text stringlengths 6 976k | token_count float64 677 677 | cluster_id int64 1 1 |
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Product Overview
PRINCIPLES OF ENGINEERING will help readers better understand the engineering concepts, mathematics, and scientific principles that form the foundation of the Project Lead the Way (PLTW) Principles Of Engineering course. Important concepts and processes are explained throughout using full-color photographs and illustrations. Appropriate for high school students, the mathematics covered includes algebra and trigonometry. Strong pedagogical features to aid comprehension include: Case Studies, boxed articles such as Fun Facts and Points of Interest, Your Turn activities, suggestions for Off-Road Exploration, connections to STEM concepts, Career Profiles, Design Briefs, and example pages from Engineers' Notebooks. Each chapter concludes with questions designed to test the reader's knowledge of information presented in the chapter, along with a hands-on challenge or exercise that compliments the content and lends itself to exploration. Key vocabulary terms are highlighted throughout the book and emphasized in margin definitions. | 677.169 | 1 |
I have used both Moore & McCabe's _Intro to the Practice of Statistics_ (IPS) and Moore's _The Basic Practice of Statistics_ (BPS). I usually abhor "big books" only 1/3 of which can be covered in a semester. But in this case, I like IPS more than BPS, although both are excellent. In the case of AP students, there is more meat in IPS to chew on, and some moderately large data sets that will require occasional use of software. (I assume that many AP teachers will prefer a TI-82 or TI-83 for small data sets and, possibly, in the first part of the course. But some reasonable software should be on the menu later on.) It is true that the reading level in IPS is greater than the reading level in BPS, but it should not be beyond legitimate AP-level students.
There are some other good possibilities:
(1) DON'T overlook the Quantitative Literacy series, authored by Richard Scheaffer, et al., at various times over the past ten years. They are available from Dale Seymour. Call 800-USA-1100.
(2) Speaking of Scheaffer, his Activity Based Statistics (a consequence of an NSF-funded project) is due to be published this spring by Springer- Verlag. This is likely to be one of the most useful resources available to AP teachers. (I have seen and used a few of the activities; many are unique, and all are thoroughly tested, and pieced together by some of the most creative people in the statistics reform movement--e.g., Ann Watkins, Gail Burrill, Jeff Witmer, etc.)
(3) Allan Rossman's _Workshop Statistics: Discovery With Data_ has just been released by Springer-Verlag. The title says it all. Get a copy.
(4) For reference, there also are:
(a) S. Chatterjee, et al., A Casebook for a First Course in Statistics and Data Analysis, Wiley, 1995.
BTW, Timothy Brown referred to Fred Djang as "the only secondary school teacher on E.T.S.'s advisory committee for developing the exam". Not true. Well, come to think of it perhaps it could be true this year, though I doubt it. But the 1994-95 Development Committee included 3 high school teachers on the 8-person committee: in addition to Djang, Chris Olsen (George Washington HS, Cedar Rapids) and Diann Resnick (Bellaire Senior HS, Bellaire, TX). And do not allow yourself to think that the college people on that committee don't know what will fly in high schools. If you ever have a chance to hear Ann Watkins (CA State U, Northridge) or Kinley Larntz (U of Minnesota) at a meeting, GO and you will see what I mean.
One more book:
If you plan to use Moore's BPS, get the supplemental _TI-82 Guide for Moore's BPS_ by Larry Morgan, which includes a set of programs MSTATPAK that supplements the -82's built-in capabilities very nicely.
'nuf said...
============================================== Bruce King Department of Mathematics and Computer Science Western Connecticut State University 181 White Street Danbury, CT 06810 (kingb@wcsub.ctstateu.edu) | 677.169 | 1 |
Welcome,
to those who have made it to find this place. There is mainly serious material on this site about science. There is a lot of information on internet about science. You can find books, lecture notes, and last but not least there is the wikipedia encyclopedia. All of this information can be useful to study. The way science is presented in these sources differs a lot. As not one source could meet my preferences, I decided to take a personal approach. The approach is characterized by giving a systematic account of the main concepts and results in a given domain. I try to avoid presenting too much details or disturbing the story by exercises and examples. I also try to put more structure than a topic oriented encyclopedia. The notes will be presented precise and concise. Main theorems are proved and sometimes an example is given. These notes would be most useful for those who had already some exposure to the theory, but need a quick review in a less cumbersome environment than a school book.
My first aim is to publish common mathematical knowledge at the university undergraduate level. This knowledge mainly developed in the 18th century by giants like Isaac Newton and Gottfried Wilhelm Leibniz is timeless and will be studied by generations of future students.
Studying science starts with two major building blocks, calculus and linear algebra. Calculus can be divided into single and multi variable calculus. In the first phase of developing this site I will focus on these domains.
For the writing of this material I use Latex in combination with MathJax . I investigated different methods to prepare these notes. But my conclusion is that the only thing needed is a free text editor. I choose for Notepad++. To save the planet please never print the information on this site.
As this site will only be updated slowly over time and is yet infancy, I refer you to the great site of Salam Kahn: Kahn Academy for all the things you misses here. | 677.169 | 1 |
Clear, rigorous definitions of mathematical terms are crucial to good scientific and technical writing-and to understanding the writings of others. Scientists, engineers, mathematicians, economists, technical writers, computer programmers, along with teachers, professors, and students, all have the need for comprehensible, working definitions of mathematical... more...
Containing more than 1,000 entries, the Dictionary of Classical and Theoretical Mathematics focuses on mathematical terms and definitions of critical importance to practicing mathematicians and scientists. This single-source reference provides working definitions, meanings of terms, related references, and a list of alternative terms and definitions.... more... | 677.169 | 1 |
Find a Tolleson Statistics covered basic mathematics, algebra, and trigonometry for many years. Also, the application of algebra to higher mathematics is a really important skill to have. Linear algebra is quickly replacing differential equations as being the most important mathematical skill to have in the engineering disciplines. | 677.169 | 1 |
Product Description
Perfect for families that already own the sold-separately Teaching Textbooks Teaching Textbooks Math 7 Extra Workbook & Answer Key, this set includes four CD-ROMs that contain step-by-step audiovisual solutions to each homework and test problem. Topics covered include basic arithmetic, including fractions, decimals, and percents; geometry (e.g. how to find the area of a circle); statistics and probability; simple graphing concept; equations and inequalities; and math in the real world. A digital gradebook grades answers as soon as they are entered and calculates percentages for each assignment. Though this CD-ROM set may technically be used without the workbook, students will then have to write out each problem; won't be able to work away from the computer; and won't receive the written summaries available in the textbook. Teaching Textbooks Grade 7.
The new hybrid CD-ROMs feature multiple improvements, including:
Easy multiple user setup (built into program)
An area where parents can access all of their students' gradebooks.
An editable gradebook where you can reset a particular lesson(s) without having to uninstall and reinstall | 677.169 | 1 |
Linear Algebra and Differential Equations Using MATLAB®, 1st Edition
These world-renowned authors integrate linear algebra and ordinary differential equations in this unique book, interweaving instructions on how to use MATLAB® with examples and theory. They use computers in two ways: in linear algebra, computers reduce the drudgery of calculations to help students focus on concepts and methods; in differential equations, computers display phase portraits graphically for students to focus on the qualitative information embodied in solutions, rather than just to learn to develop formulas for solutions. | 677.169 | 1 |
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MATH TREK Algebra 1
04/01/04
For curriculum-based algebra instruction, teachers and students can use MATH TREK Algebra 1. The multimedia program includes tutorials, assessments and student tracking. Students can use the program's scientific calculator, glossary and journal to help them complete the various exercises and activities. The assessment and student-tracking features provide immediate feedback to students so that they can stay on top of their progress. This engaging program, complete with sound, animation and graphics, can be used on stand-alone computers or a network. NECTAR Foundation, (613) 224-3031,
This article originally appeared in the 04 | 677.169 | 1 |
About the author:
Philip Klein is Professor of Computer Science
at Brown University. He was a recipient of the
National Science Foundation's Presidential
Young Investigator Award, and has received
multiple research grants from the National
Science Foundation. He has been made an ACM
Fellow in recognition of his contributions to
research on graph algorithms. He is a
recipient of Brown University's Award for
Excellence in Teaching in the Sciences.
Klein received a B.A. in Applied Mathematics
from Harvard and a Ph.D. in Computer Science
from MIT. He has been a Visiting Scientist at
Princeton's Computer Science Department, at
MIT's Mathematics Department, and at MIT's
Computer Science and Artificial Intelligence
Laboratory (CSAIL), where he is currently a
Research Affiliate.
Klein has worked at industry research labs,
including Xerox PARC and AT&T Labs, and he
has been Chief Scientist at three start-ups.
Klein was born and raised in Berkeley,
California. He started learning programming in
1974, and started attending meetings of the
Homebrew Computer Club a couple of years
later. His love for computer science has never
abated, but in a chance encounter with E. W.
Dijkstra in 1979, he was told that, if he
wanted to do computer science, he had better
learn some math.
An engaging introduction to vectors and
matrices and the algorithms that operate on
them, intended for the student who knows how to
program. Mathematical concepts and computational
problems are motivated by applications in
computer science. The reader learns bydoing,
writing programs to implement the mathematical
concepts and using them to carry out tasks and
explore the applications. Examples include:
error-correcting codes, transformations in
graphics, face detection, encryption and
secret-sharing, integer factoring, removing
perspective from an image, PageRank (Google's
ranking algorithm), and cancer detection from
cell features. A companion web site,
codingthematrix.com
provides data and support code. Most of the
assignments can be auto-graded online. Over two
hundred illustrations, including a selection of
relevantxkcdcomics. | 677.169 | 1 |
Geometry
by Ray C Jurgensen
Better than most books on the subject
This text is excellent. Easy to read, Geometry, by Ray C. Jurgensen far surpasses the content found in many of today's Geometry textbooks. Each chapter begins with a visual, tied into the subject of the text itself. Exercises are simple, yet effective in teaching the material in print form to readers .Each chapter contains an algebra review, which is a skill that is lacking in most high school geometry students. Each chapter also contains a challenge questions, along with application questions.
I firmly believe that schools and teachers should buy up these texts and use them as supplementary material for Geometry classes in high school. For the money, you cannot find a better book. I would wager, as well, that the books authored by the co-writers, Richard G. Brown, and John W. Jurgensen, would be worth investigating. Good writers flock together, and a well written math textbook is worth its weight in gold | 677.169 | 1 |
Best math tool for school and college! If you are a student, it will help you to learn calculus (analysis) and algebra!
Note: Mathematical analysis is a branch of mathematics that includes the theories of differentiation, integration, measure, limits, infinite series and analytic functions. It can be applied to physics, economics and | 677.169 | 1 |
Rather than simply a collection of problems, this book can be thought of as both a tool chest of mathematical techniques and an anthology of mathematical verse. The authors have grouped problems so as to illustrate and highlight a number of important techniques and have provided enlightening solutions in all cases. As well as this there are essays on topics that are not only beautiful but also useful. The essays are diverse and enlivened by fresh, non-standard ideas. This book not only teaches techniques but gives a flavour of their past, present and possible future implications. It is a collection of miniature mathematical works in the fullest sense.
$37Book Description
The authors have grouped problems to illustrate a number of important techniques, and provided enlightening solutions in all cases. There are numerous essays on diverse topics that are enlivened by fresh ideas. This book not only teaches techniques but gives a flavour of their past, present and possible future implications.
About the Author
Svetoslav Savchev was born in Gabrovo, Bulgaria, and graduated from Sofia University in 1980. He has worked for the journal "Mathematika" since 1981, and is currently vice editor-in-chief. He is the author of several books, and has been engaged in diverse activities for gifted students.
Titu Andreescu is the Director of the American Mathematics Competition, a program of the Mathematical Association of America (MAA). He also serves as Chair of the USA Mathematical Olympiad committee, Head Coach of the the USA International Mathematical Olympiad Team, and Director of the Mathematical Olympiad Summer Program.
Top Customer Reviews
This book is a co-work of two world-known Mathematics Olympiad Masters, Svetoslav Savchev from Bulgaria and Titu Andreescu from USA. Both are trainer/leader/problem committe member of the national/IMO team for many years, and are really excellent and famous in the olympiad field. Reader may ask that what this thin book will give in an age with numerous contest problems and with new training books appearing every year. The answer is "the Insight". It is divided into 50 sections(topics), each section is a essay containing the analyzing several(3 to 6) related problems, usuall from easier to harder. These problems are carefully selected from the huge collection of problems in numerous mathematics magazines and contests. This is the FIRST book in english to teach olympiad readers that how the problems are relating with each other, how a simple idea grows into a hard and beautiful problem, how to think in a higher point of view. However, the style of this book is rather relax and warm. This is not a textbook nor a problem book, neither suitable for training owing to its form. It's for reading, appreciating, and for those teachers/students who have been familiar enough with the standard problems and tired of training sessions. If you want to know what a mathematical olympiad is, what a beautiful cute problem is, this book is for you. This book is a pearl indeed. I really hope the authors will publish the volume 2,3,...
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Mathematical Miniatures is a wonderful book, with a new approach to an old favorite topic: problem solving. The book is comprised of 50 chapters, each one telling a mathematical story. The majority of the problems come from math Olympiads and the most clever solutions and applications are always accompanied by the name of the problem solver. One cannot help but be in awe at amazing insights in solving tremendously complicated problems, performed by high school students under the constrictive time frame of the Olympiad tests. The topics of the problems are reflective of the Math Olympiads topics, and will particularly appeal to the pure mathematician. But the beauty of the ingenious, albeit elementary mathematics will grab any passionate mathematician's attention. The sequence of 50 chapters - 50 mathematical topics is broken by nine "coffee breaks", and preceded by a warm up set. Each of the breaks proposes to the reader a handful of problems - some easy, some quite hard - all of them teasing and inviting. This is to say to the reader: don't just read about how others have cleverly solved problems, try your hand at some! I have to confess to falling for this trap several times and going through the house mumbling to myself "hmm... I used to get this in 5 minutes when I was in high school". Luckily all these problems come with solutions as well. Overall, a delightful book, recommended not only for one's mathematical reading pleasure, but also for math clubs in either high school or college.
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This book is a panorama of mathematical diamonds. That is full of various amazing problems, results, historic notes and instructive comments in mathematics. I do highly recommend this invaluable book and all other very nice books of Prof. Andreescu and his colleagues to all mathematics lovers.
A bit too over my head. Really a book for the professional, not the amateur. I love math, as explained in a college textbook, but this book was a bit too Poincaire-brief and condensed for me. I need a more verbose, baby-steps book. So that's why I'm giving it two stars. Of course, once you are a master, then less is more. I did give it away to a professional mathematician, who liked it.
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ee main syllabus by entranceindia
1.
JEE (Main) 2015 Syllabus
2.
JEE Main 2015
Syllabus
Welcome to Joint Entrance Examination (JEE Main) – 2015
JEE (Main) 2015 is for Undergraduate engineering courses to take
admission into IITs, NITs, IIITs, Other Centrally Funded Technical
Institutions, State Governments Institutions and other Institutions
JEE (Main) eligibility based on the performance of class 12th/equivalent
qualifying Examinationqualifying Examination
JEE (Main) is qualifying test for various courses like BE, B.Tech, B.Arch and
B.Planning in various institutions including IITs in India
It is conducted in two separate modes of examination i.e Online Mode &
Offline Mode
Basically there are two parts for entrance examination such as:
Paper - 1 (B.E/B.Tech) and Paper-2 (B.Arch/B.Planning)
4.
UNIT 1: SETS, RELATIONS AND FUNCTIONS:
Sets and their representation; Union, intersection and complement
of sets and their algebraic properties; Power set; Relation, Types of
relations, equivalence relations, functions;. one-one, into and onto
functions, composition of functions.
UNIT 2: COMPLEX NUMBERS AND QUADRATIC EQUATIONS:
JEE Main 2015
Mathematics Syllabus
UNIT 2: COMPLEX NUMBERS AND QUADRATIC EQUATIONS:
Complex numbers as ordered pairs of reals, Representation of
complex numbers in the form a+ib and their representation in a
plane, Argand diagram, algebra of complex numbers, modulus and
argument (or amplitude) of a complex number, square root of a
complex number, triangle inequality, Quadratic equations in real
and complex number system and their solutions. Relation between
roots and co-efficients, nature of roots, formation of quadratic
equations with given roots.
5.
.
UNIT 3: MATRICES AND DETERMINANTS:
Matrices, algebra of matrices, types of matrices, determinants and
matrices of order two and three.
Properties of determinants, evaluation of determinants, area of triangles
using determinants. Adjoint and evaluation of inverse of a square matrix
using determinants and elementary transformations, Test of consistency
JEE Main 2015
Mathematics Syllabus
using determinants and elementary transformations, Test of consistency
and solution of simultaneous linear equations in two or three variables
using determinants and matrices.
UNIT 4: PERMUTATIONS AND COMBINATIONS:
Fundamental principle of counting, permutation as an arrangement and
combination as selection, Meaning of P (n,r) and C (n,r), simple
applications.
9.
.
UNIT 11: CO-ORDINATE GEOMETRY:
Cartesian system of rectangular co-ordinates 10 in a plane, distance formula,
section formula, locus and its equation, translation of axes, slope of a line,
parallel and perpendicular lines, intercepts of a line on the coordinate axes.
Straight lines
Various forms of equations of a line, intersection of lines, angles between two
lines, conditions for concurrence of three lines, distance of a point from a line,
equations of internal and external bisectors of angles between two lines,
JEE Main 2015
Mathematics Syllabus
equations of internal and external bisectors of angles between two lines,
coordinates of centroid, orthocentre and circumcentre of a triangle, equation
of family of lines passing through the point of intersection of two lines.
Circles, conic sections
Standard form of equation of a circle, general form of the equation of a circle,
its radius and centre, equation of a circle when the end points of a diameter
are given, points of intersection of a line and a circle with the centre at the
origin and condition for a line to be tangent to a circle, equation of the tangent.
Sections of cones, equations of conic sections (parabola, ellipse and
hyperbola) in standard forms, condition for y = mx + c to be a tangent and
point (s) of tangency.
10.
.
UNIT 12 : THREE DIMENSIONAL GEOMETRY:
Coordinates of a point in space, distance between two points, section
formula, direction ratios and direction cosines, angle between two
intersecting lines.
Skew lines, the shortest distance between them and its equation.
Equations of a line and a plane in different forms, intersection of a line and
JEE Main 2015
Mathematics Syllabus
Equations of a line and a plane in different forms, intersection of a line and
a plane, coplanar lines.
UNIT 13: VECTOR ALGEBRA:
Vectors and scalars, addition of vectors, components of a vector in two
dimensions and three dimensional space, scalar and vector products,
scalar and vector triple product.
13.
.
UNIT 3: LAWS OF MOTION
Force and Inertia, Newton's First Law of motion; Momentum, Newton's
Second Law of motion; Impulse; Newton's Third Law of motion. Law of
conservation of linear momentum and its applications, Equilibrium of
concurrent forces.
Static and Kinetic friction, laws of friction, rolling friction.
Dynamics of uniform circular motion: Centripetal force and its
JEE Main 2015
Physics Syllabus
Dynamics of uniform circular motion: Centripetal force and its
applications.
UNIT 4: WORK, ENERGY AND POWER
Work done by a constant force and a variable force; kinetic and potential
energies, workenergy theorem, power.
Potential energy of a spring, conservation of mechanical energy,
conservative and non-conservative forces; Elastic and inelastic collisions
in one and two dimensions.
17.
.
UNIT 11: ELECTROSTATICS
Electric charges: Conservation of charge, Coulomb's law-forces between
two point charges, forces between multiple charges; superposition
principle and continuous charge distribution.
Electric field: Electric field due to a point charge, Electric field lines,
Electric dipole, Electric field due to a dipole, Torque on a dipole in a
uniform electric field.
Electric flux, Gauss's law and its applications to find field due to infinitely
JEE Main 2015
Physics Syllabus
Electric flux, Gauss's law and its applications to find field due to infinitely
long uniformly charged straight wire, uniformly charged infinite plane
sheet and uniformly charged thin spherical shell. Electric potential and its
calculation for a point charge, electric dipole and system of charges;
Equipotential surfaces, Electrical potential energy of a system of two point
charges in an electrostatic field.
Conductors and insulators, Dielectrics and electric polarization, capacitor,
combination of capacitors in series and in parallel, capacitance of a
parallel plate capacitor with and without dielectric medium between the
plates, Energy stored in a capacitor.
19.
.
UNIT 13: MAGNETIC EFFECTS OF CURRENT AND MAGNETISM
Biot – Savart law and its application to current carrying circular loop.
Ampere's law and its applications to infinitely long current carrying
straight wire and solenoid. Force on a moving charge in uniform magnetic
and electric fields. Cyclotron.
Force on a current-carrying conductor in a uniform magnetic field. Force
between two parallel current-carrying conductors-definition of ampere.
Torque experienced by a current loop in uniform magnetic field; Moving
JEE Main 2015
Physics Syllabus
Torque experienced by a current loop in uniform magnetic field; Moving
coil galvanometer, its current sensitivity and conversion to ammeter and
voltmeter.
Current loop as a magnetic dipole and its magnetic dipole moment. Bar
magnet as an equivalent solenoid, magnetic field lines; Earth's magnetic
field and magnetic elements.Para-, dia- and ferro- magnetic substances.
Magnetic susceptibility and permeability, Hysteresis, Electromagnets and
permanent magnets.
21.
.
UNIT 16: OPTICS
Reflection and refraction of light at plane and spherical surfaces, mirror
formula, Total internal reflection and its applications, Deviation and
Dispersion of light by a prism, Lens Formula, Magnification, Power of a
Lens, Combination of thin lenses in contact, Microscope and Astronomical
Telescope (reflecting and refracting) and their magnifyingpowers.
JEE Main 2015
Physics Syllabus
Telescope (reflecting and refracting) and their magnifyingpowers.
Wave optics: wavefront and Huygens' principle, Laws of reflection and
refraction using Huygen's principle. Interference, Young's double slit
experiment and expression for fringe width. Diffraction due to a single slit,
width of central maximum. Resolving power of microscopes and
astronomical telescopes, Polarisation, plane polarized light; Brewster's
law, uses of plane polarized light and Polaroids.
23.
.
UNIT 20: COMMUNICATION SYSTEMS
Propagation of electromagnetic waves in the atmosphere; Sky and space
wave propagation, Need for modulation, Amplitude and Frequency
Modulation, Bandwidth of signals, Bandwidth of Transmission medium,
Basic Elements of a Communication System (Block Diagram only).
UNIT 21: EXPERIMENTAL SKILLS
Familiarity with the basic approach and observations of the experiments
and activities:
JEE Main 2015
Physics Syllabus
Familiarity with the basic approach and observations of the experiments
and activities:
1. Vernier callipers-its use to measure internal and external diameter and
depth of a vessel.
2. Screw gauge-its use to determine thickness/diameter of thin
sheet/wire.
3. Simple Pendulum-dissipation of energy by plotting a graph between
square of amplitude and time.
4. Metre Scale – mass of a given object by principle of moments.
24.
.
5. Young's modulus of elasticity of the material of a metallic wire.
6. Surface tension of water by capillary rise and effect of detergents.
7. Co-efficient of Viscosity of a given viscous liquid by measuring terminal
velocity of a given spherical body.
8. Plotting a cooling curve for the relationship between the temperature of
a hot body and time.
JEE Main 2015
Physics Syllabus
a hot body and time.
9. Speed of sound in air at room temperature using a resonance tube.
10. Specific heat capacity of a given (i) solid and (ii) liquid by method of
mixtures.
11. Resistivity of the material of a given wire using metre bridge.
12. Resistance of a given wire using Ohm's law.
13. Potentiometer –
(i) Comparison of emf of two primary cells.
(ii) Determination of internal resistance of a cell.
25.
.
14. Resistance and figure of merit of a galvanometer by half deflection method.
15. Focal length of: (i) Convex mirror (ii) Concave mirror, and (iii) Convex lens
using parallax method.
16. Plot of angle of deviation vs angle of incidence for a triangular prism.
17. Refractive index of a glass slab using a travelling microscope.
18. Characteristic curves of a p-n junction diode in forward and reverse bias.
JEE Main 2015
Physics Syllabus
18. Characteristic curves of a p-n junction diode in forward and reverse bias.
19. Characteristic curves of a Zener diode and finding reverse break down voltage.
20. Characteristic curves of a transistor and finding current gain and voltage gain.
21. Identification of Diode, LED, Transistor, IC, Resistor, Capacitor from mixed
collection of such items.
22. Using multimeter to : (i) Identify base of a transistor (ii) Distinguish between
npn and pnp type transistor (iii) See the unidirectional flow of current in case of a
diode and an LED.
(iv) Check the correctness or otherwise of a given electronic component (diode,
transistor or IC).
36.
.
UNIT 12: GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF
METALS
Modes of occurrence of elements in nature, minerals, ores; Steps involved
in the extraction of metals – concentration, reduction (chemical and
electrolytic methods) and refining with special reference to the extraction
of Al, Cu, Zn and Fe; Thermodynamic and electrochemical principles
JEE Main 2015
Chemistry Syllabus
of Al, Cu, Zn and Fe; Thermodynamic and electrochemical principles
involved in the extraction of metals.
UNIT 13: HYDROGEN
Position of hydrogen in periodic table, isotopes, preparation, properties
and uses of hydrogen; Physical and chemical properties of water and
heavy water; Structure, preparation, reactions and uses of hydrogen
peroxide; Hydrogen as a fuel.
37.
.
UNIT 14: S – BLOCK ELEMENTS (ALKALI AND ALKALINE EARTH
METALS)
Group – 1 and 2 Elements
General introduction, electronic configuration and general trends in
physical and chemical properties of elements, anomalous properties of the
first element of each group, diagonal relationships.
Preparation and properties of some important compounds – sodium
carbonate and sodium hydroxide; Industrial uses of lime, limestone,
JEE Main 2015
Chemistry Syllabus
Preparation and properties of some important compounds – sodium
carbonate and sodium hydroxide; Industrial uses of lime, limestone,
Plaster of Paris and cement; Biological significance of Na, K, Mg and Ca.
UNIT 15: P – BLOCK ELEMENTS
Group – 13 to Group 18 Elements
General Introduction: Electronic configuration and general trends in
physical and chemical properties of elements across the periods and down
the groups; unique behavior of the first element in each group.
Group-wise study of the p – block elements
38.
.
Group – 14
Allotropes of carbon, tendency for catenation; Structure & properties of silicates,
and zeolites.
Group – 15
Properties and uses of nitrogen and phosphorus; Allotrophic forms of phosphorus;
Preparation, properties, structure and uses of ammonia, nitric acid, phosphine and
phosphorus halides, (PCl3, PCl5); Structures of oxides and oxoacids of
phosphorus.
JEE Main 2015
Chemistry Syllabus
phosphorus.
Group – 16
Preparation, properties, structures and uses of ozone; Allotropic forms of sulphur;
Preparation, properties, structures and uses of sulphuric acid (including its
industrial preparation); Structures of oxoacids of sulphur.
Group – 17
Preparation, properties and uses of hydrochloric acid; Trends in the acidic nature
of hydrogen halides; Structures of Interhalogen compounds and oxides and
oxoacids of halogens.
Group –18
Occurrence and uses of noble gases; Structures of fluorides and oxides of xenon. | 677.169 | 1 |
Welcome to Scribd! Start your free trial and access books, documents and more.Find out more
Academic Enhancement Center Course Syllabus Spring Term 2010
Course #
MAT 099
Course Name
Intro to Algebra
Credit
Class Schedule
Section 1: W 9:25 – 10:40 AM
3 Credits Section 2: W 1:40 – 2:55 PM
Blended Course: Computer Lab AEC and Online
Name of Instructor: Steven Diaz Phone: 305-628-6643 (office); 786-546-2415 (Cell) Email: sdiaz2@stu.edu Twitter: CafeRico IM: kaferico (Google & Yahoo) Office Hours: Monday, Tuesday, Thursday & Friday from 9 am to 3 pm. Course Description:
MAT-099 is an introductory course in Algebra. The course covers the concepts of variables, expressions, linear and quadratic equations and inequalities; algebraic manipulations, and use of graph. The goal of the course will be to ensure a solid understanding of the basic elements of algebra. Credit not applicable toward total credit graduation requirements. Eligibility to enroll in this course is based on placement examination (CPT).
New students (those who never enrolled previously in basic skills math courses) can exit this course and be placed in the next higher level math course if they score 90% in the Initial Assessment that will be administered during the first week of the course. Learning Outcomes:
By the end of this course, students should have achieved the following learning outcomes: 1. Students will demonstrate improvement of their pre-algebra skills by increasing their pretest scores by at least 25% during the post-test. 2. Students will demonstrate mastery level on each of the main concepts of the course (see below the Course Curriculum) by earning at least 70% on the assessments (i.e. quizzes) scheduled for each main concept. 3. Students will demonstrate college readiness to handle the rigor of the next mathematics course by attaining a score of at least 70% in the post-test. 4. Students will demonstrate higher confidence and motivation doing mathematics by scoring on average of at least a point higher in the Likert scale administered at the start and end of the course.
Instructional Methods
A different way of learning math with ALEKS ALEKS is a web-based assessment and learning math system that uses artificial intelligent programming to provide an individualized learning experience for every student. The instructional model of this course will mainly consist on students actively learning at their own pace with the assistance of ALEKS, the online resources available in Blackboard, and the faceto-face classroom meetings. Students must take the initiative and responsibility to use all the available resources to actively learn the course content. Instructional time will be spent less on listening class lectures and more on learning by doing and reflecting. Taking into consideration our diverse population of students and to ensure they are involved as much as possible in the learning process, this course will be based on a blended learning model. In a blended course, students complete the majority (60%) of the learning activities online (i.e. Blackboard and ALEKS), and the other learning activities (40%) takes place in the face-to-face classroom. Here is what students should expect in this course: Face-to-Face Meetings: Class will meet once a week in the classroom (math center), where students will complete learning activities led by the instructor. Students will complete a class project in every face-to-face meeting that must be published on the Internet. In addition, students must visit the math center during instructor's office hours to take the course assessments (quizzes & final exam). Asynchronous Online Meetings: Students will login at least on two different days to the course shell in Blackboard and participate in the online discussions that are posted in the discussion board. Computer assisted instruction: A learning and assessment web-based system (i.e. ALEKS) is used to help students grasp and master the course content. ALEKS will assess students' prior knowledge of the course content and create a visual representation (i.e. pie chart) of what they know and need to learn. Based on this assessment, students work on the topics they are ready to learn. Students receive immediate feedback for their performance and are continuously assessed to guarantee mastery and retention of the course content. Self-paced learning: Students complete the course content at their own pace based on their prior knowledge of the math concepts and skills covered in the course, and with the guidance of a suggested timeline available in this syllabus. Students have the opportunity to complete two courses in one academic term. Online Learning Resources: ALEKS provides detailed explanations and demonstrations of the concepts and skills covered in the course. It also provides supplementary resources such as videos, animations, Power Point presentations, math dictionary, and the course textbook (i.e. e-book). In addition, students have access in Blackboard of additional instructor-made resources (i.e. handouts, Power Points, screencasts, etc.) and math links to other Internet sites that provide tutorials, virtual manipulatives, and multimedia materials.
Available Assistance: Students have many alternatives to seek assistance to succeed in this course: (a) Visit the math center to get individual assistance from the instructor (see office hours info); (b) Visit the math center during business hours to sign up for a tutoring session; (c) Ask questions using the Question thread in the discussion board of Blackboard (questions will be answered within 24 hours); (d) Contact the instructor by phone Sunday to Thursday from 5 to 8 pm.
Course Curriculum
In this course, students will cover the following main concepts:
• • • • • • •
The main topics are represented visually with a pie chart in ALEKS. Students meet the course objectives of each main topic when they filled the pie slice that represents the main topic. In total, students must master 181 objectives (or topics) to pass this course. To see a detailed distribution of the 181 topics by main concepts, please visit the site and select the course Beginning Algebra. A dark color in the pie represents how much students have mastered of a particular main topic and a light color represents how much of the main topic students still need to master.
Course Grading Criteria:
• Your grade for this course will be based on the following components: Grading Categories
ALEKS (Fill Pie Chart) Final Exam Quizzes Class Projects/Participation TOTAL 20% 20% 25% 35% 100%
Important Notes • • • • • • • • • Students take a quiz only when they have filled a pie's slice. A quiz should be taken immediately after completing a slice of the pie. Students take quizzes in numerical order. See course outline below. Students must complete the whole pie; take all quizzes; and take Final Exam (Assessment) to have a chance passing the course. Quizzes must be taken on-site (at the math center) during the instructor's office hours. Students must retake a quiz until they earn at least 70% to pass such quiz. If an assessment pops up in the system, contact the instructor immediately. Do not take the assessment unless the instructor specified otherwise. Pie must be completed by the week prior final exam week. Incomplete grade is only granted if a student completed his/her pie and s/he took 3/4 of the quizzes by end of course.
Course Outline/Schedule:
This schedule is the suggested timeline (i.e. the slowest pace) that students should follow to complete successfully the course objectives. However, students can complete the course objectives at a faster pace. Students who fall behind the schedule are jeopardizing their chances to pass this course. ALEKS periodically prompts assessments that students must take every time. Week
•
COURSE POLICIES 1
Participation Class participation and student interaction are important components of the learning experience. In order to achieve the maximum benefits from the course, all students are expected to meet participation requirements by doing the following: Login at least on two different days to the course shell in Blackboard and actively participate in the online discussions that are posted in the discussion board. Practice Problems and Quizzes All practice problems and quizzes are done in ALEKS. Students can do the practice problems at any time and any place since ALEKS is a web-based program. Students should average at least 6 hours per week working in ALEKS to have a chance to complete the course objectives. Quizzes are taken only in the classroom during the instructor's office hours. Students must earn at least 70% to pass a quiz; otherwise, the student must retake the quiz. Attendance Educational research has proven there is a positive connection between attendance and academic success, so students are strongly urged to attend classes regularly. Face-to-Face attendance is mandatory on Wednesdays. A project will be submitted by the end of every class session. Students who miss 7 class sessions will automatically earn a No Pass (NP) grade. Contact immediately the instructor to find out how to make up an absence. Use of Computers "Computers and network systems offer powerful tools for communications among members of the St. Thomas community and of communities outside St. Thomas. When used appropriately, these tools can enhance dialogue, education, and communications. Unlawful or inappropriate use of these tools, however, can infringe on the rights of others. Activities that are expressively forbidden on St. Thomas' computers include but are not limited to the viewing, downloading or use of inappropriate materials, vandalism, virus propagation and installation of unauthorized materials. In addition, you are expected to act as a professional and use the equipment only when directed or appropriate to classroom activities. A lack of compliance with any of these directives could result in disciplinary action and dismissed of class or course. Expected Classroom Behavior Students have a responsibility to maintain both the academic and professional integrity of the school and to meet the highest standards of academic and professional conduct. Students are expected to do their own work on examinations, class preparation and assignments and to conduct themselves professionally when interacting with fellow students, faculty and staff. Academic and/or professional misconduct is subject to disciplinary action including course failure and/or probation of dismissal. No food allowed in the classroom. Dress appropriately to attend class. For additional clarification, please see Student Code of Conduct as stated in the Student Handbook. Cell Phones and Calculators Cell phones must be turned off or in vibrating mode. If a student must answer a phone call then the students must leave the classroom without disrupting the flow of the class. Students who spend a considerable amount of time attending a phone call outside the classroom will be considered absent. Calculators permitted during class only in certain circumstances the instructor considers appropriate. Assistance and Tutoring Students should take advantage of the individualized assistance from the instructor during his office hours at the Math Center (Academic Enhancement Center). One of the keys to pass this course is to ask questions without hesitation. In addition, students can sign up for tutoring sessions at the Academic Enhancement Center. Visit the center for additional info. Incomplete Grade Students will be granted an incomplete grade if s/he completed the whole pie in ALEKS by week 15 and took ¾ of the scheduled quizzes. An incomplete grade grants the student another week to complete pending assignments. •
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8
Rubric for Quizzes The following rubric (grading criteria) will be used to score quiz items. Points
1-point
Expectation
Correct answer. Work or process to support answer is logical and neatly organized. It reveals student understanding of concepts and skills. Incorrect answer. Work or process to support answer is logical and neatly organized. It reveals student understanding of concepts and skills. Minor computational or careless mistakes. Correct or Incorrect answer. Work or process to support answer is not logical or shown. It reveals student's misunderstanding of concepts and skills. Major computational mistakes.
1 1 2 3 , , , or - point 4 2 3 4
0-point
Become an Active Learner
An active learner takes control and ownership of the learning process to meet the course's goals and expectations. Active learners decide why, what and how of their learning. They do not wait for learning to happen; instead, they make it happen. The instructional model of this course requires students to become active learners to meet successfully the course objectives. The following traits are typical of active learners: 1. 2. 3. 4. 5. 6. 7. 8. 9. Identify personal goals and the steps necessary to achieve the goals. Use resources. Identify the people and tools available to aid in goal pursuit. Learn how to solve almost any problem they ever have to face. Look at situations objectively. Ask the right questions. Use time well. They organize and set priorities. Apply good reading, studying, and questioning skills to course materials. Apply good listening skills in the classroom. Find patterns and take effective notes to organize materials for studying.
10. Assess progress along the way and revise their plans.
Source:
English Second Language Learners
For students who do not speak English as their first language, the following suggestions may be helpful to succeed in this course: 1. Bring a dictionary that translates from the student's native language to English and vice versa. If a student does not have a dictionary, the following website provides word and text translation: 2. Find a classmate or group of students who speak English fluently to study for the class and to gain proficiency with the English language.
3. If there is a classmate that speaks the same native language, students can ask for clarification or assistance using their native language as long it does not disrupt the classroom learning experience. 4. The instructor of this course is bilingual (English-Spanish) and welcome students to speak Spanish during office hours or before-after class. In addition, there are many languages that have words which are pronounced and written similarly. Therefore, the instructor encourages students to sometimes use words in their native language to communicate ideas, concerns, or questions. 5. If students learned different ways or methods for simplifying or solving math problems in their countries, the instructor encourages these students to share their methods with him. In addition, ALEKS offers the option of presenting course content in Spanish for the Spanish speaker students.
Students with Disabilities
Please note that students requiring accommodations as a result of a disability must contact Maritza Rivera (e-mail: mrivera@stu.edu and phone number: 305-628-6563) at the Academic Enhancement Center. Note for Changes: The instructor reserves the right to change this syllabus at any time during the term in order to better meet the needs of this particular class group. | 677.169 | 1 |
Kaseberg/Cripe/Wildman's respected INTRODUCTORY ALGEBRA is known for an informal, interactive style that makes algebra more accessible to students while maintaining a high level of mathematical accuracy. This new edition introduces two new co-authors, Greg Cripe and Peter Wildman. The three authors have created a new textbook that introduces new pedagogy to teach students how to be better prepared to succeed in math and then life by strengthening their ability to solve critical-thinking problems. This text's popularity is attributable to the author's use of guided discovery, explorations, and problem solving, all of which help students learn new concepts and strengthen their skill retention. | 677.169 | 1 |
first calc study guides that really give students a clue.Bob Miller's student-friendlyCalc for the Cluelessfeatures quickly-absorbed, fun-to-use information and help. Students will snap upCalc for the Cluelessas they discover: Bob Miller's painless and proven techniques to learning Calculus Bob Miller's way of anticipating problems Anxiety-reducing features on every page Real-life examples that bring the math into focus Quick-take methods tht fit short study sessions (and short attention spans) The chance to have a life, rather than spend it trying to decipher calc!
Product Identifiers
ISBN-10
0070434085
ISBN-13
9780070434080
Key Details
Author
Bob Miller
Number Of Pages
150 pages
Edition Description
Revised,Student Edition of Textbook
Series
Bob Miller's Clueless
Format
Paperback
Publication Date
1997-11-22
Language
English
Publisher
McGraw-Hill Companies, The
Additional Details
Edition Number
2
Copyright Date
1998
Illustrated
Yes
Dimensions
Weight
9.5 Oz
Height
0.5 In.
Width
6.8 In.
Length
8.8 In.
Target Audience
Group
Scholarly & Professional
Classification Method
LCCN
97-043438
LC Classification Number
QA303.M68512 1998
Dewey Decimal
515
Dewey Edition
21
Publisher's Note
Publisher's Note
The book presents the material in the first semester of a standard college calculus sequence as an encouraging teacher would, explaining all topics and techniques in clear, easy-to-understand terms, answering the questions that most often crop up, anticipating students' difficulties and confusions like an experienced teacher, thereby eliminating math anxiety and bridging the ''understanding gap'' between student and standard text.
Table Of Content
Table Of Content
Student tested and approved! If you suffer from math anxiety, then sign up for private tutoring with Bob Miller! Do logarithms, sines, and cosines leave you in a cold sweat? Vectors and derivatives send stress signals to your brain? If so, then you are like the many thousands of students��students of all ages, all levels��who are anxious trying to master math. Particularly calculus. Luckily, there is a cure: Bob Miller'sCalc for the Clueless! Like the teacher you always wished you had (but never thought existed), Bob Miller brings a combination of knowledge, empathy and fun to this often troubling subject. He breaks down the learning process in an easy, nontechnical way and builds it up again using his own unique methods. "Basically, the Clueless books are my notes. Exactly the way I teach: friendly, clear...with some humor and plenty of emotion!" Meant to bridge the gap between the student, the textbook, and the teacher,Calc for the Clulessis packed with all the information you need to conquer calc. New for this second edition ofCalc Iare: Anxiety-reducing features on every page Quick-take methods that fit short study sessions (and short attention spans) More word problems Additional curve sketching problems plus added insights More complete explanation of the ellipse and hyperbola More detail on the integral theorems, with average value theorem added Plus, a new section on numerical approximation: Newton's method, trapezoidal method, parabolic method And more! "I am always delighted when a student tells me that he or she hated math...but taking a class with me has made math understandable...even enjoyable." Now it's your turn. Sharpen your #2 pencils and let Bob Miller show you how to never be clueless again!,The Beginning--Limits. The Basics. Curve Sketching Made Easy. Word Problems Made Easy. . . Well, Less Difficult. Integral Applications. Odds and Ends. | 677.169 | 1 |
...
Show More key examples and exercises from the text. New "pop-ups" reinforce key terms, definitions, and concepts while Martin-Gay presents the material. *Interactive Concept Check exercises measure students' understanding of key concepts and common trouble spots. After a student selects an answer from several multiple choice options, Martin-Gay explains why the answer was correct or incorrect. *Study Skill Builder videos reinforce the study skills-related skills and concepts found in Section 1.1, Tips for Success in Mathematics, and in the Study Skills Builder exercises found in the Student Resources section of the text. *Complete solutions on video for all exercises from the Practice Final Exam (located in the Student Resources section of the text) help students prepare for the real thing. *Chapter Test Prep Videos that help students during their most teachable moment-when they are preparing for a test. This Martin-Gay innovation provides step-by-step solutions for the exercises found in each Chapter Test. These videos are also available on YouTube(t) and in MyMathLab | 677.169 | 1 |
Most mathematicians, engineers, and many other scientists are well-acquainted with theory and application of ordinary differential equations. This book seeks to present Volterra integral and functional differential equations in that same framwork, allowing the readers to parlay their knowledge of ordinary differential equations into theory and application of the more general problems. Thus, the presentation starts slowly with very familiar concepts and shows how these are generalized in a natural way to problems involving a memory. Liapunov's direct method is gently introduced and applied to many particular examples in ordinary differential equations, Volterra integro-differential equations, and functional differential equations.
By Chapter 7 the momentum has built until we are looking at problems on the frontier. Chapter 7 is entirely new, dealing with fundamental problems of the resolvent, Floquet theory, and total stability. Chapter 8 presents a solid foundation for the theory of functional differential equations. Many recent results on stability and periodic solutions of functional differential equations are given and unsolved problems are stated.
Key Features:
Smooth
transition from ordinary differential equations to integral and functional differential equations. Unification of the theories, methods, and applications of ordinary and functional differential equations. Large collection of examples of Liapunov functions. Description of the history of stability theory leading up to unsolved problems. Applications of the resolvent to stability and periodic problems.
1. Smooth transition from ordinary differential equations to integral and functional differential equations. 2. Unification of the theories, methods, and applications of ordinary and functional differential equations. 3. Large collection of examples of Liapunov functions. 4. Description of the history of stability theory leading up to unsolved problems. 5. Applications of the resolvent to stability and periodic problems | 677.169 | 1 |
Introduction to Algebraic Curves
Overview working knowledge of elementary complex function theory and algebra together with some exposure to topology of compact surfaces) and proceeded directly to the Riemann-Roch and Abel theorems. This book differs from a number of other books on this subject in that it combines analytic and geometric methods at the outset, so that the reader can grasp the basic results of the subject. Although such modern techniques of sheaf theory, cohomology, and commutative algebra are not covered here, the book provides a solid foundation to proceed to more advanced texts in general algebraic geometry, complex manifolds, and Riemann surfaces, as well as algebraic curves. Containing numerous exercises and two exams, this book would make an excellent introductory text. | 677.169 | 1 |
GMAT Prep
One common complaint I get from students is that their algebra skills aren't where they need to be to excel on the GMAT. This complaint, invariably, is followed by a request for additional algebra drills.
If you've followed this blog for any length of time, you know that one of the themes we stress is that Quantitative Reasoning is not, primarily, a math test. Though math is certainly involved – How could it not be? – logic and reasoning are far more important factors than conventional mathematical facility. I stress this in every class I teach. So why the misconception that we need to hone our algebra chops?
I suspect that the culprit here is the explanations that often accompany official GMAC questions. On the whole, they tend to be biased in favor of purely algebraic solutions. They're always technically correct, but often suboptimal for the test-taker who needs to arrive at a solution within two minutes. Consequently, many students, after reviewing these solutions and arriving at the conclusion that they would not have been capable of the hairy algebra proffered in the official solution, think they need to work on this aspect of their prep. And for the most part it isn't true.
Here's a good example:
If x, y, and k are positive numbers such that [x/(x+y)]*10 + [y/(x+y)]*20 = k and if x < y, which of the following could be the value of k?
A) 10
B) 12
C) 15
D) 18
E) 30
A large percentage of test-takers see this question, rub their hands together, and dive into the algebra. The solution offered in the Official Guide does the same – it is about fifteen steps, few of them intuitive. If you were fortunate enough to possess the algebraic virtuosity to solve the question in this manner, you'd likely chew up 5 or 6 minutes, a disastrous scenario on a test that requires you to average 2 minutes per problem.
The upshot is that it's important for test-takers, when they peruse the official solution, not to arrive at the conclusion that they need to solve this question the same way the solution-writer did. Instead, we can use the same simple strategies we're always preaching on this blog: pick some simple numbers.
We're told that x<y, but for my first set of numbers, I like to make x and y the same value – this way, I can see what effect the restriction has on the problem. So let's say x = 1 and y = 1. Plugging those values into the equation, we get:
(1/2) * 10 + (1/2) * 20 = k
5 + 10 = k
15 = k
Well, we know this isn't the answer, because x should be less than y. So scratch off C. And now let's see what the effect is when x is, in fact, less than y. Say x = 1 and y = 2. Now we get:
(1/3) * 10 + (2/3) * 20 = k
10/3 + 40/3 = k
50/3 = k
50/3 is about 17. So when we honor the restriction, k becomes larger than 15. The answer therefore must be D or E. Now we could pick another set of numbers and pay attention to the trend, or we can employ a bit of logic and common sense. The first term in the equation x/(x+y)*10 is some fraction multiplied by 10. So this term, logically, is some value that's less than 10.
The second term in the equation is y/(x+y)*20, is some fraction multiplied by 20, this term must be less than 20. If we add a number that's less than 10 to a number that's less than 20, we're pretty clearly not going to get a sum of 30. That leaves us with an answer of 18, or D.
(Note that if you're really savvy, you'll recognize that the equation is a weighted average. The coefficients in the weighted average are 10 and 20. If x and y were equal, we'd end up at the midway point, 15. Because 20 is multiplied by y, and y is greater than x, we'll be pulled towards the high end of the range, leading to a k that must fall between 15 and 20 – only 18 is in that range.)
Takeaway: Never take a formal solution to a problem at face value. All you're seeing is one way to solve a given question. If that approach doesn't resonate for you, or seems so challenging that your conclusion is that you must purchase a host of textbooks in order to improve your formal math skills, then you haven't absorbed what the GMAT is really about. Often, the relevant question isn't, "Can you do the math?" It's, "Can you reason your way to the answer without actually doing the math?"
A recurring question from many students who are preparing for GMAT is this: When should one use the permutation formula and when should one use the combination formula?
People have tried to answer this question in various ways, but some students still remain unsure. So we will give you a rule of thumb to follow in all permutation/combination questions:
You never NEED to use the permutation formula! You can always use the combination formula quite conveniently. First let's look at what these formulas do:
Permutation: nPr = n!/(n-r)!
Out of n items, select r and arrange them in r! ways.
Combination: nCr = n!/[(n-r)!*r!]
Out of n items, select r.
So the only difference between the two formulas is that nCr has an additional r! in the denominator (that is the number of ways in which you can arrange r elements in a row). So you can very well use the combinations formula in place of the permutation formula like this:
nPr = nCr * r!
The nCr formula is far more versatile than nPr, so if the two formulas confuse you, just forget about nPr.
Whenever you need to "select," "pick," or "choose" r things/people/letters… out of n, it's straightaway nCr. What you do next depends on what the question asks of you. Do you need to arrange the r people in a row? Multiply by r!. Do you need to arrange them in a circle? Multiply by (r-1)!. Do you need to distribute them among m groups? Do that! You don't need to think about whether it is a permutation problem or a combination problem at all. Let's look at this concept more in depth with the use of a few examples.
There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can the presenters be chosen?
In this question, you simply have to choose 3 of the 8 teachers, and you know that you can do that in 8C3 ways. That is all that is required.
8C3 = 8*7*6/3*2*1 = 56 ways
Not too bad, right? Let's look at another question:
There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can all three presentations be done?
This question is a little different. You need to find the ways in which the presentations can be done. Here the presentations will be different if the same three teachers give presentations in different order. Say Teacher 1 presents, then Teacher 2 and finally Teacher 3 — this will be different from Teacher 2 presenting first, then Teacher 3 and finally Teacher 1. So, not only do we need to select the three teachers, but we also need to arrange them in an order. Select 3 teachers out of 8 in 8C3 ways and then arrange them in 3! ways:
We get 8C3 * 3! = 56 * 6 = 336 ways
Let's try another one:
Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook. How many groups of photographs are possible?
For this problem, out of 7 photographs, we just have to select 3 to make a group. This can be done in 7C3 ways:
7C3 = 7*6*5/3*2*1 = 35 ways
Here's another variation:
Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook, 1 each on the walls of his three best friends. In how many ways can he do that?
Here, out of 7 photographs, we have to first select 3 photographs. This can be done in 7C3 ways. Thereafter, we need to put the photographs on the walls of his three chosen friends. In how many ways can he do that? Now there are three distinct spots in which he will put up the photographs, so basically, he needs to arrange the 3 photographs in 3 distinct spots, which that can be done in 3! ways:
12 athletes will run in a race. In how many ways can the gold, silver and bronze medals be awarded at the end of the race?
We will start with selecting 3 of the 12 athletes who will win some position in the race. This can be done in 12C3 ways. But just selecting 3 athletes is not enough — they will be awarded 3 distinct medals of gold, silver, and bronze. Athlete 1 getting gold, Athlete 2 getting silver, and Athlete 3 getting bronze is not the same as Athlete 1 getting silver, Athlete 2 getting gold and Athlete 3 getting bronze. So, the three athletes need to be arranged in 3 distinct spots (first, second and third) in 3! ways:
Total number of ways = 12C3 * 3! ways
Note that some of the questions above were permutation questions and some were combination questions, but remember, we don't need to worry about which is which. All we need to think about is how to solve the question, which is usually by starting with nCr and then doing any other required steps. Break the question down — select people and then arrange if required. This will help you get rid of the "permutation or combination" puzzle once and for all the east coast braces for a historic winter storm (and Weezer fans can't get "My Name is Jonas"out of their heads), there's a lesson that needs to be taught from Hanover to Cambridge to Manhattan to Philadelphia to Charlottesville.
When driving in the snow:
Don't brake until you have to.
Don't make sudden turns or lane changes, and only turn if you have to.
Stay calm and leave yourself space and time to make decisions.
And those same lessons apply to GMAT Sentence Correction. Approach these questions like you would approach driving in a blizzard, and you may very well earn that opportunity to drive through blustery New England storms as you pursue your MBA. What does that mean?
1) Stay In Your Lane
Just as quick, sudden jerks of the steering wheel will doom you on snowy/icy roads, sudden and unexpected decisions on GMAT Sentence Correction will get you in trouble. Your "lane" consists of the decisions that you've studied and practiced and can calmly execute: Modifiers, Verbs (tense and agreement), Pronouns, Comparisons, Parallelism in a Series, etc. It's when you get out of that lane that you're prone to skidding well off track. For example, on this problem (courtesy the Official Guide for GMAT Review):
While Jackie Robinson was a Brooklyn Dodger, his courage in the face of physical threats and verbal attacks was not unlike that of Rosa Parks, who refused to move to the back of a bus in Montgomery, Alabama.
(A) not unlike that of Rosa Parks, who refused (B) not unlike Rosa Parks, who refused (C) like Rosa Parks and her refusal (D) like that of Rosa Parks for refusing (E) as that of Rosa Parks, who refused
Your "lane" here is to check for Modifiers (Is "who refused" correct? Is it required?) and for logical, clear meaning (it is required, because otherwise you aren't sure who refused to move to the back of the bus). But examinees are routinely baited into "jerking the wheel" and turning against the strange-but-correct structure of "not unlike." When you're taken off of your game, you often eliminate the correct answer (A) because you're turning into a decision you're just not great at making.
2) Don't Turn or Brake Until You Have To
The GMAT does test Redundancy and Pronoun Reference (among other things), but those are error types that are dangerous to prioritize – much like it's dangerous while driving in snow to decide quickly that you need to turn or hit the brakes. Too often, test-takers will slam on the Sentence Correction brakes at their first hint of, "That's redundant!" (like they would for "not unlike" above) or "There are multiple nouns – that pronoun is unclear!" and steer away from that answer choice.
The problem, as you saw above, is that often this means you're turning away from the proper path. "Not unlike" may scream "double-negative" or "redundant" to many, but it's a perfectly valid way to express the idea that the two things aren't close to identical, but they're not as different as you might think. And you don't need to know THAT, as much as you need to know that you shouldn't ever make redundancy your first decision, because if you're like most examinees you're probably not that great at you…AND you don't have to be, because the path toward your strengths will get you to your destination.
Similarly, this week the Veritas Prep Homework Help service got into an interesting email thread about why this sentence:
Based on his experience in law school, John recommended that his friend take the GMAT instead of the LSAT.
has a pronoun reference error, but this sentence:
Mothers expect unconditional love from their children, and they are rarely disappointed.
does not. And while there likely exists a technical, grammatical reason why, the GMAT reason really comes down to this: Does the problem make you address the pronoun reference? If not, don't worry about it. In other words, don't brake or turn until you have to. If you look at those sentences in GMAT problem form, you might have:
Based on his experience in law school, John recommended that his friend take the GMAT instead of the LSAT.
(A) Based on his experience in law school, John
(B) Having had a disappointing experience in law school, John
(C) Given his experience in law school, John
Here, the question forces you to deal with the pronoun problem. The major differences between the choices are that A and C involve a pronoun, and B doesn't. Here, you have to deal with that issue. But for the other sentence, you might see:
Mothers expect unconditional love from their children, and they are rarely disappointed.
(A) Mothers expect unconditional love from their children, and they are
(B) The average mother expects unconditional love from their children, and are
(C) The average mother expects unconditional love from their children, and they are
(D) Mothers, expecting unconditional love from their children, they are
Here, the only choice that doesn't include the pronoun "they" is choice B, but that choice commits a glaring pronoun (and verb) agreement error ("the average mother" is singular, but "their children" is plural…and the verb "are" is, too). So you don't need to worry about the "they" (which clearly refers to "mothers" and not "children," even though there happen to be two plural nouns in the sentence).
Grammatically, the presence of multiple nouns doesn't alone make the pronoun itself ambiguous, but strategically for the GMAT, what you really need to know is that you don't have to hit the brakes at the first sign of "unclear reference." Wait and see if the answer choices give you a chance to address that, and if they do, then make sure that those choices are free of other, more binary errors first. Don't turn or brake unless you have to.
3) Stay calm and leave yourself space to make decisions.
Just like a driver in the snow, as a GMAT test-taker you'll be nervous and antsy. But don't let that force you into rash decisions! Assess the answer choices before you try to determine whether something outside your 100% confidence interval is right or wrong in the original. You don't need to make a decision on Choice A right away, just like you don't need to change lanes simply for the sake of doing so. Have a plan and stick to it, both on the GMAT and on those snowy roads this weekend.
Today we will discuss why it is important to keep an open mind while toiling away on your GMAT studying. Don't go into test day with biases expecting that if a question tells us this, then it must ask that. GMAC testmakers are experts in surprising you and taking advantage of your preconceived notions, which is how they confuse you and convert a 600-level question to a 700-level one.
We have discussed necessary and sufficient conditions before; we have also discussed assumptions before. This question from our own curriculum is an innovative take on both of these concepts. Let's take a look.
All of the athletes who will win a medal in competition have spent many hours training under an elite coach. Michael is coached by one of the world's elite coaches; therefore it follows logically that Michael will win a medal in competition.
The argument above logically depends on which of the following assumptions?
(A) Michael has not suffered any major injuries in the past year.
(B) Michael's competitors did not spend as much time in training as Michael did.
(C) Michael's coach trained him for many hours.
(D) Most of the time Michael spent in training was productive.
(E) Michael performs as well in competition as he does in training.
First we must break down the argument into premises and conclusions:
Premises:
All of the athletes who will win a medal in competition have spent many hours training under an elite coach.
Michael is coached by one of the world's elite coaches.
Conclusion: Michael will win a medal in competition.
Read the argument carefully:
All of the athletes who will win a medal in competition have spent many hours training under an elite coach.
Are you wondering, "How does one know that all athletes who will win (in the future) would have spent many hours training under an elite coach?"
The answer to this is that it doesn't matter how one knows – it is a premise and it has to be taken as the truth. How the truth was established is none of our business and that is that. If we try to snoop around too much, we will waste precious time. Also, what may seem improbable may have a perfectly rational explanation. Perhaps all athletes who are competing have spent many hours under an elite coach – we don't know.
Basically, what this statement tells us is that spending many hours under an elite coach is a NECESSARY condition for winning. What you need to take away from this statement is that "many hours training under an elite coach" is a necessary condition to win a medal. Don't worry about the rest.
Michael is coached by one of the world's elite coaches.
It seems that Michael satisfies one necessary condition: he is coached by an elite coach.
Conclusion: Michael will win a medal in competition.
Now this looks like our standard "gap in logic". To get this conclusion, the necessary condition has been taken to be sufficient. So if we are asked for the flaw in the argument, we know what to say.
Anyway, let's check out the question (this is usually our first step):
The argument above logically depends on which of the following assumptions?
Note the question carefully – it is asking for an assumption, or a necessary premise for the conclusion to hold.
We know that "many hours training under an elite coach" is a necessary condition to win a medal. We also know that Michael has been trained by an elite coach. Note that we don't know whether he has spent "many hours" under his elite coach. The necessary condition requires "many hours" under an elite coach.
If Michael has spent many hours under the elite coach then he satisfies the necessary condition to win a medal. It is still not sufficient for him to win the medal, but our question only asks for an assumption – a necessary premise for the conclusion to hold. It does not ask for the flaw in the logic.
Focus on what you are asked and look at answer choice (C):
(C) Michael's coach trained him for many hours.
This is a necessary condition for Michael to win a medal. Hence, it is an assumption and therefore, (C) is the correct answer.
Don't worry that the argument is flawed. There could be another question on this argument which asks you to find the flaw in it, however this particular question asks you for the assumption and nothing more we celebrate Martin Luther King, Jr. this weekend, you may take some of your free time to study for the GMAT. And if you do, make sure to heed the lessons of Dr. King, particularly as you study Data Sufficiency.
If Dr. King were alive today, he would certainly be proud of the legislation he inspired to end much of the explicit bias – you can't eat here, vote there, etc. – that was part of the American legal code until the 1960s. But he would undoubtedly be dismayed by the implicit bias that still runs rampant across society.
This implicit bias is harder to detect and even harder to "fix." It's the kind of bias that, for example, the movie Freaknomics shows; often when the name at the top of a resume connotes some sort of stereotype, it subconsciously colors the way that the reader of that resume processes the rest of the information on it.
While that kind of subconscious bias is a topic for a different blog to cover, it has an incredible degree of relevance to the way that you attack GMAT Data Sufficiency problems. If you're serious about studying for the GMAT, you'll probably have long enacted your own versions of the Voting Rights Act and Civil Rights Act well before you get to test day – that is to say, you'll have figured out how to eliminate the kind of explicit bias that comes from reading a question like:
If y is an odd integer and the product of x and y equals 222, what is the value of x?
1) x > 0
2) y is a 3 digit number
Here, you'll likely see very quickly that Statement 1 is not sufficient, and come back to Statement 2 with fresh eyes. You don't know that x is positive, so you'll quickly see that y could be 111 and x could be 2, or that y could be -111 and x could be -2, so Statement 2 is clearly also not sufficient. The explicit bias that came from seeing "x is positive" is relatively easy to avoid – you know not to carry over that explicit information from Statement 1 to Statement 2.
But you also need to be just as aware of implicit bias. Try this question, as it is more likely to appear on the actual GMAT:
If y is an odd integer and the product of x and y equals 222, what is the value of x?
1) x is a prime number
2) y is a 3 digit number
On this version of the problem, people become extremely susceptible to implicit bias. You no longer get to quickly rule out the obvious "x is positive." Here, the first statement serves to pollute your mind – it is, on its own merit, sufficient (if y is odd and the product of x and y is even, the only prime number x could be is 2, the only even prime), but it also serves to get you thinking about positive numbers (only positive numbers can be prime) and integers (only integers are prime). But those aren't explicitly stated; they're just inferences that your mind quickly makes, and then has trouble getting rid of. So as you assess Statement 2, it's harder for you to even think of the possibilities that:
x could be -2 and y could be -111: You're not thinking about negatives!
x could be 2/3 and y could be 333: You're not thinking about non-integers!
On this problem, over 50% of users say that Statement 2 is sufficient (and less than 25% correctly answer A, that Statement 1 alone is sufficient), because they fall victim to that implicit bias that comes from Statement 1 whispering – not shouting – "positive integers."
Harder problems will generally prey on your more subtle bias, so you need to make sure you're giving each statement a fresh set of available options. So this Martin Luther King, Jr. weekend, applaud the progress that you have made in removing explicit bias from your Data Sufficiency regimen – you now know not to include Statement 1 directly in your assessment of Statement 2 ALONE – but remember that implicit bias is just as dangerous to your score. Pay attention to the times that implicit bias draws you to a poor decision, and be steadfast in your mission to give each statement its deserved, unbiased attention.
Today, let's take a look at an interesting number property. Once we discuss it, you might think, "I always knew that!" and "Really, what's new here?" So let me give you a question beforehand:
For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?
Think about it for a few seconds – could you come up with the answer in the blink of an eye? If yes, great! Close this window and wait for the next week's post. If no, then read on. There is much to learn today and it is an eye-opener!
Let's start by jotting down some powers of numbers:
Power of 2: 1, 2, 4, 8, 16, 32 …
Power of 3: 1, 3, 9, 27, 81, 243 …
Power of 4: 1, 4, 16, 64, 256, 1024 …
Power of 5: 1, 5, 25, 125, 625, 3125 …
and so on.
Obviously, for every power of 2, when you multiply the previous power by 2, you get the next power (4*2 = 8).
For every power of 3, when you multiply the previous power by 3, you get the next power (27*3 = 81), and so on.
Quite natural and intuitive, isn't it? Take a look at the previous question again now.
For integers x and y, 2^x + 2^y = 2^(36). What is the value of x + y?
A) 18
(B) 32
(C) 35
(D) 64
(E) 70
Which two powers when added will give 2^(36)?
From our discussion above, we know they are 2^(35) and 2^(35).
2^(35) + 2^(35) = 2^(36)
So x = 35 and y = 35 will satisfy this equation.
x + y = 35 + 35 = 70
Therefore, our answer is E.
One question arises here: Is this the only possible sum of x and y? Can x and y take some other integer values such that the sum of 2^x and 2^y will be 2^(36)?
Well, we know that no matter which integer values x and y take, 2^x and 2^y will always be positive, which means both x and y must be less than 36. Now note that no matter which two powers of 2 you add, their sum will always be less than 2^(36). For example:
2^(35) + 2^(34) < 2^(35) + 2^(35)
2^(2) + 2^(35) < 2^(35) + 2^(35)
etc.
So if x and y are both integers, the only possible values that they can take are 35 and 35.
How about something like this: 2^x + 2^y + 2^z = 2^36? What integer values can x, y and z take here?
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMATfor Veritas Prep and regularly participates in content development projects such as this blog!
Whether you were watching the College Football Playoffs or Ryan Seacrest; whether you were at a house party, in a nightclub, or home studying for the GMAT; however you rang in 2016, if 2016 is the year that you make your business school goals come true, hopefully you had one of the following thoughts immediately after seeing the number 2016 itself:
Oh, that's divisible by 9
Well, obviously that's divisible by 4
Huh, 20 and 16 are consecutive multiples of 4
2, 0, 1, 6 – that's three evens and an odd
I wonder what the prime factors of 2016 are…
Why? Because the GMAT – and its no-calculator-permitted format for the Quant Section – is a test that highly values and rewards mathematical fluency. The GMAT tests patterns in, and properties of, numbers quite a bit. Whenever you see a number flash before your eyes, you should be thinking about even vs. odd, prime vs. composite, positive vs. negative, "Is that number a square or not?" etc. And, mathematically speaking, the GMAT is a multiplication/division test more than a test of anything else, so as you process numbers you should be ready to factor and divide them at a moment's notice.
Those who quickly see relationships between numbers are at a huge advantage: they're not just ready to operate on them when they have to, they're also anticipating what that operation might be so that they don't have to start from scratch wondering how and where to get started.
With 2016, for example:
The last two digits are divisible by 4, so you know it's divisible by 4.
The sum of the digits (2 + 0 + 1 + 6) is 9, a multiple of 9, so you know it's divisible by 9 (and also by 3).
So without much thinking or prompting, you should already have that number broken down in your head. 16 divided by 4 is 4 and 2000 divided by 4 is 500, so you should be hoping that the number 504 (also divisible by 9) shows up somewhere in a denominator or division operation (or that 4 or 9 does).
So, for example, if you were given a problem:
In honor of the year 2016, a donor has purchased 2016 books to be distributed evenly among the elementary schools in a certain school district. If each school must receive the same number of books, and there are to be no books remaining, which of the following is NOT a number of books that each school could receive?
(A) 18
(B) 36
(C) 42
(D) 54
(E) 56
You shouldn't have to spend any time thinking about choices A and B, because you know that 2016 is divisible by 4 and by 9, so it's definitely divisible by 36 which means it's also divisible by every factor of 36 (including 18). You don't need to do long division on each answer choice – your number fluency has taken care of that for you.
From there, you should look at the other numbers and get a quick sense of their prime factors:
42 = 2 * 3 * 7 – You know that 2016 is divisible by 2 and 3, but what about 7?
54 = 2 * 3 * 3 * 3 – You know that 2016 is divisible by that 2 and that it's divisible by 9, so you can cover two of the 3s. But is 2016 divisible by three 3s?
56 = 2 * 2 * 2 * 7 – You know that two of the 2s are covered, and it's quick math to divide 2016 by 4 (as you saw above, it's 504). Since 504 is still even, you know that you can cover all three 2s, but what about 7?
Here's where good test-taking strategy can give you a quick leg up: to this point, a savvy 700-scorer shouldn't have had to do any real "work," but testing all three remaining answer choices could now get a bit labor intensive. Unless you recognize this: for C and E, the only real question to be asked is "Is 2016 divisible by 7?" After all, you're already accounted for the 2 and 3 out of 42, and you've already accounted for the three 2s out of 56.
7 is the only one you haven't checked for. And since there can only be one correct answer, 2016 must be divisible by 7…otherwise you'd have to say that C and E are both correct.
But even if you're not willing to take that leap, you may still have the hunch that 7 is probably a factor of 2016, so you can start with choice D. Once you've divided 2016 by 9 (here you may have to go long division, or you can factor it out), you're left with 224. And that's not divisible by 3. Therefore, you know that 2016 cannot be divided evenly into sets of 54, so answer choice D must be correct. And more importantly, good number fluency should have allowed you to do that relatively quickly without the need for much (if any) long division.
So if you didn't immediately think "divisible by 4 and 9!" when you saw the year 2016 pop up, make it your New Year's resolution to start thinking that way. When you see numbers this year, start seeing them like a GMAT expert, taking note of clear factors and properties and being ready to quickly operate on that number.
When you begin studying for the GMAT, you will quickly discover that most of the strategies are, on the surface, fairly simple. It will not come as a terribly big surprise that selecting numbers and doing arithmetic is often an easier way of attacking a problem than attempting to perform complex algebra. There is, however, a big difference between understanding a strategy in the abstract and having honed that strategy to the point that it can be implemented effectively under pressure.
Now, you may be thinking, "How hard can it possibly be to pick numbers? I see an "x" and I decide x = 5. Not so complicated." The art is in learning how to pick workable numbers for each question type. Different questions will require different types of numbers to create a scenario that truly is simpler than the algebra. The harder the problem, the more finesse that will be required when selecting numbers. Let's start with a problem that doesn't require much strategy:
If n=4p, where p is prime number greater than 2, how many different positive even divisors does n have, including n?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
Okay in this problem, "p" is a prime number greater than 2. So let's say p = 3. If n = 4p, and 4p = 4*3 = 12. Let's list out the factors of 12: 1, 2, 3, 4, 6, 12. The even factors here are 2, 4, 6, 12. There are 4 of them. So the answer is C. Not so bad, right? Just pick the first simple number that pops into your head and you're off to the races. Bring on the test!
If only it were that simple for all questions. So let's try a much harder question to illustrate the pitfalls of adhering to an approach that's overly mechanistic:
The volume of water in a certain tank is x percent greater than it was one week ago. If r percent of the current volume of water in the tank is removed, the resulting volume will be 90 percent of the volume it was one week ago. What is the value of r in terms of x?
(A) x + 10
(B) 10x + 1
(C) 100(x + 10)
(D) 100 * (x+10)/(x+100)
(E) 100 * (10x + 1)/(10x+10)
You'll notice quickly that if you simply declare that x = 10 and r =20, you may run into trouble. Say, for example, that the starting value from one week ago was 100 liters. If x = 10, a 10% increase will lead to a volume of 110 liters. If we remove 20% of that 110, we'll be removing .20*110 = 22 liters, giving us 110-22 = 88 liters. But we're also told that the resulting volume is 90% of the original volume! 88 is not 90% of 100, therefore our numbers aren't valid. In instances like this, we need to pick some simple starting numbers and then calculate the numbers that will be required to fit the parameters of the question.
So again, say the volume one week ago was 100 liters. Let's say that x = 20%, so the volume, after water is added, will be 100 + 20 = 120 liters.
We know that once water is removed, the resulting volume will be 90% of the original. If the original was 100, the volume, once water is removed, will be 100*.90 = 90 liters.
Now, rather than arbitrarily picking an "r", we'll calculate it based on the numbers we have. To summarize:
Start: 100 liters
After adding water: 120 liters
After removing water: 90 liters
We now need to calculate what percent of those 120 liters need to be removed to get down to 90. Using our trusty percent change formula [(Change/Original) * 100] we'll get (30/120) * 100 = 25%.
Thus, when x = 20, r =25. Now all we have to do is substitute "x" with "20" in the answer choices until we hit our target of 25.
Remember that in these types of problems, we want to start at the bottom of the answer choice options and work our way up:
Takeaways: Internalizing strategies is the first step in your process of preparing for the GMAT. Once you've learned these strategies, you need to practice them in a variety of contexts until you've fully absorbed how each strategy needs to be tweaked to fit the contours of the question. In some cases, you can pick a single random number. Other times, there will be multiple variables, so you'll have to pick one or two numbers to start and then solve for the remaining numbers so that you don't violate the conditions of the problem. Accept that you may have to make adjustments mid-stream. Your first selection may produce hairy arithmetic. There are no style point on the GMAT, so stay flexible, cultivate back-up plans, and remember that mental agility trumps rote memorization every time.
We know our basic probability formulas (for two events), which are very similar to the formulas for sets:
P(A or B) = P(A) + P(B) – P(A and B)
P(A) is the probability that event A will occur.
P(B) is the probability that event B will occur.
P(A or B) gives us the union; i.e. the probability that at least one of the two events will occur.
P(A and B) gives us the intersection; i.e. the probability that both events will occur.
Now, how do you find the value of P(A and B)? The value of P(A and B) depends on the relation between event A and event B. Let's discuss three cases:
1) A and B are independent events
If A and B are independent events such as "the teacher will give math homework," and "the temperature will exceed 30 degrees celsius," the probability that both will occur is the product of their individual probabilities.
Say, P(A) = P(the teacher will give math homework) = 0.4
P(B) = P(the temperature will exceed 30 degrees celsius) = 0.3
P(A and B will occur) = 0.4 * 0.3 = 0.12
2) A and B are mutually exclusive events
If A and B are mutually exclusive events, this means they are events that cannot take place at the same time, such as "flipping a coin and getting heads" and "flipping a coin and getting tails." You cannot get both heads and tails at the same time when you flip a coin. Similarly, "It will rain today" and "It will not rain today" are mutually exclusive events – only one of the two will happen.
In these cases, P(A and B will occur) = 0
3) A and B are related in some other way
Events A and B could be related but not in either of the two ways discussed above – "The stock market will rise by 100 points" and "Stock S will rise by 10 points" could be two related events, but are not independent or mutually exclusive. Here, the probability that both occur would need to be given to you. What we can find here is the range in which this probability must lie.
Maximum value of P(A and B):
The maximum value of P(A and B) is the lower of the two probabilities, P(A) and P(B).
Say P(A) = 0.4 and P(B) = 0.7
The maximum probability of intersection can be 0.4 because P(A) = 0.4. If probability of one event is 0.4, probability of both occurring can certainly not be more than 0.4.
Minimum value of P(A and B):
To find the minimum value of P(A and B), consider that any probability cannot exceed 1, so the maximum P(A or B) is 1.
Remember, P(A or B) = P(A) + P(B) – P(A and B)
1 = 0.4 + 0.7 – P(A and B)
P(A and B) = 0.1 (at least)
Therefore, the actual value of P(A and B) will lie somewhere between 0.1 and 0.4 (both inclusive).
Now let's take a look at a GMAT question using these fundamentals:
There is a 10% chance that Tigers will not win at all during the whole season. There is a 20% chance that Federer will not play at all in the whole season. What is the greatest possible probability that the Tigers will win and Federer will play during the season?
(A) 55%
(B) 60%
(C) 70%
(D) 72%
(E) 80%
Let's review what we are given.
P(Tigers will not win at all) = 0.1
P(Tigers will win) = 1 – 0.1 = 0.9
P(Federer will not play at all) = 0.2
P(Federer will play) = 1 – 0.2 = 0.8
Do we know the relation between the two events "Tigers will win" (A) and "Federer will play" (B)? No. They are not mutually exclusive and we do not know whether they are independent.
If they are independent, then the P(A and B) = 0.9 * 0.8 = 0.72
If the relation between the two events is unknown, then the maximum value of P(A and B) will be 0.8 because P(B), the lesser of the two given probabilities, is 0.8.
Since 0.8, or 80%, is the greater value, the greatest possibility that the Tigers will win and Federer will play during the season is 80%. Therefore, our answer is EHappy New Year! If you're reading this on January 1, 2016, chances are you've made your New Year's resolution to succeed on the GMAT and apply to business school. (Why else read a GMAT-themed blog on a holiday?) And if so, you're in luck: anecdotally speaking, students who study for and take the GMAT in the first half of the year, well before any major admissions deadlines, tend to have an easier time grasping material and taking the test. They have the benefit of an open mind, the time to invest in the process, and the lack of pressure that comes from needing a massive score ASAP.
This all relates to how you should approach your New Year's resolution to study for the GMAT. Take advantage of that luxury of time and lessened-pressure, and study the right way – patiently and thoroughly.
What does that mean? Let's equate the GMAT to MBA admissions New Year's resolution to the most common New Year's resolution of all: weight loss.
Someone with a GMAT score in the 300s or 400s is not unlike someone with a weight in the 300s or 400s (in pounds). There are easy points to gain just like there are easy pounds to drop. For weight loss, that means sweating away water weight and/or crash-dieting and starving one's self as long as one can. As boxers, wrestlers, and mixed-martial artists know quite well, it's not that hard to drop even 10 pounds in a day or two…but those aren't long-lasting pounds to drop.
The GMAT equivalent is sheer memorization score gain. Particularly if your starting point is way below average (which is around 540 these days), you can probably memorize your way to a 40-60 point gain by cramming as many rules and formulas as you can. And unlike weight loss, you won't "give those points" back. But here's what's a lot more like weight loss: if you don't change your eating/study habits, you're not going to get near where you want to go with a crash diet or cram session. And ultimately those cram sessions can prove to be counterproductive over the long run.
The GMAT is a test not of surface knowledge, but of deep understanding and of application. And the the problem with a memorization-based approach is that it doesn't include much understanding or application. So while there are plenty of questions in the below-average bucket that will ask you pretty directly about a rule or relationship, the problems that you'll see as you attempt to get to above average and beyond will hinge more on your ability to deeply understand a concept or to apply a concept to a situation where you might not see that it even applies.
So be leery of the study plan that nets you 40-50 points in a few weeks (unless of course that 40 takes you from 660 to 700) but then holds you steady at that level because you're only remembering and not *knowing* or *understanding*. When you're studying in January for a test that you don't need to take until the summer or fall, you have the luxury of starting patiently and building to a much higher score.
Your job this next month isn't to memorize every rule under the sun; it's to make sure you fundamentally understand the building blocks of arithmetic, algebra, logic, and grammar as it relates to meaning. Your score might not jump as high in January, but it'll be higher when decision day comes later this fall.
We know that we can perform all basic operations of addition, subtraction, multiplication and division on two equations:
a = b
c = d
When these numbers are equal, we know that:
a + c = b + d (Valid)
a – c = b – d (Valid)
a * c = b * d (Valid)
a / c = b / d (Valid assuming c and d are not 0)
When can we add, subtract, multiply or divide two inequalities? There are rules that we need to follow for those. Today let's discuss those rules and the concepts behind them.
Addition:
We can add two inequalities when they have the same inequality sign.
a < b
c < d
a + c < b + d (Valid)
Conceptually, it makes sense, right? If a is less than b and c is less than d, then the sum of a and c will be less than the sum of b and d.
On the same lines:
a > b
c > d
a + c > b + d (Valid)
Case 2: What happens when the inequalities have opposite signs?
a > b
c < d
We need to multiply one inequality by -1 to get the two to have the same inequality sign.
-c > -d
Now we can add them.
a – c > b – d
Subtraction:
We can subtract two inequalities when they have opposite signs:
a > b
c < d
a – c > b – d (The result will take the sign of the first inequality)
Conceptually, think about it like this: from a greater number (a is greater than b), if we subtract a smaller number (c is smaller than d), the result (a – c) will be greater than the result obtained when we subtract the greater number from the smaller number (b – d).
Note that this result is the same as that obtained when we added the two inequalities after changing the sign (see Case 2 above). We cannot subtract inequalities if they have the same sign, so it is better to always stick to addition. If the inequalities have the same sign, we simply add them. If the inequalities have opposite signs, we multiply one of them by -1 (to get the same sign) and then add them (in effect, we subtract them).
Why can we not subtract two inequalities when they have the same inequality sign, such as when a > b and c > d?
Say, we have 3 > 1 and 5 > 1.
If we subtract these two, we get 3 – 5 > 1 – 1, or -2 > 0 which is not valid.
If instead it were 3 > 1 and 2 > 1, we would get 1 > 0 which is valid.
We don't know how much greater one term is from the other and hence we cannot subtract inequalities when their inequality signs are the same.
Multiplication:
Here, the constraint is the same as that in addition (the inequality signs should be the same) with an extra constraint: both sides of both inequalities should be non-negative. If we do not know whether both sides are non-negative or not, we cannot multiply the inequalities.
If a, b, c and d are all non negative,
a < b
c < d
a*c < b*d (Valid)
When two greater numbers are multiplied together, the result will be greater.
Take some examples to see what happens in Case 1, or more numbers are negative:
-2 < -1
10 < 30
Multiply to get: -20 < -30 (Not valid)
-2 < 7
-8 < 1
Multiply to get: 16 < 7 (Not valid)
Division:
Here, the constraint is the same as that in subtraction (the inequality signs should be opposite) with an extra constraint: both sides of both inequalities should be non-negative (obviously, 0 should not be in the denominator). If we do not know whether both sides are positive or not, we cannot divide the inequalities.
a < b
c > d
a/c < b/d (given all a, b, c and d are positive)
The final inequality takes the sign of the numerator.
Think of it conceptually: a smaller number is divided by a greater number, so the result will be a smaller number.
Take some examples to see what happens in Case 1, or more numbers are negative.
1 < 2
10 > -30
Divide to get 1/10 < -2/30 (Not valid)
Takeaways:
Addition: We can add two inequalities when they have the same inequality signs.
Subtraction: We can subtract two inequalities when they have opposite inequality signs.
Multiplication: We can multiply two inequalities when they have the same inequality signs and both sides of both inequalities are non-negative.
Division: We can divide two inequalities when they have opposite inequality signs and both sides of both inequalities are non-negative (0 should not be in the denominator).I recently wrote about the reciprocal relationship between rate and time in "rate" questions. Occasionally, students will ask why it's important to understand this particular rule, given that it's possible to solve most questions without employing it.
There are two reasons: the first is that knowledge of this relationship can convert incredibly laborious arithmetic into a very straightforward calculation. And the second is that this same logic can be applied to other types of questions. The goal, when preparing for the GMAT, isn't to internalize hundreds of strategies; it's to absorb a handful that will prove helpful on a variety of questions.
The other night, I had a tutoring student present me with the following question:
It takes Carlos 9 minutes to drive from home to work at an average rate of 22 miles per hour. How many minutes will it take Carlos to cycle from home to work along the same route at an average rate of 6 miles per hour?
(A) 26
(B) 33
(C) 36
(D) 44
(E) 48
This question doesn't seem that hard, conceptually speaking, but here is how my student attempted to do it: first, he saw that the time to complete the trip was given in minutes and the rate of the trip was given in hours so he did a simple unit conversion, and determined that it took Carlos (9/60) hours to complete his trip.
He then computed the distance of the trip using the following equation: (9/60) hours * 22 miles/hour = (198/60) miles. He then set up a second equation: 6miles/hour * T = (198/60) miles. At this point, he gave up, not wanting to wrestle with the hairy arithmetic. I don't blame him.
Watch how much easier it is if we remember our reciprocal relationship between rate and time. We have two scenarios here. In Scenario 1, the time is 9 minutes and the rate is 22 mph. In Scenario 2, the rate is 6 mph, and we want the time, which we'll call 'T." The ratio of the rates of the two scenarios is 22/6. Well, if the times have a reciprocal relationship, we know the ratio of the times must be 6/22. So we know that 9/T = 6/22.
Cross-multiply to get 6T = 9*22.
Divide both sides by 6 to get T = 9*22/6.
We can rewrite this as T = (9*22)/(3*2) = 3*11 = 33, so the answer is B.
The other point I want to stress here is that there isn't anything magical about rate questions. In any equation that takes the form a*b = c, a and b will have a reciprocal relationship, provided that we hold c constant. Take "quantity * unit price = total cost", for example. We can see intuitively that if we double the price, we'll cut the quantity of items we can afford in half. Again, this relationship can be exploited to save time.
Take the following data sufficiency question:
Pat bought 5 lbs. of apples. How many pounds of pears could Pat have bought for the same amount of money?
(1) One pound of pears costs $0.50 more than one pound of apples.
(2) One pound of pears costs 1 1/2 times as much as one pound of apples.
Statement 1 can be tested by picking numbers. Say apples cost $1/pound. The total cost of 5 pounds of apples would be $5. If one pound of pears cost $.50 more than one pound of apples, then one pound of pears would cost $1.50. The number of pounds of pears that could be purchased for $5 would be 5/1.5 = 10/3. So that's one possibility.
Now say apples cost $2/pound. The total cost of 5 pounds of apples would be $10. If one pound of pears cost $.50 more than one pound of apples, then one pound of pears would cost $2.50. The number of pounds of pears that could be purchased for $10 would be 10/2.5 = 4. Because we get different results, this Statement alone is not sufficient to answer the question.
Statement 2 tells us that one pound of pears costs 1 ½ times (or 3/2 times) as much as one pound of apples. Remember that reciprocal relationship! If the ratio of the price per pound for pears and the price per pound for apples is 3/2, then the ratio of their respective quantities must be 2/3. If we could buy five pounds of apples for a given cost, then we must be able to buy (2/3) * 5 = (10/3) pounds of pears for that same cost. Because we can find a single unique value, Statement 2 alone is sufficient to answer the question, and we know our answer must be B.
Takeaway: Remember that in "rate" questions, time and rate will have a reciprocal relationship, and that in "total cost" questions, quantity and unit price will have a reciprocal relationship. Now the time you save on these problem-types can be allocated to other questions, creating a virtuous cycle in which your time management, your accuracy, and your confidence all improve in turn.
Today, we will take a look at the various "if/then" constructions in the GMAT Verbal section. Let us start out with some basic ideas on conditional sentences (though I know that most of you will be comfortable with these):
A conditional sentence (an if/then sentence) has two clauses – the "if clause" (conditional clause) and the "then clause" (main clause). The "if clause" is the dependent clause, meaning the verbs we use in the clauses will depend on whether we are talking about a real or a hypothetical situation.
Often, conditional sentences are classified into first conditional, second conditional and third conditional (depending on the tense and possibility of the actions), but sometimes we have a separate zero conditional for facts. We will follow this classification and discuss four types of conditionals:
1) Zero Conditional
These sentences express facts; i.e. implications – "if this happens, then that happens."
If the suns shines, the clothes dry quickly.
If he eats bananas, he gets a headache.
If it rains heavily, the temperature drops.
These conditionals establish universally known facts or something that happens habitually (every time he eats bananas, he gets a headache).
2) First Conditional
These sentences refer to predictive conditional sentences. They often use the present tense in the "if clause" and future tense (usually with the word "will") in the main clause.
If you come to my place, I will help you with your homework.
If I am able to save $10,000 by year end, I will go to France next year.
3) Second Conditional
These sentences refer to hypothetical or unlikely situations in the present or future. Here, the "if clause" often uses the past tense and the main clause uses conditional mood (usually with the word "would").
If I were you, I would take her to the dance.
If I knew her phone number, I would tell you.
If I won the lottery, I would travel the whole world.
4) Third Conditional
These sentences refer to hypothetical situations in the past – what could have been different in the past. Here, the "if clause" uses the past perfect tense and the main clause uses the conditional perfect tense (often with the words "would have").
If you had told me about the party, I would have attended it.
If I had not lied to my mother, I would not have hurt her.
Sometimes, mixed conditionals are used here, where the second and third conditionals are combined. The "if clause" then uses the past perfect and the main clause uses the word "would".
If you had helped me then, I would be in a much better spot today.
Now that you know which conditionals to use in which situation, let's take a look at a GMAT question:
Botanists have proven that if plants extended laterally beyond the scope of their root system, they will grow slower than do those that are more vertically contained.
(A) extended laterally beyond the scope of their root system, they will grow slower than do
(B) extended laterally beyond the scope of their root system, they will grow slower than
(C) extend laterally beyond the scope of their root system, they grow more slowly than
(D) extend laterally beyond the scope of their root system, they would have grown more slowly than do
(E) extend laterally beyond the scope of their root system, they will grow more slowly than do
Now that we understand our conditionals, we should be able to answer this question quickly. Scientists have established something here; i.e. it is a fact. So we will use the zero conditional here – if this happens, then that happens.
…if plants extend laterally beyond the scope of their root system, they grow more slowly than do…
So the correct answer must be (C).
A note on slower vs. more slowly – we need to use an adverb here because "slow" describes "grow," which is a verb. So we must use "grow slowly". If we want to show comparison, we use "more slowly", so the use of "slower" is incorrect here.
Let's look at another question now:
If Dr. Wade was right, any apparent connection of the eating of highly processed foods and excelling at sports is purely coincidental.
(A) If Dr. Wade was right, any apparent connection of the eating of
(B) Should Dr. Wade be right, any apparent connection of eating
(C) If Dr. Wade is right, any connection that is apparent between eating of
(D) If Dr. Wade is right, any apparent connection between eating
(E) Should Dr. Wade have been right, any connection apparent between eating
Notice the non-underlined part "… is purely coincidental" in the main clause. This makes us think of the zero conditionalSpeak like Yoda this weekend, your friends will. As today marks the release of the newest Star Wars movie, Star Wars Episode VII: The Force Awakens, young professionals around the world are lining up dressed as their favorite robot, wookie, or Jedi knight, and greeting each other in Yoda's famous inverted sentence structure. And for those who hope to awaken the force within themselves to conquer the evil empire that is the GMAT, Yoda can be your GMAT Jedi Master, too.
Learn from Yoda's speech pattern, you must.
What can Yoda teach you about mastering GMAT Sentence Correction? Beware of inverted sentences, you should. Consider this example, which appeared on the official GMAT:
Out of America's fascination with all things antique have grown a market for bygone styles of furniture and fixtures that are bringing back the chaise lounge, the overstuffed sofa, and the claw-footed bathtub.
(A) things antique have grown a market for bygone styles of furniture and fixtures that are bringing (B) things antique has grown a market for bygone styles of furniture and fixtures that is bringing (C) things that are antiques has grown a market for bygone styles of furniture and fixtures that bring (D) antique things have grown a market for bygone styles of furniture and fixtures that are bringing (E) antique things has grown a market for bygone styles of furniture and fixtures that bring
What makes this problem difficult is the inversion of the subject and verb. Much like Yoda's habit of putting the subject after the predicate, this sentence flips the subject ("a market") and the verb ("has grown"). And in doing so, the sentence gets people off track – many will see "America's fascination" as the subject (and luckily so, since it's still singular) or "all things antique" as the subject. But consider:
America's fascination is the reason for whatever is growing. "Out of America's fascination, America's fascination is growing" doesn't make any sense – the cause can't be its own effect.
So, logically, "a market" has to be the subject. But in classic GMAT style, the testmakers hide the correct answer (B) behind a strange sentence structure. Two, really – people also tend to dislike "all things antique" (preferring "all antique things" instead), but again, that's an allowable inversion in which the adjective goes after the noun.
Here is the takeaway: the GMAT will employ lots of strange sentence structures, including subject-verb inversion, a la Yoda (but only when it's grammatically warranted), so you will often need to rely on "The Force" of logic to sift through complicated sentences. Here, that means thinking through logically what the subject of the sentence should be, and also removing modifiers like "out of America's fascination…" to give yourself a more concise sentence on which to employ that logical thinking (the fascination is causing a market to develop, and that market is bringing back these old types of furniture).
Don't let the GMAT Jedi mind-trick you out of the score you deserve. See complicated sentence structures, you will, so employ the force of logic, you must.
Bryan Young served in the United States Army as an enlisted infantryman for five years, with a fifteen month tour in Iraq from 06'-07'. After leaving the military in 2008, he completed a Bachelor's Degree in Business Administration from the University of Washington. He started his career in the consumer packaged goods industry and is now looking to attend a top tier university to obtain an MBA. Along with help from Veritas Prep, he was able to raise his GMAT score from a 540 to a 690!
How did you hear about Veritas Prep?
I had been thinking about taking the GMAT for the last three years and knew that I would probably need the help of a prep course to be able to get a competitive score. Service to School, a non-profit that helps veterans make the transition from the military to undergraduate and graduate school, awarded me with a scholarship to Veritas Prep.
What was your initial Experience with the GMAT?
During my first diagnostic test, I was pretty overwhelmed. The questions were confusing and the length of the test was intimidating. Finishing the test with a 540 was a wakeup call for me. My goal was to score a 700 or higher and the score I achieved showed me just how much work I was going to need to put into the process.
How did the Veritas Prep Course help prepare you?
The resources that Veritas Prep provides are amazing. The books arrived within a few days and then I was ready to start taking the online classes. After a few classes I realized that I needed to brush up on some of the basics and was able to use their skill builders sections to get back on track. The online class format was great and helped me to learn the strategies and ask questions. Then the homework help line was where I was able to get answers on some of the more tricky questions I encountered.
Tell us about your test day experiences and how you felt throughout the experience?
The first two times I took the test I was still not as prepared as I need to be. The test day started well, but quickly went sour. I ran out of time on the integrated reasoning section and with my energy being low I wound up having my worst verbal performances.
One of the greatest aspects of Veritas Prep is that they allow you to retake the class if you feel like you need to take it again. The second time through the class helped me a lot more since I wasn't struggling with not knowing some of the basics. This helped me to fully understand the strategies for the quant section and solidify my sentence corrections skills as well. One suggestion of eating a snickers bar (or some sugary snack) made a huge difference for my energy levels and concentration on test day.
After another month and a half of studying I took the GMAT again and was excited to see the 690 with an 8 on the integrated reasoning. The score was in the range I wanted and I couldn't have been happier to be finished. Veritas Prep helped me so much throughout the year long process of beating the GMAT!
The GMAT asks a fair number of questions about the properties of two-digit numbers whose tens and units digits have been reversed. Because these questions pop up so frequently, it's worth spending a little time to gain a deeper understanding of the properties of such pairs of numbers. Like much of the content on the GMAT, we can gain understanding of these problems by simply selecting random examples of such numbers and analyzing and dissecting them algebraically.
Let's do both.
First, we'll list out some random pairs of two-digit numbers whose tens and units digits have been reversed: {34, 43}; {17, 71}; {18, 81.} Now we'll see if we can recognize a pattern when we add or subtract these figures. First, let's try addition: 34 + 43 = 77; 17 + 71 = 88; 18 + 81 = 99. Interesting. Each of these sums turns out to be a multiple of 11. This will be true for the sum of any two two-digit numbers whose tens and units digits are reversed. Next, we'll try subtraction: 43 – 34 = 9; 71 – 17 = 54; 81 – 18 = 63; Again, there's a pattern. The difference of each pair turns out to be a multiple of 9.
Algebraically, this is easy enough to demonstrate. Say we have a two-digit number with a tens digit of "a" and a units digit of "b". The number can be depicted as 10a + b. (If that isn't clear, use a concrete number to illustrate it to yourself. Let's reuse "34". In this case a = 3 and b = 4. 10a + b = 10*3 + 4 = 34. This makes sense. The number in the "tens" place should be multiplied by 10.) If the original number is 10a + b, then swapping the tens and units digits would give us 10b + a. The sum of the two terms would be (10a + b) + (10b + a) = 11a + 11b = 11(a + b.) Because "a" and "b" are integers, this sum must be a multiple of 11. The difference of the two terms would be (10a + b) – (10b + a) = 9a – 9b = 9(a – b) and this number will be a multiple of 9.
Now watch how easy certain official GMAT questions become once we've internalized these properties:
The positive two-digit integers x and y have the same digits, but in reverse order. Which of the following must be a factor of x + y?
A) 6
B) 9
C) 10
D) 11
E) 14
If you followed the above discussion, you barely need to be conscious to answer this question correctly. We just proved that the sum of two-digit numbers whose units and tens digits have been reversed is 11! No need to do anything here. The answer is D. Pretty nice.
Let's try another, slightly tougher one:
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
A) 3
B) 4
C) 5
D) 6
E) 7
This one is a little more indicative of what we're likely to encounter on the actual GMAT. It's testing us on a concept we're expected to know, but doing so in a way that precludes us from simply relying on rote memorization. So let's try a couple of approaches.
First, we'll try picking some numbers. Let's use the answer choices to steer us. Say we try B – we'll want two digits that differ by 4. So let's use the numbers 84 and 48. Okay, we can see that the difference is 84 – 48 = 36. That difference is too big, it should be 27. So we know that the digits are closer together. This means that the answer must be less than 4. We're done. The answer is A. (And if you were feeling paranoid that it couldn't possibly be that simple, you could test two numbers whose digits were 3 apart, say, 14 and 41. 41-14 = 27. Proof!)
Alternatively, we can do this one algebraically. We know that if the original two-digit numbers were 10a +b, that the new number, whose digits are reversed, would be 10b + a. If the difference between the two numbers were 27, we'd derive the following equation: (10a + b) – (10b + a) = 27. Simplifying, we get 9a – 9b = 27. Thus, 9(a – b) = 27, and a – b = 3. Also not so bad.
Takeaway: Once you've completed a few hundred practice questions, you'll begin to notice that a few GMAT strategies are applicable to a huge swath of different question types. You're constantly picking numbers, testing answer choices, doing simple algebra, or applying a basic number property that you've internalized. In this case, the relevant number property to remember is that the sum of two two-digit numbers whose units and tens digits have been reversed is always a multiple of 11, and the difference of such numbers is always a multiple of 9. Generally speaking, if you encounter a particular question type more than once in the Official Guide, it's always worth spending a little more time familiarizing yourself with it.
In Michael Lewis' Flashboys, a book about the hazards of high-speed trading algorithms, Lewis relates an amusing anecdote about a candidate interviewing for a position at a hedge fund. During this interview, the candidate receives the following question: Is 3599 a prime number? Hopefully, your testing Spidey Senses are tingling and telling you that the answer to the question is going to incorporate some techniques that will come in handy on the GMAT. So let's break this question down.
First, this is an interview question in which the interviewee is put on the spot, so whatever the solution entails, it can't involve too much hairy arithmetic. Moreover, it is far easier to prove that a large number is NOT prime than to prove that it is prime, so we should be thinking about how we can demonstrate that this number possesses factors other than 1 and itself.
Whenever we're given unpleasant numbers on the GMAT, it's worthwhile to think about the characteristics of round numbers in the vicinity. In this case, 3599 is the same as 3600 – 1. 3600, the beautiful round number that it is, is a perfect square: 602. And 1 is also a perfect square: 12. Therefore 3600 – 1 can be written as the following difference of squares:
3600 – 1 = 602 – 12
We know that x2 – y2 = (x + y)(x – y), so if we were to designate "x" as "60" and "y" as "1", we'll arrive at the following:
602 – 12 = (60 + 1)(60 – 1) = 61 * 59
Now we know that 61 and 59 are both factors of 3599. Because 3599 has factors other than 1 and itself, we've proven that it is not prime, and earned ourselves a plumb job at a hedge fund. Not a bad day's work.
But let's not get ahead of ourselves. Let's analyze some actual GMAT questions that incorporate this concept.
First:
999,9992 – 1 =
A) 1010 – 2
B) (106 – 2) 2
C) 105 (106 -2)
D) 106 (105 -2)
E) 106 (106 -2)
Notice the pattern. Anytime we have something raised to a power of 2 (or an even power) and we subtract 1, we have the difference of squares, because 1 is itself a perfect square. So we can rewrite the initial expression as 999,9992 – 12.
Using our equation for difference of squares, we get:
999,9992 – 12 = (999,999 +1)(999,999 – 1)
(999,999 + 1)(999,999 – 1) = 1,000,000* 999,998.
Take a quick glance back at the answer choices: they're all in terms of base 10, so there's a little work left for us to do. We know that 1,000,000 = 106 (Remember that the exponent for base 10 is determined by the number of 0's in the figure.) And we know that 999,998 = 1,000,000 – 2 = 106 – 2, so 1,000,000* 999,998 = 106 (106 -2), and our answer is E.
Let's try one more:
Which of the following is NOT a factor of 38 – 28?
A) 97
B) 65
C) 35
D) 13
E) 5
Okay, you'll see quickly that 38 – 28 will involve same painful arithmetic. But thankfully, we've got the difference of two numbers, each of which has been raised to an even exponent, meaning that we have our trusty difference of squares! So we can rewrite 38 – 28 as (34)2 – (24)2. We know that 34 = 81 and 24 = 16, so (34)2 – (24)2 = 812 – 162. Now we're in business.
812 – 162 = (81 + 16)(81 – 16) = 97 * 65.
Right off the bat, we can see that 97 and 65 are factors of our starting numbers, and because we're looking for what is not a factor, A and B are immediately out. Now let's take the prime factorization of 65. 65 = 13 * 5. So our full prime factorization is 97 * 13 * 5. Now we see that 13 and 5 are factors as well, thus eliminating D and E from contention. That leaves us with our answer C. Not so bad.
Takeaways:
The GMAT is not interested in your ability to do tedious arithmetic, so anytime you're asked to find the difference of two large numbers, there is a decent chance that the number can be depicted as a difference of squares.
If you have the setup (Huge Number)2 – 1, you're definitely looking at a difference of squares, because 1 is a perfect square.
If you're given the difference of two numbers, both of which are raised to even exponents, this can also be depicted as a difference of squares, as all integers raised to even exponents are, by definition, perfect squares.
I recently read Sherry Turkle's Reclaiming Conversation: The Power of Talk in a Digital Age. While the book isn't about testing advice, per se, its analysis of the costs of technology is so comprehensive that the insights are applicable to virtually every aspect of our lives.
The book's core thesis – that our smartphones and tablets are fragmenting our concentration and robbing us of a fundamental part of what it means to be human – isn't a terribly original one. The difference between Turkle's work and less effective screeds about the evils of technology is the scope of the research she provides in demonstrating how the overuse of our devices is eroding the quality of our education, our personal relationships, and our mental health.
What's amazing is that these costs are, to some extent, quantifiable. Ever wonder what the impact is of having most of our conversations mediated through screens rather than through hoary old things like facial expressions? College students in the age of smartphones score 40% lower on tests measuring indicators of empathy than college students from a generation ago. In polls, respondents who had access to smartphones by the time they were adolescents reported heightened anxiety about the prospect of face-to-face conversations in general.
Okay, you say. Disturbing as that is, those findings have to do with interpersonal relationships, not education. Can't technology be used to enhance the learning environment as well? Though it would be silly to condemn any technology as wholly corrosive, particularly in light of the fact that most schools are making a concerted effort to incorporate laptops and tablets in the classroom, Turkle makes a persuasive case that the overall costs outweigh the benefits.
In one study conducted by Pam Mueller and Daniel Oppenheimer, the researchers compared the retention rates of students who took notes on their laptops versus those who took notes by hand. The researchers' assumption had always been that taking notes on a laptop would be more beneficial, as most of us can type faster than we can write longhand. Much to their surprise, the students who took notes by hand did significantly better than those who took notes on their laptops when tested on the contents of a lecture a week later.
The reason, Mueller and Oppenheimer speculate, is that because the students writing longhand couldn't transcribe fast enough to record everything, they had to work harder to filter the information they were provided, and this additional cognitive effort allowed them to retain more. The ease of transcription – what we perceive as a benefit of technology – actually proved to be a cost. Even more disturbing, another study indicated that the mere presence of a smartphone – even if the phone is off – will cause everyone in its presence to retain less of a lecture, not just the phone's owner.
I've been teaching long enough that when I first started, it was basically unheard of for a student's attention to wander because he'd been distracted by a device. Smartphones didn't exist yet. No one brought laptops to class. Now, if I were to take a poll, I'd be surprised if there were a single student in class who didn't at least glance at a smartphone during the course of a lesson. One imagines that the same is true when students are studying on their own – a phone is nearby, just in case something important comes up. I'd always assumed the presence of these devices was relatively harmless, but if a phone that's off can degrade the quality of our study sessions, just imagine the impact of a phone that continually pings and buzzes as fresh texts, emails and notifications come in.
The GMAT is a four-hour test that requires intense focus and concentration, so anything that hampers our ability to focus is a potential drag on our scores. There's no easy solution here. I'm certainly not advocating that anyone throw away their smartphone – the fact that certain technology has costs associated with it is hardly a reason to discard that technology altogether. There are plenty of well-documented educational benefits: one can use a long train ride as an opportunity to do practice problems or watch a lecture. We can easily store data that can shed light on where we need to focus our attention in future study sessions. So the answer isn't a draconian one in which we have to dramatically alter our lifestyles. Technology isn't going anywhere – it's a question of moderation.
Takeaways: No rant about the costs of technology is going to be terribly helpful without an action plan, so here's what I suggest:
Put the devices away in class and take notes longhand. Whether you're in a GMAT prep class, or an accounting class in your MBA program, this will benefit both you and your classmates.
If you aren't using your device to study, turn it off, and make sure it's out of sight when you work. The mere visual presence of a smartphone will cause you to retain less.
Give yourself at least 2 hours of device-free time each day. This need not be when you're studying. It can also be when you're out to dinner with friends or spending time with family. In addition to improving your interpersonal relationships, conversation actually makes you smarter.
On this first Friday of November, you may start seeing some peach fuzz sprouts on the upper lips of some of your friends and colleagues. For many around the world, November means Movember, a month dedicated to the hopefully-overlapping Venn Diagram of mustaches and men's health. Why – other than the fact that this is a GMAT blog – do we mention the Venn Diagram?
Because while the Movember Foundation is committed to using mustaches as a way to increase both awareness of and funding for men's health issues (in particular prostate and testicular cancer), many young men focus solely on the mustache-growth facet of the month. And "I'm growing a mustache for Movember" without the fundraising follow-through is akin to the following quotes:
"I'm growing a mustache for Movember."
"I'm running a marathon for lymphoma research."
"I'm dumping a bucket of ice water over my head on Facebook."
"I'm taking a GMAT practice test this weekend."/"I'm going to the library to study for the GMAT."
Now, those are all noble sentiments expressed with great intentions. But another thing they all have in common is that they're each missing a critical action step in their mission to reach their desired outcome. Growing a mustache does very little to prevent or treat prostate cancer. Running a marathon isn't what furthers scientists' knowledge of lymphoma. Dumping an ice bucket over your head is more likely to cause pneumonia than to cure ALS. And taking a practice test won't do very much for your GMAT score.
Each of those actions requires a much more thorough and meaningful component. It's the fundraising behind Movember, Team in Training, and the Ice Bucket Challenge that advances those causes. It's your effort to use your mustache, sore knees, and Facebook video to encourage friends and family to seek out early diagnosis or to donate to the cause. And it's the follow-up to your GMAT practice test or homework session that helps you increase your score.
This weekend, well over a thousand practice tests will be taken in the Veritas Prep system, many by young men a week into their mustache growth. But the practice tests that are truly valuable will be taken by those who follow up on their performance, adding that extra step of action that's all so critical. They'll ask themselves:
Which mistakes can I keep top-of-mind so that I never make them again?
How could I have budgeted my time better? Which types of problems take the most time with the least probability of a right answer, and which types would I always get right if I just took the extra few seconds to double check and really focus?
Based on this test, which are the 2-3 content areas/question types that I can markedly improve upon between now and my next practice test?
How will I structure this week's study sessions to directly attack those areas?
And then they'll follow up on what they've learned, following the new week's plan of attack until it's time to again take the first step (a practice test) with the commitment to take the substantially-more-important follow-up steps that really move the needle toward success.
Taking a practice test and growing a Movember mustache are great first steps toward accomplishing noble goals, but in classic Critical Reasoning form, premise alone doesn't guarantee the conclusion. So make sure you don't leave the GMAT test center this November with an ineffective mustache and a dismal score – put in the hard work that has to accompany that first step, and this can be a Movember to Remember.
We know the formula we need to use to find the sum of n consecutive positive integers starting from 1. The formula is given as n(n+1)/2.
So the sum of first four positive integers is 4 * (4 + 1)/2 = 10.
This might seem a bit cumbersome, since it is easy to see that 1 + 2 + 3 + 4 = 10, but we know that the formula comes in very handy when n is a large number. For example, the sum of first 50 positive integers = 50 * 51/2 = 1275. Obviously, this will be a lot harder when done the "1 + 2 + 3 + 4 … + 49 + 50" way.
Now the question is, how do we adjust the same formula to find the sum of consecutive integers which do not start from 1?
Say, how do we find the sum of all positive integers from 8 to 20? The formula assumes a starting point of 1, so then we insert only the last number, n. How do we manage the 8? Let's try to figure it out
To get the sum of consecutive integers from 8 to 20, we find the sum of all integers from 1 to 20 (using the formula we know) and subtract the sum of integers from 1 to 7 out of it (using the same formula).
To generalize, the sum of all positive integers from m to n is given as:
n(n+1)/2 – (m-1)*m/2
Let's look at a question based on this concept:
If the sum of the consecutive integers from –40 to n inclusive is 356, what is the value of n?
(A) 47
(B) 48
(C) 49
(D) 50
(E) 51
If you are thinking that we haven't gone over how to adjust the formula for negative numbers, you are right, but what we have discussed is enough to solve this question.
Numbers around 0 are symmetrical. So 1 and -1 add up to equal 0. Similarly, 2 and -2 add up to equal 0, and so on…
-40, -39 … 0 … 39, 40, 41, 42, 43, 44, 45 …
The sum of all numbers from -40 to 40 will be 0. Or another way to look at it is that 0 is the mean of all numbers from -40 to 40. So the total sum of these numbers will also be 0.
The given sum is actually the sum of numbers from 41 to n only.
We know how to calculate that:
n(n+1)/2 – 40*41/2 = 356
n(n+1) = 2352
From the options, we see that n cannot be 49 or 50 because the product of 49*50 or 50*51 will end in 0, so plug in n = 48 to check whether 48*49 is equal to 2352. It is, therefore our answer is B
(Had we obtained a lower product than required, we could have said that n must be 51. Had we obtained a higher product than was required, we could have said that n is 47.)
Another method:
Use the concept of arithmetic mean and ballpark. The mean of numbers from 41 to 47 or 48 or 49… will be somewhere between 44 and 46.
Let's estimate the number of integers we need to get the sum of about 356. Each additional integer is quite large (more than 45) therefore, a difference of about 10-15 in the sum due to the various possible values of the mean will be immaterial.
45*7 = 315
45*8 = 360
This brings us very close to the value of 356.
Assuming there are 8 integers, their values will be from 41 to 48. The average of these 8 numbers will be 44.5. The total sum will be 44.5 * 8 = 356. It matches, so our answer is still Bpercentile with the proper techniques and preparation. In this "9 for 99th" videoseries, Ravi shares some of his favorite strategies to efficiently conquer the GMAT and enter that 99th percentile.
First, take a look at the previous lessons in this series: 1,2, 3, 4, 5, 6,7and8!
Lesson Nine:
Talk Like a Lawyer. When you click "Agree" on a user contract (think iTunes) or read through a GMAT question, you may just see an overkill of words. But thanks to lawyers, every word on that user agreement is carefully chosen – and that GMAT question is written the same exact way. In this final "9 for 99th" video, Ravi (a member of the bar himself) shows you how to talk and read like a lawyer, noticing those subtle word choices that can make or break your answer to those carefully-written GMAT problems you see on test day.
I recently read The Organized Mind by Daniel Levitin, a book teeming with insights about simple adjustments we can make in our daily routines to improve our productivity. I've written about this topic in the past, but it can't be emphasized enough – the primary problem most test-takers encounter is that they struggle to find enough time to study consistently.
According to GMAC, test-takers who score 700 or above spend, on average, 114 hours preparing for the exam. There's nothing magic about that number, but it does reveal that getting ready for the GMAT is an intensive ordeal. As technologies improve and our focus becomes increasingly fragmented by our proliferating gadgets, the challenge, whether we're studying for the GMAT or trying to complete a project at work, is how we can be productive and still have enough time and energy to enjoy some semblance of a personal life.
1) Sleep
First, Levitin emphasizes the importance of sleep. When we're feeling overwhelmed, our instinct is to work more and sleep less – we feel as though we need more waking hours to complete whatever tasks we have to perform. The problem with this approach is that sleep deprivation causes us to be significantly less effective and productive, so much so that the additional time we gain is more than offset by the diminished performance that results from a sleep-debt.
The statistics on the subject are nothing short of astonishing. According to economists, sleep deprivation costs U.S. businesses more than $150 billion dollars a year from accidents and lost productivity. It is also associated with increased risk for heart disease, obesity, suicide, and cancer. This is an easy fix.
Levitin recommends going to bed at the same time each night (preferably an hour earlier than you're accustomed to) and waking at the same time each morning. If it isn't possible to sleep more at night, a nap as short as 15 minutes can serve the same refreshing function. Napping has been shown to reduce our risk of developing a host of medical conditions, and the beneficial effects are so striking that many companies have designated nap rooms filled with cots.
2) Stop Multi-Tasking
Next, Levitin discusses the cognitive impact of multi-tasking. We all know that it isn't a great idea to try to study while texting or answering emails, etc., but what's striking is that the impact of allowing other activities to siphon our attention is actually quantifiable. Glenn Wilson, a British researcher from Gresham College, conducted a study in which he found that when participants were informed that they had an unread email in their inbox, their effective IQ decreased by 10 points. Moreover, he documented that the cognitive-blunting effects of multi-tasking are more pronounced than the effects of smoking marijuana.
Other studies have revealed that task-switching, in general, heightens the brain's glucose demands and amplifies anxiety, and the resulting discomfort ratchets up the desire to find some kind of distraction, such as, checking email again. Experts recommend designating two or three blocks of time a day for responding to email, and beyond that, strictly forbidding yourself to check for new messages.
A more ingenious idea comes from Lawrence Lessig, a Harvard Law professor. Lessig recommends declaring email bankruptcy, which would involve composing an automatic reply that informs whoever has contacted you that if this email requires an immediate response, they should call you, and if not, they should resend the email in a week if they haven't heard from you. This technique will allow you greater latitude in structuring your day in terms of when you respond to emails, and will, hopefully, negate the multi-tasking concerns that lead to the aforementioned IQ drop. And when you're studying for the GMAT, have a strict policy of not checking your phone or opening a new browser window.
3) Don't Procrastinate
Last, and perhaps most importantly, the book addresses the problem of procrastination. Procrastination is a universal problem and likely results from the basic architecture of the human brain, wired as it is to seek pleasure and avoid pain. Jake Eberts, a Harvard MBA and successful film producer, offers a bit of very simple but compelling advice: just get in the habit of always doing the most unpleasant thing on your agenda first. There is evidence that our willpower is gradually depleted throughout the day, so it's best to tackle the most dreaded elements of our to-do list first thing in the morning.
Takeaway: Here are three very easy things you can do, starting today, if you're having difficulty finding the time/energy to study:
1) First,sleep more. If that means a 15-minute midday nap, so be it – you will gain in productivity far more than you lose in time sacrificed.
3) Last, do the most unpleasant thing first. Whether that unpleasant thing is 25 Data Sufficiency questions, or some work-related activity, your resilience will be greatest first thing in the morning, so that's the time to tackle the task you want to do leastReading is FUNdamental: If you can read this video prompt, there are several GMAT quantitative problems that you should answer correctly…but might not on test day. As Ravi notes in this video, often students supply incorrect answers to quantitative problems not because they can't do the math, but because in doing the math they take their attention off of reading the question carefully. So heed Ravi's advice: if you're going to get a math problem wrong, get it wrong because you can't do the math, not because you can't read.
About a month ago, GMAC released the latest version of the GMAT Official Guide, 25% of which consisted of new questions. Though the GMAT tends not to change too drastically over time – how else could a school compare a score received by one candidate in 2015 to a score received by another candidate in 2010? – there can be subtle shifts of emphasis, and paying attention to the composition mix of the questions in the latest version of the Official Guide is a good way to ascertain if any such shift is in the offing.
My concern as an instructor is whether the philosophy I'm advocating and the techniques I'm teaching are as relevant for the newer questions as they have been for the older ones.
This philosophy can be summarized as follows: the GMAT is not, fundamentally, a content-based test, but rather, uses certain elements of our academic background to test how we think under pressure. Because the test is evaluating how we think, and not what we know, the cultivation of simple strategies, such as using the answer choices or picking easy numbers, is just as important as the re-mastery of the content you may have initially learned in eighth grade, but have subsequently forgotten.
Having thoroughly dissected the new questions in the latest version of the Official Guide, I can confidently report that this philosophy is more relevant than ever. Of the over 200 new quantitative questions, I didn't do extensive calculations for a single problem. If anything, the kind of fluid logic-based approach that we preach at Veritas is more critical than ever.
Take this new question, for example:
Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 4. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?
Of course, we could do this question algebraically. But if the GMAT is testing our ability to make good decisions under pressure, and if the algebra feels hard for you, then a better option is to make your life as easy as possible and select a simple number for m. If m is larger than 4, let's say that m = 5. "m" represents the number of students, so now we have 5 students and, we're told in the question stem, a total of 4 sandwiches. (The question of what kind of negligent, hard-hearted school knowingly packs only 4 sandwiches for all of its students to share will have to be addressed in another post. This question feels straight out of Oliver Twist.)
Okay. We're told that 3 of the sandwiches are divided evenly among the 5 students. (3 sandwiches)/(5 students) means each student gets 3/5 of a sandwich.
Additionally, we're told that 4 of the students don't want any part of the remaining sandwich. Because we only have 5 students and 4 of them don't want the remaining sandwich, the last student will get the entire fourth sandwich.
To summarize what we have so far: Each of the 5 students initially received 3/5 of a sandwich, and then one student received an entire additional sandwich, on top of that initial 3/5. The lucky fifth student received a total of 3/5 + 1 = 8/5 of a sandwich.
Last, we 're told that Carol ate a piece of each of the four sandwiches. But we established that only one student ate a piece of every sandwich, so Carol has to be that lucky student! Therefore, Carol ate 8/5 of a sandwich.
We're asked what fraction of a sandwich Carol ate, so the answer is simply 8/5. Now all we have to do is plug '5' in place of 'm' in each answer choice, and the one that gives us 8/5 will be our answer.
Most test-takers will simply start with A and work their way down until they find an option that works. The question-writer knows that this is how most test-takers proceed. Therefore, it's a more challenging question if the correct answer is towards the bottom of our answer choices. So let's use this logic to our advantage, start with E, and work our way up.
Answer choice E: (4m-12)/[m(m-4)]
Substituting '5' in place of 'm,' we get (4*5 – 12)/[5(5-4) = 8/5. That's it! We're done. The correct answer is E.
Takeaway: Keep reminding yourself that the GMAT (even with its new questions) is not designed to test what you know. While it is important to brush up on all of the fundamentals you acquired years before, the most successful test-takers will fluidly incorporate simple strategies when attacking complex questions, rather than simply grinding through longer calculations. Each new version of the Official Guide validates the wisdom of this approach videoseries, Ravi shares some of his favorite strategies to efficiently conquer the GMAT and enter that 99th percentile.
Read Like You Drive: very few GMAT examinees will make mistakes driving to the GMAT test center, but most test-takers will make several Reading Comprehension mistakes once they're there. As Ravi will discuss in this video, however, the two activities are much more similar than you realize: your job is to follow the signs. Certain keywords in Reading Comprehension passages will tell you when to yield, stop, turn, and pass with care, and if you're following those signs properly you can proceed much faster than your self-imposed "speed limit" (most people read the passages far too slowly – stay out of the left lane!) and save valuable time for the questions themselves.
At some point during each course I teach, I'll ask my students if they're familiar with this famous quote from Henry Ford "Whether you think you can, or you think you can't – you're right." Of course, they always know it. It's a quote so popular it's become a pedagogical cliché. Next, I'll ask them if they believe the quote is true. They usually do. I'll follow up with a series of GMAT-related questions. "Who struggles with probability questions?" "Who sees Reading Comprehension as a weakness?" Different hands go up for different questions.
They realize immediately that there's a disconnect here. Why would anyone maintain the belief that he or she struggles in a given area if he or she subscribes to the notion that the pessimistic belief is a self-fulfilling prophesy? My sense is that this disconnect is rooted in our tendency to nod politely when greeted with popular aphorisms we'd like to be true, while at some level, not really believing them.
We can pay lip service to Henry Ford all we want. Our actual belief is something more along the lines of: sure, it would be nice if you could improve your performance via thought alone, but that doesn't actually work. It's a fantasy, one that is so appealing that we'll collectively agree to pretend that it's true.
Part of my job as an instructor is to get my students to move past the cliché and somehow internalize the truth of the sentiment that our beliefs do matter. This isn't a New Age chimera that we'd like to be true. It's an area of extensive scientific research. In 2007, researchers at Stanford University conducted a study in which they tracked the development of 7th grade students who believed that intelligence was innate vs. students who believed that intelligence is a fluid phenomenon, something that can be cultivated and improved through dedicated effort.
The students who believed that intelligence is innate were deemed the "fixed mindset" group, and the group who believed that intelligence could be improved were deemed the "growth mindset" group. Most importantly, at the start of the study, these groups had similar academic background. Sure enough, over the next couple of years, there was a marked divergence in performance – the growth mindset group outperformed their fixed mindset peers by a significant margin (take a look at this study here).
One component of the growth mindset is the belief that adversity isn't evidence of an inherent shortcoming, but rather, an opportunity to learn and improve. This is absolutely essential on the GMAT. Students will, on average, take about a half-dozen practice tests. It is extremely rare that every one of those practice tests goes well.
At some point, during every class I teach, I'll get a panicked email, the general gist of which is that things had been going well, but now, after a disappointing practice test, the student has significant doubts about whether the previous successes were real. I'm often asked if it will be necessary to push the test date back. The growth mindset compels us to see this setback as a positive. Isn't it better to uncover the need for a strategic tweak on a low stakes practice test than on the official exam?
Sure enough, once my students are able to re-frame their beliefs from, "I'm just not good at X," to, "Maybe I've struggled with X in the past, but with a little practice I can actual convert this former liability into an asset," they improve. The student who struggled with probability wasn't inherently bad at probability, but had a less than stellar teacher in high school or college and never learned the underlying concepts properly. The student who struggled with Reading Comprehension simply wasn't taking notes properly.
Most importantly, the students who believed that they just weren't good at standardized tests realized that the ability to do well on standardized tests is a skill that they simply hadn't acquired yet. In the past, when they were convinced that they couldn't do well on, say, the SAT, they hadn't bothered to study, because what was the point of expending any effort if the result was going to be disappointment? Once they see that they their past struggles weren't functions of innate deficits, but rather, of self-limiting beliefs, a world of possibility opens up.
Takeaway: how we frame our thoughts with respect to academic performance is extraordinarily important. Unfortunately, our culture generally pays lip service to the growth mindset while perpetuating the notion of a fixed one. We'll thoughtlessly spout that Henry Ford quote, all the while thinking of people as high IQ or low IQ, not realizing that IQ is itself malleable (take a look at this idea here).
Think of someone you knew in high school who did unusually well on the SAT's. You probably thought, "That person is great at standardized tests," rather than "That person has been successful at cultivating a particular skill set that translated well in the domain of this one particular exam." But the latter is true. So don't set arbitrary limits of yourself, because, contrary to some our deepest intuitions, belief and performance are inextricably linked.
Once you begin studying for the GMAT, you'll realize quickly that there are different levels of mastery. There's that initial level of competence in which you learn, or relearn, many of the foundational concepts that you learned in middle school and have since forgotten. There's a more intermediate level of mastery in which you're able to blend strategic thinking with foundational concepts.
Then there's the highest level in which you achieve a kind of trance-like, fugue state that allows you to incorporate multiple strategies to break down a single complex problem and then seamlessly shift to a fresh set of strategies on the next problem, which, of course, will be testing slightly different concepts from the previous one.
It's the GMAT equivalent of becoming a Jedi who can anticipate his opponent's next light saber strike several moves in advance or becoming Neo in the Matrix, finally deciphering the structure of the streaming code that animates his synthetic world. Pick whatever sci-fi analogy you like – it's this kind of expertise that we're shooting for when we prepare for the test. The pertinent questions are then the following: how do we accomplish this level of expertise, and what does it look like once we're finally there?
Fortunately for you, dear student, our books are organized with this philosophy in mind. Once you've worked through the skill-builders and the lessons, you'll likely be at the intermediate level of competence. Then it will be through drilling with homework problems and taking practice tests that you'll achieve the level of mastery we seek. But let's take a look at a Sentence Correction question to get a sense of how our thought processes might unfold, once we're functioning in full Jedi-mode.
Unlike most severance packages, which require workers to stay until the last day scheduled to collect, workers at the automobile company are eligible for its severance package even if they find a new job before they are terminated.
(A) the last day scheduled to collect, workers at the automobile company are eligible for its severance package
(B) the last day they are scheduled to collect, workers are eligible for the automobile company's severance package
(C) their last scheduled day to collect, the automobile company offers its severance package to workers
(D) their last scheduled day in order to collect, the automobile company's severance package is available to workers
(E) the last day that they are scheduled to collect, the automobile company's severance package is available to workers
Having done hundreds of questions, you'll notice one structural clue leap immediately: "unlike." When you see words such as "like" or "unlike" you know that you're dealing with a comparison, so your first task is to make sure you're comparing appropriate items. You'll also note that the clause beginning with "which require" modifies "severance packages," so whatever is compared to these severance packages will come after the modifier.
In A, you're comparing "severance packages" to "workers." We'd rather compare severance packages to severance packages or workers to workers. No good.
That leaves us with D and E, both of which compare "severance packages" to "automobile's company severance package." Here, you're comparing one group of severance packages to another, so this is logical. But now you have to switch gears – the comparison issue allowed you to eliminate some incorrect answer choices, but you'll have to use another issue to differentiate between your remaining options.
Once we're down to two options, you can simply read the two sentences and look for differences. One difference is that E contains the word "that" in the phrase "the last day that they are scheduled to collect." Perhaps it sounds okay to your ear, but you'll recall that when "that" is used as a relative pronoun, it should touch the noun it modifies. In this case, it touches, "last day." Read literally, the phrase, "the last day that they are scheduled to collect," makes it sound as though "they" are collecting the "last day." Surely this isn't what the sentence intends to convey, so we're then left with 'D,' which is the correct answer.
Takeaway:
Notice how many disparate concepts you had to juggle here: You had to recognize the structural clue indicating that "unlike" signifies a comparison; recognize that temporarily skipping over a longer modifying phrase is an effective way to get a sense of the core clause you're evaluating; recall that once you're down to two answer choices, you can simply zero in on differences between your options; remember the rule stipulating that relative pronouns must touch what they modify; and last, you had to recognize that Sentence Correction is not only about grammar but also about logic and meaning, and all in under a minute and a half. I'd say that's pretty Jedi-like.
As we head into the Labor Day weekend here in the U.S., it seems a fitting time to talk about labor. Precious few people consider the GMAT to be a labor of love; to most aspiring (and perspiring?) MBAs, the GMAT is a lot of hard work. And while, to earn the score that you're hoping for, it's likely that you'll have to put in a good amount of sweat and a few tears (but hopefully no blood…), it's important to recognize that test day itself should not be a Labor Day!
Your hard work should take place well before you get to the test center, so that on test day you're not overworking yourself. Working too hard on test day takes time (which is a precious resource on the exam), saps your mental energy (which also tends to be in short supply as you get later into the test with only two 8-minute breaks to recharge), and leads to errors. Accordingly, here are a few tips to help you take the heavy labor out of your test day:
1. Only do the math you absolutely have to do.
The GMAT rewards efficiency and ingenuity, and has been known to set up problems that can be awful if done "by the book" but relatively smooth if you recognize common patterns. For example:
Answers are assets! If the math looks like it's going to get messy, look at the answer choices. If they're really far apart, you may be able to estimate after just a step or two. Or if the answer choices are really "clean" numbers (0, 1, 10…these are really easy numbers with which to perform calculations) you may be able to plug them into the problem and backsolve without any algebra.
Don't multiply until you've divided. Working step by step through a problem, you may see that you have to multiply, say, 51 by 18. Which is an ugly thing to have to do for two reasons: that calculation will take time by hand, and it will leave you with a new number that will be hard to work with for the following step. But the next step might be to divide by 34. If you save the multiplication (just call it (51)(18) and don't actually perform the step), then you can divide by 2 and 17. Which works out pretty cleanly: 51/17 is 3 and 18/2 is 9, so now you're just multiplying 3 by 9 and the answer has to be 27. The GMAT goes heavy on divisibility, so keep in mind that you'll do a lot of division on this test…meaning that it usually makes sense to wait to multiply until after you've seen what you'll have to divide by.
Think in terms of number properties. Often you can determine quickly whether the answer has to be even or odd, or whether it has to be positive or negative, or what the first or last digit will be. If you've made those determinations, quickly scan the answer choices and see how many fit those criteria. If only one does, you're done. And if 2-3 do but they're easier to plug in to the problem or to estimate between, then you can avoid doing the actual math.
2. Don't take too many notes.
Particularly with Reading Comprehension passages, GMAT test-takers on average take far too many notes. This hurts you for two reasons: first, it's time consuming, and on a question type that's already time consuming by nature. And second, very few of the notes that people take are useful. People tend to take notes on details – you generally write down what you don't think you'll remember – but the test will typically only ask you about one detail per passage. And the passage stays on the screen the whole time, so if you need to find a detail it's just as easy to find it on the screen as it is in your notes (plus you'll want to read the exact way that it was written, which your notes won't necessarily have). So use your time wisely: use your initial read of the passage to get a feel for the general direction of the passage, and then you'll know which area/paragraph to go back to if and when you do need to find the details.
3. Stay flexible.
The GMAT is a test that rewards "mental agility," meaning that it often designs problems that look like they should be solved one way (say, algebra) but quickly become labor-intensive that way and then reward those who are able to quickly change approaches (maybe to backsolving or picking numbers). When it looks like you've just set yourself up for a massive amount of work, take a quick step back and re-analyze. At this point are the answer choices more helpful? Should you abandon your number-picking and go back to doing the algebra? Does re-reading the question allow you to set it up differently? Generally speaking, if the math starts to get labor-intensive you're missing a better method. So let that be your catalyst for re-assessing.
As you sit down to take the GMAT (to get into a great business school to become a more valuable member of the labor force), those 4 hours you spend at the test center probably won't be a labor of love. But they shouldn't be full of labor, anyway. Heed this advice to lighten your labor and the GMAT just might feel like more of a day off than anything (like, you know, Labor Day).Practice Tests Aren't Real Tests: read the popular GMAT forums and you'll see lots of handwringing and bellyaching about practice tests scores…but not very much analysis beyond the scores themselves. In this video, Ravi (along with his alter ego Allen Iverson) talks about practice, stressing the importance of using the tests to increase your score more than to merely try to predict it. Pacing is paramount and diagnosis is divine; as Ravi will explain, practice tests are critical for learning how you would perform if that were the real thing, with the added bonus of having the opportunity to fix those things that you don't like about that practice performance.
Let's face it. Except for the folks who write the test and prepare you for the test, no one really loves the GMAT. Any anyone who tells you otherwise either scored an 800 with no prep or is lying.
But self-inflicted misery loves company, so in no particular order, let's take a look at some of the things that suck and more importantly, how to cope:
Integrated Reasoning (IR) : It was introduced a few years ago, and even though multiple surveys and studies show it does correlate well with skills needed to succeed in business and the corporate world, schools still seem to have varying opinions on its value and how best to use it in the admissions process. For now, think of IR as the appetizer or warm-up. It's tough, but it's 30 minutes and can serve as a solid warmup before tackling the tougher 'main course' of quant and verbal. You wouldn't start sprinting out of the gates in a race; treat the GMAT the same way, and if you bank some early points, that can't hurt either.
AWA: Similar to IR, it doesn't factor into your Total score, and schools differ on how they evaluate the essay. That being said, consider it a pre-pre-warm up, and more importantly, remember that schools can download a copy of your essay when they view your scores. So it's important to put forth your best effort (now is NOT the time to challenge authority and write what you truly think of the GMAT or B school admissions process) and treat it as another writing sample that schools can use to evaluate your brilliance and creativity under pressure. Also, if English isn't your first language, it's absolutely going to be leveraged as an additional writing sample.
Data Sufficiency: This isn't math, at least not in the sense that you're used to seeing. What happened to the two trains leaving from separate stations and determining where they'll meet? While that's more problem solving, data sufficiency is important for schools to gauge your decision making abilities when you have limited or inaccurate information In a perfect world, you could make informed decisions with an infinite amount of time and all of the necessary details. But the world isn't ideal, and like the cliché says, time is money. So data sufficiency quantifies what schools want to see: can you discern at what point do you have enough information to make an informed decision or at what point do you not have enough information and need to walk away.
Getting up early/Staying up late/Giving up Happy Hour aka Time Suck: We've all heard of FOMO, or "fear of missing out." You're likely going to have to make FOMO your new BFF while you're preparing. In order to get the score you want, it's important to put forth the effort. Just like training for a marathon or triathlon, you can't take shortcuts or it'll show on race day, and only you truly know the full measure of the effort you're putting forth. So before you even start studying, make sure you're mapping out a 3-4 month window where you know you can truly carve out time on a daily (regular!) basis to prepare, and more importantly, dedicate quality time to preparation.
Expenses!: The GMAT is expensive! And so is preparation! But if you think about it compared to the investment you're about to make in your future and your long-term earnings potential, $250 for the test, $20 in bus fare/gas/transportation, and $50 for a celebratory steak after you crush it is a drop in the bucket. In life, there are absolutely times you should clip coupons, look for a better value and skimp on the extras. This is not one of them. Consider the GMAT the first step in a much larger investment in yourself.
It's not rocket science (if it was, that might be the MCAT, not the GMAT), but it is important to recognize and embrace the challenges of this process. If it was easy, there would be far more individuals taking the GMAT every year (though nearly 250,000 is some decently sized competition). And one day while you're studying, you'll realize that while you don't necessarily love it, the "studying for the GMAT sucks" factor is not quite as strong as it once was. Take that as your reminder to keep your eye on the end game and keep plugging away. Your former self will thank you down the road.
Three years ago this month, the team here at Veritas Prep launched a new project to completely reinvent how we build and administer GMAT practice tests for our students. A home-built system that started with the GMAT Question Bank (launched in October, 2012) soon grew into a whole computer-adaptive testing system containing thousands of questions and employing Item Response Theory to produce some of the most authentic practice tests in the industry. We launched our new practice test in May, 2013, and five months later we made five tests available to everyone. We later added two more tests, bringing the total number to seven that anyone could get. (Veritas Prep students get five additional computer-adaptive tests, for a total of 12.)
The whole time, we kept evaluating the current bank questions (aka "items" in testing parlance), adding new ones, and measuring the ability levels of tens thousands of GMAT students. To date, we have gathered more than 12 million responses from students, and put all of that data to work to keep making our tests better and better. And we keep doing this every week.
Earlier this year, we embarked on a new chapter in the development of our computer-adaptive testing system: We began working with Dr. Larry Rudner, the former Chief Psychometrician at the Graduate Management Admission Council (GMAC), and the definitive authority on the GMAT examination. Dr. Rudner took a look at every aspect of our system, from how we manage our items, to how good each item is at helping our system measure ability levels, to how we employ Item Response Theory to produce an accurate ability level for each test taker. In the end, not only did Dr. Rudner provide us with a roadmap for how to make our tests even better, but he also gave us a great deal of praise for the system that we have now.
After months spent evaluating every aspect of their GMAT practice exams, it's clear that Veritas Prep has mastered the science of test simulation. They offer thousands of realistic questions that have been validated using Item Response Theory and a powerful computer adaptive testing algorithm that closely matches that of the real GMAT® exam. Simply stated, Veritas Prep gives students a remarkably accurate measure of how they will perform on the Official GMAT."
– Lawrence M. Rudner, PhD, MBA. Former Chief Psychometrician at GMAC and the definitive authority on the GMAT exam
Our work on our practice tests will never stop — after all, every month we add new items to our GMAT Question Bank, and many of these questions eventually make it into our computer-adaptive tests — but Dr. Rudner's endorsement is particularly satisfying given the thousands of hours that have gone into building a testing system as robust as ours. When you take this or any practice test (even the official ones from GMAC), keep in mind that it never can perfectly predict how you will perform on test day. But, with Veritas Prep's own practice tests, you have the confidence of knowing that more than three years of hard work and over 12 million responses from other students have gone into giving you as authentic a practice experience as possible.
We plan on putting this system to use in even more places, and helping even more students prepare for a wide variety of exams… That's how powerful Item Response Theory is. Stay tuned!
Finally, we love talking and writing about this stuff. If you're relatively new to studying for the GMAT or understanding how these tests work, check out some of our previous articles on computer-adaptive testing:
When I ask my students how their studying is going, the response is often to give an embarrassed smile, and admit that they just haven't found as much time as they would have liked to devote to GMAT problems. This is understandable. Most of them have full-time jobs. Many serve on the boards of non-profit organizations. Others have young families. Preparing for a test as challenging as the GMAT can often feel like taking on a part-time job, and when piled on top of an already burdensome schedule, the demands can feel overwhelming and unreasonable.
Consequently, whenever they do find time to study, they tend to cram in as much work as they can, forsaking little things like socializing, exercise, and sleep. In an earlier post, I discussed why it can be counterproductive to engage in marathon study sessions, so in this one, I want to explore strategies for consistently finding small blocks of time so that our study regimens will be less painful and more productive.
The good news is that while we all feel incredibly busy, research shows that, in actuality, we're a good deal less saturated with responsibilities than we think we are. In Overwhelmed: Work, Love, and Play When No Has the Time, Brigid Schulte discusses how our sense of having too much to do is, in a sense, a self-fulfilling prophesy. When we feel as though there's too much to do, we tend to procrastinate, and part of this procrastination involves lamenting to others about how overwhelmed we are. Of course, while we're complaining about our busy schedules, we're not exactly models of productivity, and so we fall even further behind, which compounds our overriding sense of helplessness, compelling us to complain even more, a cycle that deepens as it perpetuates itself.
So then, how do we break this cycle?
First, we need to identify the biggest productivity-killers that trigger our procrastination tendencies in the first place. It will surprise no one to hear that email is a major culprit. What is surprising, at least to me, is how much of our idea was devoted to responding to emails. According to a study conducted by Mckinsey, we spend, on average, 28% of our workdays on email.
If you're working a 10-hour day, as many of my students are, that's nearly three hours of pure email time. If they can cut this down to 2 hours, well, that's an hour of potential GMAT study time. A few simple strategies can accomplish this. This Forbes article offers some excellent advice.
The most salient recommendations are pretty simple. First, set up an auto-responder. Unless an email is urgent, the sender will not expect to hear back from you right away. Second, get in the habit of sending shorter emails. If complicated logistics are involved, make a phone call rather than going back and forth over email. Also, make judicious use of folders to prioritize which messages are most important. And last, do not, under any circumstances, send an email that is mostly about how you don't have any time to do things like, well, sending recreational emails.
Next, during those times when we'd otherwise have been on our phones complaining how much we have to do, we can instead use our phones to sneak in a bit of extra study time. Many of my students take the subway or commuter rail to work. While I don't expect anyone to crack open their GMAT books in this environment, there's no reason why they can't use a good app on their phones to sneak in a good 20-minute session each day. And if you were wondering, yes, Veritas Prep has an excellent app for precisely such occasions.
The hope is that simple strategies, like the ones outlined above, will allow you to make your study regimen both consistent and manageable, diminishing the need to over-study when you finally have a block of free time on the weekend. If you're able to do something more restorative on the weekend and feel refreshed when you begin the following work week, you'll find you'll be more productive that week and more inclined to stick with your study plan without running the risk of burnout. In time, you'll feel less busy, and paradoxically, will be able to get more done.
In the last two classes I've taught, I've had students come up to me after a session to ask about the value of brain-training exercises. The brain-training industry has been getting more attention recently as neuroscience sheds new light on how the brain works, baby-boomers worry about cognitive decline, and companies offering brain-improvement software expand. It's impossible to listen to NPR without hearing an advertisement for Lumosity, a brain-training website that now boasts 70 million subscribers. The site claims that the benefits of a regular practice range from adolescents improving their academic performance to the elderly staving off dementia.
The truth is, I never know quite what to tell these students. The research in this field, so far as I can tell is in its infancy. For years, the conventional wisdom regarding claims about brain-improvement exercises had been somewhat paradoxical. No one really believed that there was any magic regimen that would improve intelligence, and yet, most people accepted that there were tangible benefits to pursuing advanced degrees, learning another language, and generally trying to keep our brains active. In other words, we accepted that there were things we could do to improve our minds, but that such endeavors would never be a quick fix. The explanation for this disconnect is that there are two different kinds of intelligence. There is crystalized intelligence, the store of knowledge that we accumulate over a lifetime. And then there is fluid intelligence, our ability to quickly process novel stimuli. The assumption had been that crystallized intelligence could be improved, but fluid intelligence was a genetic endowment.
Things changed in 2008 with the release of a paper written by the researchers Susanne Jaeggi, martin Buschkuehl, John Jonides, and Walter Perrig. In this paper, the researches claimed to have shown that when subjects regularly played a memory game called Dual N-Back, which involved having to internalize two streams of data simultaneously, their fluid intelligence improved. This was ground-breaking.
This research has played an integral in role in facilitating the growth of the brain-training industry. Some estimates put industry revenue at over a billion dollars. There have been articles about the brain-training revolution in publications as wide-ranging as The New York Times and Wired. This cultural saturation has made it inevitable that those studying for standardized tests occasionally wonder if they're shortchanging themselves by not doing these exercises.
Unfortunately, not much research has been performed to assess the value of these brain-training exercises on standardized tests. (A few smaller studies suggest promise, but the challenge of creating a true control group makes such studies extraordinarily difficult to evaluate). Moreover, there's still debate about whether these brain-training exercises confer any benefit at all beyond helping the person training to improve his particular facility with the game he's using to train. Put another way, some say that games like Dual N-Back will improve your fluid intelligence, and this improvement translates into improvements in other domains. Others say that training with Dual N-Back will do little aside from making you unusually proficient at Dual N-Back.
It's hard to arrive at any conclusion aside from this: the debate is seriously muddled. There are claims that the research has been poorly done. There are claims that the research is so persuasive that the question has been definitively answered. Obviously, both cannot be true. My suspicion is that the better-researched exercises, such as Dual N-Back, confer some modest benefit, but that this benefit is likely to be most conspicuous in populations that are starting from an unusually low baseline.
This brings us to the relevant question: is it worth it to incorporate these brain-exercise programs into a GMAT preparation regime? The answer is a qualified 'maybe.' If you're very busy, there is no scenario in which it is worthwhile to sacrifice GMAT study time to play brain-training games that may or may not benefit you. Secondly, the research regarding the cognitive benefits of aerobic exercise, mindfulness meditation, and social interaction is far more persuasive than anything I've seen about brain-training games.
However, if you're already studying hard, working out regularly, and finding time for family and friends, and you think can sneak in another 20 minutes a day for brain-training without negatively impacting the other more important facets of your life, it can't hurt. Just know that, as with most challenging things in life, the shortcuts and hacks should always be subordinated to good, old-fashioned hard work and patience.
I like to arrive to my Monday evening classes a good half hour early so that I can spend some time talking to my students about how they spent their weekends. It helps me to get to know them, and it allows me to get a sense of the rhythm of their days. Some of my students do interesting things. They travel. They ski in the winters. They rock-climb when it's warmer. But, unfortunately, they're a minority.
The most common response is some variation of: I studied for the GMAT. Of course, they should be doing some studying, and I hope that this studying is at times enjoyable. But if an unusually satisfying Data Sufficiency problem is the highlight of your weekend, something is profoundly out of whack in your study-life balance. And yes, at times comments about studying all weekend are exaggerated for comic effect, but I think there is a distressing truth captured in these exchanges: people are so busy and overwhelmed during the week that they end up spending an unhealthy amount of time cramming for the GMAT on the weekends.
This isn't good.
It isn't good for the students' physical or psychological wellbeing; and research is beginning to show that over-studying might be bad for performance as well.
According to one study performed by Stanford University, academic performance for high school students began to deteriorate once the students' workloads exceeded two hours of homework per night. Now, there's nothing magic about the figure of two hours – one imagines that people vary in terms of stamina levels, motivation, etc. – but this notion, that doing too much work not only will fail to help you, but also will actively stymie your efforts, is one well worth considering. And though this study involved high school students, there's no reason to believe that this phenomenon wouldn't hold for adults preparing for the GMAT. When we overexert ourselves in any capacity, be it physical training, work in the office, or studying, our performance tends to suffer.
I suspect that the most important factor is that if we're studying too much, there are other beneficial things that we're not doing. Put another way, if the benefits of additional studying begin to decrease once you've been at it for a few hours, wouldn't it make more sense to use this time to engage in other activities that would not only be more enjoyable but could actually boost your score beyond what more study time could accomplish?
1. Exercise
The first, and most obvious consideration is that when we study, we're typically inactive. (My apologies to anyone who is reading this at their treadmill desk.) The research on the benefits of aerobic exercise on academic performance is unambiguous. Aerobic exercise prompts the brain to generate, not just fresh neural connections, but new neurons, a phenomenon that was considered a physical impossibility as recently as 20 years ago. Students who exercise do better, on average, than those who don't. We've been touting the benefits of exercise at Veritas Prep for years. There's no reason not to have exercise be a part of your routine. (This is to say nothing of the whole feeling better, being healthier, and living longer perk).
2. Conversation
The second, and perhaps more surprising finding, is that socialization can boost intelligence. One study, conducted by the University of Michigan, found that as little as 10 minutes of conversation can boost working memory. Moreover, they found that the total amount of socialization in one's day was positively correlated with performance on a variety of cognitive tests. (If you've been studying for Critical Reasoning, hopefully, you've taken a moment to object that correlation isn't necessarily causation. Yes, you say: it's possible that socializing causes our brains to work better; but isn't it also possible that when our brains are functioning optimally, we're more likely to seek out opportunities for socialization? Not to worry. The experiments were designed to see what happened to a given group that socialized before taking a test, and what happened to that same group when they hadn't socialized. When controlling for extraneous variables, socialization still had a robust impact on performance.)
Of course, one shouldn't take any of this research to mean that preparing for this test won't require a significant time investment. It will. But if you study so much that you stop taking care of yourself and neglect your personal relationships, you will not only make yourself unhappy, you'll be artificially limiting your intellectual potential. So yes, do those few hours of Data Sufficiency questions. Take a four-hour exam on another day. Just make sure that you're also taking time to go for a run or to play tennis or to see friends. You'll be happier and less likely to burn out. And the fact that you're also likely to do better on the exam with this approach is about as good an ancillary benefit as you're likely to find.
Veritas Prep is excited to announce a scholarship opportunity to help you achieve your target GMAT score! We've partnered with the National Society of Hispanic MBAs to offer 100 GMAT preparation courses to qualifying applicants completely free of charge!
A good GMAT score is crucial when applying to business school, and we want to help you succeed. Our GMAT courses are available in over 90 cities worldwide, and also online using new Smartboard technology.
Every GMAT course comes with the following:
36 hours of live instruction
An instructor who scored in the 99th percentile on the actual GMAT
12 lesson booklets
12 computer-adaptive practice tests
Live instructor help seven days a week
Veritas Prep GMAT on Demand pre-recorded lesson videos
3,000 GMAT practice problems and solutions
To learn more about this scholarship and how to apply, visit the NSHMBA website. The deadline to apply is May 8th, so if you're thinking about taking the GMAT, submit your application today!
We're excited to get you moving on your next step towards graduate school!
Here on the first Friday of April, we've officially ended the first quarter of the year and fiscal reports are streaming in. But who's in a hurry to finish 2015?
We're still firmly entrenched in the first third of the year, and if 2015 is the year that you plan to conquer the GMAT you're in luck. Why?
Because your GMAT study plan should include three phases:
1) Learn
One of the most common mistakes that GMAT studiers make is that they forget that they need to learn before they can execute. Are you keeping an eye on the stopwatch on every question you complete? Are you taking multiple practice tests in your first month of GMAT prep? Have you uttered the phrase "how could I ever do this in two minutes???"? If so, you're probably not paying nearly enough attention to the learning phase. In the learning phase you should:
Review core skills related to the GMAT by DOING them and not just by trying to memorize them. You were once a master of (or maybe a B-student at) factoring quadratics and identifying misplaced modifiers and completing long division. Retrain your mind to do those things well again by practicing those skills.
Learn about the GMAT question types and the strategies that will help you attack them efficiently. For this you might consider a prep course or self-study program, or you can always start by reviewing prep books and free online resources.
Take as much time as you need to complete and learn from problems. You'll learn a lot more from struggling through a problem in six minutes than you will from taking two minutes, giving up, and then reading the typewritten solution in the back of the book. Let yourself learn! Again, it's critical to learn by doing – by actively engaging with problems and talking yourself into understanding – than it is to try to memorize your way to success. The stopwatch is not your friend in the first third of your preparation!
Embrace mistakes and keep a positive attitude. The GMAT is a hard test; most people struggle with unfamiliar question formats (Data Sufficiency, anyone?) and challenging concepts (without a calculator, too). Recognize that it will take some time to learn/re-learn these skills, and that making mistakes and thinking about them is one of the best ways to learn.
2) Practice
Regardless of how you've studied, you'll need to complete plenty of practice to make sure you're comfortable implementing those strategies and using those skills on test day. Once you've developed a good sense of what the GMAT is testing and how you need to approach it, it's time to spend a few weeks devouring practice problems. Among the best sources include:
In this phase, you can start concerning yourself with the stopwatch a little and you'll want to identify weaknesses and common mistakes so that you can emphasize those. Particularly with GMAT verbal, the more official problems you see the more you develop a feel for the style of them, so it's important to emphasize practice not just for the conscious skills but also for those subconscious feelings you'll get on test day from having seen so many ways they'll ask you a question.
3) Execute
Before you take the GMAT you should have taken several practicetests. Practice tests will help you:
Work on pacing and develop a sense for how much time you'll need to complete each section. From there you can develop a pacing plan.
Determine which "silly" mistakes you tend to make under timed pressure and exam conditions, and be hyperaware of them on test day.
Develop the kind of mental stamina you'll need to hold up under a 4-hour test day. Verbal strategies can be much easier to employ in a 60-minute study session than at the end of a several-hour test! Make sure that at least a few times you take the entire test including AWA and IR for the first hour.
Continue to see new problems and hone your skills.
While it's not a terrible idea to take a practice test early in your study regimen and another partway through the Practice phase, most of your tests should come toward the end of your study process. Why? Because the learning and practice phases are so important. You can't execute until you've developed the skills and strategies necessary to do so, and you won't do nearly as effective a job of gaining and practicing those if you're not allowing yourself the time and subject-by-subject focus to learn with an open mind.
So be certain to let yourself learn with a natural progression via the GMAT Study Rule of Thirds. Learn first; then focus on practice; then emphasize execution via practice tests. Studying in thirds is the best way to ensure that you get into a school that's your first choice.
The Veritas Prep program allowed me to reach my GMAT goals and re-learn all of the quantitative skills that I had forgotten over the past several years. I am an Army veteran, six years out of college, and Veritas Prep was the perfect program to teach me the skills I needed to succeed on the GMAT. I am thankful for the quality of the curriculum, and also very appreciative of the generous scholarship from Veritas Prep through the Service2School organization. Throughout the self-study lessons, I could always count on the on-demand videos to deliver engaging, thoughtful content and guide me through the lesson of the day. I particularly enjoyed Brian's humorous references (the "alge, brah" joke stands out): The human element to the videos definitely helped me to remember many topics and leverage them on test day.
My goal was a score over 700, and I knew that I needed a structured, high-quality program to help me to get a top 10% score. After looking at several programs, Veritas Prep stood out as the one that would work for me. On day one of the program, I was contacted by Colleen Hill, who told me how to get started and offered her time for any questions I had throughout the course. I have to admit, I did not expect an actual person to contact me; it was a pleasant surprise! Upon receiving my materials in the mail and logging on to check out the online resources, I was again impressed by the quality of the materials. I found that I was more and more excited to begin the course. With everything organized and a thirty-day plan ahead of me, I began the course.
The curriculum was demanding, as I worked through it over a thirty-day period, and well-balanced to where I didn't feel that I was ever losing ground in either quant or verbal. While working through the lessons, I could also always take comfort in the fact that if I didn't understand a specific question, I could use the online homework help as a resource. Homework was challenging, which was great, and I found the explanations covered anything that I had missed when it came to why the correct answer was right, and why the wrong answers looked tempting.
When my test day finally came, I felt confident. I felt that the Veritas Prep practice tests had provided a very accurate measure of the difficulty of questions that I faced. Throughout the test, I remembered the lessons, always looking for logical ways to answer the question and leveraging a mastery of the content. I felt calm and confident throughout the test, and when I finished I had a 710. I am ecstatic at that score, especially as it is my first attempt, and I can attribute it to nothing but the exceptional quality of the Veritas Prep curriculum.
For someone who is looking for high-quality, comprehensive preparation for the GMAT, Veritas Prep should definitely be their first choice. I want to give a sincere "thank you" to Colleen and the rest of the Veritas Prep organization; you have helped me get a head start on my journey towards an MBA.
At some point during the first session of each new class I teach, I'll write my phone number on the board and mention that I take emergency calls. When I first started doing this, I figured that every now and again I'd get a call from a frantic student the night before the exam because he or she was running through some practice problems and was stumped on a concept that had previously been clear. I could then talk the student through a concept or strategy as a kind of pre-test boost. It turns out, these emergency calls happen far more often than I'd suspected, and they're never about content. They're always about anxiety. And the refrain is always the same. "When we're doing the questions in class, I understand them. When I'm working on my own with no pressure, I'm fine. But when I see the timer…" The implications are clear: the issue often isn't the content of the question, but the psychological mindset of the test-taker when he encounters it.
In fact, the link between anxiety and standardized testing is so prevalent that a Google search of 'test anxiety' yields well over 100,000,000 results. You want to make a parent nervous? Say something about Common Core. Want to freak out a high school student? Invoke the SATs. And if you're reading this article, you are likely well acquainted with the pernicious effects that the GMAT can have on the 'ol nervous system. It isn't hard to see why. These tests not only have tangible academic and professional consequences that can reverberate for years, but they shape our fundamental self-perceptions. Someone who scores in the 98th percentile on a standardized test will, no matter what he says, walk out of that test feeling different about his abilities than someone who scores in the 7th percentile, despite the fact that there are literally dozens of variables in play that have little or nothing to do with underlying intelligence. (And this supposes that there is such a thing as underlying intelligence, as opposed to a host of complexly intersecting domains of intelligence, all of which may be difficult to measure with any kind of accuracy or consistency.) This is all to say that testing anxiety is both real and inevitable. It's impossible to talk about preparation for an exam like the GMAT without addressing it.
Though this connection isn't new, much of the science behind how the brain works under pressure is quite novel, and as we learn more, this knowledge will invariably seep into how teachers and tutors prepare their students for the exam.
First, consider the physiological process by which stress makes it make more difficult to perform well on exams. We enter what psychologists call a threat state. Here is a relevant quote from Barry Mendes, an associate professor of psychology from UC San Francisco, culled from a New York Times article on the subject. (The article is itself well worth a read).
"The hallmark of a threat state is vasoconstriction — a tightening of the smooth muscles that line every blood vessel in the body. Blood pressure rises; breathing gets shallow. Oxygenated blood levels drop, and energy supplies are reduced. Meanwhile, a rush of hormones amplifies activity in the brain's amygdala, making you more aware of risks and fearful of mistakes."
And it turns out that the physiological processes in play are even more complicated than we've thought. Recent research has revealed that there is a gene that codes for the speed at which enzymes remove dopamine from various regions in the brain. Some remove dopamine quickly. Others remove it more slowly. In and of itself, this isn't terrible interesting, but what is fascinating, and relevant to this discussion, is that those who had the gene that coded for the enzymes that removed dopamine more slowly did better than the other group on IQ tests in normal conditions, but worse than the other group on tests with significant time constraints. In other words, the gene that makes you smarter in a low stress environment causes you to underperform in a stressful situation. Suddenly, we have a scientific explanation for the dozens and dozens of students I've had over the years who maintained a 3.9 GPA in college, but could not, for the life of them, understand why they struggled on standardized tests!
The implications from the above discussion may sound fairly straightforward. Stress is bad. It can hurt test performance. But it isn't that simple. It turns out that stress is one of those maddeningly elusive phenomena that we actually alter by focusing our attention on it. (Fans of quantum mechanics will recognize this as a version of the Observer Dilemma. In the quantum world, observing a particle alters the very characteristics we're attempting to observe, so there's no way to derive uncontaminated data. Scientists and philosophers have been puzzling over this for the better part of a century, and the phenomenon is no less strange now than it was when it was codified). This is best illustrated by a study conducted at Harvard. Half of the subjects were simply told that the purpose of the study was to examine the effect of anxiety on test-taking. The other students, however, were told that the anxiety during a test could actually boost performance. Sure enough, the group that was told that anxiety could boost performance did significantly better than the control group.
In other words, when we think stress is bad for us, it is. And when we think stress can be beneficial, it is. How we frame the issue in our minds has a direct and material impact on our response to trying conditions.
Moreover, there are things we can do to improve our performance in stressful situations. Pilots, for example, will practice dealing with artificial problems during test runs, and this practice yields benefits when these same problems happen during commercial flights. I'll often encourage students to create a simulated stressful environment during a practice exam so that if a similar situation should befall the student during the real test, she'll have an experience to draw on when attempting to adapt. For example, you can allow 10 minutes to elapse during a practice test so that if there is a time crunch on the real test, you'll have already practiced how to address this potential crisis.
Last, you can practice mindfulness in the weeks leading up to the exam. A study performed last year demonstrated that students who began a mindfulness practice for only two weeks demonstrated improvements in working memory and concentration, benefits that translated to significantly higher scores on standardized tests. (The students in the study took the GRE, but there's every reason to believe that mindfulness meditation would confer comparable benefits on the GMAT.) Here is an article distilling the main points of the study.
There is no avoiding stress on test day, but there is a lot we can do to reshape how we perceive this stress, and this reshaped perception can actually serve to improve our performance.
Takeaways:
Remind yourself that stress is not inherently bad. It can be a source of energy and focus that you can harness. Moreover, your belief in the bracing qualities of stress can be a self-fulfilling prophecy. Repeat that to yourself like mantra: stress can be helpful, but only if we tell ourselves so.
Simulate stressful conditions when taking practice tests so that these situations will be less alarming should they happen during the actual exam.
Consider starting a consistent mindfulness practice. The research indicating that mindfulness can boost test scores is promising, and the tangential health benefits are enormous.
For some time now, Veritas Prep team member Ravi Sreerama has been regarded as the best GMAT instructor in the industry (see for yourself!) Whether he's leading GMAT courses in Los Angeles or training students worldwide in our Next-Generation Live Online GMAT Course, Ravi keeps growing his legion of loyal followers. They want to score in the 99th percentile on the GMAT, and Ravi knows how to help them do it.
No, for the first time ever, Ravi will take his show on the road: Starting March 29, Ravi will lead a seven-day Immersion Course in New Delhi! Our Immersion Course format is entirely unique — you cover all 36 hours of the traditional Veritas Prep Full Course GMAT curriculum, but do so over seven straight days. Six of those days feature six hours of GMAT instruction each, with a break in the middle of the week.
The schedule is as follows:
Sunday: Foundations of GMAT Logic & Arithmetic
Monday: Critical Reasoning, Algebra
Tuesday: Sentence Correction, Geometry
Wednesday: Review Session and Office Hours
Thursday: Reading Comprehension, Data Sufficiency
Friday: Advanced Verbal Strategy, Statistics and Combinatorics
Saturday: Word Problems, AWA & Integrated Reasoning
Pay special attention to that Wednesday schedule — that day is dedicated to review and to office hours in which you can get one-on-one GMAT coaching from Ravi. Need to catch up? Stuck on a particular area? Have specific questions that you've been saving to ask a GMAT expert? Wednesday is when you can take advantage of Ravi being in New Delhi to brush up on the skills that matter most to you.
And, of course, you get all of the other benefits of being in a Veritas Prep Immersion Course, including the camaraderie that comes from spending seven days with a group of like-minded, ambitious GMAT students. You also receive:
36 hours of live, instructor-led class time
12 GMAT lesson booklets
12 computer adaptive practice tests
Online student account with study plan
3,000 practice problems and solutions, including video
Live homework help 7 days a week for a year
Every lesson pre-recorded in HD for review
Hurry… March 29 is coming quickly! Learn more about Ravi Sreerama's New Delhi GMAT course, and enroll as soon as you can so that you're ready when class starts on the 29th!
The GMAT presents several challenges for test takers. For many people, the issues are focused around aptitude and the ability to simply get answers right. For others, timing is a big challenge. The GMAT is as much a test of mental endurance as it is an aptitude test.
With over 90 questions in 3+ hours the GMAT requires test takers to not only answer questions correctly, but to also do so quickly. In a vacuum many test takers could answer most GMAT questions correctly under normal conditions but the time constraints imposed by the GMAT make this one of the toughest standardized tests for graduate education.
All hope is not lost however; let's discuss a few ways to prepare for the GMAT that will pay dividends on the timing front on test day.
Problem Sets
Every practice question you solve should be timed based on the average time you will have per question on the exam. Answering questions under unrealistic time scenarios does little to improve your performance especially if you are already struggling with pacing. Take the questions in sets (1, 5, 10, 20, etc.) and have your phone or stopwatch handy to make sure you are comfortable answering questions in realistic time constraints. If you are a Veritas Prep GMAT student, the Problems tab in your online account allows for the timed answering of homework questions!
Practice Exams
Too often test takers don't start taking practice exams until too close to their test date. Practice exams are an integral part of your test prep game plan. I recommend taking your practice test at a similar time of day as your test date, if possible. If your test date is Saturday morning make sure you are taking practice tests on Saturday mornings. This is a good way to get your body synced up with the physical and mental side of taking such a long and difficult test. Once you take the test make sure you are including some time for review. Getting a problem wrong can be even more valuable than getting a problem right, focus on learning from your mistakes. You should spend a considerable amount of time figuring out why you got a problem wrong so you will never get a similar problem wrong again.
Problem Recognition
For most test takers who struggle with pacing, you will also want to work on problem recognition. Pacing is about quickly identifying the question type as well as how to approach it and then answering it quickly. Spend some time finding ways to quickly identify different question types and how to approach them. Finally, be able to move on if you realize that you don't have a strong chance answering the question accurately in a reasonable amount of time. Spending an exorbitant amount of time on a question you will eventually get wrong is a death sentence on the GMAT; so don't be afraid to move on after making an educated guess.
Dozie A. is a Veritas Prep Head Consultant for the Kellogg School of Management at Northwestern University. His specialties include consulting, marketing, and low GPA/GMAT applicants. Find more of his articles here. | 677.169 | 1 |
Get a set of 5 sample papers for the CBSE Class X Mathematics SA – I exam in the form of eBook. These papers will be helpful for the students of CBSE board class 10 who are going to appear for the coming SA – I mathematics exam.
The number of students scoring a perfect 10 CGPA score in CBSE board are increasing year after year. But, scoring a perfect 10 CGPA is not a cup of tea and one can't score 10 CGPA by studying overnight or a day before the exam. Here is full proof study plan for the complete year which will help you to score a perfect 10 CGPA.
For students pursuing their school education, Class 10 and Class 12 board examination are very vital. Their performance in these two examinations will shape the path of their academic as well as professional career in the future.
CBSE Syllabus of Class 10 for the academic session 2016-17 is released by CBSE Board. With this article, students can view and download the new syllabus of CBSE Class 10 Foundation of Information Technology.
NCERT Textbooks and NCERT Solutions of CBSE Class 10th Mathematics, Science, English, Social Science and Hindi are available in cbse.jagranjosh.com. With this article, you will get the download links of NCERT Textbooks and NCERT Solutions.
NCERT textbooks,Kshitiz (Part 2) and Kratika (Part 2) are prescribed by CBSE board for Class 10 Hindi (Course A). With this article, students can get the free download for the Chapter 11 Baalgobin Bhagat.
NCERT has published two books for CBSE Class 10 Hindi (Course A) and they are, Kratika (Part 2) and Kshitiz (Part 2). You can download Chapter 6 of NCERT Hindi textbook Kshitiz (Part 2) from this article. | 677.169 | 1 |
San Leandro PrecalculusKwa D.
...In terms of applied geometry students know how to compute the surface area and volume of basic three-dimensional objects and understand how area and volume change with a change in scale. Students make conversions between different units of measurement. Undoubtedly, Precalculus course prepares for students entering into Calculus-the mathematics of changes | 677.169 | 1 |
This fifth edition of Lang's book covers all the topics traditionally taught in the first-year calculus sequence. Divided into five parts, each section of A FIRST COURSE IN CALCULUS contains examples and applications relating to the topic covered. In addition, the rear of the book contains detailed solutions to a large number of the exercises, allowing them to be used as worked-out examples -- one of the main improvements over previous editions.
Top Customer Reviews
Serge Lang's text does an effective job of teaching you the skills you need to solve challenging calculus problems, while teaching you to think mathematically. The text is principally concerned with how to solve calculus problems. Key concepts are explained clearly. Methods of solution are effectively demonstrated through examples. The challenging exercises reinforce the concepts, while enabling you to develop the skills required to solve hard problems. Answers to the majority of exercises (not just the odd-numbered ones) are provided in a hundred page appendix, making this text suitable for self-study. In some sections, such as related rates and max-min problems, Lang provides many fully worked out solutions.
As effectively as Lang conveys the key concepts and teaches you how to solve problems, he does not neglect the subject's logical development. Topics are introduced only after their logical foundations have been laid. Results are derived. Theorems are proved when Lang feels that they will add to the reader's understanding. Through his exposition and his grouping of logically related exercises, Lang teaches the reader how a mathematician thinks about the subject.
The book is divided into five sections: review of basic material, differentiation and elementary functions, integration, Taylor's formula and series, and functions of several variables. The heart of the course is the middle three sections.
Most of the topics covered in the review of basic material should be familiar to most readers.Read more ›
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This book by the late Prof. Lang covers calculus in a clear and concise manner. I own more than a few calculus books and this book is one of my favorites. The book looks like a math book in that it is not a 1200 page glossy coloring book with multi-colored inserts on every page. I think that the style of this book is a hugh improvement over most of the books on the market. I think a student who buys this book along with a good calculus study guide would be very well set.
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I had to take a refresher calculus course as a prerequisite to get into graduate school, but the assigned text (Edwards and Penney) was horrible. Like every other mass market calculus, it was filled with colorful diagrams and digressions on how to use calculators, but little in the way of explanation. Fortunately I found Lang's calculus in the university book store and it cured all of my problems. Unlike the bloated E&P, Lang's book is clear and concise. E&P covers more material to be sure, but for the essentials nothing beats Lang. After reading this book calculus became easy for me again. Which is as it should be, since calculus is a surprisingly simple subject if expalined well.
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I needed to bring my high school calculus up to speed for first year physics studies and found this to be the only book which covered the necessary ground. The material is presented in a thorough manner with the great majority of topics shown with proofs. The book is very well organized and there are abundant worked examples. Some problems are offered which deal with matters not covered in the text, but usually there is a worked example given among the answers. Lang deals with the material in a clear fashion so that the subject matter is usually not difficult to follow.On the negative side I can say that there is no human touch between the covers. His sole attempt at humor is an item following a list of problems in which he notes "relax". In the foreword he exhibits his firm belief that many freshmen arrive unprepared for college calculus, which may be true. But nowhere in the book is there a note of encouragement, so it cannot be described as reader friendly. Finally the index is pathetic--just three pages for a book of 624 pages, so that finding things can be frustrating.
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The book is OK, but I wouldn't say it's great. There are lots of exercises that ask you to do simple symbolic manipulation so you'll remember rules -- but there are too few exercises that require the reader to actually think harder and be creative. The explanations are often shallow and not as stimulating as they could be, in my opinion.
Some examples of sections that I think are not well written are the one about implicit differentiation (the discussion is too short and not clear, and there are less exercises in this section than in others); the one about rate of change (some examples are boring, like "find the rate of change of the area of a circle given the rate of change of its diameter"; he does not make it clear that he's always derives with relation to time and that, for example, the radius and height of a cylinder should be understood as functions of time, so there's a feeling of sloppiness).
It's a good book,anyway. Now, it becomes a really great book when compared to the colorful, flashy books available today.
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First, this text is not the most comprehensive and does lack rigor. Specifically the author avoids taking the examples beyond the most basic and there is no progression of building on examples to eventually illustrate the harder cases.
However, in the context of being a "first course" it fits that goal 100%. This book is why using older textbooks is actually a good idea. The organization is different than modern texts, which might prep for SAT's, but they are not always great for novice students.
The downside to older textbooks is that the format is usually different in how they approach various topics. The treatment of limits in this book is similiar to other older texts I have. Even Morris Kline's book seems to follow this pattern.
So a high school student would probably not be ready for the SAT's with this book, but they will at least learn Calculus in a very straightforward manner.
For self study or adult learners this book is better than any "dummies" book. ( not better than Kline's Calc book ) but better than stewart or larson for getting back in touch or learning Calculus for the first time.
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the last three or four years, I have recommended this book to my students' parents. I teach algebra (grade 8) and pre-algebra (grades 7 and 8) and have found that his book is written in a non-threatening, easy to understand manner. It is a perfectly wonderful reference book. Math is scary enough...this book demystifies algebra. Love it!
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This is the best reference handbook that parents and teachers will fine it useful for students. Algebra to Go, "demystifies algebra for students in 8 grades and up by explaining key and often complex math topics in a way that's clear, friendly, and understandable". It is also a great refresher for parents who may have forgotten a thing or two about Algebra..
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There are very few review books on todays new math for middle school children. This book is the perfect review and teaching aid for the math techniques used today. A must have for parents who find themselves at a loss to help their kids with math homework.
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Wonderful book if you're trying to catch up on some Algebra, I love the breakdown almost any person can follow along with basic arithmetic knowledge base. Helped me for my TEAS exam! Definite recommendation!
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ALGEBRA TO GO WAS A BOOK MY GRANDDAUGHTER SAID HER TEACHER HAD. IT HAS HELPED TO HAVE A COPY AT HOME IT IS SO MUCH EASIER TO LOOK UP JUST CERTAIN THINGS IN THE BACK AND BE ABLE TO FIND THEM SO MUCH FASTER.
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I got this for my 8th grade daughter taking Honors Algebra and it has really helped her. These days in our district, math students do not receive a text book so there is no easy way to review the lesson if something is misunderstood. My daughter could only rely on her notes which she copied from the board but could not easily understand what was written. There was no explanation to accompany homework until now. Most topics she has covered so far are in Algebra to Go. She can now do most of her homework by herself and she feels proud that she doesn't need anyone's help. I will be ordering Geometry to Go for next year. Thanks!
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What a great little book to supplement my middle schooler's math year. Every topic that he will study is in here written in very simple english that a middle school boy has no difficulty understanding. His text book is great but this little book is just perfect to cement the ideas that have already been taught in school. Plus - the ideas are presented in just a few pages so if he only needs to check on one or two things, he doesn't have to read it all.
Furtermore, my mother in law is a retired high school math teacher who tutors now and she is considering buying this for the students that she tutors. While they may be farther down the road than my middle schooler, she thinks that the basics are presented well so that the student has a good foundation for future learning.
I hope for good learning this year!
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The book is in excellent condition, no wear and tear at all. The Library stamp on the inside of the front page is the only thing that make the book an "used" book, other than that it is perfect as it would be on the shelf of a bookstore. | 677.169 | 1 |
This book is a detailed study of Gottfried Wilhelm Leibniz's creation of calculus from 1673 to the 1680s. We examine and analyze the mathematics in several of his early manuscripts as well as various articles published in the Acta Eruditorum. It studies some of the other lesser known "calculi" Leibniz created such as the Analysis Situs,...
Designed for those seeking help studying calculus in school - also valuable for adults attempting to learn/re-learn calculus. A resource for instructors supplementing their instruction. 501 Calculus Problems helps users prepare for academic exams and build problem-solving skills. Unlike textbooks, full answer explanations are provided for all problems.... more... | 677.169 | 1 |
Gauss elimination
Gauss elimination, in linear and multilinear algebra, a process for finding the solutions of a system of simultaneous linear equations by first solving one of the equations for one variable (in terms of all the others) and then substituting this expression into the remaining equations. The result is a new system in which the number of equations and variables is one less than in the original system. The same procedure is applied to another variable and the process of reduction continued until there remains one equation, in which the only unknown quantity is the last variable. Solving this equation makes it possible to "back substitute" this value in an earlier equation that contains this variable and one other unknown in order to solve for another variable. This process is continued until all the original variables have been evaluated. The whole process is greatly simplified using matrix operations, which can be performed by computers.
Learn More in these related articles:
...algebra, as large problems contain many rounding errors.Numerical analysts are generally interested in measuring the efficiency (or "cost") of an algorithm. For example, the use of Gaussian elimination to solve a linear system Ax = b containing n equations will require approximately 2n3/3 | 677.169 | 1 |
Mathematics Study Guides
Take responsibility for studying, recognizing what you do and don't know, and knowing how to get your Instructor to help you with what you don't know Attend class every day and take complete notes. Instructors formulate test questions based on material and examples covered in class as well as those in the text. Be an active participant in the classroom. Get ahead in the book; try to work some of the problems before they are covered in class. Anticipate what the Instructor's next step will be. Ask questions in class! There are usually other students wanting to know the answers to the same questions you have. Go to office hours and ask questions. The Instructor will be pleased to see that you are interested, and you will be actively helping yourself. Good study habits throughout the semester make it easier to study for tests.
Studying Math is Different from Studying Other Subjects
Math is learned by doing problems. Do the homework. The problems help you learn the formulas and techniques you do need to know, as well as improve your problem-solving prowess. A word of warning: Each class builds on the previous ones, all semester long. You must keep up with the Instructor: attend class, read the text and do homework every day. Falling a day behind puts you at a disadvantage. Falling a week behind puts you in deep trouble. A word of encouragement: Each class builds on the previous ones, all semester long. You're always reviewing previous material as you do new material. Many of the ideas hang together. Identifying and learning the key concepts means you don't have to memorize as much.
College Math is Different from High School Math
A College math class meets less often and covers material at about twice the pace that a High School course does. You are expected to absorb new material much more quickly. Tests are probably spaced farther apart and so cover more material than before. The Instructor may not even check your homework.
Take responsibility for keeping up with the homework. Make sure you find out how to do it. You probably need to spend more time studying per week - you do more of the learning outside of class than in High School. Tests may seem harder just because they cover more material
Study Time
You may know a rule of thumb about math (and other) classes: at least two hours of study time per class hour. But this may not be enough!
Take as much time as you need to do all the homework and to get complete understanding of the material. Form a study group. Meet once or twice a week (also use the phone). Go over problems you've had trouble with. Either someone else in the group will help you, or you will discover you're all stuck on the same problems. Then it's time to get help from your Instructor. The more challenging the material, the more time you should spend on it.
Mathematics: Guidelines for Study
The University of Alabama Center for Teaching and Learning 124 Osband 348-5175 As a first step to learning mathematics, it is important to realize that mathematics is easier to study in small doses. While this statement is true of almost any subject, it is particularly true of mathematics. Two hours a day is a lot more productive than 10 hours one day a week. Although you may be able to read two novels in one weekend for a literature course, it is almost impossible to catch up on two weeks of math in one weekend. The study of math is cumulative with concepts building on those previously learned. You also need "soak time", a chance to think about concepts and ideas before another is presented. Second, mathematics is not a spectator sport; it is a do-it-yourself subject. You must work the problems for yourself and recognize that there is no easy road to success. The following guidelines, however, should be helpful to you in studying math. These techniques are related to previewing, note taking, text reading, problem solving, and problem analysis. 1. Previewing Previewing is an important, but not a very time-consuming part of your study. Before class, glance over the text material that will be covered in the lecture. Get an overview of the material by reading the introductory and summary passages, the section headings and subheadings, and the diagrams. Look at the problems at the end of the section to get an overall idea of the point of the lecture. This preview should serve as a general base for anchoring the new information presented in class. 2. Note Taking In class, listen actively while taking notes. Intend to learn from the lecture. Write down explanatory remarks about the problem. Note any particular conditions of the problem, how to get from one step to another, and why the approach to the problem is taken. Try to anticipate the consequences of a theorem or the next step in a solution. During a proof, keep the conclusion in mind. If you miss or don't understand something in the lecture, jot down what you can and fill in the missing material later. As soon as possible after class, review and edit your notes. Use the margin or the back of the opposite page to summarize the materials and to list key terms or formulas. You can also use this space to take notes from the text, thus supplementing your lecture notes and creating an integrated study source. Review your notes at regular intervals, particularly as soon as possible just before and just after each class. 3. Reading the Textbook When reading the math textbook, first scan the material to obtain an overview. Then read carefully, making sure that you understand each part as you go. Since much of math is concept building, reading past a concept that is not understood may prove to be wasted time and effort. As you read, take notes on new definitions and symbols. It is especially important to translate abstract formulas into your own verbal explanations. Pay particularly close attention to derivations and sample problems. You should analyze the sample problems in the text, explaining each step in your own words and drawing diagrams to accompany these explanations. For practice, close the book and rework the examples in your own terms. Finally, note how the material relates to previous material, and stop periodically to recite the material to yourself. 4. Problem Solving Most of your study time should be spent working or studying problems. When working a problem, first read through the question to get a general overview. Second, state the unknown in your own terms and write down every piece of
information that is given. Next, devise a tentative plan to solve the problem by using one or more of the following tactics: a. b. c. d. e. f. Form relationships among all facts given. Consider formulas or definitions that might be relevant. Work backwards, asking yourself, "What do I need to know in order to find the answer?" Relate the problem to a similar textbook or lecture example. Solve a simpler version of the problem using small numbers. Break the problem into several simpler problems. Work part of the problem and see if it related to the whole. g. Check each step of the solution for correctness and clarity. Then, rewrite the solution from beginning to end, editing out blind alleys and false leads. 5. Problem Analysis After you have worked a problem, analyze it. Focus on the processes used (not the answer) and ask yourself the following questions: What concepts, formulas, and rules did I apply? What methods did I use? How does the solution compare with those in my text and notes? Can I simplify what I did? Explain each step using your own words. In this way you will sharpen your understanding of the problem and aid future study. The study tips suggested in this handout should help you to improve your performance in your math class. But remember that math courses are cumulative; if you have trouble with the material at the beginning of the course, it is likely that these problems will multiply later on. Consequently, you should seek help early if you encounter difficulty.
TEN WAYS TO REDUCE MATH ANXIETY
OK, so math makes a lot of people nervous. Is there any hope? You bet there is! Below are a few helpful hints. 1. You are not alone! Relax. Many people dislike and are nervous about math. Even mathematicians are unsure of themselves and get that sinking, panicky feeling called "math anxiety" when they first confront a new problem. 2. If you have math anxiety, admit it. If you pretend not to have it, you will not learn to overcome it or manage it. 3. If you're having math trouble, practice a little math each day. (Do you think Mozart learned how to play the piano or Michael Jackson learned how to dance just by watching?) 4. Ask questions. Some people think asking questions is a sign of weakness. It's not. It's a sign of strength. In fact, other students will be glad. (They have questions, too.) 5. Do math in a way that's natural for you. There's often more than one way to work a math problem. Maybe the teacher's way stumps you at first. Don't give up. Work to understand it your way. Then it will be easier to understand it the teacher's way. Remember, "each mind has its own method." 6. Notice your handwriting when you do math. The sloppier it gets, the more confused or angry you probably are. When it gets really sloppy, STOP. Look away for a few seconds. Then erase the messy parts. Start again. Try not to let your attitude interfere with learning math. 7. Know the basics. Be sure you know your math from earlier grades. Maybe you missed something when you moved to a new high school. Face it: Math builds on itself. You have to go back and relearn that stuff. (Don't think, "I couldn't learn it before, so I can't learn it now." Remember it's never too late to learn. Besides, you're older now. It'll be easier and quicker to learn.) 8. Don't go by memory alone. Try to understand your math. Memorizing is a real trap. When you're nervous, memory is the first thing to go. 9. Trouble with the text? Get another math book. Maybe a book in the library will explain things better. 10. Get help. Everyone needs help now and then. Try to form a study group with friends (two heads are better than one), take a review course, or work with a SLAC math tutor. SLAC also has a large math selfhelp library and educational math handouts.
STUDY SKILLS FOR MATH-RELATED COURSES
PREVIEWING Before class briefly preview the text material that will be covered in the lecture. 1. Get an overview of the material by reading the introductory and summary passages, section headings and subheadings, and diagrams. 2. Look at the problems at the end of the chapter. 3. Make note of new terms, theorems, and formulas. 4. Review (if necessary) old terms, theorems, and formulas referred to in the new material. 5. Formulate possible questions for class. Remember, the purpose of previewing is not to understand the material but to get a general idea of what the lecture will cover. This should not be a very time-consuming process. NOTE TAKING When taking notes in class, listen actively; intend to learn from the lecture. 1. Write down the instructor's explanatory remarks about the problem. a. Note how one gets from one step of the problem to another. b. Note any particular conditions of the problem. c. Note why the approach to the problem is taken. d. Note any drawings, graphs, or charts. 2. Try to anticipate the consequences of a theorem or the next step in a problem. During a proof, keep the conclusion in mind. 3. Note any concepts, rules, techniques, and problems that the instructor emphasizes. 4. Question your instructor during class about any unclear concepts or procedures. 5. If you miss something in the lecture or don't understand what's being presented, then write down what you can catch—especially key words. Be sure to skip several lines so you can fill in the missing material later. 6. As soon as possible after class, summarize, review, and edit your notes. a. Quickly read through your notes to get an overview of the material and to check for any errors or omissions. b. Fill in any information—especially explanatory remarks (see #1 above)—that you did not have time to write down or that the instructor did not provide. c. Use the margin or the back of the opposite page to summarize the material, list key terms, theorems, and formulas, and rework examples. You can also use this space to take notes from the textbook. d. Note any relationship to previous material; i.e., write down key similarities and differences between concepts in the new material and concepts in previously learned material. 7. Review your notes at regular intervals and review them with the intent to learn and retain. TEXT READING If your class lectures provide a good overall structure of the course, you can use your text to clarify and supplement your lecture notes. In order to create a single study source, insert the notes you take from the text into your lecture notes themselves as well as in the margins or the back of the opposite page. If your text provides the best overall structure of the material, then you can use your lecture notes as the supplementary source. In either case consider the following procedures: 1. Briefly preview the material. Get an overview of the content and look at the questions at the end of the chapter. 2. Read actively and read to understand thoroughly. a. Formulate questions before (from lecture notes or from previewing) and read to answer those questions. b. Know what every word and symbol means.
c. Translate abstract formulas to verbal explanations or graphic representations. d. Analyze the example problems by asking yourself these questions: What concepts, formulas, and rules were applied? What methods were used to solve the problem? Why was this method used? What was the first step? Have any steps been combined? What differences or similarities are there between the examples and homework problems? e. Further analyze the example problems by using the following procedures: Explain each step using your own words. Write these explanations on paper. Draw your own diagrams to illustrate and explain problems. For practice, write down example problems from your books. Close the book and try to work the problems. Check your work with the example to find what concepts, rules, or methods you are having trouble with. f. Check to see how the material relates to previous material. Ask yourself these questions: How was the material different from previous material? How was it the same? What totally new concepts were introduced, and how were they applied? Where does this material "fit" within the overall structure of the course? 3. Stop periodically and recall the material that you have read. 4. Review prerequisite material, if necessary. PROBLEM SOLVING Solving problems is usually the most important aspect of math-based quantitative courses. You must, therefore, spend much of your study time either working or studying problems. When working a problem, follow these steps: 1. Read through the problem at a moderate speed to get an overview of the problem. 2. Read through the problem again for the purpose of finding out what the problem is asking for (your unknown). Be able to state this in your own words. 3. If appropriate, draw a diagram and label it. 4. Read each phrase of the problem and write down (symbolically or otherwise) all information that is given. 5. Devise a tentative plan to solve the problem by using one or more of the following tactics: a. Form relationships among all facts given. (Write an equation that includes your unknown.) b. Think of every formula or definition that might be relevant to the problem. c. Work backwards; ask yourself, "What do I need to know in order to get the answer?" d. Relate the problem to a similar example from your textbook or notes. e. Solve a simpler case of the problem using extremely large or small numbers; then follow your example as if it is an example from the text. f. Break the problem into simpler problems. Work part of the problem, and see if it relates to the whole. g. Guess an answer and then try to check it to see if it is correct. The method you use to check your answer may suggest a possible plan. h. If you are making no progress, take a break and return to the problem later. 6. Once you have a plan, carry it out. If it doesn't work, try another plan. 7. Check your solution. a. Check to see if the answer is in the proper form. b. Insert your answer back into the problem. c. Make sure your answer is "reasonable." During the problem solving process, it is often helpful to say out loud all of the things you are thinking. This verbalization process can help lead you to a solution.
PROBLEM ANALYSIS After you have worked a problem, analyze it. This can help sharpen your understanding of the problem as well as aid you when working future problems. 1. Focus on the processes used (not the answer) and ask yourself these questions: a. What concept, formulas, and rules did I apply? b. What methods did I use? c. How did I begin? d. How does the solution compare with worked examples from the textbook or my notes? e. Can I do this problem another way? Can I simplify what I did? 2. Explain each step using your own words. Write these explanations on your paper. TEST PREPARATION If you have followed an approach to study as suggested in this handout, your preparation for exams should not be overly difficult. Consider these procedures: 1. Quickly review your notes to determine what topic/problems have been emphasized. 2. Look over your notes and text. Make a concept list in which you list major concepts and formulas that will be covered. 3. Review and rework homework problems, noting why the procedures were applied. 4. Note similarities and differences among problems. Do this for problems within the same chapter and for problems in different chapters. 5. Locate additional problems and use them to take a practice test. Test yourself under conditions that are as realistic as possible (e.g., no notes, time restriction, random sequence of problems, etc.). Also, try to predict test questions; make up your own problems and practice working them. TEST TAKING 1. Glance over the whole exam quickly, assessing questions as to their level of difficulty and point value. Also get a sense of how much time to spend on each question. Leave time at the end to check your work. 2. Begin to work the problems that seem easiest to you. Also give priority to those problems that are worth the most points. 3. Maximize partial credit possibilities by showing all your work. 4. If you have a lapse of memory on a certain problem, skip the problem and return to it later. TEST ANALYSIS Analyzing returned tests can aid your studying for future tests. Ask yourself the following questions: 1. Did most of the test come from the lecture, textbook, or homework? 2. How were the problems different from those in my notes, text, and homework? 3. Where was my greatest source of error (careless errors, lack of time, lack of understanding material, uncertainty of which method to choose, lack of prerequisite information, test anxiety, etc.)? 4. How can I change my studying habits to adjust for the errors I am making? IMPORTANT: The knowledge of most math/science courses is cumulative. Many concepts build on previous concepts, and a poor understanding of one concept will likely lead to a poor understanding of future concepts. Keeping up with your homework is essential. Consequently, you should seek help early if you encounter difficulty.
TWELVE MATH MYTHS
1. MEN ARE BETTER IN MATH THAN WOMEN. Research has failed to show any difference in mathematical ability between men and women. Men are reluctant to admit they have problems so they express difficulty with math by saying, "I could do it if I tried." Women are often too ready to admit inadequacy and say, "I just can't do math." 2. MATH REQUIRES LOGIC, NOT INTUITION. Few people are aware that intuition is the cornerstone of doing math and solving problems. Mathematicians always think intuitively first. Most people have mathematical intuition; they just have not learned to use or trust it. It is amazing how often the first idea you come up with turns out to be correct. 3. MATH IS NOT CREATIVE. Creativity is as central to mathematics as it is to art, literature, and music. The act of creation involves diametrical opposites—intensely and relaxing, the frustration of failure and elation of discovery, the satisfaction of seeing all the pieces fit together. It requires imagination, intellect, intuition, and an aesthetic sense about the rightness of things. 4. YOU MUST ALWAYS KNOW HOW YOU GOT THE ANSWER. Getting the answer to a problem and knowing how the answer was derived are independent processes. If you are consistently right, then you know how to do the problem. There is no need to explain it. 5. THERE IS A BEST WAY TO DO MATH PROBLEMS. A math problem may be solved by a variety of methods that express individuality and originality—but there is no best way. New and interesting techniques for doing all levels of mathematics, from arithmetic to calculus, have been discovered by students. The way math is done is very individual and personal and the best method is the one with which you feel most comfortable. 6. IT'S ALWAYS IMPORTANT TO GET THE ANSWER EXACTLY RIGHT. The ability to obtain approximate answers is often more important than getting exact answers. Feelings about the importance of the answer often are a reversion to early school years when arithmetic was taught with the idea that you were "good" when you got the right answer and "bad" when you did not. 7. IT'S BAD TO COUNT ON YOUR FINGERS. There is nothing wrong with counting on fingers as an aid to doing arithmetic. Counting on fingers actually indicates an understanding of arithmetic—more understanding than if everything were memorized. 8. MATHEMATICIANS DO PROBLEMS QUICKLY, IN THEIR HEADS. Solving new problems or learning new material is always difficult and time consuming. The only problems mathematicians do quickly are those they have solved before. Speed is not a measure of ability. It is the result of experience and practice. 9. MATH REQUIRES A GOOD MEMORY. Knowing math means that concepts make sense to you, and rules and formulas seem natural. This kind of knowledge cannot be gained through rote memorization. 10. MATH IS DONE BY WORKING INTENSELY UNTIL THE PROBLEM IS SOLVED. Solving problems requires both resting and working intensely. Going away from a problem and later returning to it allows your mind time to assimilate ideas and develop new ones. Often, upon coming back to a problem, a new insight is experienced which unlocks the solution. 11. SOME PEOPLE HAVE A "MATH MIND" AND SOME DON'T. Belief in myths about how math is done leads to a complete lack of self-confidence. But it is self-confidence that is one of the most important determining factors in mathematical performance. We have yet to encounter anyone who could not attain his or her goals once the emotional blocks were removed. 12. THERE IS A MAGIC KEY TO DOING MATH. There is no formula, rule, or general guideline that will suddenly unlock the mysteries of math. If there is a key to doing math, it is in overcoming anxiety about the subject and in using the same skills you use to do everything else.
Ten Ways to Survive the Math Blues
1. Figure out the Big Picture: Find out why you are doing this math. How does it fit with your other courses (science, geography, English, engineering)? You could do some Internet searches on the math you are studying and include "application". Get a sense of where you are going and why you are doing this. Mathematics is compulsory in most of the world – there has to be a reason… 2. Get on top of it before it gets on top of you. Yep, mathematics is one of those things that builds on prior knowledge. Yet many students learn stuff only for an examination and then promptly forget it, setting themselves up for later difficulties. Learn for the future, not for tomorrow's test. 3. Read Ahead. It is strongly advised that you read over next week's math right now. You won't understand it all, but you will have a better sense of what is coming up and how it fits with what you are doing this week. Then, when your class goes through it later, your doubts and uncertainties will reduce – and you will understand and remember it better. 4. Use more than one resource. It often happens that you can't follow the teacher's explanation and your textbook is very confusing. Borrow 2 or 3 textbooks similar to your own from your library and read what they have to say about the topic. Often they will have a diagram, a picture or an explanation that gives you the "Ahhh – I get it!" that you desire. 5. Don't join the Blame Game. Teaching mathematics is tough. Teachers really have to work hard to make math fun, interesting and engaging. It is easy to blame a teacher for a bad grade, but who is really responsible for your future? 6. Practice makes Perfect. You don't expect to be able to play guitar or drive a car without practice. Well, learning mathematics (unfortunately) involves some slogging away and doing exercises. Don't get bogged down, though – use your other resources to help you through the homework. 7. Time Management. Start homework assignments as soon as you get them. There may be some things on there that you haven't done in class yet (because maybe it is not due for a few weeks). That's good – it helps to focus your thoughts so that when you are doing that section in class, you know that it is important and you'll know what you don't know. Nobody plans to fail – but many fail to plan… 8. Don't fall into the trap of copying from a friend to survive. They probably have the wrong answer anyway. Besides, a lot of students resent being asked for their assignments for copying – they are too afraid of a ruined relationship to say no. Hey, you can do it – have the confidence in your own ability. 9. Never, never give up. Math uses a different part of the brain than most other things in school. It can be stressful when you can't figure out something. Work on something else for a while and come back to it later. 10. Keep a sense of humour! Don't lose the ability to laugh at yourself and your own mistakes. Mistakes are not the end of the world – they are the beginning of real learning!
Reasons Why People Have Math Anxiety
1. People don't try to understand; they just memorize. 2. They are underprepared - MATH IS CUMULATIVE. How to Study Math 1. Keep up - review notes after class. 2. Take good notes - put everything from the board on paper. 3. Read the text - if you don't understand, get help. 4. Get a study friend. 5. Have a set time to get math homework. Treat it as a scheduled class. The math lab is a good place to do homework. How to Study for Math Exams 1. Start at Day One - do homework. 2. Memorize formulas - use flashcards. 3. Rework problems that you missed on the homework. Math is Problem Solving 1. Read the full question. 2. Analyze and Compute. 3. Given/Find/Need: -- what's given? -- what do I need to find? -- what to I need to do? 4. Draw pictures - can simplify problem. 5. Use a calculator - do calculations twice. 6. Check your results - do the problem again another way. | 677.169 | 1 |
's Them Engaged. Keep's Them Engaged Blitzer's philosophy: present the full scope of mathematics, while always (1) engaging the student by opening their minds to learning (2) keeping the student engaged on every page (3) explaining ideas directly, simply, and clearly so they don't get "lost" when studying and reviewing. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
In the nineteenth century, French mathematician Evariste Galois developed the Galois theory of groups-one of the most penetrating concepts in modem mathematics. The elements of the theory are clearly presented in this second, revised edition of a volume of lectures delivered by noted mathematician Emil Artin. The book has been edited by Dr. Arthur N. Milgram, who has also supplemented the work with a Section on Applications. | 677.169 | 1 |
Prealgebra and Introductory Algebra, A La Carte + MathXL
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Summary
Do your students dislike carrying a big textbook around campus? We can provide an unbound, three-hole-punched version of the traditional text so that your students can carry just what they need. This unbound version comes with access to MyMaythLab or MyStatLab at a significant discount from the price of the regular text. Elayn Martin-Gaybelieveseverystudent can succeedand that is the motivating force behind her best-selling texts and acclaimed video program. With Martin-Gay you get 100% consistency in voice from text to video! Prealgebra and Introductory Algebra 2eis appropriate for a 2-sem sequence of Prealgebra (Basic Math with very early introduction to algebra) and Introductory Algebra (aka Elementary Algebra). This text was written to help students effectively make the transition from arithmetic to algebra and provide a strong foundation for success in their next, intermediate algebra course. To reach this goal, Martin-Gay introduces algebraic concepts early and repeats them as she treats traditional arithmetic topics, and then further develops their exposure to elementary-level algebra topics. The material from this text is also available split out into two separate textbooks,Prealgebra 5eandIntroductory Algebra 3e, if you prefer to use split textbooks, rather than one combined textbook for your 2-sem sequence. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
Pocket Book of Integrals and Mathematical Formulas, Second Edition, is a handy, compact reference work containing concise discussions of mathematical concepts and formulas frequently needed by students and professionals in engineering, science, applied math and statistics. One of the book's special features is its comprehensive table of integrals arranged and formatted to facilitate the rapid location of the right form. Tables of derivatives; calculus applications; graphs of important Cartesian and polar curves; and formulas from algebra, geometry, trigonometry, calculus and statistics have also been arranged with convenience in mind.
The book also lists infinite series and includes discussions and formula listings in differential equations and in advanced topics such as Fourier series, Laplace and z-transforms, vector analysis, and orthogonal polynomials. Important physical constants and numerical tables of probability distributions such as normal, Poisson, t, chi square, and F are provided. Pocket Book of Integrals and Mathematical Formulas, Second Edition, is a volume every student or professional working with engineering mathematics should have in their coat pocket.
Pocket Book of Integrals and Mathematical Formulas, Second Edition has been updated and expanded to cover even more calculus applications and additional topics, including Z-transforms, orthogonal polynomials, Bessel functions, and a summary of probability distributions.
Synopsis
This text extends the MATLAB Primer in order to represent the significant features introduced in MATLAB 5.0. This pocket book serves as an excellent resource for students and engineers requiring a high-level introduction and handy reference to MATLAB 5.0, helping readers to team efficiently and independently without having to delve through manuals. | 677.169 | 1 |
...
Show More inductive approach that includes integrated activities and tools to promote hands-on application and discovery.New! Tables provide visual connections between figures and concepts and help students better assess their level of mastery and test readiness.New! Chapter Tests have been added to the end of every chapter.New! Proofs have been varied to include written and visual proofs, as well as comparisons, to support students with different learning styles.New! Exercise sets in the Student Study Guide, with cross-references to the text, offer additional practice and review.New! Technology-related margin features encourage the use of the Geometer's Sketchpad, graphing calculators, and further explorations.New! Coverage now includes Section 2.6, Symmetry and Transformations.New! Technology Package includes the HM ClassPrep CD with computerized test bank (powered by Diploma).Updated! The number of Exercises and Explorations has been increased.Highly visual approach begins with the presentation of an idea, followed by the examination and development of a theory, verification of the theory through deduction, and finally, application of the principles to the real world.Discovery features reinforce the text's inductive approach: activities integrated throughout enable students to discover geometry concepts on their own, and section tools provide with hands-on application of geometric conceptsApplications reinforce the connection of geometry to the real world: high-interest Chapter Openers introduce the principal notion of the chapter and relate to the real world and A Perspective On... sections conclude each chapter, providing sketches that are interesting, sometimes historical, and always informative.Summaries of constructions, postulates, and theorems are provided, and an easy-to-navigate numbering system for postulates and theorems provides a user-friendly structure. In response to user feedback, paragraph proofs feature more prominently in this edition.Comprehensive appendices include Algebra Review and An Introduction to Logic. A glossary of terms, a summary of applications in the text, and selected answers are also provided in the back of the | 677.169 | 1 |
Trigonometry for Engineering Technology
Description
This book uses mechanical, civil, and architectural applications to enhance its explanations of real-world scenarios. Its open format enables it to be used as a workbook either in class or for independent study. In addition to its thorough treatment of right-angle trigonometry, the text includes oblique triangle calculations and graphing of trigonometric functions. The examples in each chapter demonstrate a specific type of problem-solving procedure and are followed by practice exercises.
Specs
Emphasizes examples from the fields of engineering technology for students who are studying to prepare for work in those fields.
Many examples and exercises include illustrations to enhance the explanation of real-world scenarios.
Each chapter includes example problems that demonstrate a problem-solving procedure for specific problem types. The example problems are followed by exercises that the students solve for practice.
Several chapters are included to supplement the right angle trigonometry topics, including Law of Sines, Law of Cosines, and graphing trigonometric functions.
An open format allows students to use the text as a workbook in class and during study | 677.169 | 1 |
Calculus III Appendix C
All professional education content courses leading to certification shall include teaching and assessment ofthe Wisconsin Content Standards in the content area.
In this column, list the Wisconsin Content Standards that are included in this course. The Standards for each content area are found in the Wisconsin Content Standards document.
In this column, indicate the nature of the performance assessments used in this course to evaluate student proficiency in each standard.
The structures within the discipline, the historical roots and evolving nature of mathematics, and the interaction between technology and the discipline.
Students are asked to compare mathematical concepts learned in Calculus I and II to their natural extensions and generalizations (evolution) in Calculus III. For example, limits, continuity, differentiability, optimization, integration, etc. are all concepts that are revisited for functions of several variables.
Facilitating the building of student conceptual and procedural understanding.
Helping all students build understanding of the discipline including:
. Confidence in their abilities to utilize mathematical knowledge.
. Awareness of the usefulness of mathematics.
. The economic implications of fine mathematical preparation.
Students are tested on their ability to apply the mathematical knowledge gained in the course to problems on homework and tests.
Exploring, conjecturing, examining and testing all aspects of problem solving.
Students asked to construct convincing mathematical arguments in class as well as on tests.
Expressing ideas orally, in writing, and visually-, using mathematical language, notation, and symbolism; translating mathematical ideas between and among contexts.
Students are challenged to show their work on homework and test problems.
Connecting the concepts and procedures of mathematics, drawing connections between mathematical strands, between mathematics and other disciplines, and with daily life.
Daily problem assignments are made to assess the student's ability to connect concepts with procedures.
Selecting appropriate representations to facilitate mathematical problem solving and translating between and among representations to explicate problem-solving situations.
For many problems, students must choose an appropriate coordinate system, order of integration, to solve some problems more efficiently.
Mathematical processes including:
. Problem solving.
. Communication.
. Reasoning and formal and informal argument.
. Mathematical connections.
. Representations.
. Technology.
Homework problems are of two types; exercise and problem-solving. Students are required to justify answers through informal arguments. Use of graphing calculators and computer algebra systems is helpful for some problems.
Number operations and relationships from both abstract and concrete perspectives identifying real world applications, and representing and connecting mathematical concepts and procedures including:
. Number sense.
. Set theory.
. Number and operation.
. Composition and decomposition of numbers, including place value, primes, factors, multiples, inverses, and the extension of these concepts throughout mathematics.
. Number systems through the real numbers, their properties and relations.
. Computational procedures.
. Proportional reasoning.
. Number theory.
Calculus III is loaded with computational procedures. Most of the course centers on how to compute various quantities.
Mathematical concepts and procedures, and the connections among them for teaching upper level number operations and relationships including:
. Advanced counting procedures, including union and intersection of sets, and parenthetical operations.
. Algebraic and transcendental numbers.
. The complex number system, including polar coordinates.
. Approximation techniques as a basis for numerical integration, fractals, and numerical-based proofs.
. Situations in which numerical arguments presented in a variety of classroom and real-world situations (e.g., political, economic, scientific, social) can be created and critically evaluated.
. Opportunities in which acceptable limits of error can be assessed (e.g., evaluating strategies, testing the reasonableness of results, and using technology to carry out computations).
Approximation techniques in the form of differentials and tangent planes appear as homework problems and test questions.
Geometry and measurement from both abstract and concrete perspectives and to identify real world applications, and mathematical concepts, procedures and connections among them including:
. Formal and informal argument.
. Names, properties, and relationships of two- and three-dimensional shapes.
. Spatial sense.
. Spatial reasoning and the use of geometric models to represent, visualize, and solve problems.
. Transformations and the ways in which rotation, reflection, and translation of shapes can illustrate concepts, properties, and relationships.
. Coordinate geometry systems including relations between coordinate and synthetic geometry, and generalizing geometric principles from a two-dimensional system to a three-dimensional system.
. Concepts of measurement, including measurable attributes, standard and non-standard units, precision and accuracy, and use of appropriate tools.
. The structure of systems of measurement, including the development and use of measurement systems and the relationships among different systems. Measurement including length, area, volume, size of angles, weight and mass, time, temperature, and money.
. Measuring, estimating, and using measurement to describe and compare geometric phenomena.
. Indirect measurement and its uses, including developing formulas and procedures for determining measure to solve problems.
Students must master quite a bit of analytic geometry: equations of lines, planes, quadric surfaces, characterize parallelism, perpendicularity, and visualize surfaces in three-dimensional contexts. Measuring areas and volumes is a fundamental problems that is dealt with in homework and test questions.
Mathematical concepts, procedures, and the connections among them for teaching upper level geometry and measurement including:
. Transformations, coordinates, and vectors and their use in problem solving. Three-dimensional geometry and its generalization to other dimensions. Topology, including topological properties and transformations.
. Opportunities to present convincing arguments by means of demonstration, informal proof, counter-examples, or other logical means to show the truth of statements and/or generalizations.
Students are tested over their understanding of cartesian, polar, cylindrical and spherical coordinate systems. They should be able to find coordinates of a given point in each of these systems and have some familiarity with transforming double and triple integrals into a given coordinate system.
Statistics and probability from both abstract and concrete perspectives and to identify real world applications, and the mathematical concepts, procedures and the connections between them including:
. Use of data to explore real-world issues.
. The process of investigation including formulation of a problem, designing a data collection plan, and collecting, recording, and organizing data.
. Probability as a way to describe chances or risk in simple and compound events.
. Outcome prediction based on experimentation or theoretical probabilities.
The principle of least-squares fit as a multidimensional optimization problem and the utility of double integrals in computing probabilities are introduced in problems.
Mathematical concepts, procedures, and the connections among them for teaching upper level statistics and probability including:
. Use of the random variable in the generation and interpretation of probability distributions.
. Descriptive and inferential statistics, measures of disbursement, including validity and reliability, and correlation.
. Probability theory and its link to inferential statistics.
. Discrete and continuous probability distributions as bases for inference.
. Situations in which students can analyze, evaluate, and critique the methods and conclusions of statistical experiments reported in journals, magazines, news media, advertising, etc.
None.
Functions, algebra, and basic concepts underlying calculus from both abstract and concrete perspectives and to identify real world applications, and the mathematical concepts, procedures and the connections among them including:
. Patterns.
. Functions as used to describe relations and to model real world situations.
. Representations of situations that involve variable quantities with expressions, equations and inequalities and that include algebraic and geometric relationships.
. Multiple representations of relations, the strengths and limitations of each representation, and conversion from one representation to another.
. Operations on expressions and solution of equations, systems of equations and inequalities using concrete, informal, and formal methods.
. Underlying concepts of calculus, including rate of change, limits, and approximations for irregular areas.
The analysis of functions of several variables is the principal objective of the course. Students are required to show mastery of elementary calculus concepts, such as, limits, differentiability, and continuity in this context. Some equation - solving required as a subtask to many problems in the homework and on tests. Some work with inequalities (making bounds: upper and lower) is also assessed on problems and tests.
Mathematical concepts, procedures, and the connections among them for teaching upper level functions, algebra, and concepts of calculus including:
. Concepts of calculus, including limits (epsilon-delta) and tangents, derivatives, integrals, and sequences and series.
. Modeling to solve problems.
. Calculus techniques including finding limits, derivatives, integrals, and using special rules. | 677.169 | 1 |
Cleve Moler
c 2011 Cleve Moler. Copyright ⃝ All rights reserved. No part of this e-book may be reproduced, stored, or transmitted in any manner without the written permission of the author. For more information, contact moler@mathworks.com. The programs described in this e-book have been included for their instructional value. These programs have been tested with care but are not guaranteed for any particular purpose. The author does not offer any warranties or representations, nor does he accept any liabilities with respect to the use of the programs. These programs should not be relied on as the sole basis to solve a problem whose incorrect solution could result in injury to person or property.
R Matlab⃝ is a registered trademark of MathWorks, Inc.TM .
For more information about relevant MathWorks policies, see:
October 4, 2011
ii
Contents
Preface
Figure 1. exmgui provides a starting point for some of the experiments.
Welcome to Experiments with MATLAB. This is not a conventional book. It is currently available only via the Internet, at no charge, from There may eventually be a hardcopy edition, but not right away. Although Matlab is now a full fledged Technical Computing Environment, it started in the late 1970s as a simple "Matrix Laboratory". We want to build on this laboratory tradition by describing a series of experiments involving applied mathematics, technical computing, and Matlab programming. iii
iv
Preface
We expect that you already know something about high school level material in geometry, algebra, and trigonometry. We will introduce ideas from calculus, matrix theory, and ordinary differential equations, but we do not assume that you have already taken courses in the subjects. In fact, these experiments are useful supplements to such courses. We also expect that you have some experience with computers, perhaps with word processors or spread sheets. If you know something about programming in languages like C or Java, that will be helpful, but not required. We will introduce Matlab by way of examples. Many of the experiments involve understanding and modifying Matlab scripts and functions that we have already written. You should have access to Matlab and to our exm toolbox, the collection of programs and data that are described in Experiments with MATLAB. We hope you will not only use these programs, but will read them, understand them, modify them, and improve them. The exm toolbox is the apparatus in our "Laboratory". You will want to have Matlab handy. For information about the Student Version, see For an introduction to the mechanics of using Matlab, see the videos at For documentation, including "Getting Started", see For user contributed programs, programming contests, and links into the world-wide Matlab community, check out To get started, download the exm toolbox, use pathtool to add exm to the Matlab path, and run exmgui to generate figure 1. You can click on the icons to preview some of the experiments. You will want to make frequent use of the Matlab help and documentation facilities. To quickly learn how to use the command or function named xxx, enter help xxx For more extensive information about xxx, use doc xxx We hope you will find the experiments interesting, and that you will learn how to use Matlab along the way. Each chapter concludes with a "Recap" section that is actually an executable Matlab program. For example, you can review the Magic Squares chapter by entering magic_recap
Preface Better yet, enter edit magic_recap
v
and run the program cell-by-cell by simultaneously pressing the Ctrl-Shift-Enter keys. A fairly new Matlab facility is the publish command. You can get a nicely formatted web page about magic_recap with publish magic_recap If you want to concentrate on learning Matlab, make sure you read, run, and understand the recaps. Cleve Moler October 4, 2011
vi
Preface
Chapter 1
Iteration
Iteration is a key element in much of technical computation. Examples involving the Golden Ratio introduce the Matlab assignment statement, for and while loops, and the plot function.
Start by picking a number, any number. Enter it into Matlab by typing x = your number This is a Matlab assignment statement. The number you chose is stored in the variable x for later use. For example, if you start with x = 3 Matlab responds with x = 3 Next, enter this statement x = sqrt(1 + x) The abbreviation sqrt √ is the Matlab name for the square root function. The quantity on the right, 1 + x, is computed and the result stored back in the variable x, overriding the previous value of x. Somewhere on your computer keyboard, probably in the lower right corner, you should be able to find four arrow keys. These are the command line editing keys. The up-arrow key allows you to recall earlier commands, including commands from
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
1
2
Chapter 1. Iteration
previous sessions, and the other arrows keys allow you to revise these commands. Use the up-arrow key, followed by the enter or return key, to iterate, or repeatedly execute, this statement: x = sqrt(1 + x) Here is what you get when you start with x = 3. x = 3 x = 2 x = 1.7321 x = 1.6529 x = 1.6288 x = 1.6213 x = 1.6191 x = 1.6184 x = 1.6181 x = 1.6181 x = 1.6180 x = √ √ √ √ √ √ These values are 3, 1 + 3, 1 + 1 + 3, 1 + 1 + 1 + 3, and so on. After 10 steps, the value printed remains constant at 1.6180. Try several other starting values. Try it on a calculator if you have one. You should find that no matter where you start, you will always reach 1.6180 in about ten steps. (Maybe a few more will be required if you have a very large starting value.) Matlab is doing these computations to accuracy of about 16 decimal digits, but is displaying only five. You can see more digits by first entering format long and repeating the experiment. Here are the beginning and end of 30 steps starting at x = 3. x = 3 1.6180
3 x = 2 x = 1.732050807568877 x = 1.652891650281070 .... x = 1.618033988749897 x = 1.618033988749895 x = 1.618033988749895 After about thirty or so steps, the value that is printed doesn't change any more. You have computed one of the most famous numbers in mathematics, ϕ, the Golden Ratio. In Matlab, and most other programming languages, the equals sign is the assignment operator. It says compute the value on the right and store it in the variable on the left. So, the statement x = sqrt(1 + x) takes the current value of x, computes sqrt(1 + x), and stores the result back in x. In mathematics, the equals sign has a different meaning. √ x= 1+x is an equation. A solution to such an equation is known as a fixed point. (Be careful not to confuse the mathematical usage of fixed point with the computer arithmetic usage of fixed point.) √ The function f (x) = 1 + x has exactly one fixed point. The best way to find the value of the fixed point is to avoid computers all together and solve the equation using the quadratic formula. Take a look at the hand calculation shown in figure 1.1. The positive root of the quadratic equation is the Golden Ratio. √ 1+ 5 ϕ= . 2 You can have Matlab compute ϕ directly using the statement phi = (1 + sqrt(5))/2 With format long, this produces the same value we obtained with the fixed point iteration, phi = 1.618033988749895
5 produce √ a figure that has three components. The first two components are graphs of x and 1 + x. The '-' argument tells the plot function to draw solid lines. The last component in the plot is a single point with both coordinates equal to ϕ. The 'o' tells the plot function to draw a circle. The Matlab plot function has many variations, including specifying other colors and line types. You can see some of the possibilities with help plot
φ
1
1
φ−1
Figure 1.3. The golden rectangle. The Golden Ratio shows up in many places in mathematics; we'll see several in this book. The Golden Ratio gets its name from the golden rectangle, shown in figure 1.3. The golden rectangle has the property that removing a square leaves a smaller rectangle with the same shape. Equating the aspect ratios of the rectangles gives a defining equation for ϕ: 1 ϕ−1 = . ϕ 1 Multiplying both sides of this equation by ϕ produces the same quadratic polynomial equation that we obtained from our fixed point iteration. ϕ2 − ϕ − 1 = 0. The up-arrow key is a convenient way to repeatedly execute a single statement, or several statements, separated by commas or semicolons, on a single line. Two more powerful constructs are the for loop and the while loop. A for loop executes a block of code a prescribed number of times. x = 3 for k = 1:31 x = sqrt(1 + x) end
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Chapter 1. Iteration
produces 32 lines of output, one from the initial statement and one more each time through the loop. A while loop executes a block of code an unknown number of times. Termination is controlled by a logical expression, which evaluates to true or false. Here is the simplest while loop for our fixed point iteration. x = 3 while x ~= sqrt(1+x) x = sqrt(1+x) end This produces the same 32 lines of output as the for loop. However, this code is open to criticism for two reasons. The first possible criticism involves the termination √ condition. The expression x ~= sqrt(1+x) is the Matlab way of writing x ̸= 1 + x. With exact arithmetic, x would never be exactly equal to sqrt(1+x), the condition would always be true, and the loop would run forever. However, like most technical computing environments, Matlab does not do arithmetic exactly. In order to economize on both computer time and computer memory, Matlab uses floating point arithmetic. Eventually our program produces a value of x for which the floating point numbers x and sqrt(1+x) are exactly equal and the loop terminates. Expecting exact equality of two floating point numbers is a delicate matter. It works OK in this particular situation, but may not work with more complicated computations. The second possible criticism of our simple while loop is that it is inefficient. It evaluates sqrt(1+x) twice each time through the loop. Here is a more complicated version of the while loop that avoids both criticisms. x = 3 y = 0; while abs(x-y) > eps(x) y = x; x = sqrt(1+x) end The semicolons at the ends of the assignment statements involving y indicate that no printed output should result. The quantity eps(x), is the spacing of the floating point numbers near x. Mathematically, the Greek letter ϵ, or epsilon, often represents a "small" quantity. This version of the loop requires only one square root calculation per iteration, but that is overshadowed by the added complexity of the code. Both while loops require about the same execution time. In this situation, I prefer the first while loop because it is easier to read and understand.
Help and Doc
Matlab has extensive on-line documentation. Statements like help sqrt help for
7 provide brief descriptions of commands and functions. Statements like doc sqrt doc for provide more extensive documentation in a separate window. One obscure, but very important, help entry is about the various punctuation marks and special characters used by Matlab. Take a look now at help punct doc punct You will probably want to return to this information as you learn more about Matlab.
Numbers
Numbers are formed from the digits 0 through 9, an optional decimal point, a leading + or - sign, an optional e followed by an integer for a power of 10 scaling, and an optional i or j for the imaginary part of a complex number. Matlab also knows the value of π . Here are some examples of numbers. 42 9.6397238 6.0221415e23 -3+4i pi
Assignment statements and names
A simple assignment statement consists of a name, an = sign, and a number. The names of variables, functions and commands are formed by a letter, followed by any number of upper and lower case letters, digits and underscores. Single character names, like x and N, and anglicized Greek letters, like pi and phi, are often used to reflect underlying mathematical notation. Non-mathematical programs usually employ long variable names. Underscores and a convention known as camel casing are used to create variable names out of several words. x = 42 phi = (1+sqrt(5))/2 Avogadros_constant = 6.0221415e23 camelCaseComplexNumber = -3+4i
Expressions
Power is denoted by ^ and has precedence over all other arithmetic operations. Multiplication and division are denoted by *, /, and \ and have precedence over addition and subtraction, Addition and subtraction are denoted by + and - and
8
Chapter 1. Iteration
have lowest precedence. Operations with equal precedence are evaluated left to right. Parentheses delineate subexpressions that are evaluated first. Blanks help readability, but have no effect on precedence. All of the following expressions have the same value. If you don't already recognize this value, you can ask Google about its importance in popular culture. 3*4 + 5*6 3 * 4+5 * 6 2*(3 + 4)*3 -2^4 + 10*29/5 3\126 52-8-2
Recap
%% Iteration Chapter Recap % This is an executable program that illustrates the statements % introduced in the Iteration chapter of "Experiments in MATLAB". % You can run it by entering the command % % iteration_recap % % Better yet, enter % % edit iteration_recap % % and run the program cell-by-cell by simultaneously % pressing the Ctrl-Shift-Enter keys. % % Enter % % publish iteration_recap % % to see a formatted report. %% Help and Documentation % help punct % doc punct %% Format format short 100/81 format long 100/81 format short
You can get started with help ^ help sin 1.2 Temperature conversion. (a) Write a Matlab statement that converts temperature in Fahrenheit, f, to Celsius, c. c = something involving f (b) Write a Matlab statement that converts temperature in Celsius, c, to Fahrenheit, f. f = something involving c
1.3 Barn-megaparsec. A barn is a unit of area employed by high energy physicists. Nuclear scattering experiments try to "hit the side of a barn". A parsec is a unit of length employed by astronomers. A star at a distance of one parsec exhibits a trigonometric parallax of one arcsecond as the Earth orbits the Sun. A barnmegaparsec is therefore a unit of volume – a very long skinny volume. A A A A A A barn is 10−28 square meters. megaparsec is 106 parsecs. parsec is 3.262 light-years. light-year is 9.461 · 1015 meters. cubic meter is 106 milliliters. milliliter is 1 5 teaspoon.
11 Express one barn-megaparsec in teaspoons. In Matlab, the letter e can be used to denote a power of 10 exponent, so 9.461 · 1015 can be written 9.461e15. 1.4 Complex numbers. What happens if you start with a large negative value of x and repeatedly iterate x = sqrt(1 + x) 1.5 Comparison. Which is larger, π ϕ or ϕπ ? 1.6 Solving equations. The best way to solve √ x= 1+x or x2 = 1 + x is to avoid computers all together and just do it yourself by hand. But, of course, Matlab and most other mathematical software systems can easily solve such equations. Here are several possible ways to do it with Matlab. Start with format long phi = (1 + sqrt(5))/2 Then, for each method, explain what is going on and how the resulting x differs from phi and the other x's. % roots help roots x1 = roots([1 -1 -1]) % fsolve help fsolve f = @(x) x-sqrt(1+x) p = @(x) x^2-x-1 x2 = fsolve(f, 1) x3 = fsolve(f, -1) x4 = fsolve(p, 1) x5 = fsolve(p, -1) % solve (requires Symbolic Toolbox or Student Version) help solve help syms syms x x6 = solve('x-sqrt(1+x)=0') x7 = solve(x^2-x-1)
12
Chapter 1. Iteration
1.7 Symbolic solution. If you have the Symbolic Toolbox or Student Version, explain what the following program does. x = sym('x') length(char(x)) for k = 1:10 x = sqrt(1+x) length(char(x)) end 1.8 Fixed points. Verify that the Golden Ratio is a fixed point of each of the following equations. ϕ= ϕ= 1 ϕ−1 1 +1 ϕ
Use each of the equations as the basis for a fixed point iteration to compute ϕ. Do the iterations converge? 1.9 Another iteration. Before you run the following program, predict what it will do. Then run it. x = 3 k = 1 format long while x ~= sqrt(1+x^2) x = sqrt(1+x^2) k = k+1 end 1.10 Another fixed point. Solve this equation by hand. 1 x= √ 1 + x2 How many iterations does the following program require? How is the final value of x related to the Golden Ratio ϕ? x = 3 k = 1 format long while x ~= 1/sqrt(1+x^2) x = 1/sqrt(1+x^2) k = k+1 end
1.12 tan(x). Figure 1.5 shows three of the many solutions to the equation x = tan x One of the solutions is x = 0. The other two in the plot are near x = ±4.5. If we did a plot over a large range, we would see solutions in each of the intervals 1 )π, (n + 1 [(n − 2 2 )π ] for integer n. (a) Does this compute a fixed point? x = 4.5
1.14 Why. The first version of Matlab written in the late 1970's, had who, what, which, and where commands. So it seemed natural to add a why command. Check out today's why command with why help why for k = 1:40, why, end type why edit why As the help entry says, please embellish or modify the why function to suit your own tastes. 1.15 Wiggles. A glimpse at Matlab plotting capabilities is provided by the function f = @(x) tan(sin(x)) - sin(tan(x)) This uses the '@' sign to introduce a simple function. You can learn more about the '@' sign with help function_handle. Figure 1.6 shows the output from the statement ezplot(f,[-pi,pi])
Figure 1.6. A wiggly function. (The function name ezplot is intended to be pronounced "Easy Plot". This pun doesn't work if you learned to pronounce "z" as "zed".) You can see that the function is very flat near x = 0, oscillates infinitely often near x = ±π/2 and is nearly linear near x = ±π . You can get more control over the plot with code like this. x = -pi:pi/256:pi; y = f(x); plot(x,y) xlabel('x') ylabel('y') title('A wiggly function') axis([-pi pi -2.8 2.8]) set(gca,'xtick',pi*(-3:1/2:3)) (a) What is the effect of various values of n in the following code? x = pi*(-2:1/n:2); comet(x,f(x)) (b) This function is bounded. A numeric value near its maximum can be found with max(y) What is its analytic maximum? (To be precise, I should ask "What is the function's supremum ?") 1.16 Graphics. We use a lot of computer graphics in this book, but studying Matlab graphics programming is not our primary goal. However, if you are curious, the
16
Chapter 1. Iteration
script that produces figure 1.3 is goldrect.m. Modify this program to produce a graphic that compares the Golden Rectangle with TV screens having aspect ratios 4:3 and 16:9. 1.17 Golden Spiral
Figure 1.7. A spiral formed from golden rectangles and inscribed quarter circles. Our program golden_spiral displays an ever-expanding sequence of golden rectangles with inscribed quarter circles. Check it out.
Chapter 2
Fibonacci Numbers
Fibonacci numbers introduce vectors, functions and recursion.
Leonardo Pisano Fibonacci was born around 1170 and died around 1250 in Pisa in what is now Italy. He traveled extensively in Europe and Northern Africa. He wrote several mathematical texts that, among other things, introduced Europe to the Hindu-Arabic notation for numbers. Even though his books had to be transcribed by hand, they were widely circulated. In his best known book, Liber Abaci, published in 1202, he posed the following problem: A man puts a pair of rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes productive? Today the solution to this problem is known as the Fibonacci sequence, or Fibonacci numbers. There is a small mathematical industry based on Fibonacci numbers. A search of the Internet for "Fibonacci" will find dozens of Web sites and hundreds of pages of material. There is even a Fibonacci Association that publishes a scholarly journal, the Fibonacci Quarterly. A simulation of Fibonacci's problem is provided by our exm program rabbits. Just execute the command rabbits and click on the pushbuttons that show up. You will see something like figure 2.1. If Fibonacci had not specified a month for the newborn pair to mature, he would not have a sequence named after him. The number of pairs would simply
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
17
18
Chapter 2. Fibonacci Numbers
Figure 2.1. Fibonacci's rabbits.
double each month. After n months there would be 2n pairs of rabbits. That's a lot of rabbits, but not distinctive mathematics. Let fn denote the number of pairs of rabbits after n months. The key fact is that the number of rabbits at the end of a month is the number at the beginning of the month plus the number of births produced by the mature pairs: fn = fn−1 + fn−2 . The initial conditions are that in the first month there is one pair of rabbits and in the second there are two pairs: f1 = 1, f2 = 2. The following Matlab function, stored in a file fibonacci.m with a .m suffix, produces a vector containing the first n Fibonacci numbers. function f = fibonacci(n) % FIBONACCI Fibonacci sequence % f = FIBONACCI(n) generates the first n Fibonacci numbers.
19 f = zeros(n,1); f(1) = 1; f(2) = 2; for k = 3:n f(k) = f(k-1) + f(k-2); end With these initial conditions, the answer to Fibonacci's original question about the size of the rabbit population after one year is given by fibonacci(12) This produces 1 2 3 5 8 13 21 34 55 89 144 233 The answer is 233 pairs of rabbits. (It would be 4096 pairs if the number doubled every month for 12 months.) Let's look carefully at fibonacci.m. It's a good example of how to create a Matlab function. The first line is function f = fibonacci(n) The first word on the first line says fibonacci.m is a function, not a script. The remainder of the first line says this particular function produces one output result, f, and takes one input argument, n. The name of the function specified on the first line is not actually used, because Matlab looks for the name of the file with a .m suffix that contains the function, but it is common practice to have the two match. The next two lines are comments that provide the text displayed when you ask for help. help fibonacci produces FIBONACCI Fibonacci sequence f = FIBONACCI(n) generates the first n Fibonacci numbers.
20
Chapter 2. Fibonacci Numbers
The name of the function is in uppercase because historically Matlab was case insensitive and ran on terminals with only a single font. The use of capital letters may be confusing to some first-time Matlab users, but the convention persists. It is important to repeat the input and output arguments in these comments because the first line is not displayed when you ask for help on the function. The next line f = zeros(n,1); creates an n-by-1 matrix containing all zeros and assigns it to f. In Matlab, a matrix with only one column is a column vector and a matrix with only one row is a row vector. The next two lines, f(1) = 1; f(2) = 2; provide the initial conditions. The last three lines are the for statement that does all the work. for k = 3:n f(k) = f(k-1) + f(k-2); end We like to use three spaces to indent the body of for and if statements, but other people prefer two or four spaces, or a tab. You can also put the entire construction on one line if you provide a comma after the first clause. This particular function looks a lot like functions in other programming languages. It produces a vector, but it does not use any of the Matlab vector or matrix operations. We will see some of these operations soon. Here is another Fibonacci function, fibnum.m. Its output is simply the nth Fibonacci number. function f = fibnum(n) % FIBNUM Fibonacci number. % FIBNUM(n) generates the nth Fibonacci number. if n <= 1 f = 1; else f = fibnum(n-1) + fibnum(n-2); end The statement fibnum(12) produces ans = 233
21 The fibnum function is recursive. In fact, the term recursive is used in both a mathematical and a computer science sense. In mathematics, the relationship fn = fn−1 + fn−2 is a recursion relation In computer science, a function that calls itself is a recursive function. A recursive program is elegant, but expensive. You can measure execution time with tic and toc. Try tic, fibnum(24), toc Do not try tic, fibnum(50), toc
To verify this claim, suppose we did not know the value of this fraction. Let x=1+ 1 1+
1 1 1+ 1+ ···
.
We can see the first denominator is just another copy of x. In other words. x=1+ 1 x
This immediately leads to x2 − x − 1 = 0 which is the defining quadratic equation for ϕ, Our exm function goldfract generates a Matlab string that represents the first n terms of the Golden Ratio continued fraction. Here is the first section of code in goldfract. p = '1'; for k = 2:n p = ['1 + 1/(' p ')']; end display(p) We start with a single '1', which corresponds to n = 1. We then repeatedly make the current string the denominator in a longer string. Here is the output from goldfract(n) when n = 7. 1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1))))))
22
Chapter 2. Fibonacci Numbers
You can see that there are n-1 plus signs and n-1 pairs of matching parentheses. Let ϕn denote the continued fraction truncated after n terms. ϕn is a rational approximation to ϕ. Let's express ϕn as a conventional fracton, the ratio of two integers ϕn = pn qn
p = 1; q = 0; for k = 2:n t = p; p = p + q; q = t; end Now compare the results produced by goldfract(7) and fibonacci(7). The first contains the fraction 21/13 while the second ends with 13 and 21. This is not just a coincidence. The continued fraction for the Golden Ratio is collapsed by repeating the statement p = p + q; while the Fibonacci numbers are generated by f(k) = f(k-1) + f(k-2); In fact, if we let ϕn denote the golden ratio continued fraction truncated at n terms, then ϕn = fn fn−1
In the infinite limit, the ratio of successive Fibonacci numbers approaches the golden ratio:
n→∞
lim
fn = ϕ. fn−1
To see this, compute 40 Fibonacci numbers. n = 40; f = fibonacci(n); Then compute their ratios. r = f(2:n)./f(1:n-1) This takes the vector containing f(2) through f(n) and divides it, element by element, by the vector containing f(1) through f(n-1). The output begins with
23 2.00000000000000 1.50000000000000 1.66666666666667 1.60000000000000 1.62500000000000 1.61538461538462 1.61904761904762 1.61764705882353 1.61818181818182 and ends with 1.61803398874990 1.61803398874989 1.61803398874990 1.61803398874989 1.61803398874989 Do you see why we chose n = 40? Compute phi = (1+sqrt(5))/2 r - phi What is the value of the last element? The first few of these ratios can also be used to illustrate the rational output format. format rat r(1:10) ans = 2 3/2 5/3 8/5 13/8 21/13 34/21 55/34 89/55 The population of Fibonacci's rabbit pen doesn't double every month; it is multiplied by the golden ratio every month.
An Analytic Expression
It is possible to find a closed-form solution to the Fibonacci number recurrence relation. The key is to look for solutions of the form fn = cρn
We've seen this equation in the chapter on the Golden Ratio. There are two possible values of ρ, namely ϕ and 1 − ϕ. The general solution to the recurrence is fn = c1 ϕn + c2 (1 − ϕ)n . The constants c1 and c2 are determined by initial conditions, which are now conveniently written f0 = c1 + c2 = 1, f1 = c1 ϕ + c2 (1 − ϕ) = 1. One of the exercises asks you to use the Matlab backslash operator to solve this 2-by-2 system of simultaneous linear equations, but it is may be easier to solve the system by hand: ϕ , 2ϕ − 1 (1 − ϕ) c2 = − . 2ϕ − 1 c1 = Inserting these in the general solution gives fn = 1 (ϕn+1 − (1 − ϕ)n+1 ). 2ϕ − 1
This is an amazing equation. The right-hand side involves powers and quotients of irrational numbers, but the result is a sequence of integers. You can check this with Matlab. n = (1:40)'; f = (phi.^(n+1) - (1-phi).^(n+1))/(2*phi-1) f = round(f) The .^ operator is an element-by-element power operator. It is not necessary to use ./ for the final division because (2*phi-1) is a scalar quantity. Roundoff error prevents the results from being exact integers, so the round function is used to convert floating point quantities to nearest integers. The resulting f begins with f = 1
Exercises
2.1 Rabbits. Explain what our rabbits simulation demonstrates. What do the different figures and colors on the pushbuttons signify? 2.2 Waltz. Which Fibonacci numbers are even? Why? 2.3 Primes. Use the Matlab function isprime to discover which of the first 40 Fibonacci numbers are prime. You do not need to use a for loop. Instead, check out help isprime help logical 2.4 Backslash. Use the Matlab backslash operator to solve the 2-by-2 system of simultaneous linear equations c1 + c2 = 1, c1 ϕ + c2 (1 − ϕ) = 1 for c1 and c2 . You can find out about the backslash operator by taking a peek at the Linear Equations chapter, or with the commands help \ help slash 2.5 Logarithmic plot. The statement semilogy(fibonacci(18),'-o') makes a logarithmic plot of Fibonacci numbers versus their index. The graph is close to a straight line. What is the slope of this line? 2.6 Execution time. How does the execution time of fibnum(n) depend on the execution time for fibnum(n-1) and fibnum(n-2)? Use this relationship to obtain an approximate formula for the execution time of fibnum(n) as a function of n. Estimate how long it would take your computer to compute fibnum(50). Warning: You probably do not want to actually run fibnum(50). 2.7 Overflow. What is the index of the largest Fibonacci number that can be represented exactly as a Matlab double-precision quantity without roundoff error? What is the index of the largest Fibonacci number that can be represented approximately as a Matlab double-precision quantity without overflowing? 2.8 Slower maturity. What if rabbits took two months to mature instead of one?
(a) Modify fibonacci.m and fibnum.m to compute this sequence. (b) How many pairs of rabbits are there after 12 months? (c) gn ≈ γ n . What is γ ? (d) Estimate how long it would take your computer to compute fibnum(50) with this modified fibnum. 2.9 Mortality. What if rabbits took one month to mature, but then died after six months. The sequence would be defined by dn = 0, n <= 0 d 1 = 1, d2 = 1 and, for n > 2, dn = dn−1 + dn−2 − dn−7 (a) Modify fibonacci.m and fibnum.m to compute this sequence. (b) How many pairs of rabbits are there after 12 months? (c) dn ≈ δ n . What is δ ? (d) Estimate how long it would take your computer to compute fibnum(50) with this modified fibnum. 2.10 Hello World. Programming languages are traditionally introduced by the phrase "hello world". An script in exm that illustrates some features in Matlab is available with hello_world Explain what each of the functions and commands in hello_world do. 2.11 Fibonacci power series. The Fibonacci numbers, fn , can be used as coefficients in a power series defining a function of x. F (x) = =
∞ ∑ n=1
fn xn
x + 2x2 + 3x3 + 5x4 + 8x5 + 13x6 + ...
29 Our function fibfun1 is a first attempt at a program to compute this series. It simply involves adding an accumulating sum to fibonacci.m. The header of fibfun1.m includes the help entries. \excise \emph{Fibonacci power series}. The Fibonacci numbers, $f_n$, can be used as coefficients in a power series defining a function of $x$. \begin{eqnarray*} F(x) & = & \sum_{n = 1}^\infty f_n x^n \\ & = & x + 2 x^2 + 3 x^3 + 5 x^4 + 8 x^5 + 13 x^6 + ... \end{eqnarray*} Our function #fibfun1# is a first attempt at a program to compute this series. It simply involves adding an accumulating sum to #fibonacci.m#. The header of #fibfun1.m# includes the help entries. \begin{verbatim} n = 1476; f = zeros(n,1); f(1) = 1; f(2) = 2; y = f(1)*x + f(2)*x.^2; t = 0; The main body of fibfun1 implements the Fibonacci recurrence and includes a test for early termination of the loop. for k = 3:n f(k) = f(k-1) + f(k-2); y = y + f(k)*x.^k; if y == t return end
30 t = y; end
Chapter 2. Fibonacci Numbers
There are several objections to fibfun1. The coefficient array of size 1476 is not actually necessary. The repeated computation of powers, x^k, is inefficient because once some power of x has been computed, the next power can be obtained with one multiplication. When the series converges the coefficients f(k) increase in size, but the powers x^k decrease in size more rapidly. The terms f(k)*x^k approach zero, but huge f(k) prevent their computation. A more efficient and accurate approach involves combining the computation of the Fibonacci recurrence and the powers of x. Let pk = fk xk Then, since fk+1 xk+1 = fk xk + fk−1 xk−1 the terms pk satisfy pk+1 = pk x + pk−1 x2 = x(pk + xpk−1 ) This is the basis for our function fibfun2. The header is essentially the same as fibfun1 function [yk,k] = fibfun2(x) % FIBFUN2 Power series with Fibonacci coefficients. % y = fibfun2(x) = sum(f(k)*x.^k). % [y,k] = fibfun2(x) also gives the number of terms required. The initialization. pkm1 = x; pk = 2*x.^2; ykm1 = x; yk = 2*x.^2 + x; k = 0; And the core. while any(abs(yk-ykm1) > 2*eps(yk)) pkp1 = x.*(pk + x.*pkm1); pkm1 = pk; pk = pkp1; ykm1 = yk; yk = yk + pk; k = k+1; end
31 There is no array of coefficients. Only three of the pk terms are required for each step. The power function ^ is not necessary. Computation of the powers is incorporated in the recurrence. Consequently, fibfun2 is both more efficient and more accurate than fibfun1. But there is an even better way to evaluate this particular series. It is possible to find a analytic expression for the infinite sum. F (x) =
∞ ∑ n=1
It is not even necessary to have a .m file. A one-liner does the job. fibfun3 = @(x) (x + x.^2)./(1 - x - x.^2) Compare these three fibfun's.
32
Chapter 2. Fibonacci Numbers
Chapter 3
Calendars and Clocks
Computations involving time, dates, biorhythms and Easter.
Calendars are interesting mathematical objects. The Gregorian calendar was first proposed in 1582. It has been gradually adopted by various countries and churches over the four centuries since then. The British Empire, including the colonies in North America, adopted it in 1752. Turkey did not adopt it until 1923. The Gregorian calendar is now the most widely used calendar in the world, but by no means the only one. In the Gregorian calendar, a year y is a leap year if and only if y is divisible by 4 and not divisible by 100, or is divisible by 400. In Matlab the following expression must be true. The double ampersands, '&&', mean "and" and the double vertical bars, '||', mean "or". mod(y,4) == 0 && mod(y,100) ~= 0 || mod(y,400) == 0 For example, 2000 was a leap year, but 2100 will not be a leap year. This rule implies that the Gregorian calendar repeats itself every 400 years. In that 400-year period, there are 97 leap years, 4800 months, 20871 weeks, and 146097 days. The 97 average number of days in a Gregorian calendar year is 365 + 400 = 365.2425. The Matlab function clock returns a six-element vector c with elements c(1) c(2) c(3) c(4) c(5) c(6) = = = = = = year month day hour minute seconds
The first five elements are integers, while the sixth element has a fractional part that is accurate to milliseconds. The best way to print a clock vector is to use fprintf or sprintf with a specified format string that has both integer and floating point fields. f = '%6d %6d %6d %6d %6d %9.3f\n' I am revising this chapter on August 2, 2011, at a few minutes after 2:00pm, so c = clock; fprintf(f,c); produces 2011 In other words, year = 2011 month = 8 day = 2 hour = 14 minute = 2 seconds = 19.470 The Matlab functions datenum, datevec, datestr, and weekday use clock and facts about the Gregorian calendar to facilitate computations involving calendar dates. Dates are represented by their serial date number, which is the number of days since the theoretical time and day over 20 centuries ago when clock would have been six zeros. We can't pin that down to an actual date because different calendars would have been in use at that time. The function datenum returns the date number for any clock vector. For example, using the vector c that I just found, the current date number is datenum(c) is 734717.585 This indicates that the current time is a little over halfway through day number 734717. I get the same result from datenum(now) or just now The datenum function also works with a given year, month and day, or a date specified as a string. For example 8 2 14 2 19.470
35 datenum(2011,8,2) and datenum('Aug. 2, 2011') both return 734717 The same result is obtained from fix(now) In two and a half days the date number will be datenum(fix(now+2.5)) 734720 Computing the difference between two date numbers gives an elapsed time measured in days. How many days are left between today and the first day of next year? datenum('Jan 1, 2012') - datenum(fix(now)) ans = 152 The weekday function computes the day of the week, as both an integer between 1 and 7 and a string. For example both [d,w] = weekday(datenum(2011,8,2)) and [d,w] = weekday(now) both return d = 3 w = Tue So today is the third day of the week, a Tuesday.
Friday the 13th
Friday the 13th is unlucky, but is it unlikely? What is the probability that the 13th day of any month falls on a Friday? The quick answer is 1/7, but that is not quite right. The following code counts the number of times that Friday occurs on the various weekdays in a 400 year calendar cycle and produces figure 3.1. (You can also run friday13 yourself.)
So the 13th day of a month is more likely to be on a Friday than any other day of the week. The probability is 688/4800 = .143333. This probability is close to, but slightly larger than, 1/7 = .142857.
Biorhythms
Biorhythms were invented over 100 years ago and entered our popular culture in the 1960s. You can still find many Web sites today that offer to prepare personalized biorhythms, or that sell software to compute them. Biorhythms are based on the notion that three sinusoidal cycles influence our lives. The physical cycle has a
37 period of 23 days, the emotional cycle has a period of 28 days, and the intellectual cycle has a period of 33 days. For any individual, the cycles are initialized at birth. Figure 3.2 is my biorhythm, which begins on August 17, 1939, plotted for an eight-week period centered around the date this is being revised, July 27, 2011. It shows that I must be in pretty good shape. Today, I am near the peak of my intellectual cycle, and my physical and emotional cycles peaked on the same day less than a week ago.
birthday: 08/17/39 100 50 0 Physical −50Emotional Intellectual −100 06/29 07/06
07/13
07/20
07/27 08/03 07/27/11
08/10
08/17
08/24
Figure 3.2. My biorhythm. A search of the United States Government Patent and Trademark Office database of US patents issued between 1976 and 2010 finds 147 patents that are based upon or mention biorhythms. (There were just 113 in 2007.) The Web site is The date and graphics functions in Matlab make the computation and display of biorhythms particularly convenient. The following code segment is part of our program biorhythm.m that plots a biorhythm for an eight-week period centered on the current date. t0 = datenum('Aug. 17, 1939') t1 = fix(now); t = (t1-28):1:(t1+28); y = 100*[sin(2*pi*(t-t0)/23) sin(2*pi*(t-t0)/28) sin(2*pi*(t-t0)/33)]; plot(t,y) You see that the time variable t is measured in days and that the trig functions take arguments measured in radians.
38
Chapter 3. Calendars and Clocks
When is Easter?
Easter Day is one of the most important events in the Christian calendar. It is also one of the most mathematically elusive. In fact, regularization of the observance of Easter was one of the primary motivations for calendar reform. The informal rule is that Easter Day is the first Sunday after the first full moon after the vernal equinox. But the ecclesiastical full moon and equinox involved in this rule are not always the same as the corresponding astronomical events, which, after all, depend upon the location of the observer on the earth. Computing the date of Easter is featured in Don Knuth's classic The Art of Computer Programming and has consequently become a frequent exercise in programming courses. Our Matlab version of Knuth's program, easter.m, is the subject of several exercises in this chapter.
Exercises
3.1 Microcentury. The optimum length of a classroom lecture is one microcentury. How long is that? 3.2 π · 107 . A good estimate of the number of seconds in a year is π · 107 . How accurate is this estimate? 3.3 datestr. What does the following program do? for k = 1:31 disp(datestr(now,k)) end
3.5 Another clock. What does the following program do? clf set(gcf,'color','white') axis off t = text(0.0,0.5,' ','fontsize',16,'fontweight','bold'); while 1 s = dec2bin(fix(86400*now)); set(t,'string',s) pause(1) end Try help dec2bin if you need help. 3.6 How long? You should not try to run the following program. But if you were to run it, how long would it take? (If you insist on running it, change both 3's to 5's.) c = clock d = c(3) while d == c(3) c = clock; end 3.7 First datenum. The first countries to adopt the Gregorian calendar were Spain, Portugal and much of Italy. They did so on October 15, 1582, of the new calendar. The previous day was October 4, 1582, using the old, Julian, calendar. So October 5 through October 14, 1582, did not exist in these countries. What is the Matlab serial date number for October 15, 1582? 3.8 Future datenum's. Use datestr to determine when datenum will reach 750,000. When will it reach 1,000,000?
41 3.9 Your birthday. On which day of the week were you born? In a 400-year Gregorian calendar cycle, what is the probability that your birthday occurs on a Saturday? Which weekday is the most likely for your birthday? 3.10 Ops per century. Which does more operations, a human computer doing one operation per second for a century, or an electronic computer doing one operation per microsecond for a minute? 3.11 Julian day. The Julian Day Number (JDN) is commonly used to date astronomical observations. Find the definition of Julian Day on the Web and explain why JDN = datenum + 1721058.5 In particular, why does the conversion include an 0.5 fractional part? 3.12 Unix time. The Unix operating system and POSIX operating system standard measure time in seconds since 00:00:00 Universal time on January 1, 1970. There are 86,400 seconds in one day. Consequently, Unix time, time_t, can be computed in Matlab with time_t = 86400*(datenum(y,m,d) - datenum(1970,1,1)) Some Unix systems store the time in an 32-bit signed integer register. When will time_t exceed 231 and overflow on such systems. 3.13 Easter. (a) The comments in easter.m use the terms "golden number", "epact", and "metonic cycle". Find the definitions of these terms on the Web. (b) Plot a bar graph of the dates of Easter during the 21-st century. (c) How many times during the 21-st century does Easter occur in March and how many in April? (d) On how many different dates can Easter occur? What is the earliest? What is the latest? (e) Is the date of Easter a periodic function of the year number? 3.14 Biorhythms. (a) Use biorhythm to plot your own biorhythm, based on your birthday and centered around the current date. (b) All three biorhythm cycles start at zero when you were born. When do they return to this initial condition? Compute the m, the least common multiple of 23, 28, and 33. m = lcm(lcm(23,28),33) Now try biorhythm(fix(now)-m)
42
Chapter 3. Calendars and Clocks
What is special about this biorhythm? How old were you, or will you be, m days after you were born? (c) Is it possible for all three biorhythm cycles to reach their maximum at exactly the same time? Why or why not? Try t = 17003 biorhythm(fix(now)-t) What is special about this biorhythm? How many years is t days? At first glance, it appears that all three cycles are reaching maxima at the same time, t. But if you look more closely at sin(2*pi*t./[23 28 33]) you will see that the values are not all exactly 1.0. At what times near t = 17003 do the three cycles actually reach their maxima? The three times are how many hours apart? Are there any other values of t between 0 and the least common multiple, m, from the previous exercise where the cycles come so close to obtaining a simultaneous maximum? This code will help you answer these questions. m = lcm(lcm(23,28),33); t = (1:m)'; find(mod(t,23)==6 & mod(t,28)==7 & mod(t,33)==8) (d) Is it possible for all three biorhythm cycles to reach their minimum at exactly the same time? Why or why not? When do they nearly reach a simultaneous minimum?
16−Sep−2010
Figure 3.3. clockex
3.15 clockex. This exercise is about the clockex program in our exm toolbox and shown in figure 3.3. (a) Why does clockex use trig functions?
43 (b) Make clockex run counter-clockwise. (c) Why do the hour and minute hands in clockex move nearly continuously while the second hand moves in discrete steps. (d) The second hand sometime skips two marks. Why? How often? (e) Modify clockex to have a digital display of your own design.
44
Chapter 3. Calendars and Clocks
Chapter 4
Matrices
Matlab began as a matrix calculator.
The Cartesian coordinate system was developed in the 17th century by the French mathematician and philosopher Ren´ e Descartes. A pair of numbers corresponds to a point in the plane. We will display the coordinates in a vector of length two. In order to work properly with matrix multiplication, we want to think of the vector as a column vector, So ( ) x1 x= x2 denotes the point x whose first coordinate is x1 and second coordinate is x2 . When it is inconvenient to write a vector in this vertical form, we can anticipate Matlab notation and use a semicolon to separate the two components, x = ( x1 ; x2 ) For example, the point labeled x in figure 4.1 has Cartesian coordinates x = ( 2; 4 ) Arithmetic operations on the vectors are defined in natural ways. Addition is defined by ( ) ( ) ( ) x1 y1 x1 + y1 x+y = + = x2 y2 x2 + y2 Multiplication by a single number, or scalar, is defined by ( ) sx1 sx = sx2
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
47 This implies that points near x are transformed to points near Ax and that straight lines in the plane through x are transformed to straight lines through Ax. Our definition of matrix-vector multiplication is the usual one involving the dot product of the rows of A, denoted ai,: , with the vector x. ( ) a1,: · x Ax = a2,: · x An alternate, and sometimes more revealing, definition uses linear combinations of the columns of A, denoted by a:,j . Ax = x1 a:,1 + x2 a:,2 For example ( )( ) ( ) ( ) ( ) 4 −3 2 4 −3 −4 =2 +4 = −2 1 4 −2 1 0 The transpose of a column vector is a row vector, denoted by xT . The transpose of a matrix interchanges its rows and columns. For example, xT = ( 2 4 ) ( ) 4 −2 T A = −3 1 Vector-matrix multiplication can be defined by xT A = AT x That is pretty cryptic, so if you have never seen it before, you might have to ponder it a bit. Matrix-matrix multiplication, AB , can be thought of as matrix-vector multiplication involving the matrixA and the columns vectors from B , or as vector-matrix multiplication involving the row vectors from A and the matrix B . It is important to realize that AB is not the same matrix as BA. Matlab started its life as "Matrix Laboratory", so its very first capabilities involved matrices and matrix multiplication. The syntax follows the mathematical notation closely. We use square brackets instead of round parentheses, an asterisk to denote multiplication, and x' for the transpose of x. The foregoing example becomes x = [2; 4] A = [4 -3; -2 1] A*x This produces x = 2 4
The matrix ( ) 1 0 I= 0 1 is the 2-by-2 identity matrix. It has the important property that for any 2-by-2 matrix A, IA = AI = A Originally, Matlab variable names were not case sensitive, so i and I were the same variable. Since i is frequently used as a subscript, an iteration index, and sqrt(-1), we could not use I for the identity matrix. Instead, we chose to use the sound-alike word eye. Today, Matlab is case sensitive and has many users whose native language is not English, but we continue to use eye(n,n) to denote the n-by-n identity. (The Metro in Washington, DC, uses the same pun – "I street" is "eye street" on their maps.)
The columns of X are the Cartesian coordinates of the 11 points shown in figure 4.2. Do you remember the "dot to dot" game? Try it with these points. Finish off by connecting the last point back to the first. The house in figure 4.2 is constructed from X by
Figure 4.3 uses matrix multiplication A*X and dot2dot(A*X) to show the effect of the resulting linear transformations on the house. All four matrices are diagonal or antidiagonal, so they just scale and possibly interchange the coordinates. The coordinates are not combined in any way. The floor and sides of the house remain at
Figure 4.3. The effect of multiplication by scaling matrices. right angles to each other and parallel to the axes. The matrix A1 shrinks the first coordinate to reduce the width of the house while the height remains unchanged. The matrix A2 shrinks the second coordinate to reduce the height, but not the width. The matrix A3 interchanges the two coordinates while shrinking one of them. The matrix A4 shrinks the first coordinate and changes the sign of the second. The determinant of a 2-by-2 matrix ( ) a1,1 a1,2 A= a2,1 a2,2 is the quantity a1,1 a2,2 − a1,2 a2,1 In general, determinants are not very useful in practical computation because they have atrocious scaling properties. But 2-by-2 determinants can be useful in understanding simple matrix properties. If the determinant of a matrix is positive, then multiplication by that matrix preserves left- or right-handedness. The first two of our four matrices have positive determinants, so the door remains on the left side of the house. The other two matrices have negative determinants, so the door is transformed to the other side of the house. The Matlab function rand(m,n) generates an m-by-n matrix with random entries between 0 and 1. So the statement
Figure 4.4 shows the effect of multiplication by these four matrices on the house. Matrices R1 and R4 have large off-diagonal entries and negative determinants, so they distort the house quite a bit and flip the door to the right side. The lines are still straight, but the walls are not perpendicular to the floor. Linear transformations preserve straight lines, but they do not necessarily preserve the angles between those lines. Matrix R2 is close to a rotation, which we will discuss shortly. Matrix R3 is nearly singular ; its determinant is equal to 0.0117. If the determinant were exactly zero, the house would be flattened to a one-dimensional straight line. The following matrix is a plane rotation. ( ) cos θ − sin θ G(θ) = sin θ cos θ We use the letter G because Wallace Givens pioneered the use of plane rotations in matrix computation in the 1950s. Multiplication by G(θ) rotates points in the plane through an angle θ. Figure 4.5 shows the effect of multiplication by the plane rotations with θ = 15◦ , 45◦ , 90◦ , and 215◦ .
10 5 0 −5 −10 −10 10 5 0 −5 −10 −10 0 G90 10 0 G15 10 10 5 0 −5 −10 −10 10 5 0 −5 −10 −10 0 G215 10 0 G45 10
You can see that G45 is fairly close to the random matrix R2 seen earlier and that its effect on the house is similar. Matlab generates a plane rotation for angles measured in radians with G = [cos(theta) -sin(theta); sin(theta) cos(theta)] and for angles measured in degrees with G = [cosd(theta) -sind(theta); sind(theta) cosd(theta)] Our exm toolbox function wiggle uses dot2dot and plane rotations to produce an animation of matrix multiplication. Here is wiggle.m, without the Handle Graphics commands. function wiggle(X) thetamax = 0.1; delta = .025; t = 0; while true theta = (4*abs(t-round(t))-1) * thetamax; G = [cos(theta) -sin(theta); sin(theta) cos(theta)] Y = G*X; dot2dot(Y); t = t + delta; end Since this version does not have a stop button, it would run forever. The variable t advances steadily by increment of delta. As t increases, the quantity t-round(t) varies between −1/2 and 1/2, so the angle θ computed by theta = (4*abs(t-round(t))-1) * thetamax; varies in a sawtooth fashion between -thetamax and thetamax. The graph of θ as a function of t is shown in figure 4.6. Each value of θ produces a corresponding plane rotation G(θ). Then
Vectors and Matrices
Here is a quick look at a few of the many Matlab operations involving vectors and matrices. Try to predict what output will be produced by each of the following statements. You can see if you are right by using cut and paste to execute the statement, or by running matrices_recap Vectors are created with square brackets. v = [0 1/4 1/2 3/4 1] Rows of a matrix are separated by semicolons or new lines. A = [8 1 6; 3 5 7; 4 9 2] There are several functions that create matrices. Z = zeros(3,4) E = ones(4,3) I = eye(4,4) M = magic(3) R = rand(2,4) [K,J] = ndgrid(1:4) A colon creates uniformly spaced vectors. v = 0:0.25:1 n = 10 y = 1:n A semicolon at the end of a line suppresses output. n = 1000; y = 1:n;
55
Matrix arithmetic
Matrix addition and subtraction are denoted by + and - . The operations A + B and A - B require A and B to be the same size, or to be scalars, which are 1-by-1 matrices. Matrix multiplication, denoted by *, follows the rules of linear algebra. The operation A * B requires the number of columns of A to equal the number of row B, that is size(A,2) == size(B,1) Remember that A*B is usually not equal to B*A If p is an integer scalar, the expression A^p denotes repeated multiplication of A by itself p times. The use of the matrix division operations in Matlab, A \ B and A / B is discussed in our "Linear Equations" chapter
Further Reading
Of the dozens of good books on matrices and linear algebra, we would like to recommend one in particular. Gilbert Strang, Introduction to Linear Algebra, Wellesley-Cambridge Press, Wellesley, MA, 2003. Besides its excellent technical content and exposition, it has a terrific cover. The house that we have used throughout this chapter made its debut in Strang's book in 1993. The cover of the first edition looked something like our figure 4.4. Chris Curtis saw that cover and created a gorgeous quilt. A picture of the quilt has appeared on the cover of all subsequent editions of the book.
Recap
%% Matrices Chapter Recap
57 % % % % % % % % % % % % % % This is an executable program that illustrates the statements introduced in the Matrices Chapter of "Experiments in MATLAB". You can access it with matrices_recap edit matrices_recap publish matrices_recap Related EXM Programs wiggle dot2dot house hand
4.9 GT . What is the effect on points in the plane of multiplication by G(θ)T ? 4.10 G. (a) What is the effect on points in the plane of multiplication by ( ) cos θ sin θ G(θ) = sin θ − cos θ (b) What is the determinant of G(θ)? (c) What happens if you modify wiggle.m to use G instead of G? 4.11 Goldie. What does the function goldie in the exm toolbox do? 4.12 Transform a hand. Repeat the experiments in this chapter with X = hand instead of X = house Figure 4.8 shows dot2dot(hand)
4.13 Mirror image. Find a 2-by-2 matrix R so that dot2dot(house) and dot2dot(R*house) as well as dot2dot(hand) and dot2dot(R*hand)
61
10 8 6 4 2 0 −2 −4 −6 −8 −10 −10 −5 0 5 10
Figure 4.8. A hand.
are mirror images of each other. 4.14 Transform your own hand. Repeat the experiments in this chapter using a plot of your own hand. Start with figure('position',get(0,'screensize')) axes('position',[0 0 1 1]) axis(10*[-1 1 -1 1]) [x,y] = ginput; Place your hand on the computer screen. Use the mouse to select a few dozen points outlining your hand. Terminate the ginput with a carriage return. You might find it easier to trace your hand on a piece of paper and then put the paper on the computer screen. You should be able to see the ginput cursor through the paper. The data you have collected forms two column vectors with entries in the range from -10 to 10. You can arrange the data as two rows in a single matrix with H = [x y]'; Then you can use dot2dot(H) dot2dot(A*H) wiggle(H) and so on. You can save your data in the file myhand.mat with save myhand H and retrieve it in a later Matlab session with load myhand
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Chapter 4. Matrices
4.15 Wiggler. Make wiggler.m, your own version of wiggle.m, with two sliders that control the speed and amplitude. In the initialization, replace the statements thetamax = 0.1; delta = .025; with thetamax = uicontrol('style','slider','max',1.0, ... 'units','normalized','position',[.25 .01 .25 .04]); delta = uicontrol('style','slider','max',.05, ... 'units','normalized','position',[.60 .01 .25 .04]); The quantities thetamax and delta are now the handles to the two sliders. In the body of the loop, replace thetamax by get(thetamax,'value'); and replace delta by get(delta,'value'); Demonstrate your wiggler on the house and the hand.
Chapter 5
Linear Equations
The most important task in technical computing.
I am thinking of two numbers. Their average is 3. What are the numbers? Please remember the first thing that pops into your head. I will get back to this problem in a few pages. Solving systems of simultaneous linear equations is the most important task in technical computing. It is not only important in its own right, it is also a fundamental, and often hidden, component of other more complicated computational tasks. The very simplest linear equations involve only one unknown. Solve 7x = 21 The answer, of course, is x= Now solve ax = b The answer, of course, is b a But what if a = 0? Then we have to look at b. If b ̸= 0 then there is no value of x that satisfies x= 0x = b
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
21 =3 7
63
64
Chapter 5. Linear Equations
The solution does not exist. On the other hand, if b = 0 then any value of x satisfies 0x = 0 The solution is not unique. Mathematicians have been thinking about existence and uniqueness for centuries. We will see that these concepts are also important in modern technical computing. Here is a toy story problem. Alice buys three apples, a dozen bananas, and one cantaloupe for $2.36. Bob buys a dozen apples and two cantaloupes for $5.26. Carol buys two bananas and three cantaloupes for $2.77. How much do single pieces of each fruit cost? Let x1 , x2 , and x3 denote the unknown price of each fruit. We have three equations in three unknowns. 3x1 + 12x2 + x3 = 2.36 12x1 + 2x3 = 5.26 2x2 + 3x3 = 2.77 Because matrix-vector multiplication has been defined the way it has, these equations can be written 3 12 1 x1 2.36 12 0 2 x2 = 5.27 0 2 3 x3 2.77 Or, simply Ax = b where A is a given 3-by-3 matrix, b is a given 3-by-1 column vector, and x is a 3-by-1 column vector with unknown elements. We want to solve this equation. If you know anything about matrices, you know that the equation can be solved using A−1 , the inverse of A, x = A−1 b This is a fine concept theoretically, but not so good computationally We don't really need A−1 , we just want to find x. If you do not know anything about matrices, you might be tempted to divide both sides of the equation by A. x= b A
This is a terrible idea theoretically – you can't divide by matrices – but it is the beginning of a good idea computationally. To find the solution to a linear system of equations with Matlab, start by entering the matrix of coefficients.
65 A = [3 12 1; 12 0 2; 0 2 3] Since all the elements of A are integers, the matrix is printed with an integer format. A = 3 12 0 12 0 2 1 2 3
Next, enter the right hand side as a column vector. b = [2.36 5.26 2.77]' The elements of b are not integers, so the default format shows four digits after the decimal point. b = 2.3600 5.2600 2.7700 Matlab has an output format intended for financial calculations, like this fruit price calculation. The command format bank changes the output to show only two digits after the decimal point. b = 2.36 5.26 2.77 In Matlab the solution to the linear system of equations Ax = b is found using the backslash operator. x = A\b Think of this as "dividing" both sides of the equation by A. The result is x = 0.29 0.05 0.89 This gives us the solution to our story problem – apples cost 29 cents each, bananas are a nickel each, and cantaloupes are 89 cents each. Very rarely, systems of linear equations come in the form xA = b where b and x are row vectors. In this case, the solution is found using the forward slash operator.
66 x = b/A
Chapter 5. Linear Equations
The two operations A\b and b/A are sometimes called left and right matrix division. In both cases, the coefficient matrix is in the "denominator". For scalars, left and right division are the same thing. The quantities 7\21 and 21/7 are both equal to 3.
Singular matrix
Let's change our story problem a bit. Assume now that Carol buys six apples and one cantaloupe for $2.77. The coefficient matrix and right hand side become A = 3 12 6 and b = 2.36 5.26 2.77 At first glance, this does not look like much of a change. However, x = A\b produces Warning: Matrix is singular to working precision. x = NaN -Inf Inf Inf and -Inf stand for plus and minus infinity and result from division of nonzero numbers by zero. NaN stands for "Not-a-Number" and results from doing arithmetic involving infinities. The source of the difficulty is that the new information about Carol's purchase is inconsistent with the earlier information about Alice's and Bob's purchases. We have said that Carol bought exactly half as much fruit as Bob. But she paid 2.77 when half of Bob's payment would have been only 2.63. The third row of A is equal to one-half of the second row, but b(3) is not equal to one-half of b(2). For this particular matrix A and vector b, the solution to the linear system of equations Ax = b does not exist. What if we make Carol's purchase price consistent with Bob's? We leave A unchanged and revise b with b(3) = 2.63 12 0 0 1 2 1
67 so b = 2.36 5.26 2.63 Now we do not have enough information. Our last two equations are scalar multiples of each other. 12x1 + 2x3 = 5.26 6x1 + x3 = 2.63 One possible solution is the solution to the original problem. x = 0.29 0.05 0.89 A*x = 2.36 5.26 2.63 But we can pick an arbitrary value for any component of the solution and then use the first equation and one of the last two equations to compute the other components. The result is a solution to all three equations. For example, here is a solution with its third component equal to zero. y = A(1:2,1:2)\b(1:2); y(3) = 0 y = 0.44 0.09 0.00 A*y = 2.36 5.26 2.63 There are infinitely many more. For this particular matrix A and vector b, the solution to Ax = b is not unique. The family of possible solutions is generated by the null vector of A. This is a nonzero vector z for which Az = 0 The general form of the solution is one particular solution, say our vector x, plus any arbitrary parameter times the null vector. For any value of t the vector y = x + tz
68 is a solution to Ay = b In Matlab z = null(A) A*z t = rand y = x + t*z A*y You can see that A*z is zero and A*y is equal to b.
Chapter 5. Linear Equations
"Their average is three."
Let's return to the question that I asked you to consider at the beginning of this chapter. I'm thinking of two numbers. Their average is three. What are the numbers? What popped into your head? You probably realized that I hadn't given you enough information. But you must have thought of some solution. In my experience, the most frequent response is "2 and 4". Let's see what Matlab responds. My problem is one linear equation in two unknowns The matrix and right hand side are A = [1/2 1/2] b = 3 We want to solve Ax = b. This is now an underdetermined system. There are fewer equations than unknowns, so there are infinitely many solutions. Backslash offers one possible solution x = A\b x = 6 0 I bet you didn't think of this solution. If we try inv(A)*b we get an error message because rectangular matrices do not have inverses. But there is something called the pseudoinverse. We can try that. x = pinv(A)*b x = 3 3
69 Did that solution occur to you? These two x's are just two members of the infinite family of solutions. If we wanted Matlab to find the [2 4]' solution, we would have to pose a problem where the solution is constrained to be a pair of integers, close to each other, but not equal. It is possible to solve such problems, but that would take us too far afield.
My Rules.
This chapter illustrates two fundamental facts about technical computing. • The hardest quantities to compute are ones that do not exist. • The next hardest are ones that are not unique.
at joints 2, 5, and 6, and we want to determine the resulting force on each member of the truss. For the truss to be in static equilibrium, there must be no net force, horizontally or vertically, at any joint. Thus, we can determine the member forces by equating the horizontal forces to the left and right at each joint, and similarly equating the vertical forces upward and downward at each joint. For the eight joints, this would give 16 equations, which is more than the 13 unknown factors to be determined. For the truss to be statically determinate, that is, for there to be a unique solution, we assume that joint 1 is rigidly fixed both horizontally and vertically and that joint 8 is fixed vertically. Resolving√ the member forces into horizontal and vertical components and defining α = 1/ 2, we obtain the following system of equations for the member forces fi : Joint 2: Joint 3: Joint 4: Joint 5: Joint 6: Joint 7: Joint 8: f2 = f6 , f3 = 10; αf1 = f4 + αf5 , αf1 + f3 + αf5 = 0; f4 = f8 , f7 = 0; αf5 + f6 = αf9 + f10 , αf5 + f7 + αf9 = 15; f10 = f13 , f11 = 20; f8 + αf9 = αf12 , αf9 + f11 + αf12 = 0; f13 + αf12 = 0.
Solve this system of equations to find the vector f of member forces.
73 5.7 Circuit. Figure 5.2 is the circuit diagram for a small network of resistors.
3
r23 2 r25
r13 i2 r12 1 i1
r34 i3 r14 r45 4 i4 vs
5
r35
Figure 5.2. A resistor network. There are five nodes, eight resistors, and one constant voltage source. We want to compute the voltage drops between the nodes and the currents around each of the loops. Several different linear systems of equations can be formed to describe this circuit. Let vk , k = 1, . . . , 4, denote the voltage difference between each of the first four nodes and node number 5 and let ik , k = 1, . . . , 4, denote the clockwise current around each of the loops in the diagram. Ohm's law says that the voltage drop across a resistor is the resistance times the current. For example, the branch between nodes 1 and 2 gives v1 − v2 = r12 (i2 − i1 ). Using the conductance, which is the reciprocal of the resistance, gkj = 1/rkj , Ohm's law becomes i2 − i1 = g12 (v1 − v2 ). The voltage source is included in the equation v3 − vs = r35 i4 . Kirchhoff 's voltage law says that the sum of the voltage differences around each loop is zero. For example, around loop 1, (v1 − v4 ) + (v4 − v5 ) + (v5 − v2 ) + (v2 − v1 ) = 0. Combining the voltage law with Ohm's law leads to the loop equations for the currents: Ri = b. Here i is the current vector, i1 i2 i = , i3 i4
You can solve the linear system obtained from the loop equations to compute the currents and then use Ohm's law to recover the voltages. Or you can solve the linear system obtained from the node equations to compute the voltages and then use Ohm's law to recover the currents. Your assignment is to verify that these two approaches produce the same results for this circuit. You can choose your own numerical values for the resistances and the voltage source.
Chapter 6
Fractal Fern
The fractal fern involves 2-by-2 matrices.
The programs fern and finitefern in the exm toolbox produce the Fractal Fern described by Michael Barnsley in Fractals Everywhere [?]. They generate and plot a potentially infinite sequence of random, but carefully choreographed, points in the plane. The command fern runs forever, producing an increasingly dense plot. The command finitefern(n) generates n points and a plot like Figure 6.1. The command finitefern(n,'s') shows the generation of the points one at a time. The command F = finitefern(n); generates, but does not plot, n points and returns an array of zeros and ones for use with sparse matrix and image-processing functions. The exm toolbox also includes fern.jpg, a 768-by-1024 color image with half a million points that you can view with a browser or a paint program. You can also view the file with F = imread('fern.png'); image(F)
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
75
76
Chapter 6. Fractal Fern
Figure 6.1. Fractal fern. If you like the image, you might even choose to make it your computer desktop background. However, you should really run fern on your own computer to see the dynamics of the emerging fern in high resolution. The fern is generated by repeated transformations of a point in the plane. Let x be a vector with two components, x1 and x2 , representing the point. There are four different transformations, all of them of the form x → Ax + b, with different matrices A and vectors b. These are known as affine transformations. The most frequently used transformation has ( ) ( ) 0.85 0.04 0 A= , b= . −0.04 0.85 1.6 This transformation shortens and rotates x a little bit, then adds 1.6 to its second component. Repeated application of this transformation moves the point up and to the right, heading toward the upper tip of the fern. Every once in a while, one of the other three transformations is picked at random. These transformations move
stands for "show graph window." It brings an existing graphics window forward, or creates a new one if necessary. clf reset resets most of the figure properties to their default values. set(gcf,'color','white','menubar','none', ... 'numbertitle','off','name','Fractal Fern') changes the background color of the figure window from the default gray to white and provides a customized title for the window. x = [.5; .5]; provides the initial coordinates of the point. h = plot(x(1),x(2),'.'); plots a single dot in the plane and saves a handle, h, so that we can later modify the properties of the plot. darkgreen = [0 2/3 0]; defines a color where the red and blue components are zero and the green component is two-thirds of its full intensity. set(h,'markersize',1,'color',darkgreen,'erasemode','none'); makes the dot referenced by h smaller, changes its color, and specifies that the image of the dot on the screen should not be erased when its coordinates are changed. A record of these old points is kept by the computer's graphics hardware (until the figure is reset), but Matlab itself does not remember them. axis([-3 3 0 10]) axis off specifies that the plot should cover the region −3 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 10, but that the axes should not be drawn.
79 stop = uicontrol('style','toggle','string','stop', ... 'background','white'); creates a toggle user interface control, labeled 'stop' and colored white, in the default position near the lower left corner of the figure. The handle for the control is saved in the variable stop. drawnow causes the initial figure, including the initial point, to actually be plotted on the computer screen. The statement p = [ .85 .92 .99 1.00];
define the four affine transformations. The statement cnt = 1; initializes a counter that keeps track of the number of points plotted. The statement tic initializes a stopwatch timer. The statement while ~get(stop,'value') begins a while loop that runs as long as the 'value' property of the stop toggle is equal to 0. Clicking the stop toggle changes the value from 0 to 1 and terminates the loop. r = rand; generates a pseudorandom value between 0 and 1. The compound if statement if r < x = elseif x = elseif x = else x = end p(1) A1*x + b1; r < p(2) A2*x + b2; r < p(3) A3*x + b3; A4*x;
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Chapter 6. Fractal Fern
picks one of the four affine transformations. Because p(1) is 0.85, the first transformation is chosen 85% of the time. The other three transformations are chosen relatively infrequently. set(h,'xdata',x(1),'ydata',x(2)); changes the coordinates of the point h to the new (x1 , x2 ) and plots this new point. But get(h,'erasemode') is 'none', so the old point also remains on the screen. cnt = cnt + 1; counts one more point. drawnow tells Matlab to take the time to redraw the figure, showing the new point along with all the old ones. Without this command, nothing would be plotted until stop is toggled. end matches the while at the beginning of the loop. Finally, t = toc; reads the timer. s = sprintf('%8.0f points in %6.3f seconds',cnt,t); text(-1.5,-0.5,s,'fontweight','bold'); displays the elapsed time since tic was called and the final count of the number of points plotted. Finally, set(stop,'style','pushbutton','string','close', ... 'callback','close(gcf)') changes the control to a push button that closes the window.
Recap
%% Fern Chapter Recap % This is an executable program that illustrates the statements % introduced in the Fern Chapter of "Experiments in MATLAB". % You can access it with % % fern_recap % edit fern_recap % publish fern_recap % % Related EXM programs %
Exercises
6.1 Fern color. Change the fern color scheme to use pink on a black background. Don't forget the stop button. 6.2 Flip the fern. Flip the fern by interchanging its x- and y -coordinates. 6.3 Erase mode. (a) What happens if you resize the figure window while the fern is being generated? Why? (b) The exm program finitefern can be used to produce printed output of the fern. Explain why printing is possible with finitefern.m but not with fern.m. 6.4 Fern stem. (a) What happens to the fern if you change the only nonzero element in the matrix A4? (b) What are the coordinates of the lower end of the fern's stem?
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6.5 Fern tip. The coordinates of the point at the upper tip end of the fern can be computed by solving a certain 2-by-2 system of simultaneous linear equations. What is that system and what are the coordinates of the tip? 6.6 Trajectories. The fern algorithm involves repeated random choices from four different formulas for advancing the point. If the k th formula is used repeatedly by itself, without random choices, it defines a deterministic trajectory in the (x, y ) plane. Modify finitefern.m so that plots of each of these four trajectories are superimposed on the plot of the fern. Start each trajectory at the point (−1, 5). Plot o's connected with straight lines for the steps along each trajectory. Take as many steps as are needed to show each trajectory's limit point. You can superimpose several plots with plot(...) hold on plot(...) plot(...) hold off
6.7 Sierpinski's triangle. Modify fern.m or finitefern.m so that it produces Sierpinski's triangle. Start at ( ) 0 x= . 0 At each iterative step, the current point x is replaced with Ax + b, where the matrix A is always ( ) 1/2 0 A= 0 1/2 and the vector b is chosen at random with equal probability from among the three vectors ( ) ( ) ( ) 0 1/2 1/4 b= , b= , and b = √ . 0 0 3/4
Chapter 7
Google PageRank
The world's largest matrix computation. (This chapter is out of date and needs a major overhaul.)
One of the reasons why GoogleTM is such an effective search engine is the PageRankTM algorithm developed by Google's founders, Larry Page and Sergey Brin, when they were graduate students at Stanford University. PageRank is determined entirely by the link structure of the World Wide Web. It is recomputed about once a month and does not involve the actual content of any Web pages or individual queries. Then, for any particular query, Google finds the pages on the Web that match that query and lists those pages in the order of their PageRank. Imagine surfing the Web, going from page to page by randomly choosing an outgoing link from one page to get to the next. This can lead to dead ends at pages with no outgoing links, or cycles around cliques of interconnected pages. So, a certain fraction of the time, simply choose a random page from the Web. This theoretical random walk is known as a Markov chain or Markov process. The limiting probability that an infinitely dedicated random surfer visits any particular page is its PageRank. A page has high rank if other pages with high rank link to it. Let W be the set of Web pages that can be reached by following a chain of hyperlinks starting at some root page, and let n be the number of pages in W . For Google, the set W actually varies with time, but by June 2004, n was over 4 billion. Let G be the n-by-n connectivity matrix of a portion of the Web, that is, gij = 1 if there is a hyperlink to page i from page j and gij = 0 otherwise. The matrix G can be huge, but it is very sparse. Its j th column shows the links on the j th page. The number of nonzeros in G is the total number of hyperlinks in W .
The quantities rj and cj are the in-degree and out-degree of the j th page. Let p be the probability that the random walk follows a link. A typical value is p = 0.85. Then 1 − p is the probability that some arbitrary page is chosen and δ = (1 − p)/n is the probability that a particular random page is chosen. Let A be the n-by-n matrix whose elements are { pgij /cj + δ : cj ̸= 0 aij = 1/n : cj = 0. Notice that A comes from scaling the connectivity matrix by its column sums. The j th column is the probability of jumping from the j th page to the other pages on the Web. If the j th page is a dead end, that is has no out-links, then we assign a uniform probability of 1/n to all the elements in its column. Most of the elements of A are equal to δ , the probability of jumping from one page to another without following a link. If n = 4 · 109 and p = 0.85, then δ = 3.75 · 10−11 . The matrix A is the transition probability matrix of the Markov chain. Its elements are all strictly between zero and one and its column sums are all equal to one. An important result in matrix theory known as the Perron–Frobenius theorem applies to such matrices. It concludes that a nonzero solution of the equation x = Ax exists and is unique to within a scaling factor. If this scaling factor is chosen so that ∑ xi = 1,
i
then x is the state vector of the Markov chain and is Google's PageRank. The elements of x are all positive and less than one. The vector x is the solution to the singular, homogeneous linear system (I − A)x = 0. For modest n, an easy way to compute x in Matlab is to start with some approximate solution, such as the PageRanks from the previous month, or x = ones(n,1)/n Then simply repeat the assignment statement x = A*x until successive vectors agree to within a specified tolerance. This is known as the power method and is about the only possible approach for very large n. In practice, the matrices G and A are never actually formed. One step of the power method would be done by one pass over a database of Web pages, updating weighted reference counts generated by the hyperlinks between pages.
85 The best way to compute PageRank in Matlab is to take advantage of the particular structure of the Markov matrix. Here is an approach that preserves the sparsity of G. The transition matrix can be written A = pGD + ez T where D is the diagonal matrix formed from the reciprocals of the outdegrees, { 1/cj : cj ̸= 0 djj = 0 : cj = 0, e is the n-vector of all ones, and z is the vector with components { δ : cj ̸= 0 zj = 1/n : cj = 0. The rank-one matrix ez T accounts for the random choices of Web pages that do not follow links. The equation x = Ax can be written (I − pGD)x = γe where γ = z T x. We do not know the value of γ because it depends upon the unknown vector x, but we can temporarily take γ = 1. As long as p is strictly less than one, the coefficient matrix I − pGD is nonsingular and the equation (I − pGD)x = e can be solved for x. Then the resulting x can be rescaled so that ∑ xi = 1.
i
until x settles down to several decimal places. It is also possible to use an algorithm known as inverse iteration. A = p*G*D + delta x = (I - A)\e x = x/sum(x) At first glance, this appears to be a very dangerous idea. Because I − A is theoretically singular, with exact computation some diagonal element of the upper triangular factor of I − A should be zero and this computation should fail. But with roundoff error, the computed matrix I - A is probably not exactly singular. Even if it is singular, roundoff during Gaussian elimination will most likely prevent any exact zero diagonal elements. We know that Gaussian elimination with partial pivoting always produces a solution with a small residual, relative to the computed solution, even if the matrix is badly conditioned. The vector obtained with the backslash operation, (I - A)\e, usually has very large components. If it is rescaled by its sum, the residual is scaled by the same factor and becomes very small. Consequently, the two vectors x and A*x equal each other to within roundoff error. In this setting, solving the singular system with Gaussian elimination blows up, but it blows up in exactly the right direction. Figure 7.1 is the graph for a tiny example, with n = 6 instead of n = 4 · 109 . Pages on the Web are identified by strings known as uniform resource locators, or URLs. Most URLs begin with http because they use the hypertext transfer protocol. In Matlab , we can store the URLs as an array of strings in a cell array. This example involves a 6-by-1 cell array. U = {' ' ' ' ' ' Two different kinds of indexing into cell arrays are possible. Parentheses denote subarrays, including individual cells, and curly braces denote the contents of the cells. If k is a scalar, then U(k) is a 1-by-1 cell array consisting of the kth cell in U, while U{k} is the string in that cell. Thus U(1) is a single cell and U{1} is the string ' Think of mail boxes with addresses on a city street. B(502) is the box at number 502, while B{502} is the mail in that box.
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alpha
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Figure 7.1. A tiny Web. We can generate the connectivity matrix by specifying the pairs of indices (i,j) of the nonzero elements. Because there is a link to beta.com from alpha.com, the (2,1) element of G is nonzero. The nine connections are described by i = [ 2 6 3 4 4 5 6 1 1] j = [ 1 1 2 2 3 3 3 4 6] A sparse matrix is stored in a data structure that requires memory only for the nonzero elements and their indices. This is hardly necessary for a 6-by-6 matrix with only 27 zero entries, but it becomes crucially important for larger problems. The statements n = 6 G = sparse(i,j,1,n,n); full(G) generate the sparse representation of an n-by-n matrix with ones in the positions specified by the vectors i and j and display its full representation. 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
89 We see that alpha has a higher PageRank than delta or sigma, even though they all have the same number of in-links. A random surfer will visit alpha over 32% of the time and rho only about 6% of the time. For this tiny example with p = .85, the smallest element of the Markov transition matrix is δ = .15/6 = .0250. A = 0.0250 0.4500 0.0250 0.0250 0.0250 0.4500 0.0250 0.0250 0.4500 0.4500 0.0250 0.0250 0.0250 0.0250 0.0250 0.3083 0.3083 0.3083 0.8750 0.0250 0.0250 0.0250 0.0250 0.0250 0.1667 0.1667 0.1667 0.1667 0.1667 0.1667 0.8750 0.0250 0.0250 0.0250 0.0250 0.0250
Notice that the column sums of A are all equal to one. The exm toolbox includes the program surfer. A statement like [U,G] = surfer(' starts at a specified URL and tries to surf the Web until it has visited n pages. If successful, it returns an n-by-1 cell array of URLs and an n-by-n sparse connectivity matrix. The function uses urlread, which was introduced in Matlab 6.5, along with underlying Java utilities to access the Web. Surfing the Web automatically is a dangerous undertaking and this function must be used with care. Some URLs contain typographical errors and illegal characters. There is a list of URLs to avoid that includes .gif files and Web sites known to cause difficulties. Most importantly, surfer can get completely bogged down trying to read a page from a site that appears to be responding, but that never delivers the complete page. When this happens, it may be necessary to have the computer's operating system ruthlessly terminate Matlab. With these precautions in mind, you can use surfer to generate your own PageRank examples. The statement [U,G] = surfer(' accesses the home page of Harvard University and generates a 500-by-500 test case. The graph generated in August 2003 is available in the exm toolbox. The statements load harvard500 spy(G) produce a spy plot (Figure 7.3) that shows the nonzero structure of the connectivity matrix. The statement pagerank(U,G) computes page ranks, produces a bar graph (Figure 7.4) of the ranks, and prints the most highly ranked URLs in PageRank order. For the harvard500 data, the dozen most highly ranked pages are
The URL where the search began, dominates. Like most universities, Harvard is organized into various colleges Harvard Medical School, the Harvard Business School, and the Radcliffe Institute. You can see that the home pages of these schools have high PageRank. With a different sample, such as the one generated by Google itself, the ranks would be different.
Further Reading
Further reading on matrix computation includes books by Demmel [?], Golub and Van Loan [?], Stewart [?, ?], and Trefethen and Bau [?]. The definitive references on Fortran matrix computation software are the LAPACK Users' Guide and Web site [?]. The Matlab sparse matrix data structure and operations are described in [?]. Information available on Web sites about PageRank includes a brief explanation at Google [?], a technical report by Page, Brin, and colleagues [?], and a comprehensive survey by Langville and Meyer [?].
Exercises
7.1 Use surfer and pagerank to compute PageRanks for some subset of the Web that you choose. Do you see any interesting structure in the results? 7.2 Suppose that U and G are the URL cell array and the connectivity matrix produced by surfer and that k is an integer. Explain what U{k}, U(k), G(k,:), G(:,k), U(G(k,:)), U(G(:,k)) are. 7.3 The connectivity matrix for the harvard500 data set has four small, almost entirely nonzero, submatrices that produce dense patches near the diagonal of the spy plot. You can use the zoom button to find their indices. The first submatrix has indices around 170 and the other three have indices in the 200s and 300s. Mathematically, a graph with every node connected to every other node is known as a clique. Identify the organizations within the Harvard community that are responsible for these near cliques. 7.4 A Web connectivity matrix G has gij = 1 if it is possible to get to page i from page j with one click. If you multiply the matrix by itself, the entries of the matrix G2 count the number of different paths of length two to page i from page j . The matrix power Gp shows the number of paths of length p. (a) For the harvard500 data set, find the power p where the number of nonzeros stops increasing. In other words, for any q greater than p, nnz(G^q) is equal to nnz(G^p). (b) What fraction of the entries in Gp are nonzero? (c) Use subplot and spy to show the nonzeros in the successive powers. (d) Is there a set of interconnected pages that do not link to the other pages? 7.5 The function surfer uses a subfunction, hashfun, to speed up the search for a possibly new URL in the list of URLs that have already been processed. Find two different URLs on The MathWorks home page that have the same hashfun value. 7.6 Figure 7.5 is the graph of another six-node subset of the Web. In this example, there are two disjoint subgraphs. (a) What is the connectivity matrix G? (b) What are the PageRanks if the hyperlink transition probability p is the default value 0.85? (c) Describe what happens with this example to both the definition of PageRank and the computation done by pagerank in the limit p → 1. 7.7 The function pagerank(U,G) computes PageRanks by solving a sparse linear
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Chapter 7. Google PageRank
alpha
delta
beta
gamma
sigma
rho
Figure 7.5. Another tiny Web. system. It then plots a bar graph and prints the dominant URLs. (a) Create pagerank1(G) by modifying pagerank so that it just computes the PageRanks, but does not do any plotting or printing. (b) Create pagerank2(G) by modifying pagerank1 to use inverse iteration instead of solving the sparse linear system. The key statements are x = (I - A)\e x = x/sum(x) What should be done in the unlikely event that the backslash operation involves a division by zero? (c) Create pagerank3(G) by modifying pagerank1 to use the power method instead of solving the sparse linear system. The key statements are G = p*G*D z = ((1-p)*(c~=0) + (c==0))/n; while termination_test x = G*x + e*(z*x) end What is an appropriate test for terminating the power iteration? (d) Use your functions to compute the PageRanks of the six-node example discussed in the text. Make sure you get the correct result from each of your three functions. 7.8 Here is yet another function for computing PageRank. This version uses the power method, but does not do any matrix operations. Only the link structure of the connectivity matrix is involved. function [x,cnt] = pagerankpow(G)
I hope you have a live Matlab and the exm functions handy. Enter the statement expgui Click on the blue line with your mouse. Move it until the green line is on top of the blue line. What is the resulting value of a? The exponential function is denoted mathematically by et and in Matlab by exp(t). This function is the solution to the world's simplest, and perhaps most important, differential equation, y ˙ = ky This equation is the basis for any mathematical model describing the time evolution of a quantity with a rate of production that is proportional to the quantity itself. Such models include populations, investments, feedback, and radioactivity. We are using t for the independent variable, y for the dependent variable, k for the proportionality constant, and dy dt for the rate of growth, or derivative, with respect to t. We are looking for a function that is proportional to its own derivative. Let's start by examining the function y ˙= y = 2t
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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Chapter 8. Exponential Function
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01. The blue curve is the graph of y = 2t . The green curve is the graph of the rate of growth, y ˙ = dy/dt. We know what 2t means if t is an integer, 2t is the t-th power of 2. 2−1 = 1/2, 20 = 1, 21 = 1, 22 = 4, ... We also know what 2t means if t = p/q is a rational number, the ratio of two integers, 2p/q is the q -th root of the p-th power of 2. √ 21/2 = 2 = 1.4142..., √ 3 25/3 = 25 = 3.1748..., √ 113 2355/113 = 2355 = 8.8250... In principal, for floating point arithmetic, this is all we need to know. All floating point numbers are ratios of two integers. We do not have to be concerned yet about the definition of 2t for irrational t. If Matlab can compute powers and roots, we can plot the graph of 2t , the blue curve in figure 8.1 What is the derivative of 2t ? Maybe you have never considered this question, or don't remember the answer. (Be careful, it is not t2t−1 .) We can plot the graph of the approximate derivative, using a step size of something like 0.0001. The following code produces figure 8.1, the graphs of both y = 2t and its approximate derivative, y ˙. t = 0:.01:2; h = .00001; y = 2.^t; ydot = (2.^(t+h) - 2.^t)/h; plot(t,[y; ydot]) The graph of the derivative has the same shape as the graph of the original function. Let's look at their ratio, y ˙ (t)/y (t).
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0.52. The ratio, y/y ˙ . plot(t,ydot./y) axis([0 2 .5 .9]) We see that the ratio of the derivative to the function, shown in figure 8.2, has a constant value, y/y ˙ = 0.6931..., that does not depend upon t . Now, if you are following along with a live Matlab, repeat the preceding calculations with y = 3t instead of y = 2t . You should find that the ratio is again independent of t. This time y/y ˙ = 1.0986.... Better yet, experiment with expgui. If we take any value a and look at y = at , we find that, numerically at least, the ratio y/y ˙ is constant. In other words, y ˙ is proportional to y . If a = 2, the proportionality constant is less than one. If a = 3, the proportionality constant is greater than one. Can we find an a so that y/y ˙ is actually equal to one? If so, we have found a function that is equal to its own derivative. The approximate derivative of the function y (t) = at is a t+ h − a t h This can be factored and written y ˙ (t) = ah − 1 t a h So the ratio of the derivative to the function is y ˙ (t) = ah − 1 y ˙ ( t) = y ( t) h The ratio depends upon h, but not upon t. If we want the ratio to be equal to 1, we need to find a so that ah − 1 =1 h
100 Solving this equation for a, we find a = (1 + h)1/h
Chapter 8. Exponential Function
The approximate derivative becomes more accurate as h goes to zero, so we are interested in the value of (1 + h)1/h as h approaches zero. This involves taking numbers very close to 1 and raising them to very large powers. The surprising fact is that this limiting process defines a number that turns out to be one of the most important quantities in mathematics e = lim (1 + h)1/h
h→0
The last line of output involves a value of h that is not zero, but is so small that it prints as a string of zeros. We are actually computing (1 + 2−51 )2 which is (1 + 1 )2251799813685248 2251799813685248
51
101 The result gives us the numerical value of e correct to 16 significant decimal digits. It's easy to remember the repeating pattern of the first 10 significant digits. e = 2.718281828... Let's derive a more useful representation of the exponential function. Start by putting t back in the picture. et = ( lim (1 + h)1/h )t
h→0
This series is a rigorous mathematical definition that applies to any t, positive or negative, rational or irrational, real or complex. The n + 1-st term is tn /n!. As n increases, the tn in the numerator is eventually overwhelmed by the n! in the denominator, so the terms go to zero fast enough that the infinite series converges. It is almost possible to use this power series for actual computation of et . Here is an experimental Matlab program. function s = expex(t) % EXPEX Experimental version of EXP(T) s = 1; term = 1; n = 0; r = 0; while r ~= s r = s; n = n + 1; term = (t/n)*term; s = s + term; end
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Chapter 8. Exponential Function
Notice that there are no powers or factorials. Each term is obtained from the previous one using the fact that tn t tn−1 = n! n (n − 1)! The potentially infinite loop is terminated when r == s, that is when the floating point values of two successive partial sums are equal. There are "only" two things wrong with this program – its speed and its accuracy. The terms in the series increase as long as |t/n| ≥ 1, then decrease after n reaches the point where |t/n| < 1. So if |t| is not too large, say |t| < 2, everything is OK; only a few terms are required and the sum is computed accurately. But larger values of t require more terms and the program requires more time. This is not a very serious defect if t is real and positive. The series converges so rapidly that the extra time is hardly noticeable. However, if t is real and negative the computed result may be inaccurate. The terms alternate in sign and cancel each other in the sum to produce a small value for et . Take, for example, t = −20. The true value of e−20 is roughly 2 · 10−9 . Unfortunately, the largest terms in the series are (−20)19 /19! and (−20)20 /20!, which are opposite in sign and both of size 4 · 107 . There is 16 orders of magnitude difference between the size of the largest terms and the size of the final sum. With only 16 digits of accuracy, we lose everything. The computed value obtained from expex(-20) is completely wrong. For real, negative t it is possible to get an accurate result from the power series by using the fact that 1 et = − t e For complex t, there is no such easy fix for the accuracy difficulties of the power series. In contrast to its more famous cousin, π , the actual numerical value of e is not very important. It's the exponential function et that's important. In fact, Matlab doesn't have the value of e built in. Nevertheless, we can use e = expex(1) to compute an approximate value for e. Only seventeen terms are required to get floating point accuracy. e = 2.718281828459045 After computing e, you could then use e^t, but exp(t) is preferable.
Logarithms
The logarithm is the inverse function of the exponential. If y = et
103 then loge (y ) = t The function loge (y ) is known as the natural logarithm and is often denoted by ln y . More generally, if y = at then loga (y ) = t The function log10 (y ) is known as the common logarithm. Matlab uses log(y), log10(y), and log2(y) for loge (y ), log10 (y ), and log2 (y ).
Exponential Growth
The term exponential growth is often used informally to describe any kind of rapid growth. Mathematically, the term refers to any time evolution, y (t), where the rate of growth is proportional to the quantity itself. y ˙ = ky The solution to this equation is determined for all t by specifying the value of y at one particular t, usually t = 0. y (0) = y0 Then y (t) = y0 ekt Suppose, at time t = 0, we have a million E. coli bacteria in a test tube under ideal laboratory conditions. Twenty minutes later each bacterium has fissioned to produce another one. So at t = 20, the population is two million. Every 20 minutes the population doubles. At t = 40, it's four million. At t = 60, it's eight million. And so on. The population, measured in millions of cells, y (t), is y (t) = 2t/20 Let k = ln 2/20 = .0347. Then, with t measured in minutes and the population y (t) measured in millions, we have y ˙ = ky, y (0) = 1 Consequently y (t) = ekt This is exponential growth, but it cannot go on forever. Eventually, the growth rate is affected by the size of the container. Initially at least, the size of the population is modelled by the exponential function.
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Suppose, at time t = 0, you invest $1000 in a savings account that pays 5% interest, compounded yearly. A year later, at t = 1, the bank adds 5% of $1000 to your account, giving you y (1) = 1050. Another year later you get 5% of 1050, which is 52.50, giving y (2) = 1102.50. If y (0) = 1000 is your initial investment, r = 0.05 is the yearly interest rate, t is measured in years, and h is the step size for the compound interest calculation, we have y (t + h) = y (t) + rhy (t) What if the interest is compounded monthly instead of yearly? At the end of the each month, you get .05/12 times your current balance added to your account. The same equation applies, but now with h = 1/12 instead of h = 1. Rewrite the equation as y (t + h) − y (t) = ry (t) h and let h tend to zero. We get y ˙ (t) = ry (t) This defines interest compounded continuously. The evolution of your investment is described by y (t) = y (0)ert Here is a Matlab program that tabulates the growth of $1000 invested at 5% over a 20 year period , with interest compounded yearly, monthly, and continuously. format bank r = 0.05; y0 = 1000; for t = 0:20 y1 = (1+r)^t*y0; y2 = (1+r/12)^(12*t)*y0; y3 = exp(r*t)*y0; disp([t y1 y2 y3]) end The first few and last few lines of output are t 0 1 2 3 4 5 .. 16 yearly 1000.00 1050.00 1102.50 1157.63 1215.51 1276.28 ....... 2182.87 monthly 1000.00 1051.16 1104.94 1161.47 1220.90 1283.36 ....... 2221.85 continuous 1000.00 1051.27 1105.17 1161.83 1221.40 1284.03 ....... 2225.54
Complex exponential
What do we mean by ez if z is complex? The behavior is very different from et for real t, but equally interesting and important. Let's start with a purely imaginary z and set z = iθ where θ is real. We then make the definition eiθ = cos θ + i sin θ This formula is remarkable. It defines the exponential function for an imaginary argument in terms of trig functions of a real argument. There are several reasons why this is a reasonable defintion. First of all, it behaves like an exponential should. We expect eiθ+iψ = eiθ eiψ This behavior is a consequence of the double angle formulas for trig functions. cos(θ + ψ ) = cos θ cos ψ − sin θ sin ψ sin(θ + ψ ) = cos θ sin ψ + sin θ cos ψ Secondly, derivatives should be have as expected. d iθ e dθ d2 iθ e dθ2 = ieiθ = i2 eiθ = −eiθ
In words, the second derivative should be the negative of the function itself. This works because the same is true of the trig functions. In fact, this could be the basis for the defintion because the initial conditions are correct. e0 = 1 = cos 0 + i sin 0 The power series is another consideration. Replace t by iθ in the power series for et . Rearranging terms gives the power series for cos θ and sin θ. For Matlab especially, there is an important connection between multiplication by a complex exponential and the rotation matrices that we considered in the chapter on matrices. Let w = x + iy be any other complex number. What is eiθ w ? Let u and v be the result of the 2-by-2 matrix multiplication ( ) ( )( ) u cos θ − sin θ x = v sin θ cos θ y Then eiθ w = u + iv This says that multiplication of a complex number by eiθ corresponds to a rotation of that number in the complex plane by an angle θ.
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Figure 8.3. Two plots of eiθ . When the Matlab plot function sees a complex vector as its first argument, it understands the components to be points in the complex plane. So the octagon in the left half of figure 8.3 can be defined and plotted using eiθ with theta = (1:2:17)'*pi/8 z = exp(i*theta) p = plot(z); The quantity p is the handle to the plot. This allows us to complete the graphic with set(p,'linewidth',4,'color','red') axis square axis off An exercise asks you to modify this code to produce the five-pointed star in the right half of the figure. Once we have defined eiθ for real θ, it is clear how to define ez for a general complex z = x + iy , ez = ex+iy = ex eiy = ex (cos y + i sin y ) Finally, setting z = iπ , we get a famous relationship involving three of the most important quantities in mathematics, e, i, and π eiπ = −1 Let's check that Matlab and the Symbolic Toolbox get this right. >> exp(i*pi) ans = -1.0000 + 0.0000i
Exercises
8.1 e cubed. The value of e3 is close to 20. How close? What is the percentage error? 8.2 expgui. (a) With expgui, the graph of y = at , the blue line, always intercepts the y -axis at y = 1. Where does the graph of dy/dx, the green line, intercept the y -axis? (b) What happens if you replace plot by semilogy in expgui? 8.3 Computing e. (a) If we try to compute (1+ h)1/h for small values of h that are inverse powers of 10, it doesn't work very well. Since inverse powers of 10 cannot be represented exactly as binary floating point numbers, the portion of h that effectively gets added to 1 is different than the value involved in the computation of 1/h. That's why we used inverse powers of 2 in the computation shown in the text. Try this: format long format compact h = 1; while h > 1.e-15 h = h/10; e = (1 + h)^(1/h); disp([h e]) end How close do you get to computing the correct value of e? (b) Now try this instead: format long format compact h = 1; while h > 1.e-15 h = h/10; e = (1 + h)^(1/(1+h-1)); disp([h e]) end How well does this work? Why? 8.4 expex. Modify expex by inserting disp([term s]) as the last statement inside the while loop. Change the output you see at the command line.
111 format compact format long Explain what you see when you try expex(t) for various real values of t. expex(.001) expex(-.001) expex(.1) expex(-.1) expex(1) expex(-1) Try some imaginary values of t. expex(.1i) expex(i) expex(i*pi/3) expex(i*pi) expex(2*i*pi) Increase the width of the output window, change the output format and try larger values of t. format long e expex(10) expex(-10) expex(10*pi*i) 8.5 Instrument expex. Investigate both the cost and the accuracy of expex. Modify expex so that it returns both the sum s and the number of terms required n. Assess the relative error by comparing the result from expex(t) with the result from the built-in function exp(t). relerr = abs((exp(t) - expex(t))/exp(t)) Make a table showing that the number of terms required increases and the relative error deteriorates for large t, particularly negative t. 8.6 Complex wiggle. Revise wiggle and dot2dot to create wigglez and dot2dotz that use multiplication by eiθ instead of multiplication by two-by-two matrices. The crux of wiggle is G = [cos(theta) sin(theta); -sin(theta) cos(theta)]; Y = G*X; dot2dot(Y); In wigglez this will become w = exp(i*theta)*z; dot2dotz(w)
112 You can use wigglez with a scaled octogon. theta = (1:2:17)'*pi/8 z = exp(i*theta) wigglez(8*z)
I first saw the wooden T puzzle shown in figure 9.1 at Puzzling World in
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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Wanaka, New Zealand. They told me that it was their most popular puzzle. I have since learned that it was a well-known toy in the 1800s and an advertising tool in the early 1900s. The underlying mathematics involves geometry, trigonometry, and arithmetic with complex numbers. The t_puzzle program in the exm toolbox demonstrates some useful programming techniques.
Figure 9.2. The four pieces.
Figure 9.2 shows the electronic version of the four pieces. They all have the same width, but different heights. One of them has an unshapely chunk cut out of it, resulting in an irregular pentagon.
Figure 9.3. The T.
It turns out to be possible to arrange the four pieces to form the capital "T" shape shown in figure 9.3, as well as the arrow and the rhombus shapes shown in figure 4.4. What happened to those all of those 45◦ angles and what happened to that chunk? If you do a Google search on "T-puzzle" you can quickly see how to solve the puzzle and form the T, but please try t_puzzle for a while before you go surfing for the solution. If you click near the center of one of the four pieces, you can move
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Figure 9.4. The arrow and the rhombus.
it horizontally and vertically. If you click near one of the vertices, you can rotate a piece about its center. If you click with the right mouse button, or, on a onebutton mouse, hold down the control key while you click, you can flip a piece over horizontally, changing its right/left-handed orientation. If you click in the subplot on the lower left of the window, you can cycle through images of the three shapes. The key to effective computation involving the T-puzzle pieces is the use of complex arithmetic. A √ complex number is formed from a pair of real numbers and the imaginary unit i = −1. For example, z = 3 + 4i The real part is 3 and the imaginary part is 4. This is the Cartesian representation of a complex number. Addition of complex numbers simply involves addition of the real parts and of the imaginary parts. (3 + 4i) + (5 − i) = (3 + 5) + (4i − i) = 8 + 3i Multiplication of complex numbers is more interesting because it makes use of the fact that i2 = −1. (3 + 4i) · (5 − i) = (3 · 5 + (4i) · (−i)) + (3 · (−i) + (4i) · 5) = (15 + 4) + (−3 + 20)i = 19 + 17i A fundamental fact involving complex numbers is Euler's formula. eiϕ = cos ϕ + i sin ϕ If you are not familiar with e or Euler's formula, see our chapter on the "Exponential Function" and the Wikipedia entry on "Euler's Identity". Or, you can just accept the formula as convenient notation.
Our T puzzle program uses the fact that multiplication by eiθ rotates a complex number through an angle θ. To see this, let w = eiθ z = reiϕ then w · z = eiθ · reiϕ = rei(θ+ϕ) In Matlab the letter i is can be used in any of three different roles. It can be an iteration counter, for i = 1:10 or a subscript, A(i,k) or the imaginary unit. z = 3 + 4i The polar form of a complex number z is obtained with the Matlab functions abs(z) and angle(z). The quantity eiϕ is written exp(i*phi). For example z = r = phi w = 3 + 4i abs(z) = angle(z) r*exp(i*phi)
The largest of the four pieces of the T puzzle can be represented in Matlab by the statement z = [0 1 1+2i 3i 0] The vector z has five complex entires. The first two elements are 0 and 1; their imaginary parts are zero. The third element is 1 + 2i; its real part is 1 and its imaginary part is 2. After that comes 3i; its real part is zero and its imaginary part is 3. The last element of z is a repeat of the first so that the line drawn by the statement line(real(z),imag(z)) returns to the origin. The result is figure 9.5. With this representation, the translations and rotations required by the graphics in t_puzzle can be programmed easily and efficiently. Translation is very easy. The statement z = z - (3-i)/2 repositions the piece in figure 9.5 at one of the corners. Do you see which corner? Rotations are also easily done. The statements
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3
2
1
0
−1
0
1
2
Figure 9.6. Rotating through multiples of nine degrees.
mu = mean(z(1:end-1)) theta = pi/20 omega = exp(i*theta) z = omega*(z - mu) + mu rotate the piece about its center through 9◦ in the counterclockwise direction. Figure 9.6 shows the result of repeating these statements several times. Let's look at each step more carefully. The statement mu = mean(z(1:end-1)) drops the last element of z because it is a repeat of the first and then computes the complex average of the coordinates of the vertices. This gives the coordinates of what we can call the center of the polygon. The angle θ = π/20 is 9◦ in radians. The statements omega = exp(i*theta) z = omega*(z - mu) + mu translate the piece so that its center is at the origin, do a complex scalar times vector multiplication to produce the rotation, and then translate the result back to its original center.
Recap
%% T Puzzle Chapter Recap % This is an executable program that illustrates the statements % introduced in the T Puzzle Chapter of "Experiments in MATLAB". % You can access it with
9.2 Stop sign. Try this for n = 8 and for other small integer values of n. Describe and explain the results. n = 8 z = exp(2*pi*i*(0:n)'/n) plot(z,'-o') axis square s = sum(z) 9.3 Strip. How many different ways are there to form the shape in figure 9.7 with the T-puzzle pieces?
Figure 9.7. A strip.
9.4 Area. If the width of each of the T-puzzle pieces is one unit, what are their areas? 9.5 Symmetry. Which one of our three shapes – T, arrow and rhombus – does not have an axis of symmetry? 9.6 Jumpy rotation. Click near a vertex of one of the T-puzzle pieces and rotate the piece slowly. You should see that the rotation is not smooth, but proceeds in discrete jumps. Why? How large are the jumps? How does t_puzzle accomplish this? 9.7 Snappy translation. Drag one of the T-puzzle pieces until it is close to, but not exactly touching, another. When you release the mouse button you sometimes see the piece snap into place. Under what circumstances does this happen? How does t_puzzle accomplish it?
121 9.8 Rotation. Reproduce figure 9.6. 9.9 Different puzzles. Do a Google search on "T-puzzle". Include the quotes and hyphen in the search string so that you get an exact match. Some of the Web pages have pieces with different sizes than the ones we have described here. (a) How many different versions of the T-puzzle are there on the Web? (b) Can you make all three of our shapes – the T, arrow, and rhombus – with the pieces shown on these Web sites. (c) Modify our t_puzzle to use the set of pieces from one of the Web sites.
Magic squares predate recorded history. An ancient Chinese legend tells of a turtle emerging from the Lo river during a flood. The turtle's shell showed a very unusual pattern – a three-by-three grid containing various numbers of spots. Of
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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course, we do not have any eye-witness accounts, so we can only imagine that the turtle looked like figure 10.1. Each of the three rows, the three columns, and the two diagonals contain a total of 15 spots. References to Lo Shu and the Lo Shu numerical pattern occur throughout Chinese history. Today, it is the mathematical basis for Feng Shui, the philosophy of balance and harmony in our surroundings and lives. An n-by-n magic square is an array containing the integers from 1 to n2 , arranged so that each of the rows, each of the columns, and the two principal diagonals have the same sum. For each n > 2, there are many different magic squares of order n, but the Matlab function magic(n) generates a particular one. Matlab can generate Lo Shu with A = magic(3) which produces A = 8 3 4 The command sum(A) sums the elements in each column to produce 15 The command sum(A')' transposes the matrix, sums the columns of the transpose, and then transposes the results to produce the row sums 15 15 15 The command sum(diag(A)) sums the main diagonal of A, which runs from upper left to lower right, to produce 15 The opposite diagonal, which runs from upper right to lower left, is less important in linear algebra, so finding its sum is a little trickier. One way to do it makes use of the function that "flips" a matrix "upside-down." 15 15 1 5 9 6 7 2
These are all the magic squares of order three. The 5 is always in the center, the other odd numbers are always in the centers of the edges, and the even numbers are always in the corners. Melancholia I is a famous Renaissance engraving by the German artist and amateur mathematician Albrecht D¨ urer. It shows many mathematical objects, including a sphere, a truncated rhombohedron, and, in the upper right hand corner, a magic square of order 4. You can see the engraving in our figure 10.2. Better yet, issue these Matlab commands
The elements of the array X are indices into the gray-scale color map named map. The image is displayed with image(X) colormap(map) axis image Click the magnifying glass with a "+" in the toolbar and use the mouse to zoom in on the magic square in the upper right-hand corner. The scanning resolution becomes evident as you zoom in. The commands load detail
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.. Figure 10.3. Detail from Melancolia.
image(X) colormap(map) axis image display the higher resolution scan of the area around the magic square that we have in figure 10.3. The command A = magic(4) produces a 4-by-4 magic square. A = 16 5 9 4 The commands sum(A), sum(A'), sum(diag(A)), sum(diag(flipud(A))) yield enough 34's to verify that A is indeed a magic square. The 4-by-4 magic square generated by Matlab is not the same as D¨ urer's magic square. We need to interchange the second and third columns. A = A(:,[1 3 2 4]) 2 11 7 14 3 10 6 15 13 8 12 1
128 changes A to A = 16 5 9 4 3 10 6 15 2 11 7 14 13 8 12 1
Chapter 10. Magic Squares
Interchanging columns does not change the column sums or the row sums. It usually changes the diagonal sums, but in this case both diagonal sums are still 34. So now our magic square matches the one in D¨ urer's etching. D¨ urer probably chose this particular 4-by-4 square because the date he did the work, 1514, occurs in the middle of the bottom row. The program durerperm interchanges rows and columns in the image produced from detail by interchanging groups of rows and columns in the array X. This is not especially important or useful, but it provides an interesting exercise. We have seen two different 4-by-4 magic squares. It turns out that there are 880 different magic squares of order 4 and 275305224 different magic squares of order 5. Determining the number of different magic squares of order 6 or larger is an unsolved mathematical problem. For a magic square of order n, the magic sum is 1∑ k µ(n) = n
n2 k=1
The integers from 1 to n2 are inserted along diagonals, starting in the middle of first row and heading in a northeasterly direction. When you go off an edge of the array, which you do at the very first step, continue from the opposite edge. When you bump into a cell that is already occupied, drop down one row and continue. We used this algorithm in Matlab for many years. Here is the code. A = zeros(n,n); i = 1; j = (n+1)/2; for k = 1:n^2 is = i; js = j; A(i,j) = k; i = n - rem(n+1-i,n); j = rem(j,n) + 1; if A(i,j) ~= 0 i = rem(is,n) + 1; j = js; end end A big difficulty with this algorithm and resulting program is that it inserts the elements one at a time – it cannot be vectorized. A few years ago we discovered an algorithm for generating the same magic squares of odd order as de La Loubere's method, but with just four Matlab matrix operations. [I,J] = ndgrid(1:n);
Both A and B are fledgling magic squares. They have equal row, column and diagonal sums. But their elements are not the integers from 1 to n2 . Each has duplicated elements between 0 and n − 1. The final statement
131 M = n*A+B+1 produces a matrix whose elements are integers between 1 and n2 and which has equal row, column and diagonal sums. What is not obvious, but is true, is that there are no duplicates. So M must contain all of the integers between 1 and n2 and consequently is a magic square. M = 17 23 4 10 11 24 5 6 12 18 1 7 13 19 25 8 14 20 21 2 15 16 22 3 9
The final result merges these two matrices to produce the magic square. The algorithm for singly even order is the most complicated and so we will give just a glimpse of how it works. If n is singly even, then n/2 is odd and magic(n) can be constructed from four copies of magic(n/2). For example, magic(10) is obtained from A = magic(5) by forming a block matrix. [ A A+75 A+50 A+25 ]
The column sums are all equal because sum(A) + sum(A+75) equals sum(A+50) + sum(A+25). But the rows sums are not quite right. The algorithm must finish by doing a few swaps of pieces of rows to clean up the row sums. For the details, issue the command. type magic
You can see the three different cases – on the left, the upper right, and the lower right. If you increase of each of the orders by 4, you get more cells, but the global shapes remain the same. The odd n case on the left reminds me of Origami.
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Further Reading
The reasons why Matlab has magic squares can be traced back to my junior high school days when I discovered a classic book by W. W. Rouse Ball, Mathematical Recreations and Essays. Ball lived from 1850 until 1925. He was a fellow of Trinity College, Cambridge. The first edition of his book on mathematical recreations was published in 1892 and the tenth edition in 1937. Later editions were revised and updated by another famous mathematician, H. S. M Coxeter. The thirteenth edition, published by Dover in 1987, is available from many booksellers, including Powells and Amazon. dp/0486253570 There are dozens of interesting Web pages about magic squares. Here are a few authors and links to their pages. Mutsumi Suzuki Eric Weisstein Kwon Young Shin Walter Trump
You will see three matrices in your workspace. You can look at all of map and caption. map caption The matrix X is 359-by-371. That's 133189 elements. Look at just a piece of it. X(101:130,101:118) The elements of X are integers in the range min(min(X)) max(max(X)) The commands image(X) axis image produce a pretty colorful display. That's because the elements of X are being used as indices into the default colormap, jet(64). You can use the intended colormap instead. colormap(map) The array map is a 64-by-3 array. Each row, map(k,:), specifies intensities of red, green and blue. The color used at point (i,j) is map(X(i,j),:). In this case, the colormap that comes with detail has all three columns equal to each other and so is the same as colormap(gray(64)) Now experiment with other colormaps colormap(hot) colormap(cool) colormap(copper) colormap(pink) colormap(bone) colormap(flag) colormap(hsv) You can even cycle through 101 colormaps. for p = 0:.001:1 colormap(p*hot+(1-p)*pink) drawnow end You can plot the three color components of a colormap like hot with
If you follow the elements in numerical order, you will be taken on a knight's tour of K. Even the step from 64 back to 1 is a knight's move. Is K a magic square? Why or why not? Try this. image(K) colormap(pink) axis square Select the data cursor icon on the figure tour bar. Now use the mouse to take the knight's tour from dark to light on the image. 10.6 ismagical. The exm function ismagical checks for four different magical properties of square arrays. Semimagic : all of the columns and all of rows have the same sum. Magic : all of the columns, all of rows and both principal diagonals have the same sum. Panmagic : all of the columns, all of rows and all of the diagonals, including the broken diagonals in both directions, have the same sum. Associative : all pairs of elements on opposite sides of the center have the same sum. For example, this matrix that has all four properties. M = 10 11 17 23 4 18 24 5 6 12 1 7 13 19 25 14 20 21 2 8 22 3 9 15 16
Here is one of the broken diagonals. Its sum is µ(5) = 65.
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Chapter 10. Magic Squares
All of the broken diagonals in both directions have the same sum, so M is panmagic. One pair of elements on opposite sides of the center is 24 and 2. Their sum is twice the center value. All pairs of elements on opposite sides of the center have this sum, so M is associative. (a) Use ismagical to verify that M has all four properties. (b) Use ismagical to investigate the magical properties of the matrices generated by the Matlab magic function. (cJ-a0,n); B = mod(I+2*J-b0,n); M = n*A + B + 1; (d2*J-a0,n); B = mod(I+3*J-b0,n); M = n*A + B + 1; 10.7 Inverse. If you have studied matrix theory, you have heard of matrix inverses. What is the matrix inverse of a magic square of order n? It turns out to depend upon whether n is odd or even. For odd n, the matrices magic(n) are nonsingular. The matrices X = inv(magic(n)) do not have positive, integer entries, but they do have equal row and column sums. But, for even n, the determinant, det(magic(n)), is 0, and the inverse does not exist. If A = magic(4) trying to compute inv(A) produces an error message. 10.8 Rank. If you have studied matrix theory, you know that the rank of a matrix is the number of linearly independent rows and columns. An n-by-n matrix is singular if its rank, r, is not equal to its order. This code computes the rank of the magic squares up to order 20, generated with
that the three different algorithms used to generate magic squares produce matrices with different rank. n odd even, not divisible by 4 divisible by 4 rank n n/2+2 3
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Chapter 11
TicTacToe Magic
Three simple games are related in a surprising way. And, the programming of the game play is instructive.
The first of the three games is Pick15. Harold Stark, who was then at the University of Michigan, told me about the game in the late 1960s. I suspect that this is the first time you have heard of it. The game involves two players. You start by listing the single digit numbers from 1 to 9. You then take turns selecting numbers from the list, attempting to acquire three numbers that add up to 15. Each number can be chosen only once. You may eventually acquire more than three numbers, but you must use exactly three of them to total 15. If neither player can achieve the desired total, the game is a draw. For example, suppose that Green and Blue are playing. They start with the list. List : 1 2 3 4 5 6 7 8 9 Green : Blue : Suppose Green has the first move and chooses 8. Then Blue chooses 4 and Green chooses 2. Now Blue should respond by choosing 5 to prevent Green from getting 2 + 5 + 8 = 15. Here is the situation after the first two rounds. List : 1 2 / 3 4 / 5 / 6 7 8 / 9
Green : 2 6 7 8 Blue : 1 4 5 Note that Green also has 7 + 8 = 15, but this does not count because there are only two numbers in the sum. Figure 11.1 shows the starting position for the Pick15 option in our Matlab tictactoe program. When you play against the computer, your moves are shown in green and the responses from the program are shown in blue.
Figure 11.1. Starting position for Pick15. Figure 11.2 shows the position for our example game after two moves from each player. Green now has to choose 6 to block Blue.
Figure 11.2. Position after two moves from each player. Figure 11.3 shows the final position for our example game. Green has won with 2+6+7 = 15.
Figure 11.3. Final position for Pick15. Please take time out from reading this chapter to try this game a few times yourself, playing against a friend on paper or against our Matlab program. I think you will find that Pick15 is more challenging than it sounds at first. The second of our three games is familiar worldwide. It is called "TicTacToe" in the United States, "Noughts and Crosses" in Great Britain, and has many other names in many other countries and languages. Our program tictactoe uses green and blue instead of X's and O's. The objective, of course, is to get your color on all three squares in a row, column, or diagonal. Figure 11.6 shows typical output.
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Figure 11.4. Typical output from a game of TicTacToe. Green has won with the third column.
Our program tictactoe uses a naive three-step strategy. • If possible, make a winning move. • If necessary, block a possible winning move by the opponent. • Otherwise, pick a random empty square. This strategy will lead to a win only when the opponent makes a error. And, even though it is possible for the player with the second move to always force a draw, this strategy will not accomplish that. Our third game, Magic15, introduces the Lo-Shu 3-by-3 magic square. Now we see that Pick15 is actually TicTacToe played on a magic square. The rows, columns and main diagonals of the magic square provide all possible ways of having three distinct numbers that sum to 15. Winning moves in Pick15 correspond to winning moves in TicTacToe. All three games are actually the same game with different displays.
Figure 11.5. Initial configuration for a game of Magic3.
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Figure 11.6. Green has won with 6+7+2 = 15 in the third column.
Game Play
The tictactoe program operates three different views of the same game. You are the green player and the computer is the blue player. The state of the game is carried in a 3-by-3 matrix X whose entries are +1 for cells occupied by green, -1 for cells occupied by blue, and 0 for as yet unoccupied cells. Obviously, the game begins with X = zeros(3,3) Here is the portion of main program that is executed when you, as the green player, click on one of the buttons or cells. It retrieves i and j, the matrix indices corresponding to the active button and then checks if the current game already has a winner. If not, it makes your move by setting X(i,j) = 1 and recursively calls tictactoe to let blue take a turn. switch case 'green' [i,j] = find(gcbo == B); if winner(X) return end X(i,j) = 1; tictactoe('blue') Here is the corresponding blue portion of the main program. If the current game does not already have a winner, it calls the strategy function to get the indices for a move and makes that move. case 'blue' if winner(X) return end [i,j] = strategy(X,-1); X(i,j) = -1; ... end
145 Briefly, then, the green moves are determined by user button clicks and the blue moves are determined by the strategy function. Here is the function that checks to see if there is a winner. It looks for any column, row, or diagonal whose elements sum to 3*p where p = 1 for green and p = -1 for blue. The only way a sum can be 3*p is if all three elements are equal to p. function p = winner(X) % p = winner(X) returns % p = 0, no winner yet, % p = -1, blue has won, % p = 1, green has won, % p = 2, game is a draw. for p = [-1 1] s = 3*p; win = any(sum(X) == s) || any(sum(X') == s) || ... sum(diag(X)) == s || sum(diag(fliplr(X))) == s; if win return end end p = 2*all(X(:) ~= 0); Here is the naive, but fairly effective strategy that the computer uses against you. It first checks to see if it can make a winning move. If not, it then checks to see if it needs to block your pending winning move. It neither of these happens, it simply chooses a random empty cell. This crude strategy makes the play interesting. It is possible for you to beat the computer if its random choices are not optimal. The exercises suggest some improvements in this strategy. function [i,j] = strategy(X,p); % [i,j] = strategy(X,p) is a better, but not perfect, move for player p. % Appear to think. pause(0.5) % If possible, make a winning move. [i,j] = winningmove(X,p); % Block any winning move by opponent. if isempty(i) [i,j] = winningmove(X,-p); end % Otherwise, make a random move. if isempty(i)
Finally, here is the function that seeks a winning move by either player. It looks for rows, columns, or diagonals with sums equal to 2*p for either value of p. function [i,j] = winningmove(X,p); % [i,j] = winningmove(X,p) finds any winning move for player p. s = 2*p; if any(sum(X) == s) j = find(sum(X) == s); i = find(X(:,j) == 0); elseif any(sum(X') == s) i = find(sum(X') == s); j = find(X(i,:) == 0); elseif sum(diag(X)) == s i = find(diag(X) == 0); j = i; elseif sum(diag(fliplr(X))) == s i = find(diag(fliplr(X)) == 0); j = 4 - i; else i = []; j = []; end The remainder of the tictactoe function is responsible for the gui. Let's see how all this works on an example. We begin with the position shown in figure 11.2, after two moves by each player. The state of the game at this point is X = 1 0 -1 0 -1 0 0 0 1
147 You see that 4 and 5 are on the antidiagonal, so green needs the 6 to prevent a win in the TicTacToe version of this game. Either way, this corresponds to (i,j) = (1,3) and setting X(i,j) = 1. X = 1 0 -1 0 -1 0 1 0 1
Now it is blue's turn. Both the first row and the third column of X have sums equal to 2, corresponding to the fact that green has two pending winning moves. With p = -1, the strategy function looks for a block by calling winningmove(X,-p). The result is (i,j) = (2,3), although it just as well could have been (i,j) = (1,2) if winningmove made its checks in another order. This leads to X = 1 0 -1 0 -1 0 1 -1 1
On the next turn, green finds that the sum along the first row is 2, and so sets X(1,2) = 1 to make the sum 3. This gives X = 1 0 -1 1 -1 0 1 -1 1
and green proclaims a win. It would be possible to implement recursive backtracking strategies like we describe in the Sudoku chapter. But there are only a few hundred possible games, so the backtracking would be exhaustive. And boring – all the games would be draws.
Exercises
11.1 Traditional. Modify tictactoe.m so that it uses traditional X's and O's. 11.2 Win. Is it possible to win against tictactoe with its naive strategy? 11.3 First move. Modify tictactoe so that the computer takes the first move. 11.4 Center square. Modify the strategy used by tictactoe.m so that, before taking a random move, it takes the center square if it is available. Does this improve the program's chances of winning or forcing a draw? 11.5 xkcd. Implement the complete tictactoc strategy available from the Web comic strip xkcd by Randal Munroe at
11.6 Computer versus computer. Modify tictactoe.m so that the computer plays against itself. Count the number of draws and wins for both sides. Run a large number of games, with and without the addition of the center square strategy.
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Chapter 12
Game of Life
Conway's Game of Life makes use of sparse matrices.
The "Game of Life" was invented by John Horton Conway, a British-born mathematician who is now a professor at Princeton. The game made its public debut in the October 1970 issue of Scientific American, in the "Mathematical Games " column written by Martin Gardner. At the time, Gardner wrote This month we consider Conway's latest brainchild, a fantastic solitaire pastime he calls "life". Because of its analogies with the rise, fall and alternations of a society of living organisms, it belongs to a growing class of what are called "simulation games" – games that resemble real-life processes. To play life you must have a fairly large checkerboard and a plentiful supply of flat counters of two colors. Of course, today we can run the simulations on our computers. The universe is an infinite, two-dimensional rectangular grid. The population is a collection of grid cells that are marked as alive. The population evolves at discrete time steps known as generations. At each step, the fate of each cell is determined by the vitality of its eight nearest neighbors and this rule: • A live cell with two live neighbors, or any cell with three live neighbors, is alive at the next step. The fascination of Conway's Game of Life is that this deceptively simple rule leads to an incredible variety of patterns, puzzles, and unsolved mathematical problems – just like real life. If the initial population consists of only one or two live cells, it expires in one step. If the initial population consists of three live cells then, because of rotational
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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Figure 12.1. A pre-block and a block. and reflexive symmetries, there are only two different possibilities – the population is either L-shaped or I-shaped. The left half of figure 12.1 shows three live cells in an L-shape. All three cells have two live neighbors, so they survive. The dead cell that they all touch has three live neighbors, so it springs to life. None of the other dead cells have enough live neighbors to come to life. So the result, after one step, is the population shown in the right half of figure 12.1. This four-cell population, known as the block, is stationary. Each of the live cells has three live neighbors and so lives on. None of the other cells can come to life.
Figure 12.2. A blinker blinking. The other three-cell initial population is I-shaped. The two possible orientations are shown in first two steps of figure 12.2. At each step, two end cells die, the middle cell stays alive, and two new cells are born to give the orientation shown in the next step. If nothing disturbs it, this blinker keeps blinking forever. It repeats itself in two steps; this is known as its period. One possible four-cell initial population is the block. Discovering the fate of the other four-cell initial populations is left to an exercise. The beginning of the evolution of the most important five-cell initial population, known as the glider, is shown in figure 12.3. At each step two cells die and two new ones are born. After four steps the original population reappears, but it has moved diagonally down and across the grid. It continues to move in this direction forever, eventually disappearing out of our field of view, but continuing to exist in the infinite universe. The fascination of the Game of Life cannot be captured in these static figures. Computer graphics lets you watch the dynamic development. We will show just more one static snapshot of the evolution of an important larger population. Figure 12.4 is the glider gun developed by Bill Gosper at MIT in 1970. The portion of the population between the two static blocks oscillates back and forth. Every 30 steps, a glider emerges. The result is an infinite stream of gliders that fly out of the field of view.
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Figure 12.3. A glider gliding.
Figure 12.4. Gosper's glider gun. Matlab is a convenient environment for implementing the Game of Life. The universe is a matrix. The population is the set of nonzero elements in the matrix. The universe is infinite, but the population is finite and usually fairly small. So we can store the population in a finite matrix, most of whose elements are zero, and increase the size of the matrix if necessary when the population expands. This is the ideal setup for a sparse matrix. Conventional storage of an n-by-n matrix requires n2 memory. But sparse storage of a matrix X requires just three vectors, one integer and one floating point vector of length nnz(X) – the number of nonzero
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elements in X – and one integer vector of length n, not n2 , to represent the start of each column. For example, the snapshot of the Gosper glider gun in figure 12.4 is represented by an 85-by-85 matrix with 68 nonzero entries. Conventional full matrix storage would require 852 = 7225 elements. Sparse matrix storage requires only 2 · 65 + 85 = 221 elements. This advantage of sparse over full storage increases as more gliders are created, the population expands, and n increases. The exm toolbox includes a program called lifex. (Matlab itself has a simpler demo program called life.) The initial population is represented by a matrix of 0's and 1's. For example, G = [1 1 1; 1 0 0; 0 1 0] produces a single glider G = 1 1 0 1 0 1 1 0 0
The universe is represented by a sparse n-by-n matrix X that is initially all zero. We might start with n = 23 so there will be a 10 cell wide border around a 3-by-3 center. The statements n = 23; X = sparse(n,n) produce X = All zero sparse: 23-by-23 The initial population is injected in the center of the universe. with the statement X(11:13,11:13) = G This produces a list of the nonzero elements X = (11,11) (12,11) (11,12) (13,12) (11,13) 1 1 1 1 1
We are now ready to take the first step in the simulation. Whether cells stay alive, die, or generate new cells depends upon how many of their eight neighbors are alive. The statements n = size(X,1); p = [1 1:n-1]; q = [2:n n];
The basic rule of Life is A live cell with two live neighbors, or any cell with three live neighbors, is alive at the next step. This is implemented with the single Matlab statement X = (X & (Y == 2)) | (Y == 3) The two characters '==' mean "is equal to". The '&' character means "and". The '|' means "or". These operations are done for all the cells in the interior of the universe. In this example, there are four cells where Y is equal to 3, so they survive or come alive. There is one cell where X is equal to 1 and Y is equal to 2, so it survives. The result is X = (11,11) (12,11) (10,12) (11,12) (12,13) 1 1 1 1 1
Our glider has taken its first step. One way to use lifex is to provide your own initial population, as either a full or a sparse matrix. For example, you create your own fleet of gliders fleet with G = [1 1 1; 1 0 0; 0 1 0]
is the home of the "Life Lexicon", maintained by Stephen Silver. Among thousands of the facts of Life, this 160-page document lists nearly 450 different initial populations, together with their history and important properties. We have included a text copy of the Lexicon with the exm toolbox in the file exm/lexicon.txt Lifex can read initial populations from the Lexicon. Calling lifex with no arguments, lifex picks a random initial population from the Lexicon. Either lifex('xyz') or lifex xyz will look for a population whose name begins with xyz. For example, the statements lifex lifex lifex lifex pre-block block blinker glider
start with the simple populations that we have used in this introduction. The statement lifex Gosper provides Gosper's glider gun. By default, the initial population is surrounded by a strip of 20 dead border cells to provide a viewing window. You can change this to use b border cells with lifex('xyz',b) If the population expands during the simulation and cells travel beyond this viewing window, they continue to live and participate even though they cannot be seen.
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Further Reading
The Wikipedia article is a good introduction. Another good introduction is available from Math.com, although there are annoying popups and ads. If you find yourself at all interested in the Game of Life, take a good look at the Lexicon, either by reading our text version or by visiting the Web page.
Recap
%% Life Chapter Recap % This is an executable program that illustrates the statements % introduced in the Life Chapter of "Experiments in MATLAB". % You can access it with % % life_recap % It does not work so well with edit and publish. % % Related EXM programs % % lifex % Generate a random initial population X = sparse(50,50); X(21:30,21:30) = (rand(10,10) > .75); p0 = nnz(X); % Loop over 100 generations. for t = 1:100 spy(X) title(num2str(t)) drawnow % Whether cells stay alive, die, or generate new cells depends % upon how many of their eight possible neighbors are alive. % Index vectors increase or decrease the centered index by one. n = size(X,1); p = [1 1:n-1];
Exercises
12.1 life vs. lifex The Matlab demos toolbox has an old program called life, without an x. In what ways is it the same as, and in what ways does it differ from, our exm gui lifex? 12.2 Four-cell initial populations. What are all of the possible four-cell initial populations, and what are their fates? You can generate one of the four-cell populations with L = [1 1 1; 1 0 0]; lifex(L,4) 12.3 Lexicon. Describe the behavior of each of these populations from the Lexicon. If any is periodic, what is its period? ants B-52 blinker puffer diehard Canada goose gliders by the dozen Kok's galaxy rabbits R2D2 spacefiller wasp washerwoman
159 12.4 Glider collisions. What happens when: A glider collides head-on with a block? A glider side-swipes a block? Two gliders collide head-on? Two gliders clip each others wings? Four gliders simultaneously leave the corners of a square and head towards its center? See also: lifex('4-8-12'). 12.5 Factory. How many steps does it take the factory to make one glider? 12.6 R-pentomino. Of all the possible five-cell initial populations, the only one that requires a computer to determine its ultimate behavior is the one that Conway dubbed the R-pentomino. It is shown in figure 12.5 and can be generated by R = [0 1 1; 1 1 0; 0 1 0]
Figure 12.5. The R-pentomino. As the simulation proceeds, the population throws off a few gliders, but otherwise remains bounded. If you make b large enough, the statement lifex(R,b) shows all of the bounded behavior. How large does this b have to be? What is the maximum population during the evolution? How many gliders are produced? How many steps does it take for the population to stabilize? How many blinkers are present in the stabilized population? What is size of the stabilized population? 12.7 Execution time. Display actual computer execution time by inserting tic and toc in lifex.m. Place the single statement tic before the start of the inner loop. Change the call of caption to caption(t,nnz(X),toc) Make the appropriate modifications in the caption subfunction at the end of lifex.m. Demonstrate your modified program on a few interesting examples. 12.8 The Ark. Run
160 lifex('ark',128)
Chapter 12. Game of Life
for a few minutes. About how much time does it take on your computer to do one step? According to the Lexicon, the ark requires 736692 steps to stabilize. About how much time will it take on your computer for the ark to stabilize? 12.9 Houses. Check out lifex(houses)
12.10 Checkerboards. This code generates an n-by-n checkerboard of 0's and 1's. [I,J] = meshgrid(1:n); C = (mod(I+J,2)==0); What are the stabilization times and final populations for n-by-n checkerboard initial populations with 3 <= n <= 30? 12.11 Early exit. Where should these code segments Xsave = X; and if isequal(X,Xsave) break end be inserted in life_recap? 12.12 Symmetry. What is the effect of inserting the statement X = (X + X' > 0) after the generation of X at the beginning of life_recap. 12.13 Live or die? Insert these statements before the first executable statement of life_recap. P = zeros(10000,2); for s = 1:10000 Insert these statements after the last executable statement. P(s,:) = [p0 p100]; end Deactivate or remove the spy, title, pause and fprintf statements. Run the resulting program and then generate histograms of the results.
X = sparse(76,76); X(34:42,20:57) = (gun=='+'); spy(X) (a) How many steps does it take for the gun to emit a glider? (b) What happens when a glider meets the boundary? How is this different from lifex('gosper')?
Benoit Mandelbrot was a Polish/French/American mathematician who has spent most of his career at the IBM Watson Research Center in Yorktown Heights, N.Y. He coined the term fractal and published a very influential book, The Fractal Geometry of Nature, in 1982. An image of the now famous Mandelbrot set appeared on the cover of Scientific American in 1985. This was about the time that computer graphical displays were becoming widely available. Since then, the Mandelbrot set has stimulated deep research topics in mathematics and has also been the basis for an uncountable number of graphics projects, hardware demos, and Web pages. To get in the mood for the Mandelbrot set, consider the region in the complex plane of trajectories generated by repeated squaring,
2 zk+1 = zk , k = 0, 1, ...
For which initial values z0 does this sequence remain bounded as k → ∞? It is easy to see that this set is simply the unit disc, |z0 | <= 1, shown in figure 13.1. If |z0 | <= 1, the sequence zk remains bounded. But if |z0 | > 1, the sequence is unbounded. The boundary of the unit disc is the unit circle, |z0 | = 1. There is nothing very difficult or exciting here. The definition is the Mandelbrot set is only slightly more complicated. It involves repeatedly adding in the initial point. The Mandelbrot set is the region in the complex plane consisting of the values z0 for which the trajectories defined by
2 zk+1 = zk + z0 , k = 0, 1, ...
Figure 13.1. The unit disc is shown in red. The boundary is simply the unit circle. There is no intricate fringe.
1.5
1
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−1
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Figure 13.2. The Mandelbrot set is shown in red. The fringe just outside the set, shown in black, is a region of rich structure.
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Figure 13.3. Two trajectories. z0 = .25-.54i generates a cycle of length four, while nearby z0 = .22-.54i generates an unbounded trajectory. Figure 13.2 shows the overall geometry of the Mandelbrot set. However, this view does not have the resolution to show the richly detailed structure of the fringe just outside the boundary of the set. In fact, the set has tiny filaments reaching into the fringe region, even though the fringe appears to be solid black in the figure. It has recently been proved that the Mandelbrot set is mathematically connected, but the connected region is sometimes so thin that we cannot resolve it on a graphics screen or even compute it in a reasonable amount of time. To see how the definition works, enter z0 = .25-.54i z = 0 into Matlab. Then use the up-arrow key to repeatedly execute the statement z = z^2 + z0 The first few lines of output are 0.2500 0.0209 -0.4057 0.0852 0.2517 ... 0.5400i 0.8100i 0.5739i 0.0744i 0.5527i
This cycle repeats forever. The trajectory remains bounded. This tells us that the starting value value, z0 = .25-.54i, is in the Mandelbrot set. The same cycle is shown in the left half of figure 13.3. On the other hand, start with z0 = .22-.54i z = 0 and repeatedly execute the statement z = z^2 + z0 You will see 0.2200 -0.0232 -0.3841 0.1136 0.2095 ... 0.5400i 0.7776i 0.5039i 0.1529i 0.5747i
Then, after 24 iterations, ... 1.5708 - 1.1300i 1.4107 - 4.0899i -14.5174 -12.0794i 6.5064e+001 +3.5018e+002i -1.1840e+005 +4.5568e+004i The trajectory is blowing up rapidly. After a few more iterations, the floating point numbers overflow. So this z0 is not in the Mandelbrot set. The same unbounded trajectory is shown in the right half of figure 13.3. We see that the first value, z0 = .25-.54i, is in the Mandelbrot set, while the second value, z0 = .22-.54i, which is nearby, is not. The algorithm doesn't have to wait until z reachs floating point overflow. As soon as z satisfies abs(z) >= 2 subsequent iterations will essential square the value of |z | and it will behave like k 22 . Try it yourself. Put these statements on one line. z0 = ... z = 0; while abs(z) < 2 z = z^2+z0; disp(z), end
167 Use the up arrow and backspace keys to retrieve the statement and change z0 to different values near .25-.54i. If you have to hit <ctrl>-c to break out of an infinite loop, then z0 is in the Mandelbrot set. If the while condition is eventually false and the loop terminates without your help, then z0 is not in the set. The number of iterations required for z to escape the disc of radius 2 provides the basis for showing the detail in the fringe. Let's add an iteration counter to the loop. A quantity we call depth specifies the maximum interation count and thereby determines both the level of detail and the overall computation time. Typical values of depth are several hundred or a few thousand. z0 = ... z = 0; k = 0; while abs(z) < 2 && k < depth z = z^2+z0; k = k + 1; end The maximum value of k is depth. If the value of k is less than depth, then z0 is outside the set. Large values of k indicate that z0 is in the fringe, close to the boundary. If k reaches depth then z0 is declared to be inside the Mandelbrot set. Here is a small table of iteration counts s z0 ranges over complex values near 0.22-0.54i. We have set depth = 512. 0.205 0.210 0.215 0.220 0.225 0.230 0.235 0.240 0.245 -0.520 -0.525 -0.530 -0.535 -0.540 -0.545 -0.550 -0.555 -0.560 512 512 512 512 512 512 33 34 62 512 512 512 512 139 199 25 20 19 512 512 512 512 113 211 21 18 17 512 512 512 512 26 21 20 18 17 512 512 35 26 24 22 20 19 18 512 36 31 28 73 25 25 21 33 44 51 74 57 56 120 63 33 162 512 512 512 512 512 512 85 512 40 512 512 512 512 512 512 512 512 344
We see that about half of the values are less than depth; they correspond to points outside of the Mandelbrot set, in the fringe near the boundary. The other half of the values are equal to depth, corresponding to points that are regarded as in the set. If we were to redo the computation with a larger value of depth, the entries that are less than 512 in this table would not change, but some of the entries that are now capped at 512 might increase. The iteration counts can be used as indices into an RGB color map of size depth-by-3. The first row of this map specifies the color assigned to any points on the z0 grid that lie outside the disc of radius 2. The next few rows provide colors for the points on the z0 grid that generate trajectories that escape quickly. The last row of the map is the color of the points that survive depth iterations and so are in the set.
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The map used in figure 13.2 emphasizes the set itself and its boundary. The map has 12 rows of white at the beginning, one row of dark red at the end, and black in between. Images that emphasize the structure in the fringe are achieved when the color map varies cyclicly over a few dozen colors. One of the exercises asks you to experiment with color maps.
Figure 13.4. Improving resolution.
Array operations.
Our script mandelbrot_recap shows how Matlab array arithmetic operates a grid of complex numbers simultaneously and accumulates an array of iteration counters, producing images like those in figure 13.4 The code begins by defining the region in the complex plane to be sampled. A step size of 0.05 gives the coarse resolution shown on the right in the figure. x = 0: 0.05: 0.80; y = x'; The next section of code uses an elegant, but tricky, bit of Matlab indexing known as Tony's Trick. The quantities x and y are one-dimensional real arrays of length n, one a column vector and the other a row vector. We want to create a twodimensional n-by-n array with elements formed from all possible sums of elements from x and y . √ zk,j = xk + yj i, i = −1, k, j = 1, ..., n This can be done by generating a vector e of length n with all elements equal to one. Then the quantity x(e,:) is a two-dimensional array formed by using x, which is the same as x(1,:), n times. Similarly, y(:,e) is a two-dimensional arrray containing n copies of the column vector y.
We see that points in the upper left of the grid, with fairly small initial z0 values, have survived 32 iterations without going outside the circle of radius two, while points in the lower right, with fairly large initial values, have lasted only one or two iterations. The interesting grid points are in between, they are on the fringe. Now comes the final step, making the plot. The image command does the job, even though this is not an image in the usual sense. The count values in c are used as indices into a 32-by-3 array of RGB color values. In this example, the jet colormap is reversed to give dark red as its first value, pass through shades of green and yellow, and finish with dark blue as its 32-nd and final value. image(c) axis image colormap(flipud(jet(depth))) Exercises ask you to increase the resolution by decreasing the step size, thereby producing the other half of figure 13.4, to investigate the effect of changing depth, and to invesigate other color maps.
Mandelbrot GUI
The exm toolbox function mandelbrot is your starting point for exploration of the Mandelbrot set. With no arguments, the statement mandelbrot provides thumbnail icons of the twelve regions featured in this chapter. The statement mandelbrot(r) with r between 1 and 12 starts with the r-th region. The statement mandelbrot(center,width,grid,depth,cmapindx) explores the Mandelbrot set in a square region of the complex plane with the specified center and width, using a grid-by-grid grid, an iteration limit of depth, and the color map number cmapindx. The default values of the parameters are center = -0.5+0i width = 3 grid = 512 depth = 256 cmapindx = 1 In other words, mandelbrot(-0.5+0i, 3, 512, 256, 1) generates figure 13.2, but with the jets color map. Changing the last argument from 1 to 6 generates the actual figure 13.2 with the fringe color map. On my laptop, these computations each take about half a second. A simple estimate of the execution time is proportional to
171 grid^2 * depth So the statement mandelbrot(-0.5+0i, 3, 2048, 1024, 1) could take (2048/512)2 · (1024/256) = 64 times as long as the default. However, this is an overestimate and the actual execution time is about 11 seconds. Most of the computational time required to compute the Mandelbrot set is spent updating two arrays z and kz by repeatedly executing the step z = z.^2 + z0; j = (abs(z) < 2); kz(j) = d; This computation can be carried out faster by writing a function mandelbrot_step in C and creating as a Matlab executable or c-mex file. Different machines and operating systems require different versions of a mex file, so you should see files with names like mandelbrot_step.mexw32 and mandelbrot_step.glnx64 in the exm toolbox. The mandelbrot gui turns on the Matlab zoom feature. The mouse pointer becomes a small magnifying glass. You can click and release on a point to zoom by a factor of two, or you can click and drag to delineate a new region. The mandelbrot gui provides several uicontrols. Try these as you read along. The listbox at the bottom of the gui allows you to select any of the predefined regions shown in the figures in this chapter. depth. Increase the depth by a factor of 3/2 or 4/3. grid. Refine the grid by a factor of 3/2 or 4/3. The depth and grid size are always a power of two or three times a power of two. Two clicks on the depth or grid button doubles the parameter. color. Cycle through several color maps. jets and hots are cyclic repetitions of short copies of the basic Matlab jet and hot color maps. cmyk cycles through eight basic colors, blue, green, red, cyan, magenta, yellow, gray, and black. fringe is a noncyclic map used for images like figure 13.2. exit. Close the gui.
The Mandelbrot set is self similar. Small regions in the fringe reveal features that are similar to the original set. Figure 13.6, which we have dubbed "Mandelbrot
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Figure 13.5. The figures in this chapter, and the predefined regions in our mandelbrot program, show these regions in the fringe just outside the Mandelbrot set.
Junior", is one example. Figure 13.7, which we call the "'Plaza", uses our flag colormap to reveal fine detail in red, white and blue. The portion of the boundary of the Mandelbrot set between the two large, nearly circular central regions is known as "The Valley of the Seahorses". Figure 13.8 shows the result of zooming in on the peninsula between the two nearly circular regions of the set. The figure can be generated directly with the command mandelbrot(-.7700-.1300i,0.1,1024,512) We decided to name the image in figure 13.9 the "West Wing" because it resembles the X-wing fighter that Luke Skywalker flies in Star Wars and because it is located near the leftmost, or far western, portion of the set. The magnification factor is a relatively modest 104 , so depth does not need to be very large. The command to generate the West Wing is mandelbrot(-1.6735-0.0003318i,1.5e-4,1024,160,1) One of the best known examples of self similarity, the "Buzzsaw", is shown in figure 13.11. It can be generated with mandelbrot(0.001643721971153+0.822467633298876i, ...
173 4.0e-11,1024,2048,2) Taking width = 4.0e-11 corresponds to a magnification factor of almost 1011 . To appreciate the size of this factor, if the original Mandelbrot set fills the screen on your computer, the Buzzsaw is smaller than the individual transistors in your machine's microprocessor. We call figure 13.12 "Nebula" because it reminds us of interstellar dust. It is generated by mandelbrot(0.73752777-0.12849548i,4.88e-5,1024,2048,3) The next three images are obtained by carefully zooming on one location. We call them the "Vortex", the "Microbug", and the "Nucleus". mandelbrot(-1.74975914513036646-0.00000000368513796i, ... 6.0e-12,1024,2048,2) mandelbrot(-1.74975914513271613-0.00000000368338015i, ... 3.75e-13,1024,2048,2) mandelbrot(-1.74975914513272790-0.00000000368338638i, ... 9.375e-14,1024,2048,2) The most intricate and colorful image among our examples is figure 13.16, the "Geode". It involves a fine grid and a large value of depth and consequently requires a few minutes to compute. mandelbrot(0.28692299709-0.01218247138i,6.0e-10,2048,4096,1) These examples are just a tiny sampling of the structure of the Mandelbrot set.
Further Reading
We highly recommend a real time fractal zoomer called "XaoS", developed by Thomas Marsh, Jan Hubicka and Zoltan Kovacs, assisted by an international group of volunteers. See If you are expert at using your Web browser and possibly downloading an obscure video codec, take a look at the Wikipedia video ... Image:Fractal-zoom-1-03-Mandelbrot_Buzzsaw.ogg It's terrific to watch, but it may be a lot of trouble to get working.
Recap
%% Mandelbrot Chapter Recap % This is an executable program that illustrates the statements
Exercises
13.1 Explore. Use the mandelbrot gui to find some interesting regions that, as far as you know, have never been seen before. Give them your own names. 13.2 depth. Modify mandelbrot_recap to reproduce our table of iteration counts for x = .205:.005:.245 and y = -.520:-.005:-.560. First, use depth = 512. Then use larger values of depth and see which table entries change. 13.3 Resolution. Reproduce the image in the right half of figure 13.4. 13.4 Big picture. Modify mandelbrot_recap to display the entire Mandelbrot set. 13.5 Color maps. Investigate color maps. Use mandelbrot_recap with a smaller step size and a large value of depth to produce an image. Find how mandelbrot computes the cyclic color maps called jets, hots and sepia. Then use those maps on your image. 13.6 p-th power. In either mandelbrot_recap or the mandelbrot gui, change the power in the Mandelbrot iteration to
p zk+1 = zk + z0
for some fixed p ̸= 2. If you want to try programming Handle Graphics, add a button to mandelbrot that lets you set p. 13.7 Too much magnification. When the width of the region gets to be smaller than about 10−15 , our mandelbrot gui does not work very well. Why? 13.8 Spin the color map. This might not work very well on your computer because it depends on what kind of graphics hardware you have. When you have an interesting region plotted in the figure window, bring up the command window, resize it so that it does not cover the figure window, and enter the command spinmap(10) I won't try to describe what happens – you have to see it for yourself. The effect is most dramatic with the "seahorses2" region. Enter help spinmap for more details.
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Figure 13.6. Region #2, "Mandelbrot Junior". The fringe around the Mandelbrot set in self similar. Small versions of the set appear at all levels of magnification.
Figure 13.7. Region #3, "Plaza", with the flag color map.
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Figure 13.8. Region #4. "Valley of the Seahorses".
Figure 13.9. Region #5. Our "West Wing" is located just off the real axis in the thin far western portion of the set, near real(z) = -1.67.
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Figure 13.10. Region #6. "Dueling Dragons".
Figure 13.11. Region #7. The "Buzzsaw" requires a magnification factor of 1011 to reveal a tiny copy of the Mandelbrot set.
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Figure 13.12. Region #8. "Nebula". Interstellar dust.
Figure 13.13. Region #9. A vortex, not far from the West Wing. Zoom in on one of the circular "microbugs" near the left edge.
Humans look for patterns, but machines use backtracking. The fascination that people have for solving Sudoku puzzles without a computer derives from the discovery and mastery of a myriad of subtle combinations and patterns that provide tips toward the solution. The Web has hundreds of sites describing these patterns, which have names like "hidden quads", "X-wing" and "squirmbag". It is not our intention to describe a Matlab program that duplicates these human pattern recognition capabilities. Instead, our program, like most other Sudoku computer codes, takes a very different approach, one that relies on the machine's almost limitless capacity to carry out brute force trial and error. We use only one pattern, singletons, together with a fundamental computer science technique, recursive backtracking. In the Sudoku world, backtracking is regarding as "guessing" and is considered bad form. But as a computational technique, it is easy to understand, straightforward to implement and guaranteed to succeed. In this chapter, we will describe five Matlab programs for experiments with Sudoku. • sudoku. An interactive program that lets you follow the computer solution process. • sudoku_basic. The same solution process as sudoku, but without the graphics interface. Much easier to read, and much faster if you just want the solution. • sudoku_all. Enumerate all solutions to a Sudoku puzzle. A valid puzzle should have only one solution.
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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Chapter 14. Sudoku • sudoku_assist. An interactive program that lets you control the solution process, as you would by hand on paper. • sudoku_puzzle. A collection of interesting puzzles.
Sudoku is actually an American invention. It first appeared, with the name Number Place, in the Dell Puzzle Magazine in 1979. The creator was probably Howard Garns, an architect from Indianapolis. A Japanese publisher, Nikoli, took the puzzle to Japan in 1984 and eventually gave it the name Sudoku, which is a kind of kanji acronym for "numbers should be single, unmarried." The Times of London began publishing the puzzle in the UK in 2004 and it was not long before it spread back to the US and around the world.
8 3 4
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Figure 14.1. A Sudoku puzzle featuring Lo-Shu, the magic square of order 3. You probably already know the rules of Sudoku. Figure 20 is an example of an initial 9-by-9 grid, with a few specified digits known as the clues. This particular puzzle reflects our interest in magic squares. It is the sum of the two matrices shown in figure 20. The matrix on the left is generated by the functions kron, which computes something known as the Kronecker product, and magic, which generates magic squares. X = kron(eye(3),magic(3)) In this case, a Kronecker product involving the identity matrix produces a block matrix with three copies of magic(3) on the diagonal. By itself, this array is not a valid Sudoku puzzle, because there is more than one way to complete it. The solution is not unique. In order to get a puzzle with a unique solution, we have added the matrix on the right in figure 20. Figure 14.3 is the final completed grid. Each row, each column, and each major 3-by-3 block, must contain exactly the digits 1 through 9. In contrast to
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Figure 14.2. Our Lo-Shu based puzzle is the sum of these two matrices.
8 3 4 2 9 7 5 6 1Figure 14.3. The completed puzzle. The digits have been inserted so that each row, each column, and each major 3-by-3 block contains 1 through 9. The original clues are shown in blue.
magic squares and other numeric puzzles, no arithmetic is involved. The elements in a Sudoku grid could just as well be nine letters of the alphabet, or any other distinct symbols.
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Figure 14.5. Candidates
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Figure 14.6. Insert singleton
Figure 14.7. Solution
Shidoku
To see how our sudoku program works, we can use Shidoku instead of Sudoku. "Shi" is Japanese for "four". The puzzles use a 4-by-4 grid and are almost trivial to solve by hand. Figure 20 is our first Shidoku puzzle and the next three figures show steps in its solution. In figure 20, the possible entries, or candidates, are shown by small digits. For example, row two contains a 3 and column one contains a 1 so the candidates in position (2,1) are 2 and 4. Four of the cells have only one candidate each. These are the singletons, shown in red. In figure 20, we have inserted the singleton 3 in the (4,1) cell and recomputed the candidates. In figure 14.7, we have inserted the remaining singletons as they are generated to complete the solution.
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Figure 14.8. diag(1:4)
Figure 14.9. No singletons.
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Figure 14.10. Backtrack step.
Figure 14.11. Solution is not unique.
The input array for figure 14.8 is generated by the Matlab statement X = diag(1:4) As figure 14.9 shows, there are no singletons. So, we employ a basic computer science technique, recursive backtracking. We select one of the empty cells and tentatively insert one of its candidates. All of the empty cells have two candidates, so we pick the first one and tentatively insert a 3 in cell (2,1). This creates a new puzzle, shown in figure 14.10. Our program is then called recursively. In this example, the new puzzle is easily solved and the result is shown in figure 14.11. However, the solution depends upon the choices that we made before the recursive call. Other choices can lead to different solutions. For this simple diagonal initial condition,
Existence and uniqueness
Mathematicians are always concerned about existence and uniqueness in the various problems that they encounter. For Sudoku, neither existence nor uniqueness can be determined easily from the initial clues. It would be very frustrating if a puzzle with no solution, or with more than one solution, were to show up in your daily newspaper. Uniqueness is an elusive property. In fact, most descriptions of Sudoku do not specify that there has to be exactly one solution. The only way that I know to check uniqueness is to exhaustively enumerate all possibilities. For our magic square puzzle in figure 20, if we were to replace the 1 in the (9,1) cell by a 5, 6 or 7, the row, column and block conditions would still be satisfied. It turns out that a 5 would produce another valid puzzle with a unique solution, a 6 would produce a puzzle with two possible solutions, and a 7 would produce a puzzle with no solution. An exercise asks you to verify these claims. It takes sudoku_all a little over half an hour on my laptop to determine that the magic square puzzle generated by X = kron(eye(3),magic(3)) without the addional entries on right in figure 20 has 283576 solutions. The puzzle generated by X = sudoku_puzzle(13) has an interesting spiral pattern. But there are over 300 solutions. An exercise asks you to find out exactly how many.
189 A number of operations on a Sudoku grid can change its visual appearance without changing its essential characteristics. All of the variations are basically the same puzzle. These equivalence operations can be expressed as array operations in Matlab. For example p = randperm(9) z = find(X > 0) X(z) = p(X(z)) permutes the digits representing the elements. Other operations include X' rot90(X,k) flipud(X) fliplr(X) X([4:9 1:3],:) X(:,[randperm(3) 4:9])
The algorithm
If we do not count the comments and GUI, sudoku.m involves less than 40 lines of code. The outline of the main program is: • Fill in all singletons. • Exit if a cell has no candidates. • Fill in a tentative value for an empty cell. • Call the program recursively.
sudoku and sudoku basic
Here is the header for sudoku_basic. function [X,steps] = sudoku_basic(X,steps) % SUDOKU_BASIC Solve the Sudoku puzzle using recursive backtracking. % sudoku_basic(X), for a 9-by-9 array X, solves the Sudoku puzzle for X % without providing the graphics user interface from sudoku.m % [X,steps] = sudoku_basic(X) also returns the number of steps. % See also sudoku, sudoku_all, sudoku_assist, sudoku_puzzle. This first bit of code checks nargin, the number of input arguments, and initializes the step counter on the first entry. Any subsequent recursive calls will have a second argument. if nargin < 2 steps = 0; end
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These comments describe the primary variables in the program. % C is the array of candidate values for each cell. % N is the vector of the number of candidates for each cell. % s is the index of the first cell with the fewest candidates. The program fills in any singletons, one at a time. The candidates are recomputed after each step. This section of code, by itself, can solve puzzles that do not require backtracking. Such puzzles can be classified as "easy". [C,N] = candidates(X); while all(N>0) & any(N==1) s = find(N==1,1); X(s) = C{s}; steps = steps + 1; [C,N] = candidates(X); end If filling in the singletons does not solve the puzzle, we reach the following section of code that implements backtracking. The backtracking generates many impossible configurations. The recursion is terminated by encountering a puzzle with no solution. if all(N>0) Y = X; s = find(N==min(N),1); for t = [C{s}] X = Y; X(s) = t; steps = steps + 1; [X,steps] = sudoku_basic(X,steps); if all(X(:) > 0) break end end end All of the bookkeeping required by backtracking is handled by the recursive call mechanism in Matlab and the underling operating system. The key internal function for sudoku is candidates. Here is the preamble. function [C,N] = candidates(X) % C = candidates(X) is a 9-by-9 cell array of vectors. % C{i,j} is the vector of allowable values for X(i,j). % N is a row vector of the number of candidates for each cell. % N(k) = Inf for cells that already have values.
% Iterate over the candidates. % Insert a tentative value.
% Recursive call. % Found a solution.
191 The following statement creates an internal function named tri that computes the indices for blocks. For example, tri(1), tri(2) and tri(3) are all equal to 1:3 because the block that includes a cell with row or column index equal to 1, 2 or 3 has row or column indices 1:3. tri = @(k) 3*ceil(k/3-1) + (1:3); Here is the core of candidates. C = cell(9,9); for j = 1:9 for i = 1:9 if X(i,j)==0 z = 1:9; z(nonzeros(X(i,:))) = 0; z(nonzeros(X(:,j))) = 0; z(nonzeros(X(tri(i),tri(j)))) = 0; C{i,j} = nonzeros(z)'; end end end For each empty cell, this function starts with z = 1:9 and uses the numeric values in the associated row, column and block to zero elements in z. The nonzeros that remain are the candidates. For example, consider the (1,1) cell in figure 20. We start with z = 1 2 3 4 5 6 7 8 9 The values in the first row change z to z = 1 0 0 0 5 6 7 8 9 Then the first column changes z to z = 1 0 0 0 5 0 7 0 9 The (1,1) block does not make any further changes, so the candidates for this cell are C{1,1} = [1 5 7 9]. The candidates function concludes with the following statements. The number of candidates, N(i,j) for cell (i,j) is the length of C{i,j}. If cell (i,j) already has a value, then X(i,j) is nonzero, C{i,j} is empty and so N(i,j) is zero. But we make these N(i,j) infinite to distinguish them from the cells that signal an impossible puzzle. N = cellfun(@length,C); N(X>0) = Inf; N = N(:)';
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An example, puzzle number one
Figure 14.12 shows the key steps in the solution of our Lo-Shu puzzle. The initial candidates include just one singleton, the red 5 in position (3,9). Five steps with five singletons allow us to fill in the cells with green 5's and 8's. We now consider the first cell with two candidates, in position (6,1). The first step in the backtracking puts a tentative 6 in position (6,1). Blue values are the initial clues, cyan values are generated by the backtracking and green values are implied by the others. After five steps we reach an impossible situation because there are no candidates for position (6,2). Try again, with a 7 in position (6,1). This puzzle is easily completed by filling in singletons. The title on the final solutions shows that it took 50 steps, including the abortive ones in the first backtrack, to reach the solution.
Exercises
14.1 xkcd. Solve the binary Sudoku puzzle by Randal Munroe in the Web comic strip xkcd at 14.2 Solve. Use sudoku to solve a puzzle from a newspaper, magazine, or puzzle book. Use sudoku_assist to solve the same puzzle by hand. 14.3 sudoku puzzle. The exm program sudoku_puzzle generates 16 different puzzles. The comments in the program describe the origins of the puzzles. How many steps are required by sudoku_basic to solve each of the puzzles? 14.4 By hand. Use sudoku_assist to solve our Lo-Shu based puzzle by hand. X = sudoku_puzzle(1); sudoku_assist(X) 14.5 Modify puzzle #1. The following program modifies the lower left hand clue in our Lo-Shu based puzzle. What happens with each of the resulting puzzles? X = sudoku_puzzle(1); for t = [0 1 5 6 7] X(9,1) = t; sudoku_all(X) end
14.7 Puzzle #3. How many solutions does sudoku_puzzle(3) have? How do they differ from each other? 14.8 Puzzle #13. How many solutions does sudoku_puzzle(13) have? In addition to the initial clues, what values are the same in all the solutions? 14.9 All solutions. What are the differences between sudoku_basic and sudoku_all? 14.10 Combine. Combine sudoku_all and sudoku_basic to create a program that returns all possible solutions and the number of steps required to find each one. Try your program on sudoku_puzzle(13). What is the average number of steps required? Hints: if S is a cell array where each cell contains one numeric value, then s = [S{:}] creates the corresponding vector. And, help mean. 14.11 Search strategy In sudoku_basic, the statement s = find(N==min(N),1); determines the search strategy. What happens when you change this to s = find(N==min(N),'last'); or s = find(X==0,'first');
14.12 Patterns. Add some human puzzle solving techniques to sudoku.m. This will complicate the program and require more time for each step, but should result in fewer total steps. 14.13 sudoku alpha. In sudoku.m, change int2str(d) to char('A'+d-1) so that the display uses the letters 'A' through 'I' instead of the digits 1 through 9. See figure 14.13. Does this make it easier or harder to solve puzzles by hand. 14.14 sudoku16. Modify sudoku.m to solve 16-by-16 puzzles with 4-by-4 blocks.
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Chapter 14. Sudoku
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Figure 14.12. The key steps in the solution of our Lo-Shu puzzle.
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Figure 14.13. Use the letters 'A' through 'I' instead of the digits 1 through 9.
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Chapter 15
Ordinary Differential Equations
Mathematical models in many different fields.
Systems of differential equations form the basis of mathematical models in a wide range of fields – from engineering and physical sciences to finance and biological sciences. Differential equations are relations between unknown functions and their derivatives. Computing numerical solutions to differential equations is one of the most important tasks in technical computing, and one of the strengths of Matlab. If you have studied calculus, you have learned a kind of mechanical process for differentiating functions represented by formulas involving powers, trig functions, and the like. You know that the derivative of x3 is 3x2 and you may remember that the derivative of tan x is 1 + tan2 x. That kind of differentiation is important and useful, but not our primary focus here. We are interested in situations where the functions are not known and cannot be represented by simple formulas. We will compute numerical approximations to the values of a function at enough points to print a table or plot a graph. Imagine you are traveling on a mountain road. Your altitude varies as you travel. The altitude can be regarded as a function of time, or as a function of longitude and latitude, or as a function of the distance you have traveled. Let's consider the latter. Let x denote the distance traveled and y = y (x) denote the altitude. If you happen to be carrying an altimeter with you, or you have a deluxe GPS system, you can collect enough values to plot a graph of altitude versus distance, like the first plot in figure 15.1. Suppose you see a sign saying that you are on a 6% uphill grade. For some
Figure 15.1. Altitude along a mountain road, and derivative of that altitude. The derivative is zero at the local maxima and minima of the altitude. value of x near the sign, and for h = 100, you will have y (x + h) − y (x) = .06 h The quotient on the left is the slope of the road between x and x + h. Now imagine that you had signs every few meters telling you the grade at those points. These signs would provide approximate values of the rate of change of altitude with respect to distance traveled, This is the derivative dy/dx. You could plot a graph of dy/dx, like the second plot in figure 15.1, even though you do not have closed-form formulas for either the altitude or its derivative. This is how Matlab solves differential equations. Note that the derivative is positive where the altitude is increasing, negative where it is decreasing, zero at the local maxima and minima, and near zero on the flat stretches. Here is a simple example illustrating the numerical solution of a system of differential equations. Figure 15.2 is a screen shot from Spacewar, the world's first video game. Spacewar was written by Steve "Slug" Russell and some of his buddies at MIT in 1962. It ran on the PDP-1, Digital Equipment Corporation's first computer. Two space ships, controlled by players using switches on the PDP-1 console, shoot space torpedoes at each other. The space ships and the torpedoes orbit around a central star. Russell's program needed to compute circular and elliptical orbits, like the path of the torpedo in the screen shot. At the time, there was no Matlab. Programs were written in terms of individual machine instructions. Floating-point arithmetic was so slow that it was desirable to avoid evaluation of trig functions in the orbit calculations. The orbit-generating program looked something like this. x = 0 y = 32768
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Figure 15.2. Spacewar, the world's first video game. The gravitational pull of the central star causes the torpedo to move in an elliptical orbit.
L: plot x y load y shift right 2 add x store in x change sign shift right 2 add y store in y go to L What does this program do? There are no trig functions, no square roots, no multiplications or divisions. Everything is done with shifts and additions. The initial value of y, which is 215 , serves as an overall scale factor. All the arithmetic involves a single integer register. The "shift right 2" command takes the contents of this register, divides it by 22 = 4, and discards any remainder. If Spacewar orbit generator were written today in Matlab, it would look something the following. We are no longer limited to integer values, so we have changed the scale factor from 215 to 1. x = 0; y = 1; h = 1/4; n = 2*pi/h; plot(x,y,'.') for k = 1:n
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Chapter 15. Ordinary Differential Equations
1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −0.6 −0.8 −1 −1 −0.5 0 0.5 1
Figure 15.3. The 25 blue points are generated by the Spacewar orbit generator with a step size of 1/4. The 201 green points are generated with a step size of 1/32. x = x + h*y; y = y - h*x; plot(x,y,'.') end The output produced by this program with h = 1/4 and n = 25 is shown by the blue dots in figure 15.3. The blue orbit is actually an ellipse that deviates from an exact circle by about 7%. The output produced with h = 1/32 and n = 201 is shown by the green dots. The green orbit is another ellipse that deviates from an exact circle by less than 1%. Think of x and y as functions of time, t. We are computing x(t) and y (t) at discrete values of t, incremented by the step size h. The values of x and y at time t + h are computed from the values at time t by x(t + h) = x(t) + hy (t) y (t + h) = y (t) − hx(t + h) This can be rewritten as x(t + h) − x(t) = y (t) h y (t + h) − y (t) = −x(t + h) h You have probably noticed that the right hand side of this pair of equations involves
203 two different values of the time variable, t and t + h. That fact turns out to be important, but let's ignore it for now. Look at the left hand sides of the last pair of equations. The quotients are approximations to the derivatives of x(t) and y (t). We are looking for two functions with the property that the derivative of the first function is equal to the second and the derivative of the second function is equal to the negative of the first. In effect, the Spacewar orbit generator is using a simple numerical method involving a step size h to compute an approximate solution to the system of differential equations x ˙ = y y ˙ = −x The dot over x and y denotes differentiation with respect to t. x ˙= dx dt
steps generates a better approximation to a circle. Actually, the fact that x(t + h) is used instead of x(t) in the second half of the step means that the method is not quite as simple as it might seem. This subtle change is responsible for the fact that the method generates ellipses instead of spirals. One of the exercises asks you to verify this fact experimentally. Mathematical models involving systems of ordinary differential equations have one independent variable and one or more dependent variables. The independent variable is usually time and is denoted by t. In this book, we will assemble all the dependent variables into a single vector y . This is sometimes referred to as the state of the system. The state can include quantities like position, velocity, temperature, concentration, and price. In Matlab a system of odes takes the form y ˙ = F (t, y ) The function F always takes two arguments, the scalar independent variable, t, and the vector of dependent variables, y . A program that evaluates F (t, y ) hould compute the derivatives of all the state variables and return them in another vector. In our circle generating example, the state is simply the coordinates of the point. This requires a change of notation. We have been using x(t) and y (t) to denote position, now we are going to use y1 (t) and y2 (t). The function F defines the velocity. ( ) y ˙ 1 (t) y ˙ (t) = y ˙ 2 (t) ( ) y2 (t) = −y1 (t) Matlab has several functions that compute numerical approximations to solutions of systems of ordinary differential equations. The suite of ode solvers includes ode23, ode45, ode113, ode23s, ode15s, ode23t, and ode23tb. The digits in the names refer to the order of the underlying algorithms. The order is related to the complexity and accuracy of the method. All of the functions automatically determine the step size required to obtain a prescribed accuracy. Higher order methods require more work per step, but can take larger steps. For example ode23 compares a second order method with a third order method to estimate the step size, while ode45 compares a fourth order method with a fifth order method. The letter "s" in the name of some of the ode functions indicates a stiff solver. These methods solve a matrix equation at each step, so they do more work per step than the nonstiff methods. But they can take much larger steps for problems where numerical stability limits the step size, so they can be more efficient overall. You can use ode23 for most of the exercises in this book, but if you are interested in the seeing how the other methods behave, please experiment. All of the functions in the ode suite take at least three input arguments. • F, the function defining the differential equations, • tspan, the vector specifying the integration interval,
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Figure 15.4. Graphs of sine and cosine generated by ode23. • y0, the vector of initial conditions. There are several ways to write the function describing the differential equation. Anticipating more complicated functions, we can create a Matlab program for our circle generator that extracts the two dependent variables from the state vector. Save this in a file named mycircle.m. function ydot = mycircle(t,y) ydot = [y(2); -y(1)]; Notice that this function has two input arguments, t and y, even though the output in this example does not depend upon t. With this function definition stored in mycircle.m, the following code calls ode23 to compute the solution over the interval 0 ≤ t ≤ 2π , starting with x(0) = 0 and y (0) = 1. tspan = [0 2*pi]; y0 = [0; 1]; ode23(@mycircle,tspan,y0) With no output arguments, the ode solvers automatically plot the solutions. Figure 15.4 is the result for our example. The small circles in the plot are not equally spaced. They show the points chosen by the step size algorithm. To produce the phase plot plot shown in figure 15.5, capture the output and plot it yourself. tspan = [0 2*pi]; y0 = [0; 1]; [t,y] = ode23(@mycircle,tspan,y0) plot(y(:,1),y(:,2)'-o')
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Chapter 15. Ordinary Differential Equations
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Figure 15.5. Graph of a circle generated by ode23. axis([-1.1 1.1 -1.1 1.1]) axis square The circle generator example is so simple that we can bypass the creation of the function file mycircle.m and write the function in one line. acircle = @(t,y) [y(2); -y(1)] The expression created on the right by the "@" symbol is known as an anonymous function because it does not have a name until it is assigned to acircle. Since the "@" sign is included in the definition of acircle, you don't need it when you call an ode solver. Once acircle has been defined, the statement ode23(acircle,tspan,y0) automatically produces figure 15.4. And, the statement [t,y] = ode23(acircle,tspan,y0) captures the output so you can process it yourself. Many additional options for the ode solvers can be set via the function odeset. For example opts = odeset('outputfcn',@odephas2) ode23(acircle,tspan,y0,opts) axis square axis([-1.1 1.1 -1.1 1.1])
207 will also produce figure 15.5. Use the command doc ode23 to see more details about the Matlab suite of ode solvers. Consult the ODE chapter in our companion book, Numerical Computing with MATLAB, for more of the mathematical background of the ode algorithms, and for ode23tx, a textbook version of ode23. Here is a very simple example that illustrates how the functions in the ode suite work. We call it "ode1" because it uses only one elementary first order algorithm, known as Euler's method. The function does not employ two different algorithms to estimate the error and determine the step size. The step size h is obtained by dividing the integration interval into 200 equal sized pieces. This would appear to be appropriate if we just want to plot the solution on a computer screen with a typical resolution, but we have no idea of the actual accuracy of the result. function [t,y] = ode1(F,tspan,y0) % ODE1 World's simplest ODE solver. % ODE1(F,[t0,tfinal],y0) uses Euler's method to solve % dy/dt = F(t,y) % with y(t0) = y0 on the interval t0 <= t <= tfinal. t0 = tspan(1); tfinal = tspan(end); h = (tfinal - t0)/200; y = y0; for t = t0:h:tfinal ydot = F(t,y); y = y + h*ydot; end However, even with 200 steps this elementary first order method does not have satisfactory accuracy. The output from [t,y]] = ode1(acircle,tspan,y0) is t = 6.283185307179587 y = 0.032392920185564 1.103746317465277 We can see that the final value of t is 2*pi, but the final value of y has missed returning to its starting value by more than 10 percent. Many more smaller steps would be required to get graphical accuracy.
Exercises
15.1 Walking to class. You leave home (or your dorm room) at the usual time in the morning and walk toward your first class. About half way to class, you realize that you have forgotten your homework. You run back home, get your homework, run to class, and arrive at your usual time. Sketch a rough graph by hand showing your distance from home as a function of time. Make a second sketch of your velocity as a function of time. You do not have to assume that your walking and running velocities are constant, or that your reversals of direction are instantaneous. 15.2 Divided differences. Create your own graphic like our figure 15.1. Make up your own data, x and y, for distance and altitude. You can use subplot(2,1,1) and subplot(2,1,2) to place two plots in one figure window. The statement d = diff(y)./diff(x) computes the divided difference approximation to the derivative for use in the second subplot. The length of the vector d is one less than the length of x and y, so you can add one more value at the end with
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Chapter 15. Ordinary Differential Equations
For more information about diff and subplot, use help diff help subplot
211 15.5 Linear system Write the system of differential equations y˙1 = y2 y˙2 = −y1 in matrix-vector form, y ˙ = Ay where y is a vector-valued function of time, ( ) y1 (t) y (t) = y2 (t) and A is a constant 2-by-2 matrix. Use our ode1 as well as ode23 to experiment with the numerical solution of the system in this form. 15.6 Example from ode23. The first example in the documentation for ode23 is y ˙ 1 = y2 y3 y ˙ 2 = −y1 y3 y ˙ 3 = −0.51 y1 y2 with initial conditions y1 (0) = 0 y2 (0) = 1 y3 (0) = 1 Compute the solution to this system on the interval 0 ≤ t ≤ 12. Reproduce the graph included in the documentation provided by the command doc ode23
15.7 A cubic system. Make a phase plane plot of the solution to the ode system
3 y˙1 = y2 3 y˙2 = −y1
The simplest model for the growth, or decay, of a population says that the growth rate, or the decay rate, is proportional to the size of the population itself. Increasing or decreasing the size of the population results in a proportional increase or decrease in the number of births and deaths. Mathematically, this is described by the differential equation y ˙ = ky The proportionality constant k relates the size of the population, y (t), to its rate of growth, y ˙ (t). If k is positive, the population increases; if k is negative, the population decreases. As we know, the solution to this equation is a function y (t) that is proportional to the exponential function y (t) = ηekt where η = y (0). This simple model is appropriate in the initial stages of growth when there are no restrictions or constraints on the population. A small sample of bacteria in a large Petri dish, for example. But in more realistic situations there are limits to growth, such as finite space or food supply. A more realistic model says that the population competes with itself. As the population increases, its growth rate decreases linearly. The differential equation is sometimes called the logistic equation. y y ˙ = k (1 − )y µ
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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Chapter 16. Predator-Prey Model
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The new parameter µ is the carrying capacity. As y (t) approaches µ the growth rate approaches zero and the growth ultimately stops. It turns out that the solution is y (t) = µηekt ηekt + µ − η
You can easily verify for yourself that as t approaches zero, y (t) approaches η and that as t approaches infinity, y (t) approaches µ. If you know calculus, then with quite a bit more effort, you can verify that y (t) actually satisfies the logistic equation. Figure 16.1 shows the two solutions when both η and k are equal to one. The exponential function y (t) = et gives the rapidly growing green curve. With carrying capacity µ = 20, the logistic function y (t) = et 20et + 19
gives the more slowly growing blue curve. Both curves have the same initial value and initial slope. The exponential function grows exponentially, while the logistic function approaches, but never exceeds, its carrying capacity. Figure 16.1 was generated with the following code. k = 1 eta = 1 mu = 20 t = 0:1/32:8;
215 y = mu*eta*exp(k*t)./(eta*exp(k*t) + mu - eta); plot(t,[y; exp(t)]) axis([0 8 0 22]) If you don't have the formula for the solution to the logistic equation handy, you can compute a numerical solution with ode45, one of the Matlab ordinary differential equation solvers. Try running the following code. It will automatically produce a plot something like the blue curve in figure 16.1. k = 1 eta = 1 mu = 20 ydot = @(t,y) k*(1-y/mu)*y ode45(ydot,[0 8],eta) The @ sign and @(t,y) specify that you are defining a function of t and y. The t is necessary even though it doesn't explicitly appear in this particular differential equation. The logistic equation and its solution occur in many different fields. The logistic function is also known as the sigmoid function and its graph is known as the S-curve. Populations do not live in isolation. Everybody has a few enemies here and there. The Lotka-Volterra predator-prey model is the simplest description of competition between two species. Think of rabbits and foxes, or zebras and lions, or little fish and big fish. The idea is that, if left to themselves with an infinite food supply, the rabbits or zebras would live happily and experience exponential population growth. On the other hand, if the foxes or lions were left with no prey to eat, they would die faster than they could reproduce, and would experience exponential population decline. The predator-prey model is a pair of differential equations involving a pair of competing populations, y1 (t) and y2 (t). The growth rate for y1 is a linear function of y2 and vice versa. y2 y ˙ 1 = (1 − )y1 µ2 y1 y ˙ 2 = −(1 − )y2 µ1 We are using notation y1 (t) and y2 (t) instead of, say, r(t) for rabbits and f (t) for foxes, because our Matlab program uses a two-component vector y. The extra minus sign in the second equation distinguishes the predators from the prey. Note that if y1 ever becomes zero, then y ˙ 2 = −y 2 and the predators are in trouble. But if y2 ever becomes zero, then y ˙ 1 = y1 and the prey population grows exponentially.
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We have a formula for the solution of the single species logistic model. However it is not possible to express the solution to this predator-prey model in terms of exponential, trigonmetric, or any other elementary functions. It is necessary, but easy, to compute numerical solutions.
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Figure 16.2. A typical solution of the predator-prey equations.
There are four parameters, the two constants µ1 and µ2 , and the two initial conditions, η1 = y1 (0) η2 = y2 (0) If we happen to start with η1 = µ1 and η2 = µ2 , then both y ˙ 1 and y ˙ 2 are zero and the populations remain constant at their initial values. In other words, the point (µ1 , µ2 ) is an equilibrium point. The origin, (0, 0) is another equilibrium point, but not a very interesting one. The following code uses ode45 to automatically plot the typical solution shown in figure 16.2. mu = [300 200]' eta = [400 100]' signs = [1 -1]' pred_prey_ode = @(t,y) signs.*(1-flipud(y./mu)).*y period = 6.5357 ode45(pred_prey_ode,[0 3*period],eta)
217 There are two tricky parts of this code. Matlab vector operations are used to define pred_prey_ode, the differential equations in one line. And, the calculation that generates figure 16.3 provides the value assigned to period. This value specifies a value of t when the populations return to their initial values given by eta. The code integrates over three of these time intervals, and so at the end we get back to where we started. The circles superimposed on the plots in figure 16.2 show the points where ode45 computes the solution. The plots look something like trig functions, but they're not. Notice that the curves near the minima are broader, and require more steps to compute, then the curves near the maxima. The plot of sin t would look the same at the top as the bottom.
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Figure 16.3. The predprey experiment.
Our Matlab program exm/predprey shows a red dot at the equilibrium point, (µ1 , µ2 ), and a blue-green dot at the initial point, (η1 , η2 ). When you drag either dot with the mouse, the solution is recomputing by ode45 and plotted. Figure 16.3 shows that two plots are produced — a phase plane plot of y2 (t) versus y1 (t) and a time series plot of y1 (t) and y2 (t) versus t. Figures 16.2 and 16.3 have the same parameters, and consequently show the same solution, but with different scaling of
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the axes. The remarkable property of the Lotka-Volterra model is that the solutions are always periodic. The populations always return to their initial values and repeat the cycle. This property is not obvious and not easy to prove. It is rare for nonlinear models to have periodic solutions. The difficult aspect of computing the solution to the predator-prey equations is determining the length of the period. Our predprey program uses a feature of the Matlab ODE solvers called "event handling" to compute the length of a period. If the initial values (η1 , η2 ) are close to the equilibrium point (µ1 , µ2 ), then the length of the period is close to a familar value. An exercise asks you to discover that value experimentally.
Exercises
16.1 Plot. Make a more few plots like figures 16.1 and 16.2, but with different values of the parameters k , η , and µ. 16.2 Decay. Compare exponential and logistic decay. Make a plot like figure 16.1 with negative k . 16.3 Differentiate. Verify that our formula for y (t) actually satisfies the logistic differential equations. 16.4 Easy as pie. In predprey, if the red and blue-green dots are close to each other, then the length of the period is close to a familar value. What is that value? Does that value depend upon the actual location of the dots, or just their relative closeness? 16.5 Period. In predprey, if the red and blue-green dots are far apart, does the length of the period get longer or shorter? Is it possible to make the period shorter than the value it has near equilibrium? 16.6 Phase. If the initial value is near the equilibrium point, the graphs of the predator and prey populations are nearly sinusoidal, with a phase shift. In other words, after the prey population reaches a maximum or minimum, the predator population reaches a maximum or minimum some fraction of the period later. What is that fraction?
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16.7 Pitstop. The predprey subfunction pitstop is involved in the "event handling" that ode45 uses to compute the period. pitstop, in turn, uses atan2 to compute angles theta0 and theta1. What is the difference between the two Matlab functions atan2, which takes two arguments, and atan, which takes only one? What happens if atan2(v,u) is replaced by atan(v/u) in predprey? Draw a sketch showing the angles theta0 and theta1. 16.8 tfinal. The call to ode45 in predprey specifies a time interval of [0 100]. What is the significance of the value 100? What happens if you change it? 16.9 Limit growth. Modify predprey to include a growth limiting term for the prey, similar to one in the logistic equation. Avoid another parameter by making the carrying capacity twice the initial value. The equations become y1 y2 )(1 − )y1 2 η1 µ2 y1 ˙ 2 = −(1 − y )y2 µ1 y ˙ 1 = (1 − What happens to the shape of the solution curves? Are the solutions still periodic? What happens to the length of the period?
Chapter 17
Orbits
Dynamics of many-body systems.
Many mathematical models involve the dynamics of objects under the influence of both their mutual interaction and the surrounding environment. The objects might be planets, molecules, vehicles, or people. The ultimate goal of this chapter is to investigate the n-body problem in celestial mechanics, which models the dynamics of a system of planets, such as our solar system. But first, we look at two simpler models and programs, a bouncing ball and Brownian motion. The exm program bouncer is a model of a bouncing ball. The ball is tossed into the air and reacts to the pull of the earth's gravitation force. There is a corresponding pull of the ball on the earth, but the earth is so massive that we can neglect its motion. Mathematically, we let v (t) and z (t) denote the velocity and the height of the ball. Both are functions of time. High school physics provides formulas for v (t) and z (t), but we choose not to use them because we are anticipating more complicated problems where such formulas are not available. Instead, we take small steps of size δ in time, computing the velocity and height at each step. After the initial toss, gravity causes the velocity to decrease at a constant rate, g . So each step updates v (t) with v (t + δ ) = v (t) − δ g The velocity is the rate of change of the height. So each step updates z (t) with z (t + δ ) = z (t) + δ v (t) Here is the core of bouncer.m.
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
generates the plot of a sphere shown in figure 17.1 and returns z0, the z -coordinates of the sphere, and h, the Handle Graphics "handle" for the plot. One of the exercises has you investigate the details of initialize_bouncer. The figure shows the situation at both the start and the end of the simulation. The ball is at rest and so the picture is pretty boring. To see what happens during the simulation, you have to actually run bouncer. The next four statements in bouncer.m are g = 9.8; c = 0.75; delta = 0.005; v0 = 21; These statements set the values of the acceleration of gravity g, an elasticity coefficient c, the small time step delta, and the initial velocity for the ball, v0. All the computation in bouncer is done within a doubly nested while loop. The outer loop involves the initial velocity v0. while v0 >= 1 ... v0 = c*v0; end To achieve the bouncing affect, the initial velocity is repeatedly multiplied by c = 0.75 until it is less than 1. Each bounce starts with a velocity equal to 3/4 of the previous one. Within the outer loop, the statements
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Figure 17.1. Initial, and final, position of a bouncing ball. To see what happens in between, run bouncer.
v = v0; z = z0; initialize the velocity v to v0 and the height z to z0. Then the inner loop while all(z >= 0) set(h,'zdata',z) drawnow v = v - delta*g; z = z + delta*v; end proceeds until the height goes negative. The plot is repeatedly updated to reflect the current height. At each step, the velocity v is decreased by a constant amount, delta*g, thereby affecting the gravitational deceleration. This velocity is then used to compute the change in the height z. As long as v is positive, the z increases with each step. When v reaches zero, the ball has reached its maximum height. Then v becomes negative and z decreases until the ball returns to height zero, terminating the inner loop. After both loops are complete, the statement finalize_bouncer activates a pushbutton that offers you the possibility of repeating the simulation. Brownian motion is not as obvious as gravity in our daily lives, but we do encounter it frequently. Albert Einstein's first important scientific paper was about Brownian motion. Think of particples of dust suspended in the air and illuminated
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by a beam of sunlight. Or, diffusion of odors throughout a room. Or, a beach ball being tossed around a stadium by the spectators. In Brownian motion an object – a dust particle, a molecule, or a ball – reacts to surrounding random forces. Our simulation of these forces uses the built-in MATLAB function randn to generate normally distributed random numbers. Each time the statement randn is executed a new, unpredictable, value is produced. The statement randn(m,n) produces an m-by-n array of random values. Each time the statement hist(randn(100000,1),60) is executed a histogram plot like the one in figure 17.2 is produced. Try executing this statement several times. You will see that different histograms are produced each time, but they all have the same shape. You might recognize the "bell-shaped curve" that is known more formally as the Gaussian or normal distribution. The histogram shows that positive and negative random numbers are equally likely and that small values are more likely than large ones. This distribution is the mathematical heart of Brownian motion.
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A simple example of Brownian motion known as a random walk is shown in figure 17.3. This is produced by the following code fragment. m = 100; x = cumsum(randn(m,1)); y = cumsum(randn(m,1));
225 plot(x,y,'.-') s = 2*sqrt(m); axis([-s s -s s]);
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Figure 17.3. A simple example of Brownian motion. The key statement is x = cumsum(randn(m,1)); This statement generates the x-coordinates of the walk by forming the successive cumulative partial sums of the elements of the vector r = randn(m,1). x1 = r1 x2 = r1 + r2 x3 = r1 + r2 + r3 ... A similar statement generates the y -coordinates. Cut and paste the code fragment into the Matlab command window. Execute it several times. Try different values of m. You will see different random walks going off in different random directions. Over many executions, the values of x and y are just as likely to be positive as negative. We want to compute an axis scale factor s so that most, but not all, of the walks stay within the plot boundaries. It turns out √ that as m, the length of the walk, increases, the proper scale factor increases like m. A fancier Brownian motion program, involving simultaneous random walks of many particles in three dimensions, is available in brownian3.m. A snapshot of the evolving motion is shown in figure 17.4. Here is the core of brownian3.m.
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n = 50; % Default number of particles P = zeros(n,3); H = initialize_graphics(P); while ~get(H.stop,'value') % Obtain step size from slider. delta = get(H.speed,'value'); % Normally distributed random velocities. V = randn(n,3); % Update positions. P = P + delta*V; update_plot(P,H); end The variable n is the number of particles. It is usually equal to 50, but some other number is possible with brownian3(n). The array P contains the positions of n particles in three dimensions. Initially, all the particles are located at the origin, (0, 0, 0). The variable H is a Matlab structure containing handles for all the user
227 interface controls. In particular, H.stop refers to a toggle that terminates the while loop and H.speed refers to a slider that controls the speed through the value of the time step delta. The array V is an n-by-3 array of normally distributed random numbers that serve as the particle velocities in the random walks. Most of the complexity of brownian3 is contained in the subfunction initialize_graphics. In addition to the speed slider and the stop button, the GUI has pushbuttons or toggles to turn on a trace, zoom in and out, and change the view point. We are now ready to tackle the n-body problem in celestial mechanics. This is a model of a system of planets and their interaction described by Newton's laws of motion and gravitational attraction. Over five hundred years ago, Johannes Kepler realized that if there are only two planets in the model, the orbits are ellipses with a common focus at the center of mass of the system. This provides a fair description of the moon's orbit around the earth, or of the earth's orbit around the sun. But if you are planning a trip to the moon or a mission to Mars, you need more accuracy. You have to realize that the sun affects the moon's orbit around the earth and that Jupiter affects the orbits of both the earth and Mars. Furthermore, if you wish to model more than two planets, an analytic solution to the equations of motion is not possible. It is necessary to compute numerical approximations. Our notation uses vectors and arrays. Let n be the number of bodies and, for i = 1, . . . , n, let pi be the vector denoting the position of the i-th body. For two-dimensional motion the i-th position vector has components (xi , yi ). For threedimensional motion its components are (xi , yi , zi ). The small system shown in figure 17.5 illustrates this notation. There are three bodies moving in two dimensions. The coordinate system and units are chosen so that initially the first body, which is gold if you have color, is at the origin, p1 = (0, 0) The second body, which is blue, is one unit away from the first body in the x direction, so p2 = (1, 0) The third body, which is red, is one unit away from the first body in the y direction, so p3 = (0, 1) We wish to model how the position vectors pi vary with time, t. The velocity of a body is the rate of change of its position and the acceleration is the rate of change of its velocity. We use one and two dots over pi to denote the velocity and acceleration vectors, p ˙i and p ¨i . If you are familiar with calculus, you realize that the dot means differentiation with respect to t. For our three body example, the first body is initially heading away from the other two bodies, so its velocity vector has two negative components, p ˙1 = (−0.12, −0.36) The initial velocity of the second body is all in the y direction, p ˙2 = (0, 0.72)
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and the initial velocity of the third body is sending it towards the second body, p ˙3 = (0.36, −0.36) Newton's law of motion, the famous F = ma, says that the mass of a body times its acceleration is proportional to the sum of the forces acting on it. Newton's law of gravitational says that the force between any two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. So, the equations of motion are ∑ pj − pi mi mj mi p ¨i = γ , i = 1, . . . , n ||pj − pi ||3
j ̸=i
Here γ is the gravitational constant, mi is the mass of the i-th body, pj − pi is the vector from body i to body j and ||pj − pi || is the length or norm of that vector, which is the distance between the two bodies. The denominator of the fraction involves the cube of the distance because the numerator contains the distance itself and so the resulting quotient involves the inverse of the square of the distance. Figure 17.6 shows our three body example again. The length of the vector r23 = p3 − p2 is the distance between p2 and p3 . The gravitation forces between the bodies located at p2 and p3 are directed along r23 and −r23 . To summarize, the position of the i-th body is denoted by the vector pi . The instantaneous change in position of this body is given by its velocity vector, denoted by p ˙i . The instantaneous change in the velocity is given by its acceleration vector, denoted by p ¨i . The acceleration is determined from the position and masses of all the bodies by Newton's laws of motion and gravitation. The following notation simplifies the discussion of numerical methods. Stack the position vectors on top of each other to produce an n-by-d array where n is the
229
1.5
1
p3
0.5
r
236. The double arrow depicts the vectors r23 = p3 − p2 and −r32 . The length of this arrow is the distance between p2 and p3 .
The masses in our three body example are m1 = 1/2, m2 = 1/3, m3 = 1/6 From these quantities, we can compute the initial value of the gravitation forces, G(P ). We will illustrate our numerical methods by trying to generate a circle. The differential equations are x ˙ = y y ˙ = −x With initial conditions x(0) = 0, y (0) = 1, the exact solution is x(t) = sin t, y (t) = cos t The orbit is a perfect circle with a period equal to 2π . The most elementary numerical method, which we will not actually use, is known as the forward or explicit Euler method. The method uses a fixed time step δ and simultaneously advances both the positions and velocities from time tk to time tk+1 = tk + δ . Pk+1 = Pk + δ Vk Vk+1 = Vk + δ G(Pk ) The forward Euler's method applied to the circle generator problem becomes xk+1 = xk + δ yk yk+1 = yk − δ xk The result for δ = 2π/30 is shown in the first plot in figure 17.7. Instead of a circle we get a growing spiral. The method is unstable and consequently unsatisfactory, particularly for long time periods. Smaller time steps merely delay the inevitable. We would see more complicated, but similar, behavior with the n-body equations. Another elementary numerical method is known as the backward or implicit Euler method. In general, it involves somehow solving a nonlinear system at each step. Pk+1 − δ Vk+1 = Pk Vk+1 − δ G(Pk+1 ) = Vk
231
1 0 −1 −1 0 1
1 0 −1 −1 0 1
1 0 −1 −1 0 1
Figure 17.7. Three versions of Euler's method for generating a circle. The first plot shows that the forward method is unstable. The second plot shows that the backward method has excessive damping. The third plot shows that symplectic method, which is a compromise between the first two methods, produces a nearly perfect circle.
For our simple circle example the implicit system is linear, so xk+1 and yk+1 are easily computed by solving the 2-by-2 system xk+1 − δ yk+1 = xk yk+1 + δ xk+1 = yk The result is shown in the second plot in figure 17.7. Instead of a circle we get a decaying spiral. The method is stable, but there is too much damping. Again, we would see similar behavior with the n-body equations. The method that we actually use is a compromise between the explicit and implicit Euler methods. It is the most elementary instance of what are known as symplectic methods. The method involves two half-steps. In the first half-step, the positions at time tk are used in the gravitation equations to update of the velocities. Vk+1 = Vk + δ G(Pk ) Then, in the second half-step, these "new" velocities are used to update the positions. Pk+1 = Pk + δ Vk+1 The novel feature of this symplectic method is the subscript k + 1 instead of k on the V term in the second half-step. For the circle generator, the symplectic method is xk+1 = xk + δ yk yk+1 = yk − δ xk+1 The result is the third plot in figure 17.7. If you look carefully, you can see that the orbit in not quite a circle. It's actually a nearly circular ellipse. And the final value does not quite return to the initial value, so the period is not exactly 2π . But the important fact is that the orbit is neither a growing nor a decaying spiral.
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1
p3
0.58. The first few steps of our example system.
There are more complicated symplectic algorithms that are much more accurate per step than this symplectic Euler. But the symplectic Euler is satisfactory for generating well behaved graphical displays. Most well-known numerical methods, including Runge-Kutta methods and traditional multistep methods, do not have this symplectic stability property and, as a result, are not as satisfactory for computing orbits over long time spans. Figure 17.8 shows the first few steps for our example system. As we noted earlier, the initial position and velocity are P = 0 1.0000 0 V = -0.1200 0 0.3600 0 0 1.0000
The three masses, 1/2, 1/3, and 1/6, are not equal, but are comparable, so all three bodies have significant affects on each other and all three move noticeable distances. We see that the initial velocity of the first body causes it to move away from the other two. In one step, its position changes from (0, 0) to small negative values, (−0.0107, −0.0653) The second body is initially at position (1, 0) with velocity (0, 1) in the positive y direction. In one step, its position changes to (0.9776, 0.1464). The x-coordinate has changed relatively little, while the y -coordinate has changed by roughly 0.72 δ . The third body moves in the direction indicated by the velocity vector in figure 17.5. After a second step we have the following values. As expected, all the trends noted in the first step continue. P = -0.0079 0.9325 0.1589 V = 0.0136 -0.2259 0.4109 -0.2779 0.7268 -0.6198
1.5
-0.1209 0.2917 0.7793
1
0.5
09. The initial trajectories of our example system.
Figure 17.9 shows an initial section of the trajectories. You should run our Experiments program orbits(3) to see the three bodies in motion. The small body and the large body orbit in a clockwise direction around each other while the medium-size body orbits in a counter-clockwise direction around the other two.
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5
0
−5 20 10 0 −10 −20 −20 0 20
Figure 17.10. The solar system, with the initial positions of all the planets and the orbits of the outer planets, Jupiter, Saturn, Uranus, and Neptune.
Our Experiments program orbits models nine bodies in the solar system, namely the sun and eight planets. Figures 17.10 and 17.11 show snapshots of the output from orbits with two different zoom factors that are necessary to span the scale of the system. The orbits for all the planets are in the proper proportion. But, obviously, the symbols for the sun and the planets do not have the same scale. Web sources for information about the solar system are provided by the University Corporation for Atmospheric Research, the Jet Propulsion Laboratory, and the US National Air and Space Museum,
Exercises
17.1 Bouncing ball. (a) What is the maximum height of the bouncing ball? (b) How many times does the ball bounce? (c) What is the effect of changing each of the four bouncer values g, c, delta, and v0. 17.2 Pluto and Ceres. Change orbits to orbits11 by adding the erstwhile planet Pluto and the recently promoted dwarf planet Ceres. See Wikipedia: and
17.3 Comet. Add a comet to orbits. Find initial conditions so that the comet has a stable, but highly elliptical orbit that extends well beyond the orbits of the planets. 17.4 Twin suns. Turn the sun in orbits into a twin star system, with two suns orbiting each other out of the plane of the planets. What eventually happens to the planetary orbits? For example, try sun1.p sun1.v sun1.m sun2.p sun2.v sun2.m = = = = = = [1 0 0]; [0 0.25 0.25]; 0.5; [-1 0 0]; [0 -0.25 -0.25]; 0.5;
Try other values as well.
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Chapter 17. Orbits
Chapter 18
Shallow Water Equations
The shallow water equations model tsunamis and waves in bathtubs.
This chapter is more advanced mathematically than earlier chapters, but you might still find it interesting even if you do not master the mathematical details. The shallow water equations model the propagation of disturbances in water and other incompressible fluids. The underlying assumption is that the depth of the fluid is small compared to the wave length of the disturbance. For example, we do not ordinary think of the Indian Ocean as being shallow. The depth is two or three kilometers. But the devastating tsunami in the Indian Ocean on December 26, 2004 involved waves that were dozens or hundred of kilometers long. So the shallow water approximation provides a reasonable model in this situation. The equations are derived from the principles of conservation of mass and conservation of momentum. The independent variables are time, t, and two space coordinates, x and y . The dependent variables are the fluid height or depth, h, and the two-dimensional fluid velocity field, u and v . With the proper choice of units, the conserved quantities are mass, which is proportional to h, and momentum, which is proportional to uh and vh. The force acting on the fluid is gravity, represented by the gravitational constant, g . The partial differential equations are: ∂h ∂ (uh) ∂ (vh) + + =0 ∂t ∂x ∂y 2 ∂ (uvh) ∂ (uh) ∂ (u2 h + 1 2 gh ) + + =0 ∂t ∂x ∂y 2 ∂ (vh) ∂ (uvh) ∂ (v 2 h + 1 2 gh ) + + =0 ∂t ∂x ∂x
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
With this notation, the shallow water equations are an instance of a hyperbolic conservation law. ∂U ∂F (U ) ∂G(U ) + + =0 ∂t ∂x ∂y One delicate aspect of this model is the boundary conditions, especially if we intend to model a real world geometry such as the Indian Ocean. For our simple experiment, we confine ourselves to a square region and specify reflective boundary conditions, u = 0 on the vertical sides of the square and v = 0 on the horizontal sides. These conditions cause any waves that reach the boundary to be reflected back into the region. More realistic models of oceans and tsunamis include terms that describe the topography of the ocean floor, the Coriolis force resulting the earth's rotation, and possibly other external forces. But the equations we are considering here are still the basis of such models.
Figure 18.1. At the beginning of a time step, the variables represent the solution at the centers of the finite difference grid.
We will use the Lax-Wendroff method to compute a numerical approximation to the solution. Introduce a regular square finite difference grid with a vector-valued solution centered in the grid cells, as shown in figure 18.1. The quantity
n Ui,j
243 represents a three component vector at each grid cell i, j that evolves with time step n. Each time step involves two stages, something like a two-stage Runge Kutta method for ordinary differential equations. The first stage is a half step; it defines 1 values of U at time step n + 2 and the midpoints of the edges of the grid, as shown in figure 18.2. Ui+ 12 = ,j
2
n+ 1
Ui,j +21
2
n+ 1
1 n n + Ui,j )− (U 2 i+1,j 1 n n = (Ui,j +1 + Ui,j ) − 2
∆t n − Fi,j ) (F n 2∆x i+1,j ∆t (Gn − Gn i,j ) 2∆y i,j +1
Figure 18.2. The first stage computes values that represent the solution at the midpoints of the edges in the finite difference grid. The second stage completes the time step by using the values computed in the first stage to compute new values at the centers of the cells, returning to figure 18.1. ∆ t n+ 1 ∆t n+ 1 n+ 1 n+ 1 n+1 n 2 2 2 Ui,j = Ui,j − (Fi+ 12 ( G − F ) − ) 1 1 − G i− 2 ,j i,j − 1 2 ,j 2 ∆x ∆y i,j + 2 Our MATLAB program, exm/waterwave, uses Lax-Wendroff to solve the shallow water equations on a square region with reflective boundary conditions. Initially, h = 1, u = 0, v = 0 over the entire region, so the solution is static. Then, at repeated intervals, a two dimensional Gaussian shaped peak is added to h, simulating an impulsive disturbance like a water drop hitting the surface. The resulting waves propagate back and forth over the region. A few snapshots of the dynamic graphic are shown in figure 18.3. The Lax-Wendroff scheme amplifies artificial, nonphysical oscillations. Eventually the numerical values overflow, producing floating point Infs and NaNs, which cause the surface plot to disappear. CLAWPACK, which stands for Conservation Law Package, is a large collection of Fortran subroutines developed by Randy LeVeque and his colleagues at the University of Washington. A version of the package specialized to modeling tsunamis has been developed by David George. See:
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Chapter 18. Shallow Water Equations
Figure 18.3. A water drop initiates a wave that reflects off the boundary.
Recap
%% Shallow Water Chapter Recap % This is an executable program that illustrates the statements % introduced in the Shallow Water Chapter of "Experiments in MATLAB". % You can access it with % % water_recap % edit water_recap % publish water_recap % % Related EXM programs % % waterwave %% Finite Differences % A simple example of the grid operations in waterwave.
Figure 19.1. The binary tree defining Morse code. A branch to the left signifies a dot in the code and a branch to the right is a dash. In addition to the root node, there are 26 nodes containing the capital letters of the English alphabet.
This chapter brings together three disparate topics: Morse code, binary trees, and cell arrays. Morse code is no longer important commercially, but it still has
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
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Chapter 19. Morse Code
some avid fans among hobbyists. Binary trees are a fundamental data structure used throughout modern computing. And cell arrays, which are unique to Matlab, are arrays whose elements are other arrays. You can get a head start on our investigation by running the exm program morse_gui Experiment with the four buttons in various combinations, and the text and code box. This chapter will explain their operation. Morse code was invented over 150 years ago, not by Samuel F. B. Morse, but by his colleague, Alfred Vail. It has been in widespread use ever since. The code consists of short dots, '.', and longer dashes, '-', separated by short and long spaces. You are certainly familiar with the international distress signal, '... --- ...', the code for "SOS", abbreviating "Save Our Ships" or perhaps "Save Our Souls". But did you notice that some modern cell phones signal '... -- ...', the code for "SMS", indicating activity of the "Short Message Service". Until 2003, a license to operate an amateur radio required minimal proficiency in Morse code. (Full disclosure: When I was in junior high school, I learned Morse code to get my ham license, and I've never forgotten it.) According to Wikipedia, in 2004, the International Telecommunication Union formally added a code for the ubiquitous email character, @, to the international Morse code standard. This was the first addition since World War I.
The Morse tree
We could provide a table showing that '.-' is the code for A, '-...' the code for B, and so on. But we're not going to do that, and our Matlab program does not start with such a table. Instead, we have figure 19.1. This is a binary tree, and for our purposes, it is the definition of Morse code. In contrast to nature, computer scientists put the root of a tree on top. Starting at this root, or any other node, and moving left along a link signifies a dot, while moving right is a dash. For example, starting at the root and moving one step to the left followed by one to the right gets us to A. So this fact, rather than a table, tells us '.-' is the code for A. The length of a Morse code sequence for a particular character is determined by the frequency of that character in typical English text. The most frequent character is "E". Consequently, its Morse sequence is a single dot and it is linked directly to the root of our tree. The least frequent characters, such as "Z" and "X", have the longest Morse sequences and are far from the root. (We will consider the four missing nodes in the tree later.) Binary trees are best implemented in Matlab by cell arrays, which are arrays whose elements are themselves other Matlab objects, including other arrays. Cell arrays have two kinds of indexing operations. Curly braces, { and }, are used for construction and for accessing individual cells. Conventional smooth parentheses, ( and ), are used for accessing subarrays. For example, C = {'A','rolling','stone','gathers','momentum','.'}
249 produces a cell array that contains six strings of different lengths. This example is displayed as C = 'A' 'rolling' 'stone' 'gathers' 'momentum' '.' The third element, denoted with curly braces by C{3}, is the string 'stone'. The third subarray, denoted with parentheses by C(3), is another cell array containing a single element, the string 'stone'. Now go back and read those last two sentences a few more times. The subtle distinction between them is both the key to the power of cell arrays and the source of pervasive confusion. Think of a string of mailboxes along C street. Assume they are numbered consecutively. Then C(3) is the third mailbox and C{3} is the mail in that box. By itself, C is the entire array of mailboxes. The expression C(1:3) is the subarray containing the first three mailboxes. And here is an unusual construction, with curly braces C{1:3} is a comma separated list, C{1}, C{2}, C{3} By itself, on the command line, this will do three assignment statements, assigning the contents of each of the first three mailboxes, one at a time, to ans. With more curly braces, {C{1:3}} is the same as C(1:3). In the computer hardware itself, there is a distinction between a memory location with a particular address and the contents of that location. This same distinction is preserved by indexing a cell array with parentheses and with braces. Did you see the "Men in Black" movies? In one of them the MIB headquarters has a bank of storage lockers. It turns out that each locker contains an entire civilization, presumably with its own lockers. At the end of the movie it is revealed that the Earth itself is a storage locker in a larger civilization. It is possible that we are all living in one element of a huge cell array. The binary tree defining Morse code is a cell array whose contents are characters and other cell arrays. Each cell represents a node in the tree. A cell, N, has three elements, The first element, N{1}, is a string with a single capital letter, X, designating the node. The second element, N{2}, is another cell array, the dot branch. The third element, N{3}, is the dash branch. There are a couple of exceptional cases. The root node does not have an associated letter, so its first element is an empty string. The U and R nodes, and the leaf nodes, have one or two empty cell arrays for branches. In principle, we could create the entire Morse binary tree with a single gigantic, but unrealistic, assignment statement. M = {'' ... {'E' ... {'I' {'S' {'U' {'A' {'R' {'W' {'T' ...
251 And finally one assignment statement at level zero to create the root node. M = {'' e t}; This function is at the heart of all our Morse code software. You can travel down this tree by first entering three commands. M = morse_tree M = M{2} M = M{3} You then can use the up-arrow on your keyboard to repeatedly select and reexecute these commands. Returning to the first command gives you a fresh tree. Executing the second command is a dot operation, moving down the tree to the left. Executing the third command is a dash operation, moving down the tree to the right. For example, the five commands M M M M M = = = = = morse_tree M{3} M{2} M{2} M{3}
correspond to the Morse sequence '-..-'. This brings you to the node 'X' {} {}
You have reached the X leaf of the tree.
Searching the tree
Returning to the "Men in Black" analogy, the Morse binary tree is a single locker. When you open that locker, you see an empty string (because the root does not have a name) and two more lockers. When you open the locker on the left, you see an 'E' and two more lockers. Now you have three choices. You can open either of the two lockers in the E locker, or you can go back to the root locker and open the locker on the right to visit the T locker. Repeatedly choosing different lockers, or different branches in the tree, corresponds to traversing the tree in different orders. Among these many possible orders, two have standard names, "depth first search" and "breadth first search". They are shown in figures 19.2 and 19.3 and you can see and hear animated versions with the morse_gui program. Depth first search visits each branch as soon as it sees it. When it has visited both branches at a node, it backs up to the first node that has an available branch. Figure 19.2 shows the progress of depth first order up to node W. E I S H V U F A R L W Nothing to the right of W has yet been visited.
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Chapter 19. Morse Code
.
.
E
I
A
S
U
R
W
H
V
F
L
.
.
Figure 19.2. A depth first search in progress. The nodes are visited from left to right.
Breadth first search takes one step along each branch at a node before it continues. The result is a top to bottom search, much like reading English language text. Figure 19.3 shows the breadth first order up to node W. E T I A N M S U R W Nothing below W has yet been visited. Depth first search uses a data structure known as a stack. Here is a code segment with a stack that simply displays the nodes of the tree in depth first order. S = {morse_tree}; while ~isempty(S) N = S{1}; S = S(2:end); if ~isempty(N) fprintf(' %s',N{1}) S = {N{2} N{3} S{:}}; end end fprintf('\n') The stack S is a cell array. Initially, it has one cell containing the tree. The while
253
.
.
E
T
I
A
N
M
S
U
R
W
.
.
Figure 19.3. A breadth first search in progress. The nodes are visited from top to bottom.
loop continues as long as the stack is not empty. Within the body of the loop a node is removed from the top of the stack and the stack shortened by one element. If the node is not empty, the single character representing the node is displayed and two new nodes, the dot and dash branches, are inserted into the top of the stack. The traversal visits recently discovered nodes before it returns to older nodes. Breadth first search uses a data structure known as a queue. Here is another code segment, this time with a queue, that displays the nodes of the tree in breadth first order. Q = {morse_tree}; while ~isempty(Q) N = Q{1}; Q = Q(2:end); if ~isempty(N) fprintf(' %s',N{1}) Q = {Q{:} N{2} N{3}}; end end fprintf('\n') This code is similar to the stack code. The distinguishing feature is that new nodes
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Chapter 19. Morse Code
are inserted at the end, rather that the beginning of the queue. A queue is employing a "First In, First Out", or FIFO, strategy, while a stack is employing a "Last In, First Out", or LIFO, strategy. A queue is like a line at a grocery store or ticket office. Customers at the beginning of the line are served first and new arrivals wait at the end of the line. With a stack, new arrivals crowd in at the start of the line. Our stack and queue codes are not recursive. They simply loop until there is no more work to be done. Here is different approach that employs recursion to do a depth first search. Actually, this does involve a hidden stack because computer systems such as Matlab use stacks to manage recursion. function traverse(M) if nargin == 0 M = morse_tree; end if ~isempty(M) disp(M{1}) traverse(M{2}) traverse(M{3}) end end % traverse
% Initial entry.
% Recursive calls.
Decode and encode
Decoding is the process of translating dots and dashes into text. Encoding is the reverse. With our binary tree, decoding is easier than encoding because the dots and dashes directly determine which links to follow. Here is a function that decodes one character's worth of dots and dashes. The function returns an asterisk if the input does not correspond to one of the 26 letters in the tree. function ch = decode(dd) M = morse_tree; for k = 1:length(dd) if dd(k) == '.' M = M{2}; elseif dd(k) == '-' M = M{3}; end if isempty(M) ch = '*'; return end end ch = M{1}; end % decode
255 Encoding is a little more work because we have to search the tree until we find the desired letter. Here is a function that employs depth first search to encode one character. A stack of dots and dashes is built up during the search. Again, an asterisk is returned if the input character is not in the tree. function dd = encode(ch) S = {morse_tree}; D = {''}; while ~isempty(S) N = S{1}; dd = D{1}; S = S(2:end); D = D(2:end); if ~isempty(N) if N{1} == ch; return else S = {N{2} N{3} S{:}}; D = {[dd '.'] [dd '-'] D{:}}; end end end dd = '*'; end % encode These two functions are the core of the decoding and encoding aspect of morse_gui.
A binary tree with four levels has 25 − 1 = 31 nodes, counting the root. Our morse_tree has a root and only 26 other nodes. So, there are four empty spots. You can see them on the dash branches of U and R and on both branches of O. So far, our definition of Morse code does not provide for the four sequences ..-.-.---. ----
How should these be decoded? We also want to add codes for digits and punctuation marks. And, it would be nice to provide for at least some of the non-English characters represented with umlauts and other diacritical marks. Morse code was invented 100 years before modern computers. Today, alphabets, keyboards, character sets and fonts vary from country to country and even from computer to computer. Our function morse_table_extended extends the tree two more levels and adds 10 digits, 8 punctuation characters, and 7 non-English characters to the 26 characters in the original morse_table. Figure 19.4 shows the result. The extend button in morse_gui accesses this extension. Some of the issues involved in representing these additional characters are pursued in the exercises.
Morse code table
We promised that we would not use a table to define Morse code. Everything has been based on the binary tree. When an actual table is desired we can generate one from the tree. Our function is named morse_code and it employs the recursive algorithm traverse from the previous section. The recursion carries along the emerging table C and a growing string dd of dots and dashes. Indexing into the table is based upon the ASCII code for characters. The ASCII standard is an 7-bit code that is the basis for computer representation of text. Seven bits provide for 27 = 128 characters. The first 32 of these are nonprinting characters that were originally used for control of teletypes and are now largely obsolete. The remaining 96 are the 52 upper and lower case letters of the English alphabet, 10 digits, and 32 punctuation marks. In Matlab the function char converts a numeric value to a character. For example char(65) ans = 'A' The function double converts a character to a floating point number and the function uint8 converts a character to an unsigned 8-bit integer. Either of these can be used as an index. For example double('A') ans = 65
257 A computer byte is 8 bits and so the ASCII standard is readily extended by another 128 characters. The actual graphic printed for some of these characters may vary from country to country and from font to font. Our function morse_code produces a table of both ASCII and Morse code from either morse_tree or morse_tree_extended. The recursive traverse algorithm is used for the depth first search. ASCII codes are used as indices into a 256-element cell array of dots and dashes. The search inserts only 26 or 51 elements into the array, so a final step creates a printable table. function C = morse_code(C,M,dd) % MORSE_CODE % C = morse_code % C = morse_code(morse_tree) % C = morse_code(morse_tree_extended) if nargin < 3 if nargin == 0 M = morse_tree; else M = C; end C = cell(256,1); dd = ''; end % Choose binary tree
You should run morse_code(morse_tree_extended) to see what output is generated on your computer.
References
[1] Wikipedia article on Morse code. See the complete Morse code tree at the end of the article.
Recap
%% Morse Code Chapter Recap % This is an executable program that illustrates the statements % introduced in the Morse Code Chapter of "Experiments in MATLAB". % You can access it with % % morse_recap % edit morse_recap
19.2 Email. Use the extend button and translate box in morse_gui to translate your email address into Morse code.
261 19.3 dash. Why didn't I use an underscore, '_', instead of a minus sign, '-', to represent a dash? 19.4 Note. What musical note is used by the sound feature in morse_gui? 19.5 Reverse order. Find this statement in function depth in morse_gui. S = {N{2} N{3} S{:}}; What happens if you interchange N{2} and N{3} like this? S = {N{3} N{2} S{:}}; 19.6 Extend. Using the tree at the end of the Wikipedia article, add more characters to morse_code_extended. If you live in a locale that has non-English characters in the alphabet, be sure to include them. 19.7 Four dashes. Why does the character with Morse code '----' cause a unique difficulty in morse_code_extended? 19.8 YouTube. Check out "Morse code" on YouTube. Be sure to listen to the "Morse code song". Who wins the Morse code versus texting contest on Jay Leno's Tonite show? 19.9 Trinary. What does this function do? Why is it named trinary. What determines how long it runs? What causes it to terminate? Why is the if statement necessary? Modify the program to make it use a depth first search. Modify the program to make it work with morse_tree_extended. function trinary T = [0 0 0 0]; Q = {morse_tree}; while any(T(1,:) < 2) p = T(1,:); y = polyval(p,3); if ~isempty(Q{1}) fprintf('%s %d%d%d%d %2d\n',Q{1}{1},p,y) Q = {Q{2:end} Q{1}{2} Q{1}{3}}; else Q = {Q{2:end} {} {}}; end T = [T(2:end,:); [T(1,2:end) 1]; [T(1,2:end) 2]]; end end % trinary 19.10 Cell arrays. Let
19.11 Fonts. Experiment with various fonts. See the Wikipedia article on the ASCII character set. You can generate a printable character table in Matlab with k = reshape([32:127 160:255],32,[])'; C = char(k) The first half of this table is standard, but the second half depends upon the fonts you are using. You can change fonts in the command window by accessing the "File" menu, then selecting "Preferences" and "Fonts". Highlight the desktop code font and use the up and down arrow keys. In my opinion the best font for the Matlab command window is Lucida Sans Typewriter You can also display the character table in the figure window with txt = text(.25,.50,C,'interp','none'); To change display fonts in the figure window, try commands like these work on your computer. set(txt,'fontname','Lucida Sans Typewriter') set(txt,'fontname','Courier New') set(txt,'fontname','Comic Sans MS') set(txt,'fontname','Wingdings') set(txt,'fontname','GiGi') Use the font names available under the command window preferences.
Chapter 20
Music
What does
12
√ 2 have to do with music?
In the theory of music, an octave is an interval with frequencies that range over a factor of two. In most Western music, an octave is divided into twelve semitones with equal frequency ratios. Since twelve semitones comprise a factor of √ two, one semitone is a factor of 12 2. And because this quantity occurs so often in this chapter, let √ 12 σ= 2 Our Matlab programs use sigma = 2^(1/12) = 1.059463094359295 Think of σ as an important mathematical constant, like π and ϕ.
Keyboard
Figure 20.1 shows our miniature piano keyboard with 25 keys. This keyboard has two octaves, with white keys labeled C D ... G A B, plus another C key. Counting both white and black, there are twelves keys in each octave. The frequency of each key is a semitone above and below its neighbors. Each black key can be regarded as either the sharp of the white below it or the flat of the white above it. So the black key between C and D is both C♯ and D♭. There is no E♯/F♭ or B♯/C♭. A conventional full piano keyboard has 88 keys. Seven complete octaves account for 7 × 12 = 84 keys. There are three additional keys at the lower left and one additional key at the upper end. If the octaves are numbered 0 through 8, then
c 2011 Cleve Moler Copyright ⃝ R is a registered trademark of MathWorks, Inc.TM Matlab⃝ October 4, 2011
263
264
Chapter 20. Music
Figure 20.1. Miniature piano keyboard.
Figure 20.2. Middle C.
a C4 = 440σ −9 ≈ 261.6256 Hz Our EXM program pianoex uses C4 as the center of its 25 keys, so the number
265 range is -12:12. The statement pianoex(0) generates and plays the sound from a sine wave with frequency C4. The resulting visual display is shown in figure 20.2. This for loop plays a two octave chromatic scale starting covering all 25 notes on our miniature keyboard. for n = -12:12 pianoex(n) end
Do Re Mi
One of the first songs you learned to sing was Do Re Mi Fa So La Ti Do If you start at C4, you would be singing the major scale in the key of C. This scale is played on a piano using only the white keys. The notes σ , the C-major scale is σ 0 σ 2 σ 4 σ 5 σ 7 σ 9 σ 11 σ 12 You can play this scale on our miniature keyboard with for n = [0 2 4 5 7 9 11 12] pianoex(n) end The number of semitones between the notes is given by the vector diff([0 2 4 5 7 9 11 12]) = [2 2 1 2 2 2 1] The sequence of frequencies in our most common scale is surprising. Why are there 8 notes in the C-major scale? Why don't the notes in the scale have uniform frequency ratios? For that matter, why is the octave divided into 12 semitones? The notes in "Do Re Me " are so familiar that we don't even ask ourselves these questions. Are there mathematical explanations? I don't have definitive answers, but I can get some hints by looking at harmony, chords, and the ratios of small integers.
Vibrations and modes
Musical instruments create sound through the action of vibrating strings or vibrating columns of air that, in turn, produce vibrations in the body of the instrument
266
Chapter 20. Music
Figure 20.3. The first nine modes of a vibrating string, and their weighted sum.
and The simplest model is a one-dimensional vibrating string, held fixed at its ends. The units of the various physical parameters can be chosen so that the length of the string is 2π . The modes are then simply the functions vk (x) = sin kx, k = 1, 2, ... Each of these functions satisfy the fixed end point conditions vk (0) = vk (2π ) = 0 The time-dependent modal vibrations are uk (x, t) = sin kx sin kt, k = 1, 2, ... and the frequency is simply the integer k . (Two- and three-dimensional models are much more complicated, but this one-dimensional model is all we need here.) Our EXM program vibrating_string provides a dynamic view. Figure 20.3 is a snapshot showing the first nine modes and the resulting wave traveling along the string. An exercise asks you to change the coefficients in the weighted sum to produce different waves.
and x = sin t, y = sin at, z = sin bt The example with a = 3/2 and b = 5/4 shown in figure 20.4. is produced by the default settings in our exm program lissajous. This program allows you to change
269 a and b by entering values in edit boxes on the figure. Entering b = 0 results in a two dimensional Lissajous figure like the one shown in figure 20.6. The simplest, "cleanest" Lissajous figures result when the parameters a and b are ratios of small integers. a= p r , b = , p, q, r, s = small integers q s
This is because the three functions p r x = sin t, y = sin t, z = sin t q s all return to zero when t = 2mπ, m = lcm(q, s) where lcm(q, s) is the least common multiple of q and s. When a and b are fractions with large numerators and denominators, the curves oscillate more rapidly and take longer to return to their starting values. In the extreme situation when a and b are irrational, the curves never return to their starting values and, in fact, eventually fill up the entire square or cube. We have seen that dividing the octave into 12 equal sized semitones results in frequencies that are powers of σ , an irrational value. The E key and G keys are four and seven semitones above C, so their frequencies are sigma^4 = 1.259921049894873 sigma^7 = 1.498307076876682 The closest fractions with small numerator and denominator are 5/4 = 1.250000000000000 3/2 = 1.500000000000000 This is why we chose 5/4 and 3/2 for our default parameters. If the irrational powers of sigma are used instead, the results are figures 20.5 and 20.7. In fact, these figures are merely the initial snapshots. If we were to let the program keep running, the results would not be pleasant.
Harmony and Intonation
Harmony is an elusive attribute. Dictionary definitions involve terms like "pleasing", "congruent", "fitting together". For the purposes of this chapter, let's say that two or more musical notes are harmonious if the ratios of their frequencies are rational
270
Chapter 20. Music
numbers with small numerator and denominator. The human ear finds such notes fit together in a pleasing manner. Strictly speaking, a musical chord is three or more notes sounded simultaneously, but the term can also apply to two notes. With these definitions, chords made from a scale with equal semitones are not exactly harmonious. The frequency ratios are powers of σ , which is irrational. Tuning a musical instrument involves adjusting its physical parameters so that it plays harmonious music by itself and in conjunction with other instruments. Tuning a piano is a difficult process that is done infrequently. Tuning a violin or a guitar is relatively easy and can be done even during breaks in a performance. The human singing voice is an instrument that can undergo continuous retuning. For hundreds of years, music theory has included the design of scales and the tuning of instruments to produce harmonious chords. Of the many possibilities, let's consider only two – equal temperament and just intonation. Equal temperament is the scheme we've been describing so far in this chapter. The frequency ratio between the notes in a chord can be expressed in terms of σ . Tuning an instrument to have equal temperament is done once and for all, without regard to the music that will be played. A single base note is chosen, usually A = 440 Hz, and that determines the frequency of all the other notes. Pianos are almost always tuned to have equal temperament. Just intonation modifies the frequencies slightly to obtain more strictly harmonious chords. The tuning anticipates the key of the music about to be played. Barbershop quartets and a capella choirs can obtain just intonation dynamically during a performance. Here is a Matlab code segment that compares equal temperament with just intonation from a strictly numerical point of view. Equal temperament is defined by repeated powers of σ . Just intonation is defined by a sequence of fractions. sigma = 2^(1/12); k = (0:12)'; equal = sigma.^k; num = [1 16 9 6 5 4 7 3 8 5 7 15 2]'; den = [1 15 8 5 4 3 5 2 5 3 4 8 1]'; just = num./den; delta = (equal - just)./equal; T = [k equal num den just delta]; fprintf('%8d %12.6f %7d/%d %10.6f %10.4f\n',T') k 0 1 2 3 4 equal 1.000000 1.059463 1.122462 1.189207 1.259921 just 1/1 16/15 9/8 6/5 5/4 1.000000 1.066667 1.125000 1.200000 1.250000 delta 0.0000 -0.0068 -0.0023 -0.0091 0.0079
The last column in the table, delta, is the relative difference between the two. We see that delta is less than one percent, except for one note. But the more important consideration is how the music sounds.
Chords
.
Figure 20.8. Dissonace and beats between two adjacent whole notes.
Chords are two or more notes played simultaneously. With a computer keyboard and mouse, we can't click on more than one key at a time. So chords are produced with pianoex by selecting the toggle switches labeled 1 through 12. The switch labeled 0 is always selected. Figure 20.8 shows the visual output generated when pianoex plays a chord involving two adjacent white keys, in this case C and D. You can see, and hear, the phenomenon known as beating. This occurs when tones with nearly equal frequencies alternate between additive reinforcement and subtractive cancellation. The relevant trig identity is a−b a+b t cos t sin at + sin bt = sin 2 2
272
Chapter 20. Music
The sum of two notes is a note with the average of the two frequencies, modulated by a cosine term involving the difference of the two frequencies. The players in an orchestra tune up by listening for beats between their instruments and a reference instrument. The most important three-note chord, or triad, is the C major fifth. If C is the lowest, or root, note, the chord is C-E-G. In just intonation, the frequency ratios are 5 3 1: : 4 2 These are the parameter values for our default Lissajous figure, shown in figure 20.4. Figures 20.9 and 20.10 show the visual output generated when pianoex plays a C major fifth with just intonation and with equal temperament. You can see that the wave forms in the oscilloscope are different, but can you hear any difference in the sound generated?
Figure 20.9. C major fifth with just intonation.
Synthesizing Music
Our pianoex program is not a powerful music synthesizer. Creating such a program is a huge undertaking, way beyond the scope of this chapter or this book. We merely want to illustrate a few of the interesting mathematical concepts involved in music. The core of the pianoex is the code that generates a vector y representing the amplitude of sound as a function of time t. Here is the portion of the code that handles equal temperament. The quantity chord is either a single note number, or a vector like [0 4 7] with the settings of the chord toggles. sigma = 2^(1/12);
piano playing middle C. This sample is loaded during the initialization of pianoex, S = load('piano_c.mat'); middle_c = double(S.piano_c)/2^15; set(gcf,'userdata',middle_c) A function from the Matlab Signal Processing Toolbox is then used to generate notes at different frequencies. middle_c = get(gcf,'userdata'); fs = 44100; t = 0:1/fs:T; y = zeros(size(t)); for n = chord y = y + resamplex(middle_c,2^(n/12),length(y)); end Figure 20.11 displays the piano simulation of the C major fith chord. You can see that the waveform is much richer than the ones obtained from superposition of sine waves.
Further Reading, and Viewing
This wonderful video shows a performance of the "Do Re Mi" song from "The Sound of Music" in the Antwerp Central Railway Station. (I hope the URL persists.) Wikipedia has hundreds of articles on various aspects of music theory. Here is one:
Exercises
20.1 Strings. The Wikipedia page has a table headed "Virtual Keyboard" that shows the frequencies of the piano keys as well as five string instruments. The open string violin frequencies are given by v = [-14 -7 0 7]' 440*sigma.^v What are the corresponding vectors for the other four string instruments? 20.2 Vibrating string. In vibrating_string.m, find the statement a = 1./(1:9) Change it to a = 1./(1:9).^2 or a = 1./sqrt(1:9) Also, change the loop control for k = 1:9 to for k = 1:2:9 or for k = 1:3:9 What effect do these changes have on the resulting wave? 20.3 Comet. Try this: a = 2/3; b = 1/2;
277 tfinal = 12*pi; t = 0:pi/512:tfinal; x = sin(t); y = sin(a*t); z = sin(b*t); comet3(x,y,z) Why did I choose this particular value of tfinal? How does this tfinal depend upon a and b? 20.4 Dissonant Lissajous. What is the Lissajous figure corresponding to two or three adjacent keys, or adjacent white keys, on the piano keyboard? 20.5 Irrational biorhythms. The biorhythms described in our "Calendars and Clocks" chapter are based on the premise that our lives are governed by periodic functions of time with periods of 23, 28 and 33 days. What is the effect of revising biorhythms.m so that the periods are irrational values near these? 20.6 Just intonation. With just intonation, the ratios of frequencies of adjacent notes are no longer equal to σ . What are these ratios? 20.7 Rational intonation. Matlab almost has the capability to discover just intonation. The Matlab function rat computes rational approximations. For example, the following statement computes the numerator n and denominator d in a rational approximation of π . [n,d] = rat(pi) n = 355 d = 113 This gives us the approximation π ≈ 355/113, which is accurate to 7 significant figures. For a less accurate approximation, specify a tolerance of 2 percent. [n,d] = rat(pi,.02) n = 22 d = 7 This gives us the familiar π ≈ 22/7. (a) Let's have rat, with a tolerance of .02, generate rational approximations to the powers of σ . We can compare the result to the rational approximations used in just intonation. In our code that compares equal temperament with just intonation, change the statements that define just intonation from num = [...]
278 den = [...] to [num,den] = rat(sigma.^k,.02);
Chapter 20. Music
You should see that the best rational approximation agrees with the one used by just intonation for most of the notes. Only notes near the ends of the scale are different. • 18/17 vs. 16/15 for k = 1 • 9/5 vs. 7/4 for k = 10 • 17/9 vs. 15/8 for k = 11 The approximations from rat are more accurate, but the denominators are primes or powers of primes and so are less likely to be compatible with other denominators. (b) Change the tolerance involved in using rat to obtain rational approximations to powers of σ . Replace .02 by .01 or .10. You should find that .02 works best. (c) Modify pianoex to also use the rational approximations produced by rat. Can you detect any difference in the sound generated by pianoex if these rat approximations are incorporated. 20.8 Musical score. Our pianoex is able to process a Matlab function that represents a musical score. The score is a cell array with two columns. The first column contains note numbers or chord vectors. The second column contains durations. If the second column is not present, all the notes have the same, default, duration. For example, here is a C-major scale. s = {0 2 4 5 7 9 11 12}' pianoex(s) A more comprehensive example is the portion of Vivaldi's "Four Seasons" in the EXM function vivaldi. type vivaldi pianoex(vivaldi) Frankly, I do not find this attempt to express music in a macine-readable form very satisfactory. Can you create something better? | 677.169 | 1 |
Details about Algebra and Trigonometry:
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Abstract
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Sharpen your skills and prepare for your precalculus exam with a wealth of essential facts in a quick-and-easy Q&A format!
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Description
This edition features the exact same content as the traditional text in a convenient, three-hole- punched, loose-leaf version. Books à la Carte also offer a great value—this format costs significantly less than a new textbook.
AGraphical Approach to Precalculus with Limits: A Unit Circle Approach illustrates how the graph of a function can be used to support the solutions of equations and inequalities involving the function. Beginning with linear functions in Chapter 1, the text uses a four-part process to analyze each type of function, starting first with the graph of the function, then the equation, the associated inequality of that equation, and ending with applications. The text covers all of the topics typically caught in a college algebra course, but with an organization that fosters students' understanding of the interrelationships among graphs, equations, and inequalities.
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Table of Contents
Chapter 1 Linear Functions, Equations, and Inequalities
1.1 Real Numbers and the Rectangular Coordinate System
1.2 Introduction to Relations and Functions
Reviewing Basic Concepts
1.3 Linear Functions
1.4 Equations of Lines and Linear Models
Reviewing Basic Concepts
1.5 Linear Equations and Inequalities
1.6 Applications of Linear Functions
Reviewing Basic Concepts
Summary
Review Exercises
Test
Chapter 2 Analysis of Graphs of Functions
2.1 Graphs of Basic Functions and Relations; Symmetry
2.2 Vertical and Horizontal Shifts of Graphs
2.3 Stretching, Shrinking, and Reflecting Graphs
Reviewing Basic Concepts
2.4 Absolute Value Functions
2.5 Piecewise-Defined Functions
2.6 Operations and Composition
Reviewing Basic Concepts
Summary
Review Exercises
Test
Chapter 3 Polynomial Functions
3.1 Complex Numbers
3.2 Quadratic Functions and Graphs
3.3 Quadratic Equations and Inequalities
Reviewing Basic Concepts
3.4 Further Applications of Quadratic Functions and Models
3.5 Higher-Degree Polynomial Functions and Graphs
Reviewing Basic Concepts
3.6 Topics in the Theory of Polynomial Functions (I)
3.7 Topics in the Theory of Polynomial Functions (II)
3.8 Polynomial Equations and Inequalities; Further Applications and Models | 677.169 | 1 |
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Synopses & Reviews
Publisher Comments
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MATLAB, Uniwersalne srodowisko do obliczen naukowo-technicznych, 3e
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Algebra Applications: Data Analysis and Probability
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Originally Released: 2008
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Product Description:
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Highlights
Description
Physics Is Frequently One Of The Hardest Subjects For Students To Tackle Because It Is A Combination Of Two Of The Toughest Subjects For Most Students: Math And Word Problems. If You Understand The Math But Don'T Do Well In Word Problems Or You Understand Word Problems But Have No Idea Where To Begin With The Math, You'Ll Have Trouble With Physics. The Physics Tutor Series Is A Complete Physics Course For The Beginning Physics Student And A Great Refresher Course For Continuing Students. The Entire Course Is A 24 Lesson Series Covering All Of The Core Topics In Detail. What Sets This Series Apart From Other Physics Teaching Tools Is That The Concepts Are Taught Entirely Thought Step-By-Step Example Problems O Increasing Difficulty. It Works By Introducing Each New Concept In An Easy To Understand Way And Using Example Problems That Are Worked Out Step-By-Step And Line By Line To Completion. If A Student Has A Problem With Coursework Or Homework, Simply Find A Similar Problem Fully Worked On In The Series And Review For The Steps Needed To Solve The Problem. Students Will Immediately Improve Their Problem-Solving Skills Which Will Help With Homework And Exams And Will Have A Reference For Many Of The Commonly Asked Problems In Physics | 677.169 | 1 |
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Synopses & Reviews
Publisher Comments
Most students taking this course do so to fulfill a requirement, but the true benefit of the course is learning how to use and understand mathematics in daily life. This quantitative reasoning text is written expressly for those students, providing them with the mathematical reasoning and quantitative literacy skills they'll need to make good decisions throughout their lives. Common-sense applications of mathematics engage students while underscoring the practical, essential uses of math.
Synopsis
This supplement provides hints for learning and studying, along with detailed, worked-out solutions to the odd-numbered unit exercises.
Synopsis
This collection of engaging articles from The New York Times explores a wide range of quantitative topics appearing in recent news items and stories. Articles are correlated to chapters in Using and Understanding Mathematics: A Quantitative Reasoning Approach, and chapter-specific references make it easy for you to relate each article to the textbook. A series of follow-up questions are also included for homework or class discussion.
Synopsis
Using and Understanding Mathematics: A Quantitative Reasoning Approach, Fifth Edition increases readers' mathematical literacy so that they better understand the mathematics used in their daily lives, and can use math effectively to make better decisions every day. Contents are organized with that in mind, with engaging coverage in sections like Taking Control of Your Finances, Dividing the Political Pie, and a full chapter about Mathematics and the Arts.
About the Author
Jeffrey Bennett specializes in mathematics and science education. He has taught at every level from pre-school through graduate school, including more than 50 college courses in mathematics, physics, astronomy, and education. His work on Using and Understanding Mathematics began in 1987, when he helped create a new mathematics course for the University of Colorado's core curriculum. Variations on this course, with its quantitative reasoning approach, are now taught at hundreds of colleges nationwide. In addition to his work in mathematics, Dr. Bennett (whose PhD is in astrophysics) has written leading college-level textbooks in astronomy, statistics, and the new science of astrobiology, as well as books for the general public. He also proposed and developed both the Colorado Scale Model Solar System on the University of Colorado at Boulder campus and the Voyage Scale Model Solar System, a permanent, outdoor exhibit on the National Mall in Washington, DC. He has recently begun writing science books for children, including the award-winning Max Goes to the Moon and Max Goes to Mars. When not working, he enjoys swimming as well as hiking the trails of Boulder, Colorado with his family.
William L. Briggs has been on the mathematics faculty at the University of Colorado at Denver for 22 years. He teaches numerous courses within the undergraduate and graduate curriculum, and has special interest in teaching calculus, differential equations, and mathematical modeling. He developed the quantitative reasoning course for liberal arts students at University of Colorado at Denver supported by his textbook Using and Understanding Mathematics. He has written two other tutorial monographs, The Multigrid Tutorial and The DFT: An Owner's Manual for the Discrete Fourier Transform, as well as Ants, Bikes, Clocks, a mathematical problem solving text for undergraduates. He is a University of Colorado President's Teaching Scholar, an Outstanding Teacher awardee of the Rocky Mountain Section of the MAA, and the recipient of a Fulbright Fellowship to Ireland. Bill lives with his wife, Julie, and their Gordon setter, Seamus, in Boulder, Colorado. He loves to bake bread, run trails, and rock climb in the mountains near his home.
Table of Contents
Prologue: Literacy and Problem Solving
Between Series, an Actress Became a Superstar (in Math); July 19, 2005
By Kenneth Chang. . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 1 Thinking Critically
Enlisting Science's Lessons to Entice More Shoppers to Spend More; September 19, 2006 | 677.169 | 1 |
Linear Programming for Beginners
Description: This book fills a gap in the linear programming literature, by explaining the steps that are illustrated but not always fully explained in every elementary operations book '" the steps that lead from the elementary and intuitive graphical method fills a gap in the linear programming literature, by explaining the steps that are illustrated but not always fully explained in every elementary operations book '" the steps that lead from the elementary and intuitive graphical method of solution to the more advanced simplex tableau method.Most of the world, even those technically trained, can get along very well by seeing a few illustrations of simple linear programming problems solved graphically, followed by instruction in the use of computer software for solving real-world problems. But there needs to be a coterie of initiates who understand the process well enough to explain it to others, to know what the pitfalls, ramifications and special cases are, and to provide further developments. I have used an informal narrative style with a number of worked out examples and detailed explanations, to put the topic within reach | 677.169 | 1 |
Beginning Algebraended for developmental math courses in beginning algebra, this text retains the hallmark features that have made the Aufmann texts market leaders: an interactive approach in an objective-based framework: a clear writing style, and an emphasis on problem-solving strategies. The acclaimed Aufmann Interactive Method, allows students to try a skill as it is introduced with matched-pair examples, offering students immediate feedback, reinforcing the concept, identifying problem areas, and, overall, promoting student success. | 677.169 | 1 |
Exploring Analytic Geometry with Mathematica®
• A large number of examples with solutions and graphics is keyed to the textual development of each topic.Preface
The study of two-dimensional analytic geometry has gone in and out of fashion several times over the past century. Mathematica provides an attractive environment for studying analytic geometry. engineers and mathematicians alike who are interested in analytic geometry can use this book and software for the study. Students. that can provide an interactive environment for studying the field. as well as producing general formulas with variables. research or just plain enjoyment of analytic geometry. • Hints are provided for improving the reader's use and understanding of Mathematica and Descarta2D. By combining the power of Mathematica with an analytic geometry software system called Descarta2D. such as Mathematica. and full solutions are illustrated using Mathematica. • More advanced topics are covered in explorations provided with each chapter. Mathematica and Descarta2D provide the following outstanding features: • The book can serve as classical analytic geometry textbook with in-line Mathematica dialogs to illustrate key concepts. however this classic field of mathematics has once again become popular due to the growing power of personal computers and the availability of powerful mathematical software systems.
v
. Mathematica also has good facilities for producing graphical plots which are useful for visualizing the graphs of two-dimensional geometry. meaning that geometry problems can be solved numerically. Mathematica supports both numeric and symbolic computations. approach to studying analytic geometry. yet powerful. the result being a simple. the author has succeeded in meshing an ancient field of study with modern computational tools. producing approximate or exact answers.
Features
Exploring Analytic Geometry with Mathematica.
• A CD-ROM is included for convenient. Sample explorations include some of the more famous theorems from analytic geometry: Carlyle's Circle • Castillon's Problem • Euler's Triangle Formula • Eyeball Theorem • Gergonne's Point • Heron's Formula • Inversion • Monge's Theorem • Reciprocal Polars • Reflection in a Point • Stewart's Theorem • plus many more.
Explorations
Each chapter in Exploring Analytic Geometry with Mathematica concludes with more advanced topics in the form of exploration problems to more fully develop the topics presented in each chapter. with complete syntax for over 100 new commands. • Complete source code for Descarta2D is provided in 30 well-documented Mathematica notebooks. There are more than 100 of these more challenging explorations. and the full solutions are provided on the CD-ROM as Mathematica notebooks as well as printed in Part VIII of the book.
. Each chapter begins with definitions of underlying mathematical terminology and develops the topic with more detailed derivations and proofs of important concepts. • The complete book is integrated into the Mathematica Help Browser for easy access and reading.
Classical Analytic Geometry
Exploring Analytic Geometry with Mathematica begins with a traditional development of analytic geometry that has been modernized with in-line chapter dialogs using Descarta2D and Mathematica to illustrate the underlying concepts. • A complete software system and mathematical reference is packaged as an affordable book. permanent storage of the Descarta2D software. The following topics are covered in 21 chapters: Coordinates • Points • Equations • Graphs • Lines • Line Segments • Circles • Arcs • Triangles • Parabolas • Ellipses • Hyperbolas • General Conics • Conic Arcs • Medial Curves • Transformations • Arc Length • Area • Tangent Lines • Tangent Circles • Tangent Conics • Biarcs.vi
Preface • A detailed reference manual provides complete documentation for Descarta2D.
125 notebooks. Maximum benefit of the book and software is gained by using it in conjunction with Mathematica. Arc Length. 30 loadable files • Exploration solutions. the complete source code for Descarta2D is provided. Ellipses. Circles. three notebooks • Complete Descarta2D source code. both in printed form in the book and as Mathematica notebook files on the CD-ROM. Hyperbolas. Conics. Biarcs) • Descarta2D Reference (philosophy and command descriptions) • Descarta2D Packages (complete source code)
.1 and Version 4. 24 interactive notebooks • Reference material for Descarta2D. In addition. manipulating and querying geometric objects in Mathematica. Medial Curves) • Geometric Functions (Transformations.0. Lines. To support the study and enhancement of the Descarta2D algorithms. Area) • Tangent Curves (Lines. Conics.
Organization of the Book
Exploring Analytic Geometry with Mathematica is a 900-page volume divided into nine parts: • Introduction (Getting Started and Descarta2D Tour) • Elementary Geometry (Points. Circles. Arcs.0.Preface
vii
Descarta2D
Descarta2D provides a full-scale Mathematica implementation of the concepts developed in Exploring Analytic Geometry with Mathematica.
CD-ROM
The CD-ROM provides the complete text of the book in Abode Portable Document Format (PDF) for interactive reading. but a passive reading and viewing of the book and notebook files can be accomplished without using Mathematica itself. These notebooks have been thoroughly tested and are compatible with Mathematica Version 3. 30 notebooks • Descarta2D packages. Triangles) • Conics (Parabolas. A reference manual section explains in detail the usage of over 100 new commands that are provided by Descarta2D for creating. the CD-ROM provides the following Mathematica notebooks: • Chapters with Mathematica dialogs.
He has managed the development of a number of commercial computer aided design systems and holds a US Patent involving the underlying data representations of geometric models.
. He has been involved in solid modeling since its inception in the early 1980's and has contributed to the theoretical foundation of the subject through several published papers. Bibliography and a detailed index). Vossler is a mechanical engineer and computer software designer with more than 20 years experience in computer aided design and geometric modeling.viii • Explorations (solution notebooks) • Epilogue (Installation Instructions.
Preface
About the Author
Donald L.
Windows and UNIX machines with Mathematica 3. This classic field of mathematics has once again become popular due to the growing power of personal computers and the availability of powerful mathematical software systems. Complete solutions are provided both as interactive Mathematica notebooks on the CD and as printed material in the book. complete with in-line Mathematica dialogs illustrating every concept as it is introduced. called Descarta2D is provided – Part VII of the book is a listing of the (30) Mathematica files notebooks supporting Descarta2D. – provides a complete computer-based environment for study of analytic geometry – all chapters and reference material are provided on the CD in addition to being printed in the book. for the underlying computer implementation. He has managed the development of a number of commercial computer aided design systems and holds a US Patent involving the underlying data representations of geometric models. Students. the source code is also in on the CD • Explorations – More than 120 challenging problems in analytic geometry are posed. which will operate on Macintosh. – Excellent theoretical presentation – Fully explained examples of all key concepts • Interactive Mathematica notebooks for the entire book. Vossler The study of two-dimensional analytic geometry has gone in and out of fashion several times over the past century. • Detailed reference manual – Complete documentation for Descarta2D – Fully integrated into the Mathematica Help Browser
About the author
Donald L. • Mathematica and Descarta2D Hints are provided to expand the reader's knowledge and understanding of Descarta2D and Mathematica . • A classic study in analytic geometry.1 or 4. the author has succeeded in meshing an ancient field of study with modern computational tools. including source code. that can provide an interactive environment for studying the field.
CD-ROM included
Full contents of book included on CD-ROM. yet powerful. • Complete software system: Descarta2D – a software system.Exploring Analytic Geometry with Mathematica
by Donald L.
. research or just plain enjoyment of analytic geometry. the result being a simple. such as Mathematica. By combining the power of Mathematica with an analytic geometry software system called Descarta2D. engineers and mathematicians alike who are interested in analytic geometry can use this book and software for the study. He has been involved in solid modeling since its inception in the early 1980's and has contributed to the theoretical foundation of the subject through several published papers.0 installed. approach to studying analytic geometry.0. Vossler is a mechanical engineer and computer software designer with more than 20 years experience in computer aided design and geometric modeling. | 677.169 | 1 |
Department Philosophy
The Math Skills Faculty strongly believe in the student's right to succeed by providing:
accurate placement,
lexible ways to move forward in the program and,
courses that satisfy varied learning styles.
The goal of the Math Skills Department at Camden County College is to provide students with the opportunity to prepare for college level mathematics. We use various modes of instruction where students utilize textbooks and technology to pursue mathematical investigations, construct mathematical ideas, and apply math to the real world. We strive to build a classroom environment where mathematical content, thought processes, and communication are cultivated and valued. | 677.169 | 1 |
Description:
This document from SpaceTEC National Aerospace Technical Education Center presents a core readiness course which will serve to prepare individuals entering the aerospace field. The document is 55 pages and contains materials on both basic and advanced math topics such as whole numbers, fractions, decimals, roots, computing area, measurement systems, and functions of numbers. | 677.169 | 1 |
I don't think either would be obsolete. Math IC seems to test very strictly a basic knowledge of high levels of algebra and geometry. Math IIC has all that, trig, limits, and all that extra stuff like matrices, series, and different kinds of probability. I'd say they don't overlap too much. | 677.169 | 1 |
Math Mammoth Metric Measuring covers measuring related topics over the span of five grades (grades 1-5). The lessons for this book have been compiled from two of my books about measuring, and are... More > presented here as a single book that only deals with metric units.
The lessons cover measuring length, weight, and volume, measuring temperature in the Celsius scale, the 24-hour clock and elapsed time, and using decimals within the metric system in general. While the book contains something for all grades from grades 1 to 5, the bulk of the lessons are targeted for second and third grade.
This book contains both explanations of the material and exercises, and is thus termed a worktext. It is designed to be very easy to teach from, requiring fairly little teacher preparation.< Less
Developed countries from the West are not the only ones who are upgrading their systems against insurance theft. With the widespread of technology and the integration of new knowledge and computer... More > geniuses, even those living in the suburbs of Africa now has their own system to secure their data and confidential information.
Axis Capital, with group of insurance and reinsurance companies in Bermuda, Australia, United Kingdom, and Singapore and in over ten states in the US, is one of the many companies reported to first integrate a tighter security system in the start of 2015 in the Asia-PacificThe focus of this book is to show the need for urban planning to increase its potential for sustainable development through applied systems thinking. By incorporating a systems approach that includes... More > enhanced information gathering and comprehensive social and environmental metrics, the planner can engage stakeholders with foresight to assist in better decision-making.
A conceptual model is included to show the intersection of an explicit information system designed for sustainable development indicators and a tacit stakeholder engagement approach.
A systems approach to urban development is not simply about eco-efficiency but a new way of seeing our planning obligations toward a vision and strategy inclusive of the diversity of people and ecosystems | 677.169 | 1 |
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Practice Makes Perfect has established itself as a reliable practical workbook series in the language-learning category. Practice Makes Perfect: Algebra, provides students with the same clear, concise approach and extensive exercises to key fields they've come to expect from the series-but now within mathematics. This book presents thorough coverage of skills, such as handling decimals and fractions, functions, and linear and quadratic equations. Practice Makes Perfect: Algebra is not focused on any particular test or exam, but complementary to most algebra curricula. Because of this approach, the book can be used by struggling students needing extra help, readers who need to firm up skills for an exam, or those who are returning to the subject years after they first studied algebra. | 677.169 | 1 |
Problem-Solving Strand for Mathematics
In this chapter, the problem-solving strategy is to Use a Graph. The examples and review questions encourage students to think about the data presented, to discuss reasonable tolerances for estimates, and to interpret graphs in real-life contexts.
Alignment with the NCTM Process Standards
The NCTM Process Standards in the use of graphs include portions of the communication, connections, and representation standards. One of the key requirements when constructing graphs is to organize and consolidate mathematical thinking in order to display the information accurately (COM.1). Graphing requires recognizing and using connections among mathematical ideas (CON.1) such as independent and dependent variables, appropriate scale when assigning values to the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}axes, or patterns and trends in the data displayed.
COM.1 - Organize and consolidate their mathematical thinking through communication.
COM.3 - Analyze and evaluate the mathematical thinking and strategies of others. | 677.169 | 1 |
I remember back in the early 60's about a mathematician who could
solve 7 equations with seven unknowns. In those days, there were no
computers readily available to solve these equations so he was
valuable to have around. Myself, I was able to solve a set of 5
equations with 5 unknowns in my linear algebra class with extreme
difficulty because I would keep making arithmetic errors. I did
manage to solve one set of five, however.
I was taught how to solve equations using matrices to solve these
equations but it was at a fairly high abstract level, so matrices
didn't help me much then.
Then, the TI-83 calculator was invented and matrices became anap! I
can easily plug in the coefficients into the TI-83 graphing calculator
and then use the inverse of the matrix to solve 5, 7 or more equations
if I want to! No problem!
Matrices don't help everybody but I suspect a few people in any
business industry uses them for initial designs. When you send people
to the moon or anywhere in space, some engineers and mathematicians
have to initially use matrices to design the equipment and plot the
journey. Once computers are programed, then you may not need them
very often.
True, not everyone will use them, but a few will always need to have
the skill in using matrices. Students who understand matrices will
appreciate this knowledge. | 677.169 | 1 |
GCSE Edexcel Maths (Revision Guide)
Description: This text has been designed to make revision as active and effective as possible. Short 20-minute 'revision sessions' break down content into manageable chunks and maximise students' concentration | 677.169 | 1 |
About the Apps
Algebra ties many parts of mathematics together with number theory, geometry, and analysis. Finding unknown quantities by using substitute letters (or variables) makes algebra challenging, interesting, and relevant to the real world.
While there are a number of quality high school Algebra apps to choose from, according to our formula these are the best.
Start your journey through algebra with Algebra Touch. You will love it. The hardest concept in algebra is how to correctly manipulate numbers, variables, and expressions in such a way as to arrive at the correct answer. Algebra Touch does it by allowing you to animate the process with guiding colors and sounds. This app is a great place to start. What you don't know already, the app will not only teach you but also reinforce essential algebraic concepts as you try to decode the results of your actions.
Elevated Math has cartoon instruction along with motivational videos as to how math is used in a variety of professions. The free video teaches the user how to break down fractions by comparing it to something that the user understands, such as music notes. Other purchased videos deal with other professions. There are two free lessons and one free video, with more that can be purchased for $1.99 each.
Mathination - Equation Solver also interacts directly with the student. Students can manipulate all parts of the exercise or equation to find the answer. It also gives students an opportunity to input their own problems. The student can easily manipulate any the part of an equation or a problem exercise to their own advantage. The price is higher than the average app price, but it's worth it.
meStudying: Algebra1 should be the next app because it not only tests the received algebraic instruction, but also teaches and reinforces troublesome concepts by providing help at the conclusion of each test question. There is nothing better or more immediate than being told if your guess was correct or incorrect, and to see why it is that way.
Algebra Explained is, in fact, a series of six apps, each designed to teach a different aspect of the subject. The first one, "Order of Operations," has fun video lectures, study cards, and end–of–unit quizzes for which kids can earn graduation caps representing mastery of a topic. The real–world applications and fun video content make this a standout app for kids who need some extra help with basic content.
Comments & Suggestions
Have a suggestion for an educational app for this category you think we should feature? Let us know. | 677.169 | 1 |
eared toward helping students learn and apply mathematics by incorporating the five-step problem-solving process, this text integrates real-world applications with proven pedagogy and an accessible writing style. The Bittinger/Ellenbogen hardback series has consistently provided teachers and students with the tools needed to succeed in developmental mathematics. With this revision, the authors have maintained of the hallmark features that have made this series so successful, including its Five-step problem solving process, student-oriented writing style, real data applications, and variety of exercises. Among the features added or revised, are the Technology Connection Boxes, Collaborative Corner Exercises, and World Wide Web integration. This series not only provides students with the tools necessary to learn and understand math, but also provides them with insights into how math works in the world around them. | 677.169 | 1 |
books.google.com - Second... Year Calculus
Second Year Calculus: From Celestial Mechanics to Special Relativity
Second Principles of Natural Philosophy in which mathematics becomes the ultimate tool for modelling physical reality, to the dawn of a radically new and often counter-intuitive age in Albert Einstein's Special Theory of Relativity in which it is the mathematical model which suggests new aspects of that reality. The development of this process is discussed from the modern viewpoint of differential forms. Using this concept, the student learns to compute orbits and rocket trajectories, model flows and force fields, and derive the laws of electricity and magnetism. These exercises and observations of mathematical symmetry enable the student to better understand the interaction of physics and mathematics.
About the author (2001)
David Bressoud is DeWitt Wallace Professor of Mathematics at Macalester College, Minnesota. He is Chair of the MAA Committee on the Undergraduate Program in Mathematics and Chair-Elect of the MAA Special Interest Group on Teaching Advanced High School Mathematics. | 677.169 | 1 |
Mathematics and Multimedia - Guillermo Bautista
Blog about mathematics, teaching, learning, and technology. Begun in October, 2009, it has included posts such as "GeoGebra Tutorial 4 - Graphs and Sliders"; "Screencasting Tutorial: Making a Math Video Lesson Using Camstudio"; "Derivative in Real Life
...more>>
Mathematics by Mr. P - Mike Poliquin
Middle school math teacher Poliquin began this blog, subtitled "making sure everyone gets it," in December, 2011. Posts have included various original problems of the week; "Eudoxus of Cnidus: the First Mathematical Superstar"; "A Quick Dip in the Deep
...more>>
mathematics.com - ENGINEERING.com
Interactive material to download or work with online; articles on current technology topics; resources and links of interest to mathematicians; technology products offered for sale. Online calculators and applets for geometry, trigonometry, calculus,Mathematics (SparkNotes.com) - WebCT.com & iTurf Inc.
Over 100 guides for mathematics ranging from pre-algebra topics to advanced work in calculus, written by students and recent graduates of Harvard University. The site also includes message boards for beginner, high school, and advanced math, calculus,
...more>>
Mathematics - Student Helpmate - Chris Divyak
Search or browse this archive of questions about algebra, calculus, geometry, statistics, trigonometry, and other college math; then pay for access to answers. To submit your own problem to Student Helpmate, type your question or upload it as a file;
...more>>
mathepower.com - Markus Hendler
A series of calculators for solving problems, for classes 1-10. Primarily in German, but with an English version. Themes include: fractions, geometry, equations, arithmetical operations, trigonometry.
...more>>
Mathepower - Markus Hendler
Online calculators for most calculations covered in the first ten years of school, from basic operations through the first stages of algebra. Available in English, German, and French.
...more>>
Math for Morons Like Us - ThinkQuest 1998
Students talk to students about math: a site designed to help you understand math concepts better. Tutorials, sample problems, and quizzes for Pre-Algebra, Algebra, Geometry, Algebra II, and Pre-Calc/Calculus, designed assuming you know some of the basic
...more>> | 677.169 | 1 |
Amazon.com description:Product Description: The authors have made this book a thorough review of basic courses in Trigonometry. In addition to traditional studies of common trigonometric functions and basic trigonometric identities, the book covers various methods for solving trigonometric equations and inequalities. It also introduces an innovative graphic approach to solving complex trigonometric inequalities and systems of trigonometric inequalities by using graphing calculators.
With a wide variety of topics, diverse exercises with answers, and logical explanations, this book will be an excellent reference for students who want to excel in Trigonometry classes and to be ready for deeper college math study | 677.169 | 1 |
unique feature of this compact student's introduction is that it presents concepts in an order that closely follows a standard mathematics curriculum, rather than structure the book along features of the software. As a result, the book provides a brief introduction to those aspects of the Mathematica software program most useful to students. The second edition of this well loved book is completely rewritten for Mathematica 6 including coverage of the new dynamic interface elements, several hundred exercises and a new chapter on programming. This book can be used in a variety of courses, from precalculus to linear algebra. Used as a supplementary text it will aid in bridging the gap between the mathematics in the course and Mathematica. In addition to its course use, this book will serve as an excellent tutorial for those wishing to learn Mathematica and brush up on their mathematics at the same time. less | 677.169 | 1 |
Product details
ISBN-13: 9780768226379
ISBN: 0768226376
Publication Date: 2003
Publisher: Carson-Dellosa Publishing, LLC
AUTHOR
School Specialty Publishing Staff, Carson-Dellosa Publishing Staff
SUMMARY
Intermediate. Based on NCTM Standards. Designed to help develop a basic understanding of various skills with practice exercises. These books progress in skill difficulty using a variety of activities. Reproducible. 6" x 9".School Specialty Publishing Staff is the author of 'Step-by-step Geometry, Intermediate ', published 2003 under ISBN 9780768226379 and ISBN 0768226 | 677.169 | 1 |
#Calculation component is a powerful calculation engine for your applications. This ActiveX component integrates expression parsing and evaluation. Generally speaking, #Calculation is very useful in two main areas: first, when a ...
... overlook the mathematical tricks behind the seemingly simple equations in this game - you may need to ... fill in the missing arithmetic symbols in the equations as quickly as you can. You will be ...
Kalkules is an universal scientific freeware calculator with an amount of untraditional functions, which can be used particularly by high school or university students. It also offers a wide range of ...
... in this game is to solve the given equations as quickly as possible. You will be given an equation at each level of the game. You need ... the answer, then click the numbers below the equation or press the corresponding keys on your keyboard ...
... available for science symbols, ions, science terms and equation formula. Interactive flowcharts can be called for mechanical and electrical units. An interactive periodic table lists elements properties. A help system ...
The program allows you to solve algebraic equations in the automatic mode. You just enter an equation in any form without any preparatory operations. Step by step Equation Wizard reduces it to a canonical form performing ...
... search for the numeric components and complete the equations in this game! You will be given a ... correct numbers on the grid to complete the equation, then click the Submit button to check if ...
... accessed by placing the mouse pointer over an equation and clicking the right mouse button. This brings ... Most of the commands operate on the clicked-on equation. In addition, there are commands to find out ...
Kalkules is an universal scientific freeware calculator with an amount of untraditional functions, which can be used particularly by high school or university students. It also offers a wide range of ...
... very few researchers using multiple regression or structural equation modeling techniques do investigate for the presence of ... the interaction simply by examining the resulting regression equation, a common way to get an intuitive feel ... | 677.169 | 1 |
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This is a fantastic book. It teaches in a very short period the basics, of many math systems. For instance, you will understand algebra after reading/studying 19 pages! I have referred to this book and referred this book since 1970 to any one that wishes to understand the basics of math and how they apply to physical tasks that we perform. I used it to understand geometry and trigonometry when I was framing houses and building stairs. It was great. I can not recommend a book more highly. Don't let it's age throw you off.
I found this book in a box of my great-grandfather's stuff, and I cannot emphasize how comprehensive it is. Methods are easy to read and understand. Lots of cheat sheet charts (i.e. tolerances, etc.) As a metalworker/construction worker and someone who is studying engineering, I highly recommend this book. It's age is irrelevant. Great for learning from scratch or refreshing your skills. Chapters include: Arithmetic, Algebra, Geometry, Trigonometry, Differential Calculus, Mechanics, Weights and Measures, Excavation and Foundations, Concrete, Brickwork, Carpentry and Building, Lathing and Plastering, Painting, Paperhanging, Glazing, Plumbing, Heating, Ventilating and Conditioning, Machine Shop Work, Sheet Metal Work, Electricity, Electronics, Print Shop, Business Mathematics, and Accounting. | 677.169 | 1 |
Copper Canyon, TX PrecalculusWilliam G.
...Topics usually covered include various properties of functions: domain and range, graphs, end behavior, piecewise and composite functions, factoring and finding the roots of higher order polynomials, exponential and logarithmic functions, rational expressions, conic sections, parametric functions... | 677.169 | 1 |
This annual anthology brings together the year's finest mathematics writing from around the world. Featuring promising new voices alongside some of the foremost names in the field, The Best Writing on Mathematics 2013 makes available to a wide audience many articles not easily found anywhere else--and you don't need to be a mathematician to enjoy... more...
This volume provides the very first transcription of correspondence between Richard Dedekind and Heinrich Weber, one of the most important instances of written dialog between mathematicians in the 19th century. Nearly every subarea of mathematics is addressed in the letters, which intensively discuss nascent developments in the field.... more...
Advanced Linear Algebra, Second Edition takes a gentle approach that starts with familiar concepts and then gradually builds to deeper results. Each section begins with an outline of previously introduced concepts and results necessary for mastering the new material. By reviewing what students need to know before moving forward, the text builds... more...
A reader-friendly introduction to modern algebra with important examples from various areas of mathematics Featuring a clear and concise approach , An Introduction to Essential Algebraic Structures presents an integrated approach to basic concepts of modern algebra and highlights topics that play a central role in various branches of mathematics.... more...
Monomial Algebras, Second Edition presents algebraic, combinatorial, and computational methods for studying monomial algebras and their ideals, including Stanley?Reisner rings, monomial subrings, Ehrhart rings, and blowup algebras. It emphasizes square-free monomials and the corresponding graphs, clutters, or hypergraphs. New to the Second Edition... more...
Introduction to Higher Algebra is an 11-chapter text that covers some mathematical investigations concerning higher algebra. After an introduction to sets of functions, mathematical induction, and arbitrary numbers, this book goes on considering some combinatorial problems, complex numbers, determinants, vector spaces, and linear equations. These... more...
Algebra and Trigonometry presents the essentials of algebra and trigonometry with some applications. The emphasis is on practical skills, problem solving, and computational techniques. Topics covered range from equations and inequalities to functions and graphs, polynomial and rational functions, and exponentials and logarithms. Trigonometric functions... more... | 677.169 | 1 |
We have been active in incorporating Mathematica, a multifunctional general computer programming environment, into the chemical engineering senior year curriculum at Colorado School of Mines (CMS). This integrated platform for symbolic, numeric, and graphical analyses has been used in process control and reaction engineering courses. Students in these classes have had mixed but generally positive reactions toward its use. In this article we will describe our experiences and provide information that allows access to example problems posted on the Internet. The use of Mathematica provides several advantages in teaching chemical engineering concepts--graphical accuracy in problem solutions presented in class, the ability to do more involved problems, and most importantly, discovery learning on the part of the student. Furthermore, the multimedia capabilities of Mathematica can be used to stimulate student interest in subjects that are inherently heavily mathematical in nature. Finally, we believe that the incorporation of this program into the curriculum provides a general engineering tool that can be useful to the students throughout their careers. | 677.169 | 1 |
Algebra Education Articles for Teachers
When the Common Core State Standards in mathematics were published, I was thrilled. Finally, here was a coherent, rigorous set of standards that provided a clear vision of mastery in the field. By outlining specific standards in this way, it gave me the ability to articulate insights and big idea...
The great physicist, Richard Feynman, once compared the scientific method to watching the gods play a cosmic game of chess. Because we cannot participate, and don't know the rules, our only chance of figuring out the game is to watch and make generalizations. Feynman had a gift for communicating ...
Most practice problems in your average algebra textbook follow a familiar pattern: they are generally one or two-step applications of the basic skill taught in the previous chapter, with the numbers shuffled around a bit. What they all have in common is that they are 'closed' problems. They have ...
It came up again a few days ago, that question every teacher dreads and every student obsesses over, especially when having a hard time on a problem or assignment: "When will we ever use this, anyway?" It is an understandable question. Abstract concepts like quadratic curves and complex numbers a... | 677.169 | 1 |
Find a White Plains, NYIn algebra, the concept of variables is introduced. Variables represent numbers and generalizations can be made using the rules of operations, i.e., addition, subtraction, multiplication, and division. The generalizations (patterns) are applied in solving linear and quadratic equations | 677.169 | 1 |
There are all too many books with titles
using the term "Precalculus" or "College Algebra with Trgonometry"
and courses with the same titles to think that this is the first time you
have seen these words. Hey, you're in some kind of a precalculus course
right now! But do you know why the subject is called "precalculus" and precisely
what it is? The answers are not as simple as saying, " it's the course you
take before you take calculus." To begin I'll try by telling you something
about the background of calculus (as I understand it) before the word "calculus"
was identified as a specific subject area of mathematics. Then the explanation
of "precalculus" will be connected to the calculus: precalculus mathematics
is mathematics that gives background for the mathematical concepts, problems,
issues and techniques that appear in the calculus course.
What is "calculus"? Quite simply a "calculus" is a method
for systematically determining a result, for arriving at a conclusion, or
(if you don't mind the redundancy) for calculating an answer [1]. In this
sense there are many calculi, such as the calculus of propositional logic,
the calculus of set operations, the calculus of probabilities, etc.
But when someone talks about "The Calculus," be it "differential" calculus,
"integral" calculus, or the calculus of infinite series, the reference is
usually to "The Calculus of Isaac Newton (1642-1727) and Gottfried Leibniz
(1646-1716)." This calculus provides procedures for solving problems
in the analysis of change: determining rates of change, predicting the
amount and explaining the quality of change, and connecting the concepts
of change with the language and symbolism of algebra that describes change.
This calculus also develops tools for solving problems of geometry: determining
a line tangent to a curve or finding the area of a planar region, predicting
the shape and explaining the graphic qualities of a curve, and connecting
these geometric concepts to the language of algebra that describes geometry.
Since its early development, calculus has grown more important. Its analysis
has been applied in many contexts: the physical sciences and engineering,
the life sciences, economics, probability. In fact calculus
has uses in practically any area of study where change is important. It
provides a theoretical basis as well as a practical tool for exact and estimated
solutions to problems in almost every scientific disciplines.
Before Newton or Leibniz, several great mathematicians had studied
many of the physical and geometric questions, from Euclid and Archimedes in
antiquity to Descartes, Fermat, and Pascal to name only a few mathematicians
living within the century before Newton. Newton and Leibniz differed from
those who had worked on these questions before them by achieving a general
overview to the problems. Their approaches, developed independently, solved
the problems using systematic techniques of calculation that depended fundamentally
on the algebraic description of the problem. With these techniques the user
could avoid repetitious conceptual analysis of each different curve
or formula. The conceptual analysis was summarized in the justifications of
the calculus rules, which were the primary achievement of the new calculus.
The results for the actual problems contained in works by Newton and Leibniz
may not have been new, but the methods were revolutionary in their generality.
Let's touch briefly on two geometric problems that were
historically important to the development of the calculus: The
tangent problem and the area problem.The Tangent Problem: You may recall a geometric
method, known at least since the time of Euclid (about 300 B.C.E.) for
drawing a line that touches a circle C only at one specific point, P, on
the circle. [ See Figure 1.] This line is called the tangent line to the
circle C at P. The construction of this line involves two simple steps:
i) draw a radius from the center of the circle, O, to the point P; and
ii) construct a line at P that is perpendicular to the radius OP. This
perpendicular line is the desired tangent line.
A similar problem for a parabola was solved by Archimedes,
(c. 287 B.C.E. - 212 B.C.E.) The problem is to find a method for drawing
a line that touches a parabola R only at one specific point, P, on the
parabola. [See Figure 2.] This line is called the tangent line to the parabola
R at P.
The general tangent problem is a little more difficult to state because
the quality of being tangent in general requires a more subtle characterization.
Ignoring the issue of what "tangent" means for now, we can describe
the general tangent problem as the problem of finding a method
for drawing a line tangent to a curve at a specific point P on the
curve.
The Area Problem: The area problem is perhaps more familiar
as a measurement problem today. No doubt you have learned many formulae
for finding the area of such common planar regions as those enclosed by
squares, rectangles, triangles, trapezoids, and circles. Earlier treatments
of area had a geometric constructive flavor with which you may not be so
familiar.
For example, the area problem for a right triangle is to give
a method of constructing a rectangular region with area equal to that of
a given right triangle. Here the geometric solution follows two steps:
i) Bisect one leg of the right triangle. ii) Form a rectangle with
one of the resulting segments and the other leg. This rectangle has the
desired property.
The statement of the general area problem in geometry is to
give a procedure for constructing a rectangle (or square) with an area
that is the same as the area of a given region in the plane.
The connection between geometry, numbers, and algebra is generally
considered one of the major contributions of the French mathematician,
scientist, and philosopher, René Descartes (1596-1650) though
some of the ideas appeared in earlier works of Oresme and Galileo. Once
measurement and algebra are added to the tools with which we analyze
the area problem, we arrive at a familiar formulation of the right triangle
area result. If the legs of the triangle have lengths a and b then the
area A of the triangle (and the rectangle) is 1/2 a b. If we use
x instead of a for the length of one leg and mx
instead of b for the other leg then we have the area determined
by the numbers m and x.
Newton and Leibniz both developed algebraic methods for a calculus
of tangents and with that they were able to explore algebraic methods to
find a calculus that would systematically resolve many problems of area,
as well as problems of volume, arc length, etc.
What then is the calculus? Briefly again, calculus is a conceptual
framework which provides systematic techniques for solving problems suitably
posed in the language of analytic geometry and algebra.
What kind of problems can be solved using calculus? Some problem
types frequently mentioned in discussions of the calculus over the centuries
of its development are quite easy to characterize. Other problems have
been developed in the past century or so as a result of the expansion of
the use of mathematics beyond the physical sciences of physics, chemistry,
and engineering, into disciplines such as biology, medicine, and economics.
The increased importance of probability and statistics in understanding
all of the sciences has also resulted in bringing other issues of
the calculus into the limelight. Here is a short list of some traditional
and non-traditional problems that use the calculus.
The Tangent Problem. Determine the line tangent to a given curve
at a given point. (Also, 1a: Define precisely the concept of "tangent.")
For example, determine the line in the plane tangent to the circle with
equation X 2 + Y 2 = 25 at the point (-4,3).
The Velocity Problem. Determine the instantaneous velocity of a
moving object. (Also, 2a: Define precisely the concept of "instantaneous
velocity.") For example, determine the instantaneous velocity of an object
moving on a straight line at time t = 5 seconds when its position at time
t seconds is t 2 + 6t meters from a given point P.
The Extremum Problems. Determine the maximum and minimum values
of a variable quantity that is dependent upon another variable quantity.
For example, suppose the variable Y depends on the variable X with Y =
X 2 - 6X. Determine any maximum and minimum values for the dependent variable
Y when the variable X is allowed to vary over real numbers between 0 and
10.
The Area Problem. Determine the area of a planar region enclosed
by a suitably defined curve. For example, determine the area of the planar
region enclosed by the X - axis, the lines X = 2, X = 5, and the
parabola with equation Y = X 2 - 6X.
The Arc Length Problem. Determine the length of a suitably defined
curve. For example, determine the length of the parabola with equation
Y = X 2 - 6X between the points (0,0) and (6,0).
The Tangent Curve Problem Reversed. Determine a curve so that the
tangent to the curve at any point on the curve is predetermined by some
description depending on the point's position in the plane. For example,
determine a curve in a coordinate plane so that the slope of the tangent
to the curve at the point (a,b) is a + b.
The Position Problem. (The Velocity Problem Reversed.) Determine
the position of an object moving on a straight line from knowledge of its
initial position and instantaneous velocity at every instant. For example,
determine the position of an object moving on a straight line at
time t = 5 seconds knowing its initial position is P on the line and its
instantaneous velocity at time t is precisely t 2 - 6t meters per second.
The Growth/Decay Problem. Given the size of a population or populations
and the rates of growth or decay of the population(s) as functions of the
time and current population sizes, describe the size of the population(s)
in the future or in the past. For example suppose a population is measured
by its biomass and it currently has a biomass of 1000 kilograms. Suppose
the population is growing at a rate that is proportional to its current
size and that one hour ago its biomass was 950 kilograms. Find
the biomass in 10 hours. Find how long it will take before the biomass
will be 2000 kilograms, 5,000 kilograms, and 10,000 kilograms.
The Probability Expected Value - Mean Problem. In an experiment
where we measure the outcome of a variable X, suppose the probability that
X< A is given by some function F(A). What is the expected value of X?
that is, what number in theory would be close to the average or mean of
the values for X if the experiment is repeated a large number of times.
For example, suppose we throw a magnetic dart at a circular dart board
of radius 2 feet and we measure X as the distance from center of the board
to where the dart lands. Suppose that the probability that X< A is given
by the function F(A)= A2/4 for 0< A < 2; 0 for A < 0; and 1 for
2 < A. What is the expected value of X? that is, what is the expected
distance of the dart from the center of the target?
Precalculus Mathematics: So now that you have a better idea
of what mathematical problems are connected to the "calculus," what
is "precalculus" mathematics? Here the simple answer we gave earlier
can be expanded. Precalculus mathematics is mathematics that gives
background for the mathematical concepts, problems, issues and techniques
that appear in the calculus course. Certainly one key background tool for
the calculus is the function concept. Being familiar with function concepts
and specific functions provides an important foundation and language for
the calculus.
To understand calculus you should have a background that allows you
to use
numbers and variables in the context of algebra,
equations and functions both algebraically and visually, and
"real world" applications that use functions to relate the quantities
involved.
So what is the sensible approach to prepare you for calculus? In
the sensible precalculus approach we will
Review and renew your understanding of numbers and variables as used in
algebra.
Review and renew your understanding of equations both algebraically and
visually.
Review, renew, and expand your understanding of functions both algebraically
and visually.
Connect "real world" applications to the equations and functions we introduce
here.
Introduce problems of the type encountered in calculus when precalculus
techniques can be used for solution.
Introduce methods from current technology that will make precalculus analysis
easier and form a foundation for later calculus use of technology.
[1] From the Complete Reference Library Dictionary, Mindscape.
The American Heritage Dictionary of the English Language. 3rd Edition,
1980, Houghton Mifflin.
Calculus a . The branch of mathematics that
deals with limits and the differentiation and integration of functions
of one or more variables.
b . A method of analysis or calculation using a special symbolic
notation.
c . The combined mathematics of differential calculus and integral
calculus. A system or method of calculation.[ Latin, small stone used in
reckoning] | 677.169 | 1 |
Gilcrest StatisticsHalston D.
...Students aren't meant to superficially memorize formulas, but to understand the how and why, the logic behind math. My | 677.169 | 1 |
advanced high school students, undergraduates, graduate students, mathematics teachers, and any lover of mathematical challenges, this two-volume set offers a broad spectrum of challenging problems — ranging from relatively simple to extremely difficult. Indeed, some rank among the finest achievements of outstanding mathematicians. Translated from a well-known Russian work entitled Non-Elementary Problems in an Elementary Exposition, the chief aim of the book is to acquaint the readers with a variety of new mathematical facts, ideas, and methods. And while the majority of the problems represent questions in higher ("non-elementary") mathematics, most can be solved with elementary mathematics. In fact, for the most part, no knowledge of mathematics beyond a good high school course is required. Volume One contains 100 problems, with detailed solutions, all dealing with probability theory and combinatorial analysis. Topics include the representation of integers as sums and products, combinatorial problems on the chessboard, geometric problems on combinatorial analysis, problems on the binomial coefficients, problems on computing probabilities, experiments with infinitely many possible outcomes, and experiments with a continuum of possible outcomes. Volume Two contains 74 problems from various branches of mathematics, dealing with such topics as points and lines, lattices of points in the plane, topology, convex polygons, distribution of objects, nondecimal counting, theory of primes, and more. In both volumes the statements of the problems are given first, followed by a section giving complete solutions. Answers and hints are given at the end of the book. Ideal as a text, for self-study, or as a working resource for a mathematics club, this wide-ranging compilation offers 174 carefully chosen problems that will test the mathematical acuity and problem-solving skills of almost any student, teacher, or mathematician.
Special Offers and Product Promotions
Editorial Reviews
Language Notes
Text: English, Russian (translation)
About the Author
Yaglom, Institute of Atmospheric Physics, Academy of Sciences, USSR.
I. M. Yaglom authored many books which have since become academic standards of reference. These include Complex Numbers in Geometry, Geometric Transformations, A Simple Non-Euclidean Geometry and its Physical Basis, and Probability and Information. He was Professor of Mathematics at Yaroslavl State University from 1974 83 and a technical consultant at the Academy of Pedagogical Sciences from 1984 88.
Top Customer Reviews
I've read the Russian original of this book. It is wonderful. I really enjoyed it, and often used the ideas I found there to solve other problems. Highly recommended, espesially for the high school students unterested in mathematics, but extremely interesting for anybody who likes math.
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It's amazing that so many results that you may have seen in more advanced books proven with more advanced techniques can also be proven with elementary methods. Some of the constructions are amazingly clever. Additionally, the authors take time to highlight the important results in elementary mathematics that are often used in more advanced areas without qualification.
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The reviewer below seems rather confused. This book has problems which COULD be solved in a mechanical way, but almost all the solutions given are simple and creative. Futhermore, there are often two solutions given. I find the problems hard, but maybe I am just not very smart. How could this book alter your perception of truth? By giving you knowledge?
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It has place in praparing students for participation in mathematical contests. It is good for induvidual work, too.
problems are ordered by topic, not by tehnique to be used, and that encourages student`s creativity. Hints, given along whit the complete solutions, help and guide them to the right solution. Also, they allow students to fine a different method of soluiton.
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Sending feedback... | 677.169 | 1 |
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MAHOBE
YEAR 11 MATHEMATICS
5
About This Book
Q&A eResources are recognised as the leading study guides for NCEA. Each
freely available title has been compiled by a team of experienced educators to
meet the study and revision needs of NCEA students. They are proving to be
valuable resources in the hands of students who want to work ahead of their
regular class programme. They also serve as effective revision programmes in the
run up to the final examinations.
This book carefully explains the mathematical concepts that will be tested in
NCEA then illustrates them with Achievement, Merit and Excellence examplars. It
allows students to practise on NCEA-type questions and provides detailed
solutions. After working through this programme, all students will be well prepared
for their final assessments.
The student who wrote the above answer on a recent assessment paper did not
use a Q&A Level 1 Mathematics eResource.
Always understand what the examiner wants! A past examination answer is
shown below. The student who wrote this answer on a recent assessment paper
did not use a Q&A Level 1 Mathematics eResource.
Expand
=
=
=
=
=
YEAR 11 MATHEMATICS
(a + b)²
(a
(a
(a
(a
(a
+
+
b)²
etc
+
+
b)²
+
b)²
b)²
b)²
MAHOBE
8
Expanding and Factorising
1
An equation or formula may contain brackets, e.g. A = 2 (a + b)
Removing the brackets from such an expression is known as expanding.
Each term inside the brackets must be multiplied by the number or variable
outside.
7(y + 7)
Substituting Values into Formulae
The process of replacing letters into a formula with numbers is known as
substitution. Some examples follow. Write out the formulae with all the values then
use a calculator.
a.
The length of a metal rafter is L (metres). The length of the rafter can change
with temperature variations.
The length can be found by the formula: L = 20 + 0.02t
t = the temperature (ºC)
Find the length of the rod when t = 29° and t = -10°.
Using t = 29°
L = 20 + 0.02 × 29
= 20.58m
Using t = -10°
Describing Linear Patterns
This section looks at how terms of a sequence are related.
a.
A matchstick pattern is shown below:
Pattern 1
i.
Pattern 2
Pattern 3
Draw a diagram of Pattern number 4.
Pattern 4
ii.
iii.
Draw a table that gives the Patterns 1 to 5 and the number of
matchsticks needed for each.
Pattern
1
2
3
4
Matches
5
9
13 17
5
21
Which pattern needs exactly 41 matchsticks?
Look for a relationship between the pattern number and the
number of matches. In this case the difference between each
number is 4. When there is a common difference the
formula for the nth term is:
nth term = dn + (a - d)
where d = common difference, a = the first term of the
sequence. In the case above nth term
Complete the table that gives the number of black tiles compared to
the number of white tiles.
white tiles
black tiles
1
2
3
4
b.
The rule for the number of white tiles (w) in terms of the number of
black tiles (b) is:
w= ........................
........................
........................
2.
Patterns can be made of matchsticks.
1
a.
2
3
4
Complete the table:
Pattern (P)
1
2
3
4
5
6
Matches (M)
b.
The rule for calculating total matches in each pattern is to multiply the
pattern number by 2 and add 1.
How many sticks will there be in pattern 10?
........................................................
Solving Linear Equations
Most equations require a number of steps before they can be solved. Each step
must be logical. Whatever you do to one side of the equation you must do to the
other. Follow the steps below to see how the following equations are solved.
a.
4x + 7 = 19
b.
5x - 8 = 15
x
c.
d.
4x + 6 = 3x + 16
5 -2=3
e.
6 - 2x = 12
f.
3(x + 2) = 5(x - 2)
The Answers
a.
Solving Factorised Equations
If two factors are multiplied together to give 0 then either one of them must be 0,
i.e. xy = 0, either x = 0 or y = 0.
Look at the examples below and see how each are solved.
a.
Simplify or Solve Rational Expressions
When fractions are added or subtracted they must have the same denominator.
= 5x + 6x
30 30
= 11x
30
Simplify:
x + x
5
6
Express:
5
3
x + x + 1 as a single fraction
5x
= 3(x + 1) +
x(x + 1)
x(x + 1)
each has a common denominator
=
3x + 3 + 5x
x(x + 1)
add the numerators
=
8x + 3
x(x + 1)
the final answer
x2 + 8x + 15
x+3
Simplify:
(x + 5)(x + 3)
x+3
= x+5
=
Solve:
4x + 1
=3
11
=>
=>
=>
4x + 1
3
=
11
1
4x + 1 = 33
4x
=>
cross multiply
= 32
x = 8
check with a calculator!
(4×8 + 1) ÷ 11 = 3
Solve:
10x + 3.5 = 16
2
=>
10x + 7
2
2
=> 10x + 7
= 16
1
= 32 multiply both sides by 2
=> 10x
= 25
=>
now solve
x = 2.5
Remember to check your answer with a calculator!
MAHOBE
YEAR 11 MATHEMATICS
33
Exercises
Simplify:
4
1.
x +
2.
3.
4.
5.
2
y
Solve:
6.
3
k=9
4
.........................
.........................
.........................
.........................
5 - 1
2b
3a
7.
m
1
8 +2= 2
.........................
.........................
.........................
.........................
3x
9x + 6
8.
2t
5 +8=4
.........................
.........................
.........................
.........................
x2 - 5x + 6
x2 - 4
9.
7e
5 - e = 10.5
.........................
.........................
.........................
.........................
-4xy × -2xy
6x2y
10.
x
x
5 + 2 = -14
.........................
.........................
.........................
.........................
YEAR 11 MATHEMATICS
MAHOBE
34
Describing Quadratic Patterns
Term (n)
Look at this sequence of numbers:
1
2
3
4
5
2,
6,
12,
20,
30,
4
The difference between each number is:
6
8
10
6
42 ...
12
The difference between these numbers is:
2
2
2
2
If the first difference between each number changes, then it could be a quadratic
sequence. When the second difference is constant, you have a quadratic
sequence - i.e., there is an n2 term.
If the second difference is 2, start with n2.
If the second difference is 4, you start with 2n2.
If the second difference is 6, you start with 3n2.
The formula for the sequence 2, 6, 12, 20, 30, 40 ... starts with n2 as the second
difference is 2. Use n2 as a starting point to calculate the formula.
Term (n) 1
2
3
4
5
6
Sequence: 2,
6,
12,
20,
30,
42
n2
4
9
16
25
36
1
The difference between the sequence and n2 is n, i.e 2-1 =1, 6-4=2.
Therefore the formula for the pattern = n2 + n
a.
Write down the next two terms of the sequence: 5, 12, 23, 38, _ , _
The first differences are: 7, 11, 15, The second difference is 4.
Continuing the sequence, the differences between each term will
be: 15 + 4 = 19 and 19 + 4 = 23
Therefore the next 2 terms in the sequence will be: 38 + 19 = 57
and 57 + 23 = 80. The sequence will be: 5, 12, 23, 38, 57, 80
b.
Find a formula for the nth term of the sequence: 5, 12, 23, 38, _ , _
The second difference is 4. Therefore the formula will start 2n2.
nth term: 1
2
3
4
5
6
Sequence: 5 12 23 38 57 80
2n2
2
8
18 32 50
72
The difference between 2n2 and original number is n + 2
Therefore the formula for the nth term is 2n2 + n + 2
MAHOBE
YEAR 11 MATHEMATICS
35
Exercises
1.
The following structures were made with slabs of wood.
a.
b.
Complete the table to give the number of slabs needed for each
structure.
Storeys (x)
1
2
3
Slabs needed
3
8
15
4
5
6
7
8
Give the rule for the relationship between the number of storeys and
slabs of wood needed.
.......................................................
.......................................................
c.
If you wanted to build a structure with 25 storeys, how many slabs of
wood would be needed?
.......................................................
2.
Look at the tile pattern below then complete the table to give the formula for
the number of white tiles. NOTE: Total tiles = grey tiles + white tiles.
Number of Tiles
Total Number
Number of
Number of
on the bottom line
of Tiles
Grey Tiles
White Tiles
n2
5n - 6
n
YEAR 11 MATHEMATICS
MAHOBE
36
3.
Sequence q = 3, 8, 15, 24, 35, ….
a.
Calculate the sixth term of the sequence.
.......................................................
.......................................................
b.
The nth term of sequence q is n2 + kn, where k represents a number.
Find the value of k.
.......................................................
.......................................................
4.
The first three terms of a sequence are: (3×4)+1, (4×5)+2, (5×6)+3.
Find the next two terms and the rule for the nth term.
............................................................
............................................................
............................................................
............................................................
5.
The first four terms of a sequence are: 4, 9, 16, 25, ....
Find the next two terms and the rule for the nth term.
............................................................
............................................................
............................................................
MAHOBE
YEAR 11 MATHEMATICS
37
Rearranging Formulae
Sometimes a formula needs to be rearranged to be more useful. A common
formula is the one that converts °F to °C, i.e °F = 1.8°C + 32.
To convert °C to °F rearrange the formula to make °C the subject.
F = 1.8C + 32
Solving Equations
An equation is the equivalent of a mathematical sentence. Within this sentence,
two expressions have the same value. If you add, subtract, multiply, or divide one
side of the equation, then you have to do exactly the same operation to the other
side of the equation.
e.g. Solve each of the following equations:
a.
3x + 4 = 25
3x = 21
subtract 4 from both sides
x
b.
Inequations
An inequation has a greater than (>) or less than (<) sign. This means that both
sides of the equation do not equal each other. Calculating values in an inequation
is much the same as with normal equations (i.e. those with = signs). But when
multiplying or dividing both sides of an inequation by a negative number then you
must change the direction of the sign. The simple example below illustrates how
multiplying or dividing by a negative number changes the "sense" of an inequation:
5 > 2
Multiply both sides by -1
=> -5 < -2
Exercises
1.
2y + 3 > 4
6.
.........................
2.
-3x + 4 < 16
.........................
7.
.........................
3.
-y
2 ≥4
8.
3 - 4x > 11
2x-9 > 7
9
....... . . . . . . . . . . . . . . . . . .
YEAR 11 MATHEMATICS
3(x - 2) ≤ 5
.........................
9.
.........................
5.
2
-2x > 3
.........................
.........................
4.
3x + 7 < 2x - 6
-x < 3x + 8
.........................
10.
5(x + 3) - 6x ≥ 12
. . . . . . . . . . . . . . . . . . . . . . ...
MAHOBE
42
Solving Quadratic Problems
Factorising a quadratic equation can make it easier to solve.
a.
The sides of an existing square warehouse are to be extended by 5 metres
and 8 metres. The area of the new extended warehouse will be 340m2.
The existing warehouse (shaded) and planned extension are shown in the
xm
diagram below.
A ball bearing rolls down a slope labeled AB. The time, t seconds, for the ball
bearing to reach B is the solution to the equation t2 + 5t = 36.
How long does it take for the ball bearing to reach B?
t2 + 5t - 36 = 0
A
(t + 9)(t - 4) = 0
T = -9, t = 4
B
It takes 4 seconds for the ball to reach B.
(It could not be a negative time).
MAHOBE
YEAR 11 MATHEMATICS
43
Exercises
1.
The diagram shows a square courtyard and square pool in one corner.
The courtyard extends 10m on two sides of the pool.
The courtyard and pool take up 225m2 .
xm
A field is 40 m longer than it is wide. The area of the field is 3200 m2
What is the length and width of the field?
.......................................................
.......................................................
.......................................................
.......................................................
.......................................................
YEAR 11 MATHEMATICS
MAHOBE
44
3.
A golf ball is hit into the air. Its flight can be calculated by the equation:
h = 40t - 8t2 where h = height from the ground and t = time in the air.
Find the time taken for the ball to reach a height of 48 metres.
Explain why there are two possible values.
............................................
............................................
.......................................................
4.
To find two positive consecutive odd integers whose product is 99 we can
use the following logic:
x is the first integer
x + 2 is the second integer
therefore x(x + 2) = 99
Continue with the logic to find the answer.
.......................................................
.......................................................
.......................................................
5.
The hypotenuse of a certain right angled triangle is 13 cm. The other two
lengths are x and (x + 7) cm.
Complete the logic below to find the lengths of the other two sides.
Using Pythagoras:
Solving Pairs of Simultaneous Equations
Some questions give two equations with two unknowns. These questions will ask
you the values of the unknowns. To solve, you can find the intersection points of
their graphs or you could use one of the following algebraic methods:
a.
Comparison: This method can be used if both equations have the same
subject. e.g. Solve for x and y when y = 90 - x and y = 63 + ½x.
90 - x = 63 + ½x
-x = -27 + ½x
-1½x = -27
subtract ½x from both sides
x = 18
b.
subtract 90 from both sides
divide both sides by -1½
y = 90 - 18
put the x value into one of the equations
y = 72
solve for y
Substitution: This method can be used if one of the equations has a single
variable as the subject. e.g. Solve the simultaneous equations: y = 3x - 9
4x - y = 13
The first equation can be substituted into the second
4x - (3x - 9) = 13
4x - 3x + 9 = 13
x + 9 = 13
x = 4
y = 3(4) - 9
put x = 4 into the other equation
y=3
c.
Elimination: Use this method if the co-efficients of either x or y are the same
in both equations. e.g. 4y - 3x = -4
8y + 3x = 28
12y
add the equations to eliminate x
= 24
y = 2
Put the solution for y (i.e. y = 2) into one of the equations:
4(2) - 3x = -4
Elton has more than twice as many CDs as Robbie. Altogether they have
56 CDs. Write a relevant equation and use it find the least number of CDs
that Elton could have.
.......................................................
.......................................................
.......................................................
3.
Elton purchases some DVDs from the mall. He buys four times as many
music DVDs as movie DVDs. The music DVDs are $2.50 each. The movie
DVDs are $1.50 each. Altogether he spends $92.
Solve the equations to find out how many music DVDs that he purchased.
S = 4V
2.5S + 1.5V = 92
.......................................................
.......................................................
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.......................................................
One of the solutions of 4x² + 8x + 3 = 0 is x = -1.5
Use this solution to find the second solution of the equation 4x² + 8x + 3 = 0
.......................................................
.......................................................
.......................................................
6.
The volume of the box shown is 60 litres.
Find the dimensions of the box.
The triangle drawn below is equilateral. The perimeter is 30 cm. Write down
two equations and solve them simultaneously to find the values of x and y.
...........................
...........................
Zahara is five years old and Maddox is four years older.
Form a relevant equation. Use it to find how many years it will take until
Zahara's and Maddox's ages in years, multiplied together make 725 years.
Let Z = Zahara's age:
Z(Z + 4) = 725
Z² + 4Z - 725 = 0
(Z + 29)(Z - 25) = 0
Z = -29 or Z = 25
Zahara will be 25 and Maddox will be 29. (25 × 29 = 725)
As Zahara is now 5 it will take another 20 years.
2.
Holmsey is using octagonal tiles to make patterns.
Pattern 1
Pattern 2
Pattern 3
Pattern 4
Holmsey has 271 octagonal tiles. He wants to use all the tiles in a pattern as
above. Write an equation to show the relationship between the pattern
number (n) and the number of tiles used (t).
Solve the equation to find the pattern number that would have 271 tiles.
Pattern number
Tiles
(n)
1
2
3
4
(t)
5
11
19
29
The first difference between the terms is: 6, 8, 10.
The second difference is 2. This means the equation will start n².
Look at the relationship between n, n² and t:
n
The design above can be modeled by the following formulae where
n = the number of squares titles on the bottom line.
Total number of square tiles = n2.
Total number of grey squares = 5n - 6.
a.
b.
Write the formula for the number of white squares.
A square courtyard is to be tiled using the design above. Each side of
the courtyard requires 25 tiles.
Give the total number of grey and white tiles required.
At the local garden centre, Mr Rose makes two rectangular garden plots.
Plot 1 is 5 metres longer than it is wide and has an area of 18.75m2.
Plot 2 is 3 metres longer than it is wide and has an area of 22.75m2.
The combined width of both gardens is 6 metres.
Find the length and width of each garden.
Show any equations you need to use.
Show all working.
Set out your work logically.
Use correct mathematical statements.
QUESTION TWO
Sue sends birthday invitations to all her friends.
Stamps cost 50 cents per envelope.
Each invitation cards costs $2.95.
Sue spends a total of $51.75.
The equation for the amount spent is 0.5x + 2.95x = 51.75 where x is the number
of cards sent. Solve the equation to find out how many friends received a birthday
invitation from Sue.
.......................................................
.......................................................
.......................................................
.......................................................
MAHOBE
QUESTION FOUR
Kate was told that two factors of 255 are 15 and 17.
Use this information to solve the quadratic equation x2 + 2x - 255 = 0
.......................................................
.......................................................
.......................................................
QUESTION FIVE
In January, Terrence's mother opened a savings account for his University study.
She made an initial deposit and then added $150 to the account every month.
After 5 months there was $950 in the account. The diagram below illustrates the
transactions.
INITIAL + $150
+ $150
DEPOSIT
+ $150
+ $150
+ $150
Total
$950
No interest is added to the account in the first 5 months. Write an expression for
the total amount of money, T, that Terrence's mother has put into the account
after m months
.......................................................
.......................................................
.......................................................
.......................................................
YEAR 11 MATHEMATICS
QUESTION SEVEN
There are V litres in Claudia's water tank. There are d "drippers" on the irrigation
hose from the tank to the garden. Each dripper uses x litres of water per day.
(a)
Write an expression to show the total amount of water, T, left in the tank
after one day.
.......................................................
(b)
At the end of the day on the 1st of April there were 150 litres of water in the
tank. The next day, 4 drippers were used to irrigate the garden and at the
end of the day there were 60 litres of water left.
Use the expression you gave above to show how much water each dripper used
on that day.
.......................................................
.......................................................
.......................................................
.......................................................
.......................................................
Amount of water, T, used by each dripper = . . . . . . . . . . . . . . . . Litres.
MAHOBE
YEAR 11 MATHEMATICS
63
QUESTION EIGHT
Graeme is designing a path around the front of his garden. His design is shown
below.
4 metres
5 metres
x metres
x metres
Garden
x metres
3x metres
The width of the path is x metres
Graeme has sufficient paving to make a path with a total area of 22 m2.
The area of the path can be written as 4x + 3x2 + (5 - 2x)x = 22.
Rewrite the equation and then solve to find the width of the path around the front
of the garden.
.......................................................
.......................................................
.......................................................
.......................................................
.......................................................
.......................................................
YEAR 11 MATHEMATICS
MAHOBE
64
QUESTION NINE
Students from Mahobe High School are about to be transported to a sports game
in two mini buses - A and B. They are all seated in the mini buses ready to depart.
•
If 3 students in bus A are moved to bus B then each bus will have the same
number of students.
•
If 2 students in bus B are moved to bus A then bus A will have twice the
number of students that are in bus B.
Use the information given to find the total number of students in the mini buses.
You must show all your working and give at least one equation that you used to
get your final answer.
The Mahobe RedBack Calculator
The NZ Centre of Mathematics recommends the Mahobe
RedBack scientific calculator for NCEA. This calculator
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take the sweat out of such calculations. Purchase it
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NCEA Level 1 English Questions & Answers
Each of the English Achievement Standards are introduced with
a wide range of writing topics along with excellent hints on how
to structure and write an outstanding essay. There are many
sample excellence essays from previous years' successful students.
This book shows you how to combine all your teacher and class
notes and practise successfully for each assessment. Q&A English
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NCEA Level 1 Science Questions & Answers
There are explanations and practice NCEA assessments of the
type that can be expected in the exam. This book allows you to
focus on the exam essentials. It will your boost confidence and
maintain motivation in the weeks leading up to the exams.
Study with this book and you will enter the exam room with
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MAHOBE
Algebra Methods
Description
This booklet and more can be downloaded for free from the New Zealand Centre of Mathematics There you can also watch over 80 videos and download over 100 similar maths revisi...
This booklet and more can be downloaded for free from the New Zealand Centre of Mathematics There you can also watch over 80 videos and download over 100 similar maths revision booklets for FREE.
This booklet is designed for the New Zealand Certificate of Educational Achievement (NCEA) Level 1 . It covers the Achievement Standard 'Algebra' and covers expanding and factorising, exponents, substituting values, linear patterns, factorising, solving equations, quadratics, rearranging formulae and simultaneous equations. There are specific achievement, merit and excellence examples and exercises along with a practise examination based on a previous external paper. | 677.169 | 1 |
Current Students
Course Search – Course Description
FAM-547 Study Algebra Online
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Study Algebra Online Discover algebra's relevance to everyday life. Become skilled at solving a variety of problems and get ahead by learning the basics of algebra before classes begin. $100 includes $72 fee. Note: Online course. Offered in partnership with ed2go. For children in grade 6 or older monitored by an adult. | 677.169 | 1 |
Details about Practical Problems in Mathematics for Drafting & CAD:
This fully updated edition of Practical Problems in Mathematics for Drafting and CAD features contemporary drafting problems, CAD drawings, and industry applications and practices-essential information designed to enhance the math skills of students concentrating their studies in the field of drafting. A comprehensive compilation of integrated math problems and CAD operations facilitates critical thinking, problem solving, and basic mathematics literacy. Real-world, everyday applications include use of a scientific calculator to solve math problems in drafting and CAD. Examples and figures cover a range of skills and levels of difficulty, and math principles provide a challenge for a variety of ability levels.
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Rent Practical Problems in Mathematics for Drafting & CAD 3rd edition today, or search our site for other textbooks by John Larkin. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Delmar Cengage Learning. | 677.169 | 1 |
Bezier and B-spline Techniques with Matlab. Edition No. 1
The aim of this book is to teach students the essential of Bezier and B-spline techniques with the aid of examples. Computer codes, which give an easy interface of Bezier and B-spline techniques to the users, are implemented as Matlab programs. The reason to choose Matlab is that it is easy to use and has a good graphical user interface. This book focuses on curves and surfaces using Bezier and B-spline techniques. It is based on the theory "Bezier and B-spline Techniques" which are known in mathematics. Interpolation and approximation methods have been illustrated intensively. Some of algorithms are represented using practical cases for example Casteljau algorithm. Students and researchers can use this book to succeed good understanding of Bezier and B-spline techniques for reliable and efficient studies in accordance within scientific applications | 677.169 | 1 |
In recent years, with the introduction of new media products, there has been a shift in the use of programming languages from FORTRAN or C to MATLAB for implementing numerical methods. This book makes use of the powerful MATLAB software to avoid complex derivations, and to teach the fundamental concepts using the software to solve practical problems. Over the years, many textbooks have been written on the subject of numerical methods. Based on their course experience, the authors use a more practical approach and link every method to real engineering and/or science problems. The main benefit is that engineers don't have to know the mathematical theory in order to apply the numerical methods for solving their real-life problems.
An Instructor's Manual presenting detailed solutions to all the problems in the book is available online Back Cover
Learn how to use MATLAB® to solve complex numerical problems
Increasingly, scientists and engineers favor MATLAB over conventional programming languages such as FORTRAN and C when they wish to solve complex problems. This book will enable readers to solve problems without needing to understand all the details of the underlying theory of numerical methods. By providing many examples of the uses of similar functions, it guides them towards the selection of the appropriate MATLAB functions for solving their problem efficiently.
The authors have incorporated existing MATLAB functions into a series of simplified, yet complete programs that may be readily adapted by students and practitioners to solve real-world engineering and science problems. Key features include:
More than 100 supplemental codes
Complete MATLAB programs to demonstrate solutions to real-life
exercises and problems
Downloadable MATLAB programs to correspond with the text
Interactive demonstration programs that course instructors can use to produce visual presentations of the solution processes of some algorithms
An overview of the Partial Differential Equation (PDE) toolbox
An appendix with MATLAB commands/functions for symbolic computation
With very little prior programming experience, students and practitioners will find this approach invaluable to quickly learn how to solve their numerical problems.
Top Customer Reviews
The authors demonstrate how Matlab can be used for advanced numerical calculations. Many Matlab books just use numerical examples that are really elementary, from a scientific or engineering viewpoint. Those texts are more about teaching Matlab. Hence they choose simple examples, to reach a broad audience. Whereas the audience for this book is rather select. Probably tending towards professionals in the physical sciences or engineering. Or possibly grad students in these fields.
The text starts with matrix [linear] algebra. Explaining how to invert a linear system of equations. Then the book goes onto finding roots of non-linear equations, using Newton-Raphson iteration. And the always useful methods of numerical differentiation and integration. Plus other topics. Each chapter has an extensive problem set.
A central theme is that you don't need much previous knowledge of Matlab. The book guides you into its abilities to help your research.
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The book is very well organized , with good descriptions and code related to the numerical methods presented. It is an excellent reference book for everyone using numerical methods and Matlab, with a slight emphasis on engineering.
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CK-12 Foundation's Single Variable Calculus FlexBook covers the following chapters: Functions, Limits, and Continuity - A review of the basics of functions is given. Students use linear approximations to study the limit process, before a more formal treatment of limits is given. Differentiation - Students explore instantaneous rate of change, and the relationship between continuity and differentiability. The Chain Rule and implicit differentiation are reviewed. Applications of Derivatives - Students gain practice with using the derivatives in related rates problems. Additional topics include The First Derivative Test, The Second Derivative Test, limits at infinity, optimization, and approximation errors. Integration - This chapter includes indefinite integrals calculus, initial value problems, definite integrals, the Fundamental Theorem of Calculus, integration by substitution, and numerical integration. Applications of Integration - This chapter includes applications of the definite integral, such as calculating areas between two curves, volumes, length of curves, and other real-world applications in physics and statistics. Transcendental Functions - This chapter includes differentiation and integration of logarithmic and exponential functions, exponential growth and decay, derivatives and integrals involving inverse trigonometric functions, and L'Hospital's Rule. Integration Techniques - Students explore integration by substitution, integration by parts, integration by partial fractions, trigonometric integrals, trigonometric substitutions, and improper integrals. Infinite Series - This chapter introduces the study of sequences and infinite series. The properties presented describe the behavior of a sequence or series, including whether a sequence approaches a number or an infinite series adds to a number.
In this lesson we will review what you have learned in previous classes about mathematical equations of relationships and corresponding graphical representations and how these enable us to address a range of mathematical applications. We will review key properties of mathematical relationships that will allow us to solve a variety of problems. We will examine examples of how equations and graphs can be used to model real-life situations.
On October 6, 1948, a U.S. Air Force B-29 Superfortress crashed soon after takeoff, killing three civilian engineers and six crew members. In June 1949, the engineers' widows filed suit against the government, determined to find out what exactly had happened to their husbands and why the three civilians had been on board the airplane in the first place. But it was the dawn of the Cold War and the Air Force refused to hand over any documents, claiming they contained classified information. The legal battle ultimately reached the Supreme Court, which in 1953 handed down a landmark decision that would, in later years, enable the government to conceal gross negligence and misconduct, block troublesome litigation, and detain criminal suspects without due-process protections.
Claim of Privilege is a mesmerizing true account of a shameful incident and its lasting impact on our nation--the gripping story of a courageous fight to right a past wrong and a powerful indictment of governmental abuse in the name of national security.
From New York Times and USA TODAY bestselling author Rachel Lee comes a sensual new series about mesmerizing vampire detectives and the love they discover, deep in the night...Jude Messenger felt the hunger deep inside him. From the moment he'd rescued Theresa Black from a late-night attack, he'd ached for things he knew he could never have. The touch of her skin. The taste of her blood. But if the vampire claimed her, one of them might not survive.So Jude did everything he could to make Terri stay away.But a demon stalked the sexy spitfire, and Jude had no choice but to protect a woman who created a longing inside him that could never be sated | 677.169 | 1 |
Understanding Elementary Algebra with Geometry A Course for College Students (with CD-ROM and iLrn™ Tutorial), 6th Edition
Hirsch and Goodman offer a mathematically sound, rigorous text to those instructors who believe students should be challenged. The text prepares students for future study in higher-level courses by gradually building students' confidence without sacrificing rigor. To help students move beyond the "how" of algebra (computational proficiency) to the "why" (conceptual understanding), the authors introduce topics at an elementary level and return to them at increasing levels of complexity. Their gradual introduction of concepts, rules, and definitions through a wealth of illustrative examples -- both numerical and algebraic--helps students compare and contrast related ideas and understand the sometimes-subtle distinctions among a variety of situations. This author team carefully prepares students to succeed in higher-level mathematics.
Additional versions of this product's ISBNs
ISBN10: 0534999727
ISBN13: 9780534999728
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Beginning graduate students in mathematics and other quantitative subjects are expected to have a daunting breadth of mathematical knowledge. But few have such a background. This book will help students to see the broad outline of mathematics and to fill in the gaps in their knowledge. The author explains the basic points and a few key results of all the most important undergraduate topics in mathematics, emphasizing the intuitions behind the subject. The topics include linear algebra, vector calculus, differential geometry, real analysis, point-set topology, probability, complex analysis, abstract algebra, and more. An annotated bibliography then offers a guide to further reading and to more rigorous foundations. This book will be an essential resource for advanced undergraduate and beginning graduate students in mathematics, the physical sciences, engineering, computer science, statistics, and economics who need to quickly learn some serious mathematics. less | 677.169 | 1 |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). | 677.169 | 1 |
Computational Science (COSC) is a scientific investigation of physical processes on the computer by the use of computational models and methods. Hypotheses are often formulated using mathematical models that can then be used to compute quantities of scientific interest and to gain new insights. COSC is now on par with the development of scientific theory and the use of experimentation in order to understand the real world. Mathematica is an ideal tool to use in this endeavor and also to teach the essential types of operations. These types of operations may be placed into three categories: Evaluation, Simulation, and Optimization. Within each category, fundamental operations will be described. One or more examples will be presented for each operation. This material is currently used to teach an undergraduate course in COSC.
Evaluation: This is the most basic category since the operations in this category are used in both simulation and optimization.
Analytic (symbolic) versus numeric computation
Data input/output and visualization
Physical units and conversions
Linear versus nonlinear systems
Data and function approximation
Computer arithmetic
Uncertainty and sensitivity in computation
Simulation: This is the most general category where models must often be implemented by writing application-specific programs.
Discrete equation models
Differential and integral equation models
Monte Carlo models
Optimization: This important category contains models that are designed to produce the best solution and typically require many controlled evaluations to determine a minimum or maximum. | 677.169 | 1 |
Math Practice Software for Standardized Tests
The Wayang Math Tutor is an intelligent electronic tutoring system designed to learn along with the student. Using interactive multimedia, the Wayang Math Tutor helps prepare middle and high school students for standardized math tests, such as the SAT, MCAS and CA-Star. Wayang provides self-paced review for students preparing for college level mathematics and adults who want to refresh their math skills.
As you progresses through the math problems presented, the tutor adapts to your individual learning pace and style. The tutor is available free to students, teachers, schools, after-school programs, and for use from home. You may log in as a guest to try at the program.*
*Guests are offered all of the same lessons and exercise as registered users, but your information will not be saved for future logins.
Using The Wayang in the Classroom
Currently, most high schools have special classes for test preparation (such as. SAT-prep or MCAS-prep classes) or remedial math classes for those students who have not passed test required for graduation. Wayang Math Tutor is an excellent tutoring tool for these kinds of classes. It's proven to increases students' scores.
Our research studies show that students improved 10% of their score after 2 hours of instruction and 20% after 3 hours using our tutoring system, on problems similar to those tested in SAT, MCAS and CA-Star. For many students, this is the difference between passing and not passing the test.
Learning Should be Fun
Help is always available, and you are never penalized for asking for help.
There is more than one hint for each problem, so you may ask for help whenever you need to.
Learn more about the Wayang Math Tutor's Game-like Interface
Parents want their children to do well
Careers that require strong math and science skills are among the fastest growing and highest paying in the nation. Along the way to their goals, standardized tests are a challenge students face. Doing well on tests such as SATs and the Massachusetts MCAS help ensure greater educational success and opportunity.
Publications
Recent Publications
Mathematics Publications
Gender and Math Learning
Emotions and Learning
Other Publications
Awards
2009 Research Award
2010 Research Award
PRESS
Funding
The research for Wayang Outpost was funded by several NSF grants: "Preparing for College: Using Technology to Support Achievement for Students with Learning Disabilities in
Mathematics", Woolf and Burleson (PIs) with Arroyo (#0931237, HRD RDE); "What kind of Math Software works for Girls?" Arroyo (PI) with Royer and Woolf (#0734060, HRD GSE/RES); "Affective Learning Companions: Modeling and supporting emotion during teaching", Woolf and Burleson (PIs) with Arroyo, Barto, and Fisher (#0705554, IIS/HCC); and a grant from the US Department of Education, Institute of Education (IES) Using Intelligent Tutoring and Universal Design To Customize The Mathematics Curriculum, to Woolf (PI) with Arroyo and Maloy (co-PIs).
Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the funding agencies.
Wayang Outpost (K-12) and Wayang Mathematics Tutor (College and Adult Education) are under development by the Center for Knowledge Communication at the University of Massachusetts Amherst. Links to the current, most stable versions are at the top of this page. If you are a researcher, or have been instructed to use a development (beta) version of Wayang, please use these links: | 677.169 | 1 |
Course 3
Course 3 provides a solid foundation in order to fully prepare students for Algebra 1.
Chapters 1-3 focus on integers, rational numbers and real numbers in order to set the stage for equations, inequalities and functions. Real-World applications to the more abstract algebraic concepts are found throughout the text. More than 5 Activity Labs per chapter ensure students receive the visual and special instruction necessary to conceptualize these abstract concepts and prepare them for advanced math courses.
Product Details
The Manipulatives Kit allows students to explore concepts in a hands-on way using a variety of measurement, geometry, algebra, and probability tools. The kit is designed for a class of 30 students.
Product Details
Product Name: Student Manipulatives Kit
System Title: PH MATH STUDENT MANIP KIT MAN N/A 2004C
Type: Ancillary, Student Resources
Medium: Other
ISBN-10: 0131156101
ISBN-13: 9780131156104
Designed for use with overhead projectors, this kit helps you demonstrate concepts in a using probability tools, including algebra tiles, tangrams, a geoboard with rubber bands, a spinner, and pattern blocks.
Product Details
Product Name: Overhead Manipulatives Kit
System Title: PH MATH OVERHEAD MANIP KIT MAN N/A 2004
Type: Student Resources, Ancillary
Medium: Other
ISBN-10: 0131158465
ISBN-13: 9780131158467
Prentice Hall Mathematics 2008 - Course 1 - Algebra Readiness Tests
Product Details
Product Name: Algebra Readiness Tests
System Title: PH MATH ALGEBRA READINESS TSTS BLM 2007C
Type: Teacher Resources
Medium: Print
ISBN-10: 0132013908
ISBN-13: 9780132013901
Online at PHSchool.com Personlized intervention for each student—
Assess—Students take Diagnostic Tests or Benchmark Tests online.
Diagnose—Based on assessment results, each student automatically recieves an assignment for any skills that have not been mastered | 677.169 | 1 |
Math 310
Our Math 310 course (Arithmetic) is the lowest level math course that is offerend here at Solano Community College. Some of the topics that are seen in Math 310 are revisted (at a quicker pace) in Math 320, our Pre-Algebra course. If you feel that you only need a small refresher of the following topics, you should consider beginning your math sequence with Math 320 (Pre-Algebra) instead of Math 310 (Arithmetic). Please note, though, that these problems are meant to be done without the use of a calculator.
1. Math 310 begins with topics relating to Whole Numbers. The following are examples of problems relating to Whole Numbers:
3.After covering Whole Numbers and Fractions, students will learn about Decimals and Percents. The following are examples of problems relating to Decimals and Percents:
2.Math 310 then heads into fractions, which is a topic that will be covered again, though quickly, in Math 320 (Pre-Algebra). The following are examples of problems relating to Fractions:
4.Other topics covered in Math 310 include Ratios and Proportion, Measurement, and Basic Geometry. A few examples of these types of problems follow. | 677.169 | 1 |
Product details
ISBN-13: 9780073312699
ISBN: 007331269X
Edition: 2
Publication Date: 2007
Publisher: McGraw-Hill College
AUTHOR
Miller, Julie, Hyde, Nancy
SUMMARY
Built by teachers, just like you, Miller/O'Neill continues to offer an enlightened approach grounded in the fundamentals of classroom experience in Beginning and Intermediate 2e. The practice of many instructors in the classroom is to present examples and have their students solve similar problems. This is realized through the Skill Practice Exercises that directly follow the examples in the textbook. Throughout the text, the authors have integrated many Study Tips and Avoiding Mistakes hints, which are reflective of the comments and instruction presented to students in the classroom. In this way, the text communicates to students, the very points their instructors are likely to make during lecture, helping to reinforce the concepts and provide instruction that leads students to mastery and success. The authors included in this edition, Problem-Recognition exercises, that many instructors will likely identify to be similar to worksheets they have personally developed for distribution to students. The intent of the Problem-Recognition exercises, is to help students overcome what is sometimes a natural inclination toward applying problem-sovling algorithms that may not always be appropriate. In addition, the exercise sets have been revised to include even more core exercises than were present in the first edition. This permits instructors to choose from a wealth of problems, allowing ample opportunity for students to practice what they learn in lecture to hone their skills and develop the knowledge they need to make a successful transition into College Algebra. In this way, the book perfectly complements any learning platform, whether traditional lecture or distance-learning; itsinstruction is so reflective of what comes from lecture, that students will feel as comfortable outside of class, as they do inside class with their instructor. For even more support, students have access to a wealth of supplements, including McGraw-Hill's online homework management system, MathZone.Miller, Julie is the author of 'Beginning and Intermediate Algebra ', published 2007 under ISBN 9780073312699 and ISBN 007331269 | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
This text introduces the vast and fascinating area of computational number theory. It treats algorithms for common number-theoretic problems in an elementary fashion, eliminating the need for an extensive prerequisite of algebra and analysis. The GP/PARI calculator is used throughout to demonstrate the working of arithmetic algorithms. The book contains detailed examples illustrating almost every algorithmic concept discussed. It also includes practical applications of arithmetic algorithms in public-key cryptography. Every chapter ends with many exercises and partial solutions are given in the appendix. | 677.169 | 1 |
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Partial differential equations (PDEs) are essential for modeling many physical phenomena. This undergraduate textbook introduces students to the topic with a unique approach that emphasizes the modern finite element method alongside the classical method of Fourier analysis.
Additional features of this new edition include broader coverage of PDE methods and applications, with new chapters on the method of characteristics, Sturm-Liouville problems, and Green s functions, and a new section on the finite difference method for the wave equation. The author continues to emphasize Fourier series and finite element methods, which were the primary scope of the first edition.
The book also features emphasis on linear algebra, particularly the idea of best approximation; realistic physical parameters and meaningful experiments for many of the examples and exercises; and tutorials for the most popular software (MATLAB, Mathematica, and Maple) that can be used to reproduce the examples and solve the exercises.
Audience: This book is written for undergraduate courses usually titled Introduction to Partial Differential Equations or Fourier Series and Boundary Value Problems.
Editorial Reviews
Review
I love this book and look forward to using it as a text in the future ... It's the first truly modern approach that I've seen in a PDE text. --Maeve McCarthy, MAA Online
Book Description
An undergraduate course on partial differential equations is found in almost every mathematics department, and this is an important offering to any student on such a course because of its fresh approach incorporating both the modern and the traditional methods of analysing and solving partial differential equations.
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I've used this book in several PDE courses aimed at engineers. It's clear, has good problems, and does an excellent job of showing the connections between partial differential equations and linear algebra. Both analytical and numerical methods are developed. I prefer this book to the competitors at this level.
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INTRODUCTION TO ALGEBRAIC THEORIES
By
A.
ADRIAN ALBERT
THE UNIVERSITY OF CHICAGO
THE UNIVERSITY OF CHICAGO PRESS CHICAGO ILLINOIS
1941 BY THE UNIVERSITY OF CHICAGO. ALL RIGHTS RESERVED. PUBLISHED JANUARY 1941. SECOND IMPRESSION APRIL 1942. COMPOSED AND PRINTED BY THE UNIVERSITY OF CHICAGO PRESS, CHICAGO, ILLINOIS, U.S.A.
COPYRIGHT
PREFACE
During recent years there has been an ever increasing interest in modern algebra not only of students in mathematics but also of those in physics, chemistry, psychology, economics, and statistics. My Modern Higher Algebra was intended, of course, to serve primarily the first of these groups, and its rather widespread use has assured me of the propriety of both its contents and its abstract mode of presentation. This assurance has been conits successful use as a text, the sole prerequisite being the subject matter of L. E. Dickson's First Course in the Theory of Equations. However, I am fully aware of the serious gap in mode of thought between the intuitive treatment of algebraic theory of the First Course and the rigorous abstract treatment of the Modern Higher Algebra, as well as the pedagogical difficulty which is a consequence.
firmed by
publication recently of more abstract presentations of the theory of equations gives evidence of attempts to diminish this gap. Another such at-
The
tempt has resulted in a supposedly less abstract treatise on modern algebra which is about to appear as these pages are being written. However, I have the feeling that neither of these compromises is desirable and that it would be far better to make the transition from the intuitive to the abstract by the
addition of a
new
mathematics
course in algebra to the undergraduate curriculum in a curriculum which contains at most two courses in algebra
and these only partly algebraic in content. This book is a text for such a course. In
terial is
fact, its only prerequisite maa knowledge of that part of the theory of equations given as a chapter of the ordinary text in college algebra as well as a reasonably complete knowledge of the theory of determinants. Thus, it would actually be pos-
a student with adequate mathematical maturity, whose only training in algebra is a course in college algebra, to grasp the contents. I have used the text in manuscript form in a class composed of third- and fourth-year
sible for
undergraduate and beginning graduate students, and they all seemed to find the material easy to understand. I trust that it will find such use elsewhere and that it will serve also to satisfy the great interest in the theory of matrices which has been shown me repeatedly by students of the social sciences. I wish to express my deep appreciation of the fine critical assistance of Dr. Sam Perils during the course of publication of this book.
UNIVERSITY OF CHICAGO
September
9,
then we shall regard f(x) and g(x) as being the
same polynomial. and multiplication as the integral operations. we shall assume then either the properties that if a and 6 are constants such that ab =
By
this
1
.
term conterm we shall mean any complex number or function independent of x. we shall merely make the unprecise assumption that our constants have the usual properties postulated in elementary algebra. We shall thus begin our exposition with a discussion of these concepts.) and g(x) are equal polynomials and write the polynomial which is the constant f(x) = g(x). For the present. We shall speak of the familiar operations of addition.CHAPTER
I
POLYNOMIALS
1. however. Our definition of a polynomial includes the use of the familiar
stant. we shall usually say merely that /(a.
+ g(x)
=
g(x)
for every polynomial g(x). A positive integral power is then
best regarded as the result of a finite repetition of the operation of multiplication. Later on in our algebraic study we shall be much more explicit about the meaning of this term. the polynomial f(x) is not the zero polynomial. We shall designate by zero and shall call this polynomial the ze^o polynomial. in the consideration of a conditional equation f(x) =
No
we seek a constant solution c such that /(c) = 0. However. confusion will arise from this usage for it will always be clear from the where context that. subtraction. This concept is frequently indicated by saying that f(x) and g(x) are identically equal and by writing f(x) = g(x). In particular. We observe that the zero polynomial has the
properties
g(x)
=
. Thus. in a discussion of polynomials f(x)
= will mean that f(x) is the zero polynomial.
A polynomial f(x)
plication of a finite
g(x)
in x is
any expression obtained number of integral operations
as the result of the apto x and constants. Polynomials in x. There are certain simple algebraic concepts with which the reader is probably well acquainted but not perhaps in the terminology and form desirable for the study of algebraic theories. If
is a second such expression and it is possible to carry out the operations indicated in the given formal expressions for/(z) and g(x) so as to obtain two identical expressions.
is
The
is most important since.
but
ex-
the zero polynomial.
. we may express f(x) as a sum of a finite number of terms of the form ax k Here k
constant called the
coefficient of
x
k
. then.
.
expressions (1) are identical.2
INTRODUCTION TO ALGEBRAIC THEORIES
if
a or 6 is zero..
the indicated integral operations in any given expression of a polynomial f(x) be carried out. We may thus speak of any such n as a virtual de-
. we may say that the expression (1) of a i = 0. If a 7* we call n the degree* of f(x).
. we have f(x) = a. q(x). a = 6* for then/(x) and g(x) are equal if and only if m n. in particular. two polynomials are equal if and only if their
with 60
. so that f(x) is the zero polynomial. Thus. if /&(x). any different formal expression of a polyg(x) = nomial in x and that f(x) g(x) in the sense defined above. But if a n = 0.. either any = a n is a constant and will be f(x) has a positive integral degree.). we shall assign to it the degree minus infinity. h(x) = re For example.+.
c8
-
c)
+
.
The constants a
unless f(x)
is
are called the coefficients of /(x)
and may be
ao
zero.
= x2 g(x) 2 2c + 3c= (cIf
2z 2 + 3z. we ob-
1.
.+&
1
= n. and 1 1 a"" such that aar
If f(x) is
a
is
a nonzero constant then a has a constant inverse
assign to a particular formal expression of a polyit occurs in f(x) by a constant c.
any polynomial of degree no may be written as an expression of the form (1) any integer n ^ no. and we may then write
(1)
coefficient is the
sum
of all their
f(x)
=
a xn
+ aixn ~ +
l
.
2c. Thus. Then it is
nomial and we replace x wherever tain a corresponding expression in
Suppose now
evident that /(c) = g(c). In other words. a n we say that the constant polynomial f(x) has degree zero.
. that is. 3 = 2z. r(x)
l)(c
2
1. If.
a non-negative integer and a is a The terms with the same exponent fc
is
may be combined into a single term whose coefficients.
+
a n -\x
+ an
.
the label
we
c which is the constant we designate by is that /(c). if g(x) pression (1) of f(x) with do 5^ polynomial and we write g(x) in the corresponding form
a second
(2)
g(x)
5^ 0.
The integer n of any expression (1) of f(x) is called the virtual degree of the expression (1). This will
*
^
Clearly
of virtual degree gree of /(a. or /(#)
called a constant polynomial in x. T(X) are poly= h(c)q(c) nomials in x such that }(x) = h(x)q(x) r(x) then f(c) r(c)
+
+
for
any
c.
=
6
zm
+ fciz".
we may always take
^
0. is polynomial unique. and are stating that for any c we have x.
What can one say about the degree of/ of/ of/* for/ k a positive integer?
. Use Ex.
Then
=
h(x). The
of
f (x)
+ g(x)
is at
most the larger of the two degrees
and
g(x). that both /and g are zero.
The
author's
.
LEMMA
g(x)
Let f(x) be nonzero and such that f(x)g(x)
degree of f (x)
=
f(x)h(x). s(x) as in
a corresponding statement about g(x)s(x) where g(x) has odd degree Ex.
/
= /(x)
a polynomial in x with real coefficients.
1. then.
and only
3.
if
State
why
it is
true that
x
is
not a factor of f(x) or g(x) then x
is
not a fac-
tor of f(x)g(x).
What
can one say about the degree of f(x)
+ g(x)
.
State the relation between the term of least degree in f(x)g(x) and those of
least degree in /(x)
7.
if
/(x)
and
g(x)
have posi-
tive leading coefficients?
2 3 3.
LEMMA 4.
2. 7 to prove that if k and only if x is a factor of /(x).
2.
EXERCISES*
1. Hint: Verify first that other-
* The early exercises in our sets should normally be taken up orally. whose almost trivial verification
we
leave to the reader.
LEMMA
sum
the
The degree of a product of two polynomials f (x) and g(x) is the and g(x).POLYNOMIALS
3
be done so as to imply that certain simple theorems on polynomials shall hold without exception.
9. The coefficient ao in (1) will be called the virtual leading coefficient of this expression of f(x) and will be called the leading coefficient oif(x) if and only if it is not zero. 4. and thus. We shall call/(x) a monic polynomial if ao = 1.
Let / and g be polynomials in x such that the following equations are satisfied Show.+/?
for
t
>
1.
Make
and
real coefficients.
State the condition that the degree of /(x)
+ g(x)
be
less
than the degree of
either /(x) or g(x). The leading coefficient of f(x) g(x) is the leading coefficients of f (x) and g(x). if f (x) and product of
of the degrees of f(x)
g(x). so is f(x)
LEMMA
stant if
A
product of two nonzero polynomials is nonzero and is a conif both factors are constants. We then have the elementary results referred to above.
6.
8. choice of oral exercises will be indicated by the language employed.
State a result about the degree and leading coefficient of any polynomial
s(x)
5.
4.
= /i
+.
g(x) are monic.
is
a positive integer then x
is
a factor of
[/(x)]*
if
(identically).
=
f(x) a polynomial.
and
g(x).
ora
j* 0. Hint: Show that / and g are nonzero polynomial solutions of these equations of least possible de= xfi as well as g = xgi. b^ CkX g(x) is
Thus a virtual degree of f(x)
.
coefficient
degree
m
1
a^ ^
1
0.a# =
2
*)0
4
=
2
if
10. r(x)
=
The Remainder Theorem
Algorithm to write
/Or)
of Algebra states that
we
use the Division
=
q(x)(x
-
c)
+r(x)
.
. r(x)
=
f(x).
then a virtual degree of s(x)
if
=
r Q (x)
m
1. This
q(x)
r Q (x).
"*
Use Ex. 11 for polynomials/.
But
Lemma
t(x)
states that
t(x)
t(x)g(x) is the
sum
of
m
and
q Q (x) 7* the degree of t(x). (6).
1. then x divides/
^
a contradiction. But then/i and g\ are also solutions grees.
Then. coefficients and not all zero. and (5) of Ex.
If Ck is the virtual leading coefficient of
m+ m. q(x)
degree
m+k>
a polynomial h(x) of virtual k l k 1. there exist unique polynomials q(x) and r(x) such that r(x)
has virtual degree
(3)
m
1.
h with complex
2. 4
a)
b)
c)
/ 2 / 2 /
2
= xg* + h* + g + (x + 2)fc
2
-
xg*
=
xh*
2
=
g. and a finite repetition b^ a Q x g(x) is n n~ l of this process yields a polynomial r(x) = f(x) + . then they are all zero: satisfying
11.
n
>
m. and
=
0.
q(x) is either zero or has degree
f (x)
n
m.zy =
c)
d)
/ 2 /
4
+ 2xfY + (* + 2zfa . g(z)
=
?o(z).
Then
Let f (x) and g(x) be polynomials of respective degrees n and m.)g(x) of b^ (a x m and leading for of virtual degree m hence n and (3) degree 1. If also f(x)
=
q Q (x)g(x)
+r
Q (x)
for r (z) of virtual
r(x) is
1.)
and
g(x) be defined respectively
(3)
by
(1)
and
(2)
with 6
^
0. a virtual degree of h(x) n'^n l 1.
12. either
n
< m and we have
with q(x}
=
0. The result of the application of the process ordinarily called long division to polynomials is a theorem which we shall call the Division Algorithm for polynomials and shall state as
Theorem
g(x)
7* 0. and h are polynomials in x with real coefficients the following equations (identically). g. Use Ex. In parts (c) and (d) complete the squares.
.
a)
6)
and
/ /
2
+ xg* = .
^
Find solutions of the equations of Ex. 9. and
=
q(x)g(x)
+ r(x)
.
For
let /(a.
if
=
the degree of s(x)
is
=
impossible. 8 to give another proof of (a). 3. Express each equation in the form a(x) apply Ex.4
INTRODUCTION TO ALGEBRAIC THEORIES
=
b(x)
wise bothf and g are not zero. The division algorithm.
to show that if/.
2.
Show by formal
m
differentiation that
if
c is
a root of multiplicity
m of f(x) =
1 of the derivative /'(x) of /(x).
Polynomial
divisibility.
numbers
3.POLYNOMIALS
so that g(x)
5
then r
the
= x = /(c).
Thus
f(x)
andg(x)are
associated if and only if each is a nonzero constant multiple of the other.
q(x)
and
h(x) are nonzero constants.
. It
for this application that
we made
the remark. and we shall say in this case that f(x) has g(x) as a factor. Hint: Write h(x) = q(x)f(x)
+
+ + b n-ic n~ for rational + r(x) and replace x by
l
c. obvious proof of this result is the use of the remark in = r c) paragraph of Section 1 to obtain /(c) = q(c)(c r. It is clear that every nonzero polynomial is associated with a monic polynomial. Observe thus that the familiar process of dividing out the leading
coefficient in
a conditional equation f(x)
=
is
equation by the equation g(x)
associated with/(x). We shall call two nonzero polynomials/(x) and g(x) associated polynomials
if
= q(x)g(x)> g(x) = f(x) divides g(x) and g(x) divides /(#).
m
Let
c
cients. Then f(x) = q(x)h(x)f(x). Thus.
Use the Division Algorithm
be a root of a polynomial /(x) of degree n and ordinary integral coeffito show that any polynomial h(c) with rational
coefficients
may
. We shall leave the statements of that theorem. g(x) T only if the polynomial r(x) of (3) is the zero polynomial. * If f(x) is a polynomial in x and c is a constant such that /(c) = then we shall
call c
a root not only of the equation /(x)
=
but also of the polynomial /(x).
. Applying Lemmas 3 and 2. and the subsequent
definitions
and theorems on the roots and corresponding factorizations of polynomials* with real or complex coefficients.
The Division Algorithm and Remainder Theorem imply the Factor Theorem a result obtained and used frequently in the study of polynomial equations. to the reader.
6
.
=0
where g(x)
that used to replace this is the monic polynomial
.
Let f(x) and g(x)
f (x)
^
by the statement that g(x) divides
we mean that
divides /(x) if and nomial q(x) such that/(x) = q(x)g(x). g(x) is a factor o/f(x). (x c) q(x) then c is a root of multiplicity What then is a necessary and sufficient condition that /(x) have multiple roots?
2.
EXERCISES
1. we have h(x)f(x).
Compute
the corresponding &< for each of
the polynomials
6 a) c
6)
c
4
+ 10c + 25c + 4c + 6c + 4c +
4 2
3 2
c)
c
6
2
2c 4
+c
3
2
1
d)
(2c
+ 3)(c + 3c)
be polynomials. Then there exists a poly-
3.
be expressed in the form 6 &ic 6 n-i. /(c)
c
+
as desired.
. so that/(x)
q(x)h(x)
=
1. The
is
fifth
has degree one and r = r(x) is necessarily a constant.
Let /(x)
=
x3
+
3x 2
+
4 in Ex.
of
polynomials
shall obtain
a property
which may best be described in terms of the concept of rational function.
.
. "n j for every
=
and/ =
n(f)
m>
n and
is
any nonzero polynomial of degree n.
+
.
finite
As in Section 1 we may express f(xi.
. x q to be any expression obtained x q and as the result of a finite number of integral operations on #1.
EXERCISES
be a polynomial in x and define m(f) = x mf(l/x) for every positive integer m.
. .
.
.
.
that m(f)
xm
/ 0. It will thus be desirable to arrange our exposition so as to precede the study
of greatest
common divisors by a discussion of the elements of the theory of polynomials and rational functions of several variables. Some of our results on polynomials extended easily to polynomials in several variables. and we shall do so.. if / 7* 0.
1.
x q ) as the
sum
of a
(4)
x$
.
in x
may be
.c.d..c. that.
If
two terms of / have the same
set of
we may combine them by adding their coefficients and thus write /as the unique sum.
. ater when we discuss the existence of a unique greatest common divisor (abbreviated.0. Show that m(f) is a polynomial in x of virtual degree m if and only if m is a virtual degree of /(x). the sum with unique coefficients. g.
.
.
Show
that m(0)
=
0.
of the term (4) and define the virtual degree in k q the virtual degree of a parXi. jX q of such a term to be k\ ticular expression of / as a sum of terms of the form (4) to be the largest of
call
.
. We define a polynomial/ = f(x\.6
INTRODUCTION TO ALGEBRAIC THEORIES
We
Two associated monic polynomials are equal. x q ) in xi.
m[m(f)}
=
/.
Let/ =
f(x)
2.
terms
(4).
constants.*)=
k
.d.
the virtual degrees of kq exponents k\.
Let g be a factor of/.
Polynomials in several variables.
.) of polynomials in x. x is not a factor of/.
1.
We
we
In our discussion of the
g.
. Prove that $
a factor of m(f) for every
m which is at
least the degree of /.
We
a the
coefficient
+
.
.
its
.
Define/
=
=
if
Show
4.
.
.
.
. .
.
(5)
/
= f(xi. number of terms of the form
azji
.
. see from this that if g(x) divides f(x) every polynomial associated with g(x) divides f(x) and that one possible way to distinguish a member of the set of all associates
shall use this property of g(x) is to assume the associate to be monic.
4.
.
k xq
..
. that is.
if /is
3.
and a virtual (2) zero.
. . xq if all terms of nomial or a form in Xi.
.
kq
T* 0.
. . x q is clearly a form of degree m + n and.
.
=
+
+
. xq The product of two forms / and g of respective degrees n and m in the same Xi. then some a kl
. cally in y Xi. g be given by .
Then g h.
.
k
ki
.
zero.
.
.
. .
.
.
. polynomial (1) of degree n = n^in x = x q with
. yx q ) = y f(xij x q ) is a form of degree k x q if and only if /(xi. Also / is the zero polynomial if and only if all its coefficients are zero.
Theorem
is not zero.
.
derivation.
.
. .
We have
2. It is a generalization of Lemma 1. in xi.
need to consider an important special x q ) a homogeneous poly(5) have the same degree
. . by Theorem 2. .
.
.POLYNOMIALS
coefficients a^ k q are constants and n/ is the degree of x q ) considered as a polynomial in Xj alone.
.
.
.
=
To
fh. continue our discussion
.
observe that a polynomial / in x\. We now use this result to obtain the second of the properties we desire.
.
.
. .
We
.
.
q
and the
.
. However.
.
.
xq
may
its coefficients
be regarded as a a a n all
.
.
Then.
+k
for a kl
^
0.
x q -\ and a not zero.
. Let f.
degree of
* fl
/
is
defined to be the
assign the
maximum sum
degree minus
fci
+
. If we x q-i prove similarly that the product of two nonzero polynomials in x\
.
.
kq
. = a 6 are then and nonzero polyof a If coefficient is &o. .
. .
Xq
and
f
be nonzero.
. is nonzero if and only if / and g are nonzero.
.
.
is
and hence have not zero. if / is given by (5) and we replace x> in (5) by + x is replaced by y k ^~ *Xj> power product x k x q ) identixq and thus that the polynomial/(i/xi.
.
.
.
.
t
.
.
. .
XQ
We have
Theorem
fg
the immediate consequence 3.
. polynomials in x\.
.
.
. Thus we shall call /(xi.
.
see that each
.
. g. leading q fg
+
nomials in Xi and ao6 ^ by Lemma 2.
. h be polynomials in
Xi.
.
.
. with 6 then a virtual degree in x q oifg is ra not n.
}
.
yx^ we
. Then we have proved that the product fg of two nonzero polynomials / and g in x\. x 2 is not zero.
=
we
.
nomial.
As before we
and have the property that nonzero constant polynomials have degree zero.
.
.
.
.
.
.
shall
type of polynomial. some of the most important simple properties of polynomials in x hold also for polynomials in several x and we shall proceed to their
infinity to the zero polynomial
v -. similarly. If / is a nonzero poly-
Here the
.
. If. Note now that a polynomial may have several different terms of the same degree and that consequently the usual definition of leading term and coefficient do not apply.
. 2.
.
. we apply the proof above to obtain ao&o ^ x q is not proved that the product fg of two nonzero polynomials in #1.
thus completed the proof of The product of any two nonzero polynomials in
Xi.
.
.
.
.)
=/o+-.
.
seen
by the reader in his earliest algebraic study to imply that every rational
x\>
.
.
.
.
5.
.+/n.
.
.
for forms g> of degree
(8)
m
fg
i
and such that g Q
^
.
.
x q ) 7* 0. .
..
. Let f and g be polynomials in xi.
tion of division
The integral operations together with the operaby a nonzero quantity form a set of what are called the
rational operations.
..
then clearly
=
h
+
.
xq
may
be expressed as a quotient
for polynomials a(xi..
closed with respect to rational operations.8
INTRODUCTION TO ALGEBRAIC THEORIES
all the terms of the same degree in a nonzero polybe grouped together into a form of this degree and then may express (5) uniquely as the sum
first
Observe nomial (5)
that
we may
(6)
/
=
/(*!.
.
. . Xq.
.*.
a(xi. By this we mean that every rational function of the elements in this set is in the set.
. x q ) and 6(xi.
+
h m+n
i
By Theorem 2 /o(7o.
.
.
A
rational function of xi.
.
.
. Thus if we call/o the leading form of/. (a/6)
(c/d)
=
(ac)/(6d).
.
.. The postulates of elementary algebra were on xi
f
.
. The coefficients of x q ) and 6(xi.
.
.
x q)
=
g
+
.
zero.
.
+ gm
0.
.
. x q ) are then called coefficients of/. we clearly have Theorem 4. coefficients has a property which we describe by saying that the set is
.
function of
.
and we may use Theorem 2
to obtain
. This may be seen to be
due to the definitions a/6 + c/d Here b and d are necessarily not
=
(ad
+ 6c)/6d.
forms of f and g.
Ao T* 0.
.
.
is
where /o is a nonzero form of the same degree n as the polynomial/ and /< a form of degree n i.
Rational functions.
xq
is
now
defined to be
any function obtained as the result of a finite number of rational operations x q and constants.
.
.
. above is evidently fundamental for the study of polynomials in several variables a study which we shall discuss only briefly in these
uct of the leading
The
result
pages.. If also
(7)
g
=
g(xi.
. Then the degree of f g is the sum of the degrees of f and g and the leading form of f g is the prodwhere the
hi are
forms of degree
m+n
and
Ao
=
.
. Let us obx q with complex serve then that the set of all rational functions in xi.
If dj(x) is the g.c.
. d(x) is then the unique 0/f(x) and g(x).
of /i(x). and hence the degree
.
dj(x)
and
//+i(x).
.
(x) is
the g.POLYNOMIALS
bd
if
9
7* 0.(*)-/(*).-+i(x) divides /i(x). f (x) is a unique polynomial.-(x). according to our definition.
the g.c. . fg = if and only
=
0.a(x) = Iand6(x) = b^ 1 is a solution of ( 10) if g(x) is given by (2) Hence. For if /(x) = 0.
.1 exists such that //~ 1
=
1.
. For every common divisor h(x) of /i(x)./.-(x)
and
//+i(aO.c. that the set of rational functions satisfies the propin Section 1 for our constants.
.d. of any number of polynomials in x not all zero to the case of two nonzero polynomials.d. d(x) and d (x) are associated monic
its
because of
We define the g.
. d(x) is a common divisor of the /i(x) of largest possible degree and is clearly the unique monic com-
mon
divisor of this degree. . Hence.
. there is no loss of generality if we assume that both f(x) and g(x) are nonzero and that the degree of is not the of than For of notation g(x) greater degree /(x). Thus the g.
. Then there exist polynomials a(x) and b(x) such that
result
(10)
is
The
d(x)
=
a(x)f(x)
+ b(x)g(x)
a monic common divisor o/f(x) and g(x). Moreover.d. and hence both
then d
(x) is
./. of /i(x)./.
.-(x)
d
(z).<x)
and d
.
.
. then g(x) divides d(x).
.
.
.
. of two polynomials and the method of its computation are essential in the study of what are called Sturm's functions and so are well known to the reader who has studied the Theory of Equations. d
(x) divides /. The existence of a g.c. If g(x) divides all the/i(x).
/.d. consistency
g. We shall repeat this material here
importance for algebraic theories. h(x) divides
d.
while
if
/
j
then /.-(x).c. of /i(x).c.fs+i(x).//+i(x).
. of
d.
erties
we assumed
or g
/
=
Observe. of polynomials f\(x).
t
of d(x) is at least that of g(x).
. then d(x) is associated with g(x).
. /. A greatest common divisor process. that is.
Moreover.d.c. then.c. .
.
6.
.d.
h l (x)
=
g(x).-+i(x)
.(x) not all zero to be any monic polynomial d(x) which divides all the fi(x). the g.
. If do(x) is a second such polynomial.c. Let f (x) and g(x) be polynomials not both zero. then d(x) and d Q (x) divide each other.
.d.d. and is such that if g(x) divides every fj(x) then g(x) divides d(x).
. We shall now study this latter problem and state the result we shall prove as Theorem 5.
above evidently reduces the problems of the existence and construction of a g.
.
polynomials and are equal.
.d.
.
and
the divisor
of /i(x).
.
we put
(11)
*.
d Q (x) divides /i(x).
b(x) = cb r (x).
An
evident proof
hi(x)
by
then (14) implies that h r (x) induction shows that h r (x) divides
both ho(x)
b*(x)
bi(x)
=
f(x)
and
=
g(x).
his geometric formulation of the analogous result on the g. observe that it not only It is therefore usually called Euclid's process.
is
=
We
the degree of hi(x) then Hi > n > while n< > conclude that our sequence must terminate with
.
.
If.(*)
.
where
hi(x)
if
n<
0.
We
enables us to prove the existence of d(x) but gives us a finite process
by
.
unless
(15)
WX)
^
-
g r_i(oO/l r_i(z)
+ hr(x)
and
(16)
for r
*. of integers.d.
where the degree of hz(x) is less than that of h^(x). who utilized it in
=
ch r (x). If we
may
At_i(x).
we
=
fc(x)M*)
+ W*)
.
Equation
(12) implies that h z (x)
= =
qi(x).
.
r
The polynomial
is
a
divisor of /(#)
and
gf(x)
and
is
associated with a monic
common
Then d(x) has the form (10) for a(x) = car (x).
Thus h r (x) hi(x) and
be replaced by h r^(x) = divides both h r ~i(x) and h r-i(x).
Equation
[qr-i(x)qr (x)
(16)
implies
that
(15)
+
l]h r (x).
Ai. .
where the degree of h z (x) is less than the degree of may apply Theorem 1 to obtain
(13)
hi(x)
hi(x). The process used above was first discovered by Euclid.c.
now. If ht(x)
^
0. Thus our division process yields a sequence of equations of the form
(14)
hi-<t(x)
=
qi-i(x)hi-i(x)
+ hi(x)
.10
INTRODUCTION TO ALGEBRAIC THEORIES
1
By Theorem
(12)
h
(x)
= q^h^x)
+ h (x)
2
.2 ()
=
ai.
Clearly also h\(x)
= =
a*(x)f(x}
a\(x)j(x)
+ b^(x)g(x) with a^(x} + bi(x)g(x) with ai(x)
and
that
Ai(x)
= =
1.
0.
1.)
=
ar (x)f(x)
+ b (x)g(x).
assume that h r (x) divides
divides
hi-.2 ()/(x)
(14)
+ 6<-i(a?)g(x)
Ai_i(x)
=
+
common
6i_i(o:)gr(a:)
then
implies
=
fe r (z)
Thus we obtain
divisor d(x)
/i r (o.$(x). We have already shown that d(x) is unique.
A r_i(z)
=
qr (x)h r (x)
>
1.
POLYNOMIALS
11
means of which d(x) may be computed. Notice finally that d(x) is computed by a repetition of the Division Algorithm on /(#), g(x) and polynomials secured from f(x) and g(x) as remainders in the application of the Division Algorithm. But this implies the result we state as Theorem 6. The polynomials a(x), b(x), and hence the greatest common
divisor d(x) of
Theorem 5
all
have coefficients which are rational functions
with rational number coefficients of the coefficients of f (x) and g(x). thus have the
We
COROLLARY. Let
If the
the coefficients of f (x)
and g(x)
numbers.
be rational numbers.
Then
the coefficients of their g.c.d. are rational
only
common divisors of f(x) and g(x)
and
are constants, then d(x)
=
1
and we
shall call f(x)
indicate this at times
by
shall also g(x) relatively prime polynomials. saying that f(x) is prime to g(x) and hence also
We
that g(x) is prime to/(x). When/(x) and g(x) are relatively prime, (10) to obtain polynomials a(x) and b(x) such that
(17)
we use
a(x)f(x)
+ g(x)b(x) =
1
.
It is interesting to observe that the polynomials a(x) and b(x) in (17) are not unique and that it is possible to define a certain unique pair and then determine all others in terms of this pair. To do this we first prove the
LEMMA
is
5.
Let f(x), g(x), and h(x) be nonzero polynomials such that f(x)
prime
to
g(x)
and
divides g(x)h(x).
Then
f(x) divides h(x).
may write g(x)h(x) = f(x)q(x) and use = [a(x)h(x) + b(x)q(x)]f(x) = h(x) b(x)g(x)]h(x)
For we
(17) to obtain [a(x)f(x)
+
as desired.
We now obtain Theorem 7. Let
Then
there exist
f (x)
of degree
n and
g(x) of degree
m
be relatively prime.
of degree at
unique polynomials a (x) of degree at most 1 such that a (x)f(x) most n b (x)g(x) =
first equation of (18) with a$(x) the remainder on division of a(x) by g(x). = a Q (x)f(x) Then a (x) has degree at most 1, a(x)f(x) b(x}g(x) = = define b Q (x) 1. b(x) c(x)f(x) and see that [b(x) c(x)f(x)]g(x)
m
12
a\(x)f(x)
[60(2)
INTRODUCTION TO ALGEBRAIC THEORIES
+ bi(x)g(x) =
bi(x)]g(x).
a$(x)j(x)
+b
Q (x)g(x)j
then /(x)
clearly
divides
By Lemma
by
degree n
&o(z)
1 is divisible
5 the polynomial 60(2) bi(x) of virtual of n and must be zero. Hence, /(x) degree
and
a (x)/(x),ai(x) a (x). This proves a (x) 6i(x),sothatai(x)/(x) 6o(#) unique. But the definition above of a$(x) as the remainder on
=
=
=
division of a(x)
by
and
g(x)
There
is
also a result
shows that then (18) holds. which is somewhat analogous to Theorem 7 for the
5 to be unary, binary, ternary, quaternary, and quinary. The terminology just described is used much more frequently in connecis
tion with theorems on forms than in the study of arbitrary polynomials.
In particular, we shall find that our principal interest
forms.
in n-ary quadratic
There are certain special forms which are quadratic in a set of variables xm 3/1, t/n and which have special importance because they Xi,
. .
.
,
.
.
.
,
,
.
xm
A bilinear form / is called symmetric if it is unaltered by the interchange
of correspondingly labeled
members
of its
two
sets of variables.
.
compare
this
. y n separately. y n in a symmetric z and yi.
.
with (22) and conclude that a quadratic form may be rej/i. This state-
ment
/ =
only
(22)
clearly has meaning only if m = ZyidijXj.
and
see that
/ may be regarded as a
. z n respectively.
. are linear in both xi. .
.
aij
=
a .
x\.
yn.
j
=
1..
.
.y
=
i
(a*y
=
a/<.
=
a..
.
. ...
.
f. y by a*.
.
. and /
a.
.
. such forms bilinear forms.
. Here again xn.
.
.
xm and yi.
A
quadratic form
/
is
evidently a
sum
of terms of the type
the type ctjx&j for
i 7* j.
. Thus skew bilinear forms are
forms of the type
(24)
f
. They may be expressed as forms
.
.Zi.
We
garded as the result of replacing the variables
.
.
.
We may write
an
=
a^ as well as
\ c/ for
i -^
a.
.
yi. n
(20)
f=
we may thus
write
so that
(21)
/
.
.
form
in yi.
We
shall call
y-i .
n)
. But / = S2/ IiXiaayj
if
=
.
final
A
m
.
.. a
-
.
.-i
j
and have djXiXj
=
a^XiXj
+
a^x^x^ so that
(23)
/=
t. bilinear form in x\.
.
we
shall obtain a theory of equivalence of quadratic
forms and shall
use the result just derived to obtain a parallel theory of symmetric bilinear
type of form of considerable interest is the skew bilinear form.
.
yn)
=
/(yi.
.i
(i.
.
. .
.
.
.
.
n)
.
. and we call a bilinear form / skew if / = f(x\.
.
.
.
y n whose
coeffi-
cients are linear forms in x\. .
j
= =
1.-l
t-1
linear
.
.
-
-
. = n.
is symmetric if and only if and hence / is symmetric if and
m=
n.14
INTRODUCTION TO ALGEBRAIC THEORIES
.
n.
x n ).
Later
forms.
.
.
.
by corresponding then the new quadratic form/(zi.
.
.
n)
. a polynomial in xi.d.
.
1.
.
.
ftiXi
+
.
ORAL EXERCISES
1.
We
ai.
.
polynomials in x as coefficients.
Hence /
j
a
.
.
. .
.
.
.
.
.
(28)
=
.
.
=
an
=
b nx n
is
the zero form.
of a finite
the g.POLYNOMIALS
where
(25)
It follows that
15
an
an
= -an
*
(t.
.
+ b n )xn
.
Use the language above to describe the following forms:
a)
ft)
c)
+ Zxy* + xl + 2xiyi + x yi + x^
z3
3
2/i
d) x\
e)
2
+ 2xfli
x#i
Xiy 2
2
2.
. x n ) is the zero polynomial.
sum
of terms an(xiyj
if
.
. Thus any polynomial in # is a linear combination of a finite number of non-negative integral powers of x with x q is a linear combination constant coefficients. .
see that
.
/
+
g
=
(a x
+
61)0:1
+
.
.
.
.
=
+
+
anx n
.
ft.
.
+
(a. .
associate (25) only with
skew
bilinear forms. x n .c.
with constant coefficients
(29)
.
.
A
linear
form
aixi
is
expressible as a
.
.
.
Linear forms.
.
.
+ an =
0.
j
=
1.
. It is important for the reader to observe thus that while (22) may be associated with both quadratic and symmetric bilinear forms we must
2.
=
#)
we
. a n The concept of linear combination has already been used without the name in several instances. xi.
.
.
number of power products x$ xj with constant coefficients.
b ny
we
. of /(x) and g(x) is a linear combination (10) of f(x) and g(x) with
.
.
Xjj
.
+
.
=
(i
=
1.
n
1.
Express the following quadratic forms as sums of the kind given by (23)
a) 2x\
ft)
:
xf
-
x\
8.
.
for
i
j j.
n)
.
x n with coefficients (27) a linear combination of xi.
. The form (27) with a\ = a^
a second form. .
.
sum
(27)
/
shall call
.
.
. If g
is
g
=
fti.
i
=
.
It is also evident that
replace the y/
.
that
is
(26)
a
is
.
.
16
Also
(30)
if
INTRODUCTION TO ALGEBRAIC THEORIES
c is
any constant. If a
any constant. we have
cf
=
(cai)xi
+
.
.
and only
if
/
=
/
+
g)
0. then.
. as such.
. u be a sequence
u
=
(ai.
.
.
.
.
.
=
by
(61.
.
. may seem to be relatively unimportant.
+
+
.
.
.
We
(31)
define
/ to be the form such that /
(
/)
=
and
see that
-/=-!.
(32)
rows and columns.
and
(35)
define the
sum
of
u and
v
u
+
v
=
(ai
+ 61.
. The sequence all of whose elements are zero will be called the zero sequence and designated by 0. The reader is already familiar with these properties which he has used in the computation of determinants by operations
on
its
Let.
(ca n )x n
.
The properties just set down are only trivial consequences of the usual properties of polynomials and.
.
.
+
=
(-a*)x n
that
is
./=
g
if
(-oi)si
g
+
(
. as the coeffiwill
and
in this formulation in
Chapter IV
be very
important for all algebraic theory.
a)
is
of
n constants a
define
called the elements of the sequence u. if
we
so desire.
.
.
.
We now consider a second sequence.
. They may be formulated abstractly.
. as properties of sequences of constants (which
cients of linear forms)
may be
thought
of.
an
+
bn)
. .
we
(33)
au
call
= ua =
(aai.
6 n)
.
Then the
(36)
linear combination
au
+ bv =
(aai
+ bbi.
Then / =
. however.
aa n )
and
(34)
au the
scalar product of
v
u by a.
.
n). It is clearly the unique sequence z with the
.
a<
=
bi (i
=
1.
.
.
.
aa n
+ bbn
)
has been uniquely defined for
all constants a and b and all sequences u and v.
If q
=
r
and the determinant
(40).
.
in x\
-
t
.
anyi
.
.
.
. .
.POLYNOMIALS
17
z = u i or every sequence u..
.
. .
consider two forms /
the same degree n.
imply that
if
x q ) of / is carried y q ) by a nonsingular linear mapping.
.
The reader should observe now that the definitions and properties derived for linear combinations of sequences are precisely those which hold for the sequences of coefficients of corresponding linear combinations of linear
forms and that the usual laws of algebra for addition and multiplication hold for addition of sequences and multiplication of sequences by constants.
.
x g ) is a form of degree n in Equivalence of forms.
.
x q we obtain the original form
We now
into g(y\.
.
is
not zero.
.
bn
a n}
. we
f(xi.
.
.
+
a ir y r
(i
=
1.
. then.
.
and we
see that the unique solution of the equation v evidently call this sequence ( w). and thus if we replace t/i. nology is justified by the fact that the equation f(xi. If / = f(xi.
.
.
. This termix q ) = g(yi. x q and we replace every x> in/ by a corresponding linear form
.
x q ) and g
=
g(x^
..
.
.
(39)
Xi
=
.
.
.
#i. yq) y q} is an identity. . .
.
. In this case y 9 as linear forms in #1.
.
.
. The statements above / is equivalent to g then g is also equivalent to /.
Then we
shall
say that /
is
equivalent to g
if
.x q ).
9.
.
.
.
.
g)
.
.
.
. y r ) of the same degree n in t/i. u is the unique sequence that u
+
+
(37)
-u = -1
w
=
(-ai.
we obtain a form g = g(y\.
+
.
. yr Then we shall say that / is carried into g (or that g is obtained from /) by
. we shall say that (39) seen that we may solve (39) for yi. constant au = We define the negative u of a sequence u to be the sequence v such v = 0.
easily
.
.
.
.
. Evidently.
.
..
-a n )
.
.
the linear mapping (39).
by the corresponding linear forms
f(xi.
.
.
.
.
.
.
=
.
.
.
. Evidently if a is the zero property that u for every u.. y q in g(y\.
.
.
.
.
it is
.
x q and
obtain a linear mapping which we may call the inverse of (39). quence
u
+x=
v is the se-
+
We
v
(38)
u =
(61
ai.
is
. Thus.
.
nonsingular.
x 2 = y*...
.
q
is
called the
identical
mapping.
.
1.18
INTRODUCTION TO ALGEBRAIC THEORIES
say simply that / and g are equivalent.
The
linear
mapping
(39) of
EXERCISES the form x< =
y* for
i
=
1. pleting the square on the term in x\ and put Xi
+
+
+
+
a) 2x1
b)
c)
+ 3xi x? + 14^10:2 + 9x1 3x1 + 18xix + 24x1
4xiX 2
2
-
d) 2xf
e)
-
XiX 2
3x1
+
2xiX 2
-
xi
3. and even of those forms only under restricted
types of linear mappings. We have now obtained the background needed for a clear understanding of matrix theory and shall proceed to its development.
6)
/
=
4x?
-
4x!X2
+ 3x1
. 3 to the following forms / to obtain equivaand their inverses to g to obtain /.2x 2
= =
2/i
2/ 2
b}
M
f
Xi
-2xi
+ ^2^2/1 +x =
2
2/2
4. What
is its effect
on any form/?
2. Apply a nonsingular linear mapping to carry each of the following forms to an cx 2) 2 .
Find the inverses of the following linear mappings:
.
lent forms g
Apply the linear mappings of Ex.
f2xi-|-
x*
'
\3xj 4.
. by coma^yl Hint: Write / = ai(xi expression of the type a\y\ cx 2 = y\.
shall usually
We shall not study the
equivalence of forms of arbitrary degree but only of the special kinds of forms described in Section 7.
.
The array
an
i
.
y n with constant coefficients
. of certain natu-
ral manipulations on the equations themselves with which the reader is shall devote this beginning chapter on matrices to that very familiar.
a in )
19
. The concept of a rectangular matrix may be thought of as arising first in connection with the study of the solution of a system
km
of
m linear equations in n unknowns y\.
chapter we shall
make a completely
explicit
statement about the nature
of these quantities. but many matrix properties are ob-
tainable
by observing the
effect.
.
on the matrix
of a system.
. y n ) according to the usual definitions and hence satisfy the properties usually assumed in algebra for rational operations. .
. The matrix of a system of linear equations.
.
(3)
The
line
Ui
=
(an.
called the coefficient matrix of the system (1).
We
study.CHAPTER
II
RECTANGULAR MATRICES AND ELEMENTARY TRANSFORMATIONS
1.
.
of coefficients arranged as they occur in (1) has the
ai2
form
a22
and
is
speak of the coefficients
.
Let us now recall some terminology with which the reader is undoubtedly
familiar.
.
.
. In a later
.
an. It is not only true that the concept of a matrix arises as above in the study of systems of linear equations. shall henceforth in s shall fc as scalar and derive our theoand (1) a/
We
rems with the understanding that they are constants (with respect to the variables y\.
20
1JNTKUJLJUUT1UJN
TU ALUEJBKA1U THEUK1ES
of coefficients in the ith equation of (1) occurs in (2) as its ith horizontal line. Thus, it is natural to call u* the ith row of the matrix A. Similarly, the coefficients of the unknowns y,- in (1) form a vertical line
(4)
which we
call
We may now
the rows of
as one
the jth column of A. speak of A as a matrix of
are 1
m
rows and n columns, as an
ra-rowed and n-columned matrix
A
by n
or, briefly, as an matrices and its columns are
m by n matrix. Then m by 1 matrices. We
shall speak of the scalars
ay as the elements of A, and they may be regarded The notation a^ which we adopt for the element of A in its ith row and jth column will be used consistently, and this usage will be of some importance in the clarity of our exposition. To avoid bulky
by one
matrices.
displayed equations we shall usually not use the notation (2) for a matrix but shall write instead
(5)
A =
(i
=
1,
]
j
=
1,
.
.
.
,
n)
If
m=
w-rowed
shall use
is a square matrix, and we shall speak of A simply as an matrix. This, too, is a concept and terminology which we square
n then A
very frequently.
ORAL EXERCISES
1.
Read
off
the elements an,
ai2, 024,
a^ in the following matrices
4050^ -1 -2 -3
c)
3147 0-1
6
1.
3
2
1
4
6
-l
-6
8/
1.
2.
Read Read
off off
the second row and the third column in each of the matrices of Ex.
3.
the systems of equations (1) with constants &,
all
zero and matrices
of coefficients as in Ex.
RECTANGULAR MATRICES
2.
21
Submatrices. In solving the system
is
(1)
by the usual methods the
equations in certain t < n of the unknowns. The corresponding coefficient matrix has s rows and t columns, and its elements lie in certain s of the rows and t of the columns of A.
reader
led to study subsystems of s
in
<
We
call
such a matrix an
s
by
of
t
submatrix
s
the elements in the remaining
m
A
is
B of A. rows and n
shall call
If s
t
<m
and
t
<
n,
columns form an
the complementary
m
up
(6)
s
by n
t
submatrix
C
and we
C
submatrix of B. Clearly, then,
the complementary submatrix of <7. time to regard a matrix as being made to It will be desirable from time
of certain of its submatrices.
B
Thus we
write
(t
A = (A)
=
l,
...,s;j=l, ...,*),
where now the symbols At, themselves represent rectangular matrices. We A it all have the same assume that for any fixed i the matrices A n, An, number of rows, and for fixed k the matrices AU, A 2 A,* have the same number of columns. It is then clear how each row of A is a 1 by t matrix whose elements are rows of AH, AH in adjacent positions and thus have columns. We for accomplished what we shall call the similarly to drawing lines mentally parallel to amounts A what of by partitioning (6) the rows and columns of A and between them and designating the arrays of elements in the smallest rectangles so formed by A</. Our principal use of (6) will be the use of the case where we shall regard A as a two by two
.
.
.
,
;t,
.
.
.
,
.
.
.
,
matrix
(A,
,
A
whose elements AI, A 2 As, A 4 are themselves rectangular matrices. Then AI and A 2 have the same number of rows, A 3 and A 4 have the same number of rows, and every row of A consists partially of a row of AI and of a corresponding row of A 2 or of a row of A 3 and a corresponding row of A 4
.
Note our usage
in (2), (5), (6), (7) of the
symbol of
equality for matrices.
always mean that two matrices are equal if and only if they are identical, that is, have the same size and equal corresponding elements.
shall
We
EXERCISES
1.
State
how
.
the columns of
A
of (7) are connected with the
columns of AI,
A
2,
A*, and
A4
2. Introduce a notation of an arbitrary six-rowed square matrix A and partition A into a three-rowed square matrix whose elements are two-rowed square matrices.
22
INTRODUCTION TO ALGEBRAIC THEORIES
A
into a two-rowed square matrix
Also partition
whose elements are three-rowed
square matrices.
3.
Write out
all
submatrices of the matrix
2
1
-1
3
4
5\
01236
7
02-1
3
-2
1
2
1
and,
4.
if
they
exist,
the complementary submatrices.
Which
of the submatrices in Ex, 3 occur in
some partitioning
of
A
as a matrix
of submatrices?
5.
rices
Partition the following matrices so that they become three-rowed square matwhose elements are two-rowed square matrices and state the results in the
(6).
notation
a)
6.
Partition the matrices of Ex. 5 into two-rowed square matrices
whose elements
are three-rowed square matrices.
7.
matrix; a one
Partition the matrices of Ex. 5 into the form (7) such that A\ is a by six matrix; a two by two matrix. Read off A 2; A 3
,
two by three and A 4 and
state their sizes.
3. Transposition. The theory of determinants arose in connection with the solution of the system (1). The reader will recall that many of the prop-
erties of
determinants were only proved as properties of the rows of a determinant, and then the corresponding column properties were merely stated as results obtained by the process of interchanging rows and columns.
We
call
rices.
Let
the induced process transposition and define it as follows for matA be an by n matrix, a notation for which is given by (5),
m
and define the matrix
(8)
,
w)
,
which we
A
A. It is an n by m matrix obtained from and columns. rows Thus, the element a*,- in the ith by interchanging row and jfth column of A occurs in A 7 as the element in its jth row and ith
shall call the transpose of
its
where A\ is a g-rowed square matrix. It is a diagonal of the square matrix Ai and is its principal diagonal. any matrix of Ex. simply.
.
t
=
1.
. If A is our m by n motion of the rigid matrix and m < n we put q = m and write (11)
The operation
A =
(A.. Give also (7).
We also observe the evident theorem which we state simply as
(10)
(AJ = A. Notice now that A' is obtained from A by using the diagonal of A as an axis for a rigid rotation of A so that each row of A becomes a column of A'. where analogous result if A =
1. Note then that in accordance with our conventions been written as
(9)
(8)
23
could have
A'
=
((a.)
(j
-
1.
....-).
We
shall pass
have now given some simple concepts in the theory of matrices and on to a study of certain fundamental operations. .
.
. . A. 1 assume that A = A'. We shall call the a.
EXERCISES
for A'.
A.
Find the form
(2) of
A
if
A =
A' and
. q = n and have
(12)
if
n
<
m.-.. In Ex.
Let
A
have the form
Give the corresponding notation
A'
if
A
is
tives of those of
3.
Let
if
also
A be a three-rowed square matrix.
)..}
.
.
.. We should also observe that if A has been partitioned so that it has the
.
m)
.
.
. we put
(A
W'
l
\
where the matrix A\ is again a #-rowed square matrix.-.
. 2.RECTANGULAR MATRICES
column.*. the diagonal of A.<=
1.
of transposition may be regarded as the result of a certain matrix which we shall now describe. The line of elements a aq of A and hence of A\ is called the principal diagonal of A an.
. 1 of Section 1. What then is the form (7) of A ? Obtain the A is the matrix whose elements are the negaA'..
.
form
(13)
(6)
then
A
1
is
the
t
by
s
matrix of matrices given by
A'
=
(G/0
(On
=
A'rfj**
1. On the other hand. a 22 or. A = A'.
n.the diagonal elements of A.
. 'Let i and r be distinct integers. for example. equation (4. The addition of a scalar multiple of one equation of (1) to another results in the addition of a corresponding multiple of a corresponding linear form to another and hence to a corresponding result on the rows of A.
*
. and the reader equations Which are permitted in this
method
familiar with the operations
on
method and which
A
are called elementary transformations on tools in the theory of matrices. & 2 .24
4. for example. multiplication by a scalar) of such sequences were defined in Section 1.
4. Lemma 4.7.
INTRODUCTION TO ALGEBRAIC THEORIES
Prove that the determinant of every three-rowed square matrix
A
with the
property that A!
5.
Solve the system
(1)
with matrix
/
1
-2
-4
1\
4=
V
for
in terms of fo.8.
Do this
system
(1)
with matrix A' and com-
Elementary transformations.9. and the operations of addition and scalar multiplication (i. Theorem 8. Let i 5^ r and B be the matrix obtained from A by interchanging its ith and rth rows (columns). c be a scalar. 2/s
Write the results as
also for the
==
x
&
A' and thus
com-
pute the matrix
pare the results. Thus. equation (10) in Chapter IV.8.10) we shall mean Section 7.
-1
2
3-2
3/
3
2/
2/1. Theorem 4. Thus we make the following DEFINITION 2. by Section 4.
B =
(&*/). The rows (columns) of an m by n matrix are sequences of n (of m) elements. Lemma 9. as. Theorem 8.e. and B be me matrix obtained by the addition to the ith row (column) of A of the multiple by c of its Tth row (column). we shall mean that theorem of the chapter in which the reference is made. We define this and the corresponding column transformation in the DEFINITION 1. Then B is said to be obtained from A by an elementary row (column) transformation of type 1. if the prefix is omitted. Then B is said to be obtained from A by an elementary row (column) transformation of type 2.
A A and will turn out to be very useful
of our transformations is the result on the rows of A of the intertwo equations of the defining system. The system
is
(1)
may
be solved by the
yield systems said to be equivalent to (1). However. 2/2.
Jt 8 .
=
A
is
zero. The resulting operations on the rows of the of the system and corresponding operations on the columns of matrix
of elimination. Our final type of transformation is induced by the multiplication of an
We shall use a corresponding notation henceforth when we make references anywhere in our text to results in previous chapters.* The
The
first
change of
left
members of (1) are linear forms.
RECTANGULAR MATRICES
25
equation of the system (1) by a nonzero scalar a. The restriction a 7* is made so that A will be obtainable from B by the like transformation for or 1 Later we shall discuss matrices whose elements are polynomials
.
in x
and use elementary transformations with polynomial
scalars a.
We
a to be a polynomial with a polynomial inverse and hence to be a constant not zero. In view of this fact we shall phrase the definition in our present environment so as to be usable in this other
shall then evidently require
situation
and hence state it as DEFINITION 3, Let the scalar a possess an inverse or 1 and the matrix B be obtained as the result of the multiplication of the ith row (column) of A by a. Then B is said to be obtained from A by an elementary row (column*) transformation of type
8.
in the theory of matrices are connected with the study of the matrices obtained from a given matrix A by the application of a finite sequence of elementary transformations, restricted by the
The fundamental theorems
particular results desired, to A. Thus, it is of basic importance to study first what occurs if we make no restriction whatever on the elementary
transformations allowed. For convenience in our discussion
the
we
first
make
DEFINITION. Let
tions.
A and Rbembyn matrices and let B be obtainable from A
many
arbitrary elementary transforma-
by the successive application of finitely
Then we
shall say that
A is
rationally equivalent to
B
and
indicate this
by writing
A ^
B.
We now
see that,
observe some simple consequences of our definition. First,
we
if is rationally equivalent to B and B is rationally equivalent to to C, the combination of the elementary transformations which carry
A
A
B
with those which carry B to C will carry equivalent to C. Observe next that every
equivalent to
c
itself.
A
to C.
Then
m by n matrix is rationally For the elementary transformations of type 2 with
=
1
A A
is
rationally
=
and of type 3 with a
are identical transformations leaving
all
matrices unaltered.
Finally, we see that if an elementary transformation carries A to B there an inverse transformation of the same type carrying B to A. In fact, the
1
is
inverse of
any transformation of type 2 defined for c is that defined for c, of type 3 defined by a is that defined for a" of type 1 is itself. But then A is rationally equivalent to B if and only if B is rationally equivalent to A.
,
verify the fact that, if we apply any elementary row transformaand then any column transformation to the result, the matrix obtained is the same as that which we obtain by applying first the column transformation and then the row transformation.
*
The reader should
tion to
A
26
INTRODUCTION TO ALGEBRAIC THEORIES
shall replace the
is rationally equivalent are rationally equivalent. have now shown that in order to prove that and are rationally both rationally equivalent to the and equivalent it suffices to prove
Thus we may and
terminology
A
to
B
in the definition above
by A
and
B
We
A
B
A
B
same matrix
C.
As a
tool in such proofs
we then prove
be
the following
LEMMA
1.
Let r
<
m,
s
<
n,
and
\
A and B
_
m by n matrices of the form
A,
/B,
r by n s rafor T by 8 rationally equivalent matrices AI and BI and B B A and and are Then matrices 2 2 rationally equivalent. tionally equivalent For it is clear that any elementary transformation on the first r rows and s
.
m
A
columns of
A
A
and the zero matrices bordering
of such transformations induced
will replace
induces a corresponding transformation on A\ and leaves A* it above unaltered. Clearly the sequence
by the transformations carrying A\
Bl <
to B\
by the matrix
A -
-\0
A*)'
by
ele-
M
r
We
similarly follow this sequence of elementary transformations
mentary transformations on the last which carry A 2 to B 2 and obtain B.
It is
m
rows and n
s
columns of
A
important also to observe that we may arbitrarily permute the rows of A by a sequence of elementary row transformations of type 2, and similarly we may permute its columns. For any permutation results from some properly chosen sequence of interchanges.
Before continuing further with the study of rational equivalence we shall introduce the familiar properties of determinants in the language of matrix theory and shall also define some important special types of matrices. We
shall
then discuss another result used for the types of proofs mentioned above.
5.
Determinants. Let
B
be the square matrix
(14)
The corresponding symbol
.....
(15)
bi
D
RECTANGULAR MATRICES
is
27
t.
called a t-rowed determinant or determinant of order
It is defined as the
sum
(16)
of the
t!
terms of the form
(-l)*^*,...^,,
. . .
where the sequence of subscripts ii, t and the permutation 1*1, of 1, 2,
.
,
it
,
.
.
,
.
.
.
it
ranges over all permutations may be carried into 1,2,...,
is
t
1)* is unique interchanges. That the sign ( son's First Course in the Theory of Equations, and
proved in L. E. Dickshall assume this result as well as all the consequent properties of determinants derived there. The determinant D will be spoken of here as the determinant of the matrix
by
i
we
B
and we
shall indicate this
by writing
(17)
DD
|fi|
equals determinant 5). Nonsquare matrices A do not have determinants, but their square submatrices have determinants called the
(read
a square matrix of n > t rows, the complementary submatrix of any t-rowed square submatrix B is an (n t) -rowed square matrix whose determinant and that of B are minors of A called complementary
minors of A.
If
A
is
minors. In particular, every element a,-/ of a matrix A defines a one-rowed square submatrix of A whose determinant is the element itself. Thus we have seen that the elements of a matrix may be regarded either as its one-rowed
square submatrices or as its one-rowed minors. We now pass to a statement of some of the most important results on determinants.
The
result
on the interchange of rows and columns of determinants menLet
'
tioned in Section 3
LEMMA
(18)
2.
may now be stated as Abe a square matrix. Then
|A'|
=
|A|.
The next three properties of determinants are those frequently used in the computation of determinants, and we shall state them now in the language we have just introduced.
LEMMA
3.
Let
B
be the matrix obtained
.
from a square matrix
A
A.
by an
by an
ele-
mentary transformation of type 1 Then |B| |A|. LEMMA 4. Let B be the matrix obtained from a square matrix
=
ele-
mentary transformation of type
2.
Then
|
B = A
|
|
.
|
LEMMA
5.
Let
B
be the matrix obtained
mentary transformation of'type The reader will recall that
to obtain
from a square matrix A by an ele3 defined for a scalar a. Then |B| =a |A|.
Lemma
3
may
be used in a simple fashion
Then B has two equal rows and by Lemma 6
will
B
B =0.
. Of particular importance is that result which might
be used to define determinants by an induction on order and which does yield the actual process ordinarily used in the expansion of a determinant.
.28
INTRODUCTION TO ALGEBRAIC THEORIES
LEMMA
6. that is. Those we shall use are.
The
and
Let
result (20) is of
fundamental importance in our theory of matrices
be applied presently together with the following parallel result. and of these
we shall use only very few.
i.
If a square matrix has two equal rows or columns.
we have
8.
Combining umns we have
(21)
this result with the corresponding property
about
col-
a ik ckq
*~1
=
k=l
(i
c 8k a k j
=
T* q.
.
n)
.
.
=
|A|
+
|B|.
its
determinant is
Finally. B.
// a square matrix has a zero row or column. j. the determinant of A is obtainable as the sum of the products of the elements a/ in any row (column) of A by their cofactors c/i. We expand B as above according to the elements of its ith row and obtain as its vanishing determinant the sum of the products of all elements in the qih row of A by the cofactors of the elements in the ith row
|
|
of A.
Thus. the
properly signed and labeled minors
(
l)
<4
"
J
'd t-/.
its
determi-
nant
is zero.
.
7. Then
(19)
|C|
are.
. Then the result we refer to states that if we i+) define c/* = ( l) 'da then
We
let
A
(20)
|
A =
|
aikCki
=
c ik a ki
(i.
n)
. also well known to the reader. s
^ j. of course.
=
1.
Another
result of this type is
LEMMA
zero. of course.
j.
.
be an n-rowed square matrix A = (o</) and define AH to be the complementary minor of aj.
LEMMA
Let A.
.
.
is the
C
be n-rowed square matrices such that the ith
row
(column) of
C
sum
of the ith
other rows (columns) of
B
and
C
row (column) of and that of B while all are the same as the corresponding rows (col-
A
umns) of A. q.
8
=
1.
There
many other properties of determinants. be the matrix obtained from a square matrix A by replacing the tth row of A by its qth row.
and.
These relations
its
will
have important
We call the matrix
(22) the adjoint of
A
and
see that
if
A=
(an)
is
an n-rowed square matrix
adjoint is the n-rowed square matrix with the cofactor of the element which appears in the jth row and ith column of A as the element in its own tth row and jth column.-/) is called a diagonal matrix matrix
A
if it is
a square matrix.
we have The determinant of a triangular matrix
is
is the
product ana22
.
Expand the determinants below and
verify the following instances of
Lem-
ma
8:
32-1
a)
1
3
1
2
3
0-1
1 1
003
-1
1
1
-3 -1 -1
3
1
-1 -1
1
0-1
1
3
1
234
6..-/)
is
triangular
if it is
true that either
a. = (a. moreover.
n)
. that is. There are certain square matrices which have speforms but which occur so frequently in the theory of matrices that they have been given special names.
Compute
the adjoint of each of the matrices
2.
EXERCISES
1.
.
(i. Clearly.
1.
. if A = then adj A = 0.
-1 -1
4
1
-3
1
234
-1
234
-1
1
Special matrices.
=
if
for all j
that an
for all j
<
i.
j
=
1.
.
. Thus a square
matrix
A =
=
(a.
It is clear that
A
is
triangular
and only
if
> i or A is
7
triangular.
|
A =
\
an.
a nn of
its
diagonal elements.
.RECTANGULAR MATRICES
The equations
square matrix
(22)
(20)
29
A
and (21) exhibit certain relations between the arbitrary and the matrix we define as
adj
A =
later consequences.
The
result
above
clearly true
if
n =
1
so that
A =
(an). The most general of these is the triangular
cial
matrix.
.
We
first
assume
it
induction
by
1 and complete our true for square matrices of order n expanding \A according to the elements of its first row or
\
column
A
in the respective cases above. a square matrix having the property that either all its elements to the right or all to the left of its diagonal are zero.
Theorem
.
The reader will
find that this usage will cause neither difficulty nor confusion. a diagonal matrix A is triangular so the product of its diagonal elements. We let A be a square matrix and partition it as in = t and the submatrices An all square matrices.
We
shall discuss the implications of this notation later.-/
=
. we shall indicate the order by a subscript and thus shall write either 7 n or I as is convenient for the n-
rowed identity matrix. Then the La(6) with s
place expansion clearly implies that if all the A/ are zero matrices for either all i > j or all i <j.
where a
= an
is
the
common
value of the diagonal elements of the matrix. .
(24) implies that
4 is necessarily square. .30
INTRODUCTION TO ALGEBRAIC THEORIES
=
its
and a/
that
If all
determinant
Clearly.
The property above and that of Theorem 1 are special instances of a more general situation.
We shall frequently feel it desirable to
of the forms
consider square matrices of either
Al
A A. Evidently Theorem 1 is the case where the A.
sidered in (24) the case where t = 2. and every
A. we call the
for all i
^ j. are one-rowed square matrices and the result con.
Any
(23)
scalar matrix
may
be indicated by
of. then |A| = |An| |A|.
It is natural to call any m by n matrix all of whose elements are zeros a zero matrix. In connection with the discussion just completed we shall define a notation which is quite useful. fact that the Laplace expansion of de-
where A! is a square matrix. Then and the reader should verify the
terminants implies that
|A|
A
-
|A
=
|
|Ai|
-|A 4 |. the diagonal elements of a diagonal matrix are equal. In any discussion of matrices we shall use the notation to represent not only the number zero but any zero matrix. Let be a square matrix partitioned as in (6)
A
and suppose that
s
=
t}
the
An are
all
square matrices.
is
matrix a scalar matrix and have \A = ajl4 The scalar matrix for which an = 1 is called the n-rowed identity matrix and will usually be designated by 7. If there be some question as to the order of I or if we are discussing
\
several identity matrices of different orders.
By an
?
r c
> =
1.
of A is the product of the determinants of its submatrices Ai. and I\ is the identity matrix of one are rationally equivalent.
.. either AI = and we have (26) for / = /i.
..
by Lemma
1. and we shall indicate this
Then
A
is
by writing
(25)
A =
diag{Ai.
A
is
rationally equivalent to a matrix
(29)
. For by elementary transformations of type
7*
we may carry any element
a pq
of
A into the element 6u of a matrix B which is rationally equivalent
1 elementary transformation of type 3 defined for a = frji we = 1.
LEMMA
matrix
9. In closing we note the following result which we referred to at the close of Section 4. the determinant
.A. or our proof shows that AI is rationally equivalent to a matrix
is
Now Ai
an
m
1
by n
A
and
A
(
But
t
then.
31
composed of zero matrices and matrices Ai = An which are what we may call its diagonal blocks.. d r \ =
to A. Moreover.}.
Every nonzero
m
by n matrix
A
is rationally equivalent to
a
/I
(26)
0\
\o
or
1
where
I is an identity matrix.RECTANGULAR MATRICES
for i 7*
j.
and then use elementary column transformations of type 2 with
dir
to replace
D
by the matrix
row.. At.
. We then apply replace B by a rationally equivalent matrix C with c\\ = of 2 with row transformations c c r \ to replace C by elementary type for the rationally equivalent matrix D = (d< ) such that du = 1.
As above.
1 matrix. Clearly.
-1\
5/
d)
/O
2\
1
c)
V-l
2
4
VO
5
3J
4. and by row transformations we may a matrix with ones on the diagonal and zeros below and then into /.
EXERCISES
1.
Carry the following matrices into rationally equivalent matrices of the
(26)
/
form
3
5
1
10
2
4\
6)
/I
1
2\
a)
I
\-l -1 -
211
O/
(2
\1
3 2
5] 3/
2023 0000
i (1 2. Hint:
The property \A\
in the first
column
carry
7.
.
is
After finitely
We
shall
show
later that the
number
of rows in the matrix I of (26)
uniquely determined by A.
/
a)
2
3-2\
-27
3
1
1
\5
/
J
.
is
^
of
A
The largest order of any nonvanishing minor of a rectangular matrix A is called the rank of A.
3
2 4
1
-x
/!
6)
\ .
1
2\
o
i
iy
Apply elementary row transformations only
arid carry
each of the following
matrices into a matrix of the form (26)
10^ 22-50 -21-52
10
00
ly
3.
*
e.
A =
(11). Some element must not be zero.
A
into
preserved by elementary transformations.
Not every matrix may be
take
carried into the
form
(26)
by row transformations
only.. The result of (20) states that every (t + l)-rowed minor of A is a sum of nuRational equivalence of rectangular matrices.32
INTRODUCTION TO ALGEBRAIC THEORIES
many
such steps we obtain
(26).g.
Show
that
if
carried into the identity matrix
the determinant of a square matrix A is not zero then A can be by elementary row* transformations alone. Apply elementary column transformations only and carry each of the following matrices into a matrix of the form (26).
|
1
I
\
|
1
I
|
|
|
. Let
and
let all (r
+
l)-rowed
Note that we are assigning zero as the rank
of the zero matrix
and that
the rank of any nonzero matrix is at least 1. and A has a Ao. then AJ = A\.
+
l)-rowed minors of
A vanish. a row erf B is in the ith row of A and a row of B is in the 0th row of B.
|
|
|
|
|
|
|
1
|
|
|
1
|
|
\
\
l
|
.4 we have B = B If. If. and the former. and our proof shows the Also there exists an r-rowed minor B 5* = B or B = B + c C in existence of a corresponding minor B =0 implies that |C| = -cr \B\ ^ 0. But then it is easy to see that C is a minor of A as well as of AO.RECTANGULAR MATRICES
merical multiples of its Crowed minors. The problem of computing the rank of a matrix
definition
A would seem from our and lemmas to involve as a minimum requirement the computation of at least one r-rowed minor of A and all (r + l)-rowed minors. then by Lemma 8 Bo = B + c C where C is a Crowed square matrix all but one of whose rows coincide with those of B. But then |B nonzero r-rowed minor |C|.
11. the
. Let then A result from the application of an elementary row transformation of either type 1 or type 3 to A. however. the correspondingly placed submatrix B of A is equal to B if no row of B is a part of the ith row of A. If A has rank r we put t = r + 1 and see of A that |B = |C| =0. and this remaining row is obtained by
ever. Bo = Of or every (r + 1) -rowed minor |B in A. By the above
|
\
. We thus clearly have
if
33
all
the latter are
zero so are
LEMMA
is at
10. A and A have the same rank. a row of B is in the ith row of A but no row is in the 0th row of A. that if an elementary row transformation be applied to the transpose A' of A to obtain Ai and the corresponding column transformation be applied to A to obtain Ao. We are thus led to study the effect of such transformations on the rank of a matrix. By Lemmas 3 and 5 every Crowed minor of A is the product by a nonzero scalar of a uniquely corresponding Crowed minor of A. then by Lemma 2.
replacing the elements of
B in the ith row of A by the correspondingly columned elements in its 0th row. Howproblem may be tremendously simplified by the application of elementary transformations. Let all (r
r. finally. If Ao results when we add to the ith row of A the product by c 7^ of its 0th row and B is a Crowed square submatrix of A. and the computations themselves generally quite complicated. Then r is the rank of A. finally.
Then
the
rank of
A
most
We may also
LEMMA
minors of
state this result as
A have a nonzero r-rowed minor A vanish. We observe. and it follows that A and Ao have the same rank. The number of determinants to be computed would then normally be rather large.
let the rank r of two by n matrices
m
A
and
B
be the same
9 to carry A into a rationally equivalent matrix (26).
r.34
INTRODUCTION TO ALGEBRAIC THEORIES
proof A' and AI have the same rank. have also the consequent COROLLARY.
We have thus proved
what we regard as the
principal result of this chapter.
respectively.
(H
0). By the above proof A and this matrix have the same rank. If
we then apply
the inverses of the column transformations in reverse order to (30).
where
G is an T-rowed matrix. I = I r
and use
Lemma
. Conversely.
and
to A.
A
<
is
rationally equivalent to
(o'
Similarly. But then Theorem 2 implies. H is an T-columned matrix. by Lemma 2 A and A! have the same minors and hence the same rank.
by a sequence of elementary row or column transformations only. as in Ex. any two rationally
equivalent matrices have the same rank. we
is
For
A
by a sequence
of elementary
first
may
obtain (30) by
row and column applying all the row
transformations and then
all
the column transformations. Every u-rowed nonsingular matrix is rationally equivalent to n-rowed identity matrix. Hence.
Theorem
they have the
Two
m by n matrices are rationally equivalent if and only if m
r is rationally equivalent to
same rank.
Theorem
3. Thus. Clearly. It is clear that the rank of the matrix in (26) is the number of rows in the matrix /. Every m by n matrix of rank r > may be carried into a
matrix of the forms
(31)
. we obtain the result of the application of the row transformations alone to A. 4 of Section 6.
But column transformations applied
to (30) clearly carry this matrix into
. Every by n matrix of rank
We
an
m by n matrix
is
(SO). In closing let us observe a result of the application to a matrix of either
the
row transformations only or column transformations only.
A matrix is called nonsingular if it is a square matrix and its determinant not zero.
I)
(30)
B is rationally equivalent to
2. A and A have the same rank. We shall prove Theorem 4. so that A\ and A( = AQ have the same rank.
and
both
G
and
H
have rank
equivalent to (30) transformations.
Compute the rank r of the following matrices by using elementary transformabut
r
tions to carry each into a matrix with all
rows
(or
columns) of zeros and an
obvious r-rowed nonzero minor.
. Hint: If necessary carry A and B into the form (30) and then apply the inverses of those transformations which carry B into (30) to carry (30) into B. it is evident that the rank of this matrix is that of (?.
B=ll
VO
/I
1
\3
3/
0. Moreover. Carry the first of each of the following pairs of matrices into the second by elementary transformations. The result for column transformations is obtained similarly.
EXERCISES
1.
-4 -2 -4 -6
I/
1\
J5=(60
\1
A =
(
2
2
1
.RECTANGULAR MATRICES
35
a matrix of the form given by the first matrix of (31).
/I
a)
1
(
1
5 5
3\ 2
/2
6)
1
(
-
1
4
14
3-2
-11
\1
-I/
3
4
16
\1
/'
c)
4
-12
2.
A =
/2
-1
3\
'
_
/I
n 6)A =
(l \2
/I
c)
-2
i\
/2
3
1
-221. G has rank r.
with n by q matrix
stitute (2) in (1)
B =
(bj k ).
carrying the y.
xm and
t/i.
.
ra.
.
m)
.
this
system was said in Section
y.
mapping (3) is usually called the product of the mappings (1) and we shall also write C = AB and call the matrix C the product
of the matrix
have now defined the product AB of an m by n matrix A and an n by q matrix B to be a certain m by q matrix C.
m)
.
by n matrix
.
if
we sub-
we obtain a
third linear
mapping
(3)
Xi
=
2*c ik z k
it is
(i
=
1.
. Moreover. .
.
y n are variables
(1)
Xi
=
atMi
(i
=
1. . .
.
.
.
.
ping
(1).
.
q)
.
Then.
.
We
36
.-.
.
.
.
A =
(a^) the matrix of the
mapping carrying the mapy\.
.
.
.
. If xi.
.
. Multiplication of matrices.
y n by a
q
(2)
yi
=
^b
ik z k
(j
=
1.
k
=
1.
.
.
n)
.
.
with
m by q matrix C =
(CM). we have defined*C so that the element dk in its ith row and fcth column is obtained
A by the matrix
B.
The
and
linear
(2). related by a system of linear equations
.
.
.CHAPTER
III
EQUIVALENCE OF MATRICES AND OF FORMS
1.
.
.9 to define a linear
Xi to the
We
call the
m
.
.
Suppose now that 21. second linear mapping
zq
are variables related to
. to the z k
.
and
easily verified
by substitution that
(4)
c ik
-
_
(i
=
1.
.
1.
.
B is an n by q matrix. they are generally not equal and may not even be matrices of the same size. Observe also that I m is the matrix of the linear transformation Xi = yi for i = 1.
.
.
A is an m by n matrix.
.
In symbols we state this result as
(8)
(AB)'
=
B'A'
.
AB
is also.
Here
trix.
matrices in the exercises below. A is m by n. But when both are defined. If A and B are n-rowed square matrices. Note the examples of noncommutative square
. (AB)' is a q by in mawhich we state is the product of the q by n matrix B' and the n by m
. and m^n.*
in the fcth
column
61*
62*
(6)
of B.6.
The transpose of a product
of two matrices is the product of
their transposes in reverse order. We have not defined and shall not define the product AB of two matrices in which the number of columns in A is not the same as the number of rows in B. we shall say that A and B are commutative if AB = BA.
(7)
if
A
is
an
m by n matrix.
Im A
a zero matrix the product then we have
.
.
+
o>i n b n k
of the products of the
elements ay in the ith row
(5)
of Ay
by the corresponding elements
?>.
= AI n = A
where 7 r represents the r-rowed identity matrix defined for every r as in Section 2. Observe that if either
A
or
B
is
Moreover. w. the fact that AB is defined need not imply that BA is defined. evidently. This latter fact is clearly so if. for example.
Finally. let us observe the following
Theorem
1.
.
Thus we have stated what we
shall
speak of as the row by column rule
for multiplying matrices.EQUIVALENCE OF MATRICES AND OF FORMS
as the
37
sum dk =
an&i*
+
+
. and thus in this case the product of (1) by (2) is immediately (2) hence (7) is trivially true. Then. B is n by m.
.
It
is
important to know that matrix multiplica-
tion has the property
(9)
(AB}C = A(BC)
for every
is
mby n matrix A.
3.
q
by
s
matrix C.
The
associative law.38
matrix
INTRODUCTION TO ALGEBRAIC THEORIES
A
f
. the powers A of any square matrix are unique.
. . Compute (3) and hence the product C = AB for the following linear changes variables (mappings) and afterward compute C by the row by column rule.
of
EXERCISES 1.
2/2
92/3
=
*i
2z 2
s3
Compute the
latter
following matrix products
is
AB.
/4
a)
3
2
1
/-I -2 -3^
1
3
3 4^
\2
1
-I/ \~1 -2
-
2
3illi
o
"I/"
t
*
A
^t
f\
i
.
I
I
Vl ==
~~~
"2*1
^"2
i~
"^"8
Ixi
=
2y l
+
-
32/2
-2/s
-ISai
[2/1=
1 2/a
+
26z 2
+
-
13* 8
Us =
2. In particular. Verify by explicit computation that EijEjk = Eik and that if j ?* g then EijE ok = 0.
2. We shall assume these two consequences of
6
.v
ci.
n by q matrix B. This result
known as the associative law for matrix multiplication and it may be shown as a consequence that no matter how we group the factors in forming a product A\ At the result is the same. Compute
also
BA
in the cases
where the
product
2\
1
defined.
We
leave the direct application of the
row by column
rule to
prove this result as an exercise for the reader.
.
|
|
9 j/9 ^ n*
o io
1
-
ith
Let the symbol EH represent the three-rowed square matrix with unity in the row and jth column and zeros elsewhere.
.
m.
.
.
I/
B
=
\0
~3\
/
/
2
-1
1
^-a. = (ft/*).
n)
. Hence. j =
1.
I
=
1.
BA =
(btftf)
(i
=
1.!
c)
A(l
2
-1
3).
.
q.
then
'
B by &<.
.
. Then that BA is defined.
. G = H.
. hu for all i and Z. where in all To prove (9) we write
A
.
B
cases
is
i
=
1.
.
.
-1 -1
2
3\
A =
3
12.
.
. Let
A =
(a/) be
an
m by n
matrix and
of
B be a diagonal matrix.EQUIVALENCE OF MATRICES AND OF FORMS
(9)
39
without further proof and refer the reader to treatises on the founda-
tions of mathematics for general discussions of such questions.-/). C = (c k i).
.
.
.
m\j =
1. k
=
1.
.
.
Then
it
clear that the element in the ith
row and
Ith
column of
G = (AB)C
is
(10)
gn
while that in the same position in
H
= A(BC)
is
(ii)
hu
y
=
i
Each of these expressions is a sum of nq terms which are respectively of the form (dijbjk^Cki and a^bjkCki).
(12)
ith diagonal element is ra-rowed so our definition of product implies that if
and designate the
B
. and (9) is proved.
.
.
n.
. But those terms in the respective sums with the same sets of subscripts are equal. gu
EXERCISES
Compute
the products
/2
o)
(AB)C and A(BC)
in the following cases.
Products by diagonal and scalar matrices. since we have already assumed that the elements of our matrices satisfy the associative law a(bc) = (ab)c.
. = (a.
s.
.
.
.
3.
Then
Let the diagonal elements of an n-rowed diagonal matrix B be the only n-rowed square matrices commutative with B are
the
We may now
n-rowed diagonal matrices. Now.
.
.
then (14) implies that a.. j
=
1.
=
0.
n)
.
.
next see that. the elements in the ith column of EjB must be zero. The only n-rowed square matrices which are commutative with every n-rowed square matrix are the scalar matrices. Then from (12) and (13) we see
that
(14)
AB = BA
if
and only
if
(&<
-
bj)av
= =
.
= l. the product of A by a diagonal matrix on the
right
is
the result of multiplying the columns of
A
in turn
by the
corre-
sponding diagonal elements of B. let m = n.
For
(15)
let
B =
(ba)
(i..
j
=
1.
As an immediate consequence
simple
of the case bi
. we have 6/ = is the matrix with 1 in its first row and jth column and zeros elsewhere.40
while
(13)
if
INTRODUCTION TO ALGEBRAIC THEORIES
B is n-rowed. prove the converse of Theorem 2
a result which
is
the
inspiration of the
name
scalar matrix.
n)
.
.-< is in the for j j* i.n)
A.
.
Theorem 4. if i 7* j and & 5^ This gives the result we shall state as
We
&/. Thus if i ^ j. and 5 is a diagonal matrix.
Every n-rowed scalar matrix
is
commutative with
all
n-rowed
square matrices. First.
Thus the product of a matrix A by a diagonal matrix B on the left is obtained as the result of multiplying the rows of A in turn by the corresponding diagonal elements of B.j.
. the
the diagonal matrix with unity in
. Since 6.
then
AB =
(aijbj)
(i
=
1.
3.
ra.
.
.
.
(i.
. If D ith column.
.
be
and suppose that
B is commutative with every n-rowed square matrix We shall select A in various ways to obtain our theorem. Equations (12) and (13) imply that the jth row of EjB is the same as that of B and the jth column of BEj is the same as that of 5.
.
..
. we let E
its
3
jth row and column and zeros elsewhere and put BEj = EjB.. A be a square matrix.
=
bn
we have
the very
Theorem
2. while all other columns of BEj are zero.
Theorem
all distinct.
and B is the scalar matrix 6/ n Let us now observe that if A is any m by n matrix and a is any scalar. fry.
0\
6)
A =
\0
-1
07
.
2.
then
(16)
is
(a! m )A
= A(al n)
by n matrix whose element in the ith row and jth column is the product by a of the corresponding element of A. This is then a type of
an
product
(17)
m
aA = Aa
like that defined in
Chapter I for sequences. we have defined (17) as the instances (16) of our matrix product (4).
.
B =
-1
o
1
Ov
/O
**
000 -2/ (2
73
d)
02 01 00-2 or
0\
0\
_ ~
/
1
0-1
1
o
o
\0
A \0
2
17
.
.
Compute
the products
rule
if
EXERCISES AB and BA by the
0\
use of (12) and (13) as well as the
row by column
71
a)
A =
\0
71
-2
37
.EQUIVALENCE OF MATRICES AND OF FORMS
41
product BDj has bu in its first row and jth column and is equal to the matrix DjB which has 6/.
Find
all
three-rowed square matrices
/I o)
B
such that
BA = AB\i
0\
4 =
-1
b)
A
.
. = bn = 6 for j = 2. However. and we shall call such a product the scalar product of a by A.
1.-in this same place.
.
. n. Hence.
INTRODUCTION TO ALGEBRAIC THEORIES
Prove by direct multiplication that
BA =
AB
if
i
2
=
1
and
4.
Let
w
be a primitive cube root of unity.
. Apply an elementary row (column) transformation to A (to B) resulting in what we shall designate by Ao(by B ). Then
(0)
(0)
(18)
Co
= AB
.
.
3 3
Compute
BAB if
Ha
where
c
ic
-d\
c)'
Hi
n
/O
-1\
o)'
c
and d are any ordinary complex numbers. 1 AB on of type by applying it to A.
. Let A be an m by n matrix. We might of course give a formula for each E and verify the statement..in the right members of (4) Thus we obtain the result of an elementary row transformation get c.
if we replace a^ by a. Elementary transformation matrices.
. Our definitions also imply that for elementary row transformations of type 3 the result stated as (18) follows
For proof we see that
we
from
(19)
.!
. 4 has the property B = al and that the A = 7.
C<>
= AB<>
.
.
q)
.*.
. We shall show that the matrix which is the result of the application of any elementary row (column) transformation to an m by n matrix A is a corresponding product EA (the product AN) where E is a uniquely determined square matrix.
. B be an n by q matrix so that C = AB is an m by q matrix.
. but it is simpler to describe E and to obtain our results by a device which is a consequence
of the following
Theorem 5.42
3.'1
(i
=
1. and then the same elementary transformation to C resulting in Co (in C ).
A
has the property
B of Ex.
and 3 are
their conjugates.
m. Prove that
BA
wAB if
5.
Show
that the matrix
matrix
6. Obtain similar results for the matrices of Ex.
.. k =
1.
4. 3.
and if Ri(a) is n-rowed.
result
and we have proved the first equation in (18).EQUIVALENCE OF MATRICES AND OF FORMS
Finally.
n. P.
./(c).
43
we
n
see that for type 2 they follow
/
from
/
n
\
n
\
1
(20)
y-i
2} (a*.2 as the
following
.
cc 9k
=
. we may interpret Theorem 2. the product AE^ is the result an elementary column transformation of type 1 on A.
R^a^R^a) = E
i
(a)^ i (a~
1
)
=
/
. First of all. and Ri(a) elementary transformation matrices of types 1. k
=
. in which 7 is the matrix called E above. The corresponding column C (0) = AB (Q) has an obvious parallel proof or may be thought of as obtained being by the process of transposition.
. Thus the elementary column transformations give rise to exactly the same set of elementary transformation matrices as were obtained from the row transformations. Here we define
EH
I.
to be the matrix obtained by interchanging the ith and jth rows of by adding c times the jth row of I to its Oh row.
and if P. then ARi(a) is the result of an elementary column transformation of type 3 on A. Similarly.7 in terms of elementary transformation matrices. then APa(c) is the result of an elementary column transformation of type 2 on A.
(22)
(PH(C)]'
=
P. Ri(a) by multiply-
ing the -ith row of / by a ? 0. P.
. We shall call EH.
E En =
i3
/
. Then we see that to apply any elementary row transformation to A we simply multiply A on the left by either EH.
q)
. Observe that
(21)
Eh =
E.
. or R<(a). and 3. 2.
=
c ik
+
1.
of
if we now assume that EH is n-rowed. P*..
.
+ ca j)b =
t
3k
I
\y-i
^a^b/^ |
/
+
c
I
\y-i
(i
^^A* /
1. respectively.
. We shall now interpret the results of Section 2.(c)
supply further
details.-(c) is n-rowed.
so that.-(c).<
= EH
.
P(-c)P w (c) = P(c)P(-c) =
/
. Finally.(c)
.
(23)
[Ri(a)Y
=
Rt(a)
. It is surely unnecessary to
We now write A = IA where / is the ra-rowed identity matrix and apply Theorem 5 to obtain A = /<>A.
By Theorem 5. that is.4.
mentary transformations carrying A into an r rows are all zero. Qt such that
.
.
^
0)
2
l
2.
.
(24)
is the
|AB|
= |A|-
|B|.
Find the elementary transformation jjrtatjafes corresponding to the elementary
of that exer-
row transformations used in Ex.30) by matrix multiplication. if
the rank of
A
is r.
A
Qt.
.Qi matrix.
transformation matrices PI.
m
.
The rank
of a product of two matrices does not exceed the rank
of either factor. and consequently B
.
EXERCISES
1.
by Theorem
2. Then r rows of the w-rowed matrix C are all zero. and the
rank of
result
C
is
the rank of
relation
on the
C and is at most r. In the theory of determinants it is the shown that symbol for the product of two determinants may be computed as the row by column product of the symbols for its factors.
.
Every nonsingular matrix
is
a product of elementary transforma-
We shall close the results with an important consequence of Theorem 2. The determinant of a product of two square matrices product of the determinants of the factors.)A(Qi
. The corresponding between the ranks of C and B is obtained similarly.
.
tion matrices. In our
The determinant
present terminology this result may be stated as Theorem 7. we
the bottom
m
obtain a rationally equivalent matrix Co = A Q B.
B =
if
(?!
.
v
Theorem 2.
.44
INTRODUCTION TO ALGEBRAIC THEORIES
LEMMA
1.
6. 2 of Sectfoi^fcfrand carry the matrices cise into the form (2.
Let
A
and
B
be
.
Express the following as products of elementary transformation matrices. and it is interesting to note that it is possible to derive the theorem as a
.
Q
t)
and only
if
A
and
B
have the same rank.
of a product.
The usual proof in determinant theory of the result is quite complicated.
is
the n-rowed identity
.
.
Theorem
For.
Thus we obtain
LEMMA 2.
by n matrices.
. tom
there exists a sequence of eleby n matrix A Q whose bot-
m
if
m
we apply these transforma-
tions to
C = AB.
.
P.3 is the case of Lemma l where = PI P. Then there exist elementary P 8 Qi.4. as desired.
.
.
6.
and hence |G = \E\ |G|. that |(?
|
| |
It follows clearly that. then Lemmas 2.
inverse
Nonsingular matrices. G|.= |/| = 1. 1. We shall. but we shall prove instead the result that if A 5^ then A" 1 is
For if
|
1
(25) holds.
.
if
.
\Ei\
|G|. give such a derivation.
Moreover.
by
A
. From our definiwe see that. and 2.. or a.
|P| and then
\B\
We apply this reto A B to obtain
|J8|
\AB\ = |Pi.20) and (2.
matrices. where adj our symbol for the adjoint matrix defined in (2.
. and Lemma 2.4. if E is an n-rowed elementary transformation matrix of 1. Hence.. then \E\ = n-rowed square matrix and <7 = EG. respectively. then EiO\ = \E t sult first to A to obtain \A\ = |Pi| . Thus.
Ei.a|G |. A ^ 0.A =
8..
|
\
|
\
A = PL.3.21) in terms of matrix product.
=
|
\A\
as desired.
. we have
Theorem
singular.P. therefore.
B
is
= 7 is the n-rowed an n-rowed square matrix which we shall
l
1
designate
(25)
by A* and thus write
1
AA~ = A.. The converse may be shown to follow from (21). respectively. .
E
t
are
.
.
I
l multiplication by the scalar \A\~ and we observe that (27) is the inis terpretation of (2. if G is an type 1.5 imply = -\G\.
A
square ?>8j^m^A has an inverse if and only
^
if
A
is
non-
we apply Theorem 7 to obtain \A JA. (23).
tions
where the Pi are elementary transformation matrices.EQUIVALENCE OF MATRICES AND OF FORMS
45
simple consequence of our theorems which were obtained independently and which we should have wished to derive even had Theorem 7 been as-
sumed. or 3.
.
. .
.
|M_i
.
I
.22).
|Pi|
.P.
\
!
|
|
\
uniquely determined as the matrix
(26)
A-*
This formula follows from the matrix equation
(27)
A(adj A)
=
(adj
A) A = |A|
. (22).
\
any elementary transformation
.. 2.
|P. Clearly. We thus let A and B be n-rowed square matrices and see that Theorem 6 states that if A = then AB = 0.| =
6. An n-rowed square matrix if there exists a matrix B such that AB = BA
A is said to have an
identity matrix. 2. by Lemma 2. let A be nonsingular so that.
.
Compute
the inverses of the following matrices
by the use
of (26).
We note also that. that
C = AB
|
the
identity matrix. This is clear since in either
l
case \A\
^
0.
if
2. that is.
Show
1. adj
A
has rank
|
|
1. if A
(28)
B are nonsingular.
Finally. and similarly
BA =
and
/.
Use Ex. shall (1) is what we called a nonsingular linear mapping in Section 1.lA~ l ) lar we have
(29)
= A(BB~ )A~ = A A" =
l
/. But this is then possible if and only if A 5^ 0.
A~
l
of (26) exists.
=
B-+A-*
1
.
(AB)-*
l
For (AB)(B.9.
that
if
A
By
is
has rank
Hint:
an n-rowed square matrix of rank n 1 ?* = 0.
if
A
is
nonsingu-
(A-O'^A')/'
1
.46
INTRODUCTION TO ALGEBRAIC THEORIES
We now prove A~~
then
B
is
unique by showing that if either AB = / or BA = / 1 necessarily the matrix A" of (26). and adj (adj A) = A if n
|
adj (adj
n~2 A A) = A
n > 2
3. Use the result above and (28) to prove that 2 and A ^ 0. PAQ(Q~ 1 adj A) (27) we have A(adj A)
0. 4.
1
A
n =
linear
q
mapping
(1)
is
with
m=
(A.
if
A" = A^I = A~ (AB) = (A~ A)B =
1
l 1
IB =
B.
1
and
(27) to
show that
|adj
A|
= lA]^
1
.)' = (A-yA'.)' is the inverse of A'.
We
use this concept later in studying the equivalence of forms.
For
=
if
I
= (AA.
then adj
for
A
=
Then the
first
n
1
rows of Q"1
adj
A
\
must be
zero. if
|
1
(2)
with
is
the product (3)
the identical mapping.
EXERCISES
1. n has an inverse mapping
is.
(The details of an instance of this method are given in the example at the end of Section 8.
We
Theorem
EXERCISES
1.
Show
is
rank 2
that the product AB of any three-rowed square matrices A and B of not zero. Equivalence of rectangular matrices.30). Compute C and verify that (AE)C = I by direct multiplication.
=
7.2 shows that the elements of P and Q in (30) may be taken to be rational functions.
We
first
DEFINITION.
is
to be regarded here as simply another
definition of equivalence given
Theorem
the
9. The reader should notice that the definition of
rational equivalence
form of the above and.
= 2. with rational coefficients. B. we may obtain P by apillustrative
plying those elementary row transformations used in that exercise to I m and similarly for Q.
AB
for the following matrices.
Observe that
P
and Q are necessarily square matrices
of
m and n rows. Hint: There exist matrices P and Q such that A = PAQ has the
form
(2. if AB = 0. Hint: If A is m by n.
Compute the ranks
A.6 PAQ has the form (2.)
2.
respectively.
exist
make the Two m by n
matrices
A and B
are called equivalent if there
nonsingular matrices
P and Q
such that
(30)
PAQ = B
.
Two
m
by n matrices are equivalent
if
and only
if they
have
same rank. we have A*B* = where B Q B and may be shown to have two rows with elements
of
QrlB has
all zero. 1 of Section 2. that above is the one which has always been given in previous expositions of matrix theory. By Lemma 2 both P and Q are products of elementary transformation matrices and therefore A and B are equivalent if and only if A and B are rationally equivalent. while the previous definition is more useful for proofs.EQUIVALENCE OF MATRICES AND OF FORMS
5. Then.
emphasize in closing that. if A and B are equivalent. Hint:
into a simpler matrix
A Q = PA by row transformations
alone. Our considerations thus far have been devised as relatively simple steps toward a goal which we may now
attain. We may now apply Lemma 1 and have the principal result of the present chapter.
Carry A B into B Q = BQ by
. of the elements of A and B. the proof of 2.30) for r
the same rank as
3.
47
Let
A
be the matrix of
-1
by
(28)
(d) above and B be the matrix of (e).
Compute matrices
such that
P and Q for each of the matrices A of Ex.
made
for the case of bi-
by the presentation of certain notations which will
simplify our
The one-rowed matrices
(31)
x'
=
(xi.
a)
A=
2
3
4
. . and thus compute the rank
stead of that of
= P(AE)Q
in-
AB.
.
.
y'
=
z'(A')-
1
= x'(A~y
of (32) for y in terms of x (or y' in terms of x').
x m)
.
xn
is
the solution
(33)
y
= A-
1
x
.
. But then the solution of (1) for yi.
have called (1) a nonsingular linear mapping singular.
y n)
let
have one-columned matrices x and y as their respective transposes.
.48
INTRODUCTION TO ALGEBRAIC THEORIES
of AoB<>
column transformations alone.
.
y'
=
(yi.
.
. and shall by n matrices
We
shall again consider
m
A
now
introduce the no-
tation
(34)
.
x
=
P'u
.
. As
we indicated at the close of Chapter I. Bilinear
4^
1-1 0-2
5/
forms.
.
y n as linear forms in
xi. A be the m by n matrix of the system of equations (1) and see that system may be expressed as either of the matrix equations
(32)
We
this
x
= Ay
.
.
.
We
if
. We now precede
the introduction of those restrictions which are
linear forms
discussion.
m=
n and
A
is
non-
.
. the problem
two forms of the same
of determining the conditions for the equivalence of
is
restricted type customarily modified by the imposition of corresponding restrictions on the linear mappings which are allowed.
. and variables Xi and y.
x'
=
y'A'
.
B
B
1-213
5-83 (3-52
8.
. It follows also that every bilinear form of rank r is equivalent to
so that
B = PAQ
A
the form
(37)
zit/i
+
.
Xi to
new
variables Uk. so that the transpose P of P is a nonsingular m-rowed 1 = Um).
(35)
y
= Q
Q and
1.
if
. we write (u\. .
v'
=
(v\ t
.
.
. and we call the m by n matrix A the matrix rankle rank of/.
.
.
.
v n ).
.
. for
r m.
carried
by these mappings
/
is
B.
. its
(31).EQUIVALENCE OF MATRICES AND OF FORMS
for a nonsingular linear
fc
49
i.
. two bilinear forms f and g are equivalent if and only if their matrices are equivalent.
.
.
.
.
A
bilinear
form /
=
Zziauj//.
.
.
m and j ~
1. Similarly. Thus.
equivalent
there exist nonsingular linear mappings (34) and (35) such um and Vi. and are yn / by say g
Here x and y are given by
.
.
The matrix
f
of
/
is
2-3
-1
3
1^
W6
W
5
. .
of/.
. square matrix and u
=
1.
.
But
if
(34) holds. .
for a nonsingular n-rowed square matrix return to the study of bilinear forms.
.
and are equivalent.
We now
. By Theorem 9 we see that two bilinear forms are equivalent if and only if they have the same rank.
. xm m n B. then x'
=
u'P and
=
(u'P)A(Qv)
=
u'(PAQ)v
.
n.
.
. for i
=
.
mapping carrying the
.
ILLUSTRATIVE EXAMPLE
We shall find nonsingular linear mappings
(34)
and
(35)
which carry the form
into a form of the type (37). Then with matrix we shall that and ?/i.
. Also let g = x'By be a bilinear form in #1. v n into which / is that the matrix of the form in ui.
.
These results complete the study of the equivalence of bilinear forms. . is
a
scalar which
may be regarded
as a one-rowed square matrix
and
is
then the
matrix product
(36)
/
=
x'Ay
.
.
+xy
r
r
.
. .
x&i 2
2x22/2
.8
+
3i.
Use the method above to
mappings which carry the
o) 2x<yi
+ 3x# . add twice the new
first
We interchange the first and second
new second row.
EXERCISES
find nonsingular linear following bilinear forms into forms of the type (37).30) f or r
=
2
by adding
five
times the
column and
V
times the second column of
/I
PA
to its third column. 2
as desired.
We then add the second row to the third row.Jtt2 + Uz = -Ui .ftt2 ~ = ^3
Vi
2/2
4W3
. add
6 times the
new
first row to the row to the third row. and obtain
which evidently has rank
the
first
2.
the second row by
/I
and obtain
0-5\
i
PA=
o
\o
-y
o/
o
The matrix P is obtained by performing the transformations above on the threerowed identity matrix.
1.50
INTRODUCTION TO ALGEBRAIC THEORIES
rows of A.
I
i 2/3
= = =
vi
t>2
+
5^3
^3
We
verify
by
direct substitution that
(-6vi
-
30t. multiply
J. and hence
/
P-f-*
\
1
-1 -I -4
0\
o
I/
.
row by
1.
Then
5\
1
Q =
\0
V
17
-
The corresponding linear mappings
f
(34)
and
(35) are given respectively
(
by
xi
I
I
Z2
Z8
= .
We
continue and carry
first
PA
into
PAQ
of the
form
(2.
mapping on
.
n so that the matrices A of forms/ = x'Ay are square matrices.
We
A
and
B
congruent
if
there exists a nonsingular matrix
P
satisfying (38).
of the type (37).
.2x yi 2xiyi Xiy + 3xi2/ + .
a)
c)
Xii/i
-
4xi2/ 2
+
Zi2/ 8
-
x 22/i
+xy
2
2
d)
e)
3
2x42/2
-
Find a nonsingular linear mapping on 2/1.
Q =
4. Hint: The matrices A of the forms of Ex.
P.
4.X + 2X32/4 + 4X42/1 2xiyi
2
Zi2/a
8
22/i
2
2
2
s
3 2/i
2
zi2/4
2
32/3
X 4t/2
+
OJ42/3
+ 5X42/4
Use the method above to find a nonsingular linear mapping on Xi.
9. Then P = A" 1 AQ = / has the unique solution
3. 2 into forms of the type (37). x8 such that it and the identity mapping on 2/1.
(38)
B = PAP'
shall call
.EQUIVALENCE OF MATRICES AND OF FORMS
c)
51
d)
+ 3xi2/ ~ + 5x + 2x y + x y + 3x . 2/3 such that it and the identical xi. y2 y* carry the following forms into forms
2.
.
Ex.
Use elementary row transformations as above
6.
. There are some important problems regarding special types of bilinear forms as well as the theory of equivalence of quadratic forms which arise when we restrict the linear mappings we use.
Use elementary column transformations
to
compute the inverses
of the fol-
lowing matrices. = P' Then/ and g and (35) are called cogredient mappings if Q (34) are clearly equivalent under cogredient mappings if and only if
We let m =
Then
. 2/2.
to
compute the inverses
of the
matrices of Section
5. x 2 x3 carry the forms of Ex. X2. 2 are nonsingular so that the corresponding matrices P have the property PA = I.
-1301 10
2-3
1
-4
1
-1\
O
I
3
-1
Congruence of square matrices.
. c 2 i congruent to A and with Ci 2 =
For either
A =
= PAP'
t
a sequence of cogredient elementary transformations of type 2 where we add c/i times the second row of C to its jth row as well as c. An gruent to A and with a. Before passing to this study we observe that are congruent if and only if A imply that A and
B
Lemma 2 and Section 7 may be carried into B by
a sequence of operations each of which consists of an elementary row transformation followed by the corresponding column transformation.
.= hiz j^ 0. P.-) con= ^22 = 0.
if
B
is
is
symmetric
if
and only
if
A
is
symmetric.. then an = an for all and consequently every diagonal element of A
10. B = P .
Hence
is zero. 2 times the first row of C to its jth row for j = 3.)
OJ
..
r.
A
then B'
if
=
PA'P'..
We
shall speak of such operations
mentary transformations of
on a matrix A as cogredient elethe three types and shall use them in our study
of the congruence of matrices.
Skew matrices and skew bilinear forms. We thus obtain a skew matrix H = (/&.' P. We now apply 1.. Multiply ^12.
.
A =
(an) is a
an
=
skew matrix. .
.
i
B
is
skew
If
and only
a>a
A
skew. en = c 22 = 0. A square matrix A is called A. where the P* are elementary transformation matrices. and so deforth as 2 (P!APDP 2 P.
\0
O/
for every P and our result is trivial. We may interchange the ith row and first row. P.
. or some an T* 0. If B = PAP' is congruent to symmetric if A' = A.
We use
this result in the proof of
Theorem
the
Two n-rowed skew
Moreover
r is
matrices are congruent if
integer 2t..52
INTRODUCTION TO ALGEBRAIC THEORIES
We
shall
not study the complicated question as to the conditions that two arbitrary square matrices be congruent but shall restrict our attention
to two special cases. PI.) = 1.'. and thus obtain the skew
.
. PjAPJ
..
same rank
an even
-It
and only if they and every skew matrix is
thus congruent to a matrix
/O
(39)
It
0\
. the ^'th and second row and thus also the corresponding columns by cogredient elementary transformations of type 1.
.. hzi = the first row and column of H by h^ and obtain a skew matrix C = (c -. Thus = P = P. .
matrix
(40)
Ac
=
\
'
E_(0 *
\1
-1\
'
A.
and
j. and skew if A' =
10. n.
.
sired.
This is true for A = H if some diagonal element of A is not zero. = 0.
+
EXERCISES
Use a method analogous to that of the exercises of Section 8 to find cogredient linear mappings carrying the following skew forms to forms of the type above.
. and any cogredient elementary transformation on the last n and columns of A induces corresponding transformations on AI.
.
a)
b)
c)
+
11.
.
. We then obtain H as the result of adding the jth row of A to its ith row and
.
.EQUIVALENCE OF MATRICES AND OF FORMS
The matrix Ao
is
53
congruent to A. the corresponding bilinear form x'Ay is a skew bilinear form. then
G and A have rank 2t.
.
a
finite
number
we may
replace
A
by
a congruent skew matrix
(41)
G =
diag
{JEi. and we shall obtain only some of its most elementary results. = a< and an = a/.
if
/
is
a skew bilinear form of rank 2t
it is
equivalent under cogredient non-
singular linear
mappings to
(xiyz
x^yi)
+
. both A and B are con-
gruent to (39).
E
t.-.
.
. a. The theory of symmetric is considerably more extensive and complicated than that of skew matrices.
0. Moreover.
.
We may evidently assume that A = A' =^ 0.
rank 2. Hence. carrying it to a congruent skew matrix H\ and carrying Ao to a congruent skew
matrix
(E
VO
It follows that. Otherwise there is some a</ ^ 0.
.
80:22/4
-
Symmetric matrices and quadratic forms.
.
Every symmetric matrix
A is congruent to
a diagonal matrix
same rank as A. two such forms are equivalent under cogredient nonsingular linear mappings if and only if they have the same rank. If A is a skew matrix. Clearly.
0}
with each
trix of
Ek =
E.
A
i
is
necessarily a
skew matrix of n
2
2 rows rows. Our
matrices
principal conclusion
may
be stated as
Theorem
of the
11. and we shall prove first that A is congruent to a symmetric matrix H = (hij) with some diagonal element ha ^ 0. after
H
of such steps. If B also is a skew maB is congruent to (41) and to A.
We call the uniquely determined symmetric matrix A the matrix of f and its rank the rank of f.
.
oj n ).7.
Theorem 11 may also be applied to quadratic forms/ = As we saw in Section 1.
. We may then use the notations developed above and see that / and g are equivalent if and only if A and B are congruent.54
INTRODUCTION TO ALGEBRAIC THEORIES
A
to its iih column.
.
+ a x*
r
.
.
=
2a.
.
x r with nonsingular matrix diag
.
. For if our nonsingular linear mapping is represented by the matrix equation x = P'u. Now in Section 1.
. / is the one-rowed square matrix
product
(44)
/
= x'Ax =
u-i
for a
symmetric matrix A.
.J and g are equivalent if and only if B = PAP'.
.
.
. follow with the corresponding column transformation.
x n with be regarded as a a r j.
.
ar
.-*
+a
.
bilinear forms
/
= x'Ay
defined
Theorem
11 then states that /is equiva-
lent under cogredient transformations to a
(43)
aiXitfi
form
a rx ryr
. and it is clear that we ultimately obtain a diagonal matrix. Then we add ~~cTiCki times the first row to its ifcth row.
. and/ = u'(PAP')u.
0}. It is a matrix equivalent to A and must have the same rank..
0.
form
in XL.
it
may
{ai.
. Thus Theorem 11 states that every quadratic form of rank r is equivalent to
(45)
ap*
+
.9 we defined the equivalence of any quadratic form / and any second quadratic
form g = y'By. A\
in the proof of
symmetric bilinear forms of rank by a symmetric matrix A of rank
r.
form
in xi.j^
as de-
permute the rows and corresponding columns of H to obtain a congruent symmetric matrix C with en = ha 7* 0.
+
The
f(xi.
.
results of
.
.
.
ha
=
a.
r. We shall
.
However.
.
The form
matrix diag
(45) is of course to be regarded as a
{ai. and obtain a symmetric matrix
We now
/On
(42)
\
\0
Aj
1 rows.
. Clearly.
.
that
is.
the jth column of
sired.
. .
. .
.
+
.
. The result above may be applied to obtain a corresponding result on
congruent to A. As is a symmetric matrix with n Theorem 10 we carry out a finite number of such steps. then x' = u'P is a consequence.
x n with nonsingular matrix A a nonsingular form. 2 are congruent
we
num-
bers as elements of
6.J
21-1
1
1
~5
3. . (c).
14/
\-l -1
O/
and
Write the symmetric bilinear forms whose matrices are those of (a). (6).
2
e)
.
What
is
the symmetric matrix
a) 3x1
6)
c)
A
of the following quadratic forms?
3
2
+ 2xiX 3x x + x\ 2xix 8x3X4 + x\ . 3 to (e).3xJ x\ +
2
0:1X2
4
-
4
2.
and
(h) of
if
Ex. 2 for quadratic forms.
allow any complex
Which
of the matrices of
Ex.
Apply the process
of Ex.
matrices
Find a nonsingular matrix P with rational elements for each of the following A such that PAP' is a diagonal matrix. Consider mappings
+
of
/ and
of g.EQUIVALENCE OF MATRICES AND OF FORMS
call
.
EXERCISES
1.
5.
.
55
a quadratic form / = x'Ax in xi.
3
7
6
1
00-2-3
I
2
-2 -1 -3 -1 -3
0\
-"0023 -l
1
6
15
3
'-3-41
-4 -5
1
1
1
/
*>
3
2
1
2
1
-
5\
1
. . and we have shown that every quadratic form of rank r in n variables may be written as a nonsingular form in r variables whose matrix
. (/). (g). Determine P by writing A =
IAI' and by applying cogredient elementary transformations.
Pf
Show
linear
= xf x\ are not equivalent under that the forms / = xf x\ and g the possible signs of values with Hint: real coefficients.
is
a diagonal matrix. 2 and use the cogredient linear mappings obtained from that exercise to obtain equivalent forms with diagonal matrices.
(d) of
4.3x^2 + 2xiX . Ex.
6. as yet. the set of all real numa. just like those of F. been necessary to emphasize it. until now. Thus all our fields contain the field
of all rational
called
numbers and are what are
will
modular fields
called nonmodular fields. c 7* are. In our discussion of the congruence and the equivalence of two matrices A and B the elements of the transformation matrices P and Q have thus far always been rational functions. with rational coefficients.56
12. of all the set of all rational functions and the set complex numbers. We now make the fol-
set of elements F is said to form a nonmodular field if F contains the set of all rational numbers and is closed with respect to rational operations such that the following properties hold:
I. given conditions that two symmetric matrices be congruent. But the reader will observe that we have not.
.
III. Now it is true that even if one were interested only in the study of matrices whose
elements are ordinary complex numbers there would be a stage of this study where one would be forced to consider also matrices whose elements are
rational functions of x.
.
INTRODUCTION TO ALGEBRAIC THEORIES
Nonmodular fields.
A
numbers
fe. c of F.
A
II.
. a fc.
+ b) + c = a + (b + c) a(b + c) = ab + ac ab = ba a + b = b + a
(a
. ab.
algebraic concept which is one of the subject the concept of a field. b. and our reason is that it is not possible to do so without some statement as to the nature of the quantities which we allow
as elements of the transformation matrices P. Examples then. The fields be defined in Chapter VI.
field of
a.
. fields in of such F.
.
lowing brief DEFINITION. bers. of the elements of A and B.
for every a.
= F(x) of all rational complex numbers. it has not.
. While we have mentioned this fact before.
Thus we
shall find it desirable to define the concept
above.
If
F
is
any
field of
functions in x with coefficients in
erties. But it is clearly possible to obtain every rational number by the application to unity of a finite number of rational operations.
(ab)c
=
a(bc)
.
of a field in such a general way as to include systems like the field defined shall do so and shall assume henceforth that what we called con-
K
We
stants in Chapter
I and scalars
thereafter are elements of
all
a fixed field F. the set F is a mathematical system having prop-
K
with respect to rational operations.
.
We
shall thus introduce
an
most fundamental concepts
of our
complex numbers is a set F of at least two distinct complex and a/c are in F for every such that a + &.
The
fields
we have already mentioned
contain the complex number
unity and are closed with respect to rational operations. with rational coefficients of any fixed complex number c.
l.
=
ab for every a and b of a
field
F.
EXERCISES
1.a ~ and a unique solution y = 6" of yb = 1 for 6^0.
1
+
1
a
+ +
=
a)
(
1
6)
a
+
Hence
=
It is
(a) =
a. of (47) to that of by = a. Then the solutions = ab~ of (46) and (47) are uniquely determined by z = a + ( b).
made
as
.
and the quotient a/6 to be a solution y
(47)
F of the equation*
. and
a
(
thus
(
1 a. Prove that F is a field.
Show
that
if a. and the existence of a solution of (46) is equivalent to that of b x = a.
'
r
"-
x
/a
b
^(b
a)
b
2.
whereas
a.
F and that there exists a unique solution x = aoix -}.
We
1
also see that the rational
number
a
1 is
defined so that
l)a = also true that
(
(
1)
1=0. In the author's Modern Higher Algebra it is shown that the solutions of (46) and (47) are unique.
the
+ bi
di
3(c
-)
a
Note that in view of III the property II implies that (b c)a = ba ca. c. and for these systems the properties just mentioned must be additional assumptions.
+
In fact. 6. (Use the definition of addition of matrices in (52). Let a. y
1
l
.
aO
=
0..
yb
=
a
Thus our hypothesis that F is closed with respect to rational operations should be interpreted to mean that any two elements a and b of F determine b and a unique product ab in F such that (46) has a a unique sum a in F and solution (47) has a solution in F if b j 0. and c range over the set of all rational numbers and F consist of all matrices of the following types.EQUIVALENCE OF MATRICES AND OF FORMS
The difference a
tion x in
(46)
57
6 is always defined in elementary algebra to be
a solu-
F
of the equation
x
+b=
in
a
.)
a
N
/a
-2b\
'~
.
+
l)a
=
a
=
1
0.
set of all matrices of the following kind
and d range over all rational numbers and i 2 form a quasi-field which is not a
/a
\c
=
field. But there are mathematical systems called quasi-fMs in which the law ab = ba
*
+
+
+
does not hold.
it
may
be concluded that the rational numbers
and
1
have the
properties
(48)
for every a of
a
+
=
al
=
a
. 6.
and of two bilinear forms under cogredient mappings. corre-
spondingly. When A = -A the matrix A is skew. Summary of results.58
INTRODUCTION TO ALGEBRAIC THEORIES
13. Moreover.
.
and these
results
may
be seen to
vary as
*
we change our assumptions on
F.
. The theory completed thus far on polynomials with constant coefficients and matrices with scalar elements may now be clarified by restating our principal results in terms of the concept of a field. the precise nature of F is again unimportant.
.
f
1
Then two corresponding quadratic forms x'Ax and x Bx
f
are equivalent in
and only if A and B are congruent in F. of two bilinear forms.
. Hence. the particular F chosen to contain the
elements of the matrices
If
is
relatively
unimportant for the theory.
We
observe
first
that
if
f(x)
and
g(x) are nonzero polynomials in x with
coefficients in a field
coefficients in F.
of
F
to the reader the explicit formulation of the definitions of equivalence in linear
. we can obtain results only
ratic
The problem
after rather complete specialization of F. we have shown that every symmetric matrix of rank r and elements in F is congruent in F to a diagonal matrix diag (a a a r 0.
.
Moreover. in F and 0} with a* ^ that correspondingly every quadratic form x'Ax is equivalent in F to
F
if
.
We leave
two forms.
.
. two bilinear forms with coefficients in F are equivalent in F if and only if they have the same rank. then we call A and B congruent in F if there exists a nonsingular matrix P with elements in F such that PAP = B.
. Let A = A' be a symmetric matrix with elements in F so that any matrix B congruent in F to A also is a symmetric matrix with elements in F. Then A and B
a
field
are equivalent in
F if and only if they have the same rank.
of finding necessary and sufficient conditions for two quadforms with coefficients in a field F to be equivalent in F is one involving the nature of F in a fundamental way. In fact.
. Similarly.
F then they have a greatest common divisor d(x) with and d(x) = a(x)f(x) + b(x)g(x) for polynomials a(x). and no simple solution of this problem exists for F an arbitrary field. and every matrix B congruent in F to A is skew.
A and B are square matrices with elements in a field F. two skew matrices with elements in F are congruent in F if and only if they have the same rank. b(x)
with coefficients in F. Since the rank of a matrix (and of a
corresponding bilinear form) is defined without reference to the nature of the field F containing its elements.
We next assume that A and B are two m by n matrices with elements in F and say that A and B are equivalent in F if there exist nonsingular matrices P and Q with elements in F such that PAQ = B. where all the forms considered have elements in F. we say that the bilinear forms x'Ay and x'By are equivalent in F under cogredient transformations* if A and B are congruent in F.
Let F be the field of either Theorem 12 or CoroUary I.
over an arbitrary
state
it.
0.
.
are congruent in F if and only if they have the same rank.
(ay)
.
. = A' of rank r is congruent to ar .j =
1. 0} >
F
A
A
. 0. If these matrices were one-rowed square matrices.
.
. For every .n)
.
. = a 6? for in F. so that
A
and
B
are
m by n matrices.
.
.
. = diag ^{ai. . Addition of matrices. Then two quadratic forms with coefficients in F are equivalent in F if and only if they have the same rank.
. We then have the obvious consequences. . m and j We now let A = (ay) and B = (6</)? where i = 1. .
59
conditions are those given in Let F be a field with the property that for every a of F there exists a quantity b such that b 2 = a.
.
numbers
if
and only
if they
COROLLARY II.
. Thus (51) will have major
importance as a formula for representing matrix computations. 0}.
=
. then 5 is 0} and hence to A.
x?
+
.
.
PA
F
P'
=
diag [I r .
.
=
.
0}.
. Hence every such form of rank r is equivalent
in
F
to
(49)
14. for and a.
.
1.
Two symmetric
matrices whose elements are complex
numhave
bers are congruent in the field of all complex the same rank.
^
P
-
67
1
.
. we should have the formula
But
it is
also true that
if
the partitioning of any
A
and
B is
carried out so
that the products in (51) have meaning and if we define the sum of two matrices appropriately. If also
.EQUIVALENCE OF MATRICES AND OF FORMS
The simplest Theorem 12.
Its
There is one other result on symmetric matrices which will be seen to have evident interest when we proof involves the computation of the product of two matrices
field
which have been partitioned into two-rowed square matrices whose elements are rectangular matrices.
.
. Then two symmetric matrices with ele-
ments in
.
COROLLARY
I.
Then we
define
(52)
S = A
+B =
SH
(i
= an
1. that is to say in F.
also congruent in
to diag {/ r
B' has rank r. The converse follows
B=
from Theorem 2. then (51) will still hold.m.
. Then if = diag {&rS 7* 6< .
+ &/
.2.
.
n.
.
+ x?
.
also holds.
Observe. #2 has rows* and n rows and g columns.
. then \A\
+ \B\ are usually not equal.
For
this equation is clearly a consequence of the corresponding property
is.
We observe also that A + (~1-A) = 0. . J5 4 has n
t
t
rows rows
. that
(55)
of
(an
+
&</)<.
A B
4
8
has
m
t
s
has n
t columns. For example.
. s and j = 1. t.
.
. .
The elements
If
of our matrices are in a field F. Then
(A
+ B)C = AC + BC
. Clearly
if
We
D
is
a matrix such that
DA
is
defined. then A B\ and \A
|
and
A
and
B
are n-rowed square mat-
+
\
if
A
and
B are
equal. B = (&.
C =
(cjk)
be any n by q matrix. that if n rices.
and
q
g columns. Then so that AI = (a*/) but now with i = 1.Cjk
have thug seen that addition of matrices has the properties (53) alfor addition of the elements of our matrices and that the assumed ways law (54). Our partitioning is now completely det columns. then
A+
=
A. we have Hence we have the properties
also
+ by = 6/ + a/. and we thus assume that Bi is a matrix of t rows and g columns./) be an mbyn matrix./*
= a^k
+ bi. however. AS has m s termined and necessarily A* has s rows and n
rows and
t
columns.
of F.
an m by n matrix.
and a^
C =
(dj) is
(bij
+ dj). We let A = (a.-.
Now
(54)
let
is
m
by n
zero matrix.
.60
INTRODUCTION TO ALGEBRAIC THEORIES
that
We have thus defined addition for any two matrices of the same shape such A + B is the matrix whose elements are the sums of correspondingly placed elements in A and B. + c*/ = an + (a</ +
6.
if
(A+B) + C = A +
the
(B
+
C)
. then
similarly
(56)
we have
D(A
+ B) = DA + DB
>
1
.
.-)
(53)
A+B = B + A.-*) be an n by q matrix. which we call the distributive law for matrix addition and multiplication. and A\ be an s by t matrix.
(51) has meaning only if B\ has t rows. + \B\ = 2|A| while |A + B\ = |2A| =
Let us now apply our definitions to derive (51).
For if AI and BI are congruent in F there exists a nonsingular matrix Pi such that PiAiP( = BI. Then P = diag {Pi.
.
B=
JE?i.
.
AB =
DJE.r } is nonsingular. 2.
.
D =
2
Ei
-
(fit.
k
=
1..
.
Then
Ho'
F
if
A
and
B
are congruent in
and only
if
AI and BI are
congruent in F. In matrix language
we have used
(58)
and
computed
as the result of partition of matrices of matrices in (59) to give (51). Theorem 13.
where we define Di.
.
.
.
obtain (51) from (57) by simply using the ranges 1. 2.
.
m
.
.
.
+ Drf*
.
(^}
.
and then have used
(57)
and addition
We shall now apply the process above to prove the following theorem on
symmetric matrices mentioned above. I n . .
may
.
ro.
s
.EQUIVALENCE OF MATRICES AND OF FORMS
and
q g columns.
61
The element in the ith row and
fcth
column of AB
may
clearly be expressed as
(i
=
1.
E =
2
(B.
.
B.
(58)
Di
=
.
.
. and A and B be the corresponding n-rowed symmetric
matrices
<>
of rank
r. g and
.
B
4)
Moreover.
q for j separately.
q)
But
this equation is equivalent to the
matrix equation given by
(57)
A = (A D2 )
Z>2.
.)
. .
Ho'S). Let Ai and BI be r-rowed nonsingular symmetric matrices with elements in F. we
. for i separately. as well as 1.
+
1.
.
s
1.
#2 by
2
. and com-
.
and
.
g
+
.
.
Conversely.
But then
|
B
1
PA
1
1
P(
J
|
5^ 0. symmetric matrix B and a skew matrix C. Hint: Put A == B (7 = 0'.p(Af( Af' \_(P l A
P'J'
~MO
P B
l
J-^A^
P(
P^
PtA
must be nonsingular since \Bi\ = and A l are congruent in F.
Computed
+ Bif
A
a)
B=
2.
Verify that (A
+ B)' =
A'
+
B' for any
m by n matrices A
and B. compute A'. g may be written as a nonsingular form g Q in r variables. and finally the original forms / and g are equivalent in F if and only if / and g Q are equivalent in F.
Theorem
same index. then / may be written as a nonsingular form / in r variables. and we
have
l
p *\
P'*
t PA p.
if
putation by the use of (51) gives PAP' a nonsingular matrix P we may write
I
PAP = B for
7
Pi
p
*}
\p*
p*)'
shall
where Pi
is
an r-rowed square matrix.
and Hence
ORAL EXERCISES
1.
+
rices
Real quadratic forms.
3.. if / and g are quadratic forms with coefficients in F so that / and g are equivalent only if they have the same rank r.
of a
Show that every n-rowed square matrix A is expressible uniquely as the sum C with B = B'.
numbers
quadratic forms with real coefficients are equivalent in if and only if they have both the same rank and the
.
14. We now call the number of positive a the index i of / and prove
15. We shall close our study of symmetric matand hence of quadratic forms with coefficients in F as well by consider= x'Ax ing the case where F is the field of all real numbers.62
INTRODUCTION TO ALGEBRAIC THEORIES
=
B. and solve.
Two
the field of all real
. Let then / = a x xf + have rank r so that we may take / + arxl for real a^ ^ 0. A! l |PJ |PJ| The result above thus states that.
.
propose to show that s = t.
.
Now. 1.
. Moreover.
.
. dt+i.
t
. Conversely.
We have now shown that two quadratic forms with real coefficients and
if
the same rank are equivalent in the field of all complex numbers but that.
\
numbers p
{PT with
(61)
1
>
^
P7
1
}
>
such that p* = diy and if P is the nonsingular matrix diag then PAP is the matrix diag {1. that = n.
.
+ ul) ^
0.
.
.
. they are inequivalent in the field of all real
numbers.
.
But
clearly
ft
=
(ttf +1
+
.
t
elements
1
and
(x\
r
t
elements
1.
.
x n).
.
We
Hence.
. we obtain as a result
(y\
+
=
.
.
Put
Xi
=
x2
=
. -1. let s > t.
+d
and
in
yn
(i
=
1.
in
.
t.
.
.
These are
exist real
linear
homogeneous equations
v\. +1
..
+ cky.
Thus.
The remaining n
certain
/(O. .
. as = t = 1. .+
*2)
.(J +l +
.
.
.
+ yj) =
yn
(yj +1
+
+ yj).
a contradiction. that by Section 14 there is no loss of generality we assume that/ = x'Ax where A is a nonsingular diagonal matrix.
.
.
.
.
/
.
.
.
= f(x
l9
.
. /
is
equivalent in
F
to
+
. there is clearly no loss of generality if we take A = is.
.
*)
.
.
n)
.
.
tO =
t>f
+
. .
.
..
>
t
unknowns.+*).
w l+1
.
have rank r
it is equivalent in F to (61) and = n and index s so that g is equiva-
F
to
(62)
(*?+. =
in s
(t
=
1. r d d r for positive d. -1}
1
.
..
.
.
0.
0.. . if 0has
lent in
the same rank and index as/. first.
.
.
. and we have the result h
+
. Then there exist real diag {di. We shall next study in some detail the important special case t = r of our discussion.
.
=
in (63)
and consider the
(64)
resulting
t
equations
AMI
t
+
.
.
.
.
+
*?)
-
(* +1
+ .
.
.
. .
. and there
numbers
v9
not
all
zero
and
satisfying these equations. .
.
+
0J
>
0.
numbers
. Our hypothesis that the form / defined by (61) and the form g defined by (62) are equivalent in the real field implies that there exist real numbers d^ such that if we substitute the linear forms
(63)
Xi
=
dnyi
+
.
.
. There is clearly no loss of generality if we assume that s ^ t and show that if s > t we arrive at a contradiction.EQUIVALENCE OF MATRICES AND OF FORMS
*
63
if
For proof we observe.
. n.
+3
.
Uj for j
equations of (63) then determine the values of x.
.
let g
hence to /. their indices are distinct.
.
=
xt
=
y.
is positive semidefinite. positive definite..
cn)
dr =
0. .
real quadratic form f(xi. . .
=
+
+
.
. yi+i
r. for all real ci not all zero.
.64
INTRODUCTION TO ALGEBRAIC THEORIES A
symmetric matrix
A
and the corresponding quadratic form /
if
=
x'Ax
are called semidefinite of rank r
A
air
is
congruent in
\
<>
F to
a matrix
. . there exist unique solutions y. and we obtain g from / by putting Xj = in / for j 9^ i*.
.
0.
j/J. .
.
>
is
= 1 and all other have n. . of A and whose corresponding columns are in the corresponding columns of A.
.
.. Then A and / are nega-
and only
if
A
and
/ are
positive.
. if /(ci. Hence.
.
.
r
=
<
= d\ + = and y/ =
+
. Conversely.
Theorem
. . if /(ci.
. We call A and/ definite
F is
if
=
n.
. every principal
is positive definite.
in F.
y
t
t
=
0.
=
t
l. if / <>
.
. c n ) = 0.
.
cn )
Now if < r.
Every principal submatrix of a positive semidefinite matrix submatrix of a positive definite matrix
For a principal submatrix B of a symmetric matrix A is defined as any im th rows m-rowed symmetric submatrix whose rows are in the tith.
. .
+
. then we put y r+ \ c n not all zero such that /(ci.
.
.
=
.
iff
+
if
cn) d\ y\ 2/ n /(ci.
f
. If f(xi. the form/ is positive definite.
. . we may take a = 1 and call A and / and call A and / negative. then g ^ Clearly. .
+
>
We
have proved
x n ) is positive semidefinite for all real Ci. . that
A
is
both semidefinite
and nonsingular. Conversely. for all real c* not all zero.
If
Ci.
equivalent to a form a(x\
is. .
for y\
. cn ) > .
.
. is positive definite if and only .
A
and only
if f (ci. d%
. x n ) is any real quadratic form.
If
the field of
all real
positive or take a
tive
=
1
numbers.
d.
. we have/ < < 0.
/(ci.
for a 5^
.
. .
=
.
cn )
^
Theorem
16.
.
. As a consequence of this result we shall prove
15.
+
+
if
!/f
(y\+i
we put
Xi
ing system
if
in (63).
.
.
.
<
.
.
Thus we may and
shall re-
our attention to positive symmetric matrices and positive quadratic forms without loss of generality. = dj of the resultof linear equations.
.
For otherwise / =
If
y\
+
. if f (ci.
./(ci.
.
cn )
.
cn)
<
then
. we have / for all dj not all zero and hence for all c not all zero. .
.
.
cn
are
any
real
numbers and
and only if the
all zero. z m ) so that g = x Q Bx Q is the quadratic form with B as matrix.
. Put XQ = (x^. .
t
.
. for all values of the x ik and for all z = a.
. and the dj may readily be seen to be all zero
c
= d
++!/?)
are
.
.
.
Thus /
is
semidefinite
if r
if it is
+
+ xj). we have seen that there exists a nonsingular transformation (63) with real d^ such that / = y\
strict
.
and we partition QQ' so an r-rowed principal submatrix of QQ'. Let
A be an by n matrix of rank r and with real elements.
is
congruent to the positive semidefinite matrix
C
of rank r and hence has
the property of our theorem. respectively.
The converse of Theorem 16 is also true.
Sec-
. We shall use the result just
obtained to prove
Theorem
Then AA' For we
is
17. a positive definite symmetric matrix. / = g ik
A
all
zero
and
for the x ik not all zero.
write
m
may
-p AA P (I'
(o
<>\o Q
>
o)
00' QQ
where
P
is
and
is
Then QQ'
that Qi
Q are nonsingular matrices of m and n rows.EQUIVALENCE OF MATRICES AND OF FORMS
hence
65
B
is
and B is Xj above
is positive definite positive semidefinite by Theorem 15.
EXERCISE
What
tion 11?
are the ranks
and
indices of the real
symmetric matrices of Ex. If = for x not all and hence for the then singular. a positive semidefinite real symmetric matrix of rank r.
2. a contradiction. zero. By Theorem 16 the ma-
trix Qi is positive definite. and we refer the reader to the author's Modern Higher Algebra for its proof.
A subset L of V n is called a linear subspace of V n if bv is in L for every a and b of F.
and we have
w
+ =
.
zero. . the quantities Ci.CHAPTER
The
(d.
over F.
1
Then
u
of F. 6 in
F and all vectors w. w of Fn The vector which we designate
all
by
(4)
O.
u+(u)
u = u
.
.
.
and
(5) of
Vn to the reader.
.
Linear spaces over a
field.
.
. We assume the laws of combination of such sequences of Section 1.
. The entire set V n will then be called the n-dimensional linear space
. (4).
of all sequences
(1)
U =
.
.
set
.
Linear subspaces.
We
the verification of the properties
2.
The
(2)
properties of a field (u
F may
u
be easily seen to imply that
+
v)
+w=
.
Cn)
may be thought of as a geometric n-dimensional space. c n the coordinates of u.
=
1
u
. v.
+ amum
.8 and call u a point or vector. and w^ in L.
+v=v+u
a(w
(3)
a(bu)
=
(o6)w
(a
+ V)u =
+ bu
. Then it is clear that L contains all linear combinations
au
+
(6)
u =
a in F.
aiUi
+
.
for
. (3).
w =
first
.
where
(5)
u =
(
Ci.
Note that the
zero vector.
+
(v
+ w)
aw
. of Vn We suppose also that the quantities Ci are in a fixed field F and call c the ith coordinate of u.
u
.
.
Vn
. and we leave
(2).
+
v)
= aw
+ aw
for all a.
of (5)
is
the quantity
and the second zero
is
the
shall use the properties just noted somewhat later in an abstract definition of the mathematical concept of linear space. every u and v of L.
cn ).is
that vector
of
whose coordinates are
u.
IV
LINEAR SPACES
1.
.
=
Q.
L-lm. e n span Vn The space spanned by the zero vector consists of the zero vector alone
..
+
. ui. Our definition of a linear space L which is a V n implies that every subspace L contains the zero vector.
.
Vn = e\F + L spanned by
+
..
. Um are not linearly independent in F. If Ui.
..
there
is
no other such expres-
sion of
is
ai
um are linearly independent if it Thus.
.
It is clear that.
F andui..
. Um
. then = Oui
. But we may actually show a finite number of vectors of Vn has a that every subspace basis in the above sense.
.
.
u m a basis
over
F
of
L and indicate
(8)
L =
UiF
+
.
+
(am
b m )um a<
&<
{ui. that the definition of linear indeIt is evident that
.
. . .
In what
. . . .
+
. ..
.
um
is
span the space
(7)
L and
so defined.
if ui. If Oum.is the vector whose jth coordinate is unity and whose other coordinates are zero.
. am um = biui
. .
.
.
. then ei.. If now
any
finite
num. Um are now seen to be linearly independent in
. .
aiUi
+
.
.
.
+
Conversely.
.
. .
um are a
set of
linearly independent vectors (of
Vn over F).
subspace of
.
DEFINITION. if
.
u m in
}
the
form
(6)
is unique.
m
.
.
shall say that
. we shall indicate.
. if and only if true that a linear combination ami (J^um = = = a 2 = .
.
.
. Then we shall
by writing
call Ui.
+
.
.
um are linearly independent and + 6mum then = (ai .
. to this definition.
Linear independence.
L = L is expressible
.
F
expression of every
u
.
.
in the
form
We
. for the time being.
. . Hence the zero vector of {MI.LINEAR SPACES
67
(6) of
We observe that the set of all linear combinations
ber
of given vectors.
independence as the spe-
cial case
u =
u
. Observe. shall write
is
L
a linear subspace L of Vn according we shall say that u\. u^.
in the
form
(6).
a
=
}
. when we call L a linear space over F that L is a linear subspace over F of some V n over F.
+
ui.
.
We now make the
.
and may be
lows
called the zero space
and designated by
L =
{0}.
{ui.bi)ui +
.
.
+ u mF
.
um
over
.
we
shall restrict
}
3.
=
.
.
. u m are linearly independent in F or that ui. first.
eJF.
(6) in at least this one way.
For
this property clearly implies linear
0.
and only
if the
they are linearly dependent in F.!!*}.
of
L =
.
fol-
our attention to the nonzero subspaces L of Vn and. .
=
.
.
. if e.
.
. am 0.
.
this
LetL-
.
& as desired.. Ui.
we
if
shall say that
A set of vectors Ui. .
.
um
be linearly inde-
pendent in F.
.
.
+
0. .
.
so + bkrvr for k = 1. But then
g k)
. now.
. 7.
. Let Abeanmbyn matrix of rank r.
P. v\F + Define B = (bki) and obtain Wk as the fcth row of the r + 1 by n matrix BG. H is an m by r matrix.
By Lemma
1
we have
LEMMA
The row spaces
2.
.
Let
P
be
an m-rowed nonsingular matrix. Thus we have
LEMMA
of
1. Any r + 1 vectors of the row space of A are linearly dependent in F.
.
b rv r sired fashion.
. .-x)
clearly 61.
. Thus.
.
. such that
(11)
PA =
(\
. they are a basis of the row space of A.
.
. By Theorem 3.
.
we
The determinant
of the r-rowed
square matrix
(12)
P = (%)
.
such that the
+ l)-rowed matrix D = (d for g.
. k = 1.6 the elementary transformation theorem as the useful
we used the matrix product equivalent of Theorem 2. and hence P is nonsingular. that Wk = bkiVi + = that Wk are any r + 1 vectors of the row space L + vr F of A.70
lar. + b rv r Hence the rows of G are linearly independent in F.
of
G
and
A
coincide.
zero rows. and by Lemma 2 there
is
.
.
.
INTRODUCTION TO ALGEBRAIC THEORIES
then the result just derived implies that the row space of A = P~ l (PA) is contained in the row space of PA and therefore that these two linear spaces are the same. and we shall now state this
LEMMA 2. Assume. r + 1.4.
P =
2
(0. + (r l)st row of D(BG) is the zero vector.
&.
+
+
=
for bi in
F
not
all zero. implies that there is no loss of generality if we permute its rows in any de-
Theorem
. if b\v\ may assume for convenience that 61 9* 0.
.
. But this is impossible since biv\ + is the first row of PG. Then
the
row spates
PA
and
A
coincide. they span the row space of G and of A.
.
of
G
span the row space of PA.
. . G is an r by n matrix. We use (11) and note that the rows of G differ from those of PA only in
Then the rows
3.
exists a nonsingular (r
r
+
1. respectively.
We
shall use this result in the proof of
The r rows of G form a basis of the row space of A.
AQ =
(H
0)
.
.
.)
.
=
(fc.
where
G and H have rank r. Then PG is r-rowed and of rank r. v r Our definition of G For we may designate the rows of G by v\.
In the proof of Theorem 3.
.6 the rank of BG is at most r.
.
.
.
. Then there exist nonsingular matrices P and Q of m and n rows.
.
r
D=
(d gk )
such
. After a permutation of the rows of A.
of the nonsinguwr+i are linearly
dependent in F. if u r are a basis of the row space of A.
.
=
row
.
0.
. u\. L = UiF + By the proof of Theorem 1 the vector u is in \Ui. then Theorem 2 implies that s < r and similarly that r < s. 2 to obtain a nonsingular r-rowed matrix
s
<
r. m. Then u k = b k \ui + + b kr u r for b ki in F.
. dr iUi
.
matrix
P
given
by
(B
is
Dif
nonsingular.
order of
L
over F. necessary.
.
If s
<
. and by Theorem 2 any n 1 vectors of eiF are linearly dependent in F. Thus any linear subspace L over F of n contains at most n linearly independent vectors.
.
.
.
r
. + urF. r+iWr+i
r
lar
matrix
D
and cannot
all
the dr +i tk form a be zero.
. The matrix G of Lemma 2 may be taken to be a submatrix of A.
.
.
exists
a
.
.
. The
. It follows that s is the rank of G. u are linearly dependent in F for every M of L. r = s is unique.
+
drr u r
u r are linearly indecontrary to our hypothesis that w 1. It follows that there
.
.
.
.
. This proves Theorem 2. we may assume that ui.
not zero.
. Any two bases of L have the same number r < n of vectorsj and we shall call r the
. We may now obtain the principal result on linear subspaces of F.
+
+
Vn
V
maximum number r < n of linearly independent vectors HI. ur in L.
.
r
.
. iWi
71
+
. This completes our proof. .
. Theorem 4. For let A be an m by n matrix whose ith row is Ui and let r be the integer
m
of
Theorem 1. But if also L = ViF + + v. The integer r of Theorem 1 is in fact the rank of the by n matrix whose ith row is Ui. pendent.
. and k = r + 1.
.F.LINEAR SPACES
dr+i.
. the vectors 101.
.
For
Vn =
+
.
e n F.
.
. u }.
the rth row of
D
is
.
. We now apply Theorem 1 to obtain Theorem 3. and that w. Every linear subspace L over F of V n over F has a basis.
.
.
+
.
.
row
of
DG =
0. and
it is
clear that
(14)
G =
Ur
then
PA
=
is
we apply Lemma
that the rth
given by (11).
.
+ d +i. the rank of A be s.
v 2
if
of Ex.
of the corresponding matrices
A are equal to the order of the subspace Li.
a linear subspace of
Vm We
of A. Thus we shall call the row space of A' the column space of A.
3.
vectors
v2 ]
if
u\. their transposes.
1
and 2
of Section 3
by the use
of elementary transformations to
compute the rank of the matrices whose rows are the given vectors.
Show
that then there exists a nonsingular matrix
P
such
PA
Give also a simple form
of Ai.\
A. the basis to
consist actually of rows of the corresponding matrix.
Section
Form the four-rowed matrices whose rows are the 3. i^} = L2 = {vi. It is
call the order of the row and column spaces and column ranks of A. v\. Show thus that LI = [HI.)
that the rows of A* are in the row space
Show
.
for P.
2.
. that the rank of the matrices formed from the first two rows of each A.
1-2 2-4
10 20
1
2301
-2 -1 -1
1
2-3-1
4-7
-2-1
11
1
2-3-1
04-12 212-2
3
3
-1001
0-11
c)
230-1
1
5100 1234
be a rectangular matrix of the form
12-1 35102 11 3-6-6 47036
4
1
7
-15 -16
4. Hint:
Mi
VO
A. the row of is its row rank. Also
A
f
Theorem 5. By Theorem 3 the rank A and A have the same rank and we have proved The row and column ranks of a matrix are equal to its rank. indeed.
and only
the ranks
is. u^.
Find a basis
of the
row space
of each of the following matrices. 3.
Solve Ex. respectively.
Let
A
Mi Ui
where A\
that
is
nonsingular.
EXERCISES
1.72
INTRODUCTION TO ALGEBRAIC THEORIES
are uniquely determined
In closing this section we note that the rows of the transpose A* of A by the columns of A and are.
Then we call (15) a one-to-one correspondence on G to G'. 4 be either symmetric or skew.
G and
G' are the
and
(16) are. In elementary algebra and analysis the sys-
tems G and G' are usually taken to be the field of all real or all complex numbers and (15) is then given by a formula y = /Or).
The concept of equivalence. In discussing the properties of mathematiand
linear spaces over
cal systems such as fields
F
it
becomes desirable
quite frequently to identify in some fashion those systems behaving exactly the same with respect to the given set of definitive properties under conshall call such systems equivalent and shall now proceed to sideration.
same system.
also that. the matrix ^5
=
0.
if
the order of A\
is
the rank of A.
Then / is given by
(15)
g->g'=f(g)
(read g corresponds to g'} such that every element g of G determines a unique corresponding element g' of (?'. This concept may be seen to be sufficiently general as to permit its extension in many
directions.
a correspondence such that every element g of G is the corresponding element f(g) of one and only one g of G. But the basic idea there is that given above of a correspondence (15) on G to (?'.
We
define this concept in terms of that of function. however. in general. it
is
f
more convenient
in general situations to say that
the range of the dependent variable.
Note.LINEAR SPACES
5.
73
Let the matrix of Ex. How/ is a function
on
G to G
or that / is a correspondence on
G to G'. Show that the choice
of
P then implies that
Show
6. and we may thus call (15) a one-to-one correspondand G' and indicate this by writing
f
g+
if
><?'. that. Let G and (?' be two systems and define a single-valued function
to G'. distinct. the functions (15) Thus we may let G be the field of all real
.
/ on
G
In elementary mathematics
it is
customary to
call
G
the range of
the independent variable
and
G
f
ever.
which
now on G
ence between
(17)
G
to G. It is clear that (15) then
(15) is
r
Suppose now that
f
defines a second one-to-one correspondence
(16)
M=
is
0'-<7.
) We then call G and G' equivalent if there exists a one-to-one correspondence between them which is preserved under the operations of their definition. have thus introduced two instances of
V
V and V
what
is
o
are equiva-
We
a very im-
portant concept in all algebra.74
INTRODUCTION TO ALGEBRAIC THEORIES
numbers and (15) be the function x 2x. A. (gh)' = g'h' for every g and h of F.
closed with respect to certain operations. The reader should observe that under our definition every mathematical system G is equivalent to itself and that if G is equivalent to a system G'.
proceed to use the concept just given in constructing the fundamental definition of this section.
(au)
=
au
for every u and v of and a of F. that of equivalence. then
G
is
equivalent to G".
Verify the statement that the field of
all
rational functions with rational coeffifield of all
cients of the
complex number
V2
is
equivalent to the
matrices
'-CD
for rational
*
a and 6 under the correspondence
A
<
>
a
+
We
of u' for the arbitrary vector of
use the notation Vo instead of to avoid confusion in the consequent usage as well as for the transpose of u. so that (16) is the function x \x. Then
we shall call V a general linear space over F. These systems consist of sets of elements #. if G and G' are distinct it is not particularly important
whether we use (15) or (16) to define our correspondence. We now see that we have defined two fields F and F' to be equivalent if there exists a one-to-one correspondence (15) between them such that (g + h)' = g' + h'. Finally.
We
linear spaces. G' of the same kind such as two fields or two linear spaces over a fixed field F. Of course.
if
G'
is
equivalent to G". then we shall say that lent over F. we might have g + h in G for every g and h in G and also g' + h' in G' for every g' and h' in G'. Let us consider two mathematical systems (?.
Let F be a fixed field and V and au are unique elements of
consist of a set of elements such that
u
+v
V
for every
u and
If
v of
is
V and
a of F. .
then G'
is
equivalent to G. and there is a one-to-one correspondence u <
that
(u
7
a second* such space UQ between V and V such
+
0)0
=u
<>
+
^o
.
. for example.
V
V
. (Thus. Let us then pass to the second case which we require for our further discussion of
.
EXERCISES
1.
(4). then the properties (2).
+
.
a)
a and
6. (4).
= ua and u v to be in V for every u.
proof of
Theorem
over
6. define au
it
a of F.
Cn)
. v.
where the equivalence between
(19)
CiVi
V and V n is
<
>
given by
. we prefer instead to define V by its equivalence to Vn This preference then requires the (somewhat trivial)
Conversely. and (5) hold be may and every u of V is uniquely expressible in the form (18) for d in F.
Thus we
justify the use of the
proving that
term linear subspace L of order r over F by L is indeed what we have called a linear space of order r over F
. (3).
F forms
Verify the statement that the set of a field equivalent to F.
6.LINEAR SPACES
2. and (5) hold for every u. v of V and shown very easily that. of V and a. Verify the statement that the mathematical system consisting of all two-rowed square matrices with elements in F and defined with respect to addition and multiplication is equivalent to the set of all matrices
fa n
ai 2
an a*
021
a
a.
field of
75
is
Verify the statement that the field of complex numbers matrices
equivalent to the
-b\
(a
\b
for real
3.
.
all
scalar matrices with elements in a field
4.
. Clearly. if V is a linear space of order n over F. if (2). .
+
Cn t>n
. w.
O22
under the correspondence indicated by the notation.
Linear spaces of
finite order.
.
.
We
shall restrict all further
study of
to be a linear spaces to linear spaces of order n over F. every quantity of V is uniquely expressible in the form
(18)
Citfi
+
. .
Every linear subspace
L
of order r over
F
of
Vn is equivalent
F
to
V
r. Moreover.
+ Cn V n
+
(ft.
. our definition implies that every two linear spaces of the
F
V F
V
same order n over F are equivalent over F.
. Our definition also implies that. then V is equivalent over F to V n However. b in F. Then
. Then we define is equivalent over to linear space of order n over a given field F if n over F. (3).
The set of all m by n matrices with elements in F.
+ u F is uniquely expressible in the form
U =
CiUi
(20)
+
.
All polynomials in independent variables x
(in x and y together).
.
_
^2.
and
the coefficient
matrix
A=
(aij)
is defined. . All polynomials in x =
coefficients in F.
.
Then
the
number
of the
vk
which are linearly independent in F is the rank of the matrix A..
+a
in
un
.
+ CrU
r
for
d
in F. L = M if and only if m = n.
linear spaces of order
.
contained in the space
of
For proof we merely observe that every vector
L =
uiF
+
.
m
EXERCISES
1.
n.
F
the
field of all
.
.
Theorem 3 should now be interpreted for arbitrary and we have a result which we state as Theorem 7. v m form a basis of L over F if and only if Vi.
in
x and
y. The set of all n-rowed diagonal matrices. ...
a) All polynomials in
6)
Find bases for the following linear spaces of polynomials with x of degree at most three.
first
row are
zero.
Thus u
in
L
uniquely determines the
d
in
F and
conversely..
finite
Verify the statement that the following sets of matrices are linear spaces of order over F and find a basis for each.
and y and degree at most two
2
c)
t*
+
t
and y
=
J
8
+
and degree at most two
numbers.
a one-to-one correspondence between
it
that
defines
an equivalence of
r.
.
.
.in
Vi
F for
which
=
anUi
+
. Moreover.
.
.
. Moreover. The set of all mbyn matrices whose elements not in the The set of all n-rowed scalar matrices. .
a)
b)
c)
d)
2.
vm
be in
L
so that
there exist quantities a.
. with
d) All polynomials in
e)
F the
field of all rational
All polynomials in a primitive cube root of unity with rational numbers.
It follows that
(21)
is
w->(ci. Let L = UiF + + unF and Vi.
.
. = n and A is nonsingular.
.<v)
r
V and V and it is trivial to verify L and V We have now seen that every linear space L of order n over F may be regarded as a linear subspace of a space M of order m over F for any m > n.76
INTRODUCTION TO ALGEBRAIC THEORIES
Vn
r
.
.
all
3.
77
numbers
+
F
m
g)
The polynomials
of (/)
but with
F the
field of all real
numbers.
+ vmF.
.
.
.
In this case the order of
L
LI
is
the
sum
. B.
. Let A = diag {1.LINEAR SPACES
the field of all rational 1) with /) All polynomials in w = u(i = and i* = v? that 2.
AB.
.
. three-rowed diagonal matrices. i. . w q ] of L will be called the sum of vm w\. {wi.
-1 (0
\0
2
01.
vm } and L 2 = Addition of linear subspaces.
Show
if
that 7.
.
if
.
5.
. then the v> and w. the subspace Lo = {wi.Li}
. . If LI = (vi. w q ) are linear subspaces over F of a space L of order n over F.
AB form
a basis of the set of
all
two-rowed square mat-
rices
5.
in
both LI and
L 2 is
that LI and
(23)
L 2 are
complementary subspaces
Lo
of their
.
.
. + and +bw = + OmVm + Jwi + a\v\ + = + (bq)w is in both LI + amvm (-bi)wi +
q
. 2}. .
.
Li and
(22)
If the
L 2 and
will
be designated generally by
Lo=
only vector which
is
{Li. and Lo has order m + If LI and Lo are linear subspaces of L and
if
L
contains LI.
. A.
Conversely.
of the orders of LI
2
and
L 2 and in fact
.
. L = wiF + + w F. 1.
For
only
if
it is
clear that
aiVi
.
.
B AB A J5 2 form
.
.
are linearly dependent in F and do = 0.
if
/O
0\
A =
then
7.
.
. .
. necessarily t>
<?
do form a basis of Lo.
A
2
. 1.
A
2
. we shall say sum and write
-
L!
+ L. w F.
the zero vector.
a basis of the set of
all
three-rowed
square matrices.
I/
2
2
A. . Hint: Prove are a basis.
7.
= viF + v F m + + WiF +
.
q
q
if
.
Show
that 7. A.
. .
v
. 1.
then Lo
= v^ =
+ +
.
q
.-
are not all zero
and therefore
that the vectors
and Wj spanning
L
if
not form a basis of
L
.
.
.
q
and
L 2 Thus v ^
t\-
implies that the a*
and
6. w.
we
shall
show
that.
.
.
Show
that. .
we may ask
.
.
. A* form a basis of the set of
4. .
-1.
2.%
in
.
. .
. Let LI of order over F be a linear subspace of L of order n
m
over F.
(-5.1.
.
Let the following vectors U{ span
Li.
.
. instead.
Then
the matrix
ft
is
nonsingular.
L
2.
.0). 2)
.1. 2)
. 4.
u3 .
.
(4.
.
.
.
.
1. 2. vectors
We put L = Vn
vi.
But then
A =
is
nonsingular.
and
A =
AoQ""
1
is
obtained by permuting the columns of A$.
(4. it is clear that the rows of are certain of the vectors e which we defined in Section 2. 2. 2.
u. u n in rem 1.
-(1.
1)
.
-1) -1)
.78
INTRODUCTION TO ALGEBRAIC THEORIES
. = * \
%=
.
The result above may be proved by the method we used to prove Theovm u\. 0.
c)
.
.
.0). -1.
theory.
.
%=
i*
(2.-.
-2.
Then G has rank m.
1.
fin
-(4.0.
. 1.
GQ =
where ft
is
nonsingular.
H
EXERCISES
1.
L
2}
to Li
and
f
to La.
vm of LI. 1)
(1.
(?.(!. Moreover.
3. 1.1).
= =
(1.
. and there exists a nonsingular
matrix
Q
such that the columns of
GQ
(ft
are a permutation of those of
ft)
. the rows of A span V n and the rows of H span the space L 2 which we have been seeking.= (1. (2. . 1.
-2.
whether a linear subspace L 2 of L exists such that L = LI + L 2 The existence of such a space is clearly a corollary of Theorem 8. Let LI be the row space of each of the by n matrices of Section 4.
let
G be the mbyn matrix whose rows are the basal
. proof using
i.
. where we apply this method to the set u let L = uiF + F. us a matrix + n However.
1)
(-2.-(!.
*s
=
(0. -1. 0. -1. Use the method above to find a basis of the corresponding V n consisting of a basis of
m
Li and of a complementary space
2.1.
-1)
-4.
-3)
(3.
.4.
Find a complement in
{
Li. 0.0). 1)
(!. 6. 3. Ex. give. 0.-(l. 3. v
span
Li.2.0)
-3.
2)
*-
-3)
-2.* = = /.
3.
.
Then
there exists
a linear subspace
L2
of
L
such that LI and
L2
are
complementary subspaces of L.2.
be linearly independent.
.
.
+
+
We now see that the system (26) is consistent if and only if A* has the same rank r as A.
.
m)
. (b k iU* + These replace the submatrix A of A * by
.
c
(i
=
1.
so that the rank of
A*
is
most
But A* has
A
as a submatrix. . if any c* 5^ 0.
. . clearly.
.
.
.
.
.80
INTRODUCTION TO ALGEBRAIC THEORIES
A*
of the system (26) to be the
Define the augmented matrix matrix
u*
m by n
(31)
A*
u
(OH.
and
(32)
see that (29)
and
(30)
imply that
6*11*
ul
=
at
+
r.
ur
u* are clearly linearly independent.
.
.
m)
. then m r of the equations (26) may be regarded as
.
matrix A* of rank r. .
. (S)A*
by
o
UT
and replace
(34)
(G
\0
Cl \
1
C.
Conversely. the has a nonzero (r l)-rowed minor.
and we choose
wi.
. if A* does have the same rank as A.
.
r)-rowed and one-columned matrix C 2 has c/bo = c* bkrCr) as its elements. This is impossible if A* is
. we have already shown that.
A*
.
./
where the (m
(ftjfcici
+
.
+b
kr u* r
(k
=
r
+
1. But.
+
.
has
to
rank
r. Moreover.
.
.
.
if
A * has the same rank r as A
t .
ain
.
.
(33)
m. .
. m. then u* We then have (29) and may apply elementary row transformations of type 1 to A* which add b kr u*) to ul for k r + 1.
.
if
we choose the notation
of the x
so that
G =
(Gi. . yn). then
(38)
Q
r
where FI and X\ have
columns.
Before solving (35)
we prove
be
LEMMA 4.
= (X G(+Xt G'
1
i
so that
. where Qi is an r by r matrix of rank r. .
serve that.
From
xQ'
this
we obtain
Q'
=
(!
^J.
.
.
cr )
.
.
.
v
=
(ci.
Thus we have
Q =
The system
(37)
may now
0)t/'
be written as y
(I T
=
t/. 2/n arbitrary. It thus remains to show that any system of r
equations in
xi.
.
.
We may write
x
=
(xi.
(36)
Let
G
an
r by
n matrix
of rank
r.
..
.
.
.
.
=
xQ'
=
( yi .
. solution yi = c for i = 1.
. r) and 2/ r +i.
.
.
x n ) and see that such a system
may
be regarded as a matrix
equation
(35)
Gx'
=
t/
. the solution of (35) is then given by z = y(Q')~ l for yi = Ci(i = 1.
But then y = xQ'
a nonsingular linear transformation and the t/ are x n Evidently.
-
-
..
diag {H.
.
x n with matrix of rank r has solutions. Obis
.
. For Lemma 2 states that G = (H 0)Qi.LINEAR SPACES
81
linear combinations of the remaining r equations and are satisfied by the solutions of these equations.. (37) has the linearly independent linear forms in Xi. 7 n _ r }.
Then
G=
(I F
0)Q
for a nonsingular n-rowed matrix Q.
.
62) with
Gi an r-rowed nonsingular matrix. r.
.
. Then singular and
is
n-rowed and nonso
is
H
H is nonsingular.
.
(35)
(7 r
0)Q 2 = (H
0).
(36) for
Q2 = Q 2 Qi.
.
. Then we shall call S a linear mapping L on the space and describe (42) as the property that S is
M
Suppose now that
so that
we
are given not only the spaces
Then a
(43)
linear
= viF + um F and L and but fixed bases mapping S uniquely determines u$ in M.
. be linear spaces of respective orders and n over Let L and consider a correspondence on L to M.
+
M
M
. as well.
. and (40) may be regarded as a correspondence u > uA defined by A whereby every u of Vm determines a unique vector uA of F n We now proceed to formulate
.1) of
x'
=
tion.
.
The system of equations a linear mapping was expressed in (3.
X.
x r as linear functions of
Xr+i.
. A and A.
.
+
uf
=
anVi
+
.
a and 6 of F.
For exercises on linear equations we
Theory of Equations.
(3.' in this equaThen we see that a linear mapping may be expressed as a matrix
Linear mappings and linear transformations.
refer the reader to the First Course in
9.
.
to u upper S) wherein every vector u of Suppose also that
L determines a unique
(au
+ 6tto) 5 =
au8
+ bu
for every
of the space
linear.
where
(41)
A
is
an
m by n matrix and
u =
(pi.
.
xw)
.
equation
(40)
v
= uA. so that
S
is
the function
u*us
u goes
us
M
.
this concept
abstractly.
. Designate the correspondence
more
M
m
F and
by the
symbol
S:
(read in
(42)
5.
.
.
. u and U Q of L.82
INTRODUCTION TO ALGEBRAIC THEORIES
(35) is given
and our solution of
(39)
by
l
X
l
.
ym)
.
.
.
.
.
.
+ ain vn
(i
=
1.
.
.
v
=
(xi.
v n F.
xn
.
. u is a vector of Vm over F v is a vector of V n over F.
. Let us interchange the roles of m and n.
.
.(Y .
.
.(
But then we have solved
the
for xi.32) as a matrix equation y'A'.
m)
.
.
Clearly.
. and hence
u\F
L=
+
.
respectively.
by
(47)
4
0)
=
i-i
y-i
. we see that S is the linear mapping
if
M
Vn
and put
(46)
u .
.
= u s in
= we assume temporarily that L = Vm and (40) and (41). if A is an m by n matrix and we define uf by (43) for the given elements an of A. Thus S determines also an m by n matrix A = (a ).1) on the space Vn
.
.
may
A. then the property that S is linear uniquely determines the mapping S. This is true since every u of L is uniquely expressible in the form
(44)
u =
in
yiUi
+
. Thus every linear mapping which is a change of be regarded as a linear mapping of the space Vm
Let us next observe the effect on the matrix defined by a linear mapping of a change of bases of the linear spaces.LINEAR SPACES
for
7
83
an in F.
for
2/<
F and
(42) implies that
(45)
us
It follows that to every linear
and M
mapping S
of
L on
M and 0wen bases of L
We
there corresponds a unique by n matrix A and conversely. But. shall call A the matrix of S with respect to the given bases of L and M. conversely.
.
+ ymum
.us
= uA
for the given matrix
variable as in (3. Define new bases of L and
M
.
w
We now
observe that
where
-
1
But
v
then.
..
. Since we are now considering only a single space. If L = F n such a
Since
M
M
M
.
.
Then P = (pki) and Q = (qu) are we may also write the second set of
equations of (47) in the form
(48)
v. u n of L of order n over F and of a second basis v n of L. we see that changes of basis in L and replace A by an equivalent matrix.
correspondence
is
given
by
u us = uA
.
n. u n of L to be the matrix A determined by (43) with Ui = Vi.
. Then we define the matrix A of a linear transformation $ on L with respect to a fixed basis ui.
. respectively.
.
and Q" 1 are arbitrary nonsingular matrices of m and n rows. as
we saw
in (3.
.
.
m
and
I
=
1.84
for k
SB 1.
.
. Let us restrict our attention to the case where v = u^ v\.
we have
(49)
where b k = Spkidarji.
.
.
apply the linearity of
S
to (47)
and obtain
wW
Substituting (43) and (48).
1
)
singular linear transformation defines a one-to-one correspondence of
L
and
itself. .
.
nonsingular and. We have defined thereby a one-to-one correspondence between the set of all n-rowed square matrices with elements in F and the set of all linear transformations on L of order n over F.33). Thus any two equivalent m by n matrices define the same mapping of L on If L and are the same space we shall henceforth call a linear mapping S of L on L a linear transformation of L.
. Since we may then solve for u
Clearly.
INTRODUCTION TO ALGEBRAIC THEORIES
.
.
=
1-1
where
JB
=
(r/j)
= Q" We
1
. the only possible meaning of the Ui and v.
.
we should and do call S a nonsingular linear transformation if A is = u8A~ we see that a nonnonsingular.in (43) can be that of a fixed basis u\.
.Hence the matrix B S with respect to our new bases is given by
i
=
(bki)
of the linear
mapping
(50)
B = PAQr
P
1
. .
.
.
.
. 1.
formation on
pass to
study those properties of a linear transnot depend on the basis of L over F we need only study those properties of square matrices A which are unchanged when we
L which do
l
.
L
defines a linear transformation of
of basis as being induced
we put
Thus we may regard a change
L when by a nonsingu-
lar linear transformation.LINEAR SPACES
85
We now
observe the effect on
0)
A
of a change of basis of L.
Define
of all vectors (points)
tions. Find the vectors of 7 4 into which (-2. 2
u =
d) 3x1
6)
-
2x1
+
2zi
+ 40:1X2 10xix 8
-4x?
+
Hxl = -6x^2
(47) of
-
18x2X3
+
1
*
Observe that a change of bases
uf
wj
0>
. (0. x%.
I
1
F4
defined for the following mat-
3. 1.
rices
Let S be the linear mapping (46) of 73 on A. Apply both S and to the vectors of Ex.
Show
A
/-4
a)
3
5\
A =
V
3 5
20
6)
A
=
-2
117
3.
We
use (47)
but note that ut
formation by
(51)
It is
= v+ and i4 = *40) so that we have P = Q.
and let C be one or the other of the curves whose coordinates satisfy the following equa(x\. 0.
/S
for the matrices of Ex.
. x$) Find the equation of the curves C s into which each S carries each C.
PAP~ We
shall call
two matrices
A
and
B
similar
if
(51) holds
and shall obtain necessary for a nonsingular matrix in our next chapter. (1.
rices
S" 1
that the linear transformations (46) of V* defined for the following matare nonsingular and find their inverse transformations.
S. similar and be tions that
P
and
sufficient condi-
A
B
EXERCISES
1. Hence a change of basis* of L with matrix P replaces the matrix A of a linear trans-
B = PAPnow
clear that in order to
1
. 0) of
F3
are
mapped by
-3
3 2
1
1\
a)
A =
I
2-6-1
-1
-5\
21
b)
A =
I
/-2 -1
3
4
2
0\
2
4)
\-l
-I/
\
-3/
0\
c)
A =
/2
\1
-2
1
3-2
-1 -I/
d)
A =
/-2
I
0-10
1 1
01
I/
\-l
2. 4). 0).
/(w) 5 have f(u) = /(w ) only if 6 = 1 for every i. We call matrices A
clearly. those
We
is an orthogonal matrix. Write B = AA' . f(u s
s
)
pfl
satisfying
(53)
AA' =
I
is
orthogonal matrices
and have shown that S
orthogonal
if
and only
if its
matrix
A uniquely only in terms of a fixed Euclidean geometry the only changes of basis allowed are those obtained by an orthogonal linear transformation. f(u)
=
uu'. c/ = for j ^ p and and see that/(w ) 2Ci6i. the norm of u (square of the length of u) to be the value f(u) = /(ci..
*
. then
BB'
and
(PAP')(PA'P')
A
..
that (/
A be a skew matrix such that / + A is nonsinguis
+ A)""^/
A)
orthogonal. all = /(w) only if other c/ = 0.. The final topic of our study of linear spaces will be a brief introduction to those linear transformations permitted in what is called Euclidean geometry. that is. have seen that S determines
basis of
Fn Now in
.+4.
S on V n which
are said
We
propose to study those linear transformations
to be length preserving and define this concept as the property that f(u) 8 f(u ) for every u of n Such transformations S will be called orthogonal. .
We
. sional linear space of vectors u = (ci.PAA'P' =
I
. Let then Vn be the n-dimenc n ) over a field F.(ft.
Show
that every orthogonal two-rowed matrix
/
A
has the form
a
=1
V
(b
.86
10. f(us ) = 2 + 26 b pq = for a p 5^ g.
for
which the matrix
P of
(51) is orthogonal.
. B = / is the identity matrix.
=
V
.
B
is
also orthogonal.
define
.
But then
PP =
1
7.
INTRODUCTION TO ALGEBRAIC THEORIES
Orthogonal linear transformations.
= us (us )' = uAAV.
P'
= P-
1
.-c/..
cn)
of the quadratic
(52)
form
/fe.
We may
by
square
Then. us = uA for an n-rowed define S matrix A.
EXERCISES
1. from which f(u) = 2.
.
What
Show
are the possible values of the determinant of
an orthogonal matrix?
2.
. Then take c p c q = 1.^-^ + ..
. so that if S is a linear transformation with orthogonal matrix and we replace A by PAP"" 1 = B.
Let I be the identity matrix.
lar.
3.) = = 2c?.
P'P = /. Put c p = 1..
fawj
0(L) and a and 6 are in F. we ^i + ^2 is in 0(L).
7.
onal (that
us = wA
if
. while.
wW =
uAA't.022
AA'
4. 1N
i
\
a)
OJ
6)
M
/3
4>
\
(4
-3J
C>
x
A
(l
\ l)
/3 (2
Orthogonal spaces. in a geometric sense. y z sin ft t/ 3/0 cos h.-)
and
0.
+
Then 0(L)
in
is
a linear subspace of
if
Vn which we shall call the space orthogonal
in
Vn to
L.
1
t
an are
1. of 0(L) is n In fact we shall prove Theorem 10. perpendicular) if uv' any orthogonal matrix and we define a linear transformation S of Vn by us = uA.
and
of the
t
such that
0?i
s*
+
2
7* 0. We now
Theorem 9. We shall call two vectors u and v of V n orthog= 0. conversely. 0(L)
and only
if
uv'
=
. = + a are derived from 6.
-i
=
XQ y
.
. Show that the corresponding
=
+
orthogonal and that every real orthogonal two-rowed matrix matrix of a rotation or its product by the matrix
matrix
is
either the
Evo
of a reflection x
5.
Wi and
have 0(01^
prove
+
bw 2 )'
=
atwj
+
w 2 are = 0.
=
yQ
.
The equations
h are x
=
is
z cos h
for rotation of axes in a real Euclidean plane through an angle sin h. we define the set to the be set of all vectors w in Vn such that tw' = for every v of L. Now
an. Hint:
put du
=
Compute
l
PAP = D =
j\
(d*.
+
.'
=u
f
=
0.
.
Find a
real orthogonal
matrix
P for each
of the following
symmetric matrices
1
A
such that
PAP'
is
a diagonal matrix.LINEAR SPACES
where a
87
=
2
s(s
+
t*
*
2
J
2
)~
1/2
. clearly.
first
note that the elements of
GH'
are the products
t^toj
. Then O(L) is the = (0 In-mXQ')" 1 o/ rank n m. Let L be a linear m. Thus orthogonal transformations on V n preserve orthogonality of vectors of V n If L = viF v m F is a linear subspace of F.
6
=
2
Z(s
+
J
2
2
)-*/
. Then. we have
11.
tr
8
wA
. if A is is.
)(s
+
)"
=
The values 021 =
form above. Let L be the row
that
subspace of order
m of Vn
.
TVien the order
G=
(I m
row space of H For proof we
space of the m by n matrix G of ran/c m so 0)Q for a nonsingular n-rowed matrix Q. For.
Hint: The result
s
is trivial if
a
2
+ #12= 1 so + 6 = (s +
2
that
2
=
J
2
range over all quantities in F A is a diagonal matrix.
dm ) has elements in F and G is an m by n matrix of where d = (di.
2. then vv = 1 + i2 = 0.
-1.88
INTRODUCTION TO ALGEBRAIC THEORIES
Vi
of the rows
(I m 0)[(0
.
.
We let HO be
=
the g
(I m
= by n matrix whose kth row is y*.
1.
(1.
But
so that
m nonzero rows. we may prove Theorem 11. the row space of H is 0(L).
But
GH =
f
/n-m)].
q
row space
m. since if F is the field of all complex numbers and
+
= (1.0). However.
u.
of the
homogeneous
linear
system Gx =
1
0. w =
vF. D has at most n = n n m.
n. 0)
ui
in-
-1. Then V n = L + 0(L). Hence 0(L) is contained in L and 0(L) = L. This is not true in general.
The sum of the
orders of
L and 0(L)
is
. 1.
v
f
.
7
EXERCISE
Let
L
be the space over the
field of all real
numbers spanned by the following
vectors u*. Let F be a field whose quantities are real numbers.
2.
u3
(3.
0. Then Gw' = Theorem 3.2.
.
i
2
L =
=
1.
-2. 3)
. and. 1) (2.17 the matrix GG' has rank m and is nonsingular. 3)
u. while also Gw = GG'd By rank m whose row space is L. Hence let w be in both L and 0(L) so that w = dG. and D must have rank Di = 0.
Let then
0(L) =
yiF
of
+
H.
=
-1. q Note that the row space of H is the set of all solutions x xn) (xi.
of
G by
the transposes of the rows Wj of H. 2.
(0. 1. and we have GH' Q The of the n rank matrix 0)Q#Q. d = 0. i). This that proves q ra.
2. For by Theorem 9 it suffices to prove that the only vector w in L and 0(L) is the zero vector.
0)
. 1. 0. 1.
f
f
. GG'd' = as desired. in
=
(0. 1)
.
u.
.
.
. clearly.
.
=
= =
(-1. 1) (1. whether V n = L 0(L). 2.
and
it is
natural to ask whether
or not they are complementary in V n that is.
then
GH' =
0.
.
+ y F of order q over F so that 0(L) contains the Evidently H has rank n m. 0)
d)
(1. and we must have g > n
.
<
. only if d = 0. 0. Find a basis of the space 0(L) in the corresponding
a)
Vn
.
.
. 3.
u3
.
4.
-1)
^-
f-l.
in-
(1. 1)
6) Ul
c)
=
(1. by q
is
g since
Q
is
nonsingular.
.
.
fr }
for monic polynomials
=
fi(x)
such that
fi
di-
in x. As we stated in that section.
Hence we
see that
are equivalent in F[x] only
if
they have the same rank.CHAPTER V
POLYNOMIALS WITH MATRIC COEFFICIENTS
1. that is. Then a ^ must be a constant polynomial. we may assume that f\ is monic and is the element in the first row and column of a matrix C = (c ) equivalent in F[x] to A.
A and B We may then
prove
LEMMA
lent
1
.
. a may
be any nonzero quantity of F.
-
X)
fi
=
diag
(fi.
Every nonzero matrix
to
A of rank r with elements in F[x] is equiva-
in F[x]
1
(S
where GI
vides
fi+i.
Matrices with polynomial elements.4.
89
.
. But
those of type 3 are restricted so that the quantity a in F[x] shall have an inverse in F[x].
We shall consider m by n matrices with elements in F[x] and define elementary transformations of three types on such matrices as in Section 2. Let
Fix]
F be a field and designate by
(1)
(read:
F
bracket x) the set of
all
polynomials in x with coefficients in F.
/i
For the elements of all matrices equivalent in Fix] to A are polynomials and in the set of all such polynomials there is a nonzero polynomial
=
/i(#) of lowest degree. The field F(x) of all rational functions of x with coefficients in F contains Flx] and it is thus clear that if A and B are equiva-
m
}
lent in F[x] they are also equivalent in F(x). We now let A and B be by n matrices with elements in F[x] and call A and B equivalent in F[x] if there exists a sequence of elementary transformations carrying A into B. we assume that in the elementary transformations of type 2 the quantities c are permitted to be any quantities of F[x\. By the Division
.
Using elementary transformations of types 3
and 1.
Then we add the first row of A i to its second row so that the submatrix in the first two rows and columns of the result is
ft
(5)
(
\fi
We
then add
a*
times the
first
column to the second column
to obtain
a matrix equivalent in F[x] to Ai and with corresponding submatrix
//. du = f\. Similarly. Then either
A2 =
may
apply the same process to
A
After a finite
number
of such steps
ultimately show that
A
is
equivalent in F[x] to a matrix (2) of
we we our lemma
0. so that these minors are also
. observe now that the elements of every <-rowed minor of a matrix
ti
=
0. = (da) with dn = for i 7* 1.
we
I'th
pass to a matrix equivalent in F[x] to
A
with
as the element in its
is
thus implies that r< in F[x] to a matrix
zero.
where
s<
and
ti
are in F[x] and the degree of
ti is
less
than the degree of /i.
or
such that
the set of
every / =
all
/<(x) is
all
monic and
is
a polynomial of least degree in
elements of
matrices equivalent in F[x] to
(4)
Write /i+i
= fiSi
+
ti.
D
column.
row
arid first
we
see that every
du
is
divisible
by /i and hence
that
A
is
equivalent in
F[x] to a matrix
(3)
Vo
where
A
2
has
w
1
rows and n
1
2.
columns.-
(6)
\f<
definition of /t thus implies that
Our
We
/ divides />i as described. But if
first
+r
for g<
q>
and
r< in
we add
times the
row
Ti
of
C
to its ith row.
A
with elements in F[x] are polynomials in
z.90
INTRODUCTION TO ALGEBRAIC THEORIES
= qifi Algorithm for polynomials we may write CH F[x] and r of degree less than the degree of f\. Our definition of /i have now shown A equivalent we Moreover.
. Hence dk divides every Mk+i so that dk divides dk+i.
91
By Section 2. or the sum the of a 2-rowed A.
.
i
r)
.POLYNOMIALS WITH MATRIC COEFFICIENTS
polynomials in
x.
.c.
.
divisible
by
0/+i for j
1
=
1. /*.
. every Crowed minor of B is either a Crowed minor of minor of A by a quantity a of F. We may state this result as LEMMA 2.
.
. it also divides every 2-rowed minor of all m by n matrices B equivalent in F[x] to A.
. Every (k + l)-rowed minor Mk+i of A may be expanded according to a row and is then a linear combination.
We
(7)
thus have d k
=
/i
. . = g r We call g the jth invariant factor of A and see that if we /r_i 02. it has the same greatest common diand hence the same g. /i = 1 we have the formula define d
It is
=
>
3
(8)
gf
=
(j
=
1. .
.
(o
then. Then d t is also the greatest common divisor of the i-rowed minors of every matrix B equivalent in F[x] to A. But A is also equivalent in F[x] to B and therefore the Crowed minors of equivalent matrices have the same common divisors. We ob-
MI + fMz where MI and Afa
are
serve also that in (2) the only nonzero fc-rowed minors are the fc-rowed
minors |diag
divides /.
Moreover.
whence
--A. with coefficients in F[x].
is
2). of fc-rowed minors of A. if the converse holds. of all fc-rowed minors of (2) is /i .
|
.
We
mentary transformations of type
0i.
.
to (2)
and
see that
A
apply elehas invariant
r. Let Abe an m by n matrix with elements in F[x] and d t be the greatest common divisor of all i-rowed minors of A. while.7 if B is obtained from A by an elementary then transformation./.+
\fi
we
<i* and since clearly every / . is
now
.
factors
0.-J for ii<z 2 < v see that the g.
. of (8)..
. product
Crowed minors of A and/ is a polynomial of F[x].d. .
then
A
is
equivalent in F[x] to
w
If. B is
equivalent in F[x] to (9) and to A.
t
customary to relabel the polynomials / and thus to write fr = gi.
-*i
*>
visors d k
equivalent in F[z] to 4.
We have
proved
.. If d in F[x] then divides every Crowed minor of A.
fk.
r
if
1.
As in our theory of the equivalence of matrices on a field we may obtain a theory of equivalence in F[x] using matrix products instead of elementary
.
Two
F[x]
if and only
m by n matrices with elements in F[x] are equivalent in if they have the same invariant factors. The invariant factors of a matrix A with elements in F[x] and rank
r are certain
i
. We call two m by n matrices A and B with elements in F[x] equivalent in F[x] if there exist elementary matrices P and Q such that PAQ = B.
transformations.
r
If
g k (x)
=
1.
. their determinants differ only by a factor in F. .
1
1
|
=
|
/
1
1.
r. Hence.= PI" adj P if |P is in F. Then.
0i.
so that a
and
6 are nonzero
But
if
P has elements in F[x]
|
so does adj P. in F) and not zero.
.
transformation matrix
is
in F.
1.
.
a
=
\P\ and b
=
\P~
l
the quantities a and 6 are in F[x]. the products P&A. In closing let us note a rather simple polynomial property of invariant factors. But.
. AQo are the matrices resulting from
lent in F[x] to identity matrices
the application of the corresponding elementary transformations to A.
. the determinant of any elementary
and B are square matrices with elements in F[x] and are equivalent in F[x].
.
Let us then
call
those gt(x)
^
1
the nontrivial invariant factors
.
.
. Then we again have the result that A and B are equivalent in F[x] if and only if they have the same invariant factors. if \A and \B\ are monic polynomials.
.
Qi(x)
such that Qi+i(x) divides gi(x) for
=
.92
INTRODUCTION TO ALGEBRAIC THEORIES
Theorem
1. in particular. ab
quantities of F.
. Thus we have proved that a square matrix P with polynomial elements is elementary if and only if its determinant
is
a constant (that
We now
is. Moreover. For under our first definition P and Q are equiva-
and hence may be expressed as products of elementary transformation matrices.
of rank r with invariant factors gi.
if
A
\
\
that
when A
is
a nonsingular matrix with
.
then g&x)
=
1
for
larger j
=
k
+
1. PAQ must have the same invariant factors as A. then the equivalence of A and B in F[x] implies that \A .
gn
as invariant factors
its
determinant
It is
if
the product g\ gn clear now that a square matrix
is
.
both
P
and P.have elements
1
in F[x}.
We
thus define a matrix
P
=
to be an elementary matrix
if
if
\.
invariant factors of P are all unity. Every m by n matrix
.
P with elements in F[x] is elementary
and thus the
and only
if
P
is
equivalent in F[x] to the identity matrix.
.
g r is equivalent in F[x] to (9). and we have the result desired. We may now redefine equivalence.
monic polynomials
1. The converse is
proved similarly.
Hence.
Hence
so will
P.\B\ and. . observe that.
. in fact. if PO and Q are elementary transformation matrices.
a)
I
x3
+
3x 2
-
3x
-
1
x3
-
x2
+ 4x -
2
/x 2
d)
x
2
+x +x
1
x
x
x2
1
+
2x
x2
\x 2
2x 2
+ 1\ +x + 2/
1
2.
Describe a process for reducing a matrix A with elements in F[x] to an equivaHow.
.
Determine the invariant factors
of the matrices of
by the use
of (8).
i
Thus
exists
an integer
t
g i+ i(x)
=
. 3 to carry this preliminary diagonal matrix into the' form (9)?
5. Let A be an m by n matrix whose elements are in F[x]...
Use elementary transformations to carry the following matrices into the
(9). of an and a*. Hint: Use the g. Express the following matrices as polynomials in x whose coefficients are matrices with constant elements.in its ith row and jth column.
6.
. then.
Reduce the matrices
of Ex.d.c.
/x(x-l)
b)
\ x(x
/x 2
c)
+ x-2
x
2
\
x
+
l)
\
+
l)j
V
+ 2*
-
3/
lent diagonal matrix.. 1 to the
form
(9)
by the use
Ex..
.
=
such that gi(x) has positive degree for
=
1.
form
\
a)
(x \Q
4.
1
the trivial invariant factors of A. process of Section 1.
Q
=
a>kj>
3.6 with/ = an.c.d.
EXERCISES
1.
1
of elementary trans-
formations.POLYNOMIALS WITH MATRIC COEFFICIENTS
of
93
there
A
}
the remaining g^x)
=
1. may we use the process of Ex. Describe a process by means of which we may use elementary row transformations involving only the fth and kth rows of A to replace A by a matrix with the g.
g r (x)
=
..
+2
divisors.
.
c.
We
shall call the rs
polynomials
the elementary divisors of A. that every g<(x) divides 0i(x).
r"i
.
.
e\j
=
^
for
i
=
2.
.
.
-x2
Let
x
+
1
~x 2
Elementary
K be the field of all complex numbers so that
ci)
ei
.
number
of invariant factors of A.)
.
.
. and thus
C\C\\ v^iuy
of a matrix
A
clear
n-(r\
&%{)
r is the
r. Those for which e^
nontrivial elementary divisors of
>
will
be called the
A.
to the
form
x
2.
where
Ci.
.
.
.Vti l
^i)
fr c
v-
* Vi cy
/
TV
^
e^ e^
1
i.
.
(r {*'
/.
rj
. Since 0+i(x) divides 0(x).
.
INTRODUCTION TO ALGEBRAIC THEORIES
Use elementary transformations to reduce the following matrices
(9).
g\(x)
=
(x
.
are the distinct complex roots of the
first
invariant factor
it is
with elements in K[x].94
7.
(x
e
c.
.
Here
.
_i = diag {0_i. d\
=
A
A
.
=
(z
c/)
for
n^ >
0.c.
.
complex numbers
the corresponding elementary divisors /. .) / with Let us then order the t exponents n. Then.
.
.
. Conversely.
.} equivalent diag {gf. q = ti t tSj and our set of polynomials may be extended to a set of
+
+
^
exactly
ts
HH =
#ij
0. .
first
has prescribed invariant factors and hence prescribed it is desirable to obtain a matrix of a form
exhibiting the elementary divisors explicitly. But then
A =
diag {^i. diag 1}.
c. in to that A.
.
Let us prove
f s be monic polynomials of F[x] which are rela2./i/2 is the only nontrivial invariant factor of 2 is equivalent in F[x] to diag (fif 2 2.POLYNOMIALS WITH MATRIC COEFFICIENTS
95
The invariant factors of A clearly determine its elementary divisors Cj where r is the uniquely as certain rs powers // of linear functions x rank of A.
./. we have r ^ t. By Theorem 2 the matrix Ai = diag {/. let us consider a set of q polynomials each a power with positive integral exponent of a monic linear factor with a complex
root.
.
.. where
0.
!.. Define gi(x) as in (10) for i = 1.
.. in Then the pairs.
.
.d. to be integers e^ satisfying t and ^ ^ Ctj ^ 0.
A
. If A has rank r.
-._i}
A =
diag {/i. 1}. In fact.
.
02j
^
._i is prime to / in to is equivalent to is F[x] diag {g.
1.
!.
We shall do this.
A
The
distinct roots in our set
hkj
may
tfyen
n *>
be labeled
Ci.
/. and we adjoin (r t)s new trivial elementary divisors // to
obtain the complete set of r invariant factors 0.
.
c.
.
the g.
. Let f i..
.
. the elementary divisors of uniquely determine its invariant factors.
A =
diag
{/i._i = diag {/i._i
=
/i
.
Then
.
.
A
whose nontrivial elementary divisors are the given hkj./2J
is
unity..
.
. It is now evident that two by n matrices with elements in K[x] are equivalent in K[x] if and only if they have the same elementary divisors.
n polynomials by adjoining t tj polynomials (x c.!}.
Theorem
.-
and our
let
tj
polynomials have the form
the
..
. only nontrivial invariant factor of the matrix tively prime fs f B } is its determinant g = f i A = diag {fi.}
is
equiv-
alent in F[x] to diag {0 f _i..
..!}.
. But 0. t be the maximum tj..
. is 5. 1}.-i.
.
is
We see now that if gi(x)
in F[x] to diag {0 t
defined
by
(10) for distinct
.
.
For each
be
number of hkj in our set./.
.
._i. . However..
The
/2 of
result is trivial for s
2
=
1.
A
t }
is
equiv-
.
. of the elements f\ and
l.
.. Then g is the only nontrivial invariant factor of A.
.
.
obtain a set of polynomials gi(x) such that g%+i(x) divides g^(x) for i 2 1. /.
./} is equivalent
. /* are relatively prime in pairs.
.. the Qi(x) are the nontrivial invariant factors of a matrix
. . Assume.
If s
=
2. then. / 1.
. F[x] equivalent . and 1J.-(z) of A.
m
The matrix
(9)
elementary divisors. 1.
.. clearly. and s is the number of distinct roots of the invariant factors
of A. Hence {</..
.
.
and our theorem
Cjj
is
proved.
Q(x).
m
with
henceforth restrict our attention to n-rowed square matrices with elements in F[x] and thus to the set of all polynomials in x with coefficients n-rowed
square matrices having elements in F. then q(x) t.
97
in F[x]
a. let
and J?o be give it in some detail. Let the elements a*.
Moreover.
while if s
t. Define
Ak =
(a^) and obtain an expression of
A
in the
form
(12)
f(x)
= A Qx>
+
. g(x) have deand leading coefficient B such that AoB 5^ 0.
then q(x)
and
us
While the proof
is
very
much
the same as that in
Theorem
1.
Matric polynomials.
nomial f(x)
ly
is
If all the
A k in
(11) are zero matrices.POLYNOMIALS WITH MATRIC COEFFICIENTS
3.-/.
. Then B^ 1 exists. of an m by n matrix A be and let s be a positive integer not less than the degree of any of the Then every a# has s as a virtual degree and we may write
Oif
(11)
for
a$
in F.
The degree of
f (x)
+
g(x) is not greater than the degree of
gree
4. the poly-
the zero polynomial. Let f (x) have degree n and leading coefficient Ao.
R(x) such that r(x) and R(x)
have virtual degree
(13)
and
f (x)
=
q(x)g(x)
+ r(x) =
=
Q(x)
g(x)Q(x)
+ R(x)
^
. if s < t.
and
such that the leading
t 1
Then
there exist unique polynomials q(x). we say that/(x) has degree s and leading coefficient Ao if ^o ^ 0.
+ A. Let us
call
these polynomials nit
rowed matric polynomials.
3.
for
s
by n matrices A k Thus f(x) is a polynomial in x of virtual degree m 6y n matrix coefficients Ak and virtual leading coefficient AQ> Moreover. Q(x) have degree s
=
0.
and we again designate
by 0. Then the degree of f (x)g(x) is m + n and the leading coefficient of f (x)g(x) is AoB
LEMMA
m
.1.
t
Let
f (x)
and g(x)
be n-rowed matric polynomials of respective
coefficient of g(x) is nonsingular. r(x). and if q\(x)g(x) has virtual deq^x) = A^B^x'"' the polynomial fi(x) = f(x)
^
.
in Chapter I we use Lemma 4 in the derivation of the Division Algorithm for matric polynomials which we state as
As
Theorem
degrees s
4. In order to be able to multiply as well as to add our matrices we shall
. We assume first that s ^ t and let A the respective leading coefficients of /(#) and g(x). Evidentf (x)
we have
LEMMA
or g(x).
.
This proves the uniqueness of g(x) and r(z).
division of f (x) by xl
C
are fn(C)
.98
gree 5
INTRODUCTION TO ALGEBRAIC THEORIES
1.
=
q$(x)g(x}
+
q(x) 0.
and
(15)
f L (C)
= CAo +0-^1
+
. Similarly.
Let us regard the first equation of (13) as the right division of f(x) by g(x).
Theorem
5. if C is an n-rowed square matrix and/(x) the polynomial /(C) might mean any one of A Q C 2
= A
. This implies a finite process by means of which we begin with i * t and leading coefficient a polynomial /i(x) of virtual degree s 1 i and then form /i+i(x) = fi(x) q i+ i(x)g(x) of virtual degree s
A^
for <7t+i(x) = A^jB^x*""*""'. and its C 0. //(x) of virtual degree t
q(x)
-
0. for ex-
ample. But leading coefficient is = r(x) r (x) is t then by Lemma 4 the degree of [qo(x) ft. and otherwise with q(x) of degree s
If also /(x)
leading coefficient
r (x) for #o(#)
A B^
and r(x) the //(x) above. the theorem in
Remainder Theorem for matric polyusual form is ambiguous since. the degree of this polynomial is h jg 0. so that J?(x)
=
in (13). of Q(x) and 12 (x) as left quotient and remainder. 1. and these matrices might all be different. respectively.
It is natural
now
to try to prove a
nomials. we call (/(x) a left
divisor of 0(x)
if
/(x)
=
gr(x)Q(x).
.
x2
.
.
=
x 2A
= xA
x. Then we shall speak
correspondingly of q(x) and r(x) as right quotient and remainder. However.
. If r(x) = 0.
Then the right and left remainders on and fi/C).+
1
.
1
. g(x)]0(x)
^
^
+
a contradiction. r(x)
= /(x)
Q
if s
1
<
The process terminates when we obtain an Thus we have the first equation of (13) with t and t. we have /(#) = q(x)g(x). C be ann-rowed square matrix with elements in F. The existence and uniqueness of Q(x) and R(x) is proved in exactly the same way except that we begin by forming /(x)
whereas
its
virtual degree
is
t
1. Also C Bo 5^ since B is nonsingular. and 0(x) is a ngrfa divisor of /(x). We shall do this and obtain a Remainder Theorem which we
state as
CA
2
Let f(x) be an n-rowed square matric polynomial (12).
CAoC.
+CA_ + A. Thus we must first define what we shall mean by/(<7). Define fn(C) (read: f right of C). the second as the left division of f(x) by g(x).
and fL (C)
(14)
(read: f
left
of C) by
f R (C)
= A
C-
+ A^*.
The second part
implies that of our
proved similarly.CD. that is. and only if f(C) = 0.
.
.
.
+
C. This statement is correct.
The
C)._i
and have
*(*)
=
q(x)(xl
-
C)
=
Then.
if
q(x)
-
x so that h(x)
unless
=
x8
-
Cx.
. if /(re) has x C is a root of f(x).
of the result just proved for matric polynomials which we state as
As a consequence
we have
the Factor Theorem
C as a right divisor if Theorem 6. if /#(C) = 0. then h R (D)
= D2 . We then 1 and r(x) r(x) where q(x) has degree s wish to draw our conclusion from /(C) = qR (C)(C C) + B.
D'
. we q(x)(xl results on the left follow
=
-
Our principal use of the result above is precisely what is usually called C as a factor.
+
(C_i
-
C.
. but let us examine it more closely.POLYNOMIALS WITH MATRIC COEFFICIENTS
field
99
C and have }(x) = q(x)g(x) + = B has elements in F.
if
(Co*
+ da-' +
. We write q(x) =
4 with g(x)
To make our proof we use Theorem
as in the case of polynomials with coefficients in a
=
xl
Cox*~
l
+
.
. The matric polynomial f (x) has xl C as a left divisor if and only if f/.
+
C.DC * D .C_i
q*(D)(DI
-
C)
= C
D-
+
(CiD-
1
-
CoD-'C)
+
.
D is any n-rowed square matrix with elements in F..C) implies that h R (D) =
qR (D)(D
*
-
C). it has xl For Theorem 5 implies that. we have
=
CoZ>
h R (D)
while
+
(Ci
-
CoC)!)-
1
+
._!x)
-
(CoC*-'
+
. then in (12) the polynomial
r(x)
=
0. then h(x) = q(x)(xl . f(x)
if
D and C are commu-
h(x)
+B
C)
+B
=
B.(C) = 0.
.
. it is nontrivial that /(C) = and follows
only from the study above where we showed that if D is any square matrix such that DC = CD.
and these matrices are equal
tative.
.
=
fe(C)(CJ
-
f(x)
C)
=
0._ 2 C)D
.
.
. However. f(x)
=
q(x)(xl
-
C).CD
E. then the trivial part of Theorem 5.
+
C_!C)
.
in general*
They
is
are equal
if
D =
/(C) =
theorem
hR (C)
+B = qR (C)(C -
if and only = and thus C.
Conversely.g.
if
have seen that/fi (C)
similarly. q M (D)(D -
C)
-
DC =
CD.
If f(x) is
n-rowed scalar matrix coefficients A^.
all
the poly-
. We call f(x) monic if a = 1. For obvi-
where /
ously this case of Theorem 4 is now the result of multiplying nomials of Theorem 1.
l) O/
6)C=(l
\0
I/
(7=12-1
\0 I/
The
characteristic matrix
and function.
the n-rowed identity matrix.100
INTRODUCTION TO ALGEBRAIC THEORIES
EXERCISES
1..
2z 4
-
x*
+ 2 -x + x 8
1
X
t
__ O'rS iJU
ff JU
\jJ(/
Q/2
2
_l
|"
(1 +
3
0*2
/v. If now g(x) is also a scalar polynomial. Q(x).)I
.(C)
by the use
of the division process
as well as
by
substitution
/O
1
0\
/O
2
0\
c)
/I
2
0\
a)C=(0
\2
4.
Use /(#)
of Ex. r(x).. l(c)
and
if
find /ft(C) and/z. and
(12) with
(16)
a matric polynomial shall call/(x) a
/(x)
is
=
(a<&
+ . the quotients q(x) and Q(x) in (13) are the same scalar polynomials and also r(x) = R(x) is scalar. then we scalar polynomial Thus A k = a k l for the a^ in F.2
y?
x
-
1
x i
2
5
2^2
T
3
2.
Express the following matrices as polynomials /(x) and g(x) with matric
coeffi-
cients
and compute
q(x).2
4 x
3a.1 by the n-rowed identity matrix.+ a.
R(x) of
(13).
A) are the
cof actors of the elements of
(in general. We now apply (3. g(A) = 0.(A) are equal for every scalar polynomial /(x).POLYNOMIALS WITH MATRIC COEFFICIENTS
If
101
any n-rowed square matrix. then q(x) We now define the minimum function of a square matrix A to be the
monic scalar polynomial of made above then implies
least degree
with
A
as a root. q(x). of the elements of B(x) is unity so that if B(x) = Q(x)q(x) for a monic scalar polynomial = 1.
A) is clearly the polynomial d n -i(x) defined f or xl A and d n (x)= \xl A \. Every square matrix is a root of its
characteristic equation.
. is a matric polynomial.
+ a. and the corresponding equation /(x) = A to obtain the characteristic equation of A. Hence B(x) A were defined in (8)._iA + aj
. The invariant factors of xl = = and by (19) \xl (xl A)B(x)d n _i(x). But then adj (x/ A) =d n _i(x)B(x). clearly. and adj (xl By the argument above we have Theorem 7.
The remark
just
Theorem
trix of A.
A) is a matric polynomial A.27) to xl
(19)
(xl
-
A)[adj (xl
-
A)]
=
[adj (xl
-
A)](xl
-
A)
= |x/.
We now say that either the polynomial /(x)
as a root
/(A) only if /(x) has xl
if
=
or the equation /(x) = has 0. Observe also that the g.
Hence.or a left-hand factor.A
Then
xl
the elements of adj (xl
|
/. where B(x) is an n-rowed square matrix with elements in F[x\.d. By Theorem 6 the matrix A is a root of /(x) if and A as either a right.c. the scalar poly-
f(x)
=
\xl
-
A\
I
the characteristic function of A.
A
We
shall call the
matrix xl
|
A
the characteristic matrix of A. By the gi(x)d-i(x)I A\I uniqueness of quotient in Theorem 4 we have
of the elements of adj (xl
(20)
The g. the de-
terminant \xl nomial
(18)
A
the characteristic determinant of A. nonscalar). the polynomials //?(A) and/i.
8.d.
g(x)
=
0i(x)7
=
(xl
-
A)J5(x)
.
. Let gi(x) be the first invariant factor of the characteristic
ma-
Then g(x)
=
gi(x)I is the
minimum function
of A.
.c. and we shall designate their common value by
is
A
(17)
/(A)
= aoA
+
.
and zero.
. as we have already ob-
served in an earlier discussion. then ~ A~~ does not exist.
A"*
it is
= -o^ (A^
l
+ M.
In closing this section we note that
(x
we
|
write f(x)
=
\xl
n
|
A\
/
=
+ aa". so that
\A\
(23)
= (-l) nan
-
It follows that. But then. If. r
|
\
(21)
P(x)
(si
-
A)
. Since A 9* 0.r(z) = 0.
. Moreover.
A A~ = A.
\A
\
^
2
0. from which g(x) = h(x) as desired.A. q(x) = 1.
+ a*) /. and since g(x) and h(x) are monic so is q(x).
then
.
. we have m > 1.
(22)
c-
\xI-A\
\Q(x)\ in F.
and we have proved
product by I o/ tfta product of the invariant factors of the characteristic matrix of A.
. A are the field of all complex numbers.
. .
and d
|
=
Hence cd
=
1. By Theorem 6 we have h(x) = (xl A)Q(x).
d
=
gl
.. It follows that
Theorem
jPAe characteristic function of
a square matrix
A is
/&#
every root of \xl
in fact
taining g\(x). and G = A m + m~ + +
and hence
l
obvious that
l
.
Q(x)
=
diag
{jfi.102
INTRODUCTION TO ALGEBRAIC THEORIES
is
For if h(x)
the
minimum function of A we may write g(x) =
and
h(x)q(x)
+
r(x) for scalar polynomials h(x)
r(x) such that the degree of r(x)
is less
than that of h(x).+
1
.and0(A) = Osothatr(A) = 0. The uniqueness in Theorem 4 then A)Q(x)q(x) states that B(x) = Q(x)q(x).
+ an-!
/)
.
. if \A\ =0.
..
. This result implies that \xl A\ is the product of g\(x) by divisors
gi(x) of g\(x}. polynomials (x \xl A\.
. the elementary divisors of xl Then whose is the are the c/)*w product c. then/(0) = -A
if
7
\
=
(-l) |A
/. A 7* and gi(x) = x m + bixm + + m the polynomial gf(a:) = g\(x)I can be the minimum function of A only if ~ b m = 0.+
l
1
.
gn ]
for elementary matrices P(x)
and Q(x). and by (20) we have g(x) = = (xl (x/ 4)5(z).gn
for c
=
|P (a:)
9. But&(A)
= 0.
fc
.
distinct roots of g\(x) as well as of
teristic roots
F
is
a root of
But
A in any field K conwe have already seen that if F is
\
xl
|
A
if
. We see now that xl A is a monic polynomial of degree n and is not = m = n in (9). Hence g(x) = h(x)q(x).
.
\
They
are called the charac-
of A.
l
M
2
.
.
6m _i/
(24)
is
a nonzero matrix with the property
AG = GA =
. then.
6.= B
1
.
Show
that
f(x)
is
any polynomial
and we
define
first
(x)
=
f(x)I t any t. Verify this for the
5.
We
have defined two n-rowed square
matrices
A
and
a nonsingular
B with elements in a field F to be similar in F if there exists matrix P with elements in F such that
PAP. P is elementary. are the minimum functions
.
if
For
PAP.
6)
a)
(o
3.
is
let
g(x)I m^ n gi(x)I m and Qi(x)I n be the
. Similarity of
square matrices.
.
Hint: Prove
by
induc-
Let
A
have the form of Ex.
What.
.
^
. A%\.
The principal result on the similarity of square matrices is then given by Theorem 10.
/O (6
1\
e>
/2 (4
3\
l)
/o
0\
6J
>
-2J
{^4i.
gi(x).
respective
minimum
and
A*
Prove that g(x)
is
the least
common
multiple of g\(x)
5.
Compute
the characteristic functions of the following matrices. Two matrices are similar in F if and only if their characteristic
matrices have the same invariant factors.
103
When
is
the
minimum
function of a matrix linear?
of the following matrices?
2.=
1
B.B = xl .
4.
Apply Ex.A)P~ == xPIP~ . 4 in the case where A\
nonsingular and
A =
z
0..B. A is in F.
.fn (A 2 )}. then/m+nCA) k tion that A = diag [Af. 3 and
functions of A. Hence xl
l
l
.
It
may
+
(
be shown that the characteristic function f(x) = x n n~* n l)*atX ( l) a n of an n-rowed square matrix
a\x
n~ l
+
+
+
A
has the
property that di is the matrices of Ex.POLYNOMIALS WITH MATRIC COEFFICIENTS
EXERCISES
1.
Let
for
A =
diag
A^} where 4i and
if
A 2 are square matrices of m
in F[x]
and n rows
f
t
respectively. .
sum
of all i-rowed principal minors of A.
/2 (l
3\
. A\.
then
\P\
P(xl
But
P has elements in F. then.
7. 6.
=
diag {fm(Ai).
. Characteristic
..
1
B
4.
....
x
62
-1
x
61
. then A\x and Q with are equivalent in F[x] if and only if there exist nonsingular matrices
elements in
+
+
F
such that
in (25).
. !.
..
BT^
Show that the hypothesis that A\ is nonsingular in Ex.
3.POLYNOMIALS WITH MATRIC COEFFICIENTS
EXERCISES
1.
yet
P
and Q do not
exist.
.
Hint: Take
P A = -Ar ^. 3 is essential by proving
/ and B\x
that Aix
J are equivalent hi
F[x].
. B is equivalent in xl nt For then xl
.
-1
x
-1
. are the nontrivial invariant factors of a matrix xl A and n is g t (x) the degree of 0(z).
105
What
are
all
possible types of invariant factors of the characteristic matrices
of square matrices
2.
. .
t ]
F[x] to diag {G i...
and Bz are n-rowed
A 2 and E\x B2 square matrices such that A\ and B\ are nonsingular.!}..
A^
Bi. and we conhas the same invariant factors as xl A Thus the problem
. 2.
B
t ]
will be similar in F to A if Bi is an n-rowed square matrix such that the only nontrivial invariant factor of the characteristic matrix of Bi is g>(x).
having
1.
Let g(x)
=
xn
(32)
010 00 000
1
(bix*-
1
+
. B = diag {xl ni BI.. if
B
6. where Gi
=
.
.
+ bn
)
and
bn
Then g(x)I
For
x
(33)
b n _i b n
and
the
is both the characteristic
minimum function
of A.
l
matrices with prescribed invariant factors.
f
clude that xl
B
diag {#. 3.
.
or 4 rows?
Give the possible elementary divisors of such matrices. g(x) is
the only nontrivial invariant factor of xl
A. then by Theorem 1 the n-rowed square matrix
.
B =
diag {Bi.
PA& =
Bi and
PA Q = B2
2
.
.
Use the proof
of
Theorem 10
to
show that
if
AI.
....
.
. If g\(x).(?}.
of constructing an n-rowed square matrix whose characteristic matrix has prescribed invariant factors is completely solved by the result we state as
A
Theorem
9.
desired.
x in the
first
row and column of (33).
t
. !.106
INTRODUCTION TO ALGEBRAIC THEORIES
& n of (33) is an (n The complementary submatrix of the element 1)1 and its derowed square triangular matrix with diagonal elements all ~ terminant is (-I)"-1 Thus the cofactor of -6n is (-l) n+ 1 (-l) n 1 = 1 and hence dn -\(x) = 1.
The
whose
construction of square matrices
characteristic matrices
simple solution. acteristic matrix xl c n But equivalent in F[x] to diag {/. For xl n The c) complementary minor of the element in the \xl A\ = (x A is a triangular matrix with diagonal nth row and first column of xl n elements all unity.
.!}.
..
. This is true if n = 1. so that Let it be true for matrices A n _i of the form (33) and of n 1 n~2 =
.
..
.. we construct matrices Aj of the form (34) for c = c..
.
+ &n-i) is the cofactor of the element
. This proves our theorem.
t
. dn-i(x) = 1. .
.
.
./* as
. ..
its
bn
=
g(x) as desired. B }.
. The matrix A = diag {Ai. It remains to prove that dn (x) = \xl A\ = g(x).
..
\
\xl
A n-i\
first
z""
(bix
+
... Let c be a complex number.
.
.. and \xl 1 rows.. .
... Thus if Ci..
c
1
A with complex number elements have prescribed elementary divisors has a see that the argument preceding Theorem 9 A
be the n-rowed square matrix
c
1
(34)
000 000
.
. and dn (x) = (x c) is the only nontrivial
. and its charrows.
c
1
c
A is (x c) n Then the only nontrivial invariant factor of xl A is a triangular matrix with diagonal elements all x c.
t
t -)
. A = x bi.. and we shall reduces the solution to the proof of Theorem 10. .!} such that / = (x then xl A is equivalent in F[x] to diag {/i.
t
tegers. and with rc A ] then has n rows./. We now expand (33) according to n~ l n~ 2 column and obtain \xl + + 6n-i)] (b\x A\ = x[x
.
.!. since then A = A\ = (61). where Bi = xl ni Ai is A = diag {Bi. and by Theorem 3 the nontrivial elementary divisors of xl A are /i.
invariant factor of A.
n are positive inc are complex numbers and ni.
however.
B =
diag {Bi.
107
Compute the
invariant factors and elementary divisors of the characteristic
matrices of the following matrices.
.
> c defines a and the correspondence c < self-equivalence or automorphism This of K. automorphism of K leaves the
elements of
its
subfield
R
unaltered. that
jR.
now
the ith nontrivial ele-
7.9. a subfield of K. Bi of the form (34).
a
=
a for every real
a.
where
R
is
the
is
field of all real
numbers.
is. a fact verified in Section 6. b
in R.
.
is
where Bi has the form (32).
topics of the theory of
their exposition
matrices other than those
to
we have
and we leave
more advanced texts.
We
then
may
call it
an automorphism over
. The quantities of the field K of all complex numbers have the form
c
=
a
+
bi
(a. 2 with the characteristic function of Bi mentary divisor of A.
Additional topics. There are
many important
discussed.POLYNOMIALS WITH MATRIC COEFFICIENTS
EXERCISES
1.
.
1. i2
=
1)
.
3.
Find a matrix
respectively.
B
t]
similar in
to each of the matrices of
Ex.
Bi
Solve Ex. The complex conc
jugate of c
=
a
bi
. and the characteristic function the ith nontrivial invariant factor of A.
.
-6
2
2
-5 -19\
1
5
1
-2
3-12
F
1
0-5
9^
of
2. Let us mention some of these topics here.
for every A.108
INTRODUCTION TO ALGEBRAIC THEORIES
a. Two matrices A and B are said to be conjunctive in K if there exists a nonsingular matrix P with elements in K such that
7
.
(J)'
=A
7
. B.
Two symmetric matrices A and B with elements in a field F are said to be orthogonally equivalent in F if there exists an orthogonal matrix P with elements in F such that PAP' = B. D be matrices of the same numbers of rows and columns. Moreover.
*
In this connection see Exs. D congruent pairs D for a nonsingular matrix P. But PP' = P'P = / so that B = PAP~ l is both similar and congruent to A. D are n-rowed square
y
matrices. where P is a unitary matrix. Then we say that two Hermitian matrices A and B are unitary equivalent if PAP7 #. B C. skew-Hermitian
We A if JF = A.
we
f
call
C and PBP =
will
A. 3 and 4 of Section
5. Then we call the pairs A. Let A. if A.
WS
The results of this theory are almost exactly the same as those of the theory of congruent matrices. a unitary matrix. in fact. Analogously. we define A to be the It is then m by n matrix whose element in its ith row and jth column is
a simple matter to verify that
Z5 = IS
. let us mention the topic of the equivalence and congruence of pairs of matrices. Both of these concepts may also be shown to be special cases of a more general concept. B equivalent* pairs if there exist nonsingular square matrices P and Q such that simultaneously PAQ = C and PBQ = D. B and C.
R=
()-*
.
If A is any m by n matrix with elements a a in K. and it is.-/. B and C. and nonsingular C. Finally. then (JHB)' = now define a matrix to be Hermitian if JF = A.
if
simultaneously
PAP' =
References to treatments of the topics mentioned above as well as others be found in the final bibliographical section of Chapter VI. Similarly. B. possible to obtain a general theory including both of the theories above as special cases. we call a matrix P with complex elements such that PP 7 = P 7 ? = /. C. We shall not state any of the results here.
.
and we are ready now to begin the study of abstract algebra.
The product ab
is
in G.
ever.CHAPTER
VI
FUNDAMENTAL CONCEPTS
1. Our first new concept is that of a set G of elements closed with respect to a single operation. respect to multiplication if for every a.
The
associative law
a (be)
=
(ab)c holds. Howtained. b. b. linear space.
There exist solutions x and y in
G of the equations
. and equivalence. correspondence. c. . we believe it desirable to precede the serious study of material such as that of the first two chapters of the Modern Higher Algebra by a brief
Modern Higher
discussion of this subject matter.
ax
=
b
ya = b
. Groups. without proofs (or exercises). It should be clear that if we do not state the nature either of the elements of G or of the operation.
desired.
III. The objective of our exposition has now been atFor our study of matrices with constant elements led us naturally to introduce the concepts of field. c of
A
G
is said to
form a group with
G
I. and we wish to define the concept that G forms a group with respect to this operation. If we wish later to consider special sets of elements with specified operations we shall then replace "product" in our definition make the
knowledge of how these concepts
by the operation
. as presented in the usual course on the theory of equations.
The reader is already familiar with the groups (with respect to ordinary multiplication) of all nonzero rational numbers. it will not matter if we indicate the operation as multiplication. of elements a. and the abstract approach of the author's
Algebra.
II. all nonzero real numbers.
109
. shall give this discussion here and shall therewith not only leave our readers with an acquaintance with the basic concepts of algebraic theory but with a
We
may lead into those branches of mathematics called the Theory of Numbers and the Theory of Algebraic Numbers.
Thus we
shall
set DEFINITION. These pages were written in order to bridge the gap between the intuitive function-theoretic study of algebra.
.
every element a of
(2)
G has a
=
1 unique inverse a" in
G
such that
aar 1
ar l a
=
e
.110
INTRODUCTION TO ALGEBRAIC THEORIES
all
and.
Then the
(3)
solutions of the equations of
Axiom
III are the unique elements
x
=
ar l b
.
of elements in a group
G
is
called its order. of all
field. Then H forms a group with respect to the same operation as does G. or it is a a finite group of order n. subgroup of a group. and
we
call
G
LEMMA
1. subfield of a field. This
finite
and we call G an infinite group. The equivalence of two groups is defined as an instance of the general definition of equivalence which we gave in Section 4. The reader may verify that an example of a finite nonabelian group
*
A
The proof
is
given in Chapter
VI
of
my
Modern Higher
Algebra.g. The concept of equivalence of two mathematical systems of the same kind as well as the
concept of subsystem (e. The products ab
=
ba
.
simple example of a finite abelian group is the set of all nth roots of unity.
groups
G
such that for every a and 6 of
G we
These are have
all
examples of
IV.
A set H of elements of a group G is called a subgroup of G if the
of
product
any two elements of H is in H. linear subspace over F of a linear space over F) are two concepts of evident funda-
mental importance which are given in algebraic theory whenever any new mathematical system is defined.
nonzero elements of any
field. H contains the identity element of G and the inverse element h~ l of every h of H. For finite groups we have the important result which we shall state without proof.
Such groups are
tiplication.
Moreover.. with respect to matrix muln-rowed square matrices with elements in a nonsingular
e called its identity element .
called commutative or abelian groups.
Every group G contains a unique element such that for every a of G
(1)
ae
=
ea
=
a
.
The number
either infinity.5.
. indeed. *
number is number n.
is
An
example of a
nondbelian (noncommutative) group
the group. Then
the order of
H
divides the order of G.
Let
H
be
a subgroup of a
finite
group G.
2.
a'TO
1
1
the identity element of [a].)'
-i.
.
.FUNDAMENTAL CONCEPTS
is
111
given by the quaternion group of order 8 whose elements are the tworowed matrices with complex elements.
a. Thus
n = mq. . and we may apply Lemma 1.
set of all
powers a
= G
e.
.
of an element a of a group
forms a subgroup of
G
which we
shall desig-
nate by
(6)
[a]
and
shall call the cyclic group generated by a.
.
The order
m is
of an element of a finite group G divides the order of G. that is. the order of the subgroup [a]. or
order
ra. Then the order of a is the least integer t such that a* = e.
Additive groups.
-AB.
f
T '
A
/
0\
(4)
Mo-. Tften
an
=
e
/or et. the elements of any linear space. a m = e. Moreover. a n =
m
)
q
=
eq
=
e. groups with respect to the operation of addition as defined for each of these mathematical systems.
or 1 a2 cr2
.
.
(7)
a2
.en/ a of G.
Let e 66
2/ie
identity element of a group
G
o/ order n.
-B. But the elements of any field.
and
[a]
consists of the
m
powers
e. Its order is called the order of
the element
a and
it
can be shown that either
a has
finite
all
the powers
(5) are distinct
and a has
distinct
infinite order.
D B
/O
-1\
o)'
-(i
[AB.
. In any field and in the set of all n-rowed square matThus we have said that the set of all nonzero
elements of a
field and the set of all nonsingular n-rowed square matrices form multiplicative groups. all form additive groups.
. a. it can be shown that a* = e if
where
e is
m
and only
since
if
m divides
m
(a
t.
.
The
(5)
-A.
We
therefore have
LEMMA
(8)
2. The reader will observe that the axioms for an additive abelian group G are those axioms
.
rices there are two operations. the set of all m by n matrices with elements in a field.
y
+
=
a
=
b
. that is.
. 2-
-(2
a).
the product of a
mands.
where.
.112
INTRODUCTION TO ALGEBRAIC THEORIES
for addition
which we gave in Section 3. Additive groups are normally assumed to be abelian.
by
-a. a. 2a.
.
3.
F is an instance of certain type of mathematical system called a Many other systems which are known to the reader are rings and we
the
shall
make
ring is an additive abelian group of at least two distinct elements such that.
c)
III.
a.
we have x
y and designate their
common
value by
Thus we
define the operation of subtraction in terms of that of addi-
In a cyclic additive group
(11)
0. a. b.
.
The
in-
a
+
(
a)
=
(
a)
+a
=
0.
II. the element
verse with respect to addition of a
is is
usually called
its zero
+
=
a
=
+
a.
m
(
a). of R.
(b
+
c)a
= ba
+
ca hold. if [a] is infinite.
associative law a(bc)
distributive laws
=
(ab)c holds.
The set consisting of all n-rowed square matrices with elements
in a field
ring. . then it may be seen that n a = q a for any integers n and q if and only if n and q are
by the
.
and we
Here
m
a does not
mean
.
Thus
a and is such that designated as the solutions of the additive formula-
tion
(9)
a
+x=
b
.
The product ab
The
is
in
R. The identity element of an additive group such that a element. . clearly.
The
a(b
+
=
ab
+ ac. for every a. (m 1) a.
. that is.
When G
b
tion.
positive integer
If [a] is
m
+
+
a
finite
equal.
(m
a)
=
.
abelian. However.12 for a field.
. c.
Rings.
of the equations of our group
(10)
Axiom
III are
z
is
= (-a)
+
6. the use of addition to designate the operation with respect to which a group G is defined is usually taken to
connote the fact that G is abelian.
if
[a]
the elements are always designated
a.
define
(
m
=
is
ra)
a
any (m
positive integer
a).
A
I. the elements of [a] are 0. and m is least positive integer such that the sum of m summands all equal to a is zero.
but means the sum a a with m sumgroup of order m.
DEFINITION.
y
6
+
(-a).
.
A ring may also contain divisors of zero. They may be found in the first chapter of the Modern Higher Algebra.FUNDAMENTAL CONCEPTS
We leave
113
to the reader the explicit formulation of the definitions of sab-
ring and equivalence of rings.
Abstract fields. If
R
is
any
ring. A ring is said to possess a unity element e if e in R has the property
=
ea
=
ae
=
a for every a of R. the set of all ordinary integers is a ring.
let c
and d range
. the ring of all n-rowed square matrices has already been seen to have such elements as well as the other properties just mentioned. that is. In nonzero elements of this ring are divisors of zero. all
rices
with elements in a
field
F
is
a noncommutative ring. However. The unity
element of the set of
n-rowed square matrices is the n-rowed identity matrix. and the unity element was always the number 1 in the other rings we have studied. The reader should also verify that all nonmodular
are rings. Observe that by making the hypothesis that R contains at least two
fields
The
elements we exclude the mathematical system consisting of zero alone from the systems we have called rings. The set of all
is
n-rowed square matrices
not a division
ring. b are in R.
4. the set of all two-rowed square matrices of the form
/O
(12)
r\
\o
with
r rational
o.
However.
we
shall designate
by
R*
*
the set of
all nonzero elements of R. The element
1
e
then has the properties of the
ordinary number
and
all
is
usually designated
by that symbol. elements a 9* 0.
may easily be seen to be a ring without a unity element. Then we shall call R a division ring if R* is a multiplicative group. Rings may now be seen to be mathematical systems R whose elements
have
ya
all the ordinary properties of numbers except that possibly the prodab ucts and ba might be different elements of R. In particular. zero element of a ring R is its identity element with respect to addition. the ring of all n-rowed square matfact. c ^ such that ac = 0. This occurs clearly if and only if the equations ax = 6. The ring of all integers is a commutative ring. the equations ax = b of
b might not have solutions in R if a ^ 0. ya = b have solutions in R* for every a and 6 of R*. A ring R is said to be commutative if ab = ba for every a and 6 of R.
. Less trivial examples are the set x q ] of all polynomials in F[x] of all polynomials in x. q
.
A =
ft
\a
*
c
is
a noncommutative division ring.
.
. noting nonsingular since A 9^ implies that dd > 0. it
generates an additive cyclic subgroup [1].
prime.
. If this cyclic group has may be shown to be equivalent to the set of all ordinary
integers.
. It is usually called the ring of real quaternions.
DEFINITION. x and coefficients (in F. then the sum a
+
is
1
p summands
such
fields
equal to the product (1
all
+1+
may F
is
+ + + + 1)0 = 0.
group. We now define fields in general. commutative ring with a unity element and withIntegral domains. out divisors of zero is called an integral domain.
under rational operations. We
.
p
-
1
. and
it
we have the property that the sum with p summands is zero.A. the set of all ordinary 1.
. the unity eley
ment
5.
where p
1
is
. be a finite group. every subfield of
F contains the subfield
generated by
F and. 1.
.
Until the present we have restricted the term "field" to mean a field containing the field of all rational numbers.
is closed with respect to rational operations. both .BjAB
of (4) as a basis over that field. 5 and d be the complex conjugate of Q of all two-rowed matrices
c
and
d.
+1+
.
element
identity element of the multiplicative group F* is then the unity 1 of F.
call
F
modular fields of
characteristic p. and [1] then of F equivalent to the field of all rational numbers. The ring Q is a linear space of
it
order 4 over the
field of all real
numbers and
has the matrices
I. cases) . The reader should verify
in particular that every
c 7*
A ^
=
cc
is
or d j
and \A\
+
this. in a field .
the order of this group.
. The whole set F is an additive group with identity element
and
1
infinite order. Any field forms a somewhat
of
F
A
trivial
example of an integral domain. It
easily be shown that
the characteristic of
subfields of a field
the same as that of
in fact. and
.
A
field is
a ring
F
such that
F*
is
a multiplicative abelian
The
0.
. The group [1] might. Its elements are then
But F
the sums
(14)
1
=
0. It is easy to show that p is a a with a follows that if a is in F. 2.114
over
all
INTRODUCTION TO ALGEBRAIC THEORIES
the set
Then
(13)
complex numbers. We a subfield generates call all such fields nonmodular fields. . however.
. the set F[XI.
of J is an a
^
Every associate of a prime is a prime.
If
u
in
J
divides
its
unity element
in J.c.
.
115
The
set of all integers
may be extended
to the field of all rational
numbers by adjoining quotients a/6.c. The problem of determining a
common
divisor) of
two
elements a and b of a unique factorization domain is solvable in terms of the factorization of a and 6.
b
and
6
are associated
and only
if
6
=
bu for a unit
Moreover. s of a and b are associates.
some
When
.
whenever
also
=
We may
#.
fact.
b divides a so does every associate of b.c. quantity p of an integral domain J is called a prime or irreducible quanis not a unit of J and the only divisors of tity of J if p p in J are units and
A
^
associates of p. Let us then formulate the problem regarding g. 's.d. Thus we are led to one of the most important problems about an inte-
gral domain.c. . of two elements a and b not both zero of J to be a common divisor d of a and b such that every common divisor of a and 6 divides d. which is neither a prime nor a unit of
A composite quantity
J. as well as the
Xij
.
.
.
J such that
is
be.
true that
qi
qa for primes
then necessarily
s
=
r
and the
qi
are associates of the p/ in
these properties hold we may call J a unique factorization integral domain.d.
pr
also ask
if it is
for a finite
number
a
of primes
. Then all g.
p of
J.
clearly
Thus u is a
also a unit. . that of determining its units. In a similar fashion we adjoin in J to imbed any integral domain J in a field. It is natural
then
to ask whether or not every composite a of
(15)
J may be
written in the form
a
=
pi
.
u. The reader is familiar with the fact that the set of all ordinary integers is such an integral domain. We call a and b relatively prime if their only common divisors are units of J.d. quotients a/6 for a and b ^ When we studied polynomials in Chapter I we studied questions about divisibility and greatest common divisor.
then u
is
called a unit of J. (greatest
author's
Modern Higher Algebra.d.
corresponding property for the set of all polynomials in x n with coefficients in a field F are derived in Chapter II of the
g. We may also study such questions about arbitrary integral domains. Every unit of J divides every a of J.
unit of
J if it has an inverse
The
inverse of a unit
Two quantities b and 6
Then
if
are said to be associated
if
if
each divides the other.
may be found by a Euclidean process.FUNDAMENTAL CONCEPTS
integers. Let us then formulate such a study.-.
. we saw that in the case of the set F[x]
the g. Then we say that a is
divisible by
b (or that b divides
a
=
a)
if
there exists an element c in
1. We define a g.c. However. This
order.
. Let J be an integral domain and a and 6 be in J.d. 6/^0. .
and if H many elements mi. We may then define for every a of R. if a c in Af. When J? is a commutative ring.
a
6.
(read:
M M
+
Let us now define the sum and product of residue
(18)
classes
by
g
+b=
a
+b
. It consists of all finite sums of elements of the form amb = {m} consists of all for a and 6 in R.c. we write {H} = {mi. contains g R if h. m show that {H} is an ideal. they have the unity element of J as a g.
. and finally whether or not may be found by the use of a Euclidean Division Algorithm. of a and b. we must ask whether there exist
x and y in
J
such that
d
is
=
ax
+
by
d
common divisor and hence a g. products
for the corresponding ideal.c. ag. A subset and a of R. This
term
is
{
de1
}
. H is any set of elements of a ring J?.
t
. am for a in R.d. We now see that one of the questions regarding an integral domain J is the question as to whether every a and b of J have a g.
If
ma of any element m of M and any element a of 1? are in M.
g
b
= a-
b
. This
M
The
ring
R
itself is
an
ideal of
R
called the unit ideal. we may designate by {H\
the set
sums of elements of the form xmy for x and y in #.
. be an ideal of R and define Let
M
(17)
o
s
6
(M)
what we
a congruent b modulo M) if a b is in M.d.d. We put into the shall call a residue class g of class every b in R such that a = 6 (M). consists of only one element m of R we write
consisting of all finite in H.
rived from
the fact that in the case
where R has
a unity
quantity R =
Evidently {0} is the zero ideal.
. It is easy to
. 6 c is It follows that c) (6 (a and b are the same resig due class) if and only if b is in g.
.
of a ring R is called an Ideals and residue class rings.116
that
INTRODUCTION TO ALGEBRAIC THEORIES
is. Clearly a = b (M) if and only if b 33 a (Af). If H consists of finitely m of jR. Moreover. . m/}. ga.
. or may be seen to be a subring of R with the property that the products
ideal of
M
M
M
M
M
am.
.c. for every g and h of Then either consists of zero alone and will be called the zero ideal of R. then (a 6 is in and b 6) = = a in M. Moreover.
y
(16)
M=
{m}
most important type of an ideal is called a principal ideal.
Let a and b ^ be integers.
. 2. etc.
then ai
+
b\
=
a
+ b.5.
this
ring. a = bqi + n with whereas r n < b This is possible only if r\ = r and q\ = q..2|&|.
We now
*
leave for the reader the application of the Euclidean process. Then let (g 1) |6| be the least multiple of |6| which exceeds a so that .
|6|.
class
is
call
call
divisorless
if
is
field.
our definitions of
sum and product
of residue
classes are unique.
the multiples
(19)
0. and have g\b\ = <?6.g\b\ = r such that < r < \b\.
are clearly a set of integers one of which exceeds* any given integer a. This coincidence occurs in most of the
topics of mathematics (in particular the Theory of Algebraic Numbers) where ideals are studied.FUNDAMENTAL CONCEPTS
may be = 06.
and we
first
shall
prove that
has the property of unique factoriare those ordinary integers u such 1 and 1 are the only units of E. aifri
It
verified readily that
It follows that
if
117
a\
=
g.
1. Then it is easy to show that if is not R the set of all the residue classes forms a ring with respect to the operations just defined. a = bq + r. -2|6|. When
all
class. and their associates 5..
The
ring of ordinary integers.
7. Every integer a is associated with 3. The process yields
We use the concept of
magnitude of integers throughout our study of the ring E. q and r such that
integers
0<r<
1
1
.
its
absolute value
a
| \
^0.
E
that uv
=
1
for
an integer
But then
Thus the primes of E are the ordinary positive prime integers 2. put 1)|6| > a.
We
observe
that the units of
v.
shall designate henceforth
E. We
note
now
that
if
b is
any
integer not zero. We
. etc..
| | | \
have thus proved the Division Algorithm for E. It
easily seen to be
an
inte-
gral domain. 61
=b
.
The set
by
it
of all ordinary integers is a ring
is
which we
zation.. We call this ring the residue class or difference ring (read: R minus
M
R M M = R the residue classes are the zero and we have not called set a M an integral domain.
-|6|. Then there exist unique 6 a = bq + r. If also < r\ < \b\. then b(q q\) = n r is divisible by 6. we the When the residue ring R ideal M a prime ideal of R. 3. to
which we used to prove Theorem
our present case. a result we state as Theorem 1.(g
+ +
We
q
= giib > Qj q = g otherwise.
in
finite
number of elements since it may be shown that any integral domain with a finite number of elements is a field. We Ma ideal of R R M M has only a a These concepts coincide the case where R
M). a g\b\ <a. 5.
. Then p divides g or h. For if a = 6c.
For let a and b be prime to m and d be the g.
.
Theorem
and b such
(20)
is
Let
f
and g
be nonzero integers.
for positive
. it does
. .
primes
qi
.
is
sometimes called the Fundain the form
Every composite integer a
is expressible
a
=
pi. But if the divisor pi of qi q8 is not equal to gi. For if p does not divide gr.
l
stages the sequence of decreasing positive integers |a|
.c.
af
+ bg
is the
a and
and
Then d
unique positive
g.
forj
=
1. But then a divisor c > 1 of d divides a or b as well as m contrary to hypothesis.
Theorem
Let
m
be
an
integer.
.118
INTRODUCTION TO ALGEBRAIC THEORIES
2. Let f.
. of p and g is either 1 or an associate of p. . and p\ = Then either we may arrange the q* so that gi pi.
|
>
.
(21). the sign
is
.
|as|
>
must terminate. and by Theorem 3 if d is prime to a it divides 6. If also
a
=
qi
.p r
. If a
it
.
. For by Theorem 2 we have af + bg
gh
=
1.
. the g.
. and we have
gi. a = p\Oz for \a^\ < \a\.
and there
exists a least divisor
Pi
But then pi is a positive prime. .
Then
the set of integers
prime
to
m
is
closed with respect to multiplication.
Then f divides h. If Oz is a prime. we write a 2 = pz with p 2 a positive prime and have (21) for r = 2.
uniquely apart from the order of the positive prime factors pi.
Then
there exist integers a
that
d
positive divisor of both f
g. If a6 is not prime to m we have d > 1.. The latter is impossible.
pr = uniquely determined by a.
>
. .. of ab and m.
We may now conclude our proof of what mental Theorem of Arithmetic.
.d. g. .
=
g.
is
composite. h be
integers such that
f
divides
gh and
is
prime
to g.
Let p be a prime divisor of gh.
|a 2
. a 2 = p 2a 3 and a = pip 2 a 3 f r a s| < |aa| After a finite number of
>
1 of a.
But by hypothesis
= /g. every divisor of b or of c is a divisor of a.
g.
.
of f
The result above implies Theorem 3.
.
Theorem
(21)
6.
g.
afh
+ fy/A = h..
. Otherwise a* is composite and has a prime divisor p% by the proof above.
.c.
Theorem
We
also
have
5.
q.
+
bq)f
is
divisible
by /.
We
A = (ah then have
4.
has divisors b such that
1
<
b
<
|a|.d.
. or pi ^ g/ s. and therefore p is prime to g.c.d.
pr
.
we ultimately obtain
r
=
s. ^ r. For by Theorem 2 there exist integers c and d such that
If
a
(23)
If 6 is in a. have proved 7.
.
. 1.
>
1.
classes
m-
I
.
.. ?.
d
>
1.. If m is a composite integer cd where
and Theorem 7 implies Theorem 8.
. and we may take p 2 = q*.
.
But a
=
to
classes a in a abelian form multiplicative group.
.
m
1.
then
m>
c. . . p2 . the elements of the residue class a are prime to m.
.
. then * We may order the p so that pi ^ p 2 ^ ^ p and similarly assume that ^ ^ 7. be
6
+ m(d
m
E
c
qc)
=
ac
ac
+ md
=
+ m(qc +
d
qc)
c
=
1. The residue
{m} defined for a prime
m
d.
the p
=
qi for
an appropriate ordering*
of the q^
8.
and therefore
and
are relatively prime.
classes of
E
modulo {m} are now the
0.
119
not divide qi and by Theorem 4 must divide q z q 8 By our hypothesis it does not divide # 2 and must divide g 3 qa This finite process leads to
. Thus the elements of the residue
.
. E {m} has divisors of zero and is not an integral domain. that every element of M
in
/i
+
r is in
. r.
=
a
+
mq.
.
m>
and d are both not the zero class.
then b
1. TTie ideals of E are principal ideals {m}
is
We
m a positive integer. Then of all integers divisible
integers
E
in
{m}
is
a ring whose zero
1
by m and whose unity element
the class
is 1.
. If h is in M. we may use Theorem 1 to write h < r < m.
The
ideals of the ring of integers. p r = ?*.
.
whose remainder
Theorem
m
on division by
all
is an integer prime to m.
. But C'd = cd = m = Q.
. and the proof just completed may be repeated.
. Let
the least positive integer in Then contains every element qm of the principal ideal {m}. = & for i = 1. 1.
and thus
Theorem The residue
(22)
in {m}.. p.
.
r
<?i
.
For
if
a
is
any
integer.
Then a
r is in
{m}
.
.
.
ac
+ md =
1
. Proceeding similarly.. Then we obtain r = 5.
a
=
we have a = mq + r for r = 0. Thus pi = gi.
class ring
Edefined for
{m}
m>
1
element
is
is
the class
1 of all
are given by (22). But mq and h are r where mq
set
.
Our
definition of
m implies that r =
.
E of all integers and m be
mq =
M be a nonzero ideal of the M M
M M..
. If m is a prime. .
a contradiction.FUNDAMENTAL CONCEPTS
. 02
and
c
{
.
. Hence if (24) and
(25)
c 3=
=
c
6. Then
ap
=
a (mod p)
for every integer a.
=
d (mod m)
hold. p 1.2. Thus f(p) = p 1 l p = a a Theorem 10 aP~ 1 is divisible by p. we will have g + c 6 d..
The
*
ring
E
{p} defined
by a
positive prime
p
is
a field*
P
whose
field
This
field is
equivalent to the subfield generated
by
its
unity quantity of any
of characteristic p.
We next have the
Theorem
(28)
11..
(26)
we have
a
+cs6+d
(mod m)
.
prime to w.
Thus the
rules (26) for
combining congruences are equivalent to the
in
defini-
tions of addition
and multiplication
E
{m}. But then a = 6 + das well as g == c and if we also have d. a
m is
We have
M
M
M
bers to write
(24)
(read:
a
a
b
(mod m)
a congruent 6 modulo m) if a 6 is divisible by m. and a is 1 and by one of the residue classes 1.
. and Theorem 8 states that E {w} is a proved Theorem 9. Thus it is customary in the Theory of Numthat is.
For /(m)
is
clearly the order of the multiplicative
group defined in Theo-
rem
8. Otherwise a is prime to p.
pbea
prime.
Our
result then follows
from
Lemma
2.
We next state the number-theoretic consequence of Theorem 8 and Lemma 2
which
is
called Euler's
Theorem
10. We observe now that a = 6 (M) for = {m} means that a b = mq.
For (28) holds if a is divisible by p.120
INTRODUCTION TO* ALGEBRAIC THEORIES
every a not in
field. Let
Fermat Theorem. An ideal of E is a prime ideal if and only ifM.. = {p} for is a divisorless ideal. Then if a is prime to m we have
af
s
1
(mod m)
. a positive prime p. a(o^~ is also di1)
visible
by
p.
ac
=
6d (mod m)
. Let f (m) be the
to
m>
(27)
and prime
Theorem and which we state as number of positive integers not greater than m. b is divisible by m.
cofc + cjc + c 2 = -c^ 1 is in F. ft = r^y* I since r is
4n. then 1.(P + 1)(P . However.
we say
K
is
a
field of degree
n over F. cubic. We call K a quad= 2. The quantities of K are then uniquely expressible in the form fc = c\ + c 2 u 2 for ci and c 2 in F. r. If fc is not in F.
. fc. fc is a root = in F. if k is in F. In particular. ^ + 1 There is a number of other results on congruences which are corollaries of theorems on rings and fields.FUNDAMENTAL CONCEPTS
nonzero elements form an abelian group
121
1.
.
Then
1)
=
=
in the field
E-
F
is
a subfield of a field
that
K which
is
+
. primitive. u are linearly dependent in F we for a 2 7* 0.
We
then have
Theorem
integer t
(29)
12.
.
9. fc2 are 2 = for c ci.
. fc is a root of a monic polynomial of degree two. The theory of linear spaces of Chapter IV implies that Ui may be taken to be any nonzero element of K. We now say that a quantity u in K generates K over F if 1. Then u generates K over F if and only if u is not in F.
{p}. u are a basis of K over F. ratic.
called a primitive root
modulo
p. u= have ai + a zu = o^ a\ is in F.1). of the monic polynomial (x fc) then Co T^ 0.
element
(30)
fc
of a quadratic field
f(x. If a linear space u\F u n F over F.
.
. or 5. The elements p of P* are the distinct roots of the equation I and are not all roots of any equation of lower degree. we shall not mention them here. rf a complete set of nonzero residue classes modulo {p Such an integer r is
P*
of order
~~ xp l
=
. u 2 over F. For if 1.
and we let r be a primitive root modulo
+
!)(
-
p. quartic. If c 0. Then if n = 1. The elements k of a quadratic field have the property that 1. Hence we may take Ui to be the unity element 1 of F. Thus evgry
)
. c 2 not all zero and linearly dependent in F. fc do not form a basis of X over F.
fc)
is
a root of an equation
=
x2
-
T(k)x
+ N(k)
=
with
(31)
T(fc)
and N(k)
in F. Let
p be a prime
of the
form 4n
+
!
Then
there exists
an
such that
t
2
+1a
(
(mod
p)
. Quadratic extensions of a field. Then it may be shown that P* is a cyclic " r p 2 are multiplicative group [r] where r is an integer such that 1. 4.
}
. Clearly. the field K is F.
For p
t*
1
=
=
r -i. t*
-
1
as desired.
. and fc is in F k = fc 1 + Ow 2 if and only if c 2 = 0. 3. However.
t
=
rn. then Ci cannot be zero.
u*
if
u generates
K
over
F we have
-
bu
+c=
. or quintic field over F according as n Let n = 2 so that K has a basis u\ = 1. and k 2 with coefficients in F.
l
.
Define a correspondence S on K to X by
k
fci
(32)
=
fci
+ to -
fc
s
=
ki
+ k us
2
. the automorphism group of K over F is the cyclic group [S] of order 2 and is the Galois group of K.
fc
But bu
5
fc
.
.
if
K were any field of degree n over F and
F
of
S and T were automorphisms
>
over
ing
K.
define ST as the result fc then k s > (fc s ) r It is easy to
k ST
of applythat the set of all
(fc
)
=
8
2
"
K
*
K
We
u)]
see
k\ +
now
that
if
fc
= =
fci
+ +
fc 2
u.
(33)
(fc
+
fc
s
)
=
ks
2
+ kj
.
that
us
is
a root in
K
of the quadratic
z equation x
bx
+
=
0. and that branch of algebra called the Galois Theory is concerned with relative properties of K and G.
(fci
+
fc 2 u)[fci
-f
fc 2
(6
=
fcifc 2 6
+
fc|u(6
u).
distinct roots in a field
(35)
K. If fc = spondence 8 in F have we k + fco = (fci + fcs) + k$ + k^u fcs + kiu for fca and ki kfi s = 5 and we have fci + fcs + (fca + fcOu (fc 2 + fcOw.
if
(36)
T(fc)
=
kk s
.
w2 =
N(k)
Hence
. Hence if is a nonmodular quadratic field over F. Then (32) is a one-to-one corres and itself if and only if u generates K over F. But b since u is not in F unless K is a modular field in which u + u = 0.
Also
fcfco
=
u.
then
fcfc^
=
c. which is not possible automorphism.
and only
if
(u
s
)
2
=
c
bu s
c. so that (fc + fc )
and
of
fc 2
K
. c
nomials in
and we propose to find the value of T(K) and N(k) as polyand the coordinates fci and fc 2 in F of k = fci + fc2W.
In either case
unaltered and
K
F
F
of
K.
In the former case
S is
defines a self-equivalence of is called an automorphism over
If
$
> the identity correspondence fc < of the elements leaving
fc.
b.
fcifcs
+
ks
(fci&4
while
kH
=
u = + kjc^u + + c) + + (fcifo + k^k^u3 + kJd(u8 But then
fc 2 fc 4
(fcifcs
fc 2 fc 4
(fcifc 4
fcifcs
)*.
(34)
if
5
(fcfco)
=
is. In our present s 2 case u s = (u s ) s = (b u) = b (b u) = u so that S is the identity = = w u if and only if 2w 6.
. the sum
us = u
or
6
u
. we would
s
first k>k and show is a group G with respect to the operation just automorphisms over F of defined.122
for b
INTRODUCTION TO ALGEBRAIC THEORIES
and
c in F.
for all
in F where us is in K.
But the quadratic equation can have only two
of the roots
is 6. In case G has order equal to the degree n of K over F it is called the Galois group of K over F.
.
N(k)
=
k\
+
ks)
k\c
+ fakj)
.
)
The
trace function
is
thus called a linear function and the norm function a multipli-
cative function.
and the polynomial of
(38)
(30) is
/(*.
fc)
=
(x
-
fc)(x
-
. a is not the square root of any quantity of F. Thus every nonmodular quadratic field is the polynomials with coefficients in F in an algebraic root u of an
2
=
a where a
is
and
F
is
nonmodular. Then (37) has the simplified form given by c =
+
+
(41)
T(k)
d? for
=
2*i
.
This
is
2
(x )x.)
*
"
(.
N(kk Q ) = N(k)
T(aifc
ai(fc
N(k Q )
2 fc )
for every ai and a 2 of F a fc s =
(aifc
and fc and fc of K. is a field. Since
in F. given. we may show that
(39)
trace of k. the nonmodular field F and the equation x 2 defined. then
if
K
is
u with coefficients in F. fc(w)
=
ring F[u] of all
2 equation x
any polynomial in F[x]. of all consist polynomials in u with coefficients
d^O. For
a 2 fc
+
2
)
ajc
+
+ ^Jc +
=
. we have u2
=
d2 (u
.
s
Note that
Let us
now assume
that the field
K is nonmodular so that K
by the root u
= F
+
uF
where u
equation
(40)
if
satisfies (31). Let us then assume with6/2) out loss of generality that u is a root of (40) so that in (31) we have 6 = 0.
whereas u
is
F and w
For
if
The
in F.
T(ajc
+
a 2 fc
)
=
aiT(fc)
+
a 2 r(fc
)
. K now
u
impossible in a field.
2 = & 2 /4 = a = (w u2 bu c 6 2 /4. Similarly.FUNDAMENTAL CONCEPTS
we have
(37)
123
T(k)
=
2fci
+kb
t
.
Now if a =
not in
d in F. we may write k(x) = ki(x ) + fci(a) + ki(a)u. it is actually the field F(u) of all
K
rational functions in
Conversely. For we may take
=
a in
F
are
<
"(?.
+ d) (u
d)
=
0.
and N(k)
is
called the
norm
of
fc.
The function T(k) is called the Moreover.
Then
K
is
also generated
%b of the
z2
=
a
(a in
F)
.
fc 2
quantities of
k(x)
is
+ d^O. then JV(Jfc) = fc
kk Q (kko) s
=
kk Q k s k$
=
(kk )(k Q k$). a.
N(k)
=
k\
-
k\a
.
s
(hkj!
=
+ asfco) = (ifc + a + + ks + 02(^0 + kfi) as
)
desired.* N(kk Q ) 2 if fc is in F.
The Theory
of Algebraic
Numbers
is
concerned principally with the integral domain consisting of the elements of degree n over the field of all rational called algebraic integers in a field numbers.
where the p< are distinct positive primes and r > I for a > 0. 2 generated by root u of u = a where a is rational and not a rational square.
.124
INTRODUCTION TO ALGEBRAIC THEORIES
identify F with the set of all two-rowed scalar matrices with elements F (so as to make K contain F). Integers of quadratic fields.
pr
.
10.
But
F(u). (fti/if ) contrary hypothesis. 2ki k\ k\a
K
y
K
of
K stating our final results as
Theorem
13.
. Hence
K
is
the field
F(u). That
K
^
fc
is that of F. We call k an integer of of if and only if of are k Thus is an T(k) N(k) (30) integers.
The integers
containing the ring
E
of
all
a quadratic field 'Kform an integral domain J ordinary integers. a = The quantities k of K have the form k = k\ + k^u where fci and k% are if the coefficients ordinary rational numbers. a 7^ d for any d of F. If
ic field is
we take 6 =
generated by
c" 1 in (43) we replace a by d.k\a = 2 2 = then we a to But For otherwise 2 7^ 0. Observe in closing that
2
if
K=
v is
such that
(43)
fc
7^
is
in F. be a quadratic field over the rational field so that is Let.
Thus we may replace the defining quantity a in F by any multiple 62a for b 7^ in F. Write a
K
K
K
integers.
. then a root of
K
F(v) for every v
=
bu
z2
=
b2 a
. if and only if ki = k = ki + k*u and N(k) = k\ . Then J consists of all linear
of
combinations
(45)
Ci
+cw
2
. then. Then every polynomial in u has the form = k = 0. unity element of over a nonmodular field fields
K
We
F
nonmodular is evident since the have thus constructed all quadratic
is
K
in terms of equations x 2
=
a for a in
F. By the use of (43) we may multiply a by an integer and hence take a inte= c2d where d has no square factor and c and d are ordinary gral. while if 1 we interpret (44) as the case a negative and r = 0. Hence every quadrata root u of the quadratic equation
(44)
x2
=
a
=
pi
. k for of take K = F + uF and have fc~ = (fc? 2 u) kla)' every
and
in
fc
1
fc
l
1
K. integer shall determine and are both We all integers ordinary integers. It is also shown easily that if K is defined by a and K by a then K and KQ are fields equivalent over F if and only if a = 62a. We shall discuss somewhat briefly the case n = 2.
m = 0.
2w
fc has the form (45) with Ci and c 2 in E.'or in either case
6
=
= mi
+mu+w
2
for
w
in (46). a contradiction. w =
It
+
form.
61. If 6 = 2 and 61 is even.
. Note that a ^
I.
If
k%a factor of 6
p were an odd prime 2 61.
This completes the proof.
Now
.
and
1.
common denominator of fci and
2fci is
fc 2 . Thus we have proved 1 that J consists of the elements (45) with w = u if a s= 2.
b 2 in
E and 6
2
the positive least
60. The elements of J are in the field K. while otherwise v? = a = 4m + 1 for an inte= J + i = 1.
We write
_ ^1 . remains to show that J is an integral domain.
then we have shown that either
fc
fc
=
61
+
b 2u with
b. kh are in J for every But the sum of fc and h of the form (45) is clearly of that fc and h of J. If b > 4 divides 2bi. T(w)
(47)
w = a + Qw.. the product
is
of the
form
2
(45)
if
w2
is
of that form. 62 = 2?w 2 1 for
. then 4 would divide 6f. b\a. 4 divides b? and b|a. and thus 2 divides 6 2 We have a contradiction and have proved that 6 = 1. and hence 2
b|a). Since a has no square factors this is possible only if p divides 62.FUNDAMENTAL CONCEPTS
for Ci and 02 in E.c. bi.
+
+ +m
+
if
a
=
1
(mod
.
But
this condition holds since
ii
w = u
then
ger m. where
125
3
w =
u
if
a
=
2 (mod 4) or a
=
(mod
4) but
(46)
w = l+
i/
a
= 1 (mod 4). ty2 . But then p would divide b?. Hence 6 is a power of 2.
and
6 2 in
1. Hence 61 = 2wi -f 1. then 2 divides bi. of
bJ7 (6?
.
w*
= m
+w
. if 6 2 were even.
A/2
-
7
>
7
fv v
__ -
1
00
00
for
bo.
However.d. 2. But
m
.
(mod
A/1
4) since
7
a has no square
__ &2
factor. and thus it suffices to show that fc A.
4).
integer. b 2
. 3 (mod 4).
= 4[mf m>i ra 2 and 6f a(^ 2 l and b\a 2 )] ordinary integers a is divisible by 4 if and only if a = 1 (mod 4).
Then
k\
1 is
the g.
an
b|a. 60 divides 2bi. so
b\
it would divide 2bi only if it divided and p 2 would divide 6? b\a as well as
-
=
that b 2 divides
divides 6 2 contrary to the definition of 6 Similarly.w . #(tiO = i(l ~ a) = -m. then 4 divides b\a. a contradiction.
+ CiC
4. and hence g ^ 4. 1.
Then one
of ci
and
c2
zero. the other
or
1. 1 so that (49) becomes
is 1
c\
+ ci =
1
.
However. Now a has no square factors. = 4 for # a and is and and this > 0.
4w2 = 9
-
8
=
=
9
.
2ci
1. u.
s -w. r(*) N(k 5 = fc fc-i. the units of J are
Then
1
=
while c 2
=
(52)
1.
=
N(k)=l.
1.
s s 1.
and we have shown that
if
a
is
negative this group
a
3
1
finite
1. But we may regard u as the ordinary positive V2
a
=
2. 2. and so is A for every integer L If h* t > 8 and have A'""* = 1. Hence we have proved that the units of J are 1. the only possible remaining cases are gr = 1.
+
c2
=
1
with any choice
1 gives
of signs. But k is in Conversely. w
.
then (48) becomes N(k)
so that
4N(k)
=
4c?
+ 4cic + c\ +c
2
2)
-
(4m
= c\ + l)c| =
4.
The
If
units of
J form a
is
multiplicative group.126
INTRODUCTION TO ALGEBRAIC THEORIES
units of
The
N(kh)
(48)
J
are elements k such that kh
=
N(K)N(h)
=
1. If g = 2 we have x\ + 1 1 only ^ #2 = 0. if (48) holds. Hence h is h for s 7* t.
3
(mod
4) so that
fc
=
Ci
+
c 2 u.
u2
=
In the remaining case a
4. k has the property kk = 1 or 8 = s 1 = = Ar or tf Hence fc* and J when k is in J since T(k ) ) (*).
If
a
s
2. 3.
N(k)
is
1 for h also in J.
then h
=
+ 2u has norm 9 -
group.
2
elm
=
But
this is
equivalent to
(50)
(2ci
-
c|a
=
determine the units of J completely and simply in case a is 4 for For both (49) and (50) have the form x\ + x\g = 1. 0. 2^1
for every
We may
Now
is
a < let a
save only a = 3. we may take a unit of J.
=
3
we
c2
use (50) and have (2ci
+c
2
2)
+ 3cl =
or 1
and
if
c2
^
we must have
c2
=
or
1.
(51)
1.
w. Then an ordinary integer which divides unity. negative.
1.
then (48)
is
equivalent to
(49)
c\
-cia=
2
1.
1
if
a
=
1
(mod
4).
-w s
. Thus (48) is a necessary and sufficient condition that k be a
unit of /.
= gives c\ for CL Clearly. c\ only if #2 1.
and the units of J are
u. 2 possible </ > ordinary integers x\ g = = 1.
s
=
Si
=
gh
+
r. Then/ =l-g N(g) Division the of a ggc. If then \t s < t < s + \ then (<-)< i while ifs + < and Hence there exist such that h% (s + 1) i.d. Let f and g ^ be in J.
Then N(s)
=
+ s u.
and the Gauss integers comprise its integral domain J. We first prove Theorem 15. ordinary integers h\
<J<s+l
|
(54)
ai^fci-fti. r = si + s\ <
2
l = h sg so that fg~ = J. It
may
similarly be shown to be infinite for
every positive a. we have = = = l with 2.
results
on primes and prime
ideals in
two
special cases. We have deteru of J and shall now study its divisibility properties.
We observe that the quotient h and the remainder r need not be unique in our present case.
|a|<i-
Put
sg
=
hi
+ hju.
gh N(g) as de-
/
=
+
sired. Every ideal of J is a principal ideal. Then the Gauss complex numbers are the elements of a quadratic field K of our study with
tional x
u-
-1. and For fg.1 is in K. The complex numbers of the form x + yi with raand y are called the Gauss complex numbers.
|*i|<J.
ft
S2
=
fe 2
-ft2. fg~\ = k\ + k^u with rational fci and fc 2 Every rational number t lies in an interval s < t < s + 1 for an ordinary integer s. existence We shall use the Algorithm to prove
+
+ 2-0-u
+
Our proof will be different and simpler than that we gave in the case of polynomials and indicated in the case of integers but has the defect of being merely existential and not constructive. mined the units 1. Then there exist elements h and r
in J such that
(53)
f
=
gh
+r
.
ft
"*
>
5*~8
>
1. By there exists an m 7*
M
M
.
< N(r) < N(g).FUNDAMENTAL CONCEPTS
and have h
units of
127
>
1
5. tf (r) N(s)N(g)
+
<
s.
Hence the
multiplicative group of the
J
is
an
infinite
group. and those for which x and y are integers are called the Gauss integers.
Gauss numbers.
M
are positive integers.
We shall not study the units of quadratic fields further but shall pass on
to
some
11. and For the norms of the nonzero elements of in such that N(m) < N(f) for every / of M. For example. Our first result is the Division Algorithm which we state as Theorem 14. if / = 2 u and g = 1 u. N(l) Ar(-u) = 1.
Then there exist b and
in J such that
(55)
is
M
M
N(r)
<
. then N(d) = p = N(g)N(h) so that
For
if
p
is
N
unit.
. then / = g h for nonunit g and h. Thus/ = mh. But must be zero. lid = = 1. For the set of all elements of the form xf + yg with x and y in J is an = {d} d in has the form (55) ideal of J. and we continue the proof of Theorem 6
M
M
. Then if p is a divisor of/ of least positive norm it is a prime divisor of /. then p = dk for N(d) > 1.
c
d
=
bf
+
eg
a common divisor of f and g. Conversely. then / divides h.
pr
f
of primes pi which are determined uniquely by
apart from their order and
unit multipliers. so is d d\ gh with N(g) ^ 1. and
follows that r
N(m).
to obtain
Theorem
(56)
17. a divisorless ideal) if and only if Let us then determine the prime quantities d = di d*u of J.
Every composite Gauss integer
f
is expressible
as a product
=
pi
. and also if p is a prime of J dividing g h. By Theorem 15 we have and divides / = 1 / + 00. A positive prime
. of f and g. We see first s = s = d zu. TTms d is a g.
in E. s s s s N(h) * 1. N(h ) = N(h).c.
Theorem
We now prove 18. c
But
=
is a dd s can
have at most two prime factors
= pp Q
for positive primes
p and p
.
let
d be a prime of J. (and. Every prime d of J is either associated with a positive prime of E or arises from the factorization in J of a positive s prime p = dd of E which is composite in J.
M
0/+lginM. The above result then implies Theorem 16. and d is
M
+
composite. then p divides g or h.
d
is
a prime. N(g) < N(f). if / is a composite integer of J.128
INTRODUCTION
TO. we have d = g h for N(g ) = N(g). N(p) = p* = N(d)N(k). N(k) > 1.
p of E is either a prime of J or the product a prime of J.d. But then p = N(d). and g = The above result may now be seen to imply that if / divides gh and is prime to g.
wA. in fact. Let f and g 6e nonzero elements of J. Moreover.
. h is a (h) gh with N(g) > 1.
N(d)
= dds
where d
is
a positive prime of E and is composite in J.
ALGEBRAIC THEORIES
and
r in
it
Theorem 14 we have / = mh + r so is r = f /.
We observe that Theorem 17 implies that an ideal M of J is a prime ideal
= {d} for a prime d of J. For if d that if d is a prime of J. m and mh are in
for h
J and
= {m}. Then
c
= N(d)
c
positive integer of
E and is either a prime or is composite.
squares congruent 0. 2 2 = = = = f or x + i/w in J. Then we have Theorem 20. we = p2 have PQ = p.
. If p divides b + u or 6 pfc 2 which is impossible.
. If t is even. Write c = f2g where f and g are positive integers and g has no square factors.
. Then cisa sum of two squares if and only if no prime factor of g has the form 4n + 3.
. then 6 by p. We note that 2= 1 and that it remains to show that p
+
1 = (l + w)(l w)is composite in = 4n + 1 is composite in J.
. or 5 modulo 8 while 4n + 3 is congruent 3 or 7 modulo 8. Since p and po are both in E.
. Thus the prime factors of c of the form 4n + 3 occur to even powers and are not factors of g. g = tf (di) N(d r ) = #(di N(di) p< 3. u) = b + 1 without dividing one of its factors 6 + u b (b + u)(b Hence p is not a prime of /. Hence. One of s and s + 1
2 (mod 4) (mod 8).
is
J
prime in J. we may write x + yu d r for primes d in J.
. p
even. But p and p are positive and must be equal.FUNDAMENTAL CONCEPTS
of E. c = N(di) N(d r).
gr
.
To prove
have
is
this result
we observe
4s
2
first
that
if
is
=
+
1.
. We use the result above to derive an interesting theorem of the Theory of Numbers.
ifc
. But a prime p of J cannot divide the product 2 u. if g = pi = for d r). and both of these integers would be divisors of p 2 But 2 2 = 4n + 3 then]V(g) = N(h) = p = x + y which is impossible.
. We call a positive integer c a sum of two squares if c = x2 + y 2 for x and y in E.
. This completes the proof. 2. Then (# + yu)(x yu) if and only if p divides either x + yu or x
. Then N(di) = p is a prime of E if and only if p 7^ 4n + 3 and otherwise N(di) p?. 0. t*
=
+ 4s +
1
=
4s(s
+
!)
an odd integer of E we + !.
We may
=
. }
.
2
=
8r
+1=
1
.
129
assume that p is associated with d and po with ds so that PQ PQ is associated with d and with p. A positive prime p of E is a prime of J if and only if p has
the
form
t
4m +
2s
t
3. we have d< in J. 1.
. and 1 == u. N(d) Therefore d is associated with the prime p of E which is prime in J.
.
.
. p r for positive primes p< of Snotof the form 4n + Conversely. We know
2
by Theorem 12 that there exists an integer b in E such that b + 1 is divisible u = p(fci + fc 2tO. = di For if c = x 2 + y 2 = (x + yu)(x yu).
We now clearly complete our study of the primes of J by proving Theorem 19. we have t = 2 55 4 of Thus sum is a two to and t (mod 8).
.
. 4. If neither have N(g} > 1.
. It follows that 2 p = 4n + 3 7* x* + y We now assume that p = gh for g and h in J and 2 = = nor h were a unit. we would have p A'Xp) N(g)N(h). For p is a prime of J and divides yu. z + t/ d r) and c JV(fc) N(fdi Note in closing that a positive prime p of the form 4n + 3 divides x* + y2 if and only if p divides both a: and y.
.
7. d
must be a unit
Number8 pp.*
y
= 4n +
y
12.
But
if
g
=
gi
+
g*u. the ideal of J consisting of all elements of J of the form
3*
+ y(l + 2u)
{d}.
serve that
6 is
bhz.
Thus
clearly
+ 2u
y
2u are primes of
J. 21.
we have
N(g)
is
=
g\
+ 5gl > 0. fc 2 =
the form k\
if
fc 2
J have
+ k^u for k\ and
. k\ = bhi. g\ <
I
But^l
=
1
impossible since g\ The units of J are 1.
in the set
E of all integers.
. It follows that the odd prime divisors of z have the form 4n + 1. and therefore 2u as divisors of ring J {3} contains 1 + 2u and 1
We let M be
(58)
If this ideal
The
ring
contains nonprincipal ideals one of which we shall exhibit.
and
3
7
We see now that 21 =
=
(1
no two of them are associated. and we have factored
21 into prime factors in J in two distinct ways.
every prime divisor p
3 of z divides both x and y. 7. y.
2u in J yet does divide their product.
1.
were a principal ideal
would
exist
a
common
divisor
d of 3 and
*
1
+ 2u. 1
* 0.130
INTRODUCTION TO ALGEBRAIC THEORIES
yu
k zu).
=
N(g)N(h) for ordi-
nary integral proper divisors N(g) and N(h) of N(k). Moreover.
We now prove
22.
We
shall close
our ex-
position with a discussion of some properties of the ring 5. the
elements
of
Oban integer of E which is a divisor in J of k\ + fc 2 w. 21.
Evidently g 2
3.c.
An
integral
domain with nonprincipal
ideals. there
(x. 1
+ 2u. 49.
Since these are nonassociated primes.
1
For further
results see L. then. The norms of the integers of our theorem are 9. For if k is a composite of
J we have
=
gh. 7. y = p& 2 for integers ki and p(ki follows. g\ + 5gl
^
1. + 2w)(l 2w). unity.d.
k
1
2u are primes of J no two
N(k)
of which are associated. 7. x = pfci. z are ordinary integers such that
x
=
fc 2
of E.
y in J)
.
=
3. then 6 must divide both k\ and fc 2 For k\ + k^u = b(hi + h^u). 2.
Theorem
The elements
3.
2. Dickson's Introduction to the Theory of
40-42. Then let x y. the principal ideal {3} of J defined for the prime 3 is not a prime ideal. E. We may show readily also that x and y cannot both be even or be odd and that z must be odd. respectively. and the only positive
proper divisors of these norms are
3. Since 5 = 3 (mod defined by a = u* = the field
J
of integers of
K
4). that if x. It
(57)
z2
+
2
2/
=
z2
. For 3 does not
divide 1
+ 2w and 1
J
the residue class
zero. z have g.
But 1 2w does not
divide
a contradiction.
Both
of these
texts as well as Chapters III." in Transactions of the American Mathematical Society. For otherwise 3 2 =
J and hence
M
M
1.
(1
Then
1
=
36
+
2u)[fc(l
-
2u)
+ (1 + + 7c]. Dickson's Modern Algebraic Theories. See also pages 74-76 and Chapter VI of L. We have shown that is not a principal ideal and does not contain Since contains 3 but not 1 it cannot contain 2. 2.
M
a divisorless ideal of J. and
Chapter VI of
J. IV. the elements of
We
have proved that
J J
M are the residue M equivalent to E
is
classes 0.
1
would be
r
in
M. Let
Since
c
M contains 3g contains 3q = the ordinary integers in M are by The ideal M contains 3 and + 2u and so contains
=
0. clude. It may be shown that every ideal of J has a basis of two elements over E.
(59)
wi
is
=
ft
3
. We shall conclude our text with the following brief bibliographical summary. XLIII (1938).
This completes our discussion of ideals and of quadratic fields. Every integer of / has the form k = ki + k^u = ki + fc 2w 2 + fo.FUNDAMENTAL CONCEPTS
1
131
or
-1
216
7. 1.
(60)
h
=
ftiWi
+ w
ft 2
We
have thus proved that consists of all quantities of the form (60) for hi and ft 2 ordinary integers.
We now proceed to
prove that
M
+
is
a
divisorless
ideal of
a prime ideal of J. 1. 386-436. H.
wz
/&
= Zu ft
(1
+ 2u) =
ft 2
w
-
1
.
2u)c for b and c in J. and of the Modern Higher Algebra inof all the material of our course. 2.h
3fti
2
ft 2
w2 =
hQ
+ h* in
E
and
M. w 2 a basis of over E.
of J. E. Note that 1 has a basis 1. Chapters II-V.
is
3}
and
is
a
field. For a discussion
V
. u over E in this sense.
If
/i
in in
M
. 1.
3<j
+
0.
M.
The theory of orthogonal and unitary equivalence of symmetric matrices is contained with generalizations in Chapter of the Modern Higher Algebra and is further generalized and connected with
'
V
the theory of equivalence of pairs of matrices in the author's paper entitled 'Symmetric and Alternate Matrices in an Arbitrary Field. Write = 3c + r for c in E and r = 0.
it
be an integer of
E
in
M so that
c
c
=
r. 2 such that 3
{
=
0.
then
=
+
Aau for
ft
and
m E. Thus we may call wi. Then k = cwi + fc 2w 2 + r = fci + fc 2
M
M
{
}
r
(M).
and
\d}
7
=
+
(1
+ 2u)7c
is
= =
{!}.
r
where
r
=
Hence
divisible
1
3.
Let us begin with references to standard topics on matrices not covered
in our introductory exposition. Wedderburn's
Lectures on Matrices. By
the proof above
+ ^2 =
for hi in E.
II. 1940). Hecke's Theorie der algebrdischen Zahlen. and its ideals on pages 106-10 of H. C. The units of the ring of integers of a quadratic field are discussed on page 233 of R. No.
trices see C.
. Weyl's Algebraic Theory of Numbers (Princeton. The theory of rings as given here is contained in the detailed discussion of Chapters I and II of the Modern Higher Algebra. H. and the theory of ideals in Chapter XI. 110]). We close with a reference to the only recent book in English on algebraic number theory. See also the much more extensive treatment in Van der Waerden's Moderne Algebra. Volume III. 5 [pp. The Theory of Matrices ("Ergebnisse der Mathematik und ihrer Grenzgebiete.132
INTRODUCTION TO ALGEBRAIC THEORIES
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"Linear algebra is a fantastic subject. On the one hand it is clean and beautiful." – Gilbert Strang This wonderful branch of mathematics is both beautiful and useful. It is the cornerstone upon which signal and image processing is built. This short chapter can not be a comprehensive survey of linear algebra; it is meant only as a brief introduction and review. The ideas and presentation order are modeled after Strang's highly recommended Linear Algebra and its Applications. At the heart of linear algebra is machinery for solving linear equations. In the simplest case, the number of unknowns equals the number of equations. For example, here are a two equations in two unknowns: 2x − y = 1
0.4 Basis 0.5 Inner Products and Projections 0.6 Linear Transforms
y
2x−y=1 (x,y)=(2,3)
x + y = 5.
(1)
x+y=5
x
There are at least two ways in which we can think of solving these equations for x and y. The first is to consider each equation as describing a line, with the solution being at the intersection of the lines: in this case the point (2, 3), Figure 0.1. This solution is termed a "row" solution because the equations are considered in isolation of one another. This is in contrast to a "column" solution in which the equations are rewritten in vector form: 2 1 x+ −1 1 y = 1 5 . (2)
Each equation now corresponds to a plane, and the row solution corresponds to the intersection of the planes (i.e., the intersection of two planes is a line, and that line intersects the third plane at a point: in this case, the point u = 1, v = 1, w = 2). In vector form, the equations take the form: 2 1 1 5 4 u + −6 v + 0 w = −2 . −2 7 2 9
(4)
The solution again amounts to finding values for u, v, and w that scale the vectors on the left so that their sum is equal to the vector on the right, Figure 0.3.
(5,−2,9)
Figure 0.3 "Column" solution
In the context of solving linear equations we have introduced the notion of a vector, scalar multiplication of a vector, and vector sum. In its most general form, a n-dimensional column vector is represented as: x1 x2 x = . , . .
Matrices, like vectors, can be added and scalar multiplied. Not surprising, since we may think of a vector as a skinny matrix: a matrix with only one column. Consider the following 3× 3 matrix: a1 a4 A = a7
a2 a5 a8
a3 a6 . a9
(13)
The matrix cA, where c is a scalar value, is given by: ca1 cA = ca4 ca7
ca2 ca5 ca8
ca3 ca6 . ca9
(14)
And the sum of two matrices, A = B + C, is given by: a1 a4 a7
a2 a5 a8
a3 b1 + c1 a6 = b4 + c4 a9 b7 + c7
b2 + c2 b5 + c5 b8 + c8
b3 + c3 b6 + c6 . b9 + c9
(15)
5
With the vector and matrix notation we can rewrite the three equations in the more compact form of Ax = b: 2 4 −2
Where the multiplication of the matrix A with vector x must be such that the three original equations are reproduced. The first component of the product comes from "multiplying" the first row of A (a row vector) with the column vector x as follows: (2 1 u v = ( 2u + 1v + 1w ) . 1) w
In its most general form the product of a m × n matrix with a n dimensional column vector is a m dimensional column vector whose ith component is:
n
aij xj ,
j=1
(19)
where aij is the matrix component in the ith row and j th column. The sum along the ith row of the matrix is referred to as the inner product or dot product between the matrix row (itself a vector) and the column vector x. Inner products are another central idea in linear algebra (more on this later). The computation for multiplying two matrices extends naturally from that of multiplying a matrix and a vector. Consider for example the following 3 × 4 and 4 × 2 matrices: a11 A = a21 a31
That is, the i, j component of the product C is computed from an inner product of the ith row of matrix A and the j th column of matrix B. Notice that this definition is completely consistent with the product of a matrix and vector. In order to multiply two matrices A and B (or a matrix and a vector), the column dimension of A must equal the row dimension of B. In other words if A is of size m × n, then B must be of size n × p (the product is of size m × p). This constraint immediately suggests that matrix multiplication is not commutative: usually AB = BA. However matrix multiplication is both associative (AB)C = A(BC) and distributive A(B + C) = AB + AC. The identity matrix I is a special matrix with 1 on the diagonal and zero elsewhere: 1 0 0 1 = . . . 0 0
I
... ... .. . ...
0 0 0 0 . . . . 0 1
(22)
Given the definition of matrix multiplication, it is easily seen that for any vector x, Ix = x, and for any suitably sized matrix, IA = A and BI = B. In the context of solving linear equations we have introduced the notion of a vector and a matrix. The result is a compact notation for representing linear equations, Ax = b. Multiplying both sides by the matrix inverse A−1 yields the desired solution to the linear equations: A−1 Ax = A−1 b Ix = A−1 b x = A−1 b (23)
A matrix A is invertible if there exists 1 a matrix B such that BA = I and AB = I, where I is the identity matrix. The matrix B is the inverse of A and is denoted as A−1 . Note that this commutative property limits the discussion of matrix inverses to square matrices. Not all matrices have inverses. Let's consider some simple examples. The inverse of a 1 × 1 matrix A = ( a ) is A−1 = ( 1/a ); but the inverse does not exist when a = 0. The inverse of a 2 × 2
1 The inverse of a matrix is unique: assume that B and C are both the inverse of matrix A, then by definition B = B(AC) = (BA)C = C, so that B must equal C.
7
matrix can be calculated as: a c b d
−1
=
1 ad − bc
d −b −c a
,
(24)
but does not exist when ad − bc = 0. Any diagonal matrix is invertible: A=
a1
..
. an
and A−1 =
1/a1 .. . 1/an
, (25)
as long as all the diagonal components are non-zero. The inverse of a product of matrices AB is (AB)−1 = B −1 A−1 . This is easily proved using the associativity of matrix multiplication. 2 The inverse of an arbitrary matrix, if it exists, can itself be calculated by solving a collection of linear equations. Consider for example a 3 × 3 matrix A whose inverse we know must satisfy the constraint that AA−1 = I: 2 4 −2
1 1 −6 0 x1 7 2
x2
x3 = e1
e2
1 0 0 e3 = 0 1 0 .(26) 0 0 1
This matrix equation can be considered "a column at a time" yielding a system of three equations Ax1 = e1 , Ax2 = e2 , and Ax3 = e3 . These can be solved independently for the columns of the inverse matrix, or simultaneously using the Gauss-Jordan method. A system of linear equations Ax = b can be solved by simply left multiplying with the matrix inverse A−1 (if it exists). We must naturally wonder the fate of our solution if the matrix is not invertible. The answer to this question is explored in the next section. But before moving forward we need one last definition. The transpose of a matrix A, denoted as At , is constructed by placing the ith row of A into the ith column of At . For example: A= 1 2 1 4 −6 0 1 and At = 2 1
the transposes: (A + B)t = At + B t. The transpose of a product of two matrices has the familiar form (AB)t = B t At . And the transpose of the inverse is the inverse of the transpose: (A−1 )t = (At )−1 . Of particular interest will be the class of symmetric matrices that are equal to their own transpose At = A. Symmetric matrices are necessarily square, here is a 3 × 3 symmetric matrix: 2 1 4 A = 1 −6 0 , 4 0 3
(28)
notice that, by definition, aij = aji . 0.3 Vector Spaces
The most common vector space is that defined over the reals, denoted as Rn . This space consists of all column vectors with n real-valued components, with rules for vector addition and scalar multiplication. A vector space has the property that the addition and multiplication of vectors always produces vectors that lie within the vector space. In addition, a vector space must satisfy the following properties, for any vectors x, y, z, and scalar c: 1. 2. 3. 4. 5. 6. 7. 8. x+y =y+x (x + y) + z = x + (y + z) there exists a unique "zero" vector 0 such that x + 0 = x there exists a unique "inverse" vector −x such that
Vector spaces need not be finite dimensional, R∞ is a vector space. Matrices can also make up a vector space. For example the space of 3 × 3 matrices can be thought of as R9 (imagine stringing out the nine components of the matrix into a column vector). A subspace of a vector space is a non-empty subset of vectors that is closed under vector addition and scalar multiplication. That is, the following constraints are satisfied: (1) the sum of any two vectors in the subspace remains in the subspace; (2) multiplication of any vector by a scalar yields a vector in the subspace. With the closure property verified, the eight properties of a vector space automatically hold for the subspace.
Example 0.1 Consider the set of all vectors in R2 whose com-
ponents are greater than or equal to zero. The sum of any two
9
vectors in this space remains in the space, but multiplication of, 1 −1 for example, the vector by −1 yields the vector 2 −2 which is no longer in the space. Therefore, this collection of vectors does not form a vector space.
Vector subspaces play a critical role in understanding systems of linear equations of the form Ax = b. Consider for example the following system: u1 u2 u3
Unlike the earlier system of equations, this system is over-constrained, there are more equations (three) than unknowns (two). A solution to this system exists if the vector b lies in the subspace of the columns of matrix A. To see why this is so, we rewrite the above system according to the rules of matrix multiplication yielding an equivalent form: u1 v1 b1 u2 + x2 v2 = b2 . x1 u3 v3 b3
v1 v2 v3
x1 x2
b1 = b2 b3
(29)
(30)
In this form, we see that a solution exists when the scaled columns of the matrix sum to the vector b. This is simply the closure property necessary for a vector subspace. The vector subspace spanned by the columns of the matrix A is called the column space of A. It is said that a solution to Ax = b exists if and only if the vector b lies in the column space of A.
Example 0.2 Consider the following over-constrained system: Ax 1 5 2 0 4 4 u v = = b b1 b2 b3
The column space of A is the plane spanned by the vectors ( 1 5 2 )t and ( 0 4 4 )t . Therefore, the solution b can not be an arbitrary vector in R3 , but is constrained to lie in the plane spanned by these two vectors.
At this point we have seen three seemingly different classes of linear equations of the form Ax = b, where the matrix A is either: 1. square and invertible (non-singular), 10
2. square but not invertible (singular), 3. over-constrained. In each case solutions to the system exist if the vector b lies in the column space of the matrix A. At one extreme is the invertible n×n square matrix whose solutions may be any vector in the whole of Rn . At the other extreme is the zero matrix A = 0 with only the zero vector in it's column space, and hence the only possible solution. In between are the singular and over-constrained cases, where solutions lie in a subspace of the full vector space. The second important vector space is the nullspace of a matrix. The vectors that lie in the nullspace of a matrix consist of all solutions to the system Ax = 0. The zero vector is always in the nullspace.
Example 0.3 Consider the following system: 1 5 2 0 4 4 1 9 6 Ax u v w = = 0 0 0 0 v w )t = ( 0 0 0 )t .
The nullspace of A contains the zero vector ( u
Notice also that the third column of A is the sum of the first two columns, therefore the nullspace of A also contains all vectors of the form ( u v w )t = ( c
3
c
−c )t (i.e., all vectors lying on a
one-dimensional line in R ).
0.4 Basis Recall that if the matrix A in the system Ax = b is invertible, then left multiplying with A−1 yields the desired solution: x = A−1 b. In general it is said that a n × n matrix is invertible if it has rank n or is full rank, where the rank of a matrix is the number of linearly independent rows in the matrix. Formally, a set of vectors u1 , u2 , ..., un are linearly independent if: c1 u1 + c2 u2 + ... + cn un = 0 (31)
(2,2)
(−1,−1)
(2,2)
(−2,0)
is true only when c1 = c2 = ... = cn = 0. Otherwise, the vectors are linearly dependent. In other words, a set of vectors are linearly dependent if at least one of the vectors can be expressed as a sum of scaled copies of the remaining vectors. Linear independence is easy to visualize in lower-dimensional subspaces. In 2-D, two vectors are linearly dependent if they lie along a line, Figure 0.4. That is, there is a non-trivial combination of the 11
(−1,2) (2,2)
(−2,0)
Figure 0.4 Linearly dependent (top/bottom) and independent (middle).
vectors that yields the zero vector. In 2-D, any three vectors are guaranteed to be linearly dependent. For example, in Figure 0.4, the vector ( −1 2 ) can be expressed as a sum of the remaining 3 linearly independent vectors: 2 ( −2 0 ) + ( 2 2 ). In 3-D, three vectors are linearly dependent if they lie in the same plane. Also in 3-D, any four vectors are guaranteed to be linearly dependent. Linear independence is directly related to the nullspace of a matrix. Specifically, the columns of a matrix are linearly independent (i.e., the matrix is full rank) if the matrix nullspace contains only the zero vector. For example, consider the following system of linear equations: u1 u2 u3
v1 v2 v3
w1 x1 0 x2 = 0 . w2 w3 x3 0
(32)
Recall that the nullspace contains all vectors x such that x1 u + x2 v + x3 w = 0. Notice that this is also the condition for linear independence. If the only solution is the zero vector then the vectors are linearly independent and the matrix is full rank and invertible. Linear independence is also related to the column space of a matrix. If the column space of a n × n matrix is all of Rn , then the matrix is full rank. For example, consider the following system of linear equations: u1 u2 u3
v1 v2 v3
w1 x1 b1 w2 x2 = b2 . w3 x3 b3
(33)
If the the matrix is full rank, then the solution b can be any vector in R3 . In such cases, the vectors u, v, w are said to span the space. Now, a linear basis of a vector space is a set of linearly independent vectors that span the space. Both conditions are important. Given an n dimensional vector space with n basis vectors v1 , ..., vn , any vector u in the space can be written as a linear combination of these n vectors: u = a1 v1 + ... + an vn . (34)
In addition, the linear independence guarantees that this linear combination is unique. If there is another combination such that: u = b1 v1 + ... + bn vn , 12 (35)
which would violate the linear independence condition. While the representation is unique, the basis is not. A vector space has infinitely many different bases. For example in R2 any two vectors that do not lie on a line form a basis, and in R3 any three vectors that do not lie in a common plane or line form a basis.
Example 0.4 The vectors ( 1 0 ) and ( 0 1 ) form the canonical
basis for R2 . These vectors are both linearly independent and span the entire vector space.
Example 0.5
The vectors ( 1
0
0 ), ( 0
1
0 ) and ( −1
0
0)
do not form a basis for R3 . These vectors lie in a 2-D plane and do not span the entire vector space.
Example 0.6 and ( 1 −1
The vectors ( 1
0
0 ), ( 0
−1
0 ), ( 0
0
2 ),
0 ) do not form a basis. Although these vectors
span the vector space, the fourth vector is linearly dependent on the first two. Removing the fourth vector leaves a basis for R3 .
For notational convenience, we will often drop the cumbersome notation of Equation (1.1), and refer to the entire sequence simply as f [x]. Discrete-time signals often arise from the periodic sampling of continuous-time (analog) signals, a process that we will cover fully in later chapters. 1.2 Discrete-Time Systems In its most general form, a discrete-time system is a transformation that maps a discrete-time signal, f [x], onto a unique g[x], and is denoted as: g[x] = T {f [x]}
Example 1.1 Consider the following system: g[x] = 1 2N + 1
N
f[x]
T
g[x]
Figure 1.2 Discrete-time system
(1.3)
f [x + k].
k=−N
In this system, the ith number in the output sequence is comf[x]
puted as the average of 2N + 1 elements centered around the ith input element. As shown in Figure 1.3, with N = 2, the output value at x = 5 is computed as 1/5 times the sum of the five input
x 3 5 7
elements between the dotted lines. Subsequent output values are computed by sliding these lines to the right.
Figure 1.3 Moving erage
Av-
Although in the above example, the output at each position depended on only a small number of input values, in general, this may not be the case, and the output may be a function of all input values.
14
1.3 Linear Time-Invariant Systems Of particular interest to us will be a class of discrete-time systems that are both linear and time-invariant. A system is said to be linear if it obeys the rules of superposition, namely: T {af1 [x] + bf2 [x]} = aT {f1 [x]} + bT {f2 [x]}, (1.4)
for any constants a and b. A system, T {·} that maps f [x] onto g[x] is shift- or time-invariant if a shift in the input causes a similar shift in the output: g[x] = T {f [x]}
Example 1.2 g[x]
The precise reason why we are particularly interested in linear time-invariant systems will become clear in subsequent chapters. But before pressing on, the concept of discrete-time systems is reformulated within a linear-algebraic framework. In order to accomplish this, it is necessary to first restrict ourselves to consider input signals of finite length. Then, any discrete-time linear system can be represented as a matrix operation of the form: g = Mf, (1.6)
where, f is the input signal, g is the output signal, and the matrix M embodies the discrete-time linear system.
Note that according to the initial definition of the system, the output signal at x = 1 is undefined (i.e., g[1] = f [0]). In the above matrix formulation we have adopted a common solution to this problem by considering the signal as wrapping around itself and setting g[1] = f [8].
Any system expressed in the matrix notation of Equation (1.6) is a discrete-time linear system, but not necessarily a time-invariant system. But, if we constrain ourselves to Toeplitz matrices, then the resulting expression will be a linear time-invariant system. A Toeplitz matrix is one in which each row contains a shifted copy of the previous row. For example, a 5 × 5 Toeplitz matrix is of the form m1 m 5 = m4 m3 m2
M
m2 m1 m5 m4 m3
m3 m2 m1 m5 m4
m4 m3 m2 m1 m5
It is important to feel comfortable with this formulation because later concepts will be built upon this linear algebraic framework. 16
m5 m4 m3 m2 m1
(1.7)
2. Linear Time-Invariant Systems
Our interest in the class of linear time-invariant systems (LTI) is motivated by the fact that these systems have a particularly convenient and elegant representation, and this representation leads us to several fundamental tools in signal and image processing. 2.1 Space: Convolution Sum In the previous section, a discrete-time signal was represented as a sequence of numbers. More formally, this representation is in terms of the discrete-time unit impulse defined as: δ[x] = 1, x = 0 0, x = 0. (2.1)
2.1 Space: Convolution Sum 2.2 Frequency: Fourier Transform
1
x
Figure 2.1 Unit impulse
Any discrete-time signal can be represented as a sum of scaled and shifted unit-impulses:
∞
Let's now consider what happens when we present a linear timeinvariant system with this new representation of a discrete-time signal: g[x] = T {f [x]} = T
∞ k=−∞
f [k]δ[x − k] .
(2.3)
17
By the property of linearity, Equation (1.4), the above expression may be rewritten as:
∞
g[x] =
k=−∞
f [k]T {δ[x − k]}.
(2.4)
Imposing the property of time-invariance, Equation (1.5), if h[x] is the response to the unit-impulse, δ[x], then the response to δ[x−k] is simply h[x−k]. And now, the above expression can be rewritten as:
∞
g[x] =
k=−∞
f [k]h[x − k].
(2.5)
Consider for a moment the implications of the above equation. The unit-impulse response, h[x] = T {δ[x]}, of a linear time-invariant system fully characterizes that system. More precisely, given the unit-impulse response, h[x], the output, g[x], can be determined for any input, f [x]. The sum in Equation (2.5) is commonly called the convolution sum and may be expressed more compactly as: g[x] = f [x] ⋆ h[x]. (2.6)
A more mathematically correct notation is (f ⋆ h)[x], but for later notational considerations, we will adopt the above notation.
Example 2.2 response: Consider the following finite-length unit-impulse h[x]
h[x] 0
The next output sum at x = −1, is computed by "sliding" the unit-impulse response along the input signal and computing a similar sum.
Since linear time-invariant systems are fully characterized by convolution with the unit-impulse response, properties of such systems can be determined by considering properties of the convolution operator. For example, convolution is commutative:
∞
f [x] ⋆ h[x] =
k=−∞
f [k]h[x − k], 18
let j = x − k
∞
∞
=
j=−∞
f [x − j]h[j] =
j=−∞
h[j]f [x − j] (2.7)
= h[x] ⋆ f [x]. Convolution also distributes over addition:
∞
f [x] ⋆ (h1 [x] + h2 [x]) =
k=−∞ ∞
f [k](h1 [x − k] + h2 [x − k]) f [k]h1 [x − k] + f [k]h2 [x − k]
∞
=
k=−∞ ∞
=
k=−∞
f [k]h1 [x − k] +
k=−∞
f [k]h2 [x − k] (2.8)
= f [x] ⋆ h1 [x] + f [x] ⋆ h2 [x].
A final useful property of linear time-invariant systems is that a cascade of systems can be combined into a single system with impulse response equal to the convolution of the individual impulse responses. For example, for a cascade of two systems: (f [x] ⋆ h1 [x]) ⋆ h2 [x] = f [x] ⋆ (h1 [x] ⋆ h2 [x]). (2.9)
f[x]
h1[x]
h2[x]
g[x]
f[x]
h1[x] * h2[x]
g[x]
Figure 2.3 Identical LTIs
This property is fairly straight-forward to prove, and offers a good exercise in manipulating the convolution sum: g1 [x] = f [x] ⋆ h1 [x]
∞
=
k=−∞
f [k]h1 [x − k] and, (2.10)
g2 [x] = (f [x] ⋆ h1 [x]) ⋆ h2 [x] = g1 [x] ⋆ h2 [x]
∞
=
j=−∞ ∞
g1 [j]h2 [x − j] substituting for g1 [x],
∞
=
j=−∞ ∞
k=−∞
=
k=−∞ ∞
f [k] f [k]
f [k]h1 [j − k] h2 [x − j]
∞
j=−∞ ∞ i=−∞
h1 [j − k]h2 [x − j] h1 [i]h2 [x − i − k]
let i = j − k,
=
k=−∞
= f [x] ⋆ (h1 [x] ⋆ h2 [x]).
(2.11)
Let's consider now how these concepts fit into the linear-algebraic framework. First, a length N signal can be thought of as a point 19
where each row contains a shifted and time-reversed copy of the unit-impulse response, h[x]. The convolution matrix can then be thought of as simply transforming the basis set. As expected, this matrix is a Toeplitz matrix of the form given earlier in Equation (1.7). The reason for the time-reversal can be seen directly from the convolution sum of Equation (2.5). More specifically, the output signal g[x] at a fixed x, is determined by summing the products of f [k]h[x − k] for all k. Note that the signal h[x − k] can be equivalently written as h[−k + x], which is a shifted (by x) and time-reversed (because of the −k) copy of the impulse response. Note also that when expressed in matrix form, it becomes immediately clear that the convolution sum is invertible, when h is not identically zero: g = M f and f = M −1 g.
f[x]
, (2.12) 0
Before pressing on, let's try to combine the main ideas seen so far into a single example. We will begin by defining a simple discretetime system, show that it is both linear and time-invariant, and compute its unit-impulse response
Example 2.3 Define the discrete-time system, T {·} as: g[x] = f [x + 1] − f [x − 1].
g[x]
f[x]
This system is linear because it obeys the rule of superposition: T {af1 [x] + bf2 [x]}
g[x]
And, in matrix form, this linear time-invariant system is given by g = M f , where:
M
=
0 1 0 −1 0 1 0 −1 0 . . . 0 0 0
0 1 0 0
0 0 1 0 0 0
... ... ... .. . ... ... ...
0 0 0 −1 0 0
0 0 0 0 −1 0
0 0 0 1 0 −1
0 0
−1 0 0 . . . .
0 1 0
2.2 Frequency: Fourier Transform
1
In the previous section the expression of a discrete-time signal as a sum of scaled and shifted unit-impulses led us to the convolution sum. In a similar manner, we will see shortly how expressing a signal in terms of sinusoids will lead us to the Fourier transform, and then to a new way to think about discrete-time systems. The basic signal used as the building block is the sinusoid: A cos[ωx + φ], −∞ < x < ∞, (2.13)
0
−1 1
0
where A is the amplitude, ω is the frequency, and φ is the phase of the sinusoid. Shown in Figure 2.6, from top to bottom, are cos[x], cos[2x], and cos[x + π/2]. Consider next the following, seemingly unrelated, complex exponential eiωx with frequency ω, and i the √ complex value −1. This function is simply a compact notation for describing a sum of the sine and cosine function: Ae
iωx
−1 1
0
−1
= A cos(ωx) + iA sin(ωx). 21
(2.14)
Figure 2.6 f [x] = A cos[ωx + φ]
The complex exponential has a special relationship with linear time-invariant systems - the output of a linear time-invariant system with unit-impulse response h[x] and a complex exponential as input is: g[x] = eiωx ⋆ h[x]
∞
that is, given a complex exponential as input, the output of a linear time-invariant system is again a complex exponential of the same frequency scaled by some amount. 3 The scaling of the complex exponential, H[w], is called the frequency response and is generally a complex-valued function expressed in terms of its real and imaginary components: H[ω] = HR [ω] + iHI [ω], (2.17)
Intuitively, this should make perfect sense. This system simply takes an input signal and outputs a delayed copy, therefore, there is no change in the magnitude of each sinusoid, while there is a phase shift proportional to the delay, x0 .
So, why the interest in sinusoids and complex exponentials? As we will show next, a broad class of signals can be expressed as a linear combination of complex exponentials, and analogous to the impulse response, the frequency response completely characterizes the system. Let's begin by taking a step back to the more familiar sinusoids, and then work our way to the complex exponentials. Any periodic discrete-time signal, f [x], can be expressed as a sum of scaled, phase-shifted sinusoids of varying frequencies: f [x] = 1 2π
π
In the language of linear algebra, the sinusoids are said to form a basis for the set of periodic signals, that is, any periodic signal can be written as a linear combination of the sinusoids. Recall 23
that in deriving the convolution sum, the basis consisted of shifted copies of the unit-impulse. But note now that this new basis is not fixed because of the phase term, φk . It is, however, possible to rewrite the Fourier series with respect to a fixed basis of zerophase sinusoids. With the trigonometric identity cos(A + B) = cos(A) cos(B)−sin(A) sin(B), the Fourier series of Equation (2.18) may be rewritten as: f [x] = = = 1 2π 1 2π 1 2π
π
ck cos[kx + φk ]
k=−π π
ck cos[φk ] cos[kx] + ck sin[φk ] sin[kx]
k=−π π
ak cos[kx] + bk sin[kx]
k=−π
(2.19)
In this expression, the constants ak and bk are the Fourier coefficients and are determined by the Fourier transform. In other words, the Fourier transform simply provides a means for expressing a signal in terms of the sinusoids. The Fourier coefficients are given by:
∞ ∞
ak =
j=−∞
f [j] cos[kj]
and
bk =
j=−∞
f [j] sin[kj]
(2.20)
Notice that the Fourier coefficients are determined by projecting the signal onto each of the sinusoidal basis. That is, consider both the signal f [x] and each of the sinusoids as T -dimensional vectors, f and b, respectively. Then, the projection of f onto b is: f0 b0 + f1 b1 + ... =
j
f j bj ,
(2.21)
where the subscript denotes the jth entry of the vector. Often, a more compact notation is used to represent the Fourier series and Fourier transform which exploits the complex exponential and its relationship to the sinusoids: eiωx = cos(ωx) + i sin(ωx), (2.22)
√ where i is the complex value −1. Under the complex exponential notation, the Fourier series and transform take the form: f [x] = 1 2π
π ∞
Comparing the Fourier transform (Equation (2.24)) with the frequency response (Equation (2.16)) we see now that the frequency response of a linear time-invariant system is simply the Fourier transform of the unit-impulse response:
∞
Both the impulse response and the magnitude of the frequency response are shown below, where for clarity, the frequency response was drawn as a continuous function.
26
Space (h[x])
0.5
Frequency (|H[ω]|)
1
0
−0.5 −1 0 1
0 −pi
0
pi
This system is an (approximate) differentiator, and can be seen from the definition of differentiation: df (x) dx =
ε→0
lim
f (x + ε) − f (x − ε) , ε
where, in the case of the system T {·}, ε is given by the distance between samples of the discrete-time signal f [x]. Let's see now if we can better understand the frequency response of this system, recall that the magnitude was given by | sin(ω)| and the phase by
π . 2
Consider now the derivative of a fixed fre-
quency sinusoid sin(ωx), differentiating with respect to x gives ω cos(ωx) = ω sin(ωx − π/2). Note that differentiation causes a phase shift of π/2 and a scaling by the frequency of the sinusoid. Notice that this is roughly in-line with the Fourier transform, the difference being that the amplitude is given by | sin(ω)| instead of ω. Note though that for small ω, | sin(ω)| ≈ ω. This discrepancy makes the system only an approximate, not a perfect, differentiator.
Linear time-invariant systems can be fully characterized by their impulse, h[x], or frequency responses, H[ω], both of which may be used to determine the output of the system to any input signal, f [x]: g[x] = f [x] ⋆ h[x] and G[ω] = F [ω]H[ω], (2.26)
where the output signal g[x] can be determined from its Fourier transform G[ω], by simply applying the inverse Fourier transform. This equivalence illustrates an important relationship between the space and frequency domains. Namely, convolution in the space domain is equivalent to multiplication in the frequency domain. This is fairly straight-forward to prove: g[x] = f [x] ⋆ h[x]
∞ ∞
Notice that this equation embodies both the Fourier transform and the Fourier series of Equation (2.24). The above form is the Fourier transform, and the Fourier series is gotten by left-multiplying with the inverse of the matrix, M −1 F = f .
28
3. Sampling: Continuous to Discrete (and back)
It is often more convenient to process a continuous-time signal with a discrete-time system. Such a system may consist of three distinct stages: (1) the conversion of a continuous-time signal to a discrete-time signal (C/D converter); (2) the processing through a discrete-time system; and (3) the conversion of the output discretetime signal back to a continuous-time signal (D/C converter). Earlier we focused on the discrete-time processing, and now we will concentrate on the conversions between discrete- and continuoustime signals. Of particular interest is the somewhat remarkable fact that under certain conditions, a continuous-time signal can be fully represented by a discrete-time signal! 3.1 Continuous to Discrete: Space
f[x]
for integer values x. In this expression, the quantity T is the sampling period. In general, continuous-time signals will be denoted with rounded parenthesis (e.g., f (·)), and discrete-time signals with square parenthesis (e.g., f [·]). This sampling operation may be considered as a multiplication of the continuous time signal with an impulse train, Figure 3.2. The impulse train is defined as:
∞
D/C
g(x)
Figure 3.1 Processing block diagram
s(x) =
k=−∞
δ(x − kT ),
(3.2)
f(x)
where δ(·) is the unit-impulse, and T is the sampling period - note that the impulse train is a continuous-time signal. Multiplying the impulse train with a continuous-time signal gives a sampled signal: fs (x) = f (x)s(x), (3.3)
x f[x]
Figure 3.2 Sampling: space
Note that the sampled signal, fs (x), is indexed on the continuous variable x, while the final discrete-time signal, f [x] is indexed on the integer variable x. It will prove to be mathematically convenient to work with this intermediate sampled signal, fs (x).
29
3.2 Continuous to Discrete: Frequency
F(w)
w −w n S(w) wn
In the space domain, sampling was described as a product between the impulse train and the continuous-time signal (Equation (3.3)). In the frequency domain, this operation amounts to a convolution between the Fourier transform of these two signals: Fs (ω) = F (ω) ⋆ S(ω) (3.4)
w −w s 0 Fs(w) ws
w w s− w n
Figure 3.3 Sampling: no aliasing
For example, shown in Figure 3.3 (from top to bottom) are the Fourier transforms of the continuous-time function, F (ω), the impulse train, S(ω), itself an impulse train, and the results of convolving these two signals, Fs (ω). Notice that the Fourier transform of the sampled signal contains multiple (yet exact) copies of the Fourier transform of the original continuous signal. Note however the conditions under which an exact replica is preserved depends on the maximum frequency response ωn of the original continuous-time signal, and the sampling interval of the impulse train, ωs which, not surprisingly, is related to the sampling period T as ωs = 2π/T . More precisely, the copies of the frequency response will not overlap if: ωn < ωs − ωn or (3.5)
F(w)
w −w n S(w) wn
ωs > 2ωn ,
w −w s 0 Fs(w) ws
The frequency ωn is called the Nyquist frequency and 2ωn is called the Nyquist rate. Shown in Figure 3.4 is another example of this sampling process in the frequency domain, but this time, the Nyquist rate is not met, and the copies of the frequency response overlap. In such a case, the signal is said to be aliased. Not surprisingly, the Nyquist rate depends on both the characteristics of the continuous-time signal, and the sampling rate. More precisely, as the maximum frequency, ωn , of the continuous-time signal increases, the sampling period, T must be made smaller (i.e., denser sampling), which in turn increases ωs , preventing overlap of the frequency responses. In other words, a signal that changes slowly and smoothly can be sampled fairly coarsely, while a signal that changes quickly requires more dense sampling. 3.3 Discrete to Continuous If the Nyquist rate is met, then a discrete-time signal fully characterizes the continuous-time signal from which it was sampled. On the other hand, if the Nyquist rate is not met, then the sampling leads to aliasing, and the discrete-time signal does not accurately represent its continuous-time counterpart. In the former case, it 30
w
Figure 3.4 Sampling: aliasing
is possible to reconstruct the original continuous-time signal, from the discrete-time signal. In particular since the frequency response of the discrete-time signal contains exact copies of the original continuous-time signals frequency response, we need only extract one of these copies, and inverse transform the result. The result will be identical to the original signal. In order to extract a single copy, the Fourier transform of the sampled signal is multiplied by an ideal reconstruction filter as shown in Figure 3.5. This filter has unit value between the frequencies −π/T to π/T and is zero elsewhere. This frequency band is guaranteed to be greater than the Nyquist frequency, ωn (i.e., ωs = 2π/T > 2ωn , so that π/T > ωn ). In the space domain, this ideal reconstruction filter has the form: sin(πx/T ) , πx/T
Fs(w)
w pi/T
Figure 3.5 Reconstruction
h(x) =
(3.6)
1
and is often referred to as the ideal sync function. Since reconstruction in the frequency domain is accomplished by multiplication with the ideal reconstruction filter, we could equivalently reconstruct the signal by convolving with the ideal sync in the space domain.
a sinusoid with frequency ω0 . We will eventually be interested in sampling this function and seeing how the effects of aliasing are manifested. But first, let's compute the Fourier transform of this signal:
∞
F (ω)
=
k=−∞ ∞
f (k)e−iωk
=
k=−∞ ∞
cos(ω0 k)(cos(ωk) − i sin(ωk))
=
k=−∞
cos(ω0 k) cos(ωk) − i cos(ω0 k) sin(ωk)
First let's consider the product of two cosines. It is easy to show from basic trigonometric identities that cos(A) cos(B) = 0 when A = B, and is equal to π when |A| = |B|. Similarly, one can show that cos(A) sin(B) = 0 for all A and B. So, the Fourier transform of cos(ω0 x) = π for |ω| = ω0 , and is 0 otherwise (see below). If the sampling rate is greater than 2ω0 , then there will be no aliasing, but if the sampling rate is less than 2ω0 , then the reconstructed signal will be of the form cos((ωs − ω0 )x), that is, the reconstructed signal will be appear as a lower frequency sinusoid - it will be aliased.
31
F(w)
w −w 0 w0
Sampling
No Aliasing
Aliasing Fs(w)
Fs(w)
w −w 0 w0 −w 0 w0
w
We will close this chapter by drawing on the linear algebraic framework for additional intuition on the sampling and reconstruction process. First we will need to restrict ourselves to the sampling of an already sampled signal, that is, consider a m-dimensional signal sub-sampled to a n-dimensional signal. We may express this operation in matrix form as follows: 1 0 g1 . . . = . . . 0 gn 1
0 0 0 0 1 0 0 0 0 0 0 0
... ... .. . ... ...
0 0 0 0 0 0 0 1 0 0 0 0
gn = Sn×m fm ,
f1 0 f2 0 . . . . . . 0 fm−1 1 fm
(3.7)
where the subscripts denote the vector and matrix dimensions, and in this example n = m/2. Our goal now is to determine when it is possible to reconstruct the signal f , from the sub-sampled signal g. The Nyquist sampling theory tells us that if a signal is band-limited (i.e., can be written as a sum of a finite number of sinusoids), then we can sample it without loss of information. We can express this constraint in matrix notation: fm = Bm×n wn , (3.8)
where the columns of the matrix B contains the basis set of sinusoids - in this case the first n sinusoids. Substituting into the above sampling equation gives: gn = Sn×m Bm×n wn = Mn×n wn . (3.9)
If the matrix M is invertible, then the original weights (i.e., the representation of the original signal) can be determined by simply left-multiplying the sub-sampled signal g by M −1 . In other 32
words, Nyquist sampling theory can be thought of as simply a matrix inversion problem. This should not be at all surprising, the trick to sampling and perfect reconstruction is to simply limit the dimensionality of the signal to at most twice the number of samples.
33
4. Digital Filter Design
Recall that the class of linear time-invariant systems are fully characterized by their impulse response. More specifically, the output of a linear time-invariant system to any input f [x] can be determined via a convolution with the impulse response h[x]: g[x] = f [x] ⋆ h[x]. (4.1)
Therefore the filter h[x] and the linear-time invariant system are synonymous. In the frequency domain, this expression takes on the form: G[ω] = F [ω]H[ω]. (4.2)
In other words, a filter modifies the frequencies of the input signal. It is often the case that such filters pass certain frequencies and attenuate others (e.g., a lowpass, bandpass, or highpass filters). The design of such filters consists of four basic steps: 1. 2. 3. 4. choose the desired frequency response choose the length of the filter define an error function to be minimized choose a minimization technique and solve
The choice of frequency response depends, of course, on the designers particular application, and its selection is left to their discretion. We will however provide some general guidelines for choosing a frequency response that is amenable to a successful design. In choosing a filter size there are two conflicting goals, a large filter allows for a more accurate match to the desired frequency response, however a small filter is desirable in order to minimize computational demands 4 . The designer should experiment with varying size filters until an equitable balance is found. With a frequency response and filter size in hand, this chapter will provide the computational framework for realizing a finite length filter that "best" approximates the specified frequency response. Although there are numerous techniques for the design of digital filters we will cover only two such techniques chosen for their simplicity and generally good performance (see Digital Filter Design by T.W. Parks and C.S. Burrus for a full coverage of many other approaches).
4
In multi-dimensional filter design, separability is also a desirable property.
34
4.1 Choosing a Frequency Response A common class of filters are bandpass in nature, that is, they pass certain frequencies and attenuate others. An ideal lowpass, bandpass, and highpass filter are illustrated in Figure 4.1 Shown is the magnitude of the frequency response in the range [0, π], since we are typically interested in designing real-valued, linear-phase filters, we need only specify one-half of the magnitude spectrum (the response is symmetric about the origin). The responses shown in Figure 4.1 are often referred to as brick wall filters because of their abrupt fall-off. A finite-length realization of such a filter produces undesirable "ringing" known as Gibbs phenomena as shown below in the magnitude of the frequency response of ideal lowpass filters of length 64, 32, and 16 (commonly referred to as the filter tap size). 64 taps
1 1
| H(w) |
Passband Stopband w 0 pi
32 taps
1
16 taps
Figure 4.1 Ideal lowpass, bandpass, and highpass
0.5
0.5
0.5
0 0
pi/2
pi
0 0
pi/2
pi
0 0
pi/2
pi
| H(w) |
These effects are particularly undesirable because neighboring frequencies may be passed or attenuated by wildly varying amounts, leading to general instabilities. To counter this problem, the designer is resigned to sacrificing the ideal response for a "softer" frequency response, Figure 4.2. Such a frequency response is amenable to a small finite-length filter free of significant ringing artifacts. The specific functional form of the soft falloff is somewhat arbitrary, however one popular form is a raised cosine. In its most general form, the frequency response takes the following form, where the bandpass nature of the response is controlled through ω0 , ω1 , ∆ω0 , and ∆ω1 .
0, 1 2 [1 − cos(π(ω − ω0 )/∆ω0 )] ,
In general, a larger transition width (i.e., ∆ω0 , and ∆ω1 ) allows for smaller filters with minimal ringing artifacts. The tradeoff, of course, is that a larger transition width moves the frequency response further from the ideal brick-wall response. In specifying only half the frequency response (from [0, π]) we are implicitly imposing a symmetric frequency response and thus assuming that the desired filter is symmetric. 35
0.5
ω0
ω1
0 0
pi
Figure 4.3 Frequency response
4.2 Frequency Sampling Once the desired frequency response has been chosen, the more difficult task of designing a finite-length filter that closely approximates this response begins. In general this problem is hard because by restricting ourselves to a small finite-length filter we are in effect asking to fit an arbitrarily complex function (the desired frequency response) with the sum of a small number of sinusoids. We begin with the most straight-forward design method - frequency sampling. Our goal is to design a filter h whose Fourier transform best approximates the specified response H, that is: F(h) = H (4.3)
In other words, the design simply involves inverse Fourier transforming the specified frequency response. This series of steps can be made more explicit and practical for computer implementation by expressing the initial constraint (Equation (4.3)) in matrix notation: Mh = H (4.5)
where H is the n-dimensional sampled frequency response, M is the n × n Fourier matrix (Equation (2.28)), and n is the chosen filter size. The filter h can be solved for by left multiplying both sides of Equation (4.5) by the inverse of the matrix F : h = M −1 H. (4.6)
Since the matrix M is square, this design is equivalent to solving for n unknowns (the filter taps) from n linearly independent equations. This fact illustrates the shortcomings of this approach, namely, that this method produces a filter with a frequency response that exactly matches the sampled response, but places no constraints on the response between sampled points. This restriction can often lead to poor results that are partially alleviated by the least-squares design presented next.
36
4.3 Least-Squares Our goal is to design a filter h that "best" approximates a specified frequency response. As before this constraint can be expressed as: M h = H, (4.7)
where M is the N × n Fourier matrix (Equation (2.28)), H is the N sampled frequency response, and the filter size is n. Note that unlike before, this equation is over constrained, having n unknowns in N > n equations. We can solve this system of equations in a least-squares sense by first writing a squared error function to be minimized: E(h) = | M h − H |2 (4.8)
Shown in Figure 4.4 are a set of lowpass, bandpass, and highpass 16-tap filters designed using this technique. In this design, the frequency response was of the form given in Equation (4.3), with start and stop bands of [ω0 , ω1 ] = [0, π/2], [π/4, 3π/4], and [π/2, π], and a transition width of ∆ω0 = ∆ω1 = π/4. The frequency response was sampled at a rate of N = 512. Finally, note that the previous frequency sampling design is equivalent to the least-squares design when the sampling of the Fourier basis is the same as the filter size (i.e., N = n). 4.4 Weighted Least-Squares One drawback of the least-squares method is that the error function (Equation (4.8)) is uniform across all frequencies. This is easily rectified by introducing a weighting on the least-squares error function: E(h) = W | M h − H |
5
0.5
0 0 1
pi/2
pi
0.5
0 0 1
pi/2
pi
0.5
0 0
pi/2
pi
2
(4.11)
Figure 4.4 Leastsquares: lowpass, bandpass, and highpass
Because of Parseval's theorem (which amounts to the orthonormality of the Fourier transform), the minimal error in the frequency domain equates to a minimal error in the space domain.
37
where W is a diagonal weighting matrix. That is, the diagonal of the matrix contains the desired weighting of the error across frequency. As before, we minimize by differentiating with respect to h: dE(h) dh = 2M t W | M h − H | = 2M t W M h − 2W M t H, then set equal to zero, and solve:
1
(4.12)
h = (M t W M )−1 M t W H.
(4.13)
0.5
Note that this solution will be equivalent to the original leastsquares solution (Equation (4.10)) when W is the identity matrix (i.e., uniform weighting).
pi/2 pi
0 0 1
0.5
0 0
Shown in Figure 4.5 is a comparison of an 8-tap lowpass filter designed with a uniform weighting, and with a weighting that 1 emphasizes the errors in the low frequency range, W (ω) = (|ω|+1)8 . Note that in the case of the later, the errors in the low frequencies are smaller, while the errors in the high frequencies have increased.
pi/2 pi
Figure 4.5 Least-squares and weighted least squares
38
5. Photons to Pixels
5.1 Pinhole Camera The history of the pinhole camera (or camera obscura) dates back as early as the fifth century B.C., and continues to be popular today among students, artists, and scientists. The Chinese philosopher Mo Ti is believed to be the first to notice that objects reflect light in all directions and that the light rays that pass through a small hole produce an inverted image. In its simplest form a pinhole camera is a light-tight box with a tiny hole in one end and a photo-sensitive material on the other. Remarkably, this simple device is capable of producing a photograph. However, the pinhole camera is not a particularly efficient imaging system (often requiring exposure times as long as several hours) and is more popular for its artistic value than for its practical value. Nevertheless, the pinhole camera is convenient because it affords a simple model of more complex imaging systems. That is, with a pinhole camera model, the projection of points from the three-dimensional world onto the two-dimensional sensor takes on a particularly simple form. Denote a point in the three-dimensional world as a column vector, P = ( X Y Z )t and the projection of this point onto the two dimensional image plane as p = ( x y )t . Note that the world and image points are expressed with respect to their own coordinate systems, and for convenience, the image coordinate system is chosen to be orthogonal to the Z-axis, i.e., the origins of the two systems are related by a one-dimensional translation along the Z−axis or optical axis. It is straight-forward to show from a similar triangles argument that the relationship between the world and image point is: dX x=− Z and dY y=− , Z (5.1)
5.1 Pinhole Camera 5.2 Lenses 5.3 CCD
Figure 5.1 Pinhole image formation
Y P Z
y
p x X
where d is the displacement of the image plane along the Z-axis 6 These equations are frequently referred to as the perspective projection equations. Although non-linear in their nature, the perspective projection equations may be expressed in matrix form
6 The value d in Equation (5.1) is often referred to as the focal length. We do not adopt this convention primarily because it is a misnomer, under the pinhole model all points are imaged in perfect focus.
An approximation to the above perspective projection equations is orthographic projection, where light rays are assumed to travel from a point in the world parallel to the optical axis until they intersect the image plane. Unlike the pinhole camera and perspective projection equations, this model is not physically realizable and is used primarily because the projection equations take on a particularly simple linear form:
Y P p Z x X y
x=X And in matrix form: x y =
and
y = Y.
(5.3)
Figure 5.3 Orthographic projection
1 0 0 1
X 0 Y 0 Z
(5.4)
Orthographic projection is a reasonable approximation to perspective projection when the difference in depth between points in the world is small relative to their distance to the image plane. In the special case when all the points lie on a single frontal-parallel d surface relative to the image plane (i.e., Z is constant in Equation (5.1)), the difference between perspective and orthographic is only a scale factor. 5.2 Lenses It is important to remember that both the perspective and orthographic projection equations are only approximations of more complex imaging systems. Commercial cameras are constructed with a variety of lenses that collect and focus light onto the image plane. That is, light emanates from a point in the world in all directions and, whereas a pinhole camera captures a single light ray, a lens collects a multitude of light rays and focuses the light to a small region on the image plane. Such complex imaging systems are often described with the simpler thin-lens model. Under the thin-lens model the projection of the central or principal ray obeys the rules of perspective projection, Equation (5.1): the point P = ( X Y Z )t is projected onto the image plane centered about the point ( x y )t = ( −dX −dY )t . If the point P Z Z 40
Y P Z
y
Figure 5.4 Thin lens
is in perfect focus, then the remaining light rays captured by the lens also strike the image plane at the point p. A point is imaged in perfect focus if its distance from the lens satisfies the following thin-lens equation: 1 1 + Z d = 1 , f (5.5)
where d is the distance between the lens and image plane along the optical axis, and f is the focal length of the lens. The focal length is defined to be the distance from the lens to the image plane such that the image of an object that is infinitely far away is imaged in perfect focus. Points at a depth of Zo = Z are imaged onto a small region on the image plane, often modeled as a blurred circle with radius r: r =
1 f
R 1 − Zo
1 1 − f Zo
−
1 , d
(5.6)
where R is the radius of the lens. Note that when the depth of a point satisfies Equation (5.5), the blur radius is zero. Note also that as the lens radius R approaches 0 (i.e., a pinhole camera), the blur radius also approaches zero for all points independent of its depth (referred to as an infinite depth of field). Alternatively, the projection of each light ray can be described in the following more compact matrix notation: l2 α2 = 1
1 −R n2 −n1 n2
0
n1 n2
l1 α1
,
(5.7)
where R is the radius of the lens, n1 and n2 are the index of refraction for air and the lens material, respectively. l1 and l2 are the height at which a light ray enters and exits the lens (the thin lens idealization ensures that l1 = l2 ). α1 is the angle between the entering light ray and the optical axis, and α2 is the angle between the exiting light ray and the optical axis. This formulation is particularly convenient because a variety of lenses can be described in matrix form so that a complex lens train can then be modeled as a simple product of matrices.
Y
Image formation, independent of the particular model, is a threedimensional to two-dimensional transformation. Inherent to such a transformation is a loss of information, in this case depth information. Specifically, all points of the form Pc = ( cX cY cZ )t , for any c ∈ R, are projected to the same point ( x y )t - the projection is not one-to-one and thus not invertible. In addition to this geometric argument for the non-invertibility of image formation, a similarly straight-forward linear algebraic argument holds. 41
y
P 1 Z
P 2
P 3 p x X
Figure 5.5 Non-invertible projection
In particular, we have seen that the image formation equations may be written in matrix form as, p = Mn×m P , where n < m (e.g., Equation (5.2)). Since the projection is from a higher dimensional space to a lower dimensional space, the matrix M is not invertible and thus the projection is not invertible. 5.3 CCD To this point we have described the geometry of image formation, how light travels through an imaging system. To complete the image formation process we need to discuss how the light that strikes the image plane is recorded and converted into a digital image. The core technology used by most digital cameras is the chargecoupled device (CCD), first introduced in 1969. A basic CCD consists of a series of closely spaced metal-oxide-semiconductor capacitors (MOS), each one corresponding to a single image pixel. In its most basic form a CCD is a charge storage and transport device: charge is stored on the MOS capacitors and then transported across these capacitors for readout and subsequent transformation to a digital image. More specifically, when a positive voltage, V , is applied to the surface of a P-type MOS capacitor, positive charge migrates toward ground. The region depleted of positive charge is called the depletion region. When photons (i.e., light) enter the depletion region, the electrons released are stored in this region. The value of the stored charge is proportional to the intensity of the light striking the capacitor. A digital image is subsequently formed by transferring the stored charge from one depletion region to the next. The stored charge is transferred across a series of MOS capacitors (e.g., a row or column of the CCD array) by sequentially applying voltage to each MOS capacitor. As charge passes through the last capacitor in the series, an amplifier converts the charge into a voltage. An analog-to-digital converter then translates this voltage into a number (i.e., the intensity of an image pixel).
Light
V
Depletion region
MOS Ground
Figure 5.6 MOS capacitor
42
6. Point-Wise Operations
6.1 Lookup Table The internal representation of a digital image is simply a matrix of numbers representing grayscale or color values. But when an image is displayed on a computer monitor we typically do not see a direct mapping of the image. An image is first passed through a lookup table (LUT) that maps the image intensity values to brightness values, Figure 6.1. If the lookup table is linear with unit slope and zero intercept then the image is directly mapped to the display, otherwise, the displayed image will not be an exact representation of the underlying image. For example, most computer monitors intentionally impose a non-linear LUT of the general form D = I α (i.e., gamma correction), where α is a real number, and D and I are the displayed and image values. A variety of interesting visual effects can be achieved by simple manipulations of the functional form of the LUT. Keep in mind though that in manipulating the lookup table, the underlying image is left untouched, it is only the mapping from pixel value to display brightness that is being effected. 6.2 Brightness/Contrast Perhaps the most common and familiar example of a LUT manipulation is to control the brightness or darkness of an image as shown in Figure 6.3. The bright and dark images of Einstein were created by passing the middle image through the LUTs shown in the same figure. The functional form of the LUT is a unit-slope line with varying intercepts: g(u) = u + b, with the image intensity values u ∈ [0, 1]. A value of b > 0 results in a brightening of the image and b < 0 a darkening of the image. Another common manipulation is that of controlling the contrast of an image as shown in Figure 6.4. The top image is said to be high contrast and the bottom image low contrast, with the corresponding LUTs shown in the same figure. The functional form of these LUTs is linear: g(u) = mu + b, where the relationship between the slope and intercept is b = 1/2(1 − m). The image contrast is increased with a large slope and negative intercept (in the limit, m → ∞ and b → −∞), and the contrast is reduced with a small slope and positive intercept (m → 0 and b → 1/2). The image is inverted with m = −1 and b = 1, that is white is mapped to black, and black is mapped to white. Of course, an image can be both contrast enhanced and brightened or darkened by simply passing the image through two (or more) LUTs. 43
Autoscaling is a special case of contrast enhancement where the minimum image intensity value is mapped to black and the maximum value is mapped to white, Figure 6.2. Autoscaling maximizes the contrast without saturating at black or white. The problem with this sort of autoscaling is that a few stray pixels can dictate the contrast resulting in a low-contrast image. A less sensitive approach is to sort the intensity values in increasing order and map the 1% and 99% intensity values to black and white, respectively. Although this will lead to a small amount of saturation, it rarely fails to produce a high-contrast image. 6.3 Gamma Correction Typically, high contrast images are visually more appealing. However a drawback of linear contrast enhancement described above is that it leads to saturation at both the low and high end of the intensity range. This may be avoided by employing a non-linear contrast adjustment scheme, also realizable as a lookup table (LUT) manipulation. The most standard approach is gamma correction, where the LUT takes the form: g(u) = uα ,
Dark
Bright
(6.1)
Figure 6.3 Brightness
where α > 1 increases contrast, and α < 1 reduces contrast. Shown in Figure 6.5 are contrast enhanced (top: α = 2) and contrast reduced (bottom: α = 1/2) images. Note that with the intensity values u scaled into the range [0, 1], black (0) and white (1) are mapped to themselves. That is, there is no saturation at the low or high end. Gamma correction is widely used in a number of devices because it yields reasonable results and is easily parameterized. One drawback to this scheme is that the gray values are mapped in an asymmetric fashion with respect to midlevel gray (0.5). This may be alleviated by employing a sigmoidal non-linearity of the form g(u) = 1 . 1 + e−αu+β (6.2)
High
In order that g(u) be bound by the interval [0, 1], it must be scaled as follows: (g(u) − c1 )/c2 , where c1 = 1/(1 + eβ ) and c2 = 1/(1+e−α+β )−c1. This non-linearity, with its two degrees of freedom, is more versatile and can produce a more balanced contrast enhancement. Shown in Figure 6.6 is a contrast enhanced image with α = 12 and β = 6.
Low
Figure 6.4 Contrast
44
6.4 Quantize/Threshold A digital image, by its very nature, is quantized to a discrete number of intensity values. For example an image quantized to 8-bits contains 28 = 256 possible intensity values, typically in the range [0, 255]. An image can be further quantized to a lower number of bits (b) or intensity values (2b ). Quantization can be accomplished by passing the image through a LUT containing a step function, Figure 6.7, where the number of steps governs the number of intensity values. Shown in Figure 6.7 is an image of Einstein quantized to five intensity values, notice that all the subtle variations in the curtain and in his face and jacket have been eliminated. In the limit, when an image is quantized to one bit or two intensity values, the image is said to be thresholded. Shown in Figure 6.8 is a thresholded image of Einstein and the corresponding LUT, a twostep function. The point at which the step function transitions from zero to one is called the threshold and can of course be made to be any value (i.e., slid left or right). 6.5 Histogram Equalize The intensity values of a typical image are often distributed unevenly across the full range of 0 to 255 (for an 8-bit image), with most the mass near mid-gray (128) and falling off on either side, Figure 6.9. An image can be transformed so that the distribution of intensity values is flat, that is, each intensity value is equally represented in the image. This process is known has histogram equalization 7 . Although it may not be immediately obvious an image is histogram equalized by passing it through a LUT with the functional form of the cumulative distribution function. More specifically, define N (u) as the number of pixels with intensity value u, this is the image histogram and a discrete approximation to the probability distribution function. Then, the cumulative distribution function is defined as:
u
Figure 6.5 Contrast: Gamma
Figure 6.6 Contrast: Sigmoid
C(y) =
i=0
N (i),
(6.3)
that is, C(u) is the number of pixels with intensity value less than or equal to u. Histogram equalization then amounts to simply inserting this function into the LUT. Shown in Figure 6.9 is Einstein before and after histogram equalization. Notice that the effect is similar to contrast enhancement, which intuitively should make sense since we increased the number of black and white pixels.
Why anyone would want to histogram equalize an image is a mystery to me, but here it is in case you do.
7
45
In all of these examples the appearance of an image was altered by simply manipulating the LUT, the mapping from image intensity value to display brightness value. Such a manipulation leaves the image content intact, it is a non-destructive operation and thus completely invertible. These operations can be made destructive by applying the LUT operation directly to the image. For example an image can be brightened by adding a constant to each pixel, and then displaying with a linear LUT. Since such an operation is destructive it may not be inveritble, for example when brightening an 8-bit image, all pixels that exceed the value 255 will be truncated to 255.
Figure 6.7 Quantize
Figure 6.8 Threshold
15000 10000 5000 0 0
100
200
15000 10000 5000 0 0
100
200
46
Figure 6.9 Histogram equalize
7. Linear Filtering
7.1 Convolution The one-dimensional convolution sum, Equation (2.5), formed the basis for much of our discussion on discrete-time signals and systems. Similarly the two-dimensional convolution sum will form the basis from which we begin our discussion on image processing and computer vision. The 1-D convolution sum extends naturally to higher dimensions. Consider an image f [x, y] and a two-dimensional filter h[x, y]. The 2-D convolution sum is then given by:
∞ ∞
In 1-D, the intuition for a convolution is that of computing inner products between the filter and signal as the filter "slides" across the signal. The same intuition holds in 2-D. Inner products are computed between the 2-D filter and underlying image as the filter slides from left-to-right/top-to-bottom. In the Fourier domain, this convolution is equivalent to multiplying the, now 2-D, Fourier transforms of the filter and image, where the 2-D Fourier transform is given by:
∞ ∞
ωx
Figure 7.1 2-D Frequency
F [ωx , ωy ] =
k=−∞ l=−∞
f [k, l]e−i(ωx k+ωy l) .
(7.2)
The notion of low-pass, band-pass, and high-pass filtering extends naturally to two-dimensional images. Shown in Figure 7.1 is a simplified decomposition of the 2-D Fourier domain parameterized by ωx and ωy ∈ [−π, π]. The inner disc corresponds to the lowest frequencies, the center annulus to the middle (band) frequencies, and the outer dark area to the highest frequencies. Two of the most common (and opposing) linear filtering operations are blurring and sharpening. Both of these operations can be accomplished with a 2-D filter and 2-D convolution, or more efficiently with a 1-D filter and a pair of 1-D horizontal and vertical convolutions. For example, a 2-D convolution with the blur filter: 0.0625 0.1250 0.0625
0.1250 0.2500 0.1250 47
0.0625 0.1250 0.0625
Figure 7.2 Low-, Band-, High-pass
can be realized by convolving in the horizontal and vertical directions with the 1-D filter: blur = ( 0.25 0.50 0.25 ) . (7.3)
That is, an outer-product of the 1-D filter with itself yields the 2-D filter - the filters are xy-separable. The separability of 2-D filters is attractive for two reasons: (1) it is computationally more efficient and (2) it simplifies the filter design. A generic blur filter may be constructed from any row of the binomial coefficients: 1 1 1 3 2 3 1 1
1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 where each row (filter) should be normalized by it's sum (i.e., blur filters should always be unit-sum so as not to increase or decrease the mean image intensity). The amount of blur is then directly proportional to the size of the filter. Blurring simply reduces the high-frequency content in an image. The opposing operation, sharpening, is meant to enhance the high-frequencies. A generic separable sharpening filter is of the form: sharp = ( 0.08 −1.00 0.08 ) . (7.4)
This filter leaves the low-frequencies intact while enhancing the contribution of the high-frequencies. Shown in Figure 7.3 are results from blurring and sharpening. 7.2 Derivative Filters
Figure 7.3 Blur Sharpen and
Discrete differentiation forms the foundation for many applications in image processing and computer vision. We are all familiar with the definition of the derivative of a continuous signal f (x): D{f (x)} = lim f (x + ε) − f (x) . ε (7.5)
ε→0
This definition requires that the signal f (x) be well defined for all x ∈ R. So, does it make sense to differentiate a discretely sampled signal, D{f [x]}, which is only defined over an integer sampling lattice? Strictly speaking, no. But our intuition may be that this is not such an unreasonable request. After all, we know how to differentiate f (x), from which the sampled signal f [x] was derived, so why not just differentiate the continuous signal f (x) and then sample the result? Surely this is what we have in mind when we 48
ask for the derivative of a sampled signal. But one should not be fooled by the seeming simplicity of our intuition, as we will soon discover the design of an accurate and efficient discrete derivative operator will prove to be sufficiently challenging. Recall from earlier chapters that under certain conditions (Nyquist theory), the relationship between the continuous and sampled signals can be expressed precisely as: f (x) = f [x] ⋆ h(x), (7.6)
On the left-hand side of the above equation is the desired quantity, the derivative of the sampled signal. On the right-hand side is a discrete convolution between two known quantities, the sampled derivative of the sync and the original sampled signal. The derivative of the sync can be expressed analytically by simply differentiating the sync function: h′ (x) = π 2 x/T 2 cos(πx/T ) − π/T sin(πx/T ) , (πx/T )2 (7.10)
0
0
where T is the sampling period at which f (x) was sampled. So, if the signal f (x) is sampled above the Nyquist rate and if it is in 49
Figure 7.4 Ideal and its derivative
sync
fact differentiable, then Equation (7.9) tells us that we can exactly compute the derivative of the sampled signal f [x], an altogether happy ending. If you are feeling a bit uneasy it is for a good reason. Although mathematically correct, we have a solution for differentiating a discretely sampled signal that is physically unrealizable. In particular the derivative of the sync, h′ (x), is spatially infinite in extent, meaning that it cannot be implemented on a finite machine. And even worse, h′ (x) falls off slowly from the origin so that truncation will cause significant inaccuracies. So we are going to have to part with mathematical perfection and design a finite-length filter. To begin we need to compute the frequency response of the ideal derivative filter. We can compute the response indirectly by first expressing f [x] in terms of its Fourier series: f [x] = 1 2π
π
Differentiation in the space domain is then seen to be equivalent to multiplying the Fourier transform F [ω] by an imaginary ramp iω. And since multiplication in the frequency domain is equivalent to convolution in the space domain, an imaginary ramp is the frequency response of the ideal derivative filter. Trying to directly design a finite length filter to this response is futile because of the discontinuity at −π/π, which of course accounts for the spatially infinite extent of h′ (x). So we are resigned to designing a filter with a periodic frequency response that "best" approximates a ramp. The simplest such approximation is that of a sinusoid where, at least in the low-frequency range, the match is reasonably good (i.e., sin(ω) = ω, for small ω). Employing the least-squares filter design technique (Equation (4.8)) we formulate a quadratic error function to be minimized: E(h) = | M h − H |2 , (7.13)
Since the desired frequency response, a sinusoid, has only two degrees of freedom, amplitude and phase, a 2-tap filter will suffice (i.e., n = 2). The resulting filter is of the form h = ( 0.5 −0.5 ). Intuitively this is exactly what we should have expected - for example, applying this filter via a convolution and evaluating at, arbitrarily, n = 0 yields: f ′ [x] = h[x] ⋆ f [x]
∞
=
k=−∞
h[x − k]f [k] (7.15)
f ′ [0] = h[1]f [−1] + h[0]f [0] = 0.5f [0] − 0.5f [−1].
Note that the derivative is being approximated with a simple twopoint difference, that is, a discrete approximation to the continuous definition in Equation (7.5). We could of course greatly improve on this filter design. But since we are really interested in multi-dimensional differentiation, let's put aside further analysis of the one-dimensional case and move on to the two-dimensional case. It has been the tendency to blindly extend the one-dimensional design to higher-dimensions, but, as we will see shortly, in higherdimensions the story becomes slightly more complicated. In the context of higher-dimensional signals we first need to consider partial derivatives. For example the partial derivative of a two dimensional signal f (x, y) in it's first argument is defined as: fx (x, y) ≡ ∂f (x, y) ∂x f (x + ε, y) − f (x, y) = lim . ε→0 ε
(7.16)
According to the Nyquist theory, the continuous and discrete signals (if properly sampled) are related by the following equality: f (x, y) = f [x, y] ⋆ h(x, y), (7.17)
) sin(πy/T ) is the two-dimensional ideal sync. where h(x, y) = sin(πx/Txy/T 2 π2 As before we apply the continuous partial differential operator to both sides:
Notice that calculating the partial derivative requires a pair of one-dimensional convolutions: a derivative filter, h′ [x], in the dimension of differentiation, and an interpolation filter, h[y], in the other dimension (for multi-dimensional signals, all remaining dimensions would be convolved with the interpolation filter). Since two-dimensional differentiation reduces to a pair of onedimensional convolutions it is tempting to simply employ the same differentiation filter used in the one-dimensional case. But since a pair of filters are now required perhaps we should give this some additional thought. In some ways the choice of filters seems trivial: chose an interpolation function h(x), differentiate it to get the derivative function h′ (x), and sample these functions to get the final digital filters h[x] and h′ [x]. So how is this different from the one-dimensional case? In the one-dimensional case only the derivative filter is employed, whereas in the two-dimensional case we require the pair of filters. And by our formulation we know that the pair of filters should satisfy the relationship that one is the derivative of the other h′ (x) = D(h(x)). And in fact this constraint is automatically enforced by the very nature in which the continuous functions are chosen, but in the final step, these functions are sampled to produce discrete filters. This sampling step typically destroys the required derivative relationship, and, although a seemingly subtle point, has dramatic effects on the accuracy of the resulting derivative operator. For example √ consider the often √ used Sobel derivative filters with h[x] = ( 1 2 1 ) /(2 + 2) and h′ [x] = ( 1 0 −1 ) /3. Shown in Figure 7.7 are the magnitudes of the Fourier transform of the derivative filter (solid line) and the interpolation filter times iω (i.e., frequency domain differentiation). If the filters obeyed the required derivative relationship than these curves would be exactly matched, which they clearly 52
f[y]
Figure 7.6 Horizontal partial differentiation
−pi
0
pi
Figure 7.7 Sobel frequency response
are not. The mismatching of the filters results in gross inaccuracies in derivative measurements. Let's see then if we can design a better set of filters. We begin by writing down the desired relationship between the derivative and interpolation filters, most conveniently expressed in the frequency domain: H ′ (ω) = iωH(ω), (7.21)
where the columns of the matrix Fm×n contain the first n Fourier basis functions (i.e., a discrete-time Fourier transform), the matrix ′ Fm×n = iωFm×n , and Wm×m is a diagonal frequency weighting matrix. Note that the dimension n is determined by the filter size and the dimension m is the sampling rate of the continuous Fourier basis functions, which should be chosen to be sufficiently large to avoid sampling artifacts. This error function can be expressed more concisely as: E(u) = |M u|2 , (7.24)
The minimal unit vector u is then simply the minimal-eigenvalue eigenvector of the matrix M t M . After solving for u, the derivative and interpolation filters can be "unpacked" and normalized so that the interpolation filter is unit sum. Below are the resulting filter values for a 3-tap and 5-tap filter pair.
53
Shown in Figure 7.8 are the matching of these filters in the frequency domain. Notice that the 5-tap filters are nearly perfectly matched. h h′ h h′
−pi 0 pi
Higher-order derivative filters can be designed by replacing the initial constraint in Equation (7.21) with H ′ (ω) = (iω)k H(ω) for a kth order derivative. A peculiar aspect of this filter design is that nowhere did we explicitly try to model a specified frequency response. Rather, the design fell naturally from the relationship between the continuousand discrete-time signals and the application of the continuous derivative operator, and in this way is quite distinct from the onedimensional case. The proper choice of derivative filters can have a dramatic impact on the applications which utilize them. For example, a common application of differential measurements is in measuring motion from a movie sequence f (x, y, t). The standard formulation for motion estimation is: fx (x, y, t)vx (x, y) + fy (x, y, t)vy (x, y) + ft (x, y, t) = 0, (7.26)
−pi
0
pi
Figure 7.8 Frequency response of matched derivative filters
where the motion vector is v = ( vx vy )t , and fx (·), fy (·), and ft (·) are the partial derivatives with respect to space and time. Shown in Figure 7.9 are the resulting motion fields for a simple translational motion with the Sobel (top panel) and matched (bottom) derivative filters used to compute the various derivatives. Although these filters are the same size, the difference in accuracy is significant.
Figure 7.9 Differential motion estimation
7.3 Steerable Filters In the previous section we showed how to compute horizontal and vertical partial derivatives of images. One may naturally wonder how to compute a derivative in an arbitrary direction. Quite remarkably it turns out that we need not design a new set of filters for each possible direction because the derivative in any direction can be synthesized from a linear combination of the horizontal and vertical derivatives. This property of derivatives has been termed steerability. There are several formulations of this property, we chose to work in the frequency domain where differentiation takes on a particularly simple form. 54
To begin, we express a two-dimensional image with respect to its Fourier series:
π π
Now, differentiation in the vertical direction v is equivalent to multiplying the Fourier transform by an imaginary vertical ramp. This trend generalizes to arbitrary directions, that is, the partial derivative in any direction α can be computed by multiplying the Fourier transform by an imaginary oriented ramp −jωα :
π π
fα (x, y) =
ωx =−π ωy =−π
−jωα F (ωx , ωy )e−j(ωx x+ωy y) ,(7.30)
where the oriented ramp can be expressed in terms of the horizontal and vertical ramps: ωα = cos(α)ωx + sin(α)ωy . (7.31)
Substituting this definition back into the partial derivative in α, Equation (7.30), gives:
π π
fα (x, y) =
ωx =−π ωy =−π π
−j[cos(α)ωx + sin(α)ωy ]F (ωx , ωy )e−j(ωx x+ωy y)
π
= cos(α)
ωx =−π ωy =−π π
−jωx F (ωx , ωy )e−j(ωx x+ωy y)
π
+ sin(α)
ωx =−π ωy =−π
−jωy F (ωx , ωy )e−j(ωx x+ωy y) (7.32)
= cos(α)fx (x, y) + sin(α)fy (x, y).
Recall that the derivative of an exponential is an exponential, so that according to the chain rule, Dx {eax } = aeax .
8
55
Notice that we obtain the horizontal and vertical derivatives when α = 0 and α = 90. This equation embodies the principle of steerability - the derivative in any direction α can be synthesized from a linear combination of the partial horizontal and vertical derivatives, fx (x, y) and fy (x, y). Pause to appreciate how remarkable this is, a pair of directional derivatives is sufficient to represent an infinite number of other directional derivatives, i.e., α can take on any real-valued number. From the previous section we know how to compute the horizontal and vertical derivatives via convolutions with an interpolation and derivative filter. To compute any other directional derivative no more convolutions are required, simply take the appropriate linear combinations of the horizontal and vertical derivatives as specified in Equation (7.32). Shown in Figure 7.10 from top to bottom is a disc f (x, y), its horizontal derivative fx (x, y), its vertical derivative fy (x, y), and its steered derivative f45 (x, y), where the steered derivative was synthesized from the appropriate linear combinations of the horizontal and vertical derivatives. The obvious benefit of steerability is that the derivative in any direction can be synthesized with minimal computational costs. Steerability is not limited to first-order derivatives. Higher-order derivatives are also steerable; the N th -order derivative is steerable with a basis set of size N + 1. For example, the second-order derivative in an arbitrary direction can be synthesized as follows: fαα = cos2 (α)fxx + 2 cos(α) sin(α)fxy + sin2 (α)fyy , (7.33)
where for notational simplicity the spatial arguments (x, y) have been dropped, and the multiple subscripts denote higher-order differentiation. Note that three partial derivatives are now needed to steer the second-order derivative. Similarly, the third-order derivative can be steered with a basis of size four: fααα = cos3 (α)fxxx + 3 cos2 (α) sin(α)fxxy +3 cos(α) sin2 (α)fxyy + sin3 (α)fyyy . (7.34) You may have noticed that the coefficients needed to steer the basis set look familiar, they are the binomial coefficients that come from a polynomial expansion. More specifically, as in Equation(7.30) the N th -order derivative in the frequency domain is computed by multiplying the Fourier transform by an imaginary oriented ramp raised to the N th power, (−jωα )N . Expressing this oriented ramp in terms of the horizontal and vertical ramps provides the basis and coefficients needed to steer derivatives of arbitrary order: (ωα )N = (cos(α)ωx + sin(α)ωy )N . 56 (7.35)
Figure 7.10 Steerability
Although presented in the context of derivatives, the principle of steerability is not limited to derivatives. In the most general case, a two-dimensional filter f (x, y) is steerable in orientation if it can be expressed as a polar-separable function, g(r)h(θ), where h(θ) is band-limited. More specifically, for an arbitrary radial component g(r), and for h(θ) expressed as:
N
h(θ) =
n=1
an cos(nθ) + bn sin(nθ)
(7.36)
then the filter is steerable with a basis size of 2N . 7.4 Edge Detection Discrete differentiation forms the foundation for many applications in computer vision. One such example is edge detection - a topic that has received an excessive amount of attention, but is only briefly touched upon here. An edge is loosely defined as an extended region in the image that undergoes a rapid directional change in intensity. Differential techniques are the obvious choice for measuring such changes. A basic edge detector begins by computing first-order spatial derivatives of an image f [x, y]: fx [x, y] = (f [x, y] ⋆ h′ [x]) ⋆ h[y] fy [x, y] = (f [x, y] ⋆ h[x]) ⋆ h [y],
′
(7.37) (7.38)
where h′ [·] and h[·] are the derivative and prefilter defined in Section 7.2. The "strength" of an edge at each spatial location is defined to be the magnitude of the gradient vector ▽[x, y] = ( fx [x, y] fy [x, y] ), defined as:
Figure 7.11 Edges
| ▽ [x, y]| =
2 fx [x, y]
+
2 fy [x, y].
(7.39)
As shown in Figure 7.11, the gradient magnitude is only the beginning of a more involved process (not discussed here) of extracting and localizing the salient and relevant edges. 7.5 Wiener Filter For any of a number of reasons a digital signal may become corrupted with noise. The introduction of noise into a signal is often modeled as an additive process, s = s+n. The goal of de-noising is ˆ to recover the original signal s from the corrupted signal s. Given ˆ a single constraint in two unknowns this problem is equivalent to my asking you "37 is the sum of two numbers, what are they?" Lacking clairvoyant powers or knowledge of how the individual numbers were selected we have little hope of a solution. But by 57
n
s
+
s
Figure 7.12 Additive noise
making assumptions regarding the signal and noise characteristics and limiting ourselves to a linear approach, a solution can be formulated known as the Wiener filter, of famed Mathematician Norbert Wiener (1894-1964). Having restricting ourselves to a linear solution, our goal is to design a filter h[x] such that: s[x] = h[x] ⋆ s[x] ˆ = h[x] ⋆ (s[x] + n[x]),
0 −pi
1
(7.40)
0
pi
that is, when the filter is convolved with the corrupted signal the original signal is recovered. With this as our goal, we reformulate this constraint in the frequency domain and construct a quadratic error functional to be minimized: E(H(ω)) = dω [H(ω)(S(ω) + N (ω)) − S(ω)]2 . (7.41)
At an intuitive level this frequency response makes sense - when the signal is strong and the noise is weak the response is close to 1 (i.e., frequencies are passed), and when the signal is weak and the noise is strong the response is close to 0 (i.e., frequencies are stopped). So we now have an optimal (in the least-squares sense) 58
frequency response in terms of the signal and noise characteristics, but of course we don't typically know what those are. But we can instantiate them by making assumptions about the general statistical nature of the signal and noise, for example a common choice is to assume white noise, N (ω) is constant for all ω, and, for natural images, to assume that S(ω) = 1/ω p . The frequency response in the top panel of Figure 7.13 was constructed under these assumptions. Shown in the bottom panel is a 7-tap filter derived from a least-squares design. This one-dimensional formulation can easily be extended to two or more dimensions. Shown in Figure 7.14 from top to bottom, is Einstein, Einstein plus noise, and the results of applying a 7 × 7 Wiener filter. Note that the noise levels are reduced but that much of the sharp image structure has also been lost, which is an unfortunate but expected side effect given that the Wiener filter is low-pass in nature.
59
8. Non-Linear Filtering
8-1 Median Filter 8-2 Dithering
8.1 Median Filter Noise may be introduced into an image in a number of different ways. In the previous section we talked about how to remove noise that has been introduced in an additive fashion. Here we look at a different noise model, one where a small number of pixels are corrupted due to, for example, a faulty transmission line. The corrupted pixels randomly take on a value of white or black, hence the name salt and pepper used to describe such noise patterns (Figure 8.1). Shown in the middle panel of Figure 8.1 is the disastrous result of applying the solution from the additive noise model (Wiener filter) to the salt and pepper noise image in the top panel. Trying to average out the noise in this fashion is equivalent to asking for the average salary of a group of eight graduate students and Bill Gates. As the income of Gates will skew the average salary so does each noise pixel when its value is so disparate from its neighbors. In such cases, the mean is best replaced with the median, computed by sorting the set of numbers and reporting on the value that lies midway. Shown in the bottom panel of Figure 8.1 is the much improved result of applying a 3 × 3 median filter to the salt and pepper noise image. More specifically, the center pixel of each 3 × 3 neighborhood is replaced with the median of the nine pixel values in that neighborhood. Depending on the density of the noise the median filter may need to be computed over a larger neighborhood. The tradeoff being that a larger neighborhood leads to a loss of detail, however this loss of detail is quite distinct from that of averaging. For example, shown in Figure 8.2 is the result of applying a 15×15 median filter. Notice that although many of the internal details have been lost the boundary contours (edges) have been retained, this is often referred to as posterization. This effect could never be achieved with an averaging filter which would indiscriminately smooth over all image structures. Because of the non-linear sorting step of a median filter it cannot be implemented via a simple convolution and is thus often more costly to implement. Outside the scope of this presentation there are a number of tricks for reducing the computational demands of a median filter.
Salt & Pepper Noise
Wiener
Median
Figure 8.1 Median filter
Figure 8.2 15 × 15 median filter
60
8.2 Dithering Dithering is a process by which a digital image with a finite number of gray levels is made to appear as a continuous-tone image. For example, shown at the top of Figure 8.3 is an 8-bit (i.e., 256 gray values) grayscale image of Richard P. Feynman. Shown below are two 1 bit (i.e., 2 gray values) images. The first was produced by thresholding, and the second by dithering. Although in both images each pixel takes on only one of two gray values (black or white), it is clear that the final image quality is critically dependent on the way in which pixels are quantized. There are numerous dithering algorithms (and even more variants within each algorithm). Sadly there are few quantitative metrics for measuring the performance of these algorithms. A standard and reasonably effective algorithm is a stochastic error diffusion algorithm based on the Floyd/Steinberg algorithm. The basic Floyd/Steinberg error diffusion dithering algorithm tries to exploit local image structure to reduce the effects of quantization. For simplicity, a 1-bit version of this algorithm is described here, the algorithm extends naturally to an arbitrary number of gray levels. This algorithm operates by scanning through the image, left to right and top to bottom. At each pixel, the gray value is first thresholded into "black" or "white", the difference between the new pixel value and the original value is then computed and distributed in a weighted fashion to its neighbors. Typically, the error is distributed to four neighbors with the following weighting:
1 16
×
• 7 3 5 1
,
where the • represents the thresholded pixel, and the position of the weights represent spatial position on a rectangular sampling lattice. Since this algorithm makes only a single pass through the image, the neighbors receiving a portion of the error must consist only of those pixels not already visited (i.e., the algorithm is casual). Note also that since the weights have unit sum, the error is neither amplified nor reduced. As an example, consider the quantization of an 8-bit image to a 1-bit image. An 8-bit image has 255 gray values, so all pixels less than 128 (mid-level gray) are thresholded to 0 (black), and all values greater than 128 are thresholded to 255 (white). A pixel at position (x, y) with intensity value 120 is thresholded to 0. The error, 120-0 = 120, is distributed to four neighbors as follows: (7/16)120 is added to the pixel at position (x + 1, y), (3/16)120 is added to the pixel at 61
Figure 8.3 Thresholding and Dithering
position (x − 1, y + 1), (5/16)120 to pixel (x, y + 1) and (1/16)120 to pixel (x + 1, y + 1). The intuition behind this algorithm is that when the pixel value of 120 is thresholded to 0 that pixel is made darker. By propagating this error the surrounding pixels are brightened making it more likely that they will be thresholded to something brighter. As a result the local neighborhood maintains it's average gray value. Qualitatively, the error diffusion algorithm reduces the effects of quantization via simple thresholding. However this algorithm does introduce correlated artifacts due to the deterministic nature of the algorithm and scanning order. These problems may be partially alleviated by introducing a stochastic process in a variety of places. Two possibilities include randomizing the error before distributing it to its neighbors (e.g., randomly scaling the error in the range 0.9 to 1.1), and alternating the scanning direction (e.g., odd lines are scanned left to right, and even lines from right to left).
62
9. Multi-Scale Transforms
63
10. Motion Estimation
10-1 Differential Motion 10-2 Differential Stereo
10.1 Differential Motion Our visual world is inherently dynamic. People, cars, dogs, etc. are (usually) moving. These may be gross motions, walking across the room, or smaller motions, scratching behind your ear. Our task is to estimate such image motions from two or more images taken at different instances in time. With respect to notation, an image is denoted as f (x, y) and an image sequence is denoted as f (x(t), y(t), t), where x(t) and y(t) are the spatial parameters and t is the temporal parameter. For example, a sequence of N images taken in rapid succession may be represented as f (x(t), y(t), t + i∆t) with i ∈ [0, N − 1], and ∆t representing the amount of time between image capture (typically on the order of 1/30th of a second). Given such an image sequence, our task is to estimate the amount of motion at each point in the image. For a given instant in space and time, we require an estimate of motion (velocity) v = ( vx vy ), where vx and vy denote the horizontal and vertical components of the velocity vector v. Shown in Figure 10.1 are a pair of images taken at two moments in time as a textured square is translating uniformly across the image. Also shown is the corresponding estimate of motion often referred to as a flow field. The flow field consists of a velocity vector at each point in the image (shown of course are only a subset of these vectors). In order to estimate motion, an assumption of brightness constancy is made. That is, it is assumed that as a small surface patch is moving, its brightness value remains unchanged. This constraint can be expressed with the following partial differential equation: ∂f (x(t), y(t), t) ∂t = 0. (10.1)
where the partials of the spatial parameters x and y with respect to time correspond to the velocity components: fx vx + fy vy + ft = 0. 64 (10.3)
The subscripts on the function f denote partial derivatives. Note again that this constraint holds for each point in space and time but that for notational simplicity the spatial/temporal parameters are dropped. This transformed brightness constancy constraint is rewritten by packing together the partial derivatives and velocity components into row and column vectors. ( fx fy ) vx vy + ft = 0 (10.4)
t fs v + ft = 0.
The space/time derivatives fs and ft are measured quantities, leaving us with a single constraint in two unknowns (the two components of the velocity vector, v). The constraint can be solved by assuming that the motion is locally similar, and integrating this constraint over a local image neighborhood. A least-squares error function takes the form:
2
If the matrix M is invertible (full rank), then the velocity can be estimated by simply left multiplying by the inverse matrix: v = −M −1 b (10.8)
The critical question then is, when is the matrix M invertible? Generally speaking the matrix is rank deficient, and hence not invertible, when the intensity variation in a local image neighborhood varies only one-dimensionally (e.g., fx = 0 or fy = 0) or zero-dimensionally (fx = 0 and fy = 0). These singularities are sometimes referred to as the aperture and blank wall problem. The motion at such points simply can not be estimated. 65
Motion estimation then reduces to computing, for each point in space and time, the spatial/temporal derivatives fx , fy , and ft . Of course the temporal derivative requires a minimum of two images, and is typically estimated from between two and seven images. The spatial/temporal derivatives are computed as follows. Given a temporal sequence of N images, the spatial derivatives are computed by first creating a temporally prefiltered image. The spatial derivative in the horizontal direction fx is estimated by prefiltering this image in the vertical y direction and differentiating in x. Similarly, the spatial derivative in the vertical direction fy is estimated by prefiltering in the horizontal x direction and differentiating in y. Finally, the temporal derivative is estimated by temporally differentiating the original N images, and prefiltering the result in both the x and y directions. The choice of filters depends on the image sequence length: an N tap pre/derivative filter pair is used for an image sequence of length N (See Section 7). 10.2 Differential Stereo Motion estimation involves determining, from a single stationary camera, how much an object moves over time (its velocity). Stereo estimation involves determining the displacement disparity of a stationary object as it is imaged onto a pair of spatially offset cameras. As illustrated in Figure 10.2, these problems are virtually identical: velocity (v) ≡ disparity (∆). Motion and stereo estimation are often considered as separate problems. Motion is thought of in a continuous (differential) framework, while stereo, with its discrete pair of images, is thought of in terms of a discrete matching problem. This dichotomy is unnecessary: stereo estimation can be cast within a differential framework. Stereo estimation typically involves a pair of cameras spatially offset in the horizontal direction such that their optical axis remain parallel (Figure 10.2). Denoting an image as f (x, y), the image that is formed by translating the camera in a purely horizontal direction is given by f (x + ∆(x, y), y). If a point in the world (X, Y, Z) is imaged to the image position (x, y), then the shift ∆(x, y) is inversely proportional to the distance Z (i.e., nearby objects have large disparities, relative to distant objects). Given this, a stereo pair of images is denoted as: fL (x + δ(x, y), y) and fR (x − δ(x, y), y), (10.9)
V
D
Figure 10.2 Motion and Stereo
where the disparity ∆ = 2δ. Our task is to determine, for each point in the image, the disparity (δ) between the left and right images. That is, to find the shift that brings the stereo pair back 66
Stereo estimation then reduces to computing, for each point in the image, spatial derivatives and the difference between the left and right stereo pair (a crude derivative with respect to viewpoint). Why, if motion and stereo estimation are similar, do the mathematical formulations look so different? Upon closer inspection they are in fact quite similar. The above formulation amounts to a constrained version of motion estimation. In particular, because of the strictly horizontal shift of the camera pair, the disparity was constrained along the horizontal direction. If we reconsider the motion estimation formulation assuming motion only along the horizontal direction, then the similarity of the formulations becomes evident. Recall that in motion estimation the brightness constancy assumption led to the following constraint: ∂f ∂x ∂f ∂y ∂f + + ∂x ∂t ∂y ∂t ∂t = 0, (10.14)
where the partial derivative of the spatial parameter x with respect to time correspond to the motion (speed) in the horizontal direction: fx vx + ft = 0. (10.16)
Unlike before, this leads to a single constraint with a single unknown which can be solved for directly: vx = − ft . fx (10.17)
This solution now looks very similar to the solution for differential stereo in Equation 10.13. In both solutions the numerator is a derivative, in one case with respect to time (motion) and in the other with respect to viewpoint (stereo). Also in both solutions, the denominator is a spatial derivative. In the stereo case, the denominator consists of the spatial derivative of the sum of the left and right image pair. This may seem odd, but recall that differentiation of a multi-dimensional function requires differentiating along the desired dimension and prefiltering along all other dimensions (in this case the viewpoint dimension). In both the differential motion and stereo formulations there exists singularities when the denominator (spatial derivative) is zero. As with the earlier motion estimation this can be partially alleviated by integrating the disparities over a local image neighborhood. However, if the spatial derivative is zero over a large area, corresponding to a surface in the world with no texture, then disparities at these points simply can not be estimated.
where the model parameters are a1 , b1 and a2 , b2 , and the system is modeled with additive noise n1 (i) and n2 (i). If we are told the model parameters, then determining which data point was generated by which model would be a simple matter of choosing, for each data point i, the model k that minimizes the error between the data and the model prediction: rk (i) = |ak x(i) + bk − y(i))|, (11.2)
for k = 1, 2 in our current example. On the other hand, if we are told which data points were generated by which model, then estimating the model parameters reduces to solving, for each model k, an over-constrained set of linear equations: xk (1) xk (2) . . .
Figure 11.1 Data two models
from
xk (n) 1
1 1 . . .
ak bk
yk (1) yk (2) = . , . .
(11.3)
yk (n)
where the xk (i) and yk (i) all belong to model k. In either case, knowing one piece of information (the model assignment or parameters) makes determining the other relatively easy. But, lacking either piece of information makes this a considerably more difficult estimation problem. The EM algorithm is an iterative two step algorithm that estimates both the model assignment and parameters. The "E-step" of EM assumes that the model parameters are known (initially, the model parameters can be assigned random values) and calculates the likelihood of each data point belonging to each model. In so doing the model assignment is made in a "soft" probabilistic fashion. That is, each data point is not explicitly assigned a single model, instead each data point i is assigned a 69
where, σN is proportional to the amount of noise in the data, and for each data point i, k wk (i) = 1. The "M-step" of EM takes the likelihood of each data point belonging to each model, and re-estimates the model parameters using weighted least-squares. That is, the following weighted error function on the model parameters is minimized: Ek (ak , bk ) =
i
wk (i)[ak x(i) + bk − y(i)]2 .
(11.7)
The intuition here is that each data point contributes to the estimation of each model's parameters in proportion to the belief that it belongs to that particular model. This quadratic error function is minimized by computing the partial derivatives with respect to the model parameters, setting the result equal to zero and solving for the model parameters. Differentiating: ∂Ek (ak , bk ) ∂ak ∂Ek (ak , bk ) ∂bk =
i
2wk (i)x(i)[ak x(i) + bk − y(i)] 2wk (i)[ak x(i) + bk − y(i)], (11.8)
=
i
setting both equal to zero yields the following set of linear equations: ak
i
wk (i)x(i)2 + bk
i
wk (i)x(i) =
i
wk (i)x(i)y(i) (11.9) wk (i)y(i).
i
ak
i
wk (i)x(i) + bk
i
wk (i) = 70
(11.10)
Rewriting in matrix form:
2 i wk (i)x(i) i wk (i)x(i) i
wk (i)x(i) i wk (i)
ak bk
=
i wk (i)x(i)y(i) i w(i)y(i)
Axk = b xk = A−1 b, (11.11)
yields a weighted least squares solution for the model parameters. Note that this solution is identical to solving the set of linear equations in Equation (11.3) using weighted least-squares. The EM algorithm iteratively executes the "E" and "M" step, repeatedly estimating and refining the model assignments and parameters. Shown in Figure 11.2 are several iterations of EM applied to fitting data generated from two linear models. Initially, the model parameters are randomly assigned, and after six iterations, the algorithm converges to a solution. Beyond the current scope are proofs on the convergence and rate of convergence of EM. At the end of this chapter is a Matlab implementation of EM for fitting multiple linear models to two-dimensional data. 11.2 Principal Component Analysis 11.3 Independent Component Analysis | 677.169 | 1 |
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How does calculus apply to real life?
A:
Quick Answer
There are many applications in which calculus is used in real life, such as calculating minimum payments due on credit cards, determining the length of cable required to connect two substations and evaluating survey data. Just as geometry is the mathematical study of shapes, calculus is the mathematical study of change.
Keep Learning
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