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Project Based Learning Pathways - David Graser
A blog about real life projects suitable for college math courses such as algebra, finite math, and business calculus. Most of these applied math projects include handouts, videos, and other resources for students, as well as a project letter. Graser,
...more>>
ProTeacher - ProTeacher
A teacher-created resource site for elementary schoolteachers, offering lessons and materials, education news, links for finding schools, news and weather, and a reference desk. An index of links to information on various academic subjects includes Mathematics
...more>>
P.T. Chang
Articles by this math professor from the University of Alaska Anchorage's Mat-Su College include "The Ostentatious World of Algebra," "In-Service Training for the Math Teacher of the 21st Century," and "Retesting in a College Remedial Mathematics Course."Pure Math Online
Pure Math Online uses live, personal tutors to help students with algebra and other math homework in real time over the Internet. Program information and signup for free demo are available on the site.
...more>>
Qdos Math - The Education Channel International, Ltd.
An interactive online program delivering elementary mathematics content. Written by teachers, the program includes assessment tests, interactive tuition/demonstration pages and follow-up exercises and activities. Qdos is available via the internet at
...more>>
QuickMath
A site for answering common algebra questions automatically: users enter their e-mail address and a mathematical expression, and decide whether they wish to expand, factorize, or simplify that expression. The answer is computed
automatically and returned
...more>>
the radical rational ... - Pam Wilson
Blog by a National Board Certified Teacher (NBCT) of Algebra II and Geometry "in search of innovative ideas with a well-balanced approach for the math classroom." Posts, which date back to July, 2011, have included "The Purpose of Life and The Quadratic
...more>>
Ralph A. Raimi Homepage - Ralph A. Raimi
The University of Rochester Professor Emeritus' notes on math education and non-technical
writings with a mathematics theme include Whatever became of the New Math?; Have we forgotten the value of memory?, a comment on the New York Regents' B-examination
...more>>
A Recursive Process - Dan Anderson
Anderson's blog, which dates back to June of 2010, has included posts such as "Robocode & Math," "Standards Based Grading," "Cake:Frosting (A look into a proper ratio of real math:cool tech)," "Paper Towels WCYDWT (What Can You Do With This?)," "Two
...more>>
Research Ideas - George Gadanidis
Disseminating math education research through performance, and studying the process and impact of research performance. Funded by the Fields Institute, Random Acts of Math brings math ideas to young children and their parents. Music videos of the Math
...more>>
Resourceaholic - Joanne Morgan
Blog of teaching ideas and resources recommended by a Key Stage 5 Coordinator in a Greater London (UK) girls' grammar school. Posts, which date back to April, 2014, have included a recurring series of "math gems" as well as "Highest Common Factor and
...more>>
Right Brained Math - Brandon Price
"Searching for divergence and creativity in all that math-y stuff." Price teaches algebra, geometry, statistics, and robotics at Austin's Khabele School, an International Baccalaureate (IB) World School. His blog, which dates back to March, 2013, has
...more>>
The Roots of the Equation - James Cleveland
Blog by a high school algebra teacher in New York City "trying to figure out how it all fits together ... [and] creating interdisciplinary problems, projects, and thinking." A Math for America (MfA) fellow, Cleveland has blogged since June, 2011, with
...more>>
Rymar - Rymar Applications
"Helping with math homework and concepts one app at a time." The Cramer's Rule app finds the solution of a system of equations with 2 or 3 unknown variables, and also reveals — via the "Help" button — all of the intermediate determinants.
...more>>
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97800714212 Yourself Calculus
While Teach Yourself Calculus is perfect for beginners who want to acquire a working knowledge of calculus, at the same time it is an excellent tool for anyone who wants to expand their knowledge beyond the basics. In a progressive, step-by-step fashion, the book builds from the ground up to offer comprehensive coverage of a range of more advanced topics such as multiple integrals. Each chapter features numerous worked examples and graded
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Applied Basic Math Worksheets - 2nd edition
Worksheets for Classroom or Lab Practice offer extra practice exercises for every section of the text, with ample space for students to show their work. These lab- and classroom-friendly workbooks also list the learning objectives and key vocabulary terms for every text section, along with vocabulary practice problems54PAPERBACK Good 032169774
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This is a 6 CD program from The Learning Company.
I created these with UltraIso.
Burn to disc or Mount to virtual drive to operate.
. It contains an alphabetical list of formulas and rules that we found particularly useful and it thoroughly explains every problem and answer presented.
However, most information is related through text with a graphic, it does not include games.
Teaching Tools:
This program presents a series of geometry topics with a corresponding multiple�choice quiz. The quizzes include math facts and hints and fully explained solutions.
The math facts section is extremely useful; it contains an alphabetical list of geometry words, formulas and rules.
Topics Covered:
This product, formally sold as The Princeton Review: Math Library, presents general math, Algebra I and II, Statistics, Geometry, Trigonometry and Calculus in six CD ROMs.
The geometry section includes 11 chapters of subjects ranging from reasoning, shapes, coordinates to transformations.
It has a good reasoning section that relates instruction on deductive, inductive, indirect, conditional statements and variants as well as Venn diagrams.
The formula resource areas include over 25 commonly used formulas, so you can easily look up the formula for finding the volume of a sphere or the slope of a line quickly. The rules area includes over 25 rules including triangle congruence theorems, circle secant theorems and parallelogram properties.
Ease of Use/Homework Help:
This product is easy to use but ran a bit slow on our system. This is the only product this high on the review list that will run on Macs.
In terms of homework help, you can quickly look up subject areas, formulas and rules on a particular subject. For our sample homework problem, we were looking for the length of the median in a trapezoid. Through this program, we looked into the trapezoid rule area and quickly found out how to solve the problem, however there was not a sample problem available or a formula. It also did not relate which theorem or property was used to solve the problem.
Help/Documentation:
This program comes with excellent product support including an inter�program user manual, online FAQs and access to telephone and email technical support.
Summary:
Mind Power Math comprehensively covers a broad range of math subjects and is a good product for older children or adults who need to renew their math skills for placement or college entrance exams.
Specializing in Homeschool, educational and early elementary software, books and videos.
**********************************************************************************
You WON'T find anyone else with my content unless they DL'd it from me and re loaded it.
Please Visit my other torrents located at
Please remember not to be dirty leecher and hit and run. I will be honest. I cannot seed all these files alone
and I need your help. Please seed to at least a ratio of 1.5 and much more if possible.
If you don't like Slow Down Load speeds, then help contribute to the total number of complete
copies by seeding to much higher ratios (say 10 or so).
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Comments
Thanks for your interest in my torrent.. please remember that I upload from home and that I have a VERY limited UL speed... I only get to a T1 once a week! Please remember that the more seeders there are the faster the DL's will be... please SEED for a while! Thank you for your understanding!
I've got 4 guys/gals at 93 percent now... it will finish today in a few hours once I get to my T1 connection... I SADLY can't get there often, please check back in 24 hours and I bet there will be a bunch of seeders (unless they are lame and hit and run like I ask them nicely NOT to) cezen
Hi, I'm new to this. When you say "seed" does that mean I just leave utorrent open on my computer so other people can dowload it from me? I want to help since you've been so helpful in uploading these homeschool torrents. Thanks!
I used UltraISO and made an image of each disc in the box. THere were six. I have NO idea where the statistics disc could have dissappeared to. Perhaps you didn't select it to DL when Utorrent popped up?
Upon further inspection, this particular set didn't come with statistics. It is a 6 disc set and doesn't include it. Perhaps this was an early release before they added statistics? or maybe in your region/country they released different stubjects? I'm not sure why you expect there to be statistics. Info please?
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A Classical Introduction to Galois Theory
Books
Galois theory is widely regarded as one of the most elegant areas of mathematics. A Classical Introduction to Galois Theory develops the topic from a historical perspective, with an emphasis on the solvability of polynomials by radicals. The book provides a gradual transition from the computational methods typical of early literature on the subject to the more abstract approach that characterizes most contemporary expositions.
The author provides an easily-accessible presentation of fundamental notions such as roots of unity, minimal polynomials, primitive elements, radical extensions, fixed fields, groups of automorphisms, and solvable series. As a result, their role in modern treatments of Galois theory is clearly illuminated for readers. Classical theorems by Abel, Galois, Gauss, Kronecker, Lagrange, and Ruffini are presented, and the power of Galois theory as both a theoretical and computational tool is illustrated through:
A study of the solvability of polynomials of prime degree
Development of the theory of periods of roots of unity
Derivation of the classical formulas for solving general quadratic, cubic, and quartic polynomials by radicals
Throughout the book, key theorems are proved in two ways, once using a classical approach and then again utilizing modern methods. Numerous worked examples showcase the discussed techniques, and background material on groups and fields is provided, supplying readers with a self-contained discussion of the topic.
A Classical Introduction to Galois Theory is an excellent resource for courses on abstract algebra at the upper-undergraduate level. The book is also appealing to anyone interested in understanding the origins of Galois theory, why it was created, and how it has evolved into the discipline it is today.
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Beginning Algebra (Student Solutions Manual) - 4th edition
ISBN13:978-0131493391 ISBN10: 0131493396 This edition has also been released as: ISBN13: 978-0131444928 ISBN10: 0131444921
The Student Solutions Manual contains detailed step-by-step solutions to odd-numbered section exercises; solutions to every (odd and even) Mental Math exercise; solutions to odd-numbered Calculator Exploration exercises; and solutions to every (odd and even) exercise found in the Chapter Reviews and Chapter Tests
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TRIO-SSS Introductory Math Courses
Math Courses 3 semester hours each
TRIO math classes help students get over the fear of math, build confidence, and prepare for their required college math courses.
AE 0131-Math I reviews fractions and decimals and introduces negative numbers and linear equations. Math I does not count toward graduation requirements, but if you NEED it to pass the next course, take it now. It's FREE.
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More About
This Textbook
Overview
Alan Tucker's newest issue of Applied Combinatorics builds on the previous editions with more in depth analysis of computer systems in order to help develop proficiency in basic discrete math problem solving. As one of the most widely used book in combinatorial problems, this edition explains how to reason and model combinatorically while stressing the systematic analysis of different possibilities, exploration of the logical structure of a problem, and ingenuity.
Editorial Reviews
From The Critics
This text for undergraduate and beginning graduate students in computer science and mathematics covers the theory and application of combinatorial reasoning. Tucker (mathematics, Stony Brook) begins with a discussion of the elements of graph theory before moving on to enumeration. The systematic analysis of different possibilities, the exploration of the logical structure of a problem, and ingenuity are emphasized throughout the text. Annotation c. Book News, Inc., Portland, OR (booknews.com)
From the Publisher
"...a well-structured text that addresses a broad range of topics... It is well presented, written clearly and easy to follow." (Times Higher Education Supplement, November 2007
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...
More About
This Book
between the characters from Math Girls, offering a fun way to learn this serious content. Each chapter comes with review problems and answers, and an appendix gives more challenging, open-ended problems for learners wanting to push the limits of their understanding. This book is most suited to middle- or high-school students who have learned basic algebra, or older readers who want to brush up on forgotten math skills.
This series came about through requests from readers who enjoyed the excitement of learning aspects of the Math Girls series, but found themselves unprepared to keep up with the mathematical content. We hope that the books in this series will help young mathematicians firm up vital math skills that will allow them to excel in more advanced
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MATLAB Student Version
04/01/03
Students in engineering, math or science have a new technical computing resource designed for their needs. The MathWorks' MATLAB Student Version includes full-featured versions of MATLAB and Simulink, the software products used by engineers, scientists and mathematicians at leading universities, research labs, technology companies and government labs. MATLAB integrates computation, data analysis, visualization and programming in one environment. Simulink is one of the leading interactive environments for modeling, simulating and analyzing dynamic systems. In addition, there is no difference between the student and professional versions of the program, which, according to the company, is important because students are learning skills with the same tools they may use in a professional arena. The program also comes with MATLAB and Simulink books to help students get started. This product has a special student price of $99. The MathWorks, (508) 647-7000
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Linear Algebra : A ModernDavid Poole's innovative book prepares students to make the transition from the computational aspects of the course to the theoretical by emphasizing vectors and geometric intuition from the start. Designed for a one- or two-semester introductory course and written in simple, "mathematical English" the book presents interesting examples before abstraction. This immediately follows up theoretical discussion with further examples and a variety of applications drawn from a number of disciplines, which reinforces the practical utility of the math, and helps students from a variety of backgrounds and learning styles stay connected to the concepts they are learning. Poole's approach helps students succeed in this course by learning vectors and vector geometry first in order to visualize and understand the meaning of the calculations that they will encounter and develop mathematical maturity for thinking abstractly.
Vectors
Introduction: The Racetrack Game
The Geometry and Algebra of Vectors
Length and Angle: The Dot Product
Exploration: Vectors and Geometry
Lines and Planes
Exploration: The Cross Product
Applications: Force Vectors
Code Vectors
Vignette: The Codabar System
Systems of Linear Equations
Introduction: Triviality
Introduction to Systems of Linear Equations
Direct Methods for Solving Linear Systems
Exploration: Lies My Computer Told Me
Exploration: Partial Pivoting
Exploration: Counting Operations: An Introduction to the Analysis of Algorithms
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This is at heart a fairly conventional transition text, but it has a number of features that encourage students to practice and improve their mathematical reading, writing, and proving skills. I doubt that it's possible for students to learn these skills only from a book: they need a careful reader to critique their work, so this book needs to be used with a teacher or other knowledgeable critic and not for individual study. The first edition appeared (and was reviewed) in 2003; this second edition has the same approach and outline, with a modest amount of additional material, and twice as many problems as before.
The book is primarily concerned with an exposition of those parts of mathematics in which students need a more thorough grounding before they can work successfully in upper-division undergraduate courses. This means that most of the book deals with logic and set theory, with some study of the real numbers and completeness. Layered on top of this conventional material are a number of features to encourage the students to build their skills in reading, writing, and proving. The book leads by example: the proofs that are given are written out fully and carefully, and are intended to be a model the students can use to construct their own proofs. The remaining material, which comprises most of the book, is presented in the form of exercises (easy, and solved in the text) and problems (harder, and including some proofs). There are several "nontheorems" and "not-a-proofs" that are fallacious and whose fallacies the student is intended to uncover.
The book ends with a series of longer projects from several areas of mathematics (not just those covered in the body of the work), that the student is supposed to figure out and write up carefully. These are very structured, in the sense that they are already broken down into a series of simpler steps. If you think (as I do) that this simplifies the task too much, you may want to supplement this text with more open-ended projects, such as those in Writing Projects for Mathematics Courses: Crushed Clowns, Cars, and Coffee to Go by Crannell et al.
Bottom line: a mathematically-conventional but pedagogically-innovative take on transition courses.
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Painless Algebra
(Barron's Painless-1 jacket DEMONSTRATES A PAINLESS WAY TO LEARN ALGEBRA, FROM INTEGERS TO QUADRATIC EQUATIONS.
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About the Book
(back cover) PAINLESS Algebra Second Edition Really. This isnat going to hurt at all . . . If you break out in a cold sweat at the very mention of algebra, this book will show you how to relax and master the subject without pain. The problems you now consider confusing are taken slowly, step by step. Simply follow the painless steps and soon youall wonder why you were ever confused in the first place! You start with integers, progress to simple equations with one variable, then move forward to gain a clear understanding of inequalities, systems of equations, exponents, and roots and radicals. Finally, you graduate to mastery of quadratic equations. Along the way, youall find fun-to-solve abrain ticklera problems in every chapter. Each abrain ticklera comes with an answer, but before you knot it, youall be figuring out all the answers for yourself. For Middle School and High School Students
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Grade 8 problems and questions on circles with answers. ... Circle problems and questions with answers for grade 8 are presented. These problems and questions deal ...
Released Test Questions Math 4 February 2009 CALIFORNIA STANDARDS TEST GRADE Introduction - Grade 4 Mathematics The following released test questions are taken from ...
Best Answer: Since the length of arc s (the length in question) is equal to the radian measure, we have: C/r = 2 C is given as 16 and so 16/r = 2 16/2 = r ...
SECONDARY SCHOOL CURRICULUM 2010 Vol 1 Main Subjects Effective from the academic session 2008-2009 of Class IX and for the Board Examination (Class X) to be held in ...
10.4.10 Contributed photoMichigan Maritime Museum Volunteer of the Year Dick Brunvand (center) enjoys the dinner and camaraderie with Bill Durant at the Museum ...
The Math Forumu0027s Internet Math Library is a comprehensive catalog of Web sites and Web pages relating to the study of mathematics. This page contains sites relating ...
8.10.11Students on deans listDavenport UniversityThe following area students were named to winter semester deans list at Davenport University in Grand Rapids ...
The City of Minnetonka website is your 24/7 connection to Minnetonka City Services.
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Tim Chartier
New Publications
Math Bytes: Google Bombs, Chocolate-Covered Pi, and Other Cool Bits in Computing, published by Princeton University Press, offers hands-on activities to engage in mathematical topics that involve computing. A handful of chocolate can create an estimate to π. Learn math behind sports ranking and bracketology, use m&ms to explore ideas from Calculus, and learn how to create a resizable computer font. Want a closer look? Get more details at Amazon and see the great reviews the book's getting.
Big Data: How Data Analytics Is Transforming the World, published by the Teaching Company, presents the fundamentals and, possibly even more, the mindset of data analytics. From infographics to sports ranking to marketing, this course covers a wide array of topics. You learn techniques so you can do data analytics on your own and also learn stories about companies like Facebook, Netflix, and JCPenney. The lectures were designed to suit beginner and more advanced mathematicians alike. Get more details at The Great Courses website.
When Life is Linear: from computer graphics to bracketology, published by the Mathematical Association of America, delves into applications of linear algebra largely in the context of computer graphics and data mining. Art Benjamin (Harvey Mudd College) calls the book "the perfect supplement to a linear algebra course." Tony DeRose (Pixar Animation Studios) calls the book a "refreshing change as it is driven by real-world applications that are inspiring and familiar..." Get more details at the MAA's website.
Introduction
I am an Associate Professor in the Department of Mathematics and Computer Science
at Davidson College. I hold both a B.S. degree in applied mathematics and a
M.S. degree in computational mathematics from Western Michigan
University. I received my Ph.D. in applied mathematics from the University of Colorado at
Boulder. During 2001-2003 I held a VIGRE postdoctoral position in the Mathematics Department
at the University of Washington. For more information on my activity, see my
curriculum vitae
(last modified January 2015).
Online Publications
The following articles were published either in the Journal of Online Mathematics and its Applications or Loci, but are all available through Loci as this journal is an aggregation of a few online journals
of the Mathematical Association of America. The following articles all contain interactive java applets.
Mountains of Fractals - Learn to produce coastlines in 2D and mountains in 2D and 3D by adapting ideas related to the construction of fractals. Introductory mathematical issues in random number generation are discussed. Java applications allow interactive exploration, (published in the Journal of Online Mathematics and its Applications).
Modeling a Changing World (co-authored with N. Dovidio) - Events in our world are often modeled with differential equations. This article develops numerical solvers for ODEs. Interactive applets explore topics including slope fields and numerically solving definite integrals and differential equations, (published in the Loci).
Integer Programming Model for the Sudoku Problem (co-authored with A. Barlett, A. Langville and T. Rankin) - Let binary integer linear programming solve Sudoku puzzles and variations on the game for you. Also, learn to prove theorems regarding puzzle creation. Java applets allow for interactive exploration, and exercises and challenge problems are included, (published in the Journal of Online Mathematics and its Applications).
Awards and Recognitions
MAA Math Ambassador -- in January 2014, Michael Pearson, Executive Director of the Mathematical Association of America, named me the inaugural MAA Math Ambassador. It is an honor and joy to share mathematics through this role.
2007 Henry L. Alder Award for Distinguished Teaching by a Beginning College or University Mathematics Faculty Member -- awarded by the
Mathematical Association of America. To read more about the award, click here.
Alfred P. Sloan Research Fellowship -- awarded by the
Alfred P. Sloan Foundation. To read more about the award, click here. To read an article about my receiving the award, click here.
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11 Actions
Which Mechanics book is the best for beginner in math major? Generally, how do you select or say filter the perfect books ? Because when I need to learn one topic, I always search the term at Amazon, and it shows tons of books, but I don't know how to find out some best books. Is there some tips or criteria for that ? Like try to use the book written by famous masters in the certain area, say Nobel prize winner. Like Abel said that, read the masters not the pupils, especially today we have too many things to learn about within limited time and energy. But it is not the case that we can find a book in a certain topic written by famous masters in it.
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books.google.com - Combinatorial design theory is a vibrant area of combinatorics, connecting graph theory, number theory, geometry, and algebra with applications in experimental design, coding theory, and numerous applications in computer science.This volume is a collection of forty-one state-of-the-art research articles... Design Theory
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Each activity requires only a few minutes and will help you pass this course and become a better math student. Many of these skills can be carried over to other disciplines and help you become a model college student. To begin, write down the following information:
a. Instructor's name
b. Instructor's office number
c. Instructor's telephone number
d. Instructor's e-mail address
e. Instructor's office hours
f. Days of the week that the class meets
g. The room number in which the
h. Is there a lab requirement for this course? How class meets often must you attend lab and where is it located
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This is the third in a series of three booklets designed to teach students about basic mathematical and physical principles such as logarithms, geometry, and size and scale. Each book is loosely based on the NASA GLAST...
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Product Description
Instead of memorizing formulas and equations, Videotext Algebra helps students to understand math through mastery learning, encouraging them to solidify each concept before moving on the next. A copy of the print materials needed for this module is included.
Module F in Videotext Algebra, this unit covers:
Parabolas (Origins, Standard Form 1, 2, 3, & 4), Intercepts)
Circles (Standard Form, Non-standard Form)
Ellipses (Standard Form, Non-standard Form)
Hyperbolas (Standard Form, Non-standard Form)
Systems (One 1st Degree, One 2nd Degree, Two 2nd Degree)
Problem Solving (Numbers, Geometric Figures)
Exponential Functions (Functions of x & y)
Logarithmic Functions(Translating, Graphs of Solution, Sets)
Operations (Properties, Finding Logarithms, Computation)
Solving Equations (Exponential, Logarithmic)
Unit Tests IX and X included
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Math for Physics Review
This course is offered every fall to review mathematical concepts and techniques to help prepare you for your first-year physics or astronomy course. Topics include linear, quadratic, and simultaneous equations, graphical representation, exponents and logarithms, trigonometric functions and identities, units, scientific notation, significant figures and vectors (different material will be covered each evening).
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Advantages to MathExcel: Better grades and retention
The MathExcel calculus class at the University of Kentucky is designed to give capable students the best chance to excel in this crucial course and have more successful college careers. MathExcel students volunteer for special sections of calculus that encourage collaborative learning in a personal and communal atmosphere. All students in calculus take the same tests and are graded on the same standards
MathExcel students get much better grades:
MathExcel students graduate from UK at a higher rate1:
Note that MathExcel Students do not have exceptional natural ability: their Math ACT score is always within two points of the rest of the class, and usually closer. MathExcel, by far the best calculus class at UK, is composed almost entirely of women, minorities, and rural students. In Fall 1995, 40 of the 68 MathExcel students were women, 10 were African-Americans, and 36 were rural students2. There was 1 white urban male. In Fall 1992 five of the top 6 students in Calculus I were in MathExcel. Four of these were women and the man was from a rural district. The top 2 overall were MathExcel women from rural districts.
These are the classes which have had at least 6 years to graduate. There were 17 MathExcel students in 1990, 52 in 1991, and 77 in 1992. The MathExcel total overall is 146.
Many of the women were from rural areas. Each year there are from 1 to 4 white males from urban schools.
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The purpose of this collection of applets and activities is to make students familiar with the basic principles of complex numbers. Combining explanatory text, exercises and interactive GeoGebra applets, this resource is suitable for both classroom lectures and distance learning.
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saxon math lesson answers
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From Wikipedia
Saxon (teaching method)
Saxon math, developed by John Saxon, is a teaching method for incremental learning of mathematics. It involves teaching a new mathematical concept every day and constantly reviewing old concepts. Early editions were deprecated for providing very few opportunities to practice the new material before plunging into a review of all previous material. Newer editions typically split the day's work evenly between practicing the new material and reviewing old material. Its primary strength is in a steady review of all previous material, which is especially important to students who struggle with retaining the math they previously learned.
In all books before Algebra 1/2 (the equivalent of a Pre-Algebra book), the book is designed for the student to complete assorted mental math problems, learn a new mathematical concept, practice problems relating to that lesson, and solve a varied number of problems which include what the students learned today and in select previous lessons -- all for one day's class. This daily cycle is interrupted for tests and additional topics. In the Algebra 1/2 book and all higher books in the series, the mental math is dropped, and tests are given more frequently.
The Saxon math program has a specific set of products to support homeschoolers, including solution keys and ready-made tests, which makes it popular among some homeschool families. It has also been adopted as an alternative to reform mathematics programs in public and private schools. Saxon teaches familiar algorithms and uses familiar terminology, unlike many reform texts, which also contributes to its popularity.
Replacing standards-based texts
By the mid 2000s, many school districts were considering abandoning experiments with reform approaches which had not produced acceptable test scores. For example, school board member Debbie Winskill in Tacoma, Washington said that the non-traditional Interactive Mathematics Program (IMP) "has been a dismal failure." Speaking to the board, Mount Tahoma High School teacher Clifford Harris noted that he taught sophomores in another district Saxon Math, and their Washington Assessment of Student Learning scores have continually climbed. Unlike IMP, Saxon program gives students plenty of chances to review material so they retain their skills, he said. In September 2006, Tacoma Public Schools introduced the Saxon books district-wide and rejected the previous IMP textbooks.
From Yahoo Answers
Question:
Answers:your teacher may be the best answer. did you understand the lesson, review the samples.
Question:
Answers:What is the question????
Question:Is there a place online where I could find the answers to the tests. My little brother lost his correction book and we need to correct one of his tests before he moves on. He is in Saxon Algebra 1/2 second edition.
I don't want to buy another book, just find the answers.
Answers:Post this question in the homeschool section of questions. I know quite a few homeschoolers use Saxon math!
Question:
Answers:Check the back of the book for the odd problems.
OR GOOGLE !!!! LOL
From Youtube
Saxon Math Upper Grades :I wish all curriculum choices were as easy as this! Saxon has proven to have the best mathematics text available today. Every school in the US that has used them has raised college board scores 20% and has tripled Calculus enrollment, has doubled Physics enrollment and has reduced the number of kids on the slow track by over 50%. The reason? Normal math books introduce a new concept with each chapter, then drop that topic as the next chapter goes on to something new. This week, it's multiplication; next week it is fractions, and I can't remember multiplication! In Saxon books, the concepts are practiced daily, over and over, until they are fully automatic and math becomes easier and more enjoyable. Upper Grades 9 - 12: includes a non-consumable hardcover Textbook, Tests and Answer Key For Pre-Algebra through Calculus. Hardcover textbook has approx 125 lessons plus additional topics. Answers to odd numbered problems are in the back of the book. Approximately 500 pages. Tests and Answer Key included. The Solutions Manual (sold separately) which gives step-by-step solutions to all the practices and problems.
(Saxon Algebra 2) Lesson 45 Practice Problem B.MOV :Check out my blog at: myjeepisfun.wordpress.com/math-homework ...where you will find all my work neatly organized with a digital copy of my work done on paper. Second Edition Book
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course is divided into four learning units and an appendix. The course begins with a unit that provides a review of...
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This course is divided into four learning units and an appendix. The course begins with a unit that provides a review of algebra specifically designed to help and prepare the student for the study of calculus. The second unit discusses functions, graphs, limits, and continuity. The third unit will introduce and explain derivatives. The fourth unit makes visual sense of derivatives by discussing derivatives and graphs. The appendix provides a large collection of reference facts, geometry, and trigonometry that will assist in solving calculus problems long after the course is over. Note that this courseMathematics 005A web hosted courseware of undergraduate single and many-variable calculus for physics and engineering students, with...
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A web hosted courseware of undergraduate single and many-variable calculus for physics and engineering students, with animated and interactive graphics. It is based on a course "Mathematical Introduction for Physicists" of the Tel-Aviv University.In addition to text, examples and exercises, the courseware takes advantage of modern technology with interactive and animated graphics (over 110) that can be projected in class, and accessed at any time by the students.Math Animated is technically based on SVG and MathML - open standards, developed by the Web Consortium, without the need of proprietary software. It runs on the most popular platforms: Windows, Mac and Linux.About 10% of the material, including text and graphics is open and does not require any registration Animated to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Math Animated
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Course HighlightsMath I is a mandatory course for students specializing in natural sciences in Junior Divisions (1st & 2nd...
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Course HighlightsMath I is a mandatory course for students specializing in natural sciences in Junior Divisions (1st & 2nd years) of the College of Arts and Sciences. Math I is offered in "Math IA" and "Math IB". Both classes cover almost the same content although Math IA focuses on theoretical aspects while Math IB emphasizes in practical computation.In Math I, students learn differential and integral calculus to acquire basic knowledge on differential and integral required for university-level study. Thus, differential and integral of high-school mathematics are prerequisite for taking this IB ( Differential and Integral Calculus) to your Bookmark Collection or Course ePortfolio
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Multivariable Calculus is an expansion of Single-Variable Calculus in that it extends single variable calculus to higher...
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Multivariable Calculus is an expansion of Single-Variable Calculus in that it extends single variable calculus to higher dimensions. This course begins with a fresh look at limits and continuity, moves to derivatives and the process of generalizing them to higher dimensions, and finally examines multiple integrals (integration over regions of space as opposed to intervals). This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Mathematics 103 online course offered by the Saylor Foundation.'Precalculus II continues the in-depth study of functions...
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This is a free online course offered by the Saylor Foundation.'Precalculus II continues the in-depth study of functions addressed in Precalculus I by adding the trigonometric functions to your function toolkit. In this course, you will cover families of trigonometric functions, as well as their inverses, properties, graphs, and applications. Additionally, you will study trigonometric equations and identities, the laws of sines and cosines, polar coordinates and graphs, parametric equations and elementary vector operations.You might be curious how the study of trigonometry, or "trig," as it is more often referred to, came about and why it is important to your studies still. Trigonometry, from the Greek for "triangle measure," studies the relationships between the angles of a triangle and its sides and defines the trigonometric functions used to describe those relationships. Trigonometric functions are particularly useful when describing cyclical phenomena and have applications in numerous fields, including astronomy, navigation, music theory, physics, chemistry, and—perhaps most importantly, to the mathematics student—calculus.In this course, you will begin by establishing the definitions of the basic trig functions and exploring their properties and then proceed to use the basic definitions of the functions to study the properties of their graphs, including domain and range, and to define the inverses of these functions and establish the properties of these. Through the language of transformation, you will explore the ideas of period and amplitude and learn how these graphical differences relate to algebraic changes in the function formulas. You will also learn to solve equations, prove identities using the trig functions, and study several applications of these functions Multivariable Calculus is the second course in the series, consisting of 26 videos, 4 Study Guides, and a set of Supplementary Notes. The series was first released in 1971-007 Calculus Revisited: Multivariable Calculus (MIT) to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material RES.18-007 Calculus Revisited: Multivariable Calculus (MIT)
Select this link to open drop down to add material RES.18-007 Calculus Revisited: Multivariable Calculus (MIT) Complex Variables, Differential Equations, and Linear Algebra is the third course in the series, consisting of 20 Videos, 3 Study Guides, and a set of Supplementary Notes. Students should have mastered the first two courses in the series (Single Variable Calculus and Multivariable Calculus) before taking this course. The series was first released in 1972, but equally valuable today for students who are learning these topicsCalculus Revisited is a series of videos and related resources that covers the materials normally found in a freshman-level...
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Calculus Revisited is a series of videos and related resources that covers the materials normally found in a freshman-level introductory calculus course. The series was first released in 1970.006 Calculus Revisited: Single Variable Calculus (MIT) to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material RES.18.006 Calculus Revisited: Single Variable Calculus (MIT)
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Geometry: Seeing, Doing, Understanding, 3rd Edition
9780716743613
ISBN:
0716743612
Edition: 3rd Pub Date: 2003 Publisher: W. H. Freeman
Summary: Jacobs innovative discussions, anecdotes, examples, and exercises to capture and hold students' interest. Although predominantly proof-based, more discovery based and informal material has been added to the text to help develop geometric intuition.
Harold R. Jacobs is the author of Geometry: Seeing, Doing, Understanding, 3rd Edition, published 2003 under ISBN 9780716743613 and 0716743612. One hundred thirty ...seven Geometry: Seeing, Doing, Understanding, 3rd Edition textbooks are available for sale on ValoreBooks.com, twelve used from the cheapest price of $35.91, or buy new starting at $119.11 [more
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Tag Archives: tex
In my Linear Algebra class we use a lot of MATLAB — including on our timed tests and all throughout our class meetings. I want to stress to students that using professional-grade technological tools is an essential part of learning a subject whose real-life applications closely involve the use of those tools. However, there are a few essential calculations in linear algebra, the understanding of which benefits from doing by hand. One of those calculations is row-reduction. Nobody does this by hand; but doing it by hand is useful for understanding elementary row operations and for getting a feel for the numerical processes that are going on under the hood. And it helps with understanding later concepts, notably that of the LU factorization of a matrix.
I have students take a mastery exam where they have to reduce a 3×5 or 4×6 matrix to reduced echelon form by hand. They are not…
Over the weekend a minor smack-talk session opened up on Twitter between Maria Andersen and about half a dozen other math people about MathType versus \(\LaTeX\). Maria is on record as being pro-MathType and yesterday she claimed that \(\LaTeX\) is "not intuitive to learn". I warned her that a pro-\(\LaTeX\) blog post was in the offing with those remarks, and so it comes to this. \(\LaTeX\) is accessible enough that every math teacher and every student in a math class at or above Calculus can (and many should) learn \(\LaTeX\) and use it for their work. I have been using \(\LaTeX\) for 15 years now and have been teaching it to our sophomore math majors for five years. I can tell you that students can learn it, and learn to love it.
Why use \(\LaTeX\) when MathType is already out there, bundled with MS Word and other office programs, tempting us with its…
There seem to be two pieces of technology that all mathematicians and other technical professionals use, regardless of how technophobic they might be: email, and \(\LaTeX\). There are ways to typeset mathematical expressions out there that have a more shallow learning curve, but when it comes to flexibility, extendability, and just the sheer aesthetic quality of the result, \(\LaTeX\) has no rival. Plus, it's free and runs on every computing platform in existence. It even runs on WordPress.com blogs (as you can see here) and just made its entry into Google Documents in miniature form as Google Docs' equation editor. \(\LaTeX\) is not going anywhere anytime soon, and in fact it seems to be showing up in more and more places as the typesetting system of choice.
But \(\LaTeX\) gets a bad rap as too complicated for normal people to use. It seems to be something people learn …
Good article here at The Productive Student giving five reasons why students should use \(\LaTeX\) as their word processor and not Microsoft Word:
1. Never worry about formatting again.
2. It looks way better. [By the way: Very nice article on LaTeX's typesetting at that link.]
3. It won't crash: LaTeX is basically a plain text file. You can edit it anywhere, in any text editor, and it basically can't crash on you. File size is very small which makes it very portable.
4. It's great for displaying equations, which is why it's the leading standard among sciencitifc scholars.
5. It fits in with the workflow of a student and allows you to do one thing well: Write.
The writer also shares some of his practices for writing papers (not necessarily math or science papers) with \(\LaTeX\), stressing \(\LaTeX\)'s ability to handle bibliographic data as the "killer feature"….
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Revamp Math Ed
The level of mathematics achievement in our public schools is eroding. Educational advancement in the U.S. has traditionally lagged Europe and Asia in the lower grades. Society accepted this and partially rationalized it because we teach all of our children rather than those showing early promise. By grade 9, our students scored well against the rest of the world.
More recently, however, this disadvantage has become general throughout our public schools. Ironically, this lag is co-dependent with a concurrent trend to teach higher-level mathematics in lower grades. How can these two trends coexist? We contend that the answer is "dummying down" of the subject material. Long division is not often taught in the lower grades. Algebra is now being taught in the seventh grade, but with little factoring. The majority of 12-year old students aren't prepared for this type of abstract thinking. Teachers certified for junior high school usually have an elementary-education background and are not trained to teach the material. The early algebra introduction is then followed by geometry in eighth grade. Again, neither the student nor the teacher is properly prepared. The result: an induction-based thought process, without formal proofs, development of analytical thinking, or logic. Then comes an Algebra II course without proper prerequisite material, followed by advanced math and calculus, again lacking the proper foundation. Textbook companies conspire in this, properly stressing PC presentation but lacking rigor.
How important is this problem? Dilution of our math curricula is certainly one of the primary reasons. Secondary mathematics is the primary resource for teaching our students to think objectively, logically, and analytically.
What is the remedy? Include more repetition of low-level math skills in lower grades. Students must be taught math skills both with and without the hand-held calculator. Math with the computer is also helpful, but only as an adjunct to traditional skills. Encourage a thorough introduction to algebra in the eighth grade. Fill the JHS void with skills in business and personal math fundamentals, such as balancing a checkbook, and engage in activities that build visual thinking skills to prepare for the abstractions to follow. Move Algebra I back to the ninth grade for most students, building on the skills introduced the previous year. This moves geometry to 10th grade, where mature minds can grapple with abstract concepts, and Algebra II to eleventh grade. This allows for both ample repetition and inclusion of some of the elements of algebra now considered "optional." Twelfth grade leaves room for trigonometry, probability and statistics, and a reasonably thorough introduction to differential calculus of, perhaps, six weeks. This recommendation is exclusive of advanced "Honors" classes. There is advantage in honors math only if it is truly "advanced". When parental wishes and other political issues temper qualification, it not only loses its purpose, but forces weaker curricula to allow slower students to catch up. Ideally, honors Algebra I in the eighth grade will eventually lead the successful student to advanced placement calculus in the twelfth and on into college. Otherwise, it has little purpose
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Calculator Two for One
Sharp doubles the potential of its EL-9900 scientific graphing calculator by equipping it with a unique two-sided keypad that has a simple side for basic operations and a complex side for advanced algebra and statistics. To change the calculator's keypad, just remove it, flip it over and snap it back in place. The EL-9900 a powerful device with 827 built-in mathematical functions, 132-by-64 dot screen and 64KB of memory yet it weighs less than 9 ounces and sells for under $100.
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On Core Mathematics
: On Core Mathematics is a complete companion program for transitioning your students to the Common Core State Standards through interactive math worksheets. This program is designed to be a supplement to your current math curriculum to help students master the Common Core Standards. On Core Mathematics offers teachers a flexible way to fill in any gaps between other math curriculums and the new standards. Teachers can selectively use only the specific lessons needed or use the entire student workbook for comprehensive coverage. On Core provides a complete Common Core Standards solution.
On Core Mathematics utilizes interactive, real-world applications that help students deepen their understanding of crucial math concepts, while addressing the Common Core Curriculum and the Standards for Mathematical Practice. On Core Mathematics is a comprehensive, ready-made resource providing instruction, practice and assessment for each Common Core State Mathematics Standard at every grade level.
On Core Mathematics for grades k-12 prepares students for assessments by developing their procedural, application, and critical thinking skills. On Core Mathematics develops student understanding of the Common Core State Standards for Mathematics in an interactive format. Each grade level includes a Student Worktext and Teacher Edition. In grades K-5, students actively participate in lessons that use math manipulatives and step-by-step example problems to develop understanding of the Common Core State Standards for Mathematics. In grades 6 through Algebra 2, students gain a deeper understanding of the Common Core State Standards for Mathematics by exploring, interacting, and reflecting on the skills and concepts presented. Published by Houghton Mifflin Harcourt.
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Education
'Complex Numbers and the Theory of Equations' presents a comprehensive understanding of the subject of equations at undergraduate level. Using a large number of solved examples and an equally ample stock of exercises with hints and answers, this book will lead students smoothly from an introductory to an advanced stage. A copious number of figures and applications are included, and have been designed to guide students toward a comprehension of the theory of equations and to aid in solving related problems.
This Glossary is a guide to the literary terms most relevant to students and readers of English literature today, thorough on the essentials and generous in its intellectual scope. The definitions are lively and precise in equipping students and general readers with a genuinely useful critical vocabulary. It identifies the thinking and controversies surrounding terms, and offers fresh insights and directions for future reading. It does this with the help of extensive cross-referencing, indexes and up-to-date bibliography (with recommended websites).
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prob
lem-solving sections appear throughout the book. Problem-solving skills are fostered with the help of an interactive, iterative presentation style: Here's the problem. How can we solve it? How can we improve the solution?
problem-solving sections appear throughout the book. Problem-solving skills are fostered with the help of an interactive, iterative presentation style: Here's the problem. How can we solve it? How can we improve the solution? ...show less
2008 Paperback Fair SOME HIGHLIGHTING AND MARKINGS MADE THROUGHOUT BOOK This book is in acceptable conditon. It is a good reading copy for personal use if you want to save some money but do not try...show more and give as a giftPanther Bookstore Milwaukee, WI
2007 Paperback Good30470270073047023
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0131861379 2005 Prentice Hall, 3rd Edition. Brand new hardcover copy, still sealed in the original publisher's shrinkwrap. "Starting with nothing more than basic high school ...algebra, this volume leads readers gradually from basic algebra to the point of actively performing mathematical research while getting a glimpse of current mathematical frontiers. Features ?Continued Fractions, Square Roots and Pell?s Equation.? The sale of this item benefits the Andover Public Library. 100% satisfaction guaranteed. We process and ship orders daily, securely and with delivery confirmation.Read moreShow Less
More About
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Overview
Starting with nothing more than basic high school algebra, this volume leads readers gradually from basic algebra to the point of actively performing mathematical research while getting a glimpse of current mathematical frontiers.
Features "Continued Fractions, Square Roots and Pell's Equation." Contains additional historical material, including material on Pell's equation and the Chinese Remainder Theorem.
Editorial Reviews
Booknews
Silverman (Brown U.) originally wrote the book as a text for a course designed to attract non-science majors with little interest in pursuing the standard calculus sequence, and convince them to study some college mathematics. He expects readers to have some facility with high school algebra and access to a calculator, though he points out that those who know how to program a computer have great fun generating reams of data and implementing assorted algorithms. He mentions concepts from calculus now and then, but does not lay them down as barriers to cross. The first edition appeared in 1997
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Math Fundamentals 16 Copies):
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About the Book
Designed to follow state and national standards for middle-school math curricula, Math Fundamentals 1, 2, 3 & 4 will work equally well together as a group of four guides or each as a standalone guide. All four 2-panel guides were written by members of the National Council of Teachers of Mathematics. Bright colors and graphic details offer proven ease of learning to this age group. Time-tested tips help students learn more effectively and avoid common pitfalls. This guide covers: Number theory, operations & measurements. Topics include: Computation, place value, Roman numerals, order of operations, scientific notation, symbols, fraction operations, divisibility rules, factorization, equivalent percents, decimals & fractions, metric system, rate conversions, time, temperature and much more!
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tudents
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The Law of Sines. The Law of Cosines. Trigonometric Form for Complex Numbers. De Moivre's Theorem and nth Roots of Complex Numbers. Vectors. The Dot Product. Chapter 8 Review Exercises. Chapter 8 Discussion Exercises.
9. SYSTEMS OF EQUATIONS AND INEQUALITIES.
Systems of Equations. Systems of Linear Equations in Two Variables. Systems of Linear Equations in More than Two Variables. Partial Fractions. Systems of Inequalities. Linear Programming. The Algebra of Matrices. The Inverse of a Matrix. Determinants. Properties of Determinants. Chapter 9 Review Exercises. Chapter 9 Discussion Exercises.
Appendix I: Using a TI-82/83 Graphing Calculator. Appendix II: Common Graphs and Their Equations. Appendix III: A Summary of Graph Transformations. Appendix IV: Graphs of the Trigonometric Functions and Their Inverses. Answers to Selected Exercises. Index of Applications. Index.
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Students can use this exercise book to learn Maple software, learn calculus, or both, as fewer than 90 Maple commands are necessary to do all of the activities. Drill exercises and rote manipulation are replaced with conceptual learning activities and an exploratory interaction with mathematics not seen in traditional courses. Calculus activities cover basic plotting, Taylor polynomials, logistic growth, and the Lagrange multiplier. Lacks an index and a
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Description
Do you want your child to look forward to learning algebra? Using Switched-On Schoolhouse 9th Grade Math from Alpha Omega Publications, your child will! How does this dynamic, computer-based homeschool curriculum work? It's very simple and easy. Just install, and in minutes your child will have a whole years' worth of algebra lessons ready at his fingertips. Must-have lessons in this Alpha Omega curriculum include engaging multimedia tools—like video clips, learning games, and animation—all integrated into content to encourage student learning! You won't find that in other traditional math textbooks.
But wait, Switched-On Schoolhouse has much more. This dynamic, cutting-edge homeschool curriculum is packed with time-saving features parents will absolutely love—like automatic grading and lesson planning, a built-in calendar, and handy message center. We've just made homeschooling easier, and much more exciting! In this math course, your child will receive an in-depth, comprehensive introduction to algebra and basic algebraic principles. Plus, Switched-On Schoolhouse 9th Grade Math has integrated, step-by-step solution keys when viewing problems from the SOS Teacher application. Come see how fun homeschooling and teaching algebra can really be. Just order Switched-On Schoolhouse 9th Grade Math
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Functions, Data, and Models
Sheldon P. Gordon and Florence S. Gordon
TEXTBOOK*
Functions, Data, and Models is a college-level algebra textbook that is written to provide the kind of mathematical knowledge and experience that students will need for courses in other fields such as biology, chemistry, business, finance, economics, and other areas that are heavily dependent on data either from laboratory experiments or from other studies. The book focuses on fundamental mathematical concepts and realistic problem-solving via mathematical modeling rather than the development of algebraic skills.
Functions, Data and Models presents college algebra in a way that differs from almost all college algebra books available today. The authors teach something new rather than covering the same ground as high school courses. By changing the content of the course, the authors are able to give students an introduction to data analysis and mathematical modeling that even students with limited algebraic skills can handle. The book contains rich exercises, many of which use real data. Also included are thought experiments or what if questions that are meant to stretch the student's mathematical thinking.
* As a textbook, Functions, Data, and Models does have DRM. Our DRM protected PDFs can be downloaded to three computers. Please note that the secure PDFs will open only on the Mac and Windows operating systems. iOS (iPad & iPhone) and Linux are not supported at this time. Click here for more information..
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Mathematics: A Practical Odyssey With Cd-rom And 1pass for Ilrn Tutorial/ Mentor/ Student Book Companion Site
Summary
Users discover the many ways in which mathematics is relevant to their lives with MATHEMATICS: A PRACTICAL ODYSSEY, 7E and its accompanying online resources. They master problem-solving skills in such areas as calculating interest and understanding voting systems and come to recognize the relevance of mathematics and to appreciate its human aspect.
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Featured Research
from universities, journals, and other organizations
Algebra Adds Value To Mathematical Biology Education
Date:
August 3, 2009
Source:
Virginia Tech
Summary:
As mathematics continues to become an increasingly important component in undergraduate biology programs, a more comprehensive understanding of the use of algebraic models is needed by the next generation of biologists to facilitate new advances in the life sciences, according to researchers.
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As mathematics continues to become an increasingly important component in undergraduate biology programs, a more comprehensive understanding of the use of algebraic models is needed by the next generation of biologists to facilitate new advances in the life sciences, according to researchers at Sweet Briar College and the Virginia Bioinformatics Institute (VBI) at Virginia Tech.
Related Articles
VBI Professor Reinhard Laubenbacher and Sweet Briar College Mathematical Sciences Professor Raina Robeva have highlighted algebraic models as one of the diverse mathematical tools needed in the professional development of up-and-coming life scientists in a new article in Science. Despite this critical need, the authors explain, algebraic models have played a less substantial role in undergraduate curricula than other methods.
Future generations of biologists will routinely use mathematical and computational approaches to develop and frame hypotheses, design experiments, and analyze results. Sound mathematical models are essential for this purpose and are currently used in the field of systems biology to understand complex biological networks. Two types of mathematical models, in particular, have been successfully used in biology to reproduce network structure and dynamics: Continuous-time models derived from differential equations (DE models) focus on the kinetics of biochemical reactions, while discrete-time algebraic models built from functions of finite-state variables focus on the logic of the connections of network variables. According to Laubenbacher and Robeva, while DE models have been included more often in undergraduate curricula integrating mathematics and biology, algebraic models should also be viewed as an important training component for students at all education levels.
"Discrete-time algebraic models created from finite-state variables, such as Boolean networks, are increasingly being used to model a variety of biochemical networks, including metabolic, gene regulatory, and signal transduction networks," says Laubenbacher. "Often, researchers do not have enough of the information required to build detailed quantitative models. Algebraic models need less information about the system to be modeled, making them useful for instances where quantitative information may be missing. All the work that goes into building them can then be used to construct detailed kinetic models, when additional information becomes available. In addition, algebraic models are much more intuitive than differential equations models, which makes them more easily accessible to life scientists."
Using algebraic models is a relatively quick, easy and reliable way for students to integrate mathematical modeling into their life sciences coursework. Creating algebraic models of biochemical networks requires only a modest mathematical background, which is usually provided in a college algebra course. Without the complexities involved in teaching students how to construct more complicated models, algebraic models make the introduction of mathematical modeling into life sciences courses more accessible for faculty members as well.
According to Robeva, "The exciting thing about algebraic models from an educational perspective is that they highlight aspects of modern-day biology and can easily fit in both the biology and mathematics curricula. At the introductory level, they provide a quick path for introducing biology students to constructing and using mathematical models in the context of contemporary problems such as gene regulation. At the more advanced level, the general study and analysis of such models often require sophisticated mathematical theories. This makes them perfect for inclusion into mathematics courses, where the biology can provide a meaningful framework for many of the abstract structures. As educators, we should actively be looking for the best ways to seize this opportunity for advancing mathematical biology." Researchers have collected feedback from more than 240 biologists in the U.S. and developed a new, detailed core concept template called BioCore Guide. The guide is intended to provide an updated ... full story
Apr. 4, 2014 — A long-standing challenge in synthetic biology has been to create gene circuits that behave in predictable and robust ways. Mathematical modeling experts and experimental biologists have now created ... full story
Sep. 16, 2013 — Researchers have found high school students in the United States achieve higher scores on a standardized mathematics test if they study from a curriculum known as integrated ... full story
Mar. 21, 2011 — The way in which teachers and textbooks use language and different metaphors in mathematics education determines how pupils develop their number sense, according to new research from
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Review your lecture notes before each lecture. (For example,
when I ask specific questions about the previous lecture, you should be
able to answer them without looking at your notes.)
You should have all electronic devices (phones, i-whatevers, MP3 players,
ear-buds, etc.) completely turned off and stowed before coming to class.
Recording devices are strictly prohibited. Using electronic
devices during class without my permission may result in their being
confiscated and in academic discipline.
Although I teach multiple sections of MATH-305, university policy
requires that you attend only the section for which you are
registered.
Consequently, you may not "float" from one section to another as a matter of
convenience.
If you miss a class, please obtain copies of the lecture notes from a classmate.
(April 13, 1743 was Thomas Jefferson's birthday. He and John Adams died within
mere
hours of one another on July 4, 1826... the 50th anniversary of the
adoption of the Declaration of Independence.)
Write the Maple code for the
Bisection Method.
April 13
NOTE: Do this immediately, and play with the code by changing the
starting interval, the tolerance, even the function. You will use this
code as the template for writing the codes for other methods and for our
first programming assignment.
Do all the examples in the 8th Maple tutorial entitled
Formatted Printing and Plot
Options.
April 14
Then change your Maple code for the bisection method so that it uses
formatted printing. From now on we will use the
printf command for printing.
Do all the examples in the second Maple tutorial entitled
Solutions of Equations.
April 16
You should complete Assignment 1 before doing this one.
Remember that these assignments will acquaint (or reacquaint)
you with Maple and prepare you for the programming assignments.
You should rework Exam 1 immediately after it
is returned.
(Recall that a score below 64% is an F. See the course policy.)
The median score was 82%!
(So half the students scored above 82 and half scored below 82.)
Write the Maple code for the trapezoidal rule.
May 4
NOTE: Do this immediately, and play with the code by changing the
number of subintervals, the limits of integration, even the integrand
f(x).
You will use this code as the template (model) for writing the codes for
other methods and for the second programming assignment.
an image of one's face taken by oneself or by another person using a
digital camera or phone, especially for posting on social networking
sites or smartphones for personal identification.
a photo ID showing only the face.
First Known Use of FACIE – 16:34 UTC,
October 12, 2014 by Kevin G. TeBeest, Michigan USA
Formerly: "profile photo" (archaic) Usage: Professor TeBeest sent a photo of himself playing his
drums
to his brother who wanted a photo ID for his smartphone.
The brother whined saying, "Send me a photo of your ugly face you stupid. .
.!"
So Professor TeBeest sent his brother a facie. Etymology: French façade
("a false, superficial, or artificial appearance or effect,"
Merriam–Webster); Italian facciata, a derivative of
faccia ("front"), from Latin facies ("face");
Geographical Use: worldwide
Not to be confused with selfie, which is a photo taken by oneself of
ones own body or part of the body, usually due to vanity.
The photo on your state driving license is an example of a facie, although
it is not a selfie.
Inform your friends and family! Let's make it
go viral.
Start using it in conversations and online and explain it when they ask you
what it means. It's fun!
Remember that:
You are responsible for successfully completing all assigned
problems in all your courses.
The exams may include problems similar to these assignments and
lecture examples and may include questions about Maple.
We must maintain a steady pace to cover the material
that constitutes Math-305. If you have difficulty with a section, be
sure to see me for help immediately.
No matter how simple a topic appears when you see my examples or read
the text, you will almost certainly have difficulty completing an exam
if you do not practice the examples and do the assignments beforehand.
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is designed for teachers to use in the classroom. Ancient Chinese math is difficult to understand, but this book is very practical in that it shows teachers important ways of thinking about math that students have not seen. For example, the Chinese used the concept of dissection of geometric figures to solve problems in surveying and geometry that use no angles, are surprisingly simple, and make students solve problems in different ways. There are a variety of problems in basic math, algebra, and number theory. The book is designed to be reproducible to use in the classroom. Highly recommended.
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Pre-Calculus Workbook For Dummies
Get the Full eBook
Synopsis
Get the confidence and math skills you need to get started withcalculus
Are you preparing for calculus? This hands-on workbook helps youmaster basic pre-calculus concepts and practice the types ofproblems you'll encounter in the course. You'll get hundreds ofvaluable exercises, problem-solving shortcuts, plenty of workspace,and step-by-step solutions to every problem. You'll also memorizethe most frequently used equations, see how to avoid commonmistakes, understand tricky trig proofs, and much more.Pre-Calculus Workbook For Dummies is the perfect tool foranyone who wants or needs more review before jumping into acalculus class. You'll get guidance and practical exercisesdesigned to help you acquire the skills needed to excel inpre-calculus and conquer the next contender-calculus.
Serves as a course guide to help you master pre-calculusconcepts
Covers the inside scoop on quadratic equations, graphingfunctions, polynomials, and more
Covers the types of problems you'll encounter in yourcoursework
With the help of Pre-Calculus Workbook For Dummies you'lllearn how to solve a range of mathematical problems as well assharpen your skills and improve your performance.
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Why have calculators? Because outside of maths exams, people are going to need to know how to deal with complicated decimal numbers and apply them to real world problems, and they will probably either use calculators or spreadsheets to do it. So having a small part of the test actually involve that doesn't seem a bad idea to me.
Yes, but then they still can punch in the acos 37. The point is to teach them to get to the point of acos 37, and that can also be done with the solution being acos 2pi or something.
I mean we could also just give them a link to wolfram alpha, and let them write down the output, but that's not the point. The point is logical, systematical thinking and learning to solve problems with given knowledge. Calculators don't help that process, they only help evaluating the exact output.
I really wish that my high school, and honestly even my undergraduate institution, had taken this sort of approach versus the number crunching nonsense.
Now that I'm taking graduate level courses, I realized how handicapped I was by my calculator. The courses feel much more difficult (actually have to work through the examples!) but they are so much more rewarding.
Applied problems can often require the calculation of a number as one step in a larger question. Yes, we could give them a link to Wolfram Alpha to get to this intermediate step. But we could also let them use a calculator. If we're not worried about their ability to calculate the answer in their head for this question, what exactly is the problem with using a calculator? People seem to be desperate to solve an "issue" without explaining what the issue actually is.
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The Third Edition of THE MATHEMATICAL PALETTE also features a robust suite of online course management, testing, and tutorial resources for instructors and students. This includes BCA/iLrn Testing and Tutorial, Personal Tutor with SMARTHINKING live online tutoring, and a Book Companion Web Site featuring online graphing calculator resources.
Additional Product Information
Features/Benefits
Each chapter is a self-contained unit, and chapters may be taught in any order according to the instructor or student's interests.
The history section at the beginning of each chapter presents material and questions that are ideal for classroom discussions, research papers, projects, speeches, and presentations.
Chapter 3, "Geometry and Art," provides a thorough, engaging, visual introduction to geometry that focuses on mathematics in art.
Artwork and photographs that relate to the mathematics being discussed make the text visually stimulating. The text includes works by Andre Derain, Mary Cassatt, Vincent van Gogh, and many more.
RESEARCH QUESTIONS and PROJECTS at the end of each history section go beyond the material and serve as an ideal source for projects, written papers, and group cooperative learning experiences.
Short history segments throughout each chapter present the development of the mathematics being discussed.
Explanations of algebra concepts are integrated into the chapters.
Problem sets, organized into Explain-Apply-Explore categories, include a variety of writing and skill-building exercises, and their unique organization ensure that students understand the mathematical concepts involved.
Summaries at the end of each chapter present terminology, formulas, and objectives, and serve as a comprehensive review of the chapter.
Consistent use of scientific calculators eliminates the time spent on paper-and-pencil computation.
Chapter-ending summaries, reviews, tests, and accurate answer sections provide students with a comprehensive review of the terminology, formulas, and objectives used in the chapter.
Table of Contents
1. NUMBERS-OLD AND NEW. Overview. A Short History of Numbers and Numerals. Chapter 1 Projects. Ancient Systems of Numeration. Hindu-Arabic System and Fractions. Numeration Systems with Other Bases. The Numbers of Technology. Types of Numbers. Summary. Review. Test. 2. LOGICAL THINKING. Overview. A Short History of Logic. Chapter Two Projects. Logic, Statements and Definitions. Inductive and Deductive Reasoning. Symbolic Logic and Truth Tables. Logic and Flowcharts. Logic and Puzzles. Chapter Two Summary. Chapter Two Review. Chapter Two Test. 3. SETS AND COUNTING. Overview. A Short History of Sets. Chapter Three Projects. Sets: Finite and Infinite. Set Operations and Venn Diagrams. Applications of Sets. Introduction to Counting. Summary. Review. Test. 4. PROBABILITY. Overview. A Short History of Probability. Chapter Four Projects. Intuitive Concepts of Probability. Calculating Probabilities. Probability and Odds. Probability of Compound Events. Conditional Probability. Expected Value. Summary. Review. Test. 5. STATISTICS AND THE CONSUMER. Overview. A Short History of Statistics. Chapter Five Projects. Arranging Information. Measures of Central Tendency. Measures of Dispersion. The Normal Distribution. Polls and the Margin of Error. Regression and Forecasting. Summary. Review. Test. 6. MODELING WITH ALGEBRA. Overview. A Short History of Algebra. Chapter Six Projects. Linear Models. Quadratic Models. Exponential Models. Logarithmic Models. Summary. Review. Test. 7. GEOMETRY AND ART. Overview. A Short History of Geometry. Chapter Seven Projects. Euclidean and Non-Euclidean Geometry. Perspective. Golden Ratios and Rectangles. Polygons and Stars. Tessellations. Fractals. Summary. Review. Test. 8. TRIGONOMETRY: A DOOR TO THE UNMEASURABLE. Overview. A Short History of Trigonometry. Chapter Eight Projects. Right Triangles, Sine, Cosine, Tangent. Solving Right Triangles. Right Triangle Applications. Laws of Sines and Cosines. Acute Triangle Applications. The Motion of a Projectile. Summary. Review. Test. 9. FINANCE MATTERS. Overview. A Short History of Interest. Projects. Percents. Simple Interest. Compound Interest. Annuities. Loans. Summary. Review. Test. 10. MATH FROM OTHER VISTAS. Overview. Differential Calculus. Integral Calculus. Pascal--Yang Hui Triangle. Voting Systems. Apportionment. Linear Programming.
What's New
History sections throughout the book have been revised to emphasize general developments in mathematics rather than detailed names and dates.
Projects at the beginning of each chapter are now divided into two groups--Research Projects and Math Projects.
Review problems, keyed to individual sections, have been added to the end-of-chapter materials.
New coverage includes the numbers of technology in Chapter 1, logic and puzzles in Chapter 2, the motion of a projectile in Chapter 8, and more.
A new Chapter 3, "Sets and Counting," expands the material on sets found in the previous edition. This chapter also includes an introduction to combinatorics.
Chapter 4, "Probability," includes more applications to non-game-related situations.
Chapter 5, "Statistics and the Consumer," now includes a section on regression and forecasting.
Chapter 6, "Modeling with Algebra," now includes many examples from sports.
Chapter 9, "Finance Matters," begins with a section on percents that includes markups, markdowns, and commissions.
More real-life problems and data are used throughout the text.
The chapter on calculus has become Chapter 10, "Math from Other Vistas," which contains views from the past (differential calculus, integral calculus, the Pascal-Yang Hui Triangle) and views of the present (voting systems, apportionment, linear programming). Resource Manual contains a brief summary of the important concepts, terms, and formulas and answers to all Explain/Apply/Explore/Review/Test problems. It also includes detailed solutions to the odd-numbered Apply and Explore problems and all chapter-ending Review and Test problems. In addition, the Instructor's Resource Manual contains material that can supplement the traditional lecture and discussion approach commonly used in mathematics classrooms. It contains suggestions for writing assignments, history discussions, guest speakers, presentations, and math projects. It also has evaluation forms for projects and presentationsList Price = $29.95
| College Bookstore Wholesale Price = $22.50
Mastering Mathematics: How to Be a Great Math Student
(ISBN-10: 0534349471 | ISBN-13: 9780534349479)
Providing solid tips for every stage of study, Mastering Mathematics stresses the importance of a positive attitude and gives students the tools to succeed in their math course. This practical guide will help students: avoid mental blocks during math exams, identify and improve areas of weakness, get the most out of class time, study more effectively, overcome a perceived "low math ability," be successful on math tests, get back on track when feeling "lost," and much more!
Written by Cynthia Arem (Pima Community College), this comprehensive workbook provides a variety of exercises and worksheets along with detailed explanations of methods to help "math-anxious" students deal with and overcome math fears.
This special version of the complete student text contains a Resource Integration Guide that correlates the supplementary items for THE MATHEMATICAL PALETTE, THIRD EDITION. With this easy-to-use tool, you can quickly compile a teaching and learning program that complements both the text and your personal teaching style. The annotated instructor's edition also includes answers printed next to all respective exercises. Graphs, tables, and other answers appear in a special answer section in the back of the textList Price = $29.95
| College Bookstore Wholesale Price = $22.50
Mastering Mathematics: How to Be a Great Math Student
(ISBN-10: 0534349471 | ISBN-13: 9780534349479)
Provides solid tips for every stage of study, stressing the importance of a positive attitude.Meet the Author
About the Author
Ronald Staszkow
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The Geometer's Sketchpad® Version 5
Special Offer Links
Dynamic Geometry® Software for Exploring Mathematics
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3 to College
The Geometer's Sketchpad is a dynamic construction, demonstration, and exploration tool
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your students can use this software program to build and investigate
mathematical models, objects, figures, diagrams, and graphs.
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Math Formulary
Education
Screenshots
Developer Notes
Math Formulary covers all mathematical formulas that are usually used in the school and the university. Where necessary graphics are included to depict and explain the topic better.
The content is organized into following groups:
- Basic Arithmetic
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About the Book
Now in its fifth edition, A Mathematics Sampler presents mathematics as both science and art, focusing on the historical role of mathematics in our culture. It uses selected topics from modern mathematics including computers, perfect numbers, and four-dimensional geometry to exemplify the distinctive features of mathematics as an intellectual endeavor, a problem-solving tool, and a way of thinking about the rapidly changing world in which we live. A Mathematics Sampler also includes unique LINK sections throughout the book, each of which connects mathematical concepts with areas of interest throughout the humanities. The original course on which this text is based was cited as an innovative approach to liberal arts mathematics in Lynne Cheney's report, "50 HOURS: A Core Curriculum for College Students," published by the National Endowment for the Humanities.
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Why? IDEA MATH provides in-depth enrichment in important mathematical areas, particularly in the fields from which contest
problems are drawn: algebra, combinatorics, geometry, and number theory. We believe students learn best by discovering,
rather than simply memorizing, essential theorems and techniques. Our process puts students at the center of the classes,
as students work together to develop their ideas. Our instructors lead this process by teaching vital ideas, giving meaningful examples,
and guiding students through the problem-solving process.
How? We apply seminar-style teaching in our classes. Class sizes are small to ensure that students receive plenty of
individual attention. Students are expected to work as a group and present their solutions and ideas to the class.
What are some unique strengths of IDEA MATH that distinguish it from other (summer/winter/online/weekend) math programs?
The unique strengths of IDEA MATH can be summarized as follows:
1. Comprehensive and well-tested teaching materials, modeled after the training materials of the Phillips Exeter Academy math team
2. Strong team of instructors, consisting of front-line teachers, coaches, and top former students
3. Emphasis on student leadership and the development of an attitude conducive to learning
In our summer and weekend programs, in order to meet the needs of the students with different mathematical backgrounds,
we always offer nine major/core series:
PC1, PC2, PC3, PC4, UC1, UC2, UC3, RC1, RP1.
In each of these programs, for the most advanced students, we also offer selected series (or selected topics from these series):
RC2, RC3, EC1, EC2, RP2, RP3, EP1, EP2, EP3.
In this IDEA MATH Summer Program at the Greater Boston Area, for interested students who have completed 9th grade and have been placed into the RC1 course or above, they can take Physics for half of the day and math for the other half. These courses are listed as:
Please visit the curriculum grid for course descriptions.
If you are not certain about the exact course you should take, make sure that the detailed academic background information
is updated in your account and register for a course that is relatively fit your background and needs. After you are enrolled (admitted
and tuition paid in full), our academic team will work with you to find the best fit course for you to study at our program.
Schedule
July 3rd (arrival day for residential students) to July 18th (departure day for residential students)
Note: We will offer a $200 tuition scholarship/discount to each residential student ($100 tuition scholarship/discount to each day student)
who meets any one (or more) of the following statuses in either 2014 or 2015 or both: MATHCOUNTS National Round participant,
USAJMO qualifier, and USAMO qualifier. A copy of proof is needed.
Day students who wish to attend supervised evening recreational and academic activities at the residential program may do so
for an additional fee. The cost will be $200 (including dinner and all activities).
* The application is ONLY condiered completed when the application fee (applied to all students) is paid in full.
** Tuition paid after this date will depend on space availability and extra charges may apply.
Please log in to your account, then click the tuition button to pay the tuition online.
Transportation arrangements to the program
For residential students, we will provide airport transportation from Boston Logan Airport to the program location between 11:00 A.M. – 5:00 P.M. on July 3rd,
2015. If students arrive outside this window and need transportation, the transportation service will be arranged with a $50 surcharge.
For day students, there will be two bus routes to the program location; Bus 1 from Lexington High School to DWC; Bus 2 from Westford Valley Plaza
to DWC. The bus fee is $100 per person for entire program and $5 per ride.
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At first glance, this is once again a reference request for "How to start machine learning".
However, my mathematical background is relatively strong and I am looking for an introduction to machineI am just finishing grade 12 pre-calculus at my school and have strong interest in math. The problem is, it seems some important elements of higher level math are not in my schools curriculum that areSuppose we have a non-linear autonomous system of two equations:
$$\begin{cases} x'(t) = F(x,y) \\ y'(t) = G(x,y) \end{cases} $$
and we obtain a fixed point for this equation, but the eigenvalues of
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Description: Learn
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modeling, and problem-solving. Offered cooperatively by Portland State
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8th Grade Algebra Ms by cby79555
8th Grade Algebra Ms. Eddy
The curriculum for 8th grade Algebra students utilizes
mathematical concepts learned in 6th and 7th grades, while
introducing new concepts needed in preparation for high
school. (This class counts as a high school credit, only for
students earning 84% or above. This A or B does count as
part of their high school GPA.)
One of the most important aspects of any middle school
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Please encourage your son or daughter to seek extra help as
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Lisa.eddy@scott.kyschools.us
Text: Algebra (Glencoe Mathematics)
Website: (wonderful examples and
practice quizzes)
Grading: Quizzes: 20-30%
Tests: 20-30%
Class work & homework: 30-50%
Projects: 0-10%
Course objectives:
To provide students with a foundation of basic skills
and operations.
To master the four content areas as approved by the
Kentucky Department of Education, KCM and NCM:
*geometry/measurement
*probability/statistics
*number sense/computation
*algebraic ideas
To develop enjoyment and appreciation of
mathematics.
To develop techniques for problem solving and
applying those techniques to a variety of situations.
To communicate mathematical ideas and information
using appropriate notation and terminology.
To recognize, make or apply the connections among
mathematical concepts to other academic disciplines
as well as the real world.
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Precalculus, CourseSmart eTextbook, 9th Edition
Description
Mike Sullivan's time-tested approach focuses students on the fundamental skills they need for the course: preparing for class, practicing with homework, and reviewing the concepts. In the Ninth Edition, Precalculus has evolved to meet today's course needs, building on these hallmarks by integrating projects and other interactive learning tools for use in the classroom or online. New Internet-based Chapter Projects apply skills to real-world problems and are accompanied by assignable MathXL exercises to make it easier to incorporate these projects into the course. In addition, a variety of new exercise types, Showcase examples, and video tutorials for MathXL exercises give instructors even more flexibility, while helping students build their conceptual understanding. CourseSmart textbooks do not include any media or print supplements that come packaged with the bound book.
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Short Course In Discrete Mathematics
9780486439464
ISBN:
0486439461
Pub Date: 2004 Publisher: Dover Pubns
Summary: What sort of mathematics do I need for computer science? In response, a pair of professors at the University of California at San Diego created this text. Explores Boolean functions and computer arithmetic; logic; number theory and cryptography; sets and functions; equivalence and order; and induction, sequences, and series. Assumes some familiarity with calculus. Original 2005 edition.
Bender, Edward A. is ...the author of Short Course In Discrete Mathematics, published 2004 under ISBN 9780486439464 and 0486439461. Two hundred seven Short Course In Discrete Mathematics textbooks are available for sale on ValoreBooks.com, fifty seven used from the cheapest price of $6.50, or buy new starting at $12.11.[read more
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Mathematics for Grob Basic Electronics
9780078271281
0078271282
Summary: Provides students with the mathematical principles needed to solve numerical problems in electricity and electronics. 13 chapters cover keeping track of the decimal point when multiplying and dividing; working with fractions; manipulating reciprocals; finding powers and roots of a number; powers of 10; logarithms; metric system; solving equations; trigonometry; binary and hexadecimal numbers; and complex numbers. ...> Bernard Grob is the author of Mathematics for Grob Basic Electronics, published 2002 under ISBN 9780078271281 and 0078271282. Five Mathematics for Grob Basic Electronics textbooks are available for sale on ValoreBooks.com, three used from the cheapest price of $20.38, or buy new starting at $116
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More About
This Textbook
Overview
The Rockswold/Krieger algebra series fosters conceptual understanding by using relevant applications and visualization to show students why math matters. It answers the common question "When will I ever use this?" Rockswold teaches students the math in context, rather than The authors believe this approach deepens conceptual understanding and better prepares students for future math courses and life.
Product Details
ISBN-13: 9780321567994
Publisher: Pearson
Publication date: 1/10/2012
Edition description: New Edition
Edition number: 1
Pages: 864
Product dimensions: 8.50 (w) x 10.80 (h) x 1.00 (d)
Meet the Author
Gary Rockswold has been a professor and teacher of mathematics, computer science, astronomy, and physical science for over 35 years. He has taught not only at the undergraduate and graduate college levels, but he has also taught middle school, high school, vocational school, and adult education. He received his BA degree with majors in mathematics and physics from St. Olaf College and his Ph.D. in applied mathematics from Iowa State University. He has been a principal investigator at the Minnesota Supercomputer Institute, publishing research articles in numerical analysis and parallel processing and is currently an emeritus professor of mathematics at Minnesota State University, Mankato. He is an author for Pearson Education and has over 10 current textbooks at the developmental and precalculus levels. His developmental coauthor and friend is Terry Krieger. They have been working together for over a decade. Making mathematics meaningful for students and professing the power of mathematics are special passions for Gary. In his spare time he enjoys sailing, doing yoga, and spending time with his family.
Additional information about Gary Rockswold can be found at
Terry Krieger has taught mathematics for 18 years at the middle school, high school, vocational, community college and university levels. His undergraduate degree in secondary education is from Bemidji State University in Minnesota, where he graduated summa cum laude. He received his MA in mathematics from Minnesota State University - Mankato. In addition to his teaching experience in the United States, Terry has taught mathematics in Tasmania, Australia and in a rural school in Swaziland, Africa, where he served as a Peace Corps volunteer. Terry is currently teaching at Rochester Community and Technical College in Rochester, Minnesota. He has been involved with various aspects of mathematics textbook publication for more than 14 years and has joined his friend Gary Rockswold as coauthor of a developmental math series published by Pearson Education. In his free time, Terry enjoys spending time with his wife and two boys, physical fitness, wilderness camping, and trout
| 677.169 | 1 |
Geared toward helping students visualize and apply mathematics, Elementary Algebra: Graphs and Models uses illustrations, graphs, and graphing technology to enhance students' mathematical skills. This is accomplished through Interactive Discoveries, Algebraic/Graphical Side-by-Sides, and the incorporation of real-data applications. In addition, students are taught problem-solving skills using the Bittinger hallmark five-step problem-solving process coupled with Connecting the Concepts and Aha! exercises. And, as you have come to expect with any Bittinger text, we bring you a complete supplements package that now includes an Annotated Instructor's Edition and MyMathLab, Addison-Wesley's online course solution.
The goal of Elementary and Intermediate Algebra: Concepts and Applications, 4e is to help todays students learn and retain mathematical concepts by preparing them for the transition from skills-oriented elementary and intermediate algebra courses to more concept-oriented college-level mathematics courses, as well as to make the transition from skill to application. This edition continues to bring your students a best-selling text that incorporates the five-step problem-solving process, real-world applications, proven pedagogy, and an accessible writing style. The Bittinger/Ellenbogen/Johnson series has consistently provided teachers and students with the tools needed to succeed in developmental mathematics. This revision has an even stronger focus on vocabulary and conceptual understanding as well as making the mathematics even more accessible to students. Among the features added are new Concept Reinforcement exercises, Student Notes that help students avoid common mistakes, and Study Summaries that highlight the most important concepts and terminology from each chapter.
KEY MESSAGE: Building on its reputation for accurate content and a unified system of instruction, the Fifth Edition of Bittinger/Ellenbogen/Johnsons Prealgebra paperback integrates success-building study tools, innovative pedagogy, and a comprehensive instructional support package with time-tested teaching techniques.
The Student Solutions Manual contains completely worked-out solutions with step-by-step annotations for all the odd-numbered exercises in the exercise sets in the text, with the exception of the thinking and writing exercises. It also includes complete, worked-out solutions to all end-of-chapter material.
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Discrete mathematics, also called finite mathematics or decision mathematics, is the study of mathematical structures that are fundamentally discrete in the sense of not supporting or requiring the notion of continuity. Objects studied in finite mathematics are largely countable sets such as integers, finite graphs, and formal languages. Concepts and notations from discrete mathematics are useful to study or describe objects or problems in computer algorithms and programming languages.״At this point only one of the planned 15 modules is currently available, that on Groups Discrete Mathematics to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Computational Discrete Mathematics
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'This review book, used in conjunction with free online YouTube videos, is designed to help students prepare for free and open online course in Elementary Algebra was produced by the WA State Board for Community & Technical...
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This free and open online course in Elementary Algebra was produced by the WA State Board for Community & Technical Colleges [ course is the study of basic algebraic operations and concepts and the structure and use of algebra. This includes the solutions to algebraic equations. factoring algebraic functions, working with rational expressions, yourThis free and open online course in Linear Algebra was produced by the WA State Board for Community & Technical Colleges...
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This free and open online course in Linear Algebra was produced by the WA State Board for Community & Technical Colleges [ course is the study of basic algebraic operations and concepts and the structure and use of algebra. This includes the solutions to algebraic equations, factoring algebraic expressions, working with rational expressions Al free and open online course in Logic was produced by the WA State Board for Community & Technical Colleges...
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This free and open online course in Logic was produced by the WA State Board for Community & Technical Colleges [ as an academic subject, is the systematic study of the standards of correct reasoning. In short, logic is the theory of reasoning. This fully online course is a comprehensive introduction to logic with an emphasis on modern logical theory Logic to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Introduction to Logic
Select this link to open drop down to add material Introduction to LogicLogic & Proofs is an introduction to modern symbolic logic. It provides a rigorous presentation of the syntax and semantics of sentential and predicate logic. However, the distinctive emphasis is on strategic argumentation. Students learn effective strategies for constructing natural deduction proofs. This learning is supported by the Carnegie Proof Lab: it provides a sophisticated interface, in which students can give arguments by strategically guided forward and backward steps and Proofs to your Bookmark Collection or Course ePortfolio
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Algebra Level 1
By Clay Cat Designs
Description
Algebra Level 1 simplistic design helps you gain confidence and accuracy in addition and subtraction of algebraic equations. Algebra is interactive fun with Algebra Level 1, and its easy to use!
Algebra Level 1's interactive interface allows you to customize the numbers you wish to focus on and choose either addition or subtraction algebraic equations. As you master one group of numbers you can increase your maximum number to make the problems more challenging.
Algebra Level 1 is ideal for the beginner or intermediate student who wishes to have a strong foundation (or review) of the basics of addition and subtraction in algebra.
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Trigonometry
Browse related Subjects ...
Read More illustrations drawn from Lance Armstrong's cycling success, the Ferris wheel, and even the human cannonball show trigonometry in action. Unique Historical Vignettes offer a fascinating glimpse at how many of the central ideas in trigonometry began. TRIGONOMETRY 6e, International Edition, uses a standard right-angle approach with an emphasis on the study skills most important for success both now and in advanced courses, such as calculus. The book's proven blend of exercises, fresh applications, and projects is combined with a simplified approach to graphing and the convenience of new Enhanced WebAssign--a leading, time-saving online homework tool--and the innovative CengageNOW teaching system. With TRIGONOMETRY 6e, International Edition, you'll find everything you need for a thorough understand of trigonometry concepts now and the solid foundation you need for future coursework and career success.
Read Less
Fair. Hardcover. All text is legible, may contain markings, cover wear, loose/torn pages or staining and much writing. SKU: 9781111826857-5 Hardcover. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9781111826857-4Very good. Hardcover. Has minor wear and/or markings. SKU: 9781111826857-3 1111826854
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1.) Create algebraic models for application-based problems by developing and solving equations and inequalities, including those involving direct, inverse, and joint variation. (Alabama)
Example: The amount of sales tax on a new car is directly proportional to the purchase price of the car. If the sales tax on a $20,500 car is $1,600, what is the purchase price of a new car that has a sales tax of $3,200?
7.) Use analytical, numerical, and graphical methods to make financial and economic decisions, including those involving banking and investments, insurance, personal budgets, credit purchases, recreation, and deceptive and fraudulent pricing and advertising. (Alabama)
Examples: Determine the best choice of certificates of deposit, savings accounts, checking accounts, or loans. Compare the costs of fixed- or variable-rate mortgage loans. Compare costs associated with various credit cards. Determine the best cellular telephone plan for a budget.
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Gauss elimination is a method of solving a system of linear equations. This site contains an interactive tool that performs...
see more
Gauss elimination is a method of solving a system of linear equations. This site contains an interactive tool that performs Gauss elimination procedure one step at a time. The result is displayed after each step. A user can also use this tool to invert a matrix Gaussian Elimination to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material Gaussian ElimMatrix Calculator is a site containing an interactive applet that let a user to input a square matrix and then with a press...
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Matrix Calculator is a site containing an interactive applet that let a user to input a square matrix and then with a press of a button compute a power of this matrix, determinant, inverse, characteristic polynomials and other useful matrix characteristics calculator to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material Matrix calculatorA collection of puzzles ranging over geometry, probability, number theory, algebra, calculus, and logic. Hints are provided,...
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A collection of puzzles ranging over geometry, probability, number theory, algebra, calculus, and logic. Hints are provided, along with fully worked solutions, and links to related mathematical Nick's Mathematical Puzzles to your Bookmark Collection or Course ePortfolio
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Synopses & Reviews
Publisher Comments:
The classical partial differential equations of mathematical physics, formulated by the great mathematicians of the 19th century, remain today the basis of investigation into waves, heat conduction, hydrodynamics, and other physical problems.
In this comprehensive treatment by a well-known Soviet mathematician, the equations are presented and explained in a manner especially designed to be accessible to the novice in the field. The reader is assumed to have no previous knowledge other than elementary analysis. From there, more advanced concepts are developed in detail and with great precision; moreover, theorems are often approached through the study of special simpler cases, before being proved in their full generality, and are applied to many particular physical problems.
After deriving the fundamental equations, the author provides illuminating expositions of such topics as Riemann's method, Lebesgue integration of multiple integrals, the equation of heat conduction, Laplace's equation and Poisson's equation, the theory of integral equations, Green's function, Fourier's method, harmonic polynomials and spherical functions, and much more.
For this third edition, various improvements in style and clarifications of the presentations were made, including a simplification of the theory of multiple Lebesgue integrals and greater precision in the proof of the Fourier method. Finally, the translation is both idiomatic as well as accurate, making the vast amount of information in this book more readily accessible to the English reader.
Synopsis:
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More About
This Textbook
Overview
This book, the first of a two-volume basic introduction to enumerative combinatorics, concentrates on the theory and application of generating functions, a fundamental tool in enumerative combinatorics. Richard Stanley covers those parts of enumerative combinatorics with the greatest applications to other areas of mathematics. The four chapters are devoted to an accessible introduction to enumeration, sieve methods--including the Principle of Inclusion-Exclusion, partially ordered sets, and rational generating functions. A large number of exercises, almost all with solutions, augment the text and provide entry into many areas not covered directly. Graduate students and research mathematicians who wish to apply combinatorics to their work will find this an authoritative reference.
This book is an introduction to enumerative combinatorics for graduate students and researchers. It concentrates on the theory and application of generating functions, a fundamental tool in enumerative combinatorics. The four chapters are devoted to enumeration, sieve methods (including the Principle of Inclusion-Exclusion), partially ordered sets, and rational generating functions. There are a large number of exercises, almost all with solutions, which greatly augment the text and provide entry into many areas not covered directly. The author stresses important connections with other areas of mathematics. This is a reissue of a book first published in 1986. The author has updated the references and included more problems. Graduate students and research mathematicians who wish to apply combinatorics to their work will find this an authoritative reference.
Meet the Author
Richard P. Stanley is a Professor of Applied Mathematics at the Massachusetts Institute of Technology. He is universally recognized as a leading expert in the field of combinatorics and its applications to a variety of other mathematical disciplines. In addition to the seminal two-volume book Enumerative Combinatorics, he is the author of Combinatorics and Commutative Algebra (1983) as well as more than 100 research articles in mathematics. Among Stanley's many distinctions are membership in the National Academy of Sciences (elected in 1995), the 2001 Leroy P. Steele Prize for mathematical exposition and the 2003 Schock 15, 2001
A Masterpiece of Mathematical Writing
It was only after having been prodded by Eratosthenes, then Librarian of Alexandria, that Archimedes was induced to write 'The Method.' Fermat is notorious for having written in the margins of Diophantus's 'Arithmetica', where there was never enough room for methods. Newton took an extra year writing 'Principia Mathematica' in order to conceal his methods. (Only after Leibniz began publishing did Newton talk openly about calculus.) Abel said about Gauss that 'he is like the fox, who effaces his tracks with his tail.' Fortunately, mathematicians of the first rank no longer deliberately hide their methods. Unfortunately, few of them seem both willing and able to write lucidly enough for nonspecialists to appreciate subtleties of approach. Richard Stanley is a refreshing exception. His two volume 'Enumerative Combinatorics' is already a classic, both for its depth and for its clarity. Reading these books, one achieves a sense, not only of 'what', but of 'why' and 'how'. Technique is generously illustrated, not only in the exposition, but in the explicit solutions of numerous well chosen exercises. These volumes comprise a masterpiece of mathematical writing.
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The main goal of this project is to improve student understanding of the geometric nature of multivariable calculus concepts,...
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The main goal of this project is to improve student understanding of the geometric nature of multivariable calculus concepts, i.e., to help them develop accurate geometric intuition about multivariable calculus concepts and the various relationships among them.To accomplish this goal, the project includes four parts:· Creating a Multivariable Calculus Visualization applet using Java and publishing it on a website: web.monroecc.edu/calcNSF· Creating a series of focused applets that demonstrate and explore particular 3D calculus concepts in a more dedicated way.· Developing a series of guided exploration/assessments to be used by students to explore calculus concepts visually on their own.· Dissemination of these materials through presentations and poster sessions at math conferences and through other publications.Intellectual Merit: This project provides dynamic visualization tools that enhance the teaching and learning of multivariable calculus. The visualization applets can be used in a number of ways:- Instructors can use them to visually demonstrate concepts and verify results during lectures.- Students can use them to explore the concepts visually outside of class, either using a guided activity or on their own.- Instructors can use the main applet (CalcPlot3D) to create colorful graphs for visual aids (color overheads), worksheets, notes/handouts, or tests. 3D graphs or 2D contour plots can be copied from the applet and pasted into a word processor like Microsoft Word.- Instructors will be able to use CalcPlot3D to create lecture demonstrations containing particular functions they specify and/or guided explorations for their own students using a scripting feature that is being integrated with this applet.The guided activities created for this project will provide a means for instructors to get their students to use these applets to actively explore and "play" with the calculus concepts.Paul Seeburger, the Principal Investigator (PI) for this grant project, has a lot of experience developing applets to bring calculus concepts to life. He has created 100+ Java applets supporting 5 major calculus textbooks (Anton, Thomas, Varberg, Salas, Hughes-Hallett). These applets essentially make textbook figures come to life. See examples of these applets at Impacts: This project will provide reliable visualization tools for educators to use to enhance their teaching in calculus and also in various Physics/Engineering classes. It is designed to promote student exploration and discovery, providing a way to truly "see" how the concepts work in motion and living color. The applets and support materials will be published and widely disseminated through the web and conference Multivariable Calculus to your Bookmark Collection or Course ePortfolio
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Demos with Positive Impact is a collection of quick classroom demos that enhance the learning of mathematics content through...
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The site includes guided interdisciplinary labs for first and second courses in statistics. As stated on the homepage for...
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The site includes guided interdisciplinary labs for first and second courses in statistics. As stated on the homepage for this site, range of more advanced topics we hope that students in any discipline can realize the intellectual content and broad applicability of statistics Stat2Labs to your Bookmark Collection or Course ePortfolio
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״The Flash mathlets are divided into two sections.The For Students and Instructors section contains Flash applets which may...
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״The Flash mathlets are divided into two sections.The For Students and Instructors section contains Flash applets which may be useful to instructors for classroom demonstrations, as well as to students for independent work and exploration. The For Developers section presents applets together with varied templates and the complete, object-oriented, easily-customizable source code that can be used by instructors for creating their own customized versions of the applets Mathlets to your Bookmark Collection or Course ePortfolio
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A searchable database of more than 1000 test questions for introductory statistics concepts. The user is prompted to select...
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A searchable database of more than 1000 test questions for introductory statistics concepts. The user is prompted to select subject material and learning outcome expectations from a variety of question formats and then downloads the items and can edit the test with a word processor Assessment Builder to your Bookmark Collection or Course ePortfolio
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Abstract: Instructors have frequently found that some content, such as mathematical formulae, chemistry laboratory...
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Abstract: Instructors have frequently found that some content, such as mathematical formulae, chemistry laboratory experiments, and business practices, are unusually difficult for students to comprehend through text-centered approaches, and that this is especially so for online students. In response, instructor-made videos (IMVs) of three to 10 minutes in length on problematic topics or subject matter areas were produced for business, chemistry, and mathematics courses. The IMVs were intended to scaffold student learning. Initial findings revealed that multimodal IMVs involving the demonstration, illustration, and presentation of key terms, knowledge, skills, and resources can help students understand important procedures, structures, or mechanisms in previously problematic content. Simply stated, IMVs can have a positive impact on student learning.Volume 8, No. 4, December 2012, pp. 298-311-Made Videos as a Scaffolding Tool to your Bookmark Collection or Course ePortfolio
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To address the critical teacher shortage in math and science, the Teaching and Learning Interchange grant produced three...
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To address the critical teacher shortage in math and science, the Teaching and Learning Interchange grant produced three products: two series of online video case studies of accomplished teaching and a resource guide explaining the techniques, ideas for engaging students, assessing learning progress, creating and maintaining a classroom conducive to learning, along with strategies for working with English Learners and Special needs students. Originally conceived for use with pre-service and intern teachers, the cases and resources have been recommended by field reviewers for use at all stages of the teacher professional continuum.The Pedagogy in Practice: Science video case studies document lessons from general science, biology and chemistry; the Pedagogy in Practice: Math video cases cover 8th grade algebra, high school algebra and geometry. Each case is supported by relevant artifacts that develop real contexts for the learning and includes teaching cues and transcripts of classroom dialog. National Board certified and Exploratorium Master teachers deconstruct their own teaching practices and share how they break subject matter standards into mind-sized learning units, then demonstrate those same learning units in action. Teacher reflections on assessment, along with links to resouces found in the Strategies in Practice Resource Guide, provide support for classroom management, lesson planning and essential strategies that describe how to adapt the instructional pedagogy modeled in the videos. All lessons are aligned simultaneously to the California standards for teaching and science subject matter
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Description! Being a student with not the best teacher, this app really helped me a lot!
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A Google User April 14, 2012
Amazing! Being a student with not the best teacher, this app really helped me a lot!
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Raising to a Power Dividing Exponents Multiplying Exponents Sets of Linear Equations
Linear Equations with One Unknown Linear Equations with Two Unknowns Linear Equations with Three Unknowns Linear Equations with Four Unknowns Solving Linear Equations by Substitution Properties of a Straight Line Non Linear Equations and Approximation and Inequalities
**REAL TEACHER TAUGHT LESSONS** Algebra This 6th Grade concisely. Algebra is taken by students who have gained skills like operation with number, rational numbers, basic equations and the basic coordinate plane.
**REAL TEACHER TAUGHT LESSONS** Algebra Readiness: This course teaches whole numbers, operations on whole numbers, rational numbers, operations on rational numbers, symbolic notation, equations and functions, and the coordinate plane. This course offers 11 full chapters in three parts with 6-8 lessons each chapter that present short easy to follow algebra readiness videos. These 5 to 10 minutes videos take students through the lesson slowly and concisely. Algebra Readiness is taken by students after Pre-Algebra if they still need to review skills before algebra or instead of Pre-Algebra for students who only need a review class before Algebra or if students want to firm skills prior to Algebra.
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High School Mathematics Extensions by txx
High School Mathematics Extensions
From Wikibooks, the open-content textbooks collection
Note: current version of this book can be found at
Contents
HIGH SCHOOL MATHEMATICS EXTENSIONS ..............................................................................................................1 CONTENTS .................................................................................................................................................................2 AUTHORS ...................................................................................................................................................................4 PRIMES AND MODULAR ARITHMETIC .........................................................................................................................5 Primes ..................................................................................................................................................................5 Introduction..........................................................................................................................................................5 Factorisation.........................................................................................................................................................6 2, 5 and 3..............................................................................................................................................................8 Theorem - Divisibility by 3..................................................................................................................................8 Finding primes .....................................................................................................................................................9 Infinitely many primes .......................................................................................................................................10 Modular Arithmetic............................................................................................................................................12 To go further ......................................................................................................................................................33 Acknowledgement..............................................................................................................................................33 Reference ...........................................................................................................................................................34 Feedback ............................................................................................................................................................34 Problem Set ........................................................................................................................................................34 Square root of minus 1 .......................................................................................................................................37 Project -- The Square Root of -1 ........................................................................................................................37 SOLUTIONS TO EXERCISES .......................................................................................................................................39 HSE Primes|Primes and Modular Arithmetic.....................................................................................................39 Diophantine equation Exercises .........................................................................................................................45 Problem set solutions .........................................................................................................................................50 DEFINITIONS ............................................................................................................................................................54 LOGIC ......................................................................................................................................................................54 Introduction........................................................................................................................................................54 Boolean algebra..................................................................................................................................................54 Propositions........................................................................................................................................................69 Logic Puzzles .....................................................................................................................................................72 Problem Set ........................................................................................................................................................74 Feedback ............................................................................................................................................................75 MATHEMATICAL PROOFS .........................................................................................................................................76 Introduction........................................................................................................................................................76 Mathematical induction......................................................................................................................................76 Proof by contradiction........................................................................................................................................79 Reading higher mathematics ..............................................................................................................................81 Axioms and Inference ........................................................................................................................................83 Problem Set ........................................................................................................................................................86 Feedback ............................................................................................................................................................87 EXERCISES ...............................................................................................................................................................88 Mathematical proofs...........................................................................................................................................88 Problem set.........................................................................................................................................................92 Mathematical Proofs Problem Set......................................................................................................................92 INFINITY AND INFINITE PROCESSES ..........................................................................................................................95 Introduction........................................................................................................................................................95 Infinite Sets ........................................................................................................................................................95 Limits Infinity got rid of ..................................................................................................................................102 Infinite series....................................................................................................................................................105
2
Feedback ..........................................................................................................................................................106 Exercises ..........................................................................................................................................................107 Infinity and infinite processes ..........................................................................................................................107 Problem set.......................................................................................................................................................109 COUNTING AND GENERATING FUNCTIONS .............................................................................................................110 Some Counting Problems.................................................................................................................................110 Generating functions ........................................................................................................................................110 Further Counting ..............................................................................................................................................116 Problem Set ......................................................................................................................................................121 Feedback ..........................................................................................................................................................125 Exercises ..........................................................................................................................................................126 Counting and Generating functions..................................................................................................................126 DISCRETE PROBABILITY ........................................................................................................................................132 Introduction......................................................................................................................................................132 Event and Probability.......................................................................................................................................132 Random Variables............................................................................................................................................136 Areas as probability..........................................................................................................................................143 Order Statistics.................................................................................................................................................143 Addition of the Uniform distribution ...............................................................................................................143 Feedback ..........................................................................................................................................................143 FINANCIAL OPTIONS ..............................................................................................................................................144 Binary tree option pricing ................................................................................................................................144 Call-Put parity ..................................................................................................................................................150 Reference .........................................................................................................................................................150 Feedback ..........................................................................................................................................................150 MATRICES .............................................................................................................................................................151 Introduction......................................................................................................................................................151 Matrix Multiplication .......................................................................................................................................153 The Identity & multiplication laws ..................................................................................................................161 Determinant and Inverses.................................................................................................................................162 Other Sections ..................................................................................................................................................168 Exercises ..........................................................................................................................................................168 Matrices............................................................................................................................................................168 Problem set.......................................................................................................................................................177 Matrices Problem Set .......................................................................................................................................177 FURTHER MODULAR ARITHMETIC .........................................................................................................................180 Introduction......................................................................................................................................................180 Modular Arithmetic with a general m ..............................................................................................................182 Two-torsion Factorisation ................................................................................................................................191 MATHEMATICAL PROGRAMMING ...........................................................................................................................193 Introduction to programming ...........................................................................................................................193 Discrete Programming .....................................................................................................................................194 Feedback ..........................................................................................................................................................196 BASIC COUNTING ...................................................................................................................................................198 Counting...........................................................................................................................................................198 PARTIAL FRACTIONS ..............................................................................................................................................201 Method of Partial Fractions..............................................................................................................................201 More on partial fraction ...................................................................................................................................204 SUMMATION SIGN ..................................................................................................................................................210 Summation Notation ........................................................................................................................................210 Operations of sum notation ..............................................................................................................................211 Beyond .............................................................................................................................................................212 COMPLEX NUMBERS ..............................................................................................................................................213 Introduction......................................................................................................................................................213
3
Arithmetic with complex numbers...................................................................................................................216 The complex root .............................................................................................................................................219 The complex plane ...........................................................................................................................................221 de Moivre's Theorem .......................................................................................................................................224 Complex root of unity ......................................................................................................................................225 PROBLEM SET ........................................................................................................................................................226 DIFFERENTIATION..................................................................................................................................................233 Differentiate from first principle ......................................................................................................................233 LICENSE .................................................................................................................................................................247 GNU Free Documentation License ..................................................................................................................247 0. PREAMBLE ................................................................................................................................................247 1. APPLICABILITY AND DEFINITIONS.....................................................................................................247 2. VERBATIM COPYING ..............................................................................................................................248 3. COPYING IN QUANTITY .........................................................................................................................248 4. MODIFICATIONS ......................................................................................................................................248 5. COMBINING DOCUMENTS.....................................................................................................................249 6. COLLECTIONS OF DOCUMENTS...........................................................................................................249 7. AGGREGATION WITH INDEPENDENT WORKS .................................................................................250 8. TRANSLATION..........................................................................................................................................250 9. TERMINATION..........................................................................................................................................250 10. FUTURE REVISIONS OF THIS LICENSE .............................................................................................250 External links ...................................................................................................................................................250
Authors
Zhuo Jia Dai, User:R3m0t, Martin Warmer, Tom Lam.
4
Primes and modular arithmetic
Primes Introduction
A prime number (or prime for short) is a natural number that can only be wholly divided by 1 and itself. For theoretical reasons, the number 1 is not considered a prime (we shall see why later on in this chapter). For example, 2 is a prime, 3 is prime, and 5 is prime, but 4 is not a prime because 4 divided by 2 equals 2 without a remainder. The first 20 primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71. Primes are an endless source of fascination for mathematicians. Some of the problems concerning primes are so difficult that even decades of work by some of the most brilliant mathematicians have failed to solve them. One such problem is Goldbach's conjecture, which states that all even numbers greater than 3 can be expressed as the sum of two primes.
Geometric meaning of primes
Let's start with an example. Given 12 pieces of square floor tiles, can we assemble them into a rectangular shape in more than one way? Of course we can, this is due to the fact that
We do not distinguish between 2×6 and 6×2 because they are essentially equivalent arrangements. But what about the number 7? Can you arrange 7 square floor tiles into rectangular shapes in more than one way? The answer is no, because 7 is a prime number.
Fundamental Theorem of Arithmetic
A theorem is a non-obvious mathematical fact. A theorem must be proven; a proposition that is generally believed to be true, but without a proof, is called a conjecture. With those definitions out of the way the fundamental theorem of arithmetic simply states that: Any natural number (except for 1) can be expressed as the product of primes in one and only one way. For example
5
Rearranging the multiplication order is not considered a different representation of the number, so there is no other way of expressing 12 as the product of primes. A few more examples
A number that can be factorised into more than 1 prime factor is a called a composite number (or composite for short). Composite is the opposite of prime. Think about it Bearing in mind the definition of the fundamental theorem of arithmetic, why isn't the number 1 considered a prime?
Factorisation
We know from the fundamental theorem of arithmetic that any integer can be expressed as the product of primes. The million dollar question is: given a number x, is there an easy way to find all prime factors of x? If x is a small number that is easy. For example 90 = 2 × 3 × 3 × 5. But what if x is large? For example x = 4539? Most people can't factorise 4539 into primes in their heads. But can a computer do it? Yes, the computer can factorise 4539 in no time. In fact 4539 = 3 × 17 × 89. There is, indeed, an easy way to factorise a number into prime factors. Just apply the method to be described below (using a computer). However, that method is too slow for large numbers: trying to factorise a number with thousands of digits would take more time than the current age of the universe. But is there a fast way? Or more precisely, is there an efficient way? There may be, but no one has found one yet. Some of the most widely used encryption schemes today (such as RSA) make use of the fact that we can't factorise large numbers into prime factors quickly. If such a method is found, a lot of internet transactions will be rendered unsafe. So if you happen to be the discoverer of such a method, don't be too forthcoming in publishing it, consult your national security agency first! Since computers are very good at doing arithmetic, we can work out all the factors of x by simply instructing the computer to divide x by 2 and then 3 then by 5 then by 7 then by 11 ... and so on, and check whether at any point the result is a whole number. Consider the following three examples of the dividing method in action. Example 1 x = 21 x / 2 = 10.5 not a whole number 6
x / 3 = 7 hence 3 and 7 are the factors of 21. Example 2 x = 153 x / 2 = 76.5 hence 2 is not a factor of 153 x / 3 = 51 hence 3 and 51 are factors of 153 51 / 3 = 17 hence 3 and 17 are factors of 153 It is clear that 3, 9, 17 and 51 are the factors of 153. The prime factors of 153 are 3, 3 and 17 (3×3×17 = 153) Example 3 2057 / 2 = 1028.5 ... 2057 / 11 = 187 187 / 11 = 17 hence 11, 11 and 17 are the prime factors of 2057.
Exercise
Factor the following numbers: 1. 2. 3. 4. 5. 6. 7. 13 26 59 82 101 121 2187 Give up if it takes too long. There is a quick way.
Fun Fact -- Is this prime?
Interestingly, due to recent developments, we can tell quickly, with the help of a computer program, whether a number is prime with 100% accuracy.
7
2, 5 and 3
The primes 2, 5, and 3 hold a special place in factorisation. Firstly, all even numbers have 2 as one of their prime factors. Secondly, all numbers whose last digit is 0 or 5 can be divided wholly by 5. The third case, where 3 is a prime factor, is the focus of this section. The underlying question is: is there a simple way to decide whether a number has 3 as one of its prime factors? Yes. See the following theorem
Theorem - Divisibility by 3
A number is divisible by 3 if and only if the sum of its digits is divisible by 3 E.g. 272 is not divisible by 3, because 2+7+2=11. And 11 isn't divisible by 3. 945 is divisible by 3, because 9+4+5 = 18. And 18 is divisible by 3. In fact 945 / 3 = 315 Is 123456789 divisible by 3? 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = (1 + 9) × 9 / 2 = 45 4+5=9 Nine is divisible by 3, therefore 45 is divisible by 3, therefore 123456789 is divisible by 3! (That this method works is a theorem, though the proof is not given here.) The beauty of the theorem lies in its recursive nature. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. How do we know whether the sum of its digits is divisible by 3? Apply the theorem again! It's too true that "... to recurse, [is] divine". Try a few more numbers yourself.
info -- Recursion
A prominent computer scientist once said "To iterate is human, to recurse, divine." But what does it mean to recurse? Before that, what is to iterate? "To iterate" simply means doing the same thing over and over again, and computers are very good at that. An example of iteration in mathematics is the exponential operation, e.g. xn means doing x times x times x times x...n times. That is an example of iteration. Thinking about iteration economically (in terms of mental resources), by defining a problem in terms of itself, is "to recurse". To recursively represent xn, we write: xn = 1 if n equals 0. if n > 0 8
What is 99? That is 9 times 9 8. But what is 98, it is 9 times 97. Repeating this way is an example of recursion.
Exercises
1. Factorise 1. 2. 3. 45 4050 2187 2. Show that the divisible-by-3 theorem works for any 3 digits numbers (Hint: Express a 3 digit number as 100a + 10b + c, where 0 ≤ a, b and c ≤ 9) 3. "
Finding primes
The prime sieve is a relatively efficient method of finding all primes less than or equal to a specified number. Let say we want to find all primes less than or equal to 50. First we write out all numbers between 0 and 51 in a table as below
Cross out 1, because it's not a prime.
Now 2 is the smallest number not crossed out in the table. We mark 2 as a prime and cross out all multiples of 2 i.e. 4, 6, 8, 10 ...
9
Now 3 is the smallest number not marked in anyway. We mark 3 as a prime and cross out all multiples of 3 i.e. 6, 9, 12, 15 ...
Continue is this way to find all the primes. When do you know you have found all the primes under 50?
Exercise
1.
The prime sieve has been applied to the table 2. Find all primes below 200
Infinitely many primes
We know some numbers can be factorised into primes. Some have only themselves as a prime factor, because they are prime. So how many primes are there? There are infinitely many! Here is a classical proof of the infinitude of primes dating back 2000 years to the ancient Greek mathematician Euclid:
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Proof of infinitude of primes
Let us first assume that there are a finite number of primes therefore there must be one prime that is greater than all others, let this prime be referred to as n. We now proceed to show the two assumptions made above will lead to non-sense, and so there are infinitely many primes. Take the product of all prime numbers to yield a number x. Thus:
Then, let y equal one more than x:
One may now conclude that y is not divisible by any of the primes up to n, since y differs from a multiple of each such prime by exactly 1. Since y is not divisible by any prime number, y must either be prime, or its prime factors must all be greater than n, a contradiction of the original assumption that n is the largest prime! Therefore, one must declare the original assumption incorrect, and that there exists an infinite number of primes. Fun Fact -- Largest known prime The largest prime known to human is 230402457-1. It has a whopping 9152052 digits! Primes of the form 2n-1 are called Mersenne primes named after the French monk/amateur mathematician.
Primes in arithmetic progression
Consider the arithmetic progression a, a + b, a + 2b, a + 3b .... if a and b share a common factor greater than 1, then a + kb for any k is not prime. But if a and b are coprimes, then there are infinitely many k's such that a + kb is prime! For example, consider a = 3, b = 4, 3, 7, 11, 15, 19, 23, 27, 31... in this rather short list, 3, 7, 11, 19, 23 and 31 are prime and they are all equal to 4k + 3 for some k. And there are infinitely many primes in this progression (see: proof by contradiction exercise Mathematical Proofs). 11
Modular Arithmetic
Introduction
Modular arithmetic connects with primes in an interesting way. Firstly recall clock-arithmetic. It is a system by which all numbers up to some positive integer, n say, are used. So if you were to start counting you would go 0, 1, 2, 3, ... , n - 1 but instead of counting n you would start over at 0. And what would have been n + 1 would be 1 and what would have been n + 2 would be 2. Once 2n had been reach the number is reset to 0 again, and so on. Very much like the clocks we have which starts at 1 and continues to 12 then back to 1 again. The sequence also continues into what would be the negitive numbers. What would have been 1 is now n - 1. For example, let's start with modulo 7 arithmetic, it's just like ordinary arithmetic except the only numbers we use are 0, 1, 2, 3, 4, 5 and 6. If we see a number outside of this range we add 7 to (or subtract 7 from) it, until it lies within that range. As mentioned above, modular arithmetic is not much different to ordinary arithmetic. For example, consider modulo 7 arithmetic
The same deal for multiplication
We have done some calculation with negative numbers. Consider 5 × -6. Since -6 does not lie in the range 0 to 6, we need to add 7 to it until it does. And -6 + 7 = 1. So in modular 7 arithmetic, -6 = 1. In the above example we showed that 5 × -6 = -30 = 5, but 5 × 1 = 5. So we didn't do ourselves any harm by using -6 instead of 1. Why? Note - Negatives: The preferred representation of -3 is 4, as -3 + 7 = 4, but using either -3 and 4 in a calculation will give us the same answer as long as we convert the final answer to a number between 0 and 6 (inclusive). Exercise Find in modulo 11 1. -1 × -5 2. 12
3×7 3. Compute all the powers of 2 21, 22, 23, ... , 210 What do you notice? Using the powers of 2 find 61, 62, 63, ... , 610 What do you notice again? 4.
i.e. find, by trial and error (or otherwise), all numbers x such that x2 = 4 (mod 11). There are two solutions, find both . 5.
i.e. find all numbers x such that x2 = 9 (mod 11). There are two solutions, find both.
Inverses
Let's consider a number n, the inverse of n is the number that when multiplied by n will give 1. Let's consider a simple example, we want to solve the following equation in modulo 7,
the (mod 7) is used to make clear that we doing arithemetic modulo 7. We want to get rid of the 5 somehow, notice that
because 3 multiplied by 5 gives 1, so we say 3 is the inverse of 5 in modulo 7. Now we multiply both sides by 3
So x = 2 modulo 7 is the required solution. Inverse is unique 13
From above, we know the inverse of 5 is 3, but does 5 have another inverse? The answer is no. In fact, in any reasonable number system, a number can have one and only one inverse. We can see that from the following proof Suppose n has two inverses b and c
From the above argument, all inverses of n must be equal. As a result, if the number n has an inverse, the inverse must be unique. From now on, we will use x-1 to denote the inverse of x if it exists. An interesting property of any modulo n arithmetic is that the number n - 1 has itself as an inverse. That is, (n - 1) × (n - 1) = 1 (mod n), or we can write (n - 1)2 = (-1)2 = 1 (mod n). The proof is left as an exercise at the end of the section. Existence of inverse Not every number has an inverse in every modulo arithmetic. For example, 3 doesn't have an inverse mod 6, i.e., we can't find a number x such that 3x = 1 mod 6 (the reader can easily check). Let's consider modulo 15 arithmetic and note that 15 is composite. We know the inverse of 1 is 1 and of 14 is 14. But what about 3, 6, 9, 12, 5 and 10? None of them has an inverse! Note that each of them shares a common factor with 15! Let's look at 3, we want to use a proof by contradiction argument to show that 3 does not have an inverse modulo 15. Suppose 3 has an inverse, which is denoted by x.
We make the jump from modular arithemetic into rational number arithmetic. If 3x = 1 in modulo 15 arithmetic, then
for some integer k. Now we divide both sides by 3, we get
But this is can't be true, because we know that x is an integer, not a fraction. Therefore 3 doesn't have an inverse in mod 15 arithmetic. To show that 10 doesn't have an inverse is harder and is left as an exercise. We will now state the theorem regarding the existence of inverses in modular arithmetic. Theorem 14
If n is prime then every number (except 0) has an inverse in modulo n arithmetic. Similarly If n is composite then every number that doesn't share a common factor with n has an inverse. It is interesting to note that division is closely related to the concept of inverses. Consider the following expresion
the conventional way to calculate the above would be to find the inverse of 3 (being 5). So
Let's write the inverse of 3 as 1/3, so we think of multiplying 3-1 as dividing by 3, we get
Notice that we got the same answer! In fact, the division method will always work if the inverse exists. Be very careful though, the expression in a different modulo system will produce the wrong answer, for example
we don't get 2, as 3-1 does not exist in modulo 9, so we can't use the division method. Exercise 1. Does 8 have an inverse in mod 16 arithemetic? If not, why not? 2. Find x mod 7 if x exists:
3. Calculate x in two ways, finding inverse and division 15
4. (Trick) Find x
5. Find all inverses mod n (n ≤ 19) This exercise may seem tedious, but it will increase your understanding of the topic by tenfold
Coprime and greatest common divisor
Two numbers are said to be coprimes if their greatest common divisor is 1. The greatest common divisor (gcd) is just what its name says it is. And there is a quick and elegant way to compute the gcd of two numbers, called Euclid's algorithm. Let's illustrate with a few examples:
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Example 1: Find the gcd of 21 and 49. We set up a 2-column table where the bigger of the two numbers is on the right hand side as follows smaller 21 larger 49
We now compute 49(mod 21) which is 7 and put it in the second row smaller column, and put 21 into the larger column. smaller 21 7 larger 49 21
Perform the same action on the second row to produce the third row. smaller 21 7 0 larger 49 21 7
Whenever we see the number 0 appear on the smaller column, we know the corresponding larger number is the the gcd of the two numbers we started with, i.e. 7 is the gcd of 21 and 49. This algorithm is called Euclid's algorithm. Example 2 Find the gcd of 31 and 101 smaller 31 8 7 1 larger 101 31 8 7 17
0 Example 3
1
Find the gcd of 132 and 200 smaller 132 68 64 4 0 Remember 1. 2. The gcd need not be a prime number. The gcd of two primes is 1 Why does the Euclid's algorithm work? It is more fun for the questioner to discover. Exercise 1. Determine whether the following sets of numbers are coprimes 1. 2. 3. 4. 5050 5051 59 78 111 369 2021 4032 larger 200 132 68 64 4
2. Find the gcd of the numbers 15, 510 and 375 info -- Algorithm An algorithm is a step-by-step description of a series of actions when performed correctly can accomplish a task. There are algorithms for finding primes, deciding whether 2 numbers are coprimes, finding inverses and many other purposes. You'll learn how to implement some of the algorithms we have seen using a computer in the chapter Mathematical Programming.
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Finding Inverses
Let's look at the idea of inverse again, but from a different angle. In fact we will provide a surefire method to find the inverse of any number. Let's consider: 5x = 1 (mod 7) We know x is the inverse of 5 and we can work out it's 3 pretty quickly. But x = 10 is also a solution, so is x = 17, 24, 31, ... 7n + 3. So there are infinitely many solutions; therefore we say 3 is equivalent to 10, 17, 24, 31 and so on. This is a crucial observation Now let's consider
A new notation is introduced here, it is the equal sign with three strokes instead of two. It is the "equivalent" sign; the above statement should read "216x is EQUIVALENT to 1" instead of "216x is EQUAL to 1". From now on, we will use the equivalent sign for modulo arithmetic and the equal sign for ordinary arithmetic. Back to the example, we know that x exists, as gcd(811,216) = 1. The problem with the above question is that there is no quick way to decide the value of x! The best way we know is to multiply 216 by 1, 2, 3, 4... until we get the answer, there are at most 816 calculations, way too tedious for humans. But there is a better way, and we have touched on it quite a few times! We notice that we could make the jump just like before into rational mathematics:
We jump into rational maths again
We jump once more
Now the pattern is clear, we shall start from the beginning so that the process is not broken:
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Now all we have to do is choose a value for f and substitute it back to find a! Remember a is the inverse of 216 mod 811. We choose f = 0, therefore e = 1, d = 13, c = 40, b = 53 and finally a = 199! If choose f to be 1 we will get a different value for a. The very perceptive reader should have noticed that this is just Euclid's gcd algorithm in reverse. Here are a few more examples of this ingenious method in action: Example 1 Find the smallest positive value of a:
Choose d = 0, therefore a = 49. Example 2 Find the smallest positive value of a:
Choose e = 0, therefore a = -152 = 669 Example 3 Find the smallest positive value of a:
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Set i = 0, then a = -21 = 34. Why is this so slow for two numbers that are so small? What can you say about the coefficients?
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Example 4 Find the smallest positive value of a:
Now d is not an integer, therefore 21 does not have an inverse mod 102. What we have discussed so far is the method of finding integer solutions to equations of the form: ax + by = 1 where x and y are the unknowns and a and b are two given constants, these equations are called linear Diophantine equations. It is interesting to note that sometimes there is no solution, but if a solution exists, it implies that infinitely many solutions exist.
Diophantine equation
In the Modular Arithmetic section, we stated a theorem that says if gcd(a,m) = 1 then a-1 (the inverse of a) exists in mod m. It is not difficult to see that if p is prime then gcd(b,p) = 1 for all b less than p, therefore we can say that in mod p, every number except 0 has an inverse. We also showed a way to find the inverse of any element mod p. In fact, finding the inverse of a number in modular arithmetic amounts to solving a type of equations called Diophantine equations. A Diophantine equation is an equation of the form ax + by = d where x and y are unknown. As an example, we should try to find the inverse of 216 in mod 811. Let the inverse of 216 be x, we can write
we can rewrite the above in every day arithmetic
which is in the form of a Diophantine equation. Now we are going to do the inelegant method of solving the above problem, and then the elegant method (using Magic Tables). 22
Both methods mentioned above uses the Euclid's algorithm for finding the gcd of two numbers. In fact, the gcd is closely related to the idea of an inverse. Let's apply the Euclid's algorithm on the two numbers 216 and 811. This time, however, we should store more details; more specifically, we want to set up an additional column called PQ which stands for partial quotient. The partial quotient is just a technical term for "how many n goes into m" e.g. The partial quotient of 3 and 19 is 6, the partial quotient of 4 and 21 is 5 and one last example the partial quotient of 7 and 49 is 7. smaller 216 163 The tables says three 216s goes into 811 with remainder 163, or symbollically: 811 = 3×216 + 163. Let's continue: smaller 216 163 53 4 1 0 larger 811 216 163 53 4 1 PQ 3 1 3 13 4 larger 811 PQ 3
Reading off the table, we can form the following expressions 811 = 3× 216 + 163 216 = 1× 163 + 53 163 = 3× 53 + 4 53 =13× 4 + 1 Now that we can work out the inverse of 216 by working the results backwards 1 = 53 - 13×4 1 = 53 - 13×(163 - 3×53) 23
1 = 40×53 - 13×163 1 = 40×(216 - 163) - 13×163 1 = 40×216 - 53×163 1 = 40×216 - 53×(811 - 3×216) 1 = 199×216 - 53×811 Now look at the equation mod 811, we will see the inverse of 216 is 199. Magic Table The Magic Table is a more elegant way to do the above calculations, let us use the table we form from Euclid's algorithm smaller 216 163 53 4 1 0 larger 811 216 163 53 4 1 PQ 3 1 3 13 4
24
Now we set up the so-called "magic table" which looks like this initially
0 1
1 0
Now we write the partial quotient on the first row: 3 1 0 1 1 0 3 1 3 4
We produce the table according to the following rule: Multiply a partial quotient one space to the left of it in a different row, add the product to the number two space to the left on the same row and put the sum in the corresponding row. It sounds more complicated then it should. Let's illustrate by producing a column: 3 1 0 1 1 0 3 1 3 1 3 4
We put a 3 in the second row because 3 = 3×1 + 0. We put a 1 in the third row because 1 = 3×0 + 1. We shall now produce the whole table without disruption: 3 1 0 1 1 0 3 4 1 1 3 1 5 4 13 19 9 53 4 81 1 21 6
I claim |199×216 - 811×53| = 1 In fact, if you have done the magic table properly and cross multiplied and subtracted the last 25
two column correctly, then you will always get 1 or -1, provided the two numbers you started with are coprimes. The magic table is just a cleaner way of doing the mathematics.
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Exercises 1. Find the smallest positive x:
2. Find the smallest positive x:
3. (a) Produce the magic table for 33a = 1 (mod 101) (b) Evaluate and express in the form p/q
What do yo notice? 4. (a) Produce the magic table for 17a = 1 (mod 317) (b) Evaluate and express in the form p/q
What do yo notice?
Chinese remainder theorem
The Chinese remainder theorem is known in China as Han Xing Dian Bing, which in its most naive translation means Han Xing counts his soldiers. The original problem goes like this: There exists a number x, when divided by 3 leaves remainder 2, when divided by 5 leaves remainder 3 and when divided by 7 leaves remaider 2. Find the smallest x. We translate the question into symbolic form:
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How do we go about finding such a x? We shall use a familar method and it is best illustrated by example: Looking at x = 2 (mod 3), we make the jump into ordinary mathematics
Now we look at the equation modulo 5
Substitute into (1) to get the following
Now look at the above modulo 7
we get
We choose b = 1 to minimise x, therefore x = 23. And a simple check (to be performed by the reader) should confirm that x = 23 is a solution. A good question to ask is what is the next smallest x that satisfies the three congruences? The answer is x = 128, and the next is 233 and the next is 338, and they differ by 105, the product of 3, 5 and 7. We will illustrate the method of solving a system of congruencies further by the following examples: Example 1 Find the smallest x that satifies:
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Solution
now substitute back into the first equation, we get
we obtain
again substituting back
Therefore 52 is the smallest x that satifies the congruencies.
Example 2
Find the smallest x that satisfies:
Solution
29
substituting back
now solve for b
again, substitue back
Therefore 269 is the smallest x that satifies the Congruences. Excercises 1. Solve for x
2. Solve for x
*Existence of a solution* The exercises above all have a solution. So does there exist a system of congruences such that no solution could be found? It certainly is possible, consider: x ≡ 5 (mod 15) x ≡ 10 (mod 21) 30
a cheekier example is: x ≡ 1 (mod 2) x ≡ 0 (mod 2) but we won't consider silly examples like that. Back to the first example, we can try to solve it by doing:
the above equation has no solution because 3 does not have an inverse modulo 21! One may be quick to conclude that if two modulo systems share a common factor then there is no solution. But this is not true! Consider:
we can find a solution
we now multiply both sides by the inverse of 5 (which is 17), we obtain
obviously, k = 3 is a solution, and the two modulo systems are the same as the first example (i.e. 15 and 21). So what determines whether a system of congruences has a solution or not? Let's consider the general case:
we have
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essentially, the problem asks us to find k and l such that the above equations are satisfied. We can approach the problem as follows
now suppose m and n have gcd(m,n) = d, and m = dmo, n = dno. We have
if (a - b)/d is an integer then we can read the equation mod mo, we have:
Again, the above only makes sense if (a - b)/d is integeral. Also if (a - b)/d is an integer, then there is a solution, as mo and no are coprimes! In summary: for a system of two congruent equations
there is a solution if and only if d = gcd(m,n) divides (a - b) And the above generalises well into more than 2 congruences. For a system of n congruences:
...
for a solution to exist, we require that if i ≠ j gcd(mi,mj) divides (ai - aj)
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Exercises Decide whether a solution exists for each of the congruencies. Explain why. 1. x ≡ 7 (mod 25) x ≡ 22 (mod 45) 2. x ≡ 7 (mod 23) x ≡ 3 (mod 11) x ≡ 3 (mod 13) 3. x ≡ 7 (mod 25) x ≡ 22 (mod 45) x ≡ 7 (mod 11) 4. x ≡ 4 (mod 28) x ≡ 28 (mod 52) x ≡ 24 (mod 32)
To go further
This chapter has been a gentle introduction to number theory, a profoundly beautiful branch of mathematics. It is gentle in the sense that it is mathematically light and overall quite easy. If you enjoyed the material in this chapter, you would also enjoy Further Modular Arithmetic, which is a harder and more rigorous treatment of the subject. Also, if you feel like a challenge you may like to try out the Probelem Set we have prepared for you. On the other hand, the project asks you to take a more investigative approach to work through some of the finer implications of the Chinese Remainder Theorem.
Acknowledgement
Acknowledgement: This chapter of the textbook owes much of its inspiration to Terry Gagen, Emeritus Associate Professor of Mathematics at the University of Sydney, and his lecture notes on "Number Theory and Algebra". Terry is a much loved figure among his students and is 33
renowned for his entertaining style of teaching.
Reference
1. Largest Known Primes--A SummaryProblem Set
1. Is there a rule to determine whether a 3-digit number is divisible by 11? If yes, derive that rule. 2. Show that p, p + 2 and p + 4 cannot all be primes. (p a positive integer) 3. Find x
4. Show that there are no integers x and y such that
5. Let p be a prime number. Show that (a)
where 34
E.g. 3! = 1*2*3 = 6
35
(b) Hence, show that
for p ≡ 1 (mod 4), i.e., show that the above when squared gives one.
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Square root of minus 1 Project -- The Square Root of -1
Notation: 1. Question 5 of the Problem Set showed that
exists for p ≡ 1 (mod 4) prime. Explain why no square root of -1 exist if p ≡ 3 (mod 4) prime. 2. Show that for p ≡ 1 (mod 4) prime, there are exactly 2 solutions to
3. Suppose m and n are integers with gcd(n,m) = 1. Show that for each of the numbers 0, 1, 2, 3, .... , nm - 1 there is a unique pair of numbers a and b such that the smallest number x that satisfies: x ≡ a (mod m) x ≡ b (mod n) is that number. E.g. Suppose m = 2, n = 3, then 4 is uniquely represented by x ≡ 0 (mod 2) x ≡ 1 (mod 3) as the smallest x that satisfies the above two congruencies is 4. In this case the unique pair of numbers are 0 and 1. 4. If p ≡ 1 (mod 4) prime and q ≡ 3 (mod 4) prime. Does
37
have a solution? Why? 5. If p ≡ 1 (mod 4) prime and q ≡ 1 (mod 4) prime and p ≠ q. Show that
has 4 solutions. 6. Find the 4 solutions to
note that 493 = 17 × 29. 7. Take an integer n with more than 2 prime factors. Consider:
Under what condition is there a solution? Explain thoroughly.
38
Solutions to exercises
HSE Primes|Primes and Modular ArithmeticFactorisation Exercises
Factorise the following numbers. (note: I know you didn't have to, this is just for those who are curious) 1. 2. 3. 4. 5. 6. 7. 101 is prime 59 is prime 13 is prime
Recursive Factorisation Exercises
Factorise using recursion. 1. 2. 3.
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Prime Sieve Exercises
1. Use the above result to quickly work out the numbers that still need to be crossed out in the table below, knowing 5 is the next prime:
The next prime number is 5. Because 5 is an unmarked prime number, and 5 * 5 = 25, cross out 25. Also, 7 is an unmarked prime number, and 5 * 7 = 35, so cross off 35. However, 5 * 11 = 55, which is too high, so mark 5 as prime ad move on to 7. The only number low enough to be marked off is 7 * 7, which equals 35. You can go no higher. 2. Find all primes below 200. The method will not be outlined here, as it is too long. However, all primes below 200 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
Modular Arithmetic Exercises
1. +5=5 2. 3. 21 = 2,22 = 4,23 = 8,24 = 16 = 5 2 = 32 = 10,26 = 64 = 9,27 = 128 = 7 28 = 256 = 3,29 = 512 = 6,210 = 1024 = 1 An easier list: 2, 4, 8, 5, 10, 9, 7, 3, 6, 1 Notice that it is not necessary to acutally compute 210 to find 210 mod 11. If you know 29 mod 11 = 6. You can find 210 mod 11 = (2*(29 mod 11)) mod 11 = 2*6 mod 11 = 12 mod 11 = 1. We can note that 29 = 6 and 210 = 1, we can calculate 62 easily: 62 = 218 = 2^8 = 3. OR by the above method 61 = 6,62 = 36 = 3,63 = 6 * 3 = 18 = 7, 64 = 6 * 7 = 42 = 9,65 = 6 * 9 = 54 = 10,66 = 6 * 10 = 60 = 5, 67 = 6 * 5 = 30 = 8,68 = 6 * 8 = 48 = 4,69 = 6 * 4 = 24 = 2,610 = 6 * 2 = 12 = 1. An easier list: 6, 3, 7, 9, 10, 5, 8, 4, 2, 1.
5
alternatively, -1 = 10, -5 = 6: 10 × 6 = 60 = 5× 11
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4.
02 = 0, 12 = 1, 22 = 4, 32 = 9, 4 = 16 = 5, 52 = 25 = 5, 62 = 36 = 3, 72 = 49 = 3, 82 = 64 = 5, 92 = 81 = 4, 102 = 100 = 1 An easier list: 0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1
2
Thus 5. x2 = -2 = 9 Just look at the list above and you'll see that
Division and Inverses Exercises
1. x = 2 -1 = 4 x = 3 -1 = 5 x = 4 -1 = 2 x = 5 -1 = 3 x = 6 -1 = 6 x = 7 - 1 = 0 - 1 therefore the inverse does not exist
2.
3.
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4. 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 13 5 0 8 11 9 6 7 9 5 4 13 13 7 11 4 3 5 7 6 4 3 9 5 10 8 3 13 2 7 5 13 16 11 12 17 2 9 15 2 2 4 5 3 7 2 2 3 3 2 4 5 3 5 2 6 7 4 3 8 7 11 2 5 3 11 4 7 8 9 5 10 11 6 9 11 3 12 13 7 5 14 15 11 8 16 17 15 14 6 9 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 mod 2 mod 3 mod 4 mod 5 mod 6 mod 7 mod 8 mod 9 mod 10 mod 11 mod 12 mod 13 mod 14 mod 15 mod 16 mod 17 mod 18 18 mod 19
12 14 10 4 5 7 8 7 3
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Coprime and greatest common divisor Exercises
1. 1.
smaller 5050 1 0
larger 5051 5050 1
5050 and 5051 are coprime 2.
smaller 59 19 2 1 0
larger 78 59 19 2 1
59 and 79 are coprime
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3.
smaller 111 36 3 0
larger 369 111 36 3
111 and 369 are not coprime 4.
smaller 2021 2011 10 1 0
larger 4032 2021 2011 10 1
2021 and 4032 are coprime
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2.We first calculate the gcd for all combinations smaller 15 0 larger 510 15
smaller 15 0
larger 375 15
smaller 375 135 105 30 15 0
larger 510 375 135 105 30 15
The gcd for any combination of the numbers is 15 so the gcd is 15 for the three numbers.
Diophantine equation Exercises
1.
There is no solution, because can never become an integer. 45
2.
We choose d=1, then x=26. 3. (a)
smaller 33 2 1 0
larger 101 33 2 1
P Q 3 1 6 2
3 0 1 1 0 3 1
16 49 16
2 101 33
(b) To be added
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4. (a) smaller 17 11 6 5 1 0 larger 317 17 11 6 5 1 PQ 18 1 1 1 5
18 0 1 1 0 18 1
1 19 1
1 37 2
1 56 3
5 317 17
(b) To be added
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Chinese remainder theorem exercises
1.
Question 1
Show that the divisible-by-3 theorem works for any 3 digits numbers (Hint: Express a 3 digit number as 100a + 10b + c, where a, b and c are ≥ 0 and < 10) Solution 1 Any 3 digits integer x can be expressed as follows x = 100a + 10b + c where a, b and c are positive integer between 0 and 9 inclusive. Now
if and only if a + b + c = 3k for some k. But a, b and c are the digits of x.
Question 2
" Solution 2 The statement is true and can be proven as in question 1. 48
Question 4
The prime sieve has been applied to the table of numbers Solution 4 The width of the grid should be 15 or a multiple of it.
Question 6
Show that n - 1 has itself as an inverse modulo n. Solution 6 (n - 1)2 = n2 - 2n + 1 = 1 (mod n) Alternatively (n - 1)2 = (-1)2 = 1 (mod n)
Question 7
Show that 10 does not have an inverse modulo 15. Solution 7 Suppose 10 does have an inverse x mod 15, 10x = 1 (mod 15) 2Ã-5x = 1 (mod 15) 5x = 8 (mod 15) 5x = 8 + 15k for some integer k x = 1.6 + 3k but now x is not an integer, therefore 10 does not have an inverse
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Problem set solutions
Question 1
Is there a rule to determine whether a 3-digit number is divisible by 11? If yes, derive that rule. Solution Let x be a 3-digit number We have
now
We can conclude a 3-digit number is divisible by 11 if and only if the sum of first and last digit minus the second is divisible by 11.
Question 2
Show that p, p + 2 and p + 4 cannot all be primes. (p a positive integer) Solution We look at the arithmetic mod 3, then p slotted into one of three categories 1st category
we deduce p is not prime, as it's a multiple of 3 2nd category
so p + 2 is not prime 3rd category
therefore p + 4 is not prime 50
Therefore p, p + 2 and p + 4 cannot all be primes.
Question 3
Find x
Solution Notice that . Then . Likewise,
and . Then
Question 4
9. Show that there are no integers x and y such that
Solution Look at the equation mod 5, we have
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but
therefore there does not exist a x such that
Question 5
Let p be a prime number. Show that (a)
where
E.g. 3! = 1Ã-2Ã-3 = 6 (b) Hence, show that
for p ≡1 (mod 4) Solution a) If p = 2, then it's obvious. So we suppose p is an odd prime. Since p is prime, some deep thought will reveal that every distinct element multiplied by some other element will give 1. Since
we can pair up the inverses (two numbers that multiply to give one), and (p - 1) has itself as an inverse, therefore it's the only element not "eliminated"
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as required. b) From part a)
since p = 4k + 1 for some positive integer k, (p - 1)! has 4k terms
there are an even number of minuses on the right hand side, so
it follows
and finally we note that p = 4k + 1, we can conclude
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Definitions Logic
Introduction
Logic is the study of the way we humans reason. In this chapter, we focus on the methods of logical reasoning, i.e. digital logic, predicate calculus, application to proofs and the (insanely) fun logical puzzles.
Boolean algebra
In the black and white world of ideals, there is absolute truth. That is to say everything is either true or false. With this philosophical backdrop, we consider the following examples: "One plus one equals two." True or false? That is (without a doubt) true! "1 + 1 = 2 AND 2 + 2 = 4." True or false? That is also true. But what about: "1 + 1 = 3 OR Sydney is in Australia" True or false? It is true! Although 1 + 1 = 3 is not true, the OR in the statement made it so that if either part of the statement is true then the whole statement is true. Now let's consider a more puzzling example "2 + 2 = 4 OR 1 + 1 = 3 AND 1 - 3 = -1" True or false? The truth or falsity of the statements depends on the order in which you evaluate the statement. If you evaluate "2 + 2 = 4 OR 1 + 1 = 3" first, the statement is false, and otherwise true. As in ordinary algebra, it is necessary that we define some rules to govern the order of evaluation, so we don't have to deal with ambiguity. Before we decide which order to evaluate the statements in, we do what most mathematician love to do -replace sentences with symbols. Let x represent the truth or falsity of the statement 2 + 2 = 4. Let y represent the truth or falsity of the statement 1 + 1 = 3. Let z represent the truth or falsity of the statement 1 - 3 = -1. Then the above example can be rewritten in a more compact way: 54
x OR y AND z To go one step further, mathematicians also replace OR by + and AND by Ã-, the statement becomes:
Now that the order of precedence is clear. We evaluate (y AND z) first and then OR it with x. The statement "x + yz" is true, or symbolically x + yz = 1 where the number 1 represents "true". There is a good reason why we choose the multiplicative sign for the AND operation. As we shall see later, we can draw some parallels between the AND operation and multiplication. The Boolean algebra we are about to investigate is named after the British mathematician George Boole. Boolean algebra is about two things -- "true" or "false" which are often represented by the numbers 1 and 0 respectively. Alternative, T and F are also used. Boolean algebra has operations (AND and OR) analogous to the ordinary algebra that we know and love.
Basic Truth tables
We have all had to memorise the 9 by 9 multiplication table and now we know it off by heart. In Boolean algebra, the idea of a truth table is somewhat similar. Let's consider the AND operation which is analogous to the multiplication. We want to consider: x AND y where and x and y each represent a true or false statement (e.g. It is raining today). It is true if and only if both x and y are true, in table form: The AND function x F F T T y F T F T x AND y F F F T 55
We shall use 1 instead of T and 0 instead of F from now on. The AND function x 0 0 1 1 y 0 1 0 1 x AND y 0 0 0 1
Now you should be able to see why we say AND is analogous to multiplication, we shall replace the AND by Ã-, so x AND y becomes xÃ-y (or just xy). From the AND truth table, we have: 0 Ã- 0 = 0 0 Ã- 1 = 0 1 Ã- 0 = 0 1 Ã- 1 = 1 To the OR operation. x OR y is FALSE if and only if both x and y are false. In table form: The OR function x 0 0 1 1 y 0 1 0 1 x OR y 0 1 1 1
We say OR is almost analoguous to addition. We shall illustrate this by replacing OR with +: 0+0=0 0+1=1 1+0=1 1 + 1 = 1 (like 1 OR 1 is 1) 56
The NOT operation is not a binary operation, like AND and OR, but a unary operation, meaning it works with one argument. NOT x is true if x is false and false if x is true. In table form: The NOT function x 0 1 NOT x 1 0
In symbolic form, NOT x is denoted x' or ~x (or by a bar over the top of x). Alternative notations:
and
Compound truth tables
The three truth tables presented above are the most basic of truth tables and they serve as the building blocks for more complex ones. Suppose we want to construct a truth table for xy + z (i.e. x AND y OR z). Notice this table involves three variables (x, y and z), so we would expect it to be bigger then the previous ones. To construct a truth table, firstly we write down all the possible combinations of the three variables: x 0 0 0 0 1 1 1 y 0 0 1 1 0 0 1 z 0 1 0 1 0 1 0
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1
1
1
There is a pattern to the way the combinations are written down. We always start with 000 and end with 111. As to the middle part, it is up to the reader to figure out. We then complete the table by hand computing what value each combination is going to produce using the expression xy + z. For example: 000 x = 0, y = 0 and z = 0 xy + z = 0 001 x = 0, y = 0 and z = 1 xy + z = 1 We continue in this way until we fill up the whole table
The procedure we follow to produce truth tables are now clear. Here are a few more examples of truth tables. Example 1 -- x + y + z
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Example 2 -- (x + yz)'
When an expression is hard to compute, we can first compute intermediate results and then the final result. Example 3 -- (x + yz')w
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Exercise Produce the truth tables for the following operations: 8. 9. 10. NAND: x NAND y = NOT (x AND y) NOR: x NOR y = NOT (x OR y) XOR: x XOR y is true if and ONLY if one of x or y is true. Produce truth tables for: 4. 5. 6. 7. 8. xyz x'y'z' xyz + xy'z xz (x + y)' 60
9. 10. 11.
x'y' (xy)' x' + y'
Laws of Boolean algebra
In ordinary algebra, two expressions may be equivalent to each other, e.g. xz + yz = (x + y)z. The same can be said of Boolean algebra. Let's construct truth tables for: xz + yz (x + y)z xz + yz
(x + y)z
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By comparing the two tables, you will have noticed that the outputs (i.e. the last column) of the two tables are the same! Definition We say two Boolean expressions are equivalent if the output of their truth tables are the same. We list a few expressions that are equivalent to each other x+0=x x Ã- 1 = x xz + yz = (x + y)z x + x' = 1 x Ã- x' = 0 x Ã- x = x x + yz = (x + y)(x + z) Take a few moments to think about why each of those laws might be true. The last law is not obvious but we can prove that it's true using the other laws:
As Dr Kuo Tzee-Char, Honorary reader of mathematics at the University of Sydney, is so fond of saying: "The only thing to remember in mathematics is that there is nothing to remember. Remember that!". You should not try to commit to memory the laws as they are stated, because some of them are so deadly obvious once you are familiar with the AND, OR and NOT operations. You should only try to remember those things that are most basic, once a high level of familiarity is developed, you will agree there really isn't anything to remember. Simplification Once we have those laws, we will want to simplify Boolean expressiosn just like we do in ordinary algebra. We can all simplify the following example with ease:
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the same can be said about:
From those two examples we can see that complex-looking expressions can be reduced very significantly. Of particular interest are expressions of the form of a sum-of-product, for example: xyz + xyz' + xy'z + x'yz + x'y'z' + x'y'z We can factorise and simplify the expression as follows xyz + xyz' + xy'z + x'yz + x'y'z' + x'y'z
It is only hard to go any further, although we can. We use the identity: x + yz = (x + y)(x + z) If the next step is unclear, try constructing truth tables as an aid to understanding.
And this is as far as we can go using the algebraic approach (or any other approach). The algebraic approach to simplication relies on the priniciple of elimination. Consider, in ordinary algebra: x+y-x We simplify by rearranging the expression as follows (x - x) + y = y 63
Although we only go through the process in our head, the idea is clear: we bring together terms that cancel themselves out and so the expression is simplified.
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De Morgan's theorems So far we have only dealt with expressions in the form of a sum of products e.g. xyz + x'z + y'z'. De Morgan's theorems help us to deal with another type of Boolean expressions. We revisit the AND and OR truth tables: x 0 0 0 1 1 1 1 1 1 1 0 0 1 0 y 0 x Ã- y 0 x+y
You would be correct to suspect that the two operations are connected somehow due to the similarities between the two tables. In fact, if you invert the AND operation, i.e. you perform the NOT operations on x AND y. The outputs of the two operations are almost the same: x 0 0 0 1 1 1 1 1 1 0 0 1 1 1 y 0 (x Ã- y)' 1 x+y
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The connection between AND, OR and NOT is revealed by reversing the output of x + y by replacing it with x' + y'. x 0 1 0 1 1 1 1 0 1 0 0 1 1 1 y 0 (x Ã- y)' 1 x' + y'
Now the two outputs match and so we can equate them: (xy)' = x' + y' this is one of de Morgan's laws. The other which can be derived using a similar process is: (x + y)' = x'y' We can apply those two laws to simplify equations: Example Express x in sum of product form 1
Example Express x in sum of product form
2
This points to a possible extension of De Morgan's laws to 3 or more 67
variables. Example Express x in sum of product form 3
Example Express x in sum of product form
4
Another thing of interest we learnt is that we can reverse the truth table of any expression by replacing each of its variables by their opposites, i.e. replace x by x' and y' by y etc. This result shouldn't have been a suprise at all, try a few examples yourself. De Morgan's laws (x + y)' = x'y' (xy)' = x' + y' Exercise 3. 1. 2. 3. 4. 5. 4. Express in simplified sum-of-product form: z = ab'c' + ab'c + abc z = ab(c + d) z = (a + b)(c + d + f) z = a'c(a'bd)' + a'bc'd' + ab'c z = (a' + b)(a + b + d)d' Show that x + yz is equivalent to (x + y)(x + z)
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Propositions
We have been dealing with propositions since the start of this chapter, although we are not told they are propositions. A proposition is simply a statement (or sentence) that is either TRUE or FALSE. Hence, we can use Boolean algebra to handle propositions. There are two special types of propositions -- tautology and contradiction. A tautology is a proposition that is always TRUE, e.g. "1 + 1 = 2". A contradiction is the opposite of a tautology, it is a proposition that is always FALSE, e.g. 1 + 1 = 3. As usual, we use 1 to represent TRUE and 0 to represent FALSE. Please note that opinions are not propositions, e.g. "George W. Bush started the war on Iraq for its oil." is just an opinion, its truth or falsity is not universal, meaning some think it's true, some do not.
Examples
• • • • • • • "It is raining today" is a proposition. "Sydney is in Australia" is a proposition. "1 + 2 + 3 + 4 + 5 = 16" is a proposition. "Earth is a perfect sphere" is a proposition. "How do you do?" is not a proposition - it's a question. "Go clean your room!" is not a proposition - it's a command. "Martians exist" is a proposition. Since each proposition can only take two values (TRUE or FALSE), we can represent each by a variable and decide whether compound propositions are true by using Boolean algebra, just like we have been doing. For example "It is always hot in Antarctica OR 1 + 1 = 2" will be evaluated as true.
Implications
Propositions of the type if something something then something something are called implications. The logic of implications are widely applicable in mathematics, computer science and general everyday common sense reasoning! Let's start with a simple example "If 1 + 1 = 2 then 2 - 1 = 1" is an example of implication, it simply says that 2 - 1 = 1 is a consequence of 1 + 1 = 2. It's like a cause and effect relationship. Consider this example: John says: "If I become a millionaire, then I will donate $500,000 to the Red Cross."
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There are four situations: 5. 6. 7. 8. John becomes a millionaire and donates $500,000 to the Red Cross John becomes a millionaire and does not donate $500,000 to the Red Cross John does not become a millionaire and donates $500,000 to the Red Cross John does not become a millionaire and does not donate $500,000 to the Red Cross
In which of the four situations did John NOT fulfill his promise? Clearly, if and only if the second situation occured. So, we say the proposition is FALSE if and only if John becomes a millionaire and does not donate. If John did not become a millionaire then he can't break his promise, because his promise is now claiming nothing, therefore it must be evaluated TRUE. If x and y are two propositions, x implies y (if x then y), or symbolically
has the following truth table: x 0 0 1 1 y 0 1 0 1 1 1 0 1
For emphasis, is FALSE if and only if x is true and y false. If x is FALSE, it does not matter what value y takes, the proposition is automatically TRUE. On a side note, the two propositions x and y need not have anything to do with each other, e.g. "1 + 1 = 2 implies Australia is in the southern hemisphere" evaluates to TRUE! If
then we express it symbolically as .
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It is a two way implication which translates to x is TRUE if and only if y is true. The if and only if operation has the following truth table: x 0 0 1 1 y 0 1 0 1 1 0 0 1
The two new operations we have introduced are not really new, they are just combinations of AND, OR and NOT. For example:
Check it with a truth table. Because we can express the implication operations in terms of AND, OR and NOT, we have open them to manipulation by Boolean algebra and de Morgan's laws. Example Is the following proposition a tautology (a proposition that's always true) 1
Solution 1
Therefore it's a tautology. Solution 2 A somewhat easier solution is to draw up a truth table of the proposition, and note that the output column are all 1s. Therefore the proposition is a tautology, because the output is 1 regardless of the inputs (i.e. x, y and z). 71
Example Show that the proposition z is a contradiction (a proposition that is always false): z = xy(x + y)' Solution
2
Therefore it's a contradiction. Back to Example 1, : . This isn't just a slab of symbols, you should be able translate it into everyday language and understand intuitively why it's true. Exercises 8. 1. 2. 9. 1. Decide whether the following propositions are true or false: If 1 + 2 = 3, then 2 + 2 = 5 If 1 + 1 = 3, then fish can't swim Show that the following pair of propositions are equivalent :
Logic Puzzles
Puzzle is an all-encompassing word, it refers to anything trivial that requires solving. Here is a collection of logic puzzles that we can solve using Boolean algebra. Example 1 We have two type of people -- knights or knaves. A knight always tell the truth but the knaves always lie. Two people, Alex and Barbara, are chatting. Alex says :"We are both knaves" Who is who? We can probably work out that Alex is a knave in our heads, but the algebraic approach to determine Alex 's identity is as follows: Let A be TRUE if Alex is a knight 72
Let B be TRUE if Barbara is a knight There are two situations, either: Alex is a knight and what he says is TRUE, OR he is NOT a knight and what he says it FALSE. There we have it, we only need to translate it into symbols: A(A'B') + A'[(A'B')'] = 1 we simplify: (AA')B' + A'[A + B] = 1 A'A + A'B = 1 A'B = 1 Therefore A is FALSE and B is TRUE. Therefore Alex is a knave and Barbara a knight. Example 2 There are three businessmen, conveniently named Abner, Bill and Charley, who order martinis together every weekend according to the following rules: 4. 5. 6. 7. 2. 3. 4. 5. If A orders a martini, so does B. Either B or C always order a martini, but never at the same lunch. Either A or C always order a martini (or both) If C orders a martini, so does A. or AB + A'B' = 1 B'C + BC' = 1 A+C=1 or CA + C'A' = 1 Putting all these into one formula and simplifying:
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Exercises
Please go to Logic puzzles.
Problem Set
1. Decide whether the following propositions are equivalent:
2. Express in simplest sum-of-product form the following proposition:
3. Translate the following sentences into symbolic form and decide if it's true: a. For all x, if x2 = 9 then x2 - 6x - 3 = 0 b. We can find a x, such that x2 = 9 and x2 - 6x - 3 = 0 are both true. 4. NAND is a binary operation: x NAND y = (xy)' Find a proposition that consists of only NAND operators, equivalent to: (x + y)w + z 5. Do the same with NOR operators. Recall that x NOR y = (x + y)' 7475
Mathematical proofs
"It is by logic that we prove, but by intuition that we discover."
Introduction
Mathematicians have been, for the past five hundred years or so, obsessed with proofs. They want to prove everything, and in the process proved that they can't prove everything (see this). This chapter will introduce the techniques of mathematical induction, proof by contradiction and the axiomatic approach to mathematics.
Mathematical induction
Deductive reasoning is the process of reaching a conclusion that is guaranteed to follow. For example, if we know • • All ravens are black birds, and For every action, there is an equal and opposite reaction then we can conclude: • • This bird is a raven, therefore it is black. This billiard ball will move when struck with a cue. Induction is the opposite of deduction. To induce, we observe how things behave in specific cases and from that we draw conclusions as to how things behave in the general case. For example:
We know it is true for all numbers, because Gauss told us. But how do we show that it's true for all positive integers? Even if we can show the identity holds for numbers from one to a trillion or any larger number we can think of, we still haven't proved that it's true for all positive integers. This is where mathematical induction comes in, it works somewhat like the dominoes. If we can show that the identity holds for some number k, and that mere fact implies that the identity also holds for k + 1, then we have effectively shown that it works for all integers. Example 1 Show that the identity
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holds for all positive integers. Solution Firstly, we show that it holds for integers 1, 2 and 3 1 = 2Ã-1/2 1 + 2 = 3Ã-2/2 1 + 2 + 3 = 4Ã-3/2 = 6 Suppose the identity holds for some number k:
This supposition is known as the induction hypothesis. We assume it is true, and aim to show that,
is also true. We proceed
which is what we have set out to show. Since the identity holds for 3, it also holds for 4, and since it holds for 4 it also holds for 5, and 6, and 7, and so on. There are two types of mathematical induction: strong and weak. In weak induction, you assume the identity holds for certain value k, and prove it for k+1. In strong induction, the identity must be true for any value lesser or equal to k, and then prove it for k+1.
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Example 2 Show that n! > 2n for n ≥ 4. Solution The claim is true for n = 4. As 4! > 24, i.e. 24 > 16. Now suppose it's true for n = k, k ≥ 4, i.e. k! > 2k it follows that (k+1)k! > (k+1)2k > 2k+1 (k+1)! > 2k+1 We have shown that if for n = k then it's also true for n = k + 1. Since it's true for n = 4, it's true for n = 5, 6, 7, 8 and so on for all n. Example 3 Show that
Solution Suppose it's true for n = k, i.e.
it follows that
We have shown that if it's true for n = k then it's also true for n = k + 1. Now it's true for n = 1 (clear). Therefore it's true for all integers.
Exercises
1. Prove that 2. Prove that for n ≥ 1,
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where xn and yn are integers. 3. Note that
Prove that there exists an explicit formula for
for all integer m. E.g.
4. The sum of all of the interior angles of a triangle is ; the sum of all the angles of a rectangle is . Prove that the sum of all the angles of a polygon with n sides, is .
Proof by contradiction
"When you have eliminated the impossible, what ever remains, however improbable must be the truth." Sir Arthur Conan Doyle The idea of a proof by contradiction is to: 6. 7. 8. First, we assume that the opposite of what we wish to prove is true. Then, we show that the logical consequences of the assumption include a contradiction. Finally, we conclude that the assumption must have been false.
Root 2 is irrational
As an example, we shall prove that is not a rational number. Recall that a rational number is a number which can be expressed in the form of p/q, where p and q are integers and q does not equal 0 (see the 'categorizing numbers' section here). First, assume that is rational:
where a and b are coprimes (i.e. both integers with no common factors). If a and b are not coprimes, we remove all common factors. In other words, a/b is in simplest form. Now, 79
continuing:
We have now found that a2 is some integer multiplied by 2. Therefore, a2 must be divisible by two. If a2 is even, then a must also be even, for an odd number squared yields an odd number. Therefore we can write a = 2c, where c is another integer.
We have discovered that b2 is also an integer multiplied by two. It follows that b must be even. We have a contradiction! Both a and b are even integers. In other words, both have the common factor of 2. But we already said that a/b is in simplest form. Since such a contradiction has been established, we must conclude that our original assumption was false. Therefore, √2 is irrational.
Contrapositive
Some propositions that take the form of if xxx then yyy can be hard to prove. It is sometimes useful to consider the contrapositive of the statement. Before I explain what contrapositive is let us see an example "If x2 is odd then x is is also odd" is harder to prove than "if x even then x2 is also even" although they mean the same thing. So instead of proving the first proposition directly, we prove the second proposition instead. If A and B are two propositions, and we aim to prove If A is true then B is true we may prove the equivalent statement If B is false then A is false instead. This technique is called proof by contrapositive. To see why those two statements are equivalent, we show the following boolean algebra 80
expressions is true (see Logic)
(to be done by the reader).
Exercises
1.Prove that there is no perfect square number for 11,111,1111,11111...... 2. Prove that there are infinitely number of k's such that, 4k + 3, is prime. (Hint: consider N = p1p2...pm + 3)
Reading higher mathematics
This is some basic information to help with reading other higher mathematical literature. ... to be expanded
Quantifiers
Sometimes we need propositions that involve some description of rough quantity, e.g. "For all odd integers x, x2 is also odd". The word all is a description of quantity. The word "some" is also used to describe quantity. Two special symbols are used to describe the quanties "all" and "some" means "for all" or "for any" means "there are some" or "there exists" Example The proposition: For all even integers x, x2 is also even. can be expressed symbolically as: 1
Example The proposition: There are some odd integers x, such that x2 is even. can be expressed symbolically as:
2
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This proposition is false. Example Consider the proposition concerning (z = x'y' + xy): For any value of x, there exist a value for y, such that z = 1. can be expressed symbolically as: 3
This proposition is true. Note that the order of the quantifiers is important. While the above statement is true, the statement
is false. It asserts that there is one value of y which is the same for all x for which z=1. The first statement only asserts that there is a y for each x, but different values of x may have different values of y. Negation Negation is just a fancy word for the opposite, e.g. The negation of "All named Britney can sing" is "Some named Britney can't sing". What this says is that to disprove that all people named Britney can sing, we only need to find one named Britney who can't sing. To express symbolically: Let p represent a person named Britney
Similarly, to disprove
we only need to find one odd number that doesn't satisfy the condition. Three is odd, but 3Ã-3 = 9 is also odd, therefore the proposition is FALSE and
is TRUE In summary, to obtain the negation of a proposition involving a quantifier, you replace the quantifier by its opposite (e.g. with ) and the quantified proposition (e.g. "x is even") by its negation (e.g. "x is odd"). Example 1
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is a true statement. Its negation is
Axioms and Inference
If today's mathematicians were to describe the greatest achievement in mathematics in the 20th century in one word, that word will be abstraction. True to its name, abstraction is a very abstract concept (see Abstraction). In this chapter we shall discuss the essence of some of the number systems we are familiar with. For example, the real numbers and the rational numbers. We look at the most fundamental properties that, in some sense, define those number systems. We begin our discussion by looking at some of the more obscure results we were told to be true • • 0 times any number gives you 0 a negative number multiplied by a negative number gives you a positive number Most people simply accept that they are true (and they are), but the two results above are simple consequences of what we believe to be true in a number system like the real numbers! To understand this we introduce the idea of axiomatic mathematics (mathematics with simple assumptions). An axiom is a statement about a number system that we assume to be true. Each number system has a few axioms, from these axioms we can draw conclusions (inferences). Let's consider the Real numbers, it has axioms Let a, b and c be real numbers For a, b, and c taken from the real numbers A1: a+b is a real number the real numbers excluding zero M1: ab (closure) M2: There exist an element, 1, such that 1a = a for all a (existence of one - an identity) 83
M3: For every a there exists a b such that ab = 1 M4: (ab)c = a(bc) (associativity of multiplication) M5: ab = ba (commutativity of multiplication) D1: a(b + c) = ab + ac (distributivity) These are the minimums we assume to be true in this system. These are minimum in the sense that everything else that is true about this number system can be derived from those axioms! Let's consider the following true identity (x + y)z = xz + yz which is not included in the axioms, but we can prove it using the axioms. We proceed:
Before we proceed any further, you will have notice that the real numbers are not the only numbers that satifies those axioms! For example the rational numbers also satify all the axioms. This leads to the abstract concept of a field. In simple terms, a field is a number system that satisfies all those axiom. Let's define a field more carefully: A number system, F, is a field if it supports + and Ã- operations such that: For a, b, and c taken from F A1: a + b is in F F with the zero removed (sometimes written F*) M1: ab is in F (closure) M2: There exist an element, 1, such that 1a = a for all a (existence of one - the identity) M3: For every a there exists a b such that ab = 1 (inverses) M4: (ab)c = a(bc) (associativity of multiplication) 84
M5: ab = ba (commutativity of multiplication) D1: a(b + c) = ab + ac (distributivity) Now, for M3, we do not let b be zero, since 1/0 has no meaning. However for the M axioms, we have excluded zero anyway. For interested students, the requirements of closure, identity, having inverses and associativity on an operation and a set are known as a group. If F is a group with addition and F* is a group with multiplication, plus the distributivity requirement, F is a field. The above axioms merely state this fact in full. Note that the natural numbers are not a field, as M3 is general not satified, i.e. not every natural number has an inverse that is also a natural number. Please note also that (-a) denotes the additive inverse of a, it doesn't say that (-a) = (-1)(a), although we can prove that they are equivalent. Example 1 Prove using only the axioms that 0 = -0, where -0 is the additive inverse of 0. Solution 1 0 = 0 + (-0) by A3: existence of inverse 0 = (-0) by A2: 0 + a = a Example 2 Let F be a field and a an element of F. Prove using nothing more than the axioms that 0a = 0 for all a. Solution 0 = 0a + (-0a) by A3 existence of inverse 0 = (0 + 0)a + (-0a) by Example 1 0 = (0a + 0a) + (-0a) by distributivity and commutativity of multiplication 0 = 0a + (0a + (-0a)) by associativity of addition 0 = 0a + 0 by A3 0 = 0a by A2. Example 3 Prove that (-a) = (-1)a. 85
Solution 3 (-a) = (-a) + 0 (-a) = (-a) + 0a by Example 2 (-a) = (-a) + (1 + (-1))a (-a) = (-a) + (1a + (-1)a) (-a) = (-a) + (a + (-1)a) (-a) = ((-a) + a) + (-1)a (-a) = 0 + (-1)a (-a) = (-1)a One wonders why we need to prove such obvious things (obvious since primary school). But the idea is not to prove that they are true, but to practise inferencing, how to logically join up arguments to prove a point. That is a vital skill in mathematics.
Exercises
1. Describe a field in which 1 = 0 2. Prove using only the axioms if u + v = u + w then v = w (subtracting u from both sides is not accepted as a solution) 3. Prove that if xy = 0 then either x = 0 or y = 0 4. In F-, the operation + is defined to be the difference of two numbers and the Ã- operation is defined to be the ratio of two numbers. E.g. 1 + 2 = -1, 5 + 3 = 2 and 9Ã-3 = 3, 5Ã-2; = 2.5. Is F- a field? 5. Explain why Z6 (modular arithmetic modular 6) is not a field.
Problem Set
1. Prove
for 2. Prove by induction that 3. Prove by induction 86
where
and and 0! = 1 by definition.
4. Prove by induction
5. Prove that if x and y are integers and n an odd integer then
is an integer.
6. Prove that (n~m) = n!/((n-m)!m!) is an integer. Where n! = n(n-1)(n-2)...1. E.g 3! = 3Ã-2Ã-1 = 6, and (5~3) = (5!/3!)/2! = 10. Many questions in other chapters require you to prove things. Be sure to try the techniques discussed in this chapter87
Exercises
Mathematical proofsMathematical induction exercises
1. Prove that 12 + 22 + ... + n2 = n(n+1)(2n+1)/6 When n=1, L.H.S. = 12 = 1 R.H.S. = 1*2*3/6 = 6/6 = 1 Therefore L.H.S. = R.H.S. Therefore this is true when n=1. Assume that this is true for some positive integer k, i.e. 12 + 22 + ... + k2 = k(k+1)(2k+1)/6
Therefore this is also true for k+1. Therefore, by the principle of mathematical induction, this holds for all positive integer n. 2. Prove that for n ≥ 1,
where xn and yn are integers. 88
When n=1,
Therefore x1=1 and y1=1, which are both integers. Therefore this is true when n=1. Assume that this is true for some positive integer k, i.e. where xk and yk are integers.
Because xk and yk are both integers, therefore xk + 5yk and xk + yk are integers also. Therefore this is true for k+1 also. Therefore, by the principle of mathematical induction, this holds for all positive integer n. 3. (The solution assume knowledge in binomial expansion and summation notation) Note that
Prove that there exists an explicit formula for
for all integer m. E.g.
It's clear that 11 + 21 + ... = (n+1)n/2. So the proposition is true for m=1. Suppose that
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has an explicit formula in terms of n for all j < k (**), we aim to prove that
also has an explicit formula. Starting from the property given, i.e.
Since we know the formula for power sum of any power less then k (**), we can solve the above equation and find out the formula for the k-th power directly. Hence, by the principle of strong mathematical induction, this proposition is true. Additional info for question 3 The method employed in question 3 to find out the general formula for power sum is called the method of difference, as shown by that we consider the sum of all difference of adjacant terms. Aside from the method above, which lead to a recursive solution for finding the general formula, there're also other methods, such as that of using generating function. Refer to the last question in the 90
generating function project page for detail.
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Problem set Mathematical Proofs Problem Set
1. For all
Therefore
When a>b and c>d, a+c>b+d ( See also Replace it if you find a better one). Therefore we have:
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3.Let us call the proposition
be P(n) Assume this is true for some n, then
Now using the identities of this function:(Note:If anyone find wikibooks ever mentioned this,include a link here!),we have:
Since for all n,
Therefore P(n) implies P(n+1), and by simple substitution P(0) is true. Therefore by the principal of mathematical induction, P(n) is true for all n. Alternate Notice that solution
letting a = b = 1, we get
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as required. 5. Let be a polynomial with x as the variable, y and n as constants.
Therefore by factor theorem(link here please), (x-(-y))=(x+y) is a factor of P(x). Since the other factor, which is also a polynomial, has integer value for all integer x,y and n(I've skipped the part about making sure all coeifficients are of integer value for this moment), it's now obivous that
is an integer for all integer value of x,y and n when n is odd.
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Infinity and infinite processes
Introduction
As soon as a child first learns about numbers, they become interested in big ones, a million, a billion, a trillion. They even make up their own, a zillion etc. One of the first mathematical questions a child asks is "what is the largest number?" This will often lead to a short explanation that there are infinitely many numbers. But there are many different types of infinity - in fact, there are infinite types of infinity! This chapter will try to explain what some of these types mean and the differences between them.
Infinite Sets
How big is infinity?
Infinity is unlike a normal number because, by definition, it is not finite. Dividing infinity by any positive number (except infinity) gives us infinity. You can also multiply it by anything except zero (or infinity) and it will not get bigger. So let's look more carefully at the different types of infinity. There was once a mathematician called Georg Cantor who created a new branch of mathematics called set theory in the late 19th century. Set theory involves collections of numbers or objects. Here's a set: {1,2,3,4,5} Is it the same size as this one? {6,7,8,9,10} Cantor's notion of sets being "the same size" does not consider whether the numbers are bigger, but whether there are the same amount of objects in it. You can easily see here that they are the same size, because you can simply count the number of members in each set. But with an infinite number of members you cannot, in a finite amount of time, count all the members of one set to see if they are the same number of them as there are in another set. In order to decide if two infinite sets have the same number of members we need to think carefully about what we do when we count. Think of a small child sharing out sweets, between her and her brother. "One for you, and one for me, two for you and two for me" and so on. She knows that they both get the same number of sweets because of the way the sharing out was done. Even if she runs out of numbers (like if she can only count up to ten) she can still distribute the candy with the even process of "another one for you and another one for 95
me". We can use the same idea to compare infinite sets. If we can find a way to pair up one member of set A with one member of set B, and if there are no members of A without a partner in B and vice versa then we can say that set A and set B have the same number of members. Example Let Set N be all counting numbers. N is called the set of natural numbers. 1,2,3,4,5,6, ... and so to infinity. Let Set B be the negative numbers -1,-2,-3, ... and so on to -infinity. Can the members of N and B be paired up? The formal way of saying this is "Can A and B be put into a one to one correspondance"? Obviously the answer is yes. 1 in set N corresponds with -1 in B. Likewise: N B 1 -1 2 -2 3 -3 and so on. So useful is the set of counting numbers that any set that can be put into a one to one correspondence with it is said to be countably infinite. Let's look at some more examples. Is the set of integers countably infinite? Integers are set N, set B and 0. ... -3,-2,-1, 0, 1, 2, 3, ... Historically this set is usually called Z. Note that N the set of natural numbers is a subset of Z. All members of N are in Z, but not all members of Z are in N. What we need to find out is if Z can be put into a one to one correspondence with N. Your first answer, given that N is a subset of Z, may be no but you would be wrong! In set theory, the order of the elements is unimportant. There is no reason why we can't rearrange the elements into any order we please as long as we don't leave any out. Z as presented above doesn't look countable, but if we rewrite it as 0, -1, 1, -2, 2, -3, 3 ..... and so on we can see that it is countable. Z N 0 1 -1 2 1 3 96
-2 4 and so on. Strange indeed! A subset of Z (namely the natural numbers) has the same number of members as Z itself? Infinite sets are not like ordinary finite sets. In fact this is sometimes used as a definition of an infinite set. An infinite set is any set which can be put into a one to one correspondence with at least one of its subsets. Rather than saying "The number of members" of a set, people sometimes use the word cardinality or cardinal value. Z and N are said to have the same cardinality. Exercises 1. Is the number of even numbers the same as the natural numbers? 2. What about the number of square numbers? 3. Is the cardinality of positive even numbers less than 100 equal to the cardinality of natural numbers less than 100? Which set is bigger? How do you know? In what ways do finite sets differ from infinite ones? 4. Using the idea of one to one correspondance prove that infinity + 1 = infinity, what about infinity + A where A is a finite set? What about infinity plus C where C is a countably infinite set?
Is the set of rational numbers bigger than N?
In this section we will look to see if we can find a set that is bigger than the countable infinity we have looked at so far. To illustrate the idea we can imagine a story. There was once a criminal who went to prison. The prison was not a nice place so the poor criminal went to the prison master and pleaded to be let out. She replied: "Oh all right - I'm thinking of a number, every day you can have a go at guessing it. If you get it correct, you can leave." Now the question is - can the criminal get himself out of jail? (Think about if for a while before you read on) Obviously it depends on the number. If the prison master chooses a natural number, then the criminal guesses 1, to first day, 2,the second day and so on until he reaches the correct number. Likewise for the integers 0 on the first day, -1 on the second day. 1 on the third day and so on. If the number is very large then it may take a long time to get out of prison but get out he will. What the prison master needs to do is choose a set that is not countable in this way. Think of a number line. The integers are widely spaced out. There are plenty of numbers inbetween the integers 0&5 for example. So we need to look at denser sets. The first set that springs to most peoples mind are the fractions. There are an infinite number of fractions between 0 and 1 so surely there are more fractions than integers? Is it possible to count fractions? Let's think about that possibility for a while. If we try to use the approach of counting all the fractions between 0 & 1 then go on to 1 - 2 and so on we will come unstuck becuase we will never finish counting 97
the ones up to 1 ( there are an infinite number of them). But does this mean that they are uncountable ? Think of the situation with the integers. Ordering them ...-2, -1, 0, 1, 2, ... renders them impossible to count, but reordering them 0, -1, 1, -2, 2, ... allows them to be counted. There is in fact a way of ordering fractions to allow them to be counted. Before we go on to it let's revert to the normal mathematical language. Mathematicians use the term rational number to define what we have been calling fractions. A rational number is any number that can be written in the form p/q where p and q are integers. So 3/4 is rational, as is -1/2. The set of all rational numbers is usally called Q. Note that Z is a subset of Q becuase any integer can be divided by 1 to make it into a rational. E.g. the number 3 can be written in the form p/q as 3/1. Now as all the numbers in Q are defined by two numbers p and q it makes sense to write Q out in the form of a table.
Note that this table isn't an exact representation of Q. It only has the positive members of Q and has a number of multiple entries.( e.g. 1/1 and 2/2 are the same number) We shall call this set Q'. It is simple enough to see that if Q' is countable then so is Q. So how do we go about counting Q'? If we try counting the first row then the second and so on we will fail because the rows are infinite in length. Likewise if we try to count columns. But look at the diagonals. In one direction they are infinite ( e.g. 1/1, 2/2, 3/3, ...) but in the other direction they are finite. So this set is countable. We count them along the finite diagonals, 1/1, 1/2, 2/1, 1/3, 2/2, 3/1.... Exercises 1. Adapt the method of counting the set Q' to show that thet Q is also countable. How will you include 0 and the negative rationals? How will you solve the problem of multiple entries representing the same number ? 2. Show that (provided that the infinites are both countable)
Can we find any sets that are bigger than N?
So far we have looked at N, Z, and Q and found them all to be the same size, even though N is a subset of Z which is a subset of Q. You might be beginning to think "Is that it? Are all infinities the same size?" In this section we will look at an set that is bigger than N. A set that cannot be put into a one to one correspondence with N no matter how it is arranged. 98
The set in question is R the real numbers. A real number is any number on the number line that is not in Q. Remember that the set Q contains all the numbers that can be written in the form p/q with p and q rational. Most numbers can never be put in this form. Examples of irrational numbers include π,e, and .
The set R is huge! Much bigger than Q. To get a feel for the different sizes of these two infinite sets consider the decimal expansions of a real number and a rational number. Rational numbers always either terminate: • 1/8 = 0.125 or repeat: • 1/9 = 0.1111111...... Imagine measuring an object such as a book. If you use a ruler you might get 10cm. If you take a bit more care to and read the mm you might get 10.2cm. You'd then have to go on to more accurate measing devices such as vernier micrometers and find that you get 10.235cm. Going onto a travelling microscope you may find its 10.235823cm and so on. In general the decimal expansion of any real measurement will be a list of digits that look completely random. Now imagine you measure a book and found it to be 10.101010101010cm. You'd be pretty surprised wouldn't you? But this is exactly the sort of result you would need to get if the book's length were rational. Rational numbers are dense (you find them all over the number line), infinite, yet much much rarer than real numbers.
How we can prove that R is bigger than Q
It's good to get a feel for the size of infinities as in the previous section. But to be really sure we have to come up with some form of proof. In order to prove that R is bigger than Q we use a classic method. We assume that R is the same size as Q and come up with a contradiction. For the sake of clarity we will restrict our proof to the real numbers between 0 and 1.We will call this set R' Clearly if we can prove that R' is bigger than Q then R must be bigger than Q also. If R' was the same size as Q it would mean that it is countable. This means that we would be able to write out some form of list of all the members of R (This is what countable means, so far we have managed to write out all our infinite sets in the form of an infinitely long list). Let's consider this list. R1 R2 R3 R4 . 99
. . Where R1 is the first number in our list, R2 is the second, and so on. Note that we haven't said what order the list is to be written. For this proof we don't need to say what the order of the list needs to be, only that the mermbers of R are listable (hence countable). Now lets write out the decimal expansion of each of the numbers in the list. 0.r11r12r13r14... 0.r21r22r23r24... 0.r31r32r33r44... 0.r41r42r43r44... . . . Here r11 means the first digit after the decimal point of the first number in the list. So if our first number happened to be 0.36921... r11 would be 3, r12 would be 6 and so on. Remember that this list is meant to be complete. By that we mean that it contains every member of R'. What we are going to do in order to prove that R is not countable is to construct a number that is not already on the list. Since the list is supposed to contain every member of R', this will cause a contradiction and therfore show that R' is unlistable. In order to construct this unlisted number we choose a decimal representation: 0.a1a2a3a4... Where a1 is the first digit after the point etc. We let a1 take any value from 0 - 9 inclusive except the digit r11. So if r11 = 3 then a1 can be 0, 1, 2, 4, 5, 6, 7, 8, or 9. Then we let a2 be any digit except r22 (the second digit of the second number on the list). Then a3 be any digit except r33 and so on. Now if this number, that we have just constructed were on the list somewhere then it would have to be equal to Rsomething. Let's see what Rsomething it might be equal to. It can't be equal to R1 because it has a different first digit (r11 and a1. Nor can it be equal to R2 because it has a different second digit, and so on. In fact it can't be equal to any number on the list becuse it differs by at least one digit from all of them. We have done what we set out to do. We have constructed a number that is in R but is not on the "list of all members of R". This means that R is bigger than any list. It is not listable. It is 100
not countable. It is a bigger infinity than Q.
Are there even bigger infinities?
There are but they are difficult to describe. The set of all the possible combinations of any number of real numbers is a bigger infinity than R. However trying to imagine such a set is mind boggling. Let's look instead at a set that looks like it should be bigger than R but turns out not to be. Remember R', which we defined earlier on as the set of all numbers on the number line between 0 and 1. Let us now consider the set of all numbers in the plane from [0,0] to [1,1]. At first sight it would seem obvious that there must be more points on the whole plane than there are in a line. But in transfinite mathematics the "obvious" is not always true and proof is the only way to go. Cantor spent three years trying to prove it true but failed. His reason for failure was the best possible. It's false.
Each point in this plane is specified by two numbers, the x coordinate and the y coordinate; x and y both belong to R. Lets consider one point in the line. 0.a1a2a3a4.... Can you think of a way of using this one number to specify a point in the plane ? Likewise can you think of a way of combining the two numbers x= 0.x1x2x3x4.... and y= 0.y1y2y3y4.... to specify a point on the line? (think about it before you read on) One way is to do it is to take a1 = x1 a2 = y1 a3 = x2 a4 = y2 . 101
. . This defines a one to one correspondence between the points in the plane and the points in the line. (Actually, for the sharp amongst you, not quite one to one. Can you spot the problem and how to cure it?) Exercises 1. Prove that the number of points in a cube is the same as the number of points on one of its sides.
Continuum hypothesis
We shall end the section on infinite sets by looking at the Continuum hypothesis. This hypothesis states that there are no infinities between the natural numbers and the real numbers. Cantor came up with a number system for transfinite numbers. He called the smallest infinity with the next biggest one and so on. It is easy to prove that the cardinality of N is (Write any smaller infinity out as a list. Either the list terminates, in which case it's finite, or it ? goes on forever, in which case it's the same size as N) but is the cardinality of the reals = To put it another way, the hypotheses states that: There are no infinite sets larger than the set of natural numbers but smaller than the set of real numbers. The hypothesis is interesting because it has been proved that "It is not possible to prove the hypothesis true or false, using the normal axioms of set theory"
Further reading
If you want to learn more about set theory or infinite sets try one of the many interesting pages on our sister project en:wikipedia. • • • • ordinal numbers Aleph numbers Set theory Hilbert's Hotel
Limits Infinity got rid of
The theory of infinite sets seems weird to us in the 21st century, but in Cantor's day it was downright unpalatable for most mathematicians. In those days the idea of infinity was too troublesome, they tried to avoid it wherever possible. 102
Unfortunately the mathematical topic called analysis was found to be highly useful in mathematics, physics, engineering. It was far too useful a field to simply drop yet analysis relies on infinity or at least infinite processes. To get around this problem the idea of a limit was invented. Consider the series
This series is called the harmonic series. Note that the terms of the series get smaller and smaller as you go further and further along the series. What happens if we let n become infinite? The term would become But this doesn't make sense. (Mathematicians consider it sloppy to divide by infinity. Infinity is not a normal number, you can't divide by it). A better way to think about it (The way you probably already do think about it, if you've ever considered the matter) is to take this approach: Infinity is very big, bigger than any number you care to think about. So let's let n become bigger and bigger and see if 1/n approaches some fixed number. In this case as n gets bigger and bigger 1/n gets smaller and smaller. So it is reasonable to say that the limit is 0. In mathematics we write this as
and it reads: the limit of 1/n as n approaches infinity is zero Note that we are not dividing 1 by infinity and getting the answer 0. We are letting the number n get bigger and bigger and so the reciprocal gets closer and closer to zero. Those 18th Century mathematicians loved this idea because it got rid of the pesky idea of dividing by infinity. At all times n remains finite. Of course, no matter how huge n is, 1/n will not be exactly equal to zero, there is always a small difference. This difference (or error) is usually denoted by ε (epsilon).
info -- infinitely small
When we talk about infinity, we think of it as something big. But there is also the infinitely small, denoted by ε (epsilon). This animal is closer to zero than any other number. Mathematicians also use the character ε to represent anything small. For example, the famous Hungarian mathematician Paul Erdos used to refer to small children as epsilons.
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Examples
Lets look at the function
What is the limit as x approaches infinity ? This is where the idea of limits really come into its own. Just replacing x with infinity gives us very little:
But by using limits we can solve it
For our second example consider this limit as x approaches infinity of x3 - x2 Again lets look at the wrong way to do it. Substituting answer zero. Now lets look at doing it the correct way, using limits into the expression gives
. Note that you cannot say that these two infinities just cancel out to give the
The last expression is two functions multiplied together. Both of these functions approach infinity as x approaches infinity, so the product is infinity also. This means that the limit does not exist, i.e. there is no finite number that the function approaches as x gets bigger and bigger. One more just to get you really familer with how it works. Calculate:
To make things very clear we shall rewrite it as
Now to calculate this limit we need to look at the properties of sin(x). Sin(x)is a function that you should already be familiar with (or you soon will be) its value oscillates between 1 and -1 104
depending on x. This means that the absolute value of sin(x) (the value ignoring the plus or minus sign) is always less than or equal to 1:
So we have 1/x which we already know goes to zero as x goes to infinity multiplied by sin(x) which always remains finite no matter how big x gets. This gives us
Exercises
Evaluate the following limits;
1.
2. 3. 4.
Infinite series
Consider the infinite sum 1/1 + 1/2 + 1/4 + 1/8 + 1/16 + .... Do you think that this sum will equal infinity once all the terms have been added ? Let's sum the first few terms.
Can you guess what
is ?
Here is another way of looking at it. Imagine a point on a number line moving along as the sum progresses. In the first term the point jumps to the position 1. This is half way from 0 to 2. In the second stage the point jumps to position 1.5 - half way from 1 to 2. At each stage in the process (shown in a different colour on the diagram) the distance to 2 is halved. The point can 105
get as close to the point 2 as you like. You just need to do the appropriate number of jumps, but the point will never actually reach 2 in a finite number of steps. We say that in the limit as n approaches infinity, Sn approaches 2.
Zeno's Paradox
The ancient Greeks had a big problem with summing infinite series. A famous paradox from the philosopher Zeno is as follows: In the paradox of Achilles and the tortoise, we imagine the Greek hero Achilles in a footrace with the plodding reptile. Because he is so fast a runner, Achilles graciously allows the tortoise a head start of a hundred feet. If we suppose that each racer starts running at some constant speed (one very fast and one very slow), then after some finite time, Achilles will have run a hundred feet, bringing him to the tortoise's starting point. During this time, the tortoise has "run" a (much shorter) distance, say one foot. It will then take Achilles some further period of time to run that distance, during which the tortoise will advance farther; and then another period of time to reach this third point, while the tortoise moves ahead. Thus, whenever Achilles reaches somewhere the tortoise has been, he still has farther to go. Therefore, Zeno says, swift Achilles can never overtake the tortoise106
Exercises Infinity and infinite processes usefull for someone and that people will correct my work if I made some mistakes
How big is infinity? exercises
1. The number of even numbers is the same as the number of natural numbers because both are countably infinite. You can clearly see the one to one correspondence. (E means even numbers and is not an official set like N) E N 2 1 4 2 6 3 8 4 2. The number of square numbers is also equal to the number of natural numbers. They are both countably infinite and can be put in one to one correspondence. (S means square numbers and is not an official set like N) S N 1 1 4 2 9 3 16 4 3. The cardinality of even numbers less than 100 is not equal to the cardinality of natural numbers less than 100. You can simply write out both of them and count the numbers. Then you will see that cardinality of even numbers less than 100 is 49 and the cardinality of natural numbers less than 100 is 99. Thus the set of natural numbers less than 100 is bigger than the set 107
of even numbers less than 100. The big difference between infinite and finite sets thus is that a finite set can not be put into one to one correspondence with any of it's subsets. While an infinite set can be put into one to one correspondence with at least one of it's subsets. 4. Each part of the sum is answered below infinity + 1 = infinity You can prove this by taking a set with a cardinality of 1, for example a set consisting only of the number 0. You simply add this set in front of the countably infinite set to put the infinite set and the inifinite+1 set into one to one correspondence. N N+1 1 0 2 1 3 2 4 3 infinity + A = infinity (where A is a finite set) You simply add the finite set in front of the infinite set like above, only the finite set doesn't need to have a cardinality of one anymore. infinity + C = infinity (where C is a countably infinite set) You take one item of each set (infinity or C) in turns, this will make the new list also countably infinite.
Is the set of rational numbers bigger than N? exercises
1. To change the matrix from Q' to Q the first step you need to take is to remove the multiple entries for the same number. You can do this by leaving an empty space in the table when gcd(topnr,bottomnr) ≠ 1 because when the gcd isn't 1 the fraction can be simplified by dividing the top and bottom number by the gcd. This will leave you with the following table.
Now we only need to add zero to the matrix and we're finished. So we add a vertical row for zero and only write the topmost element in it (0/1) (taking gcd doesn't work here because 108
gcd(0,a)=a) This leaves us with the following table where we have to count all fractions in the diagonal rows to see that Q is countably infinite.
2. To show that you have to make a table where you put one infinity in the horizontal row and one infinity in the vertical row. Now you can start counting the number of place in the table diagonally just like Q' was counted. This works because a table of size AxB contains A*B places.
Are there even bigger infinities? exercises
1. You have to use a method to map the coordinates in a plain onto a point on the line and the other way around, like the one described in the text. This method shows you that for every number on the line there is a place on the plain and for every place on the plain there is a place on the line. Thus the number of points on the line and the plain are the same. Limits Infinity got rid of exercises
1.
2.
3. 4.
Problem set
HSE PS Infinity and infinite processes
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Counting and Generating functions
Before we begin: This chapter assumes knowledge of 11. 12. 13. Ordered selection (permutation) and unordered selection (combination) covered in Basic counting, Method of Partial Fractions and, Competence in manipulating Summation Signs
Some Counting Problems
..more to come
Generating functions
..some motivation to be written Generating functions, otherwise known as Formal Power Series, are useful for solving problems like: x1 + x2 + 2x3 = m where ; n = 1, 2, 3 how many unique solutions are there if m = 55? Before we tackle that problem, let's consider the infinite polynomial: S = 1 + x + x2 + x3 + ... + xn + xn + 1... We want to obtain a closed form of this infinite polynomial. The closed form is simply a way of expressing the polynomial so that it involves only a finite number of operations.
So the closed form of 1 + x + x2 + x3 + ... 110
is
We can equate them (actually, we can't. Refer to info).
info - Infinite sums The two expressions are not equal. It's just that for certain values of x (-1 < x < 1), we can approximate the right hand side as closely as possible by adding up a large number of terms on the left hand side. For example, suppose x = 1/2, RHS = 2; we approximate the LHS using only 5 terms we get LHS equals 1 + 1/2 + 1/4 + 1/8 + 1/16 = 1.9375, which is close to 2, as you can imagine by adding more and more terms, we will get closer and closer to 2. For a more detailed discussion of the above, head to Infinity and infinite processes. Anyway we really only care about its nice algebraic properties not its numerical value. From now on we will omit the condition for equality to be true when writing out generating functions. Consider a more general case: S = A + ABx + AB2x2 + AB3x3 + ... where A and B are constants. We can derive the closed-form as follows:
The following identity as derived above is worth investing time and effort memorising.
Exercises
1. Find the closed-form: 111
(a)1 - z + z2 - z3 + z4 - z5 + ... (b)1 + 2z + 4z2 + 8z3 + 16z4 + 32z5 + ... (c)z + z2 + z3 + z4 + z5 + ... (d)3 - 4z + 4z2 - 4z3 + 4z4 - 4z5 + ... (e)1 - z2 + z4 - z6 + z8 - z10 + ... 2. Given the closed-form, find a function f(n) for the coefficients of xn:
(a)
(Hint: note the plus sign in the denominator)
(Hint: obtain the generating function for 1/(1 - z^2) first, then multiply by the (b) appropriate expression)
(c)
(Hint: break into the sum of two distinct close forms)
Method of Substitution
We are given that: 1 + z + z2 + ... = 1/(1 - z) and we can obtain many other generating functions by substitution. For example: letting z = x2 we have: 1 + x2 + x4 + ... = 1/(1 - x2) Similarly A + ABx + A(Bx)2 + ... = A/(1 - Bx) is obtained by letting z = Bx then multiplying the whole expression by A. Exercises 1. What are the coefficients of the powers of x: 1/(1 - 2x3) 2. What are the coefficients of the powers of x (Hint: take out a factor of 1/2): 1/(2 - x) 112
Linear Recurrence Relations
The Fibonacci series 1, 1, 2, 3, 5, 8, 13, 21, 34, 55... where each and every number, except the first two, is the sum of the two preceding numbers. We say the numbers are related if the value a number takes depends on the values that come before it in the sequence. The Fibonacci sequence is an example of a recurrence relation, it is expressed as:
where xn is the (n+ 1)th number in the sequence. Note that the first number in the sequence is denoted x0. Given this recurrence relation, the question we want to ask is "can we find a formula for the (n+1)th number in the sequence?". The answer is yes, but before we come to that, let's look at some examples. Example 1 The expressions
define a recurrence relation. The sequence is: 1, 1, 5, 13, 41, 121, 365... Find a formula for the (n+1)th number in the sequence. Solution Let G(z) be generating function of the sequence, meaning the coefficient of each power (in ascending order) is the corresponding number in the sequence. So the generating functions looks like this G(z) = 1 + z + 5z2 + 13z3 + 41z4 + 121z5 + ... Now, by a series of algebraic manipulations, we can find the closed form of the generating function and from that the formula for each coefficient
113
by definition xn - 2xn - 1 - 3xn - 2 = 0
by the method of partial fractions we get:
each part of the sum is in a recognisable closed-form. We can conclude that:
the reader can easily check the accuracy of the formula. Example 2
Find a non-recurrent formula for xn. Solution Let G(z) be the generating function of the sequence described above. G(z) = x0 + x1z + x2z2 + ...
114
Therefore xn = 1, for all n. Example 3 A linear recurrence relation is defined by:
Find the general formula for xn. Solution Let G(z) be the generating function of the recurrence relation. G(z)(1 - z - 6z2) = x0 + (x1 - x0)z + (x2 - x1 - 6x0)z2 + ... G(z)(1 - z - 6z2) = 1 + z2 + z3 + z4 + ... G(z)(1 - z - 6z2) = 1 + z2(1 + z + z2 + ...)
Therefore
Exercises 1. Derive the formula for the (n+1)th numbers in the sequence defined by the linear recurrence relations: 115
2. Derive the formula for the (n+1)th numbers in the sequence defined by the linear recurrence relations:
3. (Optional) Derive the formula for the (n+1)th Fibonacci numbers.
Further Counting
Consider the equation a + b = n; a, b ≥ 0 are integers For a fixed positive integer n, how many solutions are there? We can count the number of solutions: 0+n=n 1 + (n - 1) = n 2 + (n - 2) = n ... n+0=n As you can see there are (n + 1) solutions. Another way to solve the problem is to consider the generating function G(z) = 1 + z + z2 + ... + zn Let H(z) = G(z)G(z), i.e. H(z) = (1 + z + z2 + ... + zn)2 I claim that the coefficient of zn in H(z) is the number of solutions to a + b = n, a, b > 0. The reason why lies in the fact that when multiplying like terms, indices add. Consider A(z) = 1 + z + z2 + z3 + ... Let 116
B(z) = A2(z) it follows B(z) = (1 + z + z2 + z3 + ...) + z(1 + z + z2 + z3 + ...) + z2(1 + z + z2 + z3 + ...) + z3(1 + z + z2 + z3) + ... B(z) = 1 + 2z + 3z2 + ... Now the coefficient of zn (for n ≥ 0) is clearly the number of solutions to a + b = n (a, b > 0). We are ready now to derive a very important result: let tk be the number solutions to a + b = n (a, b > 0). Then the generating function for the sequence tk is T(z) = (1 + z + z2 + ... + zn + ...)(1 + z + z2 + ... + zn + ...)
i.e.
Counting Solutions to a1 + a2 + ... + am = n
Consider the number of solutions to the following equation: a1 + a2 + ... + am = n where ai ≥ 0; i = 1, 2, ... m. By applying the method discussed previously. If tk is the number of solutions to the above equation when n = k. The generating function for tk is
but what is tk? Unless you have learnt calculus, it's hard to derive a formula just by looking the equation of T(z). Without assuming knowledge of calculus, we consider the following counting problem. "You have three sisters, and you have n (n ≥ 3) lollies. You decide to give each of your sisters at least one lolly. In how many ways can this be done?" One way to solve the problem is to put all the lollies on the table in a straightline. Since there are n lollies there are (n - 1) gaps between them (just as you have 5 fingers on each hand and 4 gaps between them). Now get 2 dividers, from the (n - 1) gaps available, choose 2 and put a divider in each of the gaps you have chosen! There you have it, you have divided the n lollies 117
into three parts, one for each sister. There are
ways to do it! If you have 4 sisters,
then there are it.
ways to do it. If you have m sisters there are
ways to do
Now consider: "You have three sisters, and you have n lollies. You decide to give each of your sisters some lollies (with no restriction as to how much you give to each sister). In how many ways can this be done?" Please note that you are just solving: a1 + a2 + a3 = n where ai ≥ 0; i = 1, 2, 3. You can solve the problem by putting n + 3 lollies on the table in a straightline. Get two dividers and choose 2 gaps from the n + 2 gaps available. Now that you have divided n + 3 lollies into 3 parts, with each part having 1 or more lollies. Now take back 1 lollies from each
part, and you have solved the problem! So the number of solutions is generally, if you have m sisters and n lollies the number of ways to share the lollies is
. More
. Now to the important result, as discussed above the number of solutions to a1 + a2 + ... + am = n where ai ≥ 0; i = 1, 2, 3 ... m is
i.e.
Example 1 The closed form of a generating function T(z) is 118
and tk in the coefficient of zk is T(z). Find an explicit formula for tk. Solution
Therefore tk = k Example 2 Find the number of solutions to: a+b+c+d=n for all positive integers n (including zero) with the restriction a, b, c ,d ≥ 0. Solution By the formula
so
the number of solutions is
More Counting
We turn to a sligthly harder problem of the same kind. Suppose we are to count the number of solutions to: 2a + 3b + c = n for some integer , with a, b, also c greater than or equal zero. We can write down the closed form straight away, we note the coefficient of xn of:
119
is the required solution. This is due to, again, the fact that when multiplying powers, indices add. To obtain the number of solutions, we break the expression into recognisable closed-forms by method of partial fraction. Example 1 Let sk be the number of solutions to the following equation: 2a + 2b = n; a, b ≥ 0 Find the generating function for sk, then find an explicit formula for sn in terms of n. Solution Let T(z) be the generating functions of tk T(z) = (1 + z2 + z4 + ... + z2n + ...)2
It's not hard to see that
Example 2 Let tk be the number of solutions to the following equation: a + 2b = n; a, b ≥ 0 Find the generating function for tk, then find an explicit formula for tn in terms of n. Solution Let T(z) be the generating functions of tk T(z) = (1 + z + z2 + ... + zn + ...)(1 + z2 + z4 + ... + z2n + ...)
120
A = -1/4, B = 3/4, C = 1/4
Exercises
1. Let
be the generating functions for tk (k = 0, 1, 2 ...). Find an explicit formula for tk in terms of k. 2. How many solutions are there the following equations if m is a given constant a + b + 2c = m where a, b and c ≥ 0
Problem Set
1. A new Company has borrowed $250,000 initial capital. The monthly interest is 3%. The company plans to repay $x before the end of each month. Interest is added to the debt on the last day of the month (compounded monthly). Let Dn be the remaining debt after n months. a) Define Dn recursively. b) Find the minimum values of x. c) Find out the general formula for Dn. d) Hence, determine how many months are need to repay the debt if x = 12,000. 2. A partion of n is a sequence of positive integers (λ1,λ1,..,λr) such that λ1 ≥ λ2 ≥ .. ≥ λr and 121
λ1 + λ2 + .. + λr = n. For example, let n = 5, then (5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), (1,1,1,1,1) are all the partions of 5. So we say the number of partions of 5 is 7. Derive a formula for the number of partions of a general n.
122
3. A binary tree is a tree where each node can have up to two child nodes. The figure below is an example of a binary tree.
a) Let cn be the number of unique arrangements of a binary tree with totally n nodes. Let C(z) be a generating function of cn. (i) Define C(z) using recursion. (ii) Hence find the closed form of C(z). b) Let be a power series.
(i) By considering the n-th derivative of P(x), find a formula for pn. (ii) Using results from a) and b)(i) , or otherwise, derive a formula for cn. Hint: Instead of doing recursion of finding the change in cn when adding nodes at the buttom, try to think in the opposite way, and direction.(And no, not deleting nodes)
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Project - Exponential generating function
This project assumes knowledge of differentiation. (Optional)0. (a) (i) Differentiate log x by first principle. (ii)*** Show that the remaining limit in last part that can't be evaluated indeed converges. Hence finish the differentiation by assigning this number as a constant. (b) Hence differentiate ax. 1. Consider E(x) = ex (a) Find out the n-th derivative of E(x). (b) By considering the value of the n-th derivative of E(x) at x = 0, express E(x) in power series/infinite polynomial form. (Optional)2. (a) Find out the condition for the geometric progression(that is the ordinary generating function introduced at the begining of this chapter) to converges. (Hint: Find out the partial sum) (b) Hence show that E(x) in the last question converges for all real values of x. (Hint: For any fixed x, the numerator of the general term is exponential, while the denominator of the general term is factorial. Then what?) 3. The function E(x) is the most fundamental and important exponential generating function, it is similar to the ordinary generating function, but with some difference, most obviously having a fractorial fraction attached to each term. (a) Similar to ordinary generating function, each term of the polynomial expansion of E(x) can have number attached to it as coefficient. Now consider
Find A'(x) and compare it with A(x). What do you discover? (b) Substitute nz, where n is a real number and z is a free variable, into E(x), i.e. E(nz). What have you found?
4. Apart from A(x) defined in question 2, let 124
(a) What is A(x) multiplied by B(x)? Compare this with ordinary generating function, what is the difference? (b) What if we blindly multiply A(x) with x(or xn in general)? Will it shift coefficient like what happened in ordinary generating function? Notes: Question with *** are difficult questions, although you're not expected to be able to answer those, feel free to try your best125
Exercises Counting and Generating functions useful for someone and that people will correct my work if I made some mistakes
Generating functions exercises
1. (a)S = 1 - z + z2 - z3 + z4 - z5 + ... zS = z - z2 + z3 - z4 + z5 - ... (1 + z)S = 1
(b)S = 1 + 2z + 4z2 + 8z3 + 16z4 + 32z5 + ... 2zS = 2z + 4z2 + 8z3 + 16z4 + 32z5 + ... (1 - 2z)S = 1
(c)S = z + z2 + z3 + z4 + z5 + ... zS = z2 + z3 + z4 + z5 + ... (1 - z)S = z
(d)S = 3 - 4z + 4z2 - 4z3 + 4z4 - 4z5 + ... z(S + 1) = 4z - 4z2 + 4z3 - 4z4 + 4z5 - ... 126
S + z(S + 1) = 3 S + zS + z = 3 (1 + z)S = 3 - z
2.
(a)
S = 1 - x + x2 - x3 + x4 - x5 + ... f(n) = ( - 1)n
(b) (1 - z2)S = z3 S = z3 + z5 + z7 + z9 + ...
f(n) = 0;for n is odd 2c only contains the exercise and not the answer for the moment
(c) Linear Recurrence Relations exercises This section only contains the incomplete answers because I couldn't figure out where to go from here. 1.
127
Let G(z) be the generating function of the sequence described above. G(z) = x0 + x1z + x2z2 + ... (1 - 2z)G(z) = x0 + (x1 - 2x0)z + (x2 - 2x1)z2 + ... (1 - 2z)G(z) = 1 - z - z2 - z3 - z4 - ... (1 - 2z)G(z) = 1 - z(1 + z + z2 + ...)
xn = 1 2.
Let G(z) be the generating function of the sequence described above. G(z) = x0 + x1z + x2z2 + ... (3 + 4z - z2)G(z) = 3x0 + (3x1 + 4x0)z + (3x2 + 4x1 - x0)z2 + (3x3 + 4x2 - x1)z3 + ... (3 + 4z - z2)G(z) = 3x0 + (3x1 + 4x0)z (3 + 4z - z2)G(z) = 3 + 7z
3. Let G(z) be the generating function of the sequence described above. G(z) = x0 + x1z + x2z2 + ... (1 - z - z2)G(z) = x0 + (x1 - x0)z + (x2 - x1 - x0)z2 + (x3 - x2 - x1)z2 + ... (1 - z - z2)G(z) = 1 128
We want to factorize f(z) = z2 + z - 1 into (z - α)(z - β) , by the converse of factor theorem, if (z p) is a factor of f(z), f(p)=0. Hence α and β are the roots of the quadratic equation z2 + z - 1 = 0 Using the quadratic formula to find the roots:
In fact, these two numbers are the faomus golden ratio and to make things simple, we use the greek symbols for golden ratio from now on.
Note:
is denoted φ and
is denoted Φ
By the method of partial fraction:
129
Further Counting exercises
1. We know that
therefore
Thus Tk = ( - 1)k(k + 1) 2. a + b + c = m
Thus
*Differentiate from first principle* exercises
1.
130
131
Discrete Probability
Introduction
Probability theory is one of the most widely applicable mathematical theories. It deals with uncertainty and teaches you how to manage it. It is simply one of the most useful theories you will ever learn. Please do not misunderstand. We are not learning to predict things, rather we learn to utilise predicted chances and make them useful. Therefore, we don't care, what is the probability it will rain tomorrow?, but given the probability is 60% we can make deductions, the easiest of which is the probability it will not rain tomorrow is 40%. As suggested above, a probability is a percentage and it's between 0% and 100% (inclusive). Mathematicians like to express a probability as a proportion i.e. as a number between 0 and 1.
info - Why discrete?
Probability comes in two flavours, discrete and continuous. The continuous case is considered to be far more difficult to understand, and much less intuitive, than discrete probability and it requires knowledge of calculus. But we will touch on a little bit of the continuous case later on in the chapter.
Event and Probability
Roughly, an event is something we can assign a probability to. For example the probability it will rain tomorrow is 0.6, in here the event is it will rain tomorrow the assigned probability is 0.6. We can write P(it will rain tomorrow) = 0.6 as mathematicians like to do we can use abstract letters to represent events. In this case we choose A to represent the event it will rain tomorrow, so the above expression can be written as P(A) = 0.6 Another example a fair die will turn up 1, 2, 3, 4, 5 or 6 equally probably each time it is tossed. Let B be the event that it turns up 1 in the next toss, we write P(B) = 1/6 Misconception Please note that the probability 1/6 does not mean that it will turns up 1 in at most six tries. Its precise meaning will be discussed later on in the chapter. Roughly, it just means that on the long run (i.e. the die being tossed a large number of times), the proportion of 1's will be 132
very close to 1/6.
Impossible and Certain events
Two types of events are special. One type are the impossible events (e.g., the sum of digits of a two-digit number is greater than 18); the other type are certain to happen (e.g., a roll of a die will turn up 1, 2, 3, 4, 5 or 6). The probability of an impossible event is 0, while that of a certain event is 1. We write P(Impossible event) = 0 P(Certain event) = 1 The above reinforces a very important principle concerning probability. Namely, the range of probability is between 0 and 1. You can never have a probability of 2.5! So remember the following
for all events E.
Complement of an event
A most useful concept is the complement of an event. We use : to represent the event that the die will NOT turn up 1 in the next toss. Generally, putting a bar over a variable (that represents an event) means the opposite of that event. In the above case of a die:
it means the die will turn up 2, 3, 4, 5 or 6 in the next toss has probability 5/6. Please note that
for any event E.
Combining independent probabilities
It is interesting how independent probabilities can be combined to yield probabilities for more complex events. I stress the word independent here, because the following demonstrations will not work without that requirement. The exact meaning of the word will be discussed a little later on in the chapter, and we will show why independence is important in Exercise 10 of this section. Adding probabilities Probabilities are added together whenever an event can occur in multiple "ways." As this is a rather loose concept, the following example may be helpful. Consider rolling a single die; if we want to calculate the probability for, say, rolling an odd number, we must add up the 133
probabilities for all the "ways" in which this can happen -- rolling a 1, 3, or 5. Consequently, we come to the following calculation: P(rolling an odd number) = P(rolling a 1) + P(rolling a 3) + P(rolling a 5) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 = 50% Note that the addition of probabilities is often associated with the use of the word "or" -whenever we say that some event E is equivalent to any of the events X, Y, or Z occurring, we use addition to combine their probabilities. A general rule of thumb is that the probability of an event and the probability of its complement must add up to 1. This makes sense, since we intuitively believe that events, when welldefined, must either happen or not happen. Multiplying probabilities Probabilities are multiplied together whenever an event occurs in multiple "stages" or "steps." For example, consider rolling a single die twice; the probability of rolling a 6 both times is calculated by multiplying the probabilities for the individual steps involved. Intuitively, the first step is simply the first roll, and the second step is the second roll. Therefore, the final probability for rolling a 6 twice is as follows:
P(rolling a 6 twice) = P(rolling a 6 the first time) 1/36 2.8%
P(rolling a 6 the second time) =
=
Similarly, note that the multiplication of probabilities is often associated with the use of the word "and" -- whenever we say that some event E is equivalent to all of the events X, Y, and Z occurring, we use multiplication to combine their probabilities. Also, it is important to recognize that the product of multiple probabilities must be less than or equal to each of the individual probabilities, since probabilities are restricted to the range 0 through 1. This agrees with our intuitive notion that relatively complex events are usually less likely to occur. Combining addition and multiplication It is often necessary to use both of these operations simultaneously. Once again, consider one die being rolled twice in succession. In contrast with the previous case, we will now consider the event of rolling two numbers that add up to 3. In this case, there are clearly two steps involved, and therefore multiplication will be used, but there are also multiple ways in which the event under consideration can occur, meaning addition must be involved as well. The die could turn up 1 on the first roll and 2 on the second roll, or 2 on the first and 1 on the second. This leads to the following calculation: P(rolling a sum of 3) = P(1 on 1st roll) P(2 on 2nd roll) + P(2 on 1st roll) P(1 on 2nd roll)
134
=
+
= 1/18
5.5%
This is only a simple example, and the addition and multiplication of probabilities can be used to calculate much more complex probabilities. Exercises Let A represent the number that turns up in a (fair) die roll, let C represent the number that turns up in a separate (fair) die roll, and let B represent a card randomly picked out of a deck: 1. A die is rolled. What is the probability of rolling a 3 i.e. calculate P(A = 3)? 2. A die is rolled. What is the probability of rolling a 2, 3, or 5 i.e. calculate P(A = 2) + P(A = 3) + P(A = 5)? 3. What is the probability of choosing a card of the suit Diamonds? 4. A die is rolled and a card is randomly picked from a deck of cards. What is the probability of rolling a 4 and picking the Ace of spades, i.e. calculate P(A = 4)×P(B = Ace of spades). 5. Two dice are rolled. What is the probability of getting a 1 followed by a 3? 6. Two dice are rolled. What is the probability of getting a 1 and a 3, regardless of order? 7. Calculate the probability of rolling two numbers that add up to 7. 8. (Optional) Show the probability of C is equal to A is 1/6. 9. What is the probability that C is greater than A? 10. Gareth was told that in his class 50% of the pupils play football, 30% play video games and 30% study mathematics. So if he was to choose a student from the class randomly, he calculated the probability that the student plays football, video games or studies mathematics is 50% + 30% + 30% = 1/2 + 3/10 + 3/10 = 11/10. But all probabilities should be between 0 and 1. What mistake did Gareth make? Solutions 1. P(A = 3) = 1/6 2. P(A = 2) + P(A = 3) + P(A = 5) = 1/6 + 1/6 + 1/6 = 1/2 3. P(B = Ace of Diamonds) + ... + P(B = King of Diamonds) = 13 × 1/52 = 1/4 4. P(A = 4) × P(B = Ace of Spades) = 1/6 × 1/52 = 1/312 5. P(A = 1) × P(A = 3) = 1/36 6. P(A = 1) × P(A = 3) + P(A = 3) × P(A = 1) = 1/36 + 1/36 = 1/18 135
7. Here are the possible combinations: 1 + 6 = 2 + 5 = 3 + 4 = 7. Probability of getting each of the combinations are 1/18 as in Q6. There are 3 such combinations, so the probability is 3 × 1/18 = 1/6. 9. Since both dice are fair, C > A is just as likely as C < A. So P(C > A) = P(C < A) and P(C > A) + P(C < A) + P(A = C) = 1 But P(A = C) = 1/6 so P(C > A) = 5/12. 10. For example, some of those 50% who play football may also study mathematics. So we can not simply add them.
Random Variables
A random experiment, such as throwing a die or tossing a coin, is a process that produces some uncertain outcome. We also require that a random experiments can be repeated easily. In this section we shall start using a capital letter to represent the outcome of a random experiment. For example, let D be the outcome of a die roll, D could take the value 1, 2, 3, 4, 5 or 6, but it is uncertain. We say D is a random variable. Suppose now I throw a die, and it turns up 5, we say the observed value of D is 5. A random variable is simply the outcome of a certain random experiment. It is usually denoted by a CAPITAL letter, but its observed value is not. For example let D1,D2,...,Dn denote the outcome of n die throws, then we usually use d1,d2,...,dn to denoted the observed values of each of Di's. From here on, random variable may be abbreviated as simply rv (a common abbreviation in other probability literatures).
The Bernoulli
This section is optional and it assumes knowledge of binomial expansion. A Bernoulli experiment is basically a "coin-toss". If we toss a coin, we will expect to get a head 136
or a tail equally probably. A Bernoulli experiment is slightly more versatile than that, in that the two possible outcomes need not have the same probability. In a Bernoulli experiment you will either get a success, denoted by 1, with proability p (where p is a number between 0 and 1) or a failure, denoted by 0, with probaility 1 - p. If the random variable B is the outcome of a Bernoulli experiment, and the probability of getting a 1 is p, we say B comes from a Bernoulli distribution with success probability p and we write:
For example, if
then P(C = 1) = 0.65 and P(C = 0) = 1 - 0.65 = 0.35
Binomial Distribution
Suppose we want to repeat the Bernoulli experiment n times, then we get a binomial distribution. For example:
for i = 1, 2, ... , n. That is, there are n variables C1, C2, ... , Cn and they all come from the same Bernoulli distribution. We consider: B = C1 + C2 + ... + Cn , then B is simply the rv that counts the number of successes in n trials (experiments). Such a variables is called a binomial variable, and we write
Example 1 Aditya, Gareth, and John are equally able. Their probability of scoring 100 in an exam follows 137
a Bernoulli distribution with success probability 0.9. What is the probability of i) One of them getting 100? ii) Two of them getting 100? iii) All 3 getting 100? iv) None getting 100? Solution We are dealing with a binomial variable, which we will call B. And
i) We want to calculate P(B = 1) The probability of any of them getting 100 (success) and the other two getting below 100 (failure) is
but there are 3 possible candidates for getting 100 so
ii) We want to calculate P(B = 2) The probability is
but there are
combinations of candidates for getting 100, so
iii) To calculate
iv) The probability of "None getting 100" is getting 0 success, so 138
The above example strongly hints at the fact the binomial distribution is connected with the binomial expansion. The following result regarding the binomial distribution is provided without proof, the reader is encouraged to check its correctness.
139
If
then
This is the kth term of the binomial expansion of (p + q)n, where q = 1 - p. Exercises ...
Distribution
...
Events
In the previous sections, we have slightly abused the use of the word event. An event should be thought of as a collection of random outcomes of a certain rv. Let us introduce some notations first. Let A and B be two events, we define
to be the event of A and B. We also define
to be the event of A or B. As demonstrated in exercise 10 above,
in general. Let's see some examples. Let A be the event of getting a number less than or equal to 4 when tossing a die, and let B be the event of getting an odd number. Now P(A) = 2/3 and P(B) = 1/2 but the probability of A or B does not equal to the sum of the probabilities, as below
140
as 7/6 is greater than 1. It is not difficult to see that the event of throwing a 1 or 3 is included in both A and B. So if we simply add P(A) and P(B), some events' probabilities are being added twice! The Venn diagram below should clarify the situation a little more,
think of the blue square as the probability of B and the yellow square as the probability of A. These two probabilities overlap, and where they do is the probability of A and B. So the probability of A or B should be:
The above formula is called the Simple Inclusion Exclusion Formula. If for events A and B, we have
we say A and B are disjoint. The word means to separate. If two events are disjoint we have the following Venn diagram representing them:
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info -- Venn Diagram Traditionally, Venn Diagrams are used to illustrate sets graphically. A set being simply a collection of things, e.g. {1, 2, 3} is a set consisting of 1, 2 and 3. Note that Venn diagrams are usually drawn round. It is generally very difficult to draw Venn diagrams for more than 3 intersecting sets. E.g. below is a Venn diagram showing four intersecting sets:
Expectation
The expectation of a random variable can be roughly thought of as the long term average of the outcome of a certain repeatable random experiment. By long term average it is meant that if we perform the underlying experiment many times and average the outcomes. For example, let D be as above, the observed values of D (1,2 ... or 6) are equally likely to occur. So if you were to toss the die a large number of times, you would expect each of the numbers to turn up roughly an equal number of times. So the expectation is
. We denote the expection of D by E(D), so E(D) = 3.5 We should now properly define the expectation. Consider a random variable R, and suppose the possible values it can take are r1, r2, r3, ... , rn. We define the expectation to be E(R) = r1P(R = r1) + r2P(R = r2) + ... + rnP(R = rn) Think about it: Taking into account the expectation is the long term average of the outcomes. Can you explain why is E(R) defined the way it is? Example 1 In a fair coin toss, let 1 represent tossing a head and 0 a tail. The same coin is tossed 8 times. Let C be a random variable representing the number of heads in 8 tosses? What is the expectation of C, i.e. calculate E(C)? 142
Solution 1 ... Solution 2 ...
Areas as probability
The uniform distributions. ... ........ ...
Order Statistics
Estimate the x in U[0, x]. ...
Addition of the Uniform distribution
Adding U[0,1]'s and introduce the CLT. .... to be continued ...143
Financial Options
Binary tree option pricing
Introduction
We have all heard of at least one stock exchange. NASDAQ, Dow Jones, FTSE and Hang Sheng. Less well-known, but more useful to many people, are the future exchanges. A stock exchange allows stock brokers to trade company stocks, while the future exchanges allow more exotic derivatives to be traded. For example, financial options, which is also the focus of this chapter. An option is a contract that gives the holder the choice to buy (or sell) a certain good in some time in the future for a certain price. What are options for? Initially, they are used to protect against risk. But they are also used to take advantage foreseeable opportunities, like what Thales has done1. Thales, the great Greek philosopher, was credited with the first recorded use of an option in the western world. A popular anecdote suggests, in one particular year while still in winter, he forecasted a great harvest of olives in the coming year. He had next to no money, so he purchased the option for the use of all the olive presses in his area. Naturally, when the time to harvest came everyone wanted to use the presses he had optioned! Needless to say he made a lot of money out of it.
Basics
An option is a contract of choice. You can choose whether to exercise the option or not. If you own an option that states You may purchase 1 kg of sugar from Shop A tomorrow for $2 suppose tomorrow the market price of sugar is $3, you would want to exercise the option i.e. buy the sugar for $2. Then you would sell it for $3 on the market and make $1 in the process. But if the market price for 1 kg of sugar is $1, then you would choose not to exercise the option, because it's cheaper on the market. Let us be a little bit more formal about what an option is. In particular there are two types of option: Call Option A call option is a contract that gives the owner the option to buy an 'underlying stock' at the 'strike price', on the 'expiry date'. Put Option 144
A put option is a contract that gives the owner the option to sell an 'underlying stock' at the 'strike price', on the 'expiry date'. In the above example, the 'underlying stock' was sugar and the 'strike price' $2 and 'expiry date' is tomorrow. We shall represent an option like below {C or P, $amount, # periods to expiry} . For example {C,$3,1} represents a call option with strike price for some unspecified underlying stock expiring in one time-unit's time. A time-unit here may be a year, a month, one day or one hour. The important point is the mathematics we will present later does not really depend on what this time-unit is. Also, we need not specify the underlying stock either. Another example {P,$100,2} represents a put option with strike price $100 for some unspecified underlying stock expiring in 2 time-units' time. Now that we have a basic idea of what an option is, we can start to imagine a market place where options are traded. We assume that such a market exists. Also we assume that there is no fee of any kind to participate in a trade. Such a market is called a frictionless market. Of course, a market place where the underlying stock is traded is also assumed to exist. Info -- American or European Actually there are two major types of options: American or European. An European option allows you to exercise the option only on the 'expiry date'; while the American version allows you to exercise the option at any time prior to the 'expiry date'. We shall only discuss European options in this chapter.
Arbitrage
Another very important concept is arbitrage. In short, an arbitrage is a way to make money out of nothing. We assume that there is no free-lunch in this world, in other words our market is arbitrage-free. We will show an example of how to perform an arbitrage later on in the chapter. The real meat of this chapter is the technique used to price the options. In simple terms, we have an option, how much should it be? From this angle, we will see that the arbitrage-free requirement is a very strong one, in that it basically dictates what the price of the option should be.
145
Option's value on expiry
Pricing the option is about how much is it worth now. Of course the present value of an option depends on its possible future values. Therefore it is vital to understand how much the option is worth at expiry, when it is time to choose whether to exercise the option or not. For example, consider the option {C,$2,1} it is the call (buy) option that expires in 1 week's time (or day or year or whatever time period it is suppose to be). How much should the option be if the market price of the underlying stock on expiry is $3? What if the market price is $1? It is sensible to say the option has a value of $1 if the market price (for the underlying stock) is $3, and the option should be worthless ($0) if the market price is $1. Why do we say it is sensible to price the option as above? It is because we assume the market is arbitrage-free. Also in a market, we assume • • there is a bank that's willing to lend you money if you repay the bank in the same day you borrowed, no fee will be charged. With those assumptions, we show that if you price the option any differently, someone can make money without using any of his/her own money. For example, suppose on expiry, the market price for the underlying stock is $3 and you decide to sell the option for $0.7 (not $1 as is sensible). An intelligent buyer would do the following: Action Borrow $2.7 Purchase your option for $0.7 Purchase sugar for $2 with option Sell 1kg of sugar for $3 in market Repay bank $2.7 Money +$2.7 -$0.7 -$2 +$3 -$2.7 Balance $2.7 $2 $0 $3 $0.3
He/she made $0.3 and at no time did he/she use his/her own money (i.e. balance never less than zero)! This is a free lunch, which is contrary to the assumption of a arbitrage-free market! Exercises 1. In an arbitrage-free market, consdier an option T = {C,$100,1}. i) How much should the option be on expiry if the price of the underlying stock is $90. 146
ii) What if the underlying stock costs $110 on expiry. iii} $100? 2. Consdier an option T = {C,$10,1}. i) On expiry, would you consider buying the option if it was for sell for $2 if the underlying stock costs $8? ii) What if the underlying stock costs $7. 3. Consider the put option T = {P,$2,1}. On expiry the underlying stock costs $1. Jenny owns T, she decides on the following actions Borrow $1 Purchase the underlying stock from the market for $1 Exercise the option i.e. sell the stock for $2 Repay $1 Did she do the right thing? 4. In an arbitrage-free market, consider the put option T = {P,$2,1}. i) On expiry, how much should the option cost if the underlying stock costs $1? ii) $3? 5. Consider the put option T = {P,$2,1}. On expiry the underlying stock costs $1. And the option T is on sale for $0.5. Jenny immediately sees an arbitrage oppoturnity. Detail the actions she should take to capitalise on the arbitrage oppoturnity. (Hint: immitate the Action, Money, Balance table )
Pricing an option
Consider this hypothetical situation where a company, MassiveSoft, is in negatiation to merge with another company, Pears. The share price of MassiveSoft currently stands at $7. If the negotiation is successful, the share price will rise to $11; otherwise it will fall to $5. Experts predict the probability of a success is 90%. Consider a call option that lets you buy 1000 shares of MassiveSoft at $8 when the negotiation is finalised. How much should the option be? Since the market is arbitrage-free, the value of the option at expiry is already determined. Of course if the negotiation is successful, the option is valued at (11 - 8) × 1000 = $3000 otherwise, the option should be worthless ($0) 147
the above are the only correct values of the option at expiry or people can "rip you off".
148
Let x be the price of the option at present, we can use the following diagrams to illustrate the situation,
the diagram shows that the current price of the option should be $x, and if the negotiation is successful, it will be worth $3000, otherwise it is worthless. In similar fashion, the following diagram shows the value of the company stock now, and in the future
You may have notice that we didn't put down the probability of success or failure. Interestingly (and counter-intuitively), they don't matter! Again, the arbitrage-free principle dictates that what we have in the two diagrams above are sufficient for us to price the option! How? What is the option? It is the contract that gives you the option to buy ... Wait, wait, wait. Think of it from another angle it is a tradable object that is worth $2000 if the negotiation is successful, and $0 if otherwise This is the main idea behind how to price the option. The option must be the same price as another object that goes up to $2000 or down to $0 depending on the success of the negotiation. Hopefully, this object is something we know the price of. This idea is called constructing a replicating portfolio. A portfolio is a collection of tradable things. We want to construct a portfolio that behaves in the same way as the option. It turns out that we can construst a portfolio that behaves in the way as the option by using only two things. They are 12. 13. MassiveSoft shares and money let's assume that money is tradable in the sense that you can buy a dollar with a dollar. This concept may seem very unintuitive at first. However let's proceed with the mathematics, suppose this porfolio consists of y units of MassiveSoft shares and z units of money. If the negotiation is successful, then each share will be worth $9, and the whole portolio should be 149
worth $2000, as it behaves in the same way as the option, so we have the following
but if the negotiation is unsuccessful then the portfolio is worthless ($0) and MassiveSoft share prices will fall to $5, giving
we can easily solve the above simultaneous equations. We get
and so y = 500 and z = − $2500 So this portolio consists of 500 MassiveSoft shares and -$2500. But what is -$2500? This can be understood as an obligation to pay back some money (e.g. from borrowings) on the expiry date of the option. So the porfolio we constructed can be thought of as 500 MassiveSoft shares and an obligation to pay $2500 Now, 500 MassiveSoft shares costs $7 × 500 = 3500, so the option should be priced as 3500 2500 = $1000. Let's price a few more options. ... The famous mathematicain, John Nash, as portrayed in the movie "A beautiful mind", did some pioneering work in portfolio theory with equivalent functions.
Call-Put parity
...more to come
Reference
5. A Brief History of Options150
Matrices
Introduction
A matrix may be more popularly known as a giant computer simulation, but in mathematics it is a totally different thing. To be more precise, a matrix (plural matrices) is an array of numbers. For example, below is a typical way to write a matrix, with numbers arranged in rows and column and with round brackets around the numbers
The above matrix has 4 rows and 4 columns, so we call it a 4 × 4 (4 by 4) matrix. Also, we can have matrices of many different shapes. The shape of a matrix is the name for the dimensions of matrix (m by n, where m is the number of rows and n the number of columns). Here are some more examples of matrices This is an example of a 3 × 3 matrix:
This is an example of a 5 × 4 matrix:
This is an example of a 1 × 6 matrix:
The theory of matrices is intimately connected with that of (linear) simultaneous equations. The ancient Chinese had established a systematic way to solve simultaneous equations. The theory of simultaneous equations is furthered in the east by the Japanese mathematician, Seki and a little later by Leibniz, Newton's greatest rival. Later, Gauss (1777 - 1855), one of the three 151
pillars of modern mathematics popularised the use of Gaussian elimination, which is a simple step by step algorithm for solving any number of linear simultaneous equations. By then the use of matrices to represent simultaneous equation neatly on paper (as discussed above) has become quite common[1]. Consider the simultaneous equations: x + y = 10 x-y=4 it has the solution x = 7 and y = 3, and the usual way to solve it is to add the two equations together to eliminate the y. Matrix theory offers us another way to solve the above simultaneous equations via matrix multiplication (covered below). We will study the widely accepted way to multiply two matrices together. In theory with matrix multiplication we can solve any number of simultaneous equations, but we shall mainly restrict our attention to 2 × 2 matrices. But even with that restriction, we have opened up doors to topics simultaneous equations could never offer us. Two such examples are 9. 10. using matrices to solve linear recurrence relations which can be used to model population growth, and encrypting messages wtih matrices. We shall commence our study by learning some of the more fundamental concepts of matrices. Once we have a firm grasp of the basics, we shall move on to study the real meat of this chapter, matrix multiplication.
Elements
An element of a matrix is a particular number inside the matrix, and it is uniquely located with a pair of numbers. E.g. let the following matrix be denoted by A, or symbolically:
the (2,2)th entry of A is 5; the (1,1)th entry of A is 1, the (3,3) entry of A is 9 and the (3,2)th entry of A is 8. The (i , j)th entry of A is usually denoted ai,j and the (i , j)th entry of a matrix B is usually denoted by bi,j and so on.
Summary
• • A matrix is an array of numbers A m×n matrix has m rows and n columns 152
• •
The shape of a matrix is determined by its number of rows and columns The (i,j)th element of a matrix is located in ith row and jth column
Matrix addition & Multiplication by a scalar
Matrices can be added together. But only the matrices of the same shape can be added. This is very natural. E.g.
then
Similarly matrices can be multiplied by a number, we call the number a scalar, this is to distinguish it from a matrix. The reader need not worry about the definition here, just remember that a scalar is just a number.
in this case the scalar value is 5. In general, when we do s × A , where s is a scalar and A a matrix, we multiply each entry of A by s.
Matrix Multiplication
The widely accepted way to multiply two matrices together is definitely non-intuitive. As mentioned above, multiplication can help with solving simultaneous equations. We will now 153
give a brief outline of how this can be done. Firstly, any system of linear simultaneous equations can be written as a matrix of coefficients multiplied by a matrix of unknowns equaling a matrix of results. This description may sound a little complicated, but in symbolic form it is quite clear. The previous statement simply says that if A, x and b are matrices, then Ax = b, can be used to represent some system of simultaneous equations. The beautiful thing about matrix multiplications is that some matrices can have multiplicative inverses, that is we can multiply both sides of the equation by A-1 to get x = A-1b, which effectively solves the simultaneous equations. The reader will surely come to understand matrix multiplication better as this chapter progresses. For now we should consider the simplest case of matrix multiplication, multiplying vectors. We will see a few examples and then we will explain process of multiplication
then
Similarly if:
then
A matrix with just one row is called a row vector, similarly a matrix with just one column is called a column vector. When we multiply a row vector A, with a column vector B, we multiply the element in the first column of A by the element in the first row of B and add to that the product of the second column of A and second row of B and so on. More generally we multiply a1,i by bi,1 (where i ranges from 1 to n, the number of rows/columns) and sum up all of the products. Symbolically:
(for information on the 154
sign, see Summation_Sign)
where n is the number of rows/columns. In words: the product of a column vector and a row vector is the sum of the product of item 1,i from the row vector and i,1 from the column vector where i is from 1 to the width/height of these vectors. Note: The product of matrices is also a matrix. The product of a row vector and column vector is a 1 by 1 matrix, not a scalar.
Exercises
Multiply:
Multiplication of non-vector matrices
where A, B and C are matrices. We multiply the ith row of A Suppose with the jth column of B as if they are vector-matrices. The resulting number is the (i,j)th element of C. Symbolically:
Example 1
155
Evaluate AB = C and BA'= D, where
and
Solution
i.e.
156
i.e.
Example 2 Evaluate AB and BA where
Solution
Example 3 Evaluate AB and BA where
Solution
Example 4 Evaluate the following multiplication:
157
Solution Note that:
is a 2 by 1 matrix and
is a 1 by 2 matrix. So the multiplication makes sense and the product should be a 2 by 2 matrix.
Example 5 Evaluate the following multiplication:
Solution
Example 6 Evaluate the following multiplication:
Solution Example 7 Evaluate the following multiplication:
158
Solution Note Multiplication of matrices is generally not commutative, i.e. generally AB ≠ BA. Diagonal matrices A diagonal matrix is a matrix with zero entries everywhere except possibly down the diagonal. Multiplying diagonal matrices is really convenient, as you need only to multiply the diagonal entries together. Examples The following are all diagonal matrices
Example 1
Example 2 The above examples show that if D is a diagonal matrix then Dk is very easy to compute, all we need to do is to take the diagonal entries to the kth power. This will be an extremely useful fact later on, when we learn how to compute the nth Fibonacci number using matrices.
Exercises
1. State the dimensions of C a) C = An×pBp×m
b) 2. Evaluate. Please note that in matrix multiplication (AB)C = A(BC) i.e. the order in which you do the multiplications does not matter (proved later). a)
159
b)
160
3. Performing the following multiplications:
What do you notice?
The Identity & multiplication laws
The exercise above showed us that the matrix:
is a very special. It is called the 2 by 2 identity matrix. An identity matrix is a square matrix, whose diagonal entries are 1's and all other entries are zero. The identity matrix, I, has the following very special properties 10. 11. for all matrices A. We don't usually specify the shape of the identity because it's obvious from the context, and in this chapter we will only deal with the 2 by 2 identity matrix. In the real number system, the number 1 satisfies: r × 1 = r = 1 × r, so it's clear that the identity matrix is analogous to "1". Associativity, distributivity and (non)-commutativity Matrix multiplication is a great deal different to the multiplication we know from multiplying real numbers. So it is comforting to know that many of the laws the real numbers satisfy also carries over to the matrix world. But with one big exception, in general AB ≠ BA. Let A, B, and C be matrices. Associativity means (AB)C = A(BC) i.e. the order in which you multiply the matrices is unimportant, because the final result you get is the same regardless of the order which you do the multiplications. On the other hand, distributivity means A(B + C) = AB + AC 161
and (A + B)C = AC + BC Note: The commutative property of the real numbers (i.e. ab = ba), does not carry over to the matrix world. Convince yourself For all 2 by 2 matrices A, B and C. And I the identity matrix. 1. Convince yourself that in the 2 by 2 case: A(B + C) = AB + AC and (A + B)C = AC + BC 2. Convince yourself that in the 2 by 2 case: A(BC) = (AB)C 3. Convince yourself that:
in general. When does AB = BA? Name at least one case. Note that all of the above are true for all matrices (of any dimension/shape).
Determinant and Inverses
We shall consider the simultaneous equations: ax + by = α (1) cx + dy = β (2) where a, b, c, d, α and β are constants. We want to determine the necessary conditions for (1) and (2) to have a unique solution for x and y. We proceed: Let (1') = (1) × c Let (2') = (2) × a i.e. acx + bcy = cα (1') acx + ady = aβ (2') 162
Now let (3) = (2') - (1') (ad - bc)y = aβ - cα (3) Now y can be uniquely determined if and only if (ad - bc) ≠ 0. So the necessary condition for (1) and (2) to have a unique solution depends on all four of the coefficients of x and y. We call this number (ad - bc) the determinant, because it tells us whether there is a unique solutions to two simultaneous equations of 2 variables. In summary if (ad - bc) = 0 then there is no unique solution if (ad - bc) ≠ 0 then there is a unique solution. Note: Unique, we can not emphasise this word enough. If the determinant is zero, it doesn't necessarily mean that there is no solution to the simultaneous equations! Consider: x+y=2 7x + 7y = 14 the above set of equations has determinant zero, but there is obviously a solution, namely x = y = 1. In fact there are infiinitely many solutions! On the other hand consider also: x+y=1 x+y=2 this set of equations has determinant zero, and there is no solution at all. So if determinant is zero then there is either no solution or infinitely many solutions. Determinant of a matrix We define the determinant of a 2 × 2 matrix
to be
Inverses
It is perhaps, at this stage, not very clear what's the use of the det(A). But it's intimately connected with the idea of an inverse. Consider in the real number system a number b, it has (multiplicative) inverse 1/b, i.e. b(1/b) = (1/b)b = 1. We know that 1/b does not exist when b = 0. 163
In the world of matrices, a matrix A may or may not have an inverse depending on the value of the determinant det(A)! How is this so? Let's suppose A (known) does have an inverse B (i.e. AB = I = BA). So we aim to find B. Let's suppose further that
and
we need to solve four simultaneous equations to get the values of w, x, y and z in terms of a, b, c, d and det(A). aw + by = 1 cw + dy = 0 ax + bz = 0 cx + dz = 1 the reader can try to solve the above by him/herself. The required answer is
In here we assumed that A has an inverse, but this doesn't make sense if det(A) = 0, as we can not divide by zero. So A-1 (the inverse of A) exists if and only if det(A) ≠ 0. Summary If AB = BA = I, then we say B is the inverse of A. We denote the inverse of A by A-1. The inverse of a 2 × 2 matrix
is
provided the determinant of A, det(A) is zero. 164
Solving simultaneous equations
Suppose we are to solve: ax + by = α cx + dy = β We let
we can translate it into matrix form
i.e Aw = γ If A's determinant is not zero, then we can pre-multiply both sides by A-1 (the inverse of A)
i.e.
which implies that x and y are unique. Examples Find the inverse of A, if it exists 165
a)
b)
c)
d) Solutions
a)
b) c) No solution, as det(A) = 3ab - 3ab = 0
d) Exercises 1. Find the determinant of
. Using the determinant of A, decide whether there's a unique solution to the following simultaneous equations
2. Suppose C = AB 166
show that det(C) = det(A)det(B) for the 2 × 2 case. Note: it's true for all cases. 3. Show that if you swap the rows of A to get A' , then det(A) = -det(A' ) 4. Using the result of 2 a) Prove that if: A = P - 1BP then det(A) = det(B) b) Prove that if: Ak = 0 for some positive integer k, then det(A) = 0. 5. a) Compute A5, i.e. multiply A by itself 5 times, where
b) Find the inverse of P where
c) Verify that
d) Compute A5 by using part (b) and (c). f) Compute A10
167
Other Sections Exercises MatricesMatrix Multiplication exercises
Multiplication of non-vector matrices exercises
1. a) b) 2. 168
a)
b)
3.
The important thing to notice here is that the 2x2 matrix remains the same when multiplied with the other matrix. The matrix with only 1s on the diagonal and 0s elsewhere is known as the identity matrix, called I, and any matrix multiplied on either side of it stays the same. That is 169
NB:The remaining exercises in this section are leftovers from previous exercises in the 'Multiplication of non-vector matrices' section 3.
The important thing to notice here is that the 1 to 9 matrix remains the same when multiplied with the other matrix. The matrix with only 1s on the diagonal and 0s elsewhere is known as the identity matrix, called I, and any matrix multiplied on either side of it stays the same. That is 4. a)
b)
170
c)
d)
e) As an example I will first calculate A2
171
Now lets do the same simplifications I have done above with A5-
f)
172
Determinant and Inverses exercises
1.
The
simultaneous
equation
will
be
translated
into
the
following
matrices
Because we already know that
We can say that there is no unique solution to these simultaneous equations. 2. First calculate the value when you multiply the determinants
(ad - bc)(eh - fg) = adeh - bceh - adfg + bcfg Now let's calculate C by doing the matrix multiplication first
(ae + bg)(cf + dh) - (af + bh)(ce + dg) = aecf + bgcf + aedh + bgdh - afce - bgce - afdg - bhdg = bgcf + aedh - bgce - afdg Which is equal to the value we calculated when we multiplied the determinants, thus 173
det(C) = det(A)det(B) for the 2Ã-2 case. 3.
det(A) = ad - bc
det(A') = cb - da - det(A') = - (bc - ad) = ad - bc Thus det(A) = -det(A') is true. 4. a) A = P - 1BP det(A) = det(P - 1)det(B)det(P) = det(P - 1)det(P)det(B) = det(P - 1P)det(B) = det(I)det(B) = det(B) as det(I) = 1. thus det(A) = det(B) b) if Ak = 0 for some k it means that det(Ak) = 0. But we can write det(Ak) = det(A)k, thus det(A)k = 0. This means that det(A) = 0. 5. a) A5 =
174
b)
c)
d) A5 =
175
We see that P and it's inverse disappear when you raise the matrix to the fifth power. Thus you can see that we can calculate An very easily because you only have to raise the diagonal matrix to the n-th power. Raising diagonal matrices to a certain power is very easy because you only have to raise the numbers on the diagonal to that power. f) We use the method derived in the exercise above. A100 =
176
Problem set Matrices Problem Set 1.
Therefore the message is "iloveyou" 2. Combine the two matrices together, we have 177
178
Therefore the inverse of A is
179
Further Modular Arithmetic
In this chapter we assume the reader can find inverses and be able to solve a system of congruences (Chinese Remainder Theorem) (see: Primes and Modular Arithmetic).
Introduction
Mathematics is the queen of the sciences and number theory is the queen of mathematics. -Carl Friedrich Gauss 1777 - 1855 In the Primes and Modular Arithmetic section, we discussed the elementary properties of a prime and its connection to modular arithmetic. Our attention has been, for the most part, restricted to arithmetic mod p, where p is prime. But this need not be. In this chapter, we will start by discussing some more elementary results in arithmetic modulo p, and then we will move on to discuss those results modulo m where m is composite. In particular, we will take a closer look at the Chinese Remainder Theorem, and how it allows us to break arithmetic modulo m into components. From that point of view, the CRT is an extremely powerful tool that can help us unlock the many secrets of modulo arithmetic (with relative ease). Lastly, we will introduce the idea of an abelian group through multiplication in modular arithmetic and discuss the discrete log problem which underpins one of the most important cryptographic systems known today.
Wilson's Theorem
Wilson's theorem is a simple result that leads to a number of interesting observations in elementary number theory. It states that, if p is prime then
We know the inverse of p - 1 is p - 1, so each other number can be paired up by its inverse and eliminated. For example, let p = 7, we consider 1 × 2 × .. × 6 ≡ (2 × 4) × (3 × 5) × 1 × 6 = 6 What we have done is that we paired up numbers that are inverses of each other, then we are left with numbers whose inverse is itself. It this case, they are 1 and 6. But there is a technical difficulty. For a general prime number, p, how do we know that 1 and p - 1 are the only numbers in mod p which when squared give 1? For m not a prime, there are more than 2 solutions to x2 ≡1 (mod m), for example, let m = 15, then x = 1, 14, 4, 11 are solutions to x2 ≡1 (mod m). We can not say Wilson's theorem is true, unless we show that there can only be (at most) two 180
solutions to x2 ≡1 (mod p) when p is prime. We shall overcome this final hurdle by a simple proof by contradiction argument. You may want to skip the following proof and come up with your own justification of why Wilson's theorem is true. Let p be a prime, and x2 ≡1 (mod p). We aim to prove that there can only be 2 solutions, namely x = 1, -1
it obvious from the above that x = 1, -1 (≡p - 1) are solutions. Suppose there is another solution, x = d, and d not equal to 1 or -1. Since p is prime, we know d - 1 must have an inverse. We substitute x with d and multiply both sides by the inverse, i.e.
but we our initial assumption was that d ≠ -1. This is a contradiction. Therefore there can only be 2 solutions to x2 ≡1 (mod p).
Fermat's little Theorem
There is a remarkable little theorem named after Fermat, the prince of amateur mathematicians. It states that if p is prime and given a ≠ 0 then
This theorem hinges on the fact that p is prime. It won't work otherwise. How so? Recall that if p is prime then a ≠ 0 has an inverse. So for any b and c we must have if and only if A simple consequence of the above is that the following numbers must all be different mod p a, 2a , 3a, 4a, ..., (p-1)a and there are p - 1 of these numbers! Therefore the above list is just the numbers 1, 2, ... p - 1 in a different order. Let's see an example, take p = 5, and a = 2: 1, 2, 3, 4 multiply each of the above by 2 in mod 5, we get 2, 4, 1, 3 181
They are just the original numbers in a different order. So for any p and using Wilson's Theorem (recall: 1 × 2 × ... × (p-1) ≡ -1), , we get
on the other hand we also get
Equating the two results, we get
which is essentially Fermat's little theorem.
Modular Arithmetic with a general m
*Chinese Remainder Theorem revisited*
This section is rather theoretical, and is aimed at justifying the arithmetic we will cover in the next section. Therefore it is not necessary to fully understand the material here, and the reader may safely choose to skip the material below. Recall the Chinese Remainder Theorem (CRT) we covered in the Modular Arithmetic section. In states that the following congruences
have a solution if and only if gcd(n1,n2) divides (b - c). This deceptively simple theorem holds the key to arithmetic modulo m (not prime)! We shall consider the case where m has only two prime factors, and then the general case shall follow. Suppose m = piqj, where p and q are distinct primes, then every natural number below m (0, 1, 2, ..., m - 1) corresponds uniquely to a system of congruence mod pi and mod qj. This is due to the fact that gcd(pi,qj) = 1, so it divides all numbers. Consider a number n, it corresponds to 182
for some xn and yn. If r ≠ n then r corresponds to
Now since r and n are different, we must have either xr ≠ xn and/or yr ≠ yn For example take xn, yn for each n (0, 1, 2 ... 11) n 0 1 2 3 4 5 6 7 8 9 10 11 n (mod 22) 0 1 2 3 0 1 2 3 0 1 2 3 n (mod 3) 0 1 2 0 1 2 0 1 2 0 1 2 , then we can construct the following table showing the
Note that as predicted each number corresponds uniquely to two different systems of congruences mod 22 and mod 3.
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Exercises 1. Consider m = 45 = 32 Ã- 5. Complete the table below and verify that any two numbers must differ in at least one place in the second and third column n 0 1 2 ... 44 ? ? n (mod 32) 0 1 2 n (mod 5) 0 1 2
2. Suppose m = piqj, n corresponds to
and r corresponds to
Is it true that
and that
Arithmetic with CRT
Exercise 2 above gave the biggest indication yet as to how the CRT can help with arithmetic modulo m. It is not essential for the reader to fully understand the above at this stage. We will proceed to describe how CRT can help with arithmetic modulo m. In simple terms, the CRT helps to break a modulo-m calculation into smaller calculations modulo prime factors of m. We 184
will see what we mean very soon. As always, let's consider a simple example first. Let and we see that m has two distinct prime factors. We should demonstrate multiplication of 51 and 13 modulo 63 in two ways. Firstly, the standard way
Alternatively, we notice that
and
We can represent the two expressions above as a two-tuple (6,2). We abuse the notation a little by writing 51 = (6,2). Similarly, we write 13 = (4,6). When we do multiplication with twotuples, we multiply component-wise, i.e. (a,b) Ã- (c,d) = (ac,bd),
Now let's solve
and
we write x = 6 + 9a, which is the first congruence equation, and then
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therefore we have a = 3 + 7b, substitute back to get
which is the same answer we got from multiplying 51 and 13 (mod 63) the standard way! Let's summarise what we did. By representing the two numbers (51 and 13) as two two-tuples and multiplying component-wise, we ended up with another two-tuple. And this two-tuple corresponds to the product of the two numbers (mod m) via the Chinese Remainder Theorem. We will do two more examples. Let ways. Firstly, the standard way , and lets multiply 66 and 40 in two
and now the second way, 40 = (0,7) and 66 = (4,0) and
For the second example, we notice that there is no need to stop at just two distinct prime factors. We let , and multiply 900 and 647 (mod 975),
For the other way, we note that 900 ≡ 0 (mod 3) ≡ 0 (mod 25) ≡ 3 (mod 13), and for 647 ≡ 2 (mod 3) ≡ 22 (mod 25) ≡ 10 (mod 13),
now if we solve the following congruences
186
then we will get x ≡ 225! Why? If anything, breaking modular arithmetic in m into smaller components seems to be quite a bit of work. Take the example of multiplications, firstly, we need to express each number as a n-tuple (n is the number of distinct prime factors of m), multiply component-wise and then solve the resultant n congruences. Surely, it must be more complicated than just multiplying the two numbers and then reduce the result modulo m. So why bother studying it at all? By breaking a number m into prime factors, we have gained insight into how the arithmetic really works. More importantly, many problems in modular m can be difficult to solve, but when broken into components it suddenly becomes quite easy, e.g. Wilson's Theorem for a general m (discussed below). Exercises 1. Show that addition can also be done component-wise. 2. Multiply component-wise 32 and 84 (mod 134). ...
Euler totient
To discuss the more general form of Wilson's Theorem and Fermat's Little Theorem in mod m (not prime), it's nice to know a simple result from the famous mathematician Euler. More specifically, we want to discuss a function, called the Euler totient function (or Euler Phi), denoted ϕ. The ϕ functions does a very simple thing. For any natural number m, ϕ(m) gives the number of n < m, such that gcd(n,m) = 1. In other words, it calculates how many numbers in mod m have an inverse. We will discuss the value of ϕ (m) for simple cases first and then derive the formula for a general m from the basic results. For example, let m = 5, then ϕ (m) = 4. As 5 is prime, all non-zero natural numbers below 5 (1,2,3 and 4) are coprimes to it. So there are 4 numbers in mod 5 that have inverses. In fact, if m is prime then ϕ (m) = m - 1. We can generalise the above to m = pr where p is prime. In this case, we try to employ a counting argument to calculate ϕ (m). Note that there are pr natural numbers below m (0, 1, 2 ... pr - 1), and so ϕ (m) = pr - (number of n < m such that gcd(n,m) ≠ 1). We did that because it is easier to count the number of n 's without an inverse mod m. An element, n, in mod m does not have an inverse if and only if it shares a common factor with 187
m. But all factors of m (not equal to 1) are a multiple of p. So how many multiples of p are there in mod m? We can list them, they are
188
where the last element can be written as (pr-1 - 1)p, and so there are Therefore we can conclude
multiples of p.
We now have all the machinery necessary to derive the formula of ϕ (m) for any m. By the Fundamental Theorem of Arithmetic, any natural number m can be uniquely expressed as the product of primes, that is
where pi for i = 1, 2 ... r are distinct primes and ki are positive integers. For example 1225275 = 3Ã-52Ã-17Ã-312. From here, the reader should try to derive the following result (the CRT may help). Euler totient function ϕ
Suppose m can be uniquely expressed as below then
Wilson's Theorem
Wilson's Theorem for a general m states that the product of all the invertible element in mod m equals -1 if m has only one prime factor, or m = 2pk for some prime p equals 1 for all other cases An invertible element of mod m is a natural number n < m such that gcd(n, m) = 1. A selfinvertible element is an element whose inverse is itself. In the proof of Wilson's Theorem for a prime p, the numbers 1 to p - 1 all have inverses. This is not true for a general m. In fact it is certain that (m - 1)! ≡0 (mod m), for this reason we instead consider the product of all invertible elements in mod m. For the case where m = p is prime we also appealed to the fact 1 and p - 1 are the only elements when squared gives 1. In fact for m = pk, 1 and m - 1 (≡-1)are the only self-invertible elements (see exercise). But for a general m, this is not true. Let's take for example m = 21. In arithmetic modulo 21 each of the following numbers has itself as an inverse 1, 20, 8, 13 so how can we say the product of all invertible elements equal to 1? We use the CRT described above. Let us consider the case where m = 2pk. By the CRT, each 189
element in mod m can be represented as a two tuple (a,b) where a can take the value 0 or 1 while b can take the value 0, 1, ..., or pk - 1. Each two tuple corresponds uniquely to a pair of congruence equations and multiplication can be performed component-wise. Using the above information, we can easily list all the self-invertible elements, because (a,b)2 ≡1 means (a2,b2) = (1,1), so a is an invertible element in mod 2 and b is an invertible element in mod pk, so a ≡1 or -1, b ≡1 or -1. But in mod 2 1 ≡-1, so a = 1. Therefore, there are two elements that are self invertible in mod m = 2pk, they are (1,1) = 1, and (1, -1) = m - 1 . So in this case, the result is the same as when m has only a single prime factor. For the case where m has more than one prime factors and m≠ 2pk. Let say m has n prime factors then m can be represented as a n-tuple. Let say m has 3 distinct prime factors, then all the self-invertible elements of m are 14. 15. 16. 17. 18. 19. 20. 21. (1,1,1) (1,1,-1) (1,-1,1) (1,-1,-1) (-1,1,1) (-1,1,-1) (-1,-1,1) (-1,-1,-1) their product is (1,1,1) which corresponds to 1 in mod m. Exercise 1. Let p be a prime. Show that in arithmetic modulo pk, 1 and pk - 1 are the only self-invertible elements. ...more to come
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Fermat's Little Theorem
As mentioned in the previous section, not every element is invertible (i.e. has an inverse) mod m. A generalised version of Fermat's Last Theorem uses Euler's Totient function, it states
for all a ≠ 0 satisfying gcd(a,m) = 1. This is easy to see from the generalised version of Wilson's Theorem. We use a similar techique from the prove of Fermat's Little Theorem. We have
where the bi's are all the invertible elements mod m. By Wilson's theorem the product of all the invertible elements equals to, say, d (= 1 or -1). So we get
whihc is essentially the statement of Fermat's Little Theorem. Although the FLT is very neat, it is imprecise in some sense. For example take m = 15 = 3 Ã- 5, we know that if a has an inverse mod 15 then aϕ(15) = a8 ≡1 (mod 15). But 8 is too large, all we need is 4, by that we mean, a4 ≡1 (mod 15) for all a with an inverse (the reader can check). The Carmichael function λ(m) is the smallest number such that aλ(m) ≡1 (mod m). A question in the Problem Set deals with this function. Exercises ...more to come
Two-torsion Factorisation
It it quite clear that factorising a large number can be extremely difficult. For example, given that 76372591715434667 is the product of two primes, can the reader factorise it? Without the help of a good computer algebra software, the task is close to being impossible. As of today, there is no known efficient all purpose algorithm for factorising a number into prime factors. However, under certain special circumstances, factorising can be easy. We shall consider the two-torsion factorisation method. A 2-torsion element in modular m arithmetic is a number a such that a2 ≡1 (mod m). Let's consider an example in arithmetic modulo 21. Note that using the CRT we can represent any number in mod 21 as a two-tuple. We note that the two-torsion elements are 1 = (1,1), 13 = (1,-1), 8 = (-1,1) and 20 = (-1,-1). Of interest are the numbers 13 and 8, because 13 + 1 = (1,-1) 191
+ (1,1) = (2,0). Therefore 13 + 1 = 14 is an element sharing a common factor with 21, as the second component in the two-tuple representation of 14 is zero. Therefore GCD(14,21) = 7 is a factor of 21. The above example is very silly because anyone can factorise 21. But what about 24131? Factorising it is not so easy. But, if we are given that 12271 is a non-trivial (i.e. ≠ 1 or -1) twotorsion element, then we can conclude that both gcd(12271 + 1,24131) and gcd(12271 1,24131) are factors of 24131. Indeed gcd(12272,24131) = 59 and gcd(12270,24131) = 409 are both factors of 24131. More generally, let m be a composite, and t be a non-trivial two-torsion element mod m i.e. t ≠ 1, -1. Then gcd(t + 1,m) divides m, and gdc(t - 1,m) divides m this can be explained using the CRT. We shall explain the case where m = pq and p and q are primes. Given t is a non-trivial twotorsion element, then t has representaion (1,-1) or (-1,1). Suppose t = (-1,1) then t + 1 = (-1,1) + (1,1) = (0,2), therefore t + 1 must be a multiple of p therefore gcd(t,m) = p. In the other case where t - 1 = (-1,1) - (1,1) = (-2,0) and so gcd(t - 1,m) = q. So if we are given a non-trivial two-torsion element then we have effectively found one (and possibly more) prime factors, which goes a long way in factorising the number. In most modern public key cryptography applications, to break the system we need only to factorise a number with two prime factors. In that regard two-torsion factorisation method is frightening effectively. Of course, finding a non-trivial two-torsion element is not an easy task either. So internet banking is still safe for the moment. By the way 76372591715434667 = 224364191 Ã340395637.
Exercises
1. Given that 18815 is a two-torsion element mod 26176. Factorise 26176. ...more to come'
192
Mathematical programming
Before we begin This chapter will not attempt to teach you how to program rigorously. Therefore a basic working knowledge of the C programming language is highly recommended. It is recommended that you learn as much about the C programming language as possible before learning the materials in this chapter. Please read the first 7 lessons of "C Programming Tutorial" by About.com if you are unfamiliar with programming or the C programming language.
Introduction to programming
Programming has many uses. Some areas where programming and computer science in general are extremely important include artificial intelligence and statistics. Programming allows you to use computers flexibly and process data very quickly. When a program is written, it is written into a textual form that a human can understand. However, a computer doesn't directly understand what a human writes. It needs to be transformed into a way that the computer can directly understand. For example, a computer is like a person who reads and speaks German. You write and speak in English. The letter you write to the computer needs to be translated for the computer to speak. The program responsible for this work is referred to as the compiler. You need to compile your English-like instructions, so that the computer can understand it. Once a program has been compiled, it is hard to "un-compile" it, or transform it back into English again. A programmer writes the program (to use our analogy, in English), called source code, which is a human-readable definition of the program, and then the compiler translates this into "machine code". We recommend using the widely available gcc compiler. When we look at mathematical programming here, we will look at how we can write programs that will solve some difficult mathematical problems that would take us normally a lot of time to solve. For example, if we wanted to find an approximation to the root of the polynomial x5+x+1 - this is very difficult for a human to solve. However a computer can do this no sweat -how?
Programming language basics
We will be using the C programming language throughout the chapter, please learn about the basics of C by reading the first seven chapters of "C programming Tutorial" at About.com 193
Discrete Programming
Discrete programming deals with integers and how they are manipulated using the computer.
Understanding integral division
In C, the command int number; number = 3 / 2; will set aside some space in the computer memory, and we can refer to that space by the variable name number. In the computer's mind, number is an integer, nothing else. After number = 3 / 2; numbers equals 1, not 1.5, this is due to that fact that / when applied to two integers will give only the integral part of the result. For example: 5 / 2 equals 2 353 / 3 equals 117 99 / 7 equals 14 -19 / 2 equals -9 78 / -3 equals -29 in C. Exercises Evaluate x a) x = 7 / 2 b) x = -9 / -4 c) x = 1000 / 999 d) x = 2500 / 2501
Modelling Recursively defined functions
The factorial function n! is recursively defined: 0! = 1 n! = nÃ-(n-1)! if n ≠¥ 1
194
In C, if fact(n) is the functions as described above we want fact(0) = 1; fact(n) = n * fact(n − 1); if we should note that all recursively defined functions have a terminating condition, it is the case where the function can give a direct answer to, e.g. fact(0) = 1. We can model the factorial functions easily with the following code and then execute it: int fact (int n) { if (n == 0) return 1; if (n >= 1) return n * fact(n - 1); } The C function above models the factorial function very naturally. To test the results, we can compile the following code: #include <stdio.h> /* STanDard Input & Output Header file */ int fact (int n) { if (n == 0) return 1; if (n >= 1) return n * fact(n - 1); } void main() { int n = 5; printf("%d", fact(n)); /* printf is defined in stdio.h */ 195
} We can also model the Fibonacci number function. Let fib(n) return the (n + 1)th Fibonacci number, the following should be clear fib(0) should return 1 fib(1) should return 1 fib(n) should return fib(n - 1) + fib(n - 2); for n ≥ 2 we can model the above using C: int fib (int n) { if (n == 0 || n == 1) /* if n = 0 or if n = 1 */ return 1; if (n >= 2) return fib(n - 1) + fib(n - 2); } Again, you shall see that modelling a recursive function is not hard, as it only involves translating the mathematics in C code.
Modelling non-recursive functions
There are functions that involve only integers, and they can be modelled quite nicely using functions in C. The factorial function f(n) = n! = n(n - 1)(n - 2) ... 3Ã-2Ã-1 can be modelled quite simply using the following code
int n = 10; //get factorial for 10 int f = 1; //start f at 1 while(n > 0) //keep looping if n bigger then 0 { f = n * f; //f is now product of f and n n = n - 1; //n is one less (repeat loop) } 196
197
Basic counting
Counting
All supplementary chapters contain materials that are part of the standard high school mathematics curriculum, therefore the material is only provided for completeness and should mostly serve as revision.
Ordered Selection
Suppose there are 20 songs in your mp3 collection. The computer is asked to randomly selected 10 songs and play them in the order they are selected, how many ways are there to select the 10 songs? This type of problems is called ordered selection counting, as the order in which the things are selected is important. E.g. if one selection is 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 then 2, 1, 3, 4, 5, 6, 7, 8, 9 and 10 is considered a different selection. There are 20 ways to choose the first song since there are 20 songs, then there are 19 ways to choose the second song and 18 ways to choose the third song ... and so on. Therefore the total number of ways can be calculated by considering the following product: 20 Ã- 19 Ã- 18 Ã- 17 Ã- 16 Ã- 15 Ã- 14 Ã- 13 Ã- 12 Ã- 11 or denoted more compactly:
Here we use the factorial function, defined by 0! = 1 and words, ) In general, the number of ordered selections of m items out of n items is:
. (In other
The idea is that we cancel off all but the first m factors of the n! product.
198
Unordered Selection
Out of the 15 people in your mathematics class, five will be chosen to represent the class in a school wide mathematics competition. How many ways are there to choose the five students? This problem is called an unordered selection problem, i.e. the order in which you select the students is not important. E.g. if one selection is Joe, Lee, Sue, Britney, Justin another selection is Lee, Joe, Sue, Justin, Britney the two selections are considered equivalent. There are
ways to choose the 5 candidates in ordered selection, but there are 5! permutations of the same five candidates. (That is, 5! different permutations are actually the same combination). Therefore there are
ways of choosing 5 students to represent your class. In general, to choose (unordered selection) m candidates from n, there are
ways. We took the formula for ordered selections of m candidates from n, and then divided by m! because each unordered selection was counted as m! ordered selections.
Note:
is read "n choose m".
Examples
Example 1 How many different ways can the letters of the word BOOK be arranged? Solution 1 4! ways if the letters are all distinct. Since O is repeated twice, there are 2! permutations. Therefore there are 4!/2! = 12 ways. 199
Example 2 How many ways are there to choose 5 diamonds from a deck of cards?
Solution 2 There are 13 diamonds in the deck. So there are
ways.
Binomial expansion
The binomial expansion deals with the expansion of following expression (a + b)n Take n = 3 for example, we shall try to expand the expression manually we get (a + b)3 = (aa + ab + ba + bb)(a + b) = aaa + aab + aba + abb + baa + bab + bba + bbb = aaa + aab + aba + abb + baa + bab + bba + bbb We deliberately did not simplify the expression at any point during the expansion, we didn't even use the well known (a + b)2 = a2 + 2ab + b2. As you can see, the final expanded form has 8 terms. They are all the possible terms of powers of a and b with three factors! Since there are 3 factors in each term and all the possibles terms are in the expanded
, i.e. from expressoin. How many terms are there with only one b? The answer should be 3 possible positions, choose 1 for b. Similarly we can work out all the coefficient of like-terms. So (a + b)3 And more generally (a + b)n or more compactly using the summation sign (otherwise known as sigma notation)
200
(a + b)n
Partial fractions
Method of Partial Fractions
All supplementary chapters contain materials that are part of the standard high school mathematics curriculum, therefore the material is only provided for completeness and should mostly serve as revision.
Introduction
Before we begin, consider the following: How do we calculate this sum? At first glance it may seem difficult, but if you think carefully you will find: Thus the original problem can be rewritten as follows,
So all terms except the first and the last cancelled out, and therefore
In fact, you've just done partial fractions! Partial fractions is a method of breaking down complex fractions that involve products into sums of simpler fractions.
Method
So, how do we do partial fractions? Look at the example below:
Factorize
the 201
denominator.
202
Then we suppose we can break it down into the fractions with denominator (z-1) and (z-2) respectively. We let their numerators be a and b.
Therefore by matching coefficients of like power of z, we have:
(2)-(1):a=1 Substitute a=1 into (1):b=3 Therefore
(Need Exercises!)
203
More on partial fraction
Repeated factors
On the last section we have talked about factorizing the denominator, and have each factor as the denominators of each term. But what happens when there are repeating factors? Can we apply the same method? See the example below:
Indeed, a factor is missing! Can we multiply both the denominator and the numerator by that factor? No! Because the numerator is of degree 1, multiplying with a linear factor will make it become degree 2! (You may think:can't we set A+B+C=0? Yes, but by substituting A+B=-C, you will find out that this is impossible) From the above failed example, we see that the old method of partial fraction seems not to be working. You may ask, can we actually break it down? Yes, but before we finally attack this problem, let's look at the denominators at more detail. Consider the following example:
We can see that the power of a prime factor in the product denominator is the maximum power of that prime factor in all term's denominator. Similarly, let there be factor P1,P2,...,Pn, then we may have in general case: 204
If
we
turn
it
into
one
big
fraction,
the
denominator
will
be:
Back to our example, since the factor (x+2) has a power of 2, at least one of the term has (x + 2)2 as the denominator's factor. You may then try as follows:
But again, we can't set B=0, since that would means the latter term is 0! What is missing? To handle it properly, let's use a table to show all possible combinations of the denominator: Possible combinations of denominator Power of (x+2) 0 1 2 0 1 Power of (x-1) 0 0 0 1 1 205 Result 1 (x+2) (x+2)^2 (x-1) (x+2)(x-1) Used? Not useful Not used Used Used Not useful
2
1
(x+2)^2(x-1)
Not useful
206
So, we now know that X/(x+2) is missing, we can finally happily get the answer:
Therefore by matching coefficient of like power of x, we have As a conclusion, for a repeated factor of power n, we will have n terms with their denominator being X^n, X^(n-1), ...,X^2, X Works continuing, don't distrub :)
Alternate method for repeated factors
Other than the method suggested above, we would like to use another approach to handle the problem. We first leave out some factor to make it into non-repeated form, do partial fraction on it, then multiply the factor back, then apply partial fraction on the 2 fractions.
207
Then we do partial fraction on the latter part:
By matching coefficients of like powers of x, we have
Substitute A=4-B into (2), 2B − (4 − B) = − 1 Hence B = 1 and A = 3. We carry on:
208
Now we do partial fraction once more:
By matching coefficients of like powers of x , we have:
Substitute A=-B into (2), we have: 2B-(-B) = 1 Hence B=1/3 and A=-1/3 So finally,
209
Summation sign
Summation Notation
All supplementary chapters contain materials that are part of the standard high school mathematics curriculum, therefore the material is only provided for completeness and should mostly serve as revision. We normally use the "+" sign to represent a sum, but if the sum expression involved is complex and long, it can be confusing.
For example: Writing the above would be a tedious and messy task! To represent expression of this kind more compactly and nicely, people use the summation notation, a capital greek letter "Sigma". On the right of the sigma sign people write the expression of each term to sum, and write the upper and lower limit of the variable on top and under the sigma sign.
Example
1:
Misconception: From the above there is a common misconception that the number on top of the Sigma sign is the number of terms. This is wrong. The number on top is the number to subsitute back in the last term.
Example
2:
Tip:If the terms alternate between plus and minus, we can use the sequence
210
Exercise
14. Use the summation notation to represent the expression in the first example. Change the following into sum notation: 6. 7. 8. 9. * terms, or get more than one term in the expression) * (Hint:You need to use more than one sigma sign) Change the following sum notation into the normal representation: (Hint:reorder the
11.
12. (Need more exercise,especially "reading" sigma notation and change back into the old form)
Operations of sum notation
Although most rules related to sum makes sense in the ordinary system, in this new system of sum notation, things may not be as clear as before and therefore people summarize some rules related to sum notation (see if you can identify what they correspond to!)
•
•
•
211
• (Note:I suggest getting a visual aid on this one:showing that you can sum a two dimensional array in either direction)
•
(Index substutition)
•
(Decomposition)
•
(Factorization/Expansion)
Exercise
(put up something here please)
Beyond
"To iterate is human; to recurse, divine." When human repeated summing, they have decided to use a more advanced concept, the concept of product. And of course everyone knows we use . And when we repeat product, we use exponential. Back to topic, we now have a notation for complex sum. What about complex product? In fact, there is a notation for product also. We use the capital greek letter "pi" to denote product, and basically everything else is exactly the same as sum notation, except that the terms are not summed, but multiplied.
Example:
=
Exercise
1.It has been known that the factorial is defined inductively by:
Now try (more to go...)
to
define 212
it
by
product
notation.
Complex numbers
Introduction
All supplementary chapters contain materials that are part of the standard high school mathematics curriculum, therefore the material is only provided for completeness and should mostly serve as revision. Although the real numbers can, in some sense, represent any natural quantity, they are in another sense incomplete. We can write certain types of equations with real number coefficients which we desire to solve, but which have no real number solutions. The simplest example of this is the equation:
Your high school math teacher may have told you that there is no solution to the above equation. He/she may have even emphasised that there is no real solution. But we can, in fact, extend our system of numbers to include the complex numbers by declaring the solution to that equation to exist, and giving it a name: the imaginary unit, i. Let's imagine for this chapter that question, and i2 = - 1. exists. Hence x = i is a solution to the above
A valid question that one may ask is "Why?". Why is it important that we be able to solve these quadratics with this seemingly artificial construction? It is interesting delve a little further into the reason why this imaginary number was introduced in the first place - it turns out that there was a valid reason why mathematicians realized that such a construct was useful, and could provide deeper insight. The answer to the question lies not in the solution of quadratics, but rather in the solution of the intersection of a cubic and a line. The mathematician Cardano managed to come up with an ingenious method of solving cubics - much like the quadratic formula, there is also a formula that gives us the roots of cubic equations, although it is far more complicated. Essentially, we can express the solution of a cubic x3 = 3px + 2q in the form
An unsightly expression, indeed! You should be able to convince yourself that the line y = 3px + 2q must always hit the cubic y = x3. But try solving some equation where q2 < p3, and you run into a problem - the problem is 213
that we are forced to deal with the square root of a negative number. But, we know that in fact there is a solution for x; for example, x3 = 15x + 4 has the solution x = 4. It became apparent to the mathematician Bombelli that there was some piece of the puzzle that was missing - something that explained how this seemingly perverse operation of taking a square root of a negative number would somehow simplify to a simple answer like 4. This was in fact the motivation for considering imaginary numbers, and opened up a fascinating area of mathematics. The topic of Complex numbers is very much concerned with this number i. Since this number doesn't exist in this real world, and only lives in our imagination, we call it the imaginary unit. (Note that i is not typically chosen as a variable name for this reason.)
The imaginary unit
As mentioned above . Let's compute a few more powers of i:
As you may see, there is a pattern to be found in this.
Exercises
12. 13. 14. Compute i25 Compute i100 Compute i1000 Exercise Solutions
214
Complex numbers as solutions to quadratic equations
Consider the quadratic equation:
The x we get as a solution is what we call a complex number. (To be nitpicky, the solution set of this equation actually has two complex numbers in it; either is a valid value for x.) It consists of two parts: a real part of 3 and an imaginary part of . Let's call the real part a and the imaginary part b; then the sum is a complex number. Notice that by merely defining the square root of negative one, we have already given ourselves the ability to assign a value to a much more complicated, and previously unsolvable, quadratic equation. It turns out that 'any' polynomial equation of degree n has exactly n zeroes if we allow complex numbers; this is called the Fundamental Theorem of Algebra. We denote the real part by Re. E.g.: Re(x) = 3 and the imaginary part by Im. E.g.:
Let's check to see whether x = 3 + 2i really is solution to the equation:
Exercises
8. Convince yourself that x = 3 - 2i is also a solution to the equation. 9. Plot the points A(3, 2) and B(3, -2) on a XY plane. Draw a line for each point joining them to the origin. 215
10. Compute the length of AO (the distance from point A to the Origin) and BO. Denote them by r_1 and r2 respectively. What do you observe? 11. Compute the angle between each line and the x-axis and denote them by ϕ1 and ϕ2. What do you observe? 12. Consider the complex numbers:
Substitute z and w into the quadratic equation above using the values you have computed in Exercise 3 and 4. What do you observe? What conclusion can you draw from this?
Arithmetic with complex numbers
Addition and multiplication
Adding and multiplying two complex number together turns out to be quite straightforward. Let's illustrate with a few examples. Let x = 3 - 2i and y = 7 + 11i, and we do addition first
and now multiplication
Let's summarise the results here. • • When adding complex numbers we add the real parts with real parts, and add the imaginary parts with imaginary parts. When multiplying two complex numbers together, we use normal expansion. Whenever we see i2 we put in its place -1. We then collect like terms. But how do we calculate:
Note that the square root is only above the 5 and not the i. This is a little bit tricky, and we shall 216
cover it in the next section.
Exercises:
Compute: 6. x + y 7. x - y 8. x2 9. y2 10. xy 11. (x + y)(x - y)
Division
One way to calculate:
is to rationalise the denominator:
Utilising a similar idea, to calculate
we realise the denominator.
217
The denominator is the sum of two squares. We get:
If somehow we can always find a complex number whose product with the denominator is a real number, then it's easy to do divisions. If
and
Then zw is a real number. This is true for any 'a' and 'b' (provided they are real numbers).
Exercises
Convince yourself that the product of zw is always a real number.
Complex Conjugate
The exercise above leads to the idea of a complext conjugate. The complex congugate of a + ib is a - ib. For example, the conjugate of 2 + 3i is 2 - 3i. It is a simple fact that the product of a complex number and its conjugate is always a real number. If z is a complex number then its conjugate is denoted by . Symbolically if z = a + ib then,
The conjugate of 3 - 9i is 3 + 9i. The conjugate of 100 is 100. The conjugate of 9i - 20 is -20 - 9i. Conjugate laws Here are a few simple rules regarding the complex congugate
218
and
The above laws simply says that the sum of congugates equals the congugate of the sum; and similarly, the congugate of the product equals the product of the congugates. Consider this example: (3 + 2i) + (89 − 100i) = 92 − 98i and we can see that
which equals to
This confirms the addition conjugate law.
Exercise
Convince yourself that the multiplication law is also true.
The complex root
Now that you are equipped with all the basics of complex numbers, you can tackle the more advanced topic of root finding. Consider the question:
Express w in the form of a + ib. That is easy enough.
219
Solve (1) and (2) simultaneously to work out a and b. Observe that if, after solving for a and b, we replace them with -a and -b respectively, then they would still satisfy the two simultaneous equations above, we can see that (as expected) if w = a + ib satisfies the equation w2 = z, then so will w = -(a + ib). With real numbers, we always take the non-negative answer and call the solution which square root to take as "the" value of is very important to some calculations. . However, since there is no notion of . In fact, depends on the circumstances, and this choice "greater than" or "less than" with complex numbers, there is no such choice of
info -- Finding the square root
Finding the root of a real number is a very difficult problem to start with. Most people have no hope of finding a close estimate of without the help of a calculator. The modern method of approximating roots involves an easy to understand and ingenius piece of mathematics called the Taylor series expansion. The topic is usually covered in first year university maths as it requires an elementary understanding of an important branch of mathematics called calculus. The Newton-Raphson method of root finding is also used extensively for this purpose. Now consider the problem
Express w in the form of "a + ib". Using the methodology developed above we proceed as follows,
220
It turns out that the simultaneous equations (1) & (2) are hard to solve. Actually, there is an easy way to calculate the roots of complex numbers called the De Moivre's theorem, it allows us to calculate the nth root of any complex number with ease. But to set the method, we need understand the geometric meaning of a complex number and learn a new way to represent a complex number.
Exercises
9. Find (3 + 3i)1/2 10. Find (1 + 1i)1/2 11. Find i1/3
The complex plane
Complex numbers as ordered pairs
It is worth noting, at this point, that every complex number, a + bi, can be completely specified with exactly two real numbers: the real part a, and the imaginary part b. This is true of every complex number; for example, the number 5 has real part 5 and imaginary part 0, while the number 7i has real part 0 and imaginary part 7. We can take advantage of this to adopt an alternative scheme for writing complex numbers: we can write them as ordered pairs, in the form (a, b) instead of a+bi.
221
These should look familiar: they are exactly like the ordered pairs we use to represent poins in the plane. In fact, we can use them that way; the plane which results is called the complex plane. We refer to its x axis as the real axis, and to its y axis as the imaginary axis.
222
The complex plane
We can see from the above that a single complex number is a point in the complex plane. We can also represent sets of complex numbers; these will form regions on the plane. For example, the set
is a square of edge length 2 centered at the origin (See following diagram).
Complex-valued functions
Just as we can make functions which take real values and output real values, so we can create functions from complex numbers to real numbers, or from complex numbers to complex numbers. These latter functions are often referred to as complex-valued functions, because they evaluate to (output) a complex number; it is implicit that their argument (input) is complex as well. Since complex-valued functions map complex numbers to other complex numbers, and we have already seen that complex numbers correspond to points on the complex plane, we can see that a complex-valued function can turn regions on the complex plane into other regions. A simple example: the function f(z) = z + (0 + 1i) takes a point in the complex plane and shifts it up by 1. If we apply it to the set of points making up the square above, it will move the entire square up one, so that it "rests" on the xaxis. {To make more complicated examples, I will first have to go back and introduce the polar representation of complex numbers. Makes for much more interesting functions, :-) You can use the diagrams below or modify them to make new diagrams. I will make links to these diagrams in other places in Wikibooks:math. In the 2nd diagram showing the point, r=4 and 223
theta= 50 degrees. These types of diagrams can be used to introduce phasors, which are notations for complex numbers used in electrical engineering.}
de Moivre's Theorem
If z = reiθ then zn = rn(cos(nθ) + isin(nθ)) = r(cos(θ) + isin(θ))
224
Complex root of unity
The complex roots of unity to the nth degree is the set of solutions to the equation x^n = 1. Clearly they all have magnitude 1. They form a cyclic group under multiplication. For any given n, there are exactly n many of them, and they form a regular n-gon in the complex plane over the unit circle. A closed form solution can be given for them, by use of Euler's formula: u^n = {cos(2*pi*j/n)+i*sin(2*pi*j/n) | 0 <= j < n} The sum of the nth roots of unity is equal to 0, except for n=1, where it is equal to 1. The product of the nth roots of unity alternates between -1 and 1.
225
Problem set
Simplify: (1-i)2i Ans: 2ie-π/2 The imaginary unit 22. 23. 24. Compute i25 = i Compute i100 = 1 Compute i1000 = 1 The pattern of i1,i2,i3,... shows that i4n = 1 where n is any integer. This case applies to questions 2 and 3. For question 1, . Complex numbers as solutions to quadratic equations 15. Convince yourself that x = 3 - 2i is also a solution to the equation.
10. 11. • • • 12.
Plot the points A(3, 2) and B(3, -2) on a XY plane. Draw a line for each point joining them to the origin. Compute the length of AO (the distance from point A to the Origin) and BO. Denote them by respectively. What do you observe?
are the same Compute the angle between each line and the x-axis and denote them by . What do you
226
observe?
•
• • differ only in the sign of the number
13.
Consider the complex numbers:
Substitute z and w into the quadratic equation above using the values you have computed in Exercise 3 and 4. What do you observe? What conclusion can you draw from this?
Thus the quadratic equation will equal 0 since z and w are equal to the solutions we found when solving the equation.
Addition and multiplication
Compute: 13. • 14. x+y (3 - 2i) + (3 + 2i) = 6 x-y
227
• 15. • 16. • 17. • 18. •
3 - 2i - (3 + 2i) = -4i x2 (3 - 2i)(3 - 2i) = 9 + (2)(3)(-2i) + 4i2 = 5 - 12 i y2 (3 + 2i)(3 + 2i) = 9 + (2)(3)(2i) + 4i2 = 5 + 12 i xy (3 - 2i)(3 + 2i) = 9 + 6 i - 6i + 4i2 = 5 (x + y)(x - y) ((3 - 2i) + (3 + 2i))((3 - 2i) - (3 + 2i)) = (6)(-4i) = -24i Division Convince yourself that the product of zw is always a real number. zw = (a + bi)(a - bi) = a2 - abi + abi - b2i2 = a2 - b Complex Conjugate Convince yourself that the multiplication law is also true.
228
The complex root 1. Find (3 + 3i)1/2
Thus the solution for (3 + 3i)1/2 is: 229
2. Find (1 + 1i)1/2
230
Thus the solution for (1 + 1i)1/2 is:
3. Find i1/3
Thus the solution for i1/3 is:
231
Bases 15. 1. 2. 3. 4. 16. 1. 2. 3. 4. 17. The following numbers are written in base 2. Write them out in base 10: 101011 (base 2) = 43 (base 10) = 25 + 23 + 21 + 20= 32+8+2+1 001101 = 13 10 = 2 011 = 3 Write those numbers out in base 10 as if they were originally in base 5. 101011 (base 5) = 3256 (base 10) = 55 + 53 + 51 + 50 = 3125+125+5+1 001101 = 751 10 = 25 011 = 26 How many numbers could I write out in base 5 with only the first 4 columns? Answer: 625 = 54 (each new column multiplies the number of possibilities by 5) 13. In computing, each 1 or 0 is called a bit. They are stored in groups of 8. Each group is called a byte. How many bytes are possible? Answer: 256 = 28 (values from 00000000 through 11111111 binary, or 0 through 255 base 10) 12. Question: When editing the bytes directly, writing out 10110001 is too long and hexadecimal is used instead (in this case, B1). How many digits of hexadecimal are needed to cover all possible bytes? Answer: 2 digits (162 = number of possible 2-digit hexadecimal numbers = 256 = number of possible bytes)
232
Differentiation
Differentiate from first principle
All supplementary chapters contain materials that are part of the standard high school mathematics curriculum, therefore the material is only provided for completeness and should mostly serve as revision. This section and the *differentiation technique* section can be skipped if you are already familiar with calculus/differentiation. In calculus, differentiation is a very important operation applied to functions of real numbers. To differentiate a function f(x), we simply evaluate the limit
where the means that we let h approach 0. However, for now, we can simply think of it as putting h to 0, i.e., letting h = 0 at an appropriate time. There are various notations for the result of differentiation (called the derivative), for example
and
mean the same thing. We say, f'(x) is the derivative of f(x). Differentiation is useful for many purposes, but we shall not discuss why calculus was invented, but rather how we can apply calculus to the study of generating functions. It should be clear that if g(x) = f(x) then g'(x) = f'(x) the above law is important. If g(x) a closed-form of f(x), then it is valid to differentiate both sides to obtain a new generating function. Also if h(x) = g(x) + f(x) 233
then h'(x) = g'(x) + f'(x) This can be verified by looking at the properties of limits. Example 1 Differentiate from first principle f(x) where f(x) = x2 Firstly, we form the difference quotient
We can't set h to 0 to evaluate the limit at this point. Can you see why? We need to expand the quadratic first.
We can now factor out the h to obtain now
from where we can let h go to zero safely to obtain the derivative, 2x. So f'(x) = 2x or equivalently: (x2)' = 2x Example 2 Differentiate from first principles, p(x) = xn. We start from the difference quotient:
By the binomial theorem, we have: 234
The first xn cancels with the last, to get
Now, we bring the constant 1/h inside the brackets
and the result falls out: = nxn - 1 Important Result If p(x) = xn then p'(x) = nxn - 1 As you can see, differentiate from first principle involves working out the derivative of a function through algebraic manipulation, and for that reason this section is algebraically very difficult. Example 3 Assume that if h(x) = f(x) + g(x) then h'(x) = f'(x) + g'(x) Differentiate x2 + x5 Solution Let h(x) = x2 + x5 h'(x) = 2x + 5x4 Example 4 Show that if g(x) = AÃ-f(x) then 235
g'(x) = AÃ-f'(x) Solution
Example 5 Differentiate from first principle
Solution
236
Exercises 1. Differentiate f(z) = z3 2. Differentiate f(z) = (1 - z)2 3. Differentiate from first principle
4. Differentiate f(z) = (1 - z)3 5. Prove the result assumed in example 3 above, i.e. if f(x)=g(x)+h(x) then f′(x)=g′(x)+h′(x).
Hint: use limits.
Differentiating f(z) = (1 - z)^n
We aim to derive a vital result in this section, namely, to derive the derivative of f(z) = (1 - z)n where n ≥ 1 and n an integer. We will show a number of ways to arrive at the result. Derivation 1 Let's proceed: f(z) = (1 - z)n expand the right hand side using binomial expansion
237
differentiate both sides
now we use
and there are some cancelling
take out a common factor of -n, and recall that 1! = 0! = 1 we get
let j = i - 1, we get
but this is just the expansion of (1 - z)n-1 f'(z) = - n(1 - z)n - 1 Derivation 2 Similar to Derivation 1, we use instead the definition of a derivative:
expand using the binomial theorem
238
factorise
take the limit inside (recall that [Af(x)]' = Af'(x) )
the inside is just the derivative of zi
exactly as derivation 1, we get f'(z) = - n(1 - z)n - 1 Example Differentiate (1 - z)2 Solution 1 f(z) = (1 - z)2 = 1 - 2z + z2 f'(z) = - 2 + 2z f'(z) = - 2(1 - z) Solution 2 By the result derived above we have f'(z) = -2(1 - z)2 - 1 = -2(1 - z) Exercises Imitate the method used above or otherwise, differentiate: 1. (1 - z)3 2. (1 + z)2 3. (1 + z)3 4. (Harder) 1/(1 - z)3 (Hint: Use definition of derivative)
239
Differentiation technique
We will teach how to differentiate functions of this form:
i.e. functions whose reciprocals are also functions. We proceed, by the definition of differentiation:
Example 1
by
where g is a function of z, we get
which confirmed the result derived using a counting argument. 240
Exercises Differentiate 1. 1/(1-z)2 2. 1/(1-z)3 3. 1/(1+z)3 4. Show that (1/(1 - z)n)' = n/(1-z)n+1
Differentiation applied to generating functions
Now that we are familiar with differentiation from first principle, we should consider:
we know
differentiate both sides
therefore we can conclude that
Note that we can obtain the above result by the substituion method as well,
letting z = x2 gives you the require result.
241
The above example demonstrated that we need not concern ourselves with difficult differentiations. Rather, to get the results the easy way, we need only to differentiate the basic forms and apply the substitution method. By basic forms we mean generating functions of the form:
for n ≥ 1. Let's consider the number of solutions to a1 + a2 + a3 + ... + an = m for ai ≥ 0 for i = 1, 2, ... n. We know that for any m, the number of solutions is the coefficient to:
as discussed before. We start from:
differentiate both sides (note that 1 = 1!)
differentiate again
and so on for (n-1) times
divide both sides by (n-1)!
242
the above confirms the result derived using a counting argument.
Differentiate from first principle 1. . f'(z) = 3z2 (We know that if p(x) = xn then p'(x) = nxn − 1) 2. f(z) = (1 − z)2 = z2 − 2z + 1 f'(z) = 2z − 2 3.
4. f(z) = (1 − z)3 = − z3 + 3z2 − 3z + 1 f'(z) = − 3z2 + 6z − 3
243
5. if f(x)=g(x)+h(x) then
Differentiating f(z) = (1 - z)^n 1. f(z) = (1 − z)3 = − z3 + 3z2 − 3z + 1
2. f(z) = (1 + z)2 = z2 + 2z + 1
3. f(z) = (1 + z)3 = z3 + 3z2 + 3z + 1
244
4.
Differentiation technique 1.
We use the result of the differentation of f(z)=(1-z)^n (f'(z) = -n(1-z)n-1) 2.
245
3.
We use the result of exercise 3 of the previous section f(z)= (1+z)3 -> f'(z)=3(1+z)^2 4.
We use the result of the differentation of f(z)=(1-z)^n (f'(z) = -n(1-z)n-1)
246
License GNU Free Documentation License
Version 1.2, November 2002 Copyright (C) 2000,2001,2002 Free Software Foundation, Inc. 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. 0. PREAMBLE The 1. APPLICABILITY AND DEFINITIONS This License applies to any manual or other work, in any medium, that contains a notice placed by the copyright holder saying it can be distributed under the terms of this License. Such a notice grants a world-wide- Examples
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7. AGGREGATION WITH INDEPENDENT WORKS A 8. TRANSLATION Translation 9. TERMINATION You may not copy, modify, sublicense, or distribute the Document except as expressly provided for under this License. Any other attempt to copy, modify, sublicense or distribute the Document is void, and will automatically terminate your rights under this License. However, parties who have received copies, or rights, from you under this License will not have their licenses terminated so long as such parties remain in full compliance. 10. FUTURE REVISIONS OF THIS LICENSE The External links • • GNU Free Documentation License (Wikipedia article on the license) Official GNU FDL webpage
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A Mathematical Introduction to Robotic Manipulation presents a mathematical formulation of the kinematics, dynamics, and control of robot manipulators. It uses an elegant set of mathematical tools that emphasizes the geometry of robot motion and allows a large class of robotic manipulation problems to be analyzed within a unified framework. The foundation of the book is a derivation of robot kinematics using the product of the exponentials formula. The authors explore the kinematics of open-chain manipulators and multifingered robot hands, present an analysis of the dynamics and control of robot systems, discuss the specification and control of internal forces and internal motions, and address the implications of the nonholonomic nature of rolling contact are addressed, as well. The wealth of information, numerous examples, and exercises make A Mathematical Introduction to Robotic Manipulation valuable as both a reference for robotics researchers and a text for students in advanced robotics courses.
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is the second in a series of books that helps children learn how to read by helping them achieve fluency with the Dolch sight words. The ability to read is one of the most important skills that a child can have, because it opens the doors of knowledge. The purpose of this book is to make reading fun for the child, so that he can become fluent in reading the sight words. This book uses attractive pictures as well as encouraging words along the way to keep the child engaged. This book also contains fun and entertaining sentences that contain only the sight words that the child has learned Is Fun!: A Dolch Sight Words Book - Level 2 - Primer to your Bookmark Collection or Course ePortfolio
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This completely self-contained text proceeds from the fact that mathematics derives from the real world. For instance,...
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This completely self-contained text proceeds from the fact that mathematics derives from the real world. For instance, logical consequence is nothing but a reflection of real-world causality and statements are true or false, not because some author says so, but because the real world makes them necessarily so. Particular attention is given to the language needed to discuss and understand matters. The text is part of a package including homeworks, reviews, exams that is suitable for teaching a course in Developmental Math. The package is itself a standalone version of part of a much larger package, in progress, that should provide people with a realistic chance of going from Arithmetic to Differential Calculus in three semestersasonable Basic Algebra — With Homework, Reviews and Exams to your Bookmark Collection or Course ePortfolio
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According to Connexions, "Taking the Mystery Out of Illinois School Finance (3rd Edition) was written by Dr. Thomas A. Kersten, Professor of Educational Leadership at Roosevelt University in Chicago, Illinois. This textbook was developed over several years and is designed to explain the key principles of Illinois school finance in a way that graduate students in educational administration programs, teachers, school board members, parents, building-level school administrators, and other interested citizens can grasp the essential content without getting bogged down in excessive financial detail. Only by understanding the basics of Illinois school finance, can school administrators, board members, and other constituents make informed decisionslection Mystery Out of Illinois School Finance to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material Taking the Mystery Out of Illinois School FinanceIn "Tales of Folk and Fairies" Ms. Pyle tells 15 different children's stories from around the world; each more delightful than the last. Each story stands completely on it's own and although they were probably meant for children, adults will certainly enjoy them as well Tales of Folk and Fairies to your Bookmark Collection or Course ePortfolio
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"Curso rapido de calculo diferencial e integral" is a program that may help college and high school students understand and learn the basic concepts of differential and integral calculus. So how to solve some useful exercises using educational videos
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In this lively volume, mathematician John Allen Paulos employs his singular wit to guide us through an unlikely mathematical jungleŚthe pages of the daily newspaper. From the Senate and sex to celebrities and cults, Paulos takes stories that may not seem to involve math at all and demonstrates how mathematical na´vetÚ can put readers at a distinct... more...
What if you had to take an art class in which you were only taught how to paint a fence? What if you were never shown the paintings of van Gogh and Picasso, weren?t even told they existed? Alas, this is how math is taught, and so for most of us it becomes the intellectual equivalent of watching paint dry. In Love and Math , renowned mathematician... more...
Hailed by The Mathematical Gazette as "an extremely valuable addition to the literature of algebraic topology," this concise but rigorous introductory treatment focuses on applications to dimension theory and fixed-point theorems. The lucid text examines complexes and their Betti groups, including Euclidean space, application to dimension theory,... more...
The range of cultural interests of renowned physicist David Speiser, including the sciences, art, architecture, music, and history of science, has inspired generations of later scientists to look beyond the boundaries of their own disciplines. In this book, seventeen scholars pay tribute to his multifaceted career. more...
BASIC PRINCIPLES Planning Ahead Consistency Overall Design Page Design Design in the Small Choice of Fonts Single Column or Double Column? Foreign Alphabets TYPESETTING MATHEMATICS What is TeX? Typesetting Mathematics in TeX Typesetting English in TeX Answering the Questions from Chapter 1 in the Argot of TeX Other Methods for Typesetting Mathematics... more...
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Synopses & Reviews
Publisher Comments:
This book has been widely acclaimed for its clear, cogent presentation of the theory of partial differential equations, and the incisive application of its principal topics to commonly encountered problems in the physical sciences and engineering. It was developed and tested at Purdue University over a period of five years in classes for advanced undergraduate and beginning graduate students in mathematics, engineering and the physical sciences.
The book begins with a short review of calculus and ordinary differential equations, then moves on to explore integral curves and surfaces of vector fields, quasi-linear and linear equations of first order, series solutions and the Cauchy Kovalevsky theorem. It then delves into linear partial differential equations, examines the Laplace, wave and heat equations, and concludes with a brief treatment of hyperbolic systems of equations.
Among the most important features of the text are the challenging problems at the end of each section which require a wide variety of responses from students, from providing details of the derivation of an item presented to solving specific problems associated with partial differential equations. Requiring only a modest mathematical background, the text will be indispensable to those who need to use partial differential equations in solving physical problems. It will provide as well the mathematical fundamentals for those who intend to pursue the study of more advanced topics, including modern theory
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Calculator
External Web sites
The Museum of HP Calculators - CalculatorIllustrated information on calculators created by Hewlett-Packard (HP) between 1968 and 1986. Provides a history of calculating machines and tools, and includes technical details on early HP calculators.
Britannica Web sites
Articles from Britannica encyclopedias for elementary and high school students.
Mechanical, electromechanical, or electronic devices that perform mathematical operations automatically are called calculators. Calculators perform the basic arithmetic functions-addition, subtraction, multiplication, and division-and many can also do more complicated calculations, such as normal and inverse trigonometric functions (see trigonometry). Few inventions of recent times have had such a profound influence on daily life as the handheld, or pocket, electronic calculator. These calculators are used to save time and to reduce the chance of making mistakes and are found wherever people deal frequently with numbers-in stores, offices, banks, schools, laboratories, and homes.
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books.google.com - This... Calculus with the HP-28 and the HP-48
Discovering Calculus with the HP-28 and the HP-48
This will save valuable time on graphing and calculations. With supercalculators such as the HP-28S and the HP-48SX, students can focus on true Calculus concepts rather than on computational details.
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Buy ePub
This volume is designed for the student desiring a greater understanding of the abacus and its calculative functions. The text provides thorough explanations of the advanced operations involving negative numbers, decimals, different units of measurement, and square roots. Diagrams illustrate bead manipulation, and numerous exercises provide ample practice. Concise and easy-to-follow, this book will improve your abacus skills and help you perform calculations with greater efficiency and precision.
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MA125 Intermediate Algebra
for S2QCalculator You will need a basic scientific calculator (such as a TI-30) for this course.A graphics calculator is acceptable but not required.Please be sure that you are familiar with the basic workings of your calculator as I cannot take class time to "teach" basic calculator competency.You should bring your calculator to every class.Students who do not have a calculator with them on test days must do the test without it!
In addition the required textbook and calculator, it is to your advantage to keep an organized notebook throughout the semester that should contain all of the work you produce in this course.I suggest that your notebook be a 1½ inch (4 cm) 3-ringed binder with 5 dividers labeled and ordered as follows: COURSE INFONOTES & OUTLINESHOMEWORKQUIZZES PAPER
Note: If during the class you suspect that your grade is incorrect, you will need to produce your graded homework and quizzes to prove an error on my part; an organized notebook will help resolve any grading issues!
Mathematics is a "do" subject...it's not a spectator sport! It is imperative that you not assume that watching me in class, taking notes and completing outlines, or watching on-line videos and reading the textbook will be enough. You must practice daily, throwing yourself into the homwork assignments...this will be the only way to build your confidence and competence in math!
With that said, success in this course will depend on your individual commitment to it.My experience over the years as a student and teacher has led me to the following observation which I call the123Rule: for every 1 hour in class, you should spend 2-3 hours out of class. Given that we meet for 5 hours each class, be sure that you are spending between 10-15 hours between our Thursday classes!
Learning Outcomes: Core Learning Outcomes
State and use basic terminology and symbols of the discipline appropriately
Solve linear equations and inequalities in one variable and verify solution(s)
Manipulate and simplify exponential expressions
Perform arithmetic on and factor polynomials and solve polynomial equations
Solve "word" problems
Manipulate and simplify rational expressions
Manipulate and simplify radical expressions and translate into/ from exponential form
The Mid-Term Exam takes place on Thursday, April 11th and the Final Exam takes place on Thursday, May 9th.
Grading:
Concerning Homework Two Homework Assignments will be assigned every class meeting. Unless prior arrangements have been made with me, each homework assignment is due at the place and time indicated in class and through the on-line classroom.
Individual homework assignment grades will not be based on correct answers, but on your process and your ability to organize your work following FEA:
FEA – follow this on each homework assignment for full credit!
Each HW assignment is worth 4 points, graded on the FEA scale: Format, Effort, and Appearance and must be done on 3-holed, college-ruled paper.
Format – 1 point: Name, MATH 125, Date Assigned, Date Due in the upper right corner in descending order
Questions indicated in the upper left corner (no questions - indicate this by writing NONE there)
Concerning Quizzes Two Quizzes will be given each class except on the class days when an exam is given (when that happens, only one quiz will be given). What to prepare for on upcoming quizzes will be announced ahead of time in class and through the on-line classroom.
Each quiz will be worth 10 points and they will usually be given at the beginning and end of each class, lasting 10 - 15 minutes. Quizzes will assess your understanding of past homework and are based on process and correct answers. Unless an answer is "obvious," some work or explanation is usually required to receive full credit on quiz exercises.
Grading of quizzes will also be tied to the Learning Rubric at the end of this syllabus.
Concerning Exams The Mid-Term and Final Exams are "closed notes" and "closed book" assessments. Each exam will be 90 minutes in length and are cumulative.
Late Submission of Course Materials:
Late Homework A homework assignment not turned in on time is considered "late." You may turn in four "late" homework assignments without penalty. A "late" homework is due no later than the beginning of the next class or it becomes a "0."
After four "late" homework assignments, subsequent late or missing homework will not be accepted and will receive a "0" grade for the remainder of the semester unless I have given you instructions to the contrary.
Homework and Absences If a homework assignment is due on a day that you are absent from class (for any reason), then that homework must be turned in at the beginning of the next class or it is considered late.
If you are absent from a class when homework is assigned, then that homework is due no later than the second class following the absence or it is considered late.
Quizzes and Absences If you miss a class, or arrive late, you may make up two "missed" quizzes without penalty. Two subsequent "missed" quizzes may be made up, but only receive a maximum of 8 points. More than four "missed" quizzes without a valid reason earns a "0" on those quizzes.
Arrangements to take "missed" quizzes must be made with me before the next class meeting. Failure to do so will turn the grades for "missed" quizzes to "0."
Classroom Rules of Conduct:
My goal in this course is to provide you with a superior mathematical education!
I cannot be successful in this goal without your full cooperation. I will make every effort to make every class session, and I expect the same commitment from you.
While I may be able to understand why you must miss a class, I cannot excuse you from the material covered or from turning in homework or making up quizzes in a timely fashion.Absences lead to low scores, confusion, miscommunication, frustration, low grades, and being dropped from the course.
I will be glad to provide additional help to students who are attending class regularly and applying themselves to homework, quizzes, and exams.
Time for extra help can be obtained by making an appointment with me through the emails listed above.
Chapter 1 Provides an introduction to the properties and operations of Real Numbers. We'll also examine properties of exponents and roots, and then discuss the order in which mathematical operations are to be applied as we simplify algebraic expressions.
Chapter 2 Begins the techniques to solve linear equations in one variable followed by a discussion of formulas and how to solve problems using them. Next we will apply those skills to real-life situations as we develop a general plan for solving application problems, and specific plans for certain types of problems. We will also learn the processes of solving linear and compound inequalities and finish up the chapter by learning how to solve absolute value equations and inequalities.
Chapter 3 The Rectangular Coordinate System is introduced. We will learn how to graph lines of equations in two variables. We'll discuss how to determine slope and how to use it to graph lines and find equations of lines. We will also become acquainted with functions, and we'll finish the chapter learning how to graph linear inequalities in two variables.
Chapter 4 The techniques used to solve systems of equations are presented. First we will graphically find the solution of two intersecting lines, followed by procedures to solve a system algebraically.
Chapter 5 The properties of exponents along with the properties of polynomial expressions are covered. We will learn how to use scientific notation as a tool for expressing very large and very small numbers. We will exam methods for adding, subtracting, multiplying, and dividing polynomials. We will expand our understanding of functions to include polynomial functions.
Chapter 6 Methods of factor polynomials are presented.These methods include: factoring by grouping, factoring trinomials, factoring perfect squares and perfect cubes. We finish the chapter by focusing on a fundamental approach to factoring, and how to apply problem-solving skills when a polynomial is part of the equation.
Chapter 7 Rational expressions are introduced by examining their properties. From there we will begin the task of learning how to add, subtract, multiply, and divide rational expressions. We will also learn two methods for simplifying complex fractions and conclude the chapter by solving equations involving rational expressions.
Chapter 8 The properties of radical expressions are initially discussed before learning the procedures for simplifying, including the conversion from a radical expression to a rational exponential expression. Next we'll learn how to add, subtract, multiply, and divide radical expressions. This is followed by a presentation of methods we will use for solving equations containing radical expressions.
Chapter 9 This chapter illustrates how to transform a polynomial equation into a particular form using the technique known as "completing the square." Using that technique the important "quadratic formula" is derived which is used to solve non-factorable quadratic equations with irrational and complex number solutions Pencil & Eraser Policy:I require you to do all homework, quizzes, and exams using a pencil and eraser.You will lose 25% on any homework, quiz, or exam that wasn't done in pencil, or that has messy scratched-out work.NOTE: It is O.K. to use a pen to head your papers and a pen/highlighter to circle final answers.
Evaluate 4 out of 4 algebraic expressions
Evaluate 3 out of 4 algebraic expressions
Evaluate 2 out of 4 algebraic expressions
Evaluate 0 or 1 out of 4 algebraic expressions
Synthesis Outcomes 1
Simplify and manipulate 4 out of 4 algebraic expressions
Simplify and manipulate 3 out of 4 algebraic expressions
Simplify and manipulate 2 out of 4 algebraic expressions
Simplify and manipulate 0 or 1 algebraic expressions
Analysis Outcomes 2
Solve and check 4 out of 4 algebraic equations
Solve and check 3 out of 4 algebraic equations
Solve and check 2 out of 4 algebraic equations
Solve and check 0 or 1 out of 4 algebraic equations
Application Outcomes 3
Solve 4 out of 4 practical applications
Solve 3 out of 4 practical applications
Solve 2 out of 4 practical applications
Solve 0 or 1 practical applications
Content of Communication Outcomes 4
Graph 4 out of 4 linear equations or inequalities
Graph 3 out of 4 linear equations or inequalities
Graph 2 out of 4 linear equations or inequalities
Graph 0 or 1 linear equations or inequalities
Technical Skill in Communicating Outcomes 4
Find 4 out of 4 slopes of lines
Find 3 out of 4 slopes of lines
Find 2 out of 4 slopes of lines
Find 0 or 1 slopes of lines
First Literacy Outcomes (Formulas) 1, 2, 3
Use and evaluate 4 out
of 4 formulas
Use and evaluate 3 out
of 4 formulas
Use and evaluate 2 out
of 4 formulas
Use and evaluate 0 or 1 out
of 4 formulas
Second Literacy Outcomes (Order of Operations) 1, 2, 3
Apply order of operations to 4 out of 4 algebraic expressions
Apply order of operations to 3 out of 4 algebraic expressions
Apply order of operations to 2 out of 4 algebraic expressions
Apply order of operations to 0 or 1 out of 4 algebraic expressions
Copyright:
This material is protected by copyright and can not be reused without author permission.
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Rt-Plot Rt-Plot is a tool to generate Cartesian X/Y-plots from scientific data. You can enter and calculate tabular data. View the changing graphs, including linear and non linear regression, interpolation, differentiation and integration, during entering. Download Now!
Prime Number Spiral The Prime Number Spiral (a.k.a. the Ulam Spiral) is formed by marking the prime numbers in a spiral arrangement of the natural numbers. This is software is for exploring the Prime Number Spiral. Download Now!
School Calendar School Calendar will help you with assignment organization, project due dates, and scheduling. It can even remind you when your scheduled event is about to happen. Included are two viewing modes, search, auto-backup, iCalendar data exchange. Download Now!
Breaktru Fractions n Decimals Add, Subtract, Divide and Multiply fractions.Convert a decimal to a fraction or a fraction to a decimal. Quick and easy interface. No confusing menus. Converts decimals to fractions. Great for school or work. Handy for STOCK quote Download Now!
Archim Archim is a program for drawing the graphs of all kinds of functions. You can define a graph explicitly and parametrically, in polar and spherical coordinates, on a plane and in space (surface). Archim will be useful for teachers and students. Download Now!
Inverse Matrices The program provides detailed, step-by-step solution in a tutorial-like format to the following problem: Given a 2x2 matrix, or a 3x3 matrix, or a 4x4 matrix, or a 5x5 matrix. Find its inverse matrix by using the Gauss-Jordan elimination method. The... Download Now!
Create function graphs with the help of this tool. Create function graphs with the help of this tool. Calculus Grapher draw a graph of any function and see graphs of its derivative and integral. Don't forget to use the magnify/demagnify controls on the y-axis to adjust the scale.Non-obvious...
Graph-Calc is a full programmable graphing calculator that can simultaneously graph multiple equations. Perfect for students of trigonometry, pre-calculus, calculus, and higher math.
- Zoom in on and scroll around the graph
- Tap graph to see x and y coordinates at a particular point
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- Convenient ans button to enter previous answer
- Easily switch between radians and degrees...
Ximarc Studios Inc is proud to bring you Khan Academy Calculus 4 (videos 61-80
Ximarc Studios Inc is proud to bring you Khan Academy Calculus 6 (videos 101-120
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Preview — Project Origami
by Thomas Hull
Project Origami: Activities for Exploring Mathematics
learn about: - Solving Cubic Equations - Bucky Balls and PHiZZ units - Matrix models for folds - Gaussian Curvature and much more! These activities, which can enhance the classroom experience, also make great independent student projects and are perfect for math clubs or math circles. To provide readers of "Project Origami" with as much flexibility as possible, we have made all of the handouts in the book available online....more
Paperback, 245 pages
Published
March 16th 2006
by AK Peters
(first published March 11th 2006)
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Mathematics Masterclasses
Stretching the Imagination
Edited by Michael J. Sewell
Mathematics Masterclasses
Stretching the Imagination
Edited by Michael J. Sewell
Description
This book serves as a valuable resource for mathematics and science teachers at secondary school level, teenagers and parents. It contains written versions of Royal Institution masterclasses on a wide selection of topics in pure and applied mathematics. The masterclasses are a popular program of advanced study conducted each year for mathematically talented university-bound British youth. They serve as a unique introduction to the kinds of topics found at the undergraduate level, yet presented in a manner that is meant to stimulate interest and challenge young minds. Topics include chaos theory, meteorology, storage limitations of computers, population growth and decay, as well as the mechanics of dinosaurs. The book is well-illustrated, easy to read, and contains
worksheets with interesting problems (and solutions). The emphasis throughout is on enjoying the challenge of mathematics.
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Certainly they cannot be thoroughly understood through rote memorization.A solid algebra foundation is necessary for almost all later math classes. A student who cannot understand algebra 1 concepts will have great difficulty in their future mathematics endeavors. Algebra 1 typically includes subjects such as factoring and the quadratic formula.
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BASIC PRINCIPLES Planning Ahead Consistency Overall Design Page Design Design in the Small Choice of Fonts Single Column or Double Column? Foreign Alphabets TYPESETTING MATHEMATICS What is TeX? Typesetting Mathematics in TeX Typesetting English in TeX Answering the Questions from Chapter 1 in the Argot of TeX Other Methods for Typesetting Mathematics... more...
Krantz takes the reader on a journey around the globe and through centuries of history, exploring the many transformations that mathematical proof has undergone from its inception at the time of Euclid and Pythagoras to its versatile, present-day use. The author elaborates on the beauty, challenges and metamorphisms of thought that have accompanied... more...
This book treats all of the most commonly used theories of the integral. After motivating the idea of integral, we devote a full chapter to the Riemann integral and the next to the Lebesgue integral. Another chapter compares and contrasts the two theories. The concluding chapter offers brief introductions to the Henstock integral, the Daniell integral,... more...
This concise, well-written handbook provides a distillation of real variable theory with a particular focus on the subject's significant applications to differential equations and Fourier analysis. Ample examples and brief explanations---with very few proofs and little axiomatic machinery---are used to highlight all the major results of real... more...
This text provides a masterful and systematic treatment of all the basic analytic and geometric aspects of Bergman's classic theory of the kernel and its invariance properties. These include calculation, invariance properties, boundary asymptotics, and asymptotic expansion of the Bergman kernel and metric. Moreover, it presents a unique compendium... more...
Praise for the Second Edition: "The book is recommended as a source for middle-level mathematical courses. It can be used not only in mathematical departments, but also by physicists, engineers, economists, and other experts in applied sciences who want to understand the main ideas of analysis in order to use them to study mathematical models of... more...
? one of the difficulties that students have with university mathematics is being able to relate it to what they?ve done at school. In this respect, the work on logic, sets, proof, relations and functions plays an essential bridging role. But another problem to be addressed is to re-present mathematics as a way of knowing?rather than a static body... more...
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Wikipedia in English
This edition places emphasis on modelling and using technology in problem solving, and features applications. Step-by-step solutions are provided for every example, and this work aims to show students how the mathematical concepts have relevant, everyday applications.
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Develops student understanding of coordinate methods for representing and analyzing relations among geometric shapes, and for describing geometric change.
Topics
include:
Modeling situations with coordinates, including computer-generated graphics; distance in the coordinate plane, midpoint of a segment, and slope; designing and programming algorithms; methods for solving systems of equations; coordinate and matrix models of isometric transformations (reflections, rotations, and translations) and of size transformations; and similarity.
Unit
3
Patterns
of Association
Develops student understanding of the strength of association between two variables, how to measure the degree of the relation, and how to use this measure as a tool to create and interpret prediction lines for paired data.
Develops
student ability to recognize data patterns that involve direct
or inverse power variation, to construct and analyze those models
and combinations such as quadratic models, and to apply those
models to a variety of problems.
Topics
include:
Basic power
models with rules of the form y = axb and
combinations of power models with other simple models; analysis
of quadratic models and equations from tabular, graphic, and
symbolic viewpoints; square root and cube root relations, and
fractional power and radical expressions.
Develops student ability to model and analyze physical phenomena with triangles, quadrilaterals, and circles and to use these shapes to investigate trigonometric functions, angular velocity, and periodic change.
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Mathematics A Discrete Introduction
9780534398989
ISBN:
0534398987
Edition: 2 Pub Date: 2005 Publisher: Thomson Learning
Summary: With a wealth of learning aids and a clear presentation, this book teaches students not only how to write proofs, but how to think clearly and present cases logically beyond this course. All the material is directly applicable to computer science and engineering, but it is presented from a mathematician's perspective
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More About
This Textbook
Overview
Maintaining the standard of excellence set by the previous edition, this textbook covers the basic geometry of two- and three-dimensional spaces Written by a master expositor, leading researcher in the field, and MacArthur Fellow, it includes experiments to determine the true shape of the universe and contains illustrated examples and engaging exercises that teach mind-expanding ideas in an intuitive and informal way. Bridging the gap from geometry to the latest work in observational cosmology, the book illustrates the connection between geometry and the behavior of the physical universe and explains how radiation remaining from the big bang may reveal the actual shape of the universe
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Analytic number theory II
Number theory has its roots in ancient history but particularly since the seventeenth century, it has undergone intensive development using ideas from many branches of mathematics. In spite of the subject's maturity, there are still unsolved problems that are easy to state and understand – for example, is every even number greater than two the sum of two primes? In this intermediate-level module (and in Analytic number theory I (M823)), you'll study number theory using techniques from analysis, in particular, the convergence of series and the calculus of residues. The module is based on readings from T.M. Apostol's Introduction to Analytic Number Theory.
Register
What you will study
The Greeks were the first to classify the integers and it is to them that the first systematic study of the properties of the numbers is attributed. But after about AD 250 the subject stagnated until the seventeenth century. Since then there has been intensive development, using ideas from many branches of mathematics. There are a large number of unsolved problems in number theory that are easy to state and understand – for example:
Is every even number greater than two the sum of two primes?
Are there infinitely many 'twin primes' (primes differing by 2), such as (3, 5) or (101, 103)?
Are there infinitely many primes of the form n2 + 1?
Does there always exist a prime between n2 and (n + 1)2 for every integer n > 1?
In this intermediate-level module (and in Analytic number theory I (M823)), you will study number theory using techniques from analysis, in particular, the convergence of series and the calculus of residues. Among the results proved in this module is the prime number theorem, which estimates the number of primes up to a given value x.
This module is based on Chapters 8-14 of the set book Introduction to Analytic Number Theory by T. M. Apostol (1986, fourth edition, Springer-Verlag).
All teaching is in English and your proficiency in the English language should be adequate for the level of study you wish to take. We strongly recommend that students have achieved an IELTS (International English Language Testing System) score of at least 7. To assess your English language skills in relation to your proposed studies you can visit the IELTS website.
If you have any doubt about the suitability of the module, please speak to an adviser.
QualificationsThe material contains small print and diagrams, which may cause problems if you find reading text difficult and you may also want to use a scientific calculatorModule notes, other printed materials.
You will need
We recommend that you have access to the internet at least once a week during the module and would like to point out that vital material, such as your assignments, will be delivered onlineApostol, T.M. Introduction to Analytic Number Theory Springer £46.99 - ISBN 9780387901633 This book is Print on Demand and can be ordered through any bookseller.
Note: Presenting in October 2014 and October 2016 only.
Teaching and assessment
Support from your tutor
You will have a tutor who will help you with the study material and mark and comment on your written work, and whom you can ask for advice and guidance using a special maths eTMA processor, which is used in place of the main eTMA system, unless there are some difficulties which prevent you from doing so. In these circumstances, you must negotiate with your tutor to get their agreement to submit your assignment on paper.
You will, however, be granted the option of submitting on paper if typesetting electronically or merging scanned images of your answers to produce an electronic TMA would take you an unacceptably long time.
Students also studied
Future availability
This module is only available in alternate even-numbered years. The details given here are for the module starting in October 2016, when it will be available for the last time.
How to register
We regret that we are currently unable to accept registrations for this course. Where the course is to be presented again in the future, relevant registration information will be displayed on this page as soon as it becomes availableM829
Credits
30
OU Level
Postgraduate
SCQF level
11
FHEQ level
7
Course work includes:
4 Tutor-marked assignments (TMAs)
Examination
No residential school
Student Reviews
"A perfect conclusion to M823. Whilst the M823 stands well as a course in its own right I personally found..." Read more
"An extremely good follow up course to Analytic Number Theory I, and by the end of the course, I felt..." Read more
Course satisfaction survey
Get a prospectus
The University wishes to emphasise that, while every effort is made to regularly update this site, the material on it is subject to alteration or amendment
in the light of changes in regulations or in policy or of financial or other necessity.
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More About
This Textbook
Overview
In this book the author analyzes and compares four closely related problems, namely linear programming, integer programming, linear integration, linear summation (or counting). The focus is on duality and the approach is novel as it puts integer programming in perspective with three associated problems, and permits one to define discrete analogues of well-known continuous duality concepts, and the rationale behind them. Also, the approach highlights the difference between the discrete and continuous cases. Central in the analysis are the continuous and discrete Brion and Vergne's formulae for linear integration and counting which are not very well-known in the optimization community. This approach provides some new insights on duality concepts for integer programs, and also permits to retrieve and shed new light on some well-known results. For instance, Gomory relaxations and the abstract superadditive dual of integer programs are re-interpreted in this algebraic approach.
This book will serve graduate students and researchers in applied mathematics, optimization, operations research and computer science. Due to the substantial practical importance of some presented problems, researchers in other areas will also find this book useful.
Editorial Reviews
From the Publisher
From the reviews:
"Lasserre has produced a fascinating slim … monograph (much of the work his own) looking at the parallels between linear (respectively integer) programming on the one hand and integration (respectively integer counting) problems on the other hand. … An appendix on various transforms a hundred references and a brief index complete the work which is a welcome addition to an important set of topics." (J. Borwein, Mathematical Reviews, Issue 2010 f)
"This book is devoted to analysing four important problems: integer programming problem, linear programming problem, linear integration problem, and linear counting problem. … a very specialized book on the integer programming problem and its dual variants. … can be very helpful for researchers working in developing algorithms for the integer programming problem which is a formidable challenging problem. This is a clear and well-written book … ." (E. Almehdawe, Journal of the Operational Research Society, Vol. 61 (12),
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Highland Park, TX PrecalculusDiscrete Math -- the math of things that are not continuous, things like integers -- underpins some of the most important problems in computer science, including set theory, information theory, encryption, topology and game theory, all of which are relevant to all the major advances being made o...
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This course is designed for students who need to both prepare for the High School Equivalency Diploma (HSE) diploma or college. Students will study applied math, algebra and geometry topics. This course is designed as a math intensive over 21 classroom hours and topics will range from arithmetic to multi-step word problems in Algebra and Geometry. 7 sessions.
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Pennies For Proofs Scholarship
The University of Alaska Fairbanks Department of Mathematics and Statistics is pleased to offer the 2013 Pennies for Proofs scholarship program.
2013 Fall Competition
The Department of Mathematics and Statistics is offering $500 tuition credits for freshmen or sophomores enrolling in Math 215 (Introduction to Mathematical Proofs) in Spring 2014. Students from all majors are welcome to apply.
To be eligible to apply for this scholarship, you must
Have freshman or sophomore standing; preference will be given to first-year students.
Have at least a 3.0 average in all mathematics courses taken an UAF.
Agree to take Math 215 in Spring 2014 if you accept an award.
Scholarships will be awarded based on review of a completed application, which consists of:
About Mathematics and Proof
Mathematics has two great traditions: computation and proof. Most of your mathematics studies have been devoted to learning computation skills. Calculus is, for example, one of the finest computation tools we have. If you decide to pursue any of a number of science or engineering degrees at UAF, Calculus II is a required class and you will want to take it early in your studies.
Proofs are different. The goal of a proof is not to arrive at a number, but to show that a mathematical fact is true. For example, the early Egyptians had a rule of thumb that a triangle with side-lengths 3, 4, and 5 has a right angle. By contrast, the early Greeks had a proof that the triangles with right angles are exactly the triangles where the side lengths satisfy a2 + b2 = c2(Euclid, Book I, Propositions 47 and 48).
Here are some things you can show using proofs:
At any given time, there are two exactly opposite points on the Earth where it is exactly the same temperature.
It is impossible to comb the hairs on a sphere so that all the hairs lie flat.
Fundamental Theorem of Algebra: every polynomial equation has at least one complex solution. Maybe someone mentioned this to you in an algebra class. Is it true? How would you know?
In Math 215, Introduction to Mathematical Proofs, you learn the skills needed to put a proof together and to analyze someone else's proof. These are the skills you need to have before tackling any of results mentioned above. By learning about this side of mathematics, you'll gain a better understanding of all your mathematics courses.
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An exotic form of torture created by McGraw Hill. This system of math learning generally involves learning skills such as: typing numbers onto a calculator, realizing HomeLinks (Homework Pages) have answer keys on them, and most importantly helps your child feel like they understand what they are learning, possibly due to the fact that is because the course teaches one concept over 5 years. Generally when students move on to Algebra, they begin to realize the meaning of Everyday Mathematics. Useless crap.
See algebra. Student: Did you understand how to do the HomeLinks last night?
StudentB: No, I just looked at the key.
Student: Wh...at?! But I was looking in the Everyday Mathematics SRI that she tells us to use!
StudentB: You're an idiot.
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"The IMO Compendium" is the ultimate collection of challenging high-school-level mathematics problems and is an invaluable resource not only for high-school students preparing for mathematics competitions, but for anyone who loves and appreciates mathematics. The International Mathematical Olympiad (IMO), nearing its 50th anniversary, has... more...
In this clever guide, young readers previously daunted by math will discover they're better at it than they thought.
With clear and accessible examples, How to Be a Math Genius explores the math brain and demonstrates to readers that they use math skills all the time-they just don't know it yet. Explaining fascinating ideas in a simple and... more...
Make your brain a maths brain
Packed with things to do, Train Your Brain to be a Maths Genius, will help you calculate equations that will make your knees tremble, it will make fretting about fractions a thing of the past, and best of all - it's great fun!
Jam packed with activities, puzzles, challenges, tips and tricks to boost brain power, this... more...
This text provides an introduction to group theory with an emphasis on clear examples. The authors present groups as naturally occurring structures arising from symmetry in geometrical figures and other mathematical objects. Written in a 'user-friendly' style, where new ideas are always motivated before being fully introduced, the text will help readers... more...
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Infoplease Tools
analysis
analysis, branch of mathematics that utilizes the concepts and methods of the calculus. It includes not only basic calculus, but also advanced calculus, in which such underlying concepts as that of a limit are subjected to rigorous examination; differential and integral equations, in which the unknowns are functions rather than numbers, as in algebraic equations; complex variable analysis, in which the variables are of the form z = x + iy, where i is the imaginary unit; vector analysis and tensor analysis; differential geometry; and many other fields.
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An online course: learning units presented in worksheet format review the most important results, techniques and formulas in college and pre-college calculus. Logarithms and Exponential; Sequences; Series; Techniques of...
A tutorial on wavelet filters aimed at engineers. Focusses on "lifting," a technique for creating a general framework to design filters for every possible wavelet transform. May be read online or downloaded in...
An introduction to magnetic resonance imagining, which is based on the principles of nuclear magnetic resonance (NMR), a spectroscopic technique used by scientists to obtain microscopic chemical and physical information...
An online course: learning units presented in worksheet format review the most important results, techniques and formulas in college and pre-college differential equations. Sections include: Introduction and First...
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Algebra 1
9780078651137
ISBN:
0078651131
Pub Date: 2005 Publisher: Glencoe/McGraw-Hill School Pub Co
Summary: A flexible program with the solid content students need "Glencoe Algebra 1" strengthens student understanding and provides the tools students need to succeed--from the first day your students begin to learn the vocabulary of algebra until the day they take final exams and standardized tests.
Holliday, Berchie is the author of Algebra 1, published 2005 under ISBN 9780078651137 and 0078651131. Nine hundred nin...e Algebra 1 textbooks are available for sale on ValoreBooks.com, nine hundred four used from the cheapest price of $3.95, or buy new starting at $45.00
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About this courseEntry requirements
English reading and writing skills, and maths to at least GCSE grade C or equivalent are required. You will need to have general skills and knowledge base associated with a GCSE course or equivalent standard.
This specification is designed to:
develop the student's understanding of mathematics and mathematical processes in a way that promotes confidence and fosters enjoyment
develop abilities to reason logically and to recognise incorrect reasoning, to generalise and to construct mathematical proofs
extend their range of mathematical skills and techniques and use them in more difficult unstructured problems
use mathematics as an effective means of communication
acquire the skills needed to use technology such as calculators and computers effectively, to recognise when such use may be inappropriate and to be aware of limitations
develop an awareness of the relevance of mathematics to other fields of study, to the world of work and to society in general
On this course you will study six units:
AS Level
Unit 1 MPC1 Core 1
Unit 2 MPC2 Core 2
Unit 3 MS1B Statistics 1B
A2 Level
Unit 4 MPC3 Core 3
Unit 5 MPC4 Core 4
Unit 6 MS2B Statistics 2
Each unit has 1 written paper of 1 hour 30 minutes.
Course Content
AS Level
Unit 1 MPC1 Core 1
Co-ordinate Geometry
Quadratic functions
Differentiation
Integration
Unit 2 MPC2 Core 2
Algebra and Functions
Sequences and Series
Trigonometry
Exponentials and logarithms
Differentiation
Integration
Unit 3 MS1B Statistics 1B
Statistical Measures
Probability
Discrete Random Variables
Normal Distribution
Estimation
A2 Level
Unit 4 MPC3 Core 3
Algebra and Functions
Trigonometry
Exponentials and Logarithms
Differentiation
Integration
Numerical Methods
Unit 5 MPC4 Core 4
Algebra and Functions
Coordinate Geometry in the (x, y) plane
Sequences and Series
Trigonometry
Exponentials and Logarithms
Differentiation and Integration
Vectors
Unit 6 MS2B Statistics 2
Poisson distribution
Continuous random variables
The t-distribution
Hypothesis Testing
Chi-squared tests
AS +A2 = A Level in Maths with Statistics. Both AS and A2 level courses and examinations must be successfully completed to gain a full A Level.
AQA Specification 6360
Paper Based Version
The course comes to you as a paper-based pack delivered by courier
You will be given guidance through the Study Guide on the nuts and bolts of studying and submitting assignments
Postal assignments cannot be accepted without prior permission from the tutor
Online Version
Our online A-Level courses are fully digitised so you can study on any smart device. Your learning programme is completely flexible so you can study at a pace that suits you. All our content is broken down into bite size chunks to make your learning more manageable and effective.
Your course of study is broken down into units and sections, each of which contains lessons, activities and test papers.
Courses are delivered on our Moodle based Digital Learning Environment CloudPort allowing you to study from anywhere on any smart device as long as you have access to the Internet.
The course concludes by preparing you for examination using past papers in your chosen subject.
The course contains a number of assignments which your tutor will mark and give you valuable feedback on. We call these Tutor Marked Assignments (TMAs). You need only send the TMAs to your tutor for comment, not the self-assessment exercises which are also part of the course to help you gauge your progress.
Exams are taken at an AQA centre and we can provide an extensive list of centres for you. Please read our FAQs for further information
Our A Levels come with tutor support for up to 24 months and for this course support expires in June 2017 due to syllabus changes brought in by the Government.
You will have access to a tutor, via email, who will mark your work and guide you through the course to help you be ready for your examinations. In addition you will be supplied with a comprehensive Study Guide which will help you through the study and assessment process.
With this distance learning course, you can study the challenging and rewarding subject of A-Level Chemistry at home. Through this course, you will gain a valuable understanding of core topics, including: Foundation Chemistry; Chemistry in Action; Kinetics, Equilibria and Organic Chemistry; and Energetics, Redox and Inorganic Chemistry. Students are also provided with the theoretical basis to carry out their practical skills assessments.
With your tutor on hand to guide you every step of the way, this home study course provides a solid understanding of chemistry in order to prepare candidates for A-Level examination in this most demanding and well-respected of A-Level subjects.History is a well-respected discipline that promotes rigorous academic study and also teaches students valuable transferrable skills, such as the ability to read critically and to construct arguments. This distance learning A-Level History course offers students the opportunity to study this demanding and challenging subject at home, under the support and guidance of an experienced tutor.
Designed with the needs of distance learning students in mind, the course guides students through the four areas of study: Britain 1603-1642; The Impact of Chairman Mao; Britain 1918-1964; and Totalitarian Ideology. The course aims to encourage an evaluative, interpretative and critical approach to past events, allowing students to develop their intellectual curiosity.
This home study course is suitable for those who wish to broaden their general knowledge of the interpretation of the past events, as well as those seeking to go on to study History or related subjects at a higher level.
Through UK Distance Learning & Publishing, the fascinating subject of A-Level Biology is now available to distance learning students for study at home. Closely based on the AQA A-Level Biology specification, this comprehensive course will take you through the core topics of biological study, allowing you to gain the thorough grounding needed to prepare you for A-Level examination or to study at a higher level.
This home study course provides the opportunity for varied and in-depth study, including the topics of: Biology and Disease; The Variety of Living Organisms; Populations and Environment; and Control in Cells and in Organisms. The course also includes information on the practical skills that you will need for your assessment.
This distance learning course in A-Level Religious Studies approaches the subject of religion as an academic discipline, allowing students to develop their own views and values through a critical, reflective and evaluative approach. The course is optimized for study at home, according to the AQA A-Level specification. The units studied include: Introduction to Religious Studies; Studies in Religion; and Religion and Human Experience. Particular attention is given to the topics of World Religions: Buddhism and to the Christian New Testament.
This distance learning course is suitable for those who want to learn more about religious studies for their own interest or for those who wish to study for an A-Level qualification as preparation for study at a higher level. A-Level Religious Studies is a particularly good complement for Philosophy or History A-Levels and is suitable for those who intend to study Religious Studies or Theology at university level. The course is suitable for students or all religions or none and no prior knowledge of religious studies is required.
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College Algebra -Text Only - 7th edition wi
th withP.1 Review of Real Numbers and Their Properties P.2 Exponents and Radicals P.3 Polynomials and Special Products P.4 Factoring Polynomials P.5 Rational Expressions P.6 Errors and the Algebra of Calculus P.7 The Rectangular Coordinate System and Graphs59
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97801226765ctory Analysis, Second Edition: The Theory of Calculus
Introductory Analysis, Second Edition, is intended for the standard course on calculus limit theories that is taken after a problem solving first course in calculus (most often by junior/senior mathematics majors). Topics studied include sequences, function limits, derivatives, integrals, series, metric spaces, and calculus in n-dimensional Euclidean space * Bases most of the various limit concepts on sequential limits, which is done first * Defines function limits by first developing the notion of continuity (with a sequential limit characterization) * Contains a thorough development of the Riemann integral, improper integrals (including sections on the gamma function and the Laplace transform), and the Stieltjes integral * Presents general metric space topology in juxtaposition with Euclidean spaces to ease the transition from the concrete setting to the abstract New to This Edition * Contains new Exercises throughout * Provides a simple definition of subsequence * Contains more information on function limits and L'Hospital's Rule * Provides clearer proofs about rational numbers and the integrals of Riemann and Stieltjes * presents an appendix lists all mathematicians named in the text * Gives a glossary of symbols
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Contents
CHAPTER 4 The Chain Rule
4.1 Derivatives by the Chain Rule
4.2 Implicit Differentiation and Related Rates
4.3 Inverse Functions and Their Derivatives
4.4 Inverses of Trigonometric Functions
CHAPTER 5 Integrals
5.1 The Idea of the Integral 177
5.2 Antiderivatives 182
5.3 Summation vs. Integration 187
5.4 Indefinite Integrals and Substitutions 195
5.5 The Definite Integral 201
5.6 Properties of the Integral and the Average Value 206
5.7 The Fundamental Theorem and Its Consequences 213
5.8 Numerical Integration 220
CHAPTER 6 Exponentials and Logarithms
6.1 An Overview 228
6.2 The Exponential ex 236
6.3 Growth and Decay in Science and Economics 242
6.4 Logarithms 252
6.5 Separable Equations Including the Logistic Equation 259
6.6 Powers Instead of Exponentials 267
6.7 Hyperbolic Functions 277
CHAPTER 7 Techniques of Integration
7.1 Integration by Parts
7.2 Trigonometric Integrals
7.3 Trigonometric Substitutions
7.4 Partial Fractions
7.5 Improper Integrals
CHAPTER 8 Applications of the Integral
8.1 Areas and Volumes by Slices
8.2 Length of a Plane Curve
8.3 Area of a Surface of Revolution
8.4 Probability and Calculus
8.5 Masses and Moments
8.6 Force, Work, and Energy
CHAPTER 7
Techniques of Integration
Chapter 5 introduced the integral as a limit of sums. The calculation of areas was
started-by hand or computer. Chapter 6 opened a different door. Its new functions
ex and In x led to differential equations. You might say that all along we have been
solving the special differential equation dfldx = v(x). The solution is f = 1v(x)dx. But
the step to dyldx = cy was a breakthrough.
The truth is that we are able to do remarkable things. Mathematics has a language,
and you are learning to speak it. A short time ago the symbols dyldx and J'v(x)dx
were a mystery. (My own class was not too sure about v(x) itself-the symbol for a
function.) It is easy to forget how far we have come, in looking ahead to what is next.
I do want to look ahead. For integrals there are two steps to take-more functions
and more applications. By using mathematics we make it live. The applications are
most complete when we know the integral. This short chapter will widen (very much)
the range of functions we can integrate. A computer with symbolic algebra widens it
more.
Up to now, integration depended on recognizing derivatives. If v(x) = sec2x then
f(x) = tan x. To integrate tan x we use a substitution:,
I!&dx.=1"-
- U
- - In u = - In cos x.
What we need now ,are techniques for other integrals, to change them around until
we can attack them. Two examples are j x cos x dx and 5 , -
/ dx, which are not
immediately recognizable. With integration by parts, and a new substitution, they
become simple.
Those examples indicate where this chapter starts and stops. With reasonable effort
(and the help of tables, which is fair) you can integrate important functions. With
intense effort you could integrate even more functions. In older books that extra
exertion was made-it tended to dominate the course. They had integrals like
J(x + l)dx//-
,, which we could work on if we had to. Our time is too
valuablefor that! Like long division, the ideas are for us and their intricate elaboration
is for the computer.
Integration by parts comes first. Then we do new substitutions. Partial fractions
is a useful idea (already applied to the logistic equation y' = cy - by2). In the last
section x goes to infinity or y(x) goes to infinity-but the area stays finite. These
improper integrals are quite common. Chapter 8 brings the applications.
7.1 Integration by Parts 283
7.1 Integration by Parts
There are two major ways to manipulate integrals (with the hope of making them
easier). Substitutions are based on the chain rule, and more are ahead. Here we
present the other method, based on the product rule. The reverse of the product rule,
to find integrals not derivatives, is integration by parts.
We have mentioned Jcos2x dx and JIn dx. Now is the right time to compute
x
them (plus more examples). You will see how J In dx is exchanged for J1 dx-a
x
x x
definite improvement. Also Jxe dx is exchanged for Je dx. The difference between
the harder integral and the easier integral is a known term-that is the point.
One note before starting: Integration by parts is not just a trick with no meaning.
On the contrary, it expresses basic physical laws of equilibrium and force balance.
It is a foundation for the theory of differential equations (and even delta functions).
The final paragraphs, which are completely optional, illustrate those points too.
We begin with the product rule for the derivative of u(x) times v(x):
dv du d
u(x) + v(x)d - d (u(x)v(x)). (1)
dx dx dx
Integrate both sides. On the right, integration brings back u(x)v(x). On the left are
two integrals, and one of them moves to the other side (with a minus sign):
u(x) dx = u(x)v(x) - v(x) dx. (2)
That is the key to this section-not too impressive at first, but very powerful. It is
integration by parts (u and v are the parts). In practice we write it without x's:
7A The integration by parts formula is j u dv = uv - Jv du. (3)
The problem of integrating u dv/dx is changed into the problem of integrating
v du/dx. There is a minus sign to remember, and there is the "integrated term" u(x)v(x).
In the definite integral, that product u(x)v(x) is evaluated at the endpoints a and b:
Lb dv du
u dx u(b)v(b) - u(a)v(a) -
- v dx. (4)
a dx dx
The key is in choosing u and v. The goal of that choice is to make 5 v du easier than
j u dv. This is best seen by examples.
EXAMPLE 1 For f Inx dx choose u = In and dv = dx (so v= x):
x
In xdx = uv - v du = x ln x - x dx.
I used the basic formula (3). Instead of working with In (searching for an antideriva-
x
tive), we now work with the right hand side. There x times l/x is 1. The integral of
1 is x. Including the minus sign and the integrated term uv = x In and the constant
x
C, the answer is
J Inx dx = x Inx - x + C. (5)
For safety, take the derivative. The product rule gives Inx + x(1/x) - 1, which is Inx.
The area under y = Inx from 2 to 3 is 3 In3 - 3 - 2 In 2 + 2.
7 Techniques of Integration
To repeat: We exchanged the integral of In x for the integral of 1.
XML 5
E A P E 2 For x cos x dx choose u = x and dv = cos x dx (so v(x) = sin x):
Again the right side has a simple integral, which completes the solution:
J'xcos x d x = x sin x + c o s x + C. (7)
Note The new integral is not always simpler. We could have chosen u = cos x and
dv = x dx. Then v = fx2. Integration using those parts give the true but useless result
The last integral is harder instead of easier (x2 is worse than x). In the forward
direction this is no help. But in the opposite direction it simplifies Sf x2 sin x dx. The
idea in choosing u and v is this: Try to give u a nice derivative and du a nice integral.
E A P E 3 For J (cos x ) dx choose u = cos x and dv = cos x dx (so v = sin x):
XML ~
1
~ ( C O S) ~ = uv - J v du = cos x sin x + (sin x ) dx.
x ~ x ~
.
The integral of (sin x)' is no better and no worse than the integral of (cos x ) ~But we
~
never see (sin x ) without thinking of 1 - (cos x ) ~So substitute for (sin x ) ~ :
.
J'(cos x ) dx = cos x sin x + J' 1 dx - J (cos x)2 dx.
~
The last integral on the right joins its twin on the left, and J' 1 dx = x:
2 J (cos x ) dx = cos x sin x
~ + x.
Dividing by 2 gives the answer, which is definitely not gcos x ) ~Add any C:
.
{(cos x)' dx = f (cos x sin x + x) + C. (8)
Question Integrate (cos x)' from 0 to 2n. Why should the area be n?
Answer The definite integral is gcos x sin x + x)]:". This does give n. That area can
also be found by common sense, starting from (cos x ) + (sin x ) = 1. The area under
~ ~
~
1 is 2n. The areas under (cos x ) and (sin x ) are the same. So each one is n.
~
E A P E 4 Evaluate J tan-'x dx by choosing u = tan-'x and v = x:
XML
S tan-'x d x = uv-
S v d u = x tan-'x-
The last integral has w = 1 + x2 below and almost has dw = 2x dx above:
Substituting back into (9) gives J tan- 'x dx as x tan- 'x - f ln(1 + x2).All the familiar
inverse functions can be integrated by parts (take v = x, and add " + C" at the end).
Our final example shows how two integrations by parts may be needed, when the
first one only simplifies the problem half way.
E A P E 5 For j x2exdxchoose u = x2 and dv = exdx (so v = ex):
XML
j x2exdx= uv - v du = x2ex- ex(2xdx).
7.1 Integration by Parts 285
2
The last integral involves xex. This is better than x ex, but it still needs work:
f xexdx = uv - fv
du = xex - exdx (now u = x). (11)
Finally ex is alone. After two integrations by parts, we reach I exdx. In equation (11),
the integralof xex is xex - ex. Substituting back into (10),
f x 2exdx = x 2ex - 2[xex - ex] + C. (12)
These five examples are in the list of prime candidatesfor integration by parts:
xnex, x"sin x, x"cos x, x"ln x, exsin x, excos x, sin-'x, tan-x, ....
This concludes the presentation of the method-brief and straightforward.
Figure 7.1a shows how the areas f u dv and I v du fill out the difference between the
big area u(b)v(b) and the smaller area u(a)v(a).
U
2 v(x) 8(x) "=v(0) 6(x)
red area = large box s V(X)
- small box - gray area
= V U2 - v 1 u 1 - fvdu
2
U
1
0 X
0 vi v2 0
Fig. 7.1 The geometry of integration by parts. Delta function (area 1) multiplies v(x) at x = 0.
In the movie Stand and Deliver, the Los Angeles teacher Jaime Escalante computed
J x 2sin x dx with two integrations by parts. His success was through exercises-plus
insight in choosing u and v. (Notice the difference from f x sin x 2 dx. That falls the
other way-to a substitution.) The class did extremely well on the Advanced Place-
ment Exam. If you saw the movie, you remember that the examiner didn't believe
it was possible. I spoke to him long after, and he confirms that practice was the key.
THE DELTA FUNCTION
From the most familiar functions we move to the least familiar. The delta function is
the derivative of a step function. The step function U(x) jumps from 0 to 1 at x = 0.
We write 6(x) = dU/dx, recognizing as we do it that there is no genuine derivative at
the jump. The delta function is the limit of higher and higher spikes-from the
"burst of speed" in Section 1.2. They approach an infinite spike concentrated at a
single point (where U jumps). This "non-function" may be unconventional--it is
certainly optional-but it is important enough to come back to.
The slope dU/dx is zero except at x = 0, where the step function jumps. Thus
6(x) = 0 except at that one point, where the delta function has a "spike." We cannot
give a value for 6 at x = 0, but we know its integralacross the jump. On every interval
from - A to A, the integral of dU/dx brings back U:
6(x) dx= - dx U(x)] A = 1.
d= (13)
-A
"The area under the infinitely tall and infinitely thin spike 6(x) equals 1."
So far so good. The integral of 6(x) is U(x). We now integrate by parts for a crucial
purpose-tofindthe area under v(x)6(x). This is an ordinary function times the delta
function. In some sense v(x) times 6(x) equals v(O) times 6(x)-because away from
x = 0 the product is always zero. Thus ex6(x) equals 6(x), and sin x 6(x) = 0.
286 7 Techniques of Integration
The area under v(x)6(x) is v(0)-which integration by parts will prove:
7B The integral of v(x) times 6(x) is fA_ v(x)6(x)dx = v(0).
The area is v(0) because the spike is multiplied by v(O)-the value of the smooth
function v(x) at the spike. But multiplying infinity is dangerous, to say the least. (Two
times infinity is infinity). We cannot deal directly with the delta function. It is only
known by its integrals!As long as the applications produce integrals (as they do), we
can avoid the fact that 6 is not a true function.
The integral of v(x)6(x)= v(x)dU/dx is computed "by parts:"
=
v(x)6(x) dx v(x)U(x)] A - U(x) dx. (14)
-A - -A dx
Remember that U = 0 or U = 1. The right side of (14) is our area v(O):
A dv
v(A) . 1- 1 dx = v(A) - (v(A) - v(O))= v(O). (15)
o dx
When v(x) = 1, this answer matches f 6dx = 1. We give three examples:
S2 cos x 6(x) dx = 1 f6 5 (U(x) + 6(x))dx = 7 1_1 (6(x))2dx = c00.
A nightmare question occurs to me. What is the derivative of the deltafunction?
INTEGRATION BY PARTS IN ENGINEERING
Physics and engineering and economics frequently involve products. Work is force
times distance. Power is voltage times current. Income is price times quantity. When
there are several forces or currents or sales, we add the products. When there are
infinitely many, we integrate (probably by parts).
I start with differential equations for the displacement u at point x in a bar:
dv du
S= f(x) with v(x) = k (16)
dx dx
This describes a hanging bar pulled down by a forcef(x). Each point x moves through
a distance u(x). The top of the bar is fixed, so u(0)= 0. The stretching in the bar is
du/dx. The internal force created by stretching is v = k du/dx. (This is Hooke's law.)
Equation (16) is a balance offorces on the small piece of the bar in Figure 7.2.
0
Fig. 7.2 Difference in internal force balances external force
W
- Av =fAx or -dv/dx =f(x)
v = W at x = 1 balances hanging weight
7.1 Integration by Paits
EXAMPLE 6 Supposef(x) = F, a constant force per unit length. We can solve (16):
V(X) = - Fx + C and ku(x) = - f FX' + C x + D. (17)
The constants C and D are settled at the endpoints (as usual for integrals). At x = 0
we are given u = O so D = O . At x = 1 we are given v = W so C = W + F. Then v(x)
and u(x)give force and displacement in the bar.
To see integration by parts, multiply - dvldx =f(x) by u(x)and integrate:
] 0
f(x)u(x) dx = - ] 0 dx
u(x) dx = - u(x)v(x)]i+ ]o v(x) dx dx.
The left side is force times displacement, or external work. The last term is internal
force times stretching-or internal work. The integrated term has u(0) = 0-the fixed
support does no work. It also has -u(l)W, the work by the hanging weight. The
balance of forces has been replaced by a balance of work.
This is a touch of engineering mathematics, and here is the main point. Integration
by parts makes physical sense! When - dvldx =f is multiplied by other functions-
called test functions or virtual displacements-then equation (18) becomes the
principle of virtual work. It is absolutely basic to mechanics.
7.1 EXERCISES
Read-through questions 9 leXsinxdx 10 jexcos x dx
Integration by parts is the reverse of the a rule. It [ and 10 need two integrations. I think ex can be u or v.]
9
changes u dv into b minus c . In case u = x and 11 j eaxsin bx dx 12 jxe-"dx
1
dv = eZxdx,it changes xe2'dx to d minus e . The
definite integral ji xeZxdxbecomes f minus 9 . 13 J sin(1n x) dx 14 cos(1n x) dx
In choosing u and dv, the h of u and the i of 5
15 (In ~ ) ~ d x 16 j x 2 1 n x d x
dvldx should be as simple as possible. Normally In x goes into
i and e" goes into k . Prime candidates are u = x or
17 1sin- 'X dx 1
18 cos"(2x) dx
x 2 a n d v = s i n x o r I or m . W h e n u = x 2 w e n e e d 19 j x tan-'x dx
n integrations by parts. For 1 sin- 'x dx, the choice dv =
dx leads to o minus P .
20 1x2sin x dx (from the movie)
21 jx3cos x dx 22 j x3 sin x dx
If U is the unit step function, dU/dx = S is the unit q
function. The integral from -A to A is U(A) - U(- A) = 23 j x3exdx 24 1x sec'lx dx
r . The integral of v(x)S(x) equals s . The integral 25 1x sec2x dx 26 1x cosh x dx
jLl cos x S(x)dxequals t . In engineering, the balance of
forces -dv/dx =f is multiplied by a displacement u(x) and
integrated to give a balance of u . Compute the definite integrals 27-34.
27 ;
1 ln x dx 28 1; & dx (let u = A)
Integrate 1-16, usually by parts (sometimes twice). ;
29 1 x e""dx 30 j; ln(x2)dx
1 x sin x dx 2 jxe4"dx 31 [E x cos x dx 32 xsin x dx
3 jxe-'dx 4 x cos 3x dx 33 1 ln(x2 + 1)dx
: 34 g2 sin x dx,
x2
5 x2cos x dx (use Problem 1)
In 35-40 derive "reduction formulas" from higher to lower
powers. ,
8 j x2 e4xdx (use Problem 2) 35 xnexdx= xnex- n j xn- -'eXdx
288 7 Techniques of Integration
52 Draw the graph of v(x) if v(1) = 0 and -dv/dx =.f(x):
37 l x n c o s x dx=xnsin x - n 1xn-'sin x dx (a)f = x; (b)f = U(x - 3); (c)f = S(x - 3).
38 1xnsin x dx = 53 What integral u(x) solves k duldx = v(x) with end con-
dition u(O)=O? Find u(x) for the three v's (not f's) in
39 1(ln x)"dx = x(ln x)" - n 1(ln x)"- ldx Problem 52, and graph the three u's.
54 Draw the graph of AUlAx = [U(x + Ax) - U(x)]/Ax.
I
41 How would you compute x sin x exdx using Problem 9? What is the area under this graph?
Not necessary to do it.
Problems 55-62 need more than one integration.
I
42 How would you compute x extan- 'x dx? Don't do it.
55 Two integrations by parts lead to V = integral of v:
1
43 (a) Integrate x3sin x2dx by substitution and parts.
(b) The integral xnsin x2dx is possible if n is . I uv'dx = uv - Vu' + I Vu"dx.
Test this rule on 1x2sin x dx.
44-54 are about optional topics at the end of the section. 56 After n integrations by parts, 1u(dv/dx)dx becomes
44 For the delta function 6(x) find these integrals: uv - U'"V(~, + u ' ~ ' v ( -, + (- 1)" 1u'"'u(,- ,,dx.
~
! ,
(a) J e2xS(x)dx (b) j), v(x)6(x)dx (c) cos x 6(x)dx.
dn)s the nth derivative of u, and v(,, is the nth integral of v.
i
45 Solve dyldx = 36(x) and dyldx = 36(x) y(x). + Integrate the last term by parts to stretch this formula to
46 Strange fact: 6(2x) is diflerent from 6(x). Integrate them
n + 1 integrations.
both from -1 to 1. 57 Use Problem 56 to find [ x3exdx.
47 The integral of 6(x) is the unit step U(x). Graph the next 58 From f(x) -f(0) = [t f '(t)dt, integrate by parts (notice dt
I I
integrals R(x) = U(x)dx and Q(x) = R(x)dx. The ramp R + +
not dx) to reach f(x) =f(0) f '(0)x J","(t)(x - t)dt. Con-
and quadratic spline Q are zero at x = 0. tinuing as in Problem 56 produces Taylor's formula:
)
4,
48 In 6(x - the spike shifts to x = f. It is the derivative of 1
the shifted step U(x - 3). The integral of v(x)d(x - 3) equals f(x)=f(0)+f1(O)x+-f"(0)x2+.-+ dt.
2! n!
the value of v at x = 3. Compute
(a) 6(x - f)dx; 1;
(b) ex6(x - 4)dx; 59 What is the difference between 1; uw"dx and I; u"w dx?
(4 I! I 6(x)6(x - t)dx. ,
60 compute the areas A = [; In x dx and B = 1; eYdy. Mark
them on the rectangle with corners (0, 0), (e, 0), (e, I), (0, 1).
49 The derivative of 6(x) is extremely singular. It is a "dipole"
known by its integrals. Integrate by parts in .(b) and (c): 61 Find the mistake. I don't believe excosh x = exsinh x:
= excosh x - exsinh x + exsinh x dx.
50 Why is I!, U(x)6(x)dx equal to f? (By parts.)
62 Choose C and D to make the derivative of
51 Choose limits of integration in v(x)=J f(x)dx so that C eaXcos +
bx D eaxsinbx equal to eaXcos bx. Is this easier
dv/dx= -f(x) and v = O at x = 1. than integrating eaxcoshx twice by parts?
7.2 Trigonometric Integrals
The next section will put old integrals into new forms. For example x2 , ' -
/ dx
will become jsin20 cos20 dB. That looks simpler because the square root is gone. But
still sin20 cos28 has to be integrated. This brief section integrates any product of shes
and cosines and secants and tangents.
There are two methods to choose from. One uses integration by parts, the other
is based on trigonometric identities. Both methods try to make the integral easy (but
that may take time). We follow convention by changing the letter 8 back to x.
7.2 Trigonometric Integrals
Notice that sin4x cos x dx is easy to integrate. It is u4du. This is the goal in
Example l-to separate out cos x dx. It becomes du, and sin x is u.
EXAMPLE I j sin2xcos3x dx (the exponent 3 is odd)
Solution Keep cos x dx as du. Convert the other cos2x to 1 - sin2x:
EXAMPLE 2 5 sin5x dx (the exponent 5 is odd)
Solution Keep sin x dx and convert everything else to cosines. The conversion is
always based on sin2x + cos2x = 1:
j(l - c o ~ ~ x ) ~ s dx = !(I-
x in 2 cos2x + cos4x) sin x dx.
Now cos x is u and - sin x dx is du. We have !(- 1 + 2u2 - u4)du.
General method for 5 sinmxcosnx dx, when m or n is odd
If n is odd, separate out a single cos x dx. That leaves an even number of cosines.
Convert them to sines. Then cos x dx is du and the sines are u's.
If m is odd, separate out a single sin x dx as du. Convert the rest to cosines.
If m and n are both odd, use either method.
If m and n are both even, a new method is needed. Here are two examples.
EXAMPLE 3 5 cos2x dx (m = 0,n = 2, both even)
There are two good ways to integrate cos2x, but substitution is not one of them. If
u equals cos x, then du is not here. The successful methods are integration by parts
and double-angle formulas. Both answers are in equation (2) below-I don't see
either one as the obvious winner.
Integrating cos2x by parts was Example 3 of Section 7.1. The other approach, by
double angles, is based on these formulas from trigonometry:
cos2x = f (1 + cos 2x) sin2x = f(1- cos 2x) (1)
The integral of cos 2x is 5 sin 2x. So these formulas can be integrated directly. They
give the only integrals you should memorize-either the integration by parts form,
or the result from these double angles:
cos2x dx equals )(x + sin x cos x) or )x + 4 sin 2x (plus C). (2)
1sin2x dx equals $(x - sin x cos x) or f x - & sin 2x (plus C). (3)
EXAMPLE 4 1cos4x dx (m = 0,n = 4, both are even)
Changing cos2x to 1 - sin2x gets us nowhere. All exponents stay even. Substituting
u = sin x won't simplify sin4x dx, without du. Integrate by parts or switch to 2x.
First solution Integrate by parts. Take u = cos3x and dv = cos x dx:
1(cos3x)(cosx dx) = uv - j v du = cos3x sin x - j (sin x)(- 3 cos2x sin x dx).
The last integral has even powers sin2x and cos2x. This looks like no progress.
Replacing sin2x by 1 - cos2x produces cos4x on the right-hand side also:
J cos4x dx = cos3x sin x + 3 5 cos2x(l - cos2x)dx.
7 Techniques of Integration
Always even powers in the integrals. But now move 3 cos4x dx to the left side:
Reduction 5
4 cos4x dx = cos3x sin x + 3 cos2x dx. (4)
Partial success-the problem is reduced from cos4x to cos2x. Still an even power,
but a lower power. The integral of cos2x is already known. Use it in equation (4):
I cos4x dx = $ cos3x sin x + 3 f(x + sin x cos x) + C. (5)
Second solution Substitute the double-angle formula cos2x = 3 + 3 cos 2x:
5
cos4x dx = (f + f cos 2x)'dx = I (1 + 2 cos 2x + cos2 2x)dx.
I I
Certainly dx = x. Also 2 cos 2x dx = sin 2x. That leaves the cosine squared:
I cos22x = I f (1 + cos 4x)dx = f x + sin 4x + C.
The integral of cos4x using double angles is
$[x + sin 2x + f x + $sin 4x1 + C.
That solution looks different from equation (S), but it can't be. There all angles were
x, here we have 2x and 4x. We went from cos4x to cos22x to cos 4x, which was
integrated immediately. The powers were cut in half as the angle was doubled.
Double-angle method for I sinmxcosnx dx, when m and n are even.
Replace sin2x by f (1 - cos 2x) and cos2x by & I + cos 2x). The exponents drop to
m/2 and n/2. If those are even, repeat the idea (2x goes to 4x). If m/2 or n/2 is odd,
switch to the "general method" using substitution. With an odd power, we have du.
EXAMPLE 5 (Double angle) I sin2x cos2x dx = I i ( l - cos 2x)(1 + cos 2x)dx.
This leaves 1 - cos2 2x in the last integral. That is familiar but not necessarily easy.
We can look it up (safest) or remember it (quickest) or use double angles again:
(1-cos22x)dx=-
'I(::
4 )
x sin 4x
1 - - - - C O S ~ X dx=---
8 3 2 + C.
Conclusion Every sinmxcosnx can be integrated. This includes negative m and n-
see tangents and secants below. Symbolic codes like MACSYMA or Mathematica
give the answer directly. Do they use double angles or integration by parts?
You may prefer the answer from integration by parts (I usually do). It avoids 2x
and 4x. But it makes no sense to go through every step every time. Either a computer
does the algebra, or we use a "reduction formula" from n to n - 2:
Reduction n J cosnx dx = cosn-'x sin x + (n - 1) COS"-~X
dx. (7)
I
For n = 2 this is cos2x dx-the integral to learn. For n = 4 the reduction produces
cos2x. The integral of cos6x goes to cos4x. There are similar reduction formulas for
sinmxand also for sinmxcosnx. I don't see a good reason to memorize them.
INTEGRALS WITH ANGLES px AND qx
Instead of sin8x times cos6x, suppose you have sin 8x times cos 6x. How do you
integrate? Separately a sine and cosine are easy. The new question is the integral of
the product:
7.2 Trigonometric Intagrals
EXAMPLE 6 Find I:" sin 8x cos 6x dx. More generallyfind I:" sin px cos qx dx.
This is not for the sake of making up new problems. I believe these are the most
important definite integrals in this chapter (p and q are 0, 1,2, ...). They may be the
most important in all of mathematics, especially because the question has such a
beautiful answer. The integrals are zero. On that fact rests the success of Fourier
series, and the whole industry of signal processing.
One approach (the slow way) is to replace sin 8x and cos 6x by powers of cosines.
That involves cos14x. The integration is not fun. A better approach, which applies to
all angles px and qx, is to use the identity
sin px cos qx = f sin(p + q)x + f sin(p - q)x. (8)
Thus sin 8x cos 6x = f sin 14x + f sin 2x. Separated like that, sines are easy to
integrate:
lo2"
1 cos 14x 1 cos 2x 2"
s i n 8 x c o s 6 x d x = ------
[ I 4 2 2 0
=0.
1
Since cos 14x is periodic, it has the same value at 0 and 2n. Subtraction gives zero.
The same is true for cos 2x. The integral of sine times cosine is always zero over a
complete period (like 0 to 2n).
What about sin px sin qx and cos px cos qx? Their integrals are also zero, provided
p is dinerentfrom q. When p = q we have a perfect square. There is no negative area
to cancel the positive area. The integral of cos2px or sin2px is n.
EXAMPLE 7 I:" sin 8x sin 7x dx = 0 and I:" sin2 8x dx = n.
With two sines or two cosines (instead of sine times cosine), we go back to the
addition formulas of Section 1.5. Problem 24 derives these formulas:
sin px sin qx = - 4 cos(p + q)x + cos(p - q)x (9)
cos px cos qx = + cos(p + q)x + 9 cos(p - q)x. (10)
With p = 8 and q = 7, we get cos 15x and cos x. Their definite integrals are zero. With
x
p = 8 and q = 8, we get cos 16x and cos O (which is 1). Formulas (9) and (10) also
give a factor f . The integral of f is n:
1 sin 8x sin 7x dx = - f 1 cos 15x dx + $I:" cos x dx = 0 + 0
:" "
:
1 sin 8x sin 8x dx = - )I:"
:
" coCl6x dx + fI:" cos O dx = 0 + n
x
The answer zero is memorable. The answer n appears constantly in Fourier series.
No ordinary numbers are seen in these integrals. The case p = q = 1 brings back
cos2x dx = f + t sin 2x.
SECANTS AND T N E T
A GNS
When we allow negative powers m and n, the main fact remains true. All integrals
I sinmxcosnxdx can be expressed by known functions. The novelty for negative pow-
ers is that logarithms appear. That happens right at the start, for sin x/cos x and for
ljcos x (tangent and secant):
I tan x dx = - I duju = - lnlcos x J (here u = cos x)
I sec x dx = duju = lnlsec x + tan xl (here u = sec x + tan x).
7 Techniques of Integration
+
For higher powers there is one key identity: 1 tan2x = sec2x. That is the old
identity cos2x + sin2x = 1 in disguise (just divide by cos2x). We switch tangents to
secants just as we switched sines to cosines. Since (tan x)' = sec2x and (sec x)' =
sec x tan x, nothing else comes in.
EXAMPLE 8 [ tan2x dx = [(sec2x - 1)dx = tan x - x + C .
EXAMPLE 9 [ tan3x dx = [ tan x(sec2x - 1)dx.
The first integral on the right is [ u du = iu2, with u = tan x. The last integral is
- [ tan x dx. The complete answer is f (tan x ) + lnlcos x I + C. By taking absolute
~
values, a negative cosine is also allowed. Avoid cos x = 0.
EXAMPLE 10 Reduction
I (tan x)"dx x)"'-'
= ('an
m-1
- I(tan x)m-2dx
Same idea-separate off (tan x ) ~ sec2x - 1. Then integrate (tan x)"-'sec2x dx,
as
which is urn-'du. This leaves the integral on the right, with the exponent lowered by
2. Every power (tan x)" is eventually reduced to Example 8 or 9.
EXAMPLE II [ sec3x dx = uv - [ v du = sec x tan x - [ tan2x sec x dx
This was integration by parts, with u = sec x and v = tan x. In the integral on the
right, replace tan2x by sec2x - 1 (this identity is basic):
[ sec3x dx = sec x tan x - [ sec3x dx + [ sec x dx.
I
Bring sec3x dx to the left side. That reduces the problem from sec3x to sec x.
I believe those examples make the point-trigonometric integrals are computable.
Every product tanmxsecnx can be reduced to one of these examples. If n is even we
substitute u = tan x. If m is odd we set u = sec x. If m is even and n is odd, use a
reduction formula (and always use tan2x = sec2x- 1).
I mention very briefly a completely different substitution u = tan i x . This seems to
all students and instructors (quite correctly) to come out of the blue:
2u 1 - u2 2du
sin x = - and cos x = - and dx = - (11)
1 + u2 1 + u2 1 + u2'
The x-integral can involve sums as well as products-not only sinmxcosnx but also
+ f
1/(5 sin x - tan x). (No square roots.) The u-integral is a ratio o ordinary polynomi-
als. It is done by partial fractions.
Application o j sec x dx to distance on a map (Mercator projection)
f
The strange integral ln(sec x + tan x) has an everyday application. It measures the
distance from the equator to latitude x, on a Mercator map of the world.
All mapmakers face the impossibility of putting part of a sphere onto a flat page.
You can't preserve distances, when an orange peel is flattened. But angles can be
preserved, and Mercator found a way to do it. His map came before Newton and
Leibniz. Amazingly, and accidentally, somebody matched distances on the map with
a table of logarithms-and discovered sec x dx before calculus. You would not be
surprised to meet sin x, but who would recognize ln(sec x + tan x)?
The map starts with strips at all latitudes x. The heights are dx, the lengths are
proportional to cos x. We stretch the strips by l/cos x-then Figure 7 . 3 ~ lines up
evenly on the page. When dx is also divided by cos x, angles are preserved-a small
Trigonometric Integrals 293
A map width
Rdx Rdx
map width
Fig. 7.3 Strips at latitude x are scaled by sec x, making Greenland too large.
square becomes a bigger square. The distance north adds up the strip heights
I
dxlcos x. This gives sec x dx.
The distance to the North Pole is infinite! Close to the Pole, maps are stretched
totally out of shape. When sailors wanted to go from A to B at a constant angle with
the North Star, they looked on Mercator's map to find the angle.
7.2 EXERCISES
Read-through questions 10 Find sin2ax cos ax dx and sin ax cos ax dx.
To integrate sin4x cos3x, replace cos2x by a . Then
(sin4x- sin6x)cos x dx is b du. In terms of u = sin x the In 11-16 use the double-angle formulas (m, n even).
integral is c . This idea works for sinmxcosnx if either m 11 S",in2x dx 12 J",in4x dx
or n is d .
If both m and n are , one method is integration by
13 J cos23x dx 1
14 sin2x cos2x dx
f -I
. For sin4x dx, split off dv = sin x dx. Then v du is 15 sin2x dx + J cos2x dx 16 J sin2x cos22x dx
g . Replacing cos2x by h creates a new sin4x dx that
17 Use the reduction formula (7) to integrate cos6x.
combines with the original one. The result is a reduction to
1sin2x dx, which is known to equal I . 18 For n > 1 use formula (7) to prove
The second method uses the double-angle formula sin2x =
I . Then sin4x involves cos2 k . Another doubling
comes from cos22x = I . The integral contains the sine of
m .
19 For n = 2,4, 6, ... deduce from Problem 18 that
To integrate sin 6x cos 4x, rewrite it as isin lox + n .
The indefinite integral is 0 . The definite integral from
0 to 2 1 is
7 P . The product cos px cos qx is written as
4 cos (p + q)x + q . Its integral is also zero, except if 20 For n = 3, 5, 7, ... deduce from Problem 18 that
r when the answer is s .
With u = tan x, the integral of tangx sec2x is t . Simi-
larly J secgx (sec x tan x dx) = u . For the combination
tanmxsecnxwe apply the identity tan2x = v . After reduc-
tion we may need j tan x dx = w and J sec x dx = x . 21 (a) Separate dv = sin x dx from u = sinn- 'x and integrate
1sinnxdx by parts.
Compute 1-8 by the "general method," when m or n is odd. (b) Substitute 1 - sin2x for cosZx to find a reduction
formula like equation (7).
22 For which n does symmetry give J",osnx dx = O?
3 J sin x cos x dx 4 j cos5x dx
23 Are the integrals (a)-(f) positive, negative, or zero?
5 J sin5x cos2x dx 6 j sin3x cos3x dx
(a) J>os 3x sin 3x dx (b) j b o s x sin 2x dx
1 7
sin x cos x dx 8 1sin x cos3x dx
r :
(c) ! 2n cos x sin x dx (d) J (cos2x- sin2x)dx
J
9 Repeat Problem 6 starting with sin x cos x = $sin 2x. ;
(e) 5: cos px sin qx dx (f)
5 cos4x dx
"
294 7 Techniques of Integration
24 Write down equation (9) for p = q = 1, and (10) for p = 2, 45 j tan x sec3x dx 46 sec4x dx
q = 1. Derive (9) from the addition formulas for cos(s t) and+
cos(s - t) in Section 1.5.
49 1cot x dx 50 1csc x dx
In 25-32 compute the indefinite integrals first, then the definite
integrals.
25 jc cos x sin 2x dx 26 j",in 3x sin 5x dx
53 Choose A so that cos x - sin x = A cos(x ~14).
Then +
integrate l/(cos x - sin x).
29 :1 cos 99x cos lOlx dx 30 2
5 sin x sin 2x sin 3x dx 54 Choose A so that cos x - fi
sin x = A cos(x + n/3). Then
31 cos x/2 sin x/2 dx 32 j^, cos x dx (by parts)
x integrate l/(cos x - a
sin x)l.
33 Suppose a Fourier sine series A sin x B sin 2x + + 55 Evaluate lcos x - sin xl dx.
C sin 3x +adds up to x on the interval from 0 to n. Find
- 0 -
56 Show that a cos x + b sin x = cos (x - a) and
A by multiplying all those functions (including x) by sin x find the correct phase angle a.
and integrating from 0 to z. (B and C will disappear.)
57 If a square Mercator map shows 1000 miles at latitude
34 Suppose a Fourier sine series A sin x + B sin 2x + 30", how many miles does it show at latitude 60°?
C sin 3x + adds up to 1 on the interval from 0 to n. Find
C by multiplying all functions (including 1) by sin 3x 58 When lengths are scaled by sec x, area is scaled by
and integrating from 0 to a. (A and B will disappear.) . Why is the area from the equator to latitude x
proportional to tan x?
35 In 33, the series also equals x from -n to 0, because all
functions are odd. Sketch the "sawtooth function," which I
59 Use substitution (11) to find dx/(l + cos x).
equals x from -n to z and then has period 2n. What is the 60 Explain from areas why J^,sin2xdx = J: cos2xdx. These
sum of the sine series at x = n? "dx,
integrals add to I , so they both equal .
36 In 34, the series equals -1 from -n to 0, because sines 61 What product sin px sin qx is graphed below? Check
are odd functions. Sketch the "square wave," which is that (p cos px sin qx - q sin px cos qx)/(q2- p2) has this
alternately -1 and +1, and find A and B. derivative.
37 The area under y = sin x from 0 to n is positive. Which 62 Finish sec3x dx in Example 11. This is needed for the
frequencies p have 1; sin px dx = O? length of a parabola and a spiral (Problem 7.3.8 and
38 Which frequencies q have J; cos qx dx = O? Sections 8.2 and 9.3).
39 For which p, q is S", sin px cos qx dx = O?
40 Show that I",in px sin qx dx is always zero.
Compute the indefinite integrals 41-52.
41 sec x tan x dx 42 J tan 5x dx
43 1tan2x sec2x dx 44 1tan2x sec x dx
Trigonometric Substitutions
The most powerful tool we have, for integrating with pencil and paper and brain, is
the method of substitution. To make it work, we have to think of good substitutions-
which make the integral simpler. This section concentrates on the single most valu-
able collection of substitutions. They are the only ones you should memorize, and
two examples are given immediately.
7.3 Trigonometric Substitutions
To integrate J K i ,
(::
substitute x = sin 9. Do not set u = 1 - x2 - is missing
)
1J- dx -j (cos 0)(cos 0 40)
cos 0 d0
The expression J1 - x2 is awkward as a function of x. It becomes graceful as a
function of 8. We are practically invited to use the equation 1 - (sin 0)2 = (COS
Then the square root is simply cos 9-provided this cosine is positive.
Notice the change in dx. When x is sin 8, dx is cos 0 dO. Figure 7.4a shows the
original area with new letters. Figure 7.4b shows an equal area, after rewriting
j (COS B)(COS O dO) as 5 (cos2e)do. Changing from x to 8 gives a new height and a new
base. There is no change in area-that is the point of substitution.
,-
To put it bluntly: If we go from ,/ to cos 0, and forget the difference between
dx and dB, and just compute j cos 0 dB, the answer is totally wrong.
Fig. 7.4 Same area for Jl - x2 dx and cos28 dB. Third area is wrong: dx #dB
We still need the integral of cos20. This was Example 3 of integration by parts, and
also equation 7.2.6. It is worth memorizing. The example shows this 0 integral, and
returns to x:
EXAMPLE 1 5 cos20 dO = & sin O cos 8 + &O is after substitution
,-
/ dx = i x , , / m + 4 sin- 'x is the original problem.
We changed sin 0 back to x and cos O to -,
/. Notice that 0 is sin-'x. The answer
is trickier than you might expect for the area under a circular arc. Figure 7.5 shows
how the two pieces of the integral are the areas of a pie-shaped wedge and a triangle.
cos 0 d8
EXAMPLE 2 -0+C=sin-lx+C.
Remember: We already know sin-'x. Its derivative l/Jm was computed in
Section 4.4. That solves the example. But instead of matching this special problem
A
1
e area -8 1
= -sin-' x
1 2 2
y=dTZ?
10
I
area = ~ 1 2 I
area I x 4 - 7
2 J I
Fig. 7.5 Jm is a sum of simpler areas. Infinite graph but finite area.
dx
7 Techniques of Integration
with a memory from Chapter 4, the substitution x = sin 8 makes the solution auto-
matic. From 5 d8 = 8 we go back to sin-'x.
The rest of this section is about other substitutions. They are more complicated
than x = sin 8 (but closely related). A table will display the three main choices-sin 8,
tan 8, sec 8-and their uses.
U SIUI N
TRIGONOMETRIC S B TT TO S
After working with ,, -
/ the next step is -
/
,
. The change x = sin 8 simplified
the first, but it does nothing for the second: 4 - sin28 is not familiar. Nevertheless a
factor of 2 makes everything work. Instead of x = sin 8, the idea is to substitute x =
2 sin 8:
JF? JGGG = 2 cos 8 and dx = 2 cos 8 do.
=
Notice both 2's. The integral is 4 1cos28dB = 2 sin 8 cos 8 + 28. But watch closely.
This is not 4 times the previous 1cos28do! Since x is 2 sin 8, 8 is now sin- '(~12).
EXAMPLE 3 /
1,- dx = 4 1cos28d8 = x , / m + 2 sin- '(~12).
Based on ,/- -
/
and ,, here is the general rule for /
.
,
- Substitute
x = a sin 8. Then the a's separate out:
J ~ = , / ~ = a c o and
s ~ dx=acos8d8.
That is the automatic substitution to try, whenever the square root appears.
Here a2 = 16. Then a = 4 and x = 4 sin 8. The integral has 4 cos 8 above and below,
so it is 1dB. The antiderivative is just 8. For the definite integral notice that x = 4
means sin 8 = 1, and this means 8 = 7112.
A table of integrals would hide that substitution. The table only gives sin-'(~14).
There is no mention of 1d8 = 8. But what if 16 - x2 changes to x2 - 16?
1x=4 X
8
dx =?
EXAMPLE 5
,/F
Notice the two changes-the sign in the square root and the limits on x. Example 4
stayed inside the interval 1 1 < 4, where 16 - x2 has a square root. Example 5 stays
x
outside, where x2 - 16 has a square root. The new problem cannot use x = 4 sin 8,
because we don't want the square root of -cos28.
The new substitution is x = 4 sec 8. This turns the square root into 4 tan 8:
x = 4 sec 8 gives d x = 4 sec 8 tan 8 d8 and x2 - 16= 16sec28- 16= 16 tan2@.
This substitution solves the example, when the limits are changed to 8:
!:I3 4 sec 8 tan do -
4 tan 8
Jy3 sec8d8=ln(~ec8+tan8)]~~=ln(2+fi).
I want to emphasize the three steps. First came the substitution x = 4 sec 8. An
unrecognizable integral became sec 6 dB. Second came the new limits (8 = 0 when
13
x = 4, 8 = 7 1 when x = 8). Then I integrated sec 8.
7.3 Trigonometric Substitutions 297
Example 6 has the same x 2 - 16. So the substitution is again x = 4 sec 8:
r 16 dx fi,/2 64 sec 0 tan 0 dO i/2 cos 6 dO
EXAMPLE 6 = (x2 -- 16)3/2
8 0=,/3 (4 tan )3 /3 sin20
Step one substitutes x = 4 sec 0. Step two changes the limits to 0. The upper limit
x = oo becomes 0 = in/2, where the secant is infinite. The limit x = 8 again means 0 =
7r/3. To get a grip on the integral, I also changed to sines and cosines.
The integral of cos 6/sin20 needs another substitution! (Or else recognize
cot 0 csc 0.) With u = sin 0 we have f du/u 2 = - 1/u = - 1/sin 8:
rK/2 cos 6 dO -1 1n/ 2 2
Solution sin sin +
Jn/3 sin28 sin 8n/3 /
Warning With lower limit 0 = 0 (or x = 4) this integral would be a disaster. It divides
by sin 0, which is zero. This area is infinite.
2
(Warning) Example 5 also blew up at x = 4, but the area was not infinite. To make
the point directly, compare x-- 1/2 to x- 3/ 2 . Both blow up at x = 0, but the first one
has finite area:
dx=2 o 2 2 dx = = co.
Section 7.5 separates finite areas (slow growth of 1/ x) from infinite areas (fast
growth of x-3/2).
Last substitution Together with 16 - x 2 and x 2 - 16 comes the possibility 16 + x 2.
(You might ask about -16 - x2 , but for obvious reasons we don't take its square
root.) This third form 16 + x 2 requires a third substitution x = 4 tan 0. Then
16 + x 2 = 16 + 16 tan20 = 16 sec 20. Here is an example:
f dx f,/2 4 sec20 dO 1 /2 r
EXAMPLE 7 x=o 16 + x
2
0=o 16 sec 2 0 40 =8'
81
t
2
Table of substitutions for a - x', a2 + X2, x -
2
x = a sin 0 replaces a2 X2 by a2 cos 0 and dx by a cos 0 dO
x = a tan 0 replaces a2 + X 2 by a2 seC2O and dx by a sec20 dO
x= a sec 0 replaces x 2
-a 2
by a2 tan2
2 and dx by a sec 0 tan 0 dO
Note There is a subtle difference between changing x to sin 0 and changing sin 0 to u:
in Example 1, dx was replaced by cos 0 dO (new method)
in Example 6, cos 0 dO was already there and became du (old method).
The combination cos 0 dO was put into the first and pulled out of the second.
My point is that Chapter 5 needed du/dx inside the integral. Then (du/dx)dx
became du. Now it is not necessary to see so far ahead. We can try any substitution.
If it works, we win. In this section, x = sin 0 or sec 0 or tan 0 is bound to succeed.
dx_ xdx
d rdu
NEW = dO by trying x = tan OLD +x 2- u by seeing du
1+ X2I+X2 2u
7 Techniques of Integration
We mention the hyperbolic substitutions tanh 8, sinh 8, and cosh 8. The table below
shows their use. They give new forms for the same integrals. If you are familiar with
hyperbolic functions the new form might look simpler-as it does in Example 8.
x = a tanh8 replaces a2 - x2 by a2 sech28 and dx by a sech28 dB
x = a sinh8 replaces a2 + x2 by a2 cosh28 and dx by a cosh 8 d8
x = a cosh 8 replaces x2 - a 2 by a2 sinh28 and dx by a sinh 8 d8
EXAMPLE 8
I,/&=+ sinh 8 d8
sinh 0 = 8 C = cosh-'x + C.
dB is simple. The bad part is cosh- 'x at the end. Compare with x = sec 8:
sec 8 tan 8 d8
SJ&=j' tan0
= ln(sec 8 + tan 8) + C = ln(x + d m )+ C.
This way looks harder, but most tables prefer that final logarithm. It is clearer than
cosh-'x, even if it takes more space. All answers agree if Problem 35 is correct.
H
COMPLETING T E SQUARE
We have not said what to do for Jm -
or./, Those square roots
contain a linear term-a multiple of x. The device for removing linear terms is worth
knowing. It is called completing the square, and two examples will begin to explain it:
x2-2x+2=(x- 1=u2+ 1
The idea has three steps. First, get the x2 and x terms into one square. Here that
square was (x - 1)2= x2 - 2x + 1. Second, fix up the constant term. Here we recover
the original functions by adding 1. Third, set u = x - 1 to leave no linear term. Then
the integral goes forward based on the substitutions of this section:
The same idea applies to any quadratic that contains a linear term 2bx:
rewrite x2 + 2bx + c as (x + b)2 + C , with C = c - b2
rewrite - x2 + 2bx + c as - (x - b)2 + C , with C = c + b2
To match the quadratic with the square, we fix up the constant:
x2 + lox + 1 6 = (x + 5)2+ C leads to C = 16 - 25 = - 9
- x 2 + l o x + 1 6 = - ( x - 5)* + C leads to C = 1 6 + 2 5 = 4 1 .
EXAMPLE 9
Here u = x + 5 and du = dx. Now comes a choice-struggle on with u = 3 sec 0 or
look for du/(u2- a') inside the front cover. Then set a = 3:
Note If the quadratic starts with 5x2 or -5x2, factor out the 5 first:
+ 25 = 5(x2- 2x + 5) = (complete the square) = 5[(x - + 41.
5x2- lox
Now u = x - 1 produces 5[u2 + 41. This is ready for table lookup or u = 2 tan 8:
EXAMPLE 10
I dx -
5x2 - lox + 25 -
I du - 1 2 sec28d"
5[u2 + 41 - 5[4 sec28]
1
10
Id8,
+ '
This answer is 8/10 C. Now go backwards: 8/10 = (tan- f u)/lO = (tan- f(x - -))/lo. '
Nobody could see that from the start. A double substitution takes practice, from x
to u to 8. Then go backwards from 8 to u to x.
Final remark For u2 + aZ we substitute u = a tan 8. For u2 - a2 we substitute u =
a sec 8. This big dividing line depends on whether the constant C (after completing
the square) is positive or negative. We either have C = a* or C = - a2. The same
dividing line in the original x2 + 2bx + c is between c > b2 and c < b2. In between,
c = b2 yields the perfect square (x + b)'- and no trigonometric substitution at all.
7.3 EXERCISES
Read-through questions
The function ,-/ suggests the substitution x = a .
The square root becomes b and dx changes to c .
The integral j(1 - x2)3i2dxbecomes J d dB. The interval
3 < x < 1 changes to 8 f .
For ,/a2 - x2 the substitution is x = P with dx =
h . or x2 - a2 we use x = I with dx = 1 . Then
+
dx/(l x2) becomes j dB, because 1+ tan28 = k . The
answer is 8 = tan-'x. We already knew that I is the
derivative of tan- 'x.
+
The quadratic x2 2bx + c contains a m term 2bx. To
+
remove it we n the square. This gives (x b)2 + C with
+
C = 0 . The example x2 4x + 9 becomes P . Then
u = x + 2. In case x2 enters with a minus sign, -x2 + 4x + 9 (Important) This section started with x = sin 8 and
becomes ( q )2 + r . When the quadratic contains jd x / , / m =j dB = 8 = sin- 'JC.
4x2, start by factoring out s .
(a)Use x = cos 8 to get a different answer.
Integrate 1-20 by substitution. Change 8 back to x. (b) How can the same integral give two answers?
Compute I dx/x,/= with x = sec 0. Recompute with
x = csc 8. HOW can both answers be correct?
+
23 Integrate x/(x2 1) with x = tan 8, and also directly as a
logarithm. Show that the results agree.
24 Show that j d x / x , / a =f sec- '(x2).
Calculate the definite integrals 25-32.
&
I 8 j,-
/ dx (see 7.2.62) 25 Fa ,/- dx = area of
300 7 Techniques of Integration
+
Rewrite 43-48 as ( x + b)2 C or - ( x - b)2 + C by completing
the square.
30 1-1
-
xdx
x2+ 1
43 x 2 - 4 x + 8
45 x2 - 6x
44 - x 2 + 2 x + 8
46 - x 2 + 10
+ 2x + 1 + 4x - 12
32 jl:2Jm~ d x = area of .
47 x 2 48 x 2
49 For the three functions f ( x ) in Problems 43, 45, 47
33 Combine the integrals to prove the reduction formula integrate l / f ( x ) .
( n # 0): 50 For the three functions g(x) in Problems 44, 46, 48
d - j c d - x . integrate l / m .
n .x2+1 + +
51 For j dx/(x2 2bx c) why does the answer have different
Integrate l/cos x and 1 / ( 1 + cos x ) and J I + cos x. forms for b2 > c and b2 < c? What is the answer if b2 = c?
(a) x = i
gives d x / J x 2 - 1 = ln(sec 0 + tan 0). 52 What substitution u = x + b or u = x - b will remove the
(b)From the triangle, this answer is f = In(x + Jn). linear term?
Check that df/dx = l / J m - .
(c) Verify that coshf = i (ef + e - I ) = x. Thenf = cosh-'x,
the answer in Example 8.
(a) . =
u i
gives d x / , / x 2 + 1 = ln(sec B + tan 0).
(b)The second triangle converts this answer to g = ln(x + 53 Find the mistake. With x = sin 0 and J-x" = cos 8,
Jm). Check that dg/dx = l / J m . substituting dx = cos B dB changes
(c) Verify that sinh g = +(eg- e-g) = . so g = sinh- ' x .
u
(d)Substitute x = sinh g directly into i dx/,/+ and
integrate.
1
54 (a) If x = tan 0 then J m d x = 1 dB.
+ +
(b) Convert i[sec 0 tan 0 ln(sec 0 tan 0)] back to x.
(c) If x = sinh 0 then Jw dx = 1 dB.
+
(d)Convert i[sinh 0 cosh 0 01 back to x.
1 1 These answers agree. In Section 8.2 they will give the length
of a parabola. Compare with Problem 7.2.62.
37-42 substitute . = sinh 0. cosh 0. or tanh 0. After intee-
u
ration change back to x.
- 55 Rescale x and y in Figure 7.5b to produce the equal area
y dx in Figure 7 . 5 ~ What happens to y and what happens
.
37 1- dx
J'X - 1
dx to dx?
56 Draw y = l / J c 2 and y = l/J=
scale (1" across and up; 4" across and a" up).
to the same
57 What is wrong, if anything, with
7.4 Partial Fractions - 1
This section is about rational functions P(x)/Q(x).Sometimes their integrals are also
rational functions (ratios of polynomials). More often they are not. It is very common
for the integral of PIQ to involve logarithms. We meet logarithms immediately in the
7.4 Pattial Fractions
simple case l/(x - 2), whose integral is lnlx - 2 + C. We meet them again in a sum
1
of simple cases:
Our plan is to split PIQ into a sum like this-and integrate each piece.
Which rational function produced that particular sum? It was
This is PIQ. It is a ratio of polynomials, degree 1 over degree 3. The pieces of P are
+
collected into -4x 16. The common denominator (x - 2)(x + 2)(x) = x3 - 4x is Q.
But I kept these factors separate, for the following reason. When we start with PIQ,
and break it into a sum of pieces, thefirst things we need are the factors of Q.
In the standard problem PIQ is given. To integrate it, we break it up. The goal of
partial fractions is to find the pieces-to prepare for integration. That is the technique
to learn in this section, and we start right away with examples.
EXAMPLE 1 Suppose PIQ has the same Q but a different numerator P:
Notice the form of those pieces! They are the "partial fractions" that add to PIQ.
Each one is a constant divided by a factor of Q. We know the factors x - 2 and x + 2
and x. We don't know the constants A, B, C. In the previous case they were 1,3, - 4.
In this and other examples, there are two ways to find them.
Method 1(slow) Put the right side of (1) over the common denominator Q:
Why is A multiplied by (x + 2)(x)? Because canceling those factors will leave A/(x - 2)
+
as in equation (1). Similarly we have B/(x 2) and Clx. Choose the numbers A, B, C
so that the numerators match. As soon as they agree, the splitting is correct.
Method 2 (quicker) Multiply equation (1) by x - 2. That leaves a space:
Now set x = 2 and immediately you have A. The last two terms of (3) are zero, because
x - 2 is zero when x = 2. On the left side, x = 2 gives
Notice how multiplying by x - 2 produced a hole on the left side. Method 2 is the
"cover-up method." Cover up x - 2 and then substitute x = 2. The result is 3 =
A + 0 + 0, just what we wanted.
In Method 1, the numerators of equation (2) must agree. The factors that multiply
B and C are again zero at x = 2. That leads to the same A-but the cover-up method
avoids the unnecessary step of writing down equation (2).
302 7 Techniques of Integration
Calculation ofB Multiply equation (1) by x + 2, which covers up the (x + 2):
Now set x = - 2, so A and C are multiplied by zero:
This is almost full speed, but (4) was not needed. Just cover up in Q and give x the
right value (which makes the covered factor zero).
Calculation o C (quickest) In equation (I), cover up the factor (x) and set x = 0:
f
To repeat: The same result A = 3, B = - 1, C = 1 comes from Method 1.
EXAMPLE 2
First cover up (x - 1) on the left and set x = 1. Next cover up (x + 3) and set x = - 3:
The integral is tlnlx - 1 + ilnlx
1 + 31 + C.
EXAMPLE 3 This was needed for the logistic equation in Section 6.5:
1 A
-
~(~-by)-;
+-c -Bby'
First multiply by y. That covers up y in the first two terms and changes B to By.
Then set y = 0. The equation becomes l/c = A.
To find B, multiply by c - by. That covers up c - by in the outside terms. In the
middle, A times c - by will be zero at y = clb. That leaves B on the right equal to
l/y = blc on the left. Then A = llc and B = blc give the integral announced in
Equation 6.5.9:
f
It is time to admit that the general method o partial fractions can be very awkward.
First of all, it requires the factors of the denominator Q. When Q is a quadratic
ax2 + bx + c, we can find its roots and its factors. In theory a cubic or a quartic can
also be factored, but in practice only a few are possible-for example x4 - 1 is
(x2 - 1)(x2+ 1). Even for this good example, two of the roots are imaginary. We can
split x2 - 1 into (x + l)(x - 1). We cannot split x2 + 1 without introducing i.
The method of partial fractions can work directly with x2 + 1, as we now see.
EXAMPLE 4 dx (a quadratic over a quadratic).
This has another difficulty. The degree of P equals the degree of Q (= 2). Partial
7.4 Partial Fractions 303
jiactions cannot start until P has lower degree. Therefore I divide the leading term x2
into the leading term 3x2. That gives 3, which is separated off by itself:
Note how 3 really used 3x2 + 3 from the original numerator. That left 2x + 4. Partial
fractions will accept a linear factor 2x + 4 (or Ax + B, not just A) above a quadratic.
This example contains 2x/(x2 + I), which integrates to ln(x2 + 1). The final
4/(x2 + 1) integrates to 4 tan-'x. When the denominator is x2 + x + 1 we complete
the square before integrating. The point of Sections 7.2 and 7.3 was to make that
integration possible. This section gets the fraction ready-in parts.
The essential point is that we never have to go higher than quadratics. Every
denominator Q can be split into linear factors and quadratic factors. There is no magic
way to find those factors, and most examples begin by giving them. They go into
their own fractions, and they have their own numerators-which are the A and B
and 2x + 4 we have been computing.
The one remaining question is what to do if a factor is repeated. This happens in
Example 5.
EXAMPLE 5
The key is the new term B/(x - That is the right form to expect. With (x - l)(x - 2)
this term would have been B/(x - 2). But when (x - 1) is repeated, something new is
needed. To find B, multiply through by (x - and set x = 1:
2 x + 3 = A(x- 1)+ B becomes 5 = B when x = 1.
This cover-up method gives B. Then A = 2 is easy, and the integral is
1
2 lnlx - 1 - 5/(x - 1). The fraction 5/(x - 1)2 has an integral without logarithms.
EXAMPLE 6
This final example has almost everything! It is more of a game than a calculus
problem. In fact calculus doesn't enter until we integrate (and nothing is new there).
Before computing A, B, C, D, E, we write down the overall rules for partial fractions:
The degree of P must be less than the degree of Q. Otherwise divide their leading
terms as in equation (8) to lower the degree of P. Here 3 < 5.
Expect the fractions illustrated by Example 6. The linear factors x and x + 1
(and the repeated x2) are underneath constants. The quadratic x2 + 4 is under a
linear term. A repeated (x2 + 4)2 would be under a new Fx + G.
Find the numbers A, B, C, ... by any means, including cover-up.
Integrate each term separately and add.
We could prove that this method always works. It makes better sense to show that
it works once, in Example 6.
To find E, cover up (x - 1) on the left and substitute x = 1. Then E = 3.
To find B, cover up x2 on the left and set x = 0. Then B = 4/(0 + 4)(0 - 1) = - 1.
The cover-up method has done its job, and there are several ways to find A, C, D.
7 Techniques of Integration
Compare the numerators, after multiplying through by the common denominator Q:
The known terms on the right, from B = - 1 and E = 3, can move to the left:
We can divide through by x and x - 1, which checks that B and E were correct:
+ 4) + (Cx + D)x.
- 3x2 - 4 = A(x2
Finally x = 0 yields A = - 1. This leaves - 2x2 = (Cx + D)x. Then C = - 2 and
D=O.
You should never have to do such a problem! I never intend to do another one.
It completely depends on expecting the right form and matching the numerators.
They could also be matched by comparing coefficients of x4, x3, x2, x, 1-to give five
equations for A, B, C, D, E. That is an invitation to human error. Cover-up is the
way to start, and usually the way to finish. With repeated factors and quadratic
factors, match numerators at the end.
7.4 EXERCISES
Read-through questions Multiply by x - 1 and set x = 1. Multiply by x + 1 and set
x = - 1. Integrate. Then find A and B again by method 1-
The idea of a fractions is to express P(x)/Q(x)as a b
with numerator A(x + 1) + B(x - 1) equal to 1.
of simpler terms, each one easy to integrate. To begin, the
degree of P should be c the degree of Q. Then Q is split
Express the rational functions 3-16 as partial fractions:
into d factors like x - 5 (possibly repeated) and quadratic
factors like x2 + x + 1 (possibly repeated). The quadratic
factors have two e roots, and do not allow real linear
factors.
A factor like x - 5 contributes a fraction A/ f . Its
integral is g . To compute A, cover up h in the
denominator of P/Q. Then set x = i , and the rest of
P/Q becomes A. An equivalent method puts all fractions over
a common denominator (which is I ). Then match the
3x2 1
k . At the same point x = I this matching gives A. 9-x2+1 (divide first)
lo (x - 1)(x2+ 1)
A repeated linear factor (x - 5)2 contributes not only
A/(x - 5) but also B/ m . A quadratic factor like x2 + x + 1
+
contributes a fraction n /(x2 + x 1) involving C and D.
A repeated quadratic factor or a triple linear factor would 1 x2 + 1
bring in (Ex + F)/(x2+ x + or G/(x - 5)3. The conclusion 14 -(divide first)
l 3 X(X - 1)(x- 2)(x - 3) x+l
is that any PIQ can be split into partial o , which can
always be integrated.
1 Find the numbers A and B to split l/(.u2- x):
1
x (x- 1) (remember the
16 7
Cover up x and set x = 0 to find A. Cover up x - 1 and set 17 Apply Method 1 (matching numerators) to Example 3:
x = 1 to find B. Then integrate. 1
--- - A +-- B - A(c-by)+By
2 Find the numbers A and B to split l/(x2- 1): cy - by2 y c -by y(c - by) '
Match the numerators on the far left and far right. Why does
Ac = l? Why does - bA + B = O? What are A and B?
7.5 Improper Integrals 305
18 What goes wrong if we look for A and B so that By slibstitution change 21-28 to integrals of rational functions.
Problem 23 integrates l/sin 8 with no special trick.
Over a common denominator, try to match the numerators.
What to do first?
3x2 3x2 A Bx+C
23 IGa
sin 0 do
19 Split -- into -+-
x ~ - 1- (x-1)(x2+x+1) X-1 x2+x+l'
(a) Cover up x - 1 and set x = 1 to find A.
(b) Subtract A/(x - 1) from the left side. Find Bx + C.
(c) Integrate all terms. Why do we already know
29 Multiply this partial fraction by x - a. Then let x -+ a:
1
--- A
Q(x) - x - a
+ .*-.
20 Solve dyldt = 1- y2 by separating idyll - y2 = dt. Then
Show that A = l/Q'(a). When x = a is a double root this fails
because Q'(a) =
1 A
30 Find A in - -+ .-..Use Problem 29.
-
Integration gives 31n =t + C. With yo = 0 the con- x8-1 x-1
stant is C = . Taking exponentials gives . 31 (for instructors only) Which rational functions P/Qare the
The solution is y = . This is the S-curve. derivatives of other rational functions (no logarithms)?
1 . L
7.5 Improper Integrals1 - 1
"Zmp~oper" Jt
means that some part of y(x)dx becomes infinite. It might be b or a or
the function y. The region under the graph reaches infinitely far-to the right or left
or up or down. (Those come from b = oo and a = - oo and y + oo and y - - oo.) ,
Nevertheless the integral may "converge." Just because the region is infinite, it is not
automatic that the area is infinite. That is the point of this section-to decide when
improper integrals have proper answers.
The first examples show finite area when b = oo, then a = - m , then y = I/& at
x = 0.The areas in Figure 7.6 are 1, 1,2:
Fig. 7.6 The shaded areas are finite but the regions go to infinity.
306 7 Techniques of Integration
In practice we substitute the dangerous limits and watch what happens. When the
integral is -1/x, substituting b = oo gives "- 1/oo = 0." When the integral is ex,
substituting a = - oo gives "e-" = 0." I think that is fair, and I know it is successful.
But it is not completely precise.
The strict rules involve a limit. Calculus sneaks up on 1/oo and e-" just as it
sneaks up on 0/0. Instead of swallowing an infinite region all at once, the formal
definitions push out to the limit:
00b b b
DEFINITION y(x)dx = lim y(x)dx y(x)dx = lim y(x)dx.
a b f f - 0 a -
The conclusion is the same. The first examples converged to 1, 1, 2. Now come two
more examples going out to b = oo:
The area under 1/x is infinite: d= In x = co (1)
SX
The area under 1/xP is finite if p > 1:
"dx- x- -P 0 _-'
x, - 1 (2)
f XP 1- P p-1
The area under 1/x is like 1 + I + - + + -,which is also infinite. In fact the sum
approximates the integral-the curved area is close to the rectangular area. They go
together (slowly to infinity).
A larger p brings the graph more quickly to zero. Figure 7.7a shows a finite area
1/(p - 1)= 100. The region is still infinite, but we can cover it with strips cut out of
a square! The borderline for finite area is p = 1. I call it the borderline, but p = 1 is
strictly on the side of divergence.
The borderline is also p = 1 when the function climbs the y axis. At x = 0, the graph
of y = 1/x P goes to infinity. For p = 1, the area under 1/x is again infinite. But at x =
0 it is a small p (meaning p < 1) that produces finite area:
In
ox- =lnx 0o=0 ox- - 1-p0 -o=p ifp<l.
1 (3)
Loosely speaking "-In 0 = oo." Strictly speaking we integrate from the point x = a
f,
near zero, to get dx/x =- In a. As a approaches zero, the area shows itself as
infinite. For y = 1/x2 , which blows up faster, the area - 1/x]o is again infinite.
For y = 1/ x, the area from 0 to 1 is 2. In that case p = ½. For p = 99/100 the area
is 1/(1 - p) = 100. Approaching p = 1 the borderline in Figure 7.7 seems clear. But
that cutoff is not as sharp as it looks.
1 1 1
P
Fig. 7.7 Graphs of 1/x on both sides of p = 1. I drew the same curves!
7.5 Improper Integrals
Narrower borderline Under the graph of llx, the area is infinite. When we divide
,
by in x or (ln x ) ~the borderline is somewhere in between. One has infinite area (going
out to x = a ) , the other area is finite:
The first is dulu with u = In x. The logarithm of in x does eventually make it to
infinity. At x = 10l0, the logarithm is near 23 and ln(1n x) is near 3. That is slow!
Even slower is ln(ln(1n x)) in Problem 11. No function is exactly on the borderline.
The second integral in equation (4) is convergent (to 1). It is 1du/u2 with u = In x.
At first I wrote it with x going from zero to infinity. That gave an answer I couldn't
believe:
There must be a mistake, because we are integrating a positive function. The area
can't be zero. It is true that l/ln b goes to zero as b + oo. It is also true that l/ln a
goes to zero as a - 0. But there is another infinity in this integral. The trouble is at
,
x = 1, where In x is zero and the area is infinite.
E A P E 1 The factor e-" overrides any power xP (but only as x - a ) .
XML ,
Jr~ ' O e - ~ d x 50!
= but Jr~ - ' e - ~ d x oo.
=
It
The first integral is (50)(49)(48)--.(I). comes from fifty integrations by parts (not
recommended). Changing 50 to 3, the integral defines "i factorial." The product
*(- i)(-$).-- has no way to stop, but somehow i is ! *&.See Problem 28.
The integral ic xOe-"dx = 1 is the reason behind "zero factorial" = 1. That seems the
most surprising of all.
The area under e-"/x is (-I)! = oo. The factor e-" is absolutely no help at x = 0.
That is an example (the first of many) in which we do not know an antiderivative-
but still we get a decision. To integrate e -"/x we need a computer. But to decide that
an improper integral is infinite (in this case) or finite (in other cases), we rely on the
following comparison test:
7 6 (Corn-on test) Suppose that 0 < Nx) < v(x)..'Then the area under u(x)
i smaller than the area under Hx):
s
j'u(x)dx<ooif~u(x)dx<m iflu(x)dx=mthenjofx)dx=co.
Comparison can decide if the area is finite. We don't get the exact area, but we learn
about one function from the other. The trick is to construct a simple function (like
l/xP)which is on one side of the given function-and stays close to it:
XML
E A P E2 converges by comparison with [y $ = I.
E A P E3
XML diverges by comparison with
308 7 Techniques of Integration
EXAMPLE 4
ri dx
dx diverges by comparison with
dx
- = o.
2 fo 5x
Eo x + 4x
EXAMPLE 5 dx converges by comparison with dx = 1.
In Examples 2 and 5, the integral on the right is larger than the integral on the left.
Removing 4x and x/ increased the area. Therefore the integrals on the left are
somewhere between 0 and 1.
In Examples 3 and 4, we increased the denominators. The integrals on the right
are smaller, but still they diverge. So the integrals on the left diverge. The idea of
comparingfunctions is seen in the next examples and Figure 7.8.
EXAMPLE 6 e-xdx is below f 1 dx + e-xdx = 1 + 1.
e dxL ev dx
EXAMPLE 7 is above x In x .
J,Inx 1 x In x
EXAMPLE 8 x isbelow 'dx+ J' lo - 2 +2.
1 1
V= + ----
- I -. -11 -.
2 4-
area = o0 -
3-
red
1 2- - area
=4
- area = -oo
1-
I I
7
;X -_ ~-
1 2 e .2 .4 .6 .8
Fig. 7.8 Comparing u(x) to v(x): Se dx/ln x = oo and fo dx/lx- < 4. But oo - oo : 0.
There are two situations not yet mentioned, and both are quite common. The first is
an integral all the way from a = - oo to b = + oo. That is split into two parts, and
each part must converge. By definition, the limits at - 00 and + 00 are kept separate:
(o0 ('c 0 fb
f 0 y(x) dx = y(x) dx + y(x) dx = lim y(x) dx + lim y(x) dx.
The bell-shaped curve y = e- 2 covers a finite area (exactly i/).The region extends
to infinity in both directions, and the separate areas are •-. But notice:
0, x dx is not defined even though fb bx dx = 0 for every b.
The area under y = x is + oo00 one side of zero. The area is - oo00 the other side.
on on
We cannot accept oo - oo = 0. The two areas must be separately finite, and in this
case they are not.
7.5 Improper Integrals
EXAMPLE 9 l l x has balancing regions left and right of x = 0. Compute j?, d x / x .
This integral does not exist. There is no answer, even for the region in Figure 7 . 8 ~ .
(They are mirror images because l l x is an odd function.) You may feel that the
combined integral from -1 to 1 should be zero. Cauchy agreed with that-his
"principal value integral" is zero. But the rules say no: co - co is not zero.
7.5 EXERCISES
Read-through questions In 17-26, find a larger integral that converges or a smaller
integral that diverges.
:
An improper integral j y(x) dx has lower limit a = a or
upper limit b = b or y becomes c in the interval
a < x < b. The example jy dx/x3 is improper because d .
We should study the limit of j; dx/x3 as e . In practice
we work directly with - $x -2]y = f . For p > 1 the
improper integral g is finite. For p < 1 the improper
integral h is finite. For y = e-" the integral from 0 to co
is i .
Suppose 0 < u(x) < v(x) for all x. The convergence of i
implies the convergence of k . The divergence of
1 u(x) dx I the divergence of v(x) dx. From - co to co,
the integral of l/(ex+ e-") converges by comparison with
m . Strictly speaking we split (- co, co) into ( n , 0) and
(0, 0 ). Changing to l/(ex- e-") gives divergence, because
P . Also j'Cndxlsin x diverges by comparison with q .
The regions left and right of zero don't cancel because co - co 27 If p > 0, integrate by parts to show that
is r .
Decide convergence or divergence in 1-16. Compute the integ- The first integral is the definition of p! So the equation is p! =
rals that converge. . In particular O =
! . Another notation for
p! is T(p + 1)-using the gamma function emphasizes that p
need not be an integer.
28 Compute (- $)! by substituting x = u2:
;
1 x - 1 ' 2e - x dx = = & (known).
Then apply Problem 27 to find ($)!
29 Integrate ;
1 x2e-"dx by parts.
;
30 The beta function B(m. n) = 1 x m1 - x ) 'dx is finite
8 jYrn
sin x dx when m and n are greater than .
31 A perpetual annuity pays s dollars a year forever. With
9 n xx (by parts) 10:1 xe-.dx (by parts)
continuous interest rate c, its present value is yo = 1 se-"dt.
To receive $1000/year at c = lo%, you deposit yo =
;
.
32 In a perpetual annuity that pays once 2 year, the present
value is yo = sla + s/a2 + ... = . To receive
$1000/year at 10% (now a = 1.1) you again deposit yo =
. Infinite sums are like improper integrals.
33 The work to move a satellite (mass m) infinitely far from
"
the Earth (radius R, mass M ) is W= 1, GMm dx/x2. Evaluate
W What escape uelocity at liftoff gives an energy $mvi that
equals W?
310 7 Techniques of Integration
34 The escape velocity for a black hole exceeds the speed of *38 Compute any of these integrals found by geniuses:
light: v, > 3 lo8 m/sec. The Earth has GM = 4 *1014m3/sec2.
1 f it were compressed to radius R = , the Earth
would be a black hole.
35 Show how the area under y = 112" can be covered (draw
a graph) by rectangles of area 1 + 3 + $ + --- = 2. What is the
exact area from x = 0 to x = a? :1 xe-. cos x dx = 0 :1 cos x2dx = m.
36 Explain this paradox:
[ -- dx
S"-
-h 1 + x2 - 0 for every b but
1.
* -xdx
I x 2 diverges.
+
37 Compute the area between y = sec x and y = tan x for
39 For which p is
+
xp
- co?
x - ~
why the red area is 2, when
40 Explain from Figure 7 . 6 ~
0 < x < 7112. What is improper? Figure 7.6a has red area 1.
MIT OpenCourseWare
Resource: Calculus Online Textbook
Gilbert Strang
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