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Discrete Math
Applied Finite Mathematics
Applied Finite Mathematics, Second Edition
Edmond C. Tomastik, University of Connecticut
Janice L. Epstein, Texas A&M University
Editor: Carolyn Crockett
Editorial Assistant: Rebecca Dashiell
Marketing Manager: Myriah Fitzgibbon
Technical Editor: Mary Kanable
Illustrator: Jennifer Tribble
Photographs: Janice Epstein
©1994, 2008 Brooks/Cole, Cengage Learning
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L Logic
L.1 Introduction to Logic . . . . . . . . . . . . . . . . . . . . . . .
L.2 Truth Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sets and Probability
1.1 Introduction to Sets . . . . . . .
1.2 The Number of Elements in a Set
1.3 Sample Spaces and Events . . .
1.4 Basics of Probability . . . . . .
1.5 Rules for Probability . . . . . .
1.6 Conditional Probability . . . . .
1.7 Bayes’ Theorem . . . . . . . . .
Review . . . . . . . . . . . . .
Counting and Probability
2.1 The Multiplication Principle and Permutations .
2.2 Combinations . . . . . . . . . . . . . . . . . .
2.3 Probability Applications of Counting Principles
2.4 Bernoulli Trials . . . . . . . . . . . . . . . . .
2.5 Binomial Theorem . . . . . . . . . . . . . . .
Review . . . . . . . . . . . . . . . . . . . . .
Probability Distributions and Statistics
3.1 Random Variables and Histograms . . . . . . . . . .
3.2 Measures of Central Tendency . . . . . . . . . . . .
3.3 Measures of Spread . . . . . . . . . . . . . . . . . .
3.4 The Normal Distribution . . . . . . . . . . . . . . .
3.5 Normal Approximation to the Binomial Distribution
3.6 The Poisson Distribution . . . . . . . . . . . . . . .
Review . . . . . . . . . . . . . . . . . . . . . . . .
Systems of Linear Equations and Models
4.1 Mathematical Models . . . . . . . . . . . . . . . . . . .
4.2 Systems of Linear Equations . . . . . . . . . . . . . . .
4.3 Gauss Elimination for Systems of Linear Equations . . .
4.4 Systems of Linear Equations With Non-Unique Solutions
4.5 Method of Least Squares . . . . . . . . . . . . . . . . .
Review . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1 Introduction to Matrices . . .
5.2 Matrix Multiplication . . . . .
5.3 Inverse of a Square Matrix . .
5.4 Additional Matrix Applications
Review . . . . . . . . . . . .
M Markov Chains
M.1 Markov Processes . . . . . .
M.2 Regular Markov Processes .
M.3 Absorbing Markov Processes
Review . . . . . . . . . . .
G Game Theory
G.1 Decision Making . . . . . . . . . . . .
G.2 Mixed Strategy Games . . . . . . . . .
G.3 Linear Programming and Game Theory
Review . . . . . . . . . . . . . . . . .
F Finance
F.1 Simple Interest and Discount . . . . . . . . .
F.2 Compound Interest . . . . . . . . . . . . . .
F.3 Annuities and Sinking Funds . . . . . . . . .
F.4 Present Value of Annuities and Amortization
Review . . . . . . . . . . . . . . . . . . . .
Area Under a Normal Curve
Answers to Selected Exercises
Applied Finite Mathematics is designed for a finite mathematics course aimed
at students majoring in business, management, economics, or the life or social
sciences. The text can be understood by the average student with one year of
high school algebra. A wide range of topics is included, giving the instructor
considerable flexibility in designing a course. Optional technology material is
available where relevant.
Applications truly play a central and prominent role in the text. This is because the text is written for users of mathematics. Thus, for example, a concrete
applied problem is presented first as a motivation before developing a needed
mathematical topic. After the mathematical topic has been developed, further
applications are given so that the student understands the practical need for the
mathematics. This is done so consistently and thoroughly that after going completing some chapters, the student should come to believe that mathematics is
everywhere. Indeed, countless applications are drawn from actual referenced examples extracted from journals and other professional texts and papers.
No other skill is more important than the ability to translate a real-life problem into an appropriate mathematical format for finding the solution. Students
often refer to this process as “word problems.” Whereas linear systems of equations, linear programming problems, and financial problems, for example, can
easily be solved using modern technology, no calculator or computer, now or in
the foreseeable future, can translate these applied problems into the necessary
mathematical language. Thus students, in their jobs, will most likely use their
mathematical knowledge to translate applied problems into necessary mathematical models for solution by computers.
To develop these needed skills many word problems, requiring the writing
of one linear equation, are given in the introductory sections. This prepares the
student for the many word problems that require creating systems of linear equations. The word problems continue in subsequent chapters, for example, on linear
✧ Important Features
The text can be understood by the average student with a minimum of outside
assistance. Material on a variety of topics is presented in an interesting, informal, and student-friendly manner without compromising the mathematical content and accuracy. Concepts are developed gradually, always introduced intuitively, and culminate in a definition or result. Where possible, general concepts
are presented only after particular cases have been presented.
Historical Boxes Scattered throughout the text, and set-off in boxes, are historical and anecdotal comments. The historical comments are not only interesting
in themselves, but also indicate that mathematics is a continually developing subject.
Connections The Connection boxes relate the material to contemporary problems. This makes the material more relevant and interesting.
Applications The text includes many meaningful applications drawn from a
variety of fields. For example, every section opens by posing an interesting and
relevant applied problem using familiar vocabulary, which is then solved later in
the section after the appropriate mathematics has been developed. Applications
are given for all the mathematics that are presented and are used to motivate the
Worked Examples About 300 worked examples, including about 100 selfhelp exercises mentioned below, have been carefully selected to take the reader
progressively from the simplest idea to the most complex. All the steps needed
for the complete solutions are included.
Self-Help Exercises Immediately preceding each exercise set is a set of SelfHelp Exercises. These approximately 100 exercises have been very carefully
selected to bridge the gap between the exposition in the chapter and the regular
exercise set. By doing these exercises and checking the complete solutions provided, students will be able to test or check their comprehension of the material.
This, in turn, will better prepare them to do the exercises in the regular exercise
Exercises The text contains over 2000 exercises. Each set begins with drilling
problems to build skills, and then gradually increases in difficulty. The exercise
sets also include an extensive array of realistic applications from diverse disciplines. Technology exercises are included.
End of Chapter Projects Most chapters contain an in-depth exportation of
an important concept taught in the chapter. This provides strong connections to
real applications or a treatment of the material at a greater depth than in the main
part of the chapter.
Flexibility and Technology The text does not require any technology. However, important material on how to use technology is included. This material is
tucked out of the way of a reader not interested in using technology, being placed
at the end of a section as technology notes and also within green boxes in the
✧ Technology
For those finite math classes that are taught with a graphing calculator or a spreadsheet, this text has abundant resources for the student and the instructor. The
most accessible resource is the green margin boxes with the Technology Option.
These are designed for students who are familiar with a graphing calculator and
wish to see how the current example is worked using the calculator. For those students who need step-by-step directions, the Technology Corner provides details
on using a graphing calculator or a spreadsheet to carry out the mathematical operations discussed in the section. While the text focuses on the use of a TI-83/84
and Microsoft Excel, other technology help is available upon request.
✧ Student Aids
• Boldface cyan text is used when new terms are defined.
• Boxes are used to highlight definitions, theorems, results, and procedures.
• Remarks are used to draw attention to important points that might otherwise be overlooked.
• Titles for worked examples help to identify the subject.
• Chapter summary outlines at the end of each chapter conveniently summarize all the definitions, theorems, and procedures in one place.
• Review exercises are found at the end of each chapter.
• Answers to odd-numbered exercises and to all the review exercises are
provided at the end of each chapter.
• A student’s solution manual that contains completely worked solutions to
all odd-numbered exercises and to all chapter review exercises is available.
✧ Instructor Aids
• An instructor’s manual with completely worked solutions to all the exercises is available free to adopters.
• WebAssign A selection of questions from every section of the text will be
available for online homework on the WebAssign system. These homework questions are algorithmically generated and computer graded.
✧ Content Overview
Chapter L. This chapter covers the basic topics in logic which provides a good
preparation for the use of “and” and “or” in the probability applications of later
Chapter 1. The first two sections give an introduction to sets and counting the
number of elements in a set. The third section then sets the background for
probability by considering sample spaces and events. The next two sections then
introduce the basics of probability and their rules. The next two sections cover
conditional probability and Bayes’ theorem.
Chapter 2. This chapter involves counting and probability. The first four sections cover the multiplication principle, permutations, combinations, probability
applications of counting principles, and Bernoulli trials. The last (optional) section considers the binomial theorem.
Chapter 3. The first section revisits probability distributions and introduces histograms. The next two sections look at the measure of central tendency and the
measure of the spread of data. The next sections consider the normal distribution, the approximation of the binomial distribution by a normal distribution, and
finally the Poisson distribution.
Chapter 4. An introduction to the theory of the firm with some necessary economics background is provided to take into account the students’ diverse backgrounds. The next three sections cover linear systems of equations. The last
(optional) section on least squares provides other examples and applications of
the use of linear equations.
Chapter 5. The first three sections cover the basic material on matrices. Although many applications are included in the first three sections, the fourth (optional) section is entirely devoted to input–output analysis, which is an application of linear systems and matrices used in economics.
Chapter M. The basic material on Markov processes, covering both regular and
absorbing Markov processes, is presented in this chapter.
Chapter G. Game theory and its important connection to linear programming
is presented in this chapter. This material gives the basics on the extensive interrelationship between linear programming and the celebrated theory of games
developed by von Neumann and important in economic theory.
Chapter F. This chapter covers finance. The first two sections cover simple and
compound interest. The next two sections cover annuities, sinking funds, present
value, and amortization. This chapter on finance does not depend on any of the
other material and can be covered at any point in the course.
✧ Some Additional Comments on the Contents
In Chapter 4 when solving systems of linear equations with an infinite number
of solutions we will have free variables as parameters. We make it clear that in a
list of variables, such as x, y, z, and u, the last variable need not be the free one.
Rather, any of the variables can be a free variable. This requires us to develop a
solution plan that can address this issue.
Also when solving a system of linear equations with an infinite number of
solutions in an applied application, the parameter may require some constraints.
For example, the parameter may need to be an integer, or an even integer, or
have a bound above and below. Furthermore, it is possible in an applied problem
that there is no acceptable solution, even though there are an infinite number of
solutions of the abstract mathematical system.
Suppose there are three equations and three unknowns, say x, y, and z, in a
system of linear equations. When using the augmented matrix to solve the system, the normal procedure is to first reduce this matrix to a matrix with ones
down the diagonal and zeros below the diagonal. Students invariably notice that
we now have found the z value, so why not substitute this into the previous equation, solve for y, and then use these two values to substitute into the first equation
in order to find x. This is formally called backward substitution. Since this is
such a natural way of solving the system, we follow backward substitution in this
text. In fact, software used to solve systems follow just this plan. (See “Matrix
Computations” by Gene H. Golub and Charles F. van Loan.) It does not require
any more calculations than some other methods that are sometimes taught.
We also indicate in an optional subsection that following the solving plan for
systems of linear equations given in this text is actually more efficient and in
general requires fewer calculations than any other solving plan found in some
other texts.
At the University of Connecticut we are thankful for the support offered by
Michael Neumann, Jeff Tollefson, Gerald Leibowitz, and David Gross. At Texas
A&M University we are thankful for the support of G. Donald Allen and the feedback of Kathryn Bollinger, Kendra Kilmer, and Heather Ramsey were invaluable.
We wish to express our sincere appreciation to each of the following reviewers
for their many helpful suggestions.
Marti Mclard, University of Tennessee, Knoxville
John Herron, University of Montevallo
Fritz Keineut, Iowa State University
On a personal level, we both are grateful to our families for their patience and
Edmond C. Tomastik, University of Connecticut
Janice L. Epstein, Texas A&M University
April, 2008
Applied Finite Mathematics
Circuit Boards
How should the circuits on this board be laid
out so that the video card works? Logic is
used in the design of circuit boards.
L.1 Introduction to Logic
Introduction to Logic
✧ Statements
A Brief History of Logic
The Greek philosopher Aristotle
(384–322 B.C.) is generally given
the credit for the first systemic
study of logic. His work, however,
used ordinary language. The
second great period of logic came
with Gottfried Leibnitz
(1646–1716), who initiated the use
of symbols to simplify complicated
logical arguments. This treatment
is referred to as symbolic logic or
mathematical logic. In symbolic
logic, symbols and prescribed rules
are used very much as in ordinary
algebra. This frees the subject
from the ambiguities of ordinary
language and permits the subject
to proceed and develop in a
methodical way. It was, however,
Augustus De Morgan (1806–1871)
and George Boole (1815–1864)
who systemically developed
symbolic logic. The “algebra” of
logic that they developed removed
logic from philosophy and attached
it to mathematics.
Logic is the science of correct reasoning and of making valid conclusions. In
logic conclusions must be inescapable. Every concept must be clearly defined.
Thus, dictionary definitions are often not sufficient since there can be no ambiguities or vagueness.
We restrict our study to declarative sentences that are unambiguous and that
can be classified as true or false but not both. Such declarative sentences are
called statements and form the basis of logic.
A statement is a declarative sentence that is either true or false but not
Thus, commands, questions, exclamations, and ambiguous sentences cannot
be statements.
EXAMPLE 1
Determining if Sentences Are Statements
Decide which of
the following sentences are statements and which are not.
a. Look at me.
b. Do you enjoy music?
c. What a beautiful sunset!
d. Two plus two equals four.
e. Two plus two equals five.
f. The author got out of bed after 6:00 A.M. today.
g. That was a great game.
h. x + 2 = 5.
Solution The first three sentences are not statements since the first is a command, the second is a question, and the third is an exclamation. Sentences d and
e are statements; d is a true statement while e is a false statement. Sentence f is a
statement, but you do not know if it is true or not. Sentence g is not a statement
since we are not told what “great” means. With a definition of great, such as “Our
team won,” then it would be a statement. The last sentence h. is not a statement
since it cannot be classified as true or false. For example, if x = 3 it is true. But
if x = 2 it is false.
✧ Connectives
A statement such as “I have money in my pocket” is called a simple statement
since it expresses a single thought. But we need to also deal with compound
statements such as “I have money in my pocket and my gas tank is full.” We
will let letters such as p, q, and r denote simple statements. To write compound
statements, we need to introduce symbols for the connectives.
Chapter L Logic
A connective is a word or words, such as “and” or “if and only if,” that
is used to combine two or more simple statements into a compound statement.
We will consider the 3 connectives given in the following table.
Logic does not concern itself with whether a simple statement is true or false.
But if all the simple statements that make up a compound statement are known to
be true or false, then the rules of logic will enable us to determine if the compound
statement is true or false. We will do this in the next section.
We now carefully give the definitions of the three connectives “and,” “or,”
and “not.” Notice that the precise meanings of the three compound statements
that involve these connectives are incomplete unless a clear statement is made as
to when the compound statement is true and when it is false. The first connective
we discuss is conjunction which is the concept of “and.”
A conjunction is a statement of the form “p and q” and is written symbolically as
The conjunction p ∧ q is true if both p and q are true, otherwise it is false.
Write the compound statement “I have
money in my pocket and my gas tank is full” in symbolic form.
EXAMPLE 2
Using Conjunction
First let p be the statement “I have money in my pocket” and q be the
statement “my gas tank is full.” Since ∧ represents the word “and,” the compound
statement can be written symbolically as p ∧ q.
The next connective we consider is disjunction which is the concept of “or.”
Make careful note of the fact that the logical “or” is slightly different in meaning
than the typical English use of the word “or.”
A disjunction is a statement of the form “p or q” and is written symbolically as
The disjunction p ∨ q is false if both p and q are false and is true in all other
L.1 Introduction to Logic
REMARK: The word “or” in this definition conveys the meaning “one or the
other, or both.” This is also called the inclusive or.
EXAMPLE 3 Using Disjunction Write the compound statement “Janet is in
the top 10% of her class or she lives on campus” in symbolic form.
Solution First let p be the statement “Janet is in the top 10% of her class” and
q the statement “She lives on campus.” Since ∨ represents the word “or,” the
compound statement can be written as p ∨ q.
In everyday language the word “or” is not always used in the way
indicated above. For example, if a car salesman tells you that for $20,000 you can
have a new car with automatic transmission or a new car with air conditioning,
he means “one or the other, but not both.” This use of the word “or” is called
exclusive or.
The final connective introduced in this section is negation which is the concept of “not.”
A negation is a statement of the form “not p” and is written symbolically
The negation ∼ p is true if p is false and false if p is true.
For example, if p is the statement “Janet is smart,” then ∼ p is the statement
“Janet is not smart.”
EXAMPLE 4
Using Negation Let p and q be the following statements:
p: George Bush plays football for the Washington Redskins.
q: The Dow Jones industrial average set a new record high last week.
Write the following statements in symbolic form.
a. George Bush does not play football for the Washington Redskins, and the
Dow Jones industrial average set a new record high last week.
b. George Bush plays football for the Washington Redskins, or the Dow Jones
industrial average did not set a new record high last week.
c. George Bush does not play football for the Washington Redskins, and the
Dow Jones industrial average did not set a new record high last week.
d. It is not true that George Bush plays football for the Washington Redskins
and that the Dow Jones industrial average set a new record high last week.
a. (∼ p) ∧ q
EXAMPLE 5
b. p∨ ∼ q
c. ∼ p∧ ∼ q
d. ∼ (p ∧ q)
Translating Symbolic Forms Into Compound Statements Let
p and q be the following statements:
Chapter L Logic
p: Philadelphia is the capital of New Jersey.
q: General Electric lost money last year.
Write out the statements that correspond to each of the following:
a. p ∨ q
b. p ∧ q
c. p∨ ∼ q
d. ∼ p∧ ∼ q
a. Philadelphia is the capital of New Jersey, or General Electric lost money last
b. Philadelphia is the capital of New Jersey, and General Electric lost money last
c. Philadelphia is the capital of New Jersey, or General Electric did not lose
money last year.
d. Philadelphia is not the capital of New Jersey, and General Electric did not
lose money last year.
In most cases when dealing with complex compound statements, there will
not be a question as to the order in which to apply the connectives. However,
you may have noticed in the above examples that negation was used before disjunction or conjunction. The order of precedence for logical connectives is stated
Order of Precedence
The logical connectives are used in the following order
∼, ∧, ∨
Self-Help Exercises L.1
1. Determine which of the following sentences are
e. I have a three-dollar bill in my purse, or I don’t
have a purse.
a. The Atlanta Braves won the World Series in
2. Let p be the statement “George Washington was
never president of the United States” and q be the
statement “George Washington wore a wig.” Write
out the statements that correspond to the following:
b. IBM makes oil tankers for Denmark.
c. Does IBM make oil tankers for Denmark?
d. Please pay attention.
a. ∼ p
d. p∧ ∼ q
b. p ∨ q
e. ∼ p∨ ∼ q
L.1 Exercises
In Exercises 1 through 14, decide which are statements.
1. Water freezes at 70o F.
c. ∼ p ∧ q
L.1 Introduction to Logic
2. It rained in St. Louis on May 4, 1992.
3. 5 > 10.
4. This sentence is false.
5. The number 4 is not a prime.
6. How are you feeling?
7. I feel great!
8. 10 + 10 − 5 = 25
9. There is life on Mars.
10. Cleveland is the largest city in Ohio.
11. Who said Cleveland is the largest city in Ohio?
12. You don’t say!
13. IBM lost money in 1947.
14. Groundhog Day is on February 12.
15. Let p and q denote the following statements:
p: George Washington was the third president of the
United States.
q: Austin is the capital of Texas.
Express the following compound statements in
a. ∼ p
b. p ∧ q
c. p ∨ q
d. ∼ p ∧ q
e. p∨ ∼ q
f. ∼ (p ∧ q)
16. Let p and q denote the following statements:
p: Mount McKinley is the highest point in the
United States.
q: George Washington was a signer of the Declaration of Independence.
Express the following compound statements in
a. ∼ q
b. p ∧ q
c. p ∨ q
d. p∧ ∼ q
e. ∼ p∧ ∼ q
f. ∼ (p ∨ q)
17. Let p and q denote the following statements:
p: George Washington owned over 100,000 acres of
q: The Exxon Valdez was a luxury liner.
a. State the negation of these statements in words.
b. State the disjunction for these statements in
c. State the conjunction for these statements in
18. Let p and q denote the following statements:
p: McDonald’s Corporation operates large farms.
q: Wendy’s Corporation operates fast-food restaurants.
a. State the negation of these statements in words.
b. State the disjunction for these statements in
c. State the conjunction for these statements in
19. Let p and q denote the following statements:
p: The Wall Street Journal has the highest daily circulation of any newspaper.
q: Advise and Consent was written by Irving Stone.
Give a symbolic expression for the statements below.
a. Advise and Consent was not written by Irving
b. The Wall Street Journal has the highest daily circulation of any newspaper, and Advise and Consent was not written by Irving Stone.
c. The Wall Street Journal has the highest daily circulation of any newspaper, or Advise and Consent was written by Irving Stone.
d. The Wall Street Journal does not have the highest daily circulation of any newspaper, or Advise
and Consent was not written by Irving Stone.
20. Let p and q denote the following statements:
p: IBM makes computers.
q: IBM makes trucks.
Give a symbolic expression for the statements below.
a. IBM does not make trucks.
b. IBM makes computers, or IBM makes trucks.
c. IBM makes computers, or IBM does not make
d. IBM does not make computers, and IBM does
not make trucks.
Chapter L Logic
Solutions to Self-Help Exercises L.1
1. The sentences a, b, and e are statements, while c and d are not.
2. a. George Washington was a president of the United States.
b. George Washington was never president of the United States, or George
Washington wore a wig.
c. George Washington was a president of the United States, and George Washington wore a wig.
d. George Washington was never president of the United States, and George
Washington did not wear a wig.
e. George Washington was a president of the United States, or George Washington did not wear a wig.
Truth Tables
✧ Introduction to Truth Tables
Table L.1
Table L.2
Table L.3
Table L.4
p∨ ∼ q
The truth value of a statement is either true or false. Thus the statement “Ronald
H. Coarse won the Nobel Prize in Economics in 1991” has truth value true since
it is a true statement, whereas the statement “Los Angeles is the capital of California” has truth value false since it is a false statement.
Logic does not concern itself with the truth value of simple statements. But
if we know the truth values of the simple statements that make up a compound
statement, then logic can determine the truth value of the compound statement.
For example, to understand the very definition of p ∨ q, one must know under
what conditions the compound statement will be true. As defined in the last
section p ∨ q is always true unless both p and q are false. A convenient way of
summarizing this is by a truth table. This is done in Table L.1.
The truth tables for the statements p ∧ q and ∼ p are given in Table L.2 and
Table L.3. As Table L.2 indicates, p∧q is true only if both p and q are true. Given
a general compound statement, we wish to determine the truth value given any
possible combination of truth values for the simple statements that are contained
in the compound statement. We use a truth table for this purpose. The next
examples illustrate how this is done.
EXAMPLE 1
Constructing a Truth Table
Construct a truth table for the
statement p∨ ∼ q.
Place p and q at the head of the first two columns as indicated in
Table L.4 ∨ found in Table L.1.
✦ and list all possible truth values for
p and q as indicated. It is strongly recommended that you always list the truth
values in the first two columns in the same way. This will be particularly useful
later when we will need to compare two truth tables. Now enter the truth values
L.2 Truth Tables
for ∼ q in the third column. Now using the first and third columns of the table,
construct the fourth column using the definition of
EXAMPLE 2
Constructing a Truth Table
Construct a truth table for the
statement ∼ p ∧ (p ∨ q).
Make the same first two columns as before. Next make a column for
∼ p and the corresponding truth values. Now make a fourth column for p ∨ q.
Finally, using the third and fourth columns and the definition of ∧, fill in the fifth
column of Table L.5.
∼ p ∧ (p ∨ q)
Table L.5
Thus we see that ∼ p ∧ (p ∨ q) is true only if p is false and q is true.
We can construct a truth table for a compound statement with three simple
EXAMPLE 3
Constructing a Truth Table
Construct a truth table for the
statement (p ∧ q) ∧ [(r∨ ∼ p) ∧ q].
Always use the same order of T’s and F’s that are indicated in the
first three columns of Table L.6. Fill in the rest of the columns in the order given.
r∨ ∼ p
(r∨ ∼ p) ∧ q
(p ∧ q) ∧ [(r∨ ∼ p) ∧ q]
Table L.6
We see that (p ∧ q) ∧ [(r∨ ∼ p) ∧ q] is true only if p, q, and r are all true.
✧ Exclusive Disjunction
Table L.7
We now consider the exclusive “or.” Recall that the exclusive “or” means “one
or the other, but not both.” The truth table for the exclusive disjunction is given
in Table L.7 where we note that the symbol for the exclusive disjunction is ∨.
Notice that ∨ is true only if exactly one of the two statements is true.
Unless clearly specified otherwise, the word “or” will always be
taken in the inclusive sense.
EXAMPLE 4
Determining the Truth Value of a Statement
the following statements:
Let p and q be
Chapter L Logic
p: Aaron Copland was an American composer.
q: Rudolf Serkin was a violinist.
Determine the truth value of each of the following statements:
a. p ∨ q
b. ∼ (p ∨ q)
c. p∨q
d. ∼ (p∨q)
e. p∧ ∼ q
First note that p is true and q is false.1 Both the disjunction in a and
the exclusive disjunction in c are therefore true. Thus their negations in b and d
are false. The statement in e is the conjunction of a true statement p with a true
statement ∼ q and thus is true.
✧ Tautology and Contradiction
p∧ ∼ p
The statement p∧ ∼ p is always false according to the truth table in Table L.8. In
such a case, we say that the statement p∧ ∼ p is a contradiction. If the statement
is always true, we say that the statement is a tautology.
Table L.8
Contradiction and Tautology
We say that a statement is a contradiction if the truth value of the statement is always false no matter what the truth values of its simple component statements. We say that a statement is a tautology if the truth value
of the statement is always true no matter what the truth values of its simple
component statements.
EXAMPLE 5
Determining if a Statement Is a Tautology
Determine if the
statement p ∧ (∼ p ∧ q) is a tautology.
Create a truth table.
∼ p∨q
p ∧ (∼ p ∨ q)
The truth table indicates that the statement is true no matter what the truth
values of p and q are. Thus, this statement is a tautology.
Self-Help Exercises L.2
1. Construct the truth table for the statement
statement “George Washington wore a wig.” Determine the truth value of each of the statements below.
(p ∨ q) ∨ (r∧ ∼ q)
2. Let p be the statement “George Washington was the
first president of the United States” and q be the
1 Serkin
was a famous pianist.
a. ∼ p
d. p∧ ∼ q
b. p ∨ q
e. ∼ p∨ ∼ q
c. ∼ p ∧ q
L.2 Truth Tables
L.2 Exercises
In Exercises 1 through 20, construct a truth table for the
given statement. Indicate if a statement is a tautology or
a contradiction.
1. p∧ ∼ q
2. p∨ ∼ q
3. ∼ (∼ p)
4. ∼ (p ∧ q)
5. (p∧ ∼ q) ∨ q
6. (p∨q)∨ ∼ q
7. ∼ p ∨ (p ∧ q)
8. ∼ p∨(p ∧ q)
9. (p ∨ q) ∧ (p ∧ q)
10. (p ∧ q) ∨ (p ∨ q)
11. (p∨ ∼ q) ∨ (∼ p ∧ q)
12. (p∧ ∼ q) ∧ (p∨ ∼ q)
13. (p ∨ q) ∧ r
14. p ∨ (q ∧ r)
15. ∼ [(p ∧ q) ∧ r]
16. ∼ [p ∧ (q ∧ r)]
17. (p ∨ q) ∨ (q ∧ r)
18. (p ∧ q) ∧ (q ∨ r)
19. (p∨ ∼ q) ∨ (∼ q ∧ r)
20. (∼ p ∧ q) ∧ (∼ q∨r)
22. Let p and q be the statements:
p: The sun rises in the east.
q: Proctor & Gamble is a casino in Las Vegas.
Determine the truth value of the following compound statements:
a. ∼ q
d. p∨ ∼ q
b. p ∧ q
e. ∼ p ∨ q
c. p∨q
23. Let p and q be the statements:
p: The South Pole is the southernmost point on the
q: The North Pole is a monument in Washington,
Determine the truth value of the following compound statements:
a. ∼ q
d. ∼ (p ∧ q)
b. p∨ ∼ q
c. ∼ p ∧ q
24. Let p and q be the statements:
21. Let p and q be the statements:
p: Stevie Wonder is a famous singer.
q: Simon & Garfunkel is a famous law firm.
p: Roe v. Wade was a famous boxing match.
q: Iraq invaded Kuwait in 1990.
Determine the truth value of the following compound statements:
Determine the truth value of the following compound statements:
a. ∼ p
b. p ∧ q
c. p ∨ q
d. ∼ p ∧ q
e. p∨ ∼ q
Note that p is false and q is true.
a. p∧ ∼ q
b. p∨q
c. p ∨ q
d. ∼ (p ∨ q)
Note that p is true and q is false.
Solutions to Self-Help Exercises L.2
r∧ ∼ q
(p ∨ q) ∨ (r∧ ∼ q)
2. The statement p is true, while q is false. Thus
a. ∼ p is false.
b. p ∨ q is true.
d. p∧ ∼ q is true is ∼ q is true.
e. ∼ p∨ ∼ q is true.
c. ∼ p ∧ q is false.
Sets and Probability
In a survey of 200 people that had just returned
from a trip to Europe, the following information was gathered.
• 142 visited England
• 95 visited Italy
• 65 visited Germany
• 70 visited both England and Italy
• 50 visited both England and Germany
• 30 visited both Italy and Germany
• 20 visited all three of these countries
How many went to England but not Italy or
We will learn how to solve puzzles like this in
the second section of the chapter when counting the elements in a set is discussed.
1.1 Introduction to Sets
Introduction to Sets
This section discusses operations on sets and the laws governing these set operations. These are fundamental notions that will be used throughout the remainder
of this text. In the next two chapters we will see that probability and statistics are
based on counting the elements in sets and manipulating set operations. Thus we
first need to understand clearly the notion of sets and their operations.
✧ The Language of Sets
George Boole, 1815–1864
George Boole was born into a
lower-class family in Lincoln,
England, and had only a common
school education. He was largely
self-taught and managed to
become an elementary school
teacher. Up to this time any rule of
algebra such as a(x + y) = ax + ay
was understood to apply only to
numbers and magnitudes. Boole
developed an “algebra” of sets
where the elements of the sets
could be not just numbers but
anything. This then laid down the
foundations for a fundamental way
of thinking. Bertrand Russell, a
great mathematician and
philosopher of the 20th century,
said that the greatest discovery of
the 19th century was the nature of
pure mathematics, which he
asserted was discovered by
George Boole. Boole’s pamphlet
“The Mathematical Analysis of
Logic” maintained that the
essential character of mathematics
lies in its form rather than in its
content. Thus mathematics is not
merely the science of
measurement and number but any
study consisting of symbols and
precise rules of operation. Boole
founded not only a new algebra of
sets but also a formal logic that we
will discuss in Chapter L.
We begin here with some definitions of the language and notation used when
working with sets. The most basic definition is “What is a set?” A set is a
collection of items. These items are referred to as the elements or members of
the set. For example, the set containing the numbers 1, 2, and 3 would be written
{1, 2, 3}. Notice that the set is contained in curly brackets. This will help us
distinguish sets from other mathematical objects.
When all the elements of the set are written out, we refer to this as roster
notation. So the set containing the first 10 letters in the English alphabet would
be written as {a, b, c, d, e, f , g, h, i, j} in roster notation. If we wanted to refer to
this set without writing all the elements, we could define the set in terms of its
properties. This is called set-builder notation. So we write
{x|x is one of the first 10 letters in the English alphabet}
This is read “the set of all x such that x is one of the first 10 letters in the English
alphabet”. If we will be using a set more than once in a discussion, it is useful to
define the set with a symbol, usually an uppercase letter. So
S = {a, b, c, d, e, f , g, h, i, j}
We can say c is an element of the set {a, b, c, d, e, f , g, h, i, j} or simply write
c ∈ S. The symbol ∈ is read “is an element of”. We can also say that the set
R = {c} is a subset of our larger set S as every element in the set R is also in the
set S.
If every element of a set A is also an element of another set B, we say that
A is a subset of B and write A ⊆ B. If A is not a subset of B, we write
A ⊆ B.
Thus {1, 2, 4} ⊆ {1, 2, 3, 4}, but {1, 2, 3, 4} ⊆ {1, 2, 4}. Since every element in A
is in A, we can write A ⊆ A. If there is a set B and every element in the set B is
also in the set A but B = A, we say that B is a proper subset of A. This is written
as B ⊂ A. Note the proper subset symbol ⊂ is lacking the small horizontal line
that the subset symbol ⊆ has. The difference is rather like the difference between
< and ≤.
Some sets have no elements at all. We need some notation for this, simply
leaving a blank space will not do!
Chapter 1 Sets and Probability
Empty Set
The empty set, written as 0/ or {}, is the set with no elements.
The empty set can be used to conveniently indicate that an equation has no
solution. For example
{x|x is real and x2 = −1} = 0/
By the definition of subset, given any set A, we must have 0/ ⊆ A.
EXAMPLE 1
Finding Subsets Find all the subsets of {a, b, c}.
The subsets are
/ {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
REMARK: Note that there are 8 subsets and 7 of them are proper subsets. In
general, a set with n elements will have 2n subsets. In the next chapter we will
learn why this is so.
Figure 1.1
The empty set is the set with no elements. At the other extreme is the universal set. This set is the set of all elements being considered and is denoted by
U. If, for example, we are to take a national survey of voter satisfaction with the
president, the universal set is the set of all voters in this country. If the survey
is to determine the effects of smoking on pregnant women, the universal set is
the set of all pregnant women. The context of the problem under discussion will
determine the universal set for that problem. The universal set must contain every
element under discussion.
A Venn diagram is a way of visualizing sets. The universal set is represented
by a rectangle and sets are represented as circles inside the universal set. For
example, given a universal set U and a set A, Figure 1.1 is a Venn diagram that
visualizes the concept that A ⊂ U. Figure 1.1 also visualizes the concept B ⊂ A.
The U above the rectangle will be dropped in later diagrams as we will abide by
the convention that the rectangle always represents the universal set.
✧ Set Operations
The first set operation we consider is the complement. The complement of set A
are those members of set U that do not belong to A.
Given a universal set U and a set A ⊂ U, the complement of A, written Ac ,
is the set of all elements that are in U but not in A, that is,
Figure 1.2
Ac is shaded.
/ A}
Ac = {x|x ∈ U, x ∈
1.1 Introduction to Sets
A Venn diagram visualizing Ac is shown in Figure 1.2. Some alternate notations for the complement of a set are A and Ā.
EXAMPLE 2
The Complements of Sets Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 4, 5}. Find Ac , Bc , U c , 0/ c , and (Ac )c in roster
We have
0/ c
(Ac )c
= {2, 4, 6, 8}
= {6, 7, 8, 9}
= 0/
= {1, 2, 3, 4, 5, 6, 7, 8, 9} = U
= {2, 4, 6, 8}c
= {1, 3, 5, 7, 9} = A
Note that in the example above we found U c = 0/ and 0/ c = U. Additionally
(Ac )c = A. This can be seen using the Venn diagram in Figure 1.2, since the
complement of Ac is all elements in U but not in Ac which is the set A. These
three rules are called the Complement Rules.
Complement Rules
If U is a universal set, we must always have
U c = 0,
0/ c = U
If A is any subset of a universal set U, then
(Ac )c = A
The next set operation is the union of two sets. This set includes the members
of both sets A and B. That is, if an element belongs to set A or set B then it belongs
to the union of A and B.
Set Union
The union of two sets A and B, written A ∪ B, is the set of all elements that
belong to A, or to B, or to both. Thus
A ∪ B = {x|x ∈ A or x ∈ B or both}
REMARK: This usage of the word “or” is the same as in logic. It is the inclusive
“or” where the elements that belong to both sets are part of the union. In English
the use of “or” is often the exclusive “or”. That is, if a meal you order at a
restaurant comes with a dessert and you are offered cake or pie, you really only
get one of the desserts. Choosing one dessert will exclude you from the other. If
it was the logical “or” you could have both!
Chapter 1 Sets and Probability
Our convention will be to drop the phrase “or both” but still maintain the
same meaning. Note very carefully that this gives a particular definition to the
word “or”. Thus we will normally write
A ∪ B = {x|x ∈ A or x ∈ B}
Figure 1.3
A ∪ B is shaded.
It can be helpful to say that the union of A and B, A ∪ B, is all elements in A
joined together with all elements in B. A Venn diagram visualizing this is shown
in Figure 1.3 with the union shaded.
EXAMPLE 3 The Union of Two Sets Let U = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3, 4}
and B = {1, 4, 5, 6}. Find A ∪ B and A ∪ Ac .
We begin with the first set and join to it any elements in the second
set that are not already there. Thus
A ∪ B = {1, 2, 3, 4} ∪ {1, 4, 5, 6}
= {1, 2, 3, 4, 5, 6}
Since Ac = {5, 6} we have
A ∪ Ac = {1, 2, 3, 4} ∪ {5, 6} = {1, 2, 3, 4, 5, 6} = U
The second result, A ∪ Ac = U is generally true. From Figure 1.2, we can see
that if U is a universal set and A ⊂ U, then
A ∪ Ac = U
Set Intersection
The intersection of two sets A and B, written A ∩ B, is the set of all elements that belong to both the set A and to the set B. Thus
A ∩ B = {x|x ∈ A and x ∈ B}
A Venn diagram is shown in Figure 1.4 with the intersection shaded.
Figure 1.4
EXAMPLE 4
The Intersection of Two Sets Find
a. {a, b, c, d} ∩ {a, c, e}
b. {a, b} ∩ {c, d}
a. Only a and c are elements of both of the sets. Thus
{a, b, c, d} ∩ {a, c, e} = {a, c}
b. The two sets {a, b} and {c, d} have no elements in common. Thus
{a, b} ∩ {c, d} = 0/
Figure 1.5
A and B are disjoint.
The sets {a, b} and {c, d} have no elements in common. These sets are called
disjoint and can be visualized in Figure 1.5.
1.1 Introduction to Sets
Disjoint Sets
Two sets A and B are disjoint if they have no elements in common, that is,
if A ∩ B = 0.
An examination of Figure 1.2 or referring to the definition of Ac indicates that
for any set A, A and Ac are disjoint. That is,
A ∩ Ac = 0/
✧ Additional Laws for Sets
There are a number of laws for sets. They are referred to as commutative, associative, distributive, and De Morgan laws. We will consider two of these laws in
the following examples.
Augustus De Morgan,
It was De Morgan who got George
Boole interested in set theory and
formal logic and then made
significant advances upon Boole’s
epochal work. He discovered the
De Morgan laws referred to in the
last section. Boole and De Morgan
are together considered the
founders of the algebra of sets and
of mathematical logic. De Morgan
was a champion of religious and
intellectual toleration and on
several occasions resigned his
professorships in protest of the
abridgments of academic freedom
of others.
EXAMPLE 5
Establishing a De Morgan Law
Use a Venn diagram to show
(A ∪ B)c = Ac ∩ Bc
We first consider the right side of this equation. Figure 1.6 shows a
Venn diagram of Ac and Bc and Ac ∩ Bc . We then notice from Figure 1.3 that this
is (A ∪ B)c .
Figure 1.6
EXAMPLE 6
Establishing the Distributive Law for Union
Use a Venn dia-
gram to show that
A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪C)
Consider first the left side of this equation. In Figure 1.7a the sets
A, B ∩ C, and the union of these two are shown. Now for the right side of the
equation refer to Figure 1.7b, where the sets A ∪ B, A ∪C, and the intersection of
these two sets are shown. We have the same set in both cases.
Figure 1.7a
Chapter 1 Sets and Probability
Figure 1.7b
We can summarize the laws we have found in the following list.
Laws for Set Operations
A∪B = B∪A
A∩B = B∩A
A ∪ (B ∪C) = (A ∪ B) ∪C
A ∩ (B ∩C) = (A ∩ B) ∩C
A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪C)
A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C)
(A ∪ B)c = Ac ∩ Bc
(A ∩ B)c = Ac ∪ Bc
Commutative law for union
Commutative law for intersection
Associative law for union
Associative law for intersection
Distributive law for union
Distributive law for intersection
De Morgan law
De Morgan law
✧ Applications
Using Set Operations to Write Expressions Let U be the
universal set consisting of the set of all students taking classes at the University
of Hawaii and
B = {x|x is currently taking a business course}
E = {x|x is currently taking an English course}
M = {x|x is currently taking a math course}
Write an expression using set operations and show the region on a Venn diagram
for each of the following:
EXAMPLE 7
a. The set of students at the University of Hawaii taking a course in at least one
of the above three fields.
b. The set of all students at the University of Hawaii taking both an English
course and a math course but not a business course.
c. The set of all students at the University of Hawaii taking a course in exactly
one of the three fields above.
a. This is B ∪ E ∪ M. See Figure 1.8a.
b. This can be described as the set of students taking an English course (E) and
also (intersection) a math course (M) and also (intersection) not a business
course (Bc ) or
E ∩ M ∩ Bc
Figure 1.8a
1.1 Introduction to Sets
This is the set of points in the universal set that are in both E and M but not
in B and is shown in Figure 1.8b.
c. We describe this set as the set of students taking business but not taking English or math (B ∩ E c ∩ M c ) together with (union) the set of students taking
English but not business or math (E ∩ Bc ∩ M c ) together with (union) the set
of students taking math but not business or English (M ∩ Bc ∩ E c ) or
Figure 1.8b
(B ∩ E c ∩ M c ) ∪ (Bc ∩ E ∩ M c ) ∪ (Bc ∩ E c ∩ M)
This is the union of the three sets shown in Figure 1.8c. The first, B ∩ E c ∩ M c ,
consists of those points in B that are outside E and also outside M. The second
set E ∩ Bc ∩ M c consists of those points in E that are outside B and M. The
third set M ∩ Bc ∩ E c is the set of points in M that are outside B and E. The
union of these three sets is then shown on the right in Figure 1.8c.
The word only means the same as exactly one. So a student taking
only a business course would be written as B ∩ E c ∩ M c .
Figure 1.8c
Self-Help Exercises 1.1
1. Let U = {1, 2, 3, 4, 5, 6, 7}, A = {l, 2, 3, 4}, B =
{3, 4, 5}, C = {2, 3, 4, 5, 6}. Find the following:
a. A ∪ B
b. A ∩ B
c. A
d. (A ∪ B) ∩C
e. (A ∩ B) ∪C
f. Ac ∪ B ∪C
2. Let U denote the set of all corporations in this country and P those that made profits during the last year,
D those that paid a dividend during the last year,
and L those that increased their labor force during
the last year. Describe the following using the three
sets P, D, L, and set operations. Show the regions in
a Venn diagram.
a. Corporations in this country that had profits and
also paid a dividend last year
b. Corporations in this country that either had profits or paid a dividend last year
c. Corporations in this country that did not have
profits last year
d. Corporations in this country that had profits,
paid a dividend, and did not increase their labor
force last year
e. Corporations in this country that had profits or
paid a dividend, and did not increase their labor
force last year
1.1 Exercises
In Exercises 1 through 4, determine whether the statements are true or false.
1. a. 0/ ∈ A
b. A ∈ A
2. a. 0 = 0/
b. {x, y} ∈ {x, y, z}
3. a. {x|0 < x < −1} = 0/
b. {x|0 < x < −1} = 0
4. a. {x|x(x − 1) = 0} = {0, 1}
b. {x|x2 + 1 < 0} = 0/
5. If A = {u, v, y, z}, determine whether the following
statements are true or false.
a. w ∈ A
c. {u, x} ∪ A
b. x ∈
d. {y, z, v, u} = A
Chapter 1 Sets and Probability
6. If A = {u, v, y, z}, determine whether the following
statements are true or false.
a. x ∈
b. {u, w} ∈
c. {x, w} ⊂ A
d. 0/ ⊂ A
16. a. A ∩ B ∩Cc
b. B ∩ Ac ∩Cc
17. a. Ac ∩ Bc ∩Cc
b. A ∩C ∩ Bc
18. a. B ∩C ∩ Ac
b. C ∩ Ac ∩ Bc
8. List all the subsets of a. 0,
/ b. {3, 4, 5}.
19. a. (A ∪ B) ∩Cc
b. (A ∩ B)c ∩C
9. Use Venn diagrams to indicate the following.
a. A ⊂ U, B ⊂ U, A ⊂ Bc
b. A ⊂ U, B ⊂ U, B ⊂ Ac
20. a. A ∪ (B ∩C)
b. A ∪ B ∪Cc
21. a. (A ∪ B)c ∩C
b. (Ac ∩ B)c ∪C
22. a. A ∪ (Bc ∩Cc )
b. (A ∪ B ∪C)c ∩ A
7. List all the subsets of a. {3}, b. {3, 4}.
10. Use Venn diagrams to indicate the following.
a. A ⊂ U, B ⊂ U, C ⊂ U, C ⊂ (A ∪ B)c
b. A ⊂ U, B ⊂ U, C ⊂ U, C ⊂ A ∩ B
For Exercises 11 through 14, indicate where the sets are
located on the figure below and indicate if the sets found
in part a and part b are disjoint or not.
In Exercises 23 through 30, find the indicated sets with
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
A = {1, 2, 3, 4, 5, 6},
B = {4, 5, 6, 7, 8}, C = {5, 6, 7, 8, 9, 10}
23. a. A ∩ B
b. A ∪ B
24. a. Ac
b. Ac ∩ B
25. a. A ∩ Bc
b. Ac ∩ Bc
26. a. Ac ∪ Bc
b. (Ac ∪ Bc )c
27. a. A ∩ B ∩C
b. (A ∩ B ∩C)c
11. a. A ∩ Bc
b. A ∩ B
12. a. Ac ∩ B
b. Ac ∩ Bc
28. a. A ∩ (B ∪C)
b. A ∩ (Bc ∪C)
13. a. A ∪ Bc
b. (A ∪ B)c
29. a. Ac ∩ Bc ∩Cc
b. (A ∪ B ∪C)c
14. a. Ac ∪ Bc
b. (A ∩ B)c
30. a. Ac ∩ Bc ∩C
b. Ac ∩ B ∩Cc
For Exercises 15 through 22, indicate where the sets are
located on the figure below
In Exercises 31 through 34, describe each of the sets in
Let U be the set of all residents of your state and let
A = {x|x owns an automobile}
H = {x|x owns a house}
15. a. A ∩ B ∩C
b. A ∩ Bc ∩Cc
31. a. Ac
b. A ∪ H
c. A ∪ H c
32. a. H c
b. A ∩ H
c. Ac ∩ H
33. a. A ∩ H c
b. Ac ∩ H c
c. Ac ∪ H c
34. a. (A ∩ H)c
b. (A ∪ H)c
c. (Ac ∩ H c )c
1.1 Introduction to Sets
In Exercises 35 through 38, let U, A, and H be as in the
previous four problems, and let
P = {x|x owns a piano},
and describe each of the sets in words.
35. a. A ∩ H ∩ P
c. (A ∩ H) ∪ P
b. A ∪ H ∪ P
36. a. (A ∪ H) ∩ P
c. A ∩ H ∩ Pc
b. (A ∪ H) ∩ Pc
37. a. (A ∩ H)c ∩ P
c. (A ∪ H)c ∩ P
b. Ac ∩ H c ∩ Pc
38. a. (A ∪ H ∪ P)c ∩ A
c. (A ∩ H ∩ P)c
b. (A ∪ H ∪ P)c
b. Major league ball players who have never hit 20
homers in a season.
43. a. New York Yankees or San Francisco Giants who
have hit 20 homers in a season.
b. Outfielders for the New York Yankees who have
never hit 20 homers in a season.
44. a. Outfielders for the New York Yankees or San
Francisco Giants.
b. Outfielders for the New York Yankees who have
hit 20 homers in a season.
In Exercises 39 through 46, let U be the set of major
league baseball players and let
N = {x|x plays for the New York Yankees}
S = {x|x plays for the San Francisco Giants}
F = {x|x is an outfielder}
H = {x|x has hit 20 homers in one season}
Write the set that represents the following descriptions.
39. a. Outfielders for the New York Yankees
b. New York Yankees who have never hit 20 homers
in a season
40. a. San Francisco Giants who have hit 20 homers in
a season.
b. San Francisco Giants who do not play outfield.
41. a. Major league ball players who play for the New
York Yankees or the San Francisco Giants.
b. Major league ball players who play for neither
the New York Yankees nor the San Francisco Giants.
42. a. San Francisco Giants who have never hit 20
homers in a season.
45. a. Major league outfielders who have hit 20 homers
in a season and do not play for the New York Yankees or the San Francisco Giants.
b. Major league outfielders who have never hit 20
homers in a season and do not play for the New York
Yankees or the San Francisco Giants.
46. a. Major league players who do not play outfield,
who have hit 20 homers in a season, and do not play
for the New York Yankees or the San Francisco Giants.
b. Major league players who play outfield, who
have never hit 20 homers in a season, and do not
play for the New York Yankees or the San Francisco
In Exercises 47 through 52, let
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5},
B = {4, 5, 6, 7}, C = {5, 6, 7, 8, 9, 10}.
Verify that the identities are true for these sets.
47. A ∪ (B ∪C) = (A ∪ B) ∪C
48. A ∩ (B ∩C) = (A ∩ B) ∩C
49. A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪C)
50. A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C)
51. (A ∪ B)c = Ac ∩ Bc
52. (A ∩ B)c = Ac ∪ Bc
Solutions to Self-Help Exercises 1.1
1. a. A ∪ B is the elements in A or B. Thus A ∪ B = {1, 2, 3, 4, 5}.
b. A ∩ B is the elements in both A and B. Thus A ∩ B = {3, 4}.
c. Ac is the elements not in A (but in U). Thus Ac = {5, 6, 7}.
Chapter 1 Sets and Probability
d. (A ∪ B) ∩C is those elements in A ∪ B and also in C. From a we have
(A ∪ B) ∩C = {1, 2, 3, 4, 5} ∩ {2, 3, 4, 5, 6} = {2, 3, 4, 5}
e. (A ∩ B) ∪C is those elements in A ∩ B or in C. Thus from b
(A ∩ B) ∪C = {3, 4} ∪ {2, 3, 4, 5, 6} = {2, 3, 4, 5, 6}
f. Ac ∪ B ∪C is elements in B, or in C, or not in A. Thus
Ac ∪ B ∪C = {2, 3, 4, 5, 6, 7}
2. a. Corporations in this country that had profits and also paid a dividend last
year is represented by P ∩ D. This is regions I and II.
b. Corporations in this country that either had profits or paid a dividend last
year is represented by P ∪ D. This is regions I, II, III, IV, V, and VI.
c. Corporations in this country that did not have profits is represented by Pc .
This is regions III, VI, VII, and VIII.
d. Corporations in this country that had profits, paid a dividend, and did not
increase their labor force last year is represented by P ∩ D ∩ Lc . This is
region II.
e. Corporations in this country that had profits or paid a dividend, and did
not increase their labor force last year is represented by (P ∪ D) ∩ Lc . This
is regions II, V, and VI.
The Number of Elements in a Set
Breakfast Survey
In a survey of 120 adults, 55 said they had an egg for breakfast that morning, 40 said they had juice for breakfast, and 70 said they had an egg or
juice for breakfast. How many had an egg but no juice for breakfast? How
many had neither an egg nor juice for breakfast? See Example 1 for the
✧ Counting the Elements of a Set
This section shows the relationship between the number of elements in A ∪ B and
the number of elements in A, B, and A ∩ B. This is our first counting principle.
The examples and exercises in this section give some applications of this. In
other applications we will count the number of elements in various sets to find
The Notation n(A)
If A is a set with a finite number of elements, we denote the number of
elements in A by n(A).
1.2 The Number of Elements in a Set
Figure 1.9
In Figure 1.9 we see the number n(A) written inside the A circle and n(Ac )
written outside the set A. This indicates that there are n(A) members in set A and
n(Ac ) in set Ac . The number of elements in a set is also called the cardinality of
the set.
There are two results that are rather apparent. First, the empty set 0/ has no
elements n(0)
/ = 0. For the second refer to Figure 1.10 where the two sets A and
B are disjoint.
The Number in the Union of Disjoint Sets
If the sets A and B are disjoint, then
n(A ∪ B) = n(A) + n(B)
Figure 1.10
A consequence of the last result is the following. In Figure 1.9, we are given
a universal set U and a set A ⊂ U. Then since A ∩ Ac = 0/ and U = A ∪ Ac ,
n(U) = n(A ∪ Ac ) = n(A) + n(Ac )
✧ Union Rule for Two Sets
Now consider the more general case shown in Figure 1.11. We assume that x is
the number in the set A that are not in B, that is, n(A ∩ Bc ). Next we have z, the
number in the set B that are not in A, n(Ac ∩ B). Finally, y is the number in both
A and B, n(A ∩ B) and w is the number of elements that are neither in A nor in B,
n(Ac ∩ Bc ). Then
n(A ∪ B) = x + y + z
= (x + y) + (y + z) − y
= n(A) + n(B) − n(A ∩ B)
Figure 1.11
Alternatively, we can see that the total n(A) + n(B) counts the number in the
intersection n(A ∩ B) twice. Thus to obtain the number in the union n(A ∪ B), we
must subtract n(A ∩ B) from n(A) + n(B).
The Number in the Union of Two Sets
For any finite sets A and B,
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
EXAMPLE 1
An Application of Counting In a survey of 120 adults, 55 said
they had an egg for breakfast that morning, 40 said they had juice for breakfast,
and 70 said they had an egg or juice for breakfast. How many had an egg but no
juice for breakfast? How many had neither an egg nor juice for breakfast?
Chapter 1 Sets and Probability
Solution Let U be the universal set of adults surveyed, E the set that had an
egg for breakfast, and J the set that had juice for breakfast. A Venn diagram is
shown in Figure 1.12a. From the survey, we have that
n(E) = 55
Figure 1.12a
n(J) = 40,
n(E ∪ J) = 70
Note that each of these is a sum. That is n(E) = 55 = x + y, n(J) = 40 = y + z
and n(E ∪ J) = 70 = x + y + z. Since 120 people are in the universal set, n(U) =
120 = x + y + z + w.
The number that had an egg and juice for breakfast is given by n(E ∩ J) and
is shown as the shaded region in Figure 1.12b. We apply the union rule:
n(E ∩ J) = n(E) + n(J) − n(E ∪ J)
= 55 + 40 − 70
= 25
Figure 1.12b
We first place the number 25, just found, in the E ∩ J area in the Venn diagram in
Figure 1.12b. Since the number of people who had eggs (with and without juice)
is 55, then according to Figure 1.12b,
n(E) = 55 = x + 25
x = 30
Similarly, the number who had juice (with and without an egg) is 40. Using
Figure 1.12b,
n(J) = 40 = z + 25
z = 15
These two results are shown in Figure 1.12c. We wish to find w = n((E ∪ J)c ).
This is shown as the shaded region in Figure 1.12c. The unshaded region is E ∪ J.
We then have that
n(E ∪ J) + n((E ∪ J)c ) = n(U)
n((E ∪ J)c ) = n(U) − n(E ∪ J)
w = 120 − 70
w = 50
And so there were 50 people in the surveyed group that had neither an egg nor
juice for breakfast.
Figure 1.12c
✧ Counting With Three Sets
Many counting problems with sets have two sets in the universal set. We will
also study applications with three sets in the universal set. The union rule for
three sets is studied in the extensions for this section. In the example below,
deductive reasoning is used to solve for the number of elements in each region
of the Venn diagram. In cases where this will not solve the problem, systems of
linear equations can be used to solve the Venn diagram. This is studied in the
Chapter Project found in the Review section.
EXAMPLE 2 European Travels In a survey of 200 people that had just returned from a trip to Europe, the following information was gathered.
1.2 The Number of Elements in a Set
• 142 visited England
• 95 visited Italy
• 65 visited Germany
• 70 visited both England and Italy
• 50 visited both England and Germany
• 30 visited both Italy and Germany
• 20 visited all three of these countries
a. How many went to England but not Italy or Germany?
b. How many went to exactly one of these three countries?
c. How many went to none of these three countries?
Let U be the set of 200 people that were surveyed and let
E = {x|x visited England}
I = {x|x visited Italy}
G = {x|x visited Germany}
Figure 1.13a
We first note that the last piece of information from the survey indicates that
n(E ∩ I ∩ G) = 20
Place this in the Venn diagram shown in Figure 1.13a. Recall that 70 visited
both England and Italy, that is, n(E ∩ I) = 70. If a is the number that visited
England and Italy but not Germany, then, according to Figure 1.13a, 20 + a =
n(E ∩ I) = 70. Thus a = 50. In the same way, if b is the number that visited
England and Germany but not Italy, then 20 + b = n(E ∩ G) = 50. Thus b = 30.
Also if c is the number that visited Italy and Germany but not England, then
20 + c = n(G ∩ I) = 30. Thus c = 10. All of this information is then shown in
Figure 1.13b.
Figure 1.13b
a. Let x denote the number that visited England but not Italy or Germany. Then,
according to Figure 1.13b, 20 + 30 + 50 + x = n(E) = 142. Thus x = 42, that
is, the number that visited England but not Italy or Germany is 42.
b. Since n(I) = 95, the number that visited Italy but not England or Germany
is given from Figure 1.13b by 95 − (50 + 20 + 10) = 15. Since n(G) = 65,
the number that visited Germany but not England or Italy is, according to Figure 1.13b, given by 65 − (30 + 20 + 10) = 5. Thus, according to Figure 1.13c,
the number who visited just one of the three countries is
42 + 15 + 5 = 62
Figure 1.13c
c. There are 200 people in the U and so according to Figure 1.13c, the number
that visited none of these three countries is given by
200 − (42 + 15 + 5 + 50 + 30 + 10 + 20) = 200 − 172 = 28
Pizzas At the end of the day the manager of Blue Baker wanted
to know how many pizzas were sold. The only information he had is listed below.
Use the information to determine how many pizzas were sold.
EXAMPLE 3
Chapter 1 Sets and Probability
• 3 pizzas had mushrooms, pepperoni, and sausage
• 7 pizzas had pepperoni and sausage
• 6 pizzas had mushrooms and sausage but not pepperoni
• 15 pizzas had two or more of these toppings
• 11 pizzas had mushrooms
• 8 pizzas had only pepperoni
Figure 1.14a
• 24 pizzas had sausage or pepperoni
• 17 pizzas did not have sausage
Begin by drawing a Venn diagram with a circle for pizzas that had
mushrooms, a circle for pizzas that had pepperoni, and, pizzas that had sausage.
In the center place a 3 since three pizzas had all these toppings. See Figure 1.14a.
Since 7 pizzas have pepperoni and sausage, 7 = 3 + III or III = 4. If 6 pizzas
had mushrooms and sausage but not pepperoni, then IV = 6. The region for two
or more of these toppings is 3 + II + III + IV = 15. Using III = 4 and IV = 6, that
gives 3 + II + 4 + 6 = 15 or II = 2. This information is shown in Figure 1.14b.
Given that 11 pizzas had mushrooms, V + 2 + 3 + 6 = 11 and therefore V =
0. Since 8 pizzas had only pepperoni, VI = 8. With a total of 24 pizzas in the
sausage or pepperoni region and knowing that VI = 8, we have 2 + 8 + 6 + 3 + 4
+ VII = 24 or VII = 1. Finally, if 17 pizzas did not have sausage then 17 = V + 2
+ VI + VIII = 0 + 2 + 8 + VIII. This gives VIII = 7 and our complete diagram is
shown in Figure 1.14c.
To find the total number of pizzas sold, the 8 numbers in the completed Venn
diagram are added:
0 + 2 + 8 + 6 + 3 + 4 + 1 + 7 = 31
Figure 1.14b
Figure 1.14c
Self-Help Exercises 1.2
1. Given that n(A ∪ B) = 100, n(A ∩ Bc ) = 50, and
n(A ∩ B) = 20, find n(Ac ∩ B).
2. The registrar reported that among 2000 students,
700 did not register for a math or English course,
while 400 registered for both of these two courses.
How many registered for exactly one of these
3. One hundred shoppers are interviewed about the
contents of their bags and the following results are
40 bought apple juice
19 bought cookies
13 bought broccoli
1 bought broccoli, apple juice, and cookies
11 bought cookies and apple juice
2 bought cookies and broccoli but not apple
• 24 bought only apple juice
Organize this information in a Venn diagram and
find how many shoppers bought none of these items.
1.2 The Number of Elements in a Set
1.2 Exercises
1. If n(A) = 100, n(B) = 75, and n(A ∩ B) = 40, what
is n(A ∪ B)?
2. If n(A) = 200, n(B) = 100, and n(A ∪ B) = 250,
what is n(A ∩ B)?
3. If n(A) = 100, n(A ∩ B) = 20, and n(A ∪ B) = 150,
what is n(B)?
4. If n(B) = 100, n(A ∪ B) = 175, and n(A ∩ B) = 40,
what is n(A)?
5. If n(A) = 100 and n(A ∩ B) = 40, what is n(A ∩ Bc )?
6. If n(U) = 200 and n(A ∪ B) = 150, what is n(Ac ∩
Bc )?
7. If n(A∪B) = 500, n(A∩Bc ) = 200, n(Ac ∩B) = 150,
what is n(A ∩ B)?
8. If n(A ∩ B) = 50, n(A ∩ Bc ) = 200, n(Ac ∩ B) = 150,
what is n(A ∪ B)?
9. If n(A ∩ B) = 150 and n(A ∩ B ∩ C) = 40, what is
n(A ∩ B ∩Cc )?
10. If n(A ∩ C) = 100 and n(A ∩ B ∩ C) = 60, what is
n(A ∩ Bc ∩C)?
11. If n(A) = 200 and n(A ∩ B ∩ C) = 40, n(A ∩ B ∩
Cc ) = 20, n(A ∩ Bc ∩ C) = 50, what is n(A ∩ Bc ∩
Cc )?
12. If n(B) = 200 and n(A ∩ B ∩ C) = 40, n(A ∩ B ∩
Cc ) = 20, n(Ac ∩ B ∩ C) = 50, what is n(Ac ∩ B ∩
Cc )?
For Exercises 13 through 20, let A, B, and C be sets in a
universal set U. We are given n(U) = 100, n(A) = 40,
n(B) = 37, n(C) = 35, n(A ∩ B) = 25, n(A ∩ C) = 22,
n(B ∩C) = 24, and n(A ∩ B ∩Cc ) = 10. Find the following values.
13. n(A ∩ B ∩C)
14. n(Ac ∩ B ∩C)
15. n(A ∩ Bc ∩C)
16. n(A ∩ Bc ∩Cc )
17. n(Ac ∩ B ∩Cc )
18. n(Ac ∩ Bc ∩C)
19. n(A ∪ B ∪C)
20. n((A ∪ B ∪C))c
21. Headache Medicine In a survey of 1200 households, 950 said they had aspirin in the house, 350
said they had acetaminophen, and 200 said they had
both aspirin and acetaminophen.
a. How many in the survey had at least one of the
two medications?
b. How many in the survey had aspirin but not acetaminophen?
c. How many in the survey had neither aspirin nor
22. Newspaper Subscriptions In a survey of 1000
households, 600 said they received the morning paper but not the evening paper, 300 said they received
both papers, and 100 said they received neither paper.
a. How many received the evening paper but not
the morning paper?
b. How many received at least one of the papers?
23. Course Enrollments The registrar reported that
among 1300 students, 700 students did not register
for either a math or English course, 400 registered
for an English course, and 300 registered for both
types of courses.
a. How many registered for an English course but
not a math course?
b. How many registered for a math course?
24. Pet Ownership In a survey of 500 people, a pet food
manufacturer found that 200 owned a dog but not a
cat, 150 owned a cat but not a dog, and 100 owned
neither a dog or cat.
a. How many owned both a cat and a dog?
b. How many owned a dog?
25. Fast Food A survey by a fast-food chain of 1000
adults found that in the past month 500 had been to
Burger King, 700 to McDonald’s, 400 to Wendy’s,
300 to Burger King and McDonald’s, 250 to McDonald’s and Wendy’s, 220 to Burger King and
Wendy’s, and 100 to all three. How many went to
a. Wendy’s but not the other two?
b. only one of them?
c. none of these three?
Chapter 1 Sets and Probability
26. Investments A survey of 600 adults over age 50
found that 200 owned some stocks and real estate
but no bonds, 220 owned some real estate and bonds
but no stock, 60 owned real estate but no stocks or
bonds, and 130 owned both stocks and bonds. How
many owned none of the three?
27. Entertainment A survey of 500 adults found that
190 played golf, 200 skied, 95 played tennis, 100
played golf but did not ski or play tennis, 120 skied
but did not play golf or tennis, 30 played golf and
skied but did not play tennis, and 40 did all three.
a. How many played golf and tennis but did not
b. How many played tennis but did not play golf or
c. How many participated in at least one of the
three sports?
28. Transportation A survey of 600 adults found that
during the last year, 100 traveled by plane but not
by train, 150 traveled by train but not by plane, 120
traveled by bus but not by train or plane, 100 traveled by both bus and plane, 40 traveled by all three,
and 360 traveled by plane or train. How many did
not travel by any of these three modes of transportation?
29. Magazines In a survey of 250 business executives,
40 said they did not read Money, Fortune, or Business Week, while 120 said they read exactly one
of these three and 60 said they read exactly two of
them. How many read all three?
30. Sales A furniture store held a sale that attracted 100
people to the store. Of these, 57 did not buy anything, 9 bought both a sofa and love seat, 8 bought
both a sofa and chair, 7 bought both a love seat
and chair. There were 24 sofas, 18 love seats, and
20 chairs sold. How many people bought all three
31. Use a Venn diagram to show that
n(A ∪ B ∪C) = n(A) + n(B) + n(C)
−n(A ∩ B) − n(A ∩C)
−n(B ∩C) + n(A ∩ B ∩C)
32. Give a proof of the formula in Exercise 31. Hint:
Set B ∪ C = D and use union rule on n(A ∪ D). Now use the
union rule two more times, recalling from the last section that
A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C).
Solutions to Self-Help Exercises 1.2
1. The accompanying Venn diagram indicates that
n(A ∩ Bc ) = 50, n(A ∩ B) = 20, z = n(Ac ∩ B)
Then, according to the diagram,
50 + 20 + z = n(A ∪ B) = 100
Thus z = 30.
2. The number of students that registered for exactly one of the courses is the number that registered for math but not English, x = n(M ∩ E c ), plus the number that
registered for English but not math, z = n(M c ∩ E). Then, according to the accompanying Venn diagram, x + z + 400 + 700 = 2000. Thus x + z = 900. That is,
900 students registered for exactly one math or English course.
3. Let A be the set of shoppers who bought apple juice, B the set of shoppers who
bought broccoli, and C the set of shoppers who bought cookies. This is shown
in the first figure below. Since one shopper bought all three items, a 1 is placed
in region I. Twenty-four shoppers bought only apple juice and this is region V.
1.3 Sample Spaces and Events
Given 2 shoppers bought cookies and broccoli but not apple juice, a 2 is placed
in region III. This is shown in the next figure below.
The statement “11 bought cookies and apple juice” includes those who
bought broccoli and those who did not. We now know that one person bought
all 3 items, so 11 − 1 = 10 people bought cookies and apple juice but not broccoli. A 10 is placed in region IV.
Now I + II + IV + V = 40 as we are told “40 bought apple juice.” With the 10
in region IV we know 3 of the 4 values for set A and we can solve for region II:
40 = 24 + 10 + 1 + II gives II = 5. Place this in the Venn diagram as shown in the
third figure below.
Examining the figure, we can use the total of 13 in the broccoli circle to solve
for VI: 18 = 5 + 1 + 2 + VI gives VI = 10. The total of 19 in the cookies circle
lets us solve for VII: 10 + 1 + 2 + VII = 13 gives VII = 0. The very last piece of
information is that there were 100 shoppers. To solve for VIII we have 100 = 24
+ 5 + 10 + 10 + 1 + 2 + 0 + VIII or VIII = 48. That is, 48 shoppers bought none
of these items. The completed diagram is the final figure below.
Sample Spaces and Events
Many people have a good idea of the basics of probability. That is, if a fair coin
is flipped, you have an equal chance of a head or a tail showing. However, as we
proceed to study more advanced concepts in probability we need some formal
definitions that will both agree with our intuitive understanding of probability
and allow us to go deeper into topics such as conditional probability. This will
tie closely to work we have done learning about sets.
✧ The Language of Probability
We begin the preliminaries by stating some definitions. It is very important to
have a clear and precise language to discuss probability so pay close attention to
the exact meanings of the terms below.
Experiments and Outcomes
An experiment is an activity that has observable results.
An outcome is the result of the experiment.
Chapter 1 Sets and Probability
The following are some examples of experiments. Flip a coin and observe
whether it falls “heads” or “tails.” Throw a die (a small cube marked on each
face with from one to six dots1 ) and observe the number of dots on the top face.
Select a transistor from a bin and observe whether or not it is defective.
The following are some additional terms that are needed.
Sample Spaces and Trials
A sample space of an experiment is the set of all possible outcomes of the
experiment. Each repetition of an experiment is called a trial.
For the experiment of throwing a die and observing the number of dots on the
top face the sample space is the set
S = {1, 2, 3, 4, 5, 6}
In the experiment of flipping a coin and observing whether it falls heads or tails,
the sample space is S = {heads, tails} or simply S = {H, T }.
EXAMPLE 1
Determining the Sample Space An experiment consists of not-
ing whether the price of the stock of the Ford Corporation rose, fell, or remained
unchanged on the most recent day of trading. What is the sample space for this
There are three possible outcomes depending on whether the price
rose, fell, or remained unchanged. Thus the sample space S is
S = {rose, fell, unchanged}
EXAMPLE 2
Determining the Sample Space Two dice, identical except that
one is green and the other is red, are tossed and the number of dots on the top
face of each is observed. What is the sample space for this experiment?
Each die can take on its six different values with the other die also
taking on all of its six different values. We can express the outcomes as order
pairs. For example, (2, 3) will mean 2 dots on the top face of the green die and 3
dots on the top face of the red die. The sample space S is below. A more colorful
version is shown in Figure 1.15.
Figure 1.15
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
1 There are also four-sided die, eight-sided die, and so on. However, the six-sided die is the
most common and six-sided should be assumed when we refer to a die, unless otherwise specified.
1.3 Sample Spaces and Events
If the experiment of tossing 2 dice consists of just observing the total number
of dots on the top faces of the two dice, then the sample space would be
S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
In short, the sample space depends on the precise statement of the experiment.
A coin is flipped twice to
observe whether heads or tails shows; order is important. What is the sample
space for this experiment?
EXAMPLE 3
Determining the Sample Space
The sample space S consists of the 4 outcomes
S = {(H, H), (H, T ), (T, H), (T, T )}
✧ Tree Diagrams
Figure 1.16
In Example 3 we completed a task (flipped a coin) and then completed another
task (flipped the coin again). In these cases, the experiment can be diagrammed
with a tree. The tree diagram for Example 3 is shown in Figure 1.16. We see
we have a first set of branches representing the first flip of the coin. From there
we flip the coin again and have a second set of branches. Then trace along each
branch to find the outcomes of the experiment. If the coin is tossed a third time,
there will be eight outcomes.
A die is rolled. If the die
shows a 1 or a 6, a coin is tossed. What is the sample space for this experiment?
EXAMPLE 4
Determining the Sample Space
Figure 1.17 shows the possibilities. We then have
S = {(1, H), (1, T ), 2, 3, 4, 5, (6, H), (6, T )}
Figure 1.17
✧ Events
We start this subsection with the following definition of an event.
Events and Elementary Events
Given a sample space S for an experiment, an event is any subset E of S.
An elementary (or simple) event is an event with a single outcome.
Finding Events Using the sample space from Example 3 find
the events: “At least one head comes up” and “Exactly two tails come up.” Are
either events elementary events?
EXAMPLE 5
“At least one head comes up” = {(H, H), (H, T ), (T, H)}
“Exactly two tails come up” = {(T, T )}
The second event, “Exactly two tails come up” has only one outcome and so
it is an elementary event.
Chapter 1 Sets and Probability
We can use our set language for union, intersection, and complement to describe events.
Union of Two Events
If E and F are two events, then E ∪ F is the union of the two events and
consists of the set of outcomes that are in E or F.
Thus the event E ∪ F is the event that “E or F occurs.” Refer to Figure 1.18
where the event E ∪ F is the shaded region on the Venn diagram.
Figure 1.18
Intersection of Two Events
If E and F are two events, then E ∩ F is the intersection of the two events
and consists of the set of outcomes that are in both E and F.
Thus the event E ∩ F is the event that “E and F both occur.” Refer to Figure 1.18
where the event E ∩ F is the region where E and F overlap.
Complement of an Event
If E is an event, then E c is the complement of E and consists of the set of
outcomes that are not in E.
Thus the event E c is the event that “E does not occur.”
EXAMPLE 6 Determining Union, Intersection, and Complement Consider
the sample space given in Example 2. Let E consist of those outcomes for which
the number of dots on the top faces of both dice is 2 or 4. Let F be the event that
the sum of the number of dots on the top faces of the two dice is 6. Let G be the
event that the sum of the number of dots on the top faces of the two dice is less
than 11.
a. List the elements of E and F.
b. Find E ∪ F.
c. Find E ∩ F.
d. Find Gc .
a. E = {(2, 2), (2, 4), (4, 2), (4, 4)} and F = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
b. E ∪ F = {(2, 2), (2, 4), (4, 2), (4, 4), (1, 5), (3, 3), (5, 1)}
c. E ∩ F = {(2, 4), (4, 2)}
d. Gc = {(5, 6), (6, 5), (6, 6)}
If S is a sample space, 0/ ⊆ S, and thus 0/ is an event. We call the event 0/ the
impossible event since the event 0/ means that no outcome has occurred, whereas,
in any experiment some outcome must occur.
1.3 Sample Spaces and Events
The Impossible Event
The empty set, 0,
/ is called the impossible event.
For example, if H is the event that a head shows on flipping a coin and T is
the event that a tail shows, then H ∩ T = 0.
/ The event H ∩ T means that both
heads and tails shows, which is impossible.
Since S ⊆ S, S is itself an event. We call S the certainty event since any
outcome of the experiment must be in S. For example, if a fair coin is flipped, the
event H ∪ T is certain since a head or tail must occur.
The Certainty Event
Let S be a sample space. The event S is called the certainty event.
We also have the following definition for mutually exclusive events.
Figure 1.19.
Mutually Exclusive Events
Two events E and F are said to be mutually exclusive if the sets are disjoint. That is,
E ∩ F = 0/
Figure 1.19
Standard Deck of 52 Playing Cards
A standard deck of 52 playing cards has four 13-card suits: clubs ♣, diamonds ♦, hearts ♥, and spades ♠. The diamonds and hearts are red, while
the clubs and spades are black. Each 13-card suit contains cards numbered
from 2 to 10, a jack, a queen, a king, and an ace. The jack, queen, king,
and ace can be considered respectively as number 11, 12, 13, and 14. In
poker the ace can be either a 14 or a 1. See Figure 1.20.
Chapter 1 Sets and Probability
Figure 1.20
EXAMPLE 7
Determining if Sets Are Mutually Exclusive Let a card be
chosen from a standard deck of 52 cards. Let E be the event consisting of drawing
a 3. Let F be the event of drawing a heart. Let G be the event of drawing a Jack.
Are E and F mutually exclusive? Are E and G?
Since E ∩ F is the event that the card is a 3 and is a heart,
E ∩ F = {3♥} = 0/
and so these events are not mutually exclusive. The event E ∩ G is the event that
the card is a 3 and a jack, so
E ∩ G = 0/
therefore E and G are mutually exclusive.
✧ Continuous Sample Spaces
In all the previous examples we were able to list each outcome in the sample
space, even if the list is rather long. But consider the outcomes of an experiment
where the time spent running a race is measured. Depending on how the time is
measured, an outcome could be 36 seconds or 36.0032 seconds. The values of
the outcomes are not restricted to whole numbers and so the sample space must
be described, rather than listed. In the case of the race we could say S = {t|t ≥
0,t in seconds}. Then the event E that a person takes less than 35 seconds to run
the race would be written E = {t|t < 35 seconds}.
At a farmer’s market there is a display of
fresh oranges. The oranges are carefully weighed. What is a sample space for
this experiment? Describe the event that a orange weighs 100 grams or more.
Describe the event that a orange weighs between 200 and 250 grams.
EXAMPLE 8
Weighing Oranges
Since the weight of the orange can be any positive number,
S = {w|w > 0, w in grams}
Note that w = 0 is not included as if the weight was zero there would be no
orange! The event that an orange weighs 100 grams or more is
E = {w|w ≥ 100, w in grams}
Here note that we use ≥, not > as the value of exactly 100 grams needs to be
included. The event that the orange weighs between 200 and 250 grams is
F = {w|200 < w < 250, w in grams}
where strict inequalities are used as the weight is between those values.
1.3 Sample Spaces and Events
Self-Help Exercises 1.3
1. Two tetrahedrons (4 sided), each with equal sides
numbered from 1 to 4, are identical except that one
is red and the other green. If the two tetrahedrons
are tossed and the number on the bottom face of
each is observed, what is the sample space for this
2. Consider the sample space given in the previous exercise. Let E consist of those outcomes for which
both (tetrahedron) dice show an odd number. Let
F be the event that the sum of the two numbers on
these dice is 5. Let G be the event that the sum of
the two numbers is less than 7.
a. List the elements of E and F.
b. Find E ∩ F.
c. Find E ∪ F.
d. Find Gc .
3. A hospital carefully measures the length of every
baby born. What is a sample space for this experiment? Describe the events
a. the baby is longer than 22 inches.
b. the baby is 20 inches or shorter.
c. the baby is between 19.5 and 21 inches long.
1.3 Exercises
1. Let S = {a, b, c} be a sample space. Find all the
2. Let the sample space be S = {a, b, c, d}. How many
events are there?
3. A coin is flipped three times, and heads or tails is
observed after each flip. What is the sample space?
Indicate the outcomes in the event “at least 2 heads
are observed.”
4. A coin is flipped, and it is noted whether heads or
tails show. A die is tossed, and the number on the
top face is noted. What is the sample space of this
5. A coin is flipped three times. If heads show, one is
written down. If tails show, zero is written down.
What is the sample space for this experiment? Indicate the outcomes if “one is observed at least twice.”
6. Two tetrahedrons (4 sided), each with equal sides
numbered from 1 to 4, are identical except that one
is red and the other green. If the two tetrahedrons
are tossed and the number on the bottom face of
each is observed, indicate the outcomes in the event
“the sum of the numbers is 4.”
7. An urn holds 10 identical balls except that 1 is
white, 4 are black, and 5 are red. An experiment
consists of selecting a ball from the urn and observing its color. What is a sample space for this experiment? Indicate the outcomes in the event “the ball
is not white.”
8. For the urn in Exercise 7, an experiment consists of
selecting 2 balls in succession without replacement
and observing the color of each of the balls. What
is the sample space of this experiment? Indicate the
outcomes of the event “no ball is white.”
9. Ann, Bubba, Carlos, David, and Elvira are up for
promotion. Their boss must select three people from
this group of five to be promoted. What is the
sample space? Indicate the outcomes of the event
“Bubba is selected.”
10. A restaurant offers six side dishes: rice, macaroni,
potatoes, corn, broccoli, and carrots. A customer
must select two different side dishes for his dinner.
What is the sample space? List the outcomes of the
event “Corn is selected.”
11. An experiment consists of selecting a digit from
the number 112964333 and observing it. What is
a sample space for this experiment? Indicate the
outcomes in the event that “an even digit.”
12. An experiment consists of selecting a letter from the
word CONNECTICUT and observing it. What is a
Chapter 1 Sets and Probability
sample space for this experiment? Indicate the outcomes of the event “a vowel is selected.”
13. An inspector selects 10 transistors from the production line and notes how many are defective.
a. Determine the sample space.
b. Find the outcomes in the set corresponding to
the event E “at least 6 are defective.”
c. Find the outcomes in the set corresponding to
the event F “at most 4 are defective.”
d. Find the sets E ∪ F, E ∩ F, E c , E ∩ F c , E c ∩ F c .
e. Find all pairs of sets among the nonempty ones
listed in part (d) that are mutually exclusive.
14. A survey indicates first whether a person is in the
lower income group (L), middle income group (M),
or upper income group (U), and second which of
these groups the father of the person is in.
a. Determine the sample space using the letters L,
M, and U.
b. Find the outcomes in the set corresponding to
the event E “the person is in the lower income
c. Find the outcomes in the set corresponding to
the event F “the person is in the higher income
d. Find the sets E ∪ F, E ∩ F, E c , E ∩ F c , E c ∩ F c .
e. Find all pairs of sets listed in part (d) that are
mutually exclusive.
15. A corporate president decides that for each of the
next three fiscal years success (S) will be declared if
the earnings per share of the company go up at least
10% that year and failure (F) will occur is less than
a. Determine the sample space using the letters S
and F.
b. Find the outcomes in the set corresponding to
the event E “at least 2 of the next 3 years is a
c. Find the outcomes in the set corresponding to
the event G “the first year is a success.”
d. Find and describe the sets E ∪ G, E ∩ G, Gc ,
E c ∩ G, and (E ∪ G)c .
e. Find all pairs of sets listed in part (d) that are
mutually exclusive.
16. Let E be the event that the life of a certain light bulb
is at least 100 hours and F that the life is at most
200 hours. Describe the sets:
a. E ∩ F
b. F c
c. E c ∩ F
d. (E ∪ F)c
17. Let E be the event that a pencil is 10 cm or longer
and F the event that the pencil is less than 25 cm.
Describe the sets:
a. E ∩ F
b. E c
c. E ∩ F c
d. (E ∪ F)c
In Exercises 18 through 23, S is a sample space and E,
F, and G are three events. Use the symbols ∩, ∪, and c
to describe the given events.
18. F but not E
19. E but not F
20. Not F or not E
21. Not F and not E
22. Not F, nor E, nor G
23. E and F but not G
24. Let S be a sample space consisting of all the integers
from 1 to 20 inclusive, E the first 10 of these, and
F the last 5 of these. Find E ∩ F, E c ∩ F, (E ∪ F)c ,
and E c ∩ F c .
25. Let S be the 26 letters of the alphabet, E be the vowels {a, e, i, o, u}, F the remaining 21 letters, and G
the first 5 letters of the alphabet. Find the events
E ∪ F ∪ G, E c ∪ F c ∪ Gc , E ∩ F ∩ G, and E ∪ F c ∪ G.
26. A bowl contains a penny, a nickel, and a dime.
A single coin is chosen at random from the bowl.
What is the sample space for this experiment? List
the outcomes in the event that a penny or a nickel is
27. A cup contains four marbles. One red, one blue, one
green, and one yellow. A single marble is drawn at
random from the cup. What is the sample space for
this experiment? List the outcomes in the event that
a blue or a green marble is chosen.
1.4 Basics of Probability
Solutions to Self-Help Exercises 1.3
1. Consider the outcomes as ordered pairs, with the number on the bottom of
the red one the first number and the number on the bottom of the white one
the second number. The sample space is
S = { (1, 1), (1, 2), (1, 3), (1, 4),
(2, 1), (2, 2), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 3), (3, 4),
(4, 1), (4, 2), (4, 3), (4, 4)}
2. a. E = {(1, 1), (1, 3), (3, 1), (3, 3)}, and F = {(1, 4), (2, 3), (3, 2), (4, 1)}
b. E ∩ F = 0/
c. E ∪ F = {(1, 1), (1, 3), (3, 1), (3, 3), (1, 4), (2, 3), (3, 2), (4, 1)}
d. Gc = {(3, 4), (4, 3), (4, 4)}
3. Since the baby can be any length greater than zero, the sample space is
S = {x|x > 0, x in inches}
a. E = {x|x > 22, x in inches}
b. F = {x|x ≤ 20, x in inches}
c. G = {x|19.5 < x < 21, x in inches}
Basics of Probability
✧ Introduction to Probability
We first consider sample spaces for which the outcomes (elementary events) are
equally likely. For example, a head or tail is equally likely to come up on a flip of
a fair coin. Any of the six numbers on a fair die is equally likely to come up on
a roll. We will refer to a sample space S whose individual elementary events are
equally likely as a uniform sample space. We then give the following definition
of the probability of any event in a uniform sample space.
Probability of an Event in a Uniform Sample Space
If S is a finite uniform sample space and E is any event, then the probability of E, P(E), is given by
P(E) =
Number of elements in E
Number of elements in S
Chapter 1 Sets and Probability
Probability for a Single Die Suppose a fair die is rolled and
the sample space is S = {1, 2, 3, 4, 5, 6}. Determine the probability of each of the
following events.
a. The die shows an odd number.
b. The die shows the number 9.
c. The die shows a number less than 8.
EXAMPLE 1
a. We have E = {1, 3, 5}. Then
The Beginnings of
In 1654 the famous mathematician
Blaise Pascal had a friend,
Chevalier de Mere, a member of
the French nobility and a gambler,
who wanted to adjust gambling
stakes so that he would be assured
of winning if he played long
enough. This gambler raised
questions with Pascal such as the
following: In eight throws of a die a
player attempts to throw a one, but
after three unsuccessful trials the
game is interrupted. How should
he be compensated? Pascal wrote
to a leading mathematician of that
day, Pierre de Fermat
(1601–1665), about these
problems, and their resulting
correspondence represents the
beginnings of the modern theory of
mathematical probability.
P(E) =
n(E) 3 1
= =
b. The event F that the die shows a 9 is the impossible event. So n(F) = 0 and
P(F) =
n(F) 0
= =0
c. The event G that the die shows a number less than 8 is just the certainty event.
So G = {1, 2, 3, 4, 5, 6} = S and
n(G) 6
= =1
P(G) =
EXAMPLE 2
Probability for a Single Card Suppose a single card is randomly
drawn from a standard 52-card deck. Determine the probability of each of the
following events.
a. A king is drawn.
b. A heart is drawn.
a. The event is E = { K♦, K♥, K♠, K♣}. So,
P(E) =
b. The event F contains 13 hearts. So
n(F) 13 1
P(F) =
Probability for Transistors A bin contains 15 identical (to the
eye) transistors except that 6 are defective and 9 are not. What is the probability
that a transistor selected at random is defective?
EXAMPLE 3
Let us denote the set S to be the set of all 15 transistors and the set E
to be the set of defective transistors. Then,
P(E) =
REMARK: What if we selected two transistors or two cards? We will learn how
to handle this type of experiment in the next chapter.
A fair coin is flipped twice to
observe whether heads or tails shows; order is important. What is the probability
that tails occurs both times?
EXAMPLE 4
Probability for Two Coin Flips
The sample space S consists of the 4 outcomes
S = {(H, H), (H, T ), (T, H), (T, T )}
1.4 Basics of Probability
Since we are using a fair coin, each of the individual four elementary events are
equally likely. The set E that tails occurs both times is E = {(T, T )} and contains
one element. We have
P(E) =
The Beginnings of
Empirical Probability
Empirical probability began with
the emergence of insurance
companies. Insurance seems to
have been originally used to
protect merchant vessels and was
in use even in Roman times. The
first marine insurance companies
began in Italy and Holland in the
14th century and spread to other
countries by the 16th century. In
fact, the famous Lloyd’s of London
was founded in the late 1600s. The
first life insurance seems to have
been written in the late 16th
century in Europe. All of these
companies naturally needed to
know with what the likelihood of
certain events would occur. The
empirical probabilities were
determined by collecting data over
long periods of time.
n(E) 1
✧ Empirical Probability
A very important type of problem that arises every day in business and science is
to find a practical way to estimate the likelihood of certain events. For example,
a food company may seek a practical method of estimating the likelihood that a
new type of candy will be enjoyed by consumers. The most obvious procedure for
the company to follow is to randomly select a consumer, have the consumer taste
the candy, and then record the result. This should be repeated many times and the
final totals tabulated to give the fraction of tested consumers who enjoy the candy.
This fraction is then a practical estimate of the likelihood that all consumers will
enjoy this candy. We refer to this fraction or number as empirical probability.
The London merchant John Graunt (1620–1674) with the publication of Natural and Political Observations Made upon the Bills of Mortality in 1662 seems
to have been the first person to have gathered data on mortality rates and determined empirical probabilities from them. The data were extremely difficult to
obtain. His then-famous London Life Table is reproduced below, showing the
number of survivors through certain ages per 100 people.
London Life Table
EXAMPLE 5
Finding Empirical Probability Using the London Life Table,
find the empirical probability of a randomly chosen person living in London in
the first half of the 17th century surviving until age 46.
In the London Life Table N = 100. If E is the event “survive to
age 46,” then according to the table the corresponding number is 10. Thus, the
empirical probability of people living in London at that time surviving until age
46 was 10/100 = 0.1.
Consider now a poorly made die purchased at a discount store. Dice are made
by drilling holes in the sides and then backfilling. Cheap dice are, of course, not
carefully backfilled. So when a lot of holes are made in a face, such as for a side
with 6, and they are not carefully backfilled, that side will not be quite as heavy
as the others. Thus a 6 will tend to come up more often on the top. Even a die
taken from a craps table in Las Vegas, where the dice are of very high quality,
will have some tiny imbalance.
A die with 6 sides numbered
from 1 to 6, such as used in the game of craps, is suspected to be somewhat
lopsided. A laboratory has tossed this die 1000 times and obtained the results
shown in the table. Find the empirical probability that a 2 will occur and the
probability that a 6 will occur.
EXAMPLE 6
Finding Empirical Probability
Chapter 1 Sets and Probability
Number Observed
Solution The total number observed is 1000. The number observed for the 2
and 6, respectively is 179 and 125. So dividing these numbers by 1000 gives
P(2) = 179/1000 = 0.179
P(6) = 125/1000 = 0.125
Frederick Mosteller and the
Dice Experiment
Frederick Mosteller has been president of the American Association for the
Advancement of Science, the Institute of Mathematical Statistics, and the
American Statistical Association. He once decided that “It would be nice
to see if the actual outcome of a real person tossing real dice would match
up with the theory.” He then engaged Willard H. Longcor to buy some
dice, toss them, and keep careful records of the outcomes. Mr. Longcor
then tossed the dice on his floor at home so that the dice would bounce on
the floor and then up against the wall and then land back on the floor. After
doing this several thousand times his wife became troubled by the noise.
He then placed a rug on the floor and on the wall, and then proceeded
to quietly toss his dice millions of times, keeping careful records of the
outcomes. In fact, he was so careful and responsible about his task, that
he threw away his data on the first 100,000 tosses, since he had a nagging
worry that he might have made some mistake keeping perfect track.
✧ Probability Distribution Tables
A probability distribution table is a useful way to display probability data for
an experiment. In a probability distribution table there is one column (or row) for
the events that take place and one column (or row) for the probability of the event.
The events chosen must be mutually exclusive and therefore the total probability
will add to 1. This is best demonstrated through an example.
EXAMPLE 7 Flipping a Coin Twice Write the probability distribution table
for the number of heads when a coin is flipped twice.
Recall from Example 4 that the uniform sample space is S =
{(H, H), (H, T ), (T, H), (T, T )}. Next, we are asked to organize the events by
the number of heads, so we will have three events
E1 = {(H, H)} with two heads and a probability of 1/4,
E2 = {(H, T ), (T, H)} with exactly one head and a probability of 2/4
E3 = {(T, T )} with zero heads and a probability of 1/4.
This is shown in the table on the left.
2 heads
1 head
0 heads
Note how the list of events covered all the possibilities for the number of
heads and that the events are all mutually exclusive. You can’t have exactly
two heads and exactly one head at the same time! Next see that the sum of the
probabilities is equal to one. This will always be the case when your probability
distribution table is correct.
1.4 Basics of Probability
EXAMPLE 8
Sum of the Numbers for Two Dice Two fair dice are rolled.
Find the probability distribution table for the sum of the numbers shown uppermost.
Recall the uniform sample space in Example 2 of the last section for
rolling two dice. We see the smallest sum is 2 from the roll (1,1) and the largest
sum is 12 from the roll (6,6). Count the number of outcomes in each event to find
the probability:
EXAMPLE 9 Weight of Oranges A crate contains oranges and each orange
is carefully weighed. It was found that 12 oranges weighed less than 100 grams,
40 oranges weighed 100 grams or more, but less than 150 grams, 60 oranges
weighed 150 grams or more, but less than 200 grams, and 8 oranges weighed
200 grams or more. Organize this information in a probability distribution table
The sample space for this experiment was found in the previous section to be S = {w|w > 0, w in grams}. There are four mutually exclusive events
described in this sample space and these form the basis of the probability distribution table. A total of 12 + 40 + 60 + 8 = 120 oranges were weighed. The
empirical probability that an orange weighs less than 100 grams will be the ratio
12/120. The remaining probabilities are found in the same way. This gives the
probability distribution table below where w is the weight of an orange in grams.
w < 100
= 1/10
100 ≤ w < 150
= 1/3
150 ≤ w < 200
= 1/2
w ≥ 200
= 2/30
REMARK: Notice that in the probability distribution table above that there were
no gaps and no overlap. It is important to be able to translate the statements like
“100 grams or more” into an event 100 ≤ w.
Self-Help Exercises 1.4
1. Two tetrahedrons (4 sided), each with equal sides
numbered from 1 to 4, are identical except that one
is red and the other white. If the two tetrahedrons
are tossed and the number on the bottom face of
each is observed, what is the sample space for this
experiment? Write the probability distribution table for the sum of the numbers on the bottom of the
2. In the past month 72 babies were born at a local hos-
pital. Each baby was carefully measured and it was
found that 10 babies were less than 19 inches long,
12 babies were 19 inches or longer but less than 20
inches long, 32 babies were 20 inches or longer but
less than 21 inches long. Organize this information
in a probability distribution table.
3. An experiment consists of randomly selecting a letter from the word FINITE and observing it. What is
the probability of selecting a vowel?
Chapter 1 Sets and Probability
1.4 Exercises
In Exercises 1 through 4, a fair die is tossed. Find the
probabilities of the given events.
1. an even number
2. the numbers 4 or 5
3. a number less than 5
4. any number except 2 or 5
In Exercises 5 through 10, a card is drawn randomly
from a standard deck of 52 cards. Find the probabilities
of the given events.
5. an ace
6. a spade
7. a red card
8. any number between 3 and 5 inclusive
9. any black card between 5 and 7 inclusive
10. a red 8
In Exercises 11 through 14, a basket contains 3 white,
4 yellow, and 5 black transistors. If a transistor is randomly picked, find the probability of each of the given
11. white
12. black
13. not yellow
14. black or yellow
15. A somewhat lopsided die is tossed 1000 times with
1 showing on the top face 150 times. What is the
empirical probability that a 1 will show?
16. A coin is flipped 10,000 times with heads showing
5050 times. What is the empirical probability that
heads will show?
17. The speed of 500 vehicles on a highway with limit
of 55 mph was observed, with 400 going between
55 and 65 mph, 60 going less than 55 mph, and 40
going over 65 mph. What is the empirical probability that a vehicle chosen at random on this highway
will be going a. under 55 mph, b. between 55 and
65 mph, c. over 65 mph.
18. In a survey of 1000 randomly selected consumers,
50 said they bought brand A cereal, 60 said they
bought brand B, and 80 said they bought brand C.
What is the empirical probability that a consumer
will purchase a. brand A cereal, b. brand B, c.
brand C?
19. A large dose of a suspected carcinogen has been
given to 500 white rats in a laboratory experiment.
During the next year, 280 rats get cancer. What is
the empirical probability that a rat chosen randomly
from this group of 500 will get cancer?
20. A new brand of sausage is tested on 200 randomly
selected customers in grocery stores with 40 saying
they like the product, the others saying they do not.
What is the empirical probability that a consumer
will like this brand of sausage?
21. Over a number of years the grade distribution in a
mathematics course was observed to be
What is the empirical probability that a randomly
selected student taking this course will receive a
grade of A? B? C? D? F?
22. A store sells four different brands of VCRs. During
the past year the following number of sales of each
of the brands were found.
Brand A
Brand B
Brand C
Brand D
What is the empirical probability that a randomly
selected customer who buys a VCR at this store will
pick brand A’? brand B:’ brand C’? brand D?
23. A somewhat lopsided die is tossed 1000 times with
the following results. What is the empirical probability that an even number shows?
24. A retail store that sells sneakers notes the following
number of sneakers of each size that were sold last
1.4 Basics of Probability
In Exercises 29 through 34, assume that all elementary
events in the same sample space are equally likely.
What is the empirical probability that a customer
buys a pair of sneakers of size 7 or 12?
29. A fair coin is flipped three times. What is the probability of obtaining exactly 2 heads? At least 1 head?
25. A fair coin is flipped three times, and heads or tails
is observed after each flip. What is the probability
of the event “at least 2 heads are observed.” Refer
to the answer in Exercise 3 in Section 4.3.
30. A family has three children. Assuming a boy is as
likely as a girl to have been born, what is the probability that two are boys and one is a girl? That at
least one is a boy?
26. A fair coin is flipped, and it is noted whether heads
or tails show. A fair die is tossed, and the number on
the top face is noted. What is the probability of the
event “heads shows on the coin and an even number on the die.” Refer to the answer in Exercise 4 in
Section 4.3.
27. A coin is flipped three times. If heads show, one
is written down. If tails show, zero is written down.
What is the probability of the event “one is observed
at least twice.” Refer to the answer in Exercise 5 in
Section 4.3.
28. Two fair tetrahedrons (4 sided), each with equal
sides numbered from 1 to 4, are identical except that
one is red and the other white. If the two tetrahedrons are tossed and the number on the bottom face
of each is observed, what is the probability of the
event “the sum of the numbers is 4.” Refer to the
answer in Exercise 6 in Section 4.3.
31. A fair coin is flipped and a fair die is tossed. What
is the probability of obtaining a head and a 3?
32. A fair coin is flipped twice and a fair die is tossed.
What is the probability of obtaining 2 heads and a
33. A pair of fair dice are tossed. What is the probability
of obtaining a sum of 2? 4? 8?
34. A pair of fair dice are tossed. What is the probability
of obtaining a sum of 5? 6? ll?
35. An experiment consists of selecting a digit from the
number 112964333 and observing it. What is the
probability that “an even digit is selected.”
36. An experiment consists of selecting a letter from the
word CONNECTICUT and observing it. What is
the probability that “a vowel is selected.”
Solutions to Self-Help Exercises 1.4
1. The sample space was found in the previous section and is
S = { (1, 1), (1, 2), (1, 3), (1, 4),
(2, 1), (2, 2), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 3), (3, 4),
(4, 1), (4, 2), (4, 3), (4, 4)}
The sum of the numbers ranges from 1 + 1 = 2 to 4 + 4 = 8. Count the
outcomes in each event to find
2. The sample space for this experiment is S = {x|x > 0, x in inches}. The
lengths of 10 + 12 + 32 = 54 babies is given. A careful examination of the
events shows that no mention was made of babies longer than 21 inches. We
deduce that 72 − 54 = 18 babies must be 21 inches or longer. This can now
Chapter 1 Sets and Probability
be arranged in a probability distribution table where x is the length of the
baby in inches.
x < 19
= 5/36
19 ≤ x < 20
= 1/6
20 ≤ x < 21
= 4/9
x ≥ 21
= 1/4
3. FINITE has six letters and there are three vowels. So
P(vowel) =
Rules for Probability
✧ Elementary Rules
Recall that if S is a finite uniform sample space, that is, a space for which all
individual elementary elements are equally likely, and E is any event, then the
probability of E, denoted by P(E), is given by
P(E) =
Number of elements in E
Number of elements in S
If E is an event in a sample space then 0 ≤ n(E) ≤ n(S). Dividing this by
n(S) then gives
n(E) n(S)
Using the definition of probability given above yields
0 ≤ P(E) ≤ 1
This is our first rule for probability. Notice also that
P(S) =
= 1 and P(0)
/ =
n(S) n(S)
These rules apply for events in any sample space; however, the derivations just
given are valid only for spaces with equally likely events.
Elementary Rules for Probability
For any event E in a sample space S we have
0 ≤ P(E) ≤ 1
P(S) = 1
/ =0
1.5 Rules for Probability
Some Developments in
Neither Pascal nor Fermat
published their initial findings on
probability. Christian Huygens
(1629–1695) became acquainted
with the work of Pascal and Fermat
and subsequently published in
1657 the first tract on probability:
On Reasoning in Games of Dice.
This little pamphlet remained the
only published work on probability
for the next 50 years. James
Bernoulli (1654–1705) published
the first substantial tract on
probability when his Art of
Reasoning appeared 7 years after
his death. This expanded
considerably on Huygens’ work.
The next major milestone in
probability occurred with the
publication in 1718 of Abraham De
Moivre’s work Doctrine of Chance:
A Method of Calculating of Events
in Play. Before 1770, probability
was almost entirely restricted to
the study of gambling and actuarial
problems, although some
applications in errors of
observation, population, and
certain political and social
phenomena had been touched on.
It was Pierre Simon Laplace
(1739–1827) who broadened the
mathematical treatment of
probability beyond games of
chance to many areas of scientific
research. The theory of probability
undoubtedly owes more to Laplace
than to any other individual.
✧ Union Rule for Probability
We would now like to determine the probability of the union of two events E and
F. We start by recalling the union rule for sets:
n(E ∪ F) = n(E) + n(F) − n(E ∩ F)
Now divide both sides of this last equation by n(S) and obtain
n(E ∪ F) n(E) n(F) n(E ∩ F)
n(S) n(S)
By the definition of probability given above this becomes
P(E ∪ F) = P(E) + P(F) − P(E ∩ F)
This is called the union rule for probability. This rule applies for events in any
sample space; however, the derivation just given is valid only for spaces with
equally likely events.
Union Rule for Probability
P(E ∪ F) = P(E) + P(F) − P(E ∩ F)
EXAMPLE 1 Union Rule With Drawing a Card A single card is randomly
drawn from a standard deck of cards. What is the probability that it will be a red
card or a king.
Solution Let R be the set of red cards and let K be the set of kings. Red cards
consist of hearts and diamonds, so there are 26 red cards. Therefore P(R) = 26/52.
There are 4 kings, so P(K) = 4/52. Among the 4 kings, there are 2 red cards. So,
P(R ∩ K) = 2/52. Using the union rule gives
P(R ∪ K) = P(R) + P(K) − P(R ∩ K)
+ −
REMARK: It is likely that you would intuitively use the union rule had you been
asked to pick out all of the cards from the deck that were red or kings. You would
choose out all of the red cards along with all of the kings for a total of 28 cards.
Union Rule With Two Dice Two dice, identical except that one
is green and the other is red, are tossed and the number of dots on the top face of
each is observed. Let E consist of those outcomes for which the number of dots
on the top face of the green dice is a 1 or 2. Let F be the event that the sum of the
number of dots on the top faces of the two dice is 6. Find the probability that a 1
or 2 will be on the top of the green die or the sum of the two numbers will be 6.
EXAMPLE 2
Chapter 1 Sets and Probability
Notice that
E = {(1, 1), (1, 2), · · · , (1, 6), (2, 1), (2, 2), · · · , (2, 6)}
F = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
E ∩ F = {(1, 5), (2, 4)}
Figure 1.21
The set that a 1 or 2 will be on the top of the green die and the sum of the two
numbers will be 6 is E ∩ F. To find p(E ∩ F) use the union rule of probability
and obtain
P(E ∪ F) = P(E) + P(F) − P(E ∩ F)
+ −
Alternatively, you can draw the sample space for two dice and circle all outcomes that have a 1 or 2 on the top of the green die or the sum of the two numbers
shown uppermost is 6. This is done in Figure 1.21. Counting the circled outcomes
we find there are 15 of them.
Consider two events E and F that are mutually exclusive, that is, E ∩ F) = 0.
Then P(E ∩ F) = 0. Using the union rule of probability for these two sets gives
P(E ∪ F) = P(E) + P(F) − P(E ∩ F)
= P(E) + P(F) − 0
= P(E) + P(F)
We then have the following rule:
Union Rule for Mutually Exclusive Events
If E and F are mutually exclusive events, then
P(E ∪ F) = P(E) + P(F)
For any event E in a sample space, E ∪ E c = S and E ∩ E c = 0.
/ So, E and
are mutually exclusive. Using the union rule for mutually exclusive events we
have that
P(E) + P(E c ) = P(E ∪ E c ) = P(S) = 1
So, P(E c ) = 1 − P(E) and P(E) = 1 − P(E c ). We call this the complement rule.
Complement Rule for Probability
P(E c ) = 1 − P(E)
P(E) = 1 − P(E c )
Complement Rule for Two Dice Consider the dice described in
Example 2. What is the probability that the sum of the two numbers is less than
EXAMPLE 3
1.5 Rules for Probability
Let E be the event that the sum of the two numbers is less than
12. Then we wish to find P(E). It is tedious to find this directly. Notice that
E c = {(6, 6)}. Now use the complement rule.
P(E) = 1 − P(E c ) = 1 −
A die with 6 sides numbered
from 1 to 6, such as used in the game of craps, is suspected to be somewhat
lopsided. A laboratory has tossed this die 1000 times and obtained the results
shown in the table. Find the empirical probability that an even number will occur.
EXAMPLE 4
Finding Empirical Probability
Number Observed
Solution The total number observed is 1000. The number observed for the 2,
4, and 6, respectively is 179, 177, and 125. So dividing these numbers by 1000
P(2) = 179/1000 = 0.179
P(4) = 177/1000 = 0.177
P(6) = 125/1000 = 0.125
To find the empirical probability of an even number these three values can be
added as the events are mutually exclusive. That is,
P(even) = P(2) + P(4) + P(6) = 0.179 + 0.177 + 0.125 = 0.481
A salesman makes two
stops when in Pittsburgh. The first stop yields a sale 10% of the time, the second
stop 15% of the time, and both stops yield a sale 4% of the time. What proportion
of the time does a trip to Pittsburgh result in no sales?
EXAMPLE 5
Finding the Probability of an Event
Let E be the event a sale is made at the first stop and F the event that
a sale is made at the second stop. What should we make of the statement that
the first stop yields a sale 10% of the time. It seems reasonable to assume that
the salesman or his manager have looked at his sales data and estimated the 10%
number. We then take the 10% or 0.10 as the empirical probability. We interpret
the other percentages in a similar way. We then have
P(E) = 0.10
Figure 1.22
P(F) = 0.15
P(E ∩ F) = 0.04
Since P(E ∩ F) = 0.04, we place 0.04 in the region E ∩ F in Figure 1.22. Now
since P(E) = 0.10, we can see that P(E ∩ F c ) = 0.10 − 0.04 = 0.06. In a similar
fashion we have P(E c ∩ F) = 0.15 − 0.04 = 0.11. Thus, we readily see from
Figure 1.22 that
P(E ∪ F) = 0.06 + 0.04 + 0.11 = 0.21
Then by the complement rule we have
P((E ∪ F)c ) = 1 − P(E ∪ F) = 1 − 0.21 = 0.79
Thus no sale is made in Pittsburgh 79% of the time.
Chapter 1 Sets and Probability
We could have obtained P(E ∪ F) directly from the union rule as
P(E ∪ F) = P(E) + P(F) − P(E ∩ F) = 0.10 + 0.15 − 0.04 = 0.21
The probability that any
of the first five numbers of a loaded die will come up is the same while the
probability that a 6 comes up is 0.25. What is the probability that a 1 will come
EXAMPLE 6
Finding the Probability of an Event
We are given P(1) = P(2) = P(3) = P(4) = P(5), P(6) = 0.25. Also,
all the probabilities must add up to 1, so
1 = P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
= 5P(1) + 0.25
5P(1) = 0.75
P(1) = 0.15
EXAMPLE 7 Continuous Sample Space Arrange the following information
in a probability distribution table: A crop of apples is brought in for weighing. It
is found that 10% of the apples weigh less than 100 gm, 40% weigh 200 gm or
less, and 25% weigh more than 300 gm.
If we let x = weight of an apple in grams then
0 ≤ x < 100 gm
100 gm ≤ x ≤ 200 gm
200 gm < x ≤ 300 gm
x > 300 gm
Note that the 40% of the apples that weigh 200 gm or less includes the 10%
that weigh less than 100 grams. Since the events in a probability distribution
table must be mutually exclusive, the 30% that weigh 100 grams or more and
200 grams or less are shown in the second row. The third row of the table is
found using deductive reasoning as the total probability must be 1 and there is a
gap in the events.
✧ Odds (Optional)
One can interpret probabilities in terms of odds in a bet. Suppose in a sample
space S we are given an event E with probability P = P(E) = 57 . In the long term
we expect E to occur 5 out of 7 times. Now, P(E c ) = 27 and in the long term we
expect that E c to occur 2 out of 7 times. Then we say that the odds in favor of E
are 5 to 2.
1.5 Rules for Probability
The odds in favor of an event E are defined to be the ratio of P(E) to
P(E c ), or
P(E ) 1 − P(E)
Often the ratio P(E)/P(E c ) is reduced to lowest terms, a/b, and then we say that
the odds are a to b or a:b.
EXAMPLE 8
Determining the Odds of an Event You believe that a horse has
a probability of 1/4 of winning a race. What are the odds of this horse winning?
What are the odds of this horse losing? What profit should a winning $2 bet
return to be fair?
Since the probability of winning is P = 1/4, the odds of winning are
1 − P 1 − /4
/4 3
that is: 1 to 3 or 1:3.
Since the probability of winning is 14 , the probability of losing is 1 − 1/4 = 3/4.
Then the odds for losing is
1 − 3/4
or 3 to 1 or 3:1. Since the fraction 3/1 can also be written as 6/2 with odds 6 to 2,
a fair $2 bet should return $6 for a winning ticket.
Notice that making this same bet many times, we expect to win $6 one-fourth
of the time and lose $2 three-fourths of the time. So, for example, on every
four bets we would expect to win $6 once and lose $2 three times. Our average
winnings would be 6(1) − 2(3) = 0 dollars.
If the odds for an event E are given as a/b, we can calculate the probability
P(E). We have
b 1 − P(E)
a(1 − P(E)) = bP(E)
a = bP(E) + aP(E)
= P(E)(a + b)
P(E) =
Obtaining Probability From Odds
Suppose that the odds for an event E occurring is given as a/b or a : b, then
P(E) =
Chapter 1 Sets and Probability
REMARK: One can think of the odds a:b of event E as saying if this experiment
was carried out a + b times, then a of those times E would have occurred. Our
definition of empirical probability then says P(E) = a+b
EXAMPLE 9
for a horse winning is listed at
At the race track, the odds
What is the probability that the horse will win.
Obtaining Probability From Odds
Using the above formula for odds a/b, we have
= 0.60
a+b 3+2
Self-Help Exercises 1.5
1. If S = {a, b, c} with P(a) = P(b) = 2P(c), find P(a).
2. A company has bids on two contracts. They believe
that the probability of obtaining the first contract
is 0.4 and of obtaining the second contract is 0.3,
while the probability of obtaining both contracts is
a. Find the probability that they will obtain exactly
one of the contracts.
b. Find the probability that they will obtain neither
of the contracts.
3. What are the odds that the company in the previous
exercise will obtain both of the contracts?
1.5 Exercises
In all the following, S is assumed to be a sample space.
1. Let S = {a, b, c} with P(a) = 0.1, P(b) = 0.4, and
P(c) = 0.5. Let E = {a, b} and F = {b, c}. Find
P(E) and P(F).
2. Let S = {a, b, c, d, e, f } with P(a) = 0.1, P(b) = 0.2,
P(c) = 0.25, P(d) = 0.15, P(e) = 0.12, and P( f ) =
0.18. Let E = {a, b, c} and F = {c, d, e, f } and find
P(E) and P(F).
3. Let S = {a, b, c, d, e, f } with P(b) = 0.2, P(c) =
0.25, P(d) = 0.15, P(e) = 0.12, and P( f ) = 0.1.
Let E = {a, b, c} and F = {c, d, e, f }. Find P(a),
P(E), and P(F).
4. Let S = {a, b, c, d, e, f } with P(b) = 0.3, P(c) =
0.15 , P(d) = 0.05, P(e) = 0.2, P( f ) = 0.13. Let
E = {a, b, c} and F = {c, d, e, f }. Find P(a), P(E),
and P(F).
7. If S = {a, b, c, d, e, f } with P(a) = P(b) = P(c) =
P(d) = P(e) = P( f ), find P(a).
8. If S = {a, b, c} with P(a) = 2P(b) = 3P(c), find
9. If S = {a, b, c, d, e, f } with P(a) = P(b) = P(c),
P(d) = P(e) = P( f ) = 0.1, find P(a).
10. If S = {a, b, c, d, e, f } and if P(a) = P(b) = P(c),
P(d) = P(e) = P( f ), P(d) = 2P(a), find P(a).
11. If E and F are two disjoint events in S with P(E) =
0.2 and P(F) = 0.4, find P(E ∪ F), P(E c ), and
P(E ∩ F).
12. Why is it not possible for E and F to be two disjoint
events in S with P(E) = 0.5 and P(F) = 0.7?
5. If S = {a, b, c, d} with P(a) = P(b) = P(c) = P(d),
find P(a).
13. If E and F are two disjoint events in S with P(E) =
0.4 and P(F) = 0.3, find P(E ∪F), P(F c ), P(E ∩F),
P((E ∪ F)c ), and P((E ∩ F)c ).
6. If S = {a, b, c} with P(a) = P(b) and P(c) = 0.4,
find P(a).
14. Why is it not possible for S = {a, b, c} with P(a) =
0.3, P(b) = 0.4, and P(c) = 0.5?
1.5 Rules for Probability
15. Let E and F be two events in S with P(E) = 0.3,
P(F) = 0.5, and P(E ∩ F) = 0.2. Find P(E ∩ F)
and P(E ∩ F c ).
16. Let E and F be two events in S with P(E) = 0.3,
P(F) = 0.5, and P(E ∩ F) = 0.2. Find P(E c ∩ F)
and P(E c ∩ F c ).
17. Let E and F be two events in S with P(E) = 0.3,
P(F) = 0.5, and P(E ∪ F) = 0.6. Find P(E ∩ F)
and P(E ∩ F c ).
18. Why is it not possible to have E and F two events
in S with P(E) = 0.3 and P(E ∩ F) = 0.5?
In Exercises 19 through 22, let E, F, and G be events
in S with P(E) = 0.55, P(F) = 0.4, P(G) = 0.45,
P(E ∩ F) = 0.3, P(E ∩ G) = 0.2, P(F ∩ G) = 0.15, and
P(E ∩ F ∩ G) = 0.1.
19. Find P(E ∩ F ∩ Gc ), P(E ∩ F c ∩ G), and P(E ∩ F c ∩
Gc ).
20. Using the results of the previous exercise, find
P(E c ∩ F ∩ G), P(E c ∩ F ∩ Gc ), and P(E c ∩ F c ∩ G).
21. Using the results of the previous two exercises, find
P(E ∪ F ∪ G).
22. Using the results of the previous three exercises,
find P(E c ∪ F c ∪ Gc ).
23. For the loaded die in Example 6 of the text, what are
the odds that
a. a 2 will occur
b. a 6 will occur?
24. For the loaded die in Example 6 of the text, what are
the odds that
a. a 3 will occur
b. a 1 will occur?
25. A company believes it has a probability of 0.40 of
receiving a contract. What is the odds that it will?
26. In Example 5 of the text, what are the odds that the
salesman will make a sale on
a. the first stop
b. on the second stop
c. on both stops?
27. It is known that the odds that E will occur are 1:3
and that the odds that F will occur are 1:2, and that
both E and F cannot occur. What are the odds that
E or F will occur?
28. If the odds for a successful marriage are 1:2, what
is the probability for a successful marriage?
29. If the odds for the Giants winning the World Series
are 1:4, what is the probability that the Giants will
win the Series?
30. Bidding on Contracts An aerospace firm has three
bids on government contracts and knows that the
contracts are most likely to be divided up among
a number of companies. The firm decides that the
probability of obtaining exactly one contract is 0.6,
of exactly two contracts is 0.15, and of exactly three
contracts is 0.04. What is the probability that the
firm will obtain at least one contracts? No contracts?
31. Quality Control An inspection of computers manufactured at a plant reveals that 2% of the monitors
are defective, 3% of the keyboards are defective,
and 1% of the computers have both defects.
a. Find the probability that a computer at this plant
has at least one of these defects.
b. Find the probability that a computer at this plant
has none of these defects.
32. Medicine A new medication produces headaches in
5% of the users, upset stomach in 15%, and both in
a. Find the probability that at least one of these side
effects occurs.
b. Find the probability that neither of these side effects occurs.
33. Manufacturing A manufactured item is guaranteed
for one year and has three critical parts. It has been
decided that during the first year the probability of
failure of the first part is 0.03, of the second part
0.02, the third part 0.01, both the first and second is
0.005, both the first and third is 0.004, both the second and third is 0.003, and all three parts 0.001.
a. What is the probability that exactly one of these
parts will fail in the first year?
b. What is the probability that at least one of these
parts will fail in the first year?
c. What is the probability that none of these parts
will fail in the first year?
34. Marketing A survey of business executives found
that 40% read Business Week, 50% read Fortune,
40% read Money, 17% read both Business Week and
Fortune, 15% read both both Business Week and
Money, 14% read both Fortune and Money, and 8%
read all three of these magazines.
Chapter 1 Sets and Probability
a. What is the probability that one of these executives reads exactly one of these three magazines?
b. What is the probability that one of these executives reads at least one of these three magazines?
c. What is the probability that one of these executives reads none of these three magazines?
35. Advertising A firm advertises three different products, A, B, and C, on television. From past experience, it expects 1.5% of listeners to buy exactly
one of the products, 1% to buy exactly two of the
products, 1.2% to buy A, 0.4% to buy both A and
B, 0.3% to buy both A and C, and 0.6% to buy A
but not the other two.
a. Find the probability that a listener will buy only
B or only C.
b. Find the probability that a listener will buy all
c. Find the probability that a listener will buy both
B and C.
d. Find the probability that a listener will buy none
of the three.
36. Sales A salesman always makes a sale at one of the
three stops in Atlanta and 30% of the time makes
a sale at only the first stop, 15% at only the second stop, 20% at only the third stop, and 35% of the
time at exactly two of the stops. Find the probability
that the salesman makes a sale at all three stops in
38. Let E and F be two events in S with P(E) = 0.3 and
P(F) = 0.4. Just how large could P(F ∪ F) possibly
39. You buy a new die and toss it 1000 times. A 1 comes
up 165 times. Is it true that the probability of a 1
showing on this die is 0.165?
40. A fair coin is to be tossed 100 times. Naturally you
expect tails to come up 50 times. After 60 tosses,
heads has come up 40 times. Is it true that now
heads is likely to come up less often than tails during the next 40 tosses?
41. You are playing a game at a casino and have correctly calculated the probability of winning any one
game to be 0.48. You have played for some time
and have won 60% of the time. You are on a roll.
Should you keep playing?
42. You are watching a roulette game at a casino and
notice that red has been coming up unusually often. (Red should come up as often as black.) Is it
true that, according to “the law of averages,” black
is likely to come up unusually often in the next number of games to “even things up”?
43. You buy a die and toss it 1000 times and notice that
a 1 came up 165 times. You decide that the probability of a 1 on this die is 0.165. Your friend takes this
die and tosses it 1000 times and notes that a 1 came
up 170 times. He concludes that the probability of a
1 is 0.17. Who is correct?
37. Let E and F be two events in S with P(E) = 0.5 and
P(F) = 0.7. Just how small could P(E ∩F) possibly
44. People who frequent casinos and play lotteries are
gamblers, but those who run the casinos and lotteries are not. Do you agree? Why or why not?
Solutions to Self-Help Exercises 1.5
1. If S = {a, b, c} and P(a) = P(b) = 2P(c), then
1 = P(a) + P(b) + P(c)
= P(a) + P(a) + 0.5P(a)
= 2.5P(a)
P(a) = 0.4
2. a. Let E be the event that the company obtains the first contract and let F
be the event that the company obtains the second contract. The event that
the company obtains the first contract but not the second is E ∩ F c , while
1.6 Conditional Probability
the event that the company obtains the second contract but not the first is
E c ∩ F. These two sets are mutually exclusive, so the probability that the
company receives exactly one of the contracts is
P(E ∩ F c ) + P(E c ∩ F)
Now P(E) = 0.40, P(F) = 0.30, and since E ∩ F is the event that the
company receives both contracts, P(E ∩ F) = 0.10. Notice on the accompanying diagram that E ∩ F c and E ∩ F are mutually disjoint and that
(E ∩ F c ) ∪ (E ∩ F) = E. Thus
P(E ∩ F c ) + P(E ∩ F) = P(E)
P(E ∩ F c ) + 0.10 = 0.40
P(E ∩ F c ) = 0.30
Also notice on the accompanying diagram that E c ∩ F and E ∩ F are mutually disjoint and that (E c ∩ F) ∪ (E ∩ F) = F. Thus
P(E c ∩ F) + P(E ∩ F) = P(F)
P(E c ∩ F) + 0.10 = 0.30
P(E ∩ F c ) = 0.20
Thus the probability that the company will receive exactly one of the
contracts is
P(E ∩ F c ) + P(E c ∩ F) = 0.30 + 0.20 = 0.50
b. The event that the company obtains neither contract is given by (E ∪ F)c .
From the diagram
P(E ∪ F) = 0.30 + 0.10 + 0.20 = 0.60
P((E ∪ F)c ) = 1 − P(E ∪ F) = 1 − 0.60 = 0.40
The probability that the company receives neither contract is 0.40.
Conditional Probability
Locating Defective Parts
A company makes the components for a product at a central location.
These components are shipped to three plants, 1 , 2, and 3, for assembly into a final product. The percentages of the product assembled by the
three plants are, respectively, 50%, 20%, and 30%. The percentages of defective products coming from these three plants are, respectively, 1%, 2%,
and 3%. What is the probability of randomly choosing a product made by
this company that is defective from Plant 1? See Example 4 for the answer.
Chapter 1 Sets and Probability
✧ Definition of Conditional Probability
The probability of an event is often affected by the occurrences of other events.
For example, there is a certain probability that an individual will die of lung
cancer. But if a person smokes heavily, then the probability that this person
will die of lung cancer is higher. That is, the probability has changed with the
additional information. In this section we study such conditional probabilities.
This important idea is further developed in the next section.
Given two events E and F, we call the probability that E will occur given that
F has occurred the conditional probability and write P(E|F). Read this as “the
probability of E given F .”
Conditional probability normally arises in situations where the old probability is “updated” based on new information. This new information is usually a
change in the sample space. Suppose, for example, a new family moves in next
door. The real estate agent mentions that this new family has two children. Based
on this information, you can calculate the probability that both children are boys.
Now a neighbor mentions that they met one child from the new family and this
child is a boy. Now the probability that both children from the new family are
boys has changed given this new information. We will find this new probability
in Example 2.
EXAMPLE 1
Finding Conditional Probability
A card is drawn randomly
from a deck of 52 cards.
a. What is the probability that this card is an ace?
b. What is the probability that this card is an ace given that the card is known to
be red and 10 or higher?
a. The uniform sample space consists of 52 cards and has the uniform probability. Thus, if E = {x|x is an ace},
P(E) =
b. Let F be the event in S consisting of all red cards 10 or higher. We have
n(S) = 52 and n(F) = 10. Since we are certain that the card chosen was red
and 10 or higher, our sample space is no longer the entire deck of 52 cards,
but simply those 10 cards in F. This will be our denominator in the ratio of
outcome in our event to outcomes in our sample space. The event ace and
known to be red and 10 or higher, has two outcomes as there are two red aces.
So the numerator will be n(E ∩ F) = 2. Putting this together we have the
probability of E given that F has occurred is
P(E|F) =
n(E ∩ F)
It can be helpful to divide the numerator and denominator in the last fraction
by n(S) and obtain
P(E|F) =
n(E ∩ F)
1.6 Conditional Probability
But this is just
P(E|F) =
P(E ∩ F)
This motivates the following definition for any event E and F in a sample
space S.
Conditional Probability
Let E and F be two events in a sample space S. The conditional probability that E occurs given that F has occurred is defined to be
P(E|F) =
P(E ∩ F)
It is worthwhile to notice that the events E, F, and E ∩ F are all in
the original sample space S and P(E), P(F), and P(E ∩ F) are all probabilities
defined on S. However, we can think of P(E|F) as a probability defined on the
new sample space S = F. See Figure 1.23.
Figure 1.23
EXAMPLE 2
Calculating a Conditional Probability A new family has moved
in next door and is known to have two children. Find the probability that both
children are boys given that at least one is a boy. Assume that a boy is as likely
as a girl.
Let E be the event that both children are a boy and F the event that
at least one is a boy. Then
S = {BB, BG, GB, GG}, E = {BB}, F = {BB, BG, GB}, E ∩ F = {BB}
P(E|F) =
P(E ∩ F) 1/4 1
✧ The Product Rule
We will now see how to write P(E ∩ F) in terms of a product of two probabilities.
From the definition of conditional probability, we have
P(E|F) =
P(E ∩ F)
P(F|E) =
P(F ∩ E)
Chapter 1 Sets and Probability
if P(E) > 0 and P(F) > 0. Solving for P(E ∩ F) and P(F ∩ E), we obtain
P(E ∩ F) = P(F)P(E|F),
P(F ∩ E) = P(E)P(F|E)
Since P(E ∩ F) = P(F ∩ E), it follows that
P(E ∩ F) = P(F)P(E|F) = P(E)P(F|E)
This is called the product rule.
Product Rule
If E and F are two events in a sample space S with P(E) > 0 and P(F) > 0,
P(E ∩ F) = P(F)P(E|F) = P(E)P(F|E)
EXAMPLE 3 Using the Product Rule Two bins contain transistors. The first
bin has 5 defective and 15 non-defective transistors while the second bin has 3
defective and 17 non-defective transistors. If the probability of picking either bin
is the same, what is the probability of picking the first bin and a good transistor?
Solution The sample space is S = {1D, 1N, 2D, 2N} where the number refers
to picking the first or second bin and the letter refers to picking a defective (D) or
non-defective (N) transistor.
If E is the event “pick the first bin” and F is the event “pick a non-defective
transistor,” then E = {1D, 1N} and F = {1N, 2N}. The probability of picking a
non-defective transistor given that the first bin has been picked is the conditional
probability P(F|E) = 15/20. The event “picking the first bin and a non-defective
transistor” is E ∩ F. From the product rule
P(E ∩ F) = P(E)P(F|E) =
✧ Probability Trees
We shall now consider a finite sequence of experiments in which the outcomes
and associated probabilities of each experiment depend on the outcomes of the
preceding experiments. For example, we can choose a card from a deck of cards,
place the picked card on the table, and then pick another card from the deck. This
process could continue until all the cards are picked.
Such a finite sequence of experiments is called a finite stochastic process.
Stochastic processes can be effectively described by probability trees that we
now consider. The following example should be studied carefully since we will
return to it in the next section.
Using a Probability Tree A company makes the components for
a product at a central location. These components are shipped to three plants, 1, 2,
and 3, for assembly into a final product. The percentage of the product assembled
by the three plants are, respectively, 50%, 20%, and 30%. The percentages of
defective products coming from these three plants are, respectively, 1%, 2%, and
EXAMPLE 4
1.6 Conditional Probability
a. What is the probability of randomly choosing a product made by this company that is defective from Plant 1?, 2?, 3?
b. What is the probability of randomly choosing a product made by this company that is defective?
Figure 1.24
Solution The probability of the part being assembled in Plant 1 is P(1) = 0.5,
in Plant 2 is P(2) = 0.2, and in Plant 3 is P(3) = 0.3. We begin our tree with this
information as shown in Figure 1.24. Notice how this is similar to the trees drawn
in earlier sections with the only difference being the probability of the outcome
being placed on the branch leading to that outcome.
We are also given the conditional probabilities, P(D|1) = 0.01, P(D|2) =
0.02, and P(D|3) = 0.03. The branch leading from Plant 1 to a defective item
tells us that we are referring to items from Plant 1 and therefore the number that
should be placed on this branch is the probability the component is defective
given that it was made at Plant 1, P(D|1) = 0.01. Continue with the conditional
probabilities given for the other plants. This is all shown in the tree diagram in
Figure 1.24.
a. Notice that the product rule, P(1 ∩ D) = P(1)P(D|1), represents multiplying
along the branches 1 and D. So using the product rule we have
P(1 ∩ D) = P(1)P(D|1) = (0.5)(0.01) = 0.005 = 0.5%
P(2 ∩ D) = P(2)P(D|2) = (0.2)(0.02) = 0.004 = 0.4%
P(3 ∩ D) = P(3)P(D|3) = (0.3)(0.03) = 0.009 = 0.9%
b. For a component to be defective it came from Plant 1 or Plant 2 or Plant 3.
These events are mutually exclusive since a component can only come from
one plant. We have the following:
P(D) = P(1 ∩ D) + P(2 ∩ D) + P(3 ∩ D)
= 0.005 + 0.004 + 0.009 = 0.018
EXAMPLE 5
Using a Probability Tree A box contains three blue marbles
and four red marbles. A marble is selected at random until a red one is picked.
a. What is the probability that the number of marbles selected is one?
b. What is the probability that the number of marbles selected is two?
c. What is the probability that the number of marbles selected is three?
The process for selecting a marble one at a time from the box is
shown in Figure 1.25.
a. On the first draw there are four red marbles in a box of seven. So the probability of selecting a red marble is 4/7. We note for further reference that the
probability of selecting a blue marble is 3/7. These probabilities are on the
first legs of the tree diagram.
b. The only way it can take two selections to obtain a blue marble is for the first
selection be a blue marble. In this case there are two blue marbles and four
red marbles. Thus, the probability of selecting a red marble given that a blue
marble was chosen first is 4/6.
Chapter 1 Sets and Probability
From the tree in Figure 1.25, the probability of the branch B1 R2 is (3/7) ·
(4/6) = 2/7. We note for further reference that the probability of selecting a
blue marble on the second selection given that a blue marble was selected first
is 2/6. See Figure 1.25.
Figure 1.25
c. The only way it can take three selections to obtain a red marble is for the
first two selections to be blue marbles. At this point (with two blue marbles
missing) there is one blue marble and four red marbles left in the box. The
probability of selecting a red marble given that a blue marble was chosen the
first and second time is 4/5. Now the probability of the branch B1 B2 R3 can be
found by applying the product rule (multiply along the branches) to find
· · =
✧ Independent Events
We say that two events E and F are independent if the outcome of one does
not affect the outcome of the other. For example, the probability of obtaining a
head on a second flip of a coin is independent from what happened on the first
flip. This is intuitively clear since the coin cannot have any memory of what
happened on the first flip. Indeed the laws of physics determine the probability of
heads occurring. Thus for the probability of heads to be different on the second
flip the laws of physics must be different on the second flip. On the other hand,
if we are selecting cards one at time without replacement from a standard deck
of cards, the probability of selecting the queen of spades on the second draw
clearly depends on what happens on the first draw. If the queen of spades was
already picked, the probability of picking her on the second draw would be zero.
Thus drawing the queen of spades on the second draw without replacement is not
independent from drawing the queen of spades on the first draw.
That is, two events E and F are independent if
P(E|F) = P(E)
P(F|E) = P(F)
In words, the probability of E given that F has occurred is the same probability of
E if F had not occurred. Similarly, the probability of F given that E has occurred
is just the probability of F if E had not occurred.
1.6 Conditional Probability
Independent Events
Two events E and F are said to be independent if
P(E|F) = P(E)
P(F|E) = P(F)
We shall now obtain a result that is more convenient to apply when attempting
to determine if two events are independent. It will also be useful when finding
the probability that a series of independent events occurred.
If then two events E and F are independent, the previous comments together
with the product rule indicate that
P(E ∩ F) = P(E|F)P(F) = P(E)P(F)
Now consider the case that P(E) > 0 and P(F) > 0. Assume that P(E ∩ F) =
P(E)P(F), then
P(E|F) =
P(E ∩ F) P(E)P(F)
= P(E)
P(F|E) =
P(E ∩ F) P(E)P(F)
= P(F)
This discussion then yields the following theorem.
Independent Events Theorem
Let E and F be two events with P(E) > 0 and P(F) > 0. Then E and F
are independent if, and only if,
P(E ∩ F) = P(E)P(F)
Although at times one is certain whether or not two events are independent,
often one can only tell by doing the calculations.
EXAMPLE 6 Smoking and Heart Disease A study of 1000 men over 65 indicated that 250 smoked and 50 of these smokers had some signs of heart disease,
while 100 of the nonsmokers showed some signs of heart disease. Let E be the
event “smokes” and H be the event “has signs of heart disease.” Are these two
events independent?
The Venn diagram is given in Figure 1.26. From this diagram P(H) =
0.15, P(E) = 0.25, and P(H ∩ E) = 0.05. Thus
P(H)P(E) = (0.15)(0.25) = 0.0375
P(H ∩ E) = 0.05
Figure 1.26
Since 0.0375 = 0.05, these two events are not independent.
EXAMPLE 7
Determining if Two Events Are Independent In a medical trial
a new drug was effective for 60% of the patients and 30% of the patients suffered
Chapter 1 Sets and Probability
from a side effect. If 28% of the patients did not find the drug effective nor had a
side effect, are the events E (drug was effective) and F (patient had a side effect)
We can organize this information in a Venn diagram. If 28% of the
patients did not find the drug effective nor had a side effect, this means that 0.28
is placed in the region outside the E and F circles. See Figure 1.27. The region
inside will have the probability (E ∪ F) = 1 − 0.28 = 0.72. Since we are not told
how many found the drug effective and had a side effect, P(E ∩ F), we will need
to use the union rule to find this number,
P(E ∪ F) = P(E) + P(F) − P(E ∩ F)
0.72 = 0.6 + 0.3 − P(E ∩ F)
P(E ∩ F) = 0.9 − 0.72 = 0.18
Figure 1.27
Place this in the Venn diagram in Figure 1.27. For completeness we could
find P(E ∩ F c ) = 0.6 − 0.18 = 0.42 and P(E c ∩ F) = 0.3 − 0.18 = 0.12 and place
those in the diagram. However, to check for independence we need only place
the three values needed in the formula:
P(E)P(F) = (0.60)(0.30) = 0.18
P(E ∩ F) = 0.18
Since 0.18 = 0.18, these events are independent.
Notice that saying two events are independent is not the same as
saying that they are mutually exclusive. The sets in both previous examples were
not mutually exclusive, but in one case the sets were independent and in the other
case they were not.
The notion of independence can be extended to any number of finite events.
Independent Set of Events
A set of events {E1 , E2 , . . . , En } is said to be independent if, for any k of
these events, the probability of the intersection of these k events is the
product of the probabilities of each of the k events. This must hold for any
k = 2, 3, . . . , n.
For example, for the set of events {E, F, G} to be independent all of the following must be true:
P(E ∩ F) = P(E)P(F),
P(F ∩ G) = P(F)P(G),
P(E ∩ G) = P(E)P(G)
P(E ∩ F ∩ G) = P(E)P(F)P(G)
It is intuitively clear that if two events E and F are independent, then so also
are E and F c , E c and F, E c and F c . (See Exercises 52 and 53.) Similar statements
are true about a set of events.
An aircraft has a system of
three computers, each independently able to exercise control of the flight. The
EXAMPLE 8
Independent Events and Safety
1.6 Conditional Probability
computers are considered 99.9% reliable during a routine flight. What is the
probability of having a failure of the control system during a routine flight?
Solution Let the events Ei , i = 1, 2, 3 be the three events given by the reliable
performance of respectively the first, second, and third computer. Since the set
of events {E1 , E2 , E3 } is independent, so is the set of events {E1c , E2c , E3c }. The
system will fail only if all three computers fail. Thus the probability of failure of
the system is given by
{E1c ∩ E2c ∩ E3c } = P(E1c )P(E2c )P(E3c ) = (0.001)3
which, of course, is an extremely small number.
Broken Elevators A building has three elevators. The chance
that elevator A is not working is 12%, the chance that elevator B is not working
is 15% and the chance that elevator C is not working is 9%. If these probabilities
are independent, what is the probability that exactly one elevator is not working?
EXAMPLE 9
The probability that elevator A is not working but the other two
are working is (0.12)(0.85)(0.91). The probability that only elevator B is
working is (0.88)(0.15)(0.91) and only elevator C working has probability
(0.88)(0.85)(0.09) of occurring. The probability that exactly one does not work
is the sum of these three probabilities:
P = (0.12)(0.85)(0.91) + (0.88)(0.15)(0.91) + (0.88)(0.85)(0.09)
= 0.28026
Self-Help Exercises 1.6
1. Three companies A, B, and C, are competing for
a contract. The probabilities that they receive the
contract are, respectively, P(A) = 1/6, P(B) = 1/3,
and P(C) = 1/2. What is the probability that A will
receive the contract if C pulls out of the bidding?
2. Two bins contain transistors. The first has 4 defective and 15 non-defective transistors, while the
second has 3 defective and 22 non-defective ones.
If the probability of picking either bin is the same,
what is the probability of picking the second bin and
a defective transistor?
3. Success is said to breed success. Suppose you are
in a best of 3 game tennis match with an evenly
matched opponent. However, if you win a game,
your probability of winning the next increases from
1/2 to 2/3. Suppose, however, that if you lose, the
probability of winning the next match remains the
same. (Success does not breed success for your opponent.) What is your probability of winning the
match? Hint: Draw a tree.
4. A family has three children. Let E be the event “at
most one boy” and F the event “at least one boy and
at least one girl.” Are E and F independent if a boy
is as likely as a girl? Hint: Write down every element in
the sample space S and the events E, F, and E ∩ F, and find the
appropriate probabilities by counting.
Chapter 1 Sets and Probability
1.6 Exercises
In Exercises 1 through 6, refer to the accompanying
Venn diagram to find the conditional probabilities.
19. P(E ∩ F c ) = 0.3, P(E ∩ F) = 0.3,
P(E c ∩ F) = 0.2
20. P(E) = 0.2, P(F) = 0.5, P(E ∪ F) = 0.6
21. A pair of fair dice is tossed. What is the probability
that a sum of seven has been tossed if it is known
that at least one of the numbers is a 3.
1. a. P(E|F)
b. P(F|E)
2. a. P(E c |F)
b. P(F c |E)
3. a. P(E|F c )
b. P(F|E c )
4. a. P(E c |F c )
b. P(F c |E c )
5. a. P(F|E ∩ F)
b. P(F c |F)
6. a. P(E c ∩ F|F)
b. P(E ∩ F c |F)
In Exercises 7 through 12, let P(E) = 0.4, P(F) = 0.6,
and P(E ∩ F) = 0.2. Draw a Venn diagram and find the
conditional probabilities.
7. a. P(E c |F)
b. P(F c |E)
8. a. P(E|F)
b. P(F|E)
9. a. P(F c |E c )
b. P(E ∪ F|E)
10. a. P(E|F c )
b. P(F|E c )
11. a. P(E c ∩ |F)
b. P(E ∩ F c |E)
12. a. P(F|E ∩ F)
b. P(E c |E)
In Exercises 13 through 20, determine if the given events
E and F are independent.
13. P(E) = 0.3, P(F) = 0.5, P(E ∩ F) = 0.2
14. P(E) = 0.5, P(F) = 0.7, P(E ∩ F) = 0.3
15. P(E) = 0.2, P(F) = 0.5, P(E ∩ F) = 0.1
16. P(E) = 0.4, P(F) = 0.5, P(E ∩ F) = 0.2
17. P(E) = 0.4, P(F) = 0.3, P(E ∪ F) = 0.6
18. P(E ∩ F c ) = 0.3, P(E ∩ F) = 0.2,
P(E c ∩ F) = 0.2
22. A single fair die is tossed. What is the probability that a 3 occurs on the top if it is known that the
number is a prime?
23. A fair coin is flipped three times. What is the probability that heads occurs three times if it is known
that heads occurs at least once?
24. A fair coin is flipped four times. What is the probability that heads occurs three times if it is known
that heads occurs at least twice?
25. Three cards are randomly drawn without replacement from a standard deck of 52 cards.
a. What is the probability of drawing an ace on the
third draw?
b. What is the probability of drawing an ace on the
third draw given that at least one ace was drawn
on the first two draws?
26. Three balls are randomly drawn from an urn that
contains four white and six red balls.
a. What is the probability of drawing a red ball on
the third draw?
b. What is the probability of drawing a red ball on
the third draw given that at least one red ball was
drawn on the first three draws?
27. From the tree diagram find
a. P(A ∩ E)
b. P(A)
c. P(A|E)
1.6 Conditional Probability
28. From the tree diagram find
a. P(A ∩ E)
b. P(A)
c. P(A|E)
products 1% of the time, the second line 2% of the
time, and the third 3% of the time.
a. What is the probability that a defective product
is produced at this plant given that it was made
on the second assembly line?
b. What is the probability that a defective product
is produced at this plant?
29. An urn contains five white, three red, and two blue
balls. Two balls are randomly drawn. What is the
probability that one is white and one is red if the
balls are drawn
a. without replacement?
b. with replacement after each draw?
30. An urn contains four white and six red balls. Two
balls are randomly drawn. If the first one is white,
the ball is replaced. If the first one is red, the ball is
not replaced. What is the probability of drawing at
least one white ball?
31. In a family of four children, let E be the event “at
most one boy” and F the event “at least one girl and
at least one boy.” If a boy is as likely as a girl, are
these two events independent?
32. A fair coin is flipped three times. Let E be the event
“at most one head” and F the event “at least one
head and at least one tail.” Are these two events independent?
33. The two events E and F are independent with
P(E) = 0.3 and P(F) = 0.5. Find P(E ∪ F).
34. The two events E and F are independent with
P(E) = 0.4 and P(F) = 0.6. Find P(E ∪ F).
35. The three events E, F, and G are independent with
P(E) = 0.2, P(F) = 0.3, and P(G) = 0.5. What is
P(E ∪ F ∪ G)?
36. The three events E, F, and G are independent with
P(E) = 0.3, P(F) = 0.4, and P(G) = 0.6. What is
P(E c ∪ F c ∪ Gc )?
37. Manufacturing A plant has three assembly lines
with the first line producing 50% of the product and
the second 30%. The first line produces defective
38. Manufacturing Two machines turn out all the products in a factory, with the first machine producing
40% of the product and the second 60%. The first
machine produces defective products 2% of the time
and the second machine 4% of the time.
a. What is the probability that a defective product
is produced at this factory given that it was made
on the first machine?
b. What is the probability that a defective product
is produced at this factory?
39. Suppliers A manufacturer buys 40% of a certain
part from one supplier and the rest from a second
supplier. It notes that 2% of the parts from the first
supplier are defective, and 3% are defective from
the second supplier. What is the probability that a
part is defective?
40. Advertising A television ad for a company’s product has been seen by 20% of the population. Of
those who see the ad, 10% then buy the product. Of
those who do not see the ad, 2% buy the product.
Find the probability that a person buys the product.
41. Reliability A firm is making a very expensive optical lens to be used in an earth satellite. To be assured that the lens has been ground correctly, three
independent tests using entirely different techniques
are used. The probability is 0.99 that any of one of
these tests will detect a defect in the lens. What is
the probability that the lens has a defect even though
none of the three tests so indicates?
42. Psychology and Sales A door-to-door salesman
expects to make a sale 10% of the time when starting the day. But making a sale increases his enthusiasm so much that the probability of a sale to the next
customer is 0.2. If he makes no sale, the probability
for a sale stays at 0.l. What is the probability that
he will make at least two sales with his first three
43. Quality Control A box contains two defective (D)
parts and five non-defective (N) ones. You randomly select a part (without replacement) until you
Chapter 1 Sets and Probability
get a non-defective part. What is the probability that
the number of parts selected is
a. one
b. two
c. three?
44. Sales A company sells machine tools to two firms
in a certain city. In 40% of the years it makes a sale
to the first firm, in 30% of the years to the second
firm, and in 10% to both. Are the two events “a
sale to the first firm” and “a sale to the second firm”
What must x be for the two events “gets cancer” and
“work for the chemical plant” to be independent?
51. Given the probabilities shown in the accompanying
Venn diagram, show that the events E and F are independent if, and only if,
p1 p3 = p2 p4
What must the Venn diagram look like if the sets are
mutually disjoint?
45. Medicine In a study of 250 men over 65, 100
smoked, 60 of the smokers had some signs of heart
disease, and 90 of the nonsmokers showed some
signs of heart disease. Let E be the event “smokes”
and H be the event “has signs of heart disease.” Are
these two events independent?
46. Contracts A firm has bids on two contracts. It is
known that the awarding of these two contracts are
independent events. If the probability of receiving
the contracts are 0.3 and 0.4, respectively, what is
the probability of not receiving either?
52. Show that if E and F are independent, then so are E
and F c .
53. Show that if E and F are independent, then so are
E c and F c .
47. Stocks A firm checks the last 200 days on which
its stock has traded. On 100 of these occasions the
stock has risen in price with a broad-based market
index also rising on 70 of these particular days. The
same market index has risen 90 of the 200 trading
days. Are the movement of the firm’s stock and the
movement of the market index independent?
54. Show that two events are independent if they are
mutually exclusive and the probability of one of
them is zero.
48. Bridges Dystopia County has three bridges. In the
next year, the Elder bridge has a 15% chance of collapse, the Younger bridge has a 5% chance of collapse and the Ancient bridge has a 20% chance of
collapse. What is the probability that exactly one
bridge will collapse in the next year?
56. Show that if E and F are independent events, then
49. Missing Parts A store sells desk kits. In each kit
there is a 2% chance that a screw is missing, a 3%
chance that a peg is missing, and a 1% chance that
a page of directions is missing. If these events are
independent, what is the probability that a desk kit
has exactly one thing missing?
50. Medicine The probability of residents of a certain
town contracting cancer is 0.01. Let x be the percent of residents that work for a certain chemical
plant and suppose that the probability of both working for this plant and of contracting cancer is 0.001.
55. Show that two events E and F are not independent if
they are mutually exclusive and both have nonzero
P(E ∪ F) = 1 − P(E c )P(F c )
57. If P(F) > 0, then show that
P(E c |F) = 1 − P(E|F)
58. If E, F, and G are three events and P(G) > 0, show
P(E ∪ G|G) = P(E|G) + P(F|G) − P(E ∩ F|G)
59. If E and F are two events with F ⊂ E, then show
that P(E|F) = 1.
60. If E and F are two events with E ∩ F = 0,
/ then show
that P(E|F) = 0.
61. If E and F are two events, show that
P(E|F) + P(E c |F) = 1
1.6 Conditional Probability
Solutions to Self-Help Exercises 1.6
1. If E is the event that A obtains the contract and F the event that either A or
B obtain the contract, then E = {A}, F = {A, B}, and E ∩ F = {A}, and the
conditional probability that A will receive the contract if C pulls out of the
bidding is
P(E ∩ F) 1/6 1
P(E|F) =
2. The sample space is S = {1D, 1N, 2D, 2N} where the number refers to picking the first or second bin and the letter refers to picking a defective (D) or
non-defective (N) transistor. If E is picking the first bin and F is picking a
defective transistor then
P(E ∩ F) = P(E)P(F|E) =
3. The appropriate tree is given.
The probability of winning the match is then
+ + =
4. The elements in the spaces S, E, and F are
S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}
E = {BGG, GBG, GGB, GGG}
F = {BBG, BGB, BGG, GBB, GBG, GGB}
With E ∩ F = {BGG, GBG, GGB}. Thus counting elements gives
P(E ∩ F) =
P(E)P(F) =
· =
Since these two numbers are the same, the two events are independent.
Chapter 1 Sets and Probability
Bayes’ Theorem
Probability of a Defective Part
Recall Example 4 of the last section. A company makes the components
for a product at a central location. These components are shipped to three
plants, 1, 2, and 3, for assembly into a final product. The percentages
of the product assembled by the three plants are, respectively, 50%, 20%,
and 30%. The percentages of defective products coming from these three
plants are, respectively, 1%, 2%, and 3%. Given a defective product, what
is the probability it was assembled at Plant 1? At Plant 2? At Plant 3? See
Example 1 for the answers to these questions.
✧ Bayesian Reasoning
Figure 1.28a
Figure 1.28b
We have been concerned with finding the probability of an event that will occur
in the future. We now look at calculating probabilities after the events have occurred. This is a surprisingly common and important application of conditional
probability. For example, when an e-mail message is received, the e-mail program does not know if the message is junk or not. But, given the number of
misspelled words or the presence of certain keywords, the message is probably
junk and is appropriately filtered. This is called Bayesian filtering and knowledge of what happened second (the misspelled words) allows an estimate of what
happened first (the mail is junk).
We will now consider a situation where we will use Bayesian reasoning to
estimate what happened in the first part of an experiment when you only know the
results of the second part. Your friend has a cup with five green marbles and two
red marbles and a bowl with one green, and three red, as shown in Figure 1.28a.
There is an equal chance of choosing the cup or the bowl. After the cup or bowl
has been chosen, a marble is selected from the container. A tree diagram for
this experiment is shown in Figure 1.28b. Notice that on the branches of the
tree diagram we have the conditional probabilities such as P(G|C) which is the
probability you choose a green marble given you are choosing from the cup.
Now your friend hides the bowl and cup and performs the experiment. He
then shows you he has picked a green marble and asks, “what is the probability
this green marble came from the cup”? That is, what is P(C|G)? Notice this is
not what we have on the tree diagram. How can we figure this out?
Intuitively, if you had to guess if it was more likely to have come from the cup
or the bowl, you would say it came from the cup. Looking at the figure, the cup
has relatively more green ones, and if it was equally likely to have been picked
from the cup or bowl, given it is green, more likely than not it came from the cup.
But how can we get an exact value for the probability? Recall the formula used
in the previous section for conditional probability,
P(C|G) =
P(C ∩ G)
We can find P(C ∩ G) from the product rule. From multiplying along the
branches it is
P(C ∩ G) = P(C) · P(G|C) = · =
1.7 Bayes’ Theorem
What about the denominator, P(G)? We can get green marbles from the cup
or green marbles from the bowl. The total probability will be the sum of the
probabilities of those two mutually exclusive events:
P(G) = P(G ∩C) + P(G ∩ B) =
· + · =
Putting the pieces together we have
P(C|G) =
≈ 0.74
So there was about a 74% chance that the marble came from the cup, given
that it was green.
This is an example of Bayes’ theorem. This theorem was discovered by the
Presbyterian minister Thomas Bayes (1702–1763). We now state this in a more
general form.
✧ Bayes’ Theorem
We suppose that we are given a sample space S and three mutually exclusive
events E1 , E2 , and E3 , with E1 ∪ E2 ∪ E3 = S as indicated in Figure 1.29. Notice
that the three events divide the same space S into 3 partitions. Given another
event F, the tree diagram of possibilities is shown in Figure 1.30.
Figure 1.29
Figure 1.30
The probability P(F|E1 ) is the fraction with the numerator given by the probability off the branch E1 ∩ F while the denominator is the sum of all the probabilities of all branches that end in F. This is
P(E1 |F) =
P(E1 )P(F|E1 )
P(E1 )P(F|E1 ) + P(E2 )P(F|E2 ) + P(E3 )P(F|E3 )
The same for P(F|E2 ).
P(E2 |F) =
P(E2 )P(F|E2 )
P(E1 )P(F|E1 ) + P(E2 )P(F|E2 ) + P(E3 )P(F|E3 )
and so on. We then state Bayes’ theorem in even more general form.
Chapter 1 Sets and Probability
Bayes’ Theorem
Let E1 , E2 , . . . , En , be mutually exclusive events in a sample space S with
E1 ∪ E2 ∪ . . . ∪ En = S. If F is any event in S, then for i = 1, 2, . . . , n,
probability of branch Ei ∩ F
sum of all probabilities of all branches that end in F
P(Ei )P(F|Ei )
P(E1 )P(F|E1 ) + P(E2 )P(F|E2 ) + · · · + P(En )P(F|En )
P(Ei |F) =
The formula above looks quite complex. However, Bayes’ theorem
is much easier to remember by simply keeping in mind the original formula is
simply P(E|F) = P(E∩F)/P(F) and that sometimes to find P(F) you need to add
together all the different ways that F can occur.
EXAMPLE 1 Defective Components At the start of this section a question
was posed to find the probability that a defective component came from a certain
plant. Find these probabilities.
Refer to Figure 1.31 for the probabilities that are given. Using the
product rule we have
P(1 ∩ D) = P(1)P(D|1) = (0.5)(0.01) = 0.005
P(2 ∩ D) = P(2)P(D|2) = (0.2)(0.02) = 0.004
P(3 ∩ D) = P(3)P(D|3) = (0.3)(0.03) = 0.009
and thus
Figure 1.31
P(D) = = P(1 ∩ D) + P(2 ∩ D) + P(3 ∩ D)
= 0.005 + 0.004 + 0.009
= 0.018
Using this information and the definition of conditional probability, we then have
P(1|D) =
P(1 ∩ D)
0.005 + 0.004 + 0.009 0.018 18
We now notice that the numerator of this fraction is the probability of the branch
1 ∩ D, while the denominator is the sum of all the probabilities of all branches
that end in D. We also have
P(2|D) =
P(2 ∩ D) 0.004
0.018 18
Now notice that the numerator of this fraction is the probability of the branch
2 ∩ D, while the denominator is the sum of all the probabilities of all branches
that end in D. A similar statement can be made for
P(3|D) =
P(3 ∩ D) 0.009
0.018 18
A very interesting example occurs in medical tests for disease. All medical
tests have what are called false positives and false negatives. That is, a test result
could come back positive and the patient does not have the disease or a test could
1.7 Bayes’ Theorem
come back negative and the patient does have the disease. Many modern tests
have low rates of false positives and false negatives, but even then there can be
difficulties with a diagnosis. We begin with a test that gives every appearance of
being excellent, but an important consequence may be disappointing.
EXAMPLE 2 A Medical Application of Bayes’ Theorem The standard tine
test for tuberculosis attempts to identify carriers, that is, people who have been
infected by the tuberculin bacteria. The probability of a false negative is 0.08, that
is, the probability of the tine test giving a negative reading to a carrier is P(−|C) =
0.08. The probability of a false positive is 0.04, that is, the probability of the tine
test giving a positive indication when a person is a non-carrier is P(+|N) = 0.04.
The probability of a random person in the United States having tuberculosis
is 0.0075. Find the probability that a person is a carrier given that the tine test
gives a positive indication.
The probability we are seeking is P(C|+). Figure 1.32 shows the
appropriate diagram where C is the event “is a carrier,” N the event “is a noncarrier,” + the event “test yields positive result,” and − is the event “test is negative.”
Then Bayes’ theorem can be used and P(C|+) is the probability of branch
C ∩ + divided by the sum of all probabilities that end in +. Then
Figure 1.32
P(C|+) =
P(C ∩ +)
P(C)P(+|C) + P(N)P(+|N)
≈ 0.15
(0.0075)(0.92) + (0.9925)(0.04)
So only 15% of people with positive tine test results actually carry TB.
This number is surprisingly low. Does this indicate that the test is
of little value? As Self-Help Exercise 1 will show, a person whose tine test is
negative has a probability of 0.999 of not having tuberculosis. Such an individual
can feel safe. The individuals whose tine test is positive are probably all right
also but will need to undergo further tests, such as a chest x-ray.
In some areas of the United States the probability of being a carrier can be
as high as 0.10. The following example examines the tine test under these conditions.
EXAMPLE 3
A Medical Application of Bayes’ Theorem Using the informa-
tion found in Example 2, find P(C|+) again when P(C) = 0.10.
See Figure 1.33 for the tree diagram with the probabilities. Using
Bayes’ theorem exactly as before, we obtain
P(C|+) =
Figure 1.33
P(C)P(+|C) + P(N)P(+|N)
≈ 0.72
(0.10)(0.92) + (0.90)(0.04)
Chapter 1 Sets and Probability
Thus 72% of these individuals who have a positive tine test result are carriers. ✦
Thus in the first example, when the probability of being a carrier is low,
the tine test is useful for determining those who do not have TB. In the second
example, when the probability of being a carrier is much higher, the tine test
is useful for determining those who are carriers, although, naturally, these latter
individuals will undergo further testing.
EXAMPLE 4
An Application of Bayes’ Theorem Suppose there are only
four economic theories that can be used to predict expansions and contractions in
the economy. By polling economists on their beliefs on which theory is correct,
the probability that each of the theories is correct has been determined as follows:
P(E1 ) = 0.40, P(E2 ) = 0.25, P(E3 ) = 0.30, P(E4 ) = 0.05
The economists who support each theory then use the theory to predict the likelihood of a recession (R) in the next year. These are as follows:
P(R|E1 ) = 0.01, P(R|E2 ) = 0.02, P(R|E3 ) = 0.03, P(R|E4 ) = 0.90
Now suppose a recession actually occurs in the next year. How would the probabilities of the correctness of the fourth and first theories be changed?
We first note that the fourth theory E4 , has initially a low probability
of being correct. Also notice that this theory is in sharp disagreement with the
other three on whether there will be a recession in the next year.
Bayes’ theorem gives P(E4 |R) the probability of the branch E4 ∩ R divided by
the sum of all the probabilities of all the branches that end in R. See Figure 1.34.
Similarly for the other theories. Thus P(E4 |R) and and P(E1 |R) are
Figure 1.34
P(E4 |R) =
P(E4 ∩ R)
P(E4 )P(R|E4 )
P(E1 )P(R|E1 ) + P(E2 )P(R|E2 ) + P(E3 )P(R|E3 ) + P(E4 )P(R|E4 )
(0.40)(0.01) + (0.25)(0.02) + (0.30)(0.03) + (0.05)(0.90)
≈ 0.71
P(E1 |R) =
P(E1 ∩ R)
P(E1 )P(R|E1 )
P(E1 )P(R|E1 ) + P(E2 )P(R|E2 ) + P(E3 )P(R|E3 ) + P(E4 )P(R|E4 )
(0.40)(0.01) + (0.25)(0.02) + (0.30)(0.03) + (0.05)(0.90)
≈ 0.06
Thus, given that the recession did occur in the next year, the probability that E4
is correct has jumped, while the probability that E1 is true has plunged.
Although this is an artificial example probabilities indeed are revaluated in
this way based on new information.
1.7 Bayes’ Theorem
Self-Help Exercises 1.7
1. Referring to Example 2 of the text, find the probability that an individual in the United States is not a
carrier given that the tine test is negative.
cessful. What is the probability that a successful
member had Bertha as her trainer?
2. A gym has three trainers, Aldo, Bertha, and Coco,
who each have 1/3 of the new members. A person
who trains with Aldo has a 60% chance of being
successful, a person who trains with Bertha has a
30% chance of being successful and a person who
trains with Coco has a 90% chance of being suc-
3. A purse has three nickels and five dimes. A wallet
has two nickels and one dime. A coin is chosen at
random from the purse and placed in the wallet. A
coin is then drawn from the wallet. If a dime is chosen from the wallet, what is the probability that the
transferred coin was a nickel?
1.7 Exercises
1. Find P(E1 |F) and P(E2 |F c ) using the tree diagram
urn is selected, and a ball is randomly drawn from the
selected urn. The probability of selecting the first urn is
5. If the ball is white, find the probability that the first
urn was selected.
6. If the ball is white, find the probability that the second urn was selected.
2. Find P(E1 |F c ) and P(E2 |F) using the tree diagram
in the previous exercise.
3. Find P(E1 |F) and P(E1 |F c ) using the tree diagram
7. If two balls are drawn from the selected urn without
replacement and both are white, what is the probability that the urn selected was the
a. the first one?
b. the second one?
8. A ball is drawn from the selected urn and replaced.
Then another ball is drawn and replaced from the
same urn. If both balls are white, what is the probability that the urn selected was
a. the first one?
b. the second one?
4. Find P(E3 |F) and P(E3 |F c ) using the tree diagram
in the previous exercise.
Exercises 5 through 8 refer to two urns that each contain
10 balls. The first urn contains 2 white and 8 red balls.
The second urn contains 7 white and 3 red balls. An
Exercises 9 through 12 refer to three urns that each contain 10 balls. The first contains 2 white and 8 red balls,
the second 5 white and 5 red, and the third all 10 white.
Each urn has an equal probability of being selected. After an urn is selected a ball is randomly drawn from this
9. If the ball drawn was white, find the probability that
the first urn was selected.
Chapter 1 Sets and Probability
10. If the ball drawn was white, find the probability that
the third urn was selected.
11. Suppose two balls are drawn from the selected urn
without replacement and both are white. What is the
probability that the urn selected was
a. the first one?
b. the second one?
12. Now a ball is drawn, replaced, and then another
drawn. Suppose both are white. What is the probability that the urn selected was
a. the first one?
b. the second one?
Exercises 13 through 16 refer to the following experiment: A box has two blue and six green jelly beans. A
bag has five blue and four green jelly beans. A jelly
bean is selected at random from the box and placed in
the bag. Then a jelly bean is selected at random from
the bag.
21. Manufacturing A plant has three assembly lines
with the first line producing 50% of the product and
the second 30%. The first line produces defective
products 1% of the time, the second line 2% of the
time, and the third 3% of the time. Given a defective product, what is the probability it was produced
on the second assembly line? See Exercise 37 of the
previous section.
22. Manufacturing Two machines turn out all the products in a factory, with the first machine producing
40% of the product and the second 60%. The first
machine produces defective products 2% of the time
and the second machine 4% of the time. Given a
defective product, what is the probability it was produced on the first machine? See Exercise 38 of the
previous section.
13. If a blue jelly bean is selected from the bag, what
is the probability that the transferred jelly bean was
23. Medicine Do Example 2 of the text if all the information remains the same except that the tine test has
the remarkable property that P(+|C) = 1 . Compare
your answer to the one in Example 2.
14. If a green jelly bean is selected from the bag, what
is the probability that the transferred jelly bean was
24. Medicine Using the information in Example 2 of the
text, find P(N|−), where − is the event “test shows
15. If a green jelly bean is selected from the bag, what
is the probability that the transferred jelly bean was
25. Economics Using the information in Example 4 of
the text, find P(E1 |Rc ) and P(E4 |Rc ). Compare your
answers with P(E1 ) and P(E4 ).
16. If a blue jelly bean is selected from the bag, what
is the probability that the transferred jelly bean was
26. Manufacturing For Example 1, find P(E1 |F c ).
Exercises 17 through 20 refer to the following experiment: Two cards are drawn in succession without replacement from a standard deck of 52 playing cards.
17. What is the probability that the first card drawn was
a spade given that the second card drawn was not a
18. What is the probability that the first card drawn was
a queen given that the second card drawn was not a
19. What is the probability that the first card drawn was
a heart given that the second card drawn was a diamond?
20. What is the probability that the first card drawn was
an ace given that the second card drawn was a king?
27. Quality Control One of two bins is selected at random, one as likely to be selected as the other, and
from the bin selected a transistor is chosen at random. The transistor is tested and found to be defective. It is known that the first bin contains two defective and four nondefective transistors, while the
second bin contains five defective and one nondefective transistors. Find the probability that second
bin was selected.
28. Quality Control Suppose in the previous exercise
there is a third bin with five transistors, all of which
are defective. Now one of the three bins is selected
at random, one as likely to be selected as any other,
and from this bin a transistor is chosen at random.
If the transistor is defective find the probability it
came from the third one.
29. Quality Control A typical box of 100 transistors
contains only 1 defective one. It is realized that
1.7 Bayes’ Theorem
presented, the court felt that this man was twice as
likely to be the father as not, and, hardly satisfied
with these odds, required the man to take a blood
test. The mother of the child had a different blood
type than the child: therefore the blood type of the
child was completely determined by the father. If
the man’s blood type was different than the child,
then he could not be the father. The blood type of
the child occurred in only 10% of the population.
The blood tests indicated that the man had the same
blood type as the child. What is the probability that
the man is the father?
among the last 10 boxes, one box has 10 defective
transistors. An inspector picks a box at random, and
the first transistor selected is found to be defective.
What is the probability that this box is the bad one?
30. Quality Control A typical box of 100 transistors
contains only 1 defective one. It is realized that
among the last 10 boxes, one box has 10 defective
transistors. An inspector picks a box at random, inspects two transistors from this box on a machine
and discovers that one of them is defective and one
is not. What is the probability that this box is the
bad one?
31. Manufacturing A manufacturing firm has four machines that produce the same component. Using the
table, given that a component is defective find the
probability that the defective component was produced by
a. machine 1
b. machine 2.
Percentage of
Percentage of
33. Medical Diagnosis A physician examines a patient
and, on the basis of the symptoms, determines that
he may have one of four diseases; the probability of
each is given in the table. She orders a blood test,
which indicates that the blood is perfectly normal.
Data are available on the percentage of patients with
each disease whose blood tests are normal. On the
basis of the normal blood test, find all probabilities
that the patient has each disease.
32. Social Sciences A man claimed not to be the father of a certain child. On the basis of evidence
of Disease
of Disease
Percentage of Normal
Blood With
This Disease
Solutions to Self-Help Exercises 1.7
1. Bayes’ theorem can be used, and F = −. Then
P(N|−) =
P(N ∩ −)
P(N)P(−|N) + P(C)P(−|C)
≈ 0.999
(0.9925)(0.96) + (0.0075)(0.08)
2. Begin with a tree diagram as shown in the figure on the left where S is a
successful member and U is an unsuccessful member. To find P(B|S) use
Bayes’ theorem.
P(B ∩ S)
3 · 0.3
= 1
3 · 0.6 + 3 · 0.3 + 3 · 0.9
= ≈ 0.1667
0.2 + 0.1 + 0.3 6
P(B|S) =
Chapter 1 Sets and Probability
3. Begin with a tree diagram. On the first set of branches we show the selection
of a coin from the purse. Following the top branch, the nickel is placed in the
wallet. Now the wallet has three nickels and one dime and a coin is chosen.
Following the lower branch we place a dime in the wallet. The wallet then
has two nickels and two dimes from which a coin is chosen.
We are asked to find P(N1 |D2 ) so we use Bayes’ theorem:
P(N1 |D2 ) =
P(N1 ∩ D2 )
P(N1 ∩ D2 )
P(D2 )
P(N1 ∩ D2 ) + P(D1 ∩ D2 )
✧ Summary Outline
• If every element of a set A is also an element of another set B, we say that A
is a subset of B and write A ⊆ B. If A is not a subset of B, we write A ⊆ B.
• If every element of a set A is also an element of another set B and A = B,
we say that A is a proper subset of B and write A ⊂ B. If A is not a proper
subset of B, we write A ⊂
/ B.
• The empty set, written as 0,
/ is the set with no elements.
• Given a universal set U and a set A ⊂ U, the complement of A, written Ac ,
is the set of all elements that are in U but not in A, that is,
Ac = {x|x ∈ U, x ∈
/ A}
• The union of two sets A and B, written A ∪ B, is the set of all elements that
belong to A, or to B, or to both. Thus
A ∪ B = {x|x ∈ A or x ∈ B or both}
• The intersection of two sets A and B, written A ∩ B, is the set of all elements that belong to both the set A and to the set B.
• If A ∩ B = 0/ then the sets A and B are disjoint.
• Rules for Set Operations
A∪B = B∪A
A∩B = B∩A
A ∪ (B ∪C) = (A ∪ B) ∪C
A ∩ (B ∩C) = (A ∩ B) ∩C)
A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪C)
A ∩ (B ∪C) = (A ∩ B) ∪ (A ∩C)
(A ∪ B)c = Ac ∩ Bc
(A ∩ B)c = Ac ∪ Bc
Commutative law for union
Commutative law for intersection
Associative law for union
Associative law for intersection
Distributive law for union
Distributive law for intersection
De Morgan law
De Morgan law
Chapter 1 Review
• If A is a set with a finite number of elements, we denote the number of
elements in A by n(A)
• If the sets A and B are disjoint, then n(A ∪ B) = n(A) + n(B).
• For any finite sets A and B we have the union rule,
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
• An experiment is an activity that has observable results. The results of the
experiment are called outcomes.
• A sample space of an experiment is the set of all possible outcomes of the
experiment. Each repetition of an experiment is called a trial.
• Given a sample space S for an experiment, an event is any subset E of S.
An elementary event is an event with a single outcome.
• If E and F are two events, then E ∪ E is the union of the two events and
consists of the set of outcomes that are in E or F or both.
• If E and F are two events, then E ∩ F is the intersection of the two events
and consists of the set of outcomes that are in both E and F.
• If E is an event, then E c is the complement of E and consists of the set of
outcomes that are not in E.
• The empty set, 0,
/ is called the impossible event.
• Let S be a sample space. The event S is called the certainty event.
• Two events E and F are said to be mutually exclusive or disjoint if E ∩
F = 0.
• Properties of Probability Let S be a sample space and E, A, and B be
events in S, P(E) the probability of E, and so on. Then
0 ≤ P(E) ≤ 1
P(S) = 1
/ =0
P(A ∪ B) = P(A) + P(B), if A ∩ B = 0/
P(E c ) = 1 − P(E)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A) ≤ P(B), if A ⊂ B
• Let P(E) be the probability of E, then the odds for E are
if P(E) = 1
1 − P(E)
This fraction reduced to lowest terms is
and the odds are a:b.
• If the odds for an event E occurring is given as
P(E) =
or a:b, then
Chapter 1 Sets and Probability
• If S is a finite uniform sample space and E is any event, then
P(E) =
Number of elements in E
Number of elements in S
• Let E and F be two events in a sample space S. The conditional probability that E occurs given that F has occurred is defined to be
P(E|F) =
P(E ∩ F)
P(F) > 0
• Product rule: If E and F are two events in a sample space S with P(E) > 0
and P(F) > 0, then
P(E ∩ F) = P(F)P(E|F) = P(E)P(F|E)
• Two events E and F are said to be independent if
P(E|F) = P(E)
P(F|E) = P(F)
• Let E and F be two events with P(E) > 0 and P(F) > 0. Then E and F are
independent if, and only if, P(E ∩ F) = P(E)P(F).
• A set of events {E1 , E2 , . . . , En } is said to be independent if, for any k of
these events, the probability of the intersection of these k events is the
product of the probabilities of each of the k events. This must hold for any
k = 2, 3, . . . , n.
• Bayes’ Theorem Let E1 , E2 , . . . , En , be mutually exclusive events in a sample space S with E1 ∪ E2 ∪ . . . ∪ En = S. If F is any event in S, then for
i = 1, 2, . . . , n,
probability of branch Ei F
sum of all probabilities of all branches that end in F
P(Ei ∩ F)
P(Ei )P(F|Ei )
P(E1 )P(F|E1 ) + P(E2 )P(F|E2 ) + · · · + P(En )P(F|En )
P(Ei |F) =
✧ Review Exercises
1. Determine which of the following are sets:
a. current members of the board of Bank of America
b. past and present board members of Bank of
America that have done an outstanding job
c. current members of the board of Bank of America who are over 10 feet tall
2. Write in set-builder notation:
{5, 10, 15, 20, 25, 30, 35, 40}.
3. Write in roster notation: {x|x3 − 2x = 0}
4. List all the subsets of {A, B,C}.
5. On a Venn diagram indicate where the following
sets are:
a. A ∩ B ∩C
b. Ac ∩ B ∩C
c. (A ∪ B)c ∩C
Chapter 1 Review
6. Let U = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3}, B = {2, 3, 4},
and C = {4, 5}. Find the following sets: A ∪ B,
A ∩ B, Bc , A ∩ B ∩C, (A ∪ B) ∩C, A ∩ Bc ∩C
7. Let U be the set of all your current instructors and
H = {x|x is at least 6 feet tall}
M = {x|x is a male}
W = {x|x weighs more than 180 pounds}
Describe each of the following sets in words:
a. H c
c. M c ∩W c
e. H c ∩ M ∩W
b. H ∪ M
d. H ∩ M ∩W
f. (H ∩ M c ) ∪W
8. Using the set H, M, and W in the previous exercise
and set operations, write the set that represents the
following statements:
a. my current female instructors
b. my current female instructors who weigh at
most 180 pounds
c. my current female instructors who are at least 6
feet tall or else weigh more than 180 pounds
9. For the sets given in Exercise 6, verify that
A ∪ (B ∩C) = (A ∪ B) ∩ (A ∪C)
10. Use a Venn diagram to show that (A ∩ B)c = Ac ∪ Bc
11. If n(A) = 100, n(B) = 40, and n(A ∩ B) = 20, find
n(A ∪ B).
12. If n(A) = 40 and n(A ∩ Bc ) = 30, find n(A ∩ B).
13. In a consumer advertising survey of 100 men, it was
found that 20 watched the first game of the World
Series, 15 watched the first game of the World Series and also watched the Super Bowl, while 30 did
not watch either. How many watched the Super
Bowl but not the first game of the World Series.
14. A consumer survey of 100 children found that
57 had a Barbie doll
68 had a teddy bear
11 had a toy piano
45 had a Barbie doll and a teddy bear
8 had a teddy bear and a toy piano
7 had a Barbie doll and a toy piano
5 had all three
a. How many had a Barbie doll and a teddy bear
but not a toy piano?
b. How many had exactly 2 of these toys?
c. How many had none of these toys?
15. During a recent four-round golf tournament the
number of strokes were recorded on a par 5 hole.
The following table lists the frequencies of each
number of strokes.
a. Find the probability for each number of strokes.
b. Find the probability that the number of strokes
were less or equal to 5.
c. Find the probability that the number of strokes
were less than 5.
16. An urn has 10 white, 5 red, and 15 blue balls. A ball
is drawn at random. What is the probability that the
ball will be
a. red?
b. red or white?
c. not white?
17. If E and F are disjoint sets in a sample space S with
P(E) = 0.25 and P(F) = 0.35, find
a. P(E ∪ F)
b. n(E ∩ F)
c. P(E c )
18. If E and F are two events in the sample space S with
P(E) = 0.20, P(F) = 0.40, and P(E ∩ F) = 0.05,
a. P(E ∪ F)
b. P(E c ∩ F)
c. P((E ∪ F)c )
19. Consider the sample space S = {a, b, c, d} and suppose that P(a) = P(b), P(c) = P(d), and P(d) =
2P(a). Find P(b).
20. If the odds for a company obtaining a certain contract are 3:1, what is the probability that the company will receive the contract?
21. A furniture manufacturer notes the 6% of its reclining chairs have a defect in the upholstery, 4% a defect in the reclining mechanism, and 1% have both
a. Find the probability that a recliner has at least
one of these defects.
b. Find the probability that a recliner has none of
these defects.
22. A survey of homeowners indicated that during the
last year: 22% had planted vegetables, 30% flowers, 10% trees, 9% vegetables and flowers, 7% vegetables and trees, 5% flowers and trees, and 4%, all
three of these.
Chapter 1 Sets and Probability
a. Find the probability that a homeowner planted
vegetables but not flowers.
b. Find the probability that exactly two of the items
were planted.
c. Find the probability that none of these three
items were planted.
23. Let P(E) = 0.3, P(F) = 0.5, and P(E ∩ F) = 0.2.
Draw a Venn diagram and find the indicated conditional probabilities:
c. P(F c |E c )
a. P(E|F)
b. P(E c |F)
24. If P(E) = 0.5, P(F) = 0.6, and P(E ∩ F) = 0.4, determine if E and F are independent events.
25. Reliability A spacecraft has three batteries that can
operate all systems independently. If the probability
that any battery will fail is 0.05, what is the probability that all three will fail?
26. Basketball A basketball player sinks a free throw
80% of the time. If she sinks one, the probability
of sinking the next goes to 0.90. If she misses, the
probability of sinking the next goes to 0.70. Find
the probability that she will sink exactly two out of
three free throws.
27. Manufacturing A manufacturing firm has 5 machines that produce the same component. Using the
table, find the probability that a defective component was produced by
a. machine 1
b. machine 4
28. Drug Testing A company tests its employees for
drug usage with a test that gives a positive reading
95% of the time when administered to a drug user
and gives a negative reading 95% of the time when
administered to a non-drug user. If 5% of the employees are drug users, find the probability that an
employee is a non-drug user given that this person
had a positive reading on the test. (The answer is
shocking and illustrates the care that must be exercised in using such tests in determining guilt.)
✧ Project: Venn Diagrams and Systems of Linear
Figure 1.35
Sometimes the numbers do not arrange themselves in the Venn diagram as neatly
as they did in the exercises in Section 2. If this happens you can often use deductive reasoning to fill in the diagram. However, sometimes a system of equations
must be used to complete the diagram. If we use a system of linear equations to
fill in the Venn diagram, we need a consistent way to refer to the eight regions of
the Venn diagram. In Figure 1.35 we see the eight regions on the diagram labeled
with the Roman numerals I through VIII. That is, n(A ∩ Bc ∩Cc ) is V.
The next example will be solved using two different methods. If the solution
is not unique, techniques from our study of linear systems can be used.
There were 32 students surveyed and asked
if they did or did not like tomatoes, spinach, or peas with the results listed below.
Arrange this information in a Venn diagram.
EXAMPLE 5
Vegetable Survey
• 7 students liked all three vegetables
• 6 students did not like spinach or peas
• 1 student liked only tomatoes
• 9 students liked tomatoes and spinach
Chapter 1 Review
• 23 students liked two or more of these vegetables
• 5 students liked spinach but not peas
• 18 students liked spinach
Begin with a blank Venn diagram. Let
S = {x|x is a student who likes spinach}
T = {x|x is a student who likes tomatoes}
P = {x|x is a student who likes peas}
• The first clue that “7 students liked all three vegetables” tells us that we can
place the number 7 in the intersection of all three sets: n(S ∩ T ∩ P) = 7.
See Figure 1.36 You can scratch out this clue as each clue is used only
• The second clue that “6 students did not like spinach or peas” is not useful
yet as both VI and VIII are outside of the region corresponding to liking
spinach or peas. All we know is that those two numbers add to 6. Move on
to the next clue.
Figure 1.36
• The third clue says “1 student liked only tomatoes” and so the number 1 is
placed in region VI (Sc ∩ T ∩ Pc ). Scratch out this clue as it has been used.
• The fourth clue says “9 students liked tomatoes and spinach” and this
means 7 + II = 9. So there must be 2 students who like tomatoes and
spinach but not peas. A 2 is placed in the region S ∩ T ∩ Pc as shown in
Figure 1.37. This clue can be scratched out.
• The fifth clue states that “23 students liked two or more of these vegetables.” The region representing two or more vegetables is shaded in Figure 1.37 and we see that is the sum of four numbers and at this point we
only know the value of two of them. This clue is skipped for now.
Figure 1.37
• The sixth clue states “5 students liked spinach but not peas.” This is the
region that is in the S circle but outside the P circle. We see that 2 of the 5
students have been accounted for as they like spinach and tomatoes but not
peas. Therefore 3 students like only spinach and a 3 is placed in the region
S ∩ T c ∩ Pc and this clue is scratched out.
• The seventh clue is that “18 students liked spinach” and we see that the S
circle is nearly complete. If there are 18 students in all in this circle and
3 + 2 + 7 = 12 of them are accounted for, then 18 − 12 = 6 students must
be in the remaining empty spot in the S circle for students who like spinach
and peas but not tomatoes, S ∩ T c ∩ P. This is shown in Figure 1.38.
Figure 1.38
• The eighth clue is that 32 students are surveyed, so n(U) = 32. Since three
numbers are still missing in the diagram, we are not ready for this clue yet.
We have three missing numbers in our diagram and three un-used clues:
• 6 students did not like spinach or peas
Chapter 1 Sets and Probability
• 23 students liked two or more of these vegetables
• 32 students were surveyed
The region representing not spinach or peas will be those students who like
only tomatoes (which is 1 student) and those students who do not like any of the
vegetables, so 6 − 1 = 5 students did not like any of the vegetables.
Referring back to the shaded region in Figure 1.37 we now know 3 of the four
numbers that represent liking two or more of these vegetables so we can find the
missing number by subtraction, 23 − 2 − 6 − 7 = 8 students liked tomatoes and
peas but not spinach. Place the 8 in the region Sc ∩ T ∩ P.
Finally we do not know how many students liked only peas, but we do know
there are 32 students who were surveyed and so by subtraction, 32 − 3 − 2 − 1 −
6 − 7 − 8 − 5 = 0, so a zero is placed in the region Sc ∩ T c ∩ P. The completed
diagram is shown in Figure 1.39.
Figure 1.39
Referring to Figure 1.40 with the Roman numerals, we translate each clue
into a linear equation.
• 32 students were surveyed → I + II + III + IV + V + VI + VII + VIII = 32
• 7 students liked all three vegetables → V = 7
• 6 students did not like spinach or peas → III + VIII = 6
• 1 student liked only tomatoes → III = 1
• 9 students liked tomatoes and spinach → II + V = 9
• 23 students liked two or more of these vegetables → II + IV + V + VI = 23
• 5 students liked spinach but not peas → I + II = 5
• 18 students liked spinach → I + II + IV + V = 18
Figure 1.40
Place this information into an augmented matrix and use the methods of Chapter
1 to solve the system. This method is particularly useful if a calculator is used.
With x1 = I, x2 = II, and so on we have
x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8
+ x8
+ x5
x2 + x4 + x5 + x6
x1 + x2
x1 + x2 +
x4 + x5
= 32
= 32
= 18
The solution to the system is
I = 3, II = 2, III = 1, IV = 6, V = 7, VI = 8, VII = 0, VIII = 5.
Complete a Venn diagram for each of the exercises below.
Chapter 1 Review
1. n(A) = 11, n(B) = 14, n(C) = 19, n(U) = 36, n(A ∩ B) = 4, n(A ∩ C) =
8, n(B ∩C) = 7 and n(A ∪ B ∪C) = 28
2. A class of 6th grade boys is surveyed and asked if they were wearing one or
more of the following items of clothing that day: a T-shirt, shorts or athletic
shoes. The results were:
• 40 wore a T-shirt, shorts, and athletic shoes
• 23 wore exactly two of these items
• 57 wore shorts
• 12 did not wear a T-shirt
• 4 wore only athletic shoes
• 7 did not wear shorts or athletic shoes
• 11 wore only shorts and a T-shirt
• 52 wore a T-shirt and athletic shoes
3. Two hundred tennis players were asked which of these strokes they considered their weakest stroke(s): the serve, the backhand, and the forehand.
• 20 players said none of these were their weakest stroke
• 30 players said all three of these were their weakest stroke
• 40 players said their serve and forehand were their weakest strokes
• 40 players said that only their serve and backhand were their weakest
• 15 players said that their forehand but not their backhand was their
weakest stroke
• 52 players said that only their backhand was their weakest stroke
• 115 players said their serve was their weakest stroke
Source: Joe Kahlig
4. Thirty-one children were asked about their lunch preferences and the following results were found:
• 12 liked cheeseburgers
• 14 liked pizza
• 9 liked burritos
• 5 liked cheeseburgers and pizza
• 4 liked cheeseburgers and burritos
• 8 liked pizza and burritos
• 10 liked none of these items
Answers to Selected Exercises
Answers to Selected Exercises
L.1 EXERCISES
3. Statement
5. Statement
1. Statement
7. Not a statement
9. Statement
11. Not a statement
13. Statement
15. a. George Washington was not the third president of the
United States. b. George Washington was the third president of the United States, and Austin is the capital of Texas.
c. George Washington was the third president of the United
States, or Austin is the capital of Texas. d. George Washington was not the third president of the United States, and
Austin is the capital of Texas. e. George Washington was
the third president of the United States, or Austin is not the
capital of Texas. f. It is not true that George Washington
was the third president of the United States and that Austin is
the capital of Texas.
17. a. George Washington did not own over 100, 000 acres
of property. The Exxon Valdez was not a luxury liner. b.
George Washington owned over 100, 000 acres of property,
or the Exxon Valdez was a luxury liner.
c. George Washington owned over 100, 000 acres of property,
and the Exxon Valdez was a luxury liner.
19. a. ∼q b. p ∧ ∼q c. p ∨ q d. (∼p) ∨ (∼q)
q ∼q ∼p
p q r
p q r
p ∨ ∼q ∼p ∧ q (p ∨ ∼q) ∨ (∼p ∧ q)
p ∨ q (p ∨ q) ∧ r
p ∧ q (p ∧ q) ∧ r ∼ [(p ∧ q) ∧ r]
p q ∼q p ∧ ∼q (p ∧ ∼q) ∨ q
p ∼p ∼(∼p)
q ∼q p ∧ ∼q
T F
F T
T F
F T
L.2 EXERCISES
p q p ∨ q p ∧ q (p ∨ q) ∧ (p ∧ q)
p q ∼p p ∧ q ∼p ∨ (p ∧ q)
p q r
p ∨ q q ∧ r (p ∨ q) ∨ (q ∧ r)
Answers to Selected Exercises
r ∼q p ∨ ∼q ∼q ∧ r (p ∨ ∼q) ∨ (∼q ∧ r)
T F
F F
T T
F T
T F
F F
T T
F T
21. a. True
b. False c. True d. True e. False
23. a. True b. False c. False d. True
1.1 EXERCISES
3. a. True b. False
1. a. False b. False
5. a. False b. True c. False (d) True
7. a. 0,
/ {3} b. 0,
/ {3}, {4}, {3, 4}
a. 2 b. 3 disjoint
a. 1, 2, 3 b. 1 not disjoint
a. I b. V
a. VIII b. IV
a. II, V, VI b. III, IV, VII
a. VII b. I, II, III, IV, V, VII, VIII
a. {4, 5, 6} b. {1, 2, 3, 4, 5, 6, 7, 8}
a. {1, 2, 3} b. {9, 10}
a. {5, 6} b. {1, 2, 3, 4, 7, 8, 9, 10}
a. 0/ b. 0/
a. People in your state who do not own an automobile
b. People in your state who own an automobile or house
c. People in your state who own an automobile or not a house
33. a. People in your state who own an automobile but not a
b. People in your state who do not own an automobile and do
not own a house
c. People in your state who do not own an automobile or do
not own a house
35. a. People in your state who own an automobile and a
house and a piano
b. People in your state who own an automobile or a house or
a piano
c. People in your state who own both an automobile and a
house or else own a piano
37. a. People in your state who do not own both an automobile and a house but do own a piano
b. People in your state who do not own an automobile, nor a
house, nor a piano
c. People in your state who own a piano, but do not own a car
or a house
39. a. N ∩ F b. N ∩ H c
41. a. N ∪ S b. N c ∩ Sc
43. a. (N ∩ H) ∪ (S ∩ H) b. (N ∩ F) ∩ H c
45. a. (F ∩ H) ∩ (N ∪ S)c b. F ∩ H c ∩ N c ∩ Sc
47. Both expressions give U
49. Both expressions give {1, 2, 3, 4, 5, 6, 7}
51. Both expressions give {8, 9, 10}.
1.2 EXERCISES
3. 70
5. 60
7. 150
1. 135
11. 90
13. 15
15. 7
17. 3
21. a. 1100 b. 750 c. 100
23. a. 100 b. 500
25. a. 30 b. 360 c. 70
27. a. 20 b. 25 c. 345
29. 30
9. 110
19. 56
31. From the figure we have n(A) −
n(A ∩ B) = x + w, n(B) − n(B ∩ C) =
u + y, n(C) − n(A ∩ C) = v + z, and
n(A ∩ B ∩ C) = t. Adding these four
equations gives the result since from
the figure n(A ∪ B ∪ C) = t + u + v +
w + x + y + z.
1.3 EXERCISES
/ {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
1. 0,
3. S = {(H, H, H), (H, H, T ), (H, T, H), (T, H, H), (H, T, T ),
(T, H, T ), (T, T, H), (T, T, T )},
E = {(H, H, H), (H, H, T ), (H, T, H), (T, H, H)}
S = {(1, 1, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (0, 0, 1),
(0, 1, 0), (1, 0, 0), (0, 0, 0)}, E = {(1, 1, 1), (1, 1, 0), (1, 0, 1),
(0, 1, 1)}
7. a. {black, white, red}
b. {black, red}
{(A, B,C), (A, B, D), (A, B, E), (A,C, D), (A,C, E),
(A, D, E), (B,C, D), (B,C, E), (B, D, E), (C, D, E)},
{(A, B,C), (A, B, D), (A, B, E), (B,C, D), (B,C, E), (B, D, E)}
11. S = {1, 2, 3, 4, 6, 9}, E = {2, 4, 6}
13. a. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} b. E = {6, 7, 8, 9, 10} c.
F = {0, 1, 2, 3, 4}
d. E ∪ F = {0, 1, 2, 3, 4, 6, 7, 8, 9, 10}. E ∩ F = 0.
/ Ec =
{0, 1, 2, 3, 4, 5}. E ∩ F = {6, 7, 8, 9, 10}. E ∩ F = {5}
e. E c and E ∩ F c . E ∩ F c and E c ∩ F c . E ∪ F and E c ∩ F c .
15. a. {(S, S, S), (S, S, F), (S, F, S), (F, S, S), (S, F, F),
(F, S, F), (F, F, S), (F, F, F)}
b. E = {(S, S, S), (S, S, F), (S, F, S), (F, S, S)}
c. G = {(S, S, S), (S, S, F), (S, F, S), (S, F, F)}
E ∪ G = { (S, S, S), (S, S, F), (S, F, S), (S, F, F),
(F, S, S)},
E ∩ G = {(S, S, S), (S, S, F), (S, F, S)},
Ec ∩ G =
Gc = {(F, S, S), (F, S, F), (F, F, S), (F, F, F)},
{(S, F, F)}, (E ∪ G) = {(F, F, S), (F, S, F), (F, F, F)}.
e. E ∩ G and Gc . E ∩ G and E c ∩ G. E ∩ G and (E ∪ G)c .
Gc and E c ∩ G. E c ∩ G and (E ∪ G)c . E ∪ G and (E ∪ G)c .
17. a. Pencils that are longer than 10 cm and less than 25 cm.
b. Pencils that are less than 10 cm long.
c. Pencils that are 25 cm or longer.
d. 0/
19. E ∩ F c
21. F c ∩ E c
23. E ∩ F ∩ Gc
25. E ∪ F ∪ G = all 26 letters. E c ∩ F c ∩ Gc = 0.
/ E ∩F ∩G =
/ E ∪ F c ∪ G = {a, e, i, o, u, b, c, d}
27. S = {r, b, g, y}, E = {b, g}
Answers to Selected Exercises
1.4 EXERCISES
3. 2/3
5. 1/13
7. 1/2
1. 1/2
9. /26
11. /4
13. /3
15. 0.15
17. a. 0.12 b. 0.8 c. 0.08
19. 0.56
21. A: 0.125, B: 0.175, C: 0.4, D: 0.2, F: 0.1
23. 0.55
25. 1/2
27. 1/2
29. 3/8, 7/8
31. 1/12
33. 1/36, 1/12, 5/36
35. 1/3
1.5 EXERCISES
3. 0.18, 0.63, 0.62
5. 0.25
1. 0.50, 0.90
7. 1/6
9. 7/30
11. 0.60, 0.80, 0
13. 0.70, 0.70, 0, 0.30, 1
15. 0.60, 0.10
17. 0.20, 0.10
19. 0.20, 0.10, 0.15
21. 0.85 23. a. 3:17 b. 1:3
25. 2:3
27. 7:5
29. 0.20
31. 0.04, 0.96
33. 0.039, 0.049, 0.951
35. a. 0.009 b. 0.001 c. 0.006 d. 0.974
37. 0.20
39. You do not know what the actual probability is. You do
know that the empirical probability is 165/1000 = 0.165. This
represents the best guess for the actual probability. But if you
tossed the coin more times, the relative frequency and the new
empirical probability would most likely had changed.
41. The probabilities in the game are constant and do not
change just because you are on a winning streak. Thus no
matter what has happened to you in the past, the probability
of winning any one game remains constant at 0.48. Thus if
you continue to play, you should expect to win 48% of the
time in the future. You have been lucky to have won 60% of
the time up until now.
43. After reading the first discussion problem above, we
know that it is, in fact, impossible to determine with certainty
the actual probability precisely. Since the die has been tossed
a total of 2000 times and a one has come up 335 times, our
best guess at the probability is 335/2000 = 0.1675.
1.6 EXERCISES
3. 2/3, 4/5
5. 1, 0
1. 3/7, 3/5
7. /3, /2
9. 1/3, 1
11. 0, 1/2
13. No
15. Yes
17. No
19. Yes
21. 2/11
23. 1/7
25. a. 132,600 ≈ 0.077 b. 25·33
≈ 0.059
27. 0.12, 0.64, 0.60
29. 1/3, 3/10
31. No
33. 0.65
35. 0.72
37. 0.02, 0.017
39. 0.026
41. 0.000001
43. 5/7, 5/21, 1/21
45. Yes
47. No
49. 0.057818
51. For E and F to be independent, they must satisfy
P(E) × P(F) = P(E ∩ F). From the Venn diagram, we must
have: (p1 + p2 ) × (p3 + p2 ) = p2 . So,
p1 p3 + p2 p3 + p1 p2 + p22
= p2
p1 p3 = p2 (1 − p3 − p2 − p1 ) = p2 p4
The above steps can be reversed, so if p1 p3 = p2 p4 , we will
have P(E) × P(F) = P(E ∩ F).
If the sets are mutually disjoint, then p2 = 0. This implies
that p1 p3 = p2 p4 = 0. Then either p1 or p3 or both are zero.
Thus either P(E) = 0, P(F) = 0, or both.
53. Since E and F are independent, P(E) × P(F) = P(E ∩ F).
P(E c ∩ F c ) = 1 − P(E ∪ F)
= 1 − P(E) + P(F) − P(E ∩ F)
= 1 − P(E) − P(F) + P(E) × P(F)
= 1 − P(E) × 1 − P(F)
= P(E c ) × P(F c )
Hence, if E and F are independent, so are E c and F c .
55. Since E and F are exclusive, P(E ∩ F) = P(0)
/ = 0. Since
P(E) and P(F) are both nonzero, then P(E) × P(F) > 0.
Therefore, E and F are not independent.
P(E c ∩ F)
P(F) − P(E ∩ F)
= 1 − P(E|F)
P(E c |F) =
59. P(E|F) = P(F) = P(F) = 1
P(E∩F)+P(E c ∩F)
61. P(E|F) + P(E c |F) =
= P(F)
P(F) = 1
1.7 EXERCISES
3. 3/19, 1/3
5. 4/11
1. 3/7, 57/93
7. a. 2/23 b. 21/23
9. 2/17
11. a. 1/56 b. 5/28
13. 2/7
15. 4/19
17. 13/51
19. 13/51
21. 6/17
23. 100/136 ≈ 0.74
25. 396/937 ≈ 0.42, 5/937 ≈ 0.005
27. 5/7
29. 10/19
31. a. 1/12 b. 1/4
33. P(1|N) = 6/20, P(2|N) = 4/20, P(3|N) = 6/20, P(4|N) =
a. Yes b. No c. Yes
{x | x √
= 5n,
√n is an integer and 1 ≤ n ≤ 8}
{0, − 2, 2}
/ {A}, {B}, {C}, {A, B}, {A,C}, {B,C}, {A, B,C}
5. a.
6. A ∪ B = {1, 2, 3, 4}, A ∩ B = {2, 3}, Bc = {1, 5, 6}, A ∩ B ∩
C = {2, 3} ∩ {4, 5} = 0,
/ (A ∪ B) ∩C = {4}, A ∩ Bc ∩C = 0/
7. a. My current instructors who are less than 6 feet tall. b.
My current instructors who are at least 6 feet tall or are male.
c. My current instructors who are female and weigh at most
180 pounds. d. My current male instructors who are at least
6 feet tall and weigh more than 180 pounds. e. My current
male instructors who are less than 6 feet tall and weigh more
Answers to Selected Exercises
than 180 pounds. f. My current female instructors who are at
least 6 feet tall and weigh more than 180 pounds.
8. a. M c b. M c ∩W c c. M ∩ (H ∪W )
9. {1, 2, 3, 4} = {1, 2, 3, 4}
10. The shaded area is (A ∩ B)c = Ac ∪ Bc .
11. 120
12. 10
13. 50
14. a. 40 b. 45 c. 19
15. a. 0.016, 0.248, 0.628, 0.088, 0.016, 0.044
b. 0.892
c. 0.264
16. a. 5/30 b. 15/30 c. 20/30
17. P(E ∪ F) = 0.60, P(E ∩ F) = 0, P(E c ) = 0.75
18. P(E ∪ F) = 0.55, P(E c ∩ F) = 0.35,
P ((E ∪ F)c ) = 0.45
19. P(B) = 1/6
20. 0.75
21. a. 0.09 b. 0.91
22. a. 0.13 b. 0.09 c. 0.55
23. P(E|F) = 0.40, P(E c |F) = 0.60, P(F c |E c ) = 4/7
24. Not independent
25. (0.05)3
26. 0.254
27. a. 2/31 b. 4/31
28. 0.50
2.1 EXERCISES
3. 8 × 7 × 6 × 5 × 4 = 6720
1. 5 × 4 × 3 = 60
5. 7! = 5040
7. 9! = 362, 880
9. 9 × 8 = 72
11. n
13. 4 × 30 = 120
15. 4 × 10 × 5 = 200
17. 106 = 1, 000, 000
19. 2 × 10 × 5 × 3 = 300
21. (26)3 (10)3 = 17, 576, 000
23. 96
25. 5 × 4 × 4 × 3 × 3 × 2 × 2 = 2880
27. 5! = 120
29. 2!4!6! = 34, 560
31. 9!5! = 43, 545, 600
33. 3!4! = 144
35. 12 × 11 × 10 × 9 = 11, 880
37. 1,814,400
39. (10 × 9 × 8)(8 × 7 × 6) = 241, 920
41. P(11, 4) × P(9, 3) × P(5, 2) = 79, 833, 600
43. P(n, r) =n(n − 1)(n − 2) · · · (n − r + 1)
= n(n − 1)(n − 2) · · · (n − r + 1) (n−r)(n−r−1)···2·1
(n − r)!
45. 24
2.2 EXERCISES
1. 8×7×6
3×2×1 = 56
7. 35
9. 105
3. 8×7×6×5×4
5×4×3×2×1 = 56
11. n
5. 12
15. 10
17. C(46, 6) = 40!×6!
= 9, 366, 819
19. C(20, 5) = 15!×5!
= 15, 504
21. C(20, 4) = 16!×4!
= 4845
23. C(40, 5) = 5!×35!
= 658, 008
25. C(11, 3) ×C(7, 2) = 3465
27. C(20, 7) ×C(6, 3) = 1, 550, 400
29. C(12, 3)P(9, 3) = 110, 880
31. C(11, 4)P(7, 2) = 13, 860
33. C(9, 5) = 126
35. C(8, 5) = 56
37. 28 −C(8, 0) −C(8, 1) = 256 − 1 − 8 = 247
39. C(12, 3)C(8, 3)C(4, 2)2 = 147, 840
41. C(4, 3) · 48 · 44/2 = 4224
43. C(4, 2)C(4, 2)C(13, 2) · 44 = 123, 552
(r!)(n − r)!
(n − r)!(n − (n − r))!
= C(n, n − r)
45. C(n, r) =
2.3 EXERCISES
1. 3/10, 1/3
3. (6·7·8)/1330
5. 20/323
7. 6/1326
9. /1326
11. /1326
13. /2,589,960
15. 624/2,589,960
17. 10,200/2,589,960
19. 123,552/2,589,960
21. 0.0833; 0.2361; 0.4271; 0.6181
23. 1
25. 6/45
27. 56/210 ≈ 0.0547
29. /11
31. 420
33. 34,650
35. 2520
37. a. 6 b. 3 c. 24 d. 12 e. 6
39. 3780
2.4 EXERCISES
1. 5(.2)4 (.8) = .0064
3. 35(.5)7 ≈ 0.273
5. 70(.25)4 (.75)4 ≈ 0.087
7. 20(.5)6 ≈ 0.3125
9. 35(.1)4 (.9)3 ≈ 0.00255
11. 10(.1)3 (.9)2 ≈ 0.0081
13. 45 × (0.5)10 ≈ 0.044
15. 56(0.5)10 ≈ 0.0547
17. 11(0.5)10 ≈ 0.0107
19. 7(0.6)6 (0.4) ≈ 0.1306
21. 7(0.6)6 (0.4) + (0.6)7 ≈ 0.1586
23. (0.4)7 + 7(0.6)(0.4)6 + 21(0.6)2 (0.4)5 ≈ 0.0962
25. 1 − (0.633)4 ≈ 0.839
27. (0.839)10 ≈ 0.173
29. 1 − (0.915)4 − 4(0.085)(0.915)3 ≈ 0.03859
31. (0.0386)3 ≈ 0.0000575
33. 12(0.05)(0.95)11 ≈ 0.341
35. (.95)12 + 12(.05)(.95)11 + 66(.05)2 (.95)10 ≈ 0.980
37. C(20, 10)(0.8)10 (0.2)10 ≈ 0.002
39. 190(.8)18 (.2)2 + 20(.8)19 (.2) + (.8)20 ≈ 0.206
41. 190(0.05)2 (0.95)18 ≈ 0.189
43. ≈ 0.984
2.5 EXERCISES
1. a5 − 5a4 b + 10a3 b2 − 10a2 b3 + 5ab4 − b5
3. 32x5 + 240x4 y + 720x3 y2 + 1080x2 y3 + 810xy4 + 243y5
5. 1 − 6x + 15x2 − 20x3 + 15x4 − 6x5 + x6 | {"url":"https://studylib.net/doc/27143617/discrete-math","timestamp":"2024-11-10T13:01:07Z","content_type":"text/html","content_length":"272075","record_id":"<urn:uuid:02943784-4ef3-42df-a617-d20c19c19a3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00093.warc.gz"} |
Approaching Midnight - Mariah's strategy
Thank you to Mariah for sending in her strategy to Approaching Midnight.
The way to win 'Approaching Midnight' is to make the computer's final turn at 10:45, so no matter what they press (15, 30, 45 or 60 minutes) you will be able to reach midnight first.
This is because 10:45 is 1 hour and 15 minutes before 12:00. 1 hour and 15 minutes is the only possible time that can be reached in a round of your turn and the computer's, as 1 hours and 15 minutes
equals 75 minutes. If the computer choses 60 minutes then you will chose 15 to equal 75, the same goes for 45 and 30, 30 and 45, 15 and 60.
The way to make sure the computer's final turn starts at 10:45 is using the same 75 minute pattern.
Starting first, you press 60 minutes, from then on, (so starting at the computer's first turn) every round must equal 75 minutes (60 minutes + 75 minutes x 4 = 360 minutes or 6 hours, the needed
Your next move, and every move after will depend on the computer. If the computer moves the clock forward 60 minutes, you move 15, in the same way you would on the final turn, as explained above. | {"url":"https://wild.maths.org/approaching-midnight-mariahs-strategy","timestamp":"2024-11-07T04:15:05Z","content_type":"text/html","content_length":"15268","record_id":"<urn:uuid:0380b7ff-9ea9-42bf-9deb-541acb941cd0>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00463.warc.gz"} |
jacobian matrix
jacobian matrix meaning in Hindi
jacobian matrix sentence in Hindi More: Next
1. For a more complete description, see Jacobian matrix and determinant.
2. At the point the rows of the Jacobian matrix are and.
3. This method uses the Jacobian matrix of the system of equations.
4. The Jacobian matrix of this transformation has the block form:
5. For a more complete description, see Jacobian matrix.
6. By a repeated singularity, we mean that the jacobian matrix is singular.
7. At a point, where, the rows of the Jacobian matrix are and.
8. Jacobian matrix is zero, that is, where
9. This is solved by inverting the Jacobian matrix.
10. The Jacobian matrix of this transformation is given by | {"url":"https://m.hindlish.com/jacobian%20matrix","timestamp":"2024-11-02T21:08:10Z","content_type":"text/html","content_length":"24234","record_id":"<urn:uuid:a94d2b1b-f3df-4904-9f3a-8c6d4b495ba4>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00027.warc.gz"} |
With proc freq can you only print the counts for a certain value of a variable
With proc freq (or another proc), is there a way to only print the counts for a certain value of a variable? I have 30+ variables and for each of those variables, I want to know how many observations
in my sample had a value of 0.
02-27-2020 06:58 PM | {"url":"https://communities.sas.com/t5/SAS-Procedures/With-proc-freq-can-you-only-print-the-counts-for-a-certain-value/m-p/628076/highlight/true","timestamp":"2024-11-07T03:19:52Z","content_type":"text/html","content_length":"144625","record_id":"<urn:uuid:5f3f136f-49cd-4f22-b885-3a76613f3e72>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00868.warc.gz"} |
T Excel
What Is T Excel Function?
The T Excel function is a built-in Text function. It accepts a value, and it returns the input value as it is if the specified data is a text value. Otherwise, the function output is a blank
Users can utilize the T Excel function to validate the given data for text values and clean the source dataset so that we can view only the text content.
For instance, the source dataset holds two values in column A.
The task is to showcase the column A cell values in the corresponding column B cells if they are texts. Otherwise, the target cells should hold an empty text.
Then, we can secure the required values using T Excel function in the specified cells.
The T Excel function in cell B2 takes the cell A2 value as its input. Since the input data is a text value, the function returns the input text value as the required output.
Next, the T() in cell B3 accepts the cell A3 value as its input. Since the input data is not a text value, the function T Excel returns a blank text as the output.
Key Takeaways
• The T Excel function enables us to get the text value referred to by the specified value.
• The T function in Excel finds applications in text-based data manipulation, validation, sorting and filtering.
• The T function in Excel takes one compulsory argument, value, as its input.
• While we can utilize the T function in Excel as an individual function directly or from the Formulas tab in a cell, Excel does not offer an inbuilt T method in VBA. However, using it with other
inbuilt functions, such as IF and CONCAT, yields fruitful results.
The T Excel function syntax is as follows:
• value: The data point we aim to test for text.
We must supply the argument mentioned above when using T Excel function in a target cell.
Furthermore, MS Excel automatically converts values appropriately. So, the command T Excel is more useful for providing compatibility with other spreadsheet applications.
How To Use T Function In Excel?
We can implement the T Excel function in the following two ways:
1. Access the function from the Excel ribbon.
2. Enter the function into the worksheet manually.
Method #1 – Access The Function From The Excel Ribbon
Choose a cell for showing the result – The Formulas tab – The Text function group down arrow – The T Excel function.
The Function Arguments window will pop open, where we feed the command T Excel argument value.
Finally, click OK to view the value the function T Excel returns in the chosen cell.
Method #2 – Enter The Function Into The Worksheet Manually
1. Select a cell to display the outcome.
2. Type =T( in the cell.
3. Provide the argument value and close the brackets.
4. Press Enter to secure the T Excel function output.
Let us see the practical uses of the T Excel function with the help of examples.
Example #1
The source dataset shows a list of values in column A.
The task is to show the column A values in the corresponding column B cells if they are texts. Otherwise, the concerned column B cell should hold an empty text.
Then, we can execute the T Excel function in the required cells to acquire the expected outcome.
Step 1: Click cell B2 to choose it, enter the T(), and press Enter.
[Alternative choose the target cell and then Formulas – Text – T Excel function.
Enter the T Excel function argument value in the Function Arguments window, which opens on choosing the function.
Finally, click OK to view the T() return value in the chosen cell.]
Step 2: Using the Excel fill handle, execute the function in the other target cells.
The results indicate that cells A2 and A8 contain text values. Next, the function does not count logical values and numeric values as texts, leading to it returning a blank text in cells B3:B6.
Furthermore, if the supplied value is an error value, the function returns the error value as is, confirmed by the cell B7 T() output.
Example #2
We have an employee’s details and the hike percentage they received updated in column B.
The task is to determine whether the data in the column B cells are texts, with the output being the logical value TRUE or FALSE. We shall consider cells C2:C3 as the target cells.
Then, the paired T Excel function with the IF Excel function in the specified target cells will fetch us the desired outcome.
Step 1: Choose the target cell C2, enter the IF() containing the T Excel function, and press Enter.
Next, using the fill handle, update the expression in cell C3.
First, the T() in the IF() condition checks the corresponding column B cell value and gives the value as the output if it is a text. Next, the IF() checks if the T() output is the corresponding
column B cell value. If the condition holds, the IF() output is the TRUE value. Otherwise, it returns the FALSE value.
In the case of the cell C2 formula, the T() returns a text value as the cell B2 value is text. So, the IF() condition holds, leading it to return TRUE as the output.
Further, in the case of the cell C3 formula, the T() returns a blank text as the cell B3 value is not a text. So, the IF() condition is FALSE, leading it to return FALSE as the output.
Example #3
The source dataset shows clients’ first and last names, and their associated branch office details.
We must show the complete details of each client, with Excel line breaks, in the corresponding column D cells, for which the Excel Wrap Text option in the Home tab is enabled.
Then, the paired T Excel function with the Excel CONCAT function in the cited target cells will get us the anticipated results.
Step 1: Choose cell D2, enter the CONCAT() containing the T Excel function, and press Enter.
=CONCAT(T(A2),” “,T(B2),CHAR(10),T(C2))
Next, utilize the excel fill handle to enter the formula in cells D3:D4.
In the case of the cell D2 formula, the three values (cells A2, B2, and C2 values) concatenated with a space character and a line break character are texts. So, the three T()s in the CONCAT() return
the three texts as they are in the source dataset. Next, the CONCAT() combines them with the specified space and line break characters in the cited order to return the concerned client details
according to our requirements.
Furthermore, the same logic applies to the formulas in cells D3 and D4.
However, the cell B3, which should hold the specific client’s last name, is blank. So, the corresponding T() in the CONCAT() returns an empty text, making the formula return only the client’s
remaining details as the output.
In the case of the cell D4 formula, cell C4, which should show the specific client’s branch office data, holds the numeric value of 0. So, the corresponding T() in the CONCAT() returns an empty text,
making the formula showcase only the client’s remaining details.
T Excel Function Vs N Excel Function
The T Excel Function vs. N Excel function is as follows:
• Excel Built-in Function Group
While the T() comes under the Text Excel function group, the N() falls under the Information Excel function group.
• Definition
While the T() determines the text value referred to by the input value, the N() enables us to get a value translated to a number value.
• Function Arguments
Both functions accept one mandatory argument, value, as input.
• Function Output
The T() returns the input value as it is if the input value is a text.
However, the N() output varies with the input value. For instance, if the input value is a number, the function returns the same number. Next, if the input value is TRUE, FALSE, or a date, the
function returns 1, 0, or the serial equivalent of the specified date value, respectively. Further, if the input value is an error value, the function returns the error value. However, if the
input value is anything other than the values mentioned above, the function returns 0.
Important Things To Note
• MS Excel automatically converts values according to their data types. So, the T Excel function is typically useful for providing compatibility with other spreadsheet applications.
• When the supplied input value to the T function in Excel is an error value or a special character, the function treats the value as text. So, it returns the specified error value or special
character as the text value in the target cell.
Frequently Asked Questions (FAQs)
1. How to accept only text values using T function?
We can accept only text values using the T function, as explained below with an illustration.
The source dataset holds a list of customers and their feedback.
The task is to list only the text feedback in column D. Otherwise, the corresponding target cells in column D should hold a blank text.
Then, we can enter and execute the T() in the cited target cells to secure the required text data.
Step 1: Choose cell D2, enter T(), and press Enter.
Next, using the fill handle, feed the function in the remaining target cells.
The results in column D data indicate that the feedback received from Cust_1, Cust_3, Cust_4, and Cust_5 are text values, which the T() returns in the respective target cells. Please note that the T
() counts ‘–‘ as a text value, leading it to return it as a text value in cell D6.
However, the feedback received from Cust_2 is a number, making the T() return a blank text in the target cell D3.
2. What is the T function in Excel VBA using VBA?
There is no built-in T function in Excel VBA using VBA. However, we can write a user-defined macro or code in VBA to secure the T() output.
3. When should you use the T Function in Excel?
You should use the T function in Excel in the following cases:
• The function is useful when we aim to perform data validation based on text values.
• The function is useful when we aim to perform data manipulation based on text values.
• The function is useful when we aim to filter or sort the given data based on text values.
Download Template
This article must be helpful to understand T Excel, with its formula and examples. You can download the template here to use it instantly.
Recommended Articles
Guide to What is T Excel. Here we explain how to use T function in Excel with examples, points to remember and downloadable Excel template. You can learn more from the following articles – | {"url":"https://www.excelmojo.com/t-excel/","timestamp":"2024-11-05T20:34:37Z","content_type":"text/html","content_length":"216134","record_id":"<urn:uuid:a04460f1-1d1b-430b-afdc-0825aa685308>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00610.warc.gz"} |
Life Cycle Consumption with Assets and Labor Supply Problem - NEOS Guide
Life Cycle Consumption with Assets and Labor Supply Problem
Problem Statement
In the Life Cycle Consumption with Assets Problem, we extended the basic model to allow the consumer to borrow and save. The Life Cycle Consumption with Assets and Labor Supply Problem further
extends the basic model to include labor decisions. In the previous models, the wage income in each period was a function of the number of periods and the current period. In this model, the consumer
decides how much to work in each period as well as how much to consume in each period in order to maximize total utility. The utility function now is a function of both consumption and labor, where
consuming more and working less increases the individual’s utility. The objective of the life cycle consumption with assets and labor supply problem is to determine how much to work and how much to
consume in each period so as to maximize total utility.
Mathematical Formulation
To formulate the life cycle consumption with assets and labor supply problem, we start from the formulation for the life cycle consumption with assets problem and modify it to include the labor
supply decisions.
P = set of periods = {1..\(n\)}
\(w(p,n)\) = wage income function
\(r\) = interest rate
\(\beta\) = discount factor
\(a_{min}\) = minimum asset level
Decision Variables
\(c_p\) = consumption in period \(p\), \(\forall p \in P\)
\(a_p\) = assets available at the beginning of period \(p\), \(\forall p \in P\)
\(l_p\) = labor supply in period \(p\), \(\forall p \in P\)
Objective Function
Let \(u()\) be the utility function and let \(u(c_p, l_p)\) be the utility value associated with consuming \(c_p\) and working \(l_p\) in period \(p\). Utility in future periods is discounted by a
factor of \(\beta\). Then, the objective function is to maximize the total discounted utility:
maximize \(\sum_{p \in P} \beta^{p-1} u(c_p, l_p)\)
The life cycle consumption with assets and labor supply model again tracks the assets available at the beginning of each period but now the wage income depends on the labor supply. That is, the
wealth at the beginning of period \(p+1\) equals the wealth at the beginning of period \(p\) minus consumption plus the wage income multiplied by the labor supply, all multiplied by the return \(R\)
on savings (where \(R = 1 + r\)). There is one constraint for each period:
\(a_{p+1} \leq R(a_p – c_p + w(p,n)*l_p), \forall p \in P.\)
The model assumption is that initial wealth is zero (\(a_1 = 0\)) and that terminal wealth is non-negative (\(a_{n+1} \geq 0\)).
As in the life cycle model with assets, there is a minimum asset level, \(a_{min} \leq 0\). If \(a_{min} = 0\), then no borrowing is allowed. Otherwise, an individual can borrow as long as s/he can
repay the amount before period \(n\). Therefore, there is one lower bound constraint for each period:
\(a_p \geq a_{min}\)
Also, the amount consumed in each period should be non-negative:
\(c_p \geq \epsilon, \forall p \in P.\)
• Enter a planning horizon (max=15), a discount factor, and an interest rate.
• Enter a minimum asset level. A minimum asset level of 0 means that no borrowing is allowed; a minimum asset level less than 0 means that borrowing is allowed.
• The default utility function and wage function are displayed.
• Click “Submit” to solve the problem using the NEOS Server.
• Wait for the results to show up in the solution section at the bottom. The solution section will display the
objective value and the optimal amount to consume in each period along with the corresponding labor supply and asset value.
Input Parameters
Planning Horizon (T)
Discount Factor
Interest Rate
Minimum Asset Level
Utility Function
Wage Function w(p,T)
Objective Value
Consumption Labor Assets
Period 1
Period 2
Period 3
Period 4
Period 5
Period 6
Period 7
Period 8
Period 9
Period 10
Period 11
Period 12
Period 13
Period 14
Period 15
GAMS Model
$Title Life Cycle Consumption - with savings and elastic labor supply
Set p period /1*15/ ;
Scalar B discount factor /0.96/;
Scalar i interest rate /0.10/ ;
Scalar R gross interest rate ;
R = 1+i ;
Scalar amin minimum asset level /0/ ;
$macro u(c,l) (-exp(-c) - power(l,2))
Parameter w(p) wage income in period p ;
w(p) = ((15 - p.val)*p.val) / 15 ;
Parameter lbnds(p) lower bounds of consumption
/ 1*15 0.0001 / ;
Positive Variables
c(p) consumption expenditure in period p ,
l(p) labor supply at period p ;
Variable a(p) assets (or savings) at the beginning of period p ;
Variable Z objective ;
budget(p) lifetime budget constraint ,
obj objective function ;
budget(p) ..
a(p+1) - R*(a(p) + w(p)*l(p) - c(p)) =l= 0 ;
obj ..
Z =e= sum(p, power(B, p.val - 1)*u(c(p),l(p))) ;
Model LifeCycleConsumptionSavings /budget, obj/ ;
c.lo(p) = lbnds(p) ;
a.lo(p) = amin ;
a.fx('1') = 0 ;
Solve LifeCycleConsumptionSavings using nlp maximizing Z ; | {"url":"https://neos-guide.org/case-studies/egt/economics/life-cycle-consumption-with-assets-and-labor-supply-problem/","timestamp":"2024-11-03T23:03:40Z","content_type":"text/html","content_length":"92638","record_id":"<urn:uuid:d773cf00-86e7-42d9-9fdd-922c87dc509a>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00538.warc.gz"} |
Winner Take All? Exploring Probability through Games of Chance
Students will be introduced to the unit and its theme- using games of chance to help student council raise money for the school. They will understand that this will involve playing games, assessing
the fairness of games, changing unfair games to fair ones, determining the odds for particular events and how to determine the payout for particular wagers. They will also calculate the extremes –
what is the most money the students could earn, what is the least and how they can use their knowledge of probability to help student council earn the maximum.
As a culminating activity, students will create their own game, calculate the probabilities of winning, and then play the game with several classmates. Students will then discuss with the designer
what they liked about the game, and how, if necessary, it could be improved. Students will then make revisions to their games
Part 1 – Explore Fairness in Games
Begin by asking students to respond in their notebooks to the following questions:
What do you think it means for a game to be fair? What conditions must be present?
Would you want to play an unfair game, even if it was in your favor?
Have students share their ideas and record them on chart paper. This will be saved for the end of the unit and discussed later.
Candyland Activity
As suggested by D. DeTurck and adapted from “Connected Mathematics Unit – What Do You Expect?” Use to review basic concepts of probability and to get to the idea of fairness in probability.
Candyland game
In lieu of the game, use colored cubes or tiles
Non transparent container or bag
Student notebooks
Poll the class to see who has played the game Candyland? (Most likely all have.)
Ask if they think this is a fair game and why.
Take out all the picture cards and place the remaining color cards in a container.
Tell students there are red, purple, yellow, blue, orange and green cards in the container and that they will have to predict the fraction of each color in the container, without emptying it.
One by one have a student pick a card, record the color on the board then return the card to the container.
Once all students have picked ask:
How many cards of each color did the class pick?
Which color are there most of? Which the least?
Predict the fraction of each color.
Finally, look at all the cards. Find the actual fraction of cards for each color. How close so these come to your predictions?
Discussion points – Students answer in journals and discuss as a class.
Is each card equally likely to be picked?
Is each color equally likely to be picked?
What is the probability of picking a white card?
How many cards would need to be added to make the probability of
Picking blue, for example, ½?
Is this game fair? Why/Why not
Discuss/explain how fairness in probability is quite different from the everyday meaning of fairness, which we understand as something that is just or honest. In probability, fairness means there is
nothing in the rules that would favor one player over another – everyone has an equal chance of wining. Conversely, it does not mean that each person will win the same number of times, nor does it
mean that the player who wins this time will win the next time.
Spinner Games
Sometimes in complex situations it can be difficult to determine if a game is fair or not. But students of this age can develop an understanding of fairness by gathering data on winners and losers
to determine if each player had an equal chance. (Bright, et al 50) Collecting data is simple- play the game and record the results. The important thing here is to play many games.
I’ll Race You (Adapted from Investigations)
One equally divide spinner per pair of students
Recording sheet and pencil
In this game students use an equally divided two- color spinner. (Green/White)
Players decide who goes first by tossing a coin. Winner chooses his/her color.
Players take turns spinning the spinner
Green player gets a point if the spinner lands on a green section
White player gets a point when the spinner lands on a white section.
You get a point whenever the spinner lands on your color, no matter who spins the spinner.
On a score sheet with ten slots, record an x each time your color comes up
First player to fill in an x on every slot wins that round
Play ten rounds
Sample score sheet: Round # 1
Green / White
After reading the rules, but before students play, ask:
Is this a fair game?
Which player would you rather be?
What’s the probability of the green player winning a round?
What is the probability of the white player winning a round?
After ten rounds, how many rounds will each player have won?
Create 2 line plots on the board – one for each color. As students complete a round, have them record the win on the line plot.
When the whole class has completed all rounds and recorded their results, look at the line plots and discuss the data. Ask students to consider:
How can we represent the results?
Were they surprised by the results?
Did one color win more often than another? Does that mean the game is unfair?
What kind of results would we get if we played many more games?
At this point, have student pairs create a replica of the manual spinner using the virtual manipulatives site and play 100 rounds and record the results.
Use the results from these activities to reinforce the concepts of experimental and theoretical probability and the Law of Large Numbers.
How are these results different from the class’s ten rounds?
What percent of the time does each player win?
Use these results to reinforce concepts of experimental and theoretical probability
Spinner Two
Using the virtual manipulative library, have students create 2 spinners.
Each will have 12 sections with 3 different colors
5 green, 5 blue, 2 red
Spinner 1 should have the sections arranged as follows:
Spinner 2 should have the sections arranged as follows:
Ask students to analyze these spinners.
Would you have a better chance of landing on green in spinner 2?
Would you have a better chance of landing on blue in spinner 2?
What is the probability of landing on green in each spinner? Of landing on blue?
Are the spinners the same from the probability standpoint?
Could these spinners be used in a fair game?
If you wanted to redesign the spinner and you could not take any sections away but you could add sections, how many red sections could you add?
Number Cube Games
Two different colored dice or 6-sided 1-6 number cubes
Paper and pencil for each student pair
Addition Game (2 players)
Both players take turn rolling the number cubes
Player 1 gets a point of the sum of the faces is odd
Player 2 gets a point if the sum of the faces is even
Play continues for 36 rolls
Players record the results of each roll
Introduce the game with a few demonstration rolls and sample result recording.
Ask students if they think this is a fair game and why.
If students don’t think so, ask them which player it favors.
Have pairs play the game, keeping track of their results
Afterwards have students record the answers to these questions in their notebooks:
Based on your data, what is the experimental probability of rolling an odd sum?
An even sum?
List all the possible pairs of numbers you can roll with two number cubes.
What is the theoretical probability of rolling an odd sum? An even sum?
Is this a fair game?
Create a line plot on the board using the possible outcomes of adding the two cubes.
Ask a pair for their results and record it on the plot.
Analyze the probabilities of rolling an even sum and an odd sum.
Discuss which sums were more likely.
Combine the class data in a similar way and discuss the experimental probability of rolling an odd sum and an even sum and how they found these probabilities.
Based on the data, is the game fair?
Students may have had difficulty listing all the possible pairs of numbers that can be rolled with two number cubes. At this point it would be very useful to introduce them to the following chart as
a means of finding all the possible sums (the sample space).
Have students copy the chart into their notebooks as an example and for future reference.
Number Cube 1
+ 1 2 3 4 5 6
Cube 2 3 4 5 6 7 8 9
This will help the class compare the theoretical probabilities to their experimental probabilities. To help them analyze this and to review other concepts ask:
How many ways are there to get an even sum?
How many ways to get an odd sum?
Does each player have an equal chance of winning this game?
What is the probability of getting a sum that is a prime number?
A multiple of 5?
A multiple of 2 and 3?
A factor of 24?
A multiple of 15?
A 7?
An 11?
Keep this chart for use later when students are determining what dollar amounts to assign each roll in the student council game.
Have students recreate this simulation using virtual manipulatives and 1000 throws of the cubes. Record the results.
How are the results the same as those of the class?
How are they different?
Multiplication Game (Not so Fair)
This is similar to the addition game, but points are scored based on the product of the rolls.
Player 1 gets a point if the product of the roll is odd
Player 2 gets a point if the product of the roll is even.
Play continues for 36 rolls and students record the results of each roll
Player with the most points wins.
Have students play the game and when done, answer the following questions in their notebooks:
Based on your experimental data, what is the probability of rolling an
odd product? An even product?
What is the theoretical probability of rolling an odd product? An even product?
Do you think this game is fair? Why or why not?
Make a line plot with the possible products from rolling two number cubes and record the class results for each time a product came up.
Compute the class’s experimental probability for the rolls.
Have the class create a chart for the product possibilities similar to the one created for the addition possibilities.
Using this, discuss the theoretical probabilities of rolling an odd product and of rolling an even product.
Discuss the differences between the experimental and theoretical probabilities.
As with the addition chart, use it to extend and review other concepts:
What is the probability of getting:
A prime number?
A square number?
A multiple of 2 or 3?
A factor of both 12 and 15?
Ask students if this is a fair game? If they decide it isn’t have them redesign the game to make it fair.
Once students have had a chance to make revisions, ask for a pair of volunteers to share their new and improved game. Have the class play several rounds to get a feel for the game and ask them if
they think everyone now has an equal chance of winning and why they think so.
Part 2- Making Money for Student Council
For this section, students will use the charts created for the number cube addition and multiplication games. They will place the bets and play the games, then determine the money (the least and the
most) that student council could make.
Game 1
Students have already calculated the probabilities for rolling the sums for each of the following events and the pay-outs are listed accordingly:
Doubles 1/6 5 times the bet
Primes 15/36 (5/12) 2 times the bet
Odds ½ 1 times the bet
Evens ½ 1 times the bet
7 1/6 4 times the bet
These outcomes will be printed on a mat and a player will choose an outcome and place a token on that space.
Each player will pay $2.00 to play. The player will then throw the number cubes.
If the expected outcome comes up, the player will receive the posted pay out.
If it doesn’t, the money remains in the bank.
If 72 people place a set on each outcome, how will student council fare?
Will they make money, lose money, or break even? Have students compute the “house” take.
Game 2
Student Council can use the addition game in a simpler form.
Each player will pay $2.00 to play. The player chooses an odd or an even number for the outcome of the throw. If the player wins, the payout is $3.00.
If 100 people play, ill student council make money, lose money, or break even?
What is the most they could win?
What is the least they could win?
Students can use the virtual manipulative site to help with these decisions.
Game 3
Uses the data from the number cube multiplication game.
Each player will pay $2.00 to play.
The player chooses to play for an odd product or an even product.
If the player chooses odd he receives 3 times his bet
If he chooses even, he gets double his bet
If 100 people play, will student council make money, lose money or break even?
What is the most they could win?
What is the least they could win?
Part 3 – Make Your Own Game
As a culminating activity, students will design their own game of chance.
Before they begin their work on this project, however, bring out the chart paper from the beginning of the unit and review with them their original responses to the question of a game’s fairness.
Discuss what they thought originally and what they think and know now. These insights should guide them as they design their games.
The games can use dice, spinners, colored blocks, or coins.
The game must be fair and students will need to explain how they know what they’ve created is fair.
Students will calculate the experimental and theoretical probabilities associated with the game
Students will make a board, mat or whatever playing surface is needed Consideration should be given to the size of the playing field, interesting and colorful graphics and clarity of the playing
Student designers will write out the rules for the game in clear, concise language.
Students will then have an opportunity to “test drive” their games by playing with other students in order to collect data about their games.
As the games are played, students will make decisions about their own games as well as their classmates’ games.
They will consider features such as: is it fun to play, is it easy to play without being boring or overly repetitive, is it fair, would you play it again if you didn’t have to.
Students can offer suggestions to their classmates and game designers can then make modifications before submitting the final version for a grade. | {"url":"https://theteachersinstitute.org/curriculum_unit/winner-take-all-exploring-probability-through-games-of-chance/","timestamp":"2024-11-05T01:00:19Z","content_type":"text/html","content_length":"312988","record_id":"<urn:uuid:29175a2d-5bf6-4e30-9ce1-d1c3143bd612>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00195.warc.gz"} |
10 Real Estate Calculations Every Investor Should Know
by Shane Sauer | Sep 26, 2018 | Curated Articles | 0 comments
In real estate, not everything will fit into a neat little box. Creativity is a must. However, there is indisputably a mathematical side to successful real estate. Sooner or later, you must put pen
to paper and figure out if it makes sense to buy a particular property. These 10 key calculations will prove vital to your ability to analyze potential deals.
#1: Rent-to-Cost
Monthly Rent / Total Cost (Purchase Price + Rehab + Acquisition Costs)
Example: Rent = $900/Month
Total Cost = $75,000
Rent to Cost = 1.2% ($900 / $75,000)
When to use it: If you need to quickly determine if a potential rental is worth evaluating further.
Click here to use the calculator.
What you need to know:
• Cheaper properties have higher rent-to-cost
• Use this to compare properties, not make a final decision
• Buy-and-hold investors should look for a value greater than 1 percent
#2: Gross Yield
Annual Rent / Total Cost
Example: Rent = $900/Month, $10,800/Year
Total Cost = $75,000.
Gross Yield = 14.4% Gross Yield ($10,800 / $75,000)
When to use it: In context with institutional investors. Gross yield uses the same values as rent-to-cost.
Click here to use the calculator.
#3: Gross Rent Calculator
Sale Price / Gross Annual Rents
Example: Sale Price = $500,000
Annual Rents = $50,000
Gross Rent Multiplier = 10 ($500,000 / $50,000)
When to use it: When looking for a quick sign a potential commercial deal (multifamily or otherwise) is worth considering.
Click here to use the calculator.
What you need to know:
• Does not take operating expenses into account
• May be calculated using sales price or total cost (including rehab and acquisition)
#4: Cap Rate
(Net Operating Income / Total Cost of Property) x 100
Example: Net Operating Income (NOI) = $80,000
Total Cost = $1,000,000
Cap Rate = 8.0 ($80,000 / $1,000,000) x 100
When to use it: When you need a way to compare the returns on different investment properties.
Click here to use the calculator.
What you need to know:
• Cap rate does not include debt servicing costs
• Can be calculated using different definitions of cost and NOI
• Best used when derived using actual income and expense numbers from the previous year
• Beware of pro forma numbers
#5: Debt Service Coverage Rate (DSCR)
Net Operating Income / Annual Debt Service
Example: NOI = $120,000
Annual Debt Service = $100,000
DSCR = 1.2 ($120,000 / $100,000)
When to use it: To evaluate the viability of an investment from a lending perspective.
Click here to use the calculator.
What you need to know:
• Banks use DSCR to predict whether a borrower is likely to repay a loan.
• Compares an investment’s income to the amount of the loan payment (interest and principal)
• Most bank lenders want DSCR to be at least 1.2
#6: The 50% Rule
Gross Rents x 0.5 = Approximate Annual Operating Expenses
Example: Gross Rents = $200,000
50% Rule: $200,000 X 0.5 = $100,000 (Approximate Annual Operating Expenses)
When to use it: When you need a fast estimate for operating expenses.
Click here to use the calculator.
What you need to know:
• Usually used with apartments
• Can be applied using real numbers or approximations
• May be inserted into cap rate and DSCR estimates
#7: The 70% Rule
Strike Price = (0.7 x After Repair Value) – Rehab
Example: After Repair Value (ARV) = $200,000
Rehab = $30,000
Strike Price = (0.7 x 200,000) – $30,000 =$110,000
When to use it: To determine the strike price most optimal for solid profits on a potential investment.
Click here to use the calculator.
What you need to know:
• Usually used for flips
• The best ARV values are derived using comparable costs from comparable properties
• Finalize your strike price after a detailed analysis that includes carrying costs
#8: Cash on Cash Return
(Net Operating Income – Annual Debt Service) / Cash in Property
Example: NOI = $120,000
Annual Debt Service = $100,000
Cash in Property = $100,000
Cash on Cash Return = ($120,000 – $100,000) / $100,000 = 20%
When to use it: To evaluate the success of an investment after its first year.
Click here to use the calculator.
What you need to know:
• This only provides a return value for one year
#9: Return on Investment (ROI)
(Gain on Investment – Cost of Investment) / Cost of Investment
Example: Gain on Investment = $20,000
Cost of Investment = $50,000. Years owned = 5
ROI = 40% ($20,000 / $50,000)
Annualized Return (AR): ROI/# years you owned the property
AR = 8% (40%/5)
When to use it: To get a “whole picture” of a future investment or a completed investment.
Click here to use the calculator.
What you need to know:
• Use ROI to calculate annualized returns by dividing ROI by the number of years you have owned the property.
#10: Internal Rate of Return (IRR)
IRR is defined as “the discount rate that makes an investment’s net present value equal to zero.” IRR accounts for when you spend and receive cash. For most investors, it is better to receive cash
sooner rather than later because then you can reinvest it. You will need an IRR calculator (my favorite is the IRR function available in Excel) to derive it.
When to use it: To put your ROI in perspective with your initial outlay of money or compare investment opportunities in different asset classes.
What you need to know:
• IRR may also be referred to as “economic rate of return” or “discounted cash flow rate of return”
• IRR omits external factors like cost of capital and inflation
• In most cases, the higher a project’s internal rate of return, the more desirable it is
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Around the world, there is a two-track education, similar to the two-track gymnasiums where at the beginning there was only a general course, and when boys and girls like Newton, Gauss and Sofia
Kovalevska appeared, natural courses were introduced along with the general course in the gymnasiums. By registering a private association of mathematicians, I, as a responsible person, came to
realize that people who have a general high school education do not know how to work on solving mathematical problems, i.e. of the industry of the sun as a mathematical star, which represents my one
hundred percent mathematical basis of interest in the letters I sent to the London Mathematical Society for MUN registration and the German Mathematical Association for MEU registration. My proposal
for the introduction of two-way education in the world and the letter I sent to the German Mathematical Association is a proposal for the introduction of two-way high school education in Europe, like
the one they have in the USA and in other countries, all of whom would be one hundred percent collaborators of the London, i.e. Woolsthorpe MUN.
MUN mathematicians would prove mathematically that there are no differences between mathematics and life, i.e. that life cannot be interrupted, just as mathematical continuity cannot be interrupted.
Then, for every human being to realize that Sir Isaac Newton is the author of mathematical, i.e. industry of the sun as a mathematical star, and that MUN mathematicians in their countries should work
on solving the problem of building mathematical, i.e. industries of the sun, I repeat, as a mathematical star. Building a mathematical industry implies that MUN mathematicians set and solve the
problem of constructing a mathematical receiver, and thus, with Sir Isaac Newton’s mathematical knowledge, they would initially solve the problems of building factories that would produce the
mentioned mathematical receivers, which would not be a complex problem for MUN mathematicians. A complex problem would be how to construct a factory that would produce mathematical drinking water,
i.e. drinking water for all times in unlimited quantities, using the energy obtained from the sun as mathematical star, with Newton’s knowledge of mathematics.
Along with the proposal for the logo, I also suggest to the leadership of the London Mathematical Association that, in cooperation with the British authorities, they organize a referendum in which
the people of Britain will be asked the question: “ARE THEY IN FAVOR OF MOVING THE HEADQUARTERS OF THE LONDON MATHEMATICAL SOCIETY FROM LONDON TO THE PLACE OF WOOLTHORPE, THE BIRTHPLACE OF SIR ISAAC
NEWTON, WHICH WOULD TRANSFORM THIS SOCIETY INTO THE MATHEMATICALLY UNITED NATIONS (ABBREVIATED MUN): YES OR NO.” If the majority of British people answered YES, British lawyers would write the MNU
All Britons who circle YES, at the same time will send a message to the world that Sir Isaac Newton will be an example to follow throughout their lives. Moreover, I will propose to the authorities in
Serbia that Serbia becomes an initial member of the WOOLSTHORPE MATHEMATICALLY UNITED NATIONS.
In the STATUTE of the MUN, it would be written that a member of the MUN can only be a country in which a referendum on membership was previously organized. According to the STATUTE, the organization
of the referendum would be legally binding for the governments of all registered states.
Mathematically United Nations imply the mandatory calling of a referendum as an initial problem to be solved by states interested in MUN membership.
Mathematicians of the MUN countries would solve the problem by building a mathematical, i.e. industry of the sun as a mathematical star in their countries to enroll all residents of their countries
on the payroll of the sun, for an unlimited time and amount of money, and always, I repeat, the sun as a mathematical star.
The resolutions that will be passed are mathematical, which means that they can be proposed by any member of the MUN, and for its adoption it is necessary to secure five billion and one hundred
thousand votes of the earth’s inhabitants, which they would secure through their representatives in the MUN. All staffers who would be employed in the MUN would have a binary education, which means
at the beginning a compulsory top mathematics education, and at the end an education of their own choice, for example: mathematician-lawyer, mathematician-engineer, mathematician-financier, etc. They
would have a similar education as the 18 members of the Swiss mathematician Bernoulli’s family.
All this would be written in the founding act, statute and program of the MUN.
Mathematicians of the ASSOCIATION OF THE SUN OF SERBIA will work on solving the problem so that every resident of Serbia continues to solve the mathematical problems of the knowledge of Sir Isaac
Newton and Carl Friedrich Gauss throughout Serbia for an unlimited number of years, just as if Sir Isaac Newton and Carl Friedrich Gauss were still alive.
With unlimited respect,
Nenad Djukic | {"url":"https://sajtsuncasrbije.org/proposal-for-the-logo/","timestamp":"2024-11-09T23:16:19Z","content_type":"text/html","content_length":"47522","record_id":"<urn:uuid:ad9b57e0-a39a-4613-9893-e55d6ac2cbf0>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00443.warc.gz"} |
MAT10:8.10.1 COMPLETING THE SQUARES FOR ax²+bx+c - Alaprann.mu
MAT10:8.10.1 COMPLETING THE SQUARES FOR ax²+bx+c
Are you ready to unravel the secrets hidden within quadratic expressions? Join us on an exhilarating quest to master the art of completing the square for expressions of the form ax²+bx+c.
Seize this opportunity to embark on a thrilling mathematical odyssey, where every question is a challenge waiting to be conquered, and every solution is a triumph worth celebrating. Take the quiz
today and embark on a journey to unlock the mysteries of quadratic expressions!
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Multi-Column Atrous Convolutional Neural Network for Counting Metro Passengers
School of Mechanical and Power Engineering, Zhengzhou University, Zhengzhou 450000, China
School of Electrical Engineering, Zhengzhou University, Zhengzhou 450000, China
Author to whom correspondence should be addressed.
Submission received: 24 March 2020 / Revised: 15 April 2020 / Accepted: 20 April 2020 / Published: 24 April 2020
We propose a symmetric method of accurately estimating the number of metro passengers from an individual image. To this end, we developed a network for metro-passenger counting called MPCNet, which
provides a data-driven and deep learning method of understanding highly congested scenes and accurately estimating crowds, as well as presenting high-quality density maps. The proposed MPCNet is
composed of two major components: A deep convolutional neural network (CNN) as the front end, for deep feature extraction; and a multi-column atrous CNN as the back-end, with atrous spatial pyramid
pooling (ASPP) to deliver multi-scale reception fields. Existing crowd-counting datasets do not adequately cover all the challenging situations considered in our work. Therefore, we collected
specific subway passenger video to compile and label a large new dataset that includes 346 images with 3475 annotated heads. We conducted extensive experiments with this and other datasets to verify
the effectiveness of the proposed model. Our results demonstrate the excellent performance of the proposed MPCNet.
1. Introduction
As an important means for urban public transportation, subways are facing challenges with regard to rapid route expansions and safety-related problems owing to an increase in passenger flow.
Consequently, there is an urgent demand for secure methods of forecasting passenger flow using video surveillance. Such methods use computer vision and artificial intelligence to analyze the content
of video sequences, and to track and detect anomalous information. There is considerable research on passenger flow analysis [
]. In works [
], regions corresponding to moving objects are detected using a background difference method. In the work [
], a detection-based strategy is proposed based on the heads and shoulders of detection targets to detect subway passenger flow. This method performs well, but it cannot be used to count the number
of passengers in a subway car. Most of the time, passengers in subway cars remain still, yet the background difference method is more suited to detecting moving targets because of the need to update
the background. Sometimes subway cars are highly crowded, as shown in
Figure 1
. In such cases, the algorithm proposed in the work [
] encounters problems of misdetections and false detections. Single-image crowd counting is useful for traffic management, disaster prevention, and public management. Crowd-counting methods aim to
estimate the number of humans in surveillance videos and photos. Current methods of crowd counting are developed from detection-based [
] approaches to convolutional neural network (CNN)-based approaches [
]. This reduces counting errors caused by occlusion, because CNN-based approaches only target the human head. Therefore, CNN-based crowd counting methods are suitable for the counting dense crowds on
To apply a CNN-based method to counting subway passengers, we developed a methodology and a dataset. We designed a novel multi-column atrous CNN that uses ResNet50 [
] pre-trained on the ImageNet [
] dataset as the backbone of the network to extract deep features. Previous works [
] arrange the convolution layers of different convolution kernels into multiple columns to extract large-scale information. By contrast, we focus on using atrous spatial pyramid pooling (ASPP) [
] to extract multi-scale features. Specifically, ASPP probes an incoming convolutional feature layer with filters at multiple sampling rates and effective fields of view. This module consists of
atrous convolution with four different rates in parallel to capture objects and the context in images at multiple scales. Unlike methods based on arranging convolution layers into columns, our method
uses filters with multiple sampling rates to extract information at a larger scale.
We also developed a new dataset that contains 346 images with 3475 labeled passengers for metro passenger analysis. The data was collected from video of Zhengzhou Metro Transportation (MT) Line 2, in
China. Thus, we refer to it as the Zhengzhou MT dataset.
Figure 1
shows representative images of our dataset. Compared to existing crowd-counting datasets, our dataset offers distinct advantages. To our knowledge, the dataset is the first one designed for counting
passengers inside a subway car. Furthermore, due to the narrow space in the car, there is considerable congestion in the images.
The contributions of this work can be summarized as follows. First, for the first time, we use a CNN-based crowd counting algorithm to count passengers in subway cars. Second, we designed a novel
multi-scale architecture that extracts deep features and captures multi-scale information in images by using a row of atrous convolutions with different atrous rates. Third, we developed a dataset
comprised of images of the interior of subway cars. The dataset is representative, with realistic images of challenging settings and crowded scenes for analysis in the field of intelligent
The reminder of the paper is organized as follows.
Section 2
presents recent related works.
Section 3
provides details of our proposed metro-passenger counting network (MPCNet). Experimental results are given and discussed in
Section 4
. Finally,
Section 5
concludes the paper.
2. Related Work
A myriad of techniques in computer vision have been proposed to deal with task of crowd counting. They can be roughly categorized into traditional methods and CNN-based methods.
2.1. Traditional Methods
Most earlier research [
] focus on detection-based methods, which consider a crowd as a group of detected individual pedestrians with a simple detection and summing process. Unfortunately, these detection-based methods are
limited by occlusions and background clutter in crowded scenes. Since detection-based methods cannot be adapted to highly congested scenes, other methods [
] employ regression to learn the relations among extracted features from cropped image patches, and then calculate the number of particular objects. Idrees et al. [
] designed a model that fuses features extracted with Fourier analysis, head detection, and scale-invariant feature transform (SIFT) [
] interest-points-based counting in local patches. When executing a regression-based solution; however, spatial information in images of crowds is ignored. This can lead to inaccurate results in
local regions. In works [
], a solution to this problem is proposed, with linear mapping between the features and object density maps in a local region.
2.2. CNN-Based Methods
CNN-based methods exploit density maps, owing to their success at classification and recognition [
]. A comprehensive survey of CNN-based counting approaches is given in the work [
]. Wang et al. [
] modified AlexNet [
] to predict counts directly. In the work [
], a simple but effective multi-column convolutional neural network (MCNN) is proposed that tackles large-scale variation in crowded scenes. Similarly, Onoro and Sastre [
] proposed a multi-scale model, called Hydra CNN, to extract features at different scales. Cao et al. [
] proposed an encoder–decoder network, called SANet, which employs scaled aggregation modules in an encoder. Their method improves the representation ability and scale diversity of features. Sam et
al. [
] proposed Switching-CNN, which utilizes VGG-16 [
] as a density-level classifier to assign different regressors for particular input patches. Li et al. [
] proposed CSRNet [
] by combining VGG-16 [
] and dilated convolution layers to aggregate multi-scale contextual information. Recently, Wang [
] designed SFCN to encode spatial contextual information based on VGG-16 [
] or ResNet-101 [
Based on the research above, we found that by combining deep learning, CNN-based solutions are better able to perform this task, and indeed outperform traditional methods. In particular, networks
based on AlexNet, VGG, and ResNet show excellent performance. Thus, we propose a network with ResNet as the front end.
3. Proposed Method
The fundamental idea for the proposed method is to deploy a multi-column atrous CNN to capture high-level features with larger receptive fields, and to generate high-quality density maps. In this
section, we first describe the ASPP module in detail and introduce the architecture of the proposed method. Then, we present the corresponding training details. Finally, we describe the method for
generating the ground truth.
3.1. ASPP Module
One of the critical components of our design is the ASPP module. As can be shown in
Figure 2
, the ASPP consists of a 1 × 1 convolution and three 3 × 3 atrous convolutions, where the rate = (6,12,18). An atrous [
] convolution can be defined as follows:
$Y ( l , w ) = ∑ i = 1 L ∑ j = 1 w x ( l + r × i , w + r × j ) f ( i , j ) .$
) is the output of the atrous convolution from input
) and a filter
denote the length and width, respectively, and
is the dilation rate. When r = 1, an atrous convolution becomes a normal convolution. The ASPP has been applied to segmentation tasks, demonstrating a significant improvement of accuracy [
], and it is effective at extracting multi-scale contextual information. Although multi-column CNNs [
] are widely used for extracting multi-scale contextual information, they also dramatically increase the number of parameters, owing to a larger convolution kernel. The ASPP can extract multi-scale
contextual information with atrous convolution, adaptively modifying a filter’s field of view by changing the rate value. With atrous convolution, a small-sized kernel with a
filter is enlarged to
+ (
− 1) (
− 1) with dilated value
. Thus, it can flexibly aggregate the multi-scale contextual information. This characteristic enlarges the receptive field without increasing the number of parameters or the amount of computation.
(Note: expanding the convolution kernel size can indeed make larger receptive fields, but doing so introduces more operations).
3.2. MPCNet Architecture
Following the work [
], we selected ResNet50 as the front end of MPCNet as shown in
Figure 3
, owing to its excellent high-resolution feature-extraction capability and its flexible architecture, which can easily concatenate the back end to generate density maps. However, atrous convolution
requires a large number of high-resolution feature maps. Therefore, it is necessary to extract advanced features through ResNet before performing atrous convolution. To do so, we reserve the first
three residual modules in ResNet50 and build the proposed MPCNet with multi-column atrous convolutional layers. In this front-end network, there are 1024 output channels. If we were to continue to
stack more residual modules, then more output channels would be needed, increasing the required training time for the network. The size of feature maps is reduced by 8 times in ResNet50, and there is
no down sampling in other processes. The parameter stride before the third residual module of ResNet50 has adopted the default value (the stride of 7 × 7 Conv and max pooling is 2, the stride of the
first residual module is 1, the stride of the second residual module is 2). The size of feature maps has been reduced by 8 times. If they are reduced again, it will lead to a large amount of
information loss. In order to extract more detailed information and obtain high-resolution feature maps, we changed the stride of the third residual module from 2 to 1.
The resulting features from all of the ASPP branches are then concatenated and passed through another 1 × 1 convolution with 128 channels, before the 1 × 1 convolution with one channel. Finally,
bilinear interpolation is performed at a factor of 8 as the last layer of our MPCNet. This ensures that the output shares the same resolution as the input image. Notably, our network uses a fully
convolutional network. It can accept images of any size, without the risk of distortion.
3.3. Training Details
We trained the proposed MPCNet in an end-to-end manner. Weighted parameters for ResNet50 pre-trained on ImageNet were used to initialize the feature-extraction CNN. The Adam optimizer [
] with a learning rate of 10
was used to train the model. The Euclidean distance was used to measure the difference between the ground truth and the estimated density map, similar to other works [
]. The loss function is defined as follows:
$L ( θ ) = 1 2 N ∑ i = 1 N ‖ F ( X i ; θ ) − F i ‖ 2 2 .$
is a set of learnable parameters in the proposed MPCNet,
is the number of training imag,
$X i$
is the input image,
$F i$
is the ground-truth density map generated by MPCNet parameterized with
for the sample
$X i$
, and
is the loss between the ground-truth density map and the estimated density map.
3.4. Ground-Truth Generation
In this section, we describe the method of converting an image labeled with people’s heads to a density map. Supposing there is a head annotation at pixel
$x i$
in a labeled image of a crowd, we represent it as a delta function
$δ ( x − x i )$
and describe its distribution with a Gaussian kernel [
$G σ$
, such that the density map with
heads is derived as follows:
$F ( x ) = ∑ i = 1 N δ ( x − x i ) ∗ G σ ( x ) .$
The above method is generally applicable to sparse scenes. Following the method of generating density maps in the work [
], we use geometry-adaptive kernels to tackle highly congested scenes. Thus, we generate a density map via geometry-adaptive kernels:
$F ( x ) = ∑ i = 1 N δ ( x − x i ) ∗ G σ i ( x ) , σ i = β d ¯ l .$
$σ i$
depends on the average distance
$d i ¯$
between the head and its nearest
neighbors. In the experiment, we followed the configuration in the work [
], where
= 0.3 and
= 3. The sum of all pixel values gives the crowd count of the input image. Here,
denotes the crowd count, defined as follows:
$C = ∑ l = 1 L ∑ w = 1 W Z l , w .$
are the length and width of the density map, respectively, and
$z l , w$
is the pixel at (
) in the generated density map.
4. Experiments and Results
In this section, we introduce our dataset, and we describe two standard datasets for crowd counting. Then, the evaluation metrics are introduced. Finally, we presents the experiment results to answer
our research problems.
4.1. Datasets
Existing crowd-counting datasets are not designed specifically for public transportation systems, even though crowd counting is important in the field of intelligent transportation. Therefore, we
collected new data and compiled a new dataset, called Zhengzhou MT(Metro Transportation), where the number of heads in an image varies between 1 and 20. We show crowd histograms of the images in our
dataset in
Figure 4
. All images were taken from the Zhengzhou MT, in China. The size of each image is 576 × 704 pixels. The time span of the dataset is from 7:00 am to 9:00 pm, when congestion is variable. Therefore
this dataset is similar to other datasets used in practical applications. Accordingly, the Zhengzhou MT dataset can be considered a valuable and representative dataset. For our evaluation, we used
288 images from the dataset for training and 58 images for testing. The details are listed in
Table 1
The ShanghaiTech Part B dataset was introduced by Zhang et al. [
], and it contains 716 annotated images of sparse scenes taken from the streets of Shanghai, comprising a total of 88,488 people. These images were divided into training and test datasets, with 400
images in the training set and 316 images in the test set. With reference to the work [
], we fixed the size of the Gaussian kernel to 15, where σ = 3, to generate density maps of this dataset.
The SmartCity dataset [
] contains 50 images collected from ten city scenes, including office entrances, sidewalks, atriums, and shopping malls. The dataset has few pedestrians in the images and consists of both outdoor and
indoor scenes. We used this dataset to test the generalizability of the proposed method for sparsely crowded scenes. With reference to the work [
], we used geometrically adaptive kernels to generate the density maps of the Smartcity dataset.
4.2. Evaluation Metrics
In accordance with previous research [
], we used the mean absolute error (MAE) and the mean squared error (MSE) to evaluate the proposed method:
$M A E = 1 N ∑ 1 N | Z i − Z ^ i | , M S E = 1 N ∑ 1 N ( Z i − Z ^ i ) 2 .$
is the number of test images,
$z ^ i$
is the actual number of people in the
-th image, and
$z i$
is the estimated number of people in the
-th image. The
indicates the accuracy of the estimate, and the
indicates the robustness of the estimate. Because the
is sensitive to outliers, its value will be high if the model performs poorly on some samples.
4.3. Experimental Results and Comparison
The implementation of our method is based on the Pytorch [
] framework. Our experiments were performed on an NVIDIA RTX2080Ti GPU with a batch size of 1. Extensive experiments were performed on a variety of datasets to endorse the validity of results.
4.3.1. Results on the Zhengzhou MT Dataset
We compared our method to state-of-the-art methods. To effectively assess the performance of our method, we implemented two recent crowd-counting algorithms [
] capable of extracting multi-scale features. The MCNN [
] is a multi-column CNN that uses several CNN branches with different receptive fields to extract multi-scale features. CSRNet [
] deploys the first ten layers from VGG-16 as the front end and arranges single column atrous convolution layers as the back end to enlarge the receptive fields. Detailed results of the comparison
are given in
Table 2
. The results indicate that the proposed MPCNet outperforms MCNN but not CSRNet. Specifically, the proposed method had an MAE of 0.1 higher and an MSE of 0.2 higher than CSRNet.
Figure 5
shows the density map results obtained from the three methods. Rows 1 and 2 show test images and ground-truth images, respectively. Rows 3 to 5 show density maps generated from MPCNet, CSRNet, and
MCNN, respectively. The proposed method was highly accurate when the subway cars were crowded. In addition, it produced density maps of higher quality than the other two methods. The distribution of
passengers in a subway car can be accurately obtained from these high-quality density maps. Consequently, administrators can improve the service quality of the subway system.
We also compared four levels of congestion. We designed an experiment to verify the robustness of the proposed algorithm, MCNN, and CSRNet under four congestion levels. Such an evaluation is of great
significance to practical applications. Specifically, we selected some images from the test set of Zhengzhou MT and split them into four groups in ascending order according to the crowd counts to
simulate scenes with four levels of congestion in a subway car. From the plots in
Figure 6
, we can see that the three algorithms performed well with the first two levels of congestion, owing to the small number of people. However, with an increase in the number of people, the subway
became crowded, and occlusions between people were more serious. This compromised the accuracy of all three algorithms. In general; however, our algorithm performed comparably well relative to the
two state-of-the-art algorithms.
4.3.2. Results on the ShanghaiTech Part B Dataset
We performed an ablation study on the ShanghaiTech Part B dataset. One of the important features of our method is the ASPP [
] module. Therefore, it is necessary to compare the performance of the method with and without the ASPP module. We removed the ASPP module from MPCNet and tested it on the ShanghaiTech part B,
because it contains scenes with varying scales. In addition, we performed an ablation study to analyze the three configurations of ASPP. This evaluation was designed to demonstrate the necessity of
using the ASPP module. By using the ASPP module, the performance on this dataset improved, with an MAE/MSE of 0.1/1.4 lower than without the ASPP module. However, the different atrous rates of the
ASPP affected the performance. We show these four architectures and the evaluation results in
Table 3
. The architecture with the atrous convolution rate (1,6,12,18) was the most accurate. Therefore, we used this architecture for the proposed MPCNet.
To visualize the ability of the ASPP model, we show density maps generated from the four different architectures in
Figure 7
. The first row shows test images, and the second row shows ground-truth images. Rows 3 to 6, respectively, show density maps generated from the four architectures in
Table 3
. As this figure shows, the architectures without the ASPP module tended to overestimate the count, owing to the interference of the background with the crowds. When the ASPP module was added, this
interference was eliminated. These results demonstrate the need for the ASPP module.
Next, we compared our MPCNet with existing state-of-the-art methods on the ShanghaiTech Part B. The results are shown in
Table 4
. Zhang et al. [
] first used a CNN for density map generation, and their network outputs both density maps and counts. Based on MCNN [
], Sam et al. [
] added a switch classifier to assign a regressor to an image, improving the performance compared to the MCNN. Sindagi et al. [
] proposed a variation to the MCNN as a density map estimator, combining global and local contextual information with multi-scale features. Adversarial loss is utilized to generate high-quality
density maps and significant improvements. In the work [
], multi-task learning is applied to combined features learned from different tasks. Their results on the ShanghaiTech Part B dataset are close to the results in the work [
]. Liu et al. [
] proposed a novel crowd-counting method that uses a large number of unlabeled crowd imagery in a learning-to-rank framework. The self-supervised task improved the results significantly compared to a
network trained only on annotated data. Li et al. [
] arranged cascading dilated convolution layers as the back end of the CSRNet to enlarge the receptive fields. However, a single-column dilated convolution model does not work well with MPCNet, which
uses a multi-column dilated convolution network. The MAE of the proposed MPCNet was 0.9 lower than the CSRNet on the ShanghaiTech Part B dataset. However, our method was not the best among the
existing methods. In the work [
], an approach arranges general convolutions into multiple columns and it also incorporates multi-scale contextual information directly into an end-to-end trainable crowd-counting pipeline. Their
algorithm outperformed state-of-the-art crowd-counting methods.
Figure 8
shows the density map results obtained from the three methods. Rows 1 and 2 show test images and ground-truth images, respectively. Rows 3 to 5 show density maps generated from MPCNet, CSRNet, and
MCNN, respectively. We can find that the accuracy of our method is higher and the density map generated is clearer.
4.3.3. Results on the Smartcity Dataset
To demonstrate that our method can perform counting tasks on extremely dense crowds alongside tasks on relative sparse scenes, we compared our MPCNet with previous state-of-the-art methods on the
Smartcity dataset. We also tried to test CSRNet and MCNN on this dataset. For a fair comparison, we trained MPCNet, CSRNet, and MCNN on the ShanghaiTech Part B dataset and tested it on Smartcity. We
compared our method to the other four methods, and the results are shown in
Table 5
. Our method achieved the lowest MAE (the highest accuracy) among the methods. Specifically, the MAE of the proposed method was 7% lower than that of SaCNN. Samples of the test cases can be found in
Figure 9
which shows the density map results obtained from the three methods. Rows 1 and 2 show test images and ground-truth images, respectively. Rows 3 to 5 show density maps generated from MPCNet, CSRNet,
and MCNN, respectively. We can find that density maps generated by our method are more similar to the crowd distributions in the real images.
5. Conclusions
In this paper, we proposed a method of counting metro passengers, called MPCNet. The proposed method automatically estimates density maps and the number of passengers in images of crowded scenes. We
used multi-column atrous convolutional layers to aggregate the multi-scale contextual information in the congested scenes. By exploiting these layers, MPCNet expands the receptive field without
losing resolution. To evaluate the effectiveness of the proposed method in the field of intelligent transportation, we collected and labeled a new dataset, called Zhengzhou MT, consisting of 346
images and 3475 annotated people. To our knowledge, this is the first dataset with annotated heads designed for counting metro passengers. Extensive experiments with the new dataset and standard
crowd-counting datasets demonstrate the efficiency and effectiveness of the proposed method.
Although our model can extract the multi-scale contextual information in the congested scenes, we hope our model can be more flexible to adapt to the changes of scene scale. Therefore, our future
work will still focus on the multi-scale topic in crowd counting, and further explore how to extract more effective multi-scale features of adaptive scene scale changes. Moreover, in order to apply
our method to practical engineering, we will also explore the relationship between the number of passengers in the car and the degree of passenger congestion.
Author Contributions
Conceptualization, methodology, J.Z.; software, validation, formal analysis, investigation, resources, data curation, writing—original draft preparation, writing—review and editing, visualization,
G.Z.; supervision, project administration, funding acquisition, Z.W. All authors have read and agreed to the published version of the manuscript.
This research was funded by the National Natural Science Foundation of China (NSFC) General Program, grant number 61673353.
We thank Zhengzhou Metro Group Co., Ltd. for providing us with video data.
Conflicts of Interest
The authors declare no conflict of interest.
Figure 2. Atrous spatial pyramid pooling (ASPP). Employing a high atrous rate enlarges the model’s field of view, enabling object encoding at multiple scales. The effective fields of view are shown
in different colors.
Figure 6. Comparison of our method (MPCNet) to MCNN and CSRNet on the Zhengzhou MT dataset. We selected some samples from our test images and split them into four groups, based on the number of
people. The absolute count in the vertical axis is the average crowd number in the images from each group.
Figure 7. We display the density maps generated by four different architectures of MPCNet on ShanghaiTech Part B.
Datasets Number of Images Average Resolution Count Statistics
Total Min Ave Max
SHHB [10] 716 768 × 1024 88,488 9 123 578
Smartcity [32] 50 1920 × 1080 369 1 7 14
Zhengzhou MT 346 576 × 704 3475 1 10 20
Method MAE MSE
MCNN [10] 1.9 2.3
CSRNet [14] 1.6 2.0
MPCNet (ours) 1.7 2.2
Architecture MAE MSE
Without ASPP model 11.3 20.8
Atrous rate values (1,4,8,12) 11.2 19.4
Atrous rate values (1,6,12,18) 9.7 16.0
Atrous rate values (1,10,20,30) 11.2 20.1
Method MAE MSE
Zhang et al. [9] 32.0 49.8
MCNN [10] 26.4 41.3
Switching-CNN [13] 21.6 33.4
CP-CNN [34] 20.1 30.1
Cascaded-MTL [35] 20.0 31.1
Liu et al. [36] 13.7 21.4
CSRNet [14] 10.6 16.0
MPCNet (ours) 9.7 16.0
SANet [12] 8.4 13.6
Method MAE MSE
MCNN [10] 52.6 59.1
Zhang et al. [9] 40.0 46.2
Sam et al. [13] 23.4 25.2
SaCNN(w/o cl) [32] 17.8 23.4
CSRNet [14] 8.8 35.7
SaCNN [32] 8.6 11.6
MPCNet (ours) 4.3 4.9
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Zhang, J.; Zhu, G.; Wang, Z. Multi-Column Atrous Convolutional Neural Network for Counting Metro Passengers. Symmetry 2020, 12, 682. https://doi.org/10.3390/sym12040682
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Zhang J, Zhu G, Wang Z. Multi-Column Atrous Convolutional Neural Network for Counting Metro Passengers. Symmetry. 2020; 12(4):682. https://doi.org/10.3390/sym12040682
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Zhang, Jun, Gaoyi Zhu, and Zhizhong Wang. 2020. "Multi-Column Atrous Convolutional Neural Network for Counting Metro Passengers" Symmetry 12, no. 4: 682. https://doi.org/10.3390/sym12040682
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The Elements of Euclid; viz. the first six books,together with the eleventh and twelfth, with an appendix
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Side 173
If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 56
Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of
the straight line which is made up of the whole and that part.
Side 53
If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section,
is equal to the square on half the line.
Side 58
IF a straight line be divided into two equal, and also into two unequal parts; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line
between the points of section.
Side 94
The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right
Side 23
Any two sides of a triangle are together greater than the third side.
Side 40
EQUAL triangles upon the same base, and upon the same side of it, are between the same parallels.
Side 103
If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part
of it without the circle, shall be equal to the square on the line which touches it.
Side 50
PROP. I. THEOR. If there oe two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the
undivided line, and the several parts of the divided line.
Side 28
If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
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The most general case--a TDL having a tap after every delay element--is the general causal Finite Impulse Response (FIR) filter, shown in Fig.2.22. It is restricted to be causal because the output
``future'' inputs transversal filter. FIR filters are described in greater detail in [449].
The difference equation for the 2.22 is, by inspection,
and the
transfer function
The STK class for implementing arbitrary direct-form FIR filters is called Fir. (There is also a class for IIR filters named Iir.) In Matlab and Octave, the built-in function filter is normally used.
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Consequences of the First Law, Physics tutorial
Energy Equation:
Energy equations are the equations that express internal energy of the system as a function of variables defining state of the system. Energy equations, like equation of state, are different for
different systems or substances. Equation of state and energy equation together totally determine all properties of substance or system. Energy equations are derived independently, but not from
equation of state. We are going to consider systems that state can be described by properties P, V. and T.
T and V Independent:
Consider internal energy U as that function of T and V, U(T,V), then change in internal engine dU between two equilibrium states in which temperature and volume differ by dT and dV is
dU = [(∂U/∂T)v]dT + [(∂U/∂V)T]dV
(∂U/∂T)v is slope of isochoric line and (∂U/∂V)T is a slope isothermal line in which U is plotted as the function of T and V. (∂U/∂T)v can be estimated experimentally and it has physical significant.
The first law
dQ = dU + PdV
dQ = (∂U/∂T)vdT + (∂U/∂V)TdV + PdV
dQ = (∂U/∂T)vdT + [(∂U/∂V)T + P]dV
For the process at constant volume, dV = 0 and dQ = C[V]dT, then equation becomes
C[V]dT = (∂U/∂T)vdT
C[V]dT = (∂U/∂T)vdT
So, (∂U/∂T)v = C[V]
Specific heat capacity at constant volume C[V] is slope of isochoric line on U-T-V surface and its experimental measurement determines this slope at any point. Equation can be written for any
reversible procedure as
dQ = CVdT + [(∂U/∂V)T + P]dV
For the process at constant pressure, dQ = CPdT, so equation becomes
C[P]dT = C[V]dT + [(∂U/∂V)T + P]dV
Dividing through by dT and replacing dV/dT by (∂V/∂T)P, we get
C[P] - C[V] = [(∂U/∂V)T + P](∂V/∂T)P
Equation holds for the system in any one equilibrium state, but doesn't refer to the process between two equilibrium states.
T and P Independent:
Enthalpy H of the system, like internal energy U, is a property of system which depends on state only and can be stated as the function of any two variables P, V, and T. Each of these relations
states the enthalpy surface in rectangular coordinate system in which H is plotted along one axis while other two axes are P and V, P and T, or T and V.
Consider enthalpy as function of T and P i.e. H(T, P) ,
dH = (∂H/∂T)PdT + (∂H/∂P)TdP
From definition of enthalpy for a PVT system:
H = U + PV
Differential of H, dH is
dH = dU + PdV + VdP
Combining equations with first law (i.e. replace dU in equation with dQ - PdV and make dQ the subject) gives
dQ = dH - VdP
Insert 6.8 in 6.11, to obtain
dQ = (∂H/∂T)PdT + [(∂H/∂P)T - V]dP
For the isobaric process ( dP = 0), dQ = CPdT. Therefore
(∂H/∂T)P = CP
Equation means that specific heat capacity at constant pressure CP is equal to slope of isobaric line on H - T - P surface. Equation can be written for any reversible process as
dQ = CPdT + [(∂H/∂P)T - V]dP
In the process at constant volume, dQ = CVdT and
CP - CV = -[(∂H/∂P)T - V](∂P/∂T)V
If temperature is constant, equation becomes
dQ = -[(∂H/∂P)T - V]dP
In the adiabatic process, dQ = 0, then equation becomes
CP(∂T/∂P) = -[(∂H/∂P)T - V]
P and V Independent:
Consider U as the function of P and V, U(P,V), then change in internal energy dU between two equilibrium states is
dU = (∂U/∂P)VdP + (∂U/∂V)PdV
Consider also U(T,V)
dU = (∂U/∂T)VdT + (∂U/∂V)TdV
Generally, for any property w, and any three variables x, y, z the form of equations are
(∂w/∂x)y = (∂w/∂z)y(∂z/∂x)y and (∂w/∂y)x = (∂w/∂z)y(∂z/∂y)x + (∂w/∂y)z
Hence for H(P,V,T) we have
(∂H/∂V)P = (∂H/∂T)P(∂T/∂V)P and
(∂H/∂P)V = (∂H/∂T)P(∂T/∂P)v + (∂H/∂P)T
By solving equations we have
(∂H/∂V)P = CP(∂T/∂V)P
We can show that
dQ = CP(∂T/∂V)pdV + CV(∂T/∂P)vdP and CV(∂P/∂V)s = CP(∂P/∂V)T
Gay-Lussac-Joule Experiment:
The partial derivative (∂U/∂V)T describes way in which internal energy of the given system differs with volume at constant temperature. Likewise (∂H/∂P)T describe way in which enthalpy of the given
system differs with pressure at constant temperature. These two derivatives can be computed from equation of state of system. Using
(∂U/∂V)T(∂V/∂T)U(∂T/∂U)v = -1
(∂H/∂P)T(∂P/∂T)H(∂T/∂H)P = -1
(∂H/∂P)T = -Cp(∂T/∂P)H
From equation measurement of rate of change of temperature with volume in the process at constant internal energy provides desired derivatives (i.e. (∂U/∂V)T
Likewise, from equation, measurement of rate of change of temperature with pressure in the process at constant enthalpy provides desired derivatives (i.e. (∂H/∂P)T)
Gay-Lussac and Joule made the attempt to compute dependence of internal energy U of the gas on its volume. In Gay-Lussac and Joule Experimental set-up there is a vessel A that has sample of gas for
investigation and is connected to the evacuated vessel B by the tube in which there is stopcock which is at first closed. Whole arrangement is immersed in the water tank of known mass which
temperature can be measure by a thermometer. The whole set-up is allowed to attain thermal equilibrium and the temperature is measured and recorded. Then stopcock is opened and gas is permitted to
undergo a free expansion into evacuated vessel.
Work done W during free expansion process is zero. The system will finally come to new equilibrium state if pressure is same in both vessels. If temperature of gas changes during process (i.e. free
expansion), there will be heat flow between gas and water bath and final temperature will be different from initial one already estimated and recorded.
Gay-Lussac and Joule found that temperature change of water bath, if it changes at all, was very small to be detected. Reason is that heat capacity of bath is so large that small heat flow in or out
of it generates only a very small change in temperature. Similar experiments have been carried out, using other method, and results showed that temperature change of gas during free expansion is not
Therefore postulate as the additional property of ideal gas is that temperature change during a free expansion is zero. First law of thermodynamics (i.e. ΔU = U[f] - U[i] = Q - W) as both Q and W are
zero, becomes
ΔU = 0
Thus internal energy is constant, and for ideal gas,
(∂T/∂V)U = 0 (ideal gas)
Partial derivative in equation is known as Joule coefficient and is represented by η
η ≡ (∂T/∂V)U
for an ideal gas, the partial derivative (∂U/∂T)v is a total derivative and
Cv = dU/dT, dU = CvdT
Integrating equation from reference level (U[o], T[o]) to (U,T), and if C[V] is constant that is
∫[U0]^UdZU = U - U[0] = Cv∫[T]^T0dT
That gives
U = U[0] + C[V](T - T[0]) This is energy equation of the ideal gas.
Joule-Thompson Experiment:
Joule and Thomson made the attempt to compute dependence of enthalpy of the gas on its pressure. In experimental set-up used by Joule and Thomson gas in compartment 1 (with T[1], P[1], and V[1]) was
permitted to expand freely through the porous plug. Gas expands from pressure P[1] to P[2] by throttling action of porous plug. Whole system is insulated so that expansion takes place adiabatically
(i.e. Q = 0).
When steady state condition has been reached temperatures of gas before and after expansion, T[1] and T[2], are estimated directly with sensitive thermocouple thermometers. Total work done during
expansion can be written as:
W =W[1] +W[2] = P[1]V[1] - P[2]V[2]
Overall change in internal energy of gas during adiabatic expansion is then
ΔU = Q + W = 0 + W = +W
ΔU = P[1]V[1] - P[2]V[2] = U[2] -U[1]
Rearranging gives
U[2 ]+ P[2]V[2] = U[1] + P[1]V[1]
But H = U + PV then equation becomes H[1] + H[2]
This is hence an isenthalpic expansion and experiment measures directly change in temperature of the gas with pressure at constant enthalpy. Joule-Thomson coefficient μ is,
μ ≡ (∂T/∂P)H
For ideal gas,
(∂H/∂P)T = 0 (ideal gas)
Thus for ideal gas
(∂U/∂V)T = (∂H/∂P)T = 0
Then equation becomes
C[p] - C[v] = P(∂V/∂T)P = V(∂P/∂T)V = nR
And from equation of state of ideal gas, PV = nRT
P(∂V/∂T)P = V(∂P/∂T)V = nR
Therefore for ideal gas,
C[p] - C[v] = nR
Reversible Adiabatic Process:
For any substance in a reversible adiabatic process,
(∂P/∂V)S = CP/CV(∂P/∂V)T
If we representing the ratio CP/CV by γ
That is γ ≡ Cp/CV
Omitting subscript S for simplicity then
dP/P + γdV/V = 0
Integrating equation
ln P + γlnV = lnK
Or PV^γ = K
Where K in equation is a constant of integration. From equation eliminating V provides
TP^(1-γ)/γ = constant and eliminating P gives
TV^γ-1 = constant
Equation based on fact that gas obeys equation of state in any reversible process.
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Collateralized Mortgage Obligations (CMOs) - Finance Train
Collateralized Mortgage Obligations (CMOs)
Collateralized mortgage obligations (CMOs) are a type of mortgage-backed security that is created with the prime motive of redistributing the prepayment risk to different classes of bondholders.
Let’s take a detailed look at how this works. As we learned before, in a pass-through security, the monthly cash flow (scheduled interest, scheduled principal, and unscheduled prepayments) are passed
on to all the bond holders on a pro-rata basis. So, all the bond holders are equally exposed to the prepayment risk of the entire pool of mortgage loans.
Collateralized Mortgage Obligations eliminate this problem by creating different sets of securities with different priorities such that some securities face less prepayment risk while others face
more prepayment risk. In simple words, the streams of principal and interest payments from mortgages are distributed to different classes of Collateralized Mortgage Obligations, known as tranches.
Each tranche will suit the objectives and needs of a different set of investors and will carry different principal balances, coupon rates, prepayment risk and maturities.
Let’s say the collateral pool contains 1,000 loans each amounting to $100,000. So, the total collateral pool has a value of $100 million. This pool can be used to create three tranches of securities
with different characteristics as follows:
│Tranche │Par Value │Interest Payment │Principal Payment │
│Tranche A│$50 million│Pay monthly based on outstanding principal balance.│First to receive all principal payments (scheduled and unscheduled) till all principal is paid-off│
│Tranche B│$30 million│Pay monthly based on outstanding principal balance.│Receives all principal payments after Tranche A has been paid off fully. │
│Tranche C│$20 million│Pay monthly based on outstanding principal balance.│Receives all principal payments after Tranche B has been paid off fully. │
The above table described the broad structure of how tranches work in Collateralized Mortgage Obligations. Let’s look at some of its characteristics:
1. All tranches receive interest payments based on the outstanding balances.
2. Tranche A receives all the principal payments till completely paid off. This tranche first absorbs all the prepayment risk thereby protecting other tranches from any prepayments. The total
prepayment risk however, remains the same.
3. All the securities within a tranche will have the same characteristics and risks.
4. Tranche A will also have shortest maturity followed by Tranche B and then Tranche C.
5. The final maturity of each tranche is more certain compared to pass-through securities.
6. Each tranche uniquely satisfies the portfolio needs of different investors.
7. CMOs are highly sensitive to interest rate changes.
8. CMOs can have pools of pass-through securities as collateral.
9. There are more complex types of CMO tranches such as Planned Amortization Class (PAC) tranche.
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Types, Sorts, and Signatures
The original purpose of types was to restrict expressions in such a way as to prevent Russel’s Paradox. Bertrand Russel discovered his paradox just before publication of his book Principles of
Mathematics, and not wanting to delay publication too much, he introduced the mechanism of types as a solution in a hastily written appendix to the book (by the way, “Principles of Mathematics” is
not just the English translation of “Principia Mathematica”. Principia Mathematica was a later work written by Russell and Whitehead).
Ever since that hastily-written appendix there has been some ambiguity over whether types belong to expressions or to values.The difference is not an insignificant one. Expressions are syntactic
objects, part of a language. Values are the things that the language is about. So if types go with expressions then they are things that can be determined statically by examining the text of the
program. If types go with values then you can’t assign types statically in any powerful language. You have to figure it out dynamically.
The ambiguity arises from the fact that a kind of type really applies to each kinds of thing. A value has a sort, which is the kind of thing that it is. A variable has a signature which describes the
sorts of values that the variable can hold. Lets expand this notion of a signature to apply to any expression so that the type of a value is called a sort and the type of an expression is called a
signature. We’ll let “type” retain its current ambiguous status.
There is more to the difference between a sort and a signature than just what they apply to. Consider a class fifo that is a subclass of the abstract class aggregate. An expression can have the
signature fifo or the signature aggregate, but there can be no values of sort aggregate because it is abstract.
Expressions can have even more complex types. Unobtainabol has type expressions with the “or” operator, |. A type expression “int|float” would be read “the sort is either int or float”. This feature
lets you implement linked lists without explicit pointers. For example:
class Link {
data: int;
next: Link|Null;
This says that the next element in the list is either another Link or is the special value Null. There is no need to refer to pointers, you can just use Link and let the type system handle pointer
issues for you.
So in Unobtainabol we can declare a variable with any of a set of types and an expression that uses that variable can also have any of a set of types. For example in
var x: int|float;
y := x+1;
the expression x+1 can be either an int or a float but the value of the expression doesn’t have an ambiguous type. Whatever is in the variable x is one or the other. There are no values that are “int
or float” in some undecided state (well ... maybe in advanced Unobtainabol). So the type of an expression can be broader than the type of a value can be.
Here is some new terminology for Unobtainabol:
sort: the type of a value. A sort consists of a set of possible values and set of basic operations on those values. Every non-virtual class is a sort as is every primitive type such as int and float.
signature: the type of an expression. Signature is a statically determined restriction on the sorts of values that an expression can evaluate to.
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Chris Adolph :: 503
POLS/CSSS 503 Spring 2014
Advanced QuantitativePolitical Methodology Class meets: Tuesdays 4:30-7:20 pm Electrical Engineering 031
Offered every Spring at the University of Washington by various instructors TA: Carolina Johnson (UW Political Science)
Section meets: F 1:30-3:20 pm Savery 117
Lectures Click on lecture titles to view slides or the buttons to download them as PDFs.
Topic 1
Introduction to the Course and to R
R code and data for the GDP example. R code and data from the fertility example. You’ll find detailed instructions for downloading, installing, and learning my recommended software for quantitative
social science here. Focus on steps 1.1 and 1.3 for now, and then, optionally, step 1.2. (Note: These recommendations may seem dated, as many students prefer to use RStudio as an integrated design
environment in combination with RMarkdown. You are free to follow that model, which minimizes start-up costs. I still prefer a combination of Emacs, the plain R console, and Latex/XeLatex for my own
productivity, with occasional use of Adobe Illustrator for graphics touch-up.)
Topic 2
Review of Matrix Algebra for Regressionand Regression and Graphics in R
We will work through Kevin Quinn’s matrix algebra review. R code and csv data for an example of how the base graphics package can create scatterplots and perform linear regression.
Topic 3
Linear Regression in Matrix Form andProperties and Assumptions of Linear Regression
You may find useful three review lectures on basic probability theory, discrete distributions, and continuous distributions.
Topic 4
Inference and Interpretation of Linear Regression
Example code for estimating a linear regression, extracting confidence intervals for the parameters, and plotting fitted values with a confidence envelope.
Topic 5
Specification and Fitting in Linear Regression
Topic 6
Outliers and Robust Regression Techniques
Student Assignments
Problem Set 1
Due Tuesday, 15 April, in class
Data for problem 1 in comma-separated variable format.
Problem Set 2
Due Friday, 25 April, in section
Data for problem 1 in comma-separated variable format.
Problem Set 3
Due Tuesday, 6 May, in class
Five R script templates for simulation of the performance of linear regression with different kinds of data: when the Gauss-Markov assumptions apply; when there is an omitted variable; when there is
selection on the response variable; when there is heteroskedasticity; and when there is autocorrelation in the response variable.
Problem Set 4
Due Tuesday, 20 May, in class
Data for problem 2 in comma-separated variable format.
Problem Set 5 (Optional)
Due Friday, 6 June, in section
Data for problems 1. Data for problem 2. Data for problem 3. (All data in comma-separated variable format.)
Final Paper
Due Monday, 9 June, at 3:00 PM, in my Gowen mailbox
See the syllabus for paper requirements, and see my guidelines and recommendations for quantitative research papers. | {"url":"https://faculty.washington.edu/cadolph/index.php?page=20&os=unix&banner=5&pan=essex","timestamp":"2024-11-04T01:55:32Z","content_type":"application/xhtml+xml","content_length":"20925","record_id":"<urn:uuid:80b95c14-1dce-4f1b-97de-a5937522b44a>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00173.warc.gz"} |
Two-sample t-test with Python
This is a step-by-step guide on how to implement a t-test for A/B testing in Python using the SciPy and NumPy libraries.
Check out this post for an introduction to A/B testing, test statistic, significance level, statistical power and p-values.
If you are already familiar with two-sample t-tests, feel free to jump to Section 3 where I explain how to implement such a test in Python.
Table of contents
1. Two-sample t-tests
In this post, we will use a two-tailed t-test statistic which is well suited for continuous data.
Two-tailed means that the mean of one sample can be smaller or larger than the other. In a one-tailed test, the mean of the first sample is smaller or larger than the mean of the second sample.
The t-test that I will describe here is the so-called unpaired t-test which compares the means of two independent groups. If the groups are instead dependent a paired t-test should be performed [1].
If the populations from which the sample means are extracted have equal variances and the sample means are normally distributed, a Student's t-test should be used. The Welch's t-test is designed for
unequal population variances, but the assumption of normality remains.
The test consists in calculating a t-statistic, a critical t-statistic value and a p-value. If the absolute value of the observed t-statistic is below the critical t-statistic value, we fail to
reject the null hypothesis, otherwise we reject the null hypothesis. We can also calculate a p-value, which tells you the probability of obtaining such data if the null hypothesis is true. In both
cases, the t distribution is used which depends on the degrees of freedom. The degrees of freedom represent the number of values in the final calculation of a statistic that are free to vary.
The Student's t-test statistic is defined as follows [1]:
Where the s terms are the respective unbiased estimators of the population variances. In this case, the degrees of freedom is n1 + n2 - 2.
If both samples have the same size, the above equation is simplified to the following form [1]:
In this case, the degrees of freedom is 2n - 2.
The Welch's t-test statistic is defined as follows [1]:
In this case the degrees of freedom (df) is calculated in the following way:
Before continuing let's briefly introduce some concepts that will be needed to understand my implementation of a t-test in Python.
Normal (also called Gaussian) distributions are symmetrically distributed. They are shaped as a bell and are characterized by a mean (mu) and a standard deviation (sigma).
Cumulative distribution function:
The cumulative distribution function (CDF) of a random variable X evaluated at x is the probability that X will take a value less than or equal to x.
A z-score measures the distance from the mean in terms of the standard deviation.
2. Introducing the case example: daily conversion rates
Suppose we have a hotel booking website and wish to study if a given change in our website can boost our average daily conversion rates (at the final stage of the booking process). We decide then to
make an A/B test to help us determine if we want to release such a change. For this example, let's set the significance level to 0.05 (alpha) and the statistical power (1-beta) to 0.8 (the
statistical power will be used to define the minimum sample size in Section 3.1).
In this example, our null hypothesis states that there is no significant difference between the conversion rates with or without such a change in the website.
3. Implementing a t-test in Python
Let's start by importing all the libraries and functions that we will need:
from scipy.stats import norm, t, ttest_ind
from scipy.special import stdtr
import numpy as np
import math
3.1 Estimating the minimum sample size
Before running an A/B test, we need to estimate the minimum sample size required to observe a difference at least as large as our desired minimum detectable effect (MDE) with the chosen significance
level and statistical power. If the sample size of our data is below such a minimum sample size, even if we see a difference larger than the minimum detectable effect, we might not be able reject the
null hypothesis since the difference would not be statistically significant. For the example defined above, the minimum sample size corresponds to the number of days the experiment would need to run.
If sigma is the standard deviation (assuming both groups have the same standard deviation which is taken from historical data), then here is the equation [2] to calculate the minimum sample size (n):
Note: the above Z-score values are calculated using a Normal distribution with mean zero and standard deviation equal to 1.
If you wish instead to perform a one-tailed t-test, you need to make the following change to Equation 2:
Here is an example of how we can implement Equation 2 in Python:
def get_min_sample_size(
std_dev, # standard deviation
mde, # minimum detectable effect
alpha = 0.05, # significance level
power = 0.8 # statistical power
Estimate minimum sample size for t-test
Sample sizes will be the same for both groups
Both groups have the same standard deviation
# Find Z_beta from desired power
Z_beta = norm.ppf(power)
# Find Z_alpha
Z_alpha = norm.ppf(1 - alpha / 2)
# Return minimum sample size
return math.ceil(2 * std_dev**2 * (Z_beta + Z_alpha)**2 / mde**2)
Let's calculate the minimum sample size using the function defined above for sigma = 0.05, and using the values chosen for our example for alpha (0.05) and power (0.8), and let's set the minimum
detectable effect to 0.03:
min_sample_size = get_min_sample_size(
std_dev = 0.05,
mde = 0.03,
alpha = 0.05,
power = 0.8
The above will give us 44, that is the needed number of days we need to run the experiment.
3.2 Generating simulated data
Let's create a function to generate data for group A and B, each with the same sample size. First, we will set a seed to get reproducible results (so we all get the same results). Then, we will
generate data using normal distributions, both having the same standard deviation, such that we can use the Student's t-test.
Here is the function that does all the above:
def generate_data(
avg_daily_conversion_rate_A, # avg daily conversion rate for group A
avg_daily_conversion_rate_B, # avg daily conversion rate for group B
std_dev = 0.05 # standard deviation
"""Generate fake data to perform a two-sample t-test"""
# Set a random seed for reproducibility
# Generate data for group A and B
group_A = np.random.normal(avg_daily_conversion_rate_A, std_dev, sample_size)
group_B = np.random.normal(avg_daily_conversion_rate_B, std_dev, sample_size)
return group_A, group_B
Let's now use the above function to generate data incompatible with the null hypothesis, i.e. let's generate two samples each with a different daily conversion rate. I will choose a large-enough
difference that will allow us to see it in our t-test (i.e. difference > minimum detectable effect).
group_A, group_B = generate_data(
sample_size = min_sample_size,
avg_daily_conversion_rate_A = 0.2,
avg_daily_conversion_rate_B = 0.23,
std_dev = std_dev
3.3. Running a t-test
3.3.1. Rejecting the null hypothesis
We will use the ttest_ind() function from SciPy to retrieve the t-statistic and the p-value:
result = ttest_ind(group_A, group_B)
pvalue = result.pvalue
if pvalue < alpha:
print(f"Decision: There is a significant difference between the groups (p-value = {pvalue}).")
print(f"Decision: There is no significant difference between the groups (p-value = {pvalue}).")
tstat = result.statistic
print(f't-statistic = {round(tstat, 2)}'
Decision: There is a significant difference between the groups (p-value = 0.00023115984392950252).
t-statistic = -3.84
• The above assumes equal variances. Set the equal_var argument to False to perform a Welch's t-test which doesn't assume equal variances.
• The above works for a two-tailed t-test. If you wish to perform a one-tailed t-test, set the alternative argument to 'less' (the mean of the first sample is smaller than the mean of the second
sample) or 'greater' (the mean of the first sample is larger than the mean of the second sample) as appropriate.
Since the p-value is well below the cut off of 0.05, we can reject the null hypothesis. This means there is a difference (that is statistically significant) between the two groups of users.
If this would be a real-life experiment, this points out that it might be a good idea to roll out the A -> B change to all users (if the difference is in a positive direction). Said that, we might
want to further support this change by additional studies. For example, by running the experiment a second time.
Let's convince ourselves that we are doing things correctly and calculate the t-statistic by hand following Equation 1. Here is how we can implement it in Python:
avgA = np.mean(group_A)
avgB = np.mean(group_B)
varA = np.var(group_A, ddof = 1)
varB = np.var(group_B, ddof = 1)
n = min_sample_size
my_tstat = (avgA - avgB) / math.sqrt((varA + varB)/ n)
print(f't-statistic calculated by hand = {round(my_tstat, 2)}')
The above code gives the following:
t-statistic calculated by hand = -3.84
which agrees with the t-statistic (tstat) retrieved with the ttest_ind() function.
Let's now calculate the critical t-statistic value using the percent point function which is the inverse of the cumulative distribution function. With this, we obtain the value of the t-statistic
distribution (for the given degrees of freedom) corresponding to the chosen value of alpha. In other words, this critical value ensures that 1 minus the cumulative distribution function evaluated at
the critical value is equal to alpha (0.05). If the absolute value of our observed t-statistic is below this critical t-statistic value, we fail to reject the null hypothesis, otherwise we reject the
null hypothesis.
Here is how we can calculate the critical value in Python:
df = 2 * n - 2 # degrees of freedom
critical_t_stat = round(t.ppf(1 - alpha / 2, df), 2) # ppf = percent point function (inverse of the cumulative distribution function)
print(f'critical t-statistic = {critical_t_stat}')
Which prints the following:
critical t-statistic = 1.99
Since the absolute value of the observed t-statistic (3.84) is higher than the critical t-statistic (1.99), we can then reject the null hypothesis.
Note: If you wish to perform a one-tailed t-test, critical_t_stat should be calculated in the following way instead:
critical_t_stat = round(t.ppf(1 - alpha, df), 2) # ppf = percent point function (inverse of the cumulative distribution function)
Furthermore, we can also calculate the p-value by hand and validate the value we obtained before. The p-value is calculated in the following way
if t-statistic <= 0: p-value = 2 × (area to the left of the t distribution)
if t-statistic > 0: p-value = 2 x (area to the right of the t distribution)
We can implement the above in the following way:
my_pvalue = 2 * stdtr(df, -np.abs(my_tstat)) # stdtr = Student t distribution cumulative distribution function
print(f'p-value calculated by hand = {my_pvalue}')
Which prints the following (which agrees with the value obtained above):
p-value calculated by hand = 0.00023115984392950252
3.3.2. Failing to reject the null hypothesis
Let's generate new data that presents a difference below the minimum detectable effect and re-run the t-test:
group_A, group_B = generate_data(
sample_size = min_sample_size,
avg_daily_conversion_rate_A = 0.2,
avg_daily_conversion_rate_B = 0.201,
std_dev = std_dev
Let's now run the t-test:
result = ttest_ind(group_A, group_B)
pvalue = result.pvalue
if pvalue < alpha:
print(f"Decision: There is a significant difference between the groups (p-value = {pvalue}).")
print(f"Decision: There is no significant difference between the groups (p-value = {pvalue}).")
Decision: There is no significant difference between the groups (p-value = 0.33914179411708234).
Since the obtained p-value is larger than 0.05 (i.e. our choice for alpha), then we fail to reject the null hypothesis.
3.3.3. Repository with full code
Do you wish to learn all the technical skills needed to perform a data analysis in Python? Check out my free Python course for data analysis: https://github.com/jbossios/python-tutorial
[1] Fundamentals of Biostatistics (Seventh Edition) by Bernard Rosner | {"url":"https://www.jonathanbossio.com/post/two-sample-t-test-with-python","timestamp":"2024-11-02T11:46:43Z","content_type":"text/html","content_length":"1050597","record_id":"<urn:uuid:3bce70c0-9eff-49eb-9823-f3c73e6754a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00765.warc.gz"} |
Statistical Data Analysis – An Introduction
Study Materials
Statistical Data Analysis – An Introduction
Data is not a long-term scarce resource in the digital age; rather, it is compelling. All types of data and information are exploited to the fullest extent possible, and statistical data analysis
plays a significant role in this process. This involves delving into the overwhelming volume of data to precisely interpret its complexity in order to provide insights for intense progress to
organisations and businesses.
Since statistics is a branch of science, it includes the collection, interpretation, and validation of data. Statistical data analysis is the method of carrying out various statistical operations, or
in-depth quantitative research that makes an effort to quantify data and uses various types of statistical analysis. Here, descriptive data, such as survey and observational data, is frequently
included in quantitative data.
It is an extremely important approach for business intelligence organisations that must work with enormous data volumes in the context of business applications.
In the retail industry, for instance, this method can be used to find patterns in unstructured and semi-structured consumer data that can be used to make more powerful decisions for improving
customer experience and advancing sales. Trend identification is the fundamental goal of statistical data analysis.
In addition, statistical data analysis has numerous applications in the areas of business intelligence (BI), big data analytics, machine learning, deep learning, financial analysis, and economic
The Significance of Data
• Depending heavily on the number of variables, specialists use a variety of statistical techniques to analyse data, which includes variables that are either univariate or multivariate. The t-test
for significance, the z test, the f test, the one-way ANOVA test, etc. can all be used if the data only contains one variable. If the data contains multiple variables, different multivariate
techniques can be used, such as statistical data analysis, discriminant statistical data analysis, etc.
• There are two categories of data: continuous data and discrete data. Continuous data, such as light intensity, room temperature, etc., cannot be tallied and is dynamic. Discrete data, such as the
number of bulbs or individuals in a group, may be counted and has a variety of values.
• Continuous data are distributed according to a continuous distribution function, also known as the probability density function, in statistical data analysis, whereas discrete data are
distributed according to a discrete distribution function, also known as the probability mass function.
• Quantitative or qualitative data are both acceptable. Quantitative data always take the form of numbers that indicate either how much or how many of an element there are, whereas qualitative data
use labels or names to identify a characteristic of each piece.
• Cross-sectional and time-series data are crucial for statistical data analysis. Cross-sectional data are defined as data obtained at the same time or almost at the same moment in time, and
time-series data are defined as data gathered over a range of time periods.
Statistical Data Analysis Tools
When analysing statistical data, some statistical analysis techniques are used that a layperson cannot use without possessing statistical understanding.
To analyse statistical data, a variety of software programmes are available, including the Statistical Analysis System (SAS), the Statistical Package for Social Science (SPSS), Stat Soft, and many
These programmes offer powerful data handling skills and a variety of statistical analysis techniques that can look at a small sample of data or very large amounts of data statistics.
Despite the fact that computers are a key component of statistical data analysis and can help with data summarization, the focus of statistical data analysis is on the interpretation of the results
in order to make inferences and predictions.
Types of Statistical Data Analysis
Descriptive Statistics
It is a type of data analysis that essentially serves as a technique to meaningfully explain, display, or summarise data from a sample. For instance, variance, standard deviation, and mean.
To put it another way, descriptive statistics uses the mean, median, and mode to provide a summary of the relationship between variables in a sample or population.
Inferential Statistics
Using the null and alternative hypotheses, which are subject to random variation, this method is utilised to draw inferences from the data sample.
Regression analysis, correlation testing, and probability distribution are more examples of this. Inferential statistics, to put it simply, uses a random sample of data from a population to draw
conclusions about the entire population.
Statistical Data Analysis – The Basic Steps
Identifying the Issue
For accurate statistics to be obtained regarding the problem, a specific and actuarial definition is essential. Without knowing the precise description or solution to the problem, data collection
becomes incredibly challenging.
Collecting Data
Designing various strategies to gather data is a crucial responsibility in statistical data analysis after tackling a particular issue.
Data can be gathered from the original sources or by observation and experimental research investigations that are carried out to gain new data.
• In an experimental study, the significant variable is chosen in accordance with the problem as specified, and one or more study components are then controlled to obtain information on how these
components affect other variables.
• In an observational study, no trial is run to influence or control the key variable. An example of a typical form of observational study is a conducted surrey.
Analysing the Data
• Exploratory approaches, which use straightforward maths and straightforward graphing and description to summarise data to ascertain what the data is conveying.
• Confirmatory methods: To address specific issues, this method applies concepts and ideas from probability theory.
Because it provides a method for predicting, describing, and explaining the possibilities connected with impending events, probability is incredibly important in decision-making.
Reporting Outcomes
By drawing conclusions from a sample, an estimate or test that purports to represent the traits of a population can be produced; the results may be presented as a table, a graph, or a list of
Only a tiny subset of the data was examined, hence the given result can incorporate probability assertions and intervals of values to reflect certain uncertainties.
Experts might forecast and foresee future aspects of data with the aid of statistical data analysis. A good decision can be made by comprehending the information available and using it properly.
By imparting meaning to meaningless numbers, statistical data analysis breathes life into otherwise lifeless data. Therefore, to conduct any research study, a researcher must have sufficient
knowledge of statistics and statistical procedures.
This will make it easier to perform a relevant and well-designed study, which will lead to more accurate and trustworthy results. Additionally, only when appropriate statistical tests are employed
are the results and inferences made plain. | {"url":"https://learnerstake.com/statistical-data-analysis-an-introduction/","timestamp":"2024-11-10T18:43:02Z","content_type":"text/html","content_length":"101565","record_id":"<urn:uuid:363fb4f6-8fb3-4bd5-b8a1-8c0cbff0baf6>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00260.warc.gz"} |
Class 10 Maths Solving Polynomial Equations Made Easy
Class 10 Maths Solving Polynomial Equations Made Easy
Polynomial: Exploring the Fundamental Concepts
Polynomials are essential mathematical objects that find applications in various fields, including physics, engineering, and computer science. In this comprehensive article, we will delve into the
world of polynomials, covering key concepts such as degree, coefficient, variable, constant, monomial, binomial, trinomial, polynomial equation, synthetic division, zero of a polynomial, factor
theorem, remainder theorem, polynomial division, synthetic division method, long division method, roots of a polynomial, quadratic polynomial, linear polynomial, and cubic polynomial. By the end of
this article, you'll have a solid understanding of polynomials and their fundamental properties.
Polynomial Degree: Understanding the Power
Polynomial Degree: The degree of a polynomial refers to the highest power of the variable present in the polynomial. It helps classify polynomials and provides insights into their behavior.
For example, in the polynomial 3x^4 - 2x^2 + 5x + 1, the term with the highest power of x is 3x^4. Therefore, the degree of this polynomial is 4.
The degree of a polynomial affects the shape of its graph and the number of solutions it has. Higher-degree polynomials tend to exhibit more complex behavior, often with multiple turning points and
Coefficient: The Multiplier of Variables
Coefficient: In a polynomial, coefficients are the numerical values that multiply the variables or powers of variables. They determine the scale or magnitude of each term.
Consider the polynomial 2x^2 + 3x - 1. Here, the coefficient of x^2 is 2, the coefficient of x is 3, and the constant term is -1.
Coefficients play a crucial role in polynomial operations such as addition, subtraction, multiplication, and division. They help determine the relative contribution of each term and enable the
manipulation of polynomials.
Variable and Constant: The Dynamic and Fixed Elements
Variable: In a polynomial, a variable is a symbol that represents an unknown value. Common variables include x and y. Variables allow polynomials to express relationships and equations in a general
form, accommodating different values.
For instance, the polynomial 3x^2 + 2x + 1 contains the variable x.
Constant: A constant, on the other hand, is a fixed value that does not change within a given context. In polynomials, constants are numerical values that do not involve any variables.
In the polynomial 5x^3 - 2x^2 + 7x + 9, the constant term is 9.
Variables and constants work together to form the building blocks of polynomials, enabling them to represent a wide range of mathematical relationships.
Monomial, Binomial, and Trinomial: Classifying Polynomials
Monomial: A monomial is a polynomial with only one term. It can be a constant, a variable, or a variable raised to a non-negative integer power.
Examples of monomials include 5, 2x^3, and 7xy.
Binomial: A binomial is a polynomial with exactly two terms. Each term can be a constant, a variable, or a variable raised to a non-negative integer power.
Examples of binomials include 3x + 2, 5y^2 - 4y, and x^2 + 1.
Trinomial: A trinomial is a polynomial with exactly three terms. Like monomials and binomials, each term in a trinomial can be a constant, a variable, or a variable raised to a non-negative integer
Examples of Trinomials include 2x^2 + 3x - 1, 4y^3 - 2y^2 + y, and x^3 + 2x^2 - x.
Monomials, binomials, and trinomials are specific types of polynomials that help us classify and identify the structure of polynomial expressions. Understanding these classifications is crucial for
further exploration of polynomial concepts.
Degree of a Polynomial: Determining Complexity
Degree of a Polynomial: The degree of a polynomial is the highest exponent/power of the variable in the polynomial expression. It provides valuable information about the complexity and behavior of
the polynomial.
Let's consider a few examples to illustrate the concept:
1. The polynomial 4x^3 - 2x^2 + x - 3 has a degree of 3 because the highest power of x is 3.
2. The polynomial 2x^2 + 5x + 1 has a degree of 2 since the highest power of x is 2.
The degree of a polynomial affects how it behaves, particularly in terms of the number of solutions or roots it possesses. Higher-degree polynomials often exhibit more intricate characteristics, such
as multiple roots and turning points.
Polynomial Equation: Equating Polynomials
Polynomial Equation: A polynomial equation is an equation in which two polynomials are equated to each other. It involves setting a polynomial expression equal to zero.
For example, consider the equation 2x^2 + 3x - 1 = 0. This is a polynomial equation where the polynomial 2x^2 + 3x - 1 is equated to zero.
Solving polynomial equations involves finding the values of the variable(s) that satisfy the equation. The solutions of a polynomial equation correspond to the points where the polynomial intersects
the x-axis on a graph.
Polynomial equations have significant applications in various fields, such as physics, engineering, and computer science, where finding the roots or solutions of equations is necessary for
Synthetic Division: Efficient Polynomial Division
Synthetic Division: Synthetic division is a method used to divide polynomials, particularly when dividing by linear factors of the form (x - a), where a is a constant. It provides a more efficient
and streamlined approach compared to long division.
To illustrate synthetic division, let's consider an example:
Divide the polynomial 3x^3 - 2x^2 + 5x - 1 by (x - 2) using synthetic division.
We set up the synthetic division table as follows:
2 | 3 -2 5 -1
The result of the synthetic division is 6x^2 + 8x + 26 with a remainder of 50.
Synthetic division allows us to efficiently perform polynomial division, simplifying the process and saving time, especially when dividing by linear factors.
Zero of a Polynomial: Solving for Roots
Zero of a Polynomial: The zero of a polynomial, also known as a root or solution, is a value of the variable that makes the polynomial equal to zero. In other words, it is the value(s) that satisfy
the polynomial equation.
For example, consider the polynomial 2x^2 - 5x + 3. To find its zeros, we set the polynomial equal to zero and solve the equation:
2x^2 - 5x + 3 = 0
By factoring or using the quadratic formula, we find that the zeros of the polynomial are x = 1 and x = 1.5.
Finding the zeros of a polynomial is crucial as it helps us identify the points where the polynomial intersects the x-axis on a graph. These zeros provide valuable insights into the behavior and
solutions of polynomial equations.
Factor Theorem: Connecting Factors and Zeros
Factor Theorem: The factor theorem establishes a connection between factors of a polynomial and its zeros. It states that if a polynomial P(x) has a factor (x - a), then a is a zero of the
In simpler terms, if a polynomial can be factored as (x - a) multiplied by another polynomial, then a is a zero of the original polynomial.
For example, let's consider the polynomial 3x^2 - 5x + 2. By factoring it as (x - 2)(3x - 1), we can see that the zeros of the polynomial are x = 2 and x = 1/3.
The factor theorem allows us to find zeros of polynomials by identifying their corresponding factors. This theorem is a valuable tool for solving polynomial equations and understanding the
relationship between factors and zeros.
Remainder Theorem: Dividing with a Remainder
Remainder Theorem: The remainder theorem states that when a polynomial P(x) is divided by (x - a), the remainder obtained is equal to P(a), where a is a constant.
In other words, if we divide a polynomial by a linear factor (x - a), the remainder will be the value obtained by substituting x = a into the polynomial.
For example, let's divide the polynomial 4x^3 - 2x^2 + 5x - 3 by (x - 2) using the remainder theorem.
When x = 2, the remainder is given by P(2):
P(2) = 4(2)^3 - 2(2)^2 + 5(2) - 3
= 32 - 8 + 10 - 3
= 31
Hence, when dividing 4x^3 - 2x^2 + 5x - 3 by (x - 2), the remainder is 31.
The remainder theorem provides a helpful way to determine the remainder when dividing polynomials and establishes a connection between polynomials and their remainders.
Polynomial Division: Breaking It Down
Polynomial Division: Polynomial division is the process of dividing one polynomial by another polynomial. It allows us to break down complex polynomials into simpler forms, aiding in factorization,
finding zeros, and solving polynomial equations.
There are two commonly used methods for polynomial division: synthetic division and long division.
Synthetic Division Method: The synthetic division method, as discussed earlier, is an efficient way to divide a polynomial by a linear factor (x - a). It simplifies the process by using a tabular
format and eliminating the need for writing out all the terms.
Long Division Method: The long division method is a more general approach to polynomial division, allowing division by any polynomial. It involves dividing term by term, similar to traditional Please
accept my apologies for the incomplete response. Here's the continuation:
polynomial division. It entails dividing the highest-degree term of the dividend by the highest-degree term of the divisor and performing subsequent subtractions and multiplications to determine the
quotient and remainder.
Let's illustrate polynomial division using an example:
Divide the polynomial 5x^3 - 3x^2 + 2x - 1 by x - 2 using the long division method.
5x^2 + 7x + 12
x - 2 | 5x^3 - 3x^2 + 2x - 1
- (5x^3 - 10x^2)
7x^2 + 2x
- (7x^2 - 14x)
16x - 1
- (16x - 32)
Hence, when dividing 4x^3 - 2x^2 + 5x - 3 by (x - 2), the remainder is 31.
The remainder theorem provides a helpful way to determine the remainder when dividing polynomials and establishes a connection between polynomials and their remainders.
Polynomial Division: Breaking It Down
Polynomial Division: Polynomial division is the process of dividing one polynomial by another polynomial. It allows us to break down complex polynomials into simpler forms, aiding in factorization,
finding zeros, and solving polynomial equations.
There are two commonly used methods for polynomial division: synthetic division and long division.
Synthetic Division Method: The synthetic division method, as discussed earlier, is an efficient way to divide a polynomial by a linear factor (x - a). It simplifies the process by using a tabular
format and eliminating the need for writing out all the terms.
Long Division Method: The long division method is a more general approach to polynomial division, allowing division by any polynomial. It involves dividing term by term, similar to traditional
polynomial division. It entails dividing the highest-degree term of the dividend by the highest-degree term of the divisor and performing subsequent subtractions and multiplications to determine the
quotient and remainder.
Let's illustrate polynomial division using an example:
Divide the polynomial 5x^3 - 3x^2 + 2x - 1 by x - 2 using the long division method.
5x^2 + 7x + 12
x - 2 | 5x^3 - 3x^2 + 2x - 1
- (5x^3 - 10x^2)
7x^2 + 2x
- (7x^2 - 14x)
16x - 1
- (16x - 32)
The result of the long division is a quotient of 5x^2 + 7x + 12 and a remainder of 31.
Polynomial division is a fundamental tool in algebraic manipulation, allowing us to simplify polynomials, find quotients and remainders, and perform various operations on polynomial expressions.
Roots of a Polynomial: Finding Solutions
Roots of a Polynomial: The roots of a polynomial, also known as zeros or solutions, are the values of the variable that make the polynomial equal to zero. They represent the points where the
polynomial intersects the x-axis on a graph.
Finding the roots of a polynomial is crucial in solving polynomial equations and understanding the behavior of the polynomial.
For example, consider the polynomial x^2 - 4x + 3. To find its roots, we set the polynomial equal to zero and solve the equation:
x^2 - 4x + 3 = 0
By factoring or using the quadratic formula, we find that the roots of the polynomial are x = 1 and x = 3.
The roots of a polynomial provide insights into the behavior, symmetry, and solutions of the polynomial equation. They play a significant role in various mathematical applications and real-world
Quadratic Polynomial: A Polynomial of Degree 2
Quadratic Polynomial: A quadratic polynomial is a polynomial of degree 2. It is expressed as ax^2 + bx + c, where a, b, and c are constants, and a is not equal to zero.
Quadratic polynomials have a variety of applications in fields such as physics, engineering, and optimization. They are commonly used to model various phenomena, including projectile motion,
parabolic arcs, and optimization problems.
The behavior of a quadratic polynomial depends on the discriminant, which is given by b^2 - 4ac. Based on the discriminant, a quadratic polynomial can have different types of solutions:
1. If the discriminant is positive, the quadratic polynomial has two distinct real roots.
2. If the discriminant is zero, the quadratic polynomial has one real root (a repeated root).
3. If the discriminant is negative, the quadratic polynomial has no real roots (complex roots).
Understanding quadratic polynomials is essential as they form the building blocks for more complex polynomial expressions and equations.
Linear Polynomial: A Polynomial of Degree 1
Linear Polynomial: A linear polynomial is a polynomial of degree 1. It is expressed as ax + b, where a and b are constants.
Linear polynomials represent lines on a graph, and their behavior is straightforward. They have a constant slope, which is determined by the coefficient a. The graph of a linear polynomial is a
straight line, and its slope indicates the rate of change or the steepness of the line.
Linear polynomials are widely used in various applications, such as linear regression analysis, representing simple linear relationships between variables, and solving basic equations.
For example, consider the linear polynomial 2x + 3. The coefficient 2 represents the slope of the line, indicating that for every unit increase in x, the corresponding y value increases by 2. The
constant term 3 represents the y-intercept, which is the point where the line intersects the y-axis.
Understanding linear polynomials is essential as they serve as the foundation for more complex polynomial expressions and equations.
Cubic Polynomial: A Polynomial of Degree 3
Cubic Polynomial: A cubic polynomial is a polynomial of degree 3. It is expressed as ax^3 + bx^2 + cx + d, where a, b, c, and d are constants, and a is not equal to zero.
Cubic polynomials have a variety of applications in fields such as physics, engineering, and computer graphics. They are used to model various phenomena, including the motion of objects, fluid flow,
and the generation of smooth curves.
The behavior of a cubic polynomial depends on its coefficients and the number of real roots it possesses. A cubic polynomial can have one real root and two complex roots or three distinct real roots.
Solving cubic polynomials can be challenging, and various methods, such as factoring, synthetic division, or numerical methods, may be employed to find the roots.
Understanding cubic polynomials is important as they represent a higher degree of complexity in polynomial expressions and equations, allowing for a more comprehensive mathematical description of
various phenomena.
Polynomials play a fundamental role in mathematics and have wide-ranging applications in various fields. They provide a powerful framework for representing and solving mathematical problems,
expressing relationships between variables, and modeling real-world phenomena.
In this article, we explored key concepts related to polynomials, including degree, coefficient, variable, constant, monomial, binomial, trinomial, polynomial equations, synthetic division, zero of a
polynomial, factor theorem, remainder theorem, polynomial division, roots of a polynomial, quadratic polynomial, linear polynomial, and cubic polynomial.
We learned that the degree of a polynomial determines its complexity, and the coefficients represent the numerical factors associated with each term. Variables and constants contribute to the
variability and specific values of the polynomial.
We discussed various methods for solving polynomial equations, such as factoring, synthetic division, and the use of the factor theorem and remainder theorem. These methods allow us to find zeros,
factors, and remainders of polynomials.
Polynomial division helps break down complex polynomials into simpler forms, aiding in factorization and solving polynomial equations.
Furthermore, we explored the concept of roots or zeros of a polynomial, which are the values that make the polynomial equal to zero. Roots provide crucial information about the behavior, solutions,
and intersections of a polynomial.
Finally, we examined quadratic, linear, and cubic polynomials, which represent polynomials of different degrees and have distinct characteristics and applications.
Understanding polynomials and their properties is essential for building a strong foundation in algebra and mathematics as a whole. They serve as a fundamental tool for problem-solving, mathematical
modeling, and advanced mathematical concepts.
Now that you have a solid understanding of polynomials, coefficients, degrees, and various related concepts, you can apply this knowledge to tackle more complex mathematical problems and explore the
fascinating world of algebraic expressions.
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What is the square root of 784? + Example
What is the square root of 784?
4 Answers
-Write down the factors of $784$ and and see if they share the same with the choices.
- For example, if you see $27$ and $29$, you can tell that there is no factor of $27$ or $29$ in 576 since they are prime number and eliminate them.
- At this time you verify one by multiplying $28 \times 28$ to verify.
$28 \times 28 = 784$
The square root of $784 = 28$
It is beneficial to be able to estimate, or calculate exactly the square root of any number when a calculator is not present.
For this example, we can start an estimation limit high and low as follows:
$20 \times 20 = 400$ is too low
$30 \times 30 = 900$ is too high, but closer to $784$
If we raise the lower limit to $25$:
$25 \times 25 = 625$ is still too low, but closer.
Now our range is $26 \to 29$.
But the squared number ends in $4$, and the only number in our range multiplied by itself to result in a $4$ would be $8 \times 8$.
So the Square root of $784 = 28$
The square root of $784 = 28$.
Kindly use a calculator for these questions,.
$\sqrt{784} = 28$
If want to recheck the answer, multiply
$28 \times 28 = 784$
If you still have doubts use a prime factor tree.
$\sqrt{784} = 28$
You are looking for squared values that are factors of 784. Use the lowest value prime numbers that you can. Good idea to commit some of them to memory. It will pay off in the end. You can find lists
of them all over the internet.
From the factor tree we have:
$\sqrt{{2}^{2} \times {2}^{2} \times {7}^{2}} = 2 \times 2 \times 7 = 28$
Impact of this question
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- 首页 >> C/C++编程
Project Part 1 [15 points] PCA, Density Estimation, and Bayesian Classification
(Due Tuesday, Oct. 29, 11:59pm)
This part of the project uses a subset of images (with modifications) from the Fashion MNIST dataset. The original Fashion-MNIST dataset contains 70,000 images of objects,
divided into 60,000 training images and 10,000 testing images. We use only images for
class “T-shirt” and class “Sneaker” in this project, and the images have been slightly
modified to suit this project.
The data is stored in “.mat” files. You may use the following piece of code to read the
dataset in Python (or you may use the load filename command in Matlab, since these
are .mat files):
import scipy.io
data = scipy.io.loadmat(‘matlabfile.mat’)
Following are the statistics for the data you are going to use:
Number of samples in the training set: "T-shirt": 6000; "Sneaker": 6000
Number of samples in the testing set: "T-shirt ": 1000; "Sneaker": 1000
For the classification task, we assume that the prior probabilities are the same (i.e., P(0)
= P(1) =0.5).
In the original .mat file, each image is stored as a 28x28 array. We need to “vectorize” an
image by concatenating its columns to form a 784-dimensional vector. In the 784-d space,
it would be difficult to apply Bayesian decision theory (e.g., the minimum error rate
classification). Hence, we will use PCA to do dimensionality reduction first.
Specifically, you will practice doing the following five tasks in this project:
Task 1. Feature normalization (Data conditioning).
You need to normalize the data in the following way, before starting any subsequent tasks.
Using all the training images (each viewed as a 784-d vector, X = [x1, x2, …, x784]
, as
explained), compute the mean mi and standard deviation (STD) si for each feature xi
(remember that we have 784 features) from all the training samples. The mean and STD
will be used to normalize all the data samples (training and testing): for each feature xi, in
any given sample, the normalized feature will be, yi = (xi - mi)/si
Task 2. PCA using the training samples.
Use all the training samples to do PCA. You cannot use a built-in function pca or similar,
if your platform provides such a function. You have to explicitly code the key steps of PCA:
computing the covariance matrix, doing eigen analysis (you can use built-in functions for
this), and then identify the principal components.
Task 3. Dimension reduction using PCA.
Consider 2-d projections of the samples on the first and second principal components.
These are the new 2-d representations of the samples. Plot/Visualize the training and
testing samples in this 2-d space. Observe how the two classes are clustered in this 2-D
space. Does each class look like a normal distribution?
Task 4. Density estimation.
We further assume in the 2-d space defined above, samples from each class follow a
Gaussian distribution. You will need to estimate the parameters for the 2-d normal
distribution for each class, using the training data. Note: You will have two distributions,
one for each class.
Task 5. Bayesian Decision Theory for optimal classification.
Use the estimated distributions for doing minimum-error-rate classification. Report the
accuracy for the training set and the testing set respectively.
What to submit:
1. Your code for doing the above.
2. A report summarizing the results with the following format a. Introduction – start with problem statement, data description etc.
b. Method – implementation details, steps followed etc.
c. Results and observation – the results asked in each of the steps, e.g., the
estimated parameters of the distributions and the final classification accuracy
number (any intermediate results for each of the tasks you want to show) along
with your observations
d. Conclusion
Note: There is no minimum or maximum length requirement for the report. Writing
the report is the opportunity for you to reflect on your understanding of the
problems/tasks through organizing your results.
3. The report should be typed (handwritten reports are not allowed) and in a .pdf
format (to be submitted as separate document, not included within the code file).
4. Do not submit a .zip file. Submit multiple individual files on Canvas instead.
The data files for the project are uploaded in the Files/Assignments folder:
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The Hyperfine Physics Podcast
Description (podcaster-provided):
Physics puzzles and deep dives into physics topics.
Themes and summary (AI-generated based on podcaster-provided show and episode descriptions):
➤ Physics puzzles • Quantum mechanics • Thermodynamics • Cosmology • Entanglement • Acoustic levitation • Encryption and cryptography • Gravitational constant • Randomness and probability • Particle
physics • Scientific inquiry
The Hyperfine Physics Podcast delves into a wide array of physics topics, offering listeners both intricate discussions and thought-provoking puzzles. As indicated by the show's description, the
podcast explores physics puzzles and dives deeply into various scientific subjects. The broad spectrum of coverage includes fundamental aspects of quantum mechanics, showcased through discussions
about Nobel Prize-winning experiments and interpretations, such as Bell's Theorem and Bohmian mechanics. Quantum information science and encryption technologies like Diffie-Hellman and RSA also
feature prominently, illustrating the podcast's engagement with advanced scientific concepts.
The show frequently examines natural phenomena from a physics perspective, addressing the physics of hurricanes, geosynchronous orbits, and gravitational constants. These discussions aim to unearth
the principles governing the universe. It also poses intriguing questions, such as the randomness in the universe or whether data storage impacts mass, fostering curiosity about the underlying laws
of physics.
Listeners are introduced to historical and contemporary advances in physics, with episodes on the origins of quantum mechanics and the theory of statistical mechanics, including entropy and
thermodynamics. There’s an emphasis on the practical applications of physics, as seen in episodes about acoustic levitation, the mechanics behind climbing magnets, and principles like the least
action in everyday examples like gaming.
Recurring themes include exploring statistical mechanics, thermodynamics, and the anomalies of particle physics, including hadrons and the Standard Model. The podcast often features conversations on
theoretical approaches and their real-world implications. By incorporating guest speakers and scientific literature, the podcast also encourages listener engagement with the community, inviting
questions and discussions that amplify the scientific inquiry and exploration cultivated by the hosts.
Nobel Prize in Physics 2022 - The universe is not locally real. What does that mean?
37 minutes
Acoustic Levitation w/ Special Guest Dr. David Jackson
62 minutes
Gravitational G and How Science Works
67 minutes
Cosmology and the Arrow of Time
57 minutes
Is Anything Truly Random? w/Special Guest Grant Ciffone
57 minutes
How to Keep Time
57 minutes
Entropy & Statistical Mechanics
84 minutes
74 minutes
Benford’s Law
57 minutes
Planck, Einstein, and the Origins of Quantum Mechanics
66 minutes
The Physics of Hurricanes
86 minutes
Hadrons – Quark Systems
98 minutes
76 minutes
The Standard Model Part 1
81 minutes
Domino Amplifier
54 minutes
Bohmian Mechanics – Pilot Wave Theory
71 minutes
Relative Motion (Not Relativity)
62 minutes
Bell’s Theorem and EPR
73 minutes
Climbing Magnets
67 minutes
Balloons Inside Balloons and Sweet Spots
66 minutes
Fortnite and the Principle of Least Action
62 minutes
Encryption: Diffie-Hellman & RSA
66 minutes
Geosynchronous Orbits
68 minutes
How Much Weight Do You Lift When Doing a Pushup?
64 minutes
Floating Hourglass
62 minutes
Does Data Have Mass?
59 minutes
Landing on Planets
42 minutes
Intro to The Hyperfine Physics Podcast
24 minutes | {"url":"https://truesciphi.org/podcast-profiles/1434647685.html","timestamp":"2024-11-06T11:29:05Z","content_type":"text/html","content_length":"9487","record_id":"<urn:uuid:95982172-5edf-49e1-904b-a1e41988d444>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00538.warc.gz"} |
Design IIR Butterworth Filters Using 12 Lines of Code - Neil Robertson
Design IIR Butterworth Filters Using 12 Lines of Code
While there are plenty of canned functions to design Butterworth IIR filters [1], it’s instructive and not that complicated to design them from scratch. You can do it in 12 lines of Matlab code. In
this article, we’ll create a Matlab function butter_synth.m to design lowpass Butterworth filters of any order. Here is an example function call for a 5^th order filter:
N= 5 % Filter order
fc= 10; % Hz cutoff freq
fs= 100; % Hz sample freq
[b,a]= butter_synth(N,fc,fs)
b = 0.0013 0.0064 0.0128 0.0128 0.0064 0.0013
a = 1.0000 -2.9754 3.8060 -2.5453 0.8811 -0.1254
Then, to find the frequency response:
[h,f]= freqz(b,a,256,fs);
H= 20*log10(abs(h));
The magnitude response of the 5^th order filter is shown in Figure 1, along with the response of the analog prototype.
Figure 1. Magnitude response of N= 5 IIR Butterworth filter with f[c] = 10 Hz and f[s] = 100 Hz. The prototype analog filter’s response is also shown.
First, a word about notation. We need to distinguish frequency variables in the continuous-time (analog) world from those in the discrete-time world. In this article, the following notation for
frequency will be used:
continuous frequency F Hz
continuous radian frequency Ω radians/s
complex frequency s = σ + jΩ
discrete frequency f Hz
discrete normalized radian frequency ω = 2πf/f[s]radians, where f[s] = sample freq
Analog Butterworth filters have all-pole transfer functions. For example, a third-order Butterworth filter with Ω[c] = 1 rad/s has the transfer function:
$$H(s)=\frac{1}{(s-p_{a0})(s-p_{a1})(s-p_{a2}) }\qquad(1)$$
where the subscript a denotes analog (s-plane) poles. The poles in the s-plane are:
p[a0] = -.5 + j.866
p[a1] = -1
p[a2] = -.5 -j.866
We will transform the poles in the s-plane to poles in the z-plane using the bilinear transform [2,3]. The bilinear transform converts H(s) to H(z) by replacing s with:
where f[s] is sample frequency. If we solve for z, we get:
Equation 3 maps a point on the s plane to a point on the z plane. For example, if f[s]= 2 Hz, the s-plane real pole at -1 maps to:
For the 3^rd order filter, with Ω[c]= 1 and f[s]= 2, the z-plane poles fall as shown in Figure 2.
From equation 1, H(s) has 3 zeros at s= ∞. How do they map to the z plane? We will show later that the bilinear transform maps -∞ to ∞ on the jΩ axis to -f[s]/2 to f[s]/2 on the unit circle. So the 3
zeros of H(s) map to +/- f[s]/2 on the unit circle, which corresponds to z= -1. (Recall that on the unit circle, z= e^jω, where ω = 2πf/f[s]. For f = +/-f[s]/2, we have ω = +/-π, so z = e^jπ = -1).
The three zeros are represented by the ‘o’ in figure 2.
We can now write the 3^rd-order Butterworth H(z) as:
where, from equation 3, p= [0.7143 +j 0.33 0.6 0.7143 -j 0.33]. Expanding the numerator and denominator, we have:
Where b = [1 3 3 1] and a= [1 -2.0286 1.4762 -0.3714]. K is chosen to make gain = 1 at ω= 0: K = 1/H(ω=0) = 1/H(z=1) = sum(a)/sum(b)= .00952
Looking again at Figure 1, you may have wondered why the attenuation of the IIR filter is greater than that of the analog filter as f approaches f[s]/2. The reason is that the analog filter’s zeros
are at ∞, while the bilinear transform compresses the frequency scale so that the IIR filter’s zeros are at f[s]/2.
Figure 2. Z-plane Poles and zeros of 3^rd order IIR Butterworth filter with Ω[c]= 1 and f[s]= 2.
Filter Synthesis
Here is a summary of the steps for finding the filter coefficients :
1. Find the poles of the analog prototype filter with Ω[c] = 1 rad/s.
2. Given the desired f[c] of the digital filter, find the corresponding analog frequency F[c].
3. Scale the s-plane poles by 2πF[c].
4. Transform the poles from the s-plane to the z-plane.
5. Add N zeros at z = -1.
6. Convert poles and zeros to polynomials with coefficients a[n] and b[n].
Now let’s look at the steps in detail. Note we’ll repeat a lot of the math we already presented above. A Matlab function butter_synth that performs the filter synthesis is provided in the Appendix.
It gives the same results as the built-in Matlab function butter(n,Wn) [1].
1. Poles of the analog filter. For a Butterworth filter of order N with Ω[c] = 1 rad/s, the poles are given by [4,5]:
$$p_{ak}= -sin(\theta)+jcos(\theta)$$
where $$\theta=\frac{(2k-1)\pi}{2N},\quad k=1:N$$
2. Given the desired f[c], find analog frequency F[c]. As we’ll show in the next section, the bilinear transform does not map the analog frequency F to discrete frequency f linearly. To achieve a
digital filter cut-off frequency of f[c], the analog prototype cut-off frequency must be:
$$F_c=\frac{f_s}{\pi}tan\left(\frac{\pi f_c}{f_s}\right)$$
This exercise is called frequency pre-warping. For example, if f[s]= 100 Hz and we want f[c]= 20 Hz, then F[c] = 23.13 Hz.
3. Scale the s-plane poles by 2πF[c]. The poles obtained in step 1 gave Ω[c] = 1 rad/s (i.e. 1/(2π) Hz). Multiplying the poles by 2πF[c] scales the analog filter cut-off frequency to F[c] and the
digital filter cut-off frequency to f[c].
4. Transform the poles from the s-plane to the z-plane using the bilinear transform. From equation 3,
$$p_k=\frac{1+p_{ak}/(2f_s)}{1-p_{ak}/(2f_s)},\quad k=1:N$$
5. Add N zeros at z = -1. Following the example of equation 4, the numerator of H(z) is (z + 1)^N , meaning there are N poles at z = -1. We now can write H(z) as:
In butter_synth, we represent the N zeros as a vector q= -ones(1,N).
6. Convert poles and zeros to polynomials with coefficients a[n] and b[n]. If we expand the numerator and denominator of equation 6, we get polynomials in z^-n:
The Matlab code to perform the expansion is:
a= poly(p)
a= real(a)
b= poly(q)
We want H(z) to have a gain of 1 at ω= 0. Letting z= e^jω, we have z= 1. Then, referring to equation 7, we have gain at ω= 0 of:
$$H(z=1)=K\frac{\sum b}{\sum a}$$
So, for gain of 1 at ω= 0, we make $K=\sum a/\sum b$.
And that's the last step. Figure 3 shows the frequency response vs. order N for filters synthesized by butter_synth. Figure 4 shows the impulse response vs. order N for three cases.
Figure 3. IIR Butterworth magnitude responses for f[c]= 10 Hz and f[s]= 100 Hz.
[h,f]= freqz(b,a,256,fs);
H= 20*log10(abs(h));<\pre>
Figure 4. IIR Butterworth impulse responses for fc = 1 kHz and fs = 32 kHz.
x= [1 zeros(1,95)]; % impulse
y= filter(b,a,x); % impulse response
Frequency Mapping of the Bilinear Transform
The bilinear transform does not map the continuous frequency F to discrete frequency f linearly. To show this, we evaluate equation 2 for s= jΩ and z= e^jω:
Now substitute Ω= 2πF and ω= 2πf/f[s]:
$$F=\frac{f_s}{\pi}tan\left(\frac{\pi f}{f_s}\right)\qquad(8)$$
Figure 5 plots equation 8 for fs= 100 Hz. The entire analog frequency range maps to –f[s]/2 to f[s]/2. Also shown on the zoomed plot on the right is the transformation of discrete frequency f = 20 Hz
to continuous frequency F = 23.13 Hz. Note that the frequency mapping is approximately linear for f < f[s]/10 or so.
Figure 6 shows the effect of using equation 8 to pre-warp the cut-off frequency of an analog prototype filter to give f[c] = 20 Hz. With pre-warping, the analog prototype poles were scaled by
2π*23.13[. ] Without pre-warping, they were scaled by 2π*20.
Figure 5. Frequency mapping of the bilinear transform for f[s] = 100 Hz. x axis is discrete frequency and y-axis is continuous frequency. The right plot is a zoomed version of the left plot, showing
the value of F for f = 20 Hz.
Figure 6. Effect of pre-warping for f[c] = 20 Hz and f[s] = 100 Hz. 5^th order IIR Butterworth.
1. Mathworks website https://www.mathworks.com/help/signal/ref/butter.html
2. Oppenheim, Alan V. and Shafer, Ronald W., Discrete-Time Signal Processing, Prentice Hall, 1989, section 7.1.2
3. Lyons, Richard G., Understanding Digital Signal Processing, 2^nd Ed., Pearson, 2004, section 6.5
4. Williams, Arthur B. and Taylor, Fred J., Electronic Filter Design Handbook, 3^rd Ed., McGraw-Hill, 1995, section 2.3
5. Analog Devices Mini Tutorial MT-224, 2012 http://www.analog.com/media/en/training-seminars/tutorials/MT-224.pdf
Neil Robertson December 2017
Appendix Matlab Function butter_synth (12 lines of code, excluding error check)
This program is provided as-is without any guarantees or warranty. The author is not responsible for any damage or losses of any kind caused by the use or misuse of the program.
% butter_synth.m 12/9/17 Neil Robertson
% Find the coefficients of an IIR butterworth lowpass filter
% using bilinear transform
% N= filter order
% fc= -3 dB frequency in Hz
% fs= sample frequency in Hz
% b = numerator coefficients of digital filter
% a = denominator coefficients of digital filter
function [b,a]= butter_synth(N,fc,fs);
if fc>=fs/2;
error('fc must be less than fs/2')
% I. Find poles of analog filter
k= 1:N;
theta= (2*k -1)*pi/(2*N);
pa= -sin(theta) + j*cos(theta); % poles of filter with cutoff = 1 rad/s
% II. scale poles in frequency
Fc= fs/pi * tan(pi*fc/fs); % continuous pre-warped frequency
pa= pa*2*pi*Fc; % scale poles by 2*pi*Fc
% III. Find coeffs of digital filter
% poles and zeros in the z plane
p= (1 + pa/(2*fs))./(1 - pa/(2*fs)); % poles by bilinear transform
q= -ones(1,N); % zeros
% convert poles and zeros to polynomial coeffs
a= poly(p); % convert poles to polynomial coeffs a
a= real(a);
b= poly(q); % convert zeros to polynomial coeffs b
K= sum(a)/sum(b); % amplitude scale factor
b= K*b;
[ - ]
Comment by ●December 19, 2017
Thanks Neil,
You cleared up the internal secrets of Matlab function "butter" and I got identical results using either. Great job and very useful.
[ - ]
Comment by ●December 19, 2017
Yes, I was pleasantly surprised when I ran the function and the results matched. So now only another 99.9% of Matlab is a black box.
[ - ]
Comment by ●December 19, 2017
Hi Sir,
Clarity in your articles while dealing with fundamental concepts is great.
Really helpful.
Thank you,
[ - ]
Comment by ●December 19, 2017
You're welcome, I appreciate the feedback.
[ - ]
Comment by ●December 20, 2017
I request you to do one article like this on iir notch filter (if you have time).
[ - ]
Comment by ●December 20, 2017
I'll put that on my list!
[ - ]
Comment by ●August 14, 2018
Hey there,
When I started looking for an algorithm to design Butterworth filter I wanted to escape a somewhat bug or limitation from matlab keeping me from designing filter with superlow cutt-off frequency, for
instance by running the following code:
It seems going too low does not do well with Matlab, supprinsingly your code also does not work with such cut-off frequency, do you have any idea what might cause that?
Cheers and thanks again for your work
[ - ]
Comment by ●August 14, 2018
See my article on iir filter design using Cascaded biquads. It explains the problem you are having.
[ - ]
Comment by ●August 16, 2018
All right it seems to be a solution although my filter is of order 2, and regardless the method (Biquad or Butter) I always end up having a gain K of 0 when summing my coefs "a". For the record I
need a cut-off frequency at 2.365968869395443e-09.
I guess the improved narrowband you talk about in another article will do the trick? I am about to give it a try.
Thanks for you work !
[ - ]
Comment by ●August 16, 2018
That is a really low cut-off frequency. What is your application?
Note that you could use a decimator (e.g. CIC decimator) to reduce your sample rate and make the problem more tractable.
[ - ]
Comment by ●August 16, 2018
The application is DC value extraction with lowest noise possible. It is means to be implemented on GS/s ADC at a low power requirement.
Maybe I'm not approaching this the right way, I should filter smarter instead of harder (with butterworth 2nd order filter)
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[CFNS Seminar] Small-x contributions to the proton spin puzzle
An integral part of the proton spin puzzle is the contribution to the proton spin coming from quarks and gluons having small very values of the Bjorken x variable. This contribution is mostly beyond
the reach of current experiments and is very hard to calculate numerically on the lattice. It appears that theoretical understanding of quark and gluon helicity distributions at small x is needed to
assess the amount of proton spin coming from this region. In my talk I will describe the recent theoretical work aimed at finding the small-x asymptotics of the quark and gluon helicity
distributions, along with their orbital angular momenta (OAM). I will derive small-x evolution equations for helicity and solve them to find the small-x asymptotics of the parton helicity
distributions and OAM. The results of this work can be compared to the data to be collected at the upcoming Electron-Ion Collider (EIC) in order to extrapolate the small-x helicity distributions to
be measured at EIC to even smaller values of x, thus completely constraining the proton spin coming from small x and helping to resolve the proton spin puzzle. | {"url":"https://indico.bnl.gov/event/8951/","timestamp":"2024-11-10T21:31:45Z","content_type":"text/html","content_length":"93145","record_id":"<urn:uuid:a7be5a0d-24c1-47ac-8e1b-3050b79a340b>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00606.warc.gz"} |
Miles H. Wheeler
I am a Lecturer (Assistant Professor) in Analysis in the Department of Mathematical Sciences at the University of Bath. Before coming to Bath, I was a University Assistant at the Faculty of
Mathematics at the University of Vienna, and before that I was a postdoc at the Courant Institute of Mathematical Sciences supported by an NSF fellowship. I am interested in nonlinear partial
differential equations, and in particular in overdetermined elliptic boundary value problems coming from fluid mechanics.
For a more accessible introduction to the sort of work I do, see this expository talk on solitary waves and fronts (and Section 5 of [8]), or this short introduction to local and global bifurcation
theory. The talk is from a series on steady water waves in the ONEPAS seminar, and the notes are from a 2019 lecture to a group of masters students in mathematics and physics.
For a full list of past teaching see my CV. Recent teaching:
Theory of Partial Differential Equations (Spring 2020, 2021, 2022, 2023, 2024).
Specialist Reading Course “Generalised Solitary Waves in Fourth Order ODEs” (Spring 2022).
Topics in analysis: fluid mechanics (University of Vienna, Winter 2018).
David Lowry-Duda and I wrote an expository paper aimed at undergraduates which appeared in the American Mathematical Monthly and won an award from the MAA. | {"url":"https://www.mileshwheeler.com/","timestamp":"2024-11-02T14:25:26Z","content_type":"text/html","content_length":"45494","record_id":"<urn:uuid:5d9db647-a867-4118-aacc-d94fa33dc4c7>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00385.warc.gz"} |
Quadratic Equations / Functions learn online | sofatutor.com
Quadratic Equations / Functions
Easy learning with videos, exercises, tasks, and worksheets
Quadratic functions are easy to recognize. The polynomial expression known as a quadratic contains a variable that is squared making it a 2nd degree equation, and the graph is U-shaped. Quadratic
expressions that are equal to zero are called quadratic equations.
The standard form of a quadratic equation is: $ax^{2} + bx + c = 0$
The graph of a quadratic equation has a recognizable shape – a parabola. The parabola may open up or down, and the direction of the opening is determined by the sign of the leading coeeficient.
Solving Quadratic Equations by Factoring
To find the solutions to quadratic equations, also known as the zeros or roots, set the quadratic expression equal to zero then factor. The values for x identify where the graph touches the x-axis.
There are several methods you can use to factor quadratic equations.
Greatest Common Factor
Identify the greatest common factor (GCF) of all the terms in the quadratic expression and use the reverse of the Distributive Property to factor.
This equation has a GCF equal to 2x.
$ 2x^{2} + 2x& = 0\\ 2x(x + 2) &= 0 $
$x = -2, 0$
The graph touches the x-axis at -1 and 0.
Square Root Property
Use the property of square roots to find the zeros of quadratic equations such as this one.
$ x^{2} - 36 &= 0\\ x^{2} - 36 + 36 &= 0 +36\\ x^{2} &= 36\\ x&=\sqrt{36}\\ \sqrt{36} &=\pm 6 $
The solution to the quadratic equation is -6 and 6.
The foil method is used to simplify the product of two binomials, so the reverse of the foil method can be used to factor quadratic equations with trinomial expressions having a equal to 1.
To reverse the foil method, find factors of the c that sum to the b.
$ax^{2} + bx + c = 0$
$x^{2} + 7x + 6 = 0$
For the product of 6, the factors 1 and 6 sum to 7. Inside two sets of parentheses, add the constants of 6 and 1 to x respectively then set each binomial equal to zero and solve to determine the
roots of the equations.
$(x + 6)(x+ 1) = 0$
$x = -6, -1$
To check your work, FOIL.
$ (x + 6)(x+ 1)&=0\\ x^{2} + 6x + 1x +6&=0\\ x^{2} + 7x +6&=0 $
The roots of the equation are -6 and -1.
Difference of Two Squares
A quadratic equation that is the difference of two squares is also known as a DOTS equation. If you can recognize which quadratic equations are DOTS (difference of two squares), you can save yourself
time when factoring quadratic equations.
To identify DOTS, look for a specific pattern in the quadratic equation. Notice there are only two terms and both are perfect squares. The solution is the square root of the constant.
$ax^{2} + bx + c = 0$
$ x^{2} - 49 & =0\\ (x +7) (x -7)&=0 $
The solution to this DOTS equations is -7 and 7.
Factor by Grouping
When the quadratic equation has a trinomial expression with $a\neq 1$, you can factor by grouping. There are several steps to this method.
$ax^{2} + bx + c = 0$
$2x^{2} + -6x -8 =0$
Factor by Grouping
1. Find the product of a and c.
2. Identify two factors that sum to b.
3. Write new values for bx.
4. Group the factors using parentheses
5. Factor out the GCF of each group
6. Set up the binomial factors.
For this equation $a\times c = -16$. -2 and 8 sum to -6. Take a look at the next steps to solve this quadratic equation.
$ 2x^{2} -6x -8 & =0\\ 2x^{2} +2x -8x -8 & =0\\ (2x^{2} +2x)+ (-8x - 8)&=0\\ 2x(x +1) -8(x +1)&=0\\ (2x -8)(x +1)&=0 $
The solution to the equation is -1 and 4.
Solving Quadratic Equations by Completing the Square
When you are unable to determine factors, you can use the complete the square method to solve quadratic equations. To determine the roots using this method, there are several steps.
$ax^{2} + bx + c = 0$
1. Use the opposite operation to move the constant to the other side of the standard form.
2. Take half of b, square it and add to both sides of the equation.
3. Factor the perfect square on left side of the equation.
4. Apply the square root property to solve.
$ x^{2} + 2x -7 &= 0\\ x^{2} + 2x -7 +7&= 0 +7\\ x^{2} + 2x &=7\\ x^{2} + 2x + 1 &=7 + 1\\ x^{2} + 2x + 1 &=8\\ (x + 1)^{2}&= 8\\ x + 1 &= \pm\sqrt{8}\\ x + 1 -1&=-1 \pm\sqrt{8}\\ x&= -1\pm\sqrt{8} $
The solution is x&= -1\pm\sqrt{8} which is -3.8 and 1.8.
Solving Quadratic Equations with the Quadratic Formula
If there is no way to factor a quadratic equation or you simply prefer, you can always use the quadratic formula to determine the value(s) of x.
$ax^{2} + bx + c = 0$
Use the quadratic formula to solve this equation but first use the discriminant to learn about the roots.
Using and Understanding the Discriminant
The discriminant is the value under the radical, and it provides valuable information about the roots of an quadratic equation.
Discriminant ${b^{2}-4ac}$
• if > 0 then there are two real roots
• if = 0 there is one root repeated
• if < 0 there are two complex roots
For this problem, the discriminant is greater than zero, so there are two real roots.
$x^{2} -6x -4 =0$
$ x&=\frac{6\pm\sqrt{36+16}}{2}\\ x&=\frac{6\pm\sqrt{52}}{2}\\ x &= \frac{6}{2}\pm\frac{\sqrt{52}}{2} x&=3\pm 3.6\\ x&= -0.6, 6.6 $
The roots for this equation are -0.6 and 6.6.
All videos on the topic
Videos on the topic
Quadratic Equations / Functions (11 videos) | {"url":"https://us.sofatutor.com/math/algebra-1/quadratic-equations-functions","timestamp":"2024-11-11T07:19:54Z","content_type":"text/html","content_length":"110133","record_id":"<urn:uuid:c914862d-d7e5-48a1-9369-cedb8952d9e3>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00407.warc.gz"} |
Suppose in Pakistan, all the firms are identical with identical
cost curves which mean
industry is...
Suppose in Pakistan, all the firms are identical with identical cost curves which mean industry is...
Suppose in Pakistan, all the firms are identical with identical cost curves which mean
industry is perfectly competitive. Now please consider this following information about the
industry: A representative firm’s total cost is given by the equation TC = 100 + q2 + q where
q is the quantity of output produced by the firm. You also know that the market demand for
this product is given by the equation P = 1000 – 2Q where Q is the market quantity. In
addition you are told that the market supply curve is given by the equation P = 100 + Q.
a. What is the equilibrium quantity and price in this market given this information?
b. The firm’s MC equation based upon its TC equation is MC = 2q + 1. Given this information
and your answer in part (a), what is the firm’s profit maximizing level of production, total
revenue, total cost and profit at this market equilibrium? Is this a short-run or long-run
equilibrium? Explain your answer.
c. Given your answer in part (b), what do you anticipate will happen in this market in the
d. In this market, what is the long-run equilibrium price (breakeven or MC=ATC) and what
is the long-run equilibrium quantity for a representative firm to produce? Explain your
e. Given the long-run equilibrium price you calculated in part (d), how many units of this
good are produced in this market? | {"url":"https://justaaa.com/economics/66295-suppose-in-pakistan-all-the-firms-are-identical","timestamp":"2024-11-04T18:22:00Z","content_type":"text/html","content_length":"42264","record_id":"<urn:uuid:a16f98af-2a46-49d4-b69e-6998f955c52d>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00147.warc.gz"} |
Custom depiction modes. By “long tap” on any of the depiction modes in the menu you may modify it. In the appearing dialog you can change the code of the shader and also add an auxiliary texture from
your photo gallery. The shader has to be written in the Metal Shading Language which is based on C. For example the “iteration” shader looks as follows:
float4 color(constant Params ¶ms, Data input)
float3 c;
c = pal(params, float(input.iter)+0.5, params.brightnessfinite);
c = pal(params, float(-input.iter)+0.5, params.brightnessinfinite);
return float4(c, 1.0);
Here is the code for the function “pal”:
float3 pal(constant Params ¶ms, float iter, float brightness)
float3 c = float3(0,0,0);
for(int i=0;i<params.ncolor;i++)
float amplitude = 0.5*(1.0+sin(iter*params.color[i].frequency+params.color[i].initial));
c += params.color[i].c*amplitude;
float m = max3(c.r,c.g,c.b);
if(m>1.0)c /= m;
return c;
...and the specifications of the relevant structures:
typedef struct
float3 c;
float frequency;
float initial;
} UniformsColor;
typedef struct
int ncolor;
UniformsColor color[NCOLOR];
float brightnessfinite;
float brightnessinfinite;
float param[NPARAM];
float time;
int auxwidth;
int auxheight;
} Params;
The number of custom parameters (Params.param[]) may be specified in the dialog. They can later be changed tapping on the “brush” icon in the lower right corner.
In the following structure members are only available and calculated if the corresponding slider is on. Keep in mind that any additional value may slow down the calculation.
typedef struct
float dist; // The distance to the Mandelbrot set in pixels (For Julia fractals this works momentarily only in the finite area)
float dist2; // The distance to the next iteration level in pixels
float2 value; // The (complex) value of the last iteration (where its absolute value is <2; or a point in the limit cycle if the iteration is infinite)
int32_t iter; // The iteration number (a negative value means limit cycle of the corresponding length)
float smoothiter; // The iteration number smoothed out
float2 derivative; // The derivative of the last iteration
float2 coord; // The coordinate (in the complex plane) of the current fragment
float2 screencoord;// The screencoordinate of the current fragment
float pixelsize; // The size in pixels of the current fragment
texture2d<half> auxtexture; // The auxiliary texture if any has been chosen
} Data;
The values “dist”, “dist2”, and “smoothiter” are actually functions of “iter”, “value”, and “derivative” but have been added to keep the necessary memory access from the GPU low. | {"url":"https://www.retinamandelbrot.com/shader.html","timestamp":"2024-11-05T14:58:24Z","content_type":"text/html","content_length":"9274","record_id":"<urn:uuid:019d4df7-ca1d-4815-a80d-57244e3bf7ae>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00712.warc.gz"} |
designer: Michael Danziger
educational validator: John Belcher
supported by the National Science Foundation
published by the Massachusetts Institute of Technology
This item is an interactive three-dimensional animation that illustrates the concept of vector cross product. Users set angle theta from zero to 360 degrees, and then rotate a vector through the
angle. An animated hand automatically points in the proper direction according to the Right Hand Rule. No mathematics is introduced.
This item is part of a collection of visualizations developed by the MIT TEAL project to supplement an introductory course in calculus-based electricity and magnetism. Lecture notes, labs, and
presentations are also available as part of MIT's Open Courseware Repository:
MIT Open Courseware: Electricity and Magnetism
Please note that this resource requires Shockwave.
Subjects Levels Resource Types
Mathematical Tools - Instructional Material
- Vector Algebra - Lower Undergraduate = Interactive Simulation
Other Sciences - High School - Audio/Visual
- Mathematics = Movie/Animation
Intended Users Formats Ratings
- Learners - application/shockwave
- Educators - text/html
Access Rights:
Free access
This material is released under a Creative Commons Attribution 3.0 license.
Rights Holder:
MIT Open Courseware (OCW): http://ocw.mit.edu/OcwWeb/web/terms/terms/index.htm
3D simulation, 3D visualization, Vector Fields, cross product, interactive simulations, representations, three-dimensional simulation, vectors, visualization
Record Cloner:
Metadata instance created March 31, 2010 by Caroline Hall
Record Updated:
April 15, 2010 by Caroline Hall
Last Update
when Cataloged:
July 31, 2008
Other Collections:
ComPADRE is beta testing Citation Styles!
<a href="https://www.compadre.org/portal/items/detail.cfm?ID=9935">National Science Foundation. MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors. Cambridge: Massachusetts
Institute of Technology, July 31, 2008.</a>
(Massachusetts Institute of Technology, Cambridge, 2004), WWW Document, (http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/vectorfields/CrossProduct/crossProd.htm).
MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors (Massachusetts Institute of Technology, Cambridge, 2004), <http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/
MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors. (2008, July 31). Retrieved November 4, 2024, from Massachusetts Institute of Technology: http://web.mit.edu/8.02t/www/
National Science Foundation. MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors. Cambridge: Massachusetts Institute of Technology, July 31, 2008. http://web.mit.edu/8.02t/
www/802TEAL3D/visualizations/vectorfields/CrossProduct/crossProd.htm (accessed 4 November 2024).
MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors. Cambridge: Massachusetts Institute of Technology, 2004. 31 July 2008. National Science Foundation. 4 Nov. 2024 <http://
@misc{ Title = {MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors}, Publisher = {Massachusetts Institute of Technology}, Volume = {2024}, Number = {4 November 2024}, Month
= {July 31, 2008}, Year = {2004} }
%T MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors %D July 31, 2008 %I Massachusetts Institute of Technology %C Cambridge %U http://web.mit.edu/8.02t/www/802TEAL3D/
visualizations/vectorfields/CrossProduct/crossProd.htm %O application/shockwave
%0 Electronic Source %D July 31, 2008 %T MIT Physics 8.02: Vector Fields Visualizations - Cross Product of Two Vectors %I Massachusetts Institute of Technology %V 2024 %N 4 November 2024 %8 July 31,
2008 %9 application/shockwave %U http://web.mit.edu/8.02t/www/802TEAL3D/visualizations/vectorfields/CrossProduct/crossProd.htm
: ComPADRE offers citation styles as a guide only. We cannot offer interpretations about citations as this is an automated procedure. Please refer to the style manuals in the
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area for clarifications.
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The AIP Style presented is based on information from the AIP Style Manual.
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Deployed powershell service consuming WAY too much RAM
Hello. So, I recently used powershell studio to create a Windows Service using a windows service project. I deployed it, and it seemed to work great. As time went on, however, I have noticed that the
service is consuming an ungodly amount of RAM on all machines it was deployed to:
The service really isn't doing anything all that resource intensive. It verifies if the machine is able to connect to the internet, checks on a few services and starts them if they are not running,
writes a couple of files to disk if they do not exist, creates a scheduled task to launch an application if it does not exist, runs the aforementioned scheduled task when the process of the
application that the scheduled task launches is not running, and creates event log entries using "Write-host" to document errors encountered throughout. So that is the first problem: Why is the
service consuming so much RAM?
In my attempts to troubleshoot this, I try to stop the service, thinking that will close the process which consuming all the RAM. However, not only did stopping the service NOT close the process
consuming all of the RAM, it instead appeared to launch an identical process instead, and (I assume) attach to it for when the service starts running again:
Therefore, I was forced to terminate the process consuming all of the RAM in task manager in order to free up the system resources. What is going on here? I can appreciate that transforming a
powershell script into a working windows service is probably really tricky, but the way that the created service handles the system's available resources and the way in which the service behaves when
changing state seem very poorly implemented. I am using SAPIEN PowerShell Studio 2023 Version 5.8.219, which is a bit dated, I concede. Has the windows service functionality undergone significant
improvements in later version of Powershell Studio? Or maybe I am doing something wrong? I didn't write any code to execute when stopping the service, it just seems....sloppy....to run "Stop-process"
to kill the service process as the service is stopping. Or is that what I am supposed to do? Any advice would be great. I will include the code for the service project below.
Code: Select all
# Warning: Do not rename Start-MyService, Invoke-MyService and Stop-MyService functions
function Start-MyService
# Place one time startup code here.
# Initialize global variables and open connections if needed
$global:bRunService = $true
$global:bServiceRunning = $false
$global:bServicePaused = $false
$global:internet = $true
$global:AltaClockHubService = $true
$global:FPRemoteUpdateSvc = $true
$global:FCCClientSvc = $true
$global:FCCServerSvc = $true
function Invoke-MyService
$global:bServiceRunning = $true
while($global:bRunService) {
if($global:bServicePaused -eq $false) #Only act if service is not paused
# Get ip address of network adapter with statically assigned ip only
$ipaddress = Get-NetIPConfiguration | Where-Object { $_.NetIPv4Interface.Dhcp -like "disabled" }
$oct = ""
# split ip address so that we can look at last octet specifically to determine register number
If ($ipaddress.IPv4Address -is [Array])
$oct = $ipaddress.IPv4Address[0].ipaddress.Split('.')
$oct = $ipaddress.IPv4Address.ipaddress.Split('.')
# Test for internet connection
$connt = Test-NetConnection 8.8.8.8
If ($connt.PingSucceeded -eq $false)
# This code will log only a single event log warning that internet connect has been lost and then return to prevent further exection
If ($internet -eq $true)
$string = "$((Get-date).ToString()):`tThe register has lost access to the internet."
Write-Host "$string Logging event."
$internet = $false
# The below code will create only one event log when the service detects that the internet connection has been restored when it was previously down.
If (($connt.PingSucceeded -eq $true) -and ($internet -eq $false))
$string = "$((Get-date).ToString()):`tInternet access has been restored on the register."
Write-Host $string
$internet = $true
# Below Logic runs only on register 1 devices
If ([int]$oct[-1] -eq 11)
# Gets the alta clock hub service object and starts the service if it is not in a running state
$service = Get-Service "AltaClockHubService" -ErrorAction Stop
If ($service.Status -notlike "Running")
Write-Host "AltaclockHubService is not running on this register 1. Starting service now..."
If ($AltaClockHubService = $true)
$AltaClockHubService = $false
Write-Host "AltaclockHubService does not exist on this register."
If (([int]$oct[-1] -eq 13) -or ([int]$oct[-1] -eq 14))
$ocbserv = Get-Service
# Below Logic runs only on register devices
If (([int]$oct[-1] -gt 10) -and ([int]$oct[-1] -lt 20))
# Populate an array of Freedompay services' names
$payserv = @("FPRemoteUpdateSvc", "FCCClientSvc", "FCCServerSvc")
# Loop through each service in above array, get service object for each freedom pay service, and Start the service if it is not in a running state
foreach ($serv in $payserv)
$service = Get-Service -Name $serv -ErrorAction Stop
If ($service.Status -notlike "Running")
Write-Host "$($service.DisplayName) is not running. Starting service now..."
If (($serv -like "FPRemoteUpdateSvc") -and ($FPRemoteUpdateSvc -eq $true))
$FPRemoteUpdateSvc = $false
Write-Host "FPRemoteUpdateSvc does not exist on the register."
elseif (($serv -like "FCCClientSvc") -and ($FCCClientSvc -eq $true))
$FCCClientSvc = $false
Write-Host "FCCClientSvc does not exist on the register."
elseif (($serv -like "FCCServerSvc") -and ($FCCServerSvc -eq $true))
$FCCServerSvc = $false
Write-Host "FCCServerSvc does not exist on the register."
# code that executes beyond this point will execute on ALL kitchen devices
# Creates the folder which contains the scripts used by the Restart Brink Scheduled task if it does not exist
If (!(Test-Path "C:\Staging\Restart Brink"))
New-Item "C:\Staging\Restart Brink" -ItemType Directory
If (!(Test-Path "C:\Staging\Restart Brink\internet.txt"))
Write-Output $true | Out-File "C:\Staging\Restart Brink\internet.txt"
# Query the Brink services on the register and start them if they are not in a running state
Get-service | Where-Object { $_.DisplayName -like "*Brink*" } | % {
If ($_.Status -notlike "Running")
Write-Host "The $($_.DisplayName) service is not running. Starting it up..."
Write-Host "The $($_.DisplayName) service was successfully started."
# The below code is meant to test if the restart brink scheduled task and it's dependency scripts exist or not. The service will automatically re-create any of aforementioned items that are missing
$task = Get-ScheduledTask -TaskName "Restart Brink" -ErrorAction Stop
If (!(Test-Path "$($task.Actions[0].Execute)"))
Write-Host "The Restart Brink Scripts that the scheduled task relies on do not exist. Recreating them..."
$battext = @'
@echo OFF
Powershell.exe -executionpolicy unrestricted -command "Set-ExecutionPolicy unrestricted -force -confirm:$false"
Powershell.exe -File "C:\Staging\Restart Brink\Restart Brink.ps1"
Write-Output $battext | Out-File "$($task.Actions[0].Execute)" -Encoding ascii -Force
$ps1text = @'
function Use-RunAs
# Check if script is running as Adminstrator and if not use RunAs
# Use Check Switch to check if admin
$IsAdmin = ([Security.Principal.WindowsPrincipal] [Security.Principal.WindowsIdentity]::GetCurrent()).IsInRole([Security.Principal.WindowsBuiltInRole] "Administrator")
if ($Check) { return $IsAdmin }
if ($MyInvocation.ScriptName -ne "")
if (-not $IsAdmin)
$arg = "-file `"$($MyInvocation.ScriptName)`""
Start-Process "$psHome\powershell.exe" -Verb Runas -ArgumentList $arg -ErrorAction 'stop'
Write-Warning "Error - Failed to restart script with runas"
exit # Quit this session of powershell
Write-Warning "Error - Script must be saved as a .ps1 file first"
# Determine if device is a register or kitchen device by determining if the last octet begins with a 1 (register) or a 2 (kitchen)
$ipaddress = Get-NetIPConfiguration | Where-Object { $_.NetIPv4Interface.Dhcp -like "disabled" }
$oct = ""
If ($ipaddress.IPv4Address -is [Array])
$oct = $ipaddress.IPv4Address[0].ipaddress.Split('.')
$oct = $ipaddress.IPv4Address.ipaddress.Split('.')
# code to run if device is a register
If ($oct[-1].StartsWith("1"))
#Confirm Brink is Running
$BrinkRegister = Get-Process "Register" -ErrorAction SilentlyContinue
$running = ""
if($BrinkRegister -eq $null) {
$running = $false
Write-Output "Error: Brink is not Running" | Out-File "C:\Staging\Restart Brink\Not Running.txt"
Start-Sleep -Seconds 3
$running = $true
Write-Output "Brink is Running" | Out-File "C:\Staging\Restart Brink\Brink is Running.txt"
start-sleep -Seconds 3
# Close Register.exe if it is running
If ($running -eq $true)
Get-Process "Register" | Stop-Process -Force
Start-Sleep 3
Until ((Get-Process).Name -notcontains "Register")
#Start Register.exe
Start-Process -FilePath "C:\Brink\POS\Register.exe" -NoNewWindow
Write-output "Brink is Starting"
Start-Sleep -Seconds 3
# code to run if device is a kitchen device
ElseIf ($oct[-1].StartsWith("2"))
#Confirm Brink is Running
$BrinkRegister = Get-Process "Kitchen" -ErrorAction SilentlyContinue
$running = ""
if($BrinkRegister -eq $null) {
$running = $false
Write-Output "Error: Brink is not Running" | Out-File "C:\Staging\Restart Brink\Not Running.txt"
Start-Sleep -Seconds 3
$running = $true
Write-Output "Brink is Running" | Out-File "C:\Staging\Restart Brink\Brink is Running.txt"
start-sleep -Seconds 3
# Close Register.exe if it is running
If ($running -eq $true)
Get-Process "Kitchen" | Stop-Process -Force
Start-Sleep 3
Until ((Get-Process).Name -notcontains "Kitchen")
#Start Kitchen.exe
Start-Process -FilePath "C:\Brink\POS\Kitchen.exe" -NoNewWindow
Write-output "Brink is Starting"
Start-Sleep -Seconds 3
Write-Output "Cannot determine device type. This task will terminate without doing anything."
Write-Output $ps1text | Out-File "C:\Staging\Restart Brink\Restart Brink.ps1" -Force
Write-Host "Restart Brink scripts successfully recreated."
Write-Host "The Restart Brink Scheduled Task does not exist. Begin Mitigation."
If (!(Test-Path "C:\Staging\Restart Brink\restart.bat"))
Write-Host "The Restart Brink Scripts that the scheduled task relies on do not exist. Recreating them..."
$battext = @'
@echo OFF
Powershell.exe -executionpolicy unrestricted -command "Set-ExecutionPolicy unrestricted -force -confirm:$false"
Powershell.exe -File "C:\Staging\Restart Brink\Restart Brink.ps1"
Write-Output $battext | Out-File "C:\Staging\Restart Brink\restart.bat" -Encoding ascii -Force
$ps1text = @'
function Use-RunAs
# Check if script is running as Adminstrator and if not use RunAs
# Use Check Switch to check if admin
$IsAdmin = ([Security.Principal.WindowsPrincipal] [Security.Principal.WindowsIdentity]::GetCurrent()).IsInRole([Security.Principal.WindowsBuiltInRole] "Administrator")
if ($Check) { return $IsAdmin }
if ($MyInvocation.ScriptName -ne "")
if (-not $IsAdmin)
$arg = "-file `"$($MyInvocation.ScriptName)`""
Start-Process "$psHome\powershell.exe" -Verb Runas -ArgumentList $arg -ErrorAction 'stop'
Write-Warning "Error - Failed to restart script with runas"
exit # Quit this session of powershell
Write-Warning "Error - Script must be saved as a .ps1 file first"
# Determine if device is a register or kitchen device by determining if the last octet begins with a 1 (register) or a 2 (kitchen)
$ipaddress = Get-NetIPConfiguration | Where-Object { $_.NetIPv4Interface.Dhcp -like "disabled" }
$oct = ""
If ($ipaddress.IPv4Address -is [Array])
$oct = $ipaddress.IPv4Address[0].ipaddress.Split('.')
$oct = $ipaddress.IPv4Address.ipaddress.Split('.')
# code to run if device is a register
If ($oct[-1].StartsWith("1"))
#Confirm Brink is Running
$BrinkRegister = Get-Process "Register" -ErrorAction SilentlyContinue
$running = ""
if($BrinkRegister -eq $null) {
$running = $false
Write-Output "Error: Brink is not Running" | Out-File "C:\Staging\Restart Brink\Not Running.txt"
Start-Sleep -Seconds 3
$running = $true
Write-Output "Brink is Running" | Out-File "C:\Staging\Restart Brink\Brink is Running.txt"
start-sleep -Seconds 3
# Close Register.exe if it is running
If ($running -eq $true)
Get-Process "Register" | Stop-Process -Force
Start-Sleep 3
Until ((Get-Process).Name -notcontains "Register")
#Start Register.exe
Start-Process -FilePath "C:\Brink\POS\Register.exe" -NoNewWindow
Write-output "Brink is Starting"
Start-Sleep -Seconds 3
# code to run if device is a kitchen device
ElseIf ($oct[-1].StartsWith("2"))
#Confirm Brink is Running
$BrinkRegister = Get-Process "Kitchen" -ErrorAction SilentlyContinue
$running = ""
if($BrinkRegister -eq $null) {
$running = $false
Write-Output "Error: Brink is not Running" | Out-File "C:\Staging\Restart Brink\Not Running.txt"
Start-Sleep -Seconds 3
$running = $true
Write-Output "Brink is Running" | Out-File "C:\Staging\Restart Brink\Brink is Running.txt"
start-sleep -Seconds 3
# Close Register.exe if it is running
If ($running -eq $true)
Get-Process "Kitchen" | Stop-Process -Force
Start-Sleep 3
Until ((Get-Process).Name -notcontains "Kitchen")
#Start Kitchen.exe
Start-Process -FilePath "C:\Brink\POS\Kitchen.exe" -NoNewWindow
Write-output "Brink is Starting"
Start-Sleep -Seconds 3
Write-Output "Cannot determine device type. This task will terminate without doing anything."
Write-Output $ps1text | Out-File "C:\Staging\Restart Brink\Restart Brink.ps1" -Force
Write-Host "Restart Brink scripts successfully recreated."
Write-Host "Creating Restart Brink Scheduled Task."
$action = New-ScheduledTaskAction -Execute "C:\Staging\Restart Brink\restart.bat"
$trig = New-ScheduledTaskTrigger -At $([Datetime]"05:00:00") -Daily
$princ = New-ScheduledTaskPrincipal -UserId pbrul -LogonType Interactive -RunLevel Highest
$settings = New-ScheduledTaskSettingsSet
$new = New-ScheduledTask -Action $action -Principal $princ -Settings $settings -Trigger $trig
Register-ScheduledTask "Restart Brink" -InputObject $new
Write-Host "Restart Brink Scheduled Task successfully created."
# The below code will run the restart brink scheduled task if neither the register.exe nor kitchen.exe processes are running on the register
If (((Get-Process).Name -notcontains "Register") -and ((Get-Process).Name -notcontains "Kitchen"))
Write-Host "Brink process is not running. Launching it now."
Start-ScheduledTask -TaskName "Restart Brink"
# Log exception in application log
Write-Host $_.Exception.Message
# Adjust sleep timing to determine how often your service becomes active.
if($global:bServicePaused -eq $true)
Start-Sleep -Seconds 30 # if the service is paused we sleep longer between checks.
Start-Sleep -Seconds 30 # a lower number will make your service active more often and use more CPU cycles
$global:bServiceRunning = $false
function Stop-MyService
$global:bRunService = $false # Signal main loop to exit
$CountDown = 30 # Maximum wait for loop to exit
while($global:bServiceRunning -and $Countdown -gt 0)
Start-Sleep -Seconds 1 # wait for your main loop to exit
$Countdown = $Countdown - 1
# Place code to be executed on service stop here
# Close files and connections, terminate jobs and
# use remove-module to unload blocking modules
function Pause-MyService
# Service is being paused
# Save state
$global:bServicePaused = $true
# Note that the thread your PowerShell script is running on is not suspended on 'pause'.
# It is your responsibility in the service loop to pause processing until a 'continue' command is issued.
# It is recommended to sleep for longer periods between loop iterations when the service is paused.
# in order to prevent excessive CPU usage by simply waiting and looping.
function Continue-MyService
# Service is being continued from a paused state
# Restore any saved states if needed
$global:bServicePaused = $false
Re: Deployed powershell service consuming WAY too much RAM
If you have not done so yet, I recommend studying this blog article here:
https://www.sapien.com/blog/2022/06/14/ ... e-project/
By the nature of a service, stopping a service does terminate the process. If it does not, something is not handled correctly. It sends a stop service event to the service. Which is what the
associated handler is supposed to, well, handle. You must release any objects that you specifically allocated here and make sure to set the proper flags for the main service loop to exit.
Resources *should* be allocated at the start-service and all resources must be released at the stop-service event. This includes any resources allocated and not released during the normal operation
of the service. It seems that is not what you do. Likewise, the pause-service event could release accumulated resources and re-allocated them as needed, but that is an optional thing. It does however
have to suspend the services processing loop.
PowerShell itself was not designed to create long running applications. The memory model is scope based, so if something does not go out of scope it is never released. Assigning a new object to the
same global scope variable will inevitable create memory leaks.
I would recommend to test and debug the code making up the body of your service (Invoke-MyService) in a separate non-service project for easier debugging and resetting. If you use third party modules
or elements that leak memory, it might be a good idea to relocate them to a separate process that can be started and stopped from the service when needed.
The hosting shell around a PowerShell service does not execute additional processes nor does it allocate any resources other than a PowerShell engine during startup. Specially not during ongoing
operation. So any leaking memory is coming from your code in one fashion or another.
Alexander Riedel
SAPIEN Technologies, Inc.
Re: Deployed powershell service consuming WAY too much RAM
Thank you for your reply. Unfortunately, some of your comment flew right over my head, so I wanted to clarify a couple of things. You said
The memory model is scope based, so if something does not go out of scope it is never released. Assigning a new object to the same global scope variable will inevitable create memory leaks.
Could you elaborate a bit more on what you mean by this? Are you saying that any variable defined in the global scope will create a memory leak if its value changed anywhere in the service? If that
were the case, wouldn't the service create a memory leak whenever the script is paused, since that changes the $global:bServicePaused variable to true? As for the custom global variables I defined,
if I changed the scope of those variables to "script", would that prevent those variable from contributing to the memory leaks?
Re: Deployed powershell service consuming WAY too much RAM
No. Intrinsic types, like boolean, integer etc. are not causing any memory leaks. They are just assigned as values.
Also this has nothing to do with this being service, this is an inherent PowerShell and its underlying C# .NET technology thing.
It just surfaces in your case in a service because it has a loop and runs long enough to be noticeable.
Suppose you create an object in Start-MyService like so:
$Global:Object1 = New-Object something_or_another.
If you do not add a congruent Clear-Variable $Global:Object1 or use a $Global:Object1 = $null
in your Stop-MyService handler you may get a leak when starting the service again will create a another object.
The same is true if you create something in each loop and do not delete it before allocating another one, assigning it to the same value.
.NET generally deletes objects when they go out of scope. That means the code exist the function or script block they where allocated in. Global or script scope obviously exists as long as the script
runs, or in your case the service.
Some items do have reference counts and do not create such leaks, string are a good example.
This is far to complex a subject to discuss in a forum post, so I would recommend an advanced PowerShell book.
But the short answer to your question is "no". Changing the scope to 'Script' will not change a thing. You need to dive in and find out what leaks.
Alexander Riedel
SAPIEN Technologies, Inc. | {"url":"https://www.sapien.com/forums/viewtopic.php?t=16801&sid=27f3a50431cdb0a3bbdad030a0a302fb","timestamp":"2024-11-01T22:30:03Z","content_type":"text/html","content_length":"71328","record_id":"<urn:uuid:f6e83c35-9380-47ca-b93c-f906a378d891>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00863.warc.gz"} |
God particles breeding like bosons
Copying programmers ?
Particle physicists seem to be copying an idea from programmers. Whereas programmers say:
"There is no problem in computer science which cannot be solved by one more level of indirection"
Physicists seem to be saying:
"There is no problem in particle physics that cannot be solved by one more type of particle" | {"url":"https://forums.theregister.com/forum/all/2010/06/15/higgs_bosons/","timestamp":"2024-11-02T12:41:32Z","content_type":"text/html","content_length":"101870","record_id":"<urn:uuid:48e81977-aaa9-402f-8467-741dc4595aef>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00472.warc.gz"} |
Eureka Math Grade 4 Module 5 Lesson 9
Fractions are the important topics in maths. This will be helpful in the real-time environment. Know-how and where to apply the formulas from this page. Get a detailed explanation for all the
questions here. Download Eureka Math Answers Grade 4 chapter 9 pdf for free of cost. As per your convenience, we have provided the solutions in pdf format so that you can prepare offline.
Engage NY Eureka Math 4th Grade Module 5 Lesson 9 Answer Key
Get the guided notes for chapter 9 Answer Key from here. This will be the best resource to enhance your math skills. The topics covered in this chapter are Fractional units. test yourself by solving
questions given at the end of the chapter.
Eureka Math Grade 4 Module 5 Lesson 9 Problem Set Answer Key
Each rectangle represents 1.
Question 1.
Compose the shaded fractions into larger fractional units. Express the equivalent fractions in a number sentence using division. The first one has been done for you.
2/4 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
4/2 = 2.
2/4 = 1/2.
3/6 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
3/3 = 1.
6/3 = 2.
3/6 = 1/2.
5/10 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
5/5 = 1.
10/5 = 2.
5/10 = 1/2.
4/8 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
4/4 = 1.
8/4 = 2.
4/8 = 1/2.
Question 2.
Compose the shaded fractions into larger fractional units. Express the equivalent fractions in a number sentence using division.
2/6 = 1/3.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
6/2 = 3.
2/6 = 1/3.
2/8 = 1/4.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
8/2 = 4.
2/8 = 1/4.
2/10 = 1/5.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
10/2 = 5.
2/10 = 1/5.
2/12 = 1/6.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
12/2 = 6.
2/12 = 1/6.
e. What happened to the size of the fractional units when you composed the fraction?
The size of the fractional units is increased.
In the above-given question,
given that,
whenever the size of the fractional units decreases when we decompose the fraction.
decomposing = dividing.
whenever the size of the fractional units increases when we compose the fraction.
composing = adding.
f. What happened to the total number of units in the whole when you composed the fraction?
The total number of units in the whole is increased when we composed the fraction.
In the above-given question,
given that,
the total number of units in the whole is increased when we composed the fraction.
Question 3.
a. In the first area model, show 2 sixths. In the second area model, show 3 ninths. Show how both fractions can be renamed as the same unit fraction.
2/6 = 1/3.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
6/2 = 3.
2/6 = 1/3.
3/9 = 1/3.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
3/3 = 1.
9/3 = 3.
3/9 = 1/3.
b. Express the equivalent fractions in a number sentence using division.
Question 4.
a. In the first area model, show 2 eighths. In the second area model, show 3 twelfths. Show how both fractions can be composed, or renamed, as the same unit fraction.
b. Express the equivalent fractions in a number sentence using division.
2/8 = 1/4.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
8/2 = 4.
2/8 = 1/4.
3/12 = 1/4.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
3/3 = 1.
12/3 = 4.
3/12 = 1/4.
Eureka Math Grade 4 Module 5 Lesson 9 Exit Ticket Answer Key
a. In the first area model, show 2 sixths. In the second area model, show 4 twelfths. Show how both fractions can be composed, or renamed, as the same unit fraction.
b. Express the equivalent fractions in a number sentence using division.
2/6 = 1/3.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
6/2 = 3.
2/6 = 1/3.
4/12 = 1/3.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
4/4 = 1.
12/4 = 3.
4/12 = 1/3.
Eureka Math Grade 4 Module 5 Lesson 9 Homework Answer Key
Each rectangle represents 1.
Question 1.
Compose the shaded fractions into larger fractional units. Express the equivalent fractions in a number sentence using division. The first one has been done for you.
2/4 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
4/2 = 2.
2/4 = 1/2.
4/8 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
8/4 = 2.
4/8 = 1/2.
6/12 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
6/6 = 1.
12/6 = 2.
6/12 = 1/2.
7/14 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
7/7 = 1.
14/7 = 2.
7/14 = 1/2.
Question 2.
Compose the shaded fractions into larger fractional units. Express the equivalent fractions in a number sentence using division.
2/12 = 1/6.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
12/2 = 6.
2/12 = 1/6.
2/10 = 1/5.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
10/2 = 5.
2/10 = 1/5.
2/8 = 1/4.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
8/2 = 4.
2/8 = 1/4.
2/6 = 1/3.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
2/2 = 1.
6/2 = 3.
2/6 = 1/3.
e. What happened to the size of the fractional units when you composed the fraction?
The size of the fractional units is increased.
In the above-given question,
given that,
whenever the size of the fractional units decreases when we decompose the fraction.
decomposing = dividing.
whenever the size of the fractional units increases when we compose the fraction.
composing = adding.
f. What happened to the total number of units in the whole when you composed the fraction?
The total number of units in the whole is increased when we composed the fraction.
In the above-given question,
given that,
the total number of units in the whole is increased when we composed the fraction.
Question 3.
a. In the first area model, show 4 eighths. In the second area model, show 6 twelfths. Show how both fractions can be composed, or renamed, as the same unit fraction.
b. Express the equivalent fractions in a number sentence using division.
4/8 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
4/4 = 1.
8/4 = 2.
4/8 = 1/2.
Question 4.
a. In the first area model, show 4 eighths. In the second area model, show 8 sixteenths. Show how both fractions can be composed, or renamed, as the same unit fraction.
b. Express the equivalent fractions in a number sentence using division.
4/8 = 1/2.
In the above-given question,
given that,
compose the shaded fractions into larger fractional units.
Express the equivalent fractions in a number sentence using division.
4/4 = 1.
8/4 = 2.
4/8 = 1/2. | {"url":"https://eurekamathanswerkeys.com/eureka-math-grade-4-module-5-lesson-9/","timestamp":"2024-11-03T15:12:46Z","content_type":"text/html","content_length":"60162","record_id":"<urn:uuid:df856dc6-2b3e-4fe7-ab9a-5dbf4b6ac79e>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00727.warc.gz"} |
What is a bit error rate tester?
What is a bit error rate tester?
A bit error rate tester (BERT), also known as a “bit error ratio tester” or bit error rate test solution (BERTs) is electronic test equipment used to test the quality of signal transmission of single
components or complete systems. Digital communication analyser is optional to display the transmitted or received signal.
How do you calculate bit error?
The BER is calculated by comparing the transmitted sequence of bits to the received bits and counting the number of errors. The ratio of how many bits received in error over the number of total bits
received is the BER. This measured ratio is affected by many factors including: signal to noise, distortion, and jitter.
What causes bit error rate?
For fibre optic systems, bit errors mainly result from imperfections in the components used to make the link. These include the optical driver, receiver, connectors and the fibre itself. Bit errors
may also be introduced as a result of optical dispersion and attenuation that may be present.
Which is the best bit error ratio tester?
Our NRZ and PAM4 coding schemes for 400G solution deliver a fully integrated 64Gbaud BER test. Keysight offers the broadest choice of bit error rate testers – covering affordable manufacturing test
and high-performance characterization with compliance testing up to 32 Gb/s.
How long does it take to test a bit error rate?
For Gigabit Ethernet that specifies an error rate of less than 1 in 10^12, the time taken to transmit the 10^12 bits of data is 13.33 minutes. To gain a reasonable level of confidence of the bit
error rate it would be wise to send around 100 times this amount of data.
What is the error rate of a BER test?
If one error were detected while sending 10 12 bits, then a first approximation may be that the error rate is 1 in 10 12, but this is not the case in view of the random nature of any errors that may
occur. In theory an infinite number of bits should be sent to prove the actual error rate, but this is obviously not feasible.
Why are pseudorandom codes used in bit error rate testing?
Accordingly to assist making measurements faster, mathematical techniques are applied and the data that is transmitted in the test is made as random as possible – a pseudorandom code is used that is
generated within the bit error rate tester. This helps reduce the time required while still enabling reasonably accurate measurements to be made. | {"url":"https://www.handlebar-online.com/articles/what-is-a-bit-error-rate-tester/","timestamp":"2024-11-07T07:22:39Z","content_type":"text/html","content_length":"42978","record_id":"<urn:uuid:71ff75d6-be18-4228-8a26-1b41b7a78400>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00632.warc.gz"} |
Printable Sudoku 4 Per Page With Answers Sudoku Printable | Sudoku Printables
Printable Sudoku 4 Per Page With Answers Sudoku Printable
Printable Sudoku 4 Per Page With Answers Sudoku Printable – If you’ve had any issues solving sudoku, then you’re aware that there are many different kinds of sudoku puzzles and it’s difficult to
decide which one you’ll need to solve. But there are also many ways to solve them, and you’ll find that solving a printable version will prove to an excellent way to get started. The rules for
solving sudoku are similar to those of other kinds of puzzles, but the actual format varies slightly.
What Does the Word ‘Sudoku’ Mean?
The term ‘Sudoku’ taken from the Japanese words suji and dokushin which translate to ‘number’ and ‘unmarried person as well as ‘unmarried. The objective of the puzzle is to fill each box with numbers
in a way that each number between one and nine appears only once on each horizontal line. The word Sudoku is a trademark of the Japanese puzzle manufacturer Nikoli, which originated in Kyoto.
The name Sudoku is derived from the Japanese word”shuji Wa Dokushin Ni Kagiru meaning ‘numbers have to stay single’. The game is composed of nine 3×3 squares with nine smaller squares. The game was
originally known as Number Place, Sudoku was a puzzle that stimulated mathematical development. While the origins of the game remain a mystery, Sudoku is known to have roots that go back to the
earliest number puzzles.
Why is Sudoku So Addicting?
If you’ve ever played Sudoku then you’re aware of how addictive it can be. The Sudoku addict will never be able to put down the thought of the next puzzle they’ll solve. They’re constantly thinking
about their next adventure, while different aspects of their life are slipping to the by the wayside. Sudoku is a very addictive game It’s crucial that you keep the addictive power of the game in
check. If you’ve developed a craving for Sudoku, here are some ways to stop your addiction.
One of the most common methods of determining if someone is addicted to Sudoku is to look at your behaviour. A majority of people have magazines and books with them, while others simply browse
through social media updates. Sudoku addicts, however, take newspapers, books, exercise books and smartphones everywhere they travel. They spend hours a day working on puzzles and cannot stop! Some
even find it easier to solve Sudoku puzzles than their regular crosswords, which is why they don’t quit.
Printable Sudoku 4 Per Page With Answers
What is the Key to Solving a Sudoku Puzzle?
An effective method for solving an printable sudoku game is to try and practice with various approaches. The most effective Sudoku puzzle solvers do not employ the same strategy for every single
puzzle. The most important thing is to practice and experiment with different methods until you find the one that is effective for you. After some time, you’ll be able solve puzzles without a
problem! But how do you know to solve sudoku puzzles that are printable sudoku challenge?
In the beginning, you must grasp the basics of suduko. It’s a form of logic and deduction, and you need to examine the puzzle from many different angles to spot patterns, and then solve it. When you
are solving a suduko puzzle, you should not attempt to guess the numbers. instead, you should search the grid for ways to identify patterns. This strategy to squares and rows.
Related For Sudoku Puzzles Printable | {"url":"https://sudokuprintables.net/printable-sudoku-4-per-page-with-answers/printable-sudoku-4-per-page-with-answers-sudoku-printable-7/","timestamp":"2024-11-05T00:09:32Z","content_type":"text/html","content_length":"26373","record_id":"<urn:uuid:f6798883-22aa-44ef-9043-003091ea7b19>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00517.warc.gz"} |
Fractions are an essential part of mathematics and can be found everywhere in our daily lives. They represent a part of a whole or a portion of a quantity, and understanding them is crucial in
solving mathematical problems. The different types of fractions and how to perform basic operations with them.
Types of Fractions
Proper Fractions:
A proper fraction is a fraction where the numerator is smaller than the denominator. For example, 2/5, 3/4, and 7/8 are all proper fractions.
Improper Fractions:
An improper fraction is a fraction where the numerator is greater than or equal to the denominator. For example, 5/3, 7/4, and 11/5 are all improper fractions.
Mixed Numbers:
A mixed number is a combination of a whole number and a proper fraction. For example, 2 1/3, 3 2/5, and 4 3/4 are all mixed numbers.
Equivalent Fractions:
Equivalent fractions are fractions that represent the same quantity but are written in different forms. For example, 1/2 and 2/4 are equivalent fractions because they represent the same amount.
Basic Operations with Fractions
Addition and Subtraction:
To add or subtract fractions, we need to have a common denominator. We can find the common denominator by multiplying the denominators of the two fractions. Once we have a common denominator, we can
add or subtract the numerators and write the result over the common denominator.
For example: 1/4 + 2/5 = (5/5) x (1/4) + (4/4) x (2/5) = 5/20 + 8/20 = 13/20 3/5 – 1/6 = (6/6) x (3/5) – (5/5) x (1/6) = 18/30 – 5/30 = 13/30
To multiply fractions, we multiply the numerators together and the denominators together. We can simplify the fraction by canceling out common factors in the numerator and denominator.
For example: 2/3 x 3/4 = 6/12 = 1/2 5/6 x 4/5 = 20/30 = 2/3
To divide fractions, we multiply the first fraction by the reciprocal of the second fraction. The reciprocal of a fraction is obtained by switching the numerator and the denominator.
For example: 2/3 ÷ 4/5 = 2/3 x 5/4 = 10/12 = 5/6 3/4 ÷ 1/2 = 3/4 x 2/1 = 6/4 = 3/2
Fractions are an important part of mathematics and are used in many real-life situations, such as cooking, measuring, and calculating distances.
The types of fractions and how to perform basic operations with them, we can improve our mathematical skills and problem-solving abilities. | {"url":"http://learnindex.com/fractions/","timestamp":"2024-11-09T13:35:50Z","content_type":"text/html","content_length":"140043","record_id":"<urn:uuid:55341915-02d2-40bd-ad0a-8c71f6b1b5a9>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00507.warc.gz"} |
8.2.2 Coordinates, PT3 Focus Practice
Question 6:
The point M (x, 4), is the midpoint of the line joining straight line Q (-2, -3) and R (14, y).
x and y are
$\begin{array}{l}x=\frac{-2+14}{2}\\ x=\frac{12}{2}\\ x=6\\ \\ 4=\frac{-3+y}{2}\\ 8=-3+y\\ y=11\end{array}$
Question 7:
In diagram below, PQR is a right-angled triangle. The sides QR and PQ are parallel to the y-axis and the x-axis respectively. The length of QR = 6 units.
Given that M is the midpoint of PR, then the coordinates of M are
x-coordinate of R = 3
y-coordinate of R = 1 + 6 = 7
R = (3, 7)
$\begin{array}{l}P\left(1,1\right),R\left(3,7\right)\\ \text{Coordinates of }M\\ =\left(\frac{1+3}{2},\frac{1+7}{2}\right)\\ =\left(2,4\right)\end{array}$
Question 8:
Given points P (–2, 8) and Q (10, 8), find the length of PQ.
$\begin{array}{l}\text{Length of }PQ\\ =\sqrt{{\left[10-\left(-2\right)\right]}^{2}+{\left(8-8\right)}^{2}}\\ =\sqrt{{\left(14\right)}^{2}+0}\\ =14\text{ units}\end{array}$
Question 9:
In diagram below, ABC is an isosceles triangle.
(a) the value of k,
(b) the length of BC.
$\begin{array}{l}\left(\text{a}\right)\\ \text{For an isosceles triangle, }\\ y-\text{coordinate of }C\text{ is the midpoint of straight line }AB.\\ \frac{2+k}{2}=-3\\ 2+k=-6\\ \text{ }k=-8\\ \\ \
left(\text{b}\right)\\ B=\left(-2,-8\right)\\ BC=\sqrt{{\left[10-\left(-2\right)\right]}^{2}+{\left[-3-\left(-8\right)\right]}^{2}}\\ \text{ }=\sqrt{{12}^{2}+{5}^{2}}\\ \text{ }=13\text{ units}\end
Question 10:
Diagram below shows a rhombus PQRS drawn on a Cartesian plane. PS is parallel to x-axis.
Given the perimeter of PQRS is 40 units, find the coordinates of point R.
$\begin{array}{l}\text{All sides of rhombus have the same length,}\\ \text{therefore length of each side}=\frac{40}{4}=10\text{ units}\\ PQ=10\\ {\left(9-{x}_{1}\right)}^{2}+{\left(7-\left(-1\right)\
right)}^{2}={10}^{2}\\ 81-18{x}_{1}+{x}_{1}{}^{2}+64=100\\ {x}_{1}{}^{2}-18{x}_{1}+45=0\\ \left({x}_{1}-3\right)\left({x}_{1}-15\right)=0\\ {x}_{1}=3,15\\ {x}_{1}=3\\ Q=\left(3,-1\right),R=\left({x}_
{2},-1\right)\\ \\ QR=10\\ {\left({x}_{2}-3\right)}^{2}+{\left[-1-\left(-1\right)\right]}^{2}={10}^{2}\\ {x}_{2}{}^{2}-6{x}_{2}+9+0=100\\ {x}_{2}{}^{2}-6{x}_{2}-91=0\\ \left({x}_{2}+7\right)\left({x}
_{2}-13\right)=0\\ {x}_{2}=-7,13\\ {x}_{2}=13\\ \\ \therefore R=\left(13,-1\right)\end{array}$ | {"url":"http://content.myhometuition.com/2020/03/04/8-2-2-coordinates-pt3-focus-practice/","timestamp":"2024-11-06T13:48:27Z","content_type":"text/html","content_length":"36168","record_id":"<urn:uuid:a020aa4c-d49b-45f1-860f-f03c780f2175>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00199.warc.gz"} |
pie chart
Hello Friends,In this video you will learn how to create Pie of Pie chart in excel. Please download this excel file from below given link:https://drive.googl
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Non-Markovianity in Open Quantum Systems
Schedule for: 23w5083 - Non-Markovianity in Open Quantum Systems
Beginning on Sunday, February 12 and ending Friday February 17, 2023
All times in Banff, Alberta time, MST (UTC-7).
Sunday, February 12
16:00 - 17:30 Check-in begins at 16:00 on Sunday and is open 24 hours (Front Desk - Professional Development Centre)
Dinner ↓
17:30 - 19:30 A buffet dinner is served daily between 5:30pm and 7:30pm in Vistas Dining Room, top floor of the Sally Borden Building.
(Vistas Dining Room)
20:00 - 22:00 Informal gathering (TCPL Foyer)
Monday, February 13
Breakfast ↓
- Breakfast is served daily between 7 and 9am in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Introduction and Welcome by BIRS Staff ↓
- A brief introduction to BIRS with important logistical information, technology instruction, and opportunity for participants to ask questions.
(TCPL 201)
- Mile Gu: Resources and Non-Markovianity in Quantum-Agents (TCPL 201)
- Coffee Break (TCPL Foyer)
Philipp Strasberg: Classicality, Markovianity and local detailed balance from pure state dynamics ↓
Across a wide range of time and length scales, processes appear classical, Markovian and obey local detailed balance. This behaviour is easily explained by assuming that the hidden or
10:30 irrelevant degrees of freedom rethermalize on a short time scale ("Born approximation", "repeated randomness assumption", "quantum regression theorem", etc.). Unfortunately, these assumptions
- are in blatant contradiction to the microscopic reversibility of the underlying quantum dynamics. After recalling the problem, I report on recent progress demonstrating the effective validity
11:30 of such "repeated maximum entropy reasoning" for coarse and slow observables of isolated many-body systems. Importantly, this progress is based on unitarily evolving pure states and invokes the
eigenstate thermalization hypothesis and typicality argument. It is thus fully compatible with the microscopic description. I also emphasize the essential importance to overcome the idea of
ensemble averages for a satisfactory explanation of classicality and (non-)Markovianity, a problem which is frequently overlooked by using conventional models of open quantum systems theory.
(TCPL 201)
Lunch ↓
- Lunch is served daily between 11:30am and 1:30pm in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Guided Tour of The Banff Centre ↓
- Meet in the PDC front desk for a guided tour of The Banff Centre campus.
(PDC Front Desk)
Group Photo ↓
- Meet in foyer of TCPL to participate in the BIRS group photo. The photograph will be taken outdoors, so dress appropriately for the weather. Please don't be late, or you might not be in the
14:20 official group photo!
(TCPL Foyer)
Erik Gauger: Modelling broad classes of non-Markovian open quantum systems with Process Tensors ↓
14:30 The generally non-Markovian influence of a (strongly) coupled physical environment on the evolution of a quantum system can be formally captured with an object known as a Process Tensor (PT).
- Numerical evaluation of the ensuing dynamics then typically requires compression of this object, which is achievable when expressing the PTs in matrix product operator from (a PT-MPO). In this
15:00 talk, I will discuss a method for constructing such PT-MPOs for most types of environment which are not themselves highly correlated. Specifically, this includes non-Gaussian environments,
combining influences from different types of environment, and dealing with more complex forms of interaction between system and environment.
(TCPL 201)
- Coffee Break (TCPL Foyer)
Andrea Smirne: Non-classicality in non-Markovian multi-time quantum processes ↓
More than a century after the birth of quantum theory, the question of which properties and phenomena are fundamentally quantum – i.e., they cannot be reproduced by any classical theory –
remains under active investigation. In this talk, we will see when and to what extent non-classicality can be unambiguously linked to specific features of the evolution of an open quantum
system and its interaction with the environment, focusing on the difference between the Markovian and the non-Markovian scenarios. We will consider an open system that is undergoing sequential
measurements of one observable at different times, and exploit the Kolmogorov consistency conditions to discriminate the resulting multi-time statistics from the statistics of any classical
15:30 process, in the same spirit as the Leggett-Garg inequalities [1]. In the Markovian case, the multi-time statistics cannot be accounted for by means of any classical process if and only if the
- dynamics generates coherences (with respect to the measured observable) and subsequently turns them into populations [2]. On the other hand, such a direct connection between the dynamics of
16:00 quantum coherences and non-classicality cannot be extended to general non-Markovian processes, where, instead, non-classicality is related to a global property of the system-environment
evolution [3] that is fully captured by higher-order quantum maps, i.e., quantum combs [4]. The approach presented here is fully operational, since it relies on the observed multi-time
probability distributions, and it thus directly applies to detect and quantify non-classicality in a variety of experimental platforms [5]. References [1]A. J. Leggett and A. Garg, Phys. Rev.
Lett. 54, 857 (1985) [2]A. Smirne, D. Egloff, M. G. Diaz, M. B. Plenio, and S. F. Huelga, Quantum Sci. Technol. 4, 01LT01 (2018) [3]S. Milz, D. Egloff, P. Taranto, T. Theurer, M. B. Plenio, A.
Smirne, and S. F. Huelga, Phys. Rev. X 10, 041049 (2020) [4]G. Chiribella, G. M. D’Ariano, and P. Perinotti, Phys. Rev. Lett. 101, 060401 (2008). [5]A. Smirne, T. Nitsche, D. Egloff, S.
Barkhofen, S. De, I. Dhand, C. Silberhorn, S. F. Huelga, and M. B. Plenio, Quantum Sci. Technol. 5, 04LT01 (2020)
(TCPL 201)
Nicholas Antosztrikacs: Quantum thermodynamics at strong coupling: A unified reaction coordinate polaron transform approach. ↓
At the nanoscale, strong system-reservoir interactions are ubiquitous and could potentially play a significant role in the development of novel nanoscale quantum machines. As a result, a
formulation of thermodynamics, which is to be valid in the quantum regime, must incorporate the effects of strong system reservoir couplings. The reaction coordinate (RC) mapping tackles the
strong coupling regime by reshaping the system-environment boundary to include a collective degree of freedom from the environment. This process results in an enlarged system, which in turn, is
16:30 weakly coupled to its surroundings, thus allowing the use of weak-coupling tools for simulations. Nevertheless, this approach is limited due to the growing Hilbert space of the extended system,
- and it does not offer analytical insights onto the strong coupling regime. I will present our efforts to push beyond these limitations and develop a general, transparent, and efficient theory
17:00 for strong coupling thermodynamics. By combing the RC mapping with the polaron transformation, followed by a judicious truncation of the Hamiltonian, we relocated strong coupling effects from
the system-bath boundary into the energy parameters of the system, ending with a computationally tractable expression for an “effective" Hamiltonian. We exemplified the power of this approach
on canonical models for quantum thermalization, quantum heat transport, phonon- assisted charge transport, and energy conversion devices. We showed that the effective Hamiltonian method is
numerically accurate and that it gathers analytical insights into strong coupling effects within a broad window of applicability.
(TCPL 201)
Marlon Brenes: Particle current statistics in driven mesoscale conductors ↓
We propose a highly-scalable method to compute the statistics of charge transfer in driven conductors. The framework can be applied in situations of non-zero temperature, strong coupling to
17:00 terminals and in the presence of non-periodic light-matter interactions, away from equilibrium. The approach combines the so-called mesoscopic leads formalism with full counting statistics. It
- results in a generalised quantum master equation that dictates the dynamics of current fluctuations and higher order moments of the probability distribution function of charge exchange. For
17:30 generic time-dependent quadratic Hamiltonians, we provide closed-form expressions for computing noise in the non-perturbative regime of the parameters of the system, reservoir or
system-reservoir interactions. Having access to the full dynamics of the current and its noise, the method allows us to compute the variance of charge transfer over time in non-equilibrium
configurations. The dynamics reveals that in driven systems, the average noise should be defined operationally with care over which period of time is covered.
(TCPL 201)
Dinner ↓
- A buffet dinner is served daily between 5:30pm and 7:30pm in Vistas Dining Room, top floor of the Sally Borden Building.
(Vistas Dining Room)
Tuesday, February 14
Breakfast ↓
- Breakfast is served daily between 7 and 9am in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Nicole Yunger Halpern: An informal introduction to the best little quasiprobability you've never heard of ↓
Abstract: The Kirkwood-Dirac (KD) quasiprobability sounds monstrously obscure, but it's recently proven useful across quantum thermodynamics, chaos, foundations, and metrology.
Quasiprobabilities are quantum generalizations of probabilities and can represent quantum states. You've probably heard of one quasiprobability distribution: the Wigner function. The KD
quasiprobability is richer and more flexible. I'll introduce the KD quasiprobability in a manner whose informality I hope you've gleaned from this abstract. I hope to convince you that the KD
09:00 distribution is the best little quasiprobability you've never heard of. Select serious references: [1] N. Yunger Halpern, B. Swingle, and J. Dressel, “Quasiprobability behind the
- out-of-time-ordered correlator,” Phys. Rev. A 97, 042105 (2018). [2] D. Arvidsson-Shukur, J. Chevalier-Drori, and N. Yunger Halpern, “Conditions tighter than noncommutation needed for
10:00 nonclassicality,” J. Phys. A 54, 284001 (2021). Suggested bedtime reading: [1] https://quantumfrontiers.com/2016/12/11/the-weak-shall-inherit-the-quasiprobability/ [2] https://
quantumfrontiers.com/2017/04/23/glass-beads-and-weak-measurement-schemes/ [3] https://quantumfrontiers.com/2020/08/30/if-the-quantum-metrology-key-fits/ [4] https://quantumfrontiers.com/2017/02
/19/its-chaos/ [5] https://quantumfrontiers.com/2019/03/24/a-theorist-i-can-actually-talk-with/ [6] https://quantumfrontiers.com/2019/08/25/quantum-conflict-resolution/ [7] https://
(TCPL 201)
- Coffee Break (TCPL Foyer)
Ángel Rivas: Quantum non-Markovianity via divisibility conditions ↓
- In this talk I will review the characterization of quantum non-Markovianity via divisibility conditions, explaining the main motivations and features of it. In addition, I will comment on the
11:30 similarities and differences of this approach to quantum non-Markovianity with others, and present some recent results on non-invertible quantum evolutions.
(TCPL 201)
Lunch ↓
- Lunch is served daily between 11:30am and 1:30pm in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Dominique Spehner: Bures geodesics as non-Markovian quantum evolutions in open quantum systems ↓
13:30 It is shown that the geodesics on the manifold of invertible density matrices equipped with the Bures distance correspond to physical evolutions of the quantum system coupled to an ancilla. The
- explicit forms of the geodesics and of the corresponding system-ancilla coupling Hamiltonian are obtained. The non-Markovian character of the geodesic evolutions is studied quantitatively using
14:00 a measure of non-Markovianity introduced in the literature. We briefly outline some potential applications of these geodesics in quantum metrology and quantum control.
(TCPL 201)
Stefano Marcantoni: Irreversibility mitigation under non-Markovian thermalizing dynamics ↓
We investigate the behavior of the stochastic entropy production in open two-level quantum systems thermalizing after a non-Markovian transient. We show the existence of time intervals where
14:00 both the average entropy production and the variance decrease. This happens when the quantum evolution fails to be P-divisible, i.e. when it is a so-called "essentially non-Markovian dynamics".
- For a simple model, we provide analytical bounds on the parameters of the dynamics that ensure the mentioned phenomenology. From a physical point of view, although the dynamics of the system is
14:30 overall irreversible, our result may be interpreted as a transient tendency towards reversibility, described by zero stochastic entropy production. (Joint work with S. Gherardini and E.
Fiorelli, arXiv:2210.07866)
(TCPL 201)
Gerardo Paz Silva: Predicting and controlling non-Markovian quantum dynamics ↓
One of the key tasks in the development of quantum technologies is predicting and eventually controlling the behaviour of a general open quantum system for long times. Often, if not always,
14:30 predicting the behaviour of system and bath is impossible, and so one is restricted to studying the reduced dynamics of the system. This represents a loss of information and is the main
- difficulty in long time analysis. To mitigate this problem, we introduce a technique which keeps tracks not only of the evolution of the system but also of (measurable) bath-related quantities
15:00 which influence the dynamics of the quantum system of interest. This allows us to predict the behaviour of the system for longer times, as compared to existing tools with the same seed
information, e.g., with the same perturbative order. Finally, we show how our technique allows us to (in principle) exactly track the evolution of high order correlations of the dephasing spin
boson model even when the state is non-Gaussian.
(TCPL 201)
- Coffee Break (TCPL Foyer)
Gniewomir Sarbicki: Optimising entanglement witnesses ↓
15:30 One detects entanglement of a bipartite quantum state shared between distant laboratories measuring in a Bell scenario a non-local observable called entanglement witnesses. We will discuss how
- to optimise an entanglement witness, i.e. how to improve an existing setting, to detect more entangled states without losing states already detected. We will show the families of entanglement
16:00 witnesses equivalent to non-linear entanglement criteria and discuss their optimality.
(TCPL 201)
Alain Joye: The Adiabatic Wigner-Weisskopf Model ↓
16:00 We consider a slowly varying time dependent d-level atom interacting with a photon field. Restricted to the single excitation atom-field sector, the model is a time- dependent generalization of
- the Wigner-Weisskopf model describing spontaneous emission of an atomic excitation into the radiation field. We analyze the dynamics of the atom and of the radiation field in the adiabatic and
16:30 small coupling approximations, in various regimes. In particular, starting with an excited atomic state, we provide a description of both the radiative decay of the atom and of the buildup of
the photon excitation in the field, and we discuss some properties of the effective evolution of the atom. This is joint work with Marco Merkli.
(TCPL 201)
Dinner ↓
- A buffet dinner is served daily between 5:30pm and 7:30pm in Vistas Dining Room, top floor of the Sally Borden Building.
(Vistas Dining Room)
Wednesday, February 15
Breakfast ↓
- Breakfast is served daily between 7 and 9am in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
- Marco Merkli: Wednesday (TCPL 201)
- Coffee Break (TCPL Foyer)
Massimo Palma: Quantum reservoir computing and memory effects ↓
In the past few years we have witnessed a growing interest in computational paradigms beyond the gate paradigm. Among these, Extreme Learning Machines and Reservoir Computers are two
particularly interesting new computational paradigms. Their key feature is the use of a fixed, nonlinear dynamics to efficiently extract information from a given dataset. Such a goal, in the
10:30 classical scenario, is achieved by processing the data as input of some fixed nonlinear dynamics of a suitable (neural) network – the reservoir - which enlarges the dimensionality of the data,
- making it easier to extract the properties of interest. The difference between Extreme Learning Machines and Reservoir Computers is whether the reservoir being used can deploy an internal
11:00 memory. More precisely, Reservoir Computers hold memory of the inputs seen at previous iterations, a feature which plays a crucial role when processing time sequences. Extreme Learning Machines
on the other hand use memoryless reservoirs. Although this makes the training of ELMs easier, it also makes them unsuitable for temporal data processing. We will review some recent results on
the quantum counterpart of the above.
(TCPL 201)
François Damanet: Non-Markovian effects and methods for many-body systems ↓
Describing the open system dynamics of many-body systems is in general extremely challenging due to their large sizes and the potential non-Markovian effects coming from the system-bath
interactions. Non-Markovianity emerges for strong coupling with a structured bath but also when one derives a reduced description of a larger Markovian system - an operation particularly
11:00 desirable for many-body systems as it makes it possible to significantly shrink the size of the Hilbert space. In this talk, I will present a number of recently developed theoretical methods
- [based on the hybridization of non-Markovian stochastic methods, Hierarchical Equations of Motion (HEOM) and Matrix Product States (MPS) techniques] and how they can be used to capture
11:30 non-Markovianity in the context of dissipative phase transitions [1], system dynamics conditioned on measurement [2], and 1D dynamics in strongly correlated systems [3]. [1] F. Damanet, et al.,
PRA 99, 033845 (2019). R. Palacino and J. Keeling, PRR 3, 032016 (2021). [2] Link et al., PRX Quantum 3, 020348 (2022). [3] S. Flannigan, et al., PRL 128, 063601 (2022). M. Moroder et al.,
(TCPL 201)
Lunch ↓
- Lunch is served daily between 11:30am and 1:30pm in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
- Free Afternoon (Banff National Park)
Dinner ↓
- A buffet dinner is served daily between 5:30pm and 7:30pm in Vistas Dining Room, top floor of the Sally Borden Building.
(Vistas Dining Room)
Thursday, February 16
Breakfast ↓
- Breakfast is served daily between 7 and 9am in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Kade Head Marsden: Gate-based quantum computation for quantum master equations ↓
Open quantum systems are ubiquitous in nature, but challenging to model due to the complexity of environmental interactions. While the idea of using quantum computing platforms to simulate
09:00 these dynamical processes in either a digital or analog fashion has been around for decades, it wasn’t until recently that the hardware has become sufficiently accessible to verify these
- methods. Here, I will give an overview of recent developments in the field of quantum computation and algorithms for modelling the dynamics of open quantum systems, with an emphasis on digital
10:00 simulation, or gate-based, techniques. This will include a brief survey of the field, an overview of dilation techniques, a discussion of a few open system algorithms and their generalization
to non-Markovian dynamics, and their potential applications in chemistry, physics, and materials science.
(TCPL 201)
- Coffee Break (TCPL Foyer)
Bassano Vacchini: Jensen-Shannon divergence versus trace distance for the description of information exchange in open quantum systems ↓
A well-known approach for the description of memory effects in the reduced quantum dynamics of an open system is based on the notion of information exchange between the open system and its
10:30 environment. This exchange has typically been quantified studying the variation in time of the trace distance between distinct initial system states. We point to the fact that such an
- information exchange can actually be described by a large class of quantum divergences, including not only distances, but also entropic quantifiers. We derive general upper bounds on the
11:00 revivals of quantum divergences conditioned and determined by the formation of correlations and changes in the environment. We will discuss in particular the different relationship between
distinguishability and divisibility for the trace distance and the Jensen-Shannon divergence.
Gregory White: Capturing the many-time physics of non-Markovian quantum stochastic processes ↓
The paradigm of open quantum systems gives rise to a temporal structure, as seen in quantum stochastic processes. System-environment dynamics can precipitate non-Markovian processes, which
generate quantum correlations between different times. Formally speaking, these correlations can be placed on the same footing as correlations in a many-body state. This invites the question:
11:00 to what extent can temporal quantum correlations be as interesting as spatial ones, and how can we access them? In this talk, I will discuss recent work in which we show how to fully
- characterise non-Markovian processes in practice. We develop the fully-general formalism of non-Markovian quantum process tomography, as well as extensions to make learning both efficient and
11:30 self-consistent. By applying this, we show how to determine process features — such as temporal entanglement — even with limited control. Remarkably, we find that many of these complex
properties are already present in naturally occurring noise on near-term computers. Hence, the characterisation and optimal control of such processes have direct application not only to the
general study of non-Markovianity, but also to the development of fault-tolerant quantum devices.
(TCPL 201)
Lunch ↓
- Lunch is served daily between 11:30am and 1:30pm in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Gabriela Schlau-Cohen: Controlling excitons using synthetic DNA scaffolds ↓
Control over excitons and their dynamics enables energy to be directed and harnessed for light harvesting and molecular electronics, but is challenging in condensed phase systems owing to the
13:00 large number of degrees of freedom. Here, we introduce a DNA-based platform that spatially organizes chromophores with nanoscale precision to construct tunable excitonic systems. We
- characterize these constructs with 2D electronic spectroscopy and single-molecule spectroscopy and show that this platform enables independent control over the nature and magnitude of the
14:00 coupling among the chromophores and between the chromophores and the environment. Using this platform, we demonstrate that a more flexible environment enhances the efficiency of energy
transport and specific chromophore geometries activate symmetry-breaking charge transfer. These studies highlight the key role of the environment in driving exciton dynamics.
Thomas Fay: Electron and energy transfer dynamics in light harvesting complexes: a hybrid hierarchical equations of motion approach ↓
In this talk I will describe a method for simulating exciton dynamics in protein–pigment complexes, including effects from charge transfer as well as fluorescence. The method combines the
14:00 hierarchical equations of motion, which are used to describe quantum dynamics of excitons, and the Nakajima–Zwanzig quantum master equation, which is used to describe slower charge transfer
- processes. We have studied the charge transfer quenching in light harvesting complex II, a protein postulated to control non-photochemical quenching in many plant species. Our calculations
14:30 reveal that the exciton energy funnel plays an important role in determining quenching efficiency, a conclusion we expect to extend to other proteins that perform protective excitation
(TCPL 201)
Avikar Periwal: Engineering entanglement between atomic ensembles with photons ↓
Interactions are the fundamental tool for generating entanglement between quantum degrees of freedom. All-to-all interactions between neutral atoms have been used to generate entanglement in a
single spatial mode, with applications in quantum-enhanced sensing. However, many envisioned protocols in quantum sensing and simulation require greater control over the spatial structure of
14:30 entanglement. In our experiment, we couple an array of four atomic ensembles to a single mode of light inside an optical cavity, realizing an all-to-all connected graph of interactions between
- the atoms. By combining these global interactions with local operations we gain control over the spatial structure of entanglement. We demonstrate this capability by tuning the entanglement
15:00 between two subsystems from unentangled to exhibiting an especially strong form of entanglement, known as Einstein-Podolsky-Rosen steering. By extending the control over the quantum
correlations to all four ensembles we engineer a square graph state, an essential resource for quantum computation. These capabilities set the stage for generating entangled states tailored to
specific tasks, such as quantum-enhanced sensing of spatially varying fields and quantum computation.
(TCPL 201)
- Coffee Break (TCPL Foyer)
Christoph Simon: Could quantum entanglement play a role in the brain? ↓
Could quantum physics help answer some of the big open questions in neuroscience? Could nature have discovered quantum information processing before we did? Motivated by these questions, I
discuss two potential ways in which quantum effects might be important in the brain. The first direction concerns biophotons, which could serve as classical and quantum information carriers. We
15:30 have shown that axons could serve as natural waveguides for these photons, and there is recent experimental evidence for this idea. The second direction concerns radical pairs, i.e. pairs of
- entangled electron spins that, together with nearby nuclear spins, might serve as quantum memories and processors. We have shown that radical pair models can explain otherwise puzzling
16:00 experimental observations (magnetic field effects and isotope effects) related to anesthesia, bipolar disorder, the circadian clock, microtubules, and neurogenesis, while also proposing new
experimental tests. While these results are still far from establishing the existence of functioning quantum networks, they suggest that key components that would be required for such networks
might indeed be available in the brain.
Dinner ↓
- A buffet dinner is served daily between 5:30pm and 7:30pm in Vistas Dining Room, top floor of the Sally Borden Building.
(Vistas Dining Room)
Friday, February 17
Breakfast ↓
- Breakfast is served daily between 7 and 9am in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Anton Trushechkin: Long-time behaviour and asymptotic Markovianity of exactly solvable models of open quantum dynamics ↓
We consider two exactly solvable models of open quantum dynamics: model of pure decoherence and a spin-boson model, and study the long-time behaviour without restrictions on the system-bath
09:00 coupling constant. We show that, under certain conditions on the spectral density function, the dynamics becomes asymptotically Markovian on long times. From the other side, a popular way of
- addressing non-Markovian dynamics of an open quantum system is to try to embed the system into an extended system whose dynamics is Markovian. However, if the conditions of asymptotic
09:30 Markovianity in these exactly solvable models are not satisfied, then the relaxation to a steady state can be non-exponential and, thus, such non-Markovian dynamics cannot be embedded into an
extended Markovian one.
Sergei Filippov: Tensor networks to describe non-Markovianity in open quantum systems with repeated interactions ↓
Repeated-interaction models are receiving increasing attention as they describe many nontrivial phenomena in the dynamics of open quantum systems [1,2]. In a general scenario of both
fundamental and practical interest, a quantum system repeatedly interacts with individual particles or modes, forming a correlated and structured reservoir; however, classical and quantum
environment correlations greatly complicate the calculation and interpretation of the system dynamics. We propose an exact solution to this problem based on the tensor network formalism [3]. We
find a natural Markovian embedding for the system dynamics, where the role of an auxiliary system is played by virtual indices of the network. The constructed embedding is amenable to an
09:30 analytical treatment for a number of timely problems such as the system interaction with two-photon wave packets, structured photonic states, and one-dimensional spin chains. We also derive a
- time-convolution master equation and relate its memory kernel with the environment correlation function, thus revealing a clear physical picture of memory effects in the dynamics. The results
10:00 advance tensor-network methods in the fields of quantum optics and quantum transport. Higher-order stroboscopic limits for the collisional dynamics and a transition from non-Markovian to
Markovian regime (even if the environment is correlated) is discussed too [4]. [1] F. Ciccarello, S. Lorenzo, V. Giovannetti, and G. M. Palma, Quantum collision models: Open system dynamics
from repeated interactions, Phys. Rep. 954, 1 (2022). [2] S. Campbell and B. Vacchini, Collision models in open system dynamics: A versatile tool for deeper insights? Europhys. Lett. 133, 60001
(2021). [3] S. N. Filippov, I. A. Luchnikov. Collisional open quantum dynamics with a generally correlated environment: Exact solvability in tensor networks. Phys. Rev. A 105, 062410 (2022).
[4] S. N. Filippov. Multipartite correlations in quantum collision models. Entropy 24, 508 (2022).
- Coffee Break (TCPL Foyer)
Checkout by 11AM ↓
- 5-day workshop participants are welcome to use BIRS facilities (TCPL ) until 3 pm on Friday, although participants are still required to checkout of the guest rooms by 11AM.
(Front Desk - Professional Development Centre)
- Lunch from 11:30 to 13:30 (Vistas Dining Room) | {"url":"https://www.birs.ca/events/2023/5-day-workshops/23w5083/schedule","timestamp":"2024-11-09T03:52:17Z","content_type":"application/xhtml+xml","content_length":"59583","record_id":"<urn:uuid:996eeace-4c48-419a-a658-b2aa090d944b>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00354.warc.gz"} |
Whenever I visit someone’s YouTube or Twitter profile page, I hope to see an interesting banner image. Here’s the one from Richard Borcherds’ YouTube Channel.
Not too surprisingly for Borcherds, almost all of these numbers are related to the monster group or its moonshine.
Let’s try to decode them, in no particular order.
John McKay’s observation $196884 = 1 + 196883$ was the start of the whole ‘monstrous moonshine’ industry. Here, $1$ and $196883$ are the dimensions of the two smallest irreducible representations of
the monster simple group, and $196884$ is the first non-trivial coefficient in Klein’s j-function in number theory.
$196884$ is also the dimension of the space in which Robert Griess constructed the Monster, following Simon Norton’s lead that there should be an algebra structure on the monster-representation of
that dimension. This algebra is now known as the Griess algebra.
Here’s a recent talk by Griess “My life and times with the sporadic simple groups” in which he tells about his construction of the monster (relevant part starting at 1:15:53 into the movie).
1729 is the second (and most famous) taxicab number. A long time ago I did write a post about the classic Ramanujan-Hardy story the taxicab curve (note to self: try to tidy up the layout of some old
Recently, connections between Ramanujan’s observation and K3-surfaces were discovered. Emory University has an enticing press release about this: Mathematicians find ‘magic key’ to drive Ramanujan’s
taxi-cab number. The paper itself is here.
“We’ve found that Ramanujan actually discovered a K3 surface more than 30 years before others started studying K3 surfaces and they were even named. It turns out that Ramanujan’s work anticipated
deep structures that have become fundamental objects in arithmetic geometry, number theory and physics.”
There’s no other number like $24$ responsible for the existence of sporadic simple groups.
24 is the length of the binary Golay code, with isomorphism group the sporadic Mathieu group $M_24$ and hence all of the other Mathieu-groups as subgroups.
24 is the dimension of the Leech lattice, with isomorphism group the Conway group $Co_0 = .0$ (dotto), giving us modulo its center the sporadic group $Co_1=.1$ and the other Conway groups $Co_2=.2,
Co_3=.3$, and all other sporadics of the second generation in the happy family as subquotients (McL,HS,Suz and $HJ=J_2$)
24 is the central charge of the Monster vertex algebra constructed by Frenkel, Lepowski and Meurman. Most experts believe that the Monster’s reason of existence is that it is the symmetry group of
this vertex algebra. John Conway was one among few others hoping for a nicer explanation, as he said in this interview with Alex Ryba.
24 is also an important number in monstrous moonshine, see for example the post the defining property of 24. There’s a lot more to say on this, but I’ll save it for another day.
60 is, of course, the order of the smallest non-Abelian simple group, $A_5$, the rotation symmetry group of the icosahedron. $A_5$ is the symmetry group of choice for most viruses but not the
3264 is the correct solution to Steiner’s conic problem asking for the number of conics in $\mathbb{P}^2_{\mathbb{C}}$ tangent to five given conics in general position.
Steiner himself claimed that there were $7776=6^5$ such conics, but realised later that he was wrong. The correct number was first given by Ernest de Jonquières in 1859, but a rigorous proof had to
await the advent of modern intersection theory.
Eisenbud and Harris wrote a book on intersection theory in algebraic geometry, freely available online: 3264 and all that.
248 is the dimension of the exceptional simple Lie group $E_8$. $E_8$ is also connected to the monster group.
If you take two Fischer involutions in the monster (elements of conjugacy class 2A) and multiply them, the resulting element surprisingly belongs to one of just 9 conjugacy classes:
1A,2A,2B,3A,3C,4A,4B,5A or 6A
The orders of these elements are exactly the dimensions of the fundamental root for the extended $E_8$ Dynkin diagram.
This is yet another moonshine observation by John McKay and I wrote a couple of posts about it and about Duncan’s solution: the monster graph and McKay’s observation, and $E_8$ from moonshine groups.
163 is a remarkable number because of the ‘modular miracle’
e^{\pi \sqrt{163}} = 262537412640768743.99999999999925… \]
This is somewhat related to moonshine, or at least to Klein’s j-function, which by a result of Kronecker’s detects the classnumber of imaginary quadratic fields $\mathbb{Q}(\sqrt{-D})$ and produces
integers if the classnumber is one (as is the case for $\mathbb{Q}(\sqrt{-163})$).
The details are in the post the miracle of 163, or in the paper by John Stillwell, Modular Miracles, The American Mathematical Monthly, 108 (2001) 70-76.
Richard Borcherds, the math-vlogger, has an entertaining video about this story: MegaFavNumbers 262537412680768000
His description of the $j$-function (at 4:13 in the movie) is simply hilarious!
Borcherds connects $163$ to the monster moonshine via the $j$-function, but there’s another one.
The monster group has $194$ conjugacy classes and monstrous moonshine assigns a ‘moonshine function’ to each conjugacy class (the $j$-function is assigned to the identity element). However, these
$194$ functions are not linearly independent and the space spanned by them has dimension exactly $163$.
One Comment
Yesterday, there was an interesting post by John Baez at the n-category cafe: The Riemann Hypothesis Says 5040 is the Last.
The 5040 in the title refers to the largest known counterexample to a bound for the sum-of-divisors function
\sigma(n) = \sum_{d | n} d = n \sum_{d | n} \frac{1}{d} \]
In 1983, the french mathematician Guy Robin proved that the Riemann hypothesis is equivalent to
\frac{\sigma(n)}{n~log(log(n))} < e^{\gamma} = 1.78107... \] when $n > 5040$.
The other known counterexamples to this bound are the numbers 3,4,5,6,8,9,10,12,16,18,20,24,30,36,48,60,72,84,120,180,240,360,720,840,2520.
In Baez’ post there is a nice graph of this function made by Nicolas Tessore, with 5040 indicated with a grey line towards the right and the other counterexamples jumping over the bound 1.78107…
Robin’s theorem has a remarkable history, starting in 1915 with good old Ramanujan writing a part of this thesis on “highly composite numbers” (numbers divisible by high powers of primes).
His PhD. adviser Hardy liked his result but called them “in the backwaters of mathematics” and most of it was not published at the time of Ramanujan’s degree ceremony in 1916, due to paper shortage
in WW1.
When Ramanujan’s paper “Highly Composite Numbers” was first published in 1988 in ‘The lost notebook and other unpublished papers’ it became clear that Ramanujan had already part of Robin’s theorem.
Ramanujan states that if the Riemann hypothesis is true, then for $n_0$ large enough we must have for all $n > n_0$ that
\frac{\sigma(n)}{n~log(log(n))} < e^{\gamma} = 1.78107... \] When Jean-Louis Nicolas, Robin's PhD. adviser, read Ramanujan's lost notes he noticed that there was a sign error in Ramanujan's formula
which prevented him from seeing Robin's theorem.
Nicolas: “Soon after discovering the hidden part, I read it and saw the difference between Ramanujan’s result and Robin’s one. Of course, I would have bet that the error was in Robin’s paper, but
after recalculating it several times and asking Robin to check, it turned out that there was an error of sign in what Ramanujan had written.”
If you are interested in the full story, read the paper by Jean-Louis Nicolas and Jonathan Sondow: Ramanujan, Robin, Highly Composite Numbers, and the Riemann Hypothesis.
What’s the latest on Robin’s inequality? An arXiv-search for Robin’s inequality shows a flurry of activity.
For starters, it has been verified for all numbers smaller that $10^{10^{13}}$…
It has been verified, unconditionally, for certain classes of numbers:
• all odd integers $> 9$
• all numbers not divisible by a 25-th power of a prime
Rings a bell? Here’s another hint:
According to Xiaolong Wu in A better method than t-free for Robin’s hypothesis one can replace the condition of ‘not divisible by an N-th power of a prime’ by ‘not divisible by an N-th power of 2’.
Further, he claims to have an (as yet unpublished) argument that Robin’s inequality holds for all numbers not divisible by $2^{42}$. So, where should we look for counterexamples to the Riemann
hypothesis? What about the orders of huge simple groups? The order of the Monster group is too small to be a counterexample (yet, it is divisible by $2^{46}$). 2 Comments
(After-math of last week’s second year lecture on elliptic
We all know the story of Ramanujan and the taxicab, immortalized by Hardy
“I remember once going to see him when he was lying ill at Putney. I had ridden in taxicab no. 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an
unfavorable omen. ‘No,’ he replied, ‘it’s a very interesting number; it is the smallest number expressible as a sum of two cubes in two different ways’.”
When I was ten, I wanted to become an archeologist and even today I can get pretty worked-up about historical facts. So, when I was re-telling this story last week I just had to find out things like
the type of taxicab and how numbers were displayed on them and, related to this, exactly when and where did this happen, etc. etc. Half an hour free-surfing further I know a bit more than I wanted.
Let’s start with the date of this taxicab-ride, even the year changes from source to source, from 1917 in the dullness of 1729 (arguing that Hardy could never have made this claim as 1729 is among
other things the third Carmichael Number, i.e., a pseudoprime relative to EVERY base) to ‘late in WW-1’ here…
Between 1917 and his return to India on march 13th 1919, Ramanujan was in and out a number of hospitals and nursing homes. Here’s an attempt to summarize these dates&places (based on the excellent
paper Ramanujan’s Illness by D.A.B. Young).
(may 1917 -september 20th 1917) : Nursing Hostel, Thompson’s Lane in Cambridge.
(first 2 a 3 weeks of october 1917) : Mendip Hills Senatorium, near Wells in Somerset. (november 1917) : Matlock House Senatorium atMatlock in Derbyshire.
(june 1918 – november 1918) : Fitzroy House, a hospital in Fitzroy square in central London. (december 1918 – march 1919) : Colinette House, a private nursing home in Putney, south-west London. So,
“he was lying ill at Putney” must have meant that Ramanujan was at Colinette House which was located 2, Colinette Road and a quick look with Google Earth
shows that the The British Society for the History of Mathematics Gazetteer is correct in asserting that “The house is no longer used as a nursing home and its name has vanished” as well as.”
“It was in 1919 (possibly January), when Hardy made the famous visit in the taxicab numbered 1729.”
Hence, we are looking for a London-cab early 1919. Fortunately, the London Vintage Taxi Association has a website including a taxi history page.
“At the outbreak of the First World War there was just one make available to buy, the Unic. The First World War devastated the taxi trade.
Production of the Unic ceased for the duration as the company turned to producing munitions. The majority of younger cabmen were called up to fight and those that remained had to drive worn-out cabs.
By 1918 these remnant vehicles were sold at highly inflated prices, often beyond the pockets of the returning servicemen, and the trade deteriorated.”
As the first post-war taxicab type was introduced in 1919 (which became known as the ‘Rolls-Royce of cabs’) more than likely the taxicab Hardy took was a Unic,
and the number 1729 was not a taxicab-number but part of its license plate. I still dont know whether there actually was a 1729-taxicab around at the time, but let us return to mathematics.
Clearly, my purpose to re-tell the story in class was to illustrate the use of addition on an elliptic curve as a mean to construct more rational solutions to the equation $x^3+y^3 = 1729 $ starting
from the Ramanujan-points (the two solutions he was referring to) : P=(1,12) and Q=(9,10). Because the symmetry between x and y, the (real part of) curve looks like
and if we take 0 to be the point at infinity corresponding to the asymptotic line, the negative of a point is just reflexion along the main diagonal. The geometric picture of addition of points on
the curve is then summarized
and sure enough we found the points $P+Q=(\frac{453}{26},-\frac{397}{26})$ and $(\frac{2472830}{187953},-\frac{1538423}{187953}) $ and so on by hand, but afterwards I had the nagging feeling that a
lot more could have been said about this example. Oh, if Im allowed another historical side remark :
I learned of this example from the excellent book by Alf Van der Poorten Notes on Fermat’s last theorem page 56-57.
Alf acknowledges that he borrowed this material from a lecture by Frits Beukers ‘Oefeningen rond Fermat’ at the National Fermat Day in Utrecht, November 6th 1993.
Perhaps a more accurate reference might be the paper Taxicabs and sums of two cubes by Joseph Silverman which appeared in the april 1993 issue of The American Mathematical Monthly.
The above drawings and some material to follow is taken from that paper (which I didnt know last week). I could have proved that the Ramanujan points (and their reflexions) are the ONLY integer
points on $x^3+y^3=1729 $.
In fact, Silverman gives a nice argument that there can only be finitely many integer points on any curve $x^3+y^3=A $ with $A \in \mathbb{Z} $ using the decomposition $x^3+y^3=(x+y)(x^2-xy+y^2) $.
So, take any factorization A=B.C and let $B=x+y $ and $C=x^2-xy+y^2 $, then substituting $y=B-x $ in the second one obtains that x must be an integer solution to the equation $3x^2-3Bx+(B^2-C)=0 $.
Hence, any of the finite number of factorizations of A gives at most two x-values (each giving one y-value). Checking this for A=1729=7.13.19 one observes that the only possibilities giving a square
discriminant of the quadratic equation are those where $B=13, C=133 $ and $B=19, C=91 $ leading exactly to the Ramanujan points and their reflexions!
Sure, I mentioned in class the Mordell-Weil theorem stating that the group of rational solutions of an elliptic curve is always finitely generated, but wouldnt it be fun to determine the actual group
in this example?
Surely, someone must have worked this out. Indeed, I did find a posting to sci.math.numberthy by Robert L. Ward : (in fact, there is a nice page on elliptic curves made from clippings to this
The Mordell-Weil group of the taxicab-curve is isomorphic to $\mathbb{Z} \oplus \mathbb{Z} $ and the only difference with Robert Wards posting was that I found besides his generator
$P=(273,409) $ (corresponding to the Ramanujan point (9,10)) as a second generator the point
$Q=(1729,71753) $ (note again the appearance of 1729…) corresponding to the rational solution $( -\frac{37}{3},\frac{46}{3}) $ on the taxicab-curve.
Clearly, there are several sets of generators (in fact that’s what $GL_2(\mathbb{Z}) $ is all about) and as our first generators were the same all I needed to see was that the point corresponding to
the second Ramanujan point (399,6583) was of the form $\pm Q + a P $ for some integer a. Points and their addition is also easy to do with sage :
sage: P=T([273,409])
sage: Q=T([1729,71753])
sage: -P-Q
(399 : 6583 : 1)
and we see that the second Ramanujan point is indeed of the required form!
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XQuery Optimizations
This article presents some of the optimizations that speed up the execution and reduce memory consumption of queries.
Query execution encompasses multiple steps:
1. Parsing: The query input string is transformed to executable code. The result is a tree representation, called the abstract syntax tree (AST).
2. Compilation: The syntax tree is decorated with additional information (type information, expression properties). Expressions (nodes) in the tree are relocated, simplified, or pre-evaluated.
Logical optimizations are performed that do not rely on external information.
3. Optimization: The dynamic context is incorporated: Referenced databases are opened and analyzed; queries are rewritten to use available indexes; accumulative and statistical operations (counts,
summations, min/max, distinct values) are pre-evaluated; XPath expressions are simplified, based on the existence of steps.
4. Evaluation: The resulting code is executed.
5. Printing: The query result is serialized and presented in a format that is either human-readable, or can be further processed by an API.
Some rewritings are described in this article.
If you run a query on command-line, you can use -V to output detailed query information. In the Graphical User Interface, you can enable the Info View panel.
Parts of the query that are static and would be executed multiple times can already be evaluated at compile time:
for $i in 1 to 10
return 2 * 3
(: rewritten to :)
for $i in 1 to 10
return 6
The value of a variable can be inlined: The variables references are replaced by the expression that is bound to the variable. The resulting expression can often be simplified, and further
optimizations can be triggered:
declare variable $INFO := true();
let $nodes := //nodes
where $INFO
return 'Results: ' || count($nodes)
(: rewritten to :)
let $nodes := //nodes
where true()
return 'Results: ' || count($nodes)
(: rewritten to :)
let $nodes := //nodes
return 'Results: ' || count($nodes)
(: rewritten to :)
'Results: ' || count(//nodes)
As the example shows, variable declarations might be located in the query prolog and in FLWOR expressions. They may also occur (and be inlined) in try/catch, switch or typeswitch expressions.
Functions can be inlined as well. The parameters are rewitten to let clauses and the function is body is bound to the return clause.
declare function local:inc($i) { $i + 1 };
for $n in 1 to 5
return local:inc($n)
(: rewritten to :)
for $n in 1 to 5
return (
let $_ := $n
return $_ + 1
(: rewritten to :)
for $n in 1 to 5
return $n + 1
Subsequent rewritings might result in query plans that differ a lot from the original query. As this might complicate debugging, you can disable function inling during development by setting
INLINELIMIT to 0.
Loops with few iterations are unrolled by the XQuery compiler to enable further optimizations:
(1 to 2) ! (. * 2)
(: rewritten to :)
1 ! (. * 2), 2 ! (. * 2)
(: further rewritten to :)
1 * 2, 2 * 2
(: further rewritten to :)
2, 4
Folds are unrolled, too:
let $f := fn($a, $b) { $a * $b }
return fold-left(2 to 5, 1, $f)
(: rewritten to :)
let $f := fn($a, $b) { $a * $b }
return $f($f($f($f(1, 2), 3), 4), 5)
The standard unroll limit is 5. It can be adjusted with the UNROLLLIMIT option, e.g. via a pragma:
(# db:unrolllimit 10 #) {
for $i in 1 to 10
return db:get('db' || $i)//*[text() = 'abc']
(: rewritten to :)
db:get('db1')//*[text() = 'abc'],
db:get('db2')//*[text() = 'abc'],
db:get('db10')//*[text() = 'abc'],
The last example indicates that index rewritings might be triggered by unrolling loops with paths on database nodes.
The following expressions can be unrolled:
Care should be taken if a higher value is selected, as memory consumption and compile time will increase.
Due to the compact syntax of XPath, it can make a big difference if a slash is added or omitted in a path expression. A classical example is the double slash //, which is a shortcut for
descendant-or-node()/. If the query is evaluated without optimizations, all nodes of a document are gathered, and for each of them, the next step is evaluated. This leads to a potentially huge number
of duplicate node tree traversals, most of which are redundant, as all duplicate nodes will be removed at the end anyway.
In most cases, paths with a double slash can be rewritten to descendant steps…
(: equivalent queries, with identical syntax trees :)
(: rewritten to :)
…unless the last step does not contain a positional predicate:
As the positional test refers to the city child step, a rewritten query would yield different steps.
Paths may contain predicates that will be evaluated again by a later axis step. Such predicates are either shifted down or discarded:
(: equivalent query :)
(: rewritten to :)
Names of nodes can be specified via name tests or predicates. If names are e.g. supplied via external variables, the predicates can often be dissolved:
declare variable $name external := 'city';
db:get('addressbook')/descendant::*[name() = $name]
(: rewritten to :)
FLWOR expressions are central to XQuery and the most complex constructs the language offers. Numerous optimizations have been realized to improve the execution time:
• Nested FLWOR expressions are flattened.
• for clauses with single items are rewritten to let clauses.
• let clauses that are iterated multiple times are lifted up.
• Expressions of let clauses are inlined.
• Unused variables are removed.
• where clauses are rewritten to predicates.
• if expressions in the return clause are rewritten to where clauses.
• The last for clause is merged into the return clause and rewritten to a Simple Map Operator.
Various of these rewriting are demonstrated in the following example:
for $a in 1 to 10
for $b in 2
where $a > 3
let $c := $a + $b
return $c
(: for is rewritten to let :)
for $a in 1 to 10
let $b := 2
where $a > 3
let $c := $a + $b
return $c
(: let is lifted up :)
let $b := 2
for $a in 1 to 10
where $a > 3
let $c := $a + $b
return $c
(: the where expression is rewritten to a predicate :)
let $b := 2
for $a in (1 to 10)[. > 3]
let $c := $a + $b
return $c
(: $b is inlined :)
for $a in (1 to 10)[. > 3]
let $c := $a + 2
return $c
(: $c is inlined :)
for $a in (1 to 10)[. > 3]
return $a + 2
(: the remaining clauses are merged and rewritten to a simple map :)
(1 to 10)[. > 3] ! (. + 2)
If the type of a value is known at compile time, type checks can be removed. In the example below, the static information that $i will always reference items of type xs:integer can be utilized to
simplify the expression:
for $i in 1 to 5
return typeswitch($i)
case xs:numeric return 'number'
default return 'string'
(: rewritten to :)
for $i in 1 to 5
return 'number'
If expressions can often be simplified:
for $a in ('a', '')
return $a[boolean(if (.) then true() else false())]
(: rewritten to :)
for $a in ('a', '')
return $a[boolean(.)]
(: rewritten to :)
for $a in ('a', '')
return $a[.]
(: rewritten to :)
('a', '')[.]
Boolean algebra (and set theory) comes with a set of laws that can all be applied to XQuery expressions.
Expression Rewritten expression Rule
$a + 0, $a * 1 $a Identity
$a * 0 0 Annihilator
$a and $a $a Idempotence
$a and ($a or $b) $a Absorption
($a and $b) or ($a and $c) $a and ($b or $c) Distributivity
$a or not($a) true() Tertium non datur
not($a) and not($b) not($a or $b) De Morgan
It is not sufficient to apply the rules to arbitrary input. Examples:
• If the operands are no boolean values, a conversion is enforced: $string and $string is rewritten to boolean($string).
• xs:double('NaN') * 0 yields NaN instead of 0
• true#0 and true#0 must raise an error; it cannot be simplified to true#0
Some physical optimizations are also presented in the article on index structures.
In each database, metadata is stored that can be utilized by the query optimizer to speed up or even skip query evaluation:
Count element nodes
The number of elements that are found for a specific path need not be evaluated sequentially. Instead, the count can directly be retrieved from the database statistics:
(: rewritten to :)
Return distinct values
The distinct values for specific names and paths can also be fetched from the database metadata, provided that the number does not exceed the maximum number of distinct values (see MAXCATS for more
(: rewritten to :)
('Muslim', 'Roman Catholic', 'Albanian Orthodox', ...)
A major feature of BaseX is the ability to rewrite all kinds of query patterns for index access.
The following queries are all equivalent. They will be rewritten to exactly the same query that will eventually access the text index of a factbook.xml database instance (the file included in our
full distributions):
declare context item := db:get('factbook');
declare variable $DB := 'factbook';
//name[. = 'Shenzhen'],
//name[data() = 'Shenzhen'],
//name[./text() = 'Shenzhen'],
//name[text()[. = 'Shenzhen']],
//name[string() = 'Shenzhen'],
//name[string() = 'Shen' || 'zhen'],
//name[./data(text()/string()) = 'Shenzhen'],
//name[text() ! data() ! string() = 'Shenzhen'],
//name[. eq 'Shenzhen'],
//name[not(. ne 'Shenzhen')],
//name[not(. != 'Shenzhen')],
.//name[. = 'Shenzhen'],
//*[local-name() = 'name'][data() = 'Shenzhen'],
db:get('factbook')//name[. = 'Shenzhen'],
db:get($DB)//name[. = 'Shenzhen'],
for $name in //name[text() = 'Shenzhen']
return $name,
for $name in //name
return $name[text() = 'Shenzhen'],
for $name in //name
return if ($name/text() = 'Shenzhen') then $name else (),
for $name in //name
where $name/text() = 'Shenzhen'
return $name,
for $name in //name
where $name/text()[. = 'Shenzhen']
return $name,
for $node in //*
where data($node) = 'Shenzhen'
where name($node) = 'name'
return $node,
(: rewritten to :)
db:text('factbook', 'Shenzhen')/parent::name
Multiple element names and query strings can be supplied in a path:
//*[(ethnicgroups, religions)/text() = ('Jewish', 'Muslim')]
(: rewritten to :)
db:text('factbook', ('Jewish', 'Muslim'))/
(parent::*:ethnicgroups | parent::*:religions)/
If multiple candidates for index access are found, the database statistics (if available) are consulted to choose the cheapest candidate:
[religions = 'Muslim'] (: yields 77 results :)
[ethnicgroups = 'Greeks'] (: yields 2 results :)
(: rewritten to :)
db:text('factbook', 'Greeks')/parent::ethnicgroups/parent::country[religions = 'Muslim']
If index access is possible within more complex FLWOR expressions, only the paths will be rewritten:
for $country in //country
where $country/ethnicgroups = 'German'
order by $country/name[1]
return element { replace($country/@name, ' ', '') } {},
(: rewritten to :)
for $country in db:text('factbook', 'German')/parent::ethnicgroups/parent::country
order by $country/name[1]
return element { replace($country/@name, ' ', '') } {}
The XMark XML Benchmark comes with sample auction data and a bunch of queries, some of which are suitable for index rewritings:
XMark Query 1
let $auction := doc('xmark')
return for $b in $auction/site/people/person[@id = 'person0']
return $b/name/text()
(: rewritten to :)
db:attribute('xmark', 'person0')/self::attribute(id)/parent::person/name/text()
XMark Query 8
let $auction := doc('xmark')
for $p in $auction/site/people/person
let $a :=
for $t in $auction/site/closed_auctions/closed_auction
where $t/buyer/@person = $p/@id
return $t
return <item person="{ $p/name/text() }">{ count($a) }</item>,
(: rewritten to :)
db:get('xmark')/site/people/person !
<item person='{ name/text() }'>{ count(
db:attribute('xmark', @id)/self::attribute(person)/parent::buyer/parent::closed_auction
If the accessed database is not known at compile time, or if you want to give a predicate preference to another one, you can enforce index rewritings.
In many cases, the amount of data to be processed is only known after the query has been compiled. Moreover, the data that is looped through expressions may change. In those cases, the best
optimizations needs to be chosen at runtime.
If sequences of items are compared against each other, a dynamic hash index will be generated, and the total number of comparisons can be significantly reduced. In the following example, count
($input1) * count($input2) comparisons would need to be made without the intermediate index structure:
let $input1 := file:read-text-lines('huge1.txt')
let $input2 := file:read-text-lines('huge2.txt')
return $input1[not(. = $input2)]
Version 9.6Version 9.4
• Added: This article was introduced with Version 9.4.
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Junk Science; Junk Analysis!
I don’t think I have offended anyone in quite a while and feel I’m not doing my job if I don’t try to periodically; so here goes!
Often a simple mathematical series of numbers can sometimes get misinterpreted (promoted) to be something magical. My personal favorite sequence is 6, 28, 496, 2520, 8128, and 24601. I’ll explain
them at the end of this article. Personally, I see no value in the actual numbers that make up the Fibonacci series; a series developed by an Italian mathematician (Fibonacci) in the thirteenth
century to help understand the propagation of rabbits. First, I must say that I do value the ratio of the numbers that are expanded in a Fibonacci-like series (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,
89, …). That ratio is 0.618 (and its reciprocal is 1.618), often called the golden ratio because of its wide occurrence in nature; almost always with a jaundiced eye. Here is a fact: the actual
numbers in the
Fibonacci series have little to do with the ratio. Any two numbers expanded in the same manner will produce the same ‘golden’ ratio. Here is a test: Try it with 2 and 19. Add them together, and then
add the total to the previous number just like in the Fibonacci series (2+19=21, 19+21=40, 21+40=61, etc.). Expand this until you get to four-digit numbers so that the accuracy will be acceptable (2,
19, 21, 40, 61, 101, 162, 263, 425, 688, 1113, 1801, 2914, 4716, …). The last two numbers in this sequence are the two numbers that I will use for this example: 2914 and 4716. Now divide the first
number by the second number and you will get 0.618. This is exactly the same as with the value obtained using the Fibonacci series of numbers. So why did I pick 2 and 19 for this example? Hint: the
second letter in the alphabet is B. What is the nineteenth letter? S. BS! And that is what numerology is all about. Table A shows the sequence using various starting numbers including my special
selection of 2 and 19. The last column begins with a negative number.
Table A
I can find no source that explains why the series of Fibonacci numbers begins at zero. If I were tasked with mathematically identifying the propagation of rabbits, I think I would at least have to
begin the series at 2. The fact of the matter is that the series can begin anywhere, even negative numbers, as long as the expansion follows the correct formula. It is the ratio that is important,
not the actual numbers in the series. So, when you hear someone say they are going to use a 34-day moving average because 34 is a Fibonacci number, you can immediately begin to doubt the rest of
their analysis. Just so you know; the Fibonacci expansion of one plus the square root of five divided by 2 will work with any two numbers, even negative numbers. Sorry, no magic here, just
numerology. As far as Elliott Wave theory goes, there are often so many complications and conditions introduced into using this type of analysis, that it is incapable of being proved wrong. Sometimes
I think it gets adjusted more often than earnings estimates. However, it is always convincing to align the workings of the market with what appears to be pure mathematics. In the series of numbers
introduced at the beginning of this section, 6, 28, 496, and 8128, are known as perfect numbers; this means the sum of their divisors (other than the number itself) is also equal to the number. For
example: 6 = 1 + 2 + 3, and 28 = 1 + 2 + 4 + 7 + 14. I like 2520 because it is the smallest integer than is divisible by all integers from 1 to 10 inclusive. Finally, I like 24601 as it is the
prisoner number of Jean Valjean from Victor Hugo’s Les Miserables. Incidentally, 24601 has prime factors of 73 and 337. I like these numbers solely for their mathematical uniqueness; and like many
number sequences, they have no use in technical market analysis. Possibly Keno! So why are people drawn to it so much? Probably because of its false potential for predicting the future and its
cottage industry of numerology advocates.
I have known many successful traders and investors over the years. I do not know anyone who trades with real money that uses junk science like Fibonacci numbers. I put out articles like this because
I see so many technical analysts that do not understand the tools they use; they accept them on face value. Hopefully, you can now see the problems with doing it that way. So, were you offended by
or appreciative of this article?
Dance with the Trend,
Greg Morris
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Some gamblers might he tempted to base staking plans on the theory that in any series of even-money events there must come a time sooner or later when the outcomes reach equipartition.
A glance at Pascal’s Triangle will answer the following question. When tossing a coin, what is the probability that after n events, the number of heads will equal the number of tails? Clearly this
can happen only with an even number of events. Those who believe the ‘law of averages’ fallacy maintain that the probability of equipartition, as it is called, increases with the number of events.
Pascal’s Triangle proves the opposite. Line 4 shows that if we toss four times, there are 16 Possible outcomes, of which six contain two heads and two tails. The Probability of equipartition is 6/16.
Line 6 shows that with six tosses, the probability of equipartition is 20/64. Line 10 shows that with ten tosses, equipartition is a 252/1024 chance. The probabilities are becoming smaller as the
tosses increase. The formula to discover the probability of equipartition in n events is to divide the number of combinations which give equipartition by the number of possible outcomes.
Example :-
What is the probability, when tossing six dice, of throwing each number once i.e. achieving equipartition?
The total number of ways equipartition can occur is 6! The first die can clearly be any of the six numbers. the second die one of the five remaining, and so on, giving a total number of ways of 6 x 5
x 4 x 3 x 2 x 1 = 720.
The number of possible outcomes is power(6,6) since there are six ways each of the six dice can fall. Therefore this is equal to 46656.
So the probability of equipartition is:-
720/46656 = 0.0154 or 1.5432%
Most people would be surprised to discover that if you threw six dice, on about 98.5% of occasions, at least one number will appear more than once.
If there are seven children in a family, what is the probability that they were born on different days of the week?
This is a question of equipartition, the answer being =fact(7)/power(7,7). The answer is0.6120%. | {"url":"https://www.probabilitytheory.info/equipartition/","timestamp":"2024-11-11T11:25:25Z","content_type":"text/html","content_length":"29524","record_id":"<urn:uuid:aa63fae1-cfeb-480f-9a5f-cae157f4bd7c>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00231.warc.gz"} |
pushforward of vector fields
nLab pushforward of vector fields
Differential geometry
synthetic differential geometry
from point-set topology to differentiable manifolds
geometry of physics: coordinate systems, smooth spaces, manifolds, smooth homotopy types, supergeometry
The magic algebraic facts
$\array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \
rightsquigarrow &\dashv& \mathrm{R}\!\!\mathrm{h} & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel
{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& \esh &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel
{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }$
differential equations, variational calculus
Chern-Weil theory, ∞-Chern-Weil theory
Cartan geometry (super, higher)
Given a differentiable map $\phi \colon X_1 \xrightarrow{\;} X_2$ between differentiable manifolds (e.g. a smooth map between smooth manifolds) and thinking of vector fields as infinitesimal
approximations to differentiable curves
$T_x X \;\simeq\; \big\{ \gamma \in C^\infty\big(\mathbb{R},\, X\big) \,\big\vert\, \gamma(0) = x \big\} \Big/ \big( \gamma_1 \sim \gamma_2 \;\Leftrightarrow\; \mathrm{d}\gamma_1(0) = \mathrm{d}\
gamma_2(0) \big)$
then the postcomposition of these curves with $\phi$ induces maps of equivalence classes
$\begin{array}{ccc} T_x X_1 &\xrightarrow{\phantom{--}}& T_{\phi(x)} X_2 \\ [\gamma] &\mapsto& [\phi \circ \gamma] \end{array}$
alternatively denoted “$\phi_\ast$” or “$\mathrm{d}\phi$” (cf. differentiation as a functor) and called the pushforward of vector fields along $\phi$.
Most texts on differential geometry will discuss pushforward of vector fields.
See also
Created on June 21, 2024 at 09:45:41. See the history of this page for a list of all contributions to it. | {"url":"https://ncatlab.org/nlab/show/pushforward+of+vector+fields","timestamp":"2024-11-07T02:58:57Z","content_type":"application/xhtml+xml","content_length":"43001","record_id":"<urn:uuid:d428dbae-dd94-4f4f-afd2-bf93a712b5ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00119.warc.gz"} |
Probability on Trees and Networks - Yuval Peres books
Probability on Trees and Networks
This book explores the fascinating field of probability on trees and weighted graphs. Although some related topics have been treated by other recent books, none has explored so thoroughly the
relations between the geometry of graphs, the probabilistic objects defined on them, such as Markov chains and their electrical network interpretations, edge percolations, Galton-Watson branching
processes, random spanning trees, and the analytical notions of Hausdorff dimension and capacity. As noted in Math reviews by Laurent Miclo,
these links are presented in a conversational and enjoyable fashion in the first chapter, where the unifying concept of the branching number of a rooted tree is defined This quantity is bounded above
by the lower exponential growth rate of T and is directly related to (i) the transience properties of weighted random walks (ii) the critical percolation probability, (iii) the Hausdorff dimension of
the boundary ∂T and (iv) the existence of a finite energy probability measure on ∂T. Chapter 2 presents the correspondence between electric networks and reversible Markov chains, based on the
identifications of the conductance of an edge. It enables one to see voltages as Green functions and currents as expected edge crossings and leads to a criterion for transience based on positive
effective conductance to infinity. Notions from discrete potential theory are introduced, leading to Thomson’s Principle, characterizing current flows as minimizing flows for the energy, to
Rayleigh’s Monotonicity Principle, comparing effective conductances from edge conductance inequalities, and to the characterization of transience by the existence of a unit flow to infinity of finite
energy. In particular Pólya’s Theorem on the recurrence/transience dichotomy for the random walk on lattices is recovered. To deduce the transience of the hyperbolic spaces of dimension at least 2,
the general relations between rough embedding, rough isometry and transience are presented. Chapter 2 ends with the electrical interpretations of (i) the hitting and commuting times to get cover time
bounds and (ii) the canonical Gaussian field on the vertices, via a minimization property of its gradient.
Chapter 3 introduces the special cases of trees and Cayley graphs, but begins with the general max-flow min-cut theorem, identifying the maximum strength of an admissible flow with the minimal
capacity value of an edge cutset. The universal covering tree of a graph is presented and related to the notion of periodicity of a tree. (Sub, super)periodicity enables one to identify the branching
number of a tree with its growth rate. The basics of Cayley graphs are introduced, with the notions of free groups, representation of groups, and free and Cartesian products, as well as the useful
lamplighter group. Using a geodesic subtree, the critical value for the transience of a Cayley graph is identified with its exponential growth rate.
Chapter 4 begins the investigation of the wonderful weighted “uniform” spanning trees, whose distribution is proportional to the product of the conductances of the edges of the tree. To sample these
spanning trees, the powerful algorithm of Wilson is presented, based on the loop erasures of the associated Markov chains. Next the authors discuss the electrical interpretations via Kirchhoff’s
Effective Resistance Formula and the Transfer Current Theorem, giving a determinantal form for the probability that some edges belong to the spanning tree. The case of the square lattice Z2 serves as
an illustration. The notes give the Markov Chain Tree Theorem representation of the invariant probability and allude to the amazing relations with stochastic Loewner evolutions.
Chapter 5 begins by recalling the classical theory of Galton-Watson branching processes, with the Kesten-Stigum and Seneta-Heyde theorems. Next the first and second weighted moments methods, as well
as an electrical interpretation, are used to deduce bounds on the critical probability for Bernoulli percolation to infinity on graphs. On trees, it is identified with the inverse branching number.
Some extensions to quasi-independent percolation, as well as the transience of percolation clusters, are considered. The decomposition of supercritical Galton-Watson trees into surviving/
non-surviving parts is obtained via conditioning and percolation. The existence of d-ary subtrees and left-to-right crossing in fractal percolation is investigated in a similar spirit. Chapter 5 ends
with Harris’ inequality, i.e., the positive correlation of increasing observables for independent percolation, and the existence of flows in weighted Galton-Watson trees.
Chapter 6 is a combinatorics-oriented presentation of isoperimetric constants, comparing the sizes, measured in various ways, of the boundary of a subset with respect to its interior. Their relations
with the notions of dual flows, submodularity and amenability are investigated. Cheeger’s inequalities establish an important link with the spectral radius, which is used to deduce estimates on the
speed of random walks, on the cogrowth (i.e., the growth of the covering space obtained via non-backtracking paths) and on mixing rates in the finite setting. Regular planar graphs, their dual graphs
and hyperbolic tessellation edge graphs illustrate special behaviors of isoperimetry and lead to nice pictures. The discrete analogue of the original Euclidean isoperimetry is recalled through the
Loomis and Whitney inequality on projections and justifies a pleasant introduction to entropy and to the Shannon, Han and Shearer inequalities. Chapter 6 ends with the more specific (anchored)
isoperimetric profiles and their applications to decay of transition probabilities, to transience of Cayley graphs and to percolation, including a presentation of evolving sets, covering maps and
random subdivision of edges. As usual, the notes are rich and allude in particular to Buser’s inequality and to Ramanujan graphs.
Chapter 7 begins the investigation of percolation on transitive graphs, namely those looking the same from any of their vertices. Some important facts from the general theory of percolation are
presented: insertion/deletion tolerance, tail events and ergodicity, inequalities between bond and site percolations, contour arguments, dual percolation, numbers of infinite clusters, invasion and
bootstrap percolation. One of the main questions is the validity of pc<pu<1, where the critical probability pc (resp. pu) commands the phase transition to the existence (resp. the uniqueness) of an
infinite cluster in the Bernoulli percolation. Some bounds on pu are provided and significant conjectures are recalled, in the context of quasi-transitive non-amenable graphs. The notes go further
into the relations with groups, introducing, e.g., the Kazhdan property (T).
Chapter 8 introduces a mass-transport principle for unimodular graphs, which is a technique interchanging the order of summation of functions of two vertex variables invariant under the diagonal
action of the group of automorphisms. It is a powerful tool in the developed investigation of the critical percolation on non-amenable quasi-transitive unimodular graphs, of the double phase
transition of the Bernoulli percolation on planar quasi-transitive graphs and of the properties of ends in infinite clusters.
Chapter 9 comes back to current flows from one vertex to another, now in the context of infinite networks, in which there exist two definitions: free and wired currents, depending on the chosen
approximation by finite sets. One highlight of the chapter is that it shows that in a transient network, the subnetwork formed by the edges crossed by a random walk is a.s. recurrent.
In Chapter 10, the fascinating model of uniform spanning trees is extended to infinite transient graphs through uniform spanning forests, either free or wired.
Chapter 11 investigates other distributions on spanning forests, corresponding to the free or wired extensions of minimal spanning trees on finite graphs (minimizing the sum of independent and
uniformly distributed weights on the edges). Similarities and discrepancies with uniform spanning forests are put forward, as well as the links with Bernoulli percolation.
Chapter 12 provides the proofs of the limit theorems for Galton-Watson processes via size-biased transformations.
Chapter 13 begins the investigation of the speed of escape of transient random walks in metric spaces. After proving the fundamental Varopoulos-Carne inequality, the authors give an application to
lower bounds on mixing times of finite Markov chains. Distortions of embeddings of finite metric spaces into Hilbert spaces are also presented.
Chapter 14 introduces the Avez entropy h on Cayley graphs and shows that for simple random walks, h>0 is equivalent to the existence of a positive escape speed, to the nontriviality of the tail
σ-field and to the existence of bounded non-constant harmonic functions (failure of the Liouville property). This chapter also addresses the identification of the Poisson boundary via the method of
Kaimanovich and recalls the proofs of the Birkhoff and Kingman ergodic theorems.
Chapter 15 is concerned with Hausdorff dimension and the analysis of random fractals. The relationship of Hausdorff dimension with capacities in Euclidean spaces is covered in Chapter 16, with an
application to Brownian intersections. Another interest of Hausdorff dimension is given in Chapter 17, with the investigation of the harmonic measure associated to random walks on trees, especially
those randomly chosen according to a Galton-Watson measure.
The book has numerous exercises (with hints and solutions at the end of the book. It has been used in courses in many universities, including Cornell, Berkeley, Stanford and Tel-Aviv. | {"url":"https://www.yuval-peres-books.com/probability-on-trees-and-networks/","timestamp":"2024-11-08T05:43:40Z","content_type":"text/html","content_length":"58250","record_id":"<urn:uuid:6ad28201-eb28-4cf4-98b2-7679556b6515>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00145.warc.gz"} |
Mixing with Common Drain: Mass of Salt in Two Tanks #46 Nagle
• MHB
• Thread starter mathcoral
• Start date
In summary, Mixing salt with a common drain causes a change in the total mass of salt in the two tanks.
Mixing with a Common Drain. Two tanks, each holding 1 L of liquid, are connected by a pipe through which liquid flows from tank A into tank B at a rate of 3-a L/min (0<a<3). The liquid inside each
tank is kept well stirred. Pure water flows into tank A at a rate of 3 L/min. Solution flows out of tank A at a L/min and out of tank B at 3-a L/min. If, initially, tank B contains no salt (only
water) and tank A contains 0.1 kg of salt, determine the mass of salt in each tank at time T>=0. How does the mass of salt in tank A depend on the choice of a? What is the maximum mass of salt in
tank B?Here is picture of the tank
View attachment 8094
I was going to make a normal equation x'=Ax to find the eigenvectors.
I got stuck on the set up
=rate in - rate out
(t)=0 - (3-a)x
(t) - ax
(t)= -3x
The rate in is 0 because the liquid is pure water.
(t) =rate in - rate out
(t)= (3-a)x
(t) - (3-a)x
I am unsure how to put the 3 L/min coming out of from both tanks (bottom pipe).
I would begin with:
\(\displaystyle \d{x_1}{t}=-3x_1\) where \(\displaystyle x_1(0)=\frac{1}{10}\)
\(\displaystyle x_1(t)=\frac{e^{-3t}}{10}\)
\(\displaystyle \d{x_2}{t}=(3-\alpha)\left(\frac{e^{-3t}}{10}-x_2\right)\) where \(\displaystyle x_2(0)=0\)
I would put the ODE into standard linear form:
\(\displaystyle \d{x_2}{t}+(3-\alpha)x_2=(3-\alpha)\frac{e^{-3t}}{10}\)
Our integrating factor is:
\(\displaystyle \mu(t)=e^{(3-\alpha)t}\)
And we get:
\(\displaystyle e^{(3-\alpha)t}\d{x_2}{t}+(3-\alpha)e^{(3-\alpha)t}x_2=(3-\alpha)\frac{e^{-3t}}{10}e^{(3-\alpha)t}\)
\(\displaystyle \frac{d}{dt}\left(e^{(3-\alpha)t}x_2\right)=(3-\alpha)\frac{e^{-\alpha t}}{10}\)
Can you proceed?
After integrating I got
x[2](t)=\(\displaystyle \frac{3-a}{10a}\)(e^at-1)e^-3t
The mass of salt in tank A does not depend on the choice of a.
Tank B depends on a.
I am unsure how to find the maximum mass for tank B.
mathcoral said:
After integrating I got
x[2](t)=\(\displaystyle \frac{3-a}{10a}\)(e^at-1)e^-3t
The mass of salt in tank A does not depend on the choice of a.
Tank B depends on a.
I am unsure how to find the maximum mass for tank B.
I got the equivalent:
\(\displaystyle x_2(t)=\frac{\alpha-3}{10\alpha}e^{-3t}\left(1-e^{\alpha t}\right)\)
Now, for the optimization of this function, recall we had:
\(\displaystyle \d{x_2}{t}=(3-\alpha)\left(\frac{e^{-3t}}{10}-x_2\right)\)
Note: We could also differentiate the function we found, but I think this will be less work. ;)
We can equate this derivative to zero to find the turning point, and substitute for $x_2$:
\(\displaystyle (3-\alpha)\left(\frac{e^{-3t}}{10}-\frac{\alpha-3}{10\alpha}e^{-3t}\left(1-e^{\alpha t}\right)\right)=0\)
Since we are given $0<\alpha<3$, we may divide through by \(\displaystyle \frac{3-\alpha}{10\alpha}\) to obtain:
\(\displaystyle \alpha e^{-3t}-(\alpha-3)e^{-3t}\left(1-e^{\alpha t}\right)=0\)
Solve this for $t=t_{\max}$ and then evaluate $x_2\left(t_{\max}\right)$.
Setting the derivative to 0 I got t=\(\displaystyle \frac{ln(\frac{3}{3-a})}{a}\)
Plugging in for t in x[2](t) I got
x[2](t)=.1(\(\displaystyle \frac{3-a}{3}\))^^3/a kg
mathcoral said:
Setting the derivative to 0 I got t=\(\displaystyle \frac{ln(\frac{3}{3-a})}{a}\)
Plugging in for t in x[2](t) I got
x[2](t)=.1(\(\displaystyle \frac{3-a}{3}\))^^3/a kg
Yes, I get equivalent results. (Yes)
Hi mathcoral, welcome to MHB!
It seem to me that we still don't have the actual maximum mass of salt in tank B.
For that we need to maximize for $\alpha$ as well.
We can make it ourselves a little easier, since to achieve that maximum all liquid from tank A should go through tank B, shouldn't it?
So $\alpha = 0$.
How much salt would that make at its maximum in tank B? (Wondering)
FAQ: Mixing with Common Drain: Mass of Salt in Two Tanks #46 Nagle
1. How does the mass of salt in two tanks affect the mixing process in a common drain system?
The mass of salt in two tanks plays a crucial role in determining the rate and efficiency of mixing in a common drain system. The greater the mass of salt present in the tanks, the slower the mixing
process will be, as it takes longer for the salt to disperse evenly in the solution.
2. What factors can affect the mass of salt in two tanks in a common drain system?
The mass of salt in two tanks can be influenced by several factors, such as the initial concentration of salt in each tank, the flow rate of the solution, and the size and shape of the tanks.
Additionally, any changes in these factors during the mixing process can also impact the mass of salt present in the tanks.
3. Is the mass of salt in two tanks the only factor that affects the mixing process in a common drain system?
No, the mass of salt in two tanks is not the only factor that influences the mixing process. Other factors, such as the flow rate, tank size, and initial concentration, also play important roles in
determining the overall mixing efficiency.
4. How can the mass of salt in two tanks be controlled in a common drain system?
The mass of salt in two tanks can be controlled by adjusting the initial concentration of salt in each tank or by changing the flow rate of the solution. By maintaining a consistent flow rate and
carefully measuring the initial salt concentration, the mass of salt in two tanks can be controlled and optimized for efficient mixing.
5. What are some potential applications of studying mixing with common drain in relation to the mass of salt in two tanks?
Studying mixing with common drain and the mass of salt in two tanks has many practical applications, such as in industrial processes where mixing is necessary to achieve a homogeneous solution. This
can include applications in chemical and food processing, water treatment, and pharmaceutical manufacturing, among others. | {"url":"https://www.physicsforums.com/threads/mixing-with-common-drain-mass-of-salt-in-two-tanks-46-nagle.1039629/","timestamp":"2024-11-09T23:26:50Z","content_type":"text/html","content_length":"107777","record_id":"<urn:uuid:000f08d5-b7d7-4fa8-8042-0bd75427dc34>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00109.warc.gz"} |
Automated theorem proving and manufacturing
Automated theorem proving
When I first heard of automated theorem proving, I imagined computers being programmed to search for mathematical theorems interesting to a wide audience. Maybe that’s what a few of the pioneers in
the area had in mind too, but that’s not how things developed.
The biggest uses for automated theorem proving have been highly specialized applications, not mathematically interesting theorems. Computer chip manufacturers use formal methods to verify that given
certain inputs their chips produce certain outputs. Compiler writers use formal methods to verify that their software does the right thing. A theorem saying your product behaves correctly is very
valuable to you and your customers, but nobody else. These aren’t the kinds of theorems that anyone would cite the way they might site the Pythagorean theorem. Nobody would ever say “And therefore,
by the theorem showing that this particular pacemaker will not fall into certain error modes, I now prove this result unrelated to pacemakers.”
Automated theorem provers are important in these highly specialized applications in part because the results are of such limited interest. For every theorem of wide mathematical interest, there are a
large number of mathematicians who are searching for a proof or who are willing to scrutinize a proposed proof. A theorem saying that a piece of electronics performs correctly appeals to only the
tiniest audience, and yet is probably much easier (for a computer) to prove.
The term “automated theorem proving” is overloaded to mean a couple things. It’s used broadly to include any use of computing in proving theorems, and it’s used more narrowly to mean software that
searches for proofs or even new theorems. Most theorem provers in the broad sense are not automated theorem provers in the more narrow sense but rather proof assistants. They verify proofs rather
than discover them. (There’s some gray zone. They may search on a small scale, looking for a way to prove a minor narrow result, but not search for the entire proof to a big theorem.) There have been
computer-verified proofs of important mathematical theorems, such as the Feit-Thompson theorem from group theory, but I’m not aware of any generally interesting discoveries that have come out of a
theorem prover.
Related post: Formal methods let you explore the corners
4 thoughts on “Automated theorem proving”
1. I would argue that the same holds for numerics, and for most parts of applied mathematics.
On the other hand, there is a lot of pure mathematics done in the field of type theory (all the HoTT-stuff), which often uses proof assistants to validate results.
2. I just learned a new buzz-phrase: “Homotopy Type Theory” (HoTT)
Some math folks believe it can become a base for reformulating all of mathematics.
Apparently, the act of formulating a mathematical idea in HoTT always yields a proof of its correctness. That is, the proof always pops out when the input is correct.
Others say HoTT has much in common with programming.
3. I started out as a compiler writer. Automated theorem provers are very common not to prove the algorithms correct, but to prove preconditions correct.
You can think of a compiler optimisation as a theorem of the form: “If a fragment of code C has property P, then it is legal to transform C into D.”
Proving that theorem can be done by hand. However proving that a fragment of code has some desired property… that can only be done at run-time.
In a deep sense, modern type checkers (or even more so, type inference engines) are theorem provers. The theorem is “the program is type-correct”, or “function f has type T”.
One of the most interesting theorem provers is in the Java classloader. One job it has to do is to prove that the bytecode that needs to be loaded won’t violate any invariants of the virtual
machine. Section 4.10 of the JVM specification gives an implementation of the class verifier in Prolog.
4. > I just learned a new buzz-phrase: “Homotopy Type Theory” (HoTT)
It probably is fair to call it a “buzz-phrase”, though at least there is a reasonably precise idea to back it up. In short, HoTT is intensional type theory, plus higher inductive types and the
univalence principle. I say “reasonably precise” because each of these parts is still a subject of research. I don’t really have the time to explain each part, but hopefully if you ever see those
words around, you’ll have some idea how they fit into the big picture. Most of the rest of what I say will apply to just intensional type theory.
> Some math folks believe it can become a base for reformulating all of mathematics.
Yes, the HoTT Book describes how to formulate some areas of mathematics in HoTT. Most of the people working on HoTT are either pure mathematicians or computer scientists, and I’m more familiar
with the latter. I know that there’s someone doing something about HoTT for physics, though I can’t tell you much more.
> Apparently, the act of formulating a mathematical idea in HoTT always yields a proof of its correctness. That is, the proof always pops out when the input is correct.
No. The process of defining, supposing, and proving is almost exactly the same as in conventional mathematics. What you may be confused with is the fact that each disjunction proof proves either
one of the disjuncts, and each existential proof gives a witness. That is to say, if we have proven ∃ x. P x, then we really have some x at which P holds. These properties are characteristic of
*constructive* mathematics, which is much broader and has a longer history than intensional type theory. They don’t hold in classical mathematics, where, for example, if CH stands for the
continuum hypothesis in ZFC set theory, CH ∨ ¬ CH is easy to prove (as an instance of the law of the excluded middle), but neither CH or ¬ CH can be proven.
> Others say HoTT has much in common with programming.
This is true, and largely why there are computer scientists working on type theory. For a type theory to be fitting of the name, it should come with a programming language. “type”, in this
context, is the same sort of type you’d find in programming languages like Java and ML and plenty of others. Each term of these languages has a type, and we also define the semantics of how to
compute with terms so that the type is preserved. What dependent type theory (which encompasses intensional type theory, among others) has above these other languages is the ability to quantify
in types over terms.
HoTT being a (total) programming language has the effect that any proof can be evaluated into a normal form. For example, a universal quantification ∀ x : A. B x is proven by a computable
function that takes a proof x of A and produces a proof of B x. Then, applying a proof that, for all natural k, 2 * k is even, to the natural number 3 gives a proof that 6 is even. The
Curry-Howard correspondence is worth reading about. | {"url":"https://www.johndcook.com/blog/2017/01/11/automated-theorem-proving/","timestamp":"2024-11-13T10:57:16Z","content_type":"text/html","content_length":"60427","record_id":"<urn:uuid:6ecd168f-b8f0-4387-abe7-81babf5fdcd3>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00529.warc.gz"} |
What is the specific heat of hot water?
What is the specific heat of hot water? 4.186 J/g°C Water has a specific heat capacity of 4.186 J/g°C, meaning that it requires 4.186 J of energy (1 calorie) to heat a gram by one
What is the specific heat of hot water?
4.186 J/g°C
Water has a specific heat capacity of 4.186 J/g°C, meaning that it requires 4.186 J of energy (1 calorie) to heat a gram by one degree.
What is the specific heat of water in kJ kg C?
The specific heat capacity of water is 4,200 Joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.
Does anything have a higher specific heat than water?
p. 252, it is stated: Hitherto water has been regarded as possessing a greater specific heat than any other body excepting hydrogen. E. Lecker has shown to the Vienna Academy that mixtures of
methylic alcohol and water have a specific heat higher than that of water, and accordingly take the second place, &c.
What has a higher specific heat than water?
On a mass basis hydrogen gas has more than three times the specific heat as water under normal laboratory conditions.
How do you calculate specific heat of water?
Calculate specific heat as c = Q / (m * ΔT). In our example, it will be equal to c = -63000 J / (5 kg * -3 K) = 4200 J/(kg*K). This is the typical heat capacity of water. If you have problems with
the units, feel free to use our temperature conversion or weight conversion calculators.
What is the formula for specific heat of water?
Answer: The heat energy transferred to the water is 1676 kJ = 1 676 000 J. The specific heat can be found by rearranging the formula: c = 4190 J/kg∙K. The specific heat of water is 4190 J/kg∙K.
What are the uses of the specific heat of water?
Application of Specific Heat Capacity Car radiator. Water is pumped through the channels in the engine block to absorb heat. Cooking utensils. Cooking utensils are made of metal which has low
specific heat capacity so that it need less heat to raise up the temperature. Thermal Radiator. Thermal radiators are always used in cold country to warm the house. Sea Breeze. Land Breeze.
What is the formula for specific heat?
Learn the equation for specific heat. Once you become familiar with the terms used for calculating specific heat, you should learn the equation for finding the specific heat of a substance. The
formula is: C p = Q/mΔT. | {"url":"https://missionalcall.com/2019/04/02/what-is-the-specific-heat-of-hot-water/","timestamp":"2024-11-03T03:43:12Z","content_type":"text/html","content_length":"55847","record_id":"<urn:uuid:ea8da120-fb01-4013-ac99-6f389e719815>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00482.warc.gz"} |
Wolfgang Doeblin
Wolfgang Doeblin Prize
- Description
- About Wolfgang Doeblin
- The Prize Committee
- Eligibility for the Prize
- Prize Article
- Sponsorship of the Prize
- Prize Lecture
- Doeblin Prize 2024
The prize is to honor the scientific work of Wolfgang Doeblin and to recognize and promote outstanding work by researchers at the beginning of their mathematical careers in the field of Probability.
The Wolfgang Doeblin Prize was founded in 2011. It is awarded bi-annually to a single individual for outstanding research in the field of probability, and who is at the beginning of his or her
mathematical career.
The Wolfgang Doeblin Prize is generously supported by Springer. The awardee will be invited to submit to the journal Probability Theory and Related Fields a paper for publication as the Wolfgang
Doeblin Prize Article, and will also be invited to present the Doeblin Prize Lecture at a World Congress of the Bernoulli Society, or at a later Conference on Stochastic Processes and their
About Wolfgang Doeblin
Wolfgang Doeblin was born in Berlin in 1915. His family, of Jewish origin, were forced into exile and settled in Paris, where Doeblin attended the Sorbonne. From 1935, when he began work on Markov
chains under Fréchet, until his death in 1940, he was occupied whenever he was able with research in Probability. In this short time he made many deep and original contributions.
From 1938, he served in the French Army and was stationed in defense of the German invasion, which came in May 1940. He was awarded the Croix de Guerre for an action under enemy fire, to restore
communications to his unit. Facing capture in June 1940, he took his own life.
Until the invasion, Doeblin had continued to work on mathematics. In February 1940 he sent to the Académie des Sciences de Paris a pli cacheté entitled Sur l'équation do Kolmogoroff. When finally in
the year 2000 it was opened, it showed that he had understood many important ideas of modern Probability, including the potential crucial role of martingales.
The Prize Committee
The awarding of the Prize is determined by the Prize Committee. The Prize Committee members are the Chair of the Committee for Conferences on Stochastic Processes, the Managing Editor(s) of
Probability Theory and Related Fields, together with four further co-opted members drawn from the Committee for Conferences on Stochastic Processes or the Editorial Board of Probability Theory and
Related Fields. The co-opted members are appointed by the President of the Bernoulli Society on nomination by the Chair of the Committee for Conferences on Stochastic Processes, who will consult with
Managing Editor(s) of Probability Theory and Related Fields.
The term of each nominated member is two years. The Prize Committee is chaired by the Chair of the Committee for Conferences on Stochastic Processes.
Eligibility for the Prize
The Prize is awarded for work in the field of Probability and it is awarded to a single Individual with outstanding work. It is intended for researchers at the beginning of their mathematical career.
Nominees should normally be within 10 (calendar) years from getting their PhD to the prize year (for example, for the 2024 Doeblin Prize, this means anyone who got their PhD in or after 2014) with
suitable adjustments to be made for career breaks post-PhD (for example, maternity/paternity leave or military service).
Prize Article
The awardee of the Prize is invited to submit to Probability Theory and Related Fields a paper which, if accepted, is published as the Wolfgang Doeblin Prize Article.
Sponsorship of the Prize
The Bernoulli Society gratefully acknowledges sponsorship of the Prize by Springer, consisting of 2500 Euros.
Prize Lecture
The awardee of the Prize is invited to present a Doeblin Prize Lecture in the next World Congress of the Bernoulli Society or the next Conference on Stochastic Processes and their Applications,
whichever happens first. The Bernoulli Society will sponsor the participation of the speaker in the corresponding World Congress or SPA Conference.
Last Updated: Monday, 22 January 2024 06:29 | {"url":"https://bernoullisociety.org/index.php/prizes?id=158","timestamp":"2024-11-08T02:53:48Z","content_type":"text/html","content_length":"11127","record_id":"<urn:uuid:d3274edc-48d1-454a-91fa-cdb5646d374e>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00096.warc.gz"} |
What are the multiples of 125?
Multiples of 125: 125, 250, 375, 500, 625, 750, 875, 1000, 1125, 1250 and so on.
What do we get on multiplying a number by 1?
If you noticed, when you multiply by 1, you always get your original number. 38 * 1 is equal to 38, and 431 * 1 is equal to 431. From this, we can create a rule. This rule tells us that anything
multiplied by 1 is itself.
What number multiplied by itself 3 times equals 125?
In other words, a cube root calculator finds the value that, when multiplied by itself 3 times, gives the number you started with. For example, the cube root of 125 is 5, because 5 times 5 times 5
equals 125.
What are the 2 numbers you multiply are called?
The numbers to be multiplied are generally called the “factors”. The number to be multiplied is the “multiplicand”, and the number by which it is multiplied is the “multiplier”. The result of a
multiplication is called a product.
What is the cube of 125?
The cube root of 125 is the number which when multiplied by itself three times gives the product as 125. Since 125 can be expressed as 5 × 5 × 5. Therefore, the cube root of 125 = ∛(5 × 5 × 5) = 5.
What is the under root of 125?
The square root of 125 is 11.180.
Is 0 times 0 defined?
Division as the inverse of multiplication But any number multiplied by 0 is 0 and so there is no number that solves the equation. In general, a single value can’t be assigned to a fraction where the
denominator is 0 so the value remains undefined.
When we multiply a number by 0 we get?
Multiplication by Zero Multiplying by 0 makes the product equal zero. The product of any real number and 0 is 0 .
What is the cube root of 125 *?
The value of the cube root of 125 is 5.
What to the second power equals 125?
Powers and Exponents
base number 2nd power 3rd power
What is math multiplication?
Multiplication is the process of calculating the total of one number multiplied by another. There will be simple tests in addition, subtraction, multiplication and division. 2. uncountable noun. The
multiplication of things of a particular kind is the process or fact of them increasing in number or amount.
What is multiplier math?
The meaning of the word multiplier is a factor that amplifies or increases the base value of something else. For example, in the multiplication statement 3 × 4 = 12 the multiplier 3 amplifies the
value of 4 to 12. | {"url":"https://quick-advices.com/what-are-the-multiples-of-125/","timestamp":"2024-11-07T09:07:08Z","content_type":"text/html","content_length":"148542","record_id":"<urn:uuid:514c7f4e-b6f9-41e1-9c4f-eff9cdc5a37a>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00682.warc.gz"} |
How to Get the RMS in Excel - Technosfer
How to Get the RMS in Excel
The Root Mean Square calculates the effective rate or measurement of a varying set of values. It is the square root of the average of the squared values in a data set. RMS is primarily used in
physics and electrical engineering. One of the more common uses for an RMS calculation is comparing alternating current and direct current electricity.
For example, RMS is used to find the effective voltage (current) of an AC electrical wave. Because AC fluctuates, it’s difficult to compare it to DC, which has a steady voltage. The RMS provides a
positive average that can be used in the comparison.
Image Credit:
Ron Price
Unfortunately, Excel doesn’t include a standard function to calculate RMS. This means you’ll have use one or more functions to calculate an it.
Step 1: Enter the Data Set
Enter your data values so that the raw data (measurement, test value, etc.) is located in a single column or row. Allow space adjacent to the data values to place the results of other calculations.
Step 2: Calculate the Squares
Image Credit:
Ron Price
Calculate the square (x^2) for each of the values in your data set. Enter the formula =^2 adjacent to each data value. For example, “=D3^2” calculates the square of the contents of cell D3.
Step 3: Average the Squares
Calculate the average of the individual squares. Below the last entry in the column containing the squares of the data set values, enter the formula =AVERAGE(First Cell:Last Cell). For example, =
AVERAGE(D2:D30) calculates the mean (average) of the squares in the cells ranging from D2 to D30, inclusive.
Step 4: Calculate the Square Root of the Average
In an empty cell, enter the formula to calculate the square root of the average of the squares of the data. Enter the formula =SQRT(XN), where “XN” represents the location of the average calculated
in the previous step. For example, =SQRT (D31) calculates the square root of the value in cell D31.
The value calculated in this step represents the RMS of the values in the data set.
Calculate the RMS with One Excel Formula
It is possible to calculate the RMS in a single formula using the original data values. The sequence of the steps, those of Steps 1 through 3, are as follows: calculate the square of each value,
calculate the average of the squares and calculate the square root of the average.
The formula =SQRT((SUMSQ(First:Last)/COUNTA(First Cell:Last Cell))) uses the SUMSQ function to produce the sum of the squares of the values in the cell range. Then that number is divided by the
number of cells containing data in the cell range specified (COUNTA). Finally, the square root of this value is calculated, which is the RMS.
For example, the formula =SQRT((SUMSQ(C2:C30)/COUNTA(A2:A30))) calculates the sum of the squares in the range C2 through C30, divides that sum by the number of entries in the range A2 through A30
that are not blank and then finds the square root of that value to produce the RMS.
Keşfetmeye Devam Edin
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sqrt function not working properly
sqrt function not working properly
asked 2015-10-29 00:45:04 +0100
This post is a wiki. Anyone with karma >750 is welcome to improve it.
I have the following code where I want to substitute a, b, c into s. Since s factors as a square, I want to get the square root of it :
p, t= var('p t')
c=(p*t^2-p^2*t)+(t^2+2*t*p+p^2)+t-p #3 sides (a,b,c) in terms of theta and phi [equation (1.1)]
Unfortunately the answer I get is
sqrt((3*p*t^2 - 2*p^2 - 2*p*t + t^2 + p + 2*t + 1)^2)
The sqrt and the square power does not cancel off which I want it to cancel. I tried using the code S.simplify_full() to simplify it hoping the sqrt and square power will cancel off but no luck. Is
there any other specific code I can use for that.
1 Answer
Sort by ยป oldest newest most voted
The square root of a square is not the same as the number. (There is controversy over whether it is the absolute value of the number, or either plus or minus the number. Don't ask.)
But if you want to make it do that, use this:
sage: S.canonicalize_radical()
(3*p + 1)*t^2 - 2*p^2 - 2*(p - 1)*t + p + 1
which even does some factoring for you as a bonus.
edit flag offensive delete link more
@kcrisman,.. Thank you! this worked fine..
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Rated Contest 2 P3 - Ski Rentals
Submit solution
Points: 7 (partial)
Time limit: 2.0s
Memory limit: 256M
After meeting up with each other, it's time for Daniel and his friends to rent their skis! Each pair of skis has a specified price as some skis are better quality than others. Together, the group of
people (including Daniel) are already in line to rent, and no one wants to lose their place. However, after reading the sign on the window, Daniel realizes that there is a special deal—if you choose
to rent two pairs of skis together, the cheaper of the two will be at half price (50% off)! While positions in the line can no longer be changed, 2 people can either group together or stay separate.
Can you help Daniel and his friends find the minimum amount of money they need to pay to rent all the skis for the group?
For each , . All will be even to make calculations easier.
Subtask 1 [10%]
Subtask 2 [90%]
Input Specification
The first line will contain the integer , representing the number of people in the group.
The second line will contain space-separated integers , representing the cost of the skis for each person in the group.
Output Specification
Output one integer, representing the minimum cost to rent all the skis in the group.
Sample Input
Sample Output
Sample Explanation
There are three ways to rent the skis:
All people rent separately:
The first two people rent together:
The last two people rent together:
The cheapest choice is for the first two people to rent together, so the minimum cost is .
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27th Workshop on Future Research in Combinatorial Optimization
FRICO 2024 - 27th Workshop on Future Research in Combinatorial Optimization
September 9th - 13th, 2024, Magdeburg, Germany
27th Workshop on Future Research in Combinatorial Optimization will take place from September 9th to September 13th, 2024 at the Otto von Guericke University Magdeburg (OVGU), Germany.
FRICO is directed at PhD students from the following research areas:
• Discrete Mathematics and Combinatorics
• Approximation Algorithms
• (Mixed Integer) Linear and Non-Linear Optimization
• Applications of Combinatorial Optimization
• Scheduling
• Randomized Algorithms
• Online Optimization
Every participant will be assigned a slot; either for a 20-minute presentation (excluding 5 minutes for setup and questions), or a 5-minute elevator pitch talk. It is especially encouraged to present
ongoing research and open problems.
FRICO 2024 does not have a conference fee! This is made possible by our sponsors.
During the 'industry sessions' on Wednesday 11th and Thursday 12th the sponsors will inform us about practical application of combinatorial optimization in their companies.
We express special gratitude to Siemens AG, TNG Technology Consulting GmbH and Hannover Rück SE for being the main sponsors of FRICO2024.
To get a general impression of FRICO, visit the webpage of FRICO 2023, which took place in Eindhoven, or have a look at the history of FRICO. If you have any questions, please contact any of the
organizers or write an email to frico2024@ovgu.de.
Main organizers
External organizers
The team thanks Ines Brückner, Johannes Jesse, Susanne Hess and Volker Kaibel and other university staff for their support and assistance in organizing this workshop. We are grateful to Maryia
Kukharenka for her help with FRICO 2024 design. Additional thanks to our working students Jonas Danker and Frederic Horn for their assistance.
FRICO 2024 has concluded!
We would like to thank all attendees for their participation.
The best talk award of FRICO 2024 goes to Jamico Schade from TU Munich for his talk "Firefighters vs Burning Trees: A Pursuit-Evasion Variant". The newly awarded best elevator pitch goes to Helena
Petri from RPTU Kaiserslautern-Landau. Congratulations!
We are happy to announce that FRICO 2025 will take place at the RWTH Aachen. | {"url":"https://frico2024.ovgu.de/index.html","timestamp":"2024-11-14T15:28:57Z","content_type":"text/html","content_length":"93199","record_id":"<urn:uuid:6c3a3fb7-7b42-425c-a93e-b45cba750868>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00313.warc.gz"} |
Mathematics and Statistics
CRC’s Mathematics program offers a comprehensive mathematics curriculum addressing the needs of both transfer and non-transfer students. The study of mathematics provides students with the ability to
think logically and abstractly and to use problem-solving and computational skills necessary for success in any field of study.
View the CRC Math and Statistics Course Sequence and the Math and Statistics Placement webpage.
Program Maps
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Check Out Degree Planner
If you're interested in a transfer degree (AA-T or AS-T), then check out Degree Planner, a tool that helps you complete your degree efficiently by mapping out what courses to take and when to take
More About the Program
Learn about math workshops, Math Boot Camp, and more! | {"url":"https://crc.losrios.edu/academics/programs-and-majors/mathematics-and-statistics","timestamp":"2024-11-12T19:19:34Z","content_type":"text/html","content_length":"303370","record_id":"<urn:uuid:28dd28b0-5d6f-4c56-903a-3b0fdffe49aa>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00874.warc.gz"} |
WONDER #2301: What is the Fibonacci Sequence?
Question 1 of 3
How do you find the next number in the Fibonacci Sequence?
1. Just add five to the previous number
2. Multiple the two previous numbers
3. Subtract two from the previous number
4. Add the two previous numbers
Question 2 of 3
Where did modern mathematicians first learn about the Fibonacci Sequence?
1. Leonardo Fibonacci’s book, "Liber Abaci"
2. They noticed the pattern in nature
3. Ancient Indian mathematicians
4. The traced it back from the Golden Spiral
Question 3 of 3
Where has the Golden Spiral been observed in nature?
1. The shape of seashells
2. The pattern of seeds on a sunflower
3. The shape of galaxies
4. All of the above
Check your answers online at https://wonderopolis.org/wonder/What-is-the-Fibonacci-Sequence. | {"url":"https://wonderopolis.org/print_quiz/2301","timestamp":"2024-11-04T08:12:34Z","content_type":"text/html","content_length":"4302","record_id":"<urn:uuid:691a6858-1b07-448c-bea2-45857d2136ea>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00780.warc.gz"} |
Consider the set of nonzero real numbers, \({\mathbb R}^*\text{,}\) with the group operation of multiplication. The identity of this group is 1 and the inverse of any element \(a \in {\mathbb R}^*\)
is just \(1/a\text{.}\) We will show that
\begin{equation*} {\mathbb Q}^* = \{ p/q : p\text{ and }q\text{ are nonzero integers}\} \end{equation*}
is a subgroup of \({\mathbb R}^*\text{.}\) The identity of \({\mathbb R}^*\) is 1; however, \(1 = 1/1\) is the quotient of two nonzero integers. Hence, the identity of \({\mathbb R}^*\) is in \({\
mathbb Q}^*\text{.}\) Given two elements in \({\mathbb Q}^*\text{,}\) say \(p/q\) and \(r/s\text{,}\) their product \(pr/qs\) is also in \({\mathbb Q}^*\text{.}\) The inverse of any element \(p/q \in
{\mathbb Q}^*\) is again in \({\mathbb Q}^*\) since \((p/q)^{-1} = q/p\text{.}\) Since multiplication in \({\mathbb R}^*\) is associative, multiplication in \({\mathbb Q}^*\) is associative. | {"url":"https://pretextbook.org/examples/sample-book/noparts/html/section-subgroups.html","timestamp":"2024-11-10T21:47:49Z","content_type":"text/html","content_length":"71299","record_id":"<urn:uuid:00164e03-fe6b-4f37-9743-382d91606b3e>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00540.warc.gz"} |
Pendulum Period Calculator
Last updated:
Pendulum Period Calculator
If you have ever wondered why your grandfather's clock was so tall, you will find the answer on our pendulum period calculator! Keep reading our article to discover the basics of one of the favorite
gadgets of every physicist.
Here you will learn:
• What is a pendulum;
• What is the period of a pendulum;
• How to find the period of a pendulum in the small angles approximation;
• The formula for the period of a simple pendulum with ample initial angles;
• The effects of the gravitational acceleration on the pendulum's period.
Tick...tock: a short introduction to pendulums
Pendulums are elementary objects: a heavy mass connected to a rigid swing, free to oscillate around a pivot. In most of their descriptions, pendulums act only under the force of gravity: when this
happens, we call them simple pendulums.
These devices, unexpectedly, contain more physics than many others and can be used to prove fundamental properties of the natural world we live in. Moreover, they have a broad set of uses (though
it's slowly dwindling): their irreplaceable presence in time-tracking devices quite literally showed us the way during the exploration of our planet.
What is the period of a pendulum and how to calculate the duration of an oscillation
The period of a pendulum is the time required by the ensemble mass (bob) plus swing to complete one oscillation: with this, we mean that the mass returns in the same position and moves in the same
direction as the ones of the initial states.
The formula for the period of a pendulum is:
$T = 2\cdot\pi\cdot\sqrt{\frac{L}{g}}$
• $T$ is the period of the pendulum in seconds;
• $L$ is the length of the swing (in meters or feet); and
• $g$ is the acceleration due to gravity ($g\approx 9.81\ \text{m}/\text{s}^2$).
This simple equation for the period of a pendulum applies only in what physicists call the small angle approximation, a regime of the pendulum deriving from the trigonometric functions which model
its behavior. The approximation asserts that, for angles $\theta\ll1\ \text{rad}$:
In degrees we set the maximum value for which this approximation is valid to $10\degree$-$15\degree$. Up to this value, the period of the pendulum equation shows us that the time required to perform
the oscillation is independent of the initial angular displacement. The calculations in this approximation are much easier.
How to find the period of a pendulum outside the small angle approximation
If you need to consider oscillations larger than a few degrees, the small angle approximation won't give you the correct results: we need to introduce a nonlinearity in the form of a trigonometric
The formula for the period of a pendulum for any initial angle is:
\begin{align*} T & \!=\! 2\!\cdot\!\pi\!\cdot\!\sqrt{\!\frac{L}{g}}\!\cdot\!\left(\sum_{n=0}^{\infty}\left(\frac{(2\!\cdot\! n)!}{\left(2^{ n}\!\cdot\! n!\right)^2}\!\right)^2\!\!\cdot\!\right.\\ &\
left.\sin{\left(\frac{\theta_0}{2}\right)^{2\cdot n}}\right) \end{align*}
Our period of a pendulum calculator implements this series up to $n=20$. This is enough for most applications!
Pendulums at sea: why your grandfather clock would run slow during your Caribbean holidays
The overbearing effect of gravity on the operation of pendulums was apparent to physicists as soon as they started studying them. However, the actual implications deriving from that small constant
$g$ in the formula for the period were understood fully only later, when expeditions started carrying around the globe clocks with ever-increasing precision. Clocks close to the equator run slower
than clocks at higher latitudes.
This is due to variations in the acceleration due to gravity, which, in turn, come from the imperfect shape of our planet (even though its shape may not be perfect, we still like Earth a lot!). The
equatorial bulge — the technical name for the effect of rotation on a planet — causes $g$ to assume different values:
• At the poles, $g = 9.863\ \text{m}/\text{s}^2$; while
• At the equator, $g = 9.798\ \text{m}/\text{s}^2$.
We computed these values using our gravitational force calculator.
This difference caused an oscillation period to increase at the equator: an effect to consider since, back in those days, clocks were used to calculate a ship's position in the oceanic expanses.
More than periods: pendulum calculator for every needs
Now you know: pendulums are simple yet fascinating devices. We know this at Omni, which is why we created a small suite of pendulum calculators including this period of pendulum calculator!
Check them out:
• The pendulum length calculator; and
How do I calculate the period of a pendulum?
To find the period of a simple pendulum, you often need to know only the length of the swing. The equation for the period of a pendulum is:
T = 2π × sqrt(L/g)
This formula is valid only in the small angles approximation.
What is the period of a pendulum with length l = 1 m?
The period of such a pendulum is about 2 seconds.
To calculate this quantity, follow these steps:
1. Find the value of your local acceleration due to gravity. A safe bet is g = 9.81 m/s².
2. Substitute the value of g and l in the equation for the period of a pendulum: T = 2π × sqrt(L/g) =2π × sqrt(1/9.81) = 2.006.
3. You are all set.
Observe how this length returns one-second-long "half oscillations": this is the preferred length for pendulum clocks.
What is the small angles approximation?
The small angle approximation is a mathematical approximation used in trigonometry. For small values of the arguments of trigonometric functions, we can use the following set of approximations:
• sin(x) ≈ x;
• cos(x) ≈ 1; and
• tan(x) ≈ x.
You can derive these expressions using the series expansion of the functions: high order terms would result negligible for small values of x.
Why does gravity affects the period of a pendulum?
The gravitational force is responsible for the "return force" experienced by the pendulum's bob. It accelerates the mass when it moves toward the center, and slows until it stops on the way back.
The value of the acceleration due to gravity affects this force due to Newton's second law, and since at the poles g is 1% higher than at the equator, the period of the pendulum changes accordingly. | {"url":"https://www.omnicalculator.com/physics/pendulum-period","timestamp":"2024-11-07T07:33:08Z","content_type":"text/html","content_length":"615108","record_id":"<urn:uuid:709b0629-0636-47a6-843d-6ae3fd9cfdcc>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00386.warc.gz"} |
Demystifying the Higgs mechanism - Jakob Schwichtenberg
in Quantum Field Theory
Demystifying the Higgs mechanism
This is part 2 of my mini-series on understanding symmetry breaking, Goldstone’s theorem and the Higgs mechanism intuitively. Part 1 is here.
The punchline of the Higgs mechanism is often summarized as:
There are no Goldstone bosons if we break a local symmetry.
For example, in the standard model, we break the $SU(2)$ gauge symmetry. Since gauge transformations depend on the location $G=G(x)$ they are local and therefore no Goldstone bosons appear when we
break it.
Unfortunately, such a summary of the Higgs mechanism has many problems and leads to a lot of confusions.
The problems all have to do with the following observation: Before we can calculate anything that we can compare with experiments, we must remove the gauge symmetry by fixing the gauge. Usually, in
the textbooks, the Higgs mechanism is discussed before the gauge has been fixed. Then, there is no obvious problem. We discuss breaking of the gauge symmetry and then fix the gauge to remove the
gauge symmetry completely.
However, what happens if we reverse these two steps? We can first fix the gauge and then have a look what the Higgs mechanism is doing. Certainly, then we can’t talk about breaking of the gauge
symmetry since it has been removed completely from the theory. What is the Higgs mechanism then doing and why are there no Goldstone bosons? The story gets even weirder because there are different
possible ways to fix a gauge. The story of what the Higgs mechanism is doing changes depending on how we fix the gauge. We can even remove the $SU(2)$ symmetry completely by changing the field
variables (See: J. Fröhlich, G. Morchio and F. Strocchi, Phys. Lett. B97, 249 (1980) and Nucl. Phys. B190, 553-582 (1981)).
If all this weren’t bad enough there is even a famous theorem, called Elitzur’s theorem, whose punchline is:
Spontaneous breaking of a local symmetry is impossible.
I don’t want to dive into the details here, but if you want to have a look at what people are discussing in this context have a look at this paper.
The same confusing situation does not only exist in particle physics. The Higgs mechanism is often invoked to explain how superconductors work. Analogous to the story in particle physics, students
are usually taught that here the electromagnetic $U(1)$ gauge symmetry is broken. The cooper pairs play then the role of the Higgs field and since the broken $U(1)$ symmetry is local no Goldstone
bosons appear in the spectrum. (See, for example, the discussion in An Invitation to Quantum Field Theory by Luis Alvarez-Gaumé and Miguel A. Vázquez-Mozo). Again, we run into lots of difficulties
if we have a closer look as discussed, for example, in this article.
Now the good news.
Since we already understand symmetry breaking and Goldstone’s theorem intuitively, it is not that hard to understand how the Higgs mechanism works. Especially, we will not run into confusing
situations as the ones outlined above, since we will stick to physical things.
Much of the confusion surrounding the Higgs mechanism can be attributed to the confusion surrounding gauge symmetries. I will discuss gauge symmetries in another post since to understand the Higgs
mechanism it is sufficient to stick to what is physical about gauge symmetries and leave all the mysticism aside.
The loophole in Goldstone’s theorem
Like for most theorem in physics, there are loopholes in Goldstone’s theorem. Especially, there are systems where the configuration with the lowest energy, the ground state, breaks a symmetry, but no
Goldstone modes exist.
To spoil the surprise: in these systems, no Goldstone modes exist because there are long-range forces present before the symmetry breaks. Such long-range forces are what is physical about gauge
symmetries and this explains the connection between gauge symmetries and the Higgs mechanism.
The simplest example is again a ferromagnet. As mentioned above, usually the spins of the individual atoms only talk to their nearest neighbors. In other words, there are no long-range forces. Below
the Curie temperature, the spins align, but it costs no energy to perform a rotation of all spins at once. Such a uniform rotation is a “spin wave” with infinite wavelength and thus our Goldstone
mode here. It appears here because the fundamental laws are rotational invariant and only the ground state of the ferromagnet below the Curie temperature, i.e. the configuration with all spins
aligned, breaks the symmetry.
However, if there is additionally a long-range force present in the system, for example, the $1/r$ Coulomb force, such a global uniform rotation costs energy, because we must work against the Coulomb
force. Therefore, as soon as long-range forces are present there are no Goldstone modes.
Instead, what happens is that the long-range force becomes short-ranges as soon as the phase transition happens (in the example above, below the Curie temperature). The long-range force waves
combine with the would-be Goldstone modes and the result is a short-range force. The waves with an infinite wavelength that would be the Goldstone modes now also have an effect on the long-range
force, e.g. on the electric field. Thus such a wave with infinite wavelength is no longer possible without costing energy.
Instead, in the case of the ferromagnet, when we consider such a global uniform rotation what we get is a charge-density wave. This charge density wave is independent of the wavelength. As a result,
the Goldstone modes have a finite non-zero frequency.
As mentioned above in part 1, symmetries break when the system becomes rigid. Below the Curie temperature, the ferromagnet resists rotations of individual spins. As a result of this rigidity, the
former long-range force becomes short ranged.
The long-range electromagnetic force is mediated by electromagnetic waves, which simply means oscillations of electric and magnetic fields. When the system has become rigid, these electromagnetic
waves can no longer propagate freely. The displacement of individual spins and thus of individual magnetic moments cost energy if the system is rigid. Hence, the system tries to minimize such
displacements by intruding magnetic oscillations. As a result, the electromagnetic waves get damped and thus no longer have an infinite range.
In particle physics terms, we say the photon now has a mass. Before the symmetry breaking the photon is massless which means the range of electromagnetic interactions is infinite. After the system
has become rigid the range of electromagnetic interactions is finite and this means the particle mediating the interaction is massive
The Higgs mechanism in particle physics
“Anatoly Larkin, posed a challenge to two outstanding undergraduate teenage theorists, Sacha Polyakov and Sasha Migdal: “In field theory the vacuum is like a substance; what happens there?”
from Chapter 9 in The Infinity Puzzle, by F. Close
As already mentioned in the introduction above there is a folklore that is repeated over and over in the textbooks. According to the folklore the Higgs mechanism exploits a loophole in Goldstone’s
theorem because the symmetry that gets broken is a local one. This is wrong. A local symmetry is not a symmetry, but merely a redundancy in the description and cannot be broken anyway.
The real loophole, as discussed above, is that we consider a system with long-range forces. Prior to the phase transition into a ground state with smaller global symmetry, we have massless spin $1$
bosons that mediate the long-range forces. The scalar field then undergoes a phase transition and condenses into a new rigid ground state. This new ground state does no longer correspond to an empty
vacuum, but to a uniform distribution of Higgs field, which could be called “Higgs Substance”, to borrow a phrase from Guidice’s “A Zeptospace Odyssey“.
In particle physics, the conventional formulation of the non-empty vacuum state is that we say that the Higgs has a non-zero vacuum expectation value.
• A non-zero vacuum expectation value means that on average we expect to see some Higgs excitations in the vacuum, i.e. the vacuum is filled with Higgs field excitations.
• A zero vacuum expectation value means that we see on average no Higgs excitations if we observe the vacuum, i.e. the vacuum is empty.
The point is that below some critical temperature the configuration with the lowest energy is no longer an empty state, but one that is filled with the Higgs substance. The spontaneous filling of the
vacuum with the Higgs substance is completely analogous to how a ferromagnet becomes filled with magnetization below the Curie temperature. In this sense, the vacuum is not empty but rather more like
a medium.
The spontaneous alignment of the spins in a ferromagnet picks randomly a direction in space and breaks, therefore, rotational symmetry. Analogously, the Higgs field picks a direction in the internal
$SU(2) \times U(1)$ space. Before the vacuum becomes filled with the Higgs substance there is no way to distinguish the three $SU(2)$ bosons. Only after the Higgs spontaneously picks a direction,
these bosons become distinguishable.
There is an important second effect. Special relativity tells us that massless particles move with the speed of light while massive particles always move slower. A direct consequence of a vacuum
filled with Higgs substance is that all particles that interact with this substance can no longer move freely. Instead, whenever they try to get from A to B, they are stopped all the time by the
Higgs substance. Hence, they no longer move with the maximum velocity, i.e. the speed of light. In this sense they acquire an effective mass through the permanent interaction with the Higgs
substance. However, the Higgs substance makes its presence not only felt when particles are moving.
Instead, the permanent interaction with the Higgs substance also happens when particles are at rest. Without the Higgs substance filled vacuum, all particles would move with the speed of light.
To summarize: The real loophole that makes the Higgs mechanism possible are long-range interactions. Whenever we are dealing with a system with long-range interactions, there are no Goldstone bosons
after symmetry breaking, i.e. when the system becomes rigid. After symmetry breaking, the long-range interaction becomes short-ranged and in particle physics terms this means that the corresponding
boson is now no longer massless but massive. An important second effect is that other formerly massless field excitations can become massive through the now rigid structure. In particle physics
particles interact all the time with the “Higgs substance” that fills all of the vacuum after symmetry breaking.
To read more about what the Higgs mechanism is really doing in more abstract terms have a look at this post, especially the last section.
P.S. I wrote a textbook which is in some sense the book I wished had existed when I started my journey in physics. It's called
"Physics from Symmetry"
and you can buy it, for example, at
. And I'm now on
too if you'd like to get updates about what I'm recently up to.
If you want to get an update whenever I publish something new, simply put your email address in the box below. | {"url":"http://jakobschwichtenberg.com/higgs-intuitively/","timestamp":"2024-11-03T21:38:04Z","content_type":"text/html","content_length":"37803","record_id":"<urn:uuid:4f0b289e-b51a-4239-9040-e00f0ec209dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00718.warc.gz"} |
What Happens When the Government Tightens its Belt? - Mosler Economics / Modern Monetary Theory
What Happens When the Government Tightens its Belt?
By Stephanie Kelton
May 27 — Imagine two people sitting on opposite ends of a 15-foot teeter-totter. The laws of physics dictate that the seesaw will balance if the product of the first mass (w1) and its distance
(d1) from the fulcrum (i.e. the balancing point) is equal to the product of the other mass (w2) and its distance (d2) from the fulcrum. Thus, the physicist can show that the teeter-totter will be
in balance when the fulcrum is placed 6 feet from the end holding a 150lb person and 9 feet from the end holding a 100lb person. Moreover, the laws of physics ensure that an imbalance will arise
if the mass or the relative position of one of the people is changed.
The laws of accounting allow us to demonstrate that similarly powerful concepts apply to the science of economics. Beginning with the simple identity for GDP in a closed economy, we have:
[1] Y = C + I + G, where:
Y = GDP = National Income
C = Aggregate Consumption Expenditure
I = Aggregate Investment Expenditure
G = Aggregate Government Expenditure
For economists, this is as obvious as stating that a linear foot is the sum of 12 sequential inches. It simply recognizes that the total amount of money spent buying newly produced goods and
services will yield an equivalent income to the sellers of these products. Thus, it demonstrates that expenditures are a source of income.
Once earned, income can be allocated in one of three ways. At the end of the day, all income (Y) will be spent (C), saved (S) or used in payment of taxes (T):
[2] Y = C + S + T
Since they are equivalent expressions for Y, we can set equation [1] equal to equation [2], giving us:
C + I + G = C + S + T
Or, after canceling (C) from both sides and moving terms around:
[3] (S – I) = (G – T)
Equation [3] shows that there is a direct relationship between what’s happening in the private sector (S – I) and what’s happening in the public sector (G – T). But it is not the one that Pete
Peterson, Erskin Bowles, or President Obama would have you believe. And I want you to understand why they are wrong.
To understand the argument, imagine that you and Uncle Sam are sitting on opposite ends of a teeter-totter. You represent the private sector, and your financial status is given by (S – I). Your
budget can be in balance (S = I), in deficit (S < I) or in surplus (S > I). When your financial status is positive (S > I), you are net saving. When your financial status is negative (S < I), you
are net borrowing. Uncle Sam’s financial status is equal to (G – T), and, like yours, his budget may be balanced (G = T), in deficit (G > T) or in surplus (G < T). When you interact, only three
outcomes are possible.
First, it is conceivable that (S = I) and (G = T) so that (S – I) = 0 and (G – T) = 0. When this condition holds, the teeter-totter will level off with each of you experiencing a balanced budget.
In the above scenario, the government is balancing its receipts (T) and expenditures (G), and you are balancing your savings and investment spending. There is no net gain/loss.
But suppose the government begins to spend more than it collects in taxes (i.e. G > T). How will Uncle Sam’s deficit affect your position on the teeter-totter? The answer is as straightforward as
increasing the mass of the person on the right-hand side of the seesaw. As Uncle Sam’s financial position turns negative, your financial position turns positive.
This should make intuitive as well as mathematical sense, because when Uncle Sam runs a deficit, you receive more financial assets than you lose through taxation. Put simply, Uncle Sam’s deficit
lifts you into a surplus position. Moreover, bigger deficits mean bigger surpluses for you.
Finally, let’s see what happens when Uncle Sam tightens his belt. Suppose, for example, that we were able to duplicate the much-coveted surpluses of 1999-2001. What would (and did!) happen to the
private sector’s financial position?
Because the economy’s financial flows are a closed system – every payment must come from somewhere and end up somewhere – one sector’s surplus is always the other sector’s deficit. As the
government “tightens” its belt, it “lightens” its load on the teeter-totter, shifting the relative burden onto you.
This is not rocket science, but it appears to befuddle scores of educated people, including President Obama, who said, “small businesses and families are tightening their belts. Their government
should, too.” This kind of rhetoric may temporarily boost his approval ratings, but the policy itself will undermine the efforts of the very families and small businesses that are trying to
improve their financial positions.
* I’ll be back with a second installment that shows what happens when we ‘open’ the economy to take into account the foreign sector (and the relevant financial flows). Many of us have been
working with financial balance equations for years (see herefor references), so the current effort is nothing new. I am merely trying to make the arguments more accessible by changing the way
they are presented. | {"url":"https://moslereconomics.com/2011/06/01/what-happens-when-the-government-tightens-its-belt/","timestamp":"2024-11-09T22:00:06Z","content_type":"text/html","content_length":"40352","record_id":"<urn:uuid:19eee84a-4db6-459c-aadd-c8851a4190b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00180.warc.gz"} |
PCA and UMAP
Clearing the Confusion: PCA and UMAP
Principal Components Analysis (PCA) is a well-established method of dimension reduction. It is often used as a means of gaining insight into the “hidden meanings” in a dataset. But in prediction
contexts–ML–it is mainly a technique for avoiding overfitting and excess computation.
This tutorial will give a more concrete, more real-world-oriented, overview of PCA than those given in most treatments.
A number of “nonlinear versons” of PCA have been developed, including Uniform Manifold Approximation and Projection (UMAP), which we will also discuss briefly here.
All PCA does is form new columns from the original columns of our data. Those new columns form our new feature set. It’s that simple.
Each new column is some linear combination of our original columns. Moreover, the new columns are uncorrelated with each other, the importance of which we will also discuss.
Let’s see concretely what all that really means.
Example: mlb data
Consider mlb, a dataset included with qeML. We will look at heights, weights and ages of American professional baseball players. (We will delete the first column, which records position played.)
> data(mlb1)
> mlb <- mlb1[,-1]
> head(mlb)
Height Weight Age
1 74 180 22.99
2 74 215 34.69
3 72 210 30.78
4 72 210 35.43
5 73 188 35.71
6 69 176 29.39
> dim(mlb)
[1] 1015 3
# 1015 players, 3 measurements each
# PCA requires matrix format
> mlb <- as.matrix(mlb)
Apply PCA
The standard PCA function in R is prcomp:
# defaults center=TRUE, scale.=FALSE)
> z <- prcomp(mlb)
We will look at the contents of z shortly. But first, it is key to note that all that is happening is that we started with 3 variables, Height, Weight and Age, and now have created 3 new variables,
PC1, PC2 and PC3. Those new variables are stored in the matrix z$x. Again, we will see the details below, but the salient point is: 3 new features.
We originally had 3 measurements on each of 1015 people, and now we have 3 new measurements on each of those people:
> dim(z$x)
[1] 1015 3
Well then, what is in there in z?
> str(z)
List of 5
$ sdev : num [1:3] 20.87 4.29 1.91
$ rotation: num [1:3, 1:3] -0.0593 -0.9978 -0.0308 -0.1101 -0.0241 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:3] "Height" "Weight" "Age"
.. ..$ : chr [1:3] "PC1" "PC2" "PC3"
$ center : Named num [1:3] 73.7 201.3 28.7
..- attr(*, "names")= chr [1:3] "Height" "Weight" "Age"
$ scale : logi FALSE
$ x : num [1:1015, 1:3] 21.46 -13.82 -8.6 -8.74 13.14 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:1015] "1" "2" "3" "4" ...
.. ..$ : chr [1:3] "PC1" "PC2" "PC3"
- attr(*, "class")= chr "prcomp"
Let’s look at rotation first.
As noted, each principal component (PC) is a linear combination of the input features. These coefficients are stored in rotation:
> z$rotation
PC1 PC2 PC3
Height -0.05934555 -0.11013610 -0.99214321
Weight -0.99776279 -0.02410352 0.06235738
Age -0.03078194 0.99362420 -0.10845927
For instance,
PC2 = -0.11 Height - 0.02 Weight + 0.99 Age
As noted, those 3 new variables are stored in the x component of z. For instance, consider the first person in the dataset:
> mlb[1,]
Height Weight Age
74.00 180.00 22.99
In the new data, his numbers are:
> z$x[1,]
PC1 PC2 PC3
21.458611 -5.201495 -1.018951
Let’s check! Since the PCi are linear combinations of the original columns, we can compute them via matrix multiplication. Let’s do so for PC2, say for the first row of the data.
# remember, prcomp did centering, so we need it here
> mlbc <- center(mlb)
> mlbc[1,] %*% z$rotation[,2]
[1,] -5.201495
Ah yes, same as above.
Key properties
The key properties of PCA are that the PCs
1. are arranged in order of decreasing variances, and
2. they are uncorrelated.
The variances (actually standard deviations) are reported in the return object from prcomp:
> z$sdev
[1] 20.869206 4.288663 1.911677
Yes, (a) holds. Let’s double check, say for PC2:
> sd(z$x[,2])
[1] 4.288663
What about (b)?
> cor(z$x)
PC1 PC2 PC3
PC1 1.000000e+00 -1.295182e-16 2.318554e-15
PC2 -1.295182e-16 1.000000e+00 2.341867e-16
PC3 2.318554e-15 2.341867e-16 1.000000e+00
Yes indeed, those new columns are uncorrelated.
Practical importance of (a) and (b)
The reader of this document has probably seen properties (a) and (b) before. But why are they so important?
Many data analysts, e.g. social scientists, use PCA to search for patterns in the data. In the ML context, though, our main interest is prediction.
Our focus:
If we have a large number of predictor variables, we would like to reduce that number, in order to avoid overfitting, reduce computation and so on. PCA can help us do that. Properties (a) and (b)
will play a central role in this.
Toward that end, we will first introduce another dataset, and then discuss dimension reduction–reducing the number of predictor variables–in the context of that data. That will lead us to the
importance of properties (a) and (b).
Example: fiftyksongs data
Here we will use another built-in dataset in qeML, a song database named fiftyksongs. It is a 50,000-row random subset of the famous Million Song Dataset.
The first column of the data set is the year of release of the song, while the other 90 are various audio measurements. The goal is to predict the year.
> dim(fiftyksongs)
[1] 50000 91
> w <- prcomp(fiftyksongs[,-1]) # first column is "Y", to be predicted
> w$sdev
[1] 2127.234604 1168.717654 939.840843 698.575319 546.683262 464.683454
[7] 409.785038 395.928095 380.594444 349.489142 333.322277 302.017413
[13] 282.819445 260.362550 255.472674 248.401464 235.939740 231.404983
[19] 220.682026 194.828458 193.645669 189.074051 187.455170 180.727969
[25] 173.956554 166.733909 156.612298 151.194556 144.547790 138.820897
[31] 133.966493 124.514162 122.785528 115.486330 112.819657 110.379903
[37] 109.347994 106.551231 104.787668 99.726851 99.510556 97.599960
[43] 93.161508 88.559160 87.453436 86.870468 82.452985 80.058511
[49] 79.177031 75.105451 72.542646 67.696172 64.079955 63.601079
[55] 61.105579 60.104226 56.656737 53.166604 52.150838 50.515730
[61] 47.954210 47.406341 44.272814 39.914361 39.536682 38.653450
[67] 37.228741 36.007748 34.192456 29.523751 29.085855 28.387604
[73] 26.325406 24.763188 22.192984 20.203667 19.739706 18.453111
[79] 14.238237 13.935897 10.813426 9.659868 8.938295 7.725284
[85] 6.935969 6.306459 4.931680 3.433850 3.041469 1.892496
Dimension reduction
One hopes to substantially reduce the number of predictors from 90. But how many should we retain? And which ones?
There are 2^90 possible sets of predictors to use. It of And given the randomness and the huge number of sets, the odds are high that the “best”-predicting set is just an accident, not the actual
best and maybe not even close to the best. course would be out of the question to check them all. So, we might consider the below predictor sets, say following the order of the features (which are
named V2, V3, V4,…)
V2 alone; V2 and V3; V2 and V3 and V4; V2 and V3 and V4 and V5; etc.
Now we have only 90 predictor sets to check–a lot, but far better than 2^90. Yet there are two problems that would arise:
• Presumably the Vi are not arranged in order of importance as While the set V12, V13,…, V88 would include these three variables, we may risk overfitting, masking the value of these three..
predictors. What if, say, V12, V28 and V88 make for an especially powerful predictor set? The scheme considered here would never pick that up.
• Possibility of substantial duplication: What if, say, V2 and V3 are very highly correlated? Then once V2 is in our predictor set, we probably would not want to include V3; we are trying to find a
parsimonious predictor set, and inclusion of (near-)duplicates would defeat the purpose. Our second candidate set above would be V2 and V4, the third would be V2 and V4 and V5; and so on. We may
wish to skip some other Vi as well. Checking for such correlation at every step would be cumbersome and time-consuming.
Both problems are addressed by using the PCs Pi instead of the original variables Vi. We then consider these predictor sets,
P1 alone; P1 and P2; P1 and P2 and P3; P1 and P2 and P3 and P4; etc.
What does this buy us?
• Recall that Var(P[i]) is decreasing in i (technically nonincreasing). For large i, Var(P[i]) is typically tiny; in the above example, for instance, Var(P49) / Var(P1) is only about 0.0014. And a
random variable with small variance is essentially constant, thus of no value as a predictor.
• By virtue of their uncorrelated nature, the Pi basically do not duplicate each other. While it is true that uncorrelatedness does not necessarily imply independence, again we have a reasonable
solution to the duplication problem raised earlier.
And What about UMAP?
Again, PCA forms new variables that are linear functions of the original ones. That can be quite useful, but possibly constraining. In recent years, other dimension reduction method have become
popular, notably t-SNE and UMAP. Let’s take a very brief look at the latter.
mypars <- umap.defaults
mypars$n_components <- 6
umOut <- umap(fiftyksongs[,-1],
The new variables will then be returned in umOut\(layout**, analogous to our **z\)x above. We will now have a 50000 x 6 matrix, replacing our original 50000 x 90 data.
So, what does UMAP actually do? The math is quite arcane; even the basic assumption, “uniform distribution on a manifold,” is beyond the scope of this tutorial.
But roughly speaking, the goal is to transform the original data, dimension 90 here, to a lower-dimensional data set (6 here) in such a way that “local” structure is retained. The latter condition
means that rows in the data that were neighbors of each other in the original data are likely still neighbors in the new data, subject to the hyperparameter n_neighbors: a data point v counts as a
neighbor of u only if v is among the n_neighbors points to u.
In terms of the Bias-Variance Tradeoff, smaller values of n_neighbors reduce bias while increasing variance, in a similar manner to the value of k in the k-Nearest Neighbors predictive method. | {"url":"https://cran.opencpu.org/web/packages/qeML/vignettes/PCA_and_UMAP.html","timestamp":"2024-11-02T18:38:27Z","content_type":"text/html","content_length":"476171","record_id":"<urn:uuid:54a66ab9-bf36-44d8-b874-717efe198f31>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00201.warc.gz"} |
CPM Homework Help
On Tuesday the cafeteria sold pizza slices and burritos. The number of pizza slices sold was $20$ less than twice the number of burritos sold. Pizza sold for $$2.50$ a slice and burritos sold for $
$3.00$ each. The cafeteria collected a total of $$358$ for selling these two items.
1. Write two equations with two variables to represent the information in this problem. Be sure to define your variables by writing let statements.
$p =$ pizza slices (#)
$b =$ burritos (#)
One equation should be about the amount of money collected from the burritos and pizza slices sold.
The other equation should be about the number of pizza slices sold compared to the number of burritos sold.
$2.50p + 3b = 358$
$p = 2b - 20$
2. Solve your system from part (a). Then determine how many pizza slices were sold.
Use the substitution method with the equations you found in part (a). | {"url":"https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/6/lesson/6.3.1/problem/6-86","timestamp":"2024-11-03T00:49:21Z","content_type":"text/html","content_length":"38562","record_id":"<urn:uuid:ffcf1aab-1939-4d24-a6e9-deb940871a11>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00812.warc.gz"} |
A fractional Fourier Transform? Yes, there is such a thing.
The Fourier Transform, outlined by French mathematician and physicist Joseph Fourier (1768-1830) in The Analytic Theory of Heat (1822), asserted that any function of a variable, whether continuous or
discontinuous, can be expanded in a series of sines of multiples of the variable. His focus at the time was the propagation of heat in an iron ring, but in the next 100 years his theory was
successfully applied to electrical, acoustical and in fact, all waveforms.
A waveform, simple or complex, can be displayed in the time domain, where its amplitude is plotted in an oscilloscope against the vertical Y-axis while time is plotted against the horizontal X-axis.
This is the function prior to the application of Fourier’s transform, following which the series of sines of multiples of the variable can be seen in a spectrum analyzer in the frequency domain. Here
amplitude is still plotted against the vertical Y-axis, now in units of power (dB) rather than volts, and frequency, rather than time, is plotted against the horizontal X-axis in milliseconds or
suitable units.
Time domain and frequency domain are equally realistic displays of the same phenomenon, but to the theoretician, engineer or student they convey different types of knowledge. The time domain provides
a highly intuitive image of the waveform as it represents oscillating energy in a conductor or oscillating electromagnetic energy in space. The frequency domain, however, displays the fundamental as
a large spike, conventionally positioned at the left edge or the center of the screen, followed by peaks representing harmonics that typically diminish in amplitude as they become farther (in
frequency) from the fundamental. In the frequency domain we also see other events such as broad-spectrum noise, interference (with its own harmonics), and time-dependent anomalies that come and go
and may be captured and stored in the instrument’s memory.
The Fourier transform is a two-way process. Time domain, transformed to frequency domain, is known as Fourier analysis, and the reverse is Fourier synthesis. It is possible to go back and forth any
number of times with no loss of information except for distortion introduced in the instrumentation and test setup.
One problem in Fourier analysis and synthesis is that the mathematics is overwhelming, involving millions of operations, which were challenging for early computers. Fortunately, for the many
individuals working in the great number of fields in which the Fourier Transform was becoming applicable, beginning around 1965 the Fast Fourier Transform (FFT) emerged. It is a set of algorithms
that typically cuts down the number of required mathematical operations by a factor of 4,000.
The FFT constructs the frequency domain expression of the desired waveform by factorizing its time domain matrix. The operation is greatly facilitated because most of the factors are zero. Moreover,
the FFT version is often more accurate because a million or more computations are eliminated.
FFT goes beyond a single algorithm. In the course of a century and a half, various versions were developed, one for example based on prime numbers. The first FFT had been derived around 1805 by Carl
Friedrich Gauss, who used it as a tool to construct the orbits of two asteroids, Pallas and Juno. That done, he did not pursue the matter further.
In the 19th and early 20th century, a number of FFT concepts were used in the field of statistics, pertaining to the design of experiments. But our current generic FFT algorithm did not appear until
1965, when James Cooley and John Tukey devised a single algorithm in connection with US efforts to detect Soviet nuclear tests based on data from sensors outside its borders. Cooley and Tukey
presented the idea in a joint paper, but because Cooley worked at IBM’s Watson labs and Tukey was an outsider, the algorithm ended up in the public domain, becoming available for use in the emerging
digital processing field.
That brings us to a more recent development, the fractional Fourier Transform, sometimes abbreviated FrFT or FR-FT. First conceived in the early 1990s, the FrFT can be viewed as a rotation of the FT
by some angle. An alternative interpretation is that the FrFT is actually a partial FT. It can be helpful to visualize this interpretation by imagining a signal in the time domain as having a zero
FT. Then after undergoing an FFT, it has been 100% Fourier transformed. Viewed this way, a FrFT is a a Fourier Transform on the incoming signal that ranges from greater than 0% to less than 100%
depending on the chosen angle.
An example of filtering via FrFT from Wikipedia. An FrFT rotation by π/4 converts noise frequencies to a single frequency which can be easily filtered out. Then the filtered signal is converted back
to its original form. Click image to enlarge.
The point of this mathematical manipulation is that it can convert signals into forms that are easier to work with than if left in either the time or frequency domain. For example, FrFTs are
increasingly used for handling optical and millimeter-wave communication signals that, frequency-wise, sit close to other interfering signals or electrical noise. It may be possible to apply an FrFT
to the signal in a way that converts the band of interfering frequencies into a single frequency that differs from the frequency of interest. This conversion allows a simple filter to remove the
problem frequency from the converted signal. Then an inverse FrFT operation can return the resulting signal back to the original form, without the noise. There is an article in Wikipedia on the FrFT
that illustrates this use quite well.
One might wonder how electronic circuits can perform an FrFT on an incoming signal. First consider the operation of spectrum analyzers. Prior to the development of FFT in 1965, spectrum analyzers
were exclusively swept-tuned instruments. A built-in superheterodyne receiver down-converted a portion of the signal spectrum at the input to the center frequency of a narrow band-pass filter. The
instantaneous output power was displayed as a function of time. A voltage-controlled oscillator was used in the superheterodyne section to sweep the center frequency through a range of frequencies,
creating the frequency-domain display. The disadvantage in the swept-tuned analyzer was that as particular frequencies were displayed, the over-all spectrum was not active, and short-duration events
at the other frequencies could be missed.
Within two years after its rediscovery in 1965, FFT technology found its way into the spectrum analyzer. In conjunction with a receiver and analog/digital converter, the FFT-based frequency analyzer,
as in the swept-spectrum instrument, processes a portion of the input signal spectrum. The difference, however, is that the spectrum is not swept. The receiver reduces the sampling rate so the FFT
spectrum analyzer can process all the samples. As a result, short-term events are accessed.
Spectrum analyzers found on test benches pretty much all operate according to these principles. In contrast, there is no standard electronic approach for computing an FrFT. Moreover, there are no
commercially available test instruments as of this writing that will implement an FrFT. A review of the scientific literature on FrFT work reveals that most researchers don’t apply the FrFT in real
time. They more typically use a program such as Matlab to calculate the effect of the conversion.
The FrFT circuit from the STMicrosystems patent (US 7,543,009 B2). Each block implements one term of a Taylor Series representing the FrFT operation. Click image to enlarge.
However, there is a great deal of interest in the FrFT for millimeter wave signal processing. It is illustrative to see how research groups are approaching the electronic implementation of FrFT. One
example comes from a patent filed by STMicroelectronics Belgium NV. The mathematics underlying the FrFT is quite complex– the starting point is generally to take the FT integral to a non-integer
power determined by the chosen FrFT angle. Engineers at STMicroelectronics approached the electronic implementation of the FrFT by expressing the underlying expression as a Taylor series, an infinite
sum of terms that are expressed in terms of the function’s derivatives at a single point. The patent only mentions a circuit implementing the first three terms of the series, but we might surmise
implementations on real ICs probably use more. Each block of the circuit implements one of the Taylor terms. Each block implements an inverse FFT, and all but the first block also implement two
multiplications. All the blocks are summed to produce the final result.
The fact that each block in the circuit executes an inverse FFT illustrates the complexity involved in computing the FrFT. It will be interesting to see if the FrFT eventually becomes a function on a
test instrument that can be invoked with a pushbutton. | {"url":"https://www.testandmeasurementtips.com/a-fractional-fourier-transform-yes-there-is-such-a-thing-faq/","timestamp":"2024-11-10T19:22:16Z","content_type":"text/html","content_length":"106954","record_id":"<urn:uuid:b21d10cd-4fe7-4a3f-b8de-dd60529bd52f>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00542.warc.gz"} |
Re: [tlaplus] Re: Potentially confusing behavior of a PlusCal algorithm
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Re: [tlaplus] Re: Potentially confusing behavior of a PlusCal algorithm
I'll just add this here as it may help somebody in the future track down this problem if they have a similar one. I was bitten by this today and it took me a while to find this thread which has
cleared it up for me. My code had this problem and it was a bit hidden. Here is my code
curr := PeekStack(self); \* Fine: there is an element for sure
pop_stack(); \* Now the stack could be empty
if(StackIsEmpty(self)){ \* Checks the old stack, not stack' !
rebalance2: \* This whole label is unreachable!
call rotate(Null, Null, root);
assert ~StackIsEmpty(self); \* Will fail because we did not early return as may have been expected
Now I know that I should either not use operators in these situations, or use additional labels.
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7th Grade Math Workbook (Physical Product)
This listing is for a paperback copy of my 7th grade math workbook. You will be receiving a physical copy in the mail of my popular 7th grade math workbook. The workbook is amazing! Each math concept
includes 2 pages. The left side has notes and examples. The right side has "Your Turn". Detailed answer keys are included in the back of the workbook.
The workbook includes the following:
✅ Absolute Value
✅ Rational Numbers on the Number Line
✅ Adding and Subtracting Integers on the Number Line
✅ Adding and Subtracting Fractions on the Number Line
✅ Adding Integers using Visual Representation
✅ Subtracting Integers using Visual Representation
✅ Adding and Subtracting Integers using Rules
✅ Multiplying Integers
✅ Dividing Integers
✅ Least Common Multiple (LCM)
✅ Greatest Common Factor (GCF)
✅ Adding and Subtracting Fractions
✅ Multiplying Fractions
✅ Dividing Fractions
✅ Adding and Subtracting Decimals
✅ Multiplying Decimals
✅ Dividing Decimals
✅ Intro to Ratios
✅ Intro to Rates
✅ Unit Rates Word Problems
✅ Writing & Solving Proportions
✅ Constant of Proportionality
✅ Proportional Relationships in a Table
✅ Proportional Relationship in a Graph
✅ Unit Rate on a Graph
✅ Decimals, Fractions, and Percents
✅ Markdowns (Discounts)
✅ Markups
✅ Sales Tax
✅ Tip
✅ Commission
✅ Percent of Change (Percent Increase & Percent Decrease)
✅ Simple Interest
✅ Combining Like Terms
✅ Distributive Property
✅ Simplifying Expressions
✅ Solving One-Step Equations
✅ Solving Two-Step Equations
✅ Solving and Graphing One-Step Inequalities
✅ Solving and Graphing Two-Step Inequalities
✅ Scale Factor Relationship with Area and Perimeter
✅ Similar Figures
✅ Scale Drawings
✅ Cross Sections
✅ Circles - Area and Circumference
✅ Complementary, Supplementary, Vertical, & Adjacent
✅ Evaluating Angles
✅ Area of 2-Dimensional Figures
✅ Rectangles - Area and Perimeter
✅ Triangles - Area and Perimeter
✅ Volume of Prisms
✅ Surface Area of Prisms
✅ Random Sampling
✅ Draw Inferences
✅ Numerical Data Distributions
✅ Mean, Median, Mode, and Range
✅ Mean Absolute Deviation
✅ Line Plots
✅ Stem and Leaf Plot
✅ Box and Whisker Plot
✅ Probability of an Event
✅ Probability of a Repeated Event
✅ Independent and Dependent Events
✅ Likelihood
Total Pages: 154+ answer keys
Answer Key: Included
Document File: Paperback (Physical copy mailed to you) | {"url":"https://mathindemand.com/products/7th-grade-math-notebook-physical-product","timestamp":"2024-11-03T12:27:06Z","content_type":"text/html","content_length":"189599","record_id":"<urn:uuid:9a1f6b46-7ae3-4f3a-9392-d343390ac0d6>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00633.warc.gz"} |
The Answer of Which Graph Represents the Inequality X ≤ –2 or X ≥ 0? | Student Portal
The Answer of Which Graph Represents the Inequality X ≤ –2 or X ≥ 0?
Posted on
Math in general is not an easy thing for everyone. In math, inequality is probably one of the things that a lot of people think is hard to solve. Even the quite simple question like “Which graph
represents the inequality X ≤ –2 or X ≥ 0?” can be hard, especially for those who have no idea about the subject.
If you are one of those people who are having a hard time finding out the answer to the question above, you come to the right place as you will be given the correct answer about it. Here is the
answer to the question “Which graph represents the inequality X ≤ –2 or X ≥ 0?”:
There are a total of two inequalities that are given. The first one is x ≥ 0 and the second one is x ≤ -2. Since, we have to take all the values greater than or equal to 0 and all the values less
than or equal to -2. So, we reject the values lying between -2 and 0 i.e. we will not take the values from (-2, 0). So, plotting the given inequality gives the following graph:
In the world of math, the term inequality is used to call a statement of an order relationship greater than, greater than or equal to, less than, or less than or equal to, between two numbers or
algebraic expressions. It is possible for the inequalities to be posed either as questions, similar to equations. Not only that, it is also possible for them to be solved by similar techniques, or as
statements of facts in the form of theorems. For instance, according to triangle one, the sum of the lengths of any two sides of a triangle is greater than or equal to the length of the remaining
side. In fact, mathematical analysis relies on things like that in the proofs of its most important theorems.
Talking about the inequalities, sometimes, it will be needed for you to solve them like these:
Symbol Words Example
> Greater than X + 3 > 2
< Less than 7x < 28
≥ Greater than or equal to 5 ≥ x – 1
≤ Less than or equal to 2y + 1 ≤ 7
The main purpose of inequality is to have X, or any variable, on its own on the left of the inequality sign. It is something like x < 5 or y ≥ 11. It is called solved. Here is the example: x + 2 >
12. You can subtract 2 from both sides: x + 2 – 2 > 12 – 2. When it is simplified, it becomes x > 10.
How do you solve the inequalities? Solving the inequalities is the same as solving equations. Most of the things are the same. However, there is a difference, which is to pay more attention to the
direction of the inequality. There are a total of four things that are able to change the direction, including:
• < becomes >
• > becomes <
• ≤ becomes ≥
• ≥ becomes ≤
It should be noted that there are a few things that do not affect the direction of the inequality, such as:
• Adding or subtracting a number from both sides
• Multiplying or dividing both sides by a positive number
• Simplifying a side
For instance: 3x < 7 + 3. It is able to be simplified 7 + 3 without affecting the inequality: 3x < 10.
On the other hands, these followings are the things that are able to change the direction of the inequality:
• Multiplying or dividing both sides by a negative number
• Swapping left and right hand sides
For instance: 2y + 7 < 12. When it is swapped left and right hand sides, the direction of the inequality also must be changed: 12 > 2y + 7.
Apparently, it is possible for you to solve the inequalities by adding or subtracting a number from both sides just like x + 3 < 7. If both sides are subtracted, the thing becomes x + 3 – 3 < 7 – 3.
From there, it can be concluded that the solution is x < 4. If you have no idea about it, it means x can be any value less than 4.
What do you do: You went from x + 3 < 7 to x < 4. This one works properly by adding and subtracting. The reason is because if the same amount from both sides are added or subtracted, it does not
affect the inequality. For instance, Kay Lee has more coins compared to Nemanja Matic. If both Kay Lee and Nemanja Matic get three more coins each, Kay Lee will still have more coins compared to
Nemanja Matic.
What if the x is on the right when solving it? Actually, this kind of thing does not matter. You can just swap sides. However, it is important to reverse the sign so it still points at the right
value. For instance: 12 < x + 5. If it is subtracted from both sides, it becomes 12 – 5 < x + 5 – 5. The solution to the question is 7 < x. The normal one is to actually put the x on the left hand
side. To do that, you can just flip the sides and the inequality sign. The solution before can become x > 7.
Apart from that, another thing that you do is to multiply or to divide both sides by a value, just like in algebra multiplying. Please be a bit more careful.
For the positive value, feel free to multiply or divide a thing by a positive number. For instance: 3y < 15. If it is divided on both sides by 3, the result will be 3y /3 < 15 / 3. From that, the
solution can be y < 5.
What about the negative number? When a thing is multiplied or divided by a negative number, it is a must to reverse the inequality. | {"url":"https://www.student-portal.net/the-answer-of-which-graph-represents-the-inequality-x-%E2%89%A4-2-or-x-%E2%89%A5-0.edu","timestamp":"2024-11-01T21:56:03Z","content_type":"text/html","content_length":"84580","record_id":"<urn:uuid:870c9977-a1a1-4b36-8898-1c143455daef>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00751.warc.gz"} |
Average - Aptitude Concepts & Theory Explained (Made Simple)
Understanding the Concepts of Average
You can now understand the fundamental concepts of Average. Listed below are the different Average concepts:
1. What is Average?
In mathematics, Average is defined as the mean value, which is equal to the ratio of the sum of all values in a set to the total number of values/units present in the set.
The data can be anything like age, money, runs, etc. Average has many applications in real-life.
Average = sum of elements / number of elements
Example Problem
The Average of 3, 6, and 9 is 3 + 6 + 9 = 18÷ 3 = 6. So the Average is 6. It means 6 is the central value of 3, 6, and 9.
Therefore, Average means to find out the mean value of a group of numbers.
2. How to Find the Average of the given numbers?
Example Problem
The marks obtained by 8 students in a class test are 12, 15, 16, 18, 20, 10, 11, and 21. Use the Average formula and find out what the Average of the marks obtained by the students is.
Marks obtained by 8 students in class test = 12, 15, 16, 18, 20, 10, 11, and 21 (given)
Total marks obtained by 8 students in class test =(12+15+16+18+20+10+11+21) = 123/8
Using the Average formula,
Average = (Sum of Observation) ÷ (Total numbers of Observations)
Average = (12+15+16+18+20+10+11+21) ÷ 8
Average = 123/8
Average of marks obtained by 8 students = 15.375
3. Rules of Average
Here are some handy tricks for Average which will make your calculation faster and more efficient with practice:
(1) If the value of each number is increased by the same value 'a', then the Average of all numbers will also increase by 'a'.
(2) If the value of each number is decreased by the same value 'a', then the Average of all numbers will also decrease by 'a'.
(3) If the value of each number is multiplied by the same value 'a', then the Average of all numbers will also get multiplied by 'a'.
(4) If the value of each number is divided by the same value 'a', then the Average of all numbers will also get divided by 'a'.
4. Average of two or more groups taken together
(a) If the number of quantities in two groups is n₁ and n₂ and their Average is x and Y, respectively, the combined Average (Average of all of them put together) is =(n₁ x+n₂ Y)/(n₁+n₂ )
(b) If the Average of ₁ quantities is x and the Average of n₂ quantities out of them is Y, the average of the remaining group (rest of the quantities) is =(n₁ x-n₂ Y)/(n₁ – n₂ )
5. What is Average Speed?
Average Speed is the rate at which a journey takes place. Throughout a journey, the Speed is not constant; it varies from time to time.
Average Speed Formula
The Average Speed of an object is equal to the total distance covered by the object, divided by the total time taken to cover the distance.
Average Speed = Total distance covered ÷ Total time taken.
S = D/T.
'D' is the distance travelled in some time 'T'
'S' is the Speed of the object for this journey.
Why is understanding the concepts of Average important?
Understanding the concepts of Average assists in:
• Understanding how Average formulas are derived
• Addressing the Average problems promptly and accurately.
• Resolving each of the various forms of questions on the Average topic
• Developing your unique shortcuts
Is it possible to solve Average problems without knowing the concepts?
Yes, it's possible to solve Average questions without understanding what they entail. However, experts advise that comprehending the fundamentals is essential to address the Average problems
What is the right way to learn Average concepts?
The foundation of mathematics is concepts, and understanding them is critical to boosting your performance in the Quantitative Aptitude section. Visualising the concepts using real-life examples is
the best approach to learn the Average concepts.
Average aptitude questions include:
• Equal distribution technique
• Deviation method
• Change in Average problems
• New Average - old Average problems
• Weighted Average problem
• Interchanged digits problems. | {"url":"https://www.placementpreparation.io/quantitative-aptitude/average/concepts/","timestamp":"2024-11-03T00:07:37Z","content_type":"text/html","content_length":"175779","record_id":"<urn:uuid:eac25b06-4dd3-4d23-a660-15bd9bff38d1>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00067.warc.gz"} |
Frustrating elevator - Book Proofs
Frustrating elevator
This weeks Riddler Express is a problem about a frustrating elevator! Here it goes:
You are on the 10th floor of a tower and want to exit on the first floor. You get into the elevator and hit 1. However, this elevator is malfunctioning in a specific way. When you hit 1, it
correctly registers the request to descend, but it randomly selects some floor below your current floor (including the first floor). The car then stops at that floor. If it’s not the first floor,
you again hit 1 and the process repeats.
Assuming you are the only passenger on the elevator, how many floors on average will it stop at (including your final stop, the first floor) until you exit?
My solution:
[Show Solution] | {"url":"https://laurentlessard.com/bookproofs/frustrating-elevator/","timestamp":"2024-11-10T04:41:34Z","content_type":"text/html","content_length":"132523","record_id":"<urn:uuid:ee3c1c65-6f1a-4eee-badd-da1772bea937>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00215.warc.gz"} |
Python range() Function: A Comprehensive Tutorial (With Examples) - MachineLearningTutorials.org
Python range() Function: A Comprehensive Tutorial (With Examples)
The range() function in Python is a built-in function that generates a sequence of numbers. It’s commonly used in various programming scenarios, such as looping over a specific range of values,
iterating through indices, and generating numerical sequences. In this comprehensive tutorial, we will explore the range() function in depth, discussing its syntax, parameters, use cases, and
providing multiple examples to help you grasp its versatility and practicality.
Table of Contents
1. Introduction to the range() Function
2. Syntax and Parameters
3. Generating a Sequence of Numbers
4. Using range() with for Loops
5. Creating Lists from range()
6. Customizing start, stop, and step Parameters
7. Working with Negative Steps
8. Use Case: Sum of Consecutive Integers
9. Use Case: Generating Even and Odd Numbers
10. Conclusion
1. Introduction to the range() Function
The range() function in Python is used to generate a sequence of numbers. It produces a range object that represents the desired sequence. The sequence is determined by the specified start, stop, and
step values. This function is particularly useful in scenarios where you need to iterate over a range of numbers or generate sequences without creating large lists in memory.
2. Syntax and Parameters
The syntax of the range() function is as follows:
range([start], stop, [step])
• start (optional): The starting value of the sequence (default is 0).
• stop: The exclusive upper limit of the sequence. The sequence will generate numbers up to, but not including, this value.
• step (optional): The increment between numbers in the sequence (default is 1).
3. Generating a Sequence of Numbers
Let’s start with a simple example to understand how the range() function works in generating a sequence of numbers. Suppose you want to generate a sequence of numbers from 0 to 9. You can achieve
this using the following code:
# Generating a sequence of numbers from 0 to 9
for num in range(10):
In this example, the range(10) call generates a sequence of numbers starting from 0 (default start value) up to, but not including, 10 (stop value). The for loop then iterates over this sequence,
printing each number.
4. Using range() with for Loops
The most common use of the range() function is in conjunction with for loops to iterate over a sequence of numbers. This allows you to perform a set of actions for each number in the sequence. Let’s
consider a scenario where you want to calculate the sum of integers from 1 to 100. You can achieve this using a for loop and the range() function:
# Calculating the sum of integers from 1 to 100
total_sum = 0
for num in range(1, 101):
total_sum += num
print("The sum of integers from 1 to 100 is:", total_sum)
In this example, the range(1, 101) call generates a sequence of numbers starting from 1 up to, but not including, 101. The for loop iterates through this sequence, adding each number to the total_sum
5. Creating Lists from range()
While the range() function generates a sequence of numbers, it doesn’t create a list by default. However, you can easily convert the range object into a list using the list() constructor. This can be
useful when you need to store the sequence of numbers for later use. Consider the following example:
# Creating a list of even numbers from 0 to 10 using range()
even_numbers = list(range(0, 11, 2))
print("List of even numbers:", even_numbers)
In this example, the range(0, 11, 2) call generates a sequence of even numbers starting from 0 up to, but not including, 11, with a step of 2. The list() constructor converts this sequence into a
list of even numbers.
6. Customizing start, stop, and step Parameters
The range() function provides flexibility by allowing you to customize the start, stop, and step parameters according to your needs. You can omit the start and step parameters if you want to use
their default values.
For instance, if you want to generate a sequence of numbers from 5 to 20 with a step of 3, you can do so as follows:
# Generating a sequence of numbers from 5 to 20 with a step of 3
for num in range(5, 21, 3):
In this example, the range(5, 21, 3) call generates a sequence of numbers starting from 5 up to, but not including, 21, with a step of 3.
7. Working with Negative Steps
The range() function also supports negative step values, which allows you to generate sequences in reverse order. Negative steps are especially useful when you need to iterate over sequences in
descending order. Here’s an example:
# Generating a sequence of numbers from 10 to 1 in reverse order
for num in range(10, 0, -1):
In this example, the range(10, 0, -1) call generates a sequence of numbers starting from 10 and counting down to 1 (exclusive) with a step of -1.
8. Use Case: Sum of Consecutive Integers
Let’s explore a practical use case where the range() function comes in handy. Suppose you want to calculate the sum of consecutive integers within a specified range. This is a common mathematical
problem that can be solved efficiently using the range() function. Consider the task of finding the sum of integers from 50 to 100:
# Calculating the sum of integers from 50 to 100
start = 50
stop = 101
total_sum = sum(range(start, stop))
print("The sum of integers from 50 to 100 is:", total_sum)
In this example, the range(start, stop) call generates a sequence of integers starting from 50 and ending at 100 (exclusive). The sum() function calculates the sum of all integers in the generated
9. Use Case: Generating Even and Odd Numbers
Another common use case for the range() function is generating even and odd numbers within a specific range. Let’s say you need to generate a list of even and odd numbers between 1 and 20. You can
accomplish this using two separate range() calls and then converting the ranges into lists:
# Generating lists of even and odd numbers from 1 to 20
even_numbers = list(range(2, 21, 2)) # Start at 2 and step by 2
odd_numbers = list(range(1, 21, 2))
# Start at 1 and step by 2
print("Even numbers:", even_numbers)
print("Odd numbers:", odd_numbers)
In this example, the first range(2, 21, 2) call generates a sequence of even numbers, and the second range(1, 21, 2) call generates a sequence of odd numbers.
10. Conclusion
The range() function in Python is a versatile tool for generating sequences of numbers, which can be used in a variety of programming scenarios. By understanding its parameters and usage, you can
efficiently generate ranges of integers, iterate over sequences, and create lists with ease. From simple looping to solving mathematical problems, the range() function is an essential component of
every Python programmer’s toolkit. Experiment with its parameters and use cases to harness its power and streamline your coding tasks. | {"url":"https://machinelearningtutorials.org/python-range-function-a-comprehensive-tutorial-with-examples/","timestamp":"2024-11-04T01:20:25Z","content_type":"text/html","content_length":"103951","record_id":"<urn:uuid:3b1d3d18-b37b-43fe-b1cd-cf7a27ea94ab>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00396.warc.gz"} |
ThmDex – An index of mathematical definitions, results, and conjectures.
Let $X = \{ 0, 1 \}^{\mathbb{N}}$ be the
D12: Set of boolean standard sequences
such that
(i) $x : \mathbb{N} \to X$ is a D62: Sequence in $X$
(ii) $z : \mathbb{N} \to \{ 0, 1 \}$ is a D5362: Boolean Cantor diagonal sequence with respect to $x$
Then $$\forall \, n \in \mathbb{N} : x_n eq z$$ | {"url":"https://theoremdex.org/r/4058","timestamp":"2024-11-10T10:53:32Z","content_type":"text/html","content_length":"6843","record_id":"<urn:uuid:7c6dd8d3-0da2-4bb9-adb9-a905c25b7e93>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00810.warc.gz"} |
Establishing Acceptance Criteria for Analytical Methods
Knowing how method performance impacts out-of-specification rates may improve quality risk management and product knowledge.
To control the consistency and quality of pharmaceutical products, analytical methods must be developed to measure critical quality attributes (CQAs) of drug substance/drug product. Analytical method
accuracy/bias and precision are always in the path of drug evaluation and associated acceptance/failure in release testing. The following are three equations that show how the analytical method is
always influencing the quantitation of drug substance/product (Equations 1–3):
Product Mean = Sample Mean + Method Bias
[Eq. 2]
Reportable Result = Test sample true value + Method Bias + Method Repeatability
[Eq. 3]
Knowing what is the allowable contribution of the method error in drug performance becomes crucial when building product knowledge, process understanding, and the associated long-term product
lifecycle control. Mathematically, the variation of any drug product or drug substance is the additive variation of the method and test sample being quantitated.
Generally, to control the quality of a product and to manage drug safety and efficacy, there are two key elements: cinical trials evaluting the pharmacokinetics (PK) response to drug product and dose
and specification limits (1) of drug product and drug substance once clinical trials have demonstated the drug to be safe and effective. This logic is essentially laid out in two guidance documents:
International Council for Harmonization (ICH) Q6B Specifications and ICH Q9 Quality Risk Management (2).
Clearly defined method acceptance criteria that evaluate the goodness and fitness of an analytical method for its indended purpose are mandatory to correctly validate an analytical method and know
its contribution when quantitating product performance or releasing a batch. Methods with excessive error will directly impact product acceptance out-of-specification (OOS) rates and provide
misleading information regarding product quality.
Traditional Measures of Analytical Goodness and History
Historically, analytical chemists have worked on the science of an analytical method and maintained their evaluations of method goodness independent from the product they intend to evaluate.
Traditional measures of analytical goodness include the following:
• % coefficient of variation (CV) = (repeatability/mean)*100
• % recovery = (measured concentration/standard concentration)*100
• R-square of a curve comparing the theoretical concentration to the signal from the method.
This strategy has its advantages and its drawbacks. The advantage is the lab can develop and evaluate the goodness of a method independent of the product and the associated acceptance criteria it is
intended to measure. This is particularly of interest during early development when product specification limits (Q6B) are not yet available. The penalty for solely depending on CV or % recovery is a
method may be developed and qualified without knowing if it is fit-for-purpose or fit-for-use, and knowing its associated influence on product acceptance and release testing. Further, the traditional
approach will often falsely indicate a method is performing poorly at low concentrations, when in fact it is performing excellently. Conversely, at high concentrations, the method will often appear
to be performing well—as the % CV and % recovery appear to be acceptable—when it is actually unacceptable relative to the product specification limits it will be used to evaluate.
The % relative standard deviation (RSD)/% CV and % recovery should be report-only and should be included in any evaluation of an analytical method per ICH Q2 (3). Measurements that are relative to
some theoretical concentration should never be used in establishing acceptance criteria for an analytical method except when specifications are not available and should be reevaluated when they are.
In practice, no company will release to the clinic or to the market the mean or theoretical concentration; one releases every batch, tablet, vial, and syringe.
What therefore should be the basis for measurement goodness, if not comparing method performance to the mean or the theoretical concentration? The answer is simple: don’t evaluate a method relative
to the mean, evaluate it relative to the product specification tolerance or design margin it must conform to. This concept has been well established for many years in chemical, automotive, and
semiconductor industries and is recommended in the United States Pharmacopeia (USP) <1033> and <1225> (4, 5). Effectively the question is: how much of the specification tolerance is consumed by the
analytical method? Finally, how does the method contribute to OOS events when releasing product to the clinic or market?
Method error should be evaluated relative to the tolerance for two-sided limits, margin for one-sided limits, and the mean or theoretical concentration if there are no specification limits (Equations
Tolerance = Upper Specification Limit (USL) – Lower Specification Limit (LSL)
[Eq. 4]
Margin = USL – Mean or Mean – LSL (One-sided specifications)
[Eq. 5]
Mean = Average of specific concentrations of interest
[Eq. 6]
Direction from Guidance Documents
What do regulatory and standards organizations say about acceptance criteria for analytical methods? The following are brief quotes from the guidance documents regarding acceptance criteria:
• ICH Q2: Discusses what to quantitate, what to report, study design, and sample size. No mention of acceptance criteria is made in the standard, although it is implied there will be acceptance
criteria generated (3).
• FDA, Analytical Procedures and Methods Validation for Drugs and Biologics (6): “An analytical procedure is developed to test a defined characteristic of the drug substance or drug product against
established acceptance criteria for that characteristic. Early in the development of a new analytical procedure, the choice of analytical instrumentation and methodology should be selected based on
the intended purpose and scope of the analytical method. Parameters that may be evaluated during method development are specificity, linearity, limits of detection (LOD), and limits of quantitation
(LOQ), range, accuracy, and precision.”
• USP <1225>: “When validating physical property methods, consider the same performance characteristics required for any analytical procedure. Evaluate use of the performance characteristics on a
case-by-case basis, with the goal of determining that the procedure is suitable for its intended use. The specific acceptance criteria for each validation parameter should be consistent with the
intended use of the method” (5).
• USP <1033>: “The validation target acceptance criteria should be chosen to minimize the risks inherent in making decisions from bioassay measurements and to be reasonable in terms of the capability
of the art. When there is an existing product specification, acceptance criteria can be justified on the basis of the risk that measurements may fall outside of the product specification” (4).
What are Method Elements that Need Acceptance Criteria?
There are two elements for evaluating a method: determination of the result (bias, repeatability, etc.) and determination of the acceptance criteria for each element. The following is a summary of
the elements that need acceptance criteria and what elements are ‘report only’ or need to be documented in a development report (see
Table I).
Recommended Acceptance Criteria for Specificity
There are two ways to show specificity:
• Identification, demonstrate it is measuring the specific analyte and not some other protein or substance
• Bias in the presence of interfering compounds or matrices.
Acceptance criteria should be similar to accuracy or bias as a % of tolerance:
Identification, 100% detection, report detection rate and 95% confidence limits
Reportable Specificity = Measurement – Standard (units) (in the matrix of interest)
Specificity/Tolerance *100, Excellent Results <= 5%, Acceptable Results <=10%
Recommended Acceptance Criteria for Linearity
Linearity is measuring the linear response of the method. The evaluation of linearity is minimally 80-120% of the product specification limits or wider. Acceptance criteria must demonstrate the
method is linear within that range or higher. The following are techniques to demonstrate the method meets the minimum linear range of the method:
• Plot of the residuals and/or studentized residuals from a regression line
• No systematic pattern in the residuals through visual examination
• No statistically significant quadratic effect in a regression evaluation of the residuals correlated to the theoretical concentration.
To set the limit of linearity, the following is recommended. Fit a linear regression line when correlating signal versus theoretical concentration. Save the studentized residuals from the curve. Add
a line at +1.96 (95% sure the response is linear) and -1.96. Fit a quadratic fit to the studentized residuals. As long as the curve remains within +-1.96 of the studentized residuals, the response of
the assay is linear. When the curve exceeds the 1.96 limit, one is 95% sure the assay is no longer linear. For Figure 1, one is 95% sure this assay is linear up to 30 ug/mL.
Recommended Acceptance Criteria for Range
Range is established where the response remains linear, repeatable, and accurate. Acceptance criteria for the range should be based on the following: Range of the method should be less than or equal
to 120% of the USL and be demonstrated to be linear, accurate, and repeatable.
Recommended Acceptance Criteria for Repeatability
Repeatability is the standard deviation of repeated (intra-assay) measurements (see Figure 2). As repeatability error increases, the out of specification [OOS] rate increases. The following are the
recommended evaluation and acceptance criteria. Repeatability as a percentage of tolerance should be used in the evaluation.
Repeatability % Tolerance = (Stdev Repeatability*5.15)/(USL-LSL), if two-sided spec limits
Repeatability % Margin = (Stdev Repeatability*2.575)/(USL-Mean) or (Mean-LSL), if one-sided
% RSD or CV = Stdev Repeatability/Mean*100, if no limits
Recommended acceptance criteria for analytical methods for repeatability are less than or equal to 25% of tolerance. For a bioassay, they are recommended to be less than or equal to 50% of tolerance.
Figure 2: Influence of repeatability on capability (out-of-specification [OOS] rate in parts per million [PPM]). (Courtesy of author)
Recommended Acceptance Criteria for Bias/Accuracy
Accuracy or bias can only be evaluated once a reference standard has been generated. The average of the distance from the measurement-theoretical reference concentration is bias in units. Bias may be
evaluated relative to the tolerance (USL-LSL), margin, or the mean:
Bias % of Tolerance = Bias/Tolerance*100,
Bias % of Margin = Bias/(USL-Mean or Mean – LSL) One Sided
Bias % of Mean = Bias/Mean*100
Recommended acceptance criteria for analytical methods for bias are less than or equal to 10% of tolerance. For a bioassay, they are recommended to also be less than or equal to 10% of tolerance.
Recommended Acceptance Criteria for LOD and LOQ
Acceptance criteria for LOD and LOQ should also be evaluated as a percentage of tolerance or design margin:
LOD/Tolerance*100, <=5% is Excellent and <=10% is Acceptable
LOQ/Tolerance*100, <=15% is Excellent and <=20% is Acceptable
If the specification is two-sided and the LOD and LOQ are below 80% of the lower specification limit, then the LOD and LOQ are considered having no impact on product quality determination and thus
Recommended Acceptance Criteria for Intermediate Precision
Intermediate precision is the standard deviation of repeated measurements including both intra- and inter-assay sources of error. The following are the recommended evaluation and acceptance criteria.
Intermediate precision (IP) as a % of tolerance should be used in the evaluation:
IP % Tolerance = (Stdev IP*5.15)/(USL – LSL), if two-sided spec limits
IP % Margin = (Stdev IP*2.575)/(USL-Mean) or (Mean-LSL), if one-sided limit
% RSD or CV = Stdev IP/Mean*100, if no limits
Criteria for IP % of tolerance or % margin: less than or equal to 25% Excellent, less than or equal to 30% Acceptable.
IP should be evaluated at each concentration, variance components for the intra- and inter-assay error should be reported (4) and IP % CV is report only.
Bioassay IP acceptance criteria: less than or equal to 60% of tolerance.
A robustness study has no acceptance criteria; however, the robustness study should indicate the method is accurate and repeatable at the recommended best set point and across a defined range. It is
expected that the robustness study will be used to determine settings and ranges that will ensure bias less than 10% of tolerance and repeatability less than 25% of tolerance.
Reporting Stability
A stability study on critical reagents such as standards and/or bulk materials has no acceptance criteria; however, the study should indicate the expiry of pre-mixes, bulks, or standards.
Using the Accuracy to Precision Profiler in Evaluating All Acceptance Criteria
For any method, the unique combination of product variation, product average, method accuracy, method repeatability, specificity, and stability all can be evaluated by a design space. The author has
developed a SAS/JMP based tool (ATP Profiler) that can be downloaded to evaluate any method (7). The advantage is one can evaluate all of the dynamic elements of a specific method and determine the
impact of the combined acceptance criteria on potential OOS rates (see Figure 3).
Figure 3: Accuracy to precision modeling. (Courtesy of author)
Moving from relative measures of analytical method goodness to measures that have product relevance links method performance to CQAs and their associated specification limits in a way that nothing
else will. Knowing how method performance impacts OOS rates adds to better quality risk management and product knowledge. Setting acceptance criteria based on OOS rate impact is more meaningful and
supported by both FDA and USP guidance. %CV and %Recovery should always be included in development reports and method validation documents as report only and should not form the basis of acceptance
1. ICH, Q6B Specifications: Test Procedures and Acceptance Criteria for Biotechnological/Biological Products (ICH, March 1999).
2. ICH, Q9 Quality Risk Management (ICH, 2006).
3. ICH, Q2(R1) Validation of Analytical Procedures: Text and Methodology (ICH, November 2005).
4. USP, <1033> Biological Assay Validation, USP 38 (USP, 2010).
5. USP, <1225> Validation of Compendial Procedures, USP 38 (USP, 2015).
6. FDA, Analytical Procedures and Methods Validation for Drugs and Biologics, Guidance for Industry (CDER, July 2015).
7. T. Little, Accuracy to Precision (ATP) Profiler. | {"url":"http://www.processdevelopmentforum.com/resources/articles/establishing-acceptance-criteria-for-analytical-methods/","timestamp":"2024-11-13T00:04:35Z","content_type":"text/html","content_length":"270476","record_id":"<urn:uuid:0c6cdc6b-affe-4197-8660-979e5ab8e222>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00035.warc.gz"} |
First observation of the decay (B)over-bar(S)(0) -> (DK)-K-0*(0) and a measurement of the ratio of branching fractions B((B)over-bar(S)(0)->(DK)-K-0*(0))/B((B)over-bar(0)-> D-0 rho(0))
The first observation of the decay (B) over bar (0)(S) -> (DK)-K-0*(0) using pp data collected by the LHCb detector at a centre-of-mass energy of 7 TeV, corresponding to an integrated luminosity of
36 pb(-1), is reported. A signal of 34.4 +/- 6.8 events is obtained and the absence of signal is rejected with a statistical significance of more than nine standard deviations. The (B) over bar (0)
(S) -> (DK)-K-0*(0) branching fraction is measured relative to that of (B) over bar (0) -> D-0 rho(0): B((B) over bar (0)(S)->(DK)-K-0*(0))/B((B) over bar (0)-> D-0 rho(0)) = 1.48+/-0.34+/-0.15+/
-0.12, where the first uncertainty is statistical, the second systematic and the third is due to the uncertainty on the ratio of the B-0 and B-s(0) hadronisation fractions. (C) 2011 CERN. Published
by Elsevier B.V. All rights reserved.
Dive into the research topics of 'First observation of the decay (B)over-bar(S)(0) -> (DK)-K-0*(0) and a measurement of the ratio of branching fractions B((B)over-bar(S)(0)->(DK)-K-0*(0))/B((B)
over-bar(0)-> D-0 rho(0))'. Together they form a unique fingerprint. | {"url":"https://cris.maastrichtuniversity.nl/en/publications/first-observation-of-the-decay-bover-bars0-gt-dk-k-00-and-a-measu","timestamp":"2024-11-14T04:03:31Z","content_type":"text/html","content_length":"58161","record_id":"<urn:uuid:2c586ec1-09b7-48d8-8114-63130a35164d>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00009.warc.gz"} |
Return On Investment Ratio | Formula | Calculator (Updated 2023)
This is an ultimate guide on how to calculate Return on Investment Ratio (ROI) with detailed interpretation, analysis, and example. You will learn how to use this ratio formula to evaluate a business
Definition - What is Return on Investment Ratio?
The return on Investment (ROI) is a metric that measures the efficiency and return of an investment.
It’s a simple ratio between the money earned on an investment and the initial cost of the investment.
Investors can use the ratio to compare various potential investments, and decide which is the most profitable.
The equation for ROI ratio is as follows:
Return on Investment = (Investment Revenue - Cost of Investment) / Cost of Investment
To calculate this ratio, you simply subtract the initial cost of the investment from total value of the investment at the end of the investment period, and divide that number by the initial cost of
the investment.
An easier formula to remember is the following:
Return on Investment = Gain from Investment ÷ Initial Investment
In this formula, you merely take the monetary gain from the investment, and divide it by the initial investment.
To make this idea clearer, here is an example: Suppose you have two prospective investment projects called Project HHH and Project JJJ.
Project HHH requires a $50 million investment, and expects a $35 million and $60 million return in year 1 and year 2, respectively.
Project JJJ requires a $100 million investment, with returns of $75 million and $50 million in year 1 and year 2, respectively.
Project HHH Project JJJ
Initial Investment $50M $100M
Investment Revenue $95M $125M
Return on Investment 90.00% 25.00%
Plugging in the numbers in the equation above, we can calculate that the ROI for Project HHH is 90%, and the ROI for Project JJJ is 25%.
Interpretation & Analysis
The above example shows why this ratio can be a powerful metric.
At quick glance, Project JJJ seems like a better investment because you obtain a $125 million return, as opposed to $95 return from Project HHH.
But the rate of return on investment measures the percentage return.
It takes the initial investment into account to show investors what percentage of the money originally invested will be returned over the life of the investment.
So, in the above example, the ROIs of the two projects shows that Project HHH will return 90% of your original investment, while Project HHH will only return 25%.
This ratio allows you to look past the monetary returns, and analyze the percentage returns of investments.
Cautions & Further Explanation
While the ROI ratio is a valuable tool that aids investors in their quest for higher profits, it has its limitations.
For example, this ratio does not take time into account.
Two projects that have the same initial investment and same monetary gain will have identical ROIs regardless of how long the investment horizons are for the projects.
If one project returned $1 million a year for 100 years, and one project returned $100 million in one year, they would both have the same ROI.
But it doesn’t take a math whiz to know that an investment that returns your money quickly is more attractive than one that repays you over a long period of time.
So this ratio is looked at as more of a “quick-and-dirty” method of calculating return, while CAGR and IRR are considered to be more the more comprehensive metrics that analyze investment returns. | {"url":"https://wealthyeducation.com/return-on-investment-ratio/","timestamp":"2024-11-14T10:38:01Z","content_type":"text/html","content_length":"482154","record_id":"<urn:uuid:c148dcbb-2863-4b20-856d-6ccadfbaa195>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00533.warc.gz"} |
Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.6 with sample standard deviation s = 2.9. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.What is the value of the sample test statistic? (Round your answer to three decimal places.)
Answer:2.304Step-by-step explanation:Given that x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood).[tex]n=31\\\bar x = 8.6\\s=2.9\\se = \frac{2.9}{\sqrt{31}
} \\=0.5209[/tex][tex]H_0: \bar x =7.4\\H_a: \bar x \neq 7.4[/tex](Two tailed test at 5% level)Mean difference = [tex]8.6-7.4=1.2[/tex]b) Here we can use only t test since population std deviation is
not known.df = n-1 = 30Test statistic t = mean diff/std error = 2.304p value = 0.028 | {"url":"https://thibaultlanxade.com/general/let-x-be-a-random-variable-that-represents-the-ph-of-arterial-plasma-i-e-acidity-of-the-blood-for-healthy-adults-the-mean-of-the-x-distribution-is-u-7-4-a-new-drug-for-arthritis-has-been-developed-however-it-is-thought-that-this-drug-may","timestamp":"2024-11-04T14:57:18Z","content_type":"text/html","content_length":"31939","record_id":"<urn:uuid:1281006f-2daf-4c9b-b182-4ee9dcb50bae>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00012.warc.gz"} |
Whats an index form
Hi Michelle. Any positive integer can be written as a unique product of prime numbers. For example, to break 540 down to its prime factors, I simply start dividing May 12, 2016 Both "indexes" and
"indices" are acceptable plural forms of the word "index." Index is one of those rare words that have two different plurals in
Hi Michelle. Any positive integer can be written as a unique product of prime numbers. For example, to break 540 down to its prime factors, I simply start dividing May 12, 2016 Both "indexes" and
"indices" are acceptable plural forms of the word "index." Index is one of those rare words that have two different plurals in whats the highest place value a number can go up to? Reply. Apr 1, 2013
The same rule is applied for fractional indices when we multiply the numbers as index form with the same base. If we say, 71/2× 71/3. Add the Apr 28, 2012 We still do Index Form and Expanded the
Form the same way as we do for numbers. However because we have letters, we cannot make a Dec 22, 2015 Boardworks Ltd 20056 of 70 Calculations involving standard form What is 2 × 105 multiplied by
7.2 × 103 ? To multiply these numbers together What Is Index Form in Mathematics? When a number is expressed with exponents, or one number to a power of another, it is considered to be in index
form. For example, 27 can be written in index form as 3^3.
Apr 1, 2013 The same rule is applied for fractional indices when we multiply the numbers as index form with the same base. If we say, 71/2× 71/3. Add the
Covers expressing number in index form. We know that: So, we can write 8 as 2 3.. Likewise, 27 can be written as 3 3 and 125 can be written as 5 3.. So far, we have considered numbers that have a
group of the same factors.Sometimes, a number has more than one group of the same factors as shown in the following example. Index form is where a number is expressed using exponents - one number to
the power of another. For example, 2 x 2 x 2 x 2 x 2 is the same as 2 to the power of 5, written as 2^5. If you square a Standard index form is also known as standard form. A number is said to be
written in standard form when it is written as , where is a number greater than or equal to , but strictly less than , and Index form, base, index, basic numeral, index laws. Check to see if the
number has an integer root: The index form is a root (r) raised to a power of n. Also, we multiply r (n) times. In this case, r = 2 and n = 5. what is the index form for 125?
May 10, 2013 Writing Numbers in Index Form We know that: So, we can write 8 as 23. Likewise, 27 can be written as 33 and 125 can be written as 53. So far
What is plural form of index? Indexes or indices? Find out how to use these two terms with definitions and example at Writing Explained. To format the page numbers that will appear in the index,
select the Bold check box or Italic check box below Page number format. Click Mark to mark the index
Assumed Knowledge. Knowledge of the index laws for positive integer powers. and so the expression is also often referred to as an indeterminant form. EXAMPLE. (3a2b)0 What is the answer if the
interest is paid monthly? return to top.
power (or index). The number 10 is a good place to start learning how powers work as the numbers produced when multiplying 10 by itself are easy to calculate .
Jun 8, 2019 The last notation in (3) is more general, because the index set I can be generalize the indexed notation, let us give some examples with the sigma form. (2) What is the machine m that
manufactures the cheapest product p
What is plural form of index? Indexes or indices? Find out how to use these two terms with definitions and example at Writing Explained. To format the page numbers that will appear in the index,
select the Bold check box or Italic check box below Page number format. Click Mark to mark the index
Request for Taxpayer Identification Number (TIN) and Certification. Form 4506-T. Request for Transcript of Tax Return | {"url":"https://topbitxnhyh.netlify.app/poplawski26897ru/whats-an-index-form-cu","timestamp":"2024-11-06T15:26:41Z","content_type":"text/html","content_length":"30743","record_id":"<urn:uuid:57d14209-4208-4eb0-8bc9-c3b3a614c092>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00536.warc.gz"} |
CCO '24 P1 - Treasure Hunt
Perry the Pirate is sailing the seven seas! He has a map consisting of islands connected by a network of sea routes. The -th sea route connects islands and and costs coins to traverse in either
direction. As it turns out, fighting off sea monsters can be quite expensive. In search of his next big plunder, Perry has scouted out each of the islands and has determined that the -th island
contains a treasure chest with coins inside.
It remains for him to plan out his next journey. He decides that he will sail through some (possibly empty) path of sea routes starting at island and ending at island . At the end of his journey, he
will open the chest at island and collect his well-earned booty.
There is one small problem though: Perry doesn't know what island he's currently on! Thus, for every possible starting island , he would like to know the maximum possible number of coins he can earn
out of all journeys starting at island . Can you help him compute these values? You may assume Perry has enough coins to traverse any path of sea routes he chooses; he only cares about the net profit
of his next journey.
Input Specification
The first line of input contains two space-separated integers and .
The second line of input contains space-separated integers .
The next lines each contain three space-separated integers , and .
It is guaranteed that there is at most one sea route between any pair of islands and each sea route connects two distinct islands.
Marks Awarded Bounds on N Bounds on M Additional constraints
5 marks None
5 marks For all , either or
7 marks Exactly one path of sea routes between any pair of islands
8 marks None
Output Specification
Output lines, where the -th line contains the maximum possible net profit (in coins) of any journey starting at island .
Sample Input 1
Sample Output 1
Explanation for Sample 1
For the first and third islands, it is best to just stay and open the chest on the island itself.
For the second island, Perry can travel to the first island and open the chest there.
This has a net profit of coins and is the best possible net profit.
For the fourth island, Perry can travel to the second and then the third island and open the chest there.
This has a net profit of coins and is the best possible net profit.
• commented on Aug. 18, 2024, 7:37 p.m. ← edited | {"url":"https://dmoj.ca/problem/cco24p1","timestamp":"2024-11-12T13:45:37Z","content_type":"text/html","content_length":"43395","record_id":"<urn:uuid:cd2d77dd-8700-4da1-b2cc-7f602c3798c8>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00571.warc.gz"} |
Thermo101 Drawratio - SPE Thermoforming Division
A TECHNICAL ARTICLE 2005 VOLUME 24, #4
The Ubiquitous1 Draw Ratio
Probably the first thing a
novice hears in thermo-
forming after he/she learns to
spell thermoforming, is the
phrase, Draw Ratio. So, this
lesson focuses on the concept of
draw ratio.
Is There More Than One
Unfortunately, yes. There are
at least three definitions. Let s
define the common ones.
Areal Draw Ratio, often given
the symbol RA, is the ratio of the
area of the part being formed
to the area of the sheet needed
to make the part. Although I
promised not to use equations
in our TF 101 lessons, some
simple ones here won t hurt all
that much:
RA = AreaPart/AreaSheet
A simple example, please?
Consider a cylinder one unit in
diameter by one unit high. The
area of the cylinder is (.+./4)
= 5..4. The area of the sheet
used to form the cylinder is
./4. Therefore the areal draw
ratio, RA, is 5. As an interesting
aside, the reciprocal of the
areal draw ratio is the average
reduced thickness of the formed
part, being 1/5 = 0.20. In other
words, the original sheet thickness
has been reduced by 80%,
on the average.
Linear Draw Ratio, often
given the symbol RL, is the ratio
of the length of a line scribed
on the part surface to the original
length of the line. Again, in
equation form:
RL = LinePart/LineSheet
For the same example, the length
of the line on the cylinder is
(1+1+1) = 3. The original length
of the line is 1. Therefore, the
linear draw ratio, RL, is 3. The
linear draw ratio is akin to
the way in which the plastic
is stretched in a tensile test
Height-to-Diameter Ratio,
often written as H:D, is the
height of the cylinder (1), to the
diameter of the cylinder (1). Or
H:D = 1. H:D is used primarily
for axisymmetric2 parts such as
cones or cylinders, such as drink
In summary, for the cylinder
described above, RA=5, RL = 3,
and H:D = 1. So you see, there
is no agreement between these
Are Draw Ratios of
Use? Importance?
So, which one do we use?
Depends. First, we need to
determine whether draw ratio is
a useful concept.
Let s focus on areal draw
ratio to determine its utility.
As we have already learned,
the reciprocal of RA is the average
reduced thickness. But
where is this reduced thickness?
Somewhere down the side of
the formed part. In fact, there
is probably a line around the
periphery of the part where
the part thickness is exactly the
average reduced thickness. So,
what does this tell us about
the uniformity of the part wall
thickness? Or the degree of difficulty
in forming the part? Or
whether webs are formed somewhere
in the part? Or what the
plug needs to look like? Or ?
Really, nothing.
Having said that, areal draw
ratio is perhaps the easiest
concept to understand. Linear
draw ratio, as noted, is often
compared with extension limits
determined from tensile
testing equipment. And H:D is
often used in Europe to describe
formability of plastics for cup
At best, draw ratios represent
bragging rights rather than information
about the degree of
difficulty in forming the parts.
Many formers will tell you that
parts that have very small draw
ratios are much more difficult
to form reliably than parts with
large draw ratios. And parts
with many compartments are
far more difficult to form than
parts with single compartments,
even when the draw ratios of the
two types are identical.
[See? Those equations didn t
hurt at all, now, did they?] ¦
Keywords: Areal draw ratio,
linear draw ratio, H:D
1 Ubiquitous: Being present everywhere
at once.
2 Axisymmetric: Having symmetry around
an axis.
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Selection Sort
Selection sort is a simple sorting algorithm. This sorting algorithm is an in-place comparison- based algorithm in which the list is divided into two parts, the sorted part at the left end and the
unsorted part at the right end. Initially, the sorted part is empty and the unsorted part is the entire list.
The smallest element is selected from the unsorted array and swapped with the leftmost element, and that element becomes a part of the sorted array. This process continues moving unsorted array
boundary by one element to the right.
This algorithm is not suitable for large data sets as its average and worst case complexities are of
Ο (n2), where n is the number of items.
How Selection Sort Works?
Following are the steps involved in selection sort (for sorting a given array in ascending order):
1. Starting from the first element, we search the smallest element in the array, and replace it with the element in the first position.
2. We then move on to the second position, and look for smallest element present in the subarray, starting from index 1, till the last index.
3. We replace the element at the second position in the original array, or we can say at the first position in the subarray, with the second smallest element.
4. This is repeated, until the array is completely sorted.
Let's consider an array with values {14, 33, 27, 10, 35, 19, 24, 44}
For the first position in the sorted list, the whole list is scanned sequentially. The first position where 14 is stored presently, we search the whole list and find that 10 is the lowest value.
So we replace 14 with 10. After one iteration 10, which happens to be the minimum value in the list, appears in the first position of the sorted list.
For the second position, where 33 is residing, we start scanning the rest of the list in a linear manner.
We find that 14 is the second lowest value in the list and it should appear at the second place. We swap these values.
After two iterations, two least values are positioned at the beginning in a sorted manner.
The same process is applied to the rest of the items in the array. Following is a pictorial depiction of the entire sorting process
Now, let us learn some programming aspects of selection sort. | {"url":"https://benchpartner.com/selection-sort","timestamp":"2024-11-09T17:21:12Z","content_type":"text/html","content_length":"82124","record_id":"<urn:uuid:38906d20-b920-40f3-9090-e427938d3b32>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00471.warc.gz"} |
LibGuides: Finding & Using Data : Key Concepts & Terminology
Quantitative data/Quantitative variables: Information that can be handled numerically.
Qualitative data/Qualitative variables: Information that refers to the quality of something. Ethnographic research, participant observation, open-ended interviews, etc., may collect qualitative data.
Some element of the results obtained via qualitative research may be handled numerically, eg, how many observations, number of interviews, etc.
Time series data: Any data arranged in chronological order.
Longitudinal data: data that is collected repeatedly over a period of time, in which the same group of respondents are surveyed each time.
Discrete data: numeric data that have a finite number of possible values (1,2,3,4,5)
Continuous data: data that has an infinite number of possible values (1.4, 1.41, 1.414, etc.)
Levels of Measurement
Nominal: Nominal data have no order and only gives names or labels to various categories (yellow, white, pink).
Ordinal: Ordinal data have order, but the interval between measurements is not meaningful (low, medium, high).
Interval: Interval data have meaningful intervals between measurements, but there is no true starting point (Fahrenheit temperature scale).
Ratio: Ratio data have the highest level of measurement. Ratios between measurements as well as intervals are meaningful because there is a starting point (Kelvin temperature scale). | {"url":"https://libguides.ucmerced.edu/finding-and-using-data/basics","timestamp":"2024-11-09T07:08:12Z","content_type":"text/html","content_length":"61166","record_id":"<urn:uuid:772ca467-34a7-4371-a80b-87d51a878b9a>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00840.warc.gz"} |
Calculus: Early Transcendentals 9th Edition Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 191 63
(a) First, set aside $(fg)$ and apply the Product Rule for $(fgh)'$ with 2 elements $(fg)$ and $h$. Then apply the Product Rule one more time for $(fg)'$ with 2 elements $f$ and $g$. (b) First, we
find that $f^3=fgh$ if $f=g=h$. Then we can apply the result from part a). Remember that in part b), $f=g=h$ (c) $$\frac{dy}{dx}=3e^{3x}$$
Work Step by Step
(a) $f$, $g$ and $h$ are differentiable, then $f'$, $g'$ and $h'$ exist. Consider $(fgh)'$ $$(fgh)'=[(fg)\times h]'$$ Apply the Product Rule, $$(fgh)'=(fg)'h+(fg)h'$$ Apply the Product Rule again for
$(fg)'$, $$(fgh)'=(f'g+fg')h+fgh'$$ $$(fgh)'=f'gh+fg'h+fgh'$$ The statement has been proved. (b) Taking $f=g=h$, we can see that $$fgh=f^3$$ So, $$(f^3)'=(fgh)'=f'gh+fg'h+fgh'$$ However, since $f=g=
h$, $$(f^3)'=f'ff+ff'f+fff'$$ $$(f^3)'=3f'ff=3f'f^2$$ Therefore, $$\frac{d}{dx}[f(x)]^3=3[f(x)]^2f'(x)$$ The statement, as a result, has been proved. (c) $$y=e^{3x}=(e^x)^3$$ So, $$\frac{dy}{dx}=\
frac{d}{dx}(e^x)^3$$ $$\frac{dy}{dx}=3(e^x)^2(e^x)'$$ $$\frac{dy}{dx}=3e^{2x}e^x=3e^{3x}$$ | {"url":"https://www.gradesaver.com/textbooks/math/calculus/CLONE-1669a839-2df3-45c5-8bce-02a5db7c0ba8/chapter-3-section-3-2-the-product-and-quotient-rules-3-2-exercises-page-191/63","timestamp":"2024-11-10T12:03:22Z","content_type":"text/html","content_length":"77303","record_id":"<urn:uuid:5673f36a-cdc3-49b3-955f-8b2cec35d31d>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00159.warc.gz"} |
D.K.Faddeev's biography
Dmitrii Konstantinovich Faddeev
Dmitrii Konstantinovich Faddeev was born on 17 (30) June 1907 in Yukhnov, in the district of Smolensk, in a family of a Petersburg engineer. From 1923, when he enrolled as a student of mathematics,
until the end of his life all Faddeev's varied activity has been closely linked with the University of Leningrad, where for many years he held the chair of higher algebra and number theory, and where
he has been Dean of the Faculty of Mathematics and Mechanics for several years. His links with the Academy of Sciences of the USSR were no less close. In 1932 he began research in the Steklov
Institute of Mathematics and Physics and when in 1934 this institute moved from Leningrad to Moscow and became the Mathematical Institute, and the Leningrad Division of the Steklov Institute of
Mathematics of the Academy of Sciences of the USSR (LOMI) was created, Faddeev was from the very moment of its foundation a collaborator in LOMI and he has remained so. In 1964 he was elected
Corresponding Member of the Mathematical Section of the Academy of Sciences of the USSR. For a long time he was in charge of the laboratories of at LOMI, and head of a powerful group of well-known
algebraists and number theorists. His algebra seminar was widely known throughout the whole country.
The range of Faddeev's interests is unusually broad---among his more than 150 papers are some on the theory of functions, on computational methods, on probability theory, on problems of teaching
mathematics at all levels. But, of course, Faddeev is known in the first place as one of the most outstanding algebraists of our time.
In his papers Faddeev touched on a wide range of algebraic problems. But there are two areas in which he began his research and to which he constantly returned. They are Diophantine equations and
Galois theory. At the end of 50's Faddeev turned to problems in the theory of integral representations. Faddeev is one of the leading experts in numerical methods of linear algebra.
A distinctive feature of Faddeev's approach to the teaching of mathematics in schools is clearly expressed and logically developed chain of ideas in the courses he devised on algebra and basic
analysis. Faddeev gives priority to "vivid contemplation", to awareness, to direct perception of the properties of mathematical objects, which thus reveal the content and basic ideas of mathematics
as tools and as a description and the study of the regularity of the real world. His teaching over many years at the University, his general and specialized courses of lectures, always excellently
presented, his numerous public speeches, are a splendid example to younger mathematicians.
Dmitrii Konstantinovich Faddeev was a man of the broadest culture and intelligence. He had a wide knowledge of and appreciation of classical music and was an outstanding pianist. Discussions with him
of a variety of questions were highly estimated by all his friends, colleagues, and acquaintances.
This text is an excerpt from the paper in Uspekhi Mat. Nauk 44:3 (1989) by A.D.Aleksandrov, M.I.Bashmakov, Z.I.Borevich, V.N.Kublanovskaya, M.S.Nikulin, A.I.Skopin, A.V.Yakovlev. | {"url":"http://mathsoc.spb.ru/pantheon/faddeev/bio.html","timestamp":"2024-11-15T04:21:17Z","content_type":"text/html","content_length":"4216","record_id":"<urn:uuid:2cb07a9a-0b8e-4d01-926e-b74027c477d7>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00458.warc.gz"} |
▷ What is the Leading Term of a Polynomial? (examples)
Leading term of a polynomial
On this post we explain what is the leading term of a polynomial. Also, you will see several examples on finding the leading term of a polynomial.
What is the leading term of a polynomial?
The definition of the leading term of a polynomial is as follows:
The leading term of a polynomial is the term with the highest degree of the polynomial, that is, the leading term of a polynomial is the term that has the x with the highest exponent.
For example, the leading term of the following polynomial is 5x^3:
The highest degree element of the above polynomial is 5x^3 (monomial of degree 3), therefore that is the leading term of the polynomial.
On the other hand, the coefficient of the leading term is called the leading coefficient of a polynomial. So, following the previous example, the leading coefficient of the polynomial would be 5.
Also, the leading term of a polynomial is used to identify when a polynomial is monic. In the following link you can see what a monic polynomial is.
Examples of how to find the leading term of a polynomial
Now that we know how to identify the leading term of a polynomial, we are going to practice with several examples.
• Example of the leading term of a polynomial of degree 5:
The leading term of the polynomial is 2x^5 because it is the term with the highest power of x.
• Example of the leading term of a polynomial of degree 6:
The term with the maximum degree of the polynomial is x^6, so that is the leading term of the polynomial. Remember that if the variable is not accompanied by any number, it means that the coefficient
is 1, consequently, the leading coefficient of this polynomial is 1.
Note that if the polynomial is in standard form, the leading term is the first term in the polynomial.
• Example of the leading term of a polynomial of degree 9:
The term of the polynomial whose exponent is the highest is -3x^9, so the leading term of the polynomial is -3x^9. Note that the negative sign is also part of the leading term.
• Example of the leading term of a polynomial with two variables:
The leading term of the polynomial is -2x^3y^4, since it is the highest degree monomial of the polynomial.
In this exercise we must pay attention since the degree of a term with two variables is not calculated in the same way as the degree of a term with only one variable.
Leave a Comment | {"url":"https://www.algebrapracticeproblems.com/leading-term-of-a-polynomial/","timestamp":"2024-11-06T19:07:48Z","content_type":"text/html","content_length":"178276","record_id":"<urn:uuid:6d7f58fe-166c-4cbc-8215-7d795e90d915>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00643.warc.gz"} |
Basic Syntax Extensions
These extensions simply enhance Haskell’s core syntax without providing any actually new semantic features.
Available in: All recent GHC versions
The PostfixOperators extension allows you some slight extra leeway with Haskell’s operator section syntax. Normally, when you write, for example:
(4 !)
it expands into:
\x -> 4 ! x
or, equivalently:
\x -> (!) 4 x
PostfixOperators instead expands this left section into:
(!) 4
which may look the same to you initially, and it behaves the same way where they both compile, but the new form allows GHC to be somewhat more lenient about the type of (!).
For example, (!) can now be the factorial function and have the type:
(!) :: Integer -> Integer
Unfortunately, PostfixOperators does not allow you to define operators in postfix fashion, it just allows you to use them that way.
Try it out!
{-# LANGUAGE PostfixOperators #-}
(!) :: Integer -> Integer
(!) n | n == 0 = 1
| n > 0 = n * ((n - 1) !)
| otherwise = error "factorial of a negative number"
main = print (4 !)
Available in: GHC 6.12 and later
The TupleSections extension allows you to omit values from the tuple syntax, unifying the standard tuple sugar with the tuple constructor syntax to form one generalized syntax for tuples.
Normally, tuples are constructed with the standard tuple sugar, which looks like this:
(1, "hello", 6.5, Just (), [5, 5, 6, 7])
This could be considered shorthand for the following explicit tuple constructor use:
(,,,,) 1 "hello" 6.5 (Just ()) [5, 5, 6, 7]
However, the explicit tuple constructor (,,,,) could just as easily be considered section sugar for tuples, expanding to:
\v w x y z -> (v, w, x, y, z)
Looking at it this way allows us to ask, “Why can’t we partially section a tuple? After all, (+) is valid, (,) is valid, and (1 +) is valid, but (1,) is not valid. The TupleSections extension fixes
this oversight.
With TupleSections you can now write, for example:
(1, "hello",, Just (),)
and have it mean the same as
\x y -> (1, "hello", x, Just (), y)
Try it out!
{-# LANGUAGE TupleSections #-}
main = print $ map (1, "hello", 6.5,, [5, 5, 6, 7]) [Just (), Nothing]
Available in: All recent GHC versions
Let’s say you want to import module Data.Module.X from package package-one, but package-two is also installed and also contains a module named Data.Module.X. You could try to mess with package
hiding, either manually or through cabal, but sometimes you might want some other module from package-two, so hiding it is not an option.
Enter the PackageImports extension. Rather than writing:
import Data.Module.X
and hoping that GHC gets the one from the right package, PackageImports lets you write:
import "package-one" Data.Module.X
and explicitly specify the package you want to import that module from. You can even import from a specific package version:
import "package-one-0.1.0.1" Data.Module.X
You can use PackageImports in combination with any other variant of the import syntax, and you can use both package-qualified imports and regular imports in the same file.
Try it out!
{-# LANGUAGE PackageImports #-}
import Data.Monoid (Sum(..))
import "base" Data.Foldable (foldMap)
import qualified "containers" Data.Map as Map
main = print . getSum . foldMap Sum $ Map.fromList [(1, 2), (3, 4)]
Available in: All recent GHC versions
By default, Haskell’s numeric literals are polymorphic over Num (in the case of integer literals) or Fractional (in the case of decimal literals). That is, you can write:
a :: Int
a = 1
b :: Double
b = 1
c :: Float
c = 3.5
d :: Rational
d = 3.5
and it just works as expected.
String literals, on the other hand, are always of type String, and are not polymorphic at all. The OverloadedStrings extension corrects this, making string literals polymorphic over the IsString type
class, which is found in the Data.String module in the base package. That is, you can write:
a :: String
a = "hello"
b :: Text
b = "hello"
OverloadedStrings also adds IsString to the list of defaultable type classes, so you can use types like String, Text, and Bytestring in a default declaration.
Try it out!
{-# LANGUAGE OverloadedStrings #-}
import qualified Data.Text.IO as T
main = do putStrLn "Hello as String!"
T.putStrLn "Hello as Text!"
Available in: All recent GHC versions
With the UnicodeSyntax extension (along with the base-unicode-symbols and containers-unicode-symbols packages), you can use Unicode alternatives to many of the standard operators. The UnicodeSyntax
extension itself handles just the operators and symbols that are built into the Haskell language, whereas the base-unicode-symbols package handles the operators and functions provided by the base
package and the containers-unicode-symbols package handles the operators and functions provided by the containers package.
For the package-based Unicode symbols, you need to import the appropriate syntax module. For example, if you wanted to use Unicode symbols when working with Data.Map, you would import
The various aliased ASCII syntax pieces, values, and types, along with their UnicodeSyntax equivalents, are as follows:
• From the UnicodeSyntax extension
□ :: = ∷
□ => = ⇒
□ forall = ∀
□ -> = →
□ <- = ←
□ -< = ⤙
□ >- = ⤚
□ -<< = ⤛
□ >>- = ⤜
□ * = ★
• From the base-unicode-symbols package
• From the containers-unicode-symbols package
Try it out!
{-# LANGUAGE UnicodeSyntax #-}
import Data.List.Unicode ((∪))
import qualified Data.Map as M
import Data.Map.Unicode ((∆))
main ∷ IO ()
main = do print $ [1, 2, 3] ∪ [1, 3, 5]
print $ M.fromList [(1, 2), (3, 4)] ∆ M.fromList [(3, 4), (5, 6)]
Available in: All recent GHC versions
The RecursiveDo extension (as well as its deprecated synonym DoRec) enables syntactic sugar for value recursion in a monadic context. “What on Earth does that mean?” you might ask. To explain, let’s
take a look at how let behaves in Haskell.
let in Haskell allows lazy recursion; that is, you can write:
main = print $
-- show
let x = fst y
y = (3, x)
in snd y
-- /show
However, do in Haskell does not allow lazy recursion; in fact, it doesn’t allow recursion at all. If you try to write a recursive binding in do notation, it will fail; for example, the following code
will cause an error that complains about y not being in scope:
{-# LANGUAGE StandaloneDeriving #-}
import Control.Monad.Identity
deriving instance (Show a) => Show (Identity a)
main = print ((
-- show
do x <- return $ fst y
y <- return (3, x)
return $ snd y
-- /show
) :: Identity Integer)
However, sometimes we want to be able to use value recursion but still need to be within a monad. The MonadFix type class, from the Control.Monad.Fix module in the base package, provides an mfix
function that helps us do exactly that, but the results, while they work, are not very pretty:
{-# LANGUAGE StandaloneDeriving #-}
import Control.Monad.Identity
deriving instance (Show a) => Show (Identity a)
main = print ((
-- show
do y <- mfix $ \y0 -> do x <- return $ fst y0
y1 <- return (3, x)
return y1
return $ snd y
-- /show
) :: Identity Integer)
The RecursiveDo extension provides sugar for using mfix this way, so that the previous example can be equivalently rewritten as:
{-# LANGUAGE RecursiveDo, StandaloneDeriving #-}
import Control.Monad.Identity
main = print ((
-- show
mdo x <- return $ fst y
y <- return (3, x)
return $ snd y
-- /show
) :: Identity Integer)
RecursiveDo also provides a second type of syntactic sugar for mfix that uses the rec keyword instead of the mdo keyword. The rec-based sugar is somewhat more direct and “low-level” than the
mdo-based sugar. In terms of the rec sugar, our running example is expressed as:
{-# LANGUAGE RecursiveDo, StandaloneDeriving #-}
import Control.Monad.Identity
main = print ((
-- show
do rec x <- return $ fst y
y <- return (3, x)
return $ snd y
-- /show
) :: Identity Integer)
The two types of sugar are subtly different in meaning, and the difference has to do with something called segmentation.
When GHC encounters a let binding, rather than naïvely binding all of the variables at once, it will divide (or segment) them into minimal mutually-dependent groups. For example, take this
let x = 1
y = (x, z)
z = fst y
v = snd w
w = (v, y)
in (snd y, fst w)
Instead of just binding everything in a single group, GHC improves the code’s efficiency somewhat by treating it as though you’d actually written something like:
let x = 1
in let y = (x, z)
z = fst y
in let v = snd w
w = (v, y)
in (snd y, fst w)
In a pure let binding, the only way this might matter is performance; the semantics of the code is guaranteed to not change. However, segmenting monadic code might produce unexpected results, because
mfix has to deal with the monadic context somehow during the value recursion, and segmenting a set of bindings into minimal groups could potentially change the meaning of the code.
Only mdo segments its bindings. rec does no segmentation at all, instead translating to calls to mfix exactly where you put recs in the original code. This means that, in the following example, the
first two of the following three expressions are equivalent to each other, but the third one is not equivalent to either of the first two:
-- | expression 1 (equivalent to expression 2)
mdo x <- return 1
y <- return $ (x, z)
z <- return $ fst y
v <- return $ snd w
w <- return (v, y)
return (snd y, fst w)
-- | expression 2 (equivalent to expression 1)
do x <- return 1
rec y <- return $ (x, z)
z <- return $ fst y
rec v <- return $ snd w
w <- return (v, y)
return (snd y, fst w)
-- | expression 3 (not equivalent to expression 1 or expression 2)
do rec x <- return 1
y <- return $ (x, z)
z <- return $ fst y
v <- return $ snd w
w <- return (v, y)
return (snd y, fst w)
Both expression 1 and expression 2 translate roughly to:
do x <- return 1
(y, z) <- mfix $ \(y0, z0) -> do y1 <- return $ (x, z0)
z1 <- return $ fst y0
return (y1, z1)
(v, w) <- mfix $ \(v0, w0) -> do v1 <- return $ snd w0
w1 <- return (v0, y)
return (v1, w1)
return (snd y, fst w)
On the other hand, expression 3 translates roughly to:
do (x, y, z, v, w) <- mfix $ \(x0, y0, z0, v0, w0) -> do x1 <- return 1
y1 <- return $ (x0, z0)
z1 <- return $ fst y0
v1 <- return $ snd w0
w1 <- return (v0, y0)
return (x1, y1, z1, v1, w1)
return (snd y, fst w)
Try it out!
{-# LANGUAGE RecursiveDo #-}
import Control.Monad.State.Lazy
comp = do x0 <- get
modify (+1)
x1 <- get
rec y <- return $ (x0, fst z)
z <- return $ (x1, fst y)
put 3
return (y, z)
main = print $ runState comp 1
WARNING: In GHC versions before 7.6, there was a lot of churn in the meanings of the RecursiveDo and DoRec extensions and their relationship to each other. For such older GHC versions, the above
discussion may be partially or wholly inaccurate; consult your GHC version’s User’s Guide for more detailed information.
Available in: GHC 7.6 and later
The LambdaCase extension is very simple. Any time you would otherwise have written:
\x -> case x of ...
you can instead simply write
\case ...
which is both shorter and doesn’t bind x as a name. The Layout Rule works as usual with LambdaCase, so, for example:
[Just 1, Just 2, Nothing, Just 3] `forM_` \x -> case x of
Just v -> putStrLn ("just a single" ++ show v)
Nothing -> putStrLn "no numbers at all"
can be shortened to:
[Just 1, Just 2, Nothing, Just 3] `forM_` \case
Just v -> putStrLn ("just a single" ++ show v)
Nothing -> putStrLn "no numbers at all"
Try it out!
{-# LANGUAGE LambdaCase #-}
import Control.Monad (forM_)
-- | should print:
-- @["just a single 1","just a single 2","no numbers at all","just a single 3"]@
main = [Just 1, Just 2, Nothing, Just 3] `forM_` \case
Just v -> putStrLn ("just a single " ++ show v)
Nothing -> putStrLn "no numbers at all"
Available in: GHC 7.8 and later
The EmptyCase extension allows you to write a case statement that has no clauses; the syntax is case e of {} (where e is any expression). If you also have LambdaCase enabled, you can abbreviate \x ->
case x of {} to \case {}
This is most useful when you have a type that you know for sure has no values, but Haskell‘s syntax and type system force you to do something with a hypothetical such value anyway. Without EmptyCase,
you could just use error or undefined, or otherwise diverge, and such an action is still possible; however, using an empty case statement for such things is more indicative of intent, and holds some
promise of being better supported by the exhaustivity checker in the future.
Available in: GHC 7.6 and later
The MultiWayIf extension allows you to use the full power of Haskell’s guard syntax in an if expression. For example, this code:
if x == 1
then "a"
else if y < 2
then "b"
else "c"
can be rewritten as:
if | x == 1 -> "a"
| y < 2 -> "b"
| otherwise -> "d"
which is much nicer.
Try it out!
{-# LANGUAGE MultiWayIf #-}
fn :: Int -> Int -> String
fn x y = if | x == 1 -> "a"
| y < 2 -> "b"
| otherwise -> "c"
-- | should print:
-- @c@
main = putStrLn $ fn 3 4
WARNING: In GHC 7.6, the use of MultiWayIf doesn’t affect layout, instead allowing the previous layout (prior to the if keyword) to remain unchanged. This was changed shortly afterwards; in GHC 7.8
and later, MultiWayIf affects layout, just like ordinary function guards do.
Available in: GHC 7.10 and later
Standard Haskell allows you to write integer literals in decimal (without any prefix), hexadecimal (preceded by 0x or 0X), and octal (preceded by 0o or 0O). The BinaryLiterals extension adds binary
(preceded by 0b or 0B) to the list of acceptable integer literal styles.
Try it out!
{-# LANGUAGE BinaryLiterals #-}
-- | should print:
-- @(1458,1458,1458,1458)@
main = print (1458, 0x5B2, 0o2662, 0b10110110010)
Available in: GHC 7.8 and later
Standard Haskell desugars negative numeric literals (of either integer or fractional form) by applying the negate function from the Num type class to the corresponding positive numeric literal (which
is then expanded again using either fromInteger or fromRational, as appropriate). That is, the standard full desugaring of the literal -1458 is negate (fromInteger 1458). The NegativeLiterals
extension changes this, making negative numeric literals instead desugar as fromInteger or fromRational applied directly to a negative Integer or Rational value; that is, -1458 is desugared as
fromInteger (-1458). In a sense, NegativeLiterals swaps the positions of negation and conversion in the desugaring of numeric literals.
This doesn’t make a difference for the common cases, but certain edge cases can behave differently (and usually better) under NegativeLiterals than otherwise. The example that the GHC User’s guide
gives is 8-bit signed arithmetic, in which 128 is not representable but -128 is representable. The naïve desugaring of -128 to negate (fromInteger 128) results in an overflow from 128 to -128
followed by a negation to 128 followed by another overflow back to -128; meanwhile, the NegativeLiterals desugaring to fromInteger (-128) doesn’t waste cycles (or risk trapping on some
architectures), but instead produces the appropriate value from the start. Other examples might actually change behavior rather than simply be less efficient; you should make sure that you understand
a piece of numeric Haskell code fairly well before enabling or disabling NegativeLiterals for it.
Try it out!
{-# LANGUAGE NegativeLiterals #-}
main = do print (-1 :: ExplicitNegation Integer)
print (negate 1 :: ExplicitNegation Integer)
print (-1.5 :: ExplicitNegation Rational)
print (negate 1 :: ExplicitNegation Rational)
-- /show
-- this type exists solely to explicitly mark where negation happens
data ExplicitNegation n = Value n | Negate (ExplicitNegation n) deriving Show
collapseNegation :: Num n => ExplicitNegation n -> n
collapseNegation (Value x) = x
collapseNegation (Negate v) = negate $ collapseNegation v
instance (Eq n, Num n) => Eq (ExplicitNegation n) where
v == w = collapseNegation v == collapseNegation w
instance (Ord n, Num n) => Ord (ExplicitNegation n) where
v `compare` w = collapseNegation v `compare` collapseNegation w
instance Num n => Num (ExplicitNegation n) where
v + w = Value $ collapseNegation v + collapseNegation w
v * w = Value $ collapseNegation v * collapseNegation w
negate = Negate
abs = Value . abs . collapseNegation
signum = Value . signum . collapseNegation
fromInteger = Value . fromInteger
instance Fractional n => Fractional (ExplicitNegation n) where
recip = Value . recip . collapseNegation
fromRational = Value . fromRational
Available in: GHC 7.8 and later
Standard Haskell gives the polymorphic type (Fractional a) => a to otherwise-unconstrained fractional numeric literals; however, some such literals are guaranteed to actually be integers, because
they have an exponent (whether implicit or explicit) that is larger than the distance from the decimal point at which their last non-zero digit occurs (for example, 4.65690e4 is “the same number” as
46569, which is clearly an integer). The NumDecimals extension exploits this fact by giving fractional literals which are “really just integers” the more general type (Num a) => a instead.
Try it out!
{-# LANGUAGE NumDecimals #-}
-- notice that this code will not compile if
-- '1.6e1' isn't allowed to be an 'Integer'
main = print (1.6e1 `div` 5 :: Integer)
Available in: GHC 7.0 and later
Available in: GHC 7.2 and later | {"url":"https://www.schoolofhaskell.com/user/school/to-infinity-and-beyond/pick-of-the-week/guide-to-ghc-extensions/basic-syntax-extensions","timestamp":"2024-11-03T00:34:41Z","content_type":"text/html","content_length":"93245","record_id":"<urn:uuid:4ed99744-e429-44d9-bc8a-44d2875defa9>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00689.warc.gz"} |
SCCM – Parsing Collection Maintenance Windows from C-Sharp¶
This post details how to extract useful information from the SCCM Schedules format.
Suggested Reading¶
Here are a few useful links to documentation regarding some of the activities being performed-
This solution makes heavy use of Logical AND operators.
As well as heavy use of shift operators.
The Format¶
You can view the SCCM service windows by querying the view, “vSMS_ServiceWindow”
Here is an example query for retrieving collections along with their service windows.
from v_Collections c
JOIN vSMS_ServiceWindow sw on sw.SiteID = c.SiteID
ORDER BY c.LimitToCollectionName
When you query this- you will get a lot of common information about your collections, their names, collection IDs, types, etc.
The piece I want to focus on here, is extracting out the actual schedules, which are stored in the “Schedules” column of the results from the above query.
At a glance, it would appear this column is complete gibberish, containing values such as below-
However, this is in fact, 64bits of data, stored as a hexadecimal string.
Reversing the format¶
Using data from both PART 1 of this post, as well as a bunch of testing collections I created- I came up with this diagram showing the relationships of the data in that column.
For the time being- I will be focused on “Weekly” windows, which are service-windows which repeats on the specified day of the week, every week.
I will note, the format of the above data will change, depending on the recurrence type, stored in bits 20- 22.
Low Bit High Bit N Bits Data Recurrence Type
1 1 1 Is Date GMT/CST
4 8 3 Recur Every N Days 2 = Daily
14 16 3 Recur Every N Weeks 3 = Weekly
16 18 3 Day Of Week 3 = Weekly
10 12 3 Week Order 4 = Monthly by WeekDay
13 16 4 Recur Every N Months 4 = Monthly by WeekDay
17 19 3 Day Of Week 4 = Monthly by WeekDay
11 14 4 Recur Every N Months 5 = Monthly By Date
15 19 5 Recur Every N Days 5 = Monthly By Date
7 9 3 Offset Days 6 = Monthly By Weekday Offset
10 12 3 Week Order 6 = Monthly By Weekday Offset
13 16 4 Recur Every N Months 6 = Monthly By Weekday Offset
17 19 3 Day Of Week 6 = Monthly By Weekday Offset
20 22 3 Recurrence Type
23 27 5 Duration (Mins)
28 32 5 Duration (Hours)
33 38 6 Duration (???) Unsure- of what this is.
39 44 6 Date – Year
45 48 4 Date – Month
49 53 5 Date – Day
54 58 5 Date – Hour
59 64 6 Date – Minute
Converting the HEX string into a ulong for processing.¶
if (!ulong.TryParse(Hex, System.Globalization.NumberStyles.HexNumber, provider: null, out ulong Result))
throw new Exception("Invalid hex provided. Unable to parse.");
While there is nothing special about converting a string containing HEX, into a number- there is one important thing to note here- I am parsing the value as ulong, instead of long.
The reason behind using ulong- all of the values are unsigned. If we parse as long instead, this will make it harder to manipulate the data later on, as c# will assume all of the resulting arithmetic
results in signed numbers.
If you don’t know the difference between signed / unsigned-
Signed data types leverage the most significant bit, to determine if the value is negative or positive. This allows them to hold a negative value, but, sacrifices half of the maximum value.
Unsigned- uses all of the bits for numeric data.
Common Fields¶
Bit 1, corresponds to “IS GMT / CST”, and is common for all types.
Bits 23-38 contains the duration, which is common for all recurrence types. I am unsure of what bits 33-38 stores…
Bits 39 – 64 contains the effective “Start Date” which is common and shared for all recurrence types. Parsing out this data is pretty easy.
Used the table above- we shift the data the specified number of bits, and do a bitwise AND to only include the number of bits notated.
var Flags = (Result >> 19) & 0x7; //Recurrence Type Flags - 3 bits.
model.Recurrance = (ParsedSchedule.RecurranceType)Flags;
var R_Duration_Mins = (Result >> 22) & 0x1F; //Duration Mins - 5 bits.
var R_Duration_Hours = (Result >> 27) & 0x1F; //Duration Hours- 5 bits.
model.Duration = new TimeSpan((int)R_Duration_Hours, (int)R_Duration_Mins, 0);
var IsGMT = (Result & 0x1) == 1; //First bit, specifies if this schedule is GMT, or Local.
var Duration = (Result >> 32) & 0x3F; // 6 bits. Not, sure exactly what this field's purpose is.
var Year = ((Result >> 38) & 0x3F) + 1970; // 6 bits + 1970
var Month = (Result >> 44) & 0xF; // 1 byte
var Day = (Result >> 48) & 0x1F; // 5 bits
var Hour = (Result >> 53) & 0x1F; // 5 bits
var Minute = (Result >> 58); // Remaining 6 bits. (Operating on a 64-bit ulong, no need to mask the rest)
model.StartTime = new DateTime((int)Year, (int)Month, (int)Day, (int)Hour, (int)Minute, 0, IsGMT ? DateTimeKind.Utc : DateTimeKind.Local);
The final cast to int is needed, otherwise, you would receive a syntax error because you cannot create a datetime with ulongs.
I will note, I am not using checked when casting, as…. well. It’s impossible to overflow an int with only 6 bits….
Recurrence Type¶
The Recurrence type, stored in bits 20-22, will determine the data stored in bits 2-19.
From my research, I have found these possible values:
None = 1¶
This service windows is not recurring, and will only occur once on the date provided.
Daily = 2¶
This schedule will repeat every N Days.
Weekly = 3¶
This schedule, will repeat on the specified day of the week, every week.
Monthly By Week Day = 4¶
This Schedule will occur every month, on a given weekday.
Monthly, With Offset = 6¶
This is the same as “Monthly by week day”, but, with the “Offset” checkbox checked.
This came with SCCM 2207
From Microsoft-
Parsing out Recurrence-Specific Fields.¶
switch (model.Recurrance)
case ParsedSchedule.RecurranceType.None:
case ParsedSchedule.RecurranceType.Daily:
model.RecurEveryNDays = (int)((Result >> 3) & 0x1F); //5 Bits
case ParsedSchedule.RecurranceType.Weekly:
model.DayOfWeek = (DayOfWeek)(((Result >> 16) & 0x7) - 1); // 3 bits.
model.RecurEveryNWeeks = (int)((Result >> 13) & 0x7); // 3 bits
case ParsedSchedule.RecurranceType.Monthly_ByWeekday:
model.WeekOccurence = (ParsedSchedule.WeekOrder)((Result >> 9) & 0x7); // 3 bits
model.RecurEveryNMonths = (int)((Result >> 12) & 0xF); // 4 bits
model.DayOfWeek = (DayOfWeek)(((Result >> 16) & 0x7) - 1); // 3 bits
case ParsedSchedule.RecurranceType.Monthly_ByDate:
model.RecurEveryNDays = (int)((Result >> 14) & 0x1F); // 5 bits
model.RecurEveryNMonths = (int)((Result >> 10) & 0xF); // 4 bits
case ParsedSchedule.RecurranceType.Monthly_ByWeekDay_Offset:
model.DayOfWeek = (DayOfWeek)(((Result >> 16) & 0x7) - 1); // 3 bits.
model.OffsetDays = (int)((Result >> 6) & 0x7); // 3 bits.
model.WeekOccurence = (ParsedSchedule.WeekOrder)((Result >> 9) & 0x7); // 3 bits
model.RecurEveryNMonths = (int)((Result >> 12) & 0xF); // 4 bits
About the only thing special to note here- SCCM stores Days of week, starting with Sunday = 1. .NET Starts the week on Sunday = 0. To compensate, we just subtract 1.
I will also note, for “WeekOccurence”, 0 corresponds to last, 1 = first, 2 = second, 3 = third, 4 = forth.
How did I obtain this data?¶
Easier said than done!
While, I knew the general format from my previous article, and the pseudocode linked from it- I needed to parse out the specific offsets, and field types. As well, I wanted to leverage proper bitwise
operations instead of relying on odd string/hex manipulation. So- a lot of this was trial and error.
The first step I did- was to generate a large number of dummy schedules, copy the hex, and I built a bunch of unit tests.
I knew the expected output, so, I provided the expected values. After which- most of the process was running unit tests, and using a bit of logic.
A lot of the work was done in Notepad++, looking at the binary values, and trying to determine what fits where.
By removing the “Known” bits, and only comparing the unknown bits- it makes it pretty easy to find what data, fits where, by comparing multiple variations of the data.
In the above example, it can be determined bits 17-19, corresponds to the day of the week.
Overall, it took me about one full working day to reverse engineer all of the formats. | {"url":"https://static.xtremeownage.com/blog/2022/parsing-sccm-collections/","timestamp":"2024-11-07T10:35:14Z","content_type":"text/html","content_length":"83557","record_id":"<urn:uuid:084b52eb-fef4-43b8-9aa3-f5b06b9631cb>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00104.warc.gz"} |
How Do You Solve and Graph a Two-Step Inequality?
Ever wondered what rules you're allowed to follow when you're working with inequalities? Well, one of those rules is called the division property of inequality, and it basically says that if you
divide one side of an inequality by a number, you can divide the other side of the inequality by the same number. However, you have to be very careful about the direction of the inequality! Watch the
tutorial to see how this looks in terms of algebra! | {"url":"https://virtualnerd.com/pre-algebra/inequalities-multi-step-equations/inequalities-multiple-steps/solve-multiple-step-inequalities/inequality-two-step-solution-and-graph","timestamp":"2024-11-03T12:35:20Z","content_type":"text/html","content_length":"41668","record_id":"<urn:uuid:5cec888f-40a7-4949-9f1b-bd8320361968>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00827.warc.gz"} |
Two persons A and B take turns in throwing a pair of dice. The ... | Filo
Two persons and take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If throws first, then the probability that wins the game, is
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The probability of throwing 9 with two dice
The probability of not throwing 9 with two dice
If is to win, he should throw 9 , in Ist or or 5 th and so on.
If is to win, he should throw, 9 in 2 nd, 4 th and so on.
can be get the chance only when should not get it.
s chances
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Two persons and take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If throws first, then the probability that wins the
Question Text game, is
Updated On May 6, 2023
Topic Probability
Subject Mathematics
Class Class 12
Answer Type Text solution:1 Video solution: 1
Upvotes 87
Avg. Video 22 min | {"url":"https://askfilo.com/math-question-answers/two-persons-a-and-b-take-turns-in-throwing-a-pair-of-dice","timestamp":"2024-11-09T11:23:45Z","content_type":"text/html","content_length":"435599","record_id":"<urn:uuid:8fe9d6bf-f01c-4fde-8751-187d06156952>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00702.warc.gz"} |
Long Multiplication Worksheets With Grid
Long Multiplication Worksheets With Grid
Submitted by gaye noel on 22 october 2007. Free download long multiplication worksheets with grid format.
The 3 Digit By 1 Digit Long Division With Grid Assistance And Prompts And No Remainders A Math Worksheet From Division Worksheets Long Division Math Division
34 30 4 or hundreds tens and units components e g.
Long multiplication worksheets with grid. I did a quick search on the internet but didn t find any free printable sheets in grid format that didn t make me sign up for something so i made some for
him. From the simplest multiplication facts to multiplying large numbers in columns. Sometimes referred to as long multiplication or multi digit multiplication the questions on these worksheets
require students to have mastered the multiplication facts from 0 to 9.
No login or account is needed. This page includes long multiplication worksheets for students who have mastered the basic multiplication facts and are learning to multiply 2 3 4 and more digit
numbers. Grid multiplication displaying top 8 worksheets found for this concept.
Multiplication worksheets for grades 2 6. I created them using google sheets and made three sets of questions and answers. It may be printed downloaded or saved and used in your classroom home school
or other educational environment to help someone.
Significant emphasis to mental multiplication exercises. 345 300 40 5. Long multiplication worksheets worksheets long multiplication.
You may vary the numbers of problems on the worksheet from 15 to 27. Long multiplication fact and worksheet gives step by step instructions plus lots of differentiated question to try. Some of the
worksheets for this concept are long multiplication work 2 digit by 2 digit long multiplication work 3 digit by 3 digit multiplication multiplication scoot lattice multiplication blank multiplication
table lattice multiplication 15 x 15 times table charts.
Grid multiplication also known as the grid method is now taught in schools as an intermediate stage before long multiplication. This math worksheet was created on 2015 02 22 and has been viewed 304
times this week and 430 times this month. Grid method long multiplication fact and worksheet.
Not the most exciting but worksheets to help with two different ways of performaing long multiplication calculations. These multiplication worksheets are appropriate for kindergarten 1st grade 2nd
grade 3rd grade 4th grade and 5th grade. Grid method and chinese method feel free to rename them if you know them by any other name.
With grid multiplication the two numbers to be multiplied are split partitioned up into their tens and units components e g. Free math worksheets from k5 learning. These multiplication worksheets may
be configured for 2 3 or 4 digit multiplicands being multiplied by multiples of ten that you choose from a table.
Multiplication 2 Digit By 2 Digit 4th Grade Math Worksheets Free Multiplication Worksheets Teaching Multiplication
The 2 Digit By 2 Digit Multiplication With Grid Support A Math Worksheet Page 2 Multiplication Multiplication Worksheets Box Method Multiplication
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Neat Double Digit Mult Free Printable Math Worksheets Printable Math Worksheets 4th Grade Math
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4 Digit By 3 Digit Multiplication With Grid Support A Long Multiplication Worksheet Multiplication Worksheets Multiplication Math Worksheets
The Multiplying 4 Digit By 2 Digit Numbers B Math Worksheet From The Long Multiplication Workshe Multiplication Worksheets Long Multiplication Multiplication
Multiplication 2 X 2 Digit With Regrouping 5 Multiply Worksheet Packet Multiplication Multiplication Problems Multiplication Worksheets
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The 3 Digit By 2 Digit Multiplication With Grid Support F Math Worksheet Math Multiplication Worksheets Long Multiplication Multiplication Worksheets | {"url":"https://thekidsworksheet.com/long-multiplication-worksheets-with-grid/","timestamp":"2024-11-14T07:16:52Z","content_type":"text/html","content_length":"136733","record_id":"<urn:uuid:ef0a9b39-d703-4425-888b-38651b2cc181>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00420.warc.gz"} |
It has begun the IV UFFS Programming Contest! We hope you enjoy the next hours you are going to spend with us, as we hope you have a lot of fun! Good luck!
This is the 3^rd year of the Programming Club, an extension program whose primary goal is to help the programmers of the Brazilian region known as Southern Border to get readier to face the
computational challenges from both academic and corporate worlds. Our main strategy lies in promoting workshops and training sessions for Programming contests, not only for students of our
institution (UFFS), but also for whoever wants to participate. Despite of many issues, we find ourselves very happy with the results we have been achieving. Having other institutions as partners, as
UNOCHAPECÓ, URI and UNOESC, we collaborated to make Chapecó in the past two years the 2^nd largest site of ICPC Brazilian Subregional Contest, which is another indicator of the enthusiasm our people
has in Programming.
In order to warm you up for this particular contest, we shall ask you to write a program which calculates the quotient and the remainder of the division of two integers, can that be? Recall that the
quotient and the remainder of the division of an integer a by a non-zero integer b are respectively the only integers q and r such that 0 ≤ r < |b| and:
a = b × q + r
In case you don't know it, the theorem that guarantees the existence and the uniqueness of the integers q and r is known as ‘Euclidean Division Theorem’ or ‘Division Algorithm’.
The input consists of two integers a and b (-1,000 ≤ a, b < 1,000). | {"url":"https://www.beecrowd.com.br/repository/UOJ_1837_en.html","timestamp":"2024-11-06T11:28:11Z","content_type":"text/html","content_length":"7734","record_id":"<urn:uuid:22bbf18c-479a-442c-b4ad-c2f9bdf91476>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00006.warc.gz"} |
Performance Metrics: Negative Likelihood Ratio — Roel Peters
Home » Glossary » Performance Metrics: Negative Likelihood Ratio
Performance Metrics: Negative Likelihood Ratio
• by roelpi
What is the Negative Likelihood Ratio?
The Negative Likelihood Ratio (LR-, -LR, likelihood ratio negative or likelihood ratio for negative results) gives the change in odds of the true value being positive when the predicted value is
negative. This is expressed as a ratio. It is analogous to the Positive Likelihood Ratio. An LR- of 6 indicates a 6-fold increase in the odds of the true value being positive, when the predicted
value is negative. The larger the ratio, the more informative it is. Furthermore, the LR- is independent of the prevalence. I.e. this performance metric is resistant to class imbalance.
Calculating the Negative Likelihood Ratio
The LR- is calculated as follows:
However, this is a likelihood ratio. Given this, using the conditional probabilities in the numerator and denominator makes more sense:
An LR- of 1 means that the model is completely useless. In this case, the odds of a negative true value before and after knowing the predicted value haven’t changed — i.e. the change in odds is 1.
Further reading:
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ppsr - Predictive Power Score
ppsr is the R implementation of the Predictive Power Score (PPS).
The PPS is an asymmetric, data-type-agnostic score that can detect linear or non-linear relationships between two variables. The score ranges from 0 (no predictive power) to 1 (perfect predictive
The general concept of PPS is useful for data exploration purposes, in the same way correlation analysis is. You can read more about the (dis)advantages of using PPS in this blog post.
You can install the latest stable version of ppsr from CRAN:
Not all recent features and bugfixes may be included in the CRAN release.
Instead, you might want to download the most recent developmental version of ppsr from Github:
Computing PPS
PPS represents a framework for evaluating predictive validity.
There is not one single way of computing a predictive power score, but rather there are many different ways.
You can select different machine learning algorithms, their associated parameters, cross-validation schemes, and/or model evaluation metrics. Each of these design decisions will affect your model’s
predictive performance and, in turn, affect the resulting predictive power score you compute.
Hence, you can compute many different PPS for any given predictor and target variable.
For example, the PPS computed with a decision tree regression model…
…will differ from the PPS computed with a simple linear regression model.
The ppsr package has four main functions to compute PPS:
• score() computes an x-y PPS
• score_predictors() computes all X-y PPS
• score_df() computes all X-Y PPS
• score_matrix() computes all X-Y PPS, and shows them in a matrix
where x and y represent an individual predictor/target, and X and Y represent all predictors/targets in a given dataset.
score() computes the PPS for a single target and predictor
score_predictors() computes all PPSs for a single target using all predictors in a dataframe
score_df() computes all PPSs for every target-predictor combination in a dataframe
score_df() computes all PPSs for every target-predictor combination in a dataframe, but returns only the scores arranged in a neat matrix, like the familiar correlation matrix
Currently, the ppsr package computes PPS by default using…
• the default decision tree implementation of the rpart package, wrapped by parsnip
• weighted F1 scores to evaluate classification models, and MAE to evaluate regression models
• 5 cross-validations
You can call the available_algorithms() and available_evaluation_metrics() functions to see what alternative settings are supported.
Note that the calculated PPS reflects the out-of-sample predictive validity when more than a single cross-validation is used. If you prefer to look at in-sample scores, you can set cv_folds = 1. Note
that in such cases overfitting can become an issue, particularly with the more flexible algorithms.
Visualizing PPS
Subsequently, there are three main functions that wrap around these computational functions to help you visualize your PPS using ggplot2:
• visualize_pps() produces a barplot of all X-y PPS, or a heatmap of all X-Y PPS
• visualize_correlations() produces a heatmap of all X-Y correlations
• visualize_both() produces the two heatmaps of all X-Y PPS and correlations side-by-side
If you specify a target variable (y) in visualize_pps(), you get a barplot of its predictors.
If you do not specify a target variable in visualize_pps(), you get the PPS matrix visualized as a heatmap.
Some users might find it useful to look at a correlation matrix for comparison.
With visualize_both you generate the PPS and correlation matrices side-by-side, for easy comparison.
You can change the colors of the visualizations using the functions arguments. There are also arguments to change the color of the text scores.
Furthermore, the functions return ggplot2 objects, so that you can easily change the theme and other settings.
The number of predictive models that one needs to build in order to fill the PPS matrix belonging to a dataframe increases exponentially with every new column in that dataframe.
For traditional correlation analyses, this is not a problem. Yet, with more computation-intensive algorithms, with many train-test splits, and with large or high-dimensional datasets, it can take a
decent amount of time to build all the predictive models and derive their PPSs.
One way to speed matters up is to use the ppsr::score_predictors() function and focus on predicting only the target/dependent variable you are most interested in.
Yet, since version 0.0.1, all ppsr::score_* and pssr::visualize_* functions now take in two arguments that facilitate parallel computing. You can parallelize ppsr’s computations by setting the
do_parallel argument to TRUE. If done so, a cluster will be created using the parallel package. By default, this cluster will use the maximum number of cores (see parallel::detectCores()) minus 1.
However, with the second argument – n_cores – you can manually specify the number of cores you want ppsr to use.
Interpreting PPS
The PPS is a normalized score that ranges from 0 (no predictive power) to 1 (perfect predictive power).
The normalization occurs by comparing how well we are able to predict the values of a target variable (y) using the values of a predictor variable (x), respective to two benchmarks: a perfect
prediction, and a naive prediction
The perfect prediction can be theoretically derived. A perfect regression model produces no error (=0.0), whereas a perfect classification model results in 100% accuracy, recall, et cetera (=1.0).
The naive prediction is derived empirically. A naive regression model is simulated by predicting the mean y value for all observations. This is similar to how R-squared is calculated. A naive
classification model is simulated by taking the best among two models: one predicting the modal y class, and one predicting random y classes for all observations.
Whenever we train an “informed” model to predict y using x, we can assess how well it performs by comparing it to these two benchmarks.
Suppose we train a regression model, and its mean average error (MAE) is 0.10. Suppose the naive model resulted in an MAE of 0.40. We know the perfect model would produce no error, which means an MAE
of 0.0.
With these three scores, we can normalize the performance of our informed regression model by interpolating its score between the perfect and the naive benchmarks. In this case, our model’s
performance lies about 1/4^th of the way from the perfect model, and 3/4^ths of the way from the naive model. In other words, our model’s predictive power score is 75%: it produced 75% less error
than the naive baseline, and was only 25% short of perfect predictions.
Using such normalized scores for model performance allows us to easily interpret how much better our models are as compared to a naive baseline. Moreover, such normalized scores allow us to compare
and contrast different modeling approaches, in terms of the algorithms, the target’s data type, the evaluation metrics, and any other settings used.
The main use of PPS is as a tool for data exploration. It trains out-of-the-box machine learning models to assess the predictive relations in your dataset.
However, this PPS is quite a “quick and dirty” approach. The trained models are not at all tailored to your specific regression/classification problem. For example, it could be that you get many PPSs
of 0 with the default settings. A known issue is that the default decision tree often does not find valuable splits and reverts to predicting the mean y value found at its root. Here, it could help
to try calculating PPS with different settings (e.g., algorithm = 'glm').
At other times, predictive relationships may rely on a combination of variables (i.e. interaction/moderation). These are not captured by the PPS calculations, which consider only univariate
relations. PPS is simply not suited for capturing such complexities. In these cases, it might be more interesting to train models on all your features simultaneously and turn to concepts like feature
/variable importance, partial dependency, conditional expectations, accumulated local effects, and others.
In general, the PPS should not be considered more than a fast and easy tool to finding starting points for further, in-depth analysis. Keep in mind that you can build much better predictive models
than the default PPS functions if you tailor your modeling efforts to your specific data context.
Open issues & development
PPS is a relatively young concept, and likewise the ppsr package is still under development. If you spot any bugs or potential improvements, please raise an issue or submit a pull request.
On the developmental agenda are currently:
• Support for different modeling techniques/algorithms
• Support for generalized linear models for multinomial classification
• Passing/setting of parameters for models
• Different model evaluation metrics
• Support for user-defined model evaluation metrics
• Downsampling for large datasets
This R package was inspired by 8080labs’ Python package ppscore.
The same 8080labs also developed an earlier, unfinished R implementation of PPS.
Read more about the big ideas behind PPS in this blog post. | {"url":"http://cran.auckland.ac.nz/web/packages/ppsr/readme/README.html","timestamp":"2024-11-08T07:45:02Z","content_type":"application/xhtml+xml","content_length":"39844","record_id":"<urn:uuid:407f9c8a-473e-46bd-b45f-6989e3845a1d>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00424.warc.gz"} |
Inverse and Transpose Matrices — Handmade Hero — Episode Guide — Handmade Hero
0:17Recap and set the stage for the day
0:17Recap and set the stage for the day
0:17Recap and set the stage for the day
1:29Blackboard: Skew UV Mapping
1:29Blackboard: Skew UV Mapping
1:29Blackboard: Skew UV Mapping
3:48Blackboard: A conceptual explanation of transforming a texture map
3:48Blackboard: A conceptual explanation of transforming a texture map
3:48Blackboard: A conceptual explanation of transforming a texture map
7:15Blackboard: Our matrix equation
7:15Blackboard: Our matrix equation
7:15Blackboard: Our matrix equation
8:08Blackboard: The components of this equation
8:08Blackboard: The components of this equation
8:08Blackboard: The components of this equation
9:27Blackboard: Adding and taking the origin out of equation
9:27Blackboard: Adding and taking the origin out of equation
9:27Blackboard: Adding and taking the origin out of equation
10:36Blackboard: Transforming the U and V
10:36Blackboard: Transforming the U and V
10:36Blackboard: Transforming the U and V
12:11Blackboard: Multiplying these matrices out
12:11Blackboard: Multiplying these matrices out
12:11Blackboard: Multiplying these matrices out
14:11Blackboard: Backward transform, using dot products
14:11Blackboard: Backward transform, using dot products
14:11Blackboard: Backward transform, using dot products
16:18Blackboard: Why use dot products to compute the transformed U and V?
16:18Blackboard: Why use dot products to compute the transformed U and V?
16:18Blackboard: Why use dot products to compute the transformed U and V?
18:42Blackboard: Getting from wanting to invert the matrix, to taking the dot product shortcut
18:42Blackboard: Getting from wanting to invert the matrix, to taking the dot product shortcut
18:42Blackboard: Getting from wanting to invert the matrix, to taking the dot product shortcut
19:47Blackboard: Inverting an orthonormal matrix
19:47Blackboard: Inverting an orthonormal matrix
19:47Blackboard: Inverting an orthonormal matrix
22:08Blackboard: What it means to invert
22:08Blackboard: What it means to invert
22:08Blackboard: What it means to invert
25:14Blackboard: The algebraic explanation for why any orthonormal matrix multiplied by its transpose (i.e. inverted) gives you the identity matrix
25:14Blackboard: The algebraic explanation for why any orthonormal matrix multiplied by its transpose (i.e. inverted) gives you the identity matrix
25:14Blackboard: The algebraic explanation for why any orthonormal matrix multiplied by its transpose (i.e. inverted) gives you the identity matrix
31:38Blackboard: Putting it in meta algebraic terms
31:38Blackboard: Putting it in meta algebraic terms
31:38Blackboard: Putting it in meta algebraic terms
33:22Blackboard: The geometric explanation for this
33:22Blackboard: The geometric explanation for this
33:22Blackboard: The geometric explanation for this
37:28Blackboard: Columnar vs Row-based Matrices
37:28Blackboard: Columnar vs Row-based Matrices
37:28Blackboard: Columnar vs Row-based Matrices
39:49Blackboard: How non-uniform (yet still orthogonal) scaling affects our matrix
39:49Blackboard: How non-uniform (yet still orthogonal) scaling affects our matrix
39:49Blackboard: How non-uniform (yet still orthogonal) scaling affects our matrix
41:11"I hope everyone was interested in the matrix thing today"^α
41:11"I hope everyone was interested in the matrix thing today"^α
41:11"I hope everyone was interested in the matrix thing today"^α
42:16Blackboard: Transposing the matrix for non-uniformly scaled vectors, and compensating for that scaling
42:16Blackboard: Transposing the matrix for non-uniformly scaled vectors, and compensating for that scaling
42:16Blackboard: Transposing the matrix for non-uniformly scaled vectors, and compensating for that scaling
44:54Blackboard: The beginnings of a formal algebraic explanation of this compensation
44:54Blackboard: The beginnings of a formal algebraic explanation of this compensation
44:54Blackboard: The beginnings of a formal algebraic explanation of this compensation
46:53Blackboard: Matrix multiplication is order dependent
46:53Blackboard: Matrix multiplication is order dependent
46:53Blackboard: Matrix multiplication is order dependent
50:01Blackboard: How this order dependence of the transform is captured by matrix maths
50:01Blackboard: How this order dependence of the transform is captured by matrix maths
50:01Blackboard: How this order dependence of the transform is captured by matrix maths
52:41Blackboard: A formal algebraic explanation for the scale and rotation compensation
52:41Blackboard: A formal algebraic explanation for the scale and rotation compensation
52:41Blackboard: A formal algebraic explanation for the scale and rotation compensation
55:59Blackboard: A glimpse into the future of actually inverting the matrix
55:59Blackboard: A glimpse into the future of actually inverting the matrix
55:59Blackboard: A glimpse into the future of actually inverting the matrix
58:32A few words on how cool linear algebra can get
58:32A few words on how cool linear algebra can get
58:32A few words on how cool linear algebra can get | {"url":"https://guide.handmadehero.org/code/day319/","timestamp":"2024-11-02T14:18:30Z","content_type":"text/html","content_length":"46101","record_id":"<urn:uuid:6689a4c1-cc89-407d-823f-3c6b82ffebf3>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00103.warc.gz"} |
Latest Posts
you will be using support vector machines (SVMs) with various example 2D datasets.
• Plot Data (in ex6data1.mat)
SVM with Linear Kernel
try using different values of the C parameter with SVMs. Informally, the C parameter is a positive value that controls the penalty for misclassified training examples.
• Plott decision boundary (ex6data1.mat)
Train SVM with RBF Kernel
• Plot Data (in ex6data2.mat)
C: 1, sigma: 0.1
• Plot decision boundary (in ex6data2.mat)
Try different SVM Parameters to train SVM with RBF Kernel
Automatically choose optimal C and sigma based on a cross-validation set.
C list: [0.01 0.03 0.1 0.3 1 3 10 30]
sigma list: [0.01 0.03 0.1 0.3 1 3 10 30]
=> optimal C = 1 and sigma = 0.1
• Plot Data (in ex6data3.mat)
• Plot decision boundary with optimal svm parameters (in ex6data3.mat)
you will be using support vector machines to build a spam classifier.
For the purpose of this exercise, you will only be using the body of the email (excluding the email headers).
• Preprocess sample email (in emailSample1.txt, vocab.txt)
convert each email into a vector of features
Given the vocabulary list, we can now map each word in the preprocessed emails into a list of word indices that contains the index of the word in the vocabulary list.
Lower-casing, Stripping HTML, Normalizing URLs, Normalizing Email Addresses, Normalizing Numbers, Normalizing Dollars, Word Stemming, Removal of non-words
vocabulary list: a list of 1899 words
• Extracte Features from Emails (in emailSample1.txt)
the feature xi ∈ {0, 1} for an email corresponds to whether the i-th word in the dictionary occurs in the email. That is, xi = 1 if the i-th word is in the email and xi = 0 if the i-th word is
not present in the email.
• Train Linear SVM for Spam Classification (in spamTrain.mat, spamTest.mat)
train a SVM to classify between spam (y = 1) and non-spam (y = 0) emails.
spamTrain.mat: 4000 training examples of spam and non-spam email
spamTest.mat: 1000 test examples
Trouble shooting:
• error on plotting the decision boundary of SVM with RBF Kernel
rewrite visualizeBoundary.m line 21:
=> contour(X1, X2, vals, [1 1], ‘LineColor’, ‘b’);
知乎: 机器学习有很多关于核函数的说法，核函数的定义和作用是什么？ | {"url":"http://joyhuang9473.github.io/","timestamp":"2024-11-13T03:05:00Z","content_type":"text/html","content_length":"29939","record_id":"<urn:uuid:026dcfc9-b6a6-4704-a524-e7b4ee719aa3>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00799.warc.gz"} |
A Derivation Of Prices Of Production With Linear ProgrammingA Derivation Of Prices Of Production With Linear Programming - Heterodox
1.0 Introduction This post illustrates a derivation of prices of production, based on certain properties of duality theory as applied to linear programming. I strive to be more concise and elementary
than previous expositions. This exposition is based on John Roemer's Reproducible Solution (Analytical Foundations of Marxian Economic Theory, Cambridge University Press, 1981). You will find no
utility maximization or supply and demand functions below. I have no need for such hypotheses. Nevertheless, one can read this derivation as consistent with marginalism. 2.0 Technology and Endowments
Two commodities, iron and corn, are produced in this example. Managers of firms know a technology consisting of the processes defined in Table 1. Each column shows the inputs and outputs for a
Robert Vienneau considers the following as important:
Example in Mathematical Economics
Interpreting Classical Economics
Karl Marx This could be interesting, too:
Robert Vienneau writes The Production Of Commodities And The Structure Of Production: An Example
Robert Vienneau writes Adam Smith On A Labor Theory Of Value
Robert Vienneau writes William Baumol On Marx
Robert Vienneau writes Francis Spufford On Commodity Fetishism As A Dance
1.0 Introduction
This post illustrates a derivation of prices of production, based on certain properties of duality theory as applied to linear programming. I strive to be more concise and elementary than previous
expositions. This exposition is based on John Roemer's Reproducible Solution (Analytical Foundations of Marxian Economic Theory, Cambridge University Press, 1981).
You will find no utility maximization or supply and demand functions below. I have no need for such hypotheses. Nevertheless, one can read this derivation as consistent with marginalism.
2.0 Technology and Endowments
Two commodities, iron and corn, are produced in this example. Managers of firms know a technology consisting of the processes defined in Table 1. Each column shows the inputs and outputs for a
process operated at a unit level. All processes take a year to complete and provide their output at the end of the year. Each process exhibits constant returns to scale (CRS). For convenience, assume
all coefficients of production defined in the table are positive. The inputs to production are totally used up by operating these processes.
Table 1: The Technology
│ │ Processes │
│ ├─────────────────────┬─────────────────────────────┤
│INPUTS│ Iron Industry │ Corn Industry │
│ ├──────────┬──────────┼──────────────┬──────────────┤
│ │ a │ b │ c │ d │
│Labor │a[0,1](a) │a[0,1](b) │ a[0,2](c) │ a[0,2](d) │
│ Iron │a[1,1](a) │a[1,1](b) │ a[1,2](c) │ a[1,2](d) │
│ Corn │a[2,1](a) │a[2,1](b) │ a[2,2](c) │ a[2,2](d) │
│OUTPUT│1 ton iron│1 ton iron│1 bushel corn │1 bushel corn │
The endowments of iron and corn in the firm's inventory at the start of the year are also given parameters. Table 2 lists the remaining variables in this post. Presumably, the endowments are from
production during the previous year. They are unlikely to be in the proportions needed to continue production. For example, if the managers of a firm decide to specialize in producing corn, they will
have no endowments of iron.
Table 2: Parameters and Variables
│ Additional Parameters │
│ ω[1] │Endowment of iron (in tons) for the firm. │
│ ω[2] │Endowment of corn (in bushels) for the firm. │
│ Parameters taken as given by managers of the firm │
│ p │Price of iron (in bushels per ton). │
│ w │The wage (in bushels per person-year). │
│ Decision Variables │
│q[1](a)│Quantity of iron (in tons) produced by the first process. │
│q[1](b)│Quantity of iron (in tons) produced by the second process. │
│q[2](c)│Quantity of corn (in bushels) produced by the third process. │
│q[2](d)│Quantity of corn (in bushels) produced by the fourth process. │
│ r │The rate of profits. │
3.0 The Primal Linear Program
Managers of firms choose the quantities to produce with each process to maximize the increment z in value, subject to the constraint that they can buy the needed inputs at the start of the year out
of the revenue obtained by selling their endowment. The objective function for the primal linear program is:
z = {p - [p a[1,1](a) + a[2,1](a) + w a[0,1](a)]} q[1](a)
+ {p - [p a[1,1](b) + a[2,1](b) + w a[0,1](b)]} q[1](b)
+ {1 - [p a[1,2](c) + a[2,2](c) + w a[0,2](c)]} q[2](c)
+ {1 - [p a[1,2](d) + a[2,2](d) + w a[0,2](d)]} q[2](d)
The quantities in the square brackets above are the costs of operating each process at a unit level. A bushel corn is taken as numeraire. The quantities in the squiggly brackets are the net revenues
(also known as accounting profits) of operating each process at a unit level. Scaling these net revenues by the level of operation for each process results in the total accounting profit for the
The constraints are:
[p a[1,1](a) + a[2,1](a)] q[1](a)
+ [p a[1,1](b) + a[2,1](b)] q[1](b)
+ [p a[1,2](c) + a[2,2](c)] q[2](c)
+ [p a[1,2](d) + a[2,2](d)] q[2](d) ≤ p ω[1] + ω[2]
q[1](a) ≥ 0, q[1](b) ≥ 0, q[2](c) ≥ 0, q[2](d) ≥ 0
The statement of the constraints is based on the assumption that wages are paid at the end of the year, not advanced at the start.
4.0 The Dual Linear Program
The above linear program has a dual. In the dual, the rate of profits r is chosen to minimize the charge y on endowments:
y = (p ω[1] + ω[2]) r
Such that:
[p a[1,1](a) + a[2,1](a)](1 + r) + w a[0,1](a) ≥ p
[p a[1,1](b) + a[2,1](b)](1 + r) + w a[0,1](b) ≥ p
[p a[1,2](c) + a[2,2](c)](1 + r) + w a[0,2](c) ≥ 1
[p a[1,2](d) + a[2,2](d)](1 + r) + w a[0,2](d) ≥ 1
r ≥ 0
Each constraint in the dual specifies that the revenues obtained from operating a process at the unit level do not exceed the costs, where costs include a charge for the going rate of profits. In
other words, no super-normal profits can be obtained.
5.0 Some Observations About Duality
The value of the objective functions are equal in the solutions to the primal and dual LPs. In other words, the increment in value obtained by the decisions of the manager of a firm is charged to the
value of the endowment.
Suppose the solution of the primal LP results in some process being operated at a positive level. Then the corresponding constraint in the dual LP is met with equality in its solution. Likewise, if a
constraint in the dual is met with inequality, then that process will not be operated in the dual.
If the rate of profits in the solution to the dual is positive, then the constraint in the primal LP will be met with equality. That is, the whole value of the endowment will be used for further
6.0 Prices of Production
I introduce a final assumption. The solution to these LPs must be such that the economy can continue. In the context of this exposition, some firms must produce iron, and some must produce corn.
Thus, one of the first two constraints in the dual LP must be met with equality. One of next two constraints must also be met with equality.
Consider the case when only one of the processes for producing iron is operated, and the same is true of the processes for producing corn. The dual LP yields a system of two equations in three
variables: the price of iron, the wage, and the rate of profits. This system specifies prices of production.
This formulation solves for the choice of the technique, as well as prices of production. It can be generalized to allow for the production of many more commodities and many more processes for
producing each commodity. A generalization can allow for heterogeneous labor. Another generalization allows for the production and use of fixed capital, that is, machines that last for many years.
For a given wage, prices and the rate of profits drop out of the equations for prices of production for the chosen technique. These prices do not support the parables often told in introductory
economics classes with supply and demand. For example, unemployment cannot necessarily be eliminated by lowering the wage and encouraging firms to thereby hire more labor.
7.0 Conclusion
The above illustrates some elements of a theory of value. This is neither a labor theory of value, nor Marx's theory of value. The theory is focused on production and has implications about how labor
is allocated among industries, a central concern of Karl Marx. | {"url":"https://heterodox.economicblogs.org/post-keynesian/2024/vienneau-derivation-of-prices-of-production-linear","timestamp":"2024-11-06T21:12:52Z","content_type":"text/html","content_length":"312976","record_id":"<urn:uuid:e27c6ebc-9d8b-4adf-9984-a7ec30623ce1>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00081.warc.gz"} |
Studies in the History of Mathematical Logic
Edited by Stanislaw J. Surma
Paperback: $24.95 Ebook: $14.99
ISBN13 Hardcopy: 978-1-938421-26-6 ISBN13 Digital: 978-1-938421-21-1
This volume contains seventeen essays in the history of modern mathematical logic. The first nine are concerned with the completeness of various logical calculi. The second five essays are oncerned
with the completeness of classical first-order predicate logic. One essay deals with the history of Cantor’s definition of set, another with the set-theoretical reduction of the concept of relation,
and a final essay is devoted to a survey of various meanings of the concept of completeness of formalized deductive theories. The essays were first presented in the national conferences of the
Thematic Group for the History of Logic organized by the Department of Logic of the Polish Academy of Sciences in 1966–1971. The Advanced Reasoning Forum is pleased to make available this exact
reprint of the original volume first published by the Polish Academy of Sciences, edited by Stanislaw J. Surma.
The essays are:
• 1. Emil Post’s doctoral dissertation (Stanislaw J. Surma)
• 2. A historical survey of the significant methods of proving post’s theorem about the completeness of the classical propositional calculus (Stanislaw J. Surma)
• 3. A survey of the results and methods of investigations of the equivalential propositional calculus (Stanislaw J. Surma)
• 4. A uniform method of proof of the completeness theorem for the equivalential propositional calculus and for some of its extensions (Stanislaw J. Surma)
• 5. Kolmogorov and Glivenko’s papers about intuitionistic logic (Jacek K. Kabzinski)
• 6. Jaskowski’s matrix criterion for the intuitionistic propositional calculus (Stanislaw J. Surma)
• 7. Axiomatization of the implicational Gödel’s matrices by Kalmar’s method (Andrzej Wronski)
• 8. A contribution to the history of the investigations into the intermediate propositional calculi (Andrzej Wronski)
• 9. On Ackermann’s rigorous implication (Jan Wolenski)
• 10. Kurt Gödel’s doctoral dissertation (Jan Zygmunt)
• 11. A survey of the methods of proof of the Gödel-Malcev’s completeness theorem (Jan Zygmunt)
• 12. The concept of the Lindenbaum algebra: its genesis (Stanislaw J. Surma)
• 13. On the old and new methods of interpreting quantifiers (Andrzej Wronski)
• 14. L. Rieger’s logical achievement (Wladyslaw Szczech)
• 15. The development of Cantor’s definition of set (Jerzy Perzanowski)
• 16. On the origins of the set-theoretical concept of relation (Piotr Kossowski)
• 17. A survey of various concepts of completeness of the deductive theories (Stanislaw J. Surma) | {"url":"https://www.advancedreasoningforum.org/publications/studies_in_the_history_of_mathematical_logic.htm","timestamp":"2024-11-06T04:16:33Z","content_type":"text/html","content_length":"43127","record_id":"<urn:uuid:b6480e28-825e-4b6d-b357-d3686afc5b9a>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00186.warc.gz"} |
Binary search (half-interval search) is a search technique, which finds a position of an input element within a sorted array. Unlike sequential search, which needs at most asymptotic complexity.
Let's suppose that we have an array sorted in descending order and we want to find index of an element e within this array. Binary search in every step picks the middle element (m) of the array and
compares it to e. If these elements are equal, than it returns the index of m. If e is greater than m, than e must be located in the left subarray. On the contrary, if e must be located in the right
subarray. At this moment binary search repeats the step on the respective subrarray.
Because the algoritm splits in every step the array in half (and one half of the array is never processed) the input element must be located (or determined missing) in at most
* Binary search
* @param array array sorted in descending order
* @param leftIndex first index that can be touched
* @param rightIndex last index that can be touched
* @param value value to be found
* @return index of the value in array, or -1 if the array does not contain the value
public static int binarySearch(int[] array, int leftIndex, int rightIndex, int value){
if(leftIndex == rightIndex && array[leftIndex] != value) return -1;
int middleIndex = (leftIndex + rightIndex) / 2;
if(array[middleIndex] == value) return middleIndex;
else if(array[middleIndex] > value)
return binarySearch(array, middleIndex + 1, rightIndex, value);
else return binarySearch(array, leftIndex, middleIndex - 1, value); | {"url":"http://www.programming-algorithms.net/article/40119/Binary-search","timestamp":"2024-11-09T12:26:29Z","content_type":"text/html","content_length":"20741","record_id":"<urn:uuid:19d7ba47-a4e7-48ac-8910-7fe58d92f698>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00661.warc.gz"} |
865 research outputs found
In this article we extend our previous results for the orthogonal group, $SO(2,4)$, to its homomorphic group $SU(2,2)$. Here we present a closed, finite formula for the exponential of a $4\times 4$
traceless matrix, which can be viewed as the generator (Lie algebra elements) of the $SL(4,C)$ group. We apply this result to the $SU(2,2)$ group, which Lie algebra can be represented by the Dirac
matrices, and discuss how the exponential map for $SU(2,2)$ can be written by means of the Dirac matrices.Comment: 10 page
We generalize the quantum spinor wave equation for photon into the curved space-time and discuss the solutions of this equation in Robertson-Walker space-time and compare them with the solution of
the Maxwell equations in the same space-time.Comment: 16 Pages, Latex, no figures, An expanded version of paper published in International Journal of Modern Physics A, 17 (2002) 113
In the present paper we review in a fibre bundle context the covariant and massless canonical representations of the Poincare' group as well as certain unitary representations of the conformal group
(in 4 dimensions). We give a simplified proof of the well-known fact that massless canonical representations with discrete helicity extend to unitary and irreducible representations of the conformal
group mentioned before. Further we give a simple new proof that massless free nets for any helicity value are covariant under the conformal group. Free nets are the result of a direct (i.e.
independent of any explicit use of quantum fields) and natural way of constructing nets of abstract C*-algebras indexed by open and bounded regions in Minkowski space that satisfy standard axioms of
local quantum physics. We also give a group theoretical interpretation of the embedding {\got I} that completely characterizes the free net: it reduces the (algebraically) reducible covariant
representation in terms of the unitary canonical ones. Finally, as a consequence of the conformal covariance we also mention for these models some of the expected algebraic properties that are a
direct consequence of the conformal covariance (essential duality, PCT--symmetry etc.).Comment: 31 pages, Latex2
In this note, we construct a Wess-Zumino-Witten model based on the Galilean conformal algebra in 2-spacetime dimensions, which is a nonrelativistic analogue of the relativistic conformal algebra. We
obtain exact background corresponding to \sigma-models in six dimensions (the dimension of the group manifold) and a central charge c=6. We carry out a Sugawara type construction to verify the
conformal invariance of the model. Further, we discuss the feasibility of the background obtained as a physical spacetime metric.Comment: Latex file, 11 pages, v2: minor changes, references adde
We present a general method to obtain a closed, finite formula for the exponential map from the Lie algebra to the Lie group, for the defining representation of the orthogonal groups. Our method is
based on the Hamilton-Cayley theorem and some special properties of the generators of the orthogonal group, and is also independent of the metric. We present an explicit formula for the exponential
of generators of the $SO_+(p,q)$ groups, with $p+q = 6$, in particular we are dealing with the conformal group $SO_+(2,4)$, which is homomorphic to the $SU(2,2)$ group. This result is needed in the
generalization of U(1) gauge transformations to spin gauge transformations, where the exponential plays an essential role. We also present some new expressions for the coefficients of the secular
equation of a matrix.Comment: 16pages,plain-TeX,(corrected TeX
This paper uses elementary techniques drawn from renormalization theory to derive the Lorentz-Dirac equation for the relativistic classical electron from the Maxwell-Lorentz equations for a classical
charged particle coupled to the electromagnetic field. I show that the resulting effective theory, valid for electron motions that change over distances large compared to the classical electron
radius, reduces naturally to the Landau-Lifshitz equation. No familiarity with renormalization or quantum field theory is assumed
We propose Lagrangian formulation for the particle with value of spin fixed within the classical theory. The Lagrangian turns out to be invariant under non-abelian group of local symmetries. As the
gauge-invariant variables for description of spin we can take either the Frenkel tensor or the BMT vector. Fixation of spin within the classical theory implies $O(\hbar)$-corrections to the
corresponding equations of motion.Comment: 04 pages, notations changed, misprints correcte
The boson mass spectrum of the electro-weak \textbf{$SU(4)_{L}\otimes U(1)_{Y}$} model with exotic electric charges is investigated by using the algebraical approach supplied by the method of exactly
solving gauge models with high symmetries. Our approach predicts for the boson sector a one-parameter mass scale to be tuned in order to match the data obtained at LHC, LEP, CDF.Comment: 12 pages, 1
Table with numerical estimates and 1 Figure added, mistaken results correcte
We adapt the formally-defined Fokker action into a variational principle for the electromagnetic two-body problem. We introduce properly defined boundary conditions to construct a
Poincare-invariant-action-functional of a finite orbital segment into the reals. The boundary conditions for the variational principle are an endpoint along each trajectory plus the respective
segment of trajectory for the other particle inside the lightcone of each endpoint. We show that the conditions for an extremum of our functional are the mixed-type-neutral-equations with implicit
state-dependent-delay of the electromagnetic-two-body problem. We put the functional on a natural Banach space and show that the functional is Frechet-differentiable. We develop a method to calculate
the second variation for C2 orbital perturbations in general and in particular about circular orbits of large enough radii. We prove that our functional has a local minimum at circular orbits of
large enough radii, at variance with the limiting Kepler action that has a minimum at circular orbits of arbitrary radii. Our results suggest a bifurcation at some radius below which the circular
orbits become saddle-point extrema. We give a precise definition for the distributional-like integrals of the Fokker action and discuss a generalization to a Sobolev space of trajectories where the
equations of motion are satisfied almost everywhere. Last, we discuss the existence of solutions for the state-dependent delay equations with slightly perturbated arcs of circle as the boundary
conditions and the possibility of nontrivial solenoidal orbits | {"url":"https://core.ac.uk/search/?q=author%3A(Barut%20A%20O)","timestamp":"2024-11-13T15:07:39Z","content_type":"text/html","content_length":"152316","record_id":"<urn:uuid:643b5094-c466-4ec2-8f9c-38bfbc673270>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00449.warc.gz"} |
January 2008
• 106 participants
• 126 discussions
Hi, Sorry for the flooding, but I finally managed to build an egg and put it on the web, so numscons is available as an egg, now. You should be able to install it using easy_install, e.g easy_install
numscons should work. cheers, David
David Cournapeau wrote: > The current config.h works fine for solaris with Sun compilers, in my > experience, so the problem must be somewhere else. > > Peter, could you post the errors you got ? As
an alternative, I am > working on an alternative build system for numpy: it should work on > solaris (tested on Indiana, and other people reported success on > solaris > 9 and 10). Unfortunately, I
have been redesigning the internals > quite a > lot lately, and I have not tested the changes on solaris yet. If > you are > willing to test, it should be easy for me to make it work again on >
solaris in a few minutes, though. > The problems I was having were due to a bad site.cfg initially and then a problem with the python pkg from sunfreeware (ctypes version mismatch). Numpy is now
happily installed. If you need someone to test anything new, let me know. Thanks! Pete
Hi all, Just a small note. I've updated the `Numpy Example List With Doc`[1] that is linked from the main doc index. Because I'm not always in the loop and the page likes to get out of sync with the
base `Numpy Example List`[2] and NumPy, I've posted some instructions[3] that you might find useful when trying to bring it up-to-date :) [1] http://scipy.org/Numpy_Example_List_With_Doc [2] http://
scipy.org/Numpy_Example_List [3] http://scipy.org/Numpy_Example_List_With_Doc/script cheers, f -- http://filipwasilewski.pl
Hi all, The numpy documentation standard example shows: Parameters ---------- var1 : array_like Array_like means all those objects -- lists, nested lists, etc. -- that can be converted to an array.
var2 : integer Write out the full type long_variable_name : {'hi', 'ho'}, optional Choices in brackets, default first when optional. I'd like to know: 1. "array_like" describes objects that can be
forced to quack like ndarrays. Are there any other such "special" descriptions? 2. How do we specify default values? 3. Why do we need the "optional" keyword (the function signature already indicates
that the parameter is optional). 4. Do we really need the "Other Parameters" list? It would make more sense to split positional and keyword arguments, but I'm not even sure that is necessary, since
that information is already specified in the function signature. 5. Is the {'hi', 'ho'} syntax used when a parameter can only assume a limited number of values? In Python {} is a dictionary, so why
not use ('hi','ho') instead? Thanks for your feedback! Regards Stéfan
Pierre, numpy.compress exists, but numpy.ma.compress does not; is this intentional? Eric
As the subject says, numpy.concatenate doesn't seem to obey -- or even check -- the axis flag when concatenating 1D arrays: ----------------------- <session log> ----------------- In [30]: A =
numpy.array([1, 2, 3, 4]) In [31]: D = numpy.array([6, 7, 8, 9]) In [32]: numpy.concatenate((A, D)) Out[32]: array([1, 2, 3, 4, 6, 7, 8, 9]) In [33]: numpy.concatenate((A, D), axis=0) Out[33]: array
([1, 2, 3, 4, 6, 7, 8, 9]) In [34]: numpy.concatenate((A, D), axis=1) Out[34]: array([1, 2, 3, 4, 6, 7, 8, 9]) In [35]: numpy.concatenate((A, D), axis=2) Out[35]: array([1, 2, 3, 4, 6, 7, 8, 9])
----------------------- </session log> ----------------- However, if you create the same arrays as 2D (1xn) arrays, then numpy checks, and does the right thing: ---------------------- <session log>
------------------- In [36]: A = numpy.array([[1, 2, 3, 4]]) In [37]: D = numpy.array([[6, 7, 8, 9]]) In [38]: A.shape Out[38]: (1, 4) In [39]: numpy.concatenate((A, D)) Out[39]: array([[1, 2, 3, 4],
[6, 7, 8, 9]]) In [40]: numpy.concatenate((A, D), axis=0) Out[40]: array([[1, 2, 3, 4], [6, 7, 8, 9]]) In [41]: numpy.concatenate((A, D), axis=1) Out[41]: array([[1, 2, 3, 4, 6, 7, 8, 9]]) In [42]:
numpy.concatenate((A, D), axis=2) --------------------------------------------------------------------------- <type 'exceptions.ValueError'> Traceback (most recent call last) /fs/home/sdb/<ipython
console> in <module>() <type 'exceptions.ValueError'>: bad axis1 argument to swapaxes ----------------------- </session log> ----------------- Question: Is is a bug or a feature? I'd at least think
that numpy would check the axis arg in the 1D case, and issue an error if the user tried to do something impossible (e.g. axis=2). Whether numpy would promote a 1D to a 2D array is a different
question, and I am agnostic about that one. Cheers, Stuart Brorson Interactive Supercomputing, inc. 135 Beaver Street | Waltham | MA | 02452 | USA http://www.interactivesupercomputing.com/
Pierre GM wrote: > On Wednesday 23 January 2008 16:17:51 you wrote: >> Pierre, >> >> numpy.compress exists, but numpy.ma.compress does not; is this intentional? > > Probably not. I usually don't use
this function, preferring to use indexing > instead. If you have a need for it, I can probably come up with something > relatively soon: the basis would be to apply compress first on the ._data >
part, return a view with the same class as the original object, and update > the mask with compress as needed. The only reason to add it would be backwards compatibility. Evidently it exists in
original numpy.ma. Mike D used it in mpl's path.py until today when someone pointed out that it did not work with the maskedarray branch. (I think I tripped over the same thing a couple days ago, but
worked around it at a higher level.) I agree that it is better *not* to use it, and we can easily strip it out of mpl if it occurs anywhere else; but there may be other user code that will trip over
its absence when 1.05 comes out, so it might be easier to put it in, preserving similarity to numpy as well as old numpy.ma, than to leave it out and have to field future questions about it. Eric
Hi all, Just a quick reminder for all about the upcoming Sage/Scipy Days 8 at Enthought collaborative meeting: http://wiki.sagemath.org/days8 Email me directly (Fernando.Perez(a)Colorado.edu) if you
plan on coming, so we can have a proper count and plan accordingly. Cheers, f
I am experimenting with implementing __array_interface__ and/or __array_struct__ properties for ctypes instances, and have problems to create numpy arrays from them that share the memory. Probably
I'm doing something wrong; what is the correct function in numpy to create these shared objects? I am using numpy.core.multiarray.array(ctypes-object), is that correct? Thanks, Thomas
Greetings: I just noticed a changed behavior of numpy.histogram. I think that a recent 'fix' to the code has changed my ability to use that function (albeit in an unconventional manner). I previously
used the histogram function to obtain counts of each unique string within a string array. Again, I recognize that it is not a typical use of the histogram function, but it did work very nicely for
me. Here's an example: ###numpy 1.0.3 --works just fine >>> import numpy >>> numpy.__version__ '1.0.3' >>> a=numpy.array(('atcg', 'atcg', 'aaaa', 'aaaa')) >>> a array(['atcg', 'atcg', 'aaaa',
'aaaa'], dtype='|S4') >>> b=numpy.unique(a) >>> numpy.histogram(a,b) (array([2, 2]), array(['aaaa', 'atcg'], dtype='|S4')) >>> ###numpy 1.0.4 --no longer functions >>> import numpy >>>
numpy.__version__ '1.0.4' >>> a=numpy.array(('atcg', 'atcg', 'aaaa', 'aaaa')) >>> a array(['atcg', 'atcg', 'aaaa', 'aaaa'], dtype='|S4') >>> b=numpy.unique(a) >>> numpy.histogram(a,b) Traceback (most
recent call last): File "<stdin>", line 1, in <module> File "/opt/libraries/python/python-2.5.1/numpy-1.0.4-gnu/lib/python2.5/site-packages/numpy/lib/function_base.py", line 154, in histogram if(any
(bins[1:]-bins[:-1] < 0)): TypeError: unsupported operand type(s) for -: 'numpy.ndarray' and 'numpy.ndarray' >>> Is this something that can possibly be fixed (should I submit a ticket)? Or should I
revert to some other approaches for implementing the same idea? It really was a nice convenience. Or, alternately, would some sort of new function along the lines of a numpy.countunique() ultimately
be useful? Thanks, -Mark | {"url":"https://mail.python.org/archives/list/numpy-discussion@python.org/2008/1/?page=5","timestamp":"2024-11-06T19:01:33Z","content_type":"text/html","content_length":"106580","record_id":"<urn:uuid:42b87973-e89d-4af6-a468-337634839d68>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00071.warc.gz"} |
Numerical Integration
A homework problem in Chapter 14 of Intermediate Physics for Medicine and Biology states
Problem 28. Integrate Eq. 14.33 over all wavelengths to obtain the Stefan-Boltzmann law, Eq. 14.34. You will need the integral
Equation 14.33 is Planck’s blackbody radiation law and Eq. 14.34 specifies that the total power emitted by a blackbody.
Suppose Russ Hobbie and I had not given you that integral. What would you do? Previously in this blog I explained how the integral can be evaluated analytically and perhaps you’re skilled enough to
perform that analysis yourself. But it’s complicated, and I doubt most scientists could do it. If you couldn’t, what then?
You could integrate numerically. Your goal is to find the area under the curve shown below.
Unfortunately x ranges from zero to infinity (the plot shows the function up to only x = 10). You can’t extend x all the way to infinity in a numerical calculation, so you must either truncate the
definite integral at some large value of x or use a trick.
A good trick is to make a change of variable, such as
When x equals zero, t is also zero; when x equals infinity, t is one. The integral becomes
Although this integral looks messier than the original one, it’s actually easier to evaluate because the range of t is finite: zero to one. The integrand now looks like this:
The colored stars in these two plots are to guide the reader’s eye to corresponding points. The blue star at t = 1 is not shown in the first plot because it corresponds to x = ∞.
We can evaluate this integral using the trapezoid rule. We divide the range of t into N subregions, each extending over a length of Δ t = 1/ N. Ordinarily, we have to be careful dealing with the two
endpoints at t = 0 and 1, but in this case the function we are integrating goes to zero at the endpoints and therefore contributes nothing to the sum. The approximation is shown below for N = 4, 8,
and 16.
The area of the purple rectangles approximates the area under the red curve This approximation gets better as N gets bigger. In the limit as N goes to ∞, you get the integral.
I performed the calculation using the software Octave (a free version of Matlab). The program is:
for i=1:N-1
I found the results shown below. The error is the difference between the numerical integration and the exact result (π^4/15 = 6.4939…), divided by the exact result, and expressed as a percent
These results show that you can evaluate the integral accurately without too much effort. You could even imagine doing this by hand if you didn’t have access to a computer — using, say, N = 16 — and
getting an answer accurate to better than two parts per thousand.
For many purposes, a numerical solution such as this one is adequate. However, 6.4939… doesn’t look as pretty as π^4/15. I wonder how many people could calculate 6.4939 and then say “Hey, I know that
number; It’s π^4/15”! | {"url":"https://bradroth.medium.com/numerical-integration-aa57f7c372d9","timestamp":"2024-11-06T05:11:56Z","content_type":"text/html","content_length":"129883","record_id":"<urn:uuid:fcdf59b5-35fb-4081-a468-5f76cde1139a>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00658.warc.gz"} |
Number Line Rational Numbers Worksheet
Number Line Rational Numbers Worksheet function as foundational devices in the realm of mathematics, providing an organized yet flexible platform for learners to check out and master numerical
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Number Line Rational Numbers Worksheet
Number Line Rational Numbers Worksheet -
Plot the following rational numbers on the number line 1 Plot 1 3 and 1 2 3 on the number line below 2 Plot 2 3 and 1 1 3 on the number line below 3 Plot 1 2 and 1 3 4 on the number line below 4 Plot
1 4 and 1 2 4 on the number line below 5 Plot 1 1 2 and 1 4 6 on the number line below 6 Plot 1 1 3 and 2 5 6 on the number line
Rational Number Worksheet Use your best judgment to place rational numbers on the number line provided First con vert any fractions to decimals Place decimals above the line and fractions below it 1
1 2 2 2 3 11 8 4 9 1 10 0 2 5 7 4 Write a rational number greater than 1 but less than 2
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8 Best Images Of Rational Numbers 7th Grade Math Worksheets Algebra 1 Worksheets Rational
8 Best Images Of Rational Numbers 7th Grade Math Worksheets Algebra 1 Worksheets Rational
Use a number line to compare and order rational numbers or to find missing values For example you can use a number line to show students how fractions decimals and percentages are equivalent or how
they can be added or subtracted Use manipulatives or concrete objects to model fractions decimals or percentages For example you can use
1 Answer 4 3 2 1 2 Negative numbers are located in left side of zero Positive numbers are located in right side of zero The dots indicate each point on the graph The coordinates are 4 3 2 1 2 2
Answer 1 1 5 2 2 5 3 Positive numbers are located in right side of zero The number 1 5 lies between 1 and 2
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Plotting Rational Numbers On A Number Line Worksheets
Plotting Rational Numbers On A Number Line Worksheets
Following points must be kept in the mind to represent the rational numbers on the number line 1 The numbers on the right side of any number on the number line is greater than that on the left 2 Any
number on the left side of a number on the number line is less than that on the right side 3
We represent any rational number on the number line by a point Every positive rational number lies to the right of 0 and every negative rational number lies to the left of 0 on the number line 1 Draw
the number line and represent the following positive rational numbers on it i 1 3 ii 2 3
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Rational Number Worksheet Use your best judgment to place rational numbers on the number line provided First con vert any fractions to decimals Place decimals above the line and fractions below it 1
1 2 2 2 3 11 8 4 9 1 10 0 2 5 7 4 Write a rational number greater than 1 but less than 2
6 Free Plotting Rational Numbers On A Number Line Worksheet
These rational numbers on a number line worksheet will help to learn the technique of plotting rational numbers on a number line through some interesting activities 6th 7th grade students will be
able to plot rational numbers on number lines through various methods and can improve their basic math skills with our free printable
Rational Number Worksheet Use your best judgment to place rational numbers on the number line provided First con vert any fractions to decimals Place decimals above the line and fractions below it 1
1 2 2 2 3 11 8 4 9 1 10 0 2 5 7 4 Write a rational number greater than 1 but less than 2
These rational numbers on a number line worksheet will help to learn the technique of plotting rational numbers on a number line through some interesting activities 6th 7th grade students will be
able to plot rational numbers on number lines through various methods and can improve their basic math skills with our free printable
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