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Precalculus: Concepts Through Functions A Right Triangle - 2nd edition
Summary: Pre-calculus: Concepts Through Functions, A Right Triangle Approach to Trigonometry, Second Edition embodies Sullivan/Sullivan's hallmarks accuracy, precision, depth, strong student support, and abundant exercises while exposing readers to functions in the first chapter. To ensure that students master basic skills and develop the conceptual understanding they need for the course, this text focuses on the fundamentals: preparing for class, practicing their homework, a...show morend reviewing the concepts. After using this book, students will have a solid understanding of algebra and functions so that they are prepared for subsequent courses, such as finite mathematics, business mathematics, and engineering calculus. KEY TOPICS: Functions and Their Graphs; Linear and Quadratic Functions; Polynomial and Rational Functions; Exponential and Logarithmic Functions; Trigonometric Functions; Analytic Trigonometry; Applications of Trigonometric Functions; Polar Coordinates; Vectors; Analytic Geometry; Systems of Equations and Inequalities; Sequences; Induction; the Binomial Theorem; Counting and Probability; A Preview of Calculus: The Limit; Derivative, and Integral of a Function MARKET: For all readers interested in pre-calculus9268 +$3.99 s/h
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Math tool for high school math, middle school math teaching and studying. Function graphing and analyzing, sequence of number, analytic geometry and solid geometry. Math tool for high school math, middle school math teaching and studying. Function graphing and analyzing, sequence of number, analytic geometry and solid geometry.
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Complex Grapher Complex Grapher is a graphing calculator to create a graph of complex function. |
Introductory Algebra (11th Edition) (The Bittinger Worktext Series)
Book Description: The Bittinger Worktext Series changed the face of developmental education with the introduction of objective-based worktexts that presented math one concept at a time. This approach allowed readers to understand the rationale behind each concept before practicing the associated skills and then moving on to the next topic. With this revision, Marv Bittinger continues to focus on building success through conceptual understanding, while also supporting readers with quality applications, exercises, and new review and study materials to help them apply and retain their knowledge |
Mathematics Courses
120. Appreciation of Mathematics An exploration of topics which illustrate the power and beauty of mathematics, with a focus on the role mathematics has played in the development of Western culture. Topics differ by instructor but may include: Fibonacci numbers, mathematical logic, credit card security, or the butterfly effect. This course is designed for students who are not required to take statistics or calculus as part of their studies.
140. Statistics An introduction to statistical thinking and the analysis of data using such methods as graphical descriptions, correlation and regression, estimation, hypothesis testing, and statistical models.
160. Calculus for the Social Sciences A graphical, numerical and symbolic introduction to the theory and applications of derivatives and integrals of algebraic, exponential, and logarithmic functions, with an emphasis on applications in the social sciences. Note: A student may not receive credit for both MATH 160 and MATH 181.
181. Calculus 1 A graphical, numerical, and symbolic study of the theory and applications of the derivative of algebraic, trigonometric, exponential, and logarithmic functions, and an introduction to the theory and applications of the integral. Suitable for students of both the natural and the social sciences. Note: A student may not receive credit for both MATH 160 and MATH 181.
182. Calculus 2 A graphical, numerical, and symbolic study of the theory, techniques, and applications of integration, and an introduction to infinite series and/or differential equations. Prerequisite: MATH 181 or permission of instructor.
201. Modeling and Simulation for the Sciences A course in scientific programming, part of the interdisciplinary field of computational science. Large, open-ended, scientific problems often require the algorithms and techniques of discrete and continuous computational modeling and Monte Carlo simulation. Students learn fundamental concepts and implementation of algorithms in various scientific programming environments. Throughout, applications in the sciences are emphasized. Cross-listed as COSC 201. Prerequisites: MATH 181 or permission of instructor.
210. Multivariable Calculus A study of the geometry of three-dimensional space and the calculus of functions of several variables. Prerequisite: MATH 182.
212. Vector Calculus A study of vectors and the calculus of vector fields, highlighting applications relevant to engineering such as fluid dynamics and electrostatics. Prerequisite: MATH 182.
235. Discrete Mathematical Models An introduction to some of the important models, techniques, and modes of reasoning of non-calculus mathematics. Emphasis on graph theory and combinatorics. Applications to computing, statistics, operations research, and the physical and behavioral sciences. Prerequisite: MATH 182.
240. Differential Equations The theory and application of first- and second-order differential equations including both analytical and numerical techniques. Prerequisite: MATH 182.
250. Introduction to Technical Writing An introduction to technical writing in mathematics and the sciences with the markup language LaTeX, which is used to typeset mathematical and scientific papers, especially those with significant symbolic content.
260. Introduction to Mathematical Proof An introduction to rigorous mathematical argument with an emphasis on the writing of clear, concise mathematical proofs. Topics will include logic, sets, relations, functions, and mathematical induction. Additional topics may be chosen by the instructor. Prerequisite: MATH 182.
280. Selected Topics in Mathematics Selected topics in mathematics at the introductory or intermediate level.
310. History of Mathematics A survey of the history and development of mathematics from antiquity to the twentieth century. Prerequisite: Math 260.
410. Geometry A study of the foundations of Euclidean geometry with emphasis on the role of the parallel postulate. An introduction to non-Euclidean (hyperbolic) geometry and its intellectual implications. Prerequisite: MATH 260.
421-422. Probability and Statistics A study of probability models, random variables, estimation, hypothesis testing, and linear models, with applications to problems in the physical and social sciences. Prerequisite: MATH 210 and 260.
424. Advanced Game Theory This advanced class is intended to provide a more rigorous introduction to the main concepts and techniques of the field. These techniques will be used to investigate relevant social phenomena, such as evolutionary games, auction theory, the "prisoner's dilemma," the "tragedy of the commons," tacit collusion, competition among firms, and strategic interactions in labor, credit, and product markets. The most important classes of games will be analyzed (zero-sum games, cooperation problems, coordination games, bayesian games, signaling games, etc.), as well as the most important solution concepts (rationalizability, nash equilibrium in pure and mixed strategies, bayesian nash equilibrium, and evolutionarily stable strategies). This course also will introduce students to the main techniques of game-theoretic mathematical modeling. Cross-listed with ECO 424. Prerequisite: MATH 210.
435. Cryptology An introduction to cryptology and modern applications. Students will study various historical and modern ciphers and implement select schemes using mathematical software. Cross-listed with COSC 435. Prerequisites: MATH 220 and either MATH 235 or 260.
439. Elementary Number Theory A study of the oldest branch of mathematics, this course focuses on mathematical properties of the integers and prime numbers. Topics include divisibility, congruences, diophantine equations, arithmetic functions, primitive roots, and quadratic residues. Prerequisite: MATH 260.
441-442. Mathematical Analysis A rigorous study of the fundamental concepts of analysis, including limits, continuity, the derivative, the Riemann integral, and sequences and series. Prerequisites: MATH 210 and 260.
445. Advanced Differential Equations This course is a continuation of a first course on differential equations. It will extend previous concepts to higher dimensions and include a geometric perspective. Topics will include linear systems of equations, bifurcations, chaos theory, and partial differential equations. Prerequisite: MATH 240.
448. Functions of a Complex Variable An introduction to the analysis of functions of a complex variable. Topics will include differentiation, contour integration, power series, Laurent series, and applications. Prerequisite: MATH 260. |
was performed using a convenience sample of 90 students at a northeastern community college to determine gender differences of math anxiety and its effect on math avoidance. Four sections of an introductory English class were given a... |
Completely agree with rbednarski. We tend to reduce Algebra to a bunch of
processes and mechanisms that students repeat without thinking why they work.
Algebra is a language. A powerful one, a universal one, and independent of the
language you speak at home. That is part of its beauty.
Any language will have some rules, some "grammar" that everybody will respect in
order to be able to communicate to each other. That's all. I like the idea of
four categories. And I can't stand FOIL. There is not any operation or property
in Math called FOIL, as you know. And the order that you choose to multiply out
the terms of two binomials or polynomials in general doesn't matter. Remember
that multiplication in the set of the real numbers is commutative. All you need
to do is make sure that you multiply every term in one with every term in the
other one. So you need to understand what "term" means. Which takes you back to
the basic grammar rules of Algebra. So you go back and think about it. Terms in
Algebra are like cells in biology, or moleculae in chemistry, or words in any
language. So you start to make deeper connections between ideas, that go beyond
PEMDAS or FOIL or SOHCAHTOA or ...
Plus, when we use those stupid letters, we forget that they only make some kind
of sense in English (SOHCAHTOA also works in Spanish). I'm a bilingual Math
teacher. The universal concept is the distributive property, not the FOIL. We
should stop with all the nonsense. Think about it. If students need some kind of
mnemonic rule, they should develop their own, don't you agree? |
s a free textbook offered by BookBoon.'This book is a guide through a playlist of Calculus instructional videos. The...
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This sThe author states, "'A=B' is about identities in general, and hypergeometric identities in particular, with emphasis on...
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The author states, "'A=B' is about identities in general, and hypergeometric identities in particular, with emphasis on computer methods of discovery and proof. The book describes a number of algorithms for doing these tasks, and we intend to maintain the latest versions of the programs that carry out these algorithms on this page. So be sure to consult this page from time to time, and help yourself to the latest versions of the programs.״The entire book is available for purchase, but can also be downloaded for free at this site. This website also included some reviews of the book.
According to the author, "We study fundamental algebraic structures, namely groups, rings, fields and modules, and maps...
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According to the author, "We study fundamental algebraic structures, namely groups, rings, fields and modules, and maps between these structures. The techniques are used in many areas of mathematics, and there are applications to physics, engineering and computer science as well. In addition, I have attempted to communicate the intrinsic beauty of the subject. Ideally, the reasoning underlying each step of a proof should be completely clear, but the overall argument should be as brief as possible, allowing a sharp overview of the result.״Each chapter of the book is downloadable as a separate pdf file.
״Abstract Algebra: Theory and Applications is an open-source textbook written by Tom Judson that is designed to teach the...
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״Abstract Algebra: Theory and Applications is an open-source textbook written by Tom Judson that is designed to teach the principles and theory of abstract algebra to college juniors and seniors in a rigorous manner. Its strengths include a wide range of exercises, both computational and theoretical, plus many nontrivial applications. The first half of the book presents group theory, through the Sylow theorems, with enough material for a semester-long course. The second-half is suitable for a second semester and presents rings, integral domains, Boolean algebras, vector spaces, and fields, concluding with Galois Theory.״
'Rather than detailed explanations and worked out examples, this book uses activities intended to be done by the students in...
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'Rather than detailed explanations and worked out examples, this book uses activities intended to be done by the students in order to present the standard concepts and computational techniques of calculus. The student activities provide most of the material to be assigned as homework, but since the book does not contain the usual routine exercises, instructors wanting such exercises will need to supply their own or use a homework system such as WebWork. With this approach Active Calculusmakes it possible to teach an inquiry based learning course without severely restricting the material covered. Although this book is new, it has been class tested by the author and his colleagues both at their university and elsewhere.From the preface:Where many texts present a general theory of calculus followed by substantial collections of worked examples, we instead pose problems or situations, consider possibilities, and then ask students to investigate and explore. Following key activities or examples, the presentation normally includes some overall perspective and a brief synopsis of general trends or properties, followed by formal statements of rules or theorems. While we often offer a plausibility argument for such results, rarely do we include formal proofs.'
This is a free textbook offered by InTech.'Adaptive filtering can be used to characterize unknown systems in time-variant...
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This is a free textbook offered by InTech.'Adaptive filtering can be used to characterize unknown systems in time-variant environments. The main objective of this approach is to meet a difficult comprise: maximum convergence speed with maximum accuracy. Each application requires a certain approach which determines the filter structure, the cost function to minimize the estimation error, the adaptive algorithm, and other parameters; and each selection involves certain cost in computational terms, that in any case should consume less time than the time required by the application working in real-time. Theory and application are not, therefore, isolated entities but an imbricated whole that requires a holistic vision. This book collects some theoretical approaches and practical applications in different areas that support expanding of adaptive systems.' |
Calculus with analytic geometry
Book summary
Designed for the three-semester course for math and science majors, the Larson/Hostetler/Edwards series continues its tradition of success by being the first to offer both an Early Transcendental version as well as a new Calculus with Precalculus text. This was also the first calculus text to use computer-generated graphics (Third Edition), to include exercises involving the use of computers and graphing calculators (Fourth Edition), to be available in an interactive CD-ROM format (Fifth Edition), and to be offered as a complete, online calculus course (Sixth Edition). Every edition of the book has made the mastery of traditional calculus skills a priority, while embracing the best features of new technology and, when appropriate, calculus reform ideas. The Seventh Edition also expands its support package with an all-new set of text-specific videos.
P.S. Problem-Solving Sections, an additional set of thought-provoking exercises added to the end of each chapter, require students to use a variety of problem-solving skills and provide a challenging arena for students to work with calculus concepts.
Getting at the Concept Exercises added to each section exercise set check students' understanding of the basic concepts. Located midway through the exercise set, they are both boxed and titled for easy reference.
Review Exercises at the end of each chapter have been reorganized to provide students with a more effective study tool. The exercises are now grouped and correlated by text section, enabling students to target concepts requiring review.
The icon "IC" in the text identifies examples that appear in the Interactive Calculus 3.0 CD-ROM and Internet Calculus 2.0 web site with enhanced opportunities for exploration and visualization using the program itself and/or a Computer Algebra System.
Think About It conceptual exercises require students to use their critical-thinking skills and help them develop an intuitive understanding of the underlying theory of the calculus.
Modeling Data multi-part questions ask students to find and interpret mathematical models to fit real-life data, often through the use of a graphing utility.
Section Projects, extended applications that appear at the end of selected exercise sets. may be used for individual, collaborative, or peer-assisted assignments.
True or False? Exercises, included toward the end of many exercises sets, help students understand the logical structure of calculus and highlight concepts, common errors, and the correct statements of definitions and theorems.
Motivating the Chapter sections opening each chapter present data-driven applications that explore the concepts to be covered in the context of a real-world setting.
Hardcover, ISBN 0669095699 Publisher: D.C. Heath, 1986 Great condition for a used book! Minimal wear. Find out why millions of customers rave about Better World Books. Experience the best customer care and a 100% satisfaction guarantee.
Hardcover, ISBN 0669095699 Publisher: D.C. Heath, 1986 Shows some signs of wear, and may have some markings on the inside. Find out why millions of customers rave about Better World Books. Experience the best customer care and a 100% satisfaction guarantee. |
The two-line display scientific calculator combines statistics and advanced scientific functions and is a durable and affordable calculator for the classroom. The two-line display helps students explore math and science concepts in the classroom |
In this book we generate graphic images using the software Mathematica thus providing a gentle and enjoyable introduction to this rather technical software and its graphic capabilities. The programs we use for generating these graphics are easily adaptable to many variations. These graphic images are enhanced by introducing a variety of different... more...
Presents an introduction to MuPAD - a modern Computer Algebra System. This book shows how we can use it in various areas of mathematics. It devotes a chapter to the graphical visualization of mathematical concepts. It is a resource for conducting workshops on using Computer Algebra Systems to explore and visualize mathematical concepts. more...
Introduces the reader to Mathematica's various approximate numbers, their arithmetic and the common numerical analysis operations such as numerical integration, root-finding, equation solving, minimization, and differential equation solving. This resource is useful for practitioners, professionals, and researchers. more...
The presentation of this book is on the comprehensible application of techniques for the approximation of the mathematical problems that are frequently observed in physical sciences, engineering technology and mathematical physics. The acceptance of the technique for the solution has been justified from mathematical point of view. The Software required... more...
Need to learn MATHEMATICA? Problem SOLVED! Take full advantage of all the powerful capabilities of Mathematica with help from this hands-on guide. Filled with examples and step-by-step explanations, Mathematica Demystified takes you from your very first calculation all the way to plotting complex fractals. Using an intuitive format, this book... more...
This multi-author contributed proceedings volume contains recent advances in several areas of Computational and Applied Mathematics. Each review is written by well known leaders of Computational and Applied Mathematics. The book gives a comprehensive account of a variety of topics including - Efficient Global Methods for the Numerical Solution of Nonlinear... more...
Because of its large command structure and intricate syntax, Mathematica can be difficult to learn. Wolfram's Mathematica manual, while certainly comprehensive, is so large and complex that when trying to learn the software from scratch -- or find answers to specific questions -- one can be quickly overwhelmed. A Beginner's Guide to Mathematica offers... more...
Computing with Mathematica, 2nd edition is engaging and interactive. It is designed to teach readers how to use Mathematica efficiently for solving problems arising in fields such as mathematics, computer science, physics, and engineering. The text moves from simple to complex, often following a specific example on a number of different levels. This... more... |
bravo!
Bravo is a comprehensive calculator utility for Windows 95. It resembles a standard hand-held calculator and contains virtually all the functions you could possibly want in a calculator. It includes all the normal mathematic and trigonometric, as well as a built-in calendar. You can convert any number between 120 different measurement units grouped in 12 categories, compute all the values of a triangle (sides, angles, area, height) from any three known values, do vectorial and complex calculations, and more. Bravo is even multilingual and can be set to display in English, Italian, German, or Spanish.
Unit Conversion Tools Evaluation (Palm) A must have for every Engineer! A specialty calculator for Palm. Forget having to remember complicated formulae, just enter the values and see the solution!
Linear Systems 1.02 Linear Systems gives a complete, step-by-step solution of the following problem: Given a 2x2 linear system (two equations, two variables) or 2x3, or 3x2, or 3x3, or 3x4, or 4x3, or 4x4 linear system.
Random Number Generator PPC Random Number Generator PPC is a Windows Mobile based software that produces sequences of random numbers. Program can exclude digits and numbers from generated random sequences. User can save results to a text file.
StatCalc Download Statistics calculator and instructional aid for students and researchers.
Introduction to Integers A program deals with numbers, specificAlly integers. It includes the basic concepts of integers. Order of integers in its arrangement. Addition and subtraction of integers. Mixed operation. Multiplication and division of integrs. |
Secondary Mathematics and Additional Mathematics
Why are the books for grades 7-10 labeled 1-4?
In Singapore, students attend Primary school for 6 years, then Secondary school usually for 4 years, and then can attend junior colleges for 2-3 years. Grades 7-11 in the U.S. are roughly equivalent to Secondary 1-4 in Singapore.
Where does Additional Mathematics fit in? In Singapore, some students do a second math class in Secondary 3 and 4 called Additional Mathematics. The books we sell with Additional Mathematics in the title are used for these classes over a 2-year period. Since these books cover pre-calculus and calculus topics, they could be used after the Secondary 1-4 books.
How do these books correlate with the math sequence in the U.S.? Which is pre-algebra, algebra 1, geometry and so on?
These books have an integrated approach. They progress sequentially through various topics in algebra, geometry, and some trigonometry, with minimal review of earlier levels. Please see the
Scope and sequence. As a rough comparison:
What is the new Dimensions Math (Discovering Mathematics) Common Core Series? How does it compare? This series for 7th and 8th grade is a revision of the original Dimensions Math (Discovering Mathematics) series in order to include the Common Core Standards. A few topics were added to some chapters, and some topics were rearranged between the levels. There is some U.S. customary measurement, but metric measurement still predominates. Terms have been changed to reflect the common use in the U.S., such as gradient to slope and indices to exponents. American English spelling is used. Otherwise, it is the same as the original series. No topics were removed. It therefore follows the Singapore Mathematics Framework as well as covers the Common Core Standards.
Will the new Dimensions Math (Discovering Mathematics) Common Core Series be less advanced or rigorous than the original in order to accommodate the Common Core Standards? No, it will not. The two levels will together still include all the topics in the original. The 8th grade will in fact include a few topics from Discovering Mathematics 3 in order to cover all topics considered to be Algebra 1, such as the quadratic formula and fractional exponents (radicals). It will still include topics such as factorization of algebraic expressions and solving quadratic equations that are not part of the Common Core Standards for 8th grade but are part of the Singapore syllabus at that level.
Will there still be a teacher's edition for the new Dimensions Math (Discovering Mathematics Common Core Standard Series that includes worked out solutions? Yes, there will. It will be similar to the current one, with some notes to teachers and worked out solutions for all the problems.
How does the format of the secondary level books compare to Primary Mathematics? How do the workbooks correlate with the textbooks? What is the structure of a lesson in the textbook and how long does it take? The textbooks contain both the lesson and the practice, instead of having the practice in a separate workbook. The workbook is now a supplementary, non-consumable book with additional problems that can be done after each chapter. The lessons include some activities that are similar to the learning tasks in Primary Mathematics, but there are also more worked examples with solutions provided and more written explanations. Some lessons are meant to take several days, with exercises only at the end of the lesson.
Do you have a scope and sequence for these books? Yes. Please click here for
Scope and Sequence.
Do you have a list of contents? Sample pages? Yes. In the list of products, click on the picture of the product or on the words 'more info'. Then click on the tab that says Contents_Sample and scroll down. There are links to sample pages for each book.
Are there teacher's guides available? Tests? Answer keys? Solutions? All secondary level textbooks and workbooks have answers to most, but not all, of the problems at the back. Depending on the series, there are various teaching resources available with additional answers or solutions. See the chart below for comparison between the series. However, none of the teacher guides have detailed, already prepared daily lesson plans.
Which series should I use? It depends on your preference. New Elementary Mathematics is an older series and less modern in its format. Discovering Mathematics has fully worked solutions for all levels including the workbook problems.
Can I switch between series? You can switch between New Elementary Mathematics to Discovering Mathematics between years. You can also switch after the Dimensions Math Common Core Series 8th grade to the original Discovering Mathematics 3. There will be some repetition of two topics (fractional exponents and quadratic formula).
If my student is not able to easily answer all the problems or get most of the problems on reviews or tests correct, does that mean the program does not teach the material well? The problems have a range of difficulty level to allow all students to work to their maximum potential. If all the problems were easy, then the student is not working to his or her full potential. Some of the problems in the exercises are simply practice, but some are truly problems that allow a student to gain more depth of knowledge by reasoning through them and applying concepts in new ways. Also, in Singapore most students are expected to score between 50% and 75% on tests. Only better students will be able to score above 75%. If your student can answer all the problems easily, he or she would score well on such a test, but not being able to answer every problem easily does not necessarily indicate lack of understanding of the concepts. The grading scale in Singapore for the secondary level is:
What if I have more questions or need more help with the content or choosing which books to get? Please visit our
forums and post your questions or concerns. You can also email the curriculum advisor.
It says in the chart below that some activities in Discovering Mathematics use Geometer's Sketchpad. What is that and where can I get it? Geometer's Sketchpad is a dynamic geometry software program produced and sold by
Key Curriculum Press. (We are not associated with Key Curriculum Press and we are not responsible for any purchases made with this company.) You can use other dynamic software programs for the same activities, such as Geogebra but the steps will be different.
New Elementary Mathematics (NEM)
Discovering Mathematics original (DM)
Dimensions Math (DMCC)
Publication
date
First published 1991. New edition in 1996. This is based on the syllabus used prior to 2001 and is one of the texts used by students taking the TIMSS international test. This series is no longer used in Singapore
First published in 2008.
First published 2012. Based on the original Discovering Mathematics series.
Textbook organization
Textbook pages are black and white. Lessons in the textbook consist of explanations, worked examples, and occasional class activities which allow students to learn through discovery.
At the end of each chapter there is a summary listing the concepts learned in the chapter, but not a review exercise. Each chapter is followed by an optional "Challenger" and a "Problem Solving" exercise.
After 3-4 chapters there is a set of 5 short cumulative review exercises, and an optional Miscellaneous Exercise and Investigation for exploring some concepts in more depth. There are two practice cumulative assessments at the end of each textbook.
Textbook pages are more visually attractive than NEM and in color. There are two textbooks for each level, A and B. Lessons in the textbook consist of explanations and worked examples. Each worked example is followed by a similar question (Try It!) that students can do to see if they understood the example.
Many lessons include one or more class activities, which allow students to learn through discovery. Some of these use Geometerís Sketchpad. Some lessons are meant to take several days. Each chapter is followed by a summary listing the major concepts, a review exercise, an open-ended Extend Your Learning Curve problem, and one or two questions for journal writing.
At the end of each textbook is a list of problem solving heuristics with some examples corresponding to the topics for that level. There are no cumulative review exercises or assessments, but the chapter reviews tend to integrate concepts from earlier chapters.
See the entry under DM.
Exercise organization
Exercises vary in length and tend to be long so it is possible to select problems rather than do all of them. The type of problems depends on the lesson; some exercises are all computation or skill problems and some are all word problems. More challenging problems are marked with an asterisk.
The problems in each lesson's exercise are divided by difficulty level into Basic Practice, Further Practice, Maths@Work (application) and Brainworks (challenge). The chapter review exercises are not divided by difficulty level.
See the entry under DM.
Answers to the textbook problems.
There are answers to the regular exercises, the periodic Miscellaneous and Review exercises and the Assessments in the back of the textbook.
There are answers to the Try It! problems in the lesson, the lesson exercises except for Brainworks section, and the reviews at the back of the textbook.
See the entry under DM.
Teacher resources and fully worked solutions
The Teacherís Manual has a weekly schedule and fully worked solutions for the Challengers, Problem Solving, and Investigations. The solutions manual has fully worked solutions to the rest of the exercises. Teacher's Manual (NEMTM2) for grade 8 is no longer available.
The Teacherís Guide for each level has a weekly schedule, brief notes for the teacher, and fully worked solutions to all problems in the textbook, including class activities.
The Teaching Notes and Solutions for each level has a weekly schedule, brief notes for the teacher, and fully worked solutions to all problems in the textbook, including class activities.
Workbook
The workbook is supplementary and contains additional problems for each chapter, practice test papers every 2 chapter and 2 mid-term and final term assessments. Only the test papers have room in the book to work the problems. Answers are in the back. There are no fully worked solutions.
There is one supplementary workbook for each level. It has a set of problems for each chapter divided into Basic Practice, Further Practice, Challenging Practice, and Enrichment (even more challenging). Answers are in the back. There is a separate Teacher's Edition of the workbook with fully worked solutions.
See the entry under DM. There is a separate Workbook Solutions with fully worked solutions.
Tests
There are two sample tests at the end of each textbook and practice test papers for every 2 chapters, midterms, and semester test in the workbook. These are quite challenging, as the workbook is somewhat for enrichment.
The workbook has mid-year and end-of year sample tests. There is a separate Question Bank book for each level with test questions arranged by difficulty level. Questions will have to be selected, copied, and transferred to a test paper, since answers are included on each page in the Question Bank. |
revised, the Second Edition of the Algebra Survival Guide unleashes its power for a new generation of students. Now that the Common Core Standards have changed how math is taught, this 2nd edition aligns its content to these broad new guidelines. The new Edition also adds advanced content. In its XTREME ALGEBRA section, the new edition tackles the topics of Functions, Inequalities and the Advanced Coordinate Plane, and it teaches story problems in all three areas. These additions update the book for today's elementary and secondary students; they also provide additional support for adults taking algebra in their return to college. Plus, with its newly expanded index and glossary, the 2nd Edition makes all of its content easy to find. The book retains the cartoons, analogies and conversational format that brought out praise from all corners and garnered the book both a Parents Choice Commendation and a Golden Porch Award for pedagogical excellence.
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Sally Blakemore was born an artist and eccentric who genuinely loves people, animals, books, life, art, music and travel. Arty Projects Studio, Santa Fe, Ltd., is her award-winning book-packaging company which has produced books for Little Simon, Dutton, HarperCollins, Disney Press, Viking, Getty Press, and the Monterrey Bay Aquarium, to name just a few.
Most Helpful Customer Reviews
I purchased this book last spring in order to prepare myself to take college algebra this fall. In high school I failed algebra and I was beyond confused or befuddled. I put off going to college for 10 YEARS simply because I didn't want to do the math. I was dreading the idea of having to spend the summer studying it. I've always had a fear of math and algebra, and over time I began to hate it. Now when I say I hate it, the term hate may not be enough; my loathing was beyond measure; but from working through this book something amazing happened; I don't hate or fear algebra anymore! Seriously!!! And I understand it! And not only do I understand it; I enjoy it! If the idea of enjoying algebra makes little or no sense to you, if you have spent your time and your money struggling through books or classes that are boring, uninformative, over your head, and/or just plain stupid, then really, this book is for you! Please don't pass this book up; it will change your life. And if you could understand how much I hated algebra, you would understand how good this book really is.
This book is excellent. I took all the math offered in high school and college algebra (20 years ago!) My 7th grade
daughter is struggling with pre-algebra and I couldn't remember a thing to help her. I ordered the Survival Guide and workbook and they are both fun! My 12 year old works out of this book on her own. It really is self-teaching. Each section takes you step by step through the concepts explaining in simple terms how to do the problems. I am learning all the Algebra concepts and love the way the book is laid out. Very direct, simple, engaging & great to look at visually. I am an artist by profession and give you kudos on the graphics. I also substitute teach in our public schools and this is a great book to learn how to teach basic algebra. I work with a lot of special needs students and knowing how to explain the algebra in fun and practical ways is a blessing. I recommend this book to teachers, parents and students.
Let me start by saying that in High School, I felt like a mathmatical retard. I failed math horribly. I never got the basics. It didn't help that I was a bit lazy, but nevermind that! At age 30, and a liberal arts degree later, I decided it was time to learn Algebra properly, once and for all. This book, and the workbook that goes with it was a balanced and enlightening start that truly was painless. I have just about finished the book and will be moving on to harder algebra. I would HIGHLY recommend this book to anyone looking to simply the fog that surrounds the word "ALGEBRA".
I have been shopping for a good tutorial algebra book for several weeks so that I could help my struggling teenagers. I probably would not have shopped this book because the cover gave me the impression of chaos. I feared that the pages would be full of text that would baffle us even more. I bought this book and workbook based on the customer's reviews...and am I happy I did so. I learned not to judge a book by its cover. The book layout is easy to read with one problem per page in step-by-step detail. This is exactly what I've been seeking. Too bad Rappaport does not have a book on Geometry...that's our next level.
ps. The only reason I did not rate this book a 5 is because the contents covers ten fundamentals whereas other books cover a bit more. But the layout is worth the compromise.
I want to talk to the grownups who, like me, never got the gist of Algebra, and certainly never passed an Algebra course (Come on, soul mates, go back with me in time: remember how we sat at our desks, sweating, fighting off that sick "I know I'm not going to pass this test" stomach).
Well, fear no more--we can now emerge from the darkness that was ignorance of Algebra. We now have hope. we can now live the consummate human life, because there's a new sheriff who will help us defeat the bullies in Algebra Town (You know the ones, my friends: they sit, confidently, smugly even, figuring out how those numbers become
algebra.). The sheriff is named Josh Rappaport. But Mr. Rappaport doesn't tote a gun, no sir--his weapon doesn't mutilate, it enlightens. His weapon is the ALGEBRA SURVIVAL GUIDE ( A Conversational Handbook for the Thoroughly Befuddled). Oh my! This handbook has so charmed me, I'm reading it like a novel. Good sign, right? It's no chore reading Josh Rappaport's gem. And you will also love Sally Blakemore's illustrations. She gets it! that falling down a hole feeling ( You have to buy the handbook to see Ms. Blakemore's visual take on fear of Algebra).
So, come on grownups-who-never-passed-Algebra...wouldn't you like to live a full life? wouldn't you like to give numbers, then Algebra, a chance? I KNOW you'd like to read words like REFLEXIVE Property, COMMUTATIVE Property and ASSOCIATIVE Property without fighting the the urge to cry? Well, do yourself a favor and purchase the Algebra Survival Guide. Then, do yourself a further favor and buy the WORKBOOK. Okay? |
Help With Math:
Purveyors of MathXpert
MathXpert: software to help you learn mathematics.
AutoStep and AutoFinish
When you press the AutoStep button, MathXpert takes one more step automatically.
This screen shot shows a solution generated by pressing AutoStep four times in a row.
Instead of proceeding step-by-step, MathXpert can also present the whole solution at once
if you press AutoFinish. Red indicates the changed part of the formula.
You can press the AutoStep or AutoFinish buttons at any time, not just at the
beginning. If you have partially solved the problem, MathXpert will finish it starting
from that point. AutoStep and AutoFinish work with any problem, including ones that you
entered yourself--MathXpert actually solves the problems, rather than just presenting
solutions stored in advance. |
If this technique fails, Pólya advises: "If you can't solve a problem, then there is an easier problem you can solve: find it." Or: "If you cannot solve the proposed problem, try to solve first some related problem. Could you imagine a more accessible related problem?"
First principle: Understand the problem
"Understand the problem" is often neglected as being obvious and is not even mentioned in many mathematics classes. Yet students are often stymied in their efforts to solve it, simply because they don't understand it fully, or even in part. In order to remedy this oversight, Pólya taught teachers how to prompt each student with appropriate questions, depending on the situation, such as:
What are you asked to find or show?
Can you restate the problem in your own words?
Can you think of a picture or a diagram that might help you understand the problem?
Is there enough information to enable you to find a solution?
Do you understand all the words used in stating the problem?
Do you need to ask a question to get the answer?
The teacher is to select the question with the appropriate level of difficulty for each student to ascertain if each student understands at their own level, moving up or down the list to prompt each student, until each one can respond with something constructive.
Second principle: Devise a plan
Pólya mentions that there are many reasonable ways to solve problems. The skill at choosing an appropriate strategy is best learned by solving many problems. You will find choosing a strategy increasingly easy. A partial list of strategies is included:
Guess and check
Make an orderly list
Eliminate possibilities
Use symmetry
Consider special cases
Use direct reasoning
Solve an equation
Also suggested:
Look for a pattern
Draw a picture
Solve a simpler problem
Use a model
Work backward
Use a formula
Be creative
Use your head/noggin
Third principle: Carry out the plan
This step is usually easier than devising the plan. In general, all you need is care and patience, given that you have the necessary skills. Persist with the plan that you have chosen. If it continues not to work discard it and choose another. Don't be misled; this is how mathematics is done, even by professionals.
Fourth principle: Review/extend
Pólya mentions that much can be gained by taking the time to reflect and look back at what you have done, what worked and what didn't. Doing this will enable you to predict what strategy to use to solve future problems, if these relate to the original problem.
The book contains a dictionary-style set of heuristics, many of which have to do with generating a more accessible problem. For example:
The technique "have I used everything" is perhaps most applicable to formal educational examinations (e.g., n men digging m ditches) problems.
The book has achieved "classic" status because of its considerable influence (see the next section).
Bold text'[Italic text] ==How it works== For example: A student buys acomputer for $945. This student borrowed $500 from his best friend and spent another $445 earned from his part-time job. Now his assets are worth $945, liabilities are $500, and equity $445.
These are some simple examples, but even the most complicated transactions can be recorded in a similar way. This equation is behind debits, credits, and journal entries.
This equation is part of the transaction analysis model, for which we also write
Owners equity = Contributed Capital + Retained Earnings
Retained Earnings = Net Income − Dividends
and
Net Income = Income − Expenses
The equation resulting from making these substitutions in the accounting equation may be referred to as the expanded accounting equation, because it yields the breakdown of the equity component of the equation.
Balance sheet
An elaborate form of this equation is presented in a balance sheet which lists all assets, liabilities, and equity, as well as totals to ensure that it balances.
History
Luca Pacioli is notable for including the first published description of the method of keeping accounts that Venetianmerchants used during the Italian Renaissance, known as the double-entry accounting system. However, recently some historians and experts feel that this was already being used by the Arabs and Muslim traders with whom the Venetians would have had contact. They argue that even though Luca Pacioli formally introduced it to Europe, the credit should still go to Eastern merchants who had been using it years before. This claim is yet to be accepted by the academic community as it forces a rethink of several other aspects in this field.
From Yahoo Answers
Question:Nucore Company is thinking of purchasing a new candy-wrapping machine at a cost of $370,000. The machine should save the company approximately $70,000 in operating costs per year over its estimated useful life of 10 years. The salvage value at the end of 10 years is expected to be $15,000. (Ignore income tax effects.)
1. What is the machine's payback period?
2. Compute the net present value of the machine if the cost of capital is 12%.
3. What is the expected internal rate of return for this machine?
Can you help solve this accounting equation?
Answers:This is a corporate finance question.
Question:Journal Entries:
Make the journal entry necessary to record the transaction.
The company borrowed $125,000 in cash from Far West Bank.
I have to set this up as a chart to record activirty on an account.
I am not so sure how to do so. Could you please show me how to set up journal entries?
Answers:Debit cash in bank (I assume its deposited in bank) and credit Loans owed
Assets = liabilities plus owners equity
A debit to the left side of the equation such an asset like a bank account is adding money. A credit would be removing money from it.
A credit to the right side of the equation is also adding to those accounts. So when you add to loans owed, you are adding to loans owed or increasing loans owed.
As your company makes profits and records them, that grows your owners equity, which a right side account..which means you are adding to your ownership stake in the business.
If you pay back a portion of the loan in cash, you woud credit Cash in bank, and debit Loans owed...reducing both of those accounts.
This is just a small summary of whats involved. A class would be of use perhaps.
Question:how to record these:
1.a cash deposit of $2580 was paid by cheque and the balance was to be paid in monthly instalment?
2.transferred cash $2000 from personal account into the company account?
3. sold item with book value $9600 for cash $9600. Cash was banked into the bussiness account?
Answers:first you have left something out here -- and second you need to take you homework quesitons to the homework section!!!
Question:Use this data to compute accounts receivable turnover ratios and average collection periods for 2005 and 2006. Based on your analysis is Hickory company managing its receivables better or worse in 2006 than it did in 2005?
2005 2006
Net sales 1,425,000 1,650,000
Net receivables:
Bebinning of year375,000 333,500
End of Year 420,000 375,000
Can you help solve this accounting equation
Answers:Open your Accounting 101 textbook and read.
Hopefully you are only taking Accounting because you thought it would be an easy class, and are not an accounting or business major.
Thank You.
From Youtube
Terms, Principles, Accounting Equation - Part 2 :Discussion of the accounting equation and how to solve for unknown amounts. In addition, there is a short description of retained earnings and how it is affected by net income and dividends. Other videos in this series: Part 1 - Accounting Terms and Principles Part 3 - Revenue Recognition and Matching Principles Part 4 - The Financial Statements and how they are related |
College Algebra & Trigonometry
Course Description
Prerequisite: MATH 102 or equivalent. Continuation of MATH 102 introduces the basics of trigonometry and reviews basic properties of the complex number system. The concept of function is applied to algebraic, rational, exponential, logarithmic and trigonometric functions. Emphasis on applications of trigonometry to right and oblique triangles and vectors. Assistance available at Center for Academic Success. A scientific calculator is required. Three class hours weekly. |
Numerical Methods in Engineering with MATLAB® is a text for engineering students and a reference for practicing engineers, especially those who wish to explore the power and efficiency of MATLAB®. Examples and applications were chosen for their relevance to real world problems, and where numerical solutions are most efficient. Numerical methods are discussed thoroughly and illustrated with problems involving both hand computation and programming. MATLAB® mfiles accompany each method and are available on the book web site. This code is made simple and easy to understand by avoiding complex bookkeeping schemes, while maintaining the essential features of the method. MATLAB® was chosen as the example language because of its ubiquitous use in engineering studies and practice. Moreover, it is widely available to students on school networks and through inexpensive educational versions. Explore numerical methods with MATLAB®, a great tool for teaching scientific computation. less |
This learning object from Wisc-Online covers the right circular cylinder, examining the properties and components of the shape. The lesson uses the geometric formulas for finding the volume and surface area of the...
This is a basic course, produced by Gilbert Strang of the Massachusetts Institute of Technology, on matrix theory and linear algebra. Emphasis is given to topics that will be useful in other disciplines, including...
Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary...
This graduate-level course, created by Gilbert Strang of the Massachusetts Institute of Technology, is a continuation of Computational Science and Engineering I. Topics include numerical methods; initial-value problems;...
This course, presented by MIT and taught by Professor David Jerison, provides undergraduate level calculus instruction. The materials cover differentiation and integration of functions of one variable, with... |
More About
This Textbook
Overview
Intended for use in a beginning one-semester course in differential equations, this text is designed for students of pure and applied mathematics with a working knowledge of algebra, trigonometry, and elementary calculus. Its mathematical rigor is balanced by complete but simple explanations that appeal to readers' physical and geometric intuition.
Starting with an introduction to differential equations, the text proceeds to examinations of first- and second-order differential equations, series solutions, the Laplace transform, systems of differential equations, difference equations, nonlinear differential equations and chaos, and partial differential equations. Numerous figures, problems with solutions, and historical notes clarify the text.
Related Subjects
Meet the Author
Partial Differential Equations & Beyond Stanley J. Farlow's Partial Differential Equations for Scientists and Engineers is one of the most widely used textbooks that Dover has ever published. Readers of the many Amazon reviews will easily find out why. Jerry, as Professor Farlow is known to the mathematical community, has written many other fine texts — on calculus, finite mathematics, modeling, and other topics. We followed up the 1993 Dover edition of the partial differential equations title in 2006 with a new edition of his An Introduction toDifferential Equations and Their Applications. Readers who wonder if mathematicians have a sense of humor might search the internet for a copy of Jerry's The Girl Who Ate Equations for Breakfast (Aardvark Press, 1998).
Critical Acclaim for Partial Differential Equations for Scientists and Engineers: "This book is primarily intended for students in areas other than mathematics who are studying partial differential equations at the undergraduate level. The book is unusual in that the material is organized into 47 semi-independent lessonsrather than the more usual chapter-by-chapter approach.
"An appealing feature of the book is the way in which the purpose of each lesson is clearly stated at the outset while the student will find the problems placed at the end of each lesson particularly helpful. The first appendix consists of integral transform tables whereas the second is in the form of a crossword puzzle which the diligent student should be able to complete after a thorough reading of the text.
"Students (and teachers) in this area will find the book useful as the subject matter is clearly explained. The author and publishers are to be complimented for the quality of presentation of the material." — K. Morgan, University College, Swansea |
...However, given the right help Algebra becomes simply a game of solving little puzzles. What is the difference between solving Calculus problems and tying my shoes? Well, one I normally only do every once in a while, and the other one is Calculus |
Problem Solving in Undergraduate Mathematics
Welcome to these starting points for problem-solving in undergraduate mathematics. They were developed as part of the HE-STEM initiative to enhance the curriculum experience of STEM undergraduates including a greater emphasis on problem-solving in mathematics.
These starting points or "interactivities" have been developed by mathematics educators from Cambridge University, Liverpool Hope University and University of Manchester, in collaboration with NRICH, experts in mathematical problem-solving for learners and their teachers. They are fully functional on a range of technological platforms including tablets and handhelds. This permits highly flexible use by tutors from presentation in a lecture to small group or individual work.
Tutor notes describe the main aims for each interactivity and suggested possible uses.
We do hope you enjoy these interactivities which have been programmed by Jason Davies and released under an open source license in order to be sustainable. |
A scientific calculator can quickly perform a range of operations and functions beyond basic arithmetic. All scientific calculators perform floating point arithmetic. Most scientific calculators have trigonometric functions, logarithm and power functions, hyperbolic functions, and the factorial. Most scientific calculators support scientific notation. Some scientific calculators can perform various more advanced mathematics such as calculus. In this post, we look at scientific calculator programs for computer users, primarily for performing quick scientific or engineering calculations while working on a computer.
(1) Microsoft Windows Calculator
Microsoft Windows comes with a built-in calculator utility that has a basic and a scientific mode.
Scientific Calculator Microsoft Windows
The scientific calculator mode can be selected by selecting the Scientific Menu Item in the View Menu as shown below. The Windows Calculator from Microsoft Windows 7 Home Premium is shown.
Selecting Scientific Calculator Mode Windows Calc
(2) Macintosh OS X Calculator
Like Microsoft Windows, the Macintosh OS X operating system comes with a built-in calculator with a scientific calculator mode.
Mac OS X Scientific Calculator
The scientific calculator can be selected by selecting the Scientific menu item in the View pulldown menu as shown below.
Selecting Mac OS X Calculator Mode
(3) Unix/X Window System xcalc Utility
The X Window System comes with a scientific calculator xcalc. xcalc emulates the Texas Instruments TI-30 scientific calculator by default.
The Google search box has been able to evaluate mathematical expressions for years. It has the capabilities of a scientific calculator. In the last few weeks, Google modified its search interface to display a scientific calculator when a scientific mathematical expression is entered.
The iPhone calculator utility morphs into a scientific calculator when you hold the iPhone in landscape (not portrait) mode. Note that when the iPhone is held in portrait mode (long side vertical/short side horizontal) the iPhone Calculator is a basic arithmetic calculator.
(7) GNU Emacs Text Editor
The widely used and widely available GNU Emacs text editor has both a sophisticated calculator mode with a significant learning curve and an easy-to-use quick calculator command. This calculator has extensive scientific calculator functions bordering on a poor man's MATLAB.
MThe picture below shows the GNU Emacs calculator computing the sin(2.1*3.1) below. Note that the GNU Emacs calculator defaults to degrees rather than radians (Google, Wolfram Alpha) so the result differs from the result computed by Google and Wolfram AlphaThis is a brief post showing how to display error bars in GNU Octave, a free open-source numerical programming and analysis tool that is mostly compatible with MATLAB. Octave can display standard two dimensional plots using the plot command. Unfortunately, plot does not support error bars. Fortunately, Octave has an errorbar command which can display error bars.
This post shows how to combine the raw data plotted with the Octave plot command with a polynomial model fit to the data with the error bars on the polynomial fit results from the Octave polyfit command. The post uses the data from the US Centers for Disease Control (CDC) Autism Prevalence Summary Table for 2011, a survey of studies of the prevalence of autism spectrum disorders throughout the world.
Note, in particular, the use of the commands hold on and hold off to combine the graphical outputs of the plot and errorbar commands in a single graph. Hold on keeps the graphics and axes of the figure, so that the errorbar command does not overwrite/erase the plot. Once the figure is created with both the raw data and error bars, hold off is used to return the figure to normal behavior so an entirely new plot can be created. Also, note the syntax:
[p, s] = polyfit(x, y, n)
which returns an estimate of errors in the s structure from fitting a polynomial of degreee n to the data x and y.
The syntax:
[y, dy] = polyval(p, x, s)
evaluates the polynomial with coefficients p and error parameters s for the values x, putting the resulting values in y and the error bars (one standard deviation) in dy.
This Octave code makes the following plots with error bars displayed.
Autism in USA with Errors on Fit
The second plot shows the error bars for a fit to the worldwide autism spectrum disorder prevalence data.
Autism Prevalence Worldwide with Errors on Model Fit
Conclusion
Octave can display plots with error bars using the errorbar command or the plot and errorbar commands combined as illustrated above. In particular, the Octave plot and errorbar commands can be combined to display original data and the results of fitting a model including the error bars returned by the model, using the Octave polyfit polynomial fitting command for exampleFrom time to time, computer users may need to perform a quick calculation such as adding or multiplying two large numbers. This is particularly true if the computer user works with numbers in some way: computer graphics, finance, and many other areas. Sophisticated numerical tools such as spreadsheets, MATLAB, Mathematica, Octave, and R can be slow, cumbersome and distracting for such quick calculations. Fortunately, modern computer systems have a large number of tools that provide quick calculators. This article discusses ten widely available and widely used quick calculators.
Some quick calculators support raising a number to a power, often "**" or "^". The "^" symbol is sometimes used for the bitwise exclusive OR as in the C programming language, instead of raising a number to a power. For example,
$ python -c 'print 2**3;'
8
The Ten Quick Calculators
Microsoft Windows and MS-DOS
(1) Windows CALC COMMAND/UTILITY
The command calc (the program calc.exe) will launch a simple graphical calculator on Microsoft Windows.
DOS PROMPT>calc
MS Windows Calculator GUI
(2) MS-DOS SET /A a (op) b COMMAND
The MS-DOS SET command functions as a simple command line calculator that can perform signed integer arithmetic.
The GNU bc utility is an arbitrary precision calculator language. It is preinstalled on many Unix systems. Although it is not part of the base installation, it can be installed in the Cygwin environment. In addition to basic arithmetic, it has a small math library with a few common trigonometric and transcendental functions which can be invoked with the -l option: bc -l
bc has an annoying peculiarity, a somewhat mysterious built-in variable scale which seems to correspond to the number of digits displayed after the decimal point in the results of a division operation. By default, scale is set to zero (0). What this means is that, by default, division (and only division) gives the results of integer division; there are no decimals after the decimal point.
10/3 = 3
However, if scale is set to a positive number, the results of the division operation are reported with the requested precision.
scale = 2
10/3 = 3.33
scale = 3
10/3 = 3.333
GNU bc in console window
(5) GNU Emacs Text Editor
The widely used and widely available GNU Emacs text editor has both a sophisticated calculator mode with a significant learning curve and an easy-to-use quick calculator command.
(6) VIM Text Editor
The widely used and widely available vim text editor has a quick calculator feature.
In the VIM INSERT Mode, type Ctrl-R (nothing visible happens) followed by the equal sign "=". An equal sign will appear at the lower left corner of the VIM window. Then, enter the mathematical expression to evaluate:
= 2 + 3
Press the RETURN or ENTER key and VIM will paste the result of the calculation into the file being edited. VIM can do both integer and floating point calculation. Use simple numbers such as "2 + 3″ to get integer results. Use numbers with decimal points such as "2.1 + 3.4″ to get floating point results.
VIM Quick Calculator Step One Window Only
VIM Quick Calc Step Two Window Only
PERL, PYTHON, and RUBY
Almost all Unix systems now come with the perl programming language preinstalled. Most Unix systems now come with the python scripting language preinstalled. Many Unix systems come with the ruby scripting language preinstalled. It is easy to install perl, python, and ruby on any Unix system. All are available as native applications for MS-Windows systems as well as through the Cygwin environment which emulates Unix on MS-Windows systems. All of these scripting languages can be run at the command line or interactively as simple quick calculators.
NOTE that Python treats simple numbers such as 2 and 3 as integers. "2 / 3″ is integer division and yields zero (0). Python treats numbers with decimal points such as 2.0 and 3.0 as floating point numbers. "2.0 / 3.0″ is floating point division and yields 0.66666666.
Python also comes with an interactive GUI environment known as IDLE (after comedian Eric Idle of Monty Python fame)
IDLE Python GUI Window Only
NOTE that Python under IDLE treats simple numbers such as 2 and 3 as floating point numbers, not integers as at the command line. Sadly, computer programs often contain these inconsistencies and quirks which can sometimes bite the user, especially in mathematical or numerical projects.
NOTE that Ruby, like Python, treats simple numbers such as 2 and 3 as integers. "2 / 3″ is integer division and yields zero (0). Ruby treats numbers with decimal points such as 2.0 and 3.0 as floating point numbers. "2.0 / 3.0″ is floating point division and yields 0.66666666.
(10) GOOGLE/YAHOO/BING
Google has a calculator that evaluates mathematical expressions built into the search box. Yahoo and BING also have calculators built into their search boxes.
GOOGLE Quick Calculator Screenshot
Conclusion
In this article, ten quick calculators for computers users were presented and their basic use explained.
The quick calculators are appropriate for occasional quick calculations such as adding or multiplying two large numbers. They will work best if the computer user practices and can use the quick calculator of his/her choice quickly and easily — "second nature".
Quick calculators are often faster and less cumbersome than sophisticated numerical tools such as spreadsheets like Excel or mathematical scripting languages such as MATLAB or Octave for occasional quick calculations such as adding or multiplying two large numbers. However, sophisticated numerical and mathematical tools are better for large number crunching projectsUnderestimation of the cost, schedule, and risk of projects is common in software development and especially prevalent in mathematical software development. It is common to encounter extremely optimistic ideas about the duration and difficulty level of mathematical software projects, ironically one of the more difficult kinds of software development, as well as magical ideas about mathematics.
There is relatively little publicly available information on the scope and difficulty level of software projects of any kind. Some information is available in books and papers by various self-styled software engineering experts such as Barry Boehm, Donald Reifer, Capers Jones, and several others. These experts usually have consulting businesses and do not disclose their raw data and make limited disclosures of the results of analyses of their data.
Open source software projects can provide an excellent source of information on some aspects, such as the number of lines of computer code, of various software and mathematical software projects. This information can be independently verified by downloading the source code of an open source project and examining it, using tools like the CLOC utility to count the lines of code if needed.
Unfortunately, it is difficult to get accurate estimates of the actual effort expended on an open source project. It is difficult to verify if a contributor worked part-time, full time, or more than full time on the project. Some contributors may not be credited.
This article examines a data set of ninety-three NASA projects between the years 1971-1987 that was collected by Jairus Hihn of the NASA Jet Propulsion Laboratory (JPL). The data is the NASA 93 data set from the PROMISE Software Engineering Repository at the University of Ottawa.
The data lists the number of source lines of code (SLOC) for each project, the actual effort expended in staff months (SM), and classified the projects according to software engineering expert Barry Boehm's COCOMO I (Constructive Cost Model). The data used for Boehm's COCOMO I model is also available as a data set in the PROMISE repository.
A Note on Lines of Code
Lines of code is a very imperfect measure of the size and scope of a software project. For example, these are both one line of code in the C Programming Language:
a = 1;
and
a = (1.0/sqrt(2.0*M_PI))*exp(-(x - mean)*(x-mean)/(sigma*sigma));
There are several different definitions of lines of code used in the literature on software cost and schedule estimation. In additional, there are a range of alternatives that have been proposed to lines of codes, such as function points (currently popular).
Nonetheless, lines of code are somewhat reminiscent of Winston Churchill's quote about democracy:
It has been said that democracy is the worst form of government except all the others that have been tried.
Function points were developed for business applications and rely heavily on counting the number of inputs and outputs to a program. This often works well for business applications where the applications are often relatively simple and the complexity scales with the number of inputs and outputs. Mathematical software such as video codecs often have few inputs (one compressed file or data stream) and outputs (uncompressed video) but a very complex internal implementation (tens of thousands of lines of code). This has been recognized as a weakness of function points for some time and there are some variations such as so-called "feature points" that attempt to address this problem.
Further, methods like function points require substantial training and study to measure and learn to use. They are not relatively intuitive like lines of code. There is much more data on software projects available in lines of code than function points.
One good way to think about lines of code is that each line of code is like a single moving part in a complex machine like a grandfather clock. Some parts are simple like the first line of code above. Some parts are more complex like the second line of code above. In general, lines of code would correspond to moving parts if one tried to implement a computer program as a mechanical device like Victorian era English mathematician Charles Babbage's steam driven difference engine.
In mathematical software such as video compression, speech recognition, or other advanced applications, a line of code is usually directly equivalent to a single line of a mathematical formula or equation that a math teacher or professor might write on a blackboard or dry erase board in class. Most examples of mathematics taught in high school or college math courses cover at most a dozen blackboards. These are often building blocks of the mathematical solutions to real-world problems or cutting edge research problems. Most real-world examples of mathematical software such as video codecs such as the H.264, Flash, or Microsoft Silverlight video compression used by web sites today are many thousands of lines of code and correspond to hundreds or thousands of blackboards filled with mathematical equations and formulas.
Analysis of the NASA 93 Data
The plots below show various aspects of the NASA 93 data on the scope and effort of these software projects.
NASA 93 Software Projects
Project Size in Staff Years (Man Years)
Project Size in Thousands of Lines of Code
Project Years
NASA 93 Shown by COCOMO Mode
The COCOMO model divides software projects into three general categories or "modes". These are the embedded, semi-detached, and organic. Embedded mode projects such as flight avionics software are most similar in difficulty to mathematical software projects. Indeed, due to safety issues, flight avionics software can be more demanding, requiring higher quality, than commercial applications such as video compression for entertainment. The software productivity in lines of code per staff month is now shown for the three kinds of projects.
Software Productivity for Organic Mode Projects (NASA 93)
Software Productivity for Semi-Detached Projects (NASA 93)
Software Productivity for Embedded Projects (NASA 93)
The next plot compares the NASA 93 data to Barry Boehm's Basic COCOMO I model for Embedded Projects (red line) and to a linear fit to the NASA 93 data (green line). As can be seen, there is considerable variation between actual and estimated effort, although the models are on average roughly correct and usually within a factor of three of actual effort.
Comparison of Data to Fitted Models
The final plot shows the relative error between the actual effort and the estimated effort using the fitted model.
Relative Error (NASA 93)
Conclusion
On average, the software productivity for demanding software applications such as embedded aerospace applications tends to be quite low, in the range of two-hundred (200) lines of code per staff month (mythical man month). However, there is wide variation between actual and estimated effort. The highest productivity (defined as lines of code per staff month) among the embedded projects in the NASA 93 data set was about 700 lines of code per month, and the lowest around 50 lines of code per month. Given the difficulties in defining lines of code and measuring the quality of the delivered software, it is impossible to evaluate the significance of these variations without more detailed information on the projects.
It is important to keep in mind that numbers like two-hundred lines of code per staff month do not refer to just typing two-hundred lines of code which can take as little as a few minutes. They refer to the entire software development process, usually including requirements analysis, software design, actual coding, and especially debugging to achieve the high levels of quality required for these applications.
There are several cases where a single error in a single line of mathematical software has resulted in the loss of a multi-million dollar mission or human lives. The loss of the Mariner I probe to Mars is frequently attributed to a small error in copying a mathematical formula into the probe's computer software. In 1991 a subtle error in the mathematical software for a PATRIOT missile system resulted in an Iraqi SCUD missile penetrating to a US base in Dahran, Saudi Arabia and killing 28 soldiers. On June 4, 1995 the European Space Agency's first launch of the new Ariane 5 rocket exploded due to an error converting a 64 bit floating point number incorrectly to a 16 bit integer number in software. The loss of NASA's Mars Climate Orbiter (MCO) in 1999 has been attributed to an incorrect conversion between English units (foot-pounds) and metric units (meters-Newtons). Aviation and rocketry have especially demanding requirements for the quality of software.
While commercial applications of mathematical software such as video compression for entertainment are not always as demanding as mission-critical aerospace software, they can still be quite demanding. Viewers of compressed video such as Netflix, YouTube, BluRay, or DVD video have a pretty limited tolerance for visible artifacts and errors in the video. Almost any error in the implementation of a video codec can introduce visible artifacts or errors, so the codecs must, in general, achieve very high levels of quality, though not necessarily perfectAppendix I: Source Code for Analysis
The analysis was performed using a program written in the free open source Octave numerical programming environment which is mostly compatible with MATLAB. Here is the code. It generates additional plots beyond the ones highlighted in the body of this article. The raw data file nasa93_raw_data.txt, which is extracted from the PROMISE data file follows.
]]>Have you ever wanted to create a humorous or entertaining voice like a cartoon character's voice for a get-well video, a Valentine's video, the narration for a DVD of home videos, an advertisement for your business or some other application? This article tells how to create cartoon voices using mathematics to shift the pitch of normal voices. The article includes the Octave source code for an Octave function chipmunk that applies pitch shifting to audio.
The standard audio pitch shifting incorporated in many commonly used audio editors such as the free open-source Audacity editor is presented in detail. The article also shows the results of using a more sophisticated algorithm that produces a more natural sounding pitch-shifted voice similar to the voice of the famous cartoon character Mickey Mouse.
One of the basic concepts and methods of signal and speech processing is the Fourier transform, named after the French mathematician and physicist Joseph Fourier. The basic concept is that any real function can be represented as the sum of the trigonometric sine and cosine functions. For example, a function defined on the region can be expanded as the sum of sines and cosines:
where the coefficients and are known as Fourier coefficients. This is a continuous Fourier Transform.
There is a discrete version of the Fourier Transform, often used in digital signal processing:
where is the index of an array of discrete values such as audio samples, is the value of the th audio sample, is the index of the discrete Fourier coefficients and is the number of discrete values such as the number of audio samples in an audio "frame". The index is essentially the frequency of the Fourier component. This version of the discrete Fourier Transform uses the mathematical identity:
where
to combine the cosine and sine function components into complex functions and numbers.
In audio signal processing such as speech or music, the Fourier Transform has a straightforward meaning. The sound is broken up into a combination of frequency components. In most instrumental music, this is very simple. The music is a collection of notes or tones with specific frequencies. Percussion instruments and certain other instruments can produce more complex sounds with many frequency components. A spectrogram of a signal such as speech or music shows time on the horizontal axis and the strength of the frequency component on the vertical axis. This is the spectrogram of a pure 100 Hertz (cycles per second) tone:
Spectrogram of 100 Hz Tone
The spectrogram is generated using the specgram function in the Octave signal signal processing package by dividing the signal into a series of overlapping audio frames. Overlapping audio frames are frequently used to achieve better time resolution during signal processing in the Fourier domain. Each audio frame is windowed using the Hanning window to reduce aliasing effects.
The Fourier transform is applied to each windowed audio frame, giving a series of frequency components, which are displayed on the vertical dimension of the spectrogram. Each frequency component is a bin in frequency covering a frequency range equal to the audio sample rate divided by the number of samples in the audio frame. This frequency bin size or frequency resolution of the Fourier transform is about 20 Hz in the spectrogram above (44100 samples per second/2048 samples in an audio frame = 21.533 cycles per second). Because the 100 Hz tone in the example is not perfectly centered in the frequency bin spanning 100 Hz, the tone spreads out in the spectrogram, contributing to other bins as can be seen above. This is a limitation of the discrete Fourier transform which can lead to problems with signal processing such as pitch shifting.
Speech has a much more complex structure than a pure tone. In fact, the structure of speech remains poorly understood which is why current (2011) speech recognition systems perform poorly in realistic field conditions compared to human beings. This spectrogram shows the structure of the introduction to United States President Barack Obama's April 2, 2011 speech on the energy crisis: "Hello everybody. I'm speaking to you today from a UPS customer center in Landover, Maryland where I came to talk about an issue that is affecting families and businesses just like this one — the rising price of gas and what we can…".
President Obama on the Rising Price of Gas
The spectrogram below shows the region from 0 to 600 cycles per second (Hertz). One can see a series of bands in the spectrogram. These bands are located at integer multiples (1, 2, 3, …) of the lowest frequency band, which is often referred to as F0 in the scholarly speech literature. The bands are known as the harmonics. F0 is known as the fundamental frequency. This is the frequency of vibration of the glottis which provides the driving sound for speech and is located in the throat. The glottis vibrates at frequencies ranging from as low as 80 cycles per second (Hertz) in some men to as high as 400 cycles per second (Hertz) in some women and children. This fundamental frequency appears to be loosely correlated with the height of the speaker, higher for short speakers such as children and lower for taller women and men.
The fundamental frequency F0 fluctuates in a rhythmic pattern that is not well understood as people speak. In some languages such as Mandarin Chinese, the changing pitch conveys meaning; a word with rising pitch has a different meaning from an otherwise identical word with falling pitch. In English, a rising pitch at the end of a phrase or sentence indicates that a question is being asked. "The chair." is pronounced with falling pitch whereas "The chair?" is pronounced with a rising pitch at the end. It is difficult and even sometimes impossible to understand English if the rhythmic pattern of the fundamental frequency or pitch is abnormal.
President Obama on the Rising Price of Gas (to 600 CPS)
This spectrogram shows President Dwight David Eisenhower saying "in the councils of government we must guard against the acquisition of unwarranted influence, whether sought or unsought, by the military industrial complex" from his Farewell Address, January 17, 1961, probably his most famous phrase and his most famous speech today.
Eisenhower on the Military Industrial Complex
This spectrogram shows the spectrogram in the range 0 to 600 Hertz (cycles per second). Again, one can easily see the repeating bands.
Eisenhower on the Military Industrial Complex (to 600 CPS)
Human beings perceive something which we call "pitch" in English which appears closely related to or identical to the center frequency of the F0 band in the spectrogram. The F0 band will be higher in higher pitched speakers such as many women and most children. Both President Obama and President Eisenhower have similar pitches, varying between 200 and 75 Hertz with an average of about 150 Hertz. Nonetheless, their voices sound very different. The F0 band can be as low as 70 or 80 Hertz (cycles per second) in a few speakers. Former California governor and actor Arnold Schwarzenegger used an extremely low pitched voice while playing the Terminator, his most famous role.
In general, low pitched voices tend to convey seriousness and sometimes menace whereas high pitched voices tend to convey less seriousness, although there are exceptions. The voice of the genocidal Daleks in the BBC's Dr. Who series is both high pitched and menacing at the same time. Cartoon style voices can be created by shifting the pitch of normal speakers. This has been done for the Alvin and the Chipmunks characters created by Ross Bagdasarian Sr.. It is probable that some form of pitch shifting has been used over the years to create some of the voices of the Daleks on Dr. Who. Some robot voices have probably been created by combining pitch shifting with other audio effects.
Traditional Pitch Shifting
Pitch shifting predates the digital era. In the analog audio era, one could shift the pitch of a speaker by playing a record or tape faster or slower than normal. This shifts the pitch but also changes the tempo — speed or rate of speaking — as well. One can achieve a pure pitch shift by, for example, recording a voice performer speaking at half normal speed and then playing the recording back at twice the normal rate. In this case, the pitch will be shifted up by a factor of two and the tempo or rate of speaking will be normal. One can create the Alvin and the Chipmunks high pitched voice in this way using analog tapes or records. One can also create lower pitched voices by appropriately combining the tempo of the original voice and the playback rate of the recording. Although these voices are easily understandable, they have artificial, electronic qualities not found in normal low or high pitched speakers or voice performers intentionally creating a low or high pitched voice. The voice of Walt Disney's Mickey Mouse was performed by a series of voice artists starting with Walt Disney himself. This high pitched voice sounds much more natural than the Alvin and the Chipmunks voice.
In digital audio, it is possible to shift the pitch of the voice without changing the tempo of the speech. This can be done by manipulating the Fourier transform of the speech, the spectrogram, and converting back to the "time domain," the actual audio samples. One can simply shift the Fourier components from their original frequency bin in the spectrogram to an appropriate higher or lower frequency bin. For example, if a Fourier component is in the 100 Hz bin, one shifts this Fourier component value to the 200 Hz bin to double the pitch. This must be done for each and every non-zero Fourier component. In general, this will produce a recognizable pitch shifted voice. If the Fourier components are not centered in each bin, which is normally the situation, this pitch shifted voice will have an annoying beat or modulation. It is necessary to perform some additional mathematical acrobatics to compensate for these effects to produce a relatively smooth pitch shifted voice similar to the output of the analog processing described above.
This video is President Obama's original introduction from his April 2, 2011 speech on the energy crisis. Click on the images below to download or play the videos.
This video is President Obama speaking with his pitch doubled by shifting the Fourier components but without the mathematical acrobatics to compensate for un-centered frequency components:
This video is President Obama speaking with a chipmunked voice; his pitch has been doubled.
This video is President Obama speaking with a deep voice; his pitch has been reduced to seventy percent of normal.
Octave is a free open-source numerical programming environment that is mostly compatible with MATLAB. The Octave source code below, the Octave function chipmunk, implements the standard pitch shifting algorithm in widespread use. The Octave code requires both Octave and the Octave Forge signalsignal processing package for the specgram function which computes the spectrogram of the signal.
The videos in this article were created by downloading the original MPEG-4 videos from the White House web site and splitting the audio and video into a MS WAVE file and a sequence of JPEG still images using the FFMPEG utility. Presidential speeches and video are in the public domain in the United States. The original still images were reduced in size by half using the ImageMagickconvert utility. The audio was pitch shifted in Octave using the chipmunk function below. The new audio and video were recombined into the MPEG-4 videos in this article by again using the FFMPEG utility. Variants of this pitch shifting algorithm can be found in many programs including the widely used free open-source Audacity audio editor (the Audacity pitch shifting algorithm may be slightly different from the algorithm implemented below):
The screenshot below shows running the chipmunk function in Octave 3.2.4 on a PC under Windows XP Service Pack 2 (Click on the screenshot image to see the full size screenshot). This screenshot shows the function called from the Octave prompt using the default values of the function's arguments. The argument numberOverlaps controls the mathematics to compensate for the uncentered frequency components. If numberOverlaps is one, there is no compensation. The larger numberOverlaps, the more effective the compensation. The more overlaps, the more computer time and resources required by the pitch shifting. A value of numberOverlaps of thirty-two (32) was used to pitch shift President Obama's voice in the video above.
Running the Chipmunk Function in Octave
Although easily understandable, these pitch-shifted voices sound somewhat artificial. Indeed, this artificial quality is part of the appeal of the Alvin and the Chipmunk voice.
Pitch Shifting Gets Better
Pitch shifting algorithms have improved. It is now possible to produce voices that sound much more like natural voices at the desired new pitch, very similar to the voice of Mickey Mouse. This video is President Obama speaking with a voice similar to the voice of Mickey Mouse:
Conclusion
There are many ways to manipulate voices using mathematics. One of the most common is pitch shifting, which has been described in detail including working source code above. Traditional pitch shifting algorithms give artificial qualities to the pitch-shifted voice. There are now new, improved algorithms that can create more natural sounding pitch-shifted voices. These voices can be used for humor, entertainment, or emphasis in movies, television, video games, video advertisements for small businesses, personal and home video, and in many other applications Game of Life in Octave
16 May 2011 16:20:32 +0000
Possibly related articles:
]]>The Game of Life is a simple cellular automaton invented by mathematician John Conway. The Game of Life consists of a two dimensional grid of cells. Each cell can be "live" or "dead", often represented mathematically by 1 and 0. The grid is updated periodically in discrete time steps according to a simple rule based on the cell's neighbors, the cells immediately adjacent to the cell horizontally, vertically, and diagonally. The figure below shows a live cell (black is used for a live cell in this article) and its dead neighbors. The grid is displayed for clarity.
Cell Neighbors
The update rule for the Game of Life is:
A live cell with fewer than two live neighbors dies.
A live cell with two or three neighbors lives on to the next time step.
A live cell with more than three neighbors dies.
A dead cell with exactly three live neighbors becomes a live cell in the next time step.
The Game of Life has many interesting and entertaining properties. Amongst other features, it can implement a general purpose computer like the much more complex Pentium CPU chip. There is extensive information on the Game of Life on the Web. Interested readers are referred to this online information as well as the many traditional printed articles and books that discuss the Game of Life. This article discusses how to implement the Game of Life in Octave, a free open-source numerical programming environment that is mostly compatible with MATLAB. Octave is available in both source code and pre-compiled binaries for all three major computing platforms: MS Windows, Macintosh, and Unix/Linux. Full source code in Octave is presented and the results of several simulations of the Game of Life are shown.
Simulations of the Game of Life
There are a number of structures in the Game of Life that have interesting properties. These range from simple oscillators that cycle through a fixed set of patterns to complex self-perpetuating patterns that grow and evolve indefinitely. Several are shown below. These were simulated using an implementation of the Game of Life in Octave. The simulation generates a sequence of images showing the time evolution of the Game of Life. These image squences are either static GIF images or static JPEG images. Where possible, the images were combined into animated GIF sequences shown below using the convert utility in ImageMagick. In some cases this proved difficult and the images were converted to MPEG-4 video (MP4) using the ffmpeg command line utility.
Octave Code
The main Game of Life simulation function is simulate_life. This implementation uses a series of nested loops to implement the update rule in a way that will be familiar to users of procedural programming languages like the C programming language. Later, a function simulate_life_fast is presented which implements the update rule using the "matrix" operations in Octave with no loops over the rows and columns of the Octave matrix that represents the Game of Life universe.
The function simulate_life calls a support function cycle which uses the Octave mod function to wrap the row and column coordinates at the edges of the Game of Life universe. This closes the Game of Life universe, giving it a torus or donut shape. In the glider simulation, the glider travels off one end of the universe and reappears at the other end.
Elegant Implementation in Octave
Octave has a large number of "matrix" or "n-dimensional array" operators and functions that operate on an entire Octave matrix (two-dimensional array) or n-dimensional array without nested loops over the indices of the matrix or array. These are generally faster, more compact, and often the coding is less error prone than using nested loops. This is an implementation of the Game of Life using Octave matrix operators:
Conclusion
The Game of Life is a simple, easy to implement, entertaining cellular automaton. It is easy to implement the Game of Life in Octave (or MATLAB or SciLab). External tools such as ImageMagick convert or ffmpeg can be used to easily convert the image sequences that Octave can generate into animations in commonly used formats such as animated GIF or MPEG-4 video. Even using Octave's matrix-oriented operators to implement the Game of Life, avoidign the cumbersome and generally slow nested loops over rows and columns, the Octave implementation is still slow compared to a compiled C programming language implementation. This speed issue is probably the primary drawback to using Octave, which otherwise is very quick and convenient and has a much lower development time than low level compiled languages such as C]]>An associative array, also known as a dictionary, map, mapping, or hash table, is a powerful data structure that is built into many modern programming languages such as Python, Perl, Ruby, and many others. An associative array is a form of content addressable memory (CAM). For example, when you see someone's face, you often remember many other facts. This is John Smith. John Smith is your neighbor and lives at 123 Elm Street. John's wife is Amanda who has long blond hair and so on. As associative array associates an object, often called a key, with another object, often called a value.
In many programming languages the key is a string of characters such as "John Smith" and the value is another string such as "Address:123 Elm Street;Wife:Amanda" and so forth. In some object oriented computer programming languages, the key can be any kind of object and the value can be any kind of object such as the key <John Smith's Face Object> and the value <John Smith's Identifying Information Object>. An associative array is largely equivalent to a single table in a relational database (RDBMS). In principle, a network of associative arrays can represent complex abstract knowledge and reasoning.
Technically, associative arrays are usually implemented using a hash table. A hash table uses modulo arithmetic to map an object such as a string of characters to the numerical index of an array of values. This array of values is not a simple one dimensional array because there can be collisions where two keys, objects such as strings of characters, "hash" to the same array index. If this happens, there are various methods such as hanging a linked list of elements off the array to handle the collision. Using a hash table means that the time to look up a value is almost constant (O(1)) regardless of the size of the hash table.
The hash table may have millions of entries, but it takes the same small number of operations to look up the associated value. First, compute the array index using modulo arithmetic on the numeric value of the "key". Second, handle any collisions. The hash table should be large enough that collisions are rare. In principle, an associative array could be implemented in other ways, but hash tables of some kind are generally the fastest, most flexible way to implement an associative array. Hence, the terms associative array, dictionary, map, mapping, and hash table are often used interchangeably.
Octave is a free, open-source numerical programming environment that is mostly compatible with MATLAB. Octave is largely built around matrices (two dimensional arrays) and N (2,3,etc.) dimensional arrays. By default, matrices and N-D arrays in Octave are of the type double (usually a 64 bit IEEE-754 double precision floating point number).
This is due to the history of MATLAB (short for Matrix Laboratory) which started life as an interactive environment for performing two dimensional matrix algebra and computations. At first glance, Octave lacks associative arrays which is a significant deficiency for some types of programming including some types of mathematical programming. Octave does, in fact, have associative arrays. This article shows how to use associative arrays in Octave and use associative arrays to implement cellular automata, a type of discrete mathematical model.
Associative Arrays in Octave
While Octave lacks an explicitly identified associative array, dictionary, map, mapping, or hash table data type, Octave does have data structures or structs similar to structures in the C family of programming languages. For example, in Octave one can define a structure interactively like this:
The structure a now has two fields "key" and "key2″ with values 'value' and 'value2′. In Octave, the data structures are implemented using a hash table and can act as a fully functional associative array. Octave provides several functions to access and manipulate structures, making it easy to use a structure in Octave as an associative array.
All of these functions are useful. Most important for the purposes of this article are struct("field", value, "field", value,...) which creates a data structure explicitly, setfield(structure_name, key, value,...) which assigns a value to a key, and getfield(structure_name, key) which retrieves the value associated with key. These are used in the examples in this article to implement cellular automata.
Cellular Automata
A cellular automaton (plural, cellular automata, sometimes abbreviated as CA) is a discrete mathematical model. A typical cellular automaton consists of a grid of cells with one or more dimensions. Often, the cells have two possible values, 0 and 1, which are often displayed as black and white pixels graphically. The cellular automaton evolves over time in discrete time steps. With each time step or update, a cell changes to 0 or 1 based on the values of the cells in its neighborhood. A cellular automaton has a rule that specifies how it updates.
Cellular automata have been used in entertainment (they make pretty pictures), mathematics, physics, and a number of other fields. Probably the most well known cellular automaton is the "Game of Life", a two dimensional cellular automaton with many entertaining and interesting properties invented by the mathematician John Conway in the 1970′s.
Stephen Wolfram, the creator of the Mathematica computer algebra system and mathematical research tool, has had a long standing interest in cellular automata. His book, A New Kind of Science, speculates that the universe might be a sort of cellular automaton and be "computationally undecidable" (in layman's terms, math and science don't have all the answers). Matthew Cook, who assisted Wolfram in the research for A New Kind of Science, proved that a particularly simple cellular automaton known as "rule 110″ can function as a general purpose computer just like more complex systems such as the Pentium computer chip. An implementation of the "rule 110″ cellular automaton is one of the examples in this article.
The rule for a cellular automaton can be easily represented as an associative array that maps each possible neighborhood to a new value. This is very simple and intuitive. It is easy to implement cellular automata in programming languages with built-in associative array data types. This is illustrated in Octave in the examples in this article.
The code below tests the timing of associative arrays in Octave. As expected if a hash table is used to implement an associative array, the lookup time is largely independent of the size of the associative array in Octave, which is good. As with many features in Octave and other mathematical scripting tools, the lookup time is quite slow.
For example, on a 3GHz Macintosh running OS X, the lookup time varied between 1 and 10 milliseconds. This is much slower than a compiled implementation of a hash table in the C programming language or another fast compiled language. Although languages such as Octave are slowly closing the speed of execution gap with compiled languages such as C, the compiled languages still win handily in some cases.
There are associative arrays, also known as dictionaries, maps, mappings, or hash tables, in Octave. This is poorly documented both in the official documentation and most online information about Octave. One can perform the same tasks and implement the same algorithms with the associative arrays (data structures or structs) in Octave that one can with the explicitly identified associative array data types in Perl, Python, Ruby, Java, and many other modern programming languages. Associative arrays are very useful for implementing certain kinds of mathematics in Octave such as, but not limited to, cellular automataThis is the fourth article in an occasional series of articles about Octave, a free open-source numerical programming environment that is mostly compatible with MATLAB. This series began with the article Octave: An Alternative to the High Cost of MATLAB. This article discusses plotting and graphics in Octave. Octave has extensive plotting and graphics features including two-dimensional plots, histograms, three dimensional plots, and a range of specialized plots and charts such as pie charts. This article gives an overview of the key plotting and graphics features of Octave, discusses a few gotchas, and gives several illustrative examples.
Plotting
Octave can plot functions and data using the built-in plot functio. To illustrate thiese features, this article uses data on the price of gasoline in the United States from the United States Energy Information Administration, part of the US Department of Energy. The data is taken from the Motor Gasoline Retail Prices, U.S. City Average report released March 29, 2011. This data is currently available in Adobe Acrobat PDF file format, a comma separated value (CSV) ASCII text format, and Microsoft Excel (XLS) spreadsheet format. The data from the CSV report was imported into Microsoft Excel, exported as an ASCII tab delimited text fiel and then cut and pasted into two tab-delimited ASCII files. The EIA gas price report contains several time series in a single file (leaded gas prices, unleaded gas prices, and several more). The leaded and unleaded gas prices were extracted manually in the Notepad++ text editor. These dataq files were named: leaded.txt and unleaded.txt.
Unleaded Gas Price
unleaded = dlmread('unleaded.txt');
The plot of unleaded gasonline prices above was generated using the following Octave code:
A few comments may be helpful. dlmread is an Octave function that reads ASCII data files. It is fairly flexible and can often automatically identify the separator used in ASCII data files such as the tab or a comma. If necessary, the user can explicitly specify the separator and other parameters of the data file. Nonetheless it is common to encounter data files with various quirks. For example, the EIA gas price report contains several time series in a single file. There are many months for which either data is not available or not reported; these are indicated by a value of 10000000 for the gas price. The code above uses the Octave find function to select the valid data. Further the year and month are combined in the format YYYYMM sso January 1970 would be "197001″, February 1970 is "197002″, and so forth. Used directly in the Octave plot function, this will produce a nonsense plot that is not useful.. Thus, the example code above uses Octave fix and mod functions to compute a time in years (month numbers are converted to fractions of a calendar year). Octave has numerous advanced functions such as find , fix, rem, and so forth that can be used to clean up and reformat data as needed.
The Octave function plot handles the actually plotting of the graph. The Octae plot function is a very versatile plotting function for two dimensional data such as time series. Both the Octave user manual and the build in help (help plot) provide detailed information on the use of plot.
Histograms
Octave has built in support for histograms. A histogram is a way of displaying the frequency of occurrence of data or events. One might, for example, be interested in how often gas prices change by one percent, two percent, or ten percent in one month.
Histogram of Gas Price Changes
By default, the Octave hist function creates a histogram with ten bins, which is often not very useful. One can specify more bins easily.
Now the values in the histogram of percent changes in gas prices are normalized to 1.0. The histogram is an estimate of the probability density function for gas price changes. One might wonder whether this distribution is the Gaussian probability denisty, also known as the Normal or Bell Curve distribution. In fact, the histogram looks narrower than a typical Gaussian and also has some outliers, a long tail, which are not typical of a true Gaussian distribution. The next two figures show the Gaussian probability density function and a histogram of synthetic data generated with Gaussian statistics using the same mean and variance as the actual gas price data.
Gaussian Probability Density Function
Histogram of Synthetic Gaussian Data
It is easy to see that the Gaussian, or Nornal or Bell Curve, distribution differs from the distribution of the gas price data. The actual data has a narrower, sharper peak and long tails. Every now and then, gas prices jump shaprly. This is a pattern seen in many financial and other kinds of assets . Many popular financial models such as the Black-Scholes option piricing model assume Gaussian or near-Gaussian distributions which generally understates the risks when a fniancial asset or commodity has long non-Gaussian tails as gasoline does.
The histograms and related plots above were generated using the following Octave code:
The third argument to pie (after names ) tells Octave to "explode" the first wedge (Venezuela).
Three Dimensional Graphics
Octave has functions to plot an display three dimensional data and functions:
Built In Sombrero Plot
Built In Peaks Plot
The two 3D plots above were generated using the built-in peaksand sombrero test functions. It is possible to compute and display almost any 3D surface using the meshgrid function and 3D display functions such as plot3, surf, mesh, contour, and quiver.
Meshgrid
For people coming from another type of programming such as the C family of langauges, the meshgrid concept and function is new and may take a little getting used to. A meshgrid is basically a very simple concept which is also very powerful. Octave represents almost everything as a "matrix" or multi-dimensional array. This is the source of much of the power of Octave. One can often avoid explicitly coding loops over the elements of the Octave matrix. This speeds development and reduces errors.
A meshgrid is a two (or higher) dimenaional array in which the elements of the array are the spatial location (x or y coordinate usually) of the associated element of a spatial grid (the mesh grid). Here is some Octave code which explicitly computes and plots the sombrero function:
In this example, the Octave function meshgrid returns two arrays x and y which contin the x and y corrdinates respectively for the mesh grid elements. In this case, the x and y positions of the grid are specified by the one dimensional ticks array. The ticks run from -10 to 10.0 in steps of 0.5., a total of 41 ticks. The x array generated by meshgrid is a 41 by 51 array with the x coordinate of each element. The y array is a 41 by 41 element array with the y coordinate of each element.
The meshgrid enables one to express 3D surfaces or functions in Octave in a simple intuitive compact way:
z = sin (sqrt (x.^2 + y.^2)) ./ (sqrt (x.^2 + y.^2));
In this example, z is a two dimensional array (matrix) with the function value at the xand y coordinates specified by each grid point in the arrays xand y. Then, one can display the surface using display functions such as surf, mesh, and so forth. It usually takes some practice to get used to using meshgrid if one is not familiar with the concept.
Sombrero Meshgrid Using Surf
Sombreror Meshgrid Using Mesh
Sombrero Using Plot3
Contour Plot
Octave can create contour plots using the contourfunction.
Sombreror Meshgrid Using Contour
Vector Field Plot
Octave can create vector field plots using the quiver (as in quiver of arrows) function:
Debugging Octave
Octave has cryptic error messages. These messages almost always correctly identify the line of code that is in error. The verbal descripton of the error is often incoprehensible and may be wrong. The column number reported for the location of an error in the line is often wrong, for example indicating the start of the expression on the right hand side of an assignment statement where the problem is later in the line of code.
If a user cannot spot the error by reading the line of code, a common occurence, it is usually best to convert the line of code into several lines of code with each lnew line of code representing a sub-expression o f the original line of code. This approach will usually narrow the error/bug down to a specific symbol and identify the specific error.
Octave supports both true-matrix operations and element by element (aka element-wise) operations. For example, A*B is true-matrix multiplication if A and B are matrices. A.*B is element by element multiplication in which each element is multiplied by its corresponding element in the other matrix. It is easy to mistakenly use * where one should use .* or .* where one should use * in Octave. Pay close attention to th edistincition between the true matrix and element-by-element operators.
Conclusion
Octave has extensive built in plotting and graphics functions. There are a few weaknesses, notably some problems with the bar chart functions, at least in the Windows version of Octave 3.2.4. Users coming from a different type of programming background such as the C family of languages may need a little time and practice to adjust to the meshgrid concept. The plotting and graphics funsions of Octave are more than adequate for all common scientfic, engineeering, and general analytical tasks, both two and three dimensional]]> This article discusses the scope of mathematical programming projects using the example of several successful open source/free software projects. In the author's experience, it is common to encounter rather optimistic expectations of the cost, schedule, and risk level of mathematical research and development and programming projects; the two categories, mathematical research and development and mathematical programming, are heavily blurred together today, especially for practical and applied projects. This article provides a rough picture of the scope of mathematical programming projects based on actual historical data rather than popular culture, anecdote, or folk wisdom.
There are strong practical reasons for applied mathematical research and development and programming projects. Many potentially beneficial projects exist. These projects often suffer from the "cure for cancer" problem. With several hundred thousand people each year in the United States alone succumbing to cancer, there is little question that there is a large market for a cure for cancer. The problem is that we do not know how to cure cancer. Similarly, successful mathematical research and development and programming projects offer everything from profitable investment advice to speech recognition for mobile devices and household appliances to working fusion reactors and other new energy sources. Indeed, a cure for cancer is something that mathematical methods may offer in the future through molecular modeling or other quantitative approaches. Given the huge potential markets for successful mathematical projects, it is common to encounter individuals, organizations, and companies with great interest in particular, usually practical mathematical projects. These projects are highly unlikely to succeed without accurate ideas about the scope of the projects.
The Scope of Some Successful Free Open-Source Mathematical Software Projects
The free, open-source CLOC (Count Lines of Code) utility was used to count the number of lines of code in each project. CLOC lists the number of lines of code in each programming language in the project such as C, C++, Bourne Shell, HTML, and so forth. CLOC does not count blank lines or comment lines. Some projects include sizable amounts of installation code (in the Unix Bourne Shell for example), HTML documentation, and so forth which is counted in the total number of lines of code reported by CLOC. The actual mathematical code is typically implemented in a few languages such as C, C++, FORTRAN, or MATLAB. The term "Core Lines of Code" refers to the lines of code in these languages, as reported by CLOC, which is presumed to contain the actual mathematical software.
In general, open source projects provide a wealth of detailed information that is difficult or impossible to acquire for many commercial proprietary projects. In particular, one can see the source code, count the lines of code or other measures of size and scope, and often read comments, change logs, logs of version control systems, and so forth. Nearly all open source projects give a list of contributors somewhere in the documentation and provide rough information on the calendar duration of the project. There is usually precise information on releases and release dates. Unfortunately, it is difficult to get a reasonably exact measure of the actual effort expended on the project. Most open source projects do not publish information on exact hours worked, dollars expended, even if records exist. Several of the examples were fully or partially funded either by government funding agencies (e.g. the National Library of Medicine for the Insight Toolkit) or private sources (e.g. Intel for OpenCV), so such detailed information may be available in some cases.
The Examples
The examples were chosen as successful free open-source projects widely used within their field or application with a quality comparable to or superior to good commercial software products. Several such as FFMPEG and x264 are highly applied and used in the everyday world. Several such as the Pythia/Lund Monte Carlo are primarily scientific research tools. Some such as Octave and LAPACK span both worlds.
FFMPEG is a widely used open source audio/video encoding utility and collection of libraries. FFMPEG can encode and decode a wide range of different audio and video formats and compression schemes including h.264. It incorporates a number of other utilities and libraries. x264 is a widely used open source h.264 video encoder. The Independent JPEG Group disributes a widely used open source JPEG image encoder and decoder. Open CV is a widely used computer vision library incorporating many of the current state of the art computer vision algorithms; it is used in research and in a few commercial products. The Insight Toolkit is a toolkit of image segmentation and registration algorithms, somewhat similar to Open CV in practice, geared towward medical imaging.
The Pythia/Lund Monte Carlo is a widely used program for simulating the formation of jets of subatomic particles and other processes in experimental and theoretical particle physics, for example at the Large Hadron Collider (LHC) at CERN. Two versions, the original FORTRAN version and the more recent rewrite in C++, are listed. Electron Gamma Shower or EGS is a widely used program for simulating the interactions of electrons and photons (gamma rays and x-rays) with matter. It was originally developed for nuclear and particle phyics at the Stanford Linear Accelerator Center (SLAC), but is now widely used for medical radiation studies. LAPACK is a widely used FORTRAN library of linear algebra and other basic numerical algorithms; it is often found in other programs as well. AESCRYPT is a free open-source implementation of the Advanced Encryption Standard (AES) for data encryption. GNU Privacy Guard (GNUpg) is a free, open-source implementation of the OpenPGP encryption standard. Octave is a free, open-source numerical programming tool that is mostly compatibly with MATLAB. Octave has been discussed in previous articles by this author starting with Octave: An Alternative to the High Cost of MATLAB.
Actual Effort Estimation with Basic COCOMO
The Constructive Cost Model (COCOMO) is a software cost estimation model developed by Barry Boehm. Basic COCOMO is the original, very simple cost estimation model published by Boehm in his 1981 book Software Engineering Economics. It gives a simple, crude estimate of the effort in man-months as a function of the number of lines of code in a project. The following table gives the estimated effort in man-months/man-years from applying the "organic" Basic COCOMO model to the number of lines of code in each mathematical open source project in this article:
Program
Basic COCOMO Man-Months
Basic COCOMO Man-Years
FFMPEG 0.6.1
1,204
100
x264
201.5
16.75
IJG v8c
179.8
15
Open CV 2.2.0
2,982
248.5
Insight Toolkit
2,324
193.7
Pythia/Lund 8.145
443
36
Pythia/Lund 6.327
178
14.8
EGS
112
9.3
LAPACK 3.3.0
1,637
136.4
AESCRYPT
11
0.9
GNU Privacy Guard (GNUpg) 1.4.11
456
38
Octave 3.2.4
1,771
147.6
The following Octave/MATLAB function was used to compute the estimated man-months using the Basic COCOMO "organic" model:
While this data sample is clearly limited and a larger study is desirable, it should nonetheless be evident that successful mathematical programming projects are usually substantial. Even the smallest project on the list, the AESCRYPT encryption utility, probably took several man-months to fully develop; Basic COCOMO would estimate almost one year. Thus, expectations of a few weeks are generally unrealistic. Indeed, expectations of three calendar months, a fiscal quarter, the current fetish of American business, are usually unrealistic. On the other hand, expectations ranging from six months to several years may be realistic depending on the specific project.
In part because of heavy government funding of mathematical research and development, there are a large number of open-source, free mathematical programming projects available. This provides an excellent database of information on the size and scope of such projects, something often difficult to find for business applications where most products and projects are proprietary. Anyone considering such a mathematical project is well advised to examine comparable open source projects if they exist to determine the size and scope to the extent possible. Unfortunately, open source projects often can give only a rough measure of the actual effort (mythical man-months) used in the project. The Basic COCOMO model can provide a very rough way of estimating the actual effort of the open source project from the lines of code, but clearly a more direct way of measuring the actual effort isThis is the third in a series of articles on Octave starting with Octave, An Alternative to the High Cost of MATLAB. Octave is a free, both free as in beer and free as in speech, MATLAB compatible numerical programming tool available under the GNU General Public License. In part because MATLAB has become the de facto industry standard for numerical programming, Octave is of particular interest to individuals, companies, and organizations engaged in numerical and mathematical programming and research and development.
Octave has some limitations. The base Octave tool has no symbolic manipulation features. It is not a computer algebra system (CAS) such as Mathematica or Maple. It cannot, for example, perform symbolic integration, symbolic differentiation, factor polynomials, and so forth. Octave does have the symbolic toolbox available through the Octave Forge repository of Octave toolboxes, but the symbolic toolbox is quite limited. A better option for symbolic manipulation tasks is to use the Maxima computer algebra system in combination with Octave. Octave also lacks the ability to generate TeX or LaTex mathematical output, the de facto standard of mathematical publication. Maxima can also generate TeX output for inclusion in papers or WordPress blog posts.
Maxima
Maxima is a computer algebra system descended from MACSYMA, one of the original computer algebra systems. MACSYMA was developed at MIT in part for use in theoretical physics. Maxima is available both as source code and pre-compiled binaries for all three major computer platforms: Unix/Linux, Microsoft Windows, and Mac OS. Maxima is free software, both free as in beer and free as in speech, available under the GNU General Public License (GPL). wxMaxima is a Graphical User Interface (GUI) for Maxima available as both source code and pre-compiled binaries for all three major computer platforms. wxMaxima has human readable menu items and buttons for many common symbolic manipulation and mathematical functions. wxMaxima also has "notebooks" similar to Mathematica notebooks. There is considerable documentation on Maxima; interested readers are referred to the excellent online and published documentation on Maxima. This article is focused on using Maxima as an adjunct to Octave.
wxMaxima on a Macintosh
Maxima can perform both symbolic differentiation and integration. Symbolic differentiation is illustrated in the screen shot of Maxima above. Some optimization algorithms, used, for example, for model fitting, require the derivative of the function being optimized; the function is usually being minimized. If the function is rather complex, deriving the derivatives of the function with respect to the parameters over which the optimization is performed by hand can be time consuming, tedious, and error prone. The author has used Maxima successfully to perform the differentiation of a model function. One can then convert the Maxima output, the derivative produced by Maxima's symbolic differentiation, into an expression that can be used in Octave by using the fortran(expression) command in Maxima. The Maxima fortran command generates FORTRAN code for the Maxima expression. In many cases, the FORTRAN expressions are identical to Octave/MATLAB mathematical expressions. In some cases, the FORTRAN code generated by Maxima must be edited slightly to create valid Octave/MATLAB code.
For example, the Cauchy-Lorentz distribution is a commonly used mathematical model of a peak.
The Cauchy-Lorentz distribution is the frequency response of a forced-damped harmonic oscillator. It is widely used in physics, mathematics, and engineering under a number of different names.
Cauchy Lorentz Distribution
In using the Cauchy Lorentz distribution to model a peak in data, one typically wants to determine the values of the parameters A, mu, and W representing the magnitude of the peak (A), the position of the peak (mu), and the width of the peak (W) that best fits the data. To do this, some model fitting algorithms need the derivatives of the Cauchy Lorentz distribution with respect to each parameter A, mu, and W.
Differentiation of Cauchy Lorentz in Maxima
This is the derivative of the Cauchy Lorentz distribution with respect to the width parameter W from symbolic differentiation in Maxima:
This derivative is moderately complex. Calculating this derivative by hand is time consuming and error prone. Imagine computing the derivative of an extremely complex mathematical model with hundreds of terms by hand. The probability of error even by a highly-skilled mathematician is very high. It was for this reason that tools like Maxima were developed.
This is the FORTAN code generated by applying the Maxima fortran(expression) function to the Maxima expression for the derivative of the Cauchy Lorentz with respect to the width W
2*(x-mu)**2*A/(((x-mu)**2/W**2+1)**2*W**3)
This is actually valid Octave/MATLAB code. If the variables x, mu, A, and W are scalar variables in Octave, this FORTRAN expression will evaluate correctly. Here is the calculation in Octave when x, mu, A, and W are all scalar variables with the value 1.0.
However, in Octave the variable x is often a vector. If x is a vector, the expression above will produce an error in Octave:
octave-3.2.4.exe:25> 2*(x-mu)**2*A/(((x-mu)**2/W**2+1)**2*W**3)
error: for A^b, A must be square
octave-3.2.4.exe:25>
The reason for this error is that in Octave and MATLAB some of the operators such as * and / are not by default interpreted as element by element operators when applied to vectors and matrices. For example, the operator * is matrix multiplication by default in Octave and MATLAB. An element by element operator is an operator that is applied separately to each element in each vector or matrix that is an operand. In Octave and MATLAB, the element by element operators are .*, ./, .+, .-, and so forth. For example, if one has two vectors a and b in Octave, the operator * will give an error:
However, in Octave and MATLAB, one can multiply each element of each vector by the corresponding element of the other vector using the element by element operator .* thus:
octave-3.2.4.exe:36> a .* b
ans =
1 4 9 16 25
octave-3.2.4.exe:37>
In the output of the element by element (or elementwise) operator .*, the first element is 1*1, the second element is 2*2, and so forth.
Thus, the FORTRAN expressions generated by Maxima are not valid Octave/MATLAB code for vectors and matrices, only for scalar variables. One can convert the FORTRAN expression to a valid Octave expression for vectors by converting the non-elementwise operators to element by element operators where the operands are vectors, usually by adding a preceding dot. Here is the edited code for the example derivative:
and then this function is used to compute the value of the derivative of the Cauchy Lorentz distribution with respect to the width parameter W:
data = mydiff_edited(x, 1.0, 1.0, 1.0);
plot(data);
Plot of Vector Derivative
Maxima can also generate valid TeX code for mathematical publication through its built in tex(expression) command. This command can also be invoked through a menu item in the wxMaxima GUI (shown below).
wxMaxima with TeX Menu Item Displayed
Some TeX generated by Maxima:
tex(1/(1+x^2));
generates the TeX code:
$${{1}\over{x^2+1}}$$
which displays in WordPress after removing the $$ tags which WordPress does not need as:
All of the mathematical formulas in this article are TeX generated by Maxima in this way.
Conclusion
Octave has extensive numerical analysis and programming features. Octave has the special advantage that it is mostly compatible with MATLAB, which is currently the de facto industry standard for numerical programming. Most Octave scripts will run under MATLAB and many MATLAB scripts will run under Octave with no changes. If a user or software developer has an occasional need for symbolic manipulation features such as symbolic integration and differentiation, one can use Maxima as an adjunct to Octave. Similarly, one can use Maxima to generate TeX code for mathematical publications. If one needs to perform extensive symbolic manipulation, one may need to use Maxima or similar tools as one's primary tool.
Both Octave and Maxima have the advantage that they are free, both free as in beer and as in speech, and available as source code. There are many cases where a merger, change in corporate strategy, bankruptcy, or even the whim of an executive has resulted in a proprietary development platform being discarded or deprecated to the detriment of end users, developers, and other customers. A well known example is FoxPro, once one of the leading database programs, which Microsoft acquired and has now announced will be discontinued in favor of Microsoft's other database products. In contrast, open source development tools such as Octave and Maxima can be kept alive and indeed improved by their end users, developers, and customers if |
Mathematical Modeling For The Scientific Method - 10 edition
Summary: Part of the International Series in MathematicsMathematical Modeling For The Scientific Methodis ideal for sophomore or junior-level students that need to be grounded in math modeling for their studies in biology, engineering and/or medicine. it reviews what the scientific method is and how it is important and connected to mathematical modeling. it unites topics in statistics, linear algebra, and calculus and how they are interrelated and utilized3258.748599 +$3.99 s/h
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Introductory Mathematics 2
This course aims to consolidate and extend the topics covered in Introductory Mathematics 1. The course includes basic skills and their application to problem solving in the topics of linear function, graphing, probability and statistics.
Available in 2014
* Students will be proficient in their understanding of appropriate mathematical techniques when commencing undergraduate study. * Students will display competency, working individually and in groups. * Students will have acquired critical reasoning and problem solving skills in order to solve mathematical problems. |
How to Solve Word Problems in Geometry - 00 edition
Summary: The easiest way to solve the hardest problems! Geometry's extensive use of figures and visual calculations make its word problems especially difficult to solve. This book picks up where most textbooks leave off, making techniques for solving problems easy to grasp and offering many illustrative examples to make learning easy. Each year more than two million students take high school or remedial geometry courses. Geometry word problems are abstract and especially hard...show more to solve--this guide offers detailed, easy-to-follow solution procedures. Emphasizes the mechanics of problem-solving. Includes worked-out problems and a 50-question self-test with answers. ...show less |
The Hewlett Packard 48G calculator is a perfect choice for the purpose of surveying and engineering. This HP calculator has a separate slot for formatting different functions and storing the formulas. They have a high, powerful contrast screen with good resolution. The Hewlett Packard 48G Calculator has more space for saving the calculation results. The Hewlett Packard 48G Calculators has a built-in program, which can solve all the algebraic and trigonometric problems with 100% accuracy. This scientific calculator is mostly preferred by engineers and students who require accurate and instant results for their calculations. This HP calculator is a hand-held scientific calculator with a high battery backup.
Hewlett Packard30S is the perfect calculator for high school science and math students. This scientific calculator with its two-line display helps perform complex math functions; also it has 250 built-in functions for all your math problems. This Hewlett Packard calculator is extremely user-friendly, so you don't have to worry about confusing keys or formulae |
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Overview
The Ninth Edition of Swokowski Cole's highly respected precalculus text retains the elements that have made it so popular with instructors and students alike; the time-tested exercise sets feature a variety of applications; its exposition is clear; its uncluttered layout is appealing; and the difficulty level of problems is appropriate and consistent. With these elements the authors succeed in preparing students for calculus. PRECALCULUS: FUNCTIONS AND GRAPHS, 9/e is mathematically sound and has excellent problem sets. In this edition, all of the chapters now include numerous technology inserts and examples with specific keystrokes for the TI-83 Plus and the TI-86, ideal for students who are working with a graphing calculator for the first time. The new design of the text makes the technology inserts easily identifiable, allowing professors to skip them if desired |
Connections
lesson from Illuminations asks students to look at different classes of polynomial functions by exploring the graphs of the functions. Students should already have a grasp of linear functions, quadratic functions, and what is meant by a polynomial function. The lesson is intended for grades 9-12 and should take three class periods to complete.Wed, 19 Jan 2011 03:00:03 -0600Abstract Algebra: Online Study Guide
site from Northern Illinois University provides online notes for students using the Abstract Algebra textbook (which is also available online). The materials cover the topics of integers, functions, groups, polynomials, communicative rings, fields, structure of groups, Galois theory and unique factorization. Indexes of definitions and theorems are also available on the site.Thu, 6 Jan 2011 03:00:02 -0600Polynomial Puzzler
algebra lesson helps students explore polynomials by solving puzzles. The activity explains the relationship between expanding and factoring polynomials, as well as factoring trinomials, and multiplying monomials and binomials. The lesson includes an activity sheet, downloadable in PDF format. The material is appropriate for grades 9-12 and should require 1 class period to complete.Fri, 31 Dec 2010 03:00:03 -0600Taylor Polynomials I
by Lang Moore and David Smith for the Connected Curriculum Project, this is a module using differentiation to find coefficients of polynomial approximations to functions that are not polynomials. This is one of a much larger set of learning modules hosted by Duke University.Tue, 15 Jun 2010 03:00:01 -0500Taylor Polynomials II
by Lang Moore and David Smith of the Connected Curriculum Project, this module continues the study of polynomial approximations to functions, concentrating on the region of convergence. This is one within a much larger set of learning modules hosted by Duke University.Wed, 19 May 2010 03:00:02 -0500LessonCorner: Math Worksheets
excellent site, from LessonCorner, helps educators create customized math worksheets for students of all levels. Creating worksheets is free, and visitors can also name them what they wish and print them to handout to students, and there are a number different kinds of math problems to choose from, from the very simple such as telling time, to more complex algebraic problems like adding and multiplying polynomials. The site owners also provide answer keys and take recommendations for new topics of worksheets. The site could be a helpful addition to any math course from elementary to higher education.Wed, 8 Apr 2009 12:50:16 -0500Function Institute
(x,y) functions: linear (slope-intercept, point-slope, and general forms), polynomial (definition, roots, graphs), and exponential (definition, exponential growth, radioactive decay, money matters - simple, compound, and continuous interest, effective annual rate, ordinary annuity, and loans). From the Mathematics area of Zona Land: Education in Physics and Mathematics.Fri, 19 Sep 2008 03:00:05 -0500Elementary Algebra
by HippoCampus, a project of the Monterey Institute for Technology and Education, this free online course "is a study of the basic skills and concepts of elementary algebra, including language and operations on sets, operations on signed numbers, simple linear equations and inequalities in one variable, operations on polynomials (including beginning techniques of factoring), integer exponents, brief introduction to radicals, introduction to graphing, and applications." The course has seven chapters: Basic algebra principles; Linear equations and set theory; Inequalities & absolute values; Graphs of linear equations; Exponents, monomials, and polynomials; Factoring polynomials and solving quadratic equations; and Rational and radical expressions and equations. Each is broken into two or three lessons containing objectives, readings, multimedia components, and sample problems. The Topic View section of the site provides the concepts taught in the course in either alphabetical or sequential order for educators looking for more specific and targeted supporting materials for an introductory algebra classroom.Thu, 17 Jul 2008 03:00:02 -0500Introductory Algebra: Algebra 1B
by HippoCampus, a project of the Monterey Institute for Technology and Education, this free online course follows up on a previous course, Algebra 1A, which "develops algebraic fluency by providing students with the skills needed to solve equations and perform important manipulations with numbers, variables, equations, and inequalities. In addition, the course develops proficiency with operations involving monomial and polynomial expressions." Along with providing a syllabus, the Course View section of the site is broken into three units: Exponents, monomials, and polynomials; Relations, functions, & quadratic equations; and Rational & radical expressions & equations. Each unit has five lessons, and each lesson has objectives, readings, multimedia components, assessments, and answers. Also, for instructors looking for more targeted teaching tools, the Topic View of the course presents both a sequential and alphabetical list of individual concepts covered in the course.Wed, 16 Jul 2008 03:00:03 -0500Mathematics and Student Life Skills
course, designed for Miami Dade Community College, integrates arithmetic and beginning algebra for the undergraduate student. By applying math to real-life situations most students experience during college, the instructors attempt to make math both fun and applicable. The instructors specifically wish to dissipate the anxiety many college students feel when approaching math at an advanced level. Students can use the information provided on this website to help apply mathematical concepts to their own lives, while instructors can use the assignments, syllabus, and lecture notes to create their own relevant assignments in a mathematics course.Mon, 5 May 2008 03:00:13 -0500Art and Algebra
by artist Cynthia Wilson at Spokane Falls Community College, this lesson combines art, geometry, and algebra to create two-dimensional models for abstract paintings. On this page, visitors will find a very brief description, along with materials for creating an in-class project complete with objectives, materials, procedures, and a handout for students. In it, students are asked to analyze the use of geometric shapes in the works of Piet Mondrian and Frank Lloyd Wright and create their own compositions by tiling. It is an excellent resource, which allows educators to illustrate to students the importance of mathematics in other disciplines.Wed, 23 Apr 2008 03:00:02 -0500Algebra Review in Ten Lessons
University of Akron has created these excellent algebra tutorials that review some of the main topics in the discipline. There are ten lessons, which focus on topics like radicals and exponents, basic algebra, expansion, polynomials, functions, and trig curves. Each lesson has a table of contents and interactive resources like quizzes, in-line examples, and exercises. Words that appear in green or brown are hyperlinks; click on them to learn more about that topic. Tutorials are viewed as a PDF file, and users must have Acrobat Reader 3.0 or greater to access them. This is a perfect resource for anyone who needs to refresh their knowledge of basic algebra concepts, and is also great for those who are just learning about the subject.Wed, 9 Apr 2008 03:00:02 -0500Mathematics with Alice
with Alice to a wonderland of math! This website utilizes Lewis Carroll's bright universe and most-recognizable character in order to teach mathematical concepts. Many students may feel as though they have stepped through the looking glass when attempting to learn math, but here Alice can help them find their way back again by explaining unfamiliar mathematical terms and concepts in a clear, demonstrative fashion. Despite the use of a children's book as the organizing theme, this website remains true to the historical Carroll and is designed with community college students in mind. Instructors will also find Alice's assistance helpful either as an assignment or inspiration for their classes.Fri, 4 Apr 2008 03:00:02 -0500Polynomials, Rational Functions
page reserved for the analytic study of polynomial functions studied in calculus classes. History, applications and related fields and subfields; textbooks, reference works, and tutorials; software and tables; selected topics; other web sites with this focus.Fri, 21 Dec 2007 03:00:02 -0600College Algebra Online Tutorials
introduction to this site remarks, "If you need help in college algebra, you have come to the right place." Their statement is accurate, as the staff members at the West Texas A&M University's Virtual Math Lab have done a fine job creating a series of online algebra tutorials for students and anyone else who might be returning to the world of algebra. First-time visitors should look at their online guide to the tutorials to learn how their tutorials are organized. After that, they should feel free to browse through any of the 59 tutorials offered here. Each tutorial contains information about learning objectives, full explanations, and numerous examples of how to correctly solve problems.Mon, 10 Dec 2007 03:00:01 -0600 |
Xtreme Calculations is an advanced calculator software, which has almost all features related to math. It also has an interface similar to the Windows desktop, making it easier to navigate. Apart from math features like solving quadratic, cubic and biquadratic equations and many more math related features, XC also provides a Leisure "app" which includes a few games and e-mailIn 2012, Chicago teachers built a grassroots movement through education and engagement of an entire union membership, taking militant action in the face of enormous structural barriers and a hostile Democratic Party leadership. The teachers won massive concessions from the city and have become a new model for school reform led by teachers themselves, rather than by billionaires. Strike for America is the story of this movement, and how it has become the defining struggle for the labor movement today |
Calculus I
Calculus is the study of rates of change of functions (the derivative), accumulated areas under curves (the integral), and how these two ideas are (surprisingly!) related. The concepts and techniques involved apply to medicine, economics, engineering, physics, chemistry, biology, ecology, geology, and many other fields. Such applications appear throughout the course, but we will focus on understanding concepts deeply and will approach functions from graphical, numeric, symbolic, and descriptive points of view. Conference work will explore additional mathematical topics. This seminar is intended for students planning further study in mathematics or in science, medicine, engineering, economics, or any technical field, as well as students who seek to enhance their logical thinking and problem-solving skills. Facility with high-school algebra and basic geometry are prerequisites for this course. Prior exposure to trigonometry and/or pre-calculus is highly recommended. |
This is an introductory, but also a sufficient book, on MATLAB programming for science and engineering students. Today, it is impossible to avoid the use of MATLAB for effective teaching and learning... More > in science and engineering. This practice and exposure is also important in professional practice after graduation. Holistic MATLAB integrates the potency of MATLAB through symbolic, numeric, graphic, and text programming. It is developed through examples from many disciplines in science and engineering. The development of code is supported by detailed explanation and intuitive exploration is always encouraged throughout the book using quizzes. It is based on the integration of MATLAB in more than dozen courses that the author teaches in a prominent engineering school. It is written in simple style so that the students can self-learn.< Less |
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Math for Healthcare Professionals: Dosage Calculations and Fundamentals of Medication Administration builds upon a student's existing basic math skills and teaches them the more complex solving calculations that are representative of today's healthcare practice. The foundation for learning to calculate dosages and intravenous administration will be based on the ratio and proportion model, which also adapts to the dimensional Analysis method—an increasingly popular method as dosage calculations are universalized.
Math for Healthcare Professionals contains diagnostic test questions at the beginning of each chapter, sample questions throughout each chapter, and diagnostic test questions at the end of the each chapter to analyze proficiency of that subject matter. The print book includes a CD-ROM with additional |
plain-English guide to the basics of trigFrom sines and cosines to logarithms, conic sections, and polynomials, this friendly guide takes the torture out of trigonometry, explaining basic concepts in plain English, offering lots of easy-to-grasp example problems, and adding a dash of humor and fun. It also explains the "why" of trigonometry, using real-world examples that illustrate the value of trigonometry in a variety of careers.Mary Jane Sterling (Peoria, IL) has taught mathematics at Bradley University in Peoria for more than 20 years. She is also the author of the highly successful Algebra For Dummies (0-7645-5325-9). |
First Course of Collegiate Mathematics
9780894645921
ISBN:
0894645927
Publisher: Krieger Publishing Company
Summary: Intended for all students in their freshman year of college, this book is organized around eight fundamental mathematical processes: conjecture, logical argumentation, formal demonstration, algorithmic thinking, correspondence, enumeration, limiting processes, and approximation. Topically, the book cuts across several traditional branches of mathematics including algebra, trigonometry, number theory, and analysis. Bo...th formal demonstrations and problem solving with extensive applications to the physical sciences are stressed. Use of the microcomputer as a working tool is also emphasized throughout the book. This text would be useful in courses such as general college mathematics, introduction to mathematical thinking, and pre-calculus.
Dence, Joseph B. is the author of First Course of Collegiate Mathematics, published under ISBN 9780894645921 and 0894645927. Seven First Course of Collegiate Mathematics textbooks are available for sale on ValoreBooks.com, and seven used from the cheapest price of $0.06 |
Maple is the highly regarded application that lets you quickly and accurately create and simplify all types of
mathematical expressions, from simple equations to complex quadratics. In this primer, we will cover all that is
needed to start using the remarkable program. |
How is that you can walk into a classroom and gain an overall sense of thequality of math instruction taking place there? What contributes to gettingthat sense? In Math Sense, Chris Moynihan explores some of the componentsthat comprise the look, sound, and feel of effective teaching and learning.Does the landscape of the classroom feature such items... more...
How can we solve the national debt crisis? Should you or your child take on a student loan? Is it safe to talk on a cell phone while driving? Are there viable energy alternatives to fossil fuels? What could you do with a billion dollars? Could simple policy changes reduce political polarization? These questions may all seem very different, but theyWhile computational technologies are transforming the professional practice of mathematics, as yet they have had little impact on school mathematics. This pioneering text develops a theorized analysis of why this is and what can be done to address it. It examines the particular case of symbolic calculators (equipped with computer algebra systems) in... more... |
ISBN: 0071464700
Description
This work teaches business-management students all the basic mathematics used in a retail business and follows the standard curriculum of Business Math courses.
From the Back Cover
Make It Your Business to Learn Business Math. Here's the fast and easy way!
Keep your business running in tip-top shape with a firm grasp of the business math required for day-to-day transactions. Whether you need to calculate sales tax or keep records of inventory, experienced math instructor Allan G. Bluman provides a painless and effective approach to mastering the mathematical skills necessary for today's business world.
With Business Math Demystified, you master the subject one simple step at a time -- at your own speed. This unique self-teaching guide offers a quiz at the end of each chapter to pinpoint weaknesses and a 75-question final exam to reinforce the entire book.
Become a savvy business owner or manager with Business Math Demystified:
Follow the standard curriculum of business math courses
Easily master basic business mathematics concepts
Supplement your college studies with this self-study guide and decipher complex terms in an easy-to-read format
Learn how to calculate the markup, sales tax, and discount on items sold
Get through lengthy computations with the sections on using a scientific calculator
Understand depreciation, inventory, promissory notes, financial statements, stocks and bonds, and more
Whether you want help in operating a profitable retail business or need to get through that business math course, Business Math Demystified is the perfect shortcutBusiness Math Demystified (Repost |
books.google.com - From... notions of algebra
Basic notions of algebra
From of sureness of foot and lightness of touch in the exposition... which transports the reader effortlessly across the whole spectrum of algebra...Shafarevich's book - which reads as comfortably as an extended essay - breathes life into the skeleton and will be of interest to many classes of readers; certainly beginning postgraduate students would gain a most valuable perspective from it but... both the adventurous undergraduate and the established professional mathematician will find a lot to enjoy..." Math. Gazette
From inside the book
Review: Basic Notions of Algebra
User Review - Joecolelife - Goodreads
The author explains not just the definitions but also the 'philosophy' of algebra with highly non- trivial examples from geometry,analysis & topology not to mention algebra itself. Examples motivate ...Read full review
Review: Basic Notions Of Algebra
User Review - Joecolelife - Goodreads
An off-hand account of algebra by one of the best authorities of the subject. Recommended as a "serious" pass-time.Read full review |
I am selling my calculus book used for calculus 1-4. I will also include the solution manual for the $125 price. Both books are in like new condition. The Calculus book will include a binder. My email is: regnierz@msoe.edu |
Name and Form — Locating the Patterns of Orderliness That Connect a Function with Its Graph and Describe Numerical Relationships
A
mathematical function quantifies the relationship between two related
quantities and can be used to model change. Functions and their graphs
are essential to all branches of mathematics and their applications.
Topics: domain and range, average rate of change, graphs, functions
(linear, exponential, logarithmic, and quadratic), and applications. (4
credits) Prerequisite: MATH 153 |
entals
This book is a response to those instructors who feel that calculus textbooks are too big. In writing the book James Stewart asked himself:What is ...Show synopsisThis book is a response to those instructors who feel that calculus textbooks are too big. In writing the book James Stewart asked himself:What is essential for a three-semester calculus course for scientists and engineers? Stewart's ESSENTIAL CALCULUS: EARLY TRANSCENDENTALS offers a concise approach to teaching calculus, focusing on major concepts and supporting those with precise definitions, patient explanations, and carefully graded problems. ESSENTIAL CALCULUS: EARLY TRANSCENDENTALS is only 850 pages-two-thirds the size of Stewart's other calculus texts (CALCULUS, Fifth Edition and CALCULUS, EARLY TRANSCENDENTALS, Fifth Edition)-yet it contains almost all of the same topics. The author achieved this relative brevity mainly by condensing the exposition and by putting some of the features on the website ... Despite the reduced size of the book, there is still a modern flavor: Conceptual understanding and technology are not neglected, though they are not as prominent as in Stewart's other books. ESSENTIAL CALCULUS: EARLY TRANSCENDENTALS has been written with the same attention to detail, eye for innovation, and meticulous accuracy that have made Stewart's textbooks the best-selling calculus texts in the world |
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A group Finnish mathematics students, teachers and researchers have made history during the last weekend of September, producing an open-license High School Mathematics book in a three-day booksprint, for the first time in the world.
"Usually writing a school book is a solitary process requiring at least a year of time, for which the author will receive a small financial compensation. Now, the same was achieved with inspired creative force over one weekend and everyone will get to benefit from the fruit of our labour," says project leader Vesa Linja-aho.
As far as as is known, this was the first one-weekend schoolbook sprint in the world. A similar approach has previously been used to produce manuals for open-source computer programs.
The new mathematics book consists of just over hundred pages and three sections: reading areas, applications, functions and equations. An electronic version of the book, as well as its LaTeX source code, is available for download at . A printed version will be available this year. Before printing, the book will be proofread by the community.
All 30 000 Finnish High school students can benefit from this book. The price for a similar mathematics book is around 15 euros, hence the possible value of the book project can be be up to half a million euros.
"The most important aspect of this project is openness and transparency. All development and evolution require copying and revising. This does not occur in current textbooks, which cannot be altered due to copyright reasons," says project member Tommi Sottinen, professor of business mathematics at the University of Vaasa.
The book will be published under open Creative Commons license, so anyone can copy and edit the book freely.
Over 30 math teachers, researchers and students took part in the project, but representatives of the end users were also present. High School seniors Tiina Salola and Anni Saarelainen tested the book as it was written. "We like the book very much as it presents complicated things clearly. The book starts from basics and does not make assumptions on the knowledge level of the student," said Salola and Saarelainen.
The project team hope that this will the beginning of a new way of producing learning materials.
"High-quality free material decreases the cost of studying. Currently even in Finland, where education is free, an individual student will pay over a thousand euros for books during their study years. Reduction in these costs will promote educational equality", says Linja-aho. |
Foundations of Mathematical & Computational Economics
9780324235838
ISBN:
0324235836
Edition: 1 Pub Date: 2006 Publisher: Thomson Learning
Summary: Economics doesn't have to be a mystery anymore. FOUNDATIONS OF MATHEMATICAL AND COMPUTATION ECONOMICS shows you how mathematics impacts economics and econometrics using easy-to-understand language and plenty of examples. Plus, it goes in-depth into computation and computational economics so you'll know how to handle those situations in your first economics job. Get ready for both the test and the workforce with this ...economics textbook.
Dadkhah, Kamran is the author of Foundations of Mathematical & Computational Economics, published 2006 under ISBN 9780324235838 and 0324235836. Four hundred eighty three Foundations of Mathematical & Computational Economics textbooks are available for sale on ValoreBooks.com, one hundred twenty eight used from the cheapest price of $8.40, or buy new starting at $90.32.[read more]
Ships From:Philadelphia, PAShipping:Standard, Second DayComments:Book in 'Good' Condition and will show signs of use, and may contain writing, underlining, &a... [more]24235836 Brand new book. Ship from multiple locations, including USA, UK, ASIA. 3-5 business days Express Delivery to USA/UK/Europe/Asia/Worldwide. Tracking number will be [more]
0324235836 Brand new book. Ship from multiple locations, including USA, UK, ASIA. 3-5 business days Express Delivery to USA/UK/Europe/Asia/Worldwide. Tracking number will be provided. Satisfaction guaranteed.[less] |
COURSE DESCRIPTION
What's the sure road to success in calculus? The answer is simple: Precalculus. Traditionally studied after Algebra II, this mathematical field covers advanced algebra, trigonometry, exponents, logarithms, and much more. These interrelated topics are essential for solving calculus problems, and by themselves
are powerful methods for describing the real world, permeating all areas of science and every branch of mathematics. Little wonder, then, that precalculus is a core course in high schools throughout the country and an important review subject in college.
Unfortunately, many students struggle in precalculus because they fail to see the links between different topics—between one approach to finding an answer and a startlingly different, often miraculously simpler, technique. As a result, they lose out on the enjoyment and fascination of mastering an amazingly useful tool box of problem-solving strategies.
And even if you're not planning to take calculus, understanding the fundamentals of precalculus can give you a versatile set of skills that can be applied to a wide range of fields—from computer science and engineering to business and health care.
Mathematics Describing the Real World: Precalculus and Trigonometry is your unrivaled introduction to this crucial subject, taught by award-winning Professor Bruce Edwards of the University of Florida. Professor Edwards is coauthor of one of the most widely used textbooks on precalculus and an expert in getting students over the trouble spots of this challenging phase of their mathematics education.
"Calculus is difficult because of the precalculus skills needed for success," Professor Edwards points out, adding, "In my many years of teaching, I have found that success in calculus is assured if students have a strong background in precalculus."
A Math Milestone Made Clear
In 36 intensively illustrated half-hour lectures, supplemented by a workbook with additional explanations and problems, Mathematics Describing the Real World takes you through all the major topics of a typical precalculus course taught in high school or college. Those who will especially benefit from Professor Edwards's lucid and engaging approach include
high school and college students currently enrolled in precalculus who feel overwhelmed and want coaching from an inspiring teacher who knows where students stumble;
parents of students, who may feel out of their depth with the advanced concepts taught in precalculus;
those who have finished Algebra II and are eager to get a head start on the next milestone on the road to calculus;
beginning calculus students who want to review and hone their skills in crucial precalculus topics;
anyone motivated to learn precalculus on his or her own, whether as a home-schooled pupil or as an adult preparing for a new career.
The Powerful Tools of Precalculus
With precalculus, you start to see all of mathematics as a unified whole—as a group of often radically different techniques for representing data, analyzing problems, and finding solutions. And you discover that these techniques are ultimately connected in a beautiful way. Perceiving these connections helps you choose the best tool for a given problem:
Algebraic functions: Including polynomial functions and rational functions, these equations relate the input value of a variable to a single output value, corresponding to countless everyday situations in which one event depends on another.
Trigonometry: Originally dealing with the measurement of triangles, this subject has been vastly enriched by the concept of the trigonometric function, which models many types of cyclical processes, such as waves, orbits, and vibrations.
Exponential and logarithmic functions: Often involving the natural base, e, these functions are built on terms with exponents and their inverse, logarithms, and describe phenomena such as population growth and the magnitude of an earthquake on the Richter scale.
Complex numbers: Seemingly logic-defying, complex numbers are based on the square root of –1, designated by the symbol i. They are essential for solving many technical problems and are the basis for the beautiful patterns in fractal geometry.
Vectors: Quantities like velocity have both magnitude and direction. Vectors allow the direction component to be specified in a form that allows addition, multiplication, and other operations that are crucial in fields such as physics.
Matrices: A matrix is a rectangular array of numbers with special rules that permit two matrices to be added or multiplied. Practically any situation where data are collected in columns and rows can be treated mathematically as a matrix.
In addition, Professor Edwards devotes two lectures to conic sections, slicing a cone mathematically into circles, parabolas, ellipses, and hyperbolas. You also learn when it's useful to switch from Cartesian to polar coordinates; how infinite sequences and series lead to the concept of the limit in calculus; and two approaches to counting questions: permutations and combinations. You close with an introduction to probability and a final lecture that features an actual calculus problem, which your experience in precalculus makes ... elementary!
Real-World Mathematics
Believing that students learn mathematics most effectively when they see it in the context of the world around them, Professor Edwards uses scores of interesting problems that are fun, engaging, and often relevant to real life. Among the many applications of precalculus that you'll encounter are these:
Public health: A student with a new strain of flu arrives at college. How long before every susceptible person is infected? An exponential function called the logistic growth model shows how quickly an epidemic spreads.
Surveying: Suppose you have to measure the diagonal width of a marsh without getting wet. It's a simple matter of walking two sides of a triangle on dry land and then using trigonometry to determine the length of the third side that spans the marsh.
Astronomy: One of the most famous cases involving the sine and cosine functions that model periodic phenomena occurred in 1967, when astronomer Jocelyn Bell detected a radio signal from space at 1.3373-second intervals. It proved to be the first pulsar ever observed.
Acoustics: The special properties of an ellipse explain why a person standing at a given spot in the U.S. Capitol's Statuary Hall can hear a whisper from someone standing 85 feet away.
Computer graphics: How do you make an object appear to rotate on a computer screen? Matrix algebra allows you to move each pixel in an image by a specified angle by multiplying two matrices together.
Probability: Have you ever forgotten your four-digit ATM PIN number? What is the probability that you can guess it? A simple calculation shows that you would have to punch numbers nonstop for many hours before being assured of success.
An Adventure in Mathematical Learning
A three-time Teacher of the Year in the College of Liberal Arts and Sciences at the University of Florida, Professor Edwards has a time-tested approach to making difficult material accessible. In Mathematics Describing the Real World, he enlivens his lectures with study tips and a feature he calls "You Be the Teacher," in which he puts you in the professor's shoes by asking how you would design a particular test problem or answer one of the frequently asked questions he gets in the classroom. For example, are all exponential functions increasing? After you hear Professor Edwards's explanation, you'll know that when someone uses the term "exponentially," you should ask, "Do you mean exponential growth or decay?"—for it can go in either direction. He also gives valuable tips on using graphing calculators, pointing out their amazing capabilities—and pitfalls.
LECTURES
36Lectures
Precalculus is important preparation for calculus, but it's also a useful set of skills in its own right, drawing on algebra, trigonometry, and other topics. As an introduction, review the essential concept of the function, try your hand at simple problems, and hear Professor Edwards's recommendations for approaching the course.
The most common type of algebraic function is a polynomial function. As examples, investigate linear and quadratic functions, probing different techniques for finding roots, or "zeros." A valuable tool in this search is the intermediate value theorem, which identifies real-number roots for polynomial functions.
Step into the strange and fascinating world of complex numbers, also known as imaginary numbers, where i is defined as the square root of -1. Learn how to calculate and find roots of polynomials using complex numbers, and how certain complex expressions produce beautiful fractal patterns when graphed.
Investigate rational functions, which are quotients of polynomials. First, find the domain of the function. Then, learn how to recognize the vertical and horizontal asymptotes, both by graphing and comparing the values of the numerator and denominator. Finally, look at some applications of rational functions.
Discover how functions can be combined in various ways, including addition, multiplication, and composition. A special case of composition is the inverse function, which has important applications. One way to recognize inverse functions is on a graph, where the function and its inverse form mirror images across the line y = x.
You have already used inequalities to express the set of values in the domain of a function. Now study the notation for inequalities, how to represent inequalities on graphs, and techniques for solving inequalities, including those involving absolute value, which occur frequently in calculus.
Explore exponential functions—functions that have a base greater than 1 and a variable as the exponent. Survey the properties of exponents, the graphs of exponential functions, and the unique properties of the natural base e. Then sample a typical problem in compound interest.
A logarithmic function is the inverse of the exponential function, with all the characteristics of inverse functions covered in Lecture 5. Examine common logarithms (those with base 10) and natural logarithms (those with base e), and study such applications as the "rule of 70" in banking.
Learn the secret of converting logarithms to any base. Then review the three major properties of logarithms, which allow simplification or expansion of logarithmic expressions—methods widely used in calculus. Close by focusing on applications, including the pH system in chemistry and the Richter scale in geology.
Practice solving a range of equations involving logarithms and exponents, seeing how logarithms are used to bring exponents "down to earth" for easier calculation. Then try your hand at a problem that models the heights of males and females, analyzing how the models are put together.
Finish the algebra portion of the course by delving deeper into exponential and logarithmic equations, using them to model real-life phenomena, including population growth, radioactive decay, SAT math scores, the spread of a virus, and the cooling rate of a cup of coffee.
Trigonometry is a key topic in applied math and calculus with uses in a wide range of applications. Begin your investigation with the two techniques for measuring angles: degrees and radians. Typically used in calculus, the radian system makes calculations with angles easier.
The Pythagorean theorem, which deals with the relationship of the sides of a right triangle, is the starting point for the six trigonometric functions. Discover the close connection of sine, cosine, tangent, cosecant, secant, and cotangent, and focus on some simple formulas that are well worth memorizing.
Trigonometric functions need not be confined to acute angles in right triangles; they apply to virtually any angle. Using the coordinate plane, learn to calculate trigonometric values for arbitrary angles. Also see how a table of common angles and their trigonometric values has wide application.
The graphs of sine and cosine functions form a distinctive wave-like pattern. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. Such behavior occurs throughout nature and led to the discovery of rapidly rotating stars called pulsars in 1967.
Continue your study of the graphs of trigonometric functions by looking at the curves made by tangent, cosecant, secant, and cotangent expressions. Then bring several precalculus skills together by using a decaying exponential term in a sine function to model damped harmonic motion.
For a given trigonometric function, only a small part of its graph qualifies as an inverse function as defined in Lecture 5. However, these inverse trigonometric functions are very important in calculus. Test your skill at identifying and working with them, and try a problem involving a rocket launch.
An equation that is true for every possible value of a variable is called an identity. Review several trigonometric identities, seeing how they can be proved by choosing one side of the equation and then simplifying it until a true statement remains. Such identities are crucial for solving complicated trigonometric equations.
In calculus, the difficult part is often not the steps of a problem that use calculus but the equation that's left when you're finished, which takes precalculus to solve. Hone your skills for this challenge by identifying all the values of the variable that satisfy a given trigonometric equation.
Study the important formulas for the sum and difference of sines, cosines, and tangents. Then use these tools to get a preview of calculus by finding the slope of a tangent line on the cosine graph. In the process, you discover the derivative of the cosine function.
Return to the subject of triangles to investigate the law of sines, which allows the sides and angles of any triangle to be determined, given the value of two angles and one side, or two sides and one opposite angle. Also learn a sine-based formula for the area of a triangle.
Given three sides of a triangle, can you find the three angles? Use a generalized form of the Pythagorean theorem called the law of cosines to succeed. This formula also allows the determination of all sides and angles of a triangle when you know any two sides and their included angle.
Vectors symbolize quantities that have both magnitude and direction, such as force, velocity, and acceleration. They are depicted by a directed line segment on a graph. Experiment with finding equivalent vectors, adding vectors, and multiplying vectors by scalars.
Apply your trigonometric skills to the abstract realm of complex numbers, seeing how to represent complex numbers in a trigonometric form that allows easy multiplication and division. Also investigate De Moivre's theorem, a shortcut for raising complex numbers to any power.
Embark on the first of four lectures on systems of linear equations and matrices. Begin by using the method of substitution to solve a simple system of two equations and two unknowns. Then practice the technique of Gaussian elimination, and get a taste of matrix representation of a linear system.
Deepen your understanding of matrices by learning how to do simple operations: addition, scalar multiplication, and matrix multiplication. After looking at several examples, apply matrix arithmetic to a commonly encountered problem by finding the parabola that passes through three given points.
Get ready for applications involving matrices by exploring two additional concepts: the inverse of a matrix and the determinant. The algorithm for calculating the inverse of a matrix relies on Gaussian elimination, while the determinant is a scalar value associated with every square matrix.
Use linear systems and matrices to analyze such questions as these: How can the stopping distance of a car be estimated based on three data points? How does computer graphics perform transformations and rotations? How can traffic flow along a network of roads be modeled?
In the first of two lectures on conic sections, examine the properties of circles and parabolas. Learn the formal definition and standard equation for each, and solve a real-life problem involving the reflector found in a typical car headlight.
Continue your survey of conic sections by looking at ellipses and hyperbolas, studying their standard equations and probing a few of their many applications. For example, calculate the dimensions of the U.S. Capitol's "whispering gallery," an ellipse-shaped room with fascinating acoustical properties.
How do you model a situation involving three variables, such as a motion problem that introduces time as a third variable in addition to position and velocity? Discover that parametric equations are an efficient technique for solving such problems. In one application, you calculate whether a baseball hit at a certain angle and speed will be a home run.
Take a different mathematical approach to graphing: polar coordinates. With this system, a point's location is specified by its distance from the origin and the angle it makes with the positive x axis. Polar coordinates are surprisingly useful for many applications, including writing the formula for a valentine heart!
Get a taste of calculus by probing infinite sequences and series—topics that lead to the concept of limits, the summation notation using the Greek letter sigma, and the solution to such problems as Zeno's famous paradox. Also investigate Fibonacci numbers and an infinite series that produces the number e.
Counting problems occur frequently in real life, from the possible batting lineups on a baseball team to the different ways of organizing a committee. Use concepts you've learned in the course to distinguish between permutations and combinations and provide precise counts for each.
What are your chances of winning the lottery? Of rolling a seven with two dice? Of guessing your ATM PIN number when you've forgotten it? Delve into the rudiments of probability, learning basic vocabulary and formulas so that you know the odds.
In a final application, locate a position on the surface of the earth with a two-dimensional version of GPS technology. Then close by finding the tangent line to a parabola, thereby solving a problem in differential calculus and witnessing how precalculus paves the way for the next big mathematical adventure.VIDEO OR AUDIO?
Because of the highly visual nature of the subject matter, this course is available exclusively on video. |
MATH 96
Preparation for Elementary Statistics
This single one-semester course can be used to replace the traditional pathway to statistics (Math 88, Math 90, and Math 103 or Math 110).
Math 96 offers a streamlined approach to preparing you for transfer-level statistics at Cuyamaca College (Math 160), but it's nothing like your typical high school algebra course. We will study only the core concepts from arithmetic, pre-algebra, algebra, and introductory statistics that are needed to understand the basics of college-level statistics. We'll do a lot of writing in this course to answer questions such as, "Are adult's breakfast cereals healthier than children's breakfast cereals?" and "Do banks practice discriminatory lending practices?"
This is a new pathway to statistics designed for students who do NOT plan to major in math, science, engineering or business. Furthermore upon successful completion of Math 96, it is assumed that you will enroll in Math 160 Elementary Statistics at Cuyamaca College. Successful completion of Math 96 qualifies you to take transfer-level Math 160 at Cuyamaca College only and does not qualify you to take any other math course. At this time Math 96 does not transfer to other colleges. However, Math 160 meets the degree requirements at Cuyamaca College and transfers to the four-year institutions. So once you successfully complete Math 96 and Math 160, you could be done with the required math for your two-year and/or four-year degrees.
For financial aid purposes, Math 96 will count towards your financial aid if you are receiving financial aid at Cuyamaca College. It will NOT count towards your financial aid if you are receiving financial aid at Grossmont College. If you plan to take this course and have applied for financial aid, please contact an advisor at the financial aid office of either college to help you understand your financial aid options. |
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This is the definitive app for calculus!Simply insert your function into The Calculus Curve Sketching App.Features:- derivative- second derivative- third derivative- antiderivatives- zeros of the func...
Study Guide To Calculus is the complete guide, covering the basics to the advanced, full of pictures and useful reference guides to get you all the information you need to have a compete understanding...
Sci Calculus is a professional scientific and graphics calculator with many useful features. In addition to the classic functions of a scientific calculator this application is also able to calculate ...
Why study harder when you can study smarter?This powerful application contains a rich collection of examples, tutorials, and solvers for the following topics:- Completing the Square- Quadratic Functio...
Calculus has been known to bring students to tears. Now you have an expert in your corner. This application contains a rich collection of examples, tutorials and solvers, crafted by a professional m...
Forget taking down calculus formulas on a paper! Calculus Quick Reference lists down all the important formulas and evaluation techniques used in calculus which makes it easier for you to memorize and...
Use the power of your brain and see how good you can get in this Mental Calculus game== Main Screen ==Change player name or access high score by pressing menuChoose what type of calculus method you wa...
PreCalculus Buddy is a reference manual for students in technical and engineering programs. The app covers hundreds of definitions and rules, and has an user friendly and intuitive interface. Many of ...
You will use it from high school all the way to graduate school and beyond.FeaturesIncludes both Calculus I and IIClear and concise explanationsDifficult concepts are explained in simple termsIllustra...
Calculus may not seem very important to you but the lessons and skills you learn will be with for your whole lifetime!Calculus is the mathematical study of continuous change. It helps you practice and...
For a $Pi donation, you can gain access to the beta release channel for Calculus Tools. New feature updates will be pushed out more quickly, but they may not be entirely stable or fully functional.Ne...
Excel HSC Mathematics Quick Study is the perfect tool for studying and revising on the go! This app is designed specifically for the HSC Mathematics course. There are two parts to the app: 1. HSC stud...
Genius Kids Maths is a funny mental math game for kids of all ages.- The children choose their options by themselves - math drills with the 4 operations (addition, subtraction, multiplication and divi...
This is an informative app consisting of interesting facts, tricks and tips in mathematics.Must have for any school/college going student. It'll help you solve calculations quickly. You can learn tr |
Students explore the concept of average value of a function. In this average value of a function instructional activity, students use their Ti-Nspire to determine the average value of a quadratic function. Students take the integrals of velocity functions using trapezoids.
Students explore the concept of arc length. In this arc length lesson, students find the arc length of a sine curve over a specific interval using integrals. Students use their Ti-Nspire to determine the integrals of sine curves.
In this calculus activity, students use integration to solve word problems they differentiate between integration and anti derivatives, and between definite and indefinite integrals. There are 3 questions with an answer keyEleventh and twelfth graders find the volume of figures using cross sections. They use their Ti-Nspire to find the volume of a solid formed by cross sections of a function. Pupils find the integrals of the shapes using cross sections.
Students approximate the arc length of a cycloid. In this arc length of a cycloid lesson, students solve a problem from an episode of NUMB3RS involving the rotation of a tire. Students take th integral of the the cosine function. Students plot a cycloid as a tire rotates. Students approximate the arc length of a curve by drawing segments spanning the arc.
Twelfth graders investigate the capabilities of the TI-89. In this calculus lesson, 12th graders explore the parametric equation for a circle, for arc length of curves, and for trajectories. Students investigate the symbolic and graphical representation of vectors. Students use polar functions of investigate the area bounded by a curve. Students investigate a 3D graphing application.
Twelfth graders explore an application of the definite integrals. In this calculus instructional activity, 12th graders graph three functions on the same domain and each goes through the same three points. Students use the symbolic capacity of their calculator and calculus to find the shortest of each of the paths through these points. Students explore the piece-wise linear function which also goes through the three points.
Twelfth graders investigate an application of definite integrals. In this Calculus lesson plan, 12th graders use the symbolic capacity of the TI-89 calculator and the concept of the definite integral to explore the area bounded by a function and the x-axis.
Sal continues his discussion of decay by showing students the math involved in determining how much a substance is left after one half-life, two half-lives, and even three half-lives have gone by. He sets up a general function of time that can be used to determine the remaining amount of a substance after 10 minutes, or three billion years have elapsed!
Twelfth graders investigate differential equations. In this calculus lesson, 12th graders are presented with a step-by-step illustrated review of the process used in solving differential equations and an application problem. Students solve differential equations and application of differential equations.
Twelfth graders investigate an application of definite integrals. In this calculus lesson, 12th graders explore the use of definite integral to represent the area bounded by two curves. Students rely on the symbolic capacity of the TI-89 to assist them in solving the problem.
In this Calculus worksheet, students use the TI-89 calculator to compute derivatives and anti-derivatives in order to solve a crossword puzzle. The two page worksheet contains twenty-one problems. Answers are not provided.
Watch this video with your junior engineers or your higher-level mathematicians. In it, Drew builds half of an arch bridge with building blocks. There happens to be a magnetic board behind the blocks, on which he affixes a magnet for each block. When the blocks are removed, what appears to be a graph of an integral function, a natural logarithm, is left behind. Really revealing! |
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Description
Beginning with the origin of the four color problem in 1852, the field of graph colorings has developed into one of the most popular areas of graph theory. Introducing graph theory with a coloring theme, Chromatic Graph Theory explores connections between major topics in graph theory and graph colorings as well as emerging topics.
This self-contained book first presents various fundamentals of graph theory that lie outside of graph colorings, including basic terminology and results, trees and connectivity, Eulerian and Hamiltonian graphs, matchings and factorizations, and graph embeddings. The remainder of the text deals exclusively with graph colorings. It covers vertex colorings and bounds for the chromatic number, vertex colorings of graphs embedded on surfaces, and a variety of restricted vertex colorings. The authors also describe edge colorings, monochromatic and rainbow edge colorings, complete vertex colorings, several distinguishing vertex and edge colorings, and many distance-related vertex colorings.
With historical, applied, and algorithmic discussions, this text offers a solid introduction to one of the most popular areas of graph theory.
Reviews
… The book is written in a student-friendly style with carefully explained proofs and examples and contains many exercises of varying difficulty. … The book is intended for standard courses in graph theory, reading courses and seminars on graph colourings, and as a reference book for individuals interested in graphs colourings.
—Zentralblatt MATH 1169
… well-conceived and well-written book … written in a reader-friendly style, and there is a sufficient number of exercises at the end of each chapter.
—Miklós Bóna, University of Florida, MAA Online, January 2009
Contents
The Origin of Graph Colorings
Introduction to Graphs
Fundamental Terminology
Connected Graphs
Distance in Graphs
Isomorphic Graphs
Common Graphs and Graph Operations
Multigraphs and Digraphs
Trees and Connectivity
Cut Vertices, Bridges, and Blocks
Trees
Connectivity and Edge-Connectivity
Menger's Theorem
Eulerian and Hamiltonian Graphs
Eulerian Graphs
de Bruijn Digraphs
Hamiltonian Graphs
Matchings and Factorization
Matchings
Independence in Graphs
Factors and Factorization
Graph Embeddings
Planar Graphs and the Euler Identity
Hamiltonian Planar Graphs
Planarity versus Nonplanarity
Embedding Graphs on Surfaces
The Graph Minor Theorem
Introduction to Vertex Colorings
The Chromatic Number of a Graph
Applications of Colorings
Perfect Graphs
Bounds for the Chromatic Number
Color-Critical Graphs
Upper Bounds and Greedy Colorings
Upper Bounds and Oriented Graphs
The Chromatic Number of Cartesian Products
Coloring Graphs on Surfaces
The Four Color Problem
The Conjectures of Hajós and Hadwiger
Chromatic Polynomials
The Heawood Map-Coloring Problem
Restricted Vertex Colorings
Uniquely Colorable Graphs
List Colorings
Precoloring Extensions of Graphs
Edge Colorings of Graphs
The Chromatic Index and Vizing's Theorem
Class One and Class Two Graphs
Tait Colorings
Nowhere-Zero Flows
List Edge Colorings
Total Colorings of Graphs
Monochromatic and Rainbow Colorings
Ramsey Numbers
Turán's Theorem
Rainbow Ramsey Numbers
Rainbow Numbers of Graphs
Rainbow-Connected Graphs
The Road Coloring Problem
Complete Colorings
The Achromatic Number of a Graph
Graph Homomorphisms
The Grundy Number of a Graph
Distinguishing Colorings
Edge-Distinguishing Vertex Colorings
Vertex-Distinguishing Edge Colorings
Vertex-Distinguishing Vertex Colorings
Neighbor-Distinguishing Edge Colorings
Colorings, Distance, and Domination
T-Colorings
L(2, 1)-Colorings
Radio Colorings
Hamiltonian Colorings
Domination and Colorings
Epilogue
Appendix: Study Projects
General References
Bibliography
Index (Names and Mathematical Terms)
List of Symbols
Exercises appear at the end of each chapter.
Related Subjects
Name: Chromatic Graph Theory (Hardback) – Chapman and Hall/CRC
Description: By Gary Chartrand, Ping ZhangSeries Editor: Kenneth H. Rosen. Beginning with the origin of the four color problem in 1852, the field of graph colorings has developed into one of the most popular areas of graph theory. Introducing graph theory with a coloring theme, Chromatic Graph Theory explores connections...
Categories: Combinatorics, Discrete Mathematics, Algorithms & Complexity |
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Starting at $1229Blackboard, Access Code Card, Exc with Mini-Excursions
Excursions in Modern Mathematics, Books a la Carte Edition
Excursns In Modrn Mathemtcs W/ Mini-Excursns
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Author Biography
Peter Tannenbaum has bachelor's degrees in Mathematics and Political Science and a Ph. D. in Mathematics, all from the University of California, Santa Barbara. He has held faculty positions at the University of Arizona, Universidad Simon Bolivar (Venezuela), and is currently professor of mathematics at the California State University, Fresno. His current research interests are in the interface between mathematics, politics and behavioral economics. He is also involved in mathematics curriculum reform and teacher preparation. His hobbies are travel, foreign languages and sports. He is married to Sally Tannenbaum, a professor of communication at CSU Fresno, and is the father of three (twin sons and a daughter).
Table of Contents
Part 1. The Mathematics of Social Choice
1. The Mathematics of Voting: The Paradox of Democracy
1.1 Preference Ballots and Preference Schedules
1.2 The Plurality Method
1.3 The Borda Count Method
1.4 The Plurality-with-Elimination Method (Instant Runoff Voting)
1.5 The Method of Piecewise Comparisons
1.6 Rankings
Profile: Kenneth J. Arrow
Key Concepts
Exercises
Projects and Papers
References and Further Readings
2. The Mathematics of Power: Weighted Voting
2.1 An Introduction to Weighted Voting
2.2 The Banzhaf Power Index
2.3 Applications of the Banzhaf Power Index
2.4 The Shapely-Shubik Power Index
2.5 Applications of the Shapely-Shubik Power Index
Profile: Lloyd S. Shapely
Key Concepts
Exercises
Projects and Papers
References and Further Readings
3. The Mathematics of Sharing: Fair-Division Games
3.1 Fair-Division Games
3.2 Two Players: The Divider-Chooser Method
3.3 The Lone-Divider Method
3.4 The Lone-Chooser Method
3.5 The Last-Diminisher Method
3.6 The Method of Sealed Bids
3.7 The Method of Markers
Profile: Hugo Steinhaus
Key Concepts
Exercises
Projects and Papers
References and Further Readings
4. The Mathematics of Apportionment: Making the Rounds
4.1 Apportionment Problems
4.2 Hamilton's Method and the Quota Rule
4.3 The Alabama and Other Paradoxes
4.4 Jefferson's Method
4.5 Adams's Method
4.6 Webster's Method
Historical Note: A Brief History of Apportionment in the United States
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 1: Apportionment Today
Part 2. Management Science
5. The Mathematics of Getting Around: Euler Paths and Circuits
5.1 Euler Circuit Problems
5.2 What is a Graph?
5.3 Graph Concepts and Terminology
5.4 Graph Models
5.5 Euler's Theorems
5.6 Fleury's Algorithm
5.7 Eulerizing Graphs
Profile: Leonard Euler
Key Concepts
Exercises
Projects and Papers
References and Further Readings
6. The Mathematics of Touring: The Traveling Salesman Problem
6.1 Hamilton Circuits and Hamilton Paths
6.2 Complete Graphs
6.3 Traveling Salesman Problems
6.4 Simple Strategies for Solving TSPs
6.5 The Brute-Force and Nearest-Neighbor Algorithms
6.6 Approximate Algorithms
6.7 The Repetitive Nearest-Neighbor Algorithm
6.8 The Cheapest Link Algorithm
Profile: Sir William Rowan Hamilton
Key Concepts
Exercises
Projects and Papers
References and Further Readings
7. The Mathematics of Networks: The Cost of Being Connected
7.1 Trees
7.2 Spanning Trees
7.3 Kruskal's Algorithm
7.4 The Shortest Network Connecting Three Points
7.5 Shortest Networks for Four or More Points
Profile: Evangelista Torricelli
Key Concepts
Exercises
Projects and Papers
References and Further Readings
8. The Mathematics of Scheduling: Chasing the Critical Path
8.1 The Basic Elements of Scheduling
8.2 Directed Graphs (Digraphs)
8.3 Scheduling with Priority Lists
8.4 The Decreasing-Time Algorithm
8.5 Critical Paths
8.6 The Critical-Path Algorithm
8.7 Scheduling with Independent Tasks
Profile: Ronald L. Graham
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 2: A Touch of Color
Part 3. Growth And Symmetry
9. The Mathematics of Spiral Growth: Fibonacci Numbers and the Golden Ratio
9.1 Fibonacci's Rabbits
9.2 Fibonacci Numbers
9.3 The Golden Ratio
9.4 Gnomons
9.5 Spiral Growth in Nature
Profile: Leonardo Fibonacci
Key Concepts
Exercises
Projects and Papers
References and Further Readings
10. The Mathematics of Money: Spending it, Saving It, and Growing It
10.1 Percentages
10.2 Simple Interest
10.3 Compound Interest
10.4 Geometric Sequences
10.5 Deferred Annuities: Planned Savings for the Future
Key Concepts
Exercises
Projects and Papers
References and Further Readings
11. The Mathematics of Symmetry: Beyond Reflection
11.1 Rigid Motions
11.2 Reflections
11.3 Rotations
11.4 Translations
11.5 Glide Reflections
11.6 Symmetry as a Rigid Motion
11.7 Patterns
Profile: Sir Roger Penrose
Key Concepts
Exercises
Projects and Papers
References and Further Readings
12. The Geometry of Fractal Shapes: Naturally Irregular
12.1 The Koch Snowflake
12.2 The Sierpinski Gasket
12.3 The Chaos Game
12.4 The Twisted Sierpinski Gasket
12.5 The Mandelbrot Set
Profile: Benoit Mandelbrot
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 3: The Mathematics of Population Growth: There is Strength in Numbers |
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various areas of calculus. Featuring many diagrams and illustrations, this text highlights the problem-solving skills you need to know. It helps you to shorten your study time, increase your test scores, and get your best possible final grade. Full description
3000 Solved Problems in Linear Algebra Helping you master linear algebra, this guide also ... more
helps you cut study time and hone problem-solving skills. It includes: 3000 solved problems with complete solutions; an index to help you locate the types of problems you want to solve; problems like those you'll find on your exams; and techniques for choosing the correct approach to problems. Full description
category books comics magazines about speedy hen ltd by continuing with this checkout and ordering from speedy hen you are accepting our current terms and conditions details of which can be found by clicking here author s s a nasar content note illustrations country of publication united states date of publication 01 01 1988 format paperback genre level 1 adult non fiction specialist genre level 2 engineerin |
Mathematics for Elementary Teachers-Activity Manual - 3rd edition
Summary: An integral part of the text written by Beckmann herself, the Activities Manual contains fully integrated activities getting students engaged in exploring, discussing, and ultimately reaching a true understanding of mathematics. The manual is included with every new copy of the text.
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$31.6130351 2011 Trade paperback 3rd ed. Good. Trade paperback (US). Glued binding. 496 p. Audience: General/trade. I have for sale a GOOD CONDITION softbound textbook of 469 pages titled "MATHEMATICS FOR...show more ELEMENTARY TEACHERS: ACTIVITY MANUAL" Third Edition, written by Sybilla Beckmann with a copyright year of 2011 by Pearson Education ( ISBN 0-321-64696-7) (30351) This textbook shows moderate cover, corner and edge wear. There is some wear to the outside cover corners, resulting in some minor upturning of the tips of some of the internal page corners near the cover. There is a small amount of front cover curl, as often happens with softcover textbooks. Fanning through the pages, I found a minor amount of markings and/or highlighting, but certainly not excessive |
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It conveys a lively and modern picture of mathematics; emphasises the close links between different branches of mathematics; responds to the enormous advances in mathematics due to the ever-growing influence of computer science and the role of modern PCs; contains a wealth of material ranging from elementary facts to modern and highly-sophisticated results and methods; and includes a comprehensive bibliography of all contemporary literature in the main fields of mathematics.
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"With so much mathematics in compact form, this book will be useful as a quick reference for those working in such fields as physics, engineering, and economics, as well as for mathematicians."-- |
IntroductionIn Differential Equations with Linear Algebra, we endeavor to introduce studentsto two interesting and important areas of mathematics that enjoy powerfulinterconnections and applications. Assuming that students have completed asemester of multivariable calculus, the text presents an introduction to criticalthemes and ideas in linear algebra, and then, in its remaining seven chapters,investigates differential equations while highlighting the role that linearity playsin their study. Throughout the text, we strive to reach the following goals: • To motivate the study of linear algebra and differential equations through interesting applications in order that students may see how theoretical results can answer fundamental questions that arise in physical situations. • To demonstrate the fact that linear algebra and differential equations can be presented as two parts of a mathematical whole that is coherent and interconnected. Indeed, we regularly discuss how the structure of solutions to linear differential equations and systems of equations exemplify important ideas in linear algebra, and how linear algebra often answers key questions regarding differential equations. • To present an exposition that is intended to be read and understood by students. While certainly every textbook is written with students in mind, often the rigor and formality of standard mathematical presentation takes over, and books become difficult to read. We employ an examples-first philosophy that uses an intuitive approach as a lead-in to more general, theoretical results. xi
xii Introduction • To develop in students a deep understanding of what may be their first exposure to post-calculus mathematics. In particular, linear algebra is a fundamental subject that plays a key role in the study of much higher level mathematics; through its study, as well as our investigations of differential equations, we aim to provide a foundation for further study in mathematics for students who are so interested.Whether designed for mathematics or engineering majors, many universitiesoffer a hybrid course in linear algebra and differential equations, and this textis written for precisely such a class. At other institutions, linear algebra anddifferential equations are treated in two separate courses; in settings where linearalgebra is a prerequisite to the study of differential equations, this text may alsobe used for the differential equations course, with its first chapter on linearalgebra available as a review of previously studied material. More details on theways the book can be implemented in these courses follows shortly in the sectionHow to Use this Text. An overriding theme of the book is that if a differentialequation or system of such equations is linear, then we can usually solve itexactly.Linear algebra and systems firstIn most other texts that present the subjects of differential equations and linearalgebra, the presentation begins with first-order differential equations, followedby second- and higher order linear differential equations. Following these topics,a modest amount of linear algebra is introduced before beginning to considersystems of linear differential equations. Here, however, we begin on the veryfirst page of the text with an example that shows the natural way that systemsof linear differential equations arise, and use this example to motivate theneed to study linear algebra. We then embark on a one-chapter introductionto linear algebra that aims not only to introduce such important conceptsas linear combinations, linear independence, and the eigenvalue problem,but also to foreshadow the use of such topics in the study of differentialequations. Following chapter 1, we consider first-order differential equations brieflyin chapter 2, using the study of linear first-order equations to highlight someof the key ideas already encountered in linear algebra. From there, we quicklyproceed to an in-depth presentation of systems of linear differential equationsin chapter 3. In that setting, we show how the eigenvalues of an n × n matrix Anaturally provide the general solution to systems of linear differential equationsin the form x = Ax. Moreover, we include examples that show how anysingle higher order linear differential equation may be converted to a system ofequations, thus providing further motivation for why we choose to study systemsfirst. Through this approach, we again strive to emphasize critical connectionsbetween linear algebra and differential equations and to demonstrate the mostimportant ideas that arise in the study of each. In the remainder of the text, the
Introduction xiiirole of linear algebra is continually emphasized, even in the study of nonlinearequations and systems.Features of the textInstructors and students alike will find several consistent features in thepresentation. • Each chapter begins with one or two motivating problems that present a natural situation—often a physical application—in which linear algebra or differential equations arises. From such problems, we work to develop related ideas in subsequent sections that enable us to solve the original problem. In discussing the motivating problems, we also endeavor to use our intuition to predict the solution(s) we expect to find, and then later test our results against these predictions. • In almost every section of the text, we use an examples-first approach. By this we mean that we introduce a certain type of problem that we are interested in solving, and then consider a relatively simple one that can be solved by intuition or ideas studied previously. From the solution of an elementary example, we then discuss how this approach can be generalized or modified to solve more complex examples, and then ultimately prove or state theorems that provide general results that enable the solution of a wide range of problems. With this philosophy, we strive to demonstrate how the general theory of mathematics comes from experimenting and investigating through individual examples followed by looking for overall trends. Moreover, we often use this approach to foreshadow upcoming ideas: for example, while studying linear algebra, we look ahead to a handful of fundamental differential equations. Similarly, early on in our investigations of the Laplace transform, we regularly attempt to demonstrate through examples how the transform will be used to solve initial-value problems. • While there are many formal theoretical results that hold in both linear algebra and differential equations, we have endeavored to emphasize intuition. Specifically, we use the aforementioned examples-first approach to solve sample problems and then present evidence as to why the details of the solution process for a small number of examples can be generalized to an overall structure and theory. This is in contrast to many books that first present the overall theory, and then demonstrate the theory at work in a sequence of subsequent examples. In addition, we often eschew formal proofs, choosing instead to present more heuristic or intuitive arguments that offer evidence of the truth of important theorems. • Wherever possible, we use visual reasoning to help explain important ideas. With over 100 graphics included in the text, we have provided
xiv Introduction figures that help deepen students' understanding and offer additional perspective on essential concepts. By thinking graphically, we often find that an appropriate picture sheds further light on the solution to a problem and how we should expect it to behave, thus adding to our intuition and understanding. • With computer algebra systems (CASs), such as Maple and Mathematica, approaching their twentieth year of existence, these technologies are an important part of the landscape of the teaching and learning of mathematics. Especially in more sophisticated subjects with computationally complicated problems, these tools are now indispensable. We have chosen to integrate instructional support for Maple directly within the text, while offering similar commentary for Mathematica, MATLAB, and SAGE on our website, differentialequations/. For each, students can find directions for how to effectively use computer algebra systems to generate important graphs and execute complicated or tedious calculations. Many sections of the text are followed by a short subsection on "Using Maple to . . .." Parallel sections for the other CASs, numbered similarly, can be found on the website. • Each chapter ends with a section titled For further study. In this setting, rather than a full exposition, a sequence of leading questions is presented to guide students to discover some key ideas in more advanced problems that arise naturally from the material developed to date. These sections can be used as a basis for instructor-led in-class discussions or as the foundation for student projects or other assignments. Interested students can also pursue these topics on their own.How to use this textThere are two courses for which this text is well-suited: a hybrid course in linearalgebra and differential equations, or a course in differential equations thatrequires linear algebra as a prerequisite. We address each course separately withsome suggestions for instructors.Linear algebra and differential equationsFor a hybrid course in the two subjects, instructors should begin with chapter 1on linear algebra. There, in addition to an introduction to many essentialideas in the subject, students will encounter a handful of examples on lineardifferential equations that foreshadow part of the role of linear algebra in thefield of differential equations. The goal of the chapter on linear algebra is tointroduce important ideas such as linear combinations, linear independenceand span, matrix algebra, and the eigenvalue problem. At the close of chapter 1
Introduction xvwe also introduce abstract vector spaces in anticipation of the structural rolethat vector spaces play in solving linear systems of differential equations andhigher order linear differential equations. Instructors may choose to move onfrom chapter 1 upon completing section 1.10 (the eigenvalue problem), as thisis the last topic that is absolutely essential for the solution of linear systems ofdifferential equations in chapter 3. Discussion of ideas like basis, dimension,and vector spaces of functions from the final two sections of chapter 1 can occuralongside the development of general solutions to systems of linear differentialequations or higher order linear differential equations. Over the past decade or two, first-order differential equations have becomea standard topic that is normally discussed in calculus courses. As such,chapter 2 can be treated lightly at the instructor's discretion. In particular, itis reasonable to expect that students are familiar with direction fields, separabledifferential equations, Euler's method, and several fundamental applications,such as Newton's law of Cooling and the logistic differential equation. It isless likely that students will have been exposed to integrating factors as asolution technique for linear first-order equations and the solution methodsfor exact equations. In any case, chapter 2 is not one on which to linger.Instructors can choose to selectively discuss a small number of sections in class,or assign the pages there as a reading assignment or project for independentinvestigation. Chapter 3 on systems of linear differential equations is the heart of thetext. It can be begun immediately following section 1.10 in chapter 1. Here wefind not only a large number of rich ideas that are important throughout thestudy of differential equations, but also evidence of the essential role that linearalgebra plays in the solution of these systems. As is noted on several occasionsin chapter 3, any higher order linear differential equation may be converted toa system of first-order equations, and thus an understanding of systems enablesone to solve these higher order equations as well. Thus, the material in chapter 4may be de-emphasized. Instructors may choose to provide a brief overview, inclass, of how the ideas in solving linear systems translate naturally to the higherorder case, or may choose to have students investigate these details on their ownthrough a sequence of reading and homework assignments or a group project.Section 4.5 on beats and resonance is one to discuss in class as these phenomenaare fascinating and important and the perspective of higher order equations is amore natural context in which to consider their solution. The Laplace transform is a topic that affords discussion of a variety ofimportant ideas: linear transformations, differentiation and integration, directsolution of initial-value problems, discontinuous forcing functions, and more.In addition, it can be viewed as a gateway to more sophisticated mathematicaltechniques encountered in more advanced courses in mathematics, physics,and engineering. Chapter 5 is written with the goal of introducing studentsto the Laplace transform from the perspective of how it can be used to solveinitial-value problems. This emphasis is present throughout the chapter, andculminates in section 5.5.
xvi Introduction Finally, a course in both linear algebra and differential equations shouldnot be considered complete until there has been at least some discussionof nonlinearity. Chapter 6 on nonlinear higher order equations and systemsoffers an examination of this concept from several perspectives, all of whichare related to our previous work with linear differential equations. Directionfields, approximation by linear systems, and an introduction to numericalapproximation with Euler's method are natural topics with which to round outthe course. Due to the time required to introduce the subject of linear algebrato students, the final two chapters of the text (on numerical methods and seriessolutions) are ones we would normally not expect to be considered in a hybridcourse.Differential equations with a linear algebra prerequisiteFor a differential equations course in which students have already taken linearalgebra, chapter 1 may be used as a reference for students, or as a source of reviewas needed. The comments for the hybrid course above for chapters 2–5 hold fora straight differential equations class as well, and we would expect instructorsto use the time not devoted to the study of linear algebra to focus more onthe material on nonlinearity in chapter 6, numerical methods in chapter 7, andseries solutions in chapter 8. The first several sections of chapter 7 may be treatedany time after first-order differential equations have been discussed; only thefinal section in that chapter is devoted to systems and higher order equationswhere the methods naturally generalize work with first-order equations. In addition to spending more time on the final three chapters of the text,instructors of a differential equations-only course can take advantage of themany additional topics for consideration in the For further study sections thatclose each chapter. There is a wide range of subjects from which to choose, boththeoretical and applied, including discrete dynamical systems, how raindropsfall, matrix exponentials, companion matrices, Laplace transforms of periodicpiecewise continuous forcing functions, and competitive species.AppendicesFinally, the text closes with five appendices. The first three—on integrationtechniques, polynomial zeros, and complex numbers—are intended as a reviewof familiar topics from courses as far back in students' experience as high schoolalgebra. The instructor can refer to these topics as necessary and encouragestudents to read them for review. Appendix D is different in that it aims toconnect some key ideas in linear algebra and differential equations through amore sophisticated viewpoint: linear transformations of vector spaces. Someof the material there is appropriate for consideration following chapter 1,but it is perhaps more suited to discussion after the Laplace transform hasbeen introduced. Finally, appendix E contains answers to nearly all of theodd-numbered exercises in the text.
Introduction xviiAcknowledgmentsWe are grateful to our institutions for the time and support provided to workon this manuscript; to several anonymous reviewers whose comments haveimproved it; to our students for their feedback in classroom-testing of the text;and to all students and instructors who choose to use this book. We welcomeall comments and suggestions for improvement, while taking full responsibilityfor any errors or omissions in the text. Matt Boelkins/J. L. Goldberg/Merle Potter
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Differential Equations with Linear Algebra
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1Essentials of linear algebra1.1 Motivating problemsThe subjects of differential equations and linear algebra are particularlyimportant because each finds a wide range of applications in fundamentalphysical problems. We consider two situations that involve systems of equationsto motivate our work in this chapter and much of the remainder of the text. The pollution of bodies of water is an important issue for humankind.Environmental scientists are particularly interested in systems of rivers andlakes where they can study the flow of a given pollutant from one body of waterto another. For example, there is great concern regarding the presence of avariety of pollutants in the Great Lakes (Lakes Michigan, Superior, Huron, Erie,and Ontario), including salt due to snow melt from highways. Due to the largenumber of possible ways for salt to enter and exit such a system, as well as themany lakes and rivers involved, this problem is mathematically complicated. Butwe may gain a feel for how one might proceed by considering a simple system oftwo tanks, say A and B, where there are independent inflows and outflows fromeach, as well as two pipes with opposite flows connecting the tanks as picturedin figure 1.1. We will let x1 denote the amount of salt (in grams) in A at time t (inminutes). Since water flows into and out of the tank, and each such flow carriessalt, the amount of salt x1 is changing as a function of time. We know fromcalculus that dx1 /dt measures the rate of change of salt in the tank with respectto time, and is measured in grams per minute. In this basic model, we can seethat the rate of change of salt in the tank will be the difference between the netrate of salt flowing in and the net rate of salt flowing out. 3
4 Essentials of linear algebra A B Figure 1.1 Two tanks with inflows, outflows, and connecting pipes. As a simplifying assumption, we will suppose that the volume of solution ineach tank remains constant and all inflows and outflows happen at the identicalrate of 5 liters per minute. We will further assume that the tanks are uniformlymixed so that the salt concentration in each is identical throughout the tank ata given time t . Let us now suppose that the volume of tank A is 200 liters; as we just noted,the pipe flowing into A delivers solution at a rate of 5 liters per minute. Moreover,suppose that this entering water is contaminated with 4 g of salt per liter. Ananalysis of the units on these quantities shows that the rate of inflow of saltinto A is 5 liters 4 g g · = 20 (1.1.1) min liter minThere is one other inflow to consider, that being the pipe from B, which we willconsider momentarily after first examining the behavior of the outflow. For the solution exiting the drain from A at a rate of 5 liters/min, observeits concentration is unknown and depends on the amount of salt in the tank attime t . In particular, since there are x1 g of salt in the tank at time t , and thisis distributed over the volume of 200 liters, we can say (using the simplifyingassumption that the tank's contents stay uniformly mixed) that the rate ofoutflow of salt in each of the exiting pipes is 5 liters x1 g x1 g · = (1.1.2) min 200 liters 40 minSince there are two such exit flows, this means that the combined rate of outflowof salt from A is twice this amount, or x1 /20 g/min. Finally, there is one last inflow to consider. Note that solution from B isentering A at a rate of 5 liters per minute. If we assume that B has a (constant)volume of 400 liters, this flow has a salt concentration of x2 g/400 liters. Thusthe rate of salt entering A from B is 5 liters x2 g x2 g · = (1.1.3) min 400 liters 80 min
Motivating problems 5Combining the rates of inflow (1.1.1) and (1.1.3) and outflow (1.1.2), whereinflows are considered positive and outflows negative, leads us to the differentialequation dx1 x2 x1 = 20 + − (1.1.4) dt 80 20 Since we have two tanks in the system, there is a second differential equationto consider. Under the assumptions that B has a volume of 400 liters, the pipeentering B carries a concentration of salt of 7 g/liter, and the net rates of inflowand outflow match those into A, a similar analysis to the above reveals that dx2 x1 x2 = 35 + − (1.1.5) dt 40 40Together, these two DEs form a system of DEs, given by dx1 x2 x1 = 20 + − (1.1.6) dt 80 20 dx2 x1 x2 = 35 + − dt 40 40 Systems of DEs are therefore, seen to play a key role in environmentalprocesses. Indeed, they find application in studying the vibrations of mechanicalsystems, the flow of electricity in circuits, the interactions between predatorsand prey, and much more. We will begin our examination of the mathematicsinvolved with systems of differential equations in chapter 3. An important question related to the above system of DEs leads us to amore familiar mathematical situation, one that is the foundation of much of thesubject of linear algebra. For the system of tanks above, we might ask, "underwhat circumstances is the amount of salt in the two tanks not changing?" Insuch a situation, neither x1 nor x2 varies, so the rate of change of each is zero,and therefore dx1 dx2 = =0 dt dtSubstituting these values into the system of DEs, we see that this results in thesystem of linear equations x2 x1 0 = 20 + − (1.1.7) 80 20 x1 x2 0 = 35 + − 40 40Multiplying both sides of the first equation by eighty and the second by fortyand rearranging terms, we find an equivalent system to be 4x1 − x2 = 1600 x1 − x2 = −1400 Geometrically, this system of linear equations represents the set of all pointsthat simultaneously lie on each of the two lines given by the respective equations.
6 Essentials of linear algebraThe solution of such 2 × 2 systems is typically discussed in introductory algebraclasses where students learn how to solve systems like these with the methodsof substitution and elimination. Doing so here leads to the unique solutionx1 = 1000, x2 = 2400; one interpretation of this ordered pair is that the systemof two tanks has an equilibrium state where, if the two tanks ever reach thislevel of salinity, that salinity will then stay constant. With further study oflinear algebra and DEs, we will be able to show that over time, regardless ofhow much salt is initially in each tank, the amount of salt in A will approach1000 g, while that in B will approach 2400 g. We will thus call the equilibriumpoint stable. Electrical circuits are another physical situation where systems of linearequations naturally arise. Flow of electricity through a collection of wires issimilar to the flow of water through a sequence of pipes: current measures theflow of electrons (charge carriers) past a given point in the circuit. Typically,we think about a battery as a source that provides a flow of electricity, wiresas a collection of paths along which the electricity may flow, and resistorsas places in the circuit where electricity is converted to some sort of outputsuch as heat or light. While we will discuss the principles behind the flowof electricity in more detail in section 3.8, for now a basic understanding ofKirchoff's laws enables us to see an important application of linear systemsof equations. In a given loop or branch j of a circuit, current is measured in amperes (A)and is denoted by the symbol Ij . Resistances are measured in ohms ( ), and theenergy produced by the battery is measured in volts. As shown in figure 1.2, weuse arrows in the circuit to represent the direction of flow of the current; when 10V + − I3 6Ω I3 I2 I2 a b 2Ω I1 4Ω 3Ω I1 + − 5V Figure 1.2 A simple circuit with two loops, two batteries, and four resistors.
Motivating problems 7this flow is away from the positive side of a battery (the circles in the diagram),then the voltage is taken to be positive. Otherwise, the voltage is negative. Two fundamental laws govern how the currents in various loops of thecircuit behave. One is Kirchoff's current law, which is essentially a conservationlaw. It states that the sum of all current flowing into a node equals the sum ofthe current flowing out. For example, in figure 1.2 at junction a, I1 + I3 = I2 (1.1.8)Similarly, at junction b, we must have I2 = I1 + I3 . This equation is identicalto (1.1.8) and adds no new information about the currents. Ohm's law governs the flow of electricity through resistors, and states thatthe voltage drop across a resistor is proportional to the current. That is, V = IR,where R is a constant that is the amount of resistance, measured in ohms. Forinstance, in the circuit given in figure 1.2, the voltage drop through the 3-resistor on the bottom right is V = 3 . Kirchoff's voltage law states that, in anyclosed loop, the sum of the voltage drops must be zero. Since the battery that ispresent maintains a constant voltage, it follows that in the bottom loop of thegiven circuit, 4I1 + 2I2 + 3I1 = 5 (1.1.9)Similarly, in the upper loop, we have 6I3 + 2I2 = 10 (1.1.10)Finally, in the outer loop, taking into account the direction of flow of electricityby regarding opposing flows as having opposing signs, we observe 6I3 − 4I1 − 3I1 = −5 + 10 (1.1.11)Taking (1.1.8) through (1.1.11), combining like terms, and rearranging each sothat indices are in increasing order, we have the system of linear equations I 1 − I2 + I3 = 0 7I1 + 2I2 =5 (1.1.12) 2I2 + 6I3 = 10 −7I1 + 6I3 = 5We will call the system (1.1.12) a 4 × 3 system to represent the fact that it is acollection of four linear equations in three unknown variables. Its solution—theset of all possible values of (I1 , I2 , I3 ) that make all four equations simultaneouslytrue—provides the current in each loop of the circuit. In this first chapter, we will develop our understanding of the more generalsituation of systems of linear equations with m linear equations in n unknownvariables. This problem will lead us to consider important ideas from the theoryof matrices that play key roles in a variety of applications ranging from computergraphics to population dynamics; related ideas will find further applications inour subsequent study of systems of differential equations.
8 Essentials of linear algebra1.2 Systems of linear equationsLinear equations are the simplest of all possible equations and are involvedin many applications of mathematics. In addition, linear equations play afundamental role in the study of differential equations. As such, the notionof linearity will be a theme throughout this book. Formally, a linear equation invariables x1 , . . . , xn is one having the form a1 x1 + a2 x2 + · · · + an xn = b (1.2.1)where the coefficients a1 , . . . , an and the value b are real or complex numbers.For example, 2x1 + 3x2 − 5x3 = 7is a linear equation, while 2 x1 + sin x2 − x3 ln x1 = 5is not. Just as the equation 2x1 + 3x2 = 7 describes a line in the x1 –x2 plane, thelinear equation 2x1 + 3x2 − 5x3 = 7 determines a plane in three-dimensionalspace. A system of m linear equations in n unknown variables is a collection of mlinear equations in n variables, say x1 , . . . , xn . We often refer to such a system asan "m × n system of equations." For example, x1 + 2x2 + x3 = 1 (1.2.2) x1 + x2 + 2x3 = 0is a system of two linear equations in three unknown variables. A solution to thesystem is any point (x1 , x2 , x3 ) that makes both equations simultaneously true;the solution set for (1.2.2) is the collection of all such solutions. Geometrically,each of these two equations describes a plane in three-dimensional space, asshown in figure 1.3, and hence the solution set consists of all points that lie onboth of the planes. Since the planes are not parallel, we expect this solution set to 10 x3 x1+ 2x2+ x3 = 1 5 x1+ x2+ 2x3 = 0 2 x2 x1 2 Figure 1.3 The intersection of the planes x1 + 2x2 + x3 = 1 and x1 + x2 + 2x3 = 0.
Systems of linear equations 9form a line in R3 . Note that R denotes the set of all real numbers; R3 representsfamiliar three-dimensional Euclidean space, the set of all ordered triples withreal entries. The solution set for the system (1.2.2) may be determined using elementaryalgebraic steps. We say that two systems are equivalent if they share the samesolution set. For example, if we multiply both sides of the first equation by −1and add this to the second equation, we eliminate x1 in the second equation andget the equivalent system x1 + 2x2 + x3 = 1 −x2 + x3 = −1Next, we multiply both sides of the second equation by −1 to get x1 + 2x2 + x3 = 1 x 2 − x3 = 1Finally, if we multiply the second equation by −2 and add it to the first equation,it follows that x1 + 3x3 = −1 (1.2.3) x2 − x3 = 1This shows that any solution (x1 , x2 , x3 ) of the original system must satisfy the(simpler) equivalent system of equations x1 = −1 − 3x3 and x2 = 1 + x3 . Saiddifferently, any point in R3 of the form (−1 − 3x3 , 1 + x3 , x3 ), where x3 ∈ R (herethe symbol '∈' means is an element of ), is a solution to the system. Replacingx3 by the parameter t , we recognize that the solution to the system is the lineparameterized by (−1 − 3t , 1 + t , t ), t ∈ R (1.2.4)which is the intersection of the two planes with which we began, as seen infigure 1.3. Note that this shows there are infinitely many solutions to the givensystem of equations; a particular example of such a solution may be found byselecting any value of t (i.e., any point on the line). We can also check that theresulting point makes both of the original equations true. It is not hard to see in the 2 × 2 case that any linear system has either nosolution (the lines are parallel), a unique solution (the lines intersect once), orinfinitely many solutions (the two equations represent the same line). Thesethree options (no solution, exactly one solution, or infinitely many) turn out tobe the only possible cases for any m × n system of linear equations. A systemwith at least one solution is said to be consistent, while a system with no solutionis called inconsistent. In our work above from (1.2.2) to (1.2.3) in reducing the given system ofequations to a simpler equivalent one, it is evident that the coefficients of thesystem played the key role, while the variables x1 , x2 , and x3 (and the equals sign)were essentially placeholders. It proves expedient to therefore change notationand collect all of the coefficients into a rectangular array (called a matrix) andeliminate the redundancy of repeatedly writing the variables. Let us reconsider
10 Essentials of linear algebraour above work in this light, where we will now refer to rows in the coefficientmatrix rather than equations in the original system. When we create a right-mostcolumn consisting of the constants from the right-hand side of each equation,we often say we have an augmented matrix. From the 'simplest' version of the system at (1.2.3), the correspondingaugmented matrix is 1 0 3 −1 0 1 −1 1The 0's represent variables that have been eliminated in each equation. Fromthis, we see that our goal in working with a matrix that represents a systemof equations is essentially to introduce as many zeros as possible throughoperations that do not change the solution set of the system. We now repeat theexact same steps we took with the system above, but translate our operations tobe on the matrix, rather than the equations themselves. We begin with the augmented matrix 1 2 1 1 1 1 2 0To introduce a zero in the bottom left corner, we add −1 times the first row tothe second row, to yield a new row 2 and the updated matrix 1 2 1 1 0 −1 1 −1The '0' in the second entry of the first column shows that we have eliminatedthe presence of the x1 variable in the second equation. Next, we can multiplyrow 2 by −1 to obtain an updated row 2 and the augmented matrix 1 2 1 1 0 1 −1 1Finally, if we multiply row 2 by −2 and add this to row 1, we find a new row 1and the matrix 1 0 3 −1 0 1 −1 1At this point, we have introduced as many zeros as possible1 , and have arrivedat our goal of the simplest possible equivalent system. We can reinterpret thematrix as a system of equations: the first row implies that x1 + 3x3 = −1, whilethe second row implies x2 − x3 = 1. This leads us to find, as we did above,that any solution (x1 , x2 , x3 ) of the original system must be of the form (−1 −3x3 , 1 + x3 , x3 ), where x3 ∈ R.1 Any additional row operations to introduce zeros in the third or fourth columns will replace thezeros in columns 1 or 2 with nonzero entries.
Systems of linear equations 11 We will commonly need to refer to the number of rows and columns in amatrix. For example, the matrix 1 0 3 −1 0 1 −1 1has two rows and four columns; therefore, we say this is a 2 × 4 matrix. Ingeneral, an m × n matrix has m rows and n columns. Observe that if wehave a 2 × 3 system of equations, its corresponding augmented matrix will be2 × 4. The above example demonstrates the general fact that there are basicoperations we can perform on an augmented matrix that, at each stage, resultin the matrix representing an equivalent system of equations; that is, theseoperations do not change the solution to the system, but rather make the solutionmore easily obtained. In particular, we may 1. Replace one row by the sum of itself and a multiple of another row; 2. Interchange any two rows; or 3. Scale a row by multiplying every entry in a given row by a fixed nonzero constant.These three types of operations are typically called elementary row operations.Two matrices are row equivalent if there is a sequence of elementary rowoperations that transform one matrix into the other. When matrices are usedto represent systems of linear equations, as was done above, it is always the casethat row-equivalent matrices correspond to equivalent systems. We desire to use elementary row operations systematically to produce rowequivalent matrices from which we may easily interpret the solution to a systemof equations. For example, the solution to the system represented by ⎡ ⎤ 1 0 0 −5 ⎣0 1 0 6⎦ (1.2.5) 0 0 1 −3is easy to obtain (in particular, x1 = −5, x2 = 6, x3 = −3), while the solution for ⎡ ⎤ 3 −2 4 −39 ⎣−1 2 7 −4⎦ 6 9 −3 33is not, even though the two matrices are equivalent. Therefore, we desire eachvariable in the system to be represented in its corresponding augmented matrixas infrequently as possible. Essentially our goal is to get as many columns of thematrix as possible to have one entry that is 1, while all the rest of the entries inthat column are 0.
12 Essentials of linear algebra A matrix is said to be in reduced row echelon form (RREF) if and only ifthe following characteristics are satisfied: • All nonzero rows are above any rows with all zeros • The first nonzero entry (or leading entry) in a given row is 1 and is in a column to the right of the first nonzero entry in any row above it • Every other entry in a column with a leading 1 is 0For example, the matrix in (1.2.5) is in RREF, while the matrix ⎡ ⎤ 1 −2 4 −5 ⎣0 2 7 6⎦ 0 0 −3 −3is not, since two of the rows lack leading 1's, and columns 2 and 3 lack zeros inthe entries above the lowest nonzero locations. Each leading 1 in RREF is said to be in a pivot position, the column in whichthe 1 lies is termed a pivot column, and the leading 1 itself is called a pivot.Rows with all zeros do not contain a pivot position. The process by which rowoperations are applied to a matrix to convert it to RREF is usually called Gauss–Jordan elimination. We will also say that we "row-reduced" a given matrix.While this process can be described in a somewhat cumbersome algorithm, it isbest demonstrated with a few examples. By working through the details of thefollowing problems (in particular by deciding which elementary row operationswere performed at each stage), the reader will not only learn the basics of rowreduction, but also will see and understand the key possibilities for the solutionset of a system of linear equations.Example 1.2.1 Solve the system of equations 3x1 + 2x2 − x3 = 8 x1 − 4x2 + 2x3 = −9 (1.2.6) −2x1 + x2 + x3 = −1Solution. We begin with the corresponding augmented matrix ⎡ ⎤ 3 2 −1 8 ⎣ 1 −4 2 −9⎦ −2 1 1 −1and then perform a sequence of row operations. The arrows below denote thefact that one or more row operations have been performed to produce a rowequivalent matrix. We find that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 3 2 −1 8 1 −4 2 −9 1 −4 2 −9 ⎣ 1 −4 2 −9⎦ → ⎣ 3 2 −1 8⎦ → ⎣0 14 −7 35⎦ → −2 1 1 −1 −2 1 1 −1 0 −7 5 −19
14 Essentials of linear algebradid not contribute any restrictions on the system. Moreover, as the matrix isnow in RREF, we can see that the simplest equivalent system is given by thetwo equations x1 + x3 = 3 and x2 − x3 = −1. In other words, x1 = 3 − x3 andx2 = −1 + x3 . Since the variable x3 has no restrictions on it, we call x3 a freevariable. This implies that the system under consideration has infinitely manysolutions, each having the form (3 − t , −1 + t , t ), where t ∈ R (1.2.8) In the next section, we will begin to emphasize the role that vectors play insystems of linear equations. For example, the ordered triple (3 − t , −1 + t , t )in (1.2.8) may be viewed as a vector in R3 . In addition, the representation (1.2.8)of the set of all solutions involving the parameter t is often called the parametricvector form of the solution. As we saw in the very first system of equationsdiscussed in this section, example 1.2.2 shows that the three planes given in thesystem (1.2.7) meet in a line.Example 1.2.3 Solve the system of equations x1 + 2x2 − x3 = 1 x1 + x2 =2 3x1 + x2 + 2x3 = 7Solution. Observe that the only difference between this example and theprevious one is that the "8" in the third equation has been replaced with "7."We proceed with identical row operations to those above and find that⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 2 −1 1 1 2 −1 1 1 2 −1 1 1 0 1 3⎣1 1 0 2⎦ → ⎣0 −1 1 1⎦ → ⎣0 1 −1 −1⎦ → ⎣0 1 −1 −1⎦ 3 1 2 7 0 −5 5 4 0 −5 5 4 0 0 0 −1In this case, the final row of the reduced matrix corresponds to the equation0x1 + 0x2 + 0x3 = −1. Since there are no points (x1 , x2 , x3 ) that make thisequation true, it follows that there can be no points which simultaneously satisfyall three equations in the system. Said differently, the three planes given in theoriginal system of equations do not meet at a single point, nor do they meet ina line. Therefore, the system has no solution; recall that we call such a systeminconsistent. Note that the only difference between example 1.2.2 and example 1.2.3 is oneconstant in the righthand side in the equation of one of the planes. This changedthe result dramatically, from the case where the system had infinitely manysolutions to one where no solutions were present. This is evident geometricallyif we think about a situation where three planes meet in a line, and then we alterthe equation of one of the planes to shift it to a new plane parallel to its originallocation: the three planes will no longer have any points in common.
Systems of linear equations 15Algebraically, we can see what is so special about the one constant we changed(8 to 7) if we replace this value with an arbitrary constant, say k, and performrow operations: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 2 −1 1 1 2 −1 1 1 0 1 3 ⎣1 1 0 2⎦ → ⎣0 1 −1 −1⎦ → ⎣0 1 −1 −1⎦ 3 1 2 k 0 −5 5 k −3 0 0 0 k −8This shows that for any value of k other than 8, the resulting system of linearequations will be inconsistent, therefore having no solutions. In the case thatk = 8, we see that a free variable arises and then the system has infinitely manysolutions. Overall, the question of consistency is an important one for any linearsystem of equations. In asking "is this system consistent?" we investigate whetheror not the system has at least one solution. Moreover, we are now in a positionto understand how RREF determines the answer to this question. We note fromconsidering the RREF of a matrix that there are two overall cases: either thesystem contains an equation of the form 0x1 + · · · + 0xn = b, where b is nonzero,or it has no such equation. In the former case, the system is inconsistent andhas no solution. In the latter case, it will either be that every variable is uniquelydetermined, or that there are one or more free variables present, in which casethere are infinitely many solutions to the system. This leads us to state thefollowing theorem.Theorem 1.2.1 For any linear system of equations, there are only three possiblecases for the solution set: there are no solutions, there is a unique solution, orthere are infinitely many solutions.This central fact regarding linear systems will play a key role in our studies.1.2.1 Row-reduction using MapleObviously one of the problems with the process of row reducing a matrixis the potential for human arithmetic errors. Soon we will learn how to usecomputer software to execute all of these computations quickly; first, though,we can deepen our understanding of how the process works, and simultaneouslyeliminate arithmetic mistakes, by using a computer algebra system in a step-by-step fashion. Our software of choice is Maple. For now, we only assume that theuser is familiar with Maple's interface, and will introduce relevant commandswith examples as we go. We will use the LinearAlgebra package in Maple, which is loaded usingthe command > with(LinearAlgebra):(The symbol '>' is called a Maple prompt; the program makes this available tothe user automatically, and it should not be entered by the user.) To demonstrate
16 Essentials of linear algebravarious commands, we will revisit the system from example 1.2.1. The readershould explore this code actively by entering and experimenting on his or herown. Recall that we were interested in row-reducing the augmented matrix ⎡ ⎤ 3 2 −1 8 ⎣ 1 −4 2 −9⎦ −2 1 1 −1 We enter the augmented matrix, say A, column-wise in Maple with thecommand > A := <<3,1,-2>|<2,-4,1>|<-1,2,1>|<8,-9,-1>>;We first want to swap rows 1 and 2; this is accomplished by entering > A1 := RowOperation(A,[1,2]);Note that this stores the result of this row operation in the matrix A1, which isconvenient for use in the next step. After executing the most recent command,the following matrix will appear on the screen: ⎡ ⎤ 1 −4 2 −9 A1 := ⎣ 3 2 −1 8⎦ −2 1 1 −1To perform row-replacement, our next step is to add (−3) · R1 to R2 (whererows 1 and 2 are denoted R1 and R2 ) to generate a new second row; similarly,we will add 2 · R1 to R3 for an updated row 3. The commands that accomplishthese steps are > A2 := RowOperation(A1,[2,1],-3); > A3 := RowOperation(A2,[3,1],2);and lead to the following output: ⎡ ⎤ 1 −4 2 −9 A3 := ⎣0 14 −7 35⎦ 0 −7 5 −19Next, we will scale row 2 by a factor of 1/14 using the command > A4 := RowOperation(A3,2,1/14);to find that ⎡ ⎤ 1 −4 2 −9 A4 := ⎣0 5⎦ 1 −1 2 2 0 −7 5 −19
20 Essentials of linear algebra32. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) Two lines must either intersect or be parallel. (b) A system of three linear equations in three unknown variables can have exactly three solutions. (c) If the RREF of a matrix has a row of all zeros, then the corresponding system must have a free variable present. (d) If a system has a free variable present, then the system has infinitely many solutions. (e) A solution to a 4 × 3 linear system is a list of four numbers (x1 , x2 , x3 , x4 ) that simultaneously makes every equation in the system true. (f) A matrix with three columns and four rows is 3 × 4. (g) A consistent system is one with exactly one solution.33. Suppose that we would like to find a quadratic function p(t ) = a2 t 2 + a1 t + a0 that passes through the three points (1, 4), (2, 7), and (3, 6). How does this problem lead to a system of linear equations? Find the function p(t ). (Hint: p(1) = 4 implies that 4 = a2 12 + a1 1 + a0 .)34. Find a quadratic function p(t ) = a2 t 2 + a1 t + a0 that passes through the three points (−1, 1), (2, −1), and (5, 4). How does this problem involve a system of linear equations?35. For the circuit shown at the left in figure 1.5, set up and solve a system of linear equations whose solution is the respective currents I1 , I2 , and I3 .36. For the circuit shown at the right in figure 1.5, set up and solve a system of linear equations whose solution is the respective currents I1 , I2 , and I3 . 20V I3 + − 2ΩI3 2Ω 4Ω I3 I3 I2 I2 I2 I2 + − 5Ω 3Ω 6VI1 1Ω I1 I1 1Ω I1 + − + − 10V 8VFigure 1.5 Circuits for use in exercises 35 and 36.
Linear combinations 211.3 Linear combinationsAn important theme in mathematics that is especially present in linear algebrais the value of considering the same idea from a variety of different perspectives.Often, we can make statements that on the surface may seem unrelated, whenin fact they ultimately mean the same thing, and one of the statements is mostadvantageous for solving a particular problem. Throughout our study of linearalgebra, we will see that the subject offers a wide variety of perspectives andterminology for addressing the central concept: systems of linear equations. Inthis section, we take another look at the concept of consistency, but do so in adifferent, geometric light.Example 1.3.1 Consider the system of equations x1 − x2 = 1 x1 + x2 = 3 (1.3.1) x1 + 2x2 = 4Rewrite the system in vector form and explore how two vectors are beingcombined to form a third, particularly in terms of the geometry of R3 . Thensolve the system.Solution. In multivariable calculus, we learn to think of vectors in R3 verymuch like we think of points. For example, given the point (a , b , c), we maywrite v = a , b , c or v = ai + bj + ck to denote the vector v that emanatesfrom (0, 0, 0) and ends at (a , b , c). (Here i, j, and k represent the standard unitcoordinate vectors: i is the vector from (0, 0, 0) to (1, 0, 0), j to (0, 1, 0), and k to(0, 0, 1).) In linear algebra, we will prefer to take the perspective of writing such anordered triple as a matrix with only one column, also known as a column vector,in the form ⎡ ⎤ a v = ⎣b⎦ (1.3.2) cTo save space, we will sometimes use the equivalent notation2 v = [a b c ]T .Recall that two vectors are equal if and only if their corresponding entries areequal, that a vector may be multiplied by a scalar, and that any two vectors ofthe same size may be added.2 The 'T ' stands for transpose, and the transpose of a matrix is achieved by turning every columninto a row.
22 Essentials of linear algebra We can now re-examine the system of equations (1.3.1) in the light ofequality among vectors. In particular, observe that it is equivalent to say ⎡ ⎤ ⎡ ⎤ x1 − x2 1 ⎣ x1 + x2 ⎦ = ⎣ 3 ⎦ (1.3.3) x1 + 2x2 4since two vectors are equal if and only if their corresponding entries areequal. Recalling further that vectors are added component-wise, we canrewrite (1.3.3) as ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x1 −x2 1 ⎣ x1 ⎦ + ⎣ x 2 ⎦ = ⎣ 3 ⎦ (1.3.4) x1 2x2 4Finally, we observe in (1.3.4) that the first vector on the left-hand side hasa common factor of x1 in each component, and the second vector similarlycontains x2 . Since a scalar multiple of a vector is computed component-wise,here we can rewrite the equation once more, now in the form ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 1 x 1 ⎣ 1 ⎦ + x2 ⎣ 1 ⎦ = ⎣ 3 ⎦ (1.3.5) 1 2 4Equation (1.3.5) is equivalent to the original system (1.3.1), but is now beingviewed in a very different way. Specifically, this last equation asks if there arevalues of x1 and x2 for which x 1 v 1 + x2 v 2 = bwhere ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 1 v1 = ⎣ 1 ⎦ , v2 = ⎣ 1 ⎦ , and b = ⎣ 3 ⎦ (1.3.6) 1 2 4If we plot the vectors v1 , v2 , and b, an interesting situation comes to light, asseen in figure 1.6. In particular, it appears as if all three vectors lie in the sameplane. Moreover, if we think about the parallelogram law of vector addition andstretch the vector v1 by a factor of 2, we see the image in figure 1.7. This showsgeometrically that it appears b = 2v1 + v2 ; a quick check of the vector arithmeticconfirms that this is in fact the case. In other words, the unique solution to thesystem (1.3.1) is x1 = 2 and x2 = 1. Among the many important ideas in example 1.3.1, perhaps most significantis the way we were able to re-cast a problem about a system of linear equationsas a question involving vectors. In particular, we saw that it was equivalent toask if there exist constants x1 and x2 such that x 1 v 1 + x2 v 2 = b (1.3.7)
24 Essentials of linear algebra In light of this new terminology of linear combinations, in example 1.3.1we saw that the question "is there a solution to the linear system (1.3.1)?" isequivalent to asking "is the vector b a linear combination of the vectors v1and v2 ?" If we now consider the more general situation of a system of linearequations, say a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 . . . am1 x1 + am2 x2 + · · · + amn xn = bmit follows (as in section 1.2) that we can view this system in terms of theaugmented matrix [a1 a2 · · · an b]where a1 is the vector in Rm representing the first column of the augmentedmatrix, and so on. Now, however, we have the additional perspective, as inexample 1.3.1, that the columns of the augmented matrix A are precisely thevectors being used to form a linear combination in an attempt to construct b.That is, the general m × n linear system above asks the question, "is b a linearcombination of a1 , . . . , an ?" We make the connection between linear combinations and augmentedmatrices more explicit by defining matrix–vector multiplication in terms of linearcombinations.Definition 1.3.2 Given an m × n matrix A with columns a1 , . . . , an that arevectors in Rm , if x is a vector in Rn , then we define the product Ax by theequation ⎡ ⎤ x1 ⎢ x2 ⎥ ⎢ ⎥ Ax = [a1 a2 · · · an ] ⎢ . ⎥ = x1 a1 + x2 a2 + · · · + xn an (1.3.9) ⎣ . ⎦ . xn That is, the matrix–vector product of A and x is the vector Ax obtainedby taking the linear combination of the column vectors of A according to theweights prescribed by the entries in x. Certainly we must have the same numberof entries in x as columns in A, or Ax will not be defined. The following examplehighlights how to compute and interpret matrix–vector products.Example 1.3.2 Let a1 = [1 − 4 2]T and a2 = [−3 1 5]T , and let A be thematrix whose columns are a1 and a2 . Compute Ax, where x = [−5 2]T , andinterpret the result in terms of linear combinations.
Linear combinations 25Solution. By definition, we have that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −3 1 −3 −11 −5 Ax = ⎣ −4 1⎦ = −5 ⎣ −4 ⎦ + 2 ⎣ 1 ⎦ = ⎣ 22 ⎦ 2 2 5 2 5 0The above computations show clearly that the vector Ax = [−11 22 0]T is alinear combination of a1 and a2 .Following a few more computational examples in homework exercises, thereader will quickly see how to compute the product Ax whenever it is defined;usually we skip past the intermediate stage of writing out the explicit linearcombination of the columns and simply write the resulting vector. Matrix–vector multiplication also has several important general properties, some ofwhich will be explored in the exercises. For now, we simply list these propertieshere for future reference: for any m × n matrix A, vectors x, y ∈ Rn , and c ∈ R, • A(x + y) = Ax + Ay • A(cx) = c(Ax)The first property shows that matrix multiplication distributes over addition;the second demonstrates that a scalar multiple can be taken either before or aftermultiplying the vector x by A. These two properties of matrix multiplication areoften referred to as being properties of linearity—note the use of only scalarmultiplication and vector addition in each, and the linear appearance of eachequation.3 Finally, note that it is also the case that A0n = 0m , where 0n is thevector in Rn with all entries being zero, and 0m is the corresponding zero vectorin Rm . There is one more important perspective that this new matrix–vectorproduct notation permits. Recall that, in example 1.3.1, we learned that thequestion "is b a linear combination of a1 and a2 ?" is equivalent to asking "isthere a solution to the system of linear equations whose augmented matrix hascolumns a1 , a2 , and b?" Now, in light of matrix–vector multiplication, we alsosee that the question "is b a linear combination of a1 and a2 ?" may be rephrasedas asking "does there exist a vector x such that Ax = b?" That is, are thereweights x1 and x2 (the entries in vector x) such that b is a linear combination ofthe columns of A? In particular, we may now adopt the perspective that we desire to solve theequation Ax = b for the unknown vector x, where A is a matrix whose entriesare known, and b is a vector whose entries are known. This equation is strikinglysimilar to the most elementary of equations encountered in algebra, ones such as2x = 7. Therefore, we see that the linear equation Ax = b, involving matrices andvectors, is of fundamental importance as it is another way of expressing questions3 A deeper discussion of the notion of linear transformations can be found in appendix D.
26 Essentials of linear algebraregarding linear combinations and solutions of systems of linear equations. Insubsequent sections, we will explore this equation from several perspectives.1.3.1 Markov chains: an application of matrix–vector multiplicationPeople are often distributed naturally among various groupings. For example,much political discussion in the United States is centered on three classificationsof voters: Democrat, Republican, and Independent. A similar situation can beconsidered with regard to peoples' choices for where to live: urban, suburban, orrural. In each case, the state of the population at a given time is its distributionamong the relevant categories. Furthermore, in each of these situations, it is natural to assume that if weconsider the state of the system at a given point in time, its state depends on thesystem's state in the preceding year. For example, the percentage of Democrats,Republicans, and Independents in the year 2020 ought to be connected to therespective percentages in 2019. Let us assume that a population of voters (of constant size) is considered inwhich every-one must classified as either D, R, or I (Democrat, Republican, orIndependent). Suppose further that a study of voter registrations over manyyears reveals the following trends: from one year to the next, 95 percentof Democrats keep their registration the same. For the remaining 5 percentwho change parties, 2 percent become Republicans and 3 percent becomeIndependents. Similar data for Republicans and Independents is given in thefollowing table. Future party (↓)/current party (→) D(%) R(%) I(%) Democrat 95 3 7 Republican 2 90 13 Independent 3 7 80 If we let Dn , Rn , and In denote the respective numbers of registeredDemocrats, Republicans, and Independents in year n, then the table showsus how to determine the respective numbers in year n + 1. For example, Dn+1 = 0.95Dn + 0.03Rn + 0.07In (1.3.10)since 95 percent of the Democrats in year n stay registered Democrats, and3 percent of Republicans and 7 percent of Independents change to Democrats.Similarly, we have Rn+1 = 0.02Dn + 0.90Rn + 0.13In (1.3.11) In+1 = 0.03Dn + 0.07Rn + 0.80In (1.3.12)
Linear combinations 27If we combine (1.3.10), (1.3.11), and (1.3.12) in a single vector equation, then ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ Dn+1 0.95 0.03 0.07 ⎣ Rn+1 ⎦ = Dn ⎣ 0.02 ⎦ + Rn ⎣ 0.90 ⎦ + In ⎣ 0.13 ⎦ (1.3.13) In+1 0.03 0.07 0.80Here we find that linear combinations of vectors have naturally arisen. Note,for example, that the vector [0.03 0.90 0.07]T is the Republican vector, andrepresents the likelihood that a Republican in a given year will be in oneof the three parties in the following year. More specifically, we observe thatprobabilities are involved: a Republican has a 3 percent likelihood of registeringas a Democrat in the following year, a 90 percent likelihood of staying aRepublican, and 7 percent chance of becoming an Independent. The sum ofthe entries in each column vector is 1. If we use the vector x (n) to represent ⎡ ⎤ Dn x (n) = ⎣ Rn ⎦ Inand use matrix–vector multiplication to represent the linear combination ofvectors in (1.3.13), then (1.3.13) is equivalently expressed by the equation x (n+1) = Mx (n) (1.3.14)where M is the matrix ⎡ ⎤ 0.95 0.03 0.07 M = ⎣ 0.02 0.90 0.13 ⎦ 0.03 0.07 0.80The matrix M is often called a transition matrix since it shows how the populationtransitions from state n to state n + 1. We observe that in order for such amatrix to represent the probabilities that groups in a particular set of states willtransition to another set of states, the columns of the matrix M must be non-negative and add to 1. Such a matrix is called a stochastic matrix or a Markovmatrix. Finally, we call any system such as the one with three classifications ofvoters, where the state of the system in a given observation period results fromapplying probabilities to a previous state, a Markov chain or Markov process. We see, for example, that if we had a group of 250 000 voters that at yearn = 0 was distributed among Democrats, Republicans, and Independents bythe vector (with entries measured in thousands) x (0) = [120 110 20]T then wecan easily compute the projected distribution of voters in subsequent years. Inparticular, (1.3.14) implies ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 118.70 117.80 117.18x (1) = Mx (0) = ⎣ 104⎦, x (2) = Mx (1) = ⎣ 99.52⎦, x (3) = Mx (2) = ⎣ 96.18⎦ 27.3 32.68 36.65
28 Essentials of linear algebraInterestingly, if we continue the sequence, we eventually find that there is verylittle variation from one vector x (n) to the next. For example, ⎡ ⎤ ⎡ ⎤ 116.67 116.79 x (17) = ⎣ 85.95 ⎦ ≈ x (18) = ⎣ 85.76 ⎦ 47.42 47.44In fact, as we will learn in our later study of eigenvectors, there exists a vectorx ∗ called the steady-state vector for which x ∗ = Mx ∗ . This shows that the systemcan reach a state in which it does not change from one year to the next. Another example is instructive.Example 1.3.3 Geographers studying a metropolitan area have observed atrend that while the population of the area stays roughly constant, people withinthe city and its suburbs are migrating back and forth. In particular, suppose that85 percent of people whose homes are in the city keep their residence from oneyear to the next; the remainder move to the suburbs. Likewise, while 92 percentof people whose homes are in suburbs will live there the next year, the other8 percent will move into the city. Assuming that in a given year there are 230 000 people living in the city and270 000 people in the surrounding suburbs, predict the population distributionover the next 3 years.Solution. If we let Cn and Sn denote the populations of the city and suburbsin year n, the given information tells us that the following relationships hold: Cn+1 = 0.85Cn + 0.08Sn Sn+1 = 0.15Cn + 0.92SnUsing the notation Cn x (n) = Snwe can model the changing distribution of the population between the city andsuburbs with the Markov process x (n+1) = Mx (n) , where M is the Markov matrix 0.85 0.08 M= 0.15 0.92In particular, starting with x (0) = [230 270]T , we see that 217.10 207.17 199.52 x (1) = , x (2) = , x (3) = 282.90 292.83 300.48 As with voter distribution, this example is oversimplified. For instance,we have not taken into account members of the population who move intoor away from the metropolitan area. Nonetheless, the basic ideas of Markovprocesses are important in the study of systems whose current state depends onpreceding ones, and we see the key role matrices and matrix multiplication playin representing them.
30 Essentials of linear algebra 5. Recall from multivariable calculus that given vectors x , y ∈ R3 , the dot product of x and y, x · y, is computed by taking x · y = x1 y1 + x2 y2 + x3 y3 How can matrix–vector multiplication (when defined) be viewed as the result of computing several appropriate dot products? Explain. 6. For the system of equations given below, determine a vector equation with an equivalent solution. What is the system asking in regard to linear combinations of certain vectors? x1 + 2x2 = 1 x1 + x2 = 0 In addition, determine a matrix A and vector b so that the equation Ax = b is equivalent to the given system of equations. 7. For the system of differential equations (1.1.6) (also given below) from the introductory section, how can we rewrite the system in matrix–vector notation? dx1 x 1 x2 = 20 − + dt 20 80 dx2 x 1 x2 = 35 + − dt 40 40 Hint: recall that if x(t ) is a vector function, we write x (t ) or dx /dt for the vector [dx1 /dt dx2 /dt ]T . 8. Determine if the vector b = [−3 1 5 9. Determine if the vector b = [0 7 410. We know from our work in this section that the matrix equation Ax = b corresponds both to a vector equation and a system of linear equations. What is the augmented matrix that represents this system of equations?In exercises 11–15, let A be the stated matrix and b the given vector. Solvethe linear equation Ax = b by converting the equation to a system of linearequations and row-reducing appropriately. If the system has more than onesolution, express the solution in parametric vector form. Finally, write a sentencein each case that explains how the vector b is related to linear combinations ofthe columns of A. 4 5 −1 1311. A = , b= 3 1 2 −4
32 Essentials of linear algebra24. Let A be an m × n matrix, x and y ∈ Rn , and c ∈ R. Show that (a) A(x + y) = Ax + Ay (b) A(cx) = c(Ax)25. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) To compute the product Ax, the vector x must have the same number of entries as the number of rows in A. (b) A linear combination of three vectors in R3 produces another vector in R3 . (c) If b is a linear combination of v1 and v2 , then there exist scalars c1 and c2 such that c1 v1 + c2 v2 = b. (d) If A is a matrix and x and b are vectors such that Ax = b, then x is a linear combination of the columns of A. (e) The equation Ax = 0 can be inconsistent.26 1 1 88 Given that the population of 250 million is initially distributed in 100 million urban, 100 million suburban, and fifty million rural, predict the population distribution in each of the following five years.27. Car-owners can be grouped into classes based on the vehicles they own. A study of owners of sedans, minivans, and sport utility vehicles shows that the likelihood that an owner of one of these automobiles will replace it with another of the same or different type is given by the table Future vehicle (↓)/ Sedan(%) Minivan(%) SUV(%) current vehicle (→) Sedan 91 3 2 Minivan 7 95 8 SUV 2 2 90
The span of a set of vectors 33 If there are currently 100 000 sedans, 60 000 minivans, and 80 000 SUVs among the owners being studied, predict the distribution of vehicles among the population after each owner has replaced her vehicle 3 times.1.4 The span of a set of vectorsIn section 1.3, we saw that the question "is b a linear combination of a1 anda2 ?" provides an important new perspective on solutions of linear systems ofequations. It is natural to slightly rephrase this question and ask more generally"which vectors b may be written as linear combinations of a1 and a2 ?" Weexplore this question further through the following sequence of examples.Example 1.4.1 Describe the set of all vectors in R2 that may be written as alinear combination of the vector a1 = [2 1]T .Solution. Since we have just one vector a1 , any linear combination of a1 hasthe form ca1 , which of course is a scalar multiple of a1 . Geometrically, thevectors that are linear combinations of a1 are stretches of a1 , which lie on theline through (0, 0) in the direction of a1 , as shown in figure 1.8. In this first example, we see a visual way to interpret the question aboutlinear combinations: essentially we want to know "which vectors can we createusing only linear combinations of a1 ?" The answer is not surprising: only vectorsthat lie on the line through the origin in the direction of a1 . Next, we consider how the situation changes when we consider two parallelvectors. x2 4 a1 x1 −4 4 −4 Figure 1.8 The set of all linear combinations of a1 in example 1.4.1.
34 Essentials of linear algebraExample 1.4.2 Describe the set of all vectors in R2 that may be written as alinear combination of the vectors a1 = [2 1]T and a2 = [−1 − 1 ]T . 2Solution. Observe first that − 1 a1 = a2 . Here we are considering the set of all 2vectors y of the form 2 −1 y = c1 + c2 1 −1 2In figure 1.9, we observe that the vectors a1 and a2 point in opposing directions.When we take a linear combination of these vectors to form y, we are addinga stretch of c1 units of the first to a stretch of c2 units of the second. Becausethe two directions are parallel, this leaves the resulting vector as a stretch ofone of the two original vectors, and therefore on the line through the originin their direction. This may also be seen algebraically since − 1 a1 = a2 implies 2y = c1 a1 + c2 a2 = c1 a1 − 1 c2 a1 = (c1 − 1 c2 )a1 . 2 2 We note particularly that since the two given vectors a1 and a2 are parallel,any linear combination of them is actually a scalar multiple of a1 . Thus, theresulting set of all linear combinations is identical to what we found with thesingle vector given in example 1.4.1. Finally, we consider the situation where we consider all linear combinationsof two non-parallel vectors. x2 4 a1 x1 −4 a2 4 −4 Figure 1.9 The set of all linear combinations of a1 and a2 in example 1.4.2.Example 1.4.3 Describe the set of all vectors in R2 that may be written as alinear combination of the vectors a1 = [2 1]T and a2 = [1 2]T .
The span of a set of vectors 35 x2 4 a1+ a2 −2a1+ 2a2 a2 a1 x1 −4 4 7/3 a1− 5/3 a2 −4 Figure 1.10 Linear combinations of a1 and a2 from example 1.4.3.Solution. Algebraically, we are again considering the set of all vectors y suchthat y = c1 a1 + c2 a2 . A visual way to think about how the set of all such vectors ylooks is found in the question, "which vectors can we create by taking a stretchof a1 and adding this to a stretch of a2 ?" If we consider a plot of the given two vectors a1 and a2 and think of the"grid" that is formed by considering all of their stretches and the sums of theirstretches, we have the picture shown in figure 1.10. The fact that a1 and a2 arenot parallel enables us to "get off the line" that each one generates through theorigin. For example, if we simply take the sum of these two vectors and sety = a1 + a2 , by the parallelogram law of vector addition we arrive at the newvector [3 3]T shown in figure 1.10. Two other linear combinations are shownas well, and from here it is not hard to visualize the fact that we can createany vector in the plane using linear combinations of the non-parallel vectors a1and a2 . In other words, the set of all linear combinations of a1 and a2 is R2 . It is also possible to verify our findings in example 1.4.3 algebraically. Wewill explore this further in the exercises and in section 1.5. Certainly we are not limited to considering linear combinations of only twovectors. We therefore introduce a more formal perspective and terminology todescribe the phenomena examined in the above examples.Definition 1.4.1 Given a set of vectors S = {v1 , . . . , vk }, vi ∈ Rm , the span of S,denoted Span(S) or Span{v1 , . . . , vk }, is the set of all linear combinations of thevectors v1 , . . . , vk . Equivalently, Span(S) is the set of all vectors y of the form y = c1 v1 + · · · + ck vk
36 Essentials of linear algebrawhere c1 , . . . , ck are scalars. We also say that Span(S) is the subset of Rm spannedby the vectors v1 , . . . , vk . For any single nonzero vector v1 ∈ Rm , Span{v1 } consists of all vectors thatlie on the line through the origin in Rm in the direction of v1 . For two non-parallel vectors v1 , v2 ∈ Rm , Span{v1 , v2 } is the plane through the origin thatcontains both the vectors v1 and v2 . Next, let us recall that our interest in linear combinations was motivated bya desire to look at systems of linear equations from a new perspective. How isthe concept of span related to linear systems? We begin to answer this questionby considering the special situation where b = 0. A system of linear equations that can be represented in matrix form bythe equation Ax = 0 is said to be homogeneous; the case when b = 0 is termednonhomogeneous. We also call the equation Ax = 0 a homogeneous equation.By the definition of matrix–vector multiplication, it is immediately clear thatA0 = 0 (note that these two zero vectors may be of different sizes), and thusany homogeneous equation has at least one solution and is guaranteed to beconsistent. We will usually call the solution x = 0 the trivial solution. Underwhat circumstances will a homogeneous system have nontrivial solutions? Howis this question related to the span of a set of vectors? The following exampleprovides insight into these questions.Example 1.4.4 Solve the homogeneous system of linear equations given by theequation Ax = 0 where A is the matrix ⎡ ⎤ 1 1 1 1 ⎢2 1 −1 3⎥ A=⎢ ⎣1 0 −2 2⎦ ⎥ 8 5 −1 11If more than one solution exists, express the solution in parametric vector form.Solution. To begin, we augment the matrix A with a column of zeros torepresent the vector 0 in the system given by Ax = 0. We then row-reduce thisaugmented matrix to find ⎡ ⎤ ⎡ ⎤ 1 1 1 1 0 1 0 −2 2 0 ⎢2 1 −1 3 0⎥ ⎢0 1 3 −1 0⎥ ⎢ ⎥ ⎢ ⎥ ⎣1 0 −2 2 0⎦ → ⎣0 0 0 0 0⎦ 8 5 −1 11 0 0 0 0 0 0We observe that the system has two free variables, and therefore infinitely manysolutions. In particular, these solutions must satisfy the equations x1 − 2x3 + 2x4 = 0 x2 + 3x3 − x4 = 0
The span of a set of vectors 37where x3 and x4 are free. Equivalently, using these equations and vector additionand scalar multiplication, it must be the case that any solution x to Ax = 0 hasthe form ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x1 2x3 − 2x4 2 −2 ⎢ x ⎥ ⎢ −3x3 + x4 ⎥ ⎢ ⎥ ⎢ ⎥ x=⎢ 2⎥=⎢ ⎥ = x3 ⎢ −3 ⎥ + x4 ⎢ 1 ⎥ (1.4.1) ⎣ x3 ⎦ ⎣ x3 ⎦ ⎣ 1⎦ ⎣ 0⎦ x4 x4 0 1where x3 , x4 ∈ R. Note particularly that this shows that every solution x to theoriginal homogeneous equation Ax = 0 can be expressed as a linear combinationof the two vectors on the rightmost side of (1.4.1). Moreover, it is also the casethat every linear combination of these two vectors is a solution to the equation.In light of the terminology of span, we can say that the set of all solutions to thehomogeneous equation Ax = 0 is Span{v1 , v2 }, where ⎡ ⎤ ⎡ ⎤ 2 −2 ⎢ −3 ⎥ ⎢ 1⎥ v1 = ⎢ ⎥ ⎣ 1 ⎦ , v2 = ⎣ 0 ⎦ ⎢ ⎥ 0 1 In this section, we have seen that the set of all linear combinations of aset of vectors can be interpreted geometrically, particularly in the case whenwe only have one or two vectors present, by thinking about lines and planes. Inaddition, the span of a set of vectors arises naturally in considering homogeneousequations in which infinitely many solutions are present. In that situation, theset of all solutions can be expressed as the span of a set of k vectors, wherek is the number of free variables that arise in row-reducing the augmentedmatrix.Exercises 1.4 In exercises 1–6, solve the homogeneous equation Ax = 0,given the matrix A. If infinitely many solutions exist, express the solution set asthe span of the smallest possible set of vectors. 1 −3 2 1. A = −4 1 0 ⎡ ⎤ −4 2 2. A = ⎣ 1 −3⎦ 6 5 −5 8 3. A = 10 −16 −4 2 4. A = 2 −1
38 Essentials of linear algebra ⎡ ⎤ 3 1 −1 5. A = ⎣ 1 3 1⎦ −1 1 3 ⎡ ⎤ 1 −1 2 6. A = ⎣ 4 −2 6⎦ −7 3 −10 7. Let A be an m × n matrix where n > m. Is it possible that Ax = 0 has only the trivial solution? Explain why or why not. 8. Let A be an m × n matrix where n ≤ m. Is it guaranteed that Ax = 0 will have only the trivial solution? Explain why or why not. 9. Determine if the vector b = [11 − 4]T is in the span of the vectors a1 = [3 − 2]T and a2 = [−9 6]T . Justify your answer carefully.10. Determine if the vector b = [−17 31]T is in the span of the vectors a1 = [1 0]T and a2 = [0 1]T . What do you observe?11. Determine if the vector b = [9 17 11]T is in the span of the vectors a1 = [−1 2 1]T , a2 = [3 1 1]T , and a3 = [1 5 3]T . Justify your answer.12. Explain why the vector b = [3 2]T does not lie in the span of the set S, where S = {v } and v = [1 1]T .13. Describe geometrically the set W = Span{v1 , v2 }, where v1 = [1 1 1]T and v2 = [−3 0 2]T .14. Can every vector b ∈ R3 be found in W = Span{v1 , v2 }, where v1 = [1 1 1]T and v2 = [−3 0 2]T ? If so, explain why. If not, find a vector not in W and justify your answer.15. Show that every point (vector) that lies on the line with equation 2x1 − 3x2 = 0 also lies in the set W = Span{v1 }, where v1 = [3 2]T .16. Show that every point (vector) that lies on the plane with equation −x + y + z = 0 also lies in the set W = Span{v1 , v2 }, where v1 = [1 − 1 2]T and v2 = [2 1 1]T .17. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) The span of a single nonzero vector in R2 can be thought of as a line through the origin. (b) The span of any two nonzero vectors in R3 can be viewed as a plane through the origin in R3 . (c) If Ax = b holds true for a given matrix A and vectors x and b, then x lies in the span of the columns of A.
Systems of linear equations revisited 39 (d) It is possible for a homogeneous equation Ax = 0 to be inconsistent. (e) The number of free variables present in the solution to Ax = 0 is the same as the number of pivot columns in the matrix A.1.5 Systems of linear equations revisitedFrom our initial work with row-reducing a system of linear equations to ourrecent discussions of linear combinations and span, we have seen already thatthere are several perspectives from which to view a system of linear equations.One is purely algebraic: "is there at least one ordered list (x1 , . . . , xn ) that makesevery equation in a given system true?" Here we are viewing the system inthe form a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 . . .=. . . am1 x1 + am2 x2 + · · · + amn xn = bmIn light of linear combinations, we can rephrase this question geometrically as"is the vector b a linear combination of the vectors a1 , . . . , an ?", where ai is theith column of the coefficient matrix of the system. From this standpoint, askingif the system has a solution can be thought of in terms of the question, "doesthe vector b belong to the span of the columns of A?" Finally, through matrixmultiplication, we can also express this system of equations in its simplest form:Ax = b. From all of this, we know that the question, "Does Ax = b have at leastone solution?" is one of fundamental importance. We have also seen that in the special case of the homogeneous equationAx = 0, the answer to the above questions is always affirmative, since settingx = 0 guarantees that we have at least one solution. In what follows, wefurther explore the nonhomogeneous case Ax = b, with particular emphasison understanding characteristics of the matrix A that enable us to answer thequestions in the preceding paragraph. We begin by revisiting example 1.4.2 from a more algebraic perspective.Example 1.5.1 For which vectors b is the equation Ax = b consistent, if A isthe matrix whose columns are the vectors a1 = [2 1]T and a2 = [−1 − 1 ]T ? 2Solution. By the definition of matrix multiplication, this question is equivalentto asking, "which vectors b are linear combinations of the columns of A?" Thisquestion may be equivalently rephrased as "which vectors b are in the span ofthe columns of A?" We have already answered this question from a geometricperspective in example 1.4.2, where we saw that since a1 and a2 are parallel,it follows that every vector in R2 that lies on the line through the origin in
40 Essentials of linear algebrathe direction of a1 can be written as a linear combination of the two vectors.Nonetheless, it is insightful to explore algebraically why this is the case.Letting b be the vector whose entries are b1 and b2 and writing the equationAx = b in the form of an augmented matrix, we row-reduce and find that 2 −1 b1 1 −1 b2 → 2 1 − 2 b2 1 0 0 b1 − 2b2The second row in the augmented matrix represents the equation 0x1 + 0x2 = b1 − 2b2Observe that if b1 − 2b2 = 0, this equation cannot possibly be true, and thereforethe system would be inconsistent. Said differently, the only way for Ax = b tobe consistent is for b1 − 2b2 = 0. That is, if b is a vector such that b1 = 2b2 , or 2b2 b= b2then Ax = b is consistent. This makes sense geometrically, since the span of thecolumns of A is all the stretches of the vector a1 = [2 1]T .An important lesson to take from example 1.5.1 is that the equation Ax = bdiscussed there is not consistent for every choice of b. In fact, the equation is onlyconsistent for very limited choices of b. For example, if b = [6 3]T , the equationis consistent, but if b = [6 k ]T for any k = 3, the equation is inconsistent.Moreover, we should observe that for the matrix in this example, A does nothave a pivot position in every row. This is what ultimately leads to the algebraicequation 0x1 + 0x2 = b1 − 2b2 , and the potential inconsistency of Ax = b. At this point in our work, it is important that we begin to generalize ourobservations in order to apply them in new, but similar, circumstances. Weagain emphasize that it is a noteworthy characteristic of linear algebra that thediscipline often offers great flexibility through the large number of ways to saythe same thing; at times, one way of stating a fact can give more insight thanothers, and therefore it is important to be well versed in shifting among multipleperspectives. The following theorem is of the form "the following statements areequivalent"; this means that if any one of the statements is true, all the others areas well. Likewise, if any one statement is false, every statement in the theoremmust be false. This theorem formalizes our findings in the example above, and, in somesense, our work in the first several sections of the text.Theorem 1.5.1 Let A be an m × n matrix and b a vector in Rm so that theequation Ax = b represents a system of m linear equations in n unknownvariables. The following statements are equivalent: a. The equation Ax = b is consistent b. The vector b is a linear combination of the columns of A
Systems of linear equations revisited 41 c. The vector b is in the span of the columns of A d. When the augmented matrix [A b] is row-reduced, there are no rows where the first n entries are zero and the last entry is nonzero. The following example demonstrates how we can use theorem 1.5.1 toanswer questions about span and linear combinations.Example 1.5.2 Does the vector b = [1 − 7 − 14]T belong to the span of thevectors a1 = [1 3 4]T , a2 = [2 1 − 1]T , and a3 = [0 5 9]T ? Does the resultchange if we ask the same question about the vector c = [1 − 7 − 13]T ?Solution. By theorem 1.5.1, we know that it is equivalent to ask if the equationAx = b is consistent, where b is the given vector and A is the matrix whosecolumns are a1 , a2 , and a3 . To answer that question, we consider the augmentedmatrix [A | b] and row-reduce: ⎡ ⎤ ⎡ ⎤ 1 2 0 1 1 0 2 −3 ⎣3 1 5 −7⎦ → ⎣0 1 −1 2⎦ 4 −1 9 −14 0 0 0 0Because this system of equations is consistent, it follows that b is indeed a linearcombination of the columns of A and therefore b lies in the span of a1 , a2 ,and a3 . If we instead consider the vector c stated in the example and proceedsimilarly, row-reduction shows that ⎡ ⎤ ⎡ ⎤ 1 2 0 1 1 0 2 0 ⎣3 1 5 −7⎦ → ⎣0 1 −1 0⎦ 4 1 9 −13 0 0 0 1which implies that the system is inconsistent and therefore c is not a linearcombination of the columns of A, or equivalently, c does not lie in the span ofa1 , a2 , and a3 . At this point, it is natural to think the situations in examples 1.5.1 and 1.5.2are somewhat dissatisfying: sometimes Ax = b is consistent, and sometimes not,all depending on our choice of b. A natural question to ask is, "are there matricesA for which Ax = b is consistent for every choice of b?" With that question, weare certainly interested in the properties of the matrix A that make this situationoccur. We next revisit example 1.4.3 and explore these issues further.Example 1.5.3 For which vectors b is the equation Ax = b consistent, if A isthe matrix whose columns are the vectors a1 = [2 1]T and a2 = [1 2]T ?
42 Essentials of linear algebraSolution. Proceeding as in the previous example, we row reduce theaugmented matrix form of the equation and find that 3 b1 − 3 b2 2 1 2 1 b1 1 0 → 1 2 b2 0 1 − 1 b 1 + 2 b2 3 3Algebraically, this shows that regardless of the entries we select for the vectorb, we can always find a solution to the equation Ax = b. In particular, x isthe vector in R2 whose components are x1 = 2 b1 − 1 b2 and x2 = − 1 b1 + 2 b2 . 3 3 3 3Thus the equation Ax = b is consistent for every b in R2 . Note that this isnot surprising, given our work in example 1.4.3, where we found that froma geometric perspective, every vector b ∈ R2 could be written as a linearcombination of a1 and a2 . This example simply confirms that finding, but nowfrom an algebraic point of view. In terms of a key property of the matrix in example 1.5.3, we see that A hasa pivot position in every row. In particular, there is no row in RREF(A) wherewe encounter all zeros, and thus it is impossible to ever encounter an equationof the form 0 = k, where k = 0. This is, therefore, one property of the matrix Athat guarantees consistency for every choice of b. We generalize our findings in this example in the following theorem, whichis similar to theorem 1.5.1, but now focuses solely the matrix A and no longerrequires a vector b to be initially chosen.Theorem 1.5.2 Let A be an m × n matrix. The following statements areequivalent: a. The equation Ax = b is consistent for every b ∈ Rm b. Every vector b ∈ Rm is a linear combination of the columns of A c. The span of the columns of A is Rm d. A has a pivot position in every row. That is, when the matrix A is row-reduced, there are no rows of all zeros. Our next example shows how we can apply theorem 1.5.2 to answer generalquestions about the span of a set of vectors and the consistency of related systemsof equations.Example 1.5.4 Does the vector b = [1 − 7 − 13]T belong to the span of thevectors a1 = [1 3 4]T , a2 = [2 1 − 1]T , and a3 = [0 5 10]T ? Can every vectorin R3 be found in the span of the vectors a1 , a2 , and a3 ?Solution. Just as in example 1.5.2, we know by theorem 1.5.1 that it isequivalent to ask if the equation Ax = b is consistent, where b is the givenvector and A is the matrix whose columns are a1 , a2 , and a3 . We thus consider
Systems of linear equations revisited 43the augmented matrix [A | b] and row-reduce: ⎡ ⎤ ⎡ ⎤ 1 2 0 1 1 0 0 −5 ⎣3 1 5 −7⎦ → ⎣0 1 0 3⎦ 4 −1 10 −13 0 0 1 1Because this system of equations is consistent, it follows that b is indeed a linearcombination of the columns of A and therefore b lies in the span of a1 , a2 , anda3 . But by theorem 1.5.2 we can now make a much more general observation.Because we see that the coefficient matrix A has a pivot in every row, it followsthat regardless of which vector b we choose in R3 , we can write that vectoras a linear combination of the columns of A. That is, the vectors a1 , a2 , anda3 span all of R3 and the equation Ax = b will be consistent for every choiceof b. This example demonstrates that it is in some sense ideal if a matrix A hasa pivot in every row. As we proceed with further study of linear algebra, wewill focus more and more on properties of the coefficient matrix and theirimplications for related systems of equations. We conclude this section byexamining a key link between homogeneous and nonhomogeneous equationsin order to foreshadow an essential concept in our pending study of differentialequations.Example 1.5.5 Solve the nonhomogeneous system of linear equations givenby the equation Ax = b where A and b are ⎡ ⎤ ⎡ ⎤ 1 1 1 1 1 ⎢2 1 −1 3⎥ ⎢ −8 ⎥ A=⎢ ⎥ ⎢ ⎣1 0 −2 2⎦ , b = ⎣ −9 ⎦ ⎥ 8 5 −1 11 −22If more than one solution exists, express the solution in parametric vector form.Solution. Note that the coefficient matrix A is identical to the one inexample 1.4.4, so that here we are simply considering a related nonhomogeneousequation. We augment the matrix A with b and then row reduce to find ⎡ ⎤ ⎡ ⎤ 1 1 1 1 1 1 0 −2 2 −9 ⎢2 1 −1 3 −8⎥ ⎢ 3 −1 10⎥ ⎢ ⎥ → ⎢0 1 ⎥ ⎣1 0 −2 2 −9⎦ ⎣0 0 0 0 0⎦ 8 5 −1 11 −22 0 0 0 0 0As we found with the homogeneous equation, the system is consistent and hastwo free variables, and therefore infinitely many solutions. These solutions mustsatisfy the equations x1 = −9 + 2x3 − 2x4 x2 = 10 − 3x3 + x4
44 Essentials of linear algebrawhere x3 and x4 are free. Equivalently, it must be the case that any solution xhas the form ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x1 −9 + 2x3 − 2x4 −9 2 −2 ⎢ x ⎥ ⎢ 10 − 3x3 + x4 ⎥ ⎢ 10 ⎥ ⎢ −3 ⎥ ⎢ 1⎥ x=⎢ 2⎥=⎢ ⎣ x3 ⎦ ⎣ ⎥=⎢ ⎥ ⎢ ⎥ ⎦ ⎣ 0 ⎦ + x3 ⎣ 1 ⎦ + x 4 ⎣ 0 ⎦ ⎢ ⎥ x3 x4 x4 0 0 1where x3 , x4 ∈ R. Observe that if we let xp = [−9 10 0 0]T and let xh be anyvector of the form ⎡ ⎤ ⎡ ⎤ 2 −2 ⎢ −3 ⎥ ⎢ ⎥ xh = t ⎢ ⎥+s⎢ 1⎥ ⎣ 1⎦ ⎣ 0⎦ 0 1then any solution to the equation Ax = b has the form x = xp + xh . Moreover,it is now apparent that this vector xh is the same general solution vector thatwe found for the corresponding homogeneous equation in example 1.4.4. Inaddition, it is straightforward to check that Axp = b. Thus, we see that the generalsolution to the nonhomogeneous equation contains the general solution to thecorresponding homogeneous equation. It appears from example 1.5.5 that if we have a solution, say xp , toa nonhomogeneous equation Ax = b, we may add any solution xh to thehomogeneous equation Ax = 0 to xp and still have a solution to Ax = b. To seewhy any vector of the form xp + xh is a solution to Ax = b, let us assume that xpis a solution to Ax = b, and xh is a solution to Ax = 0. We claim that x = xp + xhis also a solution to Ax = b. This holds since Ax = A(xp + xh ) = Axp + Axh = b+0 =b (1.5.1) Clearly, this shows that the solution to the corresponding homogeneousequation plays a central role in the solution of nonhomogeneous equations.One observation we can make is that in the event we can find a single particularsolution xp to the nonhomogeneous equation, if the corresponding homoge-neous equation has at least one free variable, then we know that there must beinfinitely many solutions to the nonhomogeneous equation as well. We couldeven take the perspective that, in order to solve a nonhomogeneous equation, wesimply need to do two things: find one particular solution to Ax = b, and thencombine that particular solution with the general solution to the correspondinghomogeneous equation Ax = 0. While this is not so useful with systems of linearalgebraic equations, it turns out that this approach of solving the homogeneousequation first is essential in the solution of differential equations.
Systems of linear equations revisited 45 The following example shows how the same structure is present in a classof differential equations that we will discuss in detail in section 2.3.Example 1.5.6 Consider the differential equations y + 3y = 0 and y + 3y = 6.Compare and contrast the solutions to these two equations.Solution. The first equation, y + 3y = 0, we will call a homogeneous linearfirst-order differential equation. Note that it asks a straightforward question:what function y(t ) is such that the function's derivative plus 3 times itself is thezero function? Said differently, we seek a function y such that y = −3y. Fromour experience with exponential functions in calculus, we know that if y = e −3t ,then y = −3e −3t . The same is true for functions like y = 2e −3t and y = −5e −3t ;indeed, we see that for any constant C, the function y = Ce −3t satisfies thedifferential equation. (It also turns out that these are the only functions thatsatisfy the differential equation.) If we next consider the related differential equation y + 3y = 6 – one thatwe will call a nonhomogeneous linear first-order differential equation—we see thatthere is one obvious solution to the equation. In particular, if we let y(t ) be theconstant function y(t ) = 2, then y (t ) = 0 and this function clearly makes thedifferential equation true since 3 × 2 = 6. Now, we should wonder if we have found all of the possible solutions toy + 3y = 6. The answer is no: as we will see in section 2.3, it turns out that thegeneral solution y to this differential equation is y(t ) = 2 + Ce −3tWe can verify that this is the case by direct substitution. Note that y = −3Ce −3tand therefore y + 3y = −3Ce −3t + 3(2 + Ce −3t ) = −3Ce −3t + 6 + 3Ce −3t = 6Observe the structure of this solution function: if we let yp = 2, we have aparticular solution to the nonhomogeneous equation. Further, letting yh =Ce −3t , this is the general solution to the related homogeneous equation. Thisdemonstrates that the overall solution to the nonhomogeneous equation is y = yp + yh = 2 + Ce −3tExercises 1.5 For each of the following m × n matrices A in exercises 1–8,determine whether the equation Ax = b is consistent for every choice of b ∈ Rm .If not, describe the set of all b ∈ Rm for which the equation is consistent. In eachcase, explain your reasoning fully. 4 −1 1. A = 1 −4 4 −1 2. A = −12 3
48 Essentials of linear algebra27. Suppose that A is a 6 × 9 matrix that has a pivot in every row. What can you say about the consistency of Ax = b for every b ∈ R6 ? Why?28. Suppose that A is a 3 × 4 matrix and that the span of the columns of A is R3 . What can you say about the consistency of Ax = b for every b ∈ R3 ? Why?29. If possible, give an example of a 3 × 2 matrix A such that the span of the columns of A is R3 . If finding such a matrix is impossible, explain why.30. Suppose that A is a 4 × 3 matrix for which the homogeneous equation Ax = 0 has only the trivial solution. Will the equation Ax = b be consistent for every b ∈ R4 ? Explain. For the vectors b for which Ax = b is indeed a consistent equation, how many solution vectors x does each equation have? Why?31. Suppose that A is a 3 × 4 matrix for which the homogeneous equation Ax = 0 has exactly one free variable present. Will the equation Ax = b be consistent for every b ∈ R3 ? Explain. For the vectors b for which Ax = b is indeed a consistent equation, how many solution vectors x does each equation have? Why?32. Suppose that A is a 4 × 5 matrix for which the homogeneous equation Ax = 0 has exactly two free variables present. Will the equation Ax = b be consistent for every b ∈ R4 ? Explain. For the vectors b for which Ax = b is indeed a consistent equation, how many solution vectors x does each equation have? Why?33. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) If Ax = b is consistent for at least one vector b, then A has a pivot in every row. (b) If A is a 4 × 3 matrix, then it is possible for the columns of A to span R4 . (c) If A is a 3 × 3 matrix with exactly two pivot columns, then the columns of A do not span R3 . (d) If A is a 3 × 4 matrix, then the columns of A must span R3 . (e) If y and z are solutions to the equation Ax = 0, then the vector y + z is also a solution to Ax = 0. (f) If y and z are solutions to the equation Ax = b, where b = 0, then the vector y + z is also a solution to Ax = b.34. Solve the linear first-order differential equation y + y = 3 by first finding all functions yh that satisfy the homogeneous equation y + y = 0 and then determining a constant function yp that is a solution to y + y = 3. Verify by direct substitution that y = yh + yp is a solution to the given equation.35. Solve the linear first-order differential equation y − 5y = 6 by first finding all functions yh that satisfy the homogeneous equation y − 5y = 0 and
Linear independence 49 then determining a constant function yp that is a solution to y − 5y = 6. Verify by direct substitution that y = yh + yp is a solution to the given equation.1.6 Linear independenceIn theorem 1.5.2, we found that when solving Ax = b, an ideal situation occurswhen A has a pivot position in every row. Equivalently, this means that theequation Ax = b is guaranteed to have at least one solution for every vectorb ∈ Rm (when A is m × n), or that every b ∈ Rm can be written as a linearcombination of the columns of A. In other words, regardless of the choice of b,the equation Ax = b is always consistent. Because the equation is consistent, weare guaranteed that at least one solution x exists. In what follows, we exploreconditions that imply not only that at least one solution exists, but in fact thatonly one solution exists. First, we consider the simpler situation of homogeneousequations. In section 1.4, we discovered that the equation Ax = 0 is always consistent.Because x = 0 always makes this equation true, we know that we at least havethe trivial solution present. It is natural to ask: under what conditions on Ais the trivial solution the only solution to the homogeneous equation Ax = 0?Geometrically, we are asking whether or not a nontrivial linear combination ofthe columns of A can be formed that leads to the zero vector. We revisit an earlier example to further explore these issues.Example 1.6.1 Does the equation Ax = 0 have nontrivial solutions if A is thematrix whose columns are a1 = [2 1]T and a2 = [−1 − 1 ]T ? Discuss the 2geometric implications of your conclusions.Solution. We first consider the corresponding augmented matrix and rowreduce, finding that 2 −1 0 1 −1 0 2 → 1 −1 0 2 0 0 0This shows that any vector x = [x1 x2 ]T that satisfies x1 = 1 x2 will be a solution 2to Ax = 0. The presence of the free variable x2 implies that there are infinitelymany nontrivial solutions to this equation. If we interpret the matrix–vector product Ax as the linear combinationAx = x1 a1 + x2 a2 , then the equation 1 x2 a1 + x2 a2 = 0 2implies geometrically that the zero vector (on the right) may be expressedas a nontrivial linear combination of a1 and a2 . For example, a1 + 2a2 = 0.
50 Essentials of linear algebra x2 4 a1 x1 −4 a2 4 −4 Figure 1.11 Linear combinations of a1 and a2 from example 1.6.1.Indeed, if we consider figure 1.11 this conclusion is evident: if we add onelength of a1 to two lengths of a2 , we end up at 0. Another way to express the equation a1 + 2a2 = 0 is to write a1 = −2a2 . Inthis setting, we can see that a1 depends on a2 , and that the relationship is givenby a linear equation. We hence say that a1 and a2 are linearly dependent vectors. The situation in example 1.6.1, where the vectors a1 and a2 are parallel is incontrast to that of example 1.4.3, where we instead considered the non-parallelvectors a1 = [2 1]T and a2 = [1 2]T ; in that setting, if we solve the associatedhomogeneous equation Ax = 0, we find that 2 1 0 1 0 0 → 1 2 0 0 1 0In this case, the only solution to Ax = 0 is the trivial solution, x = 0. Thegeometry of the situation also informs us: if we desire a linear combination ofthe vectors a1 and a2 (as shown in figure 1.12) that results in the zero vector, wesee that the only way to accomplish this is to take 0a1 + 0a2 . Said differently, ifwe take any nontrivial linear combination c1 a1 + c2 a2 , we end up at a locationother than the origin. When a1 and a2 in example 1.6.1 were parallel, we said that a1 and a2 werelinearly dependent. In the current context, where a1 and a2 are not parallel, itmakes sense to say that a1 and a2 are linearly independent, since neither dependson the other. Of course, in linear algebra we often consider sets of more than two vectors.The next definition formalizes what the terms linearly dependent and linearlyindependent mean in a more general context. Observe that the key criterion is
Linear independence 51 x2 4 a2 a1 x1 −4 4 −4 Figure 1.12 Linear combinations of a1 and a2 from example 1.4.3.a geometric one: can we form a nontrivial linear combination of vectors thatresults in 0?Definition 1.6.1 Given a set S = {v1 , . . . , vk } where each vector vi ∈ Rm , theset S is linearly dependent if there exists a nontrivial solution x to the vectorequation x1 v1 + x2 v2 + · · · + xk vk = 0 (1.6.1)If (1.6.1) has only the trivial solution, then we say the set S is linearlyindependent. Note that (1.6.1) also takes us back to the fundamental questions aboutany linear system of equations: "does at least one solution exist?" (Yes; the zerovector is always a solution.) And "is that solution unique?" (Maybe; only ifthe vectors are linearly independent and the zero vector is the only solution.)The latter question addresses the fundamental issue of linear independence. Weconsider an example to demonstrate how we interpret the language of this mostrecent definition as well as how we will generally respond to the question ofwhether or not a set of vectors is linearly independent.Example 1.6.2 Determine whether the set S = {v1 , v2 , v3 } is linearlyindependent or linearly dependent if ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 0 v1 = ⎣ 1 ⎦ , v2 = ⎣ 0 ⎦ , v3 = ⎣ 1 ⎦ −1 1 1
52 Essentials of linear algebraSolution. By definition, the linear independence of the set S rests on whetheror not nontrivial solutions exist to the vector equation x1 v1 + x2 v2 + x3 v3 = 0.Letting A = [v1 v2 v3 ], we know that this question is equivalent to determiningwhether or not Ax = 0 has a nontrivial solution. Considering the augmentedmatrix [A 0] and row-reducing, we find ⎡ ⎤ ⎡ ⎤ 1 −1 0 0 1 0 0 0 ⎣ 1 0 1 0⎦ → ⎣0 1 0 0⎦ (1.6.2) −1 1 1 0 0 0 1 0It follows that Ax = 0 has only the trivial solution, and therefore the set S islinearly independent. Geometrically, this means that if we take any nontrivialcombination of v1 , v2 , and v3 , the result is a vector that is not the zero vector.From example 1.6.2, we see how we will normally test a set of vectors for linearindependence: we take advantage of our understanding of linear combinationsand matrix multiplication and convert the vector equation x1 v1 + x2 v2 + · · · +xk vk = 0 to the matrix equation Ax = 0, where A is the matrix with columnsv1 , . . . , vk . Row-reducing, we can test whether or not nontrivial solutions existto Ax = 0 by examining pivot locations in the matrix A. Several facts about linear dependence and independence will prove to beuseful in many aspects of our upcoming work. We simply state them here, andleave their verification to the exercises at the end of this section: • Any set containing the zero vector is linearly dependent. • Any set {v1 } consisting of a single nonzero vector is linearly independent. • Any set of two vectors {v1 , v2 } is linearly independent whenever v1 is not a scalar multiple of v2 . • The columns of a matrix A are linearly independent if and only if the equation Ax = 0 has only the trivial solution. The concepts of linear independence and span both involve linearcombinations of a set of vectors. Furthermore, there are many important andnatural connections between span and linear independence. The next exampleextends the previous one and lays the foundation for a discussion of severalgeneral results.Example 1.6.3 Let the vectors v1 , v2 , v3 , and v4 be given by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 0 5 v1 = ⎣ 1 ⎦ , v2 = ⎣ 0 ⎦ , v3 = ⎣ 1 ⎦ , v4 = ⎣ 6 ⎦ −1 1 1 −1Let R = {v1 , v2 }, S = {v1 , v2 , v3 }, and T = {v1 , v2 , v3 , v4 }. Which of the sets R,S, and T are linearly independent? Which of the sets R, S, and T span R3 ?
Linear independence 53Solution. We have already seen in example 1.6.2 that the set S is linearlyindependent. Moreover, we saw that when we let A = [v1 v2 v3 ] and row-reducethe augmented matrix for the equation Ax = 0, it follows that ⎡ ⎤ ⎡ ⎤ 1 −1 0 0 1 0 0 0 ⎣ 1 0 1 0⎦ → ⎣0 1 0 0⎦ −1 1 1 0 0 0 1 0Not only does this show that the vectors in set S are linearly independent (Ax = 0has only the trivial solution because A has a pivot in every column so there areno free variables present), but also, by theorem 1.5.2, the vectors in S span R3since A has a pivot in every row. Since the vectors in S span R3 , this means thatwe can write every vector in R3 as a linear combination of the three vectors in S.Moreover, since A has a pivot in every column, it will also follow that every suchlinear combination is unique: every vector in R3 can be written in exactly oneway as a linear combination of v1 , v2 , and v3 . What happens if we remove v3 from S and instead consider the set R ={v1 , v2 }? To answer the question of linear independence, we ask if there is anontrivial solution to the vector equation x1 v1 + x2 v2 = 0. Equivalently, we letB be the 3 × 2 matrix whose columns are v1 and v2 and solve Bx = 0. Doing so,we find that ⎡ ⎤ ⎡ ⎤ 1 −1 0 1 0 0 ⎣ 1 0 0⎦ → ⎣0 1 0 ⎦ −1 1 0 0 0 0so only the trivial solution exists and thus the set R is linearly independent. Noteagain that this is due to the fact that B has a pivot in every column. This shouldnot be surprising, since we removed a vector from the linearly independent set Sto get the set R: if the vectors in S do not depend on one another, neither shouldthe vectors in R. On the other hand, we can also say by theorem 1.5.2 that the set R does notspan R3 , since B does not have a pivot position in every row. For example, thevector b = [0 1 1]T cannot be written as a linear combination of v1 and v2 .This can be seen by row-reducing the augmented matrix that represents Bx = b,where we find that ⎡ ⎤ ⎡ ⎤ 1 −1 0 1 0 0 ⎣ 1 0 1⎦ → ⎣0 1 0⎦ −1 1 1 0 0 1The last equation tells us that 0x1 + 0x2 = 1, which is impossible, and thus bcannot be written as a linear combination of the vectors in R. Finally, we consider the set T = {v1 , v2 , v3 , v4 }. To test if T is linearlyindependent, we let C be the matrix whose columns are v1 , v2 , v3 , and v4 ,
54 Essentials of linear algebraand consider the equation Cx = 0, which corresponds to the equation x1 v1 +x2 v2 + x3 v3 + x4 v4 = 0. Row-reducing, ⎡ ⎤ ⎡ ⎤ 1 −1 0 5 0 1 0 0 2 0 ⎣ 1 0 1 6 0⎦ → ⎣0 1 0 −3 0⎦ −1 1 1 −1 0 0 0 1 4 0Note that the variable x4 is free, since C does not have a pivot in its fourthcolumn. This shows that any vector x with entries x1 , x2 , x3 , and x4 such thatx1 = −2x4 , x2 = 3x4 , and x3 = −4x4 will be a solution to the equation Cx = 0.For example, taking x4 = 1, it follows that ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 −1 0 5 0 −2 ⎣ 1 ⎦ + 3 ⎣ 0 ⎦ − 4 ⎣ 1 ⎦ + 1 ⎣ 6 ⎦ = ⎣ 0 ⎦ −1 1 1 −1 0Thus, the set T is linearly dependent. We can also see from our computationsthat the set T does indeed span R3 , since the matrix C has a pivot position inevery row. This result should be expected: we have already shown that everyvector in R3 can be written as a linear combination of the vectors in S, and theset T contains all three vectors in S. There are many important generalizations we can make from example 1.6.3.For instance, from an algebraic perspective we see that we can easily answerquestions about the linear independence and span of the columns of a matrixsimply by considering the location of pivots in the matrix. In particular, thecolumns of A are linearly independent if and only if A has a pivot in everycolumn, while the columns of A span Rm if and only if A has a pivot in everyrow. We state these results formally in the two following theorems.Theorem 1.6.1 Let A be an m × n matrix. The following statements areequivalent: a. The columns of A span Rm . b. A has a pivot position in every row. c. The equation Ax = b is consistent for every b ∈ Rm . In the next theorem, note particularly the change in emphasis in state-ment (b) from rows to columns when considering pivot positions in the matrix.Theorem 1.6.2 Let A be an m × n matrix. The following statements areequivalent: a. The columns of A are linearly independent. b. A has a pivot position in every column. c. The equation Ax = 0 has only the trivial solution.
Linear independence 55 At this point, it appears ideal if a set is linearly independent or spans Rm .The best scenario, then, is the case when a set has both of these propertiesand forms a linearly independent spanning set. In this case, for the matrixwhose columns are the vectors in the set, we need the matrix to have apivot in every column, as well as in every row. As we saw in example 1.6.3with the set S and the corresponding matrix A, this can only happen whenthe number of vectors in the set S matches the number of entries in eachvector. In other words, the corresponding matrix A must be square. Obviouslyif a square matrix has a pivot in every row, it must also have a pivotin every column, and vice versa. We close our current discussion with animportant result that links the concepts of linear independence and span inthe columns of a square matrix; theorem 1.6.3 is a consequence of the twopreceding ones.Theorem 1.6.3 Let A be an n × n matrix. The following statements areequivalent: a. The columns of A are linearly independent. b. The columns of A span Rn . c. A has a pivot position in every column. d. A has a pivot position in every row. e. For each b ∈ Rn , the equation Ax = b has a unique solution.Theorem 1.6.3 shows that square matrices play a particularly important role inlinear algebra, an idea that will further demonstrate itself when we study thenotion of the inverse of a matrix in the following section. We conclude this section with a look ahead to our study of linear differentialequations, in which the concepts of linear independence and span will also finda prominent role.Example 1.6.4 Consider the differential equation y + y = 0. Explain why thefunction y = c1 cos t + c2 sin t is a solution to the differential equation.Solution. In our upcoming study of differential equations, we will call theequation y + y = 0 a linear second-order homogeneous equation with constantcoefficients. Equations of this form will be considered in chapter 3 and be thefocus of chapter 4. For now, we can intuitively understand why y = c1 cos t + c2 sin t is asolution to the equation. Note that in order to solve the equation y + y = 0, wemust find all functions y such that y = −y. From our experience in calculus,we know that d d [sin t ] = cos t and [cos t ] = − sin t dt dt
Linear independence 5710. Suppose that S is a set of three vectors in R5 . Is it possible for S to span R5 ? Why or why not?11. Suppose that S is a set of two vectors in R3 . Is S linearly independent, linearly dependent, or not necessarily either? Explain your answer.12. Let S be a set of four vectors in R3 . Is it possible for S to be linearly independent? Is it possible for S to span R3 ? Why or why not?13. Let S be a set of five vectors in R4 . Must S span R4 ? Is it possible for S to be linearly independent? Explain.14. If A is an m × n matrix, for what relationship between n and m are the columns of A guaranteed to not span Rm ? For what relationship between n and m will the columns have to be linearly dependent?15. Prove that any set that contains the zero vector must be linearly dependent.16. Explain why any set consisting of a single nonzero vector must be linearly independent.17. Show that any set of two vectors, {v1 , v2 }, is linearly independent if and only if v1 is not a scalar multiple of v2 .18. Explain why the columns of a matrix A are linearly independent if and only if the equation Ax = 0 has only the trivial solution.19. Let v1 = [−1 2 1]T , v2 = [3 1 1]T , and v3 = [5 3 k ]T . For what value(s) of k is {v1 , v2 , v3 } linearly independent? For what value(s) of k is v3 in the span of {v1 , v2 }? How are these two questions related?20. Consider the set S = {v1 , v2 , v3 } where v1 = [1 0 0]T , v2 = [0 1 0]T , and v3 = [0 0 1]T . Explain why S spans R3 , and also why S is linearly independent. In addition, determine the weights x1 , x2 , and x3 that allow you to write the vector [−27 13 91]T as a (unique) linear combination of v1 , v2 , v3 . What do you observe?21. Let A be a 4 × 7 matrix. Suppose that when solving the homogeneous equation Ax = 0 there are three free variables present. Do the columns of A span R4 ? Explain. Are the columns of A linearly dependent, linearly independent, or is it impossible to say? Justify your answer.22. Suppose that A is a 9 × 6 matrix and that A has six pivot columns. Are the columns of A linearly dependent, linearly independent, or is it impossible to say? Do the columns of A span R9 , or is it impossible to tell? Justify your answers.23. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) If the system represented by Ax = 0 has a free variable present, then the columns of the matrix A are linearly independent vectors.
58 Essentials of linear algebra (b) If a matrix has more columns than rows, then the columns of the matrix must be linearly dependent. (c) If an m × n matrix A has a pivot in every column, then the columns of A span Rm . (d) If A is an m × n matrix that is not square, it is possible for its columns to be both linearly independent and span Rm .24. Consider the linear second-order homogeneous differential equation y + y = 0. Show by direct substitution that y1 = e t and y2 = e −t are solutions to the differential equation. In addition, show by substitution that any linear combination y = c1 e t + c2 e −t is also a solution.25. We have seen that the general solution to the linear second-order differential equation y + y = 0 is given by y(t ) = c1 sin(t ) + c2 cos(t )26. It can be shown that the solution to the linear second-order differential equation y − y = 0 is given by y(t ) = c1 e t + c2 e −t1.7 Matrix algebraFor a given system of linear equations, we are now interested in solving thevector equation Ax = b, where A is a known m × n matrix, b ∈ Rm is given,and we seek x ∈ Rn . It is natural to compare this equation to an elementarylinear equation such as 2x = 7. The key algebraic step in solving 2x = 7 isto divide both sides of the equation by 2. Said differently, we multiply bothsides by the multiplicative inverse of the number 2. In anticipation of a newapproach to solving the vector equation Ax = b, we carefully state the detailsrequired to solve 2x = 7. In particular, from the equation 2x = 7, it follows that2 (2x) = 2 (7), so that ( 2 · 2)x = 2 . Thus, 1 · x = 2 , so x = 2 . From a sophisticated1 1 1 7 7 7perspective, to solve the equation 2x = 7, we need to be able to multiply, to havea multiplicative identity (that is, the number 1), and to be able to compute amultiplicative inverse (here, the number 1 ). 2
Matrix algebra 59 In this section, we lay the foundation for similar ideas that provide analternate way to solve the equation Ax = b: essentially we are interested indetermining whether we can find a matrix B so that when we compute BA theresult is the matrix equivalent of "1". To do this, we will first have to learn whatit means to multiply two matrices; a simpler (and still important) place to beginis with the addition of matrices and multiplication of matrices by scalars. We already know how to add vectors and multiply them by scalars; similarprinciples hold for matrices. Two matrices can be added (or subtracted) if andonly if they have an identical number of rows and columns. When addition(subtraction) is defined, the result is computed component-wise. Furthermore,the multiple of a matrix by a scalar c ∈ R is attained by multiplying every entryof the matrix by the same constant c. The following example demonstrates thesebasic facts.Example 1.7.1 Let A and B be the matrices 1 3 −4 −6 10 −1 A= , B= 0 −7 2 3 2 11Compute A + B and −3A.Solution. Since A and B are both 2 × 3, their sum is defined and is given by 1 3 −4 −6 10 −1 −5 13 −5 A+B = + = 0 −7 2 3 2 11 3 −5 13The scalar multiple of a matrix is always defined, and −3A is given by −3 −9 12 −3A = 0 21 −6Matrix addition, when defined, has all of the expected properties of addition.In particular, A + B = B + A, so order does not matter, and we say matrixaddition is commutative. Since A + (B + C) = (A + B) + C, the way we groupmore than two matrices to add also does not matter and we say matrix additionis associative. There is even a matrix that acts like the number 0. If Z is a matrixof the same number of rows and columns as A such that every entry in Z is zero,then it follows that A + Z = Z + A = A. We call this zero matrix the additiveidentity. The next natural operation to consider, of course, is multiplication. Whatdoes it mean to multiply two matrices? And when does it even make senseto multiply two matrices? We know for matrix–vector multiplication that theproduct Ax computes the vector b that is the unique linear combination ofthe columns of A having the entries of the vector x as weights. Moreover, thisproduct is only defined when the number of entries in x matches the number ofcolumns of A. If we now consider a matrix B, we can naturally think about thematrix product AB by considering the columns of B, say b1 , . . . , bk . In particular,we make the following definition.
60 Essentials of linear algebraDefinition 1.7.1 If A is an m × n matrix, and B is a matrix whose columnsare b1 , . . . , bk such that the matrix–vector product Abj is defined for eachj = 1, . . . , k, then we define the matrix product AB by AB = [Ab1 Ab2 · · · Abk ] (1.7.1) Note particularly that since A has n columns, in order for Abj to be definedeach bj must belong to Rn . This in turn implies that the matrix B must havedimensions n × k. Specifically, the number of rows in B must equal the numberof columns in A. We explore matrix multiplication and its properties in the nextexample.Example 1.7.2 Let A and B be the matrices 1 3 −4 −6 10 A= , B= 0 −7 2 3 2Compute the matrix products AB and BA, or explain why they are not defined.Solution. First we consider AB. To do so, we would have to compute both Ab1and Ab2 , where b1 and b2 are the columns of B. But neither of these products isdefined, since A has three columns and B has just two rows. Thus, AB is not defined. On the other hand, BA is defined. For instance, we can compute the firstcolumn of BA by taking Ba1 , where we see that −6 10 1 −6 Ba1 = = 3 2 0 3Similar computations for Ba2 and Ba3 show that −6 −88 44 BA = 3 −5 −8There are several important observations to make based on example 1.7.2. Oneis that if A is m × n and B is n × k so that the product AB is defined, then theresulting matrix AB is m × k. This is true since the columns of AB are each ofthe form Abj , thus being linear combinations of the columns of A, which havem entries, so that AB has m rows. Moreover, we have to consider each of theproducts Ab1 , . . . , Abk , therefore giving AB k columns. Furthermore, we clearly see that order matters in matrix multiplication.Specifically, given matrices A and B for which AB is defined, it is not evenguaranteed that BA is defined, much less that AB = BA. Even when bothproducts are defined, it is possible (even typical) that AB = BA. Formally, wesay that matrix multiplication is not commutative. This fact will be exploredfurther in the exercises. It is, however, the case that matrix multiplication (formatrices of the appropriate sizes) is both associative and distributive. That is,A(BC) = (AB)C and A(B + C) = AB + AC, again provided the sizes of thematrices make the relevant products and sums defined.
Matrix algebra 61 Now, we should not forget our motivation for considering matrix multi-plication: we want to develop an alternative approach to solving equations ofthe form Ax = b by multiplying A by another matrix B so that the product BAis the matrix equivalent of the number 1 (while simultaneously multiplying b bythe same matrix B). What is the matrix equivalent of the number 1? We considerthis question and more in the following example.Example 1.7.3 Consider the matrices 5 11 1 0 A= and I2 = −3 −7 0 1Compute AI2 and I2 A. What is special about the matrix I2 ?Solution. Using the rules for matrix multiplication, we observe that 5 11 1 0 5 11 AI2 = = =A −3 −7 0 1 −3 −7and similarly 1 0 5 11 5 11 I2 A = = =A 0 1 −3 −7 −3 −7Thus, we see that multiplying the matrix A by I2 has no effect on the matrix A.The matrix I2 in example 1.7.3 is important because it has the propertythat I2 A = A for any matrix A with two rows (not simply the matrix A inexample 1.7.3) and AI2 = A for any A with two columns. We can similarly showthat if I3 is the matrix ⎡ ⎤ 1 0 0 I3 = ⎣0 1 0⎦ 0 0 1then I3 A = A for any matrix A with three rows, and AI3 = A for any matrix Awith three columns. Similar results hold for corresponding matrices In of largersize; each of these matrices acts like the number 1, since multiplying othermatrices by In has no effect on the given matrix. Matrices which when multiplied by other matrices do not change the othermatrices, are called identity matrices. More formally, the n × n identity matrix Inis the square matrix whose diagonal entries all equal 1, and whose off-diagonalentries are all 0. (The diagonal entries in a matrix are those whose row andcolumn indices are the same.) Often, when the context is clear, we will writesimply I, rather than In . We also note that In is the only matrix that is n × n andacts as a multiplicative identity. Finally, it is evident that for any m × n matrixA, Im A = AIn = A. In the next section, we will explore the notion of the inverseof a matrix, and there see that identity matrices play a central role. One final algebraic operation with matrices merits formal introductionhere. Given a matrix A, its transpose, denoted AT , is the matrix whose columns
62 Essentials of linear algebraare the rows of A. That is, taking the transpose of a matrix replaces its rows withits columns, and vice versa. For example, if A is the 2 × 3 matrix 1 3 −4 A= 0 −7 2then its transpose AT is the 3 × 2 matrix ⎡ ⎤ 1 0 AT = ⎣ 3 −7⎦ −4 2Note that this is the same notation we regularly use to express a column vector inthe form b = [1 2 3]T . In the case that A is a square matrix, taking its transposeresults in swapping entries across its diagonal. For example, if ⎡ ⎤ 5 −2 7 A = ⎣ 0 −3 −1⎦ −4 8 −6then ⎡ ⎤ 5 0 −4 AT = ⎣−2 −3 8⎦ 7 −1 −6The transpose operator has several nice algebraic properties, some of which willbe explored in the exercises. For example, for matrices for which the appropriatesums and products are defined, (A + B)T = AT + BTand (AB)T = BT AT For a square matrix such as 3 −1 A= −1 2it happens that AT = A. Any square matrix A for which AT = A is said tobe symmetric. It turns out that symmetric matrices have several especially niceproperties in the context of more sophisticated concepts that arise later in thetext, and we will revisit them at that time.1.7.1 Matrix algebra using MapleWhile it is important that we first learn to add and multiply matrices by handto understand how these processes work, just like with row-reduction it isreasonable to expect that we will often use available technology to performtedious computations like multiplying a 4 × 5 and 5 × 7 matrix. Moreover, inreal-world applications, it is not uncommon to have to deal with matrices that
Matrix algebra 63have thousands of rows and thousands of columns, or more. Here we introducea few Maple commands that are useful in performing some of the algebraicmanipulations we have studied in this section. Let us consider some of the matrices defined in earlier examples: 1 3 −4 −6 10 −6 10 −1 A= , B= , C= 0 −7 2 3 2 3 2 11After defining each of these three matrices with the usual commands in Maple,such as > A := <<1,0>|<3,-7>|<-4,2>>;we can execute the sum of A and C and the scalar multiple −3B with thecommands > A + C; > -3*B;for which Maple will report the outputs −5 13 −5 18 −30 and 3 −5 13 −9 −6We have previously seen that to compute a matrix–vector product, the periodis used to indicate multiplication, as in > A.x;. The same syntax holds formatrix multiplication, where defined. For example, if we wish to compute theproduct BA, we enter > B.A;which yields the output −6 −88 44 3 −5 −8 If we try to have Maple compute an undefined product, such as AB throughthe command > A.B;, we get the error message Error, (in LinearAlgebra:-MatrixMatrixMultiply) first matrix column dimension (3) <> second matrix row dimension (2) In the event that we need to execute computations involving an identitymatrix, rather than tediously enter all the 1's and 0's, we can use the built-inMaple command IdentityMatrix(n); where n is the number of rows andcolumns in the matrix. For example, entering > Id := IdentityMatrix(4);
Matrix algebra 65 3. Discuss the differences between multiplying two square matrices versus multiplying non-square matrices. That is, under what circumstances can two square matrices be multiplied? How does the situation change for non-square matrices? In addition, if the product AB is defined, is BA? 4. Give an example of 2 × 2 matrices A and B for which AB = BA. 5. Give an example of 2 × 2 matrices A and B for which AB = BA. 6. If A is m × n and B is n × k, and neither A nor B is square, can AB ever equal BA? Explain.In exercises 7–9, let A be the given matrix. If possible, find a matrix B such thatBA = I2 ; if B exists, determine whether BA = AB. 2 0 7. A = 0 5 2 4 8. A = 0 5 1 −1 9. A = −1 2In exercises 10 and 11, for the given matrix A, answer each of the followingquestions:(a) Are the columns of A linearly independent?(b) Do the columns of A span R2 ?(c) How many pivot positions does A have?(d) Solve the equation Ax = 0 by row reducing by hand. Is A row equivalent to an important matrix?(e) If possible, determine a 2 × 2 matrix B such that BA = I2 . −1 210. A = 2 −3 −1 211. A = 2 −412. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) If A and B are matrices of the same size, then the products AB and BA are always defined. (b) If A and B are matrices such that the products AB and BA are both defined, then AB = BA. (c) If A and B are matrices such that AB is defined, then (AB)T = AT BT .
66 Essentials of linear algebra (d) If A and B are matrices such that A + B is defined, then (A + B)T = AT + BT .13. Compute the prescribed algebraic computations in exercise 1 using a computer algebra system.14. Compute the prescribed algebraic computations in exercise 2 using a computer algebra system.1.8 The inverse of a matrixWe have observed repeatedly that linear algebra is a subject centered on oneidea—systems of linear equations—viewed from several different perspectives.Continuing with this theme, we have recently considered an alternative methodfor solving the equation Ax = b by attempting to find a matrix B such thatBA = I, where I is the appropriate identity matrix. If we can in fact find such amatrix B, it follows that B(Ax) = Bb (1.8.1)By the associativity of matrix multiplication and the defining property of B, itfollows that B(Ax) = (BA)x = Ix = x (1.8.2)Equations (1.8.1) and (1.8.2) together imply that x = Bb. Thus, the existence ofsuch a matrix B shows us how we can solve Ax = b by multiplication. It turns outthat from a computational point of view, row-reduction is a superior approachto solving Ax = b; nonetheless, the perspective that it may be possible to solvethe equation through the use of a multiplicative inverse has many importanttheoretical applications. In addition, similar ideas will be encountered in ourstudy of differential equations. Our work in section 1.7 showed that if A and B are not square matrices,it is never the case that AB and BA are equal. Thus it is only possible to find amatrix B such that AB = BA = I if A is square (though even then it is not alwaysthe case that such a matrix B exists). Moreover, as we know from theorem 1.6.3,some square matrices have the important property that the equation Ax = b hasa unique solution for every possible choice of b. For the next few sections, we therefore focus our attention almost exclusivelyon square matrices. Here, our emphasis is on the questions "when does a matrixB exist such that AB = BA = I?" and "when such a matrix B exists, how can wefind it?" The next definition formalizes the notion of the inverse of a matrix.Definition 1.8.1 If A is an n × n matrix, we say that A is invertible if and onlyif there exists an n × n matrix B such that AB = BA = In (1.8.3)
The inverse of a matrix 67When A is invertible, we call B the inverse of A and write B = A−1 (read "B isA-inverse"). If A is not invertible, A is often called a singular matrix, and thussaying "A is invertible" is equivalent to saying "A is nonsingular." It can be shown (see exercise 19) that if A is an invertible n × n matrix,then its inverse is unique (i.e., a given matrix cannot have two distinct inverses).In addition, we note from our discussion above in (1.8.1) and (1.8.2) that ifA is invertible, then the equation Ax = b has a solution for every b ∈ Rn . Inparticular, that solution is x = A−1 b. Moreover, since Ax = b has a solutionfor every b ∈ Rn , we know from theorem 1.6.1 that A has a pivot position inevery row. From this, the fact that A is square, and theorem 1.6.3, it follows thatAx = b has a unique solution for every b ∈ Rn . We state this result formally inthe following theorem.Theorem 1.8.1 If A is an n × n invertible matrix, then the equation Ax = bhas a unique solution for every b ∈ Rn .Before beginning to explore how to find the inverse of a matrix, as well as whenthe inverse even exists, we consider an example to see how we may check if twomatrices are inverses and how to apply an inverse to solve a related equation.Example 1.8.1 Let A and B be the matrices 4 5 2/3 −5/3 A= , B= 1 2 −1/3 4/3Show that A and B are inverses, and then use this fact to solve Ax = b, whereb = [−7 3]T , without using row reduction.Solution. The reader should verify that the following matrix products indeedhold: 4 5 2/3 −5/3 1 0 AB = = 1 2 −1/3 4/3 0 1and 2/3 −5/3 4 5 1 0 BA = = −1/3 4/3 1 2 0 1This shows that indeed B = A−1 . Note, equivalently, that A = B−1 . Now, we caneasily solve the equation Ax = b where b is the given vector: 2/3 −5/3 −7 −29/3 x = A− 1 b = = −1/3 4/3 3 19/3Of course, what is not clear in example 1.8.1 is how, given the matrix A, onemight determine the entries in the inverse matrix B = A−1 . We now explore thisin the 3 × 3 case for a general matrix A, and along the way learn conditions thatguarantee that A−1 exists.
68 Essentials of linear algebra Given a 3 × 3 matrix A, we seek a matrix B such that AB = I3 . Let thecolumns of B be b1 , b2 , and b3 , and the columns of I3 be e1 , e2 , and e3 . Thecolumn-wise definition of matrix multiplication then tells us that the followingthree vector equations must hold: Ab1 = e1 , Ab2 = e2 , and Ab3 = e3 (1.8.4)For the unique inverse matrix B to exist, it follows that each of these equationsmust have a unique solution. Clearly if A has a pivot position in every row (or,equivalently, the columns of A span R3 ), then by theorem 1.6.3 it follows thatwe can find unique vectors b1 , b2 , and b3 that make these three equations hold.Thus, any one of the conditions in theorem 1.6.3 will guarantee that B = A−1exists. Moreover, if A−1 exists, we know from theorem 1.8.1 that every conditionin theorem 1.6.3 also holds. Momentarily, let us assume that A is indeed invertible. If we proceed to findthe matrix B by solving the three equations in (1.8.4), we see that row-reductionprovides an approach for producing all three vectors at once. To find thesevectors one at a time, it would be necessary to row-reduce each of the threeaugmented matrices [A e1 ], [A e2 ], and [A e3 ] (1.8.5)In each case, the exact same elementary row operations will be applied to A andthus be applied, respectively, to the vectors e1 , e2 , and e3 . As such, we may doall of them at once by considering the augmented matrix [A e1 e2 e3 ] (1.8.6)Note particularly that the form of the augmented matrix in (1.8.6) is [A I3 ]. Ifwe now row-reduce this matrix, and A has a pivot in every row, it follows thatwe will be able to read the coefficients of A−1 from the result. This process is bestilluminated by an example, so we now explore how these computations lead usto A−1 in a concrete situation.Example 1.8.2 Find the inverse of the matrix ⎡ ⎤ 2 1 −2 A=⎣ 1 1 −1⎦ −2 −1 3Solution. Following the discussion above, we augment A with the 3 × 3identity matrix and row-reduce. It follows that ⎡ ⎤ ⎡ ⎤ 2 1 −2 1 0 0 1 0 0 2 −1 1 ⎣ 1 1 −1 0 1 0⎦ → ⎣0 1 0 −1 2 0⎦ −2 −1 3 0 0 1 0 0 1 1 0 1These computations demonstrate two important things. The first is that therow reduction of A in the first three columns of the augmented matrix shows
The inverse of a matrix 69that A has a pivot position in every row, and therefore A is invertible. Moreover,the row-reduced form of [A I3 ] tells us that A−1 is the matrix ⎡ ⎤ 2 −1 1 A−1 = ⎣−1 2 0⎦ 1 0 1Again, we observe from our preceding discussion and example 1.8.2 that we havefound an algorithm for finding the inverse of a square matrix A. We augment Awith the corresponding identity matrix and row-reduce. Provided that A has apivot in every row, we find by row-reducing that [A I] → [I A−1 ]That is, row-reduction of an invertible matrix A augmented with the identitymatrix leads us directly to the inverse, A−1 . Next, we examine what happens in the event that a square matrix is notinvertible.Example 1.8.3 Find the inverse of the matrix 2 1 A= −6 −3provided the inverse exists. If the inverse does not exist, explain why.Solution. We augment A with the 2 × 2 identity matrix and row-reduce,finding that 2 1 1 0 1 1 0 −1 2 6 → −6 −3 0 1 0 0 1 1 3Again, we see at least two key facts from these computations: A does not havea pivot position in every row, and thus A is not invertible. In particular, recallthat we are solving two vector equations simultaneously in these computations:Ab1 = e1 and Ab2 = e2 . If we consider the first of these and observe the row-reduction 2 1 1 1 1 0 → 2 −6 −3 0 0 0 1we see that this system of equations is inconsistent—the last row of theaugmented matrix is equivalent to the equation 0b11 + 0b12 = 1, where b =[b11 b12 ]T . This is yet another way of saying that A does not have an inverse.The above two examples together show us, in general, how we answer twoquestions at once: does the square matrix A have an inverse? And if so, what isA−1 ? In a computational sense, we can simply row-reduce A augmented withthe appropriate identity matrix and then observe if A has a pivot position inevery row. If A is row equivalent to the appropriately sized identity matrix, thenA is invertible and A−1 will be revealed through the row-reduction.
70 Essentials of linear algebra We close this section with a formal statement of a theorem that summarizesour discussion. Note particularly how this result extends theorem 1.6.3 anddemonstrates the theme of linear algebra: one idea from several perspectives.We will refer to this result as The Invertible Matrix Theorem.Theorem 1.8.2 (The Invertible Matrix Theorem) Let A be an n × n matrix.The following In addition to being of great theoretical significance, inverse matricesfind many key applications. We investigate one such use in the followingsubsection.1.8.1 Computer graphicsLinear algebra is the engine that drives computer animations. While animatedmovies originally were constructed by artists hand-drawing thousands of similarsketches that were photographed and played in sequence, today such filmsare created entirely with computers. Once a figure has been constructed,moving the image around the screen is essentially an exercise in matrixmultiplication. Every pixel in an image on a computer screen can be represented throughcoordinates. For an elementary example, consider an animated figure which, ata given point in time, has its hand located at the point (3, 4). To see how a basicanimation can be built, assume further that the figure's elbow is at the origin(0, 0), and that an animator wishes to make the hand wave back and forth. Thisenables us to represent the forearm of the figure with the vector v = [3 4]T . If we now consider the matrix √ 3/2 √1/2 − R= 1/2 3/2and apply the matrix R to the vector v, we see that the product is √ √ 3/2 √1/2 3 − 3 3 − 4/2 √ 0.598 Rv = = ≈ 1/2 3/2 4 3 + 4 3/2 4.964
The inverse of a matrix 71 Rv 5 v 3 Figure 1.13 The vectors v = [3 4] and Rv = [0.598 4.964]T .Thus, the figure's hand is now located at the point (0.598, 4.964). In fact,the hand has been rotated 30◦ counterclockwise about the origin, as shownin figure 1.13. The matrix R is known as a rotation matrix; its impact on any vector is torotate the vector 30◦ counterclockwise about the origin. One way to see why thisis so is to compute the vectors Re1 and Re2 , where e1 and e2 are the columnsof the 2 × 2 identity matrix. Since each of those two vectors is rotated 30◦ whenmultiplied by R, the same thing happens to any vector in R2 , because any suchvector may be written as a linear combination of e1 and e2 . Not only do computer animations show one application of matrix–vectormultiplication, but they also demonstrate the need for inverse matrices. Forinstance, suppose we knew that the matrix R had been applied to some unknownvector v and that the result was 2 Rv = 5That is, a hand located at some unknown point v was waved and had been movedto the new point (2, 5). An animator might want to wave the hand back so thatit ended up at its original location, which is again represented by the vector v.To do so, he must answer the question "for which vector v is Rv = [2 5]T ?" We now know that one way to solve for v is to use the inverse of R. Thematrix R is clearly invertible because its columns are linearly independent; wecan compute R −1 in the standard way to find that √ 3/2 √1/2 R −1 = −1/2 3/2
72 Essentials of linear algebraWe can solve for v by computing 2 v = R −1 (Rv) = R −1 5so that √ −1 2 3/2 √1/2 2 4.232 v=R = ≈ 5 −1/2 3/2 5 3.330Of course, in actual animations, we would not wave the hand by a single 30◦rotation, but rather through a sequence of consecutive small rotations, forinstance, 1-degree rotations. Again, computers enable us to do thousands ofsuch computations almost instantly and make amazing animations possible. We consider an additional example to see the role of matrices to store dataas well as matrices and their inverses to transform the data.Example 1.8.4 Consider the matrix 0 1 B= 1 0Let v1 = [2 1]T , v2 = [3 3]T , and v3 = [4 0]T be the vertices of a triangle inthe plane. Compute Bv1 , Bv2 , and Bv3 . Sketch a picture of the new triangle thathas resulted from applying the matrix B to the vertices (2, 1), (3, 3), and (4, 0).What is the impact of the matrix B on each point? Finally, determine the inverseof B. What do you observe?Solution. We observe first that 0 1 2 1 0 1 3 3 Bv1 = = , Bv2 = = , and 1 0 1 2 1 0 3 3 0 1 4 0 Bv1 = = 1 0 0 4From these calculations, we see that multiplying by B moves a given point to anew point that corresponds to the one found by switching the coordinates ofthe given point. Geometrically, the matrix B accomplishes a reflection acrossthe line y = x in the plane, as we can see in figure 1.14.Moreover, if we think about how we might undo reflection across the line y = x,it is clear that to restore a point to its original location, we need to reflect thepoint back across the line. Said differently, the inverse of the matrix B must bethe matrix itself. We can confirm that B−1 = B by computing the product 0 1 0 1 BB = =I 1 0 1 0It is noteworthy that the calculations of Bv1 , Bv2 , and Bv3 can be simplified intoa single matrix product if we let T = [v1 v2 v3 ]. That is, the matrix T holds the
The inverse of a matrix 73 5 (0,4) (3,3) (1,2) (2,1) (4,0) 5 Figure 1.14 The triangle with vertices v1 = [2 1]T , v2 = [3 3]T , and v3 = [4 0]T and its image under multiplication by the matrix B.coordinates of the three points in the given triangle; the product BT is then theimage of the triangle under multiplication by the matrix B. A more complicatedpolygonal figure than a triangle would be stored in a matrix with additionalcolumns. Of course, the actual work of computer animations is much more com-plicated than what we have presented here. Nonetheless, matrix multiplicationis the platform on which the entire enterprise of animated films is built. Inaddition to achieving rotations and reflections, matrices can be used to dilate(or magnify) images, to shear images, and even to translate them (provided thatwe are clever about the coordinate system we use to represent points). Finally,matrices are even essential to the storage of images, as each column of a matrixcan be viewed as a data point in an image. More about the application of matricesand their inverses to computer graphics can be learned in one of the projectsfound at the end of this chapter. In addition, a deeper discussion of the notionof linear transformations (of which reflection and rotation matrices are a part)can be found in appendix D.1.8.2 Matrix inverses using MapleCertainly we can use Maple's row-reduction commands to find inverses ofmatrices. However, an even simpler command exists that enables us to avoidhaving to enter the corresponding identity matrix. Let us consider the twomatrices from examples 1.8.2 and 1.8.3. Let ⎡ ⎤ 2 1 −2 A=⎣ 1 1 −1⎦ −2 −1 3If we enter the command > MatrixInverse(A);
The inverse of a matrix 75 Then execute computations to find the explicit weights by which b is a linear combination of the columns of A. ⎡ ⎤ 1 0 0 9. Let E be the elementary matrix given by E = ⎣0 0 1⎦. Note that E is 0 1 0 obtained by interchanging rows 2 and 3 of the 3 × 3 identity matrix. Choose a 3 × 3 matrix A, and compute EA. What is the effect on A of multiplication by E?10. Without doing any row-reduction, determine E−1 where E is the matrix defined in exercise 9. (Hint: E−1 EI = I. Think about the impact that E has on I, and then what E−1 must accomplish.) ⎡ ⎤ 1 0 011. Let E be the elementary matrix given by E = ⎣0 c 0⎦. Note that E is 0 0 1 obtained by scaling the second row of the 3 × 3 identity matrix by the constant c. Choose a 3 × 3 matrix A, and compute EA. What is the effect on A of multiplication by E?12. Without doing any row reduction, determine E−1 where E is the matrix defined in exercise 11. What do you observe? ⎡ ⎤ 1 0 013. Let E be the elementary matrix given by E = ⎣ 0 1 0⎦. Note that E is a 0 1 obtained by applying the row operation of taking a times row 1 of the 3 × 3 identity matrix and adding it to row 3 to form a new row 3. Choose a 3 × 3 matrix A, and compute EA. What is the effect on A of multiplication by E?14. Without doing any row reduction, determine E−1 where E is the matrix defined in exercise 13. (Hint: E−1 EI = I. Think about the impact that E has on I, and then what E−1 must accomplish.) √ √ 1/√2 −1/√215. Let A = . Compute A−1 . What do you observe about the 1/ 2 1/ 2 relationship between A and A−1 ? cos θ − sin θ16. Let θ be any real number and A = . Compute AT and sin θ cos θ AT A. What do you observe about the relationship between A and AT ?17. Let A and B be invertible n × n matrices with inverses A−1 and B−1 , respectively. Show that AB is also an invertible matrix by finding (AB)−1 in terms of A−1 and B−1 .18. Let A be an invertible matrix. Explain why A−1 is also invertible, and find (A−1 )−1 .19. Show that if A is an invertible n × n matrix, then its inverse is unique. (Hint: suppose that both B and C are inverses of A. What can you say about AB and AC?)
76 Essentials of linear algebra20. For real numbers a and b, the Zero Product Property states that "if a · b = 0, then a = 0 or b = 0." Said differently, if a = 0 and b = 0, then a · b = 0. Let 0 be the 2 × 2 zero matrix (i.e., all entries are zero). Does the Zero Product Property hold for matrices? That is, can you find two nonzero matrices A and B such that AB = 0? Can you find such matrices where none of the entries in A or B are zero? If so, what kind of matrices are A and B?21. Does there exist a 2 × 2 matrix A, none of whose entries are zero, such that A2 = 0?22. Does there exist a 2 × 2 matrix A other than the identity matrix such that A2 = I? What is special about such a matrix?23. Let D be a diagonal matrix, P an invertible matrix, and A = PDP−1 . Using the expression PDP−1 for A, compute and simplify the matrix A2 = A · A. Do likewise for A3 = A · A · A. What will be the simplified form of An in terms of P, D, and P−1 ? a b24. Let A be the matrix . Find conditions on a, b, c, and d that c d guarantee that Ax = 0 has infinitely many solutions. What must therefore be true about a, b, c, and d in order for A to be invertible? √25. Let A = √1/2 3/2 and v1 , v2 , v3 be the vectors that emanate from − 3/2 1/2 the origin to the vertices of the triangle given by (2, 1), (3, 3), and (4, 0). Compute the new triangle that results from applying the matrix A to the given vertices, and sketch a picture of the original triangle and the resulting image. What is the effect of multiplying by A?26. Suppose that A in exercise 25 was applied to a different set of three unknown vectors x1 , x2 , and x3 . The resulting output from these products is −4 0 2 Ax1 = , Ax2 = , and Ax3 = 2 3 1 In other words, the new image after multiplying by A is the triangle whose vertices are (−4, 2), (0, 3), and (2, 1). Determine the exact vectors x1 , x2 , and x3 and sketch the original triangle that was mapped to the triangle with vertices (−4, 2), (0, 3), and (2, 1).27. Consider the matrix 0 −1 B= 1 0 Let v1 = [2 1]T , v2 = [3 3]T , and v3 = [4 0]T . Compute Bv1 , Bv2 , and Bv3 . Sketch a picture of the new triangle that has resulted from applying the matrix B to the vertices (1, 1), (2, 3), and (4, 0). What is the geometric effect of the matrix B on each point?
The inverse of a matrix 7728. Determine the inverse of B in exercise 27. What do you observe?29. An unknown 2 × 2 matrix C is applied to the two vectors v1 = [1 1]T and v2 = [2 3]T , and the results are Cv1 = [0.1 0.7]T and Cv2 = [−0.1 1.8]T . Determine the entries in the matrix C.30. Suppose that a computer graphics programmer decides to use the matrix √ √ 1/√2 1/√2 A= 1/ 2 1/ 2 Why is the programmer's choice a bad one? What will be the result of applying this matrix to any collection of points?31 3 1 88 Given that the population of 250 million in a certain year is distributed among 100 million urban, 100 million suburban, and 50 million rural, determine the population distribution in each of the preceding two years.32. Car-owners can be grouped into classes based on the vehicles they own. A study of owners of sedans, minivans, and sport-utility vehicles shows sUV 2 2 90 If there are currently 100 000 sedans, 60 000 minivans, and 80 000 SUVs among the owners being studied, determine the distribution of vehicles among the population before each current owner replaced his or her previous vehicle.
78 Essentials of linear algebra33. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) If A is a matrix with a pivot in every row, then A is invertible. (b) If A is an invertible matrix, then its columns are linearly independent. (c) If Ax = b has a unique solution, then A is an invertible matrix. (d) If A and B are invertible matrices, then (AB)−1 exists and (AB)−1 = A−1 B−1 . (e) If A is a square matrix row equivalent to the identity matrix, then A is invertible. (f) If A is a square matrix and Ax = b has a solution for a given vector b, then Ax = c has a solution for every choice of c. (g) If R is a matrix that reflects points across a line through the origin, then R −1 = R. (h) If A and B are 2 × 2 matrices with all nonzero entries, then AB cannot equal the 2 × 2 zero matrix.1.9 The determinant of a matrixThe Invertible Matrix Theorem (theorem 1.8.2) tells us that there are severaldifferent ways to determine whether or not a matrix is invertible, and hencewhether or not an n × n system of linear equations has a unique solution. Thereis at least one more useful way to characterize invertibility, and that is throughthe concept of a determinant. As seen in exercise 24 of section 1.8, it may beshown through row-reduction that the general 2 × 2 matrix a b c dis invertible if and only if ad − bc = 0. We call the quantity (ad − bc) thedeterminant of the matrix A, and write4 det(A) = ad − bc. Note that thisexpression provides a condition on the entries of matrix A that determineswhether or not A is invertible. We can explore similar ideas for larger matrices. For example, if we take anarbitrary 3 × 3 matrix ⎡ ⎤ a11 a12 a13 A = ⎣a21 a22 a23 ⎦ a31 a32 a33and row-reduce in order to explore conditions under which the matrix has apivot position in every row, it turns out to be necessary that the quantity D = a11 a22 a33 − a11 a23 a32 − a12 a21 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a314 Some authors use the notation |A| instead of det(A).
80 Essentials of linear algebraNext, to determine whether or not A is invertible, we row-reduce A to see if Ahas a pivot position in every row. Doing so, we find that ⎡ ⎤ ⎡ ⎤ 2 −1 1 1 0 1 ⎣ 1 1 2⎦ → ⎣0 1 1⎦ −3 0 −3 0 0 0Thus, we see that A does not have a pivot in every row, and therefore A is notinvertible. Of course, we should note that the primary motivation for the conceptof the determinant comes from the question, "is A invertible?" Indeed, onereason the 3 × 3 matrix in the above example is not invertible is preciselybecause its determinant is zero. Later in this section, we will formally establishthe connection between the value of the determinant and the invertibility of ageneral n × n matrix. It is clear at this point that determinants of most n × n matrices withn ≥ 3 require a substantial number of computations. Certain matrices, however,have particularly simple determinants to calculate, as the following exampledemonstrates.Example 1.9.2 Compute the determinant of the matrix ⎡ ⎤ 2 −2 7 A = ⎣0 −5 3⎦ 0 0 4In addition, determine if A is invertible.Solution. Again using the definition, we see that −5 3 0 3 0 −5 det(A) = 2 det − (−2) det + 7 det 0 4 0 4 0 0 = 2(−5 · 4 − 2 · 0) + 2(0 − 0) + 7(0 − 0) = 2(−5)(4) = −40Note particularly that the determinant of A is the product of its diagonal entries.Moreover, A clearly has a pivot position in every row, and so by this fact(or equivalently by the nonzero determinant of A) we see that A is invertible.In general, the determinant of any triangular matrix (one where all entrieseither below or above the diagonal are zero) is simply the product of its diagonalentries. There are other interesting properties that the determinant has, severalof which are explored in the next example for the 2 × 2 case.Example 1.9.3 Let a b A= c d
The determinant of a matrix 81be an arbitrary 2 × 2 matrix. Explore the effect of elementary row operations onthe determinant of A.Solution. First, let us consider a row swap, calling A1 the matrix c d A1 = a bWe observe immediately that det(A) = ad − bc and det(A1 ) = cb − ad =− det(A). We next consider scaling; let A2 be the matrix whose first row is [ka kb ], ascaled version of row 1 in A. We see that det(A2 ) = kad − kbc = k(ad − bc) =k · det(A). Finally, replacing, say, row 2 of A by the sum of k times row 1 with itself,we arrive at the matrix a b A3 = c + ka d + kbThen det(A3 ) = a(d + kb) − b(c + ka) = ad + kab − bc − kab = ad − bc = det(A). Thus, we see that for the 2 × 2 case, swapping rows in a matrix changesonly the sign of the determinant, scaling a row by a nonzero constant scales thedeterminant by the same constant, and executing a row replacement does notchange the value of the determinant at all. These demonstrate the effect that thethree elementary row operations from the process of row-reduction have on a2 × 2 matrix A.Given that the general definition of the determinant is recursive, it should notbe surprising that the properties witnessed in example 1.9.3 can be shown tohold for n × n matrices. We state this result formally as our next theorem.Theorem 1.9.1 Let A be an n × n matrix and k a nonzero constant. Then a. If two rows of A are exchanged to produce matrix B, then det(B) = − det(A). b. If one row of A is multiplied by k to produce B, then det(B) = k det(A). c. If B results from a row replacement in A, then det(B) = det(A). Theorem 1.9.1 enables us to more clearly see the link between invertibilityand determinants. Through a finite number of row interchanges and rowreplacements, any square matrix A may be row-reduced to upper triangularform U (where we have all subdiagonal zeros, but we do not necessarily scale toget 1's on the diagonal). It follows from theorem 1.9.1 that det(A) = (−1)k det(U),where k is the number of row interchanges needed. Note that since U istriangular, its determinant is the product of its diagonal entries, and these entries
82 Essentials of linear algebralie in the pivot locations of A. Thus, A has a pivot in every row if and only if thisdeterminant is nonzero. Specifically, we have shown that A is invertible if andonly if det(A) = 0. To conclude this section, we note that linear algebra has once again affordedan alternate perspective on the problem of solving an n × n system of linearequations, and we can now add an additional statement involving determinantsto the Invertible Matrix Theorem.Theorem 1.9.2 (Invertible Matrix Theorem) Let A be an n × n matrix. Thefollowing1.9.1 Determinants using MapleObviously for most square matrices of size greater than 3 × 3, the computationsnecessary to find determinants are tedious and present potential for error.As with other concepts that require large numbers of arithmetic operations,Maple offers a single command that enables us to take advantage of theprogram's computational powers. Given a square matrix A of any size, we simplyenter > Determinant(A);As we explore properties of determinants in the exercises of this section,it will prove useful to be able to generate random matrices. Within theLinearAlgebra package in Maple, one accomplishes this for a 3 × 3 matrixwith the command > RandomMatrix(3);For example, if we wanted to consider the determinant of a random matrix Awe could enter the code > A := RandomMatrix(3); > det(A);See exercise 11 for a particular instance where this code will be useful.
The determinant of a matrix 83Exercises 1.9 Compute (by hand) the determinant of each of the followingmatrices in exercises 1–7, and hence state whether or not the matrix is invertible. 2 1 1. A = 2 2 2 4 2. A = 1 2 ⎡ ⎤ 2 1 −3 3. A = ⎣2 2 5⎦ 2 3 −1 ⎡ ⎤ 2 1 3 4. A = ⎣2 2 4⎦ 2 3 5 ⎡ ⎤ −3 1 0 5 ⎢ 0 2 −4 0⎥ 5. A = ⎢ ⎣ 0 ⎥ 0 −7 11⎦ 0 0 0 6 ⎡ ⎤ a a d 6. A = ⎣b b e ⎦ c c f 7. In , where In is the n × n identity matrix. 1 2 8. For which value(s) of h is the matrix invertible? Explain your −3 h answer in at least two different ways. 2−z 1 9. For which value(s) of z is the matrix invertible? Why? 1 2−z10. For which value(s) of z do nontrivial solutions x to the equation 2−z 1 x = 0 exist? For one such value of z, determine a nontrivial 1 2−z solution x to the equation.11. In a computer algebra system, devise code that will generate two random 3 × 3 matrices A and B, and that subsequently computes det(A), det(B), and det(AB). What theorem do you conjecture is true about the relationship between det(AB) and the individual determinants det(A) and det(B)?
84 Essentials of linear algebra12. In a computer algebra system, devise code that will generate a random 3 × 3 matrix A and that subsequently computes its transpose AT , as well as det(A) and det(AT ). What theorem do you conjecture is true about the relationship between det(A) and det(AT )?13. Use the formula conjectured in exercise 11 above to show that if A is 1 invertible, then det(A−1 ) = . (Hint: AA−1 = I.) det(A)14. What can you say about the determinant of any square matrix in which one of the columns (or rows) is zero? Why?15. What can you say about the determinant of any square matrix where one of the columns (or rows) is repeated in the matrix? Why?16. Suppose that A is a n × n matrix and that Ax = 0 has infinitely many solutions. What can you say about det(A)? Why?17. Suppose that A2 is not invertible. Can you determine if A is invertible or not? Explain.18. Two matrices A and B are said to be similar if there exists an invertible matrix P such that A = PBP−1 . What can you say about the determinants of similar matrices?19. Let A be an arbitrary 2 × 2 matrix of the form a b c d where a = 0 and A is assumed to be invertible. Working by hand, row reduce the augmented matrix [A I2 ] and hence determine a formula for A−1 in terms of the entries of A. What role does det(A) play in the formula for A−1 ?20. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) Swapping the rows in a square matrix A does not change the value of det(A). (b) If A is a square matrix with a pivot in every column, then det(A) = 0. (c) The determinant of any diagonal matrix is the product of its diagonal entries. (d) If A is an n × n matrix and Ax = b has a unique solution for every b ∈ Rn , then det(A) = 0.1.10 The eigenvalue problemAnother powerful characteristic of linear algebra is the way the subject oftenallows us to better understand an infinite collection of objects in terms of theproperties of a small, finite number of elements in the set. For example, if we have
The eigenvalue problem 85a set of three linearly independent vectors that spans R3 , then every vector in R3may be understood as a unique linear combination of the three special vectorsin the linearly independent spanning set. Thus, in some ways it is sufficient tounderstand these three vectors, and to use that knowledge to better understandthe rest of the vectors in R3 . In a similar way, as we will see in this section, foran n × n matrix A there are up to n important vectors (called eigenvectors) thatenable us to better understand a variety of properties of the matrix. The process of matrix multiplication enables us to associate a function withany given matrix A. For example, if A is a 2 × 2 matrix, then we may define afunction T by the formula T (x) = Ax (1.10.1)Note that the domain of the function T is R2 , the set of all vectors with twoentries. Moreover, note that every output of the function T is also a vector inR2 . We therefore use the notation T : R2 → R2 . This is analogous to familiarfunctions like f (x) = x 2 , where for every real number input we obtain a realnumber output (f : R → R); the difference here is that for the function T , forevery vector input we get a vector output. In what follows, we go in search ofspecial input vectors to the function T for which the corresponding output isparticularly simple to compute. The next example will highlight the propertiesof the vector(s) we seek.Example 1.10.1 Explore the geometric effect of the matrix 2 1 A= 1 2on the vectors u = [1 0]T and v = [1 1]T from the perspective of the functionT (x) = Ax.Solution. We first compute T (u) = Au = [2 1]T . In figure 1.15, we see a plotof the vector u on the left, and T (u) on the right. This shows that the geometriceffect of T on u is to rotate u and stretch it. For the vector v, we observethat T (v) = Av = [3 3]T . Graphically, as shown in figure 1.16, it is clear thatT (v) is simply a stretch of v by a factor of 3. Said slightly differently, we mightwrite that 3 1 T (v) = Av = =3 = 3v 3 1This shows that the result of the function T (and hence the matrix A) beingapplied to the vector v is particularly simple: v is only stretched by T .For any n × n matrix A, there is an associated function T : Rn → Rn definedby T (x) = Ax. This function takes a given vector in Rn and maps it to acorresponding vector in Rn ; in every case, we may view this output as resultingfrom the input vector being stretched and/or rotated. Input vectors that are
86 Essentials of linear algebra 3 3 T(u) u T−3 3 −3 3 −3 −3Figure 1.15 The vectors u and T (u) in example 1.10.1. T(v) 3 3 v T−3 3 −3 3 −3 −3Figure 1.16 The vectors v and T (v).only stretched have corresponding outputs that are simplest to determine: theinput vector is simply multiplied by a scalar. To put this another way, for thesestretched-only vectors, multiplying them by A is equivalent to multiplying themby a constant. Such vectors prove to be important for a host of reasons, and arecalled the eigenvectors of a matrix A.Definition 1.10.1 For a given n × n matrix A, a nonzero vector v is said to bean eigenvector of A if and only if there exists a scalar λ such that Av = λv (1.10.2)The scalar λ is called the eigenvalue corresponding to the eigenvector v.
The eigenvalue problem 87 In example 1.10.1, we found that the vector v = [1 1]T is an eigenvectorof the given matrix A with corresponding eigenvalue 3 since Av = 3v. What isnot yet clear is how we even begin to find eigenvectors and eigenvalues. We willsoon see that some of the many different perspectives we can take on systems oflinear equations will help us solve this problem. In general, given an n × n matrix A, we seek eigenvectors v that are, bydefinition, nonzero and satisfy the equation Av = λv. In one sense, what makesthis problem challenging is that neither v nor λ is initially known. We thusexplore some different perspectives on the problem to see if we can highlightthe role of either v or λ. Early in this chapter, we spent significant effortstudying homogeneous equations and the circumstances under which they havenontrivial solutions. Here, the eigenvector problem can be rephrased in a similarlight. Subtracting λv from both sides of (1.10.2), we equivalently seek λ and vsuch that Av − λv = 0 (1.10.3)Viewing λv as (λI)v, we can factor (1.10.3) and write (A − λI)v = 0 (1.10.4)Now the question becomes, "for which values of λ does (1.10.4) have a nontrivialsolution?" At this point, we recall theorem 1.6.2, which tells us that the equationBx = 0 has only the trivial solution if and only if the matrix B has a pivot inevery column. To have a nontrivial solution, we therefore want A − λI to nothave a pivot in every column. In (1.10.4), the matrix A − λI is square, so by theInvertible Matrix Theorem such a nontrivial solution exists if and only if A − λIis not invertible. This last observation brings us, finally, to determinants. As we saw inSection 1.9, a matrix is invertible if and only if its determinant is nonzero.Therefore, a nontrivial solution to (1.10.4) exists whenever λ is such thatdet(A − λI) = 0. In the next example, we explore how this equation enablesus to find the eigenvalues of a matrix A, and hence the eigenvectors as well.Example 1.10.2 Find the eigenvalues and eigenvectors of the matrix 2 1 A= 1 2Solution. As seen in our preceding discussion, by the definition of eigenvaluesand eigenvectors, λ is an eigenvalue of A if and only if the equation (A −λI)v = 0has a nontrivial solution. Note first that A − λI is the matrix A with the scalar λsubtracted from each diagonal entry since 2 1 λ 0 2−λ 1 A − λI = − = 1 2 0 λ 1 2−λ
88 Essentials of linear algebraWe next compute det(A − λI) so that we can see which values of λ make thisdeterminant zero. In particular, we have 2−λ 1 det(A − λI) = det 1 2−λ = (2 − λ)2 − 1 = λ2 − 4λ + 3 (1.10.5)Thus, in order for det(A − λI) = 0, λ must satisfy the equation λ2 − 4λ +3 = 0. Factoring, (λ − 3)(λ − 1) = 0, and therefore λ = 3 and λ = 1 areeigenvalues of A. The value λ = 3 is not surprising, given our earlier discoveriesin example 1.10.1.Next, we proceed to find the eigenvectors that correspond to each eigenvalue.Beginning with λ = 3, we seek nonzero vectors v that satisfy Av = 3v, orequivalently (A − 3I)v = 0This problem is a familiar one: solving a homogeneous system of linear equationsfor which infinitely many solutions exist. Augmenting A − 3I with a column ofzeros and row-reducing, we find that −1 1 0 1 −1 0 → 1 −1 0 0 0 0Note that from the very definition of an eigenvector, by which we seek anontrivial solution to (A − λI)v = 0, it must be the case at this point that thematrix A − λI does not have a pivot in every row. Interpreting the row-reducedmatrix with the free variable v2 , we find that the vector v = [v1 v2 ]T must satisfyv1 − v2 = 0. Thus, any vector v of the form v2 1 v= = v2 v2 1is an eigenvector of A that corresponds to the eigenvalue λ = 3. In particular,we observe that any scalar multiple of the vector v = [1 1]T is an eigenvector ofA with associated eigenvalue 3. We say that the set of all eigenvectors associatedwith eigenvalue 3 is the eigenspace corresponding to λ = 3. It now only remains to find the eigenvectors associated with λ = 1. Weproceed in the same manner as above, now solving the homogeneous equation(A − 1I)v = 0. Row-reducing, we find that 1 1 0 1 1 0 → 1 1 0 0 0 0
The eigenvalue problem 89and therefore the eigenvector v must satisfy v1 + v2 = 0 and have the form −v2 −1 v= = v2 v2 1Here, any scalar multiple of v = [−1 1]T is an eigenvector of A correspondingto λ = 1.There are several important general observations to be made from exam-ple 1.10.2. One is that for any 2 × 2 matrix, the matrix will have 0, 1, or2 real eigenvalues. This comes from the fact that det(A − λI) is a quadraticfunction in the variable λ, and therefore can have up to two real zeros. Whileit is possible to consider complex eigenvalues, we will wait until these arisein our study of systems of differential equations to address them in detail. Inaddition, we note that there are infinitely many eigenvectors associated with eacheigenvalue. Often we will be interested in finding representative eigenvectors—ones for which all others with the same eigenvalue are linear combinations.Finally, it is worthwhile to note that the two representative eigenvectors foundin example 1.10.2, corresponding respectively to the two distinct eigenvalues, arelinearly independent. More on why this is important will be discussed at the endof this section; for now, we remark that it is possible to show that eigenvectorscorresponding to distinct eigenvalues are always linearly independent. This factwill be proved in exercise 16. The observations in the preceding paragraph generalize to the case ofn × n matrices. It may be shown that det(A − λI) is a polynomial of degree nin λ. This function is usually called the characteristic polynomial; the equationdet(A − λI) = 0 is typically referred to as the characteristic equation. Becausethe characteristic polynomial has degree n, it follows that A has up to n realeigenvalues5 . Next we consider two additional examples that demonstrate some more ofthe possibilities and important ideas that arise in trying to find the eigenvaluesand eigenvectors of a given matrix.Example 1.10.3 Determine the eigenvalues and eigenvectors of the matrix √ √ 1/√2 −1/√2 R= 1/ 2 1/ 2In addition, explore the geometric effect of the function T (v) = Rv on vectorsin R2 .5 See appendix C for a review and discussion of important properties of roots of polynomialequations.
92 Essentials of linear algebraThis leads us to see that the corresponding eigenvector has form ⎡ ⎤ 5 v = v3 ⎣ −2 ⎦ 1Therefore, we see that for this matrix A, the matrix has two distinct eigenvalues(−4 and 3), and each of these eigenvalues has only one associated linearlyindependent eigenvector. That is, every eigenvector of A associated with λ = −4is a scalar multiple of [−2 8 1]T while every eigenvector associated with λ = 3 3is a scalar multiple of [5 − 2 1]T .In the three preceding examples, we have seen that an n × n matrix has up ton real eigenvalues. It turns out that there are also up to n linearly independenteigenvectors of the matrix. For many reasons, the best possible scenario iswhen a matrix has n linearly independent eigenvectors, such as the matrix Ain example 1.10.2. In that 2 × 2 situation, A had two distinct real eigenvalues,and two corresponding linearly independent eigenvectors. One reason that thisis so useful is that the eigenvectors are not only linearly independent, but alsospan R2 . If we call the two eigenvectors found in example 1.10.2 u and v,corresponding to λ = 3 and μ = 1, respectively, then, since these two vectorsare linearly independent in R2 and span R2 , we can write every vector in R2uniquely as a linear combination of u and v. In particular, given a vector x, there exist coefficients α and β such that x = αu + β vIf we are interested in computing Ax, we can do so now solely by knowinghow A acts on the eigenvectors. Specifically, if we apply the linearity of matrixmultiplication and the definition of eigenvectors, we have Ax = A(α u + β v) = α Au + β Av = αλu + βμvThis then reduces matrix multiplication essentially to scalar multiplication. In conclusion, we have seen in this section that via matrix multiplication,every matrix can be viewed as a function in the way that, through multiplication,it stretches and rotates vectors. Those vectors that are only stretched arecalled eigenvectors, and the factor by which the matrix stretches them arecalled eigenvalues. By knowing the eigenvalues and eigenvectors, we can betterunderstand how A acts on an arbitrary vector, and, with some more sophisticatedapproaches, even further understand key properties of the matrix. Some of theseproperties will be studied in detail later in this text when we consider systems ofdifferential equations.
The eigenvalue problem 931.10.1 Markov chains, eigenvectors, and GoogleIn a Markov process such as the one discussed in subsection 1.3.1 that representsthe transition of voters from one classification to another, it is natural to wonderwhether or not there is a distribution of voters for which the total number ineach category will remain constant from one year to the next. For example, forthe Markov process represented by x (n+1) = Mx (n) (1.10.6)where M is the matrix ⎡ ⎤ 0.95 0.03 0.07 M = ⎣0.02 0.90 0.13⎦ 0.03 0.07 0.80we can ask: is there a voter distribution x such that Mx = x? In light of ourmost recent work with eigenvalues and eigenvectors, we see that this questionis equivalent to asking if the matrix M has λ = 1 as an eigenvalue with somecorresponding eigenvector that can represent a voter distribution. If we compute the eigenvalues and eigenvectors of M, we find that theeigenvalues are λ = 1.000, 0.911, 0.739. The eigenvector corresponding to λ = 1is v = [0.770 0.558 0.311]T . Scaling v so that the sum of its entries is 250, wesee that the eigenvector v = [117.450 85.113 47.437]Trepresents the distribution of a population of 250 000 people in such a way thatthe total number of Democrats, Republicans, and Independents does not changefrom one year to the next, under the hypothesis that voters change categoriesannually according to the likelihoods expressed in the Markov matrix M. Thiseigenvector is sometimes also called a stationary vector. Remarkably, we can also note that in our earlier computations insubsection 1.3.1 for this Markov chain, we observed that the sequence of vectorsx (1) , x (2) , . . . , x (20) , . . . was approaching a single vector. In fact, the limiting valueof this sequence is the eigenvector v = [117.450 85.113 47.437]T . That thisphenomenon occurs is the result of the so-called Power method, a rudimentarynumerical technique for computing an eigenvalue–eigenvector pair of a matrix.More about this concept can be studied in the project on discrete dynamicalsystems found in section 1.13.3.Example 1.10.5 Find the stationary vector from the matrix in example 1.3.3.Solution. Under the assumptions stated in example 1.3.3, we saw that themigration of citizens from urban to suburban areas of a metropolitan area, orvice versa, were modeled by the Markov process x (n+1) = Mx (n) where M is thematrix 0.85 0.08 M= 0.15 0.92
94 Essentials of linear algebraSolving the equation x = Mx by writing (M − I)x = 0, we see that we need tofind the eigenvector of x that corresponds to λ = 1. Doing so, we find that theeigenvector is 0.4706 v= 0.8824Scaling this vector so that the sum of its entries is one, we see that the populationstabilizes when it is distributed with 34.78 percent in the city and 65.22 percentin the suburbs, in accordance with the vector [0.3478 0.6522]T .One of the most stunning applications of eigenvalues and eigenvectors can befound on the World Wide Web. In particular, the idea of finding a stationaryvector that satisfies Mx = x is at the center of Google's Page Rank Algorithmthat it uses to index the importance of billions of pages on the Internet. What isparticularly challenging about this problem is the fact that the stochastic matrixM used by the algorithm is a square matrix that has one column for every pageon the World Wide Web that is indexed by Google! In early 2007, this meantthat M was a matrix with 25 billion columns. Nonetheless, properties of thematrix M and sophisticated numerical algorithms make it possible for moderncomputers to quickly find the stationary vector of M and hence provide the userwith the results we have all grown accustomed to in using Google.61.10.2 Using Maple to find eigenvalues and eigenvectorsDue to its reliance upon determinants and the solution of polynomialequations, the eigenvalue problem is computationally difficult for any caselarger than 3 × 3. Sophisticated algorithms have been developed to computeeigenvalues and eigenvectors efficiently and accurately. One of these is the so-called QR algorithm, which through an iterative technique produces excellentapproximations to eigenvalues and eigenvectors simultaneously. While Maple implements these algorithms and can find both eigenvaluesand eigenvectors, it is essential that we not only understand what the programis attempting to compute, but also how to interpret the resulting output.As always, in what follows we are working within the LinearAlgebrapackage. Given an n × n matrix A, we can compute the eigenvalues of A with thecommand > Eigenvalues(A);6 A detailed description of how the Page Rank Algorithm works and the role that eigenvectors playmay be read at
The eigenvalue problem 95Doing so for the matrix 2 1 A= 1 2from example 1.10.2 yields the Maple output 3 1Despite the vector format, the program is telling us that the two eigenvaluesof the matrix A are 3 and 1. If we desire the eigenvectors, too, we can use thecommand > Eigenvectors(A);which leads to the output 3 1 −1 , 1 1 1Here, the first vector tells us the eigenvalues of A. The following matrix holds thecorresponding eigenvectors in its columns; the vector [1 1]T is the eigenvectorcorresponding to λ = 3 and [−1 1]T corresponds to λ = 1. Maple is extremely powerful. It is not at all bothered by complex numbers.So, if we enter a matrix like the one in example 1.10.3 that has no real eigenvalues,Maple will find complex eigenvalues and eigenvectors. To see how this appears,we enter the matrix √ √ 1/√2 −1/√2 R= 1/ 2 1/ 2and execute the command > Eigenvectors(R);The resulting output is √ √ 2 2+ 1 1 2I 2 I −I √ √ , 1 1 2 2− 1 1 2I 2Note that here Maple is using 'I ' to denote not the identity matrix, but rather√ −1. Just as we saw in example 1.10.3, R does not have any real eigenvalues. Wecan use familiar properties of complex numbers (most importantly, I 2 = 1) toactually check that the equation Ax = λx holds for the listed complex eigenvaluesand complex eigenvectors above. However, at this point in our study, thesecomplex eigenvectors are of less importance, so we defer further details on themuntil later work with systems of differential equations. One final example is relevant here to see how Maple deals with repeatedeigenvalues and missing eigenvectors. If we enter the 3 × 3 matrix A from
The eigenvalue problem 97 9. A 2 × 2 matrix A has eigenvalues 5 and −1 and corresponding eigenvectors u = [0 1]T and v = [1 0]T . Use this information to compute Ax, where x is the vector x = [−5 4]T .10. A 2 × 2 matrix A has eigenvalues −3 and −2 and corresponding eigenvectors u = [−1 1]T and v = [1 1]T . Use this information to compute Ax, where x is the vector x = [−3 5]T .11. Consider the matrix ⎡ ⎤ −2 1 1 A = ⎣ 1 −2 1⎦ 1 1 −2 (a) Determine the eigenvalues and eigenvectors of A. (b) Does R3 have a linearly independent spanning set that consists of eigenvectors of A?12. Consider the matrix 3 −1 A= −1 3 (a) Determine the eigenvalues and eigenvectors of A, and show that A has two linearly independent eigenvectors. (b) Let P be the matrix whose columns are two In particular, explain how A10 can be easily computed by using the diagonal matrix D along with P and P−1 .13. Consider the matrix ⎡ ⎤ 3 −1 1 A = ⎣−1 3 −1⎦ 1 −1 3 (a) Determine the eigenvalues and eigenvectors of A, and show that A has three linearly independent eigenvectors. (b) Let P be the matrix whose columns are three
98 Essentials of linear algebra14. Prove that an n × n matrix A is invertible if and only if A has no eigenvalue equal to zero.15. Show that if A, B, and P are square matrices (with P invertible) such that B = PAP−1 , then A and B have the same eigenvalues. (Hint: consider the characteristic equation for PAP−1 .)16. Prove that if A is a 2 × 2 matrix and v and u are eigenvectors of A corresponding to distinct eigenvalues λ and μ, then v and u are linearly independent. (Hint: suppose to the contrary that v and u are linearly dependent.)17. For a differentiable function y, denote the derivative of y with respect to x by D(y). Now consider the function y = e 7x , and compute D(y). For what value of λ is D(y) = λy? Explain how this value behaves like an eigenvalue of the operator D. What is the corresponding eigenvector? How does the problem change if we consider y = e rx for any other real value of r?18. For a vector-valued function x(t ), let the derivative of x with respect to t be denoted by D(x). For the function e −2t x(t ) = −3e −2t compute D(x). For what value(s) of λ is D(x) = λx? Explain how it appears from your work that the operator D has an eigenvalue-eigenvector pair.19 90 3 2 Suburban 7 96 10 Rural 3 1 88 Given that a population of 250 million is present, is there a stationary vector that reveals a population which does not change from year to year?20. Car-owners can be grouped into classes based on the vehicles they own. A study of owners of sedans, minivans, and sport utility vehicles shows
Generalized vectors 99 SUV 2 2 90 If there are currently 100 000 vehicles in the population under study, is there a stationary vector that represents a distribution in which the number of owners of each type of vehicle will not change as they replace their vehicles?21. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) If x is any vector and λ is a constant such that Ax = λx, then x is an eigenvector of A. (b) If Ax = 0 has nontrivial solutions, then λ = 0 is an eigenvalue of A. (c) Every 3 × 3 matrix has three real eigenvalues. (d) If A is a 2 × 2 matrix, then A can have up to two real linearly independent eigenvectors.1.11 Generalized vectorsThroughout our work with vectors in Rn , we have regularly used several keyalgebraic properties they possess. For example, any two vectors u and v can beadded to form a new vector u + v, any single vector can be multiplied by a scalarto determine a new vector cu, and there is a zero vector 0 with the propertythat for any vector v, v + 0 = v. Of course, we use other algebraic properties ofvectors as well, often implicitly. Other sets of mathematical objects behave in ways that are algebraicallysimilar to vectors. The purpose of this section is to expand our perspectiveon what familiar mathematical entities might also reasonably be called vectors;much of this expanded perspective is in anticipation of our pending work withdifferential equations and their solutions. We motivate our study with severalfamiliar examples, and then summarize a collection of formal properties that allthese examples share.Example 1.11.1 Let M2×2 denote the collection of all 2 × 2 matrices with realentries. Show that if A and B are any 2 × 2 matrices and c ∈ R, then A + B andcA are also 2 × 2 matrices. In addition, show that there exists a "zero matrix" Zsuch that A + Z = A for every matrix A.
100 Essentials of linear algebraSolution. Let a11 a12 b b A= and B = 11 12 a21 a22 b21 b22By the definition of matrix addition, a11 + b11 a12 + b12 A+B = a21 + b21 a22 + b22and thus we see that A + B is also a 2 × 2 matrix. Recall that it only makes sensefor matrices of the same size to be added; here we are simply pointing out theobvious fact that the sum of two matrices of the same size is yet another matrixof the same size. In the same way, ca11 ca12 cA = ca21 ca22which shows that not only is the scalar multiple defined, but also that cA is a2 × 2 matrix. Finally, if we let Z be the 2 × 2 matrix all of whose entries are zero, 0 0 Z= 0 0then our work with matrix sums shows us immediately that A + Z = A for everypossible 2 × 2 matrix A.Certainly, we can see that there is nothing particularly special about the 2 × 2case in this example; the same properties will hold for Mm×n for any positiveinteger values of m and n. Mathematicians often use the language "M2×2 is closed under additionand scalar multiplication" and "M2×2 contains a zero element " to describethe observations we made in example 1.11.1. Specifically, to say that a set isclosed under an operation means simply that if we perform the operation onan appropriate number of elements from the set, the result is another elementin the set. We next consider several more examples of sets that demonstrate theproperties of being closed and having a zero element.Example 1.11.2 Let P2 denote the set of all polynomials of degree 2 or less.That is, P2 is the set of all functions of the form p(x) = a2 x 2 + a1 x + a0where a0 , a1 , a2 ∈ R. Show that P2 is closed under addition and scalarmultiplication, and that P2 contains a zero element.Solution. Before we formally address the stated tasks, let us remind ourselveshow we add polynomial functions. If we are given, say, f (x) = 2x 2 − 5x + 11 andg (x) = 4x − 3, we compute (f + g )(x) = f (x) + g (x) = 2x 2 − 5x + 11 + 4x − 3.We can then add like terms to simplify and find that (f + g )(x) = 2x 2 − x + 8.
Generalized vectors 101Similarly, if we wanted to compute (−3f )(x), we have (−3f )(x) = −3f (x) =−3(2x 2 − 5x + 11) = −6x 2 + 15x − 33. We now show that P2 is indeed closed under the operations of additionand scalar multiplication. Given two arbitrary elements of P2 , say f (x) = a2 x 2 +a1 x + a0 and g (x) = b2 x 2 + b1 x + b0 , it follows upon adding and combininglike terms that (f + g )(x) = (a2 + b2 )x 2 + (a1 + b1 )x + (a0 + b0 )which is obviously a polynomial of degree 2 or lower, and thus f + g is anelement of P2 . In the same way, for any real value c, (cf )(x) = ca2 x 2 + ca1 x + ca0which also belongs to P2 . Finally, it is evident that if we let z(x) = 0x 2 + 0x + 0(i.e., z(x) is the zero function), then (f + z)(x) = f (x) for any choice of f in P2 .Here, too, we should observe that while these properties hold for P2 , there isnothing special about the 2. In fact, Pn (the set of all polynomials of degree n orless) has the exact same properties. Even P, the set of all polynomials, behavesin the same manner.Example 1.11.3 From calculus, consider the set C [−1, 1] of all continuousfunctions on the interval [−1, 1]. That is, C [−1, 1] = {f | f is continuous on [−1, 1]}.Show that C [−1, 1] is closed under addition and scalar multiplication, and alsothat C [−1, 1] contains a zero element.Solution. Two standard facts from calculus tell us that the sum of any twocontinuous functions is also a continuous function and that a constant multipleof a continuous function is also a continuous function. Thus C [−1, 1] isclosed under addition and scalar multiplication. Furthermore, the zero functionz(x) = 0 is itself continuous, which shows that C [−1, 1] indeed has a zero element.One of the principal reasons that we are shifting our attention from vectors in Rnto this more generalized concept of vector where the objects under considerationare often functions is the fact that our focus in subsequent chapters will be solvingdifferential equations. The solution to a differential equation is a function thatmakes the equation true. Moreover, we will also see that for certain importantclasses of differential equations, there are multiple solutions to the equation andthat often these solution sets are closed under addition and scalar multiplicationand also contain the zero function. From each of the above examples, we see that Rn has many importantproperties that we can consider in a broader context. We therefore introducethe notion of a vector space, which is a set of objects that have defined operationsof addition and scalar multiplication that satisfy the list of ten rules below. Theconcept of a vector space is a generalization of Rn .
102 Essentials of linear algebra While many of the rules are technical in nature, the most important onesto verify turn out to be the three that we have focused on so far: being closedunder addition, closed under scalar multiplication, and having a zero element.All three sets described in the above examples are vector spaces, as is Rn .Definition 1.11.1 A vector space is a nonempty set V of objects, on whichoperations of addition and scalar multiplication are defined, where the objectsin V (called vectors) adhere to the following ten rules: 1. For every u and v in V , the sum u + v is in V (V is "closed under vector addition") 2. For every u and v in V , u + v = v + u ("vector addition is commutative") 3. For every u, v , w in V , (u + v) + w = v + (u + w) ("vector addition is associative") 4. There exists a zero vector 0 in V such that u + 0 = u for every u ∈ V (0 is called the additive identity of V ) 5. For every u ∈ V , there is a vector −u such that u + (−u) = 0 (−u is called the additive inverse of u) 6. For every u ∈ V and every scalar c, the scalar multiple cu ∈ V (V is "closed under scalar multiplication") 7. For every u and v in V and every scalar c, c(u + v) = cu + cv ("scalar multiplication is distributive over vector addition") 8. For every u ∈ V and scalars c and d, (c + d)u = cu + du 9. For every u ∈ V and scalars c and d, c(du) = (cd)u10. For every u ∈ V , 1u = u Sometimes we can take a sub-collection (i.e., a subset) of the vectors in avector space, and that smaller set itself acts like a vector space. For example, theset of all polynomial functions is a vector space. If we take just the polynomialsof degree 2 or less (as in example 1.11.2 above), that subset is itself a vectorspace. This leads us to introduce the notion of a subspace.Definition 1.11.2 Given a vector space V , let H be a subset of V (i.e.,every object in H is also in V .) There are then operations of addition andscalar multiplication on objects in H : specifically, the same addition and scalarmultiplication as on the objects in V . We say H is a subspace of V if and only ifall three of the following conditions hold: 1. H is closed under addition 2. H is closed under scalar multiplication 3. H contains the zero element of V
Generalized vectors 103 We close this section with two important examples of subspaces. The firstis a subspace of Rn associated with a given matrix A. The second is a subspaceof the set of all continuous functions on [−1, 1].Example 1.11.4 Recall the matrix A from example 1.10.4 in section 1.10, ⎡ ⎤ 5 6 2 A = ⎣0 −1 −8⎦ 1 0 −2Show that the set of all eigenvectors that correspond to a given eigenvalue of Aforms a subspace of R3 .Solution. In example 1.10.4, we saw that the eigenvalues of A are λ = −4 (withmultiplicity 1) and λ = 3 (with multiplicity 2). In addition, the correspondingeigenvectors are v = [−2 8 1]T for λ = −4 and v = [5 − 2 1]T for λ = 3. In 3particular, recall that every scalar multiple of vλ=−4 is also an eigenvector of Acorresponding to λ = −4. We now show that the set of all these eigenvectorscorresponding to λ = −4 is a subspace of R3 . Let Eλ=−4 denote the set of all vectors v such that Av = −4v. First, certainlyit is the case that A0 = −40. This shows that the zero element of R3 is anelement of Eλ=−4 . Furthermore, we have already seen that every scalar multipleof an eigenvector is itself an eigenvector, and thus Eλ=−4 is closed under scalarmultiplication. Finally, suppose we have two vectors x and y such that Ax = −4xand Ay = −4y. Observe that by properties of linearity, A(x + y) = Ax + Ay = −4x − 4y = −4(x + y)which shows that (x + y) is also an eigenvector of A corresponding to λ = −4.Therefore, Eλ=−4 is closed under addition. This shows that Eλ=−4 is indeed a subspace of R3 . In a similar fashion, Eλ=3is also a subspace of R3 .Our observations for the eigenspaces of the 2 × 2 matrix A in example 1.11.4hold in general for any n × n matrix A: the set of all eigenvectors correspondingto a given eigenvalue of A forms a subspace of Rn .Example 1.11.5 Show that the set of all linear combinations of the sine andcosine functions is a subspace of the vector space C of all continuous functions.Solution. We let C denote the vector space of all continuous functions, andnow let H be the subset of C which is defined to be all functions that are linearcombinations of sin t and cos t . That is, a typical element of H is a function f ofthe form f (t ) = c1 sin t + c2 cos t
104 Essentials of linear algebrawhere c1 and c2 are any real scalars. We need to show that the set H contains thezero function from C , that H is closed under scalar multiplication, and that His closed under addition. First, if we choose c1 = c2 = 0, the function z(t ) = 0 sin t + 0 cos t = 0 is thefunction that is identically zero, which is the (continuous) zero function from C .Next, if we take a function from H , say f (t ) = c1 sin t + c2 cos t , and multiply itby a scalar k, we get kf (t ) = k(c1 sin t + c2 cos t ) = (kc1 ) sin t + (kc2 ) cos twhich is of course another element in H , so H is closed under scalarmultiplication. Finally, if we consider two elements f and g in H , given byf (t ) = c1 sin t + c2 cos t and g (t ) = d1 sin t + d2 cos t , then it follows that f (t ) + g (t ) = (c1 sin t + c2 cos t ) + (d1 sin t + d2 cos t ) = (c1 + d1 ) sin t + (c2 + d2 ) cos tso that H is closed under addition, too. Thus, H is a subspace of C .In fact, it turns out that the subspace considered in example 1.11.5 contains allof the solutions to a familiar differential equation. We will revisit this issue inexample 1.11.7. It is also instructive to consider an example of a set that is not asubspace.Example 1.11.6 Consider the vector space C [−1, 1] of all continuous functionson the interval [−1, 1]. Let H be the set of all functions with the property thatf (−1) = f (1) = 2. Determine whether or not H is a subspace of C [−1, 1].Solution. The set H does not satisfy any of the three required propertiesof subspaces, so any one of these suffices to show that H is not a subspace. Inparticular, the zero function z(t ) = 0 does not have the property that z(−1) = 2,and thus the zero function from C [−1, 1] does not lie in H , so H is not a subspace. We could also observe that any scalar multiple of a function whose value att = −1 and t = 1 is 2 will result in a new function whose value at these points isnot 2; similarly, the sum of two functions whose values at t = −1 and t = 1 are 2will lead to a new function whose values at these points is 4. These facts togethershow that H is not closed under scalar multiplication, nor under addition.As we have already mentioned, we are considering this generalization of theterm vector to include mathematical objects like functions because this structureunderlies the study of differential equations, and this vector space perspectivewill help us to better understand a variety of key ideas when we are solvingimportant problems later on. To foreshadow these coming ideas, we presentan example of an elementary differential equation that shows how the setof solutions to the equation is in fact the subspace of continuous functionsconsidered in example 1.11.5.
Generalized vectors 105Example 1.11.7 Consider the differential equation y +y = 0Show that y1 = sin t and y2 = cos t are solutions to this differential equation,and that every function of the form y = c1 y1 + c2 y2 is a solution as well.Solution. This example is very similar to example 1.6.4. Because of itsimportance, we discuss the current problem in full detail here as well. For any equation, a solution is an object that makes the equation true. Inthe above differential equation, y represents a function. The equation asks "forwhich functions y is the sum of y and its second derivative equal to zero?" Observe first that if we let y1 = sin t , then y1 = cos t , so y1 = − sin t , andtherefore y1 + y1 = − sin t + sin t = 0. In other words, y1 is a solution to thedifferential equation. Similarly, for y2 = cos t , y2 = − sin t and y2 = − cos t , sothat y2 + y2 = − cos t + cos t = 0. Thus, y2 is also a solution to the differentialequation. Now, consider any function y of the form y = c1 y1 + c2 y2 . That is, let y be anylinear combination of the two solutions we have already found. We then have y = c1 sin t + c2 cos tso that, using standard properties of the derivative (properties which are linearin nature), it follows that y = c1 cos t − c2 sin tand y = −c1 sin t − c2 cos tWe, therefore see that y + y = (−c1 sin t − c2 cos t ) + (c1 sin t + c2 cos t ) = −c1 sin t + c1 sin t − c2 cos t + c2 cos t =0so that y is indeed also a solution of y + y = 0.In example 1.11.7, we find a large number of connections to our work insystems of linear equations and linear algebra: properties of linearity, linearcombinations of vectors, homogeneous equations, infinitely many solutions,and more. In particular, the set of all solutions to the differential equation inexample 1.11.7 is precisely the subspace of continuous functions examined inexample 1.11.5. Certainly, we will revisit these topics in greater detail as weprogress in our study of differential equations.Exercises 1.11 In exercises 1–16, determine whether or not the set H is asubspace of the given vector space V . If H is a subspace, show that it satisfies the
Generalized vectors 10719. Explain why for any set of vectors {u, v } in Rn , Span{u, v } is a subspace of Rn . Similarly, explain why Span {v1 , . . . , vk } is a subspace of Rn for any set {v1 , . . . , vk }. ⎧⎡ ⎤ ⎫ ⎨ 2a + b ⎬20. Let V = R3 and H = ⎣ a − b ⎦ : a , b ∈ R . Determine vectors u and ⎩ ⎭ 3a + 5b v so that H can be expressed as the set Span{u, v }, and hence explain why H is a subspace of R3 . ⎧⎡ ⎤ ⎫ ⎨ 2a + b ⎬21. Let V = R3 and H = ⎣ −2 ⎦ : a , b ∈ R . Explain why H is not a ⎩ ⎭ 3a + 5b subspace of R3 .22. Let A be an m × n matrix. The null space of the matrix A, denoted Nul(A) is the set of all solutions to the equation Ax = 0. Explain why Nul(A) is a subspace of Rn .23. Let A be an m × n matrix. The column space of the matrix A, denoted Col(A) is the set of all linear combinations of the columns of A. Explain why Col(A) is a subspace of Rm .In exercises 24–27, use the definitions of the null space Nul(A) and columnspace Col(A) of a matrix given in exercises 22 and 23. 2 1 −124. Let A = . Is the vector v = [−2 1 1]T in Nul(A)? Justify your 1 3 4 answer clearly. In addition, describe all vectors that belong to Nul(A) as the span of a finite set of vectors. ⎡ ⎤ 1 −225. Let A = ⎣ 3 1⎦. Is the vector v = [−2 1 1]T in Col(A)? Justify your −4 0 answer. Is the vector u = [−1 4 − 4]T in Col(A)? In addition, describe all vectors that belong to Col(A) as the span of a finite set of vectors.26. Given a matrix A and a vector v, is it easier to determine whether v lies in Nul(A) or Col(A)? Why?27. Given a matrix A and a vector v, is it easier to describe Nul(A) or Col(A) as the span of a finite set of vectors? Why?28. Consider the differential equation y = 3y. Explain why any function of the form y = Ce 3t is a solution to this equation. Is the set of all these solutions a subspace of the vector space of continuous functions?29. Consider the differential equation y = 3y − 3. Explain why any function of the form y = Ce 3t + 1 is a solution to this equation. Is the set of all these solutions a subspace of the vector space of continuous functions?
108 Essentials of linear algebra30. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) If H is a subspace of a vector space V , then H is itself a vector space. (b) If H is a subset of a vector space V , then H is a subspace of V . (c) The set of all linear combinations of any two vectors in R3 is a subspace of R3 . (d) Every nontrivial subspace of a vector space has infinitely many elements.1.12 Bases and dimension in vector spacesIn section 1.11, we saw that some common sets we encounter in mathematics arevery similar to Rn . For instance, the set M2×2 of all 2 × 2 matrices, the set P2 of allpolynomials of degree 2 or less, and the set C [−1, 1] of all continuous functionson [−1, 1] are sets that contain a zero element, are closed under addition, andare closed under scalar multiplication. In addition, because they each satisfy theother required seven characteristics we noted, these sets are all vector spaces. Wespecifically observe that this enables us to take linear combinations of elementsof a vector space, because addition and scalar multiplication are defined andclosed in these collections of objects. Every vector space has further characteristics that are similar to Rn .For example, it is natural to discuss now-familiar concepts such as linearindependence and span in the context of the more generalized notion of vector.As we will see, the definitions of these terms in the setting of vector spaces arealmost identical to those we encountered earlier in Rn . Moreover, just as we canfrequently describe sets in Rn in terms of a small number of special vectors, wewill find that this often occurs in general vector spaces. We begin by updating two key definitions.Definition 1.12.1 In a vector space V , given a set S = {v1 , . . . , vk } where eachvector vi ∈ V , the set S is linearly dependent if there exists a nontrivial solutionto the vector equation x1 v1 + x2 v2 + · · · + xk vk = 0 (1.12.1)If (1.12.1) has only the trivial solution (x1 = · · · = xk = 0), then we say the set Sis linearly independent. The only difference between this definition and definition 1.6.1 that weencountered in section 1.6 is that Rn has been replaced by V . Just as withvectors in Rn , it is an equivalent formulation to say that a set S in a vector spaceV is linearly independent if and only if no vector in the set may be written as alinear combination of the other vectors in the set. We can also define the span of a set of vectors in a vector space V .
Bases and dimension in vector spaces 109Definition 1.12.2 In a vector space V , given a set of vectors S = {v1 , . . . , vk },vi ∈ V , the span of S, denoted Span(S) or Span{v1 , . . . , vk }, is the set of all linearcombinations of the vectors v1 , . . . , vk . Equivalently, Span(S) is the set of allvectors y of the form y = c1 v1 + · · · + ck vk ,where c1 , . . . , ck are scalars. We also say that Span(S) is the subset of V spannedby the vectors v1 , . . . , vk .In example 1.6.3 in section 1.6, we studied three sets R, S, and T in R3 . Rcontained two vectors and was linearly independent but did not span R3 ; Scontained three vectors, was linearly independent, and spanned R3 ; and Tconsisted of four vectors, was linearly dependent, and spanned R3 . In thatsetting, we came to see that the set S was in some ways the best of the three:it had both key properties of being linearly independent and a spanning set. Inother words, the set had enough vectors to span R3 , but not so many vectors asto generate redundancy by being linearly dependent. Through the next definition, we will now call such a set a basis, even in thegeneralized setting of vector spaces and subspaces.Definition 1.12.3 Let V be a vector space and H a subspace of V . A set B ={v1 , v2 , . . . , vk } of vectors in H is called a basis of H if and only if B is linearlyindependent and Span(B) = H . That is, B is a basis of H if and only if it is alinearly independent spanning set. Several examples now follow that use the terminology of linear indepen-dence, span, and basis in the context of different vector spaces.Example 1.12.1 In the vector space P of all polynomials, consider the subspaceH = P2 of all polynomials of degree 2 or less. Show that the set B = {1, t , t 2 } isa basis for H . Is the set {1, t , t 2 , 4 − 3t } also a basis for H ?Solution. To begin, we observe that every element of H = P2 is a polynomialfunction of the form p(t ) = a0 + a1 t + a2 t 2 . In particular, every element ofP2 is a linear combination of the functions 1, t , and t 2 , and therefore the setB = {1, t , t 2 } spans H . In addition, to determine whether the set B is linearly independent, weconsider the equation c 0 + c 1 t + c2 t 2 = 0 (1.12.2)and ask whether or not this equation has a nontrivial solution. Keeping inmind that the '0' on the right-hand side represents the zero function in P2 , thefunction that is everywhere equal to zero, we can see that if at least one of c0 ,c1 , or c2 is nonzero, we will be guaranteed to have either a nonzero constantfunction, a linear function, or a quadratic function, thus making c0 + c1 t + c2 t 2
110 Essentials of linear algebranot identically zero. This shows that (1.12.2) has only the trivial solution, andtherefore the set B = {1, t , t 2 } is linearly independent. Having shown that B isa linearly independent spanning set for H = P2 , we can conclude that B is abasis for H . On the other hand, the set {1, t , t 2 , 4 − 3t } is not a basis for H since we canobserve that the element 4 − 3t is a linear combination of the elements 1 and t :4 − 3t = 4 · 1 − 3 · t . This shows that the set {1, t , t 2 , 4 − 3t } is linearly dependentand thus cannot be a basis.Example 1.12.2 Consider the set H of all functions of the form y = c1 sin t +c2 cos t . In the vector space C of all continuous functions, explain why the setB = {sin t , cos t } is a basis for the subspace H .Solution. First, we recall that H is indeed a subspace of C [−1, 1] due to ourwork in example 1.11.5. By the definition of H (the set of all functions of the form y = c1 sin t +c2 cos t ), we see immediately that B is a spanning set for H . In addition, it isclear that the functions sin t and cos t are not scalar multiples of one another:any scalar multiple of sin t is simply a vertical stretch of the function, whichcannot result in cos t . This tells us that the set B = {sin t , cos t } is also linearlyindependent, and therefore is a basis for H .Example 1.12.3 In R3 , consider the set B = {e1 , e2 , e3 }, where e1 = [1 0 0]T ,e2 = [0 1 0]T , and e3 = [0 0 1]T . Explain why B is a basis for R3 . Is the set S = {v1 , v2 , v3 }, where v1 = [1 2 − 1]T , v2 = [−1 1 3]T , andv3 = [0 3 1]T also a basis for R3 ?Solution. First, we observe that while the formal definition of a basis refersto the basis of a subspace H of a vector space V , since every vector space is asubspace of itself, it follows that we can also discuss a basis for a vector space. Considering the set B = {e1 , e2 , e3 }, we observe that the vectors in this setare the columns of the 3 × 3 identity matrix. By the Invertible Matrix Theorem,it follows that the set B is linearly independent because I3 has a pivot in everycolumn. Likewise, the set B spans R3 since I3 has a pivot in every row. As alinearly independent spanning set in R3 , B is indeed a basis. For the set S whose elements are the columns of the matrix ⎡ ⎤ 1 −1 0 A=⎣ 2 1 3⎦ −1 3 1we again use the Invertible Matrix Theorem to determine whether or not S is abasis for R3 . Row-reducing A, it is straightforward to see that A is row equivalentto the identity matrix, and therefore is invertible. In particular, A has a pivot in
Bases and dimension in vector spaces 111every column and every row, and thus the columns of A are linearly independentand span R3 . It follows that S is also a basis for R3 .The basis B = {e1 , e2 , e3 } consisting of the columns of the 3 × 3 identity matrixis often referred to as the "standard basis of R3 ." In addition, by our work inexample 1.12.3, we can see the role that the Invertible Matrix Theorem plays indetermining whether a set of vectors in Rn is a basis or not. Specifically, sincewe know that it is logically equivalent for the columns of a square matrix A to belinearly independent and to be a spanning set for Rn , it follows that a matrix Ais invertible if and only if its columns form a basis for Rn . We therefore updatethe Invertible Matrix Theorem with an additional statement as follows.Theorem 1.12.1 (Invertible Matrix Theorem) Let A be an n × n matrix. Thefollowing statements are equivalent: a. A is invertible. b. The columns of A are linearly independent i. The columns of A form a basis for Rn . Our next example demonstrates how certain families of vectors naturallyform subspaces of Rn and how vector arithmetic can be used to determine abasis for the subspace they form. ⎡ ⎤ 3a + b − c ⎢ 4a − 5b + c ⎥Example 1.12.4 Consider the set W of all vectors of the form ⎢ ⎥ ⎣ a + 2b − 3c ⎦. a −bShow that W is a subspace of R 4 and determine a basis for this subspace.Solution. First, we observe that a typical element v of W is a vector of the form ⎡ ⎤ 3a + b − c ⎢ 4a − 5b + c ⎥ v=⎢ ⎣ a + 2b − 3c ⎦ ⎥ a −b
112 Essentials of linear algebraUsing properties of vector addition and scalar multiplication, we can write ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 3 1 −1 ⎢4⎥ ⎢ −5 ⎥ ⎢ 1⎥ v = a⎢ ⎥+b⎢ ⎣1⎦ ⎥ ⎢ ⎣ 2 ⎦ + c ⎣ −3 ⎦ ⎥ 1 −1 −1From this, we observe that W may be viewed as the span of the set S ={w1 , w2 , w3 }, where ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 3 1 −1 ⎢4⎥ ⎢ −5 ⎥ ⎢ ⎥ w1 = ⎢ ⎥ , w2 = ⎢ ⎥ , w3 = ⎢ 1 ⎥ ⎣1⎦ ⎣ 2⎦ ⎣ −3 ⎦ 1 −1 −1As seen in exercise 19 in section 1.11, the span of any set of vectors in Rngenerates a subspace of Rn ; it follows that W is a subspace of R4 . Moreover, wecan observe that S = {w1 , w2 , w3 } is a linearly independent set since ⎡ ⎤ ⎡ ⎤ 3 1 −1 1 0 0 ⎢4 −5 1⎥ ⎢ ⎥ ⎢ ⎥ → ⎢0 1 0⎥ ⎣1 2 −3 ⎦ ⎣0 0 1⎦ 1 −1 −1 0 0 0Since S both spans the subspace W and is linearly independent, it follows thatS is a basis for W .In example 1.12.4 we used the fact that the span of any set in Rn is a subspaceof Rn . This result extends to general vector spaces and is stated formally in thefollowing theorem.Theorem 1.12.2 In any vector space V , the span of any set of vectors forms asubspace of V . It is not hard to prove this result. Since the span of a set contains all linearcombinations of the set, it must contain the zero combination and be closedunder both vector addition and scalar multiplication. One of the reasons that a basis for a subspace is important is that a basistells us the minimum number of vectors needed to fully describe every elementof the subspace. More specifically, given a basis B for a subspace W , we knowthat we can write every element of W uniquely as a linear combination of theelements in the basis. Note that a subspace does not have a unique basis; forexample, in example 1.12.3, we saw two different bases for R3 . Furthermore, in R3 we have seen that the standard basis (and one exampleof another basis) has three elements. By the Invertible Matrix Theorem, it isclear that every basis of R3 consists of three vectors since we are requiredto have a set that is both linearly independent and spans R3 . Likewise, anybasis of Rn will have n elements. It can be shown that even in vector spaces
Bases and dimension in vector spaces 113other than Rn , any two bases of a subspace are guaranteed to have the samenumber of elements. Therefore, this number of elements in a basis can be usedto identify a fundamental property of any subspace: the minimum number ofelements needed to describe all of the elements in the space. We call this numberthe dimension of the subspace.Definition 1.12.4 Given a subspace W in a vector space V and a basis B forW , the number of elements in B is the dimension of W . Equivalently, if B has kelements, we write dim(W ) = k. Thus we naturally use the language that "R3 is three-dimensional" andsimilarly that "Rn has dimension n." Similarly, we can say dim(P2 ) = 3 (seeexample 1.12.1), and that the dimension of the vector space of all linearcombinations of the functions sin t and cos t is two (see example 1.12.2). In closing, it is worth recalling example 1.6.3 in section 1.6, where weconsidered three sets R, S, and T in R3 . R contained two vectors and waslinearly independent but did not span R3 ; S contained three vectors, was linearlyindependent, and spanned R3 ; and T consisted of four vectors, was linearlydependent, and spanned R3 . Since the set S has both key properties of beinglinearly independent and a spanning set, we can say that the set S is a basis forR3 , which further reflects the fact that dim(R3 ) = 3.Exercises 1.12 In the vector space V given in each of exercises 1–7, determinea basis for the subspace H and hence state the dimension of H . ⎧ ⎡ ⎤ ⎫ ⎨ 2 ⎬ 1. V = R3 , H = t ⎣ 0 ⎦ : t ∈ R ⎩ ⎭ −1 2. V = P2 , H = at 2 : a ∈ R ⎧⎡ ⎤ ⎫ ⎪ 2a + 3b ⎪ ⎪ ⎪ ⎨⎢ ⎥ ⎬ 3. V = R 4 , H = ⎢ a − 4b ⎥ : a , b ∈ R ⎪⎣ −3a + 2b ⎦ ⎪ ⎪ ⎪ ⎩ a −b ⎭ 4. V = P (the vector space of all polynomials), H = Pn (the subspace of all polynomials of degree n or less) 2 −1 5. V = R2 , H = x : Ax = 0 where A = −6 3 1 −3 2 −1 6. V = R4 , H = x : Ax = 0 where A = −2 5 0 4 a 0 7. V = M2×2 , H = A ∈ M2×2 : A = b c
114 Essentials of linear algebra 8. Determine whether or not the following set S is a basis for R3 . If not, is some subset of S a basis for R3 ? Explain. ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ ⎨ 1 0 1 2 ⎬ S = ⎣ 0 ⎦,⎣ 1 ⎦,⎣ 1 ⎦,⎣ 1 ⎦ ⎩ ⎭ 1 1 1 3 9. Is the set S = {[1 2]T , [2 1]T } a basis for R2 ? Justify your answer.10. Is the set S = {[1 2]T , [−4 − 8]T } a basis for R2 ? Justify your answer.11. Is the set S = {[1 2 1 1]T , [2 1 1 − 1]T , [−1 1 3 1]T , [2 4 5 1]T } a basis for R4 ? Justify your answer.12. Is the set S = {[1 2 1 1]T , [2 1 1 − 1]T , [−1 1 3 1]T , [2 4 5 0]T } a basis for R4 ? Justify your answer.13. Can a set with three vectors be a basis for R4 ? Why or why not?14. Can a set with seven vectors be a basis for R6 ? Why or why not?15. Not every vector space has a basis with finitely many elements. If there is not a finite basis, then we say that the vector space is infinite dimensional. Explain why the vector space P of all polynomial functions is an infinite dimensional vector space.16. Let V be the vector space V = C [−1, 1] and H the subset defined by H = f ∈ C [−1, 1] : f is differentiable Explain why H is an infinite dimensional subspace of V and why we cannot explicitly write down the elements in a basis for H .17. Recall from exercises 22 and 23 in section 1.11 that the null space of a matrix A is the subspace of all solutions to the equation Ax = 0 and that the column space of A is the space spanned by the columns of A. By exploring several different examples of matrices A of your choice, discuss how the dimensions of the null and column spaces are related to the number of pivot columns in the matrix. In particular, explain what you can say about the relationship between the sum of the dimensions of the null and column spaces and the number of columns in the matrix A.18. Decide whether each of the following sentences is true or false. In every case, write one sentence to support your answer. (a) Any set of five vectors is a basis for R5 . (b) If S is a linearly independent set of six vectors in R6 , then S is a basis for R6 . (c) If the determinant of a 3 × 3 matrix A is zero, then the columns of A form a basis for R3 . (d) If A is an n × n matrix whose columns span Rn , then the columns of A form a basis for Rn .
For further study 1151.13 For further study1.13.1 Computer graphics: geometry and linear algebra at workIn modern computer graphics, images consisting of sets of pixels are movedaround the screen through mathematical computations that rely on linearalgebra. If we focus on two-dimensional objects, there are several basic movesthat we must be able to perform: translation, rotation, reflection, and dilation.In what follows, we explore the role that linear algebra plays in the geometry oflinear transformations and computer graphics.(a) In section 1.8.1 we began to develop an understanding of how matrix multiplication can be used to move a two-dimensional image around the plane. If you have not already read this section, do so now. If we take the perspective that a given point in the plane is stored in the vector v, then for any 2 × 2 matrix A, the matrix A moves the vector via multiplication to the new location Av. If we have a finite set of points (which together constitute an image), we can store the points in a matrix M whose columns represent the individual points), and the new image which results from multiplication by A is given by AM. Consider the triangle with vertices (0, 0), (3, 1), and (2, 2), stored in the matrix 0 3 2 M= 0 1 2 Choose three different matrices A and compute AM. Then explain why it is impossible to use multiplication by a 2 × 2 matrix to translate the triangle so that all three of its vertices appear in new locations.(b) Due to our discovery in (a) that a simple translation is impossible using 2 × 2 matrices, we introduce the notion of homogeneous coordinates; instead of representing points in the two-dimensional plane as [x y ]T , we move to a plane in three-dimensional space where the third coordinate is always 1. That is, instead of [x y ]T we use [x y 1]T . Consider the matrix A given by ⎡ ⎤ 1 0 a A = ⎣0 1 b ⎦ (1.13.1) 0 0 1 and the triangle from (a) which can be represented in homogeneous coordinates by the matrix ⎡ ⎤ 0 3 2 M = ⎣0 1 2⎦ 1 1 1
116 Essentials of linear algebra Compute AM. What has happened to each vertex of the triangle represented by M? Explain in terms of the parameters a and b in A.(c) Using a = 2 and b = −1 in (1.13.1) along with the triangle M from above, compute AM in order to determine the translation of the triangle 2 units in the x-direction and −1 units in the y-direction. Sketch both the original triangle and its image under this translation.(d) In order to view some more sophisticated graphics, we use Maple in our computations that follow. Rather than performing operations on a triangle, we will use the syntax > with(plots): with(LinearAlgebra): > setoptions(scaling=constrained, axes=boxed, tickmarks=[5,5]): > X := cos(t)*(1+sin(t))*(1+0.3*cos(8*t))* (1+0.1*cos(24*t)): > Y := sin(t)*(1+sin(t))*(1+0.3*cos(8*t))* (1+0.1*cos(24*t)): > plot([X,Y,t=0..2*Pi], color=blue, thickness=3); which generates a parametric curve whose plot is the leaf shown in figure 1.18. Input these commands in Maple, as well as the syntax > leaf := plot([X,Y,t=0..2*Pi], color=grey, thickness=1): to store the image of the original leaf in leaf. 2.0 1.0 0.0 −1.0 0.0 1.0 Figure 1.18 A Maple leaf.
For further study 117 Finally, for a given matrix A of the form ⎡ ⎤ a11 a12 a13 A = ⎣a21 a22 a23 ⎦ 0 0 1 and a vector Z = [X Y 1], compute AZ (by hand) to show how AZ depends on the entries in A.(e) By our work in (c) and (d), if we now let ⎡ ⎤ 1 0 2 A = ⎣0 1 −1⎦ 0 0 1 the product AZ should result in translation of the leaf by the vector [2 − 1]T . To test this, we define the matrix A in Maple by > A := <<1,0,0>|<0,1,0>|<2,-1,1>>; and compute the coordinates in the new image by > Xnew := A[1,1]*X + A[1,2]*Y + A[1,3]*1: > Ynew := A[2,1]*X + A[2,2]*Y + A[2,3]*1: > image1 := plot([Xnew,Ynew,t=0..2*Pi], thickness=3, color=blue): The last command above plots the resulting image and stores it in image1. Display both the original leaf and the new image with the command > display(leaf, image1); and show that this results indeed in the translated leaf as shown in figure 1.19.(f) In section 1.8.1, we learned that a matrix of the form cos θ − sin θ R= sin θ cos θ is known as a rotation matrix and, through multiplication, rotates any vector by θ radians counterclockwise about the origin. To work with a rotation matrix in homogeneous coordinates, we update the matrix as follows: ⎡ ⎤ cos θ − sin θ 0 R = ⎣ sin θ cos θ 0⎦ 0 0 1 Let us say that we wanted to perform two operations on the leaf. First, we wish to translate the leaf as above along the vector [2 − 1]T , and then we
118 Essentials of linear algebra 1 −1 −1 1 3 Figure 1.19 The original leaf and its transla- tion by [2 − 1]T . want to rotate the resulting image π/4 radians clockwise about the origin. We can accomplish this through two matrices by computing their product, as the following discussion shows. From (e), we know that using the matrix > Translation := <<1,0,0>|<0,1,0>|<2,-1,1>>; leads to the desired translation. Likewise, the matrix > Rotation := <<1/sqrt(2),-1/sqrt(2), 0>|<1/sqrt(2),1/sqrt(2),0>|<0,0,1>>; will produce the sought rotation. Explain why the matrix > A := Rotation.Translation; will produce the combined translation and rotation, and plot the resulting figure by updating your computations for Xnew and Ynew and using the syntax > image2 := plot([Xnew,Ynew,t=0..2*Pi], thickness=4, color=black): > display(leaf, image1, image2);(g) What is the result of applying the matrix ⎡ ⎤ 1 0 0 ⎢ ⎥ A = ⎣0 1 0⎦ 2 0 0 1
For further study 119 on the leaf? What kind of geometric transformation is performed by this matrix? What matrix would keep the height of the leaf constant but stretch its width by a factor of 2?(h) It can be shown that to reflect an image across a line through the origin that forms an angle α with the positive x-axis, the necessary matrix is ⎡ ⎤ cos 2α sin 2α 0 A = ⎣ sin 2α − cos 2α 0⎦ 0 0 1 By finding the appropriate value of α , find the matrix that will reflect an image across the line y = x and compute and plot the image of the original leaf under this reflection. (i) Exercises for further practice and investigation: 1. Find the image of the original leaf under rotation about the origin by 2π/3 radians, followed by a reflection across the y-axis. 2. Find the image of the original leaf under rotation about the point (−3, 1) by −π/6 radians. (Hint: To rotate about a point other than the origin, first translate that point to the origin, then rotate, then translate back.) 3. Find the image of the original leaf under translation along the vector [3 2]T , followed by reflection across the line y = x /2.1.13.2 Bézier curvesIn what follows7 , we explore the use of a specific type of parametric curves,called Bézier curves (pronounced "bezzy-eh"), which have a variety of importantapplications. These curves were originally developed by two automobileengineers in France in the 1960s, P. Bézier and P. de Casteljau, who were workingto develop mathematical formulas to graph the smooth, wiggle-free curves thatformed the shape of a car's body. Today, Bézier curves find their way into ourlives every day: they are used to create the letters that appear in typeset fonts.The principles that govern these curves involve fundamental mathematics fromlinear algebra and calculus.(a) In calculus, we study parametric curves given in the form x = f (t ), y = g (t ), where f and g are each functions of the parameter t . Another way to denote this situation is to write P(t ) = (f (t ), g (t )) where t belongs to some interval of real numbers. Note that P(t ) is essentially a vector; the graph of P(t ) is the parametric curve traced out by7 The material in this project has been adapted from Steven Janke's chapter "Designer Curves" inApplications of Calculus, MAA Notes Number 29, Philip Straffin, Ed.
120 Essentials of linear algebra the vector over time. It will be most convenient if we simply write this as P(t ) = (x(t ), y(t )) in what follows. In this problem we begin to consider some special formulas for x(t ) and y(t ). To parameterize the line between the points P0 (1, 3) and P1 (3, 7), we can think about wanting to make x go from 1 to 3, and y go from 4 to 7. Indeed, we want these to occur simultaneously as t goes from 0 to 1. Consider the parameterization: x = x(t ) = 1 + t (3 − 1) = t · 3 + (1 − t ) · 1 y = y(t ) = 3 + t (7 − 3) = t · 7 + (1 − t ) · 3 0≤t ≤1 Observe that when t = 0, x = 1 and y = 3, and when t = 1, x = 3 and y = 7. Show that the curve parameterized by these two equations is indeed the line segment between P0 and P1 . For instance, you might use algebra to eliminate the variable t , thereby deducing a relationship between x and y.(b) We can think about the equations for x and y in (a) in a more compact manner. Consider the following vector notation to replace the previous equations: x(t ) 3 1 P(t ) = =t + (1 − t ) (1.13.2) y(t ) 7 3 This is sometimes referred to as taking a convex combination of the points (1, 3) and (3, 7), because t and 1 + t are both nonnegative and sum to 1. Using the above style, write the parametric equations for the line segment that passes between the general points P0 (x0 , y0 ) and P1 (x1 , y1 ).(c) An even more concise notation is to simply write P(t ) = (1 − t )P0 + tP1 . We will now use this notation to combine two or more of these parameterizations for line segments in a way that constructs curves that can be "controlled" in very interesting ways. Consider three points, labeled P0 , P1 , and P2 . In the most recent form of P(t ) given above at (1.13.2), write parameterizations for the two line segments from P0 to P1 and from P1 to P2 , as pictured below. Call the first parameterization P (1) (t ) and the second parameterization P (2) (t ). In addition, determine the parameterizations P (1) (t ) and P (2) (t ) for the specific set of points P0 (2, 3), P1 (4, 7), and P2 (7, 1). Show your work, and write each out in the expanded form where you have an expression for x(t ) and another for y(t ).(d) From the two line-segment parameterizations in (c), we will now create a new parametric plot by taking similar combinations of P (1) (t ) and P (2) (t ).
For further study 121 P2 P0 P (2) (1) P P1 Figure 1.20 Theline seg- ments from P0 to P1 and P1 to P2 . Consider the function Q(t ) defined as follows: Q(t ) = (1 − t ) · P (1) (t ) + t · P (2) (t ) (1.13.3) First, substitute in (1.13.3) your expressions for P (1) (t ) and P (2) (t ) from (c) that involve the general points P0 , P1 , and P2 . Simplify the result as much as possible in order to write the formula for Q in the following form: Q(t ) = a0 (t )P0 + a1 (t )P1 + a2 (t )P2 where a0 (t ), a1 (t ), and a2 (t ) are polynomial functions of t . Then, using the specific parameterizations for P (1) (t ) and P (2) (t ) for the points P0 (2, 3), P1 (4, 7), and P2 (7, 1), determine the parametric equations for x(t ) and y(t ) that make up the function Q(t ). For each of these three parameterizations (P (1) , P (2) , and Q), use Maple to sketch a plot8 and describe the results in detail. For example, how does Q(t ) look in comparison to the two line segments? What kind of functions make up the components x(t ) and y(t ) in Q? What is true about Q(0) relative to the points P0 , P1 , and P2 ? Q(1)? What direction is a particle moving along Q(t ) headed as t starts out away from 0? As t gets near to 1? (e) It turns out that we will have even more freedom and control in drawing curves if we start with four control points, P0 , P1 , P2 , and P3 . The development here is similar to what was done above, just using a greater number of points. First, parameterize the segments from P0 to P1 (with P (1) (t )), P1 to P2 (with P (2) (t )), and from P2 to P3 (with P (3) (t )). The usual formulas apply8 The Maple syntax to plot a parametric curve (f (t ), g (t )) on the interval [a , b ] is> plot([f(t),g(t),t=a..b]);.
122 Essentials of linear algebra here; write down the basic form of each P (j) (t ), j = 1, 2, 3, in terms of the various points Pi . Then combine, as in (d) above, the parameterizations for the first two segments to get a new function Q (1) ; also combine the parameterizations for the second two segments to get Q (2) . These Q parameterizations are written as Q (1) (t ) = (1 − t ) · P (1) (t ) + t · P (2) (t ) Q (2) (t ) = (1 − t ) · P (2) (t ) + t · P (3) (t ) Finally, combine Q (1) and Q (2) to get a new parametric function that we call B(t ) according to the natural formula B(t ) = (1 − t ) · Q (1) (t ) + t · Q (2) (t ) By substituting appropriately for Q (1) (t ) and Q (2) (t ) and then replacing these with the appropriate P (j) (t ) functions, show that B(t ) = P0 (1 − t )3 + 3P1 t (1 − t )2 + 3P2 t 2 (1 − t ) + P3 t 3 . B(t ) is called a cubic Bézier curve. By finding and using appropriate t values, show that the points P0 and P3 both lie on the curve given by B(t ).(f) Write the formulas for x(t ) and y(t ) that give the parameterizations for the cubic Bézier curve that has the four control points P0 (2, 2), P1 (5, 10), P2 (40, 20), and P3 (10, 5). Use Maple to plot each of the parametric curves given by P (j) (t ), j = 1 . . . 3, Q (1) (t ), Q (2) (t ), and B(t ) in the same window. Discuss how the various curves combine to form others.(g) For the general Bézier curve with control points P0 (x0 , y0 ), P1 (x1 , y1 ), P2 (x2 , y2 ), and P3 (x3 , y3 ), derive the equation for the tangent line to the curve at the point (x0 , y0 ), and prove that the point (x1 , y1 ) lies on this tangent line. (Hint: to determine the slope of the tangent line, use the chain rule in the standard way for finding dy /dx for a parametric curve.)(h) Laser printers and the program Postscript use Bézier curves to construct the fonts that we use to represent letters. For example, a picture of the letter g is shown below that reveals the control points and Bézier curves required to accomplish this. In Maple, use two or more Bézier curves to sketch a reasonable representation of the letter S. (You need not try to emulate the thickness of the 'g' that is shown above.) Then, use an appropriate number of Bézier curves to create an approximation of the lowercase letter 'a,' in the form shown here in quotes. State the control points required for the various curves.
For further study 123 Figure 1.21 The letter g. (i) Discuss the role that vectors and linear combinations play in the development of Bézier curves.1.13.3 Discrete dynamical systemsA linear discrete dynamical system is a model that represents changes in a systemfrom time k to time k + 1 by the rule x (k +1) = Ax (k)A discrete dynamical system is similar to a Markov chain, but we no longerrequire that the columns of the matrix A sum to 1. A key issue in either scenariois the long term behavior of the quantity x (k) being modeled. In what follows, weexplore the role of eigenvalues and eigenvectors in determining this long-termbehavior and study an important application of these ideas.(a) To begin investigating the long-term behavior of the system, we will assume that A is an n × n matrix with n real linearly independent eigenvectors v1 , . . . , vn . Furthermore, assume that the corresponding real eigenvalues of A satisfy the relationship |λ1 | > |λ2 | ≥ · · · ≥ |λn | Consider an initial vector x (0) . Explain why there exist constants c1 , . . . , cn such that x (0) = c1 v1 + c2 v2 + · · · + cn vn and show that Ax (0) = c1 λ1 v1 + c2 λ2 v2 + · · · + cn λn vn
124 Essentials of linear algebra Furthermore, show that x (k) = Ak x (0) = c1 λk v1 + c2 λk v2 + · · · + cn λk vn 1 2 n (1.13.4)(b) In (1.13.4), divide both sides by λk . What can you conclude about 1 (λ2 /λ1 )k as k → ∞? Why can you make similar conclusions about (λj /λ1 )k for j = 3 . . . n? Hence explain why for large k k 1 Ak x (0) ≈ c1 v1 λ1 and thus why Ak x (0) is an approximate eigenvector of A corresponding to v1 .(c) In studying a population like spotted owls, mathematical ecologists often pay close attention to the various numbers of a species at different stages of life. For example, for spotted owls there are three pronounced groupings: juveniles (under 1 year), subadults (1 to 2 years old), and adults (2 years and older). The owls mate during the latter two stages, breed as adults, and can live for up to 20 years. A critical time in the life cycle and survival of these owls is when the juvenile leaves the nest to build a home of its own.9 Let the number of spotted owls in year k be represented by the vector ⎡ ⎤ jk x (k) = ⎣ sk ⎦ ak where jk is the number of juveniles, sk the number of subadults, and ak the number of adults. Using field data, mathematical ecologists have determined10 that a particular spotted owl population is modeled by the discrete dynamical system ⎡ ⎤ 0 0 0.33 x (k +1) = ⎣0.18 0 0⎦ x (k) 0 0.71 0.94 What does this model imply about the percent of juveniles that survive to become subadults? About the percent of subadults that survive to become adults? About the percent of adults that survive from one year to the next? What percent of adults produce juvenile offspring in a given year?(d) Assume that in a given region, ecologists have measured the present populations as follows: j0 = 200, s0 = 45, and a0 = 725. Use the model stated in (c) to determine the population x (k) = [jk sk ak ]T for k = 1, . . . , 20. Do you think the spotted owl will become extinct? Give a 9 To read more about the issue of spotted owl survival, see the introduction to chapter 5 ofDavid C. Lay's Linear Algebra and its Applications.10 R. H. Lamberson et al., "A Dynamic Analysis of the Viability of the Northern Spotted Owl in aFragmented Forest Environment," Conservation Biology 6 (1992), 505–512.
For further study 125 convincing argument using not only your computations of the population vectors but also the results of (b).(e) Say that r is the fraction of juveniles that survive from one year to the next (that is, replace 0.18 in the matrix of the model with r) . By experimenting with different values of r, determine the minimum fraction of juveniles that must survive from one year to the next in order for the spotted owl population not to become extinct. How does your answer depend on the eigenvalues of the matrix?(f) Let A be the n × n matrix of a discrete dynamical system and assume that A has n real linearly independent eigenvectors. Let x (0) be an initial vector and let ρ (A) denote the maximum absolute value of an eigenvalue of A. Show that the following are true: (i) If ρ (A) < 1, then limk →∞ Ak x (0) = 0. (ii) If ρ (A) = 1 and λ = 1 is the unique eigenvalue having this maximum absolute value, then limk →∞ Ak x (0) is an eigenvector of A. (iii) If ρ (A) > 1, then there exist choices of x (0) for which Ak x (0) grows without bound.
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2First-order differential equations2.1 Motivating problemsDifferential equations arise naturally in many problems encountered whenmodeling physical phenomena. To begin our study of this subject, we introducetwo fundamental examples that demonstrate the central role that differentialequations play in our world. In section 1.1, we discussed how the amount of salt present in a system oftwo tanks can be modeled through a system of differential equations. Here, aneven simpler situation is considered: our goal is to predict the amount of saltpresent in a city's water reservoir at time t , given a set of determining conditions. Suppose that the reservoir is filled to its capacity of 10 000 m3 , and thatmeasurements indicate an initial concentration of salt of C0 = 0.02 g/m3 . Notethat it follows there are A0 = 200 g of salt initially present. As the city drawsthis solution from the reservoir for use, new solution (water with some saltconcentration) from the local treatment facility flows into the reservoir so thatthe volume of water present in the tank stays constant. Let us assume that theconcentration of salt in the inflowing solution is 0.01 g/m3 , and that the rateof this inflow is 1000 m3 /day. Since the city is also assumed to be drawingsolution at an equal rate from the reservoir, the outflow also occurs at a rate of1000 m3 /day. We are interested in several key questions. How much salt is in the tank attime t ? What is the concentration of salt in the water being used by the city attime t ? What happens to these values over time? We will let A(t ) denote the amount of salt in the tank at time t . Theinstantaneous rate of change dA /dt of A(t ) is given by the difference between 127
128 First-order differential equationsthe rate at which salt is entering the tank and the rate at which salt isleaving. Exploring the given information regarding inflow and outflow, we candetermine these rates precisely. Since solution is entering the reservoir at 1000 m3 /day containing aconcentration of 0.01 g/m3 , it follows that salt is entering the tank at a rate of m3 g g 1000 · 0.01 3 = 10 day m dayFor salt leaving the reservoir, the situation is slightly more complicated. Since wedo not know the exact amount of salt present in the reservoir at time t , we denotethis by A(t ). Assuming that the solution in the reservoir is uniformly mixed, theconcentration of salt in the outflowing solution is the ratio of the amount A(t )of salt to the volume of the tank. That is, the outflowing concentration is A(t ) g 10 000 m3Since this outflow is occurring at a rate of 1000 m3 /day, it follows that salt isleaving the tank at a rate of m3 A(t )g A(t ) g 1000 · = day 10 000 m3 10 day It now follows that the instantaneous rate of change dA /dt of salt in thetank in grams per day is given by the difference of the rate of salt entering andthe rate of salt leaving the tank. Specifically, dA A(t ) = 10 − (2.1.1) dt 10Note carefully what this last equation is saying: A(t ) is an unknown function,but we have an equation that relates this unknown function to its derivative.Such an equation is called a differential equation. The solution to this equationis a function A(t ) that makes the equation true. If we can solve the equation forA(t ), we then will be able to predict the amount of salt in the tank at any time t .Determining such solutions and their long-term behaviors is the main focus ofthis chapter. Another important application of differential equations involves pop-ulation growth. Consider a population P(t ) of animals. As likelihood ofreproduction depends on the number of animals present, it is natural to assumethat the rate of change of P(t ) is directly proportional to P(t ). Phrased in termsof the derivative, this assumption means that dP = kP(t ) (2.1.2) dtwhere k is some positive constant. Observe that (2.1.2) is a differential equationinvolving the function P. It is a standard exercise in calculus to show thatfunctions of the form P(t ) = P0 e ktare solutions to (2.1.2).
Definitions, notation, and terminology 129 Because the function P(t ) = P0 e kt exhibits unbounded growth over time, itturns out that this exponential growth model is not realistic beyond a relativelyshort period of time. A related, but more sophisticated, model of populationgrowth is the logistic differential equation dP P(t ) = kP(t ) 1 − dt Awhere the constant k is considered the reproductive rate of the population andthe constant A is the surrounding environment's carrying capacity. For example,if a population had a relative growth rate of k = 0.02 and a carrying capacity ofA = 100, the population function would satisfy the differential equation dP P(t ) = 0.02P(t ) 1 − dt 100The logistic model, usually credited to the Dutch mathematician Pierre Verhulst,accounts not only for reproductive growth, but also for mortality by consideringenvironmental limitations on maximum population. The logistic equation ismore challenging to solve; we will do so in section 2.7. In addition to mixing problems and models of population growth, differ-ential equations enjoy widespread applications in other physical phenomena.Differential equations are also mathematically interesting in and of themselves,and in upcoming sections we will study not only their applications, but also theirkey properties and characteristics to better understand the subject as a whole.2.2 Definitions, notation, and terminologyAs we have seen with the examples dA A = 10 − (2.2.1) dt 10 dP P = 0.02P 1 − (2.2.2) dt 100 y +y = 0 (2.2.3)a differential equation is an equation relating an unknown function to one ormore of its derivatives. Usually we will suppress the notation "A(t )" and insteadsimply write "A," as in (2.2.1). We will interchangeably use the notations yand dy /dt to represent the first derivative; similarly, y = d 2 y /dt 2 . Other bookssometimes employ the notations y = D(y) = y and y = D 2 (y) = y . ˙ ¨ A solution of a differential equation is a differentiable function that satisfiesthe equation on some interval (a , b) of values for the independent variable.For example, the function y = sin t is a solution to (2.2.3) on (−∞, ∞) sincey = − sin t , and − sin t + sin t = 0 for all values of t . Given any differential equation, we are interested in determining all of itssolutions. But many, if not most, differential equations are difficult or impossible
130 First-order differential equationsto solve. For example, the equation y + ty = t(which is only a slightly modified version of (2.2.3)) has no solution in termsof elementary functions.1 In such situations, we may turn to qualitative orapproximation methods that may enable us to analyze how a solution shouldbehave, while perhaps not being able to determine an explicit formula for thefunction. Equations (2.2.1), (2.2.2), and (2.2.3) are often called ordinary differentialequations, in contrast to partial differential equations such as ∂ 2u ∂ 2u + =0 ∂x2 ∂y2where the solution function u(x , y) has two independent variables x and y. Ourfocus will be on ordinary differential equations, as partial differential equationsare beyond the scope of this text. The order of a differential equation is the orderof the highest derivative present. For example, (2.2.1) and (2.2.2) are first-orderdifferential equations since they only involve first derivatives. Equation (2.2.3)is second-order. For now, we limit our attention to first-order equations; higherorder equations will be discussed in detail in subsequent chapters. It is important to note that every student of calculus learns to solve a certainclass of differential equations through integration. For example, the problem,"find a function y whose derivative is te t " can be restated as a differentialequation. In particular, this problem can be stated as the differential equation dy = te t (2.2.4) dtIntegrating both sides with respect to t and using integration by parts on theright, it follows that y(t ) = te t − e t + Cis a solution for any choice of the constant C. Here we see an importantfact: differential equations typically have a family of infinitely many solutions.Determining all possible members of that family, like determining all solutionsto systems of linear equations in linear algebra, will be a central component ofour work. Calculus students also know that if we are given one more piece ofinformation about the function y along with (2.2.4), it is possible to uniquelydetermine the integration constant, C. For example, had the problem aboveread, "find a function y whose derivative is te t such that y(0) = 5," we couldintegrate to find y = te t − e t + C, just as we did previously, and then use theinitial condition y(0) = 5 to see that C must satisfy the equation 5 = 0 · e0 − e0 + C1 This fact is not obvious.
Definitions, notation, and terminology 131and thus C = 6. When we are given a differential equation of order n alongwith n initial conditions, we say that we are solving an initial-value problem.2In the given example, y = te t − e t + 6 is the solution to the stated initial-valueproblem. Based on the example above and our experience in calculus, it is clear thatintegration is an obvious (and often effective) approach to solving differentialequations of the form dy = f (t ) dtwhere f (t ) is a given function. If we can integrate f symbolically, then thedifferential equation is solved. Even if f (t ) cannot by integrated symbolicallywith respect to t , we can still use techniques like numerical integration tosuccessfully attack the problem. The situation grows more complicated whenwe want to solve differential equations that also involve the unknown functiony, such as dy = te y dtIn what follows in this chapter, we seek to classify first-order equations intotypes that can be solved in a straightforward way by symbolic means (ofteninvolving integration), as well as to develop methods that can be used to generateapproximate solutions in situations where a symbolic solution is either difficultor impossible to attain. Throughout, the general form of the equations we areconsidering will be y = f (t , y), where the function f (t , y) represents somecombination of the independent variable t and the unknown function y. It is also important to note that a wide range of first-order initial-valueproblems are guaranteed to have unique solutions. This is stated formally in thefollowing theorem, whose proof may be found in more advanced texts.Theorem 2.2.1 Consider the initial-value problem given by y = f (t , y),y(t0 ) = y0 . If the function f (t , y) is continuous on a rectangle that includes(t0 , y0 ) in its interior and the partial derivative3 fy (t , y) is continuous onthat same rectangle, then there exists an interval containing t0 on which theinitial-value problem has a unique solution. Often the dependent variable, or unknown function y, in a differentialequation will model an important quantity in some physical problem: theamount of salt in a tank at time t , the number of members of a populationat a given time, or the position of a mass attached to a spring. As such, wewill place particular emphasis on the graph of the solution function in order tobetter understand what the differential equation is telling us about the physicalsituation it models.2 We often use the abbreviation IVP to stand for the phrase "initial-value problem."3 We typically use the notation fy (t , y) = ∂ f /∂ y.
132 First-order differential equations Just as geometry and graphical interpretations shaped our understandingof linear algebra in chapter 1, these perspectives will prove extremely helpful inour study of differential equations. We begin our explorations of these graphicalinterpretations through the reservoir problem from section 2.1 and the earlierexample y = te t . So far in our references to derivatives in the reservoir and populationmodels, we have viewed the derivative as measuring the instantaneous rate ofchange of a quantity that is varying. From a more geometric point of view, wealso know that the derivative of a function measures the slope of the tangentline to the function's graph at a given point. For example, with the differentialequation dA A = 10 − (2.2.5) dt 10we can say that if, at some time t , the amount of salt A is A = 20, then dA /dt =10 − 20/10 = 8. Thus, if A(t ) is a solution to the differential equation, it followsthat at any time where A(t ) = 20, A (t ) = 8. Graphically, this means that at sucha point, the slope of the tangent line to the curve must be 8. Since we are interested in the function A(t ) over an interval of t -values, wealso expect that A(t ) will take on a wide range of values. As such, it is natural tocompute the slope of the tangent line determined by (2.2.5) for a large numberof different values of A and t . Obviously computers are best suited to such atask, and, as we will see in the introduction to Maple commands at the end ofthis section, Maple and other computer algebra systems provide tools for doingso. Computing values of dA /dt over a grid of t and A values, we can plot asmall portion of each corresponding tangent line at the point (t , A), and see theresulting slope field (or direction field). The slope field for (2.2.5) is shown infigure 2.1. A(t) 200 150 100 50 t 10 20 30 40 50 Figure 2.1 The slope field for (2.2.5); the graph of the solution cor- responding to an initial condition A(0) = 200 is included.
Definitions, notation, and terminology 133Observe that a slope field provides an intuitive way to understand theinformation a first-order differential equation possesses: the slope at each pointgives the direction of the solution at that point. Indeed, we use arrows insteadof small lines in order to indicate the flow of the solution as time increases. Inessence, the slope field is a map that the solution must navigate based on theinitial point from which the function starts. For example, if we use the initialcondition A(0) = 200 (as was given in the original example in section 2.1), wecan start a graph at the point (0, 200) and follow the map. Doing so yields thecurve shown in figure 2.1. Note particularly how we can clearly see the slope of the solution curvefitting with the slopes present in the direction field. Moreover, observe that thedirection field provides an immediate overall sense of how every solution to thedifferential equation behaves: for any solution A(t ), A(t ) → 100 as t → ∞. Thismakes sense physically, too, since the saltwater solution entering the reservoirhas concentration 0.01 g/m3 . Over time, the concentration of solution in thereservoir should tend to that level, and with 10 000 m3 of solution present in thereservoir, we expect the amount of salt to approach 100 g. Another example of a differential equation's slope field provides furtherinsights. For the differential equation dy = te t (2.2.6) dtits slope field for the window −2 ≤ t ≤ 1 and −2 ≤ y ≤ 2 is given in figure 2.2. We noted earlier that the general solution to this equation is y = te t − e t + C.Moreover, given any initial condition, we can determine C. For example, ify(0) = 1/2, C = 3/2. Likewise, if y(0) = 0, C = 1, and if y(0) = −1, C = 0. Ifwe plot the corresponding three functions with the slope field, then (as shownin figure 2.2) the three members of the family of all solutions to the originaldifferential equation appear as shown. In integral calculus, students learn about families of antiderivatives4 andhow two members of such a family differ only by a constant. Here, we see thisfact graphically in the slope field of figure 2.2, and can add the perspective thatthere exists a family of solutions to a certain differential equation. In upcomingsections, we will learn new techniques for how to determine solutions analyticallyin various circumstances, while not losing sight of the fact that every first-orderdifferential equation can be interpreted graphically through a direction field. Finally, there is an important type of first-order differential equation (DE)for which solutions can be determined algebraically. A first-order DE is said tobe autonomous if it can be written in the form y = f (y). That is, the independentvariable t is not involved explicitly in f (y). For example, the equation y = 1 − y2 (2.2.7)is autonomous.4 An antiderivative F of a function f is a function that satisfies F = f .
134 First-order differential equations y(t) 1.6 0.8 t −2.0 −1.5 −1.0 −0.5 0.5 1.0 −0.8 −1.6 Figure 2.2 The slope field for (2.2.6) along with three solution functions for the initial conditions y(0) = 1/2, y(0) = 0, and y(0) = −1. In addition, a solution y to a DE is called an equilibrium or constant solutionif the function y is constant. In (2.2.7), both y = 1 and y = −1 are equilibriumsolutions to the DE above. Such a solution is stable if all solutions with initialconditions y(t0 ) = y0 with y0 close to the equilibrium solution result in theoverall solution to the IVP tending toward the equilibrium solution. Otherwise,the equilibrium solution is called unstable. We close this section with an example regarding an autonomous differentialequation.Example 2.2.1 Consider the differential equation y = (y 2 − 1)(y − 3)2 .Determine all equilibrium solutions to the equation, as well as whether or noteach is stable or unstable. Finally, plot the direction field for the equation andinclude plots of the equilibrium solutions.Solution. To find the equilibrium solutions, we assume that y is a constantfunction, and therefore y = 0. Solving the algebraic equation 0 = (y 2 − 1)(y − 3)2we find that y = −1, y = 1, and y = 3 are the equilibrium solutions of thegiven DE. We can decide the stability of each equilibrium solution by studying thesign of y near the equilibrium value; note that (y − 3)2 is always nonnegative.To consider the stability of y = −1, observe that when y < −1, y = (y + 1)(y − 1)(y − 3)2 > 0, since the first two terms are both negative and the third ispositive. When y > −1 (and y < 1), it follows y = (y + 1)(y − 1)(y − 3)2 < 0
Definitions, notation, and terminology 135 y(t) 3 2 1 t −0.5 0.5 −1 Figure 2.3 The slope field for y = (y 2 − 1) (y − 3)2 along with its three equilibrium solutions.since the middle term is negative while the other two are positive. Hence, if asolution starts just below y = −1, that solution will increase toward −1, whereasif a solution starts just above y = −1, it will decrease to −1. This makes theequilibrium y = −1 stable. These observations are easiest to make visually in the direction field. As seenin figure 2.3, the constant solution y = −1 is stable, since any solution with aninitial condition just above or just below y = −1 will tend to y = −1. However,the solution at y = 1 is unstable, since any solution with an initial value justabove or just below y = 1 will tend away from 1 (and tend toward y = 3 ory = −1, respectively). Finally, although solutions just below y = 3 tend to 3,any solution that begins just above y = 3 will increase away from that constantsolution, and hence y = 3 is also unstable.52.2.1 Plotting slope fields using MapleJust as our work in linear algebra required the use of Maple's LinearAlgebra package, to take advantage of the software's support for the studyof differential equations we use the DEtools package, loading it with thecommand > with(DEtools):5 Some authors call a solution such as y = 3 in this example semi-stable, since there is stability on oneside and instability on the other.
136 First-order differential equationsTo plot the direction field associated with a given differential equation, itis convenient to first define the equation itself in Maple. This is accom-plished (for the equation from the reservoir problem) through the followingcommand: > Eq1 := diff(A(t), t) = 10-1/10*A(t);Note that the differential equation of interest is now stored in "Eq1". The slopefield may now be generated by the command > DEplot(Eq1, A(t), t = 0 .. 50, A(t) = 0 .. 200, color = grey, arrows=large);This command produces the slope field of figure 2.1, but without any particularsolution satisfying an initial value included. It is important to note that the rangeof t and A(t ) values is extremely important. Without a well-chosen windowselected by the user, the plot Maple generates may not be very insightful.For example, if the above command were changed so that the range of A(t )values is 0 .. 10, almost no information can be gained from the slopefield. As such, we will strive to learn to analyze the expected behavior of adifferential equation from its form so that we can choose windows well inrelated plots; we may often have to experiment and explore to find graphs thatare useful. Finally, if we are interested in one or more related initial-value problems,a variation of the DEplot command enables us to sketch the graph of eachcorresponding solution. For example, the command > DEplot(Eq1, A(t), t = 0 .. 50, A(t) = 0 .. 200, color = grey, arrows=large, [[0,200]]);will generate not only the slope field, but also the graph of the solution A(t )that satisfies A(0) = 200, as shown in figure 2.1. Additional curves for differentinitial conditions may be plotted by listing the other conditions to be satisfied:for example, in the stated command above we could replace [[0,200]] with[[0,200], [0,100], [0,0]] to include the plots of the three solutioncurves that respectively satisfy A(0) = 200, A(0) = 100, and A(0) = 0.Exercises 2.2 1. Consider the differential equation y = 4y. (a) What is the order of this equation? (b) Show via substitution that the function y = e 2t is a solution to this equation. (c) Are there any other functions of the form y = e rt (r = 2) that are also solutions to the equation? If so, which? Justify your answer.
Definitions, notation, and terminology 1372. For a ball thrown straight up from an initial height s(0) = 4 meters at an initial velocity of s (0) = 10 m/s, we know that after being thrown, the only force acting on the ball is gravity, provided we neglect air resistance. Knowing that acceleration due to gravity is constant at −9.81 m/s2 , it follows that s (t ) = −9.81. Use the given information to determine s(t ), the function that tells us the height of the ball at time t . Then determine the maximum height the ball reaches, as well as the time the ball lands.3. In the differential equation dA /dt = 10 − A /10 from the reservoir problem, explain why the function A(t ) = 100 is an equilibrium solution to the equation. Is it stable or unstable? Why?4. Consider the logistic differential equation dP P = 0.02P 1 − dt 100 Use Maple to plot the direction field for this equation. Print the output and, by hand, sketch the solutions that correspond to the initial conditions P(0) = 10, P(0) = 75, and P(0) = 125. What is the long-term behavior of every solution P(t ) for which P(0) > 0? Are there any constant (or equilibrium) solutions to the equation? Explain what these observations tell you about the behavior of the population being modeled.5. For the logistic differential equation dP P = 0.001P 1 − dt 25 how should the direction field appear? Use the constant/equilibrium solutions to the equation as well as the long-term behavior of the population to help you sketch, by hand, the direction field for this DE.6. By constructing tangent lines over a grid with at least sixteen vertices, sketch a direction field by hand for each of the following differential equations. (a) y = 1 − y (b) y = 1 (t − y) 2 (c) y = 1 (t + y) 2 (d) y = 1 − t7. Without using Maple to plot direction fields, match each of the following differential equations with its corresponding direction field. Write at least one sentence to explain the reasoning behind each of your choices. dy dy dy dy (a) = y −t (b) = ty (c) =y (d) =t dt dt dt dt
140 First-order differential equationsare all first-order DEs. Recall that in section 1.12 we discussed linear combina-tions of generalized vectors. Here we can view y and y as functions that belongto a vector space, and thus think about whether a certain combination of y andy is a linear combination or not. We say that any differential equation of theform a1 (t )y + a0 (t )y = b(t ) (2.3.4)is a linear first-order differential equation, since a linear combination of y and y isbeing formed. Any other first-order differential equation is said to be nonlinear.If we stipulate that a1 (t ) = 0, we can divide through by a1 (t ) and hence write y + p(t )y = f (t ) (2.3.5)as the standard form for a linear first-order equation. We call f (t ) theforcing function. Above, note that (2.3.1) and (2.3.3) are nonlinear equations,while (2.3.2) is linear. The simplest linear first-order differential equations are those for which theforcing function is zero. We naturally call the equation y + p(t )y = 0 (2.3.6)a homogeneous linear first-order DE. We consider a particular example thatshows how every such homogeneous DE may be solved.Example 2.3.1 Solve the differential equation y + (1 + 3t 2 )y = 0. In addition,solve the initial-value problem that is given by the same DE and the initialcondition y(0) = 4.Solution. We will use integration to solve for y. Rearranging the givenequation, we observe that y = −(1 + 3t 2 )y . Dividing both sides by y, we findthat y = −(1 + 3t 2 ) yKeeping in mind the fact that y and y are each unknown functions of t , weintegrate both sides of the previous equation with respect to t : y dt = −(1 + 3t 2 ) dt yWe recognize from the chain rule that the left-hand side is ln y. Thus, integratingthe polynomial in t on the right yields ln y = −t − t 3 + CWe note that while an arbitrary constant arises on each side of the equation whenintegrating, it suffices to simply include one constant on the right. Finally, wesolve for y using properties of the natural logarithm and exponential functionsto find that y = e −t −t +C = e C e −t −t 3 3
Linear first-order differential equations 141Since C is a constant, so is e C , and thus we write y = Ke −t −t 3Observe that we have found an entire family of functions that solve the originaldifferential equation: regardless of the constant K , the above function y is asolution. If we consider the stated initial-value problem and apply the giveninitial condition y(0) = 4, we immediately see that K = 4, and the solution tothe initial-value problem is y = 4e −t −t 3The solution method in example 2.3.1 can be generalized to apply to anyhomogeneous linear first-order DE. Using the notation p(t ) to replace thefunction 1 + 3t 2 , which is the coefficient of y, the same steps above may be usedto find the solution to the standard homogeneous linear first-order differentialequation. We state this result in the following theorem.Theorem 2.3.1 For any homogeneous linear first-order differential equationof the form y + p(t )y = 0,the general solution is y = Ke −P(t ) , where P is any antiderivative of p. Moreover,for the initial condition y(t0 ) = y0 , if p(t ) is continuous on an interval containingt0 , then the solution to the corresponding initial-value problem is unique. The uniqueness of the solution to the initial-value problem follows fromtheorem 2.2.1. But perhaps the most important lesson to learn from this result isthat a homogeneous linear first-order DE can always be solved. This is analogousto our experience with homogeneous linear systems of algebraic equations inchapter 1. In particular, note that by taking K = 0, the zero function (y = 0)is always a solution to y + p(t )y = 0; in addition, the homogeneous linearfirst-order DE has infinitely many solutions. This is very similar to how, for agiven matrix A, the homogeneous equation Ax = 0 always has the zero vectoras a solution and, in the case where A is singular, Ax = 0 has infinitely manysolutions. Having now completely addressed the case of a homogeneous linear first-order DE, we turn to the nonhomogeneous case. In particular, we are interestedin solving the equation y + p(t )y = f (t ) (2.3.7)where f (t ) is not identical to zero. Recalling the product rule from calculus, d [v(t ) · y ] = v(t )y + v (t )y (2.3.8) dtwe observe that the left-hand side of (2.3.7), y + p(t )y, looks similar to theright-hand side of (2.3.8). If we multiply both sides of (2.3.7) by an unknown
Linear first-order differential equations 143There are several important observations to make from our work inexample 2.3.2. First, the parentheses at (2.3.14) are essential. Without them,e −2t is not multiplied by the entire antiderivative, and the function y would nolonger be a solution to the given DE. A second is that if we had instead solved the corresponding homogeneousdifferential equation y + 2y = 0, we would have found the so-called com-plementary solution yh = Ce −2t . Moreover, by observing that y = 4 − 2y =2(2 − y), if we consider the function yp = 2, it is apparent that yp is asolution to the nonhomogeneous equation y + 2y = 4. In addition, if weomit the constant of integration C in (2.3.14), it follows that the methodderived in (2.3.13) can be viewed as producing a so-called particular solutionyp that is a solution to the given nonhomogeneous linear first-order differentialequation. Thus we see that the method derived in (2.3.13) and implemented tofind (2.3.15) ultimately expresses the solution to the original nonhomogeneouslinear first-order DE in the form y = yp + yhwhere yp is a particular solution to the nonhomogeneous equation, while yh isthe complementary solution, the solution to the corresponding homogeneousequation. This situation reminds us of one way to view the general solution to a systemof linear equations given by Ax = b, where in (1.5.1) in section 1.5 we foundthat x = xp + xh . A further discussion of this property of linear first-order DEswill occur in theorem 2.3.2 to close the current section. Before doing so, weconsider another example.Example 2.3.3 Solve the nonhomogeneous first-order linear differentialequation y + y tan t = cos tIn addition, solve the initial-value problem (IVP) that is given by the same DEand the initial condition y(π/3) = 1.Solution. We first determine the integrating factor v(t ). Since p(t ) = tan t , itfollows that P(t ) = tan t dt = − ln(cos t )Thus, v(t ) = e − ln(cos t ) . Applying the integrating factor and using properties ofexponential and logarithmic functions, we now observe that y = e ln(cos t ) cos t · e − ln(cos t ) dt 1 = cos t cos t dt cos t
Applications of linear first-order differential equations 1472.4 Applications of linear first-order differential equationsA large number of important physical situations can be modeled by linearfirst-order differential equations. In this section we introduce several suchapplications through examples and explore further scenarios in the exercises.2.4.1 Mixing problemsRecall that in section 2.1, we encountered a problem where a saltwater solutionwas entering and exiting a city's water reservoir. Specifically, in (2.1.1) weencountered the DE dA A(t ) = 10 − dt 10This equation, rewritten in the form 1 A + A = 10 10is a linear first-order DE that we now can easily solve. With p(t ) = 1/10, theintegrating factor is v(t ) = e t /10 , and therefore A = e −t /10 e t /10 · 10 dt (2.4.1) = e −t /10 (100e t /10 + C) (2.4.2) −t /10 = 100 + Ce (2.4.3)From this result, we can also confirm our previous observation that as t → ∞,A(t ) → 100, for any solution A(t ) to the differential equation. Moreover, if weconsider the initial condition A(0) = 200 stated along with the original problemin section 2.1, it follows that A(t ) = 100 + 100e −t /10 Certainly we can consider a wide range of variations on this mixingproblem by changing concentrations, flow rates, and tank volumes. In everysuch scenario, the most important thing to keep in mind is that the rate ofchange of salt (or whatever quantity is under consideration) is the differencebetween the rate of salt entering and the rate exiting. Furthermore, an analysisof units is often very helpful. We consider one more example to demonstratewhat can occur when the entering and exiting solutions are flowing at differentrates.Example 2.4.1 Consider a tank in which 1 g of chlorine is initially present in100 m3 of a solution of water and chlorine. A chlorine solution concentrated at0.03 g/m3 flows into the tank at a rate of 1 m3 /min, while the uniformly mixedsolution exits the tank at 2 m3 /min. At what time is the maximum amount ofchlorine present in the tank, and how much is present?
148 First-order differential equationsSolution. To answer the questions posed, we set up and solve an IVP. We letA(t ) denote the amount of chlorine in the tank (in grams) at time t (in minutes).We note from the inflow that the rate at which chlorine is entering the tank isgiven by m3 g rate in = 1 · 0.03 3 (2.4.4) min mFor the exiting flow, we must compute the concentration of chlorine present inthe solution leaving the tank. This concentration is given by the ratio of amountpresent in grams to the total volume of solution in the tank at time t . In thisproblem, note that the volume is changing as a function of time. In particular,since solution enters at 1 m3 /min and exits at 2 m3 /min, it follows that thevolume V (t ) of solution present in the tank is decreasing at a rate of 1 m3 /min.With 100 m3 initially present, we observe that V (t ) = 100 − t is the volume ofsolution in the tank at time t . Thus, the concentration of chlorine in the solutionexiting the tank at time t is given by m3 A(t ) g 2 · A(t ) g rate out = 2 · = (2.4.5) min V (t ) m3 100 − t minIt follows from (2.4.4) and (2.4.5) that the overall instantaneous rate of changeof chlorine in the tank with respect to time is dA 2A = rate in − rate out = 0.03 − dt 100 − tNote that we also have the initial condition A(0) = 1. Rearranging the differentialequation, we see that we must solve the nonhomogeneous linear first-orderequation 2 A + A = 0.03 (2.4.6) 100 − tApplying the approach discussed in section 2.3, followed by the initial condition,it can be shown that the solution to (2.4.6) is A(t ) = 3 − 0.03t − 0.0002(100 − t )2From the quadratic nature of this solution, as well as from the direction fieldshown in figure 2.4, we can see that this function has a maximum value. Itis a straightforward exercise to show that this maximum of A(t ) occurs whent = 25 min and that the maximum is A = 1.125 g.2.4.2 Exponential growth and decayA radioactive substance emits particles; in doing so, the substance decreases itsmass. This process is known as radioactive decay. For example, the radioactiveisotope carbon-14 emits particles and loses half its mass over a period of5730 years. For any such isotope, the instantaneous rate of decay is proportionalto the mass of the substance present at that instant. Thus, assuming an initial
Applications of linear first-order differential equations 149 A(t) 2.0 1.0 t 50 100 Figure 2.4 Direction field for (2.4.6) with solution corresponding to the initial condition A(0) = 1.mass M0 is present, it follows that the mass M (t ) of the substance at time t mustsatisfy the initial-value problem M = −kM , M (0) = M0 (2.4.7)for some positive constant k. Note that the minus sign is present in (2.4.7) sincethe mass M (t ) is decreasing. It follows from our work with homogeneous linearfirst-order DEs in section 2.3 that the solution to this equation is M (t ) = M0 e −kt (2.4.8)Similarly, experiments show that a population with zero death rate (e.g., a colonyof bacteria with sufficient food and no predators) grows at a rate proportionalto the size of the population at time t . In particular, if P(t ) is the populationpresent at time t and P0 is the initial population, then P satisfies the initial-value problem P = kP, P(0) = P0 , for some positive constant k. Here, itfollows that P(t ) = P0 e kt (2.4.9) Problems involving radioactive decay and exponential population growthare very similar and should be familiar to students from past courses in calculusand precalculus. We include one example here for review and several more inthe exercises at the end of the section.Example 2.4.2 A radioactive isotope initially has 40 g of mass. After 10 daysof radioactive decay, its mass is 39.7 g. What is the isotope's half-life? At whattime t will 1 g remain?Solution. Because the isotope decays radioactively, we know that its massM (t ) must have the form M (t ) = M0 e −kt . To answer the questions posed, wemust first determine the constant k. In the given problem, we know that M0 = 40
150 First-order differential equationsand that M (10) = 39.7. It follows that 39.7 = 40e −10kDividing both sides of the equation by 40, taking natural logs, and solving for k,we find that 1 39.7 k = − ln 10 40 To compute the half life, we now solve the equation M0 = M0 e −kt 2for t . In particular, we have 1 39.7 20 = 40e 10 ln 40 tDividing by 40 and taking natural logs, 1 1 39.7 ln = ln t 2 10 40so 1 ln 2 t= 1 39.7 10 ln 40Thus the half-life of the isotope is approximately 921 days. Finally, to determine when 1 g of the substance will remain, we simply solvethe equation 1 39.7 1 = 40e 10 ln 40 tDoing so shows that t ≈ 4900 days.2.4.3 Newton's law of CoolingSuppose that T (t ) is the temperature of a body immersed in a coolersurrounding medium such as air or water. Sir Isaac Newton postulated (andexperiments confirm) that the body will lose heat at a rate proportional tothe difference between its present temperature and the temperature of itssurroundings. If we assume that the temperature of the surrounding medium isconstant, say Tm , and that the warmer body's initial temperature is T (0) = T0 ,then Newton's law of Cooling can be expressed through the initial-value problem T = −k(T − Tm ), T (0) = T0 (2.4.10)Written in the standard form of a nonhomogeneous linear first-order DE, wefind that T satisfies the IVP T + kT = kTm , T (0) = T0 (2.4.11)
Applications of linear first-order differential equations 151Solving this problem in the standard way reveals that the temperature of thecooling body must satisfy T (t ) = (T0 − Tm )e −kt + Tm (2.4.12)We consider an example with some particular details given in order to analyzethe behavior of the temperature function.Example 2.4.3 A can of soda at room temperature 70◦ F is placed in arefrigerator that maintains a constant temperature of 40◦ F. After 1 hour inthe refrigerator, the temperature of the soda is 58◦ F. At what time will the soda'stemperature be 41◦ F?Solution. Let T (t ) denote the temperature of the soda at time t in degreesF; note that T0 = 70. Since the surrounding temperature is 40, T satisfies theinitial-value problem T = −k(T − 40), T (0) = 70and therefore by (2.4.12) T has the form T (t ) = 30e −kt + 40In particular, note that the temperature is decreasing exponentially as timeincreases and tending towards 40◦ F, the temperature of the refrigerator, ast → ∞. To determine the constant k, we use the additional given information thatT (1) = 58, and therefore 58 = 30e −k + 40It follows that e −k = 3/5, and thus k = ln(5/3). To now answer the originalquestion, we solve the equation 41 = 30e − ln(5/3)t + 40and find that t = ln(30)/ ln(5/3) ≈ 6.658 h.Exercises 2.4 1. A population of bacteria is growing at a rate proportional to the number of cells present at time t . If initially 100 million cells are present and after 6 hours 300 million cells are present, what is the doubling time of the population? At what time will 100 billion cells be present? 2. The half-life of a radioactive element is 2000 years. What percentage of its original mass is left after 10 000 years? After 11 000 years? 3. The evaporation rate of moisture from a sheet hung on a clothesline is proportional to the sheet's moisture content. If one half of the moisture evaporates in the first 30 min, how long will it take for 95 percent of the moisture to evaporate?
152 First-order differential equations 4. A population of 200 million people is observed to grow at a rate proportional to the population present and to be increasing at a rate of 2 percent per year. How long will it take for the population to triple? 5. In a certain lake, wildlife biologists determine that the walleye population is growing very slowly. In particular, they conclude that the population growth is modeled by the differential equation P = 0.002P, where P is measured in thousands of walleye, and time t is measured in years. The biologists estimate that the initial population of walleye in the lake is 100 000 fish. To enhance the fishery, the department of conservation begins planting walleye fingerlings in the lake at a rate of 5000 walleye per year. (a) Write an IVP that the population P(t ) of walleye in the lake in year t will satisfy under the assumption that walleye are being added to the lake at a rate of 5000 fish per year. (b) Solve the IVP stated in (a). (c) In 20 years, how many more walleye will be in the lake than if the biologists had not planted any fish? 6. Solve the IVP A = 0.03 − 2/(100 − t ) A, A(0) = 1, in order to verify the stated solution in example 2.4.1. 7. Brine (saltwater) is entering a 25 m3 tank at flow rate of 0.25 m3 /min and at a concentration of 6 g/m3 . The uniformly mixed solution exits the tank at a rate of 0.25 m3 /min. Assume that initially there are 15 m3 of solution in the tank at a concentration of 3 g/m3 . (a) State an IVP that is satisfied by A(t ), the amount of salt in grams in the tank at time t . (b) What will happen to the amount of salt in the tank as t → ∞? Why? (c) Plot a direction field for the IVP stated in (a), including a plot of the solution. (d) At exactly what time will there be 75 g of salt present in the tank? 8. Brine is entering a 25-m3 tank at flow rate of 0.5 m3 /min and at a concentration of 6 g/m3 . The uniformly mixed solution exits the tank at a rate of 0.25 m3 /min. Assume that initially there are 5 m3 of solution in the tank at a concentration of 25 g/m3 . (a) State an IVP that is satisfied by the amount of salt A(t ) in grams in the tank at time t . (b) Solve the IVP stated in (a). For what values of t is this problem valid? Why? (c) At exactly what time will the least amount of salt be present in the tank? How much salt will there be at that time? (d) Plot a direction field for the IVP stated in (a), including a plot of the solution. Discuss why this direction field and the solution make sense in the physical context of the problem.
Applications of linear first-order differential equations 153 9. A body of water is polluted with mercury. The lake has a volume of 200 million cubic meters and mercury is present in a concentration of 5 grams per million cubic meters. Health officials state that any level above 1 g per million cubic meters is considered unsafe. If water unpolluted by mercury flows into the lake at a rate of 0.5 million cubic meters per day, and uniformly mixed lake water flows out of the lake at the same rate, how long will it take for the lake to reach a mercury concentration that is considered safe?10. An average person takes eighteen breaths per minute and each breath exhales 0.0016 m3 of air that contains 4 percent more carbon dioxide (CO2 ) than was inhaled. At the start of a seminar containing 300 participants, the room air contains 0.4 percent CO2 . The ventilation system delivers 10 m3 of fresh air per minute to the room whose volume is 1500 m3 . Find an expression for the concentration level of CO2 in the room as a function of time; assume that air is leaving the room at the same rate that it enters.11. Solve the general Newton's law of Cooling IVP T = −k(T − Tm ), T (0) = T0 in order to verify the solution stated in (2.4.12).12. A potato at room temperature of 72◦ F is placed in an oven set at 350◦ F. After 30 min, the potato's temperature is 105◦ F. At what time will the potato reach a temperature of 165◦ F?13. An object at a temperature of 80◦ C is placed in a refrigerator maintained at 5◦ C. If the temperature of the object is 75◦ C at 20 min after it is placed in the refrigerator, determine the time (in hours) the object will reach 10◦ C.14. An object at a temperature of 9◦ C is placed in a refrigerator that is initially at 5◦ C. At the same time the object is placed in the refrigerator, the refrigerator's thermostat is adjusted in order to raise the temperature inside from 5◦ C to 10◦ C; the function that governs the temperature of the 10 refrigerator is R(t ) = . 1 + e −0.75t (a) Using the refrigerator's temperature constant k from exercise 13, modify Newton's law of Cooling appropriately to state an IVP whose solution is the temperature of the object. (b) Plot a direction field for the IVP from (a) and sketch an approximate solution to the IVP. (c) Discuss the qualitative behavior of the solution to the IVP. Estimate the minimum temperature the object achieves.15. On a cold, winter evening with an outdoor temperature of 4◦ F, a home's furnace fails at 10 pm. At the time of the furnace failure, the indoor temperature was 68◦ F. At 2 am, the indoor temperature was 60◦ F.
154 First-order differential equations Assuming the outside temperature remains constant, at what time will the homeowner have to begin to worry about pipes freezing due to an indoor temperature below 32◦ F?2.5 Nonlinear first-order differential equationsSo far in our work with differential equations, we have seen that linear first-order differential equations have many interesting properties. One is that anyIVP that corresponds to a linear first-order DE (with reasonably well-behavedfunctions p(t ) and f (t )) is guaranteed to have a unique solution. In addition,through our development of integrating factors, we have a method by which wecan always (at least in theory) determine a solution for the differential equation. Any differential equation that is not linear is called nonlinear. Thus,nonlinear differential equations constitute every other type of equation wecan conceive. Unfortunately, nonlinear equations are (in general) far moredifficult to solve than linear ones. We will limit ourselves in this section toconsidering a few relatively common special cases of nonlinear first-orderdifferential equations that can be solved analytically. In section 2.6, we willconsider qualitative and approximation techniques that enable us to gainvaluable information from a nonlinear initial-value problem, even in the eventthat we cannot solve it explicitly.2.5.1 Separable equationsIn example 2.3.1 in section 2.3, we solved the differential equation y = −(1 +t 2 )y. While this equation is linear, our method provides insight into how toapproach a class of nonlinear equations whose structure is similar. We begin byconsidering a slightly modified example.Example 2.5.1 Solve the nonlinear first-order differential equation y = −(1 + t 2 )y 2 (2.5.1)Solution. Following our approach in example 2.3.1, we can separate thevariables y and t algebraically to arrive at the equation dy y −2 = −1 − 3t 2 dtIntegrating both sides of this equation with respect to t , dy (y(t ))−2 dt = (−1 − 3t 2 ) dt (2.5.2) dtThe left-hand side may be simplified to y −2 dy. Thus, evaluating each integralin (2.5.2), we find that −y −1 = −t − t 3 + C (2.5.3)
Nonlinear first-order differential equations 155We note again that since an arbitrary constant of integration arises on each side,it suffices to include just one. It is essential here to observe that by successfullyintegrating, we have removed the presence of y in the equation, and now haveonly an algebraic, rather than differential, equation in t and y. Solving (2.5.3)algebraically for y, it follows 1 y= t+ 3t − C 1 3The strategy of example 2.5.1 may be applied to any differential equation ofthe form y = f (t , y) where f (t , y) can be decomposed into a product of twofunctions of t and y only. That is, if we can write f (t , y) = g (t ) · h(y)then we are able to separate the variables in the equation, writing all of they-terms on one side (multiplied by y ), and writing all of the t -terms on theother. Any differential equation of the form y = g (t ) · h(y) is said to be separable.We attempt to solve a separable differential equation by separating the variablesand writing 1 y = g (t ) (2.5.4) h(y)Writing y in the alternate notation dy /dt , we have 1 dy = g (t ) (2.5.5) h(y) dtHence when we integrate both sides of (2.5.5) with respect to t , we find 1 dy = g (t ) dt h(y)Now, all of this work is only useful if we arrive at integrals we can actually √evaluate. For example, if the left-hand side is sin y dy, we are really no closerto solving for y than we were when considering the initial differential equation. In section 2.6, we will address ways to approximate the solution of suchequations that we seem unable to solve analytically. For now, we consider a fewexamples of separable equations that we can solve, with more to follow in theexercises.Example 2.5.2 Find a family of solutions to the differential equation y = e t +2y tand a solution to the corresponding initial-value problem with the conditionthat y(1) = 1.
156 First-order differential equationsSolution. First, we may write e t +2y = e t e 2y . Thus, we have y = e t e 2y tSeparating the variables, it follows that dy e −2y = te t dtIntegrating both sides with respect to t , we may now write e −2y dy = te t dtUsing integration by parts on the right and evaluating both integrals, we have 1 − e −2y = (t − 1)e t + C 2To now solve algebraically for y, we first multiply both sides by −2. Since C isan arbitrary constant, −2C is just another constant, one that we will denote byC1 . Hence e −2y = −2(t − 1)e t + C1Taking logarithms and solving for y, we can conclude that 1 y = − ln(−2(t − 1)e t + C1 ) 2is the family of functions that provides the general solution to the original DE. To solve the corresponding IVP with y(1) = 1, we observe that 1 1 y(1) = − ln(−2(1 − 1)e t + C1 ) = − ln(C1 ) = 1 2 2so ln(C1 ) = −2, and therefore C1 = e −2 . The solution to the IVP is 1 y = − ln(−2(t − 1)e 1 + e −2 ) 2Example 2.5.3 Is the following differential equation linear or nonlinear? ty + y 2 = 4Classify the equation, and solve it to find a general family of solutions.Solution. We note that the given equation is nonlinear due to the presence ofy 2 in the equation; said differently, the left-hand side is not a linear combinationof y and y . To separate the variables, we first write ty = 4 − y 2Dividing both sides by t (4 − y 2 ), it follows that 1 dy 1 = 4−y 2 dt t
158 First-order differential equationsof the form d φ = 0, from which it follows that φ (t , y) = K , for some constantK . Assuming that we can find the function φ (t , y), we have then transformedthe original differential equation in t and y to an algebraic equation in t and y,one that we can hopefully solve for y. In the current example, let us suppose that such a function φ (t , y) exists,and therefore that ∂φ = ty 2 (2.5.7) ∂tand ∂φ = 2 + t 2y (2.5.8) ∂yIntegrating both sides of (2.5.7) with respect to t , it follows that 1 φ (t , y) = t 2 y 2 + g (y) 2The function g (y) arises since the partial derivative with respect to t of anyfunction of only y is zero. For φ to satisfy the condition in (2.5.8), we see thatwe must take the partial derivative with respect to y of our most recent resultand set this equal to 2 + t 2 y. Doing so, we find that ∂φ = t 2 y + g (y) = 2 + t 2 y ∂yTherefore, g (y) = 2, so g (y) = 2y, and we have found that 1 φ (t , y) = t 2 y 2 + 2y 2Since it is the case that d φ = 0, we know that φ (t , y) = K , and therefore t andy are related by the algebraic equation 1 2 2 t y + 2y = K 2From the quadratic formula, it follows that √ −2 ± 4 + 2Kt 2 y= t2and we have solved the original equation. The choice of "+" or "−" in thesolution would depend on the value given in an initial condition. There are several important lessons to learn from this example. One is someterminology. If a differential equation can be written in the form M (t , y)dt + N (t , y)dy = 0 (2.5.9)and there exists a function φ (t , y) such that φt (t , y) = M (t , y) and φy (t , y) =N (t , y), then since the differential equation is of the form d φ (t , y) = 0, we saythat the equation is exact. So, certainly a first check of whether an equation might be exact consistsin trying to write it in the form of (2.5.9). Still, there is the issue of whether or
Nonlinear first-order differential equations 159not φ exists. If φ does exist, and we further assume that M (t , y) and N (t , y)have continuous first-order partial derivatives, then it follows from Clairaut'sTheorem in multivariable calculus that My (t , y) = φty = φyt = Nt (t , y)Thus, if (2.5.9) is exact, then it must be the case that My = Nt . Said differently,if My = Nt , then the differential equation is not exact. In fact, it turns out thatif My = Nt , then the equation is guaranteed to be exact, but this result is muchmore difficult to prove. As a consequence of this, it suffices for us to check ifMy = Nt as a first step; if so, the equation is indeed exact and we then proceedto try to find the function φ in order to solve the differential equation. If not,another approach is needed. An example is instructive.Example 2.5.4 Solve the differential equation t y + ln(ty) + 1 = 0 ySolution. We begin by observing that this equation is neither linear norseparable. Thus, writing the derivative in differential notation, we have t dy + ln(ty) + 1 = 0 y dtand then rearranging algebraically, t (ln(ty) + 1)dt + dy = 0 (2.5.10) yLetting M (t , y) = ln(ty) + 1 and N (t , y) = t /y, we observe that 1 1 1 My = t = and Nt = ty y yand therefore, My = Nt . Hence the differential equation is exact and we canassume that a function φ exists such that φt = M (t , y) and φy = N (t , y). Since the latter equation is more elementary, we consider φy = t /y, andintegrate both sides with respect to y. Doing so, we find that φ (t , y) = t ln y + h(t ) (2.5.11)From (2.5.10), φ must also satisfy φt = ln(ty) + 1, so we take the partial derivativeof both sides of (2.5.11) with respect to t to find that φt = ln y + h (t ) = ln(ty) + 1From this and properties of the logarithm, we observe that ln y + h (t ) = ln t + ln y + 1and thus h (t ) = ln t + 1. It follows (integrating by parts and simplifying) thath(t ) = t ln t . Thus, we have demonstrated that the original equation is indeed
162 First-order differential equations41. (y + t )y + y = t , y(0) = 142. y sin 2t + 2y cos 2t = 0, y(π/4) = 1/2 √43. Consider the IVP y = y, y(0) = 0. Show that this IVP has more than one solution. Does this result contradict theorem 2.2.1?2.6 Euler's methodWhile we have learned to solve certain classes of differential equationsexplicitly—including linear first-order, separable, and exact equations—wemust also develop the ability to estimate solutions to initial-value problems thatwe cannot solve analytically. Direction fields will play a key role in motivatingour work, as we see in the following introductory example. Consider the initial-value problem dy + y 2 = t , y(0) = 1 (2.6.1) dtThis DE is not linear due to the presence of y 2 . In addition, since we canwrite y = t − y 2 , we see that the right-hand side may not be expressed as aproduct of two functions that each involve just one of the variables t and y.Thus, the equation is not separable. Finally, writing the equation in the formdy + (y 2 − t )dt = 0, it is straightforward to check that this equation is notexact. While it may seem frustrating to not be able to use any of the solutionmethods we have discussed so far, it is important to realize that many differentialequations cannot be solved explicitly by analytic techniques. As such, we mustexplore how we can use our understanding of derivatives to estimate certainvalues of the solution to an IVP. For the given DE, writing y = t − y 2 , we can generate the direction field thatis shown in figure 2.5. For the initial condition y(0) = 1, visually estimating howthe solution y(t ) will flow through the direction field, we can roughly estimatethat y(1/2) ≈ 0.75. But if we think about the calculus underpinnings of slopefields, we can be much more precise in our estimate. Recall that a direction field for a DE y = f (t , y) is created by observing thatthe slope of the tangent line to the solution curve y(t ) at the point (t0 , y0 ) isf (t0 , y0 ). In the current example, we know that the solution to the IVP must passthrough the point (t0 , y0 ) = (0, 1). At this point, the slope of the tangent line tothe solution curve is m = 0 − 12 = −1; note also that m ≈ y / t , where y isthe exact change in y from t = 0 to t = 1/2, due to the fact that the tangent lineapproximates the solution curve for values near the point of tangency. Thus, aswe step from t0 = 0 to t = 1/2, a change of 1/2 in the t -direction will generatean approximate change y = t · m = 1/2 · (−1) = −1/2 in y. Therefore, fromour original y-value of 1, a change of −1/2 leads us to the approximation thaty(1/2) ≈ 1/2.
Euler's method 163 y(t) 2 1 t −1 1 2 3 −1 Figure 2.5 The direction field for (2.6.1). y(t) 2.0 1.5 1.0 0.5 t 0.5 1.0 1.5 2.0 Figure 2.6 Taking one step to esti- mate y(0.5) in (2.6.1). Graphically, this estimation approach amounts to following the tangentline to the solution curve for some prescribed change in t . We can see this infigure 2.6, where it is immediately evident that our estimate is too small. Incalculus, we learn that while the tangent line approximation to a differentiablefunction is good near the point of tangency, the approximation gets poorerand poorer the further we move from the point of tangency. Thus, a naturalapproach to the estimation problem at hand is to take a smaller step, thensearch the direction field for a new direction to follow, and then take anothersmall step. In this situation, we are much like a hiker lost in the woods who isattempting to navigate by compass: just as the hiker is best served by checking acompass frequently, so are we best served by checking slopes frequently.
164 First-order differential equations y(t) 2.0 1.5 1.0 0.5 t 0.5 1.0 1.5 2.0 Figure 2.7 Two steps of size 0.25 to estimate y(0.5) in (2.6.1). So, rather than stepping the full distance of 1/2 from t = 0 to t = 1/2, letus first step to t = 1/4, find an estimate to y(1/4), and then proceed from thereto estimate y(1/2). Starting at (0, 1), we know that the slope of the tangent lineto the solution curve at this point is m0 = f (0, 1) = −1. Stepping t = 0.25, itfollows that we experience a change in y along the tangent line of y = m0 t =−1(0.25) = −0.25. Thus, we have that y(0.25) ≈ y(0) + y = 1 − 0.25 = 0.75. Now we repeat this process from the point (0.25, 0.75). At this point, theslope of the tangent line to the solution curve is m1 = f (0.25, 0.75) = 0.25 −(0.75)2 = −0.3125. Taking a step of t = 0.25, it follows that the change iny along the tangent line will be y = m1 t = −0.3125(0.25) = −0.078125.Thus we have that y(0.5) ≈ 0.75 − 0.078125 = 0.671875. We record our workgraphically in figure 2.7, where our improved approximation is apparent, thoughthe estimate is still too small.It is evident from our work in this first example that we can significantlyimprove our ability to estimate an initial-value problem's solution at varioust -values by developing an iterative process that uses reasonably small step sizes.In particular, we want to imitate the way in which we took two steps, but ratherbe able to take n steps using a step-size of t = h. Throughout, the key idea isalways that we are estimating the solution function by determining its tangentline at a given point, and then following the tangent line for the determined stepsize. We observe that when moving along any line from a given point (told , yold )to a new point (tnew , ynew ), it follows that ynew = yold + y y = yold + · t t = yold + m · t (2.6.2)
Euler's method 165Another essential observation to make is that the slope m at each step of ourapproximation is given by m = y = f (t , y) in the differential equation that weare attempting to solve. In particular, if we have some approximation at time tkgiven by yk , the slope of the tangent line to the solution curve at this point isgiven by f (tk , yk ). Therefore, using this value for m in (2.6.2) and letting h = tbe the step size, we now have ynew = yold + hf (told , yold ) (2.6.3)Hence, starting from the initial condition (t0 , y0 ), we are able to generate thesequence of points (t1 , y1 ), . . . , (tn , yn ), where for each n ≥ 0, tn+1 = tn + h and yn+1 = yn + hf (tn , yn ) (2.6.4)The value yn is an approximation of the exact solution value y(tn ) at each step,so that yn ≈ y(tn ) for each n ≥ 1. This method of approximating the solution toan initial-value problem is known as Euler's method.Example 2.6.1 For the initial-value problem dy + y 2 = t , y(0) = 1 dtthat we have just considered, apply Euler's method to estimate the value ofy(1/2) using h = 0.1.Solution. At the end of this section, the implementation of Euler's method ina spreadsheet such as Excel will be discussed. Here, we simply report the resultsof such a computer implementation. If we use a step size of h = 0.1, we see thatwe will take five steps to move from t0 = 0 to t5 = 0.5, the point at which weseek to approximate y. Doing so yields the output shown in table 2.1. With just five steps, we can see in the direction field in figure 2.8, togetherwith a piecewise linear plot of the approximate solution, that we have anapparently good estimate in the above table for how the actual solution tothis IVP behaves on this interval.In the example we have been considering with various step sizes, oneshortcoming is that we do not have a precise sense of how accurate our Table 2.1 Euler's method applied to the IVP y = t − y 2 , y(0) = 1, using h = 0.1 tn yn 0 1 0.1 0.9 0.2 0.829 0.3 0.7802759 0.4 0.749392852 0.5 0.733233887
166 First-order differential equations y(t) 2.0 1.5 1.0 0.5 t 0.5 1.0 1.5 2.0 Figure 2.8 Five steps of size h = 0.1 to estimate y(0.5).approximations are. One way to explore this issue is to apply Euler's method toan IVP that we can solve exactly, and then compare our estimates with actualsolution values. We do so in the following example.Example 2.6.2 Solve the IVP y = y − t , y(0) = 0.5 exactly, and use Euler'smethod with the step sizes h = 0.2 and h = 0.1 to estimate the value of y(1).Hence analyze the effect that step size has on error in the method.Solution. We first observe that y = y − t is a linear first-order DE. Applyingour work from section 2.3, we can determine that the solution to this equationis y = 1 + t + Ce t . The initial condition y(0) = 0.5 then implies that C = −1/2,so that the solution to the IVP is et y(t ) = 1 + t − 2If we apply Euler's method with h = 0.2 and take 5 steps to determine ynat each, and also evaluate y(tn ) at each stage, the resulting output is shownin table 2.2. Here, we observe the obvious pattern that the further we step away fromthe initial condition, the greater the error we encounter. This is a naturalconsequence of the use of linear approximations. To get a further sense of how the error at a given step depends on step size,we now apply the same method with h = 0.1. Doing so produces the results intable 2.3. For ease of display and comparison to the case where h = 0.2, we onlyreport the results from every other step. By comparing the approximations in the preceding two tables at thecommon values of t = 0.2, 0.4, 0.8, 1 we can see that cutting the step size inhalf appears to have reduced the error by a factor of approximately 2.
168 First-order differential equationsIn a given row of the spreadsheet, we will view the data (as labeled in the cellsbelow) step number n, step size h, t -value tn , approximate current y-value yn ,slope f (tn , yn ), and updated y-value yn+1 . We will demonstrate the development of such an Excel spreadsheet for theparticular example y = t − y 2 , y(0) = 1 using a step size of h = 0.1. To begin,we establish names for the various columns, say in cells A1, B1, C1, D1, E1,and F1, as shown below by entering the text "n", "h", etc., in the respective cellsshown below. A B C D E F 1 n h t n y n f(t n,y n) y n+1In row 2, we now enter the given data at step zero. In particular, in cell A2 weenter the step number ("0"), in B2 the chosen step size ("0.1"), in C2 thestarting t -value ("0"), in D2 the starting y-value ("1"), and in E2, we apply thefunction f (t , y) to get the slope at the point at this step. That is, since in this IVPf (t , y) = t − y 2 , we enter in E2 the command "=C2 - D2ˆ2". We now alsohave enough information entered to compute y1 in cell F2. Using the rule fromEuler's method, we know y1 = y0 + hf (t0 , y0 ). In our spreadsheet, this implieswe must enter "=D2 + B2*E2". Doing so, the result (y1 = 0.9) appears in cellF2. Now our spreadsheet should appear as shown. A B C D E F 1 n h t n y n f(t n,y n) y n+1 2 0 0.1 0 1 −1 0.9In row 3, we may now build subsequent entries based on existing data. Toincrease the step number, in A3 we enter "=A2 + 1". Since the step-sizestays constant throughout, in B3 we input "=B2". Because the next t -valuewill be the preceding t -value plus the step size (t1 = t0 + h), we enter inC3 the command "=C2 + B2". We also have the next y-value, so in D3we enter "=F2" to have this data available in the given row. The slope atstep 1 is computed according to the same rule (given by f (t , y)) as it was atstep 0. Hence in cell E3 we simply paste a copy of cell E2, which ensuresthat Excel uses the same computations, but updates them for the currentstep. Equivalently, we can directly enter in E3 the text "=C3 - D3ˆ2".Cell F3 computes the newest y-value: the same rule as in step 0 must befollowed, so we can copy and paste cell F2 into F3, or equivalently enter inF3 "=D3 + B3*E3".
Euler's method 169 At this stage, we see on the screen the following. A B C D E F1 n h t n y n f(t n,y n) y n+12 0 0.1 0 1 −1 0.93 1 0.1 0.1 0.9 −0.71 0.829Now we can harness the power of Excel to compute as many subsequent steps aswe like. By using the mouse to highlight row 3 (cells A3 through F3), and thenplacing the cursor on the bottom right corner of cell F3, we can then click anddrag downward to fill subsequent rows with similar calculations. For example,doing so through row 5 (i.e., down to F7) yields the following table. A B C D E F1 n h t n y n f(t n,y n) y n+12 0 0.1 0 1 -1 0.93 1 0.1 0.1 0.9 -0.71 0.8294 2 0.1 0.2 0.829 -0.487241 0.78027595 3 0.1 0.3 0.7802759 -0.30883048 0.7493928526 4 0.1 0.4 0.749392852 -0.161589647 0.7332338877 5 0.1 0.5 0.733233887 -0.037631934 0.729470694Besides the ease of iteration past the first two rows, there are further advantagesExcel offers. One is that changing one appropriately-chosen cell will update allof our computations. For example, if we are interested in the change inducedby a different step size, say h = 0.05, all we need to do is enter "0.05" in cellB2, and every other cell will update accordingly. In addition, if we desire to seethe graphical results of our work, we can use Excel's Chart Wizard. To plot our approximations, we can simultaneously highlight the t andy columns in our chart above (cells C2 through C7 and D2 through D7), andthen go to Insert menu and select Chart (alternatively, we may click on the ChartWizard icon on the toolbar). In the prompt window that arises, we choose "XY(Scatter)" and select one of the graph style options at the right by clicking onthe desired one. By clicking "Next" in a few subsequent windows (in whichadvanced users can avail themselves of more options), we eventually get to afinal window where our graph appears and the option to "Finish." Clicking on"Finish," the graph will appear in the spreadsheet and may be moved around
170 First-order differential equations 1.2 1 0.8 0.6 Series1 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 Figure 2.9 An Excel plot of an approximate solution to the IVP y = t − y 2 , y(0) = 1, for 0 ≤ t ≤ 0.5.by clicking and dragging it accordingly. We see the resulting plot displayed as infigure 2.9.Exercises 2.6 1. Consider the IVP y = t /y, y(1) = 3 (where we assume that y is always positive). (a) Program Excel to use Euler's method to determine an estimate of the value of y(3). Do so using a step size of h = 0.2 2. Consider the IVP y = (1 − t )(1 + y), y(0) = 2. (a) Program Excel to use Euler's method to determine an estimate to the value of y(1.6). Do so using step sizes of h = 0.2 and h = 0.1. Show the results in a table and create an appropriate plot of the approximate solution.
Euler's method 171 3. Consider the IVP y = (t − y)2 /4, y(0) = 1/2. (a) Program Excel to use Euler's method to determine an estimate to the value of y(1.5). Do so using step sizes of h = 0.1 and h = 0.05. Show the results in a table and create an appropriate plot of the approximate solution. (b) Explain why you cannot solve the given IVP explicitly. (c) Use a computer algebra system appropriately to plot a direction field for the given differential equation. By hand, sketch a solution that satisfies the above IVP. Compare your work in (a) to the direction field. 2 4. Consider the IVP y = e t − y, y(1) = 4, t > 0. t (a) Program Excel to use Euler's method to determine an estimate to the value of y(2.2). Do so using step sizes of h = 0.1 and h = 0.05In each of exercises 5–10, find an approximate solution to the stated IVP byusing Euler's method with h = 0.1 on the interval [0, 1]. In addition, find anexact solution and compare the values and plots of the approximate and exactsolutions. 5. y + 2ty = 0, y(0) = −2 6. y = 2y − 1, y(0) = 2 7. y − y = 0, y(0) = 2
172 First-order differential equations 8. (y )2 + 2y = 0, y(0) = 2 9. y y2 = 8, y(0) = 110. (t + 1)yy = −1 − y 2 , y(0) = 2In each of exercises 11–14, find an approximation solution to the stated IVP byusing Euler's method with h = 0.1 on the interval [0, 1]. In addition, explainwhy it is not possible to solve the IVP exactly by established methods.11. (y )2 − 2y 2 = t , y(0) = 212. y − sin y = 2e t , y(0) = 013. y + y 3 = t 3 , y(0) = 214. (t + 1)yy = −1 − y 2 − t 2 , y(0) = 22.7 Applications of nonlinear first-order differential equationsIn this section, we explore two examples of nonlinear differential equations.It is important to recall that if an equation is nonlinear, it is possible that wemay not be able to solve for the solution function explicitly. Regardless, we canuse direction fields to qualitatively understand the behavior of solution curves;furthermore, if we are unable to find an exact solution function, we may employEuler's method to generate approximate solutions.2.7.1 The logistic equationWe have recently learned that if a population is assumed to grow at a constantrelative growth rate (or in a way such that the rate of change of the populationis proportional to the size of the population), then the population functionsatisfies the initial-value problem P = kP , P(0) = P0This leads to the familiar population model P(t ) = P0 e kt , which is also studiedin algebra and calculus courses. While this model is a natural one, it is alsounrealistic: over significant periods of time, the function P will grow to valuesthat become unreasonable since the function exhibits unbounded growth. Therefore, we now explore a more plausible population model. Let usassume we know that a given population P has the tendency over time tolevel off at a value A. The value A is often called the carrying capacity of thepopulation; as the name indicates, it is the maximum population sustainable bythe surrounding environment. It is natural to further assume that if P is closeto, but less than A, then dP /dt will be small and positive, indicating that thepopulation will be growing slowly. Similarly, if P is close to, but greater than A,we will want dP /dt to be negative and close to zero, so that the population willbe decreasing slowly.
Applications of nonlinear first-order differential equations 173 At the same time, we want to maintain the natural inherent exponentialcharacteristic of growth, so when P is relatively small (in comparison to A), wewould like for dP /dt to be approximately kP for some appropriate constant k.The combination of all these criteria led Dutch mathematician Pierre Verhulst(1804–1849) to propose the differential equation dP P = kP 1 − (2.7.1) dt Aas a more realistic model of population growth, where k and A are positiveconstants. Equation (2.7.1) is known as the logistic differential equation. That the logistic equation may be solved in general (to determine an explicitsolution P involving k and A) will be shown in the exercises. We consider herea specific example where k and A are given to provide further insight into thebehavior of solutions to this equation.Example 2.7.1 A population P(t ) exhibits logistic growth according to themodel dP P = 0.05P 1 − , P(0) = 10 dt 75(a) Determine the values of P for which P is an increasing function(b) Plot the direction field for the differential equation(c) Determine the value(s) of P for which P is increasing most rapidly(d) Solve the IVP explicitly for PSolution.(a) To determine where P is increasing, we require that dP /dt > 0. If P < 0, note that (1 − P /75) > 0, which makes dP /dt < 0, so we need P > 0 and (1 − P /75) > 0 to make dP /dt positive. This occurs on the interval 0 < P < 75, so for these P values, P is an increasing function of t . We note further that if P > 75 or P < 0, then dP /dt < 0 and P is a decreasing function. Finally, it is evident that both P = 0 and P = 75 are equilibrium solutions, which makes sense given the physical interpretation of the population model.(b) Using familiar commands in Maple, we can plot the direction field for this differential equation. Note in advance the behavior we expect from our work above: two equilibrium solutions at 0 and 75, plus certain increasing and decreasing behavior. Finally, note that our analysis of the equation suggests a good range of values to select for P when plotting, say, P = −10 . . . 100. As always, some experimentation with t may be necessary to get a useful plot. The plot is shown in figure 2.10.
174 First-order differential equations P(t) 100 75 50 25 t 25 50 75 100 Figure 2.10 The slope field for dP /dt = 0.05P(1 − P /75).(c) To decide where P is increasing most rapidly, we seek the maximum value of P . Graphically, we can observe in figure 2.10 that this appears to occur approximately halfway between P = 0 and P = 75. This is reasonable in light of the physical meaning of the logistic equation, since at this point the population has accumulated some substantial numbers to increase its growth rate, while not being close enough to the carrying capacity to have its growth slowed. We can determine this point of greatest increase in P analytically as well. Note that P = 0.05P(1 − P /75) = 0.05P − 0.0006P 2 , so that P is determined by a quadratic function of P. We have already observed that this quadratic function has zeros at the equilibrium solutions (P = 0 and P = 75), and furthermore, we know that every quadratic function achieves is extremum (a maximum in this case, since the function g (P) = 0.05P − 0.0006P 2 is concave down) at the midpoint of its zeros. Hence, P is maximized precisely when P = 75/2.(d) Our final task is to solve the given initial-value problem explicitly for P. We first solve the differential equation dP = 0.05P (1 − P /75) dt for P. Note that this equation is separable and nonlinear. Separating variables, we first write dP = 0.05dt (2.7.2) P(1 − P /75) Because the left-hand side is a rational function of P, we may use the method of partial fractions to integrate the left-hand side of (2.7.2). Observe that 1 75 = P(1 − P /75) P(75 − P)
Applications of nonlinear first-order differential equations 175Now, letting 75 A B = + P(75 − P) P 75 − Pit follows that A = 1 and B = 1, so that (2.7.2) may now be written as 1 1 − dP = 0.05 dt (2.7.3) P P − 75Integrating both sides of (2.7.3), we find that P must satisfy the equation ln |P | − ln |P − 75| = 0.05t + CUsing a standard property of logarithms, the left-hand side may beexpressed as ln |P |/|P − 75|, and hence using the definition of the naturallogarithm, it follows that P = e 0.05t +C = Ke 0.05t P − 75where K = e C . Since K is an arbitrary constant, the sign of K will absorbthe ± that arises from the presence of the absolute value signs, and thus wemay write P = Ke 0.05t P − 75Multiplying both sides by P − 75 and expanding, we see that P = PKe 0.05t − 75Ke 0.05tand gathering all terms involving P on the left, P(1 − Ke 0.05t ) = −75Ke 0.05tThus, it follows that −75Ke 0.05t P= 1 − Ke 0.05tMultiplying the top and bottom of the right-hand side by −1/(Ke 0.05t ), itfollows that 75 P= 1 − Me −0.05twhere M = 1/K . In this final form, it is evident that as t → ∞, P(t ) → 75,which fits with the given carrying capacity in the original problem. At thispoint, we can use the initial condition P(0) = 10 to solve for M ; doing soresults in the equation 10 = 75/(1 − M ), which yields that M = −13/2,and thus 75 P= 13 −0.05t 1+ 2 eA plot of this function (shown in figure 2.11), along with comparison toour work throughout this example, demonstrates that our solution iscorrect.
176 First-order differential equations y 80 40 t 50 100 150 Figure 2.11 The solution P = 75/ (1 + 13 e −0.05t ) to the IVP dP /dt = 2 0.05P(1 − P /75), P(0) = 10. For the general logistic differential equation dP P = kP 1 − dt Aan argument similar to the one we just completed can be used to show that thesolution to this equation is A P(t ) = , 1 + Me −ktwhere M is a constant that may be determined by an initial condition. This factwill be shown in exercise 1 for this section.2.7.2 Torricelli's lawSuppose that a water tank has a hole in its base with area a, through which wateris flowing. Let h(t ) be the depth of the water and V (t ) be the volume of waterin the tank at time t . At what rates are h(t ) and V (t ) changing? Evangelista Torricelli (1608–1647) discovered what has come to be knownas Torricelli's law, which describes the way water in an open tank will flowthrough a small hole in the bottom. To develop this law, let us consider6how water molecules will rearrange themselves as water exits the tank and therelationship between the potential and kinetic energy of a small mass m of water.The potential energy lost as a small mass m of water falls from a height h > 0 ismgh, where g is the gravitational constant; at the same time, the kinetic energygained as an equal mass m exits the tank is 1 mv 2 , where v is the velocity at 2which the water is flowing. Equating the potential and kinetic energy, we find6 Our approach follows that of R. D. Driver in "Torricelli's law: An Ideal Example of an ElementaryODE," Amer. Math. Monthly, 105(5) (May 1998), pp. 453–455.
Applications of nonlinear first-order differential equations 177that mgh = 1 mv 2 , so that 2 v = 2ghThis model assumes that no friction is present; a slightly more realistic modeltakes a fraction of this velocity, depending on the viscosity. For simplicity, wewill consider the ideal case where friction is not considered. If we now consider the water exiting the tank, it follows that the rate ofchange dV /dt of volume in the tank is determined by the product of the area aof the hole and the exiting water's velocity v. In other words, dV = −av = −a 2gh (2.7.4) dtAt this point, observe that we have related the rate of change of volume to theheight of the water in the tank at time t . Instead, we desire to either relatedV /dt and V or dh /dt and h. Of course, height and volume are related. If weassume that A(y) denotes the tank's cross sectional area at height y, then integralcalculus tells us that the volume of the tank up to height h is given by h V (h) = A(y)dy 0Furthermore, by the Fundamental Theorem of Calculus, differentiating V (h)implies dV /dh = A(h), and thus by the chain rule, dV dV dh dh = = A(h) dt dh dt dtUsing this new expression for dV /dt in (2.7.4), it follows that dh A(h) = −a 2gh (2.7.5) dtwhich is a differential equation in h. In particular, this nonlinear equationpredicts, given a tank of a particular shape (as determined by A(h)) with ahole of area a, the behavior of the function h(t ) that describes the height of thewater at time t . We explore this further in the following example.Example 2.7.2 For a cylindrical tank of height 2 m and radius 0.3 m, filledto the top with water, how long does it take the tank to drain once a hole ofdiameter 4 cm is opened?Solution. In this situation, the cross sectional area A(h) of the tank at heighth is constant because each is a circle of radius 0.3, so that A(h) = 0.09π . Inaddition, the area of the hole in square meters is a = π (0.02)2 = 0.0004π , andthe gravitational constant is g = 9.8 m/s2 . Since we have already established thatA(h)dh /dt = −a 2gh, we therefore conclude that h satisfies the equation dh √ 0.09π = −0.0004π 19.6h dt
Applications of nonlinear first-order differential equations 179 (d) Solve the initial-value problem explicitly for P to show that A P(t ) = 1 + Me −kt and determine M in terms of A and P0 .2. The growth of an animal population is governed by the equation 500 dP = 50 − P P dt where P(t ) is the number of animals in the herd at time t . The initial population is known to be 125. Determine the solution P(t ), sketch its graph, and decide whether there will ever be more than 125 or fewer than 50 animals present.3. Consider the differential equation dP /dt = −0.02P 2 + 0.08P. (a) What are the equilibrium solutions to this equation? (b) Determine whether each equilibrium solution is stable or unstable. (c) At what value of P is the function growing most rapidly? (d) Under the initial condition P(0) = 0.25, determine the time at which P(t ) = 3.4. Consider a fish population that grows according to the model dP = 0.05P − 0.000005P 2 dt where t is measured in years, and P is measured in thousands. (a) Determine the population of fish at time t if initially P(0) = 1000. What is the carrying capacity of the population? (b) Suppose that the fish population is established as growing according to the above model in the absence of fish being removed from the lake. Suppose that harvesting begins at a rate of 20 000 fish per year. How does the differential equation governing the fish population change? Explain. (c) Plot a direction field for the updated differential equation you found in part (b). Discuss the new equilibrium solutions for the fish population. Can you solve the IVP with P(0) = 1000? (d) How would the DE change if wildlife biologists began planting 30 000 fish per year in the lake, and no harvesting occurred?5. Solve the initial-value problem dP = 6 − 7P + P 2 , P(0) = 2 dt Sketch your solution curve P(t ) and explain why it makes sense in light of the equilibrium solutions to the given equation and your understanding of where dP /dt is positive and negative.
180 First-order differential equations 6. A cruise ship leaves port with 2500 vacationers aboard. At the time the boat leaves the dock, ten recent visitors of an amusement park are sick with the flu. Let S(t ) denote the number of people at time t who have had the flu at some time since leaving port. (a) Assuming that the rate at which the flu virus spreads is directly proportional to the product of the number of people who have had the flu times the number of people not yet infected, write a differential equation whose solution is the function S(t ). Explain why the differential equation is a logistic equation. (b) Solve the differential equation you found in (a). Assume that four days into the trip, 150 people have been sick with the flu. Clearly show how all constants are identified, and sketch a graph of your solution curve. (c) How many people have been sick seven days into the trip? How long would the boat have to stay at sea for half the vacationers to get ill? 7. A cylindrical tank of height 4 m and radius 1 m is full of water. A small hole of diameter 1 cm is opened in the bottom of the tank. Use Torricelli's law to determine how long it will take for all the water to drain from the tank. 8. A cylindrical tank of height 1.2 m and radius 30 cm is originally full of water. A small hole is opened in the bottom of the tank, and after 15 min, the water in the tank has dropped 10 cm. According to Torricelli's law, how large is the hole and how long will it take the tank to drain? √ 9. Consider a tank that is generated by taking the curve x = y and revolving it about the y-axis. Assume that the tank is full of water to a depth of 1.2 m and that a hole of diameter 1 cm is opened in the bottom. Use Torricelli's law to determine how long it will take for all the water to drain from the tank.10. Suppose a hemispherical bowl has top radius of 30 cm and at time t = 0 is full of water. At that moment a circular hole with diameter 1.2 mm is opened in the bottom of the tank. Use Torricelli's law to determine how long it will take for all the water to drain from the tank.11. For an open cylindrical tank, Torricelli's law tells us that if a small hole is opened, the height of the water at time t obeys the IVP dh √ = −k h , h(t0 ) = h0 dt where k is a constant that depends on the radius of the tank and the radius of the hole. In this exercise, we will take k = 1. (a) Explain why theorem 2.2.1 does not guarantee a unique solution to the IVP dh √ = − h , h(1) = 0 dt (b) Explain why it is physically impossible to determine the height of the water at time t < 1 in a tank which satisfies h(1) = 0.
For further study 181 (c) Show that for any c < 1, the function 2 1 2c − 2t 1 if t < c h(t ) = 0 if t ≥ c is a solution to the IVP in (a). (d) Explain how the result of (c) can be interpreted physically in light of the time when the tank becomes empty. Compare your findings to those in (a) and (b).2.8 For further study2.8.1 Converting certain second-order DEs to first-order DEsLinear second-order differential equations such as y + p(t )y + q(t )y = f (t ) (2.8.1)will be the focus of upcoming work in chapters 3 and 4. But there are somesecond-order equations we can solve at present. For example, if q(t ) = 0in (2.8.1), then we can perform a process called reduction of order to convert theequation to a first-order one.(a) Consider the second-order equation y + p(t )y = f (t ). Using the substitution u = y , convert the equation to a new first-order DE involving the function u.(b) Use a standard solution technique to state the solution u to the differential equation in (a) in terms of p(t ) and f (t ). (Your answer will involve integrals.)(c) Explain how you would use your result in (b) to find the solution y to the original DE.(d) Use reduction of order to solve each of the following second-order IVPs. (i) y + 2y = 4, y(0) = 2, y (0) = 1 (ii) y + tan(t )y = t , y(0) = 1, y (0) = 0 (iii) y + 2t 1+t 2 y = t 2, y(0) = 0, y (0) = 1 (iv) y + 4−t y = 4 − t , 1 y(0) = 1, y (0) = 1(e) Reduction of order can be performed on certain nonlinear differential equations as well. For instance, suppose that we have an equation of form y = g (y )h(t ) (2.8.2) Show that the substitution u = y converts (2.8.2) to a first-order equation in u. Explain how you would approach solving the new equation in u.
182 First-order differential equations (f) Solve each of the following second-order IVPs. (i) y = (y )2 t 2 , y(0) = 1, y (0) = 0 t + t (y )2 (ii) y = , y(0) = 2, y (0) = 1 y (iii) y = e 2t +y , y(0) = 0, y (0) = 0 (iv) y = y, y(0) = 3, y (0) = 52.8.2 How raindrops fallThe following questions and discussion are based on the article "FallingRaindrops" by Walter J. Meyer7 . When a raindrop falls, various forces act upon it. We explore several differ-ent models that show the importance of adjusting assumptions appropriately tomatch physical conditions. Let us first assume that the only force acting uponthe raindrop is the acceleration due to gravity. Under this assumption, Galileo(1564–1642) hypothesized that the falling raindrop would gain an extra 32 ft/sin velocity for every second for which it falls. In other words, the acceleration ofthe raindrop is constant and equal to 32 ft/sec2 .(a) Let y(t ) denote the distance (in feet) traveled by the rain drop after it has been falling for t seconds. Write an initial-value problem involving y(t ) based on the above assumption. Solve this IVP; be sure to introduce appropriate initial conditions based on the context of the problem.(b) Assuming that the raindrop starts from rest at an elevation of 3000 ft, how long does it take the raindrop to fall to earth? What is the raindrop's velocity when it hits the ground? Why is this model unrealistic? (c) We next must attempt to account for the air resistance the raindrop encounters through a slightly more sophisticated model. For a raindrop having diameter d ≤ 0.00025 ft, this model, sometimes known as Stoke's law, states that the acceleration of the raindrop due to gravity is opposed by an acceleration directly proportional to the velocity of the raindrop at that instant. Suppose that the constant of proportionality is given by c /d 2 , where c ≈ 3.29 × 10−6 ft2 /s is an experimentally determined constant. Write a new IVP (again involving y(t ) and its relevant derivatives) for the raindrop having diameter d. Do not yet attempt to solve this equation. Leave d as an unknown constant.(d) Letting v = y and using the fact that the raindrop starts from rest, convert the IVP in (c) to a first-order IVP involving v. Using d = 0.00012 ft (which can be considered a drizzle), produce a slope field corresponding to the7 See Applications of Calculus, MAA Notes Number 29, pp. 101–111.
For further study 183 differential equation in v. On this slope field, sketch a graphical approximation of the solution to the stated IVP. Describe the behavior of the raindrop's velocity based on the slope field you constructed in the problem above.(e) In the model in (d), we will say that the long-term limiting velocity of the raindrop is its terminal velocity, denoted vterm . Calculate this terminal velocity by using the IVP to answer the following questions: What is the initial velocity of the raindrop? What is the equilibrium solution of the differential equation? What happens to the velocity of the raindrop if it ever reaches the equilibrium value? Why, in view of the differential equation, must the velocity of the raindrop increase from its initial value to the equilibrium value? (f) Use your result from (e) to determine the terminal velocities for raindrops having diameters of 0.00009, 0.00012, and 0.00015 ft, respectively. Graph vterm as a function of d, and comment on the phenomena observed.(g) Solve the IVP from (d) explicitly for v. Graph your solution, and then use your solution to calculate vterm as well.(h) Assuming that a raindrop of diameter 0.00012 ft starts from rest at 3000 ft, how long does it take the raindrop to fall to the ground? What is its velocity at the instant it hits the ground? Do your answers surprise you? Is it raining hard or barely raining when raindrops are this size? (i) When the diameter of the raindrop becomes too large, the force of air resistance on the raindrop becomes so appreciable that Stoke's model loses accuracy as well. This leads to a third model, known as the velocity-squared model. This model states that when a raindrop has diameter d ≥ 0.004 ft, the acceleration due to gravity is opposed by an acceleration directly proportional to the square of the velocity of the raindrop at that instant. Here the constant of proportionality is given by k /d, where k ≈ 0.00046. (j) Repeat questions (c), (d), and (e) for the velocity-squared model. Compare your findings with those of Stoke's model. For example, how do the terminal velocities of small raindrops compare with those of large raindrops? For which type of raindrop, small or large, does the terminal velocity increase more rapidly as a function of diameter?(k) Finally, explicitly solve the IVP arising from the velocity-squared model for the velocity function v(t ). Graph your solution v(t ) for an appropriate choice of d and compare the result to the results in (j).2.8.3 Riccati's equationThe Ricatti equation y + p(t )y + q(t )y 2 = f (t ) (2.8.3)
184 First-order differential equationsand its study are attributed to the Italian mathematician Jacobo Riccati(1667–1748). Observe that this nonlinear equation is a modification of thestandard linear first-order equation y + p(t )y = f (t ). Through the followingsteps, we will use a change of variables to transform the Riccati equation into alinear, second-order differential equation.(a) We consider a change of variables to convert (2.8.3) from being a differential equation in y to a new equation in v. Let v be a function that satisfies the relationship v = q(t )y(t )v(t ) (i) Differentiate v = qyv with respect to t to show that v = (qyv) = q yv + qy v + qyv (2.8.4) (ii) Show that q yv = q v /q.(b) Multiply both sides of the Riccati equation (2.8.3) by qv and use (i) and (ii) to show that the left-hand side may be written q vqy + vqpy + vq 2 y 2 = v + p − v (2.8.5) q(c) Use your work in (b) to show that the Riccati equation may now be re-expressed as the second-order equation in v given by q v + p− v − vqf = 0 (2.8.6) q(d) Explain how you would solve the Riccati equation in the special case when f (t ) = 0. Note particularly that to solve (2.8.6) with f (t ) = 0, you must reduce the order of the equation through an appropriate substitution, say u = v . See section 2.8.1 for further details on this technique. In addition, note that your goal is to find the solution y to the original equation (2.8.3). Be sure to explain how the functions v and u are used in this process.(e) Solve the following differential equations, each of which is a Riccati equation. (i) y + 2y + 4y 2 = 0 (ii) y + 1 y + t 2 y 2 = 0 t (iii) y + y tan t + y 2 cos t = 02.8.4 Bernoulli's equationThe Bernoulli brothers, James (1654–1705) and John (1667–1748), contributedto the solution of y + p(t )y = q(t )y n , n = 1 (2.8.7)
3Linear systems of differential equations3.1 Motivating problemsIn section 1.1, we considered how the amount of salt present in a system oftwo tanks can be modeled through a system of differential equations. In thatparticular example, we assumed that the volume of solution in each tank (as seenin figure 3.1) remains constant and all inflows and outflows happen at theidentical rate of 5 liter/min, and further that that the tanks are uniformly mixedso that the salt concentration in each is identical throughout each tank at a giventime t . With the additional premises that the volume of solution in tank A is200 liters and the independent inflow entering A carries water contaminatedwith 4g/liter of salt, we can develop a differential equation that models x1 (t ),the amount of salt (in grams) in tank A at time t . Likewise, by presuming thattank B holds solution of volume 400 liters and the inflow entering B carriesa concentration of salt of 7g/liter, a similar analysis produces a differentialequation whose solution is x2 (t ), the amount of salt (in grams) in tank B attime t . In particular, we found in (1.1.6) that the following system of differentialequations arose: dx1 x1 x2 =− + + 20 dt 20 80 (3.1.1) dx2 x1 x2 = − + 35 dt 40 40With our experience in linear algebra, we can now represent this system inmatrix notation. In particular, if we simultaneously consider the amounts of 187
188 Linear systems of differential equations A B Figure 3.1 Two tanks with inflows, outflows, and connecting pipes.salt x1 (t ) and x2 (t ) as entries in the vector function x1 (t ) x(t ) = x2 (t )we know that dx1 /dt x (t ) = (3.1.2) dx2 /dtMoreover, in (3.1.1) we recognize the familiar form of a matrix product in theterms involving x1 and x2 . Specifically, −x1 /20 + x2 /80 −1/20 1/80 x1 = (3.1.3) x1 /40 − x2 /40 1/40 −1/40 x2With the observations from (3.1.2) and (3.1.3) substituted into (3.1.1) andreplacing the quantities 20 and 35 with the appropriate vector, we may nowwrite the system of differential equations in the form −1/20 1/80 20 x = x+ (3.1.4) 1/40 −1/40 35Letting A be the matrix of coefficients that multiplies the vector x and b thevector [20 35]T , we can also write the system in (3.1.4) in the simplified form x = Ax + b (3.1.5)This form reminds us of the familiar nonhomogeneous linear first-orderdifferential equation with constant coefficients, for instance, an equation such as y = 2y + 5 (3.1.6)In this chapter, we will study similarities between (3.1.5) and (3.1.6) withthe specific goal of learning how to completely solve nonhomogeneouslinear systems of differential equations with constant coefficients such as thesystem (3.1.4). We will be especially interested in the role that linear algebraplays in identifying certain characteristics of the coefficient matrix A that enableus to find all solutions to the system. Before we proceed to an in-depth study of linear systems of differential equa-tions, at least one more motivating example is appropriate. A spring-mass system
Motivating problems 189 y y −y(t) −y(t)displacement −y(t) mass t t equilibriumFigure 3.2 A spring-mass system shown at two different points in time; −y(t ) denotesthe displacement of the mass from equilibrium (where displacements below the t -axisare considered positive).is a physical situation that models vibrations; for example, such a system arisesany time a mass attached to a spring is set in motion. We choose to envision thissituation vertically, as seen in figure 3.2, though one can also imagine the massresting on a table and moving horizontally. We consider some of the physics of basic springs and motion under theinfluence of gravity in order to develop a differential equation that describes thespring-mass system. Initially, the mass will stretch the spring from its naturallength. Hooke's law states that the force necessary to stretch a spring a distancex from its natural length is given by the equation F (x) = kxwhere k is the spring constant. Assume that the mass stretches the spring adistance L0 . Then from Hooke's law, when the system is in equilibrium, we seethat the force Fs exerted by the spring must be Fs = −kL0Here the minus sign indicates that the force is opposing the natural downwarddisplacement of the spring. Note particularly that we view the downwarddirection as positive. We also know that gravity acts on the mass with forceFg given by Fg = mg If the system is in static equilibrium, we know that the sum of the two forcesis zero. In other words, Fg + Fs = 0and therefore mg = kL0Once the system is set in motion by some initial force or displacement, wetrack the location of the mass at time t with a function y(t ). In particular,y(t ) represents the displacement of the mass from the equilibrium position attime t ; note that y = 0 is the equilibrium position of the system. We continue to
190 Linear systems of differential equationsdesignate the downward direction as positive, so y(t ) > 0 means that the massis below the equilibrium position, while y(t ) < 0 means the mass is above theequilibrium position. We can see the role y(t ) plays in figure 3.2 as it tracksthe displacement of the mass from equilibrium and thus traces out a curve withrespect to time. We can now use Newton's second law to obtain a differential equation thatgoverns the system. The forces that act on the mass are: • Gravity, with Fg = mg . • The spring force Fs . Note now that at a given time t the displacement of the spring from its natural length is L0 + y(t ), so that by Hooke's law we have Fs = −k(L0 + y). • A possible damping force Fd . Motion may be damped due to air resistance, friction, or some sort of external damping system (usually called a dashpot). We assume that damping forces are directly proportional to the velocity of the mass. Under this assumption, it follows that Fd = −cy . Again, the minus sign indicates that this force opposes the motion of the mass. The positive constant c is called the damping constant. • Finally, there may be an external driving force present (such as the periodic force that drives a piston in an engine). We call this a forcing function F (t ); the role of forcing functions will be considered in detail later on in this chapter. Newton's second law demands that the resultant force (that is, the sum of allthe forces) on the mass must be equal to ma, where a is the body's acceleration(which is also y ). Summing all the aforementioned forces and equating theresult with ma = my , we find my = Fg + Fs + Fd + F (t ) (3.1.7)Using the formulas we developed earlier and substituting in (3.1.7) yields my = mg − k(L0 + y) − cy + F (t ) (3.1.8)Now recall that mg − kL0 = 0, rearrange (3.1.8), and divide by m. This leadsus to the standard form of the differential equation that governs a spring masssystem, c k 1 y + y + y = F (t ) (3.1.9) m m mNote that (3.1.9) is a nonhomogeneous linear second-order differential equation. To see how such a second-order linear differential equation is linkedto a system of linear differential equations, let's consider the specific examplewhere c = 1, m = 1, k = 6, and F (t ) = 0, which results in the equation y + y + 6y = 0 (3.1.10)
The eigenvalue problem revisited 191If we introduce the functions x1 and x2 through the substitutions y = x1 andy = x2 , then x1 (t ) represents the displacement of the mass at time t and x2 (t )is the velocity of the mass at time t . Observe first that x 1 = x2 (3.1.11)Moreover, since x2 = y , we can rewrite (3.1.10) as x2 + x2 + 6x1 = 0.Equivalently, x2 = −6x1 − x2 (3.1.12)Thus (3.1.11) and (3.1.12) generate the system of differential equations x 1 = x2 (3.1.13) x2 = −6x1 − x2which may also be expressed in matrix form as 0 1 x = x (3.1.14) −6 −1We have therefore shown that the linear second-order differential equa-tion (3.1.9) that describes a spring-mass system may be converted to the systemof linear first-order equations (3.1.14) through the substitution x1 = y, x2 = y . In fact, any linear higher order differential equation may be convertedthrough a similar substitution to a system of linear first-order equations.Therefore, by learning to understand and solve systems of linear equations,we will be able to determine the behavior of higher order linear equations aswell. It is this fact that motivates us to study systems of linear equations prior tothe study of higher order single equations.3.2 The eigenvalue problem revisitedAs we begin our study of linear systems of first-order differential equations, weare ultimately interested in two main questions: the first asks, for a linear systemx = Ax such as 2 3 x = x 2 1how can we explicitly solve the system for x(t )? In addition, what is the long-term behavior of the solution x(t ) to such a system? How does its graphappear? We start our investigation by thinking carefully about the meaningof the matrix equation x = Ax and compare our experience with the singlefirst-order differential equation x = ax. Note that we naturally begin with thehomogeneous system x = Ax; later we will consider nonhomogeneous systemsof the form x = Ax + b. In every case, we seek a vector function x(t ) that solvesthe given system. An elementary example is instructive.
192 Linear systems of differential equationsExample 3.2.1 Solve the linear system x = Ax, where −3 0 A= 0 −1Explain the role that the eigenvalues and eigenvectors of A play in the generalsolution, and graph and discuss the solution curves for different choices of initialconditions.Solution. First, we observe that the system x1 −3 0 −3 0 x1 =x = x= (3.2.1) x2 0 −1 0 −1 x2tells us that we seek two functions x1 (t ) and x2 (t ) such that x1 = −3x1 andx2 = −x2 . Because the matrix of the system is diagonal, the problem is especiallysimple. In particular, the system is uncoupled, which means that the differentialequation for x1 does not involve x2 and the equation for x2 does not involve x1 . From our experience with linear first-order equations, we know that thegeneral solution to x1 = −3x1 is x1 (t ) = c1 e −3t and that the solution to x2 = −x2is x2 (t ) = c2 e −t . Writing the solution to the system as a single vector, we have x1 c e −3t x= = 1 −t (3.2.2) x2 c2 eRewriting x in another form sheds further insight on the key components of thissolution. Writing x as the sum of two vectors, we find c1 e −3t 0 1 0 x= + = c1 e −3t + c2 e −t (3.2.3) 0 c2 e −t 0 1Here, we can make a key observation about the eigenvalues and eigenvectors ofA: because A is diagonal, its eigenvalues are its diagonal entries, λ1 = −3 andλ2 = −1. Moreover, its corresponding eigenvectors may be easily confirmed tobe the vectors 1 0 v1 = and v2 = 0 1Thus, in (3.2.3), we see the interesting fact that the solution has the form x =c1 e λ1 t v1 + c2 e λ2 t v2 ; the eigenvalues and eigenvectors therefore play a centralrole in the system's behavior. Finally, we explore the solutions to several related initial-value problemsfor select initial conditions. If we have the initial condition x(0) = [4 0]T , wesee in (3.2.3) that c1 = 4 and c2 = 0, so that the solution to the IVP is 1 x(t ) = 4e −3t 0Two key observations can be made about this solution curve: one is that itsgraph is a straight line, since for every value of t , x is a scalar multiple of the
The eigenvalue problem revisited 193vector [1 0]T . Note particularly that the direction of this line is given by theeigenvector corresponding to λ1 = −3. The other important fact is that e −3t → 0as t → ∞, and therefore x(t ) → 0, so that the solution approaches the origin astime increases without bound. For the initial condition x(0) = [0 5]T , it follows from (3.2.3) that c1 = 0and c2 = 5, and thus the solution to this IVP is 0 x(t ) = 5e −t 1Similar observations about the behavior of this solution may be made to thosenoted above for the first chosen initial condition: this solution curve is linearand approaches the origin as t → ∞. Finally, if we consider an initial condition that does not correspond to aneigenvector of the system, such as x(0) = [4 5]T , (3.2.3) tells us that c1 = 4 andc2 = 5, and thus 1 0 x = 4e −3t + 5e −t 0 1This last solution's graph is not a straight line. As seen in figure 3.3, which showsthe three different solutions based on the differing initial conditions, we see theconsistent behavior that every solution tends to the origin as t → ∞, as well asthat the eigenvectors play a key role in how these graphs appear. We will discussthis graphical perspective further in sections 3.4 and 3.5. The long-term behavior of the solutions to the system (3.2.1) inexample 3.2.1 suggests that every solution tends to the zero vector. In fact, theorigin itself is a solution, a so-called constant or equilibrium solution. That is, if x2 solution through (0,5) 5 solution through (4,5) solution through (4,0) x1 5 Figure 3.3 Plots of solutions to three IVPs for the system in example 3.2.1. Arrows indicate the direction of flow along the solution curve as time increases.
194 Linear systems of differential equationswe consider whether there is any constant vector x that is a solution to x = Ax,it follows that x = 0, and thus x must satisfy Ax = 0. From our work withhomogeneous linear equations, we know that x = 0 is always a solution to thisequation, and thus the zero vector is a constant solution to every homogeneouslinear system of first-order differential equations. In sections 3.4 and 3.5 we willinvestigate the so-called stability of this equilibrium solution. There is a second perspective from which we can see how eigenvectorsand eigenvalues arise in the solution of linear systems of differential equations.After constant solutions, the next simplest type of solutions to such a systemare straight-line solutions. In other words, solutions whose graph is a straightline in space form a particularly important type of solution to a system. In thepreceding example, we saw two such straight-line solutions: each occurred inthe direction of an eigenvector and passed through the origin. In search of a general straight-line solution to x = Ax, we know that anysuch solution must have the form x(t ) = f (t )v, where f (t ) is a scalar functionand v is a constant vector. This form guarantees that x(t ) traces out a path thatis a straight line through 0 in the direction of v. In order for x(t ) to satisfy thesystem, we observe that since x (t ) = f (t )v, the equation f (t )v = A(f (t )v) (3.2.4)must hold. Moreover, since f (t ) is a scalar, the linearity of matrix multiplicationallows us to rewrite (3.2.4) as f (t )v = f (t )Av (3.2.5)Equation (3.2.5) is strongly reminiscent of the equation we use to defineeigenvalues and eigenvectors: Ax = λx. In fact, if f (t ) = λf (t ), then (3.2.5)implies that λf (t )v = f (t )AvFurther, if f (t ) = 0, then λv = Av, and λ and v must be an eigenvalue-eigenvector pair of A. It is therefore natural for us to want f to satisfy the single differentialequation f (t ) = λf (t ). From our work in chapter 2, we know that f (t ) = Ce λtis the general solution to this equation. Substituting this form for f in (3.2.5),we now observe that λe λt v = e λt Av (3.2.6)and since e λt is never zero, we can simplify (3.2.6) to λv = Av (3.2.7)which is satisfied precisely when v is an eigenvector of A with correspondingeigenvalue λ. Our most recent work has demonstrated that if x(t ) is a function of theform x(t ) = e λt v that is a solution to x = Ax, then (λ, v) is an eigenpair ofthe coefficient matrix A. In fact, the converse also holds (as will be shown in
The eigenvalue problem revisited 195the exercises), so that the following result is true for any n × n system of linearfirst-order differential equations.Theorem 3.2.1 Let A be an n × n matrix. The vector function x(t ) = e λt v is asolution to the homogeneous linear system of first-order differential equationsgiven by x = Ax if and only if v is an eigenvector of A with correspondingeigenvalue λ. We close this section with one more example to demonstrate theorem 3.2.1and one of its important consequences.Example 3.2.2 Consider the system of differential equations given by x1 = −2x1 − 2x2 x2 = −4x1Write the system in the form x = Ax and show that A has two real eigenvalueswith corresponding linearly independent eigenvectors. Verify by substitutionthat for each eigenvalue-eigenvector pair, x(t ) = e λt v is a solution of the system.In addition, show that any linear combination of such solutions is also a solutionto the system.Solution. First, we observe that the system can be expressed in the formx = Ax by using the matrix −2 −2 A= −4 0We briefly review the process of determining the eigenvalues and eigenvectorsof a matrix A; in most future occurrences, we will use Maple to determine thisinformation using the commands introduced in section 1.10.2.Since the eigenvalues are the roots of the characteristic equation, we solvedet(A − λI) = 0. Doing so, 0 = det(A − λI) −2 − λ −2 = det = −λ(−2 − λ) − 8 −4 −λ = λ2 + 2λ − 8 = (λ + 4)(λ − 2)so the eigenvalues of A are λ = −4 and λ = 2. To find the eigenvector v that corresponds to λ = −4, we solve the equation(A − (−4I))v = 0. Row-reducing the appropriate augmented matrix yields 2 −2 0 1 −1 0 → −4 4 0 0 0 0
The eigenvalue problem revisited 197Example 3.2.2 provides the foundation for much of our study of linear systemsof differential equations. It shows that when we can find real eigenvaluesand eigenvectors, these lead us directly to solutions of the system. In addition,any linear combination of such solutions is also a solution to the system; westate this formally in the next theorem.Theorem 3.2.2 If (λ1 , v1 ), (λ2 , v2 ), . . . , (λk , vk ) are eigenpairs of an n × nmatrix A and c1 , . . . , ck are any scalars, then x(t ) = c1 e λ1 t v1 + c2 e λ2 t v2 + · · · + ck e λk t vkis a solution to x = Ax. In upcoming sections, we will determine whether we have found all ofthe solutions to a given system, address some subtle issues that arise when wecannot find enough real eigenvalues and eigenvectors, and better understand thegraphical and long-term behavior of solutions. The exercises in this section willhelp further illuminate the roles of eigenvalues and eigenvectors as well as someof the issues that arise when there is an insufficient number of real eigenvectorsfor a given system's matrix.Exercises 3.2In exercises 1–7, compute by hand the eigenvalues and eigenvectors of the givenmatrix. 1 4 1. A = 2 3 0 4 2. A = 1 0 0 3 3. A = 3 8 2 2 4. A = −1 −1 ⎡ ⎤ 2 2 0 5. A = ⎣1 2 1⎦ 1 2 1 ⎡ ⎤ 3 0 1 6. A = ⎣0 2 0⎦ 5 0 −1 ⎡ ⎤ 2 1 0 7. A = ⎣0 2 1⎦ 0 0 2
198 Linear systems of differential equations 8. Consider the system of differential equations given by x1 = −2x1 + 3x2 x2 = x1 − 4x2 (a) Determine a matrix A so that the system may be written in the form x = Ax. (b) Determine all constant (equilibrium)1 2]T . Discuss the graphical behavior of this solution. 9. Consider the system of differential equations given by x1 = −x1 + 2x2 x2 = −7x1 + 8−2 0]T . Discuss the graphical behavior of this solution.10. Consider the system of differential equations given by x1 = 2x1 + 3x2 x2 = −4 Explain how you could find this same general solution without determining eigenvalues and eigenvectors. (Hint: focus on x2 (t ) first.) (g) Solve the initial-value problem x = Ax, x(0) = [0 1]T . Discuss the graphical behavior of this solution.
The eigenvalue problem revisited 19911. Consider the system of differential equations given by x1 = −2x1 + x2 x2 = −2 your straight-line solutions from (d). (f) Attempt to solve the initial-value problem x = Ax, x(0) = [1 1]T . What does this tell you about the proposed general solution in (e)?12. Consider the system of differential equations given by x1 = 2x1 + 9x2 x2 = −x1 − 2 Are there any straight-line solutions to x = Ax. Why or why not?13. Consider the system of differential equations given by x1 = −3x1 + x2 x2 = 3x1 − x2 (a) Determine a matrix A so that the system may be written in the form x = Ax. (b) Determine all constant solutions to x = Ax. Compare and contrast your findings with preceding exercises. (c) Compute the eigenvalues and eigenvectors of A. (d) Determine all straight-line solutions to x = Ax. How many such solutions exist? (e) Find a more general solution to x = Ax by taking all possible linear combinations of your straight-line solutions from (d). (f) Solve the initial-value problem x = Ax, x(0) = [3 0]T . Discuss the graphical behavior of this solution.14. Consider the system of differential equations given by x1 = 3x1 + x2 + x3 x2 = x1 + 3x2 + x3 x3 = x1 + x2 + 3x3
The eigenvalue problem revisited 201set up, but do not solve a system of differential equations or initial-value problemwhose solution would give the amount of salt in each tank at time t . Write eachsystem in matrix form.23. A system of two tanks is24. Suppose that in exercise 23 all of the given information remains the same except for the fact that instead of saltwater flowing into each tank, pure water flows in; that is, the concentration of salt in the entering solution is 0 g/liter for each tank.25. In a closed system of two tanks (i.e., 326. In a closed system of three tanks (i.e., one for which there are no input flows and no output flows), the following information is given. Tank A Tank B Tank C Tank Volume 100 liters 150 liters 125 liters Rates of outflows to B: 3 liters/min to C: 1 liter/min to A: 4 liters/min to other tanks Rates of outflows to C: 4 liters/min to A: 3 liters/min to B: 1 liter/min to other tanks
202 Linear systems of differential equations27. Show that if (λ, v) is an eigenpair of the matrix A, then x(t ) = e λt v is a solution to the homogeneous system of linear differential equations given by x = Ax.3.3 Homogeneous linear first-order systemsIn preceding sections, we have encountered examples of systems of two(or three) linear differential equations in two (or three) unknown functions.More generally, a linear system of n differential equations in n unknown functions(or simply, a linear system) is a collection of differential equations for which weseek unknown functions x1 (t ), . . . , xn (t ) when given n equations with coefficientfunctions aij (t ) and bi (t ) in the form dx1 = a11 (t )x1 + a12 (t )x2 + · · · + a1n (t )xn + b1 (t ) dt dx2 = a21 (t )x1 + a22 (t )x2 + · · · + a2n (t )xn + b2 (t ) dt . . . . . . dxn = an1 (t )x1 + an2 (t )x2 + · · · + ann (t )xn + bn (t ) dtIt will be convenient to write the above system in matrix form. If we let x denotethe vector function whose entries are x(t ) = [xi (t )], A(t ) the n × n matrix offunctions whose entries are A = [aij (t )], and b(t ) the vector of functions whoseentries are b = [bi (t )], then the above system can be rewritten simply as x (t ) = A(t )x(t ) + b(t ) (3.3.1)In much of our work, we will suppress the independent variable t and writex = Ax + b. Moreover, it will most often be the case that, as in examples 3.2.1and 3.2.2, the matrix A has all constant entries. Indeed, from this point on,unless otherwise noted, we will assume the matrix A has constant entries. In the event that b = 0, we say that the linear system is homogeneous. Ifb is nonzero, the system is nonhomogeneous. We have already encounteredin theorems 3.2.1 and 3.2.2 the important facts that for any homogeneousfirst-order linear system x = Ax, every solution of the form x(t ) = e λt v requires(λ, v) to be an eigenpair of A, and that any linear combination of such solutionsis also a solution to the system. Just as with individual differential equations, to each system of equationswe can associate an initial-value problem. Using the matrix notation (3.3.1), if
Homogeneous linear first-order systems 203we assume that we also have the initial condition x(t0 ) = x0 , then we have thestandard initial-value problem x (t ) = A(t )x(t ), x(t0 ) = x0 (3.3.2) We next consider a theoretical result (whose proof we omit) that willframe our overall work with systems. The following theorem is analogous tothe earlier result we encountered in theorem 2.2.1 regarding the existence of aunique solution to the initial-value problem associated with a single first-orderdifferential equation.Theorem 3.3.1 In (3.3.2), let the entries of the matrix A(t ) be continuousfunctions on a common interval I that contains the value t0 . Then there exists aunique solution x(t ) to (3.3.2) on the interval I . In particular, we note that in examples where the matrix A has constantcoefficients, the entries are continuous functions, so that the IVP x = Ax,x(0) = x0 is guaranteed to have a unique solution. We now examine this resultmore closely through a particular example, revisiting a problem we consideredin the preceding section.Example 3.3.1 Determine the unique solution to the IVP given by −2 −2 −5 x = x , x(0) = (3.3.3) −4 0 3Solution. We note, by theorem 3.3.1, that a unique solution exists. Moreover,from our work in example 3.2.2, every function of the form 1 1 x(t ) = c1 e −4t + c2 e 2t (3.3.4) 1 −2is a solution to the system x = Ax. We now explore whether we can findconstants c1 and c2 in order that the function x(t ) will satisfy the given initialcondition in (3.3.3). The initial condition in (3.3.3) and (3.3.4) together imply −5 1 1 = x(0) = c1 e 0 + c2 e 0 3 1 −2or equivalently 1 1 −5 c1 + c2 = (3.3.5) 1 −2 3We note that since the vectors [1 1]T and [1 − 2]T (which are eigenvectorsof A) are linearly independent and span R2 , we are guaranteed a uniquesolution to (3.3.5). Row-reducing the system (3.3.5), we find 1 1 −5 1 0 −7 3 → 1 −2 3 0 1 −8 3
204 Linear systems of differential equationsThus, we have shown 7 1 8 1 x(t ) = − e −4t − e 2t 3 1 3 −2is the unique solution to the given initial-value problem.One especially important observation from example 3.3.1 can be made regardingthe point at which we solved for the constants c1 and c2 : we were guaranteednot only that a solution existed, but also that it was unique, due to the fact thattwo linearly independent eigenvectors of the 2 × 2 matrix A were present in thegeneral solution (3.3.4). Indeed, if we imagine wanting to solve any similar IVPwith the freedom to choose any initial vector x(0), it will be necessary that x(0)can be written as a linear combination of the vectors v1 and v2 , whenever thegeneral solution has form x(t ) = c1 e λ1 t v1 + c2 e λ2 t v2This situation is indicative of the general fact that for all 2 × 2 linear systemsof DEs, we must have two parts to the general solution, in order to be ableto uniquely determine the constants c1 and c2 . Note further that for thesolutions x1 (t ) = e λ1 t v1 and x2 (t ) = e λ2 t v2 we encountered above, x1 (0) = v1and x2 (0) = v2 are linearly independent and form a basis for R2 . Thislinear independence of the constant vectors v1 and v2 turns out to have animportant analog in the linear independence of certain solutions to the systemof differential equations. More generally, we can consider these same issues for an n × n homogeneoussystem. Because theorem 3.3.1 guarantees the existence of a unique solution tothe corresponding IVP for every initial condition x(0) ∈ Rn , when we thinkabout the structure of the general solution, it is natural to think this solutionwill have form x(t ) = c1 x1 (t ) + c2 x2 (t ) + · · · + cn xn (t )where {x1 (0), x2 (0), . . . , xn (0)} form a basis for Rn .These observations, together with our earlier work in theorem 3.2.2 that showedthat every linear combination of solutions to the general homogeneous linearsystem of DEs (3.3.1) is also a solution to (3.3.1), help explain why the set of allsolutions to x = Ax, where A is a matrix with constant coefficients, is a vectorspace of dimension n. We state this formally in the following result.Theorem 3.3.2 The set of all solution vectors to the homogeneous linearsystem x = Ax, where A is an n × n matrix with constant coefficients, forms avector space of dimension n. Theorem 3.3.2 shows us that in order to solve an n × n system ofhomogeneous first-order DEs, we must find n linearly independent solutions tothe system. Said differently, the general solution to x = Ax will have form x(t ) = c1 x1 (t ) + c2 x2 (t ) + · · · + cn xn (t ) (3.3.6)
208 Linear systems of differential equations = e −t (3e 7t ) − 3e 2t (−2e 4t ) + e 5t (e t ) = 10e 6t = 0Since W [x1 , x2 , x3 ] = 0 for at least one t -value (in fact, for all t ), it follows bytheorem 3.3.5 that the functions x1 , x2 , and x3 are linearly independent.In conclusion, we now know that when we encounter a homogeneous system ofn linear first-order differential equations in n unknown functions, the set of allsolutions to the system forms an n-dimensional vector space. Hence, we seek nlinearly independent solutions to the system x = Ax. Such a set x1 , . . . , xn of nlinearly independent solution vectors to this system is called a fundamental set.Moreover, given a set of fundamental solutions x1 , . . . , xn to x = Ax, on someinterval I , the general solution to the system is x(t ) = c1 x1 + · · · + cn xn We have also seen that if an n × n matrix A has n linearly independentreal eigenvectors, then these eigenvectors and their corresponding eigenvaluesgenerate a fundamental set for the system x = Ax. In subsequent sections wewill find that, even in the case when an insufficient number of real eigenvectorsexists, the eigenvalue problem enables us to build a fundamental set. Moreover,we will investigate how fundamental solutions allow us to fully understand thegraphical behavior of solutions and the stability of equilibrium solutions to thesystem.Exercises 3.3 1. If x = Ax represents the system of differential equations given by a 4 × 4 matrix A with constant entries, how many linearly independent solutions to the system do we need to find in order to determine the general solution? What if A is 7 × 7? 2. Consider the second-order differential equation y + y = 0. Using the substitutions y = x1 and y = x2 , convert the given second-order differential equation to a system of first-order equations. What is the dimension of the solution space to the system? What does this tell you about the dimension of the solution space to the original second-order equation? 3. Consider the third-order differential equation y + 3y + 3y + y = 0. Using the substitutions y = x1 , y = x2 , and y = x3 , convert the given differential equation to a system of first-order equations. What is the dimension of the solution space to the system? What does this tell you about the dimension of the solution space to the original third-order equation?
210 Linear systems of differential equations (c) Solve the IVP with the initial condition x(0) = [3 2]T . (d) Explain how you could solve the original system given in this problem without using eigenvalues and eigenvectors.13. Let x = Ax be given by the matrix 0 −1 A= 1 0 (a) Compute the eigenvalues and eigenvectors of A. Explain why the eigenvalues and eigenvectors do not produce any real linearly independent solutions to the system. (b) Verify through direct substitution that x1 (t ) = [cos t sin t ]T and x2 (t ) = [− sin t cos t ]T are solutions to the given system x = Ax. (c) Show that the solutions you verified in (b) are linearly independent, and hence state the general solution to the system. (d) Solve the IVP with the initial condition x(0) = [3 2]T .14. Let x = Ax be given by the matrix ⎡ ⎤ 5 6 2 A = ⎣0 −1 −8⎦ 1 0 −2 (a) Compute the eigenvalues and eigenvectors of A. Explain why your work determines two linearly independent solutions to the system, but that one additional linearly independent solution remains to be found. (b) Verify through direct substitution that x3 (t ) = te 3t [5 − 2 1]T + e 3t [1 1/2 0]T is a solution to the given system x = Ax. (c) Show that the set of three solutions from (a) and (b) is linearly independent, and hence state the general solution to the system. (d) Solve the IVP with the initial condition x(0) = [3 2 1]T .15. Consider the second-order differential equation y + y = 0. Convert this equation to a system of first-order equations and solve the system. Use your work to state the general solution y to the original equation. (Hint: See exercise 13.)16. Convert the second-order differential equation y + 3y + 2y = 0 to a system of first-order equations and solve the system. Use your work to state the general solution y to the original equation.17. Convert the third-order differential equation y − y = 0 to a system of first-order equations and solve the system. Use your work to state the general solution y to the original equation.
Systems with all real linearly independent eigenvectors 2113.4 Systems with all real linearly independent eigenvectorsIn this section, we closely examine the graphical and long-term behavior ofsolutions to 2 × 2 systems in the case where the coefficient matrix A has two real,linearly independent eigenvectors. We do so through a sequence of examplesthat demonstrate a variety of possibilities that naturally lead to discussion of thestability of equilibrium solutions. We first review the graphical behavior of vector functions, a subjectnormally encountered in multivariable calculus. For the system x = Ax inthe case where A is 2 × 2, every solution x(t ) is a vector function whose outputlies in R2 . In particular, the graph of x(t ) is the curve that is traced out by thevectors x(t ) at various times t . For example, if 1 0 e −t x(t ) = e −t + et = (3.4.1) 0 1 etis a function we have found by solving a system of differential equations, thenevaluating x(t ) at t = −1, 0, and 1 yields the vectors 2.719 1 0.368 x(−1) ≈ , x(0) = , and x(1) ≈ (3.4.2) 0.368 1 2.719Plotting these vectors helps indicate how x(t ) traces out the parametric curvegiven by (x1 (t ), x2 (t )) = (e −t , e t ), shown at left in figure 3.4. In addition, it is important to recall the meaning of x (t ), the derivative ofa vector function. The direction of the vector x (t ) indicates the instantaneousdirection of motion of a particle traveling along the curve traced out by x(t ),while the magnitude of x (t ) determines the instantaneous speed of the particleat time t . For our purposes, the direction of motion is most important because x2 x2 4 4 (0.368, 2.719) (0.368, 2.719) (1,1) (1,1) (2.719, 0.368) (2.719, 0.368) x1 x1−4 4 −4 4 −4 −4Figure 3.4 At left, the solution curve x(t ) given in (3.4.1). At right, the solution curvex(t ) given in (3.4.1), along with corresponding scaled derivative vectors at times t = −1,t = 0, and t = 1.
212 Linear systems of differential equationsthis indicates a flow along the solution curve as time increases. Thus, ratherthan plotting the vector x (t ) at various times, we plot scaled versions of it, eachemanating from the tip of x(t ). For example, since −e −t x (t ) = (3.4.3) etit follows that −2.719 −1 −0.368 x (−1) ≈ , x (0) = , and x (1) ≈ (3.4.4) 0.368 1 2.719Plotting scaled versions of each of these vectors emanating from x(−1), x(0),and x(1), respectively, we see the updated image at the right in figure 3.4. These plots of the derivative vectors and the flow of the solution curveremind us of our earlier work with slope fields for single differential equations.Indeed, since a solution curve such as x(t ) will always be the result of solvingsome differential equation x = Ax, we realize that we have a formula for x , justas we had a formula for y in examples like y = −2y. In the example discussedabove, we can view x(t ) as being the solution to the system x = Ax where A isthe matrix −1 0 A= (3.4.5) 0 1so that x (t ) satisfies the equation x1 (t ) −x1 (t ) = x (t ) = Ax(t ) = (3.4.6) x2 (t ) x2 (t )In particular, (3.4.6) indicates how, for any point (x1 , x2 ) in the plane, we caneasily compute x at that point, and hence know the direction of the flow ofthe solution curve that passes through that point. Using a computer to conductsuch computations at points sampled throughout the plane (with each resultingvector scaled to be of equal length), we get a picture of the so-called directionfield for the system, shown at left in figure 3.5, which is analogous to a directionfield for a single differential equation. If we now superimpose our plot of the solution curve in figure 3.4 in thedirection field, now shown on the right in figure 3.5, we see clearly the role thatthe derivative x and the direction field play in determining the graph of thesolution x, as well as the typical behavior of a solution as time increases. The x1 –x2 plane is usually called the phase plane; note that the independentvariable t is implicit in the flow, while the behavior of the curve relative to thecoordinate axes demonstrates the interrelationship between the componentsx1 (t ) and x2 (t ) of the solution x(t ). Sample solution curves, such the one plottedin figure 3.5, are typically called trajectories. Each distinct trajectory is a solutionto an initial-value problem; the one in figure 3.5 can be viewed as the solutionto x = Ax , x(0) = [1 1]T .
Systems with all real linearly independent eigenvectors 213 x2 x2 4 4 x1 x1−4 4 −4 4 −4 −4Figure 3.5 At left, the direction field for the system x = Ax given by (3.4.5). At right,the solution to (3.4.5) that is given by (3.4.1). We will now explore the direction field, phase plane, and trajectories forseveral examples of 2 × 2 systems of linear differential equations for which thecoefficient matrix has two real linearly independent eigenvectors. An importanttheme throughout will be the long-range behavior of solutions x(t ) as t → ∞.In addition, we will study the equilibrium solutions of each system; a solutionx(t ) is an equilibrium or constant solution if and only if x(t ) is constant for allvalues of t .Example 3.4.1 Consider the system of differential equations given by x = 3 2Ax where A = . Compute the eigenvalues and eigenvectors of A and 2 3state the general solution to the system. In addition, determine all equilibriumsolutions of the system. Finally, plot the direction field for the system, sketchseveral trajectories, and discuss the long-term behavior of solutions relative tothe equilibrium solution(s).Solution. The Maple command > Eigenvectors(A) produces the output 5 1 −1 1 1 1so that A has eigenvalues λ1 = 5 and λ2 = 1, with corresponding eigenvectorsv1 = [1 1]T and v2 = [−1 1]T . We therefore know that the general solution tox = Ax is 1 −1 x(t ) = c1 e 5t + c2 e t 1 1To find the equilibrium solution(s), we seek all constant vectors x that satisfyx = Ax. In this situation, since x is constant with respect to t , we know that
214 Linear systems of differential equationsx = 0, so therefore we must solve the system of linear equations given by Ax = 0where 3 2 A= 2 3Since det(A) = 0, it follows that A is an invertible matrix, so the only solutionto Ax = 0 is x = 0. Thus the system has the origin as its only equilibriumsolution. At the end of this section, in subsection 3.4.1, we will show how to useMaple to plot direction fields for systems. In this and subsequent examples,well simply provide these plots for discussion. In figure 3.6, we see not only thedirection field generated by the system, but also the plots of several trajectories,which are natural to sketch (even by hand, once the direction field is provided)by following the map that the direction field provides. Note particularly the straight-line solutions that follow the eigenvectorsv1 = [1 1]T and v2 = [−1 1]T . Moreover, since both eigenvalues are positive,the respective scalar functions e 5t and e t both increase without bound as t → ∞.This explains why the flow along each straight-line solution is away from theorigin. Indeed, every solution besides the zero solution flows away from theequilibrium solution at the origin. In chapter 2, we considered single autonomous differential equations suchas y = 2y − 4. When we found equilibrium solutions to such equations, wealso classified their stability based on the behavior exhibited in the directionfield. We do likewise with equilibrium solutions for systems. In example 3.4.1, x2 4 x1 −4 4 −4 Figure 3.6 The direction field for the system x = Ax of example 3.4.1 along with several trajectories.
Systems with all real linearly independent eigenvectors 215we found that x = 0 is the only equilibrium solution of the system, and thatevery non-constant solution flows away from 0. This shows that 0 is an unstableequilibrium, and in this case we naturally call 0 a repelling node. We next explore the behavior of a system where both eigenvalues arenegative.Example 3.4.2 Consider the system of differential equations given by x = Ax −2 2where A = . Compute the eigenvalues and eigenvectors of A, and 1 −3state the general solution to the system. In addition, determine all equilibriumsolutions to the system. Finally, plot the direction field for the system, sketchseveral trajectories, and discuss the long-term behavior of solutions relative tothe equilibrium solution(s).Solution. Using Maple, we find that A has eigenvalues λ1 = −1 and λ2 = −4,with corresponding eigenvectors v1 = [2 1]T and v2 = [−1 1]T . The generalsolution to x = Ax is therefore 2 −1 x(t ) = c1 e −t + c2 e −4t 1 1To find the equilibrium solution, we set x = 0. Solving the system of linearequations given by Ax = 0, we see that since A is an invertible matrix, the onlysolution to Ax = 0 is x = 0, so the system has the origin as its only equilibriumsolution. Plotting the direction field and several trajectories, as shown in figure 3.7,we observe that all solutions flow towards the equilibrium solution at the origin.This makes sense due to the presence of the scalar functions e −4t and e −t inthe general solution, as each approaches 0 as t → ∞, and thus it follows thatx(t ) → 0 as t → ∞. Moreover, note the two straight-line solutions that showflow along stretches of the two eigenvectors v1 = [2 1]T and v2 = [−1 1]T .Because every non-constant solution to the system in example 3.4.2 approachesthe equilibrium solution at 0, we say that the origin is a stable equilibrium.Moreover, based on the patterns in the flow, we use the terminology that 0 is anattracting node. We study the third case for a 2 × 2 linear system of differential equationswith two real, nonzero eigenvalues in the next example: the eigenvalues haveopposing signs. 3 −2Example 3.4.3 Let A = and consider the system of differential 2 −2
216 Linear systems of differential equations x2 4 x1 −4 4 −4 Figure 3.7 The direction field for the system x = Ax in example 3.4.2 along with several trajectories.Solution. We find that A has eigenvalues λ1 = 2 and λ2 = −1, withcorresponding eigenvectors v1 = [2 1]T and v2 = [1 2]T . It follows that thegeneral solution to x = Ax is 2 1 x(t ) = c1 e 2t + c2 e − t 1 2Since A is an invertible matrix, the only solution to Ax = 0 is x = 0, so the originis only equilibrium solution of the system. As figure 3.8 shows, the direction field and various trajectories exhibit adifferent type of behavior around the origin. In particular, solutions that donot lie on either eigenvector appear to initially flow toward the origin, and thenturn away and tend toward the straight-line solution associated with the positiveeigenvalue. More specifically, it appears that solutions that do not pass througha point on the line in the direction of the eigenvector [1 2]T are eventuallyattracted to stretches of the eigenvector [2 1]T . This is reasonable since in thegeneral solution, e −t will tend to 0 as t → ∞, leaving the function c1 e 2t [2 1]Tto dominate.Since some solutions that pass through points near the origin tend away fromthe origin as t → ∞, the origin is an unstable equilibrium in example 3.4.3.Moreover, as the trajectories remind us of the contour plot in multivariablecalculus of a surface whose graph looks like a saddle, we say in this context aswell that the origin is a saddle point. The preceding examples demonstrate the three possible cases for a 2 × 2system with real, nonzero eigenvalues: both positive, both negative, or opposites.Our next example investigates the situation when one eigenvalue is zero.
Systems with all real linearly independent eigenvectors 217 x2 4 x1 −4 4 −4 Figure 3.8 The direction field for the system x = Ax of example 3.4.3 along with several trajectories. −3 1Example 3.4.4 For the matrix A = and the corresponding system 3 −1of differential equations x = Ax, find the general solution of the system anddetermine all equilibrium solutions. Furthermore, plot the direction field for thesystem along with sketches of several trajectories; discuss the long-term behaviorof solutions relative to the equilibrium solution(s).Solution. We first do the standard computations to find that A has eigenvaluesλ1 = −4 and λ2 = 0, with corresponding eigenvectors v1 = [−1 1]T andv2 = [1 3]T . Thus, the general solution to x = Ax is −1 1 x(t ) = c1 e −4t + c2 1 3We immediately notice something different about x(t ). In particular, becausethe second eigenvalue is 0, the scalar function e 0t has no effect on the generalsolution. Furthermore, with e −4t the only part of x(t ) that changes with t , wecan see that for any nonzero constant c1 and any c2 , the graph of x(t ) is alwaysa straight line where the direction is given by the eigenvector corresponding tothe nonzero eigenvalue. In addition, the presence of a zero eigenvalue has a significant impact onthe system's equilibrium solutions. The fact that the columns of A are scalarmultiples of each other leads us to see immediately that A is not invertible;this can be equivalently deduced from the fact that A has a zero eigenvalue.The singularity of A further implies that the homogeneous equation Ax = 0has infinitely many solutions. In particular, row-reducing the appropriate
218 Linear systems of differential equationsaugmented matrix, we find that −3 1 0 1 −1/3 0 → 3 −1 0 0 0 0This implies that any constant vector x of the form 1 x = x1 3satisfies the equation x = Ax, and therefore is an equilibrium solution. Noteespecially that x = x1 [1 3]T is an eigenvector associated with λ = 0, and thusevery eigenvector associated with the zero eigenvalue is an equilibrium solutionto the system. The interesting behaviors that we have discussed algebraically are seenin figure 3.9. Specifically, every non-constant solution is a straight linesolution in the direction of the eigenvector [−1 1]T that is drawn toward anequilibrium point that lies on the eigenvector [1 3]T corresponding to the zeroeigenvalue.The flows in figure 3.9, as well as the long-term behavior of the function e −4t inthe general solution x(t ), clearly demonstrate that every equilibrium solutionto the system is stable. Moreover, we say that each such equilibrium point is anattracting node. There are two important observations to make in closing. One is that westill must address the situations where A lacks two real linearly independenteigenvectors; we will do so in the next section. In addition, examples 3.4.1–3.4.4 x2 4 x1 −4 4 −4 Figure 3.9 The direction field for the system x = Ax of example 3.4.4 along with several trajectories.
Systems with all real linearly independent eigenvectors 219indicate that plotting a direction field is perhaps best left to a computer; however,in the case where A has two real, linearly independent eigenvectors, it is astraightforward exercise use the eigenvectors to plot these straight-line solutionsby hand and to use the signs of the corresponding eigenvalues to understandthe flows along the straight line solutions. Then, it is not difficult to imagine theoverall appearance of the direction field and sketch several probable trajectoriesby hand, thus fully understanding the graphical behavior of all solutions to thesystem.3.4.1 Plotting direction fields for systems using MapleWe again use the DEtools package, and load it with the command > with(DEtools):To plot the direction field associated with a given system of differentialequations, we first define the system itself, similar to how we defined a singledifferential equation in order to plot its slope field. We do this through the 3 2following command for the system with coefficient matrix A = from 2 3example 3.4.1. > sys := diff(x(t),t)= 3*x(t)+2*y(t), diff(y(t),t)= 2*x(t)+3*y(t);The system of differential equations of interest is now stored in "sys". Whilewe typically use x1 (t ) and x2 (t ) to represent the component functions in ourdiscussion of the theory and solution of systems, in working with Maple it isoften simpler to use x(t ) and y(t ). The direction field may now be generated bythe command > DEplot([sys], [x(t),y(t)], t=-1..1, x=-4..4, y=-4..4, arrows=large, color=gray);This command produces the output shown at left in figure 3.10.From here, it is a straightforward exercise to sketch trajectories by hand. Ofcourse, Maple has the capacity to include trajectories that pass through anyinitial conditions we choose. For example, if we are interested in the variousinitial conditions x(0) = (2, 2), (0, 4), (4, 0), and (−1, 1), we can modify theearlier DEplot command to > DEplot([sys], [x(t),y(t)], t=-1.6..3.6, x=-4..4, y=-4..4, arrows=large, color=gray, [[x(0)=-2,y(0)=0], [x(0)=0,y(0)=-2], [x(0)=2,y(0)=0], [x(0)=0,y(0)=2],
220 Linear systems of differential equations x2 x2 4 4 x1 x1 −4 4 −4 4 −4 −4Figure 3.10 At left, the direction field for the system x = Ax of example 3.4.1. At right,the same direction field with several trajectories. [x(0)=0.1,y(0)=0.1], [x(0)=-0.1,y(0)=-0.1], [x(0)=0.1,y(0)=-0.1], [x(0)=-0.1,y(0)=0.1]]);The results of this most recent DEplot command are shown at rightin figure 3.10. As always, the user can experiment some with the window in which the plotis displayed: the range of x- and y-values can affect how clearly the direction fieldis revealed, and the range of t -values determines how much of each trajectory isplotted.Exercises 3.4 1. Consider the system of differential equations x = Ax given by 2 −1 A= 3 −2 2. Consider the system of differential equations x = Ax given by 3 1 A= 1 3 (a) Determine the general solution to the system x = Ax. (b) Classify the stability of all equilibrium solutions to the system.
Systems with all real linearly independent eigenvectors 221 (c) Sketch all straight-line solutions to the system and hence plot several nonlinear trajectories in the phase plane. 3. Consider the system of differential equations x = Ax given by −3 2 A= 2 −3 4. Consider the system of differential equations x = Ax given by −2 0 A= 0 −2 (a) Determine the general solution to the system x = Ax. (b) Classify the stability of all equilibrium solutions to the system. (c) Sketch the straight-line solutions to the system that correspond to the two linearly independent eigenvectors. Why is every solution to this system also a straight-line solution? 5. Consider the system of differential equations x = Ax given by −2 2 A= 1 −1 (a) Determine the general solution to the system x = Ax. (b) Classify the stability of all equilibrium solutions to the system. (c) Why is every non-constant solution to this system also a straight-line solution? How are these straight-line solutions related to the eigenvectors of the system?In exercises 6–9, let x(t ) be the stated general solution to some system x = Ax.State the straight-line solutions to the system, classify the stability of the origin,and sketch some sample trajectories. 1 3 6. x(t ) = c1 e −2t + c2 e −5t 3 1 −1 1 7. x(t ) = c1 e 4t + c2 e −3t 2 2 2 1 8. x(t ) = c1 e 2t + c2 −1 1
222 Linear systems of differential equations 1 −1 9. x(t ) = c1 e 0.1t + c2 e 10t 1 110. For the system x = Ax whose general solution is given in exercise 6, determine a possible matrix A for the system. (Hint: If A is a matrix with all real linearly independent eigenvectors and those eigenvectors are the columns of a matrix P, then A satisfies the equation AP = PD, where D is the diagonal matrix whose entries are the eigenvalues of A in order corresponding to the eigenvectors in the columns of P.)11. For the system x = Ax whose general solution is given in exercise 7, determine a possible matrix A for the system.12. Consider the four systems of equations given by x = Ax where A is given by the matrices I, II, III, and IV below. Match each system with one of the four direction field plots (a), (b), (c), and (d) given below. Write one sentence for each to explain the reasoning behind your choice. 5 3 2 −4 2 7 2 3 I. A = II. A = III. A = IV. A = 3 5 −1 2 7 2 3 −6 x2 x2 4 4(a) x1 (b) x1 −4 4 −4 4 −4 −4 x2 x2 4 4(c) x1 (d) x1 −4 4 −4 4 −4 −4
When a matrix lacks two real linearly independent eigenvectors 223In exercises 13–17, solve the IVP x = Ax with the given matrix A and statedinitial condition. 2 −113. A = , x(0) = [1 2] 3 −2 3 114. A = , x(0) = [−3 1]T 1 3 −3 215. A = , x(0) = [1 − 2]T 2 −3 −2 016. A = , x(0) = [−2 − 2]T 0 −2 −2 217. A = , x(0) = [1 4]T 1 −1In exercises 18–22, use the standard substitution to convert the given second-order differential equation to a system of two linear first-order equations. Solvethe system to hence determine the solution y to the second-order equation.18. y − y − 6y = 019. y − 6y + 5y = 020. y + 4y = 021. y + 3y + 2y = 022. y + y = 03.5 When a matrix lacks two real linearly independent eigenvectorsWe have seen repeatedly, both in theory and in specific examples, that when a2 × 2 matrix A has two real linearly independent eigenvectors, we can determinethe general solution to x = Ax and its graphical behavior. In this section,we address two remaining cases: when A has a repeated eigenvalue and onlyone associated real linearly independent eigenvector, and when A has complexeigenvalues and eigenvectors. In each case, we work through preliminary exam-ples to discover general patterns and principles, expand these principles withappropriate theorems, and explore and discuss graphical behavior along the way.Example 3.5.1 Consider the system of differential equations given by x = Ax −2 1where A = . Compute the eigenvalues and eigenvectors of A and 0 −2
224 Linear systems of differential equationsexplain why this alone does not lead to the general solution of the system. Bynoting that the system is partially coupled, solve the system and determine asecond real, linearly independent solution. Finally, state the general solution.Solution. By inspection, since A is a triangular matrix, we see that λ = −2is a repeated eigenvalue of A with multiplicity 2. From this, we deduce thatv1 = [1 0]T is a corresponding eigenvector, and therefore one solution tox = Ax is x1 = c1 e −2t [1 0]T . However, A lacks a second linearly independenteigenvector associated with λ = −2; therefore, we need to find a second reallinearly independent solution to the system in order to determine the generalsolution to x = Ax. In this example, we are fortunate that the system is onlypartially coupled and that therefore we may solve the system directly by usingtechniques for single differential equations from chapter 2. In particular, noting that the second equation in the system is x2 = −2x2 ,it follows immediately that the solution to this single differential equation isx2 (t ) = ce −2t . Substituting this result into the equation x1 = −2x1 + x2 , itremains for us to solve the single nonhomogeneous linear first-order differentialequation x1 = −2x1 + ce −2tApplying our understanding of such equations from section 2.3, via theintegrating factor v(t ) = e 2t we know that 1 x1 (t ) = e 2t · ce −2t dt = e −2t (ct + k) e 2tTo summarize, with x1 (t ) and x2 (t ) as the components of x(t ), we have foundthat a solution to the system is x1 (t ) x(t ) = x2 (t ) e −2t (ct + k) = (3.5.1) ce −2tIf we factor this expression to write x(t ) as a linear combination of two vectorsin order to more clearly identify the role of the constants in (3.5.1), we see e −2t te −2t x(t ) = k + c −2t (3.5.2) 0 eIn this form, two key observations can be made. First, each individual vectorin (3.5.2) may be verified to be a solution to the given system. Moreover, thesetwo vectors are linearly independent. Hence, (3.5.2) is the general solution tothe given system.While it is good that we were able to solve the system in example 3.5.1, it is stillunclear how we will proceed in similar circumstances when neither equation inthe system may be solved by techniques for single first-order equations. That is,
When a matrix lacks two real linearly independent eigenvectors 225if the equation for x1 involves x2 and the equation for x 2 involves x1 , butthe system's matrix has only one linearly independent eigenvector, we cannotemploy the approach used in example 3.5.1. However, the general form of thesolution (3.5.2) can help us guess an appropriate form of the needed secondlinearly independent solution in the more general case. Recall that we know that whenever (λ, v) is a real eigenpair of A, the functionx(t ) = e λt v is a solution to x = Ax, and moreover x(t ) is a straight-line solutionto the system. In example 3.5.1, we found that for the given matrix, which had arepeated eigenvalue and only one associated linearly independent eigenvector,the scalar function te λt arose in the solution. If we recall that our original workwith e λt v arose from guessing that a function of the form f (t )v was a solutionto x = Ax, example 3.5.1 now suggests that in the case where we are missingan eigenvector, we consider a vector function that somehow involves the scalarfunction te λt as a second linearly independent solution to x = Ax. A closer lookat (3.5.2) suggests the form of this second solution we seek. In particular, recalling that the matrix A in example 3.5.1 had v1 = [1 0]Tas the eigenvector corresponding to λ = −2, rewriting (3.5.2) reveals the role v1plays in the general solution. Specifically, 1 1 0 x(t ) = ke −2t + cte −2t + ce −2t (3.5.3) 0 0 1and since x1 (t ) = e −2t [1 0]T is the standard solution that arises through theeigenpair, we see from (3.5.3) that the second linearly independent solution 1 0 x2 (t ) = te −2t + e −2t 0 1has the form te −2t v + e −2t u, where u is not an eigenvector of A correspondingto λ = −2. This suggests a form for the second solution when this case arises ingeneral. We now consider this situation for an arbitrary matrix with the appropriateproperties. Let A be a 2 × 2 matrix with a single real, repeated eigenvalue λ withonly one linearly independent eigenvector v. Note specifically that we knowAv = λv and x1 (t ) = e λt v is a solution to x = Ax. Now consider a secondfunction x2 (t ) = te λt v + e λt u (3.5.4)where u is an unknown constant vector and (λ, v) remains an eigenpair of A.We seek conditions on u that will make x2 (t ) a solution to x = Ax; as wehave previously encountered in several instances, direct substitution into thedifferential equation reveals the constraints on u. First, differentiating (3.5.4) gives x2 (t ) = (λte λt + e λt )v + λe λt u (3.5.5)Next, observe that multiplying x2 (t ) by A yields Ax2 (t ) = A(te λt v + e λt u) = te λt (Av) + e λt (Au) (3.5.6)
226 Linear systems of differential equationsIn order for x2 (t ) to be a solution to x = Ax, it follows from (3.5.5) and (3.5.6)that we require the equality (λte λt + e λt )v + λe λt u = te λt (Av) + e λt (Au) (3.5.7)to hold. Using the fact that Av = λv and expanding, we find λte λt v + e λt v + λe λt u = λte λt v + e λt (Au) (3.5.8)With λte λt v present on both sides of (3.5.8), we can simplify the equality to e λt v + λe λt u = e λt (Au) (3.5.9)Since e λt is never zero, we observe from (3.5.9) that u must satisfy the equation v + λu = Au (3.5.10)In other words, (A − λI)u = v, where (as we assumed earlier) v is an eigenvectorof A that corresponds to the eigenvalue λ. In particular, note that v satisfiesthe equation (A − λI)v = 0. We summarize our work above in the followingtheorem.Theorem 3.5.1 If A is a 2 × 2 matrix with repeated eigenvalue λ and only onecorresponding linearly independent eigenvector v, then the general solution tox = Ax is given by x(t ) = c1 e λt v + c2 e λt (t v + u)where u satisfies the equation (A − λI)u = v.The vector u is often called a generalized eigenvector of A corresponding to λ.We now demonstrate the role of theorem 3.5.1 in the following example. 1 4Example 3.5.2 Let A = and consider the system of differential −1 5Solution. We find that A has a single repeated eigenvalue λ = 3 with just onecorresponding linearly independent eigenvector v = [2 1]T . Thus, one linearlyindependent solution to x = Ax is x1 (t ) = e 3t v. Applying theorem 3.5.1, wedetermine a second linearly independent solution to the system. Specifically,we first solve the vector equation (A − 3I)u = v. To do so, we row-reduce theappropriate augmented matrix and find −2 4 2 1 −2 −1 → −1 2 1 0 0 0It follows that the vector u must have components u1 and u2 that satisfy theequation u1 = 2u2 − 1, where u2 is a free variable. Since we only need one
When a matrix lacks two real linearly independent eigenvectors 227 x2 4 x1 −4 4 −4 Figure 3.11 The direction field for the system x = Ax of example 3.5.2 along with several trajectories.such vector u, we choose u2 = 0 and thus u1 = −1. From theorem 3.5.1, itnow follows that a second linearly independent solution to x = Ax is givenby the function x2 (t ) = e 3t (t v + u). In particular, the general solution tox = A x is 2 2 −1 x(t ) = c1 e 3t + c2 e 3t t + 1 1 0We note further that since A is an invertible matrix, the only solution toAx = 0 is x = 0, so the origin is the only equilibrium solution of thesystem. As figure 3.11 shows, the direction field and several trajectories exhibitbehavior consistent with the fact that the system has just one straight-line solution, the one that corresponds to the single linearly independenteigenvector of A. Note as well that since the system's only eigenvalue ispositive, every non-constant solution flows away from the origin as t → ∞.In example 3.4.3, the origin is obviously an unstable equilibrium solution.Because there is only one linearly independent eigenvector for the system, wecall the origin a degenerate node, and in this case where λ = 3 > 0 and all thetrajectories flow away from the origin, this degenerate node is also called arepelling node. We now consider an example that reveals the other possible situation thatcan arise when a matrix A lacks two real linearly independent eigenvectors: whenA has no real eigenvalues and no real eigenvectors.
228 Linear systems of differential equationsExample 3.5.3 Consider the system x = Ax given by the matrix 0 −1 A= 1 0Compute the eigenvalues and eigenvectors of A and explain why this does notlead directly to the general solution of the system. In addition, plot the directionfield for the system to confirm these observations from a graphical perspective.Using familiarity with solutions to single differential equations and the form ofthe equations for the given system, determine the general solution to the system.Solution. The eigenvalues of the matrix A are computed using thecharacteristic equation −λ −1 det(A − λI) = det = λ2 + 1 = 0 1 −λ √We see that λ2 = −1, so that λ = ±i, where i is the complex number2 i = −1. To determine the eigenvector associated with λ = i, we solve (A − iI)v = 0.Row-reducing the appropriate matrix with complex entries just as we would amatrix with real entries, we observe −i − 1 0 1 −i 0 1 −i 0 → → 1 −i 0 −i −1 0 0 0 0where the first step was achieved by swapping the two rows, while the last stepwas achieved by computing the row replacement iR1 + R2 → R2 . It follows thatany eigenvector v associated with λ = i must have components v1 and v2 thatsatisfy v1 = iv2 . Choosing v 2 = 1, we see that an eigenvector v corresponding toλ = i is v = [i 1]T . Similar computations with λ = −i show that a correspondingeigenvector is v = [−i 1]T . While we might suggest at this point that i x(t ) = e it 1is a solution to x = Ax, such a solution involves the complex number i, andis not a real solution to the system. A plot of the direction field for the systemreveals further why no real solutions arise directly from the eigenvectors. Inparticular, if we examine figure 3.12, the direction field and various trajectoriesexhibit behavior consistent with the fact that the system has no straight-linesolutions due to the fact that it has no real eigenpairs: every trajectory appearsto be circular. In this example, we will suspend our work with eigenvalues and eigenvectorsand see whether we can determine a solution to the system more directly. If weexamine the two equations given in the system x = Ax, we observe that we2 A review of key concepts with complex numbers may be found in appendix B.
When a matrix lacks two real linearly independent eigenvectors 229 x2 4 x1 −4 4 −4 Figure 3.12 The direction field for the system x = Ax of example 3.5.3.are trying to solve the two equations x1 = −x2 and x2 = x1 simultaneously. Inparticular, we seek two functions x1 (t ) and x2 (t ) such that the derivative of thefirst is the opposite of the second and the derivative of the second is the first. Thisis a familiar scenario encountered in calculus and we recognize that x1 (t ) = cos tand x2 (t ) = sin t form a pair of such functions. Further consideration revealsthat the choices x1 (t ) = − sin t and x2 (t ) = cos t also satisfy the system. Our recent observations show that the vector functions cos t − sin t x1 (t ) = and x2 (t ) = sin t cos teach form a real solution to x = Ax; moreover, it is clear that x1 (t ) andx2 (t ) are not scalar multiples of one another, and thus these are two linearlyindependent solutions to the system. Therefore, theorem 3.3.2 implies that thegeneral solution to the given system is cos t − sin t x(t ) = c1 + c2 (3.5.11) sin t cos tThe presence of the sine and cosine functions in the entries of x will also lead tothe circular trajectories we expect from the direction field in figure 3.12.Example 3.5.3 shows several new phenomena. In every preceding example wehave considered for 2 × 2 systems x = Ax, eigenpairs have directly provided atleast one real solution to the system. But for the latest system we examined,the eigenpairs appeared to not produce any solutions to the system at all.Moreover, for the first time in our work with linear systems, the sine and cosine
230 Linear systems of differential equationsfunctions arose. An important question to consider at this point is whether thecomplex eigenpair i λ = i, v = (3.5.12) 1can be linked to the general solution that we found in (3.5.11). It turns outthat the key idea lies in understanding how the exponential function e z behaveswhen the input z is a complex number. The great Swiss mathematician Leonhard Euler (1707–1783) is creditedwith discovering Euler's formula, which states that for any real number t , e it = cos t + i sin t (3.5.13)In exercise 14 in this section, one way to derive Euler's formula through Taylorseries for the exponential and trigonometric functions is explored. For now, wewill simply accept (3.5.13) and put it to use. Using the first complex eigenpair found in example 3.5.3, let us considerthe standard form of a potential solution to x = Ax, x(t ) = e λt v, using theeigenpair identified in (3.5.12). Here, since the solution we are considering is infact complex, we will use the notation z(t ). Using Euler's formula and complexarithmetic, observe that i z(t ) = e it 1 i = (cos t + i sin t ) 1 i cos t − sin t = (3.5.14) cos t + i sin tWhen working with complex numbers, it is often useful to identify the real andimaginary parts of the numbers. That is, for a complex number z = a + ib wherea and b are real, we call a the real part of z, and b the imaginary part of z. Thesame distinctions hold for vectors with complex entries. Considering (3.5.14),if we separate this vector into its real and imaginary parts, we may write − sin t cos t z(t ) = +i (3.5.15) cos t sin tIf we now compare the general solution to x = Ax that we found in (3.5.11)to (3.5.15) above, we can make a critical observation. The two linearlyindependent solutions to the system seen in (3.5.11) are in fact the real andcomplex parts of the vector z(t ) which arose from considering z(t ) = e λt vwhere (λ, v) was a complex eigenpair of A. That this fact holds in general is ournext stated theorem.Theorem 3.5.2 If A is a real 2 × 2 matrix with a complex eigenvalue λ = a + iband corresponding eigenvector v = p + iq, where a, b, p, and q are real, then
232 Linear systems of differential equations x2 4 x1 −4 4 −4 Figure 3.13 The direction field for the system x = Ax of example 3.5.4 along with several trajectories.real eigenvectors and therefore no straight-line solutions. Moreover, since thereal part of λ = −1 + 2i is negative, the role of e −t in the general solution (3.5.16)draws every solution to 0 and thus the origin is a stable equilibrium.In cases such as the one in example 3.5.4 where there are no straight-linesolutions and every nonconstant solution tends to 0 as t → ∞, we naturallysay that 0 is a spiral sink. Note that this case corresponds to the situation wherethe real part of a complex eigenvalue is negative. If the real part a of λ = a + bi ispositive, then we will have e at present in the general solution, and this will driveevery solution away from the origin. We therefore call 0 a spiral source and notethat this equilibrium solution is unstable. Finally, in the event that a = 0 in thecomplex eigenvalue λ = a + bi, as it was in example 3.5.3, then all nonconstantsolutions will orbit the origin while neither being drawn toward or repelled fromthe equilibrium solution. See, for example, figure 3.12. Such an equilibrium iscalled a center and is considered stable. In our discussions in this section we have addressed the two possible casesfor a 2 × 2 matrix A which lacks two linearly independent eigenvectors. Ourwork extends naturally to the case of more general n × n systems where then × n matrix A may or may not have n real linearly independent eigenvectors.Of course, in the case where A has a full set of n real linearly independenteigenvectors, the eigenpairs allow the general solution to the system to bedetermined. In cases where some of the eigenvalues are complex, or repeatedwith missing eigenvectors, we can work with each individual eigenvalue to buildreal linearly independent solutions in ways similar to our preceding work. Someexamples are explored in the exercises that follow.
234 Linear systems of differential equationsthe matrix A. For each, classify the stability of the origin as an equilibriumpoint of the system given by x = Ax. 8. p(λ) = λ2 − 4 9. p(λ) = λ2 + 410. p(λ) = λ2 + λ + 111. p(λ) = λ2 − 10λ + 912. p(λ) = λ2 − 2λ + 513. p(λ) = λ2 + 3λ + 214. Recall or look up the formulas for the Taylor series about a = 0 for each of the functions e x , sin x, and cos x. Assuming that the Taylor series for e x is valid for complex numbers x, compute e ib and compare the result to the expansions for cos b and i sin b to show that e ib = cos b + i sin b In addition, show that e a +ib = e a (cos b + i sin b)In exercises 15–19, a matrix A is given. For each, consider the system ofdifferential equations x = Ax and respond to (a) - (d).(a) Determine the general solution to the system x = Ax.(b) Classify the stability of all equilibrium solutions to the system.(c) How many straight-line solutions does this system of equations have? Why?(d) Use a computer algebra system to plot the direction field for this system and sketch several trajectories by hand. 0 −215. A = 2 0 2 −316. A = 3 2 −2 117. A = 0 −2 −4 518. A = −5 4 7 −119. A = 4 11
236 Linear systems of differential equations3.6 Nonhomogeneous systems: undetermined coefficientsSo far in our studies of systems of linear differential equations, we have focusedalmost exclusively on the case where the system is homogeneous and can berepresented in the form x = Ax. We now begin to investigate nonhomogeneoussystems, which are systems of the form x = Ax + b where b = 0. In section 3.1, we encountered a system of two tanks where we wereinterested in the amount of salt in each tank at time t . With the amount ofsalt in the two tanks represented respectively by x1 (t ) and x2 (t ), we saw thatthese component functions had to satisfy the system of differential equationsgiven by −1/20 1/80 x1 20 x = + (3.6.1) 1/40 −1/40 x2 35and that this system is naturally represented in the form x = Ax + b (3.6.2) In our most recent work with the homogeneous equation x = Ax, we notedseveral times the analogy to solving the single first-order differential equationx = ax. In particular, we observed the key role that e λt plays in the process ofsolving homogeneous systems of equations, much like e at does in the solutionof a single homogeneous linear first-order equation. We next naturally consider the linear first-order analogy of (3.6.2),a nonhomogeneous equation such as y = 2y + 5 (3.6.3)In section 2.3, we made the observation in theorem 2.3.3 that for any linearfirst-order differential equation in the form y + p(t )y = f (t )if yp is any solution to the nonhomogeneous equation and yh is a solution tothe corresponding homogeneous equation, then y = yp + yh is a solution to thenonhomogeneous equation. In our studies of linear algebra in chapter 1, we made a similar observationin section 1.5: if we have a solution xp to the nonhomogeneous equation Ax = b,and we add to xp any solution xh to the homogeneous equation Ax = 0, the result(x = xp + xh ) is also a solution to Ax = b. See (1.5.1) to revisit the details of thisdiscussion. Note that in this purely linear algebra context, x is a vector whoseentries are constant. These two preceding observations for linear first-order differential equa-tions and systems of linear algebraic equations are now applied to thenonhomogeneous system of linear first-order differential equations, x = Ax + b.We note specifically that in this context, x(t ) is a function of t . Let's return to
Nonhomogeneous systems: undetermined coefficients 237the known situation of the homogeneous system x = Ax and denote its solutionby xh (t ). In addition, suppose we are able to determine a single solution xp (t )to the nonhomogeneous equation x = Ax + b. We claim that the functionx(t ) = xh (t ) + xp (t ) is the general solution of the nonhomogeneous equation.To see this, we substitute directly into x = Ax + b and verify that the equationis satisfied. By properties of linearity, observe that x (t ) = xh (t ) + xp (t ) (3.6.4)and furthermore Ax + b = A(xh + xp ) + b = Axh + Axp + b (3.6.5)By how we defined xh (t ) and xp (t ), we know that xh (t ) = Axh (t ) andxp (t ) = Axp (t ) + b, and thus (3.6.5) implies Ax + b = xh (t ) + xp (t ) (3.6.6)From (3.6.4) and (3.6.6), we see that x = xh + xp is indeed a solution tox = Ax + b. In fact, we have found the general solution to the nonhomogeneoussystem, as stated in the following theorem.Theorem 3.6.1 Let A be an n × n matrix with constant coefficients. If xh isthe general solution to the homogeneous system x = Ax and xp is any solutionto the nonhomogeneous system x = Ax + b, then x = xh + xp is the generalsolution to x = Ax + b. Theorem 3.6.1 provides an approach that will guide us throughout ourefforts to solve nonhomogeneous systems of differential equations. First, wesolve the associated homogeneous system to find xh , a process we are familiarwith. We usually call xh the complementary solution to the equation x = Ax + b.Next, we must find a so-called particular solution xp to the nonhomogeneoussystem x = Ax + b. Although a more sophisticated approach will be introducedin the next section, for now we will investigate a few examples in which theprocess of finding such a particular solution xp is relatively straightforward.Example 3.6.1 From the system of two tanks discussed in sections 1.1 and 3.1,consider the nonhomogeneous system of linear differential equations given by −1/20 1/80 20 x = x+ (3.6.7) 1/40 −1/40 35By solving the associated homogeneous system and determining a particularsolution to the nonhomogeneous system, find the general solution to the givensystem. In addition, plot an appropriate direction field and discuss the long-term behavior of solutions and their meaning in the context of the salt ineach tank. Determine and sketch the solution to the IVP with initial conditionx(0) = [2000 1000]T .
238 Linear systems of differential equationsSolution. We begin by solving x = Ax, where −1/20 1/80 A= 1/40 −1/40The eigenvalues of A are approximately λ1 = −0.158 and λ2 = −0.592,with corresponding eigenvectors approximated by v1 = [0.366 1.000]T andv2 = [−1.366 1.000]T . It follows that the general solution xh is 0.366 −1.366 xh (t ) = c1 e −0.158t + c2 e −0.592t 1.000 1.000Next, we must determine a particular solution xp to the nonhomogeneousequation x = Ax + b. In this particular example, b is a constant vector.Therefore, it is natural to guess that a constant vector xp will satisfy thenonhomogeneous equation. More than this, we should recall from earlierdiscussions of the problem leading to the given system that the vector xrepresents the amounts of salt in two connected tanks as streams of inflowdeliver salt, each at a constant rate. Our intuition suggests that over time thetwo tanks should approach a stable equilibrium, and hence an equilibrium (andtherefore constant) solution should be present. Therefore, we assume that xp is a constant vector and observe that thisimmediately implies that xp = 0. Substituting into x = Ax + b, it follows thatxp must satisfy the system of linear equations 0 = Axp + b or Axp = −b. With thegiven entries of A and b, this leads us to row reduce the appropriate augmentedmatrix and find that −1/20 1/80 −20 1 0 1000 → 1/40 −1/40 −35 0 1 2400This shows xp = [1000 2400]T is a particular solution to x = Ax + b, and, morespecifically, is an equilibrium solution of the system. Moreover, it now followsthat the general solution to the system is given by 0.366 −1.366 1000 x(t ) = xh (t ) + xp (t ) = c1 e −0.158t + c2 e −0.592t + 1.000 1.000 2400 (3.6.8)If we add the initial condition that x(0) = [2000 1000] T , we can solve for theconstants c1 and c2 , and plot the appropriate corresponding trajectory, as shownin figure 3.14. In both (3.6.8) and figure 3.14 we can see how the long-termbehavior of every solution tends to the equilibrium solution. Moreover, in thedirection field we can also recognize the straight-line solutions that correspondto lines in the direction of each eigenvector but that now pass through theequilibrium solution (1000, 2400).From example 3.6.1, we observe that in cases where we want to solve x = Ax + band b is itself a constant vector, xp may be determined by assuming that xp is aconstant vector and solving 0 = Axp + b. If xp is not constant, then the situationis more complicated, as we discover in the following example.
Nonhomogeneous systems: undetermined coefficients 239 x2 5000 3000 equilibrium solution solution through (1000, 2400) (2000, 1000) 1000 x1 1000 2000 Figure 3.14 The direction field for the system x = Ax + b of example 3.6.1.Example 3.6.2 Find the general solution of the nonhomogeneous systemgiven by 2 −1 cos 2t x = x+ (3.6.9) 3 −2 0 2 −1Solution. Since the eigenvalues of A = are λ1 = −1 and λ2 = 1 with 3 −2corresponding eigenvectors v1 = [1 3]T and v2 = [1 1]T , it follows that thecomplementary solution to the related homogeneous system is 1 1 xh = c1 e −t + c2 e t 3 1 To determine the particular solution xp to the given nonhomogeneoussystem, we need to find a vector function x(t ) that simultaneously satisfiesthe system (3.6.9). Due to the presence of cos 2t in the vector b, it is naturalto guess that the components of xp will somehow involve cos 2t . In addition,since xp plays a role in the system, we must account for the possibility that thederivative of cos 2t may also arise; moreover, since Ax will also be computed,linear combinations of vectors that involve the entries in x will be present.Therefore, we make the reasonable guess that xp has the form a cos 2t + b sin 2t xp = (3.6.10) c cos 2t + d sin 2tand attempt to determine values for the undetermined coefficients a , b , c , andd that make xp a solution to the system.We accomplish this by direct substitution into (3.6.9). First, observe that −2a sin 2t + 2b cos 2t xp = (3.6.11) −2c sin 2t + 2d cos 2t
Nonhomogeneous systems: undetermined coefficients 241call the process of finding xp through a guess involving unknown constants themethod of undetermined coefficients. To gain a better sense of the guesses that are involved in using undeterminedcoefficients, we turn to the following example.Example 3.6.3 For nonhomogeneous linear systems of the form x = Ax + bwhere A is a matrix with constant entries, state the natural guess to use for xpwhen the vector b is e −t 1 t2 e −3t (a) b = (b) b = (c) b = (d) b = 2e −t t 0 −2Solution. (a) With b = [e −t 2e −t ]T , it is natural to expect that any particular solution must involve e −t in its components. Specifically, we make the guess that Ae −t xp = Be −t and substitute directly into x = Ax + b in order to attempt to find values of A and B for which xp satisfies the given system.3(b) Given b = [1 t ]T , we must account for the fact that xp and its derivative can involve constant and linear functions of t . In particular, we suppose that At + B xp = Ct + D and substitute appropriately in an effort to determine A, B, C, and D.(c) For b = [t 2 0]T , with one quadratic term present in b, it is necessary to include quadratic terms in each entry of xp . But since the derivative of xp will be taken, linear terms must be included as well. Finally, once linear terms are included, for the same reason we must permit the possibility that constant terms can be present in xp . Therefore, we guess the form At 2 + Bt + C xp = Dt 2 + Et + F(d) With b = [e −3t − 2]T having both an exponential and constant term present, we account for both of these scalar functions and their derivative by assuming that Ae −3t + B xp = Ce −3t + D3 It is possible that the guess can fail to work, in which case a modified form for x is required. One psetting where this may occur is when λ = −1 is an eigenvalue of A, whereby a vector involving e −talready appears in the complementary solution xh . See exercise 8 for further investigation of thisissue.
242 Linear systems of differential equationsThe method of undetermined coefficients is not foolproof: it is certainly possibleto guess incorrectly (as noted in the footnote related to part (a) of example 3.6.3).If our guess is incorrect, an inconsistent linear system of algebraic equations willarise, which tells us we need to modify our guess. Besides the possibility ofguessing incorrectly, it can also be the case that the computations involved indetermining xp are very cumbersome. In the next section, we consider a differentapproach, one that parallels our solution of single linear first-order differentialequations of the form y + p(t )y = f (t ), that provides, at least in theory, analgorithmic approach to solving any nonhomogeneous system x = Ax + bwhere the matrix A has real, constant entries. Finally, we note that the presence of nonconstant entries in the vector b ina nonhomogeneous system x = Ax + b makes it impossible to plot a directionfield for the system. In particular, when we sketch direction fields, we rely onthe fact that regardless of time, t , the direction vector x to the solution curve xis dependent only on the location (x1 , x2 ), and not on t . When b is nonconstantand a function of t , this is no longer the case and we therefore are left with onlyalgebraic approaches to the problem. If b is constant, then we can generate thedirection field for the system, such as the one shown in figure 3.14.Exercises 3.6 In each of exercises 1–4, show by direct substitution that thegiven particular solution xp is indeed a solution to the stated nonhomogeneoussystem of equations. Hence determine the general solution to the stated system. −1 3 5 −4 1. x = x+ , xp = 2 −3 −1 −3 1 −2 e 2t −1/3 2. x = x+ , xp = e 2t −2 1 0 2/3 2 1 sin t −2/5 −3/10 3. x = x+ , xp = sin t + cos t 1 2 0 1/10 1/5 −3 1 e 2t + 1 1 3/14 4. x = x+ , xp = + e 2t 1 −1 1 2 1/14 5. Consider the system of differential equations 1 1 1 x = x+ 4 1 −3 (a) Explain why it is reasonable to assume that xp is a constant vector, and use this assumption to determine a particular solution to the given nonhomogeneous system. (b) Determine the complementary solution xh to the associated homogeneous system, x = Ax. (c) State the general solution to the system. (d) Is there an equilibrium solution to this system? If so, is it stable? Explain.
Nonhomogeneous systems: undetermined coefficients 2436. Consider the system of differential equations 1 1 e 4t x = x+ 4 1 0 (a) Explain why it is reasonable to assume that xp is a vector of the form ae 4t xp = be 4t Then use this assumption to determine a particular solution to the given nonhomogeneous system. (b) Determine the complementary solution xh to the associated homogeneus system, x = Ax. (c) State the general solution to the system.7. Consider the system of differential equations 1 1 e −2t + 1 x = x+ 4 1 2e −2t + 3 (a) Explain why it is reasonable to assume that xp is a vector of the form ae −2t + b xp = ce −2t + d Use this assumption to determine a particular solution to the given nonhomogeneous system. (b) Determine the complementary solution xh to the associated homogeneus system, x = Ax. (c) State the general solution to the system.8. Consider the system of differential equations 1 1 e −t x = x+ 4 1 0 (a) Explain why it is reasonable to assume that xp is a vector of the form ae −t xp = be −t (b) Show that the form of xp above does not result in a particular solution to the system. (c) By assuming that xp is a vector of the form ae −t + bte −t xp = ce −t + dte −t determine a particular solution to the given nonhomogeneous system (d) Determine the complementary solution xh to the associated homogeneus system, x = Ax. (e) State the general solution to the system.
246 Linear systems of differential equationsNow observe that the right side of the above equation—the overall vectorformulation of x—can be expressed as a matrix product. In particular, we write x= C (3.7.2)where C is the vector whose entries are the arbitrary constants c1 , . . . , cn thatarise in the formulation of the general solution x, and (t ) is the matrix whosecolumns are the n linearly independent solutions to x = Ax. We call (t ) thefundamental solution matrix of the system. At this point, it is essential to make two observations about (t ). The firstis that (t ) is nonsingular for every relevant value of t . This holds because thecolumns of (t ) are linearly independent since, by definition, they are linearlyindependent solutions of x = Ax. Second, we note that (t ) = A (t ). Sincethe derivative of (t ) is taken component-wise, this equation is simply thematrix way to say that each column of (t ) satisfies the homogeneous systemof equations x = Ax. Now, recall (3.7.2) where we expressed the complementary solution inthe form xh = (t )C. As we now seek a particular solution xp to thenonhomogeneous equation, it is natural to suppose that xp has the form xp (t ) = (t )u(t ) (3.7.3)where u(t ) is a function yet to be determined. We now substitute this guess forxp into x = Ax + b(t ) to see what conditions u must satisfy. For ease of display,in what follows we suppress the "(t )" notation in each of the functions , u, u ,and b. By the product rule, xp = ( u) = u + uand so substituting into the system x = Ax + b(t ), we have u + u = A u+b (3.7.4)Recalling our observation above that = A , we can substitute in (3.7.4)to find u +A u = A u+b (3.7.5)We next subtract A u from both sides of (3.7.5) to deduce that u =b (3.7.6)Since we are interested in determining the unknown function u, and we knowthat is nonsingular, we may now write −1 u = b (3.7.7)and, therefore, u must have the form −1 u(t ) = (t )b(t ) dt (3.7.8)
Nonhomogeneous systems: variation of parameters 247Finally, recalling the supposition we made in (3.7.3) that xp = u, (3.7.8) nowimplies −1 xp (t ) = (t ) (t )b(t ) dt (3.7.9)It is remarkable how this form of xp aligns with our experience with a single linearfirst-order differential equation and the form of its solution given by (3.7.1). Wesummarize our above work in the following theorem.Theorem 3.7.1 If A is an n × n matrix with constant entries, (t ) isthe fundamental solution matrix of the homogeneous system of differentialequations x = Ax, and b(t ) is a continuous vector function, then a particularsolution xp to the nonhomogeneous system x = Ax + b(t ) is given by −1 xp (t ) = (t ) (t )b(t ) dt (3.7.10)The approach to finding a particular solution given in theorem 3.7.1 is oftencalled variation of parameters. We next consider an example to see theorem 3.7.1at work.Example 3.7.1 Find the general solution of the nonhomogeneous systemgiven by 2 −1 0 x = x+ t 3 −2 4Solution. From our determination of the eigenvalues and eigenvectors of thesame coefficient matrix in example 3.6.2, the complementary solution is 1 1 xh = c1 e −t + c2 e t 3 1Therefore, the fundamental matrix is e −t e t (t ) = 3e −t e tAccording to (3.7.10), we next need to compute −1 . While the inverse of thismatrix of functions may be computed by row-reducing [ | I] in the usual way,because of the function coefficients in it is much easier to use a shortcut forcomputing the inverse of a 2 × 2 matrix that we established in exercise 19 ofsection 1.9. Specifically, if a b A= c dis an invertible matrix, then 1 d −b A−1 = det(A) −c a
250 Linear systems of differential equations At each stage in applying variation of parameters it is essential to simplify. Inparticular, −1 (t ) should be simplified as much as possible before computing −1 (t )b(t ), and similarly, −1 (t )b(t ) dt should be simplified as much aspossible before computing (t ) −1 (t )b(t ) dt . One option, of course, isto use a computer algebra system to avoid the more tedious aspects of thecomputations. We offer some suggestions for how to use Maple to assist in thecomputations in the following subsection.3.7.1 Applying variation of parameters using MapleHere we address how Maple can be used to execute the computations in aproblem such as the one posed in example 3.7.2, where we are interested insolving the nonhomogeneous linear system of equations given by 2 −1 1/(e t + 1) x = x+ 3 −2 1As usual, we load the Linear Algebra package. > with(LinearAlgebra):Because we already know how to find the complementary solution, we focuson determining xp by variation of parameters. First, we use the complementarysolution, 1 1 xh = c1 e −t + c2 e t 3 1to define the fundamental matrix (t ): > Phi := <<exp(-t),3*exp(-t)>|<exp(t),exp(t)>>;We next use the MatrixInverse command to find −1 by entering > MatrixInverse(Phi);The resulting output is − 1 e −t 2 1 1 1 2 e −t 3 1 2 et − 1 e1t 2We can simplify this result using negative exponents; Maple can do so throughthe following command, through which we also store −1 in PhiInv: > PhiInv := simplify(MatrixInverse(Phi));Next, in order to compute −1 (t )b(t ), we must enter the function b(t ). Weenter > b := <<1/(exp(t)+1),1>>;
Applications of linear systems 253 1 1 2t + 2 e t cos t e t sin t 11. x = x+ , (t ) = −1 1 0 −e t sin t e t cos t ⎡ ⎤ ⎡ t⎤ ⎡ 2t ⎤ 2 1 0 e e te 2t 0 12. x = ⎣0 2 0⎦ x + ⎣ 1⎦ , (t ) = ⎣ 0 e 2t 0⎦ 0 0 1 0 0 0 e −t 3.8 Applications of linear systems In this section, we consider three fundamental physical problems that may be modeled and studied using linear systems of differential equations. 3.8.1 Mixing problems Through our study of the motivating example provided at the start of chapter 1 and reconsidered at the beginning of the current chapter, we have seen that mixing problems naturally lead to nonhomogeneous linear systems of differential equations. Below, we examine a slightly more complicated example. Consider a system of three tanks connected in such a way that each of the tanks has an independent inflow that delivers salt solution to it, each has an independent outflow (drain), and each tank is connected to the other two with both outflow and inflow pipes. The relevant information about each tank is given in table 3.2. We set up a system of differential equations whose solution represents the amount of salt in each tank at time t and state the system in matrix form. For tank A, we denote the amount of salt (in grams) in the tank at time t (in minutes) by x1 (t ). Similarly, we let x2 (t ) and x3 (t ) represent the amount of salt in tanks B and C. A careful check of the given data shows that for each tank the total ratesTable 3.2Saltwater mixing in three tanks A, B, and C Tank A Tank B Tank CTank volume 50 liters 100 liters 200 litersRate of inflow to the tank 2 liters/min 4 liters/min 5 liters/minConcentration of salt in inflow 0.25 g/liter 2 g/liter 0.9 g/literRate of drain outflow 2 liters/min 4 liters/min 5 liters/minRates of outflows to other tanks to B: 3 liters/min to C: 1 liter/min to A: 4 liters/minRates of outflows to other tanks to C: 4 liters/min to A: 3 liters/min to B: 1 liter/min
254 Linear systems of differential equationsof inflow and outflow of solution balance so that the volume of solution in eachtank is constant. From the given information on the independent inflow to the tank, weknow that tank A gains salt at a rate of g liters g 0.25 ·2 = 0.5 (3.8.1) liter min minFurthermore, tank A also gains salt from the two inflows that come from tanksB and C. For tank B, which contains 100 liters of solution, solution flows to Aat a rate of 3 liters/min with a concentration of x2 (t )/100 g/liter, so that salt isgained by tank A at a rate of x2 g liters 3x2 g ·3 = (3.8.2) 100 liter min 100 minSimilarly, the flow from tank C to tank A results in A gaining salt at a rate of x3 g liters x3 g ·4 = (3.8.3) 200 liter min 50 min Tank A is also losing salt through its three outflows: a drain, flow to tankB, and flow to tank C. Since the concentration of solution in tank A at timet is x1 (t )/50 g/liter, it follows that each outflow carries this concentration ofsalt, doing so at respective rates of 2 liters/min, 3 liters/min, and 4 liters/min.This shows that solution is leaving tank A at a cumulative rate of 9 liters/min,therefore causing the rate at which salt is lost from tank A to be x1 g liters 9x1 g ·9 = (3.8.4) 50 liter min 50 minCombining the rates of inflow and outflow in (3.8.1), (3.8.2), (3.8.3), and (3.8.4),it follows that x1 (t ) satisfies the differential equation 3x2 4x3 9x1 x1 = 0.5 + + − (3.8.5) 100 200 50 Similar reasoning shows that x2 (t ) and x3 (t ) satisfy the differentialequations 3x1 x3 8x2 x2 = 8 + + − (3.8.6) 50 200 100and 4x1 x2 10x3 x3 = 4.5 + + − (3.8.7) 50 100 200Rearranging (3.8.5), (3.8.6), and (3.8.7) and writing the system they generate inmatrix form, we see ⎡ ⎤ ⎡ ⎤ −9/50 3/100 1/50 0.5 x = ⎣ 3/50 −2/25 1/200⎦ x + ⎣ 8⎦ (3.8.8) 2/25 1/100 −1/20 4.5
256 Linear systems of differential equations L k1 k2 m1 equilibrium m2 equilibrium Figure 3.15 Two masses m1 and m2 joined by two springs, at equilibrium. Next, we consider the more complicated case of a system involving twomasses and two springs, but omit damping and driving forces. In particular,suppose that a mass m1 is attached to a spring with spring constant k1 and thatfrom m1 a second spring with constant k2 and mass m2 is attached, as shownin figure 3.15. While we represent the masses with boxes, for our theoreticalwork we assume we are working with point-masses, where all of the mass isconcentrated at a single point. We can envision these points as lying at thecenters of the respective boxes in figure 3.15. To omit damping, we assume that the surface on which the masses rest isfrictionless. In addition, once the masses are set in motion by some collectionof initial displacements and velocities, we let x1 (t ) denote the displacement ofm1 from its equilibrium position and x2 (t ) the displacement of m2 from itsequilibrium position and set the system in motion, as shown in figure 3.16. We seek a system of first-order differential equations that models thissituation. Note that m1 has two springs attached to it, so each spring exertsforces on m1 . One is F1 = −k1 x1 , which is the force the first spring exerts tooppose the displacement of the first mass. Next, observe that when the system isat equilibrium, the distance between the two masses is some constant L. Oncethe system is set in motion, the distance between the two masses is L + x2 − x1 .As such, the second spring is being stretched a length of x2 − x1 beyond whereit is when the system is at equilibrium. On mass m1 this exerts a force in theopposite direction of F1 , specifically the force F2 = k2 (x2 − x1 ) on m1 . On thesecond mass m2 there is only this same force exerted by the second spring, butin the opposite direction as on m1 . In particular, F3 = −k2 (x2 − x1 ) acts on m2 . L x1 x2 Figure 3.16 Two masses m1 and m2 and two springs displaced from equilibrium.
Applications of linear systems 257 Now, because we have omitted damping and forcing, these are the onlyforces acting on m1 and m2 . Newton's second law tells us that the sum of allforces acting on an object must equal the object's mass times its acceleration. Inparticular, we have m1 x1 = −k1 x1 + k2 (x2 − x1 ) m2 x2 = −k2 (x2 − x1 )Dividing through by m1 and m2 , respectively, these observations lead us to thesystem of linear second-order differential equations k1 k2 x1 = − x1 + (x2 − x1 ) m1 m1 (3.8.11) k2 x2 = − (x2 − x1 ) m2To study the behavior of this system with the techniques that we have developed,we must convert each of the second-order equations to a system of two first-order equations. Before doing so, we introduce specific numerical values for themasses and spring constants to simplify our work. We let k1 = 2 and k2 = 1, andm1 = 2 and m2 = 4. This yields the system x1 = −x1 + 0.5(x2 − x1 ) (3.8.12) x2 = −0.25(x2 − x1 )Using the substitutions y1 = x1 , y2 = y1 = x1 , y3 = x2 , y4 = y3 = x2 , it followsthat (3.8.12) results in the system of four first-order equations given by y1 = y2 y2 = −y1 + 0.5(y3 − y1 ) (3.8.13) y3 = y4 y4 = −0.25(y3 − y1 )Letting y be the vector [y1 y2 y3 y4 ]T , we can write (3.8.13) in matrix form, ⎡ ⎤ 0 1 0 0 ⎢−1.5 0 0.5 0⎥ y =⎢⎣ ⎥y (3.8.14) 0 0 0 1⎦ 0.25 0 −0.25 0From this, we can now analyze the overall behavior of the coupled spring-masssystem. In particular, the eigenvalues and eigenvectors of the coefficient matrixin (3.8.14) will enable us to find the general solution y. Given initial conditions,we can fully describe the functions yi (t )—particularly y1 and y3 , which representthe respective displacements of the masses in the system—and understand thebehavior of the system over time. This problem and others like it are exploredfurther in the exercises at the end of this section.
258 Linear systems of differential equations3.8.3 RLC circuitsThe flow of electricity through a circuit, much like the flow of water in apipe, naturally involves relationships with rates of change. As such, the studyof electrical current involves differential equations. Here, we explore somefundamental properties of electricity and how these lead to such equations. Throughout what follows, we will make use of the analogy that the flow ofcharge carriers in an electrical circuit is like the flow of particles in a movingstream of water. Just as we consider flow of water in a pipe to be the numberof water particles flowing past a given point during a certain time interval, thecurrent I (t ) in a circuit at time t is proportional to the number of positivecharge carriers that move past any given point per second in the conductor.Note particularly that current measures a rate of change of charge. Current is measured in amperes(amp), the base unit through which all otherunits will be defined. One ampere corresponds to 6.2420 × 1018 charge carriersper second moving past a given point. The unit of charge is a coulomb, which isthe amount of charge that flows through a cross section of a wire in one secondwhen a one amp current is flowing. In other words, 1 amp = 1 coulomb/s Here, we begin to see how derivatives and integrals are involved in the studyof electricity. The current I (t ) at time t is by definition a rate of change of charge.Thus, by the Fundamental Theorem of Calculus, the total amount of charge thatflows past a given point on a time interval [t0 , t1 ] is given by t1 I (s) ds (3.8.15) t0If we let Q(t ) measure the total accumulated charge at a given point in the circuitfrom time t0 up to time t , then we have t Q(t ) = Q(t0 ) + I (s) ds (3.8.16) t0and therefore Q (t ) = I (t ). As current flows through a circuit, the charge carriers and elements in thecircuit exchange energy. We, therefore, define a potential function V throughouta circuit. The energy (per coulomb of charge) that has been exchanged by thecharge carriers as they flow from point a to point b is computed as Vab = Va − Vbwhere Va and Vb are the values of the potential function at points a and b in thecircuit. The difference Vab is called the voltage drop from a to b and is measured injoules per coulomb, which are also known as volts. If we again think of the flowof water through a pipe, the concept of voltage drop is analogous to the changein water pressure between points a and b. Batteries, for example, maintain avoltage drop between two terminals; the energy provided by a battery's internalchemicals produces a constant amount of energy per coulomb as charge carriers
Applications of linear systems 259move throughout the battery, which raises the function V by the voltage ratingof the battery. As current flows through a circuit, energy is lost. This makes the potentialV at one end lower than the potential at the other. Over a portion of a circuit,say from a to b, where a substantial amount of energy is lost, we say that such aportion is called a resistor. Good examples of resistors are light bulbs and heatingelements, because they show how electrical energy can be converted into lightand heat. The voltage drop across a resistor and the current flowing through it aremodeled by Ohm's law, which says that the potential difference Vab between theendpoints a and b of a resistor is proportional to the current flowing throughthe resistor. In other words, Vab = IR (3.8.17)where R is a constant called the resistance. The unit of resistance is the ohm,which is equal to one volt per ampere, or one volt-second per coulomb. A changing electrical current I (t ) in a segment of a circuit will create achanging magnetic field that results in a voltage drop between the ends of asegment. When this effect is large, such as in a coil between points b and c (theeffect can be magnified by different geometrical arrangements of the circuit),the device that induces the effect is called an inductor. Faraday's law tells us whathappens with the voltage drops across inductors. In particular, the voltage dropacross an inductor is proportional to the rate of change of the current, or, inother words dI Vbc = L (3.8.18) dtwhere L is a constant called the inductance. Note specifically that Faraday's lawregards the rate of change of current. Inductance is measured in henries. Finally, if a circuit is broken and we include two plates separated by aninsulating material (such as air), and the terminals of the circuit are connectedto a voltage source (such as a battery), then charges will build up on the plates.In the ongoing analogy to water, this is similar to a tank used to store water toprovide a source of pressure. We call the set of plates a capacitor, and speak ofthe total charge Q(t ) on the capacitor. From (3.8.16), since we know that current I is the rate of change of chargeQ, if we know an initial charge Q(t0 ), then given a current I (t ) we can find thecharge Q(t ) by the relationship t Q(t ) = Q(t0 ) + I (s) ds (3.8.19) t0Finally, Coulomb's law states that the voltage drop Vcd across a capacitor betweenpoints c and d is proportional to the charge on the capacitor, or 1 1 t Vcd = Q(t ) = Q(t0 ) + I (s)ds (3.8.20) C C t0where C is called the capacitance of the capacitor and is measured in farads.
260 Linear systems of differential equations All three of the laws (3.8.17), (3.8.18), and (3.8.20) are based onexperimental observations of circuits. Similarly, Kirchoff's law is a conservationlaw that tells us what we can expect for the voltage drops across various parts ofa circuit. Simply stated, Kirchoff's law says that if we pick a sequence of pointsin a closed circuit, then the sum of the voltage drops across these segments iszero. Specifically, for points a1 , a2 , . . . , an , Va1 a2 + Va2 a3 + · · · + Van−1 an + Van a1 = 0 (3.8.21) A final necessary law for us to consider is Kirchoff's current law, which tellsus that at each point of a circuit, the sum of currents flowing into a pointequals the sum of the currents flowing out. For a simple RLC circuit with oneloop, Kirchoff's current law guarantees that we can use a single function I (t ) tomodel the current at any point at a given time t ; for circuits with multiple loops,multiple functions I (t ) are needed. Now we are prepared to see how these fundamental laws of electricity leadto a second-order differential equation, and hence a 2 × 2 system of first-orderDEs. Let us consider an RLC circuit that consists of a resistor, inductor, andcapacitor, along with some energy (voltage) source E(t ), arranged in series, asshown in figure 3.17. Kirchoff's law leads us directly to second-order differentialequations that determine the behavior of the current I (t ) in the circuit and thecharge Q(t ) on the capacitor.By Ohm's law, we know thatVab = IR. Similarly, Faraday's law implies that Vbc = tL dI and Coulomb's law tells us that Vcd = C Q(t ) = C Q(t0 ) + t0 I (s) ds . dt 1 1Finally, we know from the voltage source that Vda = −E(t ). Kirchoff's law nowyields the equation Vab + Vbc + Vcd + Vda = 0, or 1 RI (t ) + LI (t ) + Q(t ) = E(t ) (3.8.22) C L b c R I(t) C + − a d E(t) Figure 3.17 An RLC circuit with resistance R, inductance L, capacitance C, and energy source E(t ).
Applications of linear systems 261Recalling that Q (t ) = I (t ), we may rewrite (3.8.22) in two different ways. If wedifferentiate both sides of (3.8.22), and rearrange the terms in decreasing orderof derivatives, it follows immediately that the current I (t ) must satisfy the linearsecond-order differential equation 1 LI (t ) + RI (t ) + I (t ) = E (t ) (3.8.23) CIf instead we substitute Q for I in (3.8.22), then we see that Q is the solution tothe linear second-order differential equation 1 LQ (t ) + RQ (t ) + Q(t ) = E(t ) (3.8.24) CWe can therefore study the behaviors of different RLC circuits based on the givenresistance, inductance, capacitance, and supplied voltage. Moreover, as we wellknow, any such linear second-order differential equation may be converted to asystem of first-order equations. For example, letting x1 = I and x2 = I , we canconvert (3.8.23) to the system of equations x1 = x2 1 R 1 x2 = − x1 − x2 + E (t ) CL L LExample 3.8.1 Determine all solutions I (t ) for an RLC circuit when L = 20 H,R = 80 , C = 10−2 F, and the external voltage is given by the functionE(t ) = 50 sin 2t .Solution. From (3.8.23) and the given information, we can immediatelydetermine the second-order differential equation that I (t ) satisfies. In particular,since E(t ) = 50 sin 2t , we have E (t ) = 100 cos 2t , and using the values for L, C,and R, I (t ) is a solution to the equation 20I + 80I + 100I = 100 cos 2t (3.8.25)Using the substitution x1 = I and x2 = I and multiplying both sides of (3.8.25)by 1/20, the system becomes x1 = x2 x2 = −5x1 − 4x2 + 5 cos 2tFrom this, we can write the system in matrix form as 0 1 0 x = x+ (3.8.26) −5 −4 5 cos 2tFor the coefficient matrix A in (3.8.26), we compute the eigenvalues andeigenvectors in order to find the complementary solution xh of the system.
264 Linear systems of differential equationsof the original equation (3.8.25) and c1 e −2t (−2 cos t + sin t ) + c2 e −2t (− cos t −2 sin t ) the transient solution. Overall, we have now seen several examples of important phenomenagoverned by linear systems of differential equations. Further examples will beconsidered in the exercises.Exercises 3.8 1. In a closed system of two tanks (i.e, 1 2. Consider a system of two tanks Initially, Tank A has 20 g of salt present in its solution, and Tank B has 75 g of salt present in its solution. Set up and solve an initial-value problem whose solution will determine the amount of salt in each tank at time t . Discuss the graphical behavior of the solution x(t ) (whose components are the amount of salt in each
Applications of linear systems 265 tank at time t ). Is there an equilibrium solution to the system? If so, what is it?3. Suppose that in exercise 2 all of the given information remains the same except for the fact that instead of saltwater flowing into each tank, pure water flows in. How do the results of your work in exercise 2 change?4. In a closed system of three tanks (that is, one for which there are no input flows and no output flows), the following information is given. Tank A Tank B Tank C Tank volume 100 liters 150 liters 125 liters Rates of outflows to B: 3 liters/min to C: 1 liters/min to A: 4 liters/min to other tanks Rates of outflows to C: 4 liters/min to A: 3 liters/min to B: 1 liter/min to other tanks5. In a system of three tanks of saltwater, the following information is given. Tank A Tank B Tank C Tank volume 400 liters 200 liters 300 liters Rate of inflow 7 liters/min 0 liters/min 0 liters/min to the tank Concentration of 10 g/liter n/a n/a salt in inflow Rate of drain outflow 0 liters/min 0 liters/min 7 liters/min Rates of outflows to B: 7 liters/min to C: 7 liters/min to A: 0 liters/min to other tanks Rates of outflows to C: 0 liters/min to A: 0 liters/min to B: 0 liters/min to other tanks
266 Linear systems of differential equations Each tank is full; tank A contains solution whose initial concentration is 20 g/liter. Tank B contains solution whose initial concentration is 50 g/liter. Tank C contains pure water. Without setting up a system of differential equations, first use your intuition to describe what you think will be the behavior of the functions x1 (t ), x2 (t ), and x3 (t ) that measure the amount of salt in each of the three respective tanks at time t . Then, set up and solve an initial-value problem whose solution will tell you the amount of salt in each tank at time t . Discuss the graphical behavior of each component of the solution x(t ) and compare it to your intuitive expectations. Is there an equilibrium solution to the system? If so, what is it? 6. In a system of three tanks of saltwater interconnected with pipes of inflow and outflow to and from each, the following information is given. Tank A Tank B Tank C Tank volume 400 liters 800 liters 500 liters Rate of inflow 5 liters/min 10 liters/min 5 liters/min to the tank Concentration of 25 g/liter 15 g/liter 40 g/liter salt in inflow Rate of drain outflow 4 liters/min 7 liters/min 9 liters/min Rates of outflows to B: 6 liters/min to C: 5 liters/min to A: 4 liters/min to other tanks Rates of outflows to C: 4 liters/min to A: 5 liters/min to B: 1 liter/min to other tanks Assume that the system is such that initially there is a concentration of 10 g/liter of salt in each of the three tanks. Set up and solve an initial-value problem whose solution will tell you the amount of salt in each tank at time t . Discuss the graphical behavior of each component of the solution x(t ). Is there an equilibrium solution to the system? If so, what is it? 7. Recall that for a spring-mass system of mass m, spring constant k, and damping constant c, the displacement y(t ) of the mass from equilibrium is governed by the linear second-order differential equation c k 1 y + y + y = F (t ) m m m For a mass of 0.5 kg with spring constant k = 2 N/m in an undamped, unforced system, assume the mass is displaced 0.4 m from equilibrium and released (i.e., y(0) = 0.4 and y (0) = 0).
Applications of linear systems 267 8. For a mass of 0.5 kg with spring constant k = 2 N/m and damping constant c = 0.5 N·s/m in an unforced system, assume the mass is 9. For a mass of 0.5 kg with spring constant k = 2 N/m and damping constant c = 0.5 N·s/min a forced system with forcing function F (t ) = cos 2t N, assume the mass is initially Use variation of parameters to solve the system in (b), and graph the component function x1 (t ). Discuss the long-term behavior of the spring-mass system.10. In section 3.8.2, we considered a system of two masses attached to two springs in parallel, where a mass m1 is attached to a spring with spring constant k1 and from m1 a second spring with constant k2 and mass m2 is attached. See figure 3.16. If we assume that the surface on which the masses rest is frictionless and let let x1 (t ) denote the displacement of m1 from its equilibrium position and x2 (t ) the displacement of m2 from its equilibrium position and se |
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Show More, without first gaining a proper understanding of the underlying theory, limits the ability to use matrices and to apply them to new problems. This book explains matrices in the detail required by engineering or science students, and it discusses linear systems of ordinary differential equations. These students require a straightforward introduction to linear algebra illustrated by applications to which they can relate. It caters of the needs of undergraduate engineers in all disciplines, and provides considerable detail where it is likely to be helpful. According to the author the best way to understand the theory of matrices is by working simple exercises designed to emphasize the theory, that at the same time avoid distractions caused by unnecessary numerical calculations. Hence, examples and exercises in this book have been constructed in such a way that wherever calculations are necessary they are straightforward. For example, when a characteristic equation occurs, its roots (the eigenvalues of a matrix) can be found by inspection. The author of this book is Alan Jeffrey, Emeritus Professor of mathematics at the University of Newcastle upon Tyne. He has given courses on engineering mathematics at UK and US Universities |
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Mathematics
Mathematics provides the logical and analytical tools for tackling many of the important problems of our time. By its very nature, mathematics provides the means to break many problems into manageable pieces that can be analyzed and solved. In fact, mathematical approaches have been central to solving problems and modeling phenomena in a wide array of disciplines. Probability and statistical analysis are fundamental for mapping and analyzing the human genome. Advanced mathematical theories provide the keys to analyzing the risk of rare events, a basic problem of the financial markets. In physics, geometry finds applications to particle physics, to string theory, and to cosmology. In neuroscience, exciting new research into the structure and functioning of the brain relies heavily on the insights provided by mathematical modeling. These are but a few of the contemporary problems relying on mathematical analysis. Mathematical thinking is grounded in rigor and abstraction, but draws its vitality from questions arising in the natural world as well as applications to industry and technology.
Mathematics majors acquire solid foundations in differential and integral calculus, as well as basic concepts of algebra and modern geometry. Students are introduced to classical subjects such as complex and real analysis, abstract algebra, number theory, and topology. Students interested in applications of mathematics to social and physical sciences may pursue courses in numerical methods, theoretical mechanics, probability, dynamical systems, and differential equations.
Mathematics majors at NYUAD attain a breadth of knowledge within the field, pursue their own interests in math electives, explore the role of mathematics as an applied discipline, and undertake a capstone project. The major offers a rigorous and broad foundation in mathematics through seven required courses: Calculus; Linear Algebra; Multivariable Calculus; Ordinary Differential Equations; Real Analysis 1; Introduction to Probability and Statistics; Abstract Algebra 1.
Students select three electives. To attain greater depth in analysis, algebra or calculus, students choose Real Analysis 2, Abstract Algebra 2 or Vector Analysis. The second elective must be a course in applied mathematics, such as Discrete Mathematics, Numerical Methods, Cryptography, Introduction to Mathematical Modeling or Introduction to Game Theory. The third elective may be any other course in mathematics.
Mathematics majors must also complete a concentration in one of the following areas, which use mathematics or mathematical modeling: Computer Science, Economics or the Natural Sciences
Requiring mathematics majors to complete a concentration provides them with a basic knowledge of how math is applied to a specific discipline and is intended to foster the requisite capstone projects in which math majors work closely with students from other areas to solve problems and answer questions.
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A fundamental understanding of mathematical functions is critical before engaging in the rigors of calculus. This course examines single variable functions, including their algebraic and geometric properties. The course begins with a rigorous exploration of the following question: What is a function, and how can it be represented geometrically as a graph? The course delves into standard function manipulations and examines a range of mathematical functions, including polynomial, rational, trigonometric, exponential, and logarithmic functions. Placement into Mathematical Functions is decided by discussion with mentors and the results of a mathematics placement examination.
Students in the NYUNY Mathematics Dept: This course is equivalent toMATH-UA 009 Algebra and Calculus
This course presents the foundations of calculus by examining functions and their derivatives and integrals, with an emphasis on proofs and theorems and an introduction to basic mathematical analysis. While the derivative measures the instantaneous rate of change of a function, the definite integral measures the total accumulation of a function over an interval. Indeed, the relationship between differentiation (finding a derivative) and integration (determining an integral) is described in the Fundamental Theorem of Calculus. In addition to two weekly lectures, students attend a weekly discussion section that provides opportunities for rigorous analysis of proofs and theorems associated with the material. This course is primarily intended for students considering Mathematics as a major or for students who seek an in-depth understanding of the arguments that support calculus. Placement into Calculus is decided by discussion with mentors and the results of a mathematics placement examination. With permission of the program in mathematics, Calculus with Applications may substitute for Calculus.
This is a required course for Economics majors until the class of 2017.
Students in the NYUNY Mathematics Dept: This course is equivalent to MATH-UA 121 Calculus I
This course presents the foundations of calculus by examining functions and their derivatives and integrals with a special emphasis placed on the utilitarian nature of the subject material. Applications to natural science, engineering, and the social sciences, particularly economics, are emphasized. Since the derivative measures the instantaneous rate of change of a function and the definite integral measures the total accumulation of a function over an interval, these two ideas form the basis for nearly all mathematical formulas in science, engineering, economics, and other fields. This course also provides instruction in how to model situations in order to solve problems. Applications include graphing, and maximizing and minimizing functions. In addition to two weekly lectures, students attend a weekly discussion section focused on applications of calculus in science, engineering, or social science, depending on their primary interest. Placement into Calculus with Applications is decided by discussion with mentors and the results of a mathematics placement examination.
May not be taken if MATH-AD 110 Calculus is completed
This is a required course for Economics majors until the class of 2017. This course can be substituted with MATH-AD 110 Calculus.
Please note: During Summer 2014, classes will be held on Thursday May 22 and Saturday May 24, and then on Sunday through Thursday from May 28-June 19. Exams will be held on Saturday May 21.
This course explores functions of several variables and has applications to science and engineering as well as economics. Specific topics include: vectors in the plane and space; partial derivatives with applications; double and triple integrals; spherical and cylindrical coordinates; surface and line integrals; and divergence, gradient, and curl. In addition, the theorems of Gauss and Stokes are rigorously introduced.
Students in the NYUNY Mathematics Dept: This course is equivalent toMATH-UA 123 Calculus III
Please note: During Summer 2014, classes will be held on Thursday May 22 and Saturday May 24, and then on Sunday through Thursday from May 28-June 19. Exams will be held on Saturday June 21. Duration: 4 weeks
In many applications of mathematics a response of some systems is nearly a linear function of the input. These linear systems, which arise in elasticity, in electrical engineering, and in economics for example, involve linear equations in many unknowns. The associated matrix algebra is a rich and beautiful field of mathematics. It is also central to the analysis of linear ordinary and partial differential equations. The material in this course includes systems of linear equations, Gaussian elimination, matrices, determinants, Cramer's rule, vectors, vector spaces, basis and dimension, linear transformations, eigenvalues, eigenvectors, and quadratic forms.
Ordinary differential equations arise in virtually all fields of applied mathematics. Newton's equations of motion, the rate equations of chemical reactions, the currents flowing in electric circuits, all can be expressed as ordinary differential equations. The solutions of these equations usually evolve a combination of analytic and numerical methods. The course studies first- and second-order equations, solutions using infinite series, Laplace transforms, linear systems, numerical methods.
MATH-AD 116 is a corequisite.
Students in the NYUNY Mathematics Dept: This course is equivalent to MATH-UA 262 Ordinary Differential Equations.
An introduction to discrete mathematics, emphasizing proof and abstraction, as well as the applications to the computational sciences. Topics include: sets, relations, and functions; graphs and trees; algorithms, proof techniques, and order of magnitude analysis; Boolean algebra and combinatorial circuits; Formal logic, formal languages, and automata; combinatorics, probability, and statistics.
This course comprises a combination of the theory of probability and the mathematical foundations with techniques of modern statistical analysis. It is designed to acquaint the student with both probability and statistics in the context of their applications to the sciences. In probability: mathematical treatment of chance; combinatorics; binomial, Poisson, and Gaussian distributions; law of large numbers and the normal distribution; application to coin-tossing, radioactive decay, and so on. In statistics: sampling; normal and other useful distributions; testing of hypotheses; confidence intervals; correlation and regression; and applications to scientific, industrial, and financial data.
Students in the NYUNY Mathematics Dept: This course is equivalent to MATH-UA 235 Probability and Statistics.
Algebra is a part of every field of mathematics, and has applications in the discrete systems of computer science. Fractions, together with their familiar laws of addition, multiplication, and division, provide an example of algebra. The complex numbers form another. This course introduces more general algebras, and their properties and applications. Topics considered in this course include groups, homomorphisms, automorphisms and permutation groups. Rings, ideals and quotient rings, Euclidean rings, and polynomial rings are also considered.
Students in the NYUNY Mathematics Dept: This course is equivalent to MATH-UA 343 Algebra I.
One of the most remarkable applications of abstract algebra is the solution of algebraic equations: for example, to finding the roots of a polynomial. This course develops the ideas needed to study this problem, culminating in the celebrated theory of Galois. The topics include extension fields and roots of polynomials, constructions with straight edge and compass. Unique factorization in rings, elements of Galois theory.
Many of the complex systems of natural science can be formulated as a dynamical system—one whose changes are determined only by the current state. These systems are typically nonlinear, and often exhibit the random behavior associated with chaos. Topics of the course include: dynamics of maps and of first-order and second-order differential equations; stability, bifurcations, limit cycles, dissection of systems with fast and slow time scales. The geometric viewpoint, including phase planes, are stressed. Chaotic behavior is introduced in the context of one-variable maps (the logistic), fractal sets, etc. Applications are drawn from physics and biology.
Complex analysis is a powerful tool with diverse applications in mathematics, science, and engineering. Functions of a complex variable arise in elasticity, electrical engineering, and in fluid dynamics, to name a few examples. The geometrical content of analysis in the complex plane is especially appealing. Topics include: complex numbers and complex functions; differentiation and the Cauchy-Riemann equations, Cauchy's theorem, and the Cauchy integral formula; singularities, residues, Taylor and Laurent series; fractional linear transformations and conformal mapping; analytic continuation; and applications to fluid flow.
Often the most difficult part of the task of the applied mathematician is the formulation of an analyzable model in the face of a perplexing phenomenon or data set. This course is designed to give students an introduction to all aspects of this process. The course consists of several modules, each a self-contained problem, taken from biology, economics, and other areas of science. In the process the student experiences the formulation and analysis of a model and its validation by numerical simulation and comparison with data. The mathematical tools to be developed include dimensional analysis, optimization, simulation, probability, and elementary differential equations. The necessary mathematical and scientific background is developed as needed. Students participate in formulating models as well as in analyzing them.
Numerical analysis explores how mathematical problems can be analyzed and solved with a computer. As such, the subject has very broad applications in mathematics, physics, engineering, finance, and the life sciences. This course gives an introduction to this subject for Mathematics majors. Theory and practical examples using Matlab is combined to study of topics ranging from simple root-finding procedures to differential equations and the finite element method.
Perhaps the purest of pure mathematics, number theory nevertheless finds important application to cryptography and computer science generally. The recent solution of Fermat's last theorem brought attention to the subject. In mathematics, number theory is associated with many outstanding problems, including the famous Riemann hypothesis. Topics to be covered include divisibility theory and prime numbers, linear and quadratic congruences, the classical number-theoretic functions, continued fractions, and diophantine equations.
Many laws of physics are formulated as partial differential equations, e.g., the propagation of sound waves, the diffusion of a gas, and the flow of a fluid. This course discusses the simplest examples of such laws as embodied in the wave equation, the diffusion equation, and Laplace's equation. The course also discusses nonlinear conservation laws and the theory of shock waves. Applications to physics, chemistry, biology, and population dynamics are given.
This course is designed as a review of the calculus of several variables with emphasis on vector methods. Topics to be treated include: functions of several variables; partial derivatives, chain rule, change of variables, Lagrange multipliers; inverse and implicit function theorems; vector calculus (divergence, gradient, and curl); theorems of Gauss, Green, and Stokes with applications to fluids, gravity, electromagnetism, and the like. The course also treats an introduction to differential forms and degree and fixed points of mappings with applications.
Analysis builds a more rigorous foundation for calculus and prepares the way for more advanced courses. The emphasis is on the careful formulation of the concepts of calculus, and the formulation and proof of key theorems. The goal is to understand the need for and the nature of a mathematical proof. The course studies the real number system, the convergence of sequences and series, functions of one real variable, continuity, connectedness, compactness, and metric spaces.
The second part of the analysis series is devoted to the calculus of functions of several variables. The transition from a single variable to many variables involves important new concepts, which are essential to understanding applications to the natural world. Topics include the rigorous study of functions of several variables, limits and continuity, differentiable functions, the implicit function theorem, transformation of multiple integrals,
This course is intended for students who are highly motivated and seek the opportunity to conduct field research with a faculty sponsor from the NYUAD program in Mathematics. Students with the necessary background in course work and who, in the opinion of a faculty sponsor, possess intellectual independence and ability may register for this course. The student must approach a faculty member in his or her field of interest to obtain sponsorship. Typically, this course is only open to students with a minimum overall GPA of 3.3 and a minimum major GPA of 3.5, and registration requires permission of the sponsoring faculty member. Forms for Directed Study in Research in Mathematics are available from the Office of the Dean of Science.
Topology is a major branch of mathematics, which is concerned with the geometry of sets of points in space of arbitrary dimension. One aspect of the subject deals with the classification of sets based upon their structure, not their specific shape. Topology has applications in physics, biology, and dynamical systems. The material includes metric spaces, topological spaces, compactness, connectedness, covering spaces,
Intermediate-level course on the principles and applications of dynamics. Topics include the Lagrangian and Hamiltonian formulations of mechanics, conservation laws, central force motion, rotational kinematics and dynamics, normal modes and small oscillations, and chaos theory.
Prerequisites include MATH-AD 116 or MATH-AD 121
Students in the NYUNY Physics Dept: This course is equivalent to PHYS-UA 120 Dynamics
This course introduces the basic concepts of elementary game theory in a way that allows students to use them in solving simple problems. Topics include: the basics of cooperative and non-cooperative game theory; basic solution concepts such as Nash equilibrium and the core; and the extensions of these solutions to dynamic games and situations of incomplete information. Students are exposed to a variety of simple games with varied and useful applications: zero-sum games; the Prisoner's Dilemma; coordination games; the Battle of the Sexes; repeated games; and elementary signaling games. The course relies on a wide array of example applications of game theory in the social sciences.
Students who have a strong mathematical background but who do not have the required Calculus pre-requisite are encouraged to contact Professor Chacon for a possible pre-requisite waiver.
Students in the NYUNY Politics Dept: This course is equivalent to POL-UA 840 Intro to Game Theory
Foundations of Science 1: Energy and Matterprovides a comprehensive introduction to these two fundamental concepts that are so famously unified in the equality E=mc2. Following an introduction to the physical sciences, the course focuses on velocity, acceleration, forces, and energy, while simultaneously introducing students to atoms and molecules. Chemical reactions are examined, and the energy changes associated with them are investigated via a thorough analysis of the three laws of thermodynamics. Laboratory exercises focus on the guiding principles of the scientific method and an introduction to experimental design, and scientific presentation, including technical writing. Weekly discussion sections are designed to hone proficiency at solving problems in a collaborative, team environment.
Foundations of Science 2: Forces and Interactions introduces students to fundamental forces, including gravity and electrical forces. Concurrently, atomic theory, the theory of molecular bonding, and atomic and molecular structures and shapes, in which forces and energy play a role, are investigated. Students apply these concepts to understanding molecules related to the life sciences. Laboratory exercises focus on acquisition of data and analysis with a continued emphasis on technical presentation. Weekly discussion sections are designed to hone proficiency at solving problems in a collaborative, team environment.
focuses on changes in systems in the physical and living worlds. Capacitors, current, and basic circuits are explored with an eye toward understanding their applications to chemical reactions and the behavior of living cells. The rates and directions of chemical reactions are explored as chemical kinetics and chemical equilibrium are investigated with a special focus on acid-base chemistry. These fundamental physical and chemical principles are used to describe basic cellular monomers and polymers including DNA, RNA, and protein, and the sequence of events that leads to information flow and its regulation in the cell nucleus. They are also applied to macroscopic systems found in the biosphere. Laboratory exercises focus on fundamental protocols and tools needed to sharpen basic laboratory skills. Weekly discussion sections are designed to hone proficiency at solving problems in a collaborative, team environment.
Foundations of Science 4: Form and Function explores a question applicable to all branches of science: How does the form or shape of a physical entity set its function? This leads to another question: If a specific function is desired, can a form or shape be engineered or modified to execute or improve that function? The course examines the form/function concept in magnetic and electrical fields, the behavior and design of small molecules, and the activity of proteins as the workhorse in biological systems. Laboratory exercises require students to design experiments related to crystals and crystallography, and to examine chemical forms at the macroscopic and microscopic levels. Weekly discussion sections are designed to hone proficiency at solving problems in a collaborative, team environmentConcentration in Applied Mathematics
Mathematics is often associated with science, particularly physics and chemistry, but it is indeed the language and tool of the contemporary life sciences, including ecology and environmental studies, as well as the world of business and the economy. The concentration in Applied Mathematics at NYU Abu Dhabi is designed to prepare students in science and the social sciences with the critical quantitative tools and reasoning skills needed to solve problems in those disciplines.
Mathematics Faculty
2012 - 2013 Academic Year |
Product Description
This teacher's guide accompanies Singapore Math's sold-separately Discovering Mathematics Textbook 3A. This guide contains a weekly schedule lists the objectives for each lesson, brief notes on teaching for each chapter, and fully worked solutions for all questions and problems in the textbook. 164 pages, softcover. Grades 9-10. |
Essentials of Pre - Calculus - 98 edition
Summary: REA's Essentials provide quick and easy access to critical information in a variety of different fields, ranging from the most basic to the most advanced. As its name implies, these concise, comprehensive study guides summarize the essentials of the field covered. Essentials are helpful when preparing for exams, doing homework and will remain a lasting reference source for students, teachers, and professionals. Pre-Calculus reviews sets, numbers, operations and prope...show morerties,coordinate geometry, fundamental algebraic topics, solving equations and inequalities, functions, trigonometry, exponents and logarithms,conic sections, matrices and determinants |
This algebra lesson from Illuminations has students collect data for rolling objects of differing sizes in order to further understand periodic phenomena. They will then create two sinusoidal graphs of the data....
This interdisciplinary lesson uses musical terms and concepts to teach algebra and geometry. Students will analyze musical scales and frequencies generated by a geometric sequence, and relate sine waves to musical...
This algebra unit from illuminations provides an in depth exploration of exponential models in context. The model of light passing through water is used to demonstrate exponential functions and related mathematical ties earth science concepts in with algebra. The forest-fire danger rating index is applied to a mathematical model. Students will learn real-world meaning of the intercepts and slope in... |
This is a lecture based course. Students will develop an understanding of the material by working exercises. Some instructors may require students to make presentations.
Goals and Objectives of the Course
This course is primarily for mathematics majors. Its goals and objectives are:
1. to develop an understanding of the structure of the real number system,
2. to develop an understanding of the theory of calculus,
3. to improve the student's skills in the mathematical methods, including the logical and orderly presentation of proofs.
Assessment Measures
Assessment measures include a final examination and may include any of the following:
in-class tests, quizzes, exercises, in-class presentations and research papers.
Other Course Information
The instructor may require that the students use a computer in exercises designed to enhance the understanding of theoretical material. |
, Hybrid
Reflecting Cengage Learning's commitment to offering flexible teaching solutions and value for students and instructors, this new hybrid edition ...Show synopsisReflecting Cengage Learning's commitment to offering flexible teaching solutions and value for students and instructors, this new hybrid edition features the instructional presentation found in the printed text while delivering end-of-section exercises online in Enhanced WebAssign. The result - a briefer printed text that engages students online! Do your students attempt to memorize facts and mimic examples to make it through algebra? James Stewart, author of the worldwide, best-selling calculus texts, saw this scenario time and again in his classes. So, along with longtime coauthors Lothar Redlin and Saleem Watson, he wrote College Algebra, specifically to help students learn to think mathematically and to develop genuine problem-solving skills. Comprehensive and evenly-paced, the text has helped hundreds of thousands of students. Incorporating technology, real-world applications, and additional useful pedagogy, the sixth edition promises to help more students than ever build conceptual understanding and a core of fundamental skills33600435-5Good. Paperback. Missing components. May include moderately...Good. Paperback. Missing components. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9781133600435 NO ACCESS CARD! ! NO SUPPLEMENTS! ! TEXT ONLY! ! ! This item may not include any CDs, Infotracs, Access cards or other supplementary material 1133600433 *VALUE PRICED USED BOOK* RETURNS ARE NO...Good. 1133600433 ALIBRIS.
Description:New. 1133600433 *NEW! * RETURNS ARE NO PROBLEM! We LOVE happy...New. 1133600433 *NEW! * RETURNS ARE NO PROBLEM! We LOVE happy customers. All of our orders are sent with tracking information when possible. ALIBRIS |
Sophus Lie
algebraBranch of mathematics in which arithmetical operations and formal manipulations are applied to abstract symbols rather than specific numbers. The notion that there exists such a distinct subdiscipline...
differential equationMathematical statement containing one or more derivatives —that is, terms representing the rates of change of continuously varying quantities. Differential equations are very common in science and engineering,...
group theoryIn modern algebra, a system consisting of a set of elements and an operation for combining the elements, which together satisfy certain axioms. These require that the group be closed under the operation...
mathematicsThe science of structure, order, and relation that has evolved from elemental practices of counting, measuring, and describing the shapes of objects. It deals with logical reasoning and quantitative calculation,... |
Learn Mathematics On Your Smartphone [NOOK Book] For easy reading, a comprehensive list of hundreds of topics each with a graphic image and explanatory text ...
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Clive W. Humphris was born in Maldon Essex UK in 1946 leaving school at fifteen to begin an apprenticeship as an instrument maker with Marconi in Chelmsford working in both electronics and mechanical engineering fields. At a very early age he developed a passion for electronics and engineering that was to go on to provide an interesting and enjoyable career in both the retail television electronics being involved at a senior management level in the launch of Colour Television, Video Recorders and use of Computers in schools and later in the commercial sector worked with the main television broadcast companies in and around London for national televised events.
Leaving school with little more than a basic education has proved to be a massive advantage in being able to understand the needs and learning difficulties of many students particularly adults and led to the production of training materials that have a very strong practical application. A Further Education Teachers Certificate and two post graduate diplomas in Management Studies and Industrial Relations and Personnel Management ensured the author linked a practical technical background with strong interest in how people learn and develop with a belief that a poor early education does not need to restrict opportunities providing the individual is prepared to put in a little effort..
The early nineties were the start of the computer based education and began to feature distance learning where STEM subjects (Science, Technology, Engineering and Maths) being practical, but with a need for an academic understanding lent themselves to the emerging computer market where the PC graphics and were to be the start of the author spending the next twenty years developing computer based training packages. These packages now provide the core material for all his eBooks and Paperback publications available on all the available eReaders. Downloaded and activated with the payment eBook receipt at no additional cost thereby enabling students, teachers and hobbyists to enjoy comprehensive education software at a tiny fraction of the previous published price. An unlimited user networked version of the software is also available as part of the 'Teachers Pack' series of December 29, 2013
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OUTCASTS
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Andy bio update
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Elements of Set Theory
Description: This is an introductory undergraduate textbook in set theory. In mathematics these days, essentially everything is a set. Some knowledge of set theory is necessary part of the background everyone needs for further study of mathematics. It is alsoMore objects. This book starts with material that nobody can do without. There is no end to what can be learned of set theory, but here is a beginning |
1. Maths Coverage Mangahigh is a comprehensive and powerful maths teaching resource offering full coverage of the UK National Curriculum with more than 400 different challenges ranging from addition to quadratic factorisation
Statistics is a mathematical science pertaining to the collection, analysis, interpretation, and presentation of data. It is applicable to a wide variety of academic disciplines, from the physical and social sciences to the humanities; it is also used and misused for making informed decisions in all areas of business and government. (wikipedia.org) its include histogram graphics theory, probability, ANOVA, Summarizing data, deviation, Correlation and causation, etc. Download Basic Statistics Formula ListBasic Statistics 1it include : Measures of Location, Sets, Probability, Measures of spread, deviation, Normal density function, Confidence intervals for the mean, Hypothesis test for the mean , and Simple linear regression Basic Practice of Statistics it include : Exploring Data: Distributions,...
Mathematical Formula Table and List |
A mathematical puzzle where equations are all mixed up. One solution per puzzle. ... the correct equations to solve the puzzle. Equations run in various formats like boxes, crosses, diagonals, ... diagonals, mazes. Equations go in all directions - follow the arrows. ... to reinforce math or just use your logic ....
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- lacks dynamic feedback. Flash Math duplicates the positive aspect of flash cards and ... Also, Flash Math scales problems according to childrens proficiency so that ... less intricate equations . The problems presented are for children in grades ...
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IO Software Equations is the advanced equations editor which combines the power of features with ... possible combinations, equation library, real time Visual and LaTex editing, using ... own scientific fonts , a fully configurable Speed Bar for fast symbol ... symbol access, Equations ! can be used as standalone or together with ... ... |
Math 60330
Basic Geometry and Topology
Fall Semester, 2010
General Information
Text: There is no required textbook. Here are notes for the point set topology part of the course (covering the first fives lectures). See the syllabus for information about additional references. There are some books for the course on the reference shelf of the math library.
Exams: The midterm as well as the final exam will both consist of two parts: a take-home and an in-class part, each worth half of the total points. The in-class part, similar to the quizzes, is about definitions and theorems, possibly requiring some basic calculations, but involving no proofs. The take-home part, similar to homework problems, will involve more extensive calculations and proofs.
Homework: Homework is an integral part of the course. You are permitted,
in fact encouraged, to work together and help one another with homework, although what you turn in should be written by you. Providing detailed arguments in your homework is important, since learning how to write mathematics in a rigorous and yet concise and readable way is an essential part of graduate school in mathematics. Homework problems will be posted on the course web-page and will be due on a to be determined day each week. After the due date I will provide a detailed solution to each homework set online.
Pre/post lecture preparation: As with most math classes, the material in one class
often depends heavily on stuff covered in previous classes.
Hence it is a good idea to look over your notes before the next class. I'm very happy to discuss questions concerning the material of previous classes at the beginning of each class.
Quizzes: There will be weekly quizzes on a to be determined day of the week testing basic understanding of definitions and theorems.
Course website: |
Product Description
Students will develop foundational math skills needed for higher education and practical life skills with ACE's Math curriculum. PACE 1088 covers identifying four types of prisms, and learning formulas to determine the surface area and volume of a prism, cylinder, and sphere |
Mathematical Circles Topics - Tom Davis
Mathematical Circles are for students of high school age or younger who want to increase their abilities to reason about mathematical problems. Students who study in circles learn to do mathematical olympiad-style problems (essay or proofs, not quick
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Mathematical Interactivities - David Hellam
Interactive mathematical challenges, mostly written in Flash, covering number, algebra, data, shape, and space for K-12 students. Site also includes a forum for teachers to discuss this sort of resource, its development, and its use in the classroom.
...more>>
Mathematics and Multimedia - Guillermo Bautista
Blog about mathematics, teaching, learning, and technology. Begun in October, 2009, it has included posts such as "GeoGebra Tutorial 4 - Graphs and Sliders"; "Screencasting Tutorial: Making a Math Video Lesson Using Camstudio"; "Derivative in Real Life
...more>>
Mathematics by Mr. P - Mike Poliquin
Middle school math teacher Poliquin began this blog, subtitled "making sure everyone gets it," in December, 2011. Posts have included various original problems of the week; "Eudoxus of Cnidus: the First Mathematical Superstar"; "A Quick Dip in the Deep
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mathematics.com - ENGINEERING.com
Interactive material to download or work with online; articles on current technology topics; resources and links of interest to mathematicians; technology products offered for sale. Online calculators and applets for geometry, trigonometry, calculus,
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Mathematics Curriculum Notes - Alan Selby
A model for mathematics instruction from primary school to college. The first part describes Pattern Based Reason, its origins, benefits and limitations in many subjects. In mathematics education there are two barriers to comprehension to be lowered or
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Mathematics (SparkNotes.com) - WebCT.com & iTurf Inc.
Over 100 guides for mathematics ranging from pre-algebra topics to advanced work in calculus, written by students and recent graduates of Harvard University. The site also includes message boards for beginner, high school, and advanced math, calculus,
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Mathematics - Student Helpmate - Chris Divyak
Search or browse this archive of questions about algebra, calculus, geometry, statistics, trigonometry, and other college math; then pay for access to answers. To submit your own problem to Student Helpmate, type your question or upload it as a file;
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mathepower.com - Markus Hendler
A series of calculators for solving problems, for classes 1-10. Primarily in German, but with an English version. Themes include: fractions, geometry, equations, arithmetical operations, trigonometry.
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Mathepower - Markus Hendler
Online calculators for most calculations covered in the first ten years of school, from basic operations through the first stages of algebra. Available in English, German, and French.
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Math for Morons Like Us - ThinkQuest 1998
Students talk to students about math: a site designed to help you understand math concepts better. Tutorials, sample problems, and quizzes for Pre-Algebra, Algebra, Geometry, Algebra II, and Pre-Calc/Calculus, designed assuming you know some of the basic
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Math Fundamentals Problem of the Week - Math Forum
Math problems for students working with concepts of number, operation, and measurement, as well as introductory geometry, data, and probability. The goal is to challenge students with non-routine problems and encourage them to put their solutions into
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MathGifs
"Illuminating mathematics through animations." This blog, which dates back to September, 2013, has included illustrated posts such as "Reflective Property of the Ellipse," "Rotations Through Translations," "Fourier Pumpkins," "Sonic Booms," and "The Mandelbrot
...more>> |
This is a free textbook offered by BookBoon.'The success of Group Theory is impressive and extraordinary. It is, perhaps, the...
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This is a free textbook offered by BookBoon.'The success of Group Theory is impressive and extraordinary. It is, perhaps, the most powerful and influential branch of all Mathematics. Its influence is strongly felt in almost all scientific and artistic disciplines (in Music, in particular) and in Mathematics itself. Group Theory extracts the essential characteristics of diverse situations in which some type of symmetry or transformation appears. Given a non-empty set, a binary operation is defined on it such that certain axioms hold, that is, it possesses a structure (the group structure). The concept of structure, and the concepts related to structure such as isomorphism, play a decisive role in modern Mathematics.The general theory of structures is a powerful tool. Whenever someone proves that his objects of study satisfy the axioms of a certain structure, he immediately obtains all the valid results of the theory for his objects. There is no need to prove each one of the results in particular. Indeed, it can be said that the structures allow the classification of the different branches of Mathematics (or even the different objects in Music (! )).The present text is based on the book in Spanish "Teoría de Grupos: un primer curso" by Emilio Lluis-Puebla, published by the Sociedad Matemática Mexicana This new text contains the material that corresponds to a course on the subject that is offered in the Mathematics Department of the Facultad de Ciencias of the Universidad Nacional Autónoma de México plus optional introductory material for a basic course on Mathematical Music Theory.This text follows the approach of other texts by Emilio Lluis-Puebla on Linear Algebra and Homological Algebra. A modern presentation is chosen, where the language of commutative diagrams and universal properties, so necessary in Modern Mathematics, in Physics and Computer Science, among other disciplines, is introduced.This work consists of four chapters. Each section contains a series of problems that can be solved with creativity by using the content that is presented there; these problems form a fundamental part of the text. They also are designed with the objective of reinforcing students' mathematical writing. Throughout the first three chapters, representative examples (that are not numbered) of applications of Group Theory to Mathematical Music Theory are included for students who already have some knowledge of Music Theory.In chapter 4, elaborated by Mariana Montiel, the application of Group Theory to Music Theory is presented in detail. Some basic aspects of Mathematical Music Theory are explained and, in the process, some essential elements of both areas are given to readers with different backgrounds. For this reason, the examples follow from some of the outstanding theoretical aspects of the previous chapters; the musical terms are introduced as they are needed so that a reader without musical background can understand the essence of how Group Theory is used to explain certain pre-established musical relations. On the other hand, for the reader with knowledge of Music Theory only, this chapter provides concrete elements, as well as motivation, to begin to understand Group Theory.'
A comprehensive website that provides information on all aspects of sound and music recording. Individual sections provide...
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A comprehensive website that provides information on all aspects of sound and music recording. Individual sections provide tips for recording a variety of musical instruments. There are sections covering the use of mixing effects such as reverb, compression, and equalization handy glossary is aimed at students studying Edexcel's AS or A2 level Music Technology courses. It can be used simply to look up terms as they crop up, or flicked through whenever there are five minutes to spare. Some of the benefits of this book are: Concise, accurate descriptions of more than 230 music technology termsDescriptions of the main effects, their parameters and usesDescriptions of the main microphone types and polar patternsCross-referenced and clickable to enable easy browsingPlenty of illustrations and diagrams'
An interactive music theory quiz that helps students improve their ability to identify scale types (Major, minor, mixolydian...
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An interactive music theory quiz that helps students improve their ability to identify scale types (Major, minor, mixolydian etc.). The scale is given on the staff and students must choose from one of the eight types, then click the 'Submit' button. A 'Show Answer' button can help them. |
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BJU Math 9: Algebra 1, Tests
This set of tests will help assess student knowledge, and provide a helpful way to assign grades. Quizzes, chapter tests, and exams are provided for one student. Glue binding for easy removal of tests.
This resource is also known as Bob Jones Algebra 1 Tests, Grade 9, 2nd Edition.
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The answer key is not included. The stock number for the answer key is WW123174. |
This book, which grew out of lectures given over the course of several years at Kharkov University for students in the Faculty of Mechanics and Mathematics, is devoted to classical integral transforms, principally the Fourier transform, and their applications. The author develops the general theory of the Fourier transform for the space \(L^1(E_n)\) of integrable functions of \(n\) variables. His proof of the inversion theorem is based on the general Bochner theorem on integral transforms, a theorem having other applications within the subject area of the book. The author also covers Fourier-Plancherel theory in \(L^2(E_n)\). In addition to the general theory of integral transforms, connections are established with other areas of mathematical analysis--such as the theory of harmonic and analytic functions, the theory of orthogonal polynomials, and the moment problem--as well as to mathematical physics.
Reviews
""This book is remarkable for its rigor, brevity, and systematic expression which, together with the problems proposed in each chapter, make it extremely useful for students, mathematicians, and physicists.""
-- Mathematical Reviews
Table of Contents
Averaging operators and the Bochner theorem
The Fourier transform in \(L^1\)
The inversion theorem in \(L^1\). The Poisson integral
Harmonic functions. The Dirichlet problem for a ball and a half-space
The Fourier transform in \(L^2\)
Hermite functions
Spherical functions
Positive definite functions
The Hankel transform
Orthogonal polynomials and the moment problem
The class \(H^2\). The Paley-Wiener theorem
Boundary properties of functions analytic in the upper half-plane and the Hilbert transform
The Poisson summation formula and some of its applications
Applications of the Laplace and Fourier transforms to the solution of boundary value problems in mathematical physics |
Suitable for the GCSE Modular Mathematics, this book covers different concepts through artwork and diagrams.
Synopsis:
This book is revised in-line with the 2007 GCSE Modular Mathematics specification. This Student Book is delivered in colour giving clarity to different concepts through artwork and diagrams. Worked examples, practice exercises and examiners tips ensure students are fully prepared for their exams. It is written by an experienced author team, including Senior Examiners, which means you can trust that the 2007 specification is covered to ensure exam success |
Discrete Mathematics
Description: The widespread use of computers and the rapid growth in computer science have led to a new emphasis on discrete mathematics, a discipline which deals with calculations involving a finite number of steps. This book provides a well-structuredMore...
The widespread use of computers and the rapid growth in computer science have led to a new emphasis on discrete mathematics, a discipline which deals with calculations involving a finite number of steps. This book provides a well-structured introduction to discrete mathematics, taking a self-contained approach that requires no ancillary knowledge of mathematics, avoids unnecessary abstraction, and incorporates a wide rage of topics, including graph theory, combinatorics, number theory, coding theory, combinatorial optimization, and abstract algebra. Amply illustrated with examples and exercises |
academics
Mathematics
Pitzer's mathematics courses are designed to serve three purposes: general education; service to courses in social, behavioral and natural sciences; and the basis for the mathematics major.
Pitzer Advisers: D. Bachman, J. Grabiner, J. Hoste.
General Education in Mathematics
What is mathematics? What are its major methods and conclusions? How is it related to other subjects? What do modern mathematicians do? Several Pitzer courses specifically address these questions. These courses (described below) are: MATH001 PZ, Mathematics, Philosophy and the "Real World"; MATH 010 PZ The Mathematical Mystery Tour; MATH 015 PZ Mathematics for Teachers I: Number and Operation; MATH 016 PZ Mathematics for Teachers II: Geometry and Data. These courses cover mathematical material that is exciting and sophisticated and yet accessible to students with a standard high school education in mathematics. As such they offer students an excellent opportunity to break fresh ground in kinds of mathematics they are not likely to have seen before. All of these courses meet Pitzer's Educational Objective in Formal Reasoning.
The Precalculus and Calculus Sequences
MATH 025 PZ, Precalculus, is designed to prepare students for Calculus I. The course reviews linear, quadratic and polynomial functions, before introducing the exponential, logarithmic and trigonometric functions. These are the functions most widely used in the quantitative social sciences and natural sciences. MATH 025 PZ does not fulfill the Quantitative Reasoning Requirement.
MATH 030 PZ, MATH 031 PZ and MATH 032 PZ comprise the calculus sequence. The calculus, since it studies motion and change, is the key mathematical tool in understanding growth, decay and motion in the physical, biological, and social sciences. Pitzer offers MATH 030 PZ, MATH 031 PZ and MATH 032 PZ each year. Calculus is also offered at the other Claremont Colleges.
We also offer more advanced courses as part of The Claremont Colleges' Intercollegiate Mathematics program. |
School Textbooks & Study Guides: Maths, Science & Technical
New Maths Frameworking
From 13 To 14
English
UK School Key Stage 3
0007268076
Detailed item information
Description
Year 9 Practice Book 3 is aimed at students working at levels 6-8, helping them to progress with confidence, succeed in the National Tests and take the next step towards GCSE. With hundreds of levelled practice questions corresponding to topics covered in Year 9 Pupil Book 3, it is ideal for extra class work, homework and catch-up classes.
Key Features
Author(s)
Brian Speed, Keith Gordon, Kevin Evans, Trevor Senior
Publisher
HarperCollins Publishers
Date of Publication
09/06/2008
Language(s)
English
Format
Paperback
ISBN-10
0007268076
ISBN-13
9780007268078
Genre
School Textbooks & Study Guides: Maths, Science & Technical
Series Title
New Maths Frameworking
Series Part/Volume Number
No. 41
Publication Data
Place of Publication
London
Country of Publication
United Kingdom
Imprint
Collins Educational
Dimensions
Weight
280 g
Width
192 mm
Height
265 mm
Pagination
64
Editorial Details
Edition Statement
2nd Revised edition
Age Details
Educational Level
UK School Key Stage 3
Interest Age
From 13 To 14
Description
Author Biography
Collectively, Evans, Gordon, Speed and Senior have over 100 years of teaching experience, both in the classroom and leading maths departments. They all currently hold senior positions within examining bodies and have been extensively involved in the development and piloting of new specifications.
Copyright in bibliographic data and cover images is held by Nielsen Book Services Limited or by the publishers or by their respective licensors: all rights reserved.
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Calculus : Single and Multivariable - 5th edition
Summary: Calculus teachers recognize Calculus as the leading resource among the ''reform'' projects that employ the rule of four and streamline the curriculum in order to deepen conceptual understanding. The fifth edition uses all strands of the ''Rule of Four'' - graphical, numeric, symbolic/algebraic, and verbal/applied presentations - to make concepts easier to understand. The book focuses on exploring fundamental ideas rather than comprehensive coverage of multiple similar cases that are ...show morenot fundamentally unique. Readers will also gain access to WileyPLUS, an online tool that allows for extensive drills and practice. Calculus teachers will build on their understanding in the field and discover new ways to present concepts to their students229.40 +$3.99 s/h
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College Algebra Solver
College algebra covers a range of topics like linear equations, quadratic equations, matrices, logarithms and polynomials. It is preferred that college level students have a sound understanding of algebra 1 and algebra 2. But even the best of us stumble at times, so there is no reason to let your apprehensions about algebra close that path for you. Help is available to all who ask and its much closer than you think. Free online college algebra solvers are readily available and cover all the topics that are covered in college algebra courses. Starting with basic refresher lessons and moving on to more advanced topics, free college algebra solvers have proven to be a great help to college students everywhere.
Free College Algebra Solver
So how exactly do college algebra problem solvers work and what can you expect from them? One of the best things about them is that you can start from whichever level of algebra you are comfortable with. Most online college algebra solvers cover high school algebra as well so that you can go back from time to time to anything that needs more clarity. Free online college algebra solvers have step by step solutions to your problems. College algebra equation solvers give step by step solutions to each problem. They also explain how each step leads to the next and what operations or rules were involved. Both college algebra equation solvers and college algebra word problem solvers provide practice questions and worksheets, mock test and exam prep help to make learning Algebra easy.
Solved Examples
Question 1: The sum of the square of the two consecutive positive even integers is 164. Find the integers. Solution:
Let two consecutive even integers, x and x + 2 Step 1: sum of the square of the two consecutive positive even integers = 164 |
Introduction Infinite Sequences Infinite Series and Convergence Taylor Series and Taylor Polynomials The Integral Test Comparison Tests for Positive-Term Series Alternating Series and Absolute Convergence Power Series Power Series Computations Series Solutions of Differential Equations
12. Vectors and Matrices
Vectors in the Plane Rectangular Coordinates and Three-Dimensional Vectors The Cross Product of Vectors Lines and Planes in Space Curves and Motions in Space Curvature and Acceleration Cylinders and Quadric Surfaces Cylindrical and Spherical Coordinates
Double Integrals Double Integrals over More General Regions Area and Volume by Double Integration Double Integrals in Polar Coordinates Applications of Double Integrals Triple Integrals Integration in Cylindrical and Spherical Coordinates Surface Area Change of Variables in Multiple Integrals |
Intermediate Algebra - 2nd edition
Summary: This student-focused text addresses individual learning styles through the use of a complete study system that starts with a learning styles inventory and presents targeted learning strategies designed to guide students toward success in this and future college-level courses.
Students who approach math with trepidation will find that Intermediate Algebra, Second Edition, builds competence and confidence. The study system, introduced at the outset and used c...show moreonsistently throughout the text, transforms the student experience by applying time-tested strategies to the study of mathematics. Learning strategies dovetail nicely into the overall system and build on individual learning styles by addressing students' unique strengths. The authors talk to students in their own language and walk them through the concepts, showing students both how to do the math and the reasoning behind it. Tying it all together, the use of the Algebra Pyramid as an overarching theme relates specific chapter topics to the 'big picture' of algebra200630.47 +$3.99 s/h
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Books Revisited Chatham, NJ
Possible retired library copy, some have markings or writing.
$125.93 +$3.99 s/h
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Penntext Downingtown, PA
May have minimal notes/highlighting, minimal wear/tear. Please contact us if you have any Questions |
Introduction to General Topology
9780201576986
ISBN:
0201576988
Edition: 1 Publisher: Benjamin-Cummings Publishing Company
Summary: An Introduction to General Topology is a clear, well motivated, rigorous introduction to point-set topology. Through careful exposition, it helps students think abstractly, providing an excellent foundation and preparation for further studies in topology and advanced mathematics study. The text contains few illustrations, encouraging students to draw their own and motivating them to think abstractly.
Cain, G...eorge L. is the author of Introduction to General Topology, published under ISBN 9780201576986 and 0201576988. Twenty Introduction to General Topology textbooks are available for sale on ValoreBooks.com, fifteen used from the cheapest price of $24.70, or buy new starting at $136.15 |
Hi this question is kind of a natural offshoot to this question My topic covers very conventional topics like :
Cartesian and Polar Coordinates in 3 Dim, second Degree eqns in 3 vars, reduction to canonical forms, straight lines, shortest distance between 2 skew lines, Plane, sphere, cone, cylinder, paraboloid, ellipsoid,hyperboloid of one and two sheets and their properties.
I had been working on my own for some time, but a paucity of time and the book being slightly dry has led me to this point.
Is there any repository of video lectures to help me through stuff like ellipsoid, paraboloid etc... I am having a bit of difficulty in putting together the large themes (no book does that, its my own approach: you can say a mental archive/model of the big ideas. I have often noticed doing this leads me to internalise things better) in my brain, and I think listening to a video lecture might help(coming second to a one to one class). |
Understanding Intermediate Algebra A Course for College Students
9780534417956
0534417957
Summary: Dr. Arthur Goodman (Ph.D., Yeshiva University) currently teaches in the mathematics department at Queens College of the City University of New York.
Hirsch, Lewis is the author of Understanding Intermediate Algebra A Course for College Students, published 2005 under ISBN 9780534417956 and 0534417957. Six hundred twenty Understanding Intermediate Algebra A Course for College Students textbooks are available f...or sale on ValoreBooks.com, one hundred twenty three used from the cheapest price of $24.00, or buy new starting at $190.69Lewis Hirsch and Arthur Goodman strongly believe that students can understand what they are learning in algebra and why. The authors meticulously explain why things are done [more]
Lewis Hirsch and Arthur Goodman strongly believe that students can understand what they are learning in algebra and why. The authors meticulously explain why things are done in a certain way, illustrate how and why concepts are related and demonstrat534421618-2-1-1 Orders ship the same or next bu [more]
ALTERNATE EDITION: Missing components. Instructor Edition: Same as student edition with additional notes or answers. Almost new condition. SKU:9780534421618-2-1-1 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions.[less] |
Dacula ExcelI use math jargon but try to make sure the student not only understands what (s)he is doing, but why and how it relates to other topics they have learned. I think it is important for students to understand the interconnectedness of mathematical principles as well as the "method" behind solving p... |
Mathematics
Mathematics
The Mathematics Department offers a broad, in-depth curriculum for exploring all aspects of mathematics – including quantities, changes, abstraction, structure and space – through small, engaging classroom settings. Within the department there is a great sense of discovery and collaboration as students and faculty are active in research, journal publication, conference presentations, mathematics competitions and many have received distinguished awards and recognition.
Research
is an integral part of the program and is incorporated into coursework as well as ongoing, unique research solutions outside of the classroom. In addition to collaborative research projects with faculty, students are highly active in the Math Club where they have the opportunity to discuss mathematics, solve problems and participate in social activities. A dedicated tutoring facility, great professor accessibility and the encouragement of open discussions in the classroom all contribute to a nurturing and supportive learning environment that lends a deep exploration of mathematics. Graduates are poised to excel in graduate programs and have become highly successful alumni who are skilled educators, actuaries, engineers, financial mathematicians and more.
Interested in helping share the fun of math and science with K12 students? This summer we have internships available for students to help assist and teach at science and math summer camps. The internships for College for Kids lasts approximately two and a half weeks and pays $1000. Preference is given to science and math [...]
Math major Josiah Reiswig scored in the top 45 percent of the students in the nation in the Putnam Competition. The Putnam Competition consists of two three-hour blocks. In each block, students are given six extremely difficult problems to solve. The competition is the largest, most competitive mathematics competition in North America and attracts participantsMike |
More About
This Textbook
Overview
This is a new edition of the precalculus text developed by the Consortium based at Harvard University and funded by a National Science Foundation Grant. The text is thought-provoking for well-prepared students while still accessible to students with weaker backgrounds. It provides numerical and graphical approaches as well as algebraic approaches to give students another way of mastering the material. This approach encourages students to persist, thereby lowering failure rates. A large number of real-world examples and problems enable students to create mathematical models that will help them understand the world in which they live.
The focus is on those topics that are essential to the study of calculus and these topics are treated in depth.
Linear, exponential, power, and periodic functions are introduced before polynomial and rational functions to take advantage of their use to model physical phenomena.
Building on the Consortium's Rule of Four: Each function is represented symbolically, numerically, graphically, and verbally where appropriate |
Fairfax Station ACT Math...This may have been as an official position, but usually just happened by itself since I never minded interruptions to help them and never criticized them for what they did not know - we all have many things we don't know.I purchased my first Mathematica license in 1989 and have been using Mat... |
Product Details
Complex Numbers and Geometry by Liang-shin Hahn
Provides a self-contained introduction to complex numbers and college geometry written in an informal style with an emphasis on the motivation behind the ideas... The author engages the reader with a casual style, motivational questions, interesting problems and historical notes. -Mathematical Reviews
The purpose of this book is to demonstrate that complex numbers and geometry can be blended together beautifully, resulting in easy proofs and natural generalizations of many theorems in plane geometry--such as the Napoleon theorem, the Ptolemy-Euler theorem, the Simson theorem, and the Morley theorem. Beginning with a construction of complex numbers, readers are taken on a guided tour that includes something for everyone, even those with advanced degrees in mathematics. Yet, the entire book is accessible to a talented high-school student.The book is self- contained-no background in complex numbers is assumed-and can be covered at a leisurely pace in a one semester course. Many of the chapters can be read independently. Over 100 exercises are included. The book would be suitable as a text for a geometry course, or for a problem solving seminar, or as enrichment for the student who wants to know more. |
Genre: Traditional story Learning Objectives: Language Comprehension Strand 7: Explore how particular words are used, including words and expressions with similar meanings. Writing Opportunities Strand 9: Sustain form in narrative, including use of person and time (e. g. writing about story characters).
Stewart's SINGLE VARIABLE CALCULUS WITH VECTOR FUNCTIONS has the mathematical precision, accuracy, clarity of exposition and outstanding examples and problem sets that characterized all of James Stewart's texts. In this new text, Stewart focuses on problem solving, using the pedagogical system that has worked so well for students in a wide variety of academic settings |
Groups and Geometry (Oxford Science Publications)
Book Description: This volume presents the Oxford Mathematical Institute notes for the enormously successful advanced undergraduate and first-year graduate student course on groups and geometry. The book's content closely follows the Oxford syllabus but covers a great deal more material than did the course itself. The book is divided into two parts: the first covers the fundamentals of groups, and the second covers geometry and its symbiotic relationship with groups. Both parts contain a number of useful examples and exercises. This book will be welcomed by student and teacher alike as a lucidly written text on an important topic |
Essential Mathematical Methods for Physicists
9780120598779
ISBN:
0120598779
Pub Date: 2003 Publisher: Academic Pr
Summary: This new adaptation of Arfken and Weber's bestselling Mathematical Methods for Physicists, Fifth Edition, is the most modern collection of mathematical principles for solving physics problems. Additional explanations and examples provide models and context for methods applicable to a wide range of physics theories and applications. Features: · Many detailed, worked-out examples illustrate how to use and apply mathema...tical techniques to solve physics problems · Frequent and thorough explanations help readers understand, recall, and apply the theory · Introductions and review material provide context and emphasis on key ideas · Many routine exercises reinforce basic, foundational concepts and computations "True to the title, this new text achieves a comprehensive coverage of the "essential" topics in mathematical physics at the undergraduate level. This new version is filled with enlightening examples, which is the key to undergraduate teaching . More importantly, many examples are real problems from various fields of physics. Illustration of the mathematical principles behind the solution of these problems further enhances the connection between this course and other courses in a physics curriculum. " David Hwang, University of California at Davis
Weber, Hans J. is the author of Essential Mathematical Methods for Physicists, published 2003 under ISBN 9780120598779 and 0120598779. Seven hundred twenty nine Essential Mathematical Methods for Physicists textbooks are available for sale on ValoreBooks.com, one hundred sixteen used from the cheapest price of $26.73, or buy new starting at $39.00Corners/spine lightly worn/bent, covers lightly scuffed, copyright page lightly stained, front fly/first few pages lightly stained/creased at right top corner, text pages clean/tight/bright |
books.google.com - This text is designed for an introductory probability course at the university level for undergraduates in mathematics, the physical and social sciences, engineering, and computer science. It presents a thorough treatment of probability ideas and techniques necessary for a firm understanding of the subject.... to Probability |
Descriptive analysis and presentation of either single-variable data or bivariate data, probability, probability distributions, normal probability distributions, sample variability, statistical inferences involving one and two populations, analysis of variance, linear correlation and regression analysis. Statistical computer software will be extensively integrated as a tool in the description and analysis of data.
Math 8A prepares the student for the study of calculus by providing important skills in algebraic manipulation, interpretation, and problem solving at the college level. Topics will include basic algebraic concepts, complex numbers, equations and inequalities of the first and second degree, functions, and graphs, linear and quadratic equations, polynomial functions, exponential and logarithmic functions, systems of equations, matrices and determinants, right triangle trigonometry, and the Law of Sines and Cosines.
This course is a standard beginning algebra course, including algebraic expressions, linear equations and inequalities in one variable, graphing, equations and inequalities in two variables, integer exponents, polynomials, rational expressions and equations, radicals and rational exponents, and quadratic equations. Mathematics 205, 205A and 205B, and 206 have similar course content. This course may not be taken by students who have completed Mathematics 205B or 206 with a grade of "C" or better. This course may be taken for Mathematics 205B credit (2.5 units) by those students who have successfully completed Mathematics 205A with a grade of "C" or better.
Prerequisite: Math 205 or Math 205A and Math 205B with a grade of 'C' or better.
Transferable: No
This course introduces the vocabulary and principles of Euclidean Geometry. Methods of proof including inductive and deductive reasoning will be developed. Concepts of congruence and similarity, angles, lines, polygons, and circles will be covered. Additional topics such as solid geometry, analytical geometry, transformations, and basic trigonometry may be included as time allows.
This course covers operations with integers, fractions and decimals and associated applications, percentages, ratio, and geometry and measurement, critical thinking and applications. Elementary algebra topics such as variables, expressions, and solving equations are introduced.
Sect#
Type
Room
Instructor
Units
Days
Time Start-End
Footnotes
8107
L/L
SS110
FULLER G
3.0
DAILY
0800A - 1020A
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Geometry with Geometry Explorer
Book Description: Geometry with Geometry Explorer combines a discovery-based geometry text with powerful integrated geometry software. This combination allows for the deep exploration of topics that would be impossible without well-integrated technology, such as hyperbolic geometry, and encourages the kind of experimentation and self-discovery needed for students to develop a natural intuition for various topics in geometry |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). |
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PROFESSOR STRANG: OK,
let's start with a
10
00:00:24 --> 00:00:26
review and preview.
11
00:00:26 --> 00:00:30
I put a P up there because
we're really looking into
12
00:00:30 --> 00:00:36
the Fourier part that just
started this morning.
13
00:00:36 --> 00:00:41
And there'll be some homework
from these early sections about
14
00:00:41 --> 00:00:44
the Fourier stuff, so we maybe
we should just do a few
15
00:00:44 --> 00:00:51
of those problems or
discuss here today.
16
00:00:51 --> 00:00:54
Just in advance.
17
00:00:54 --> 00:00:57
Can I say one thing about
MATLAB and the MATLAB
18
00:00:57 --> 00:00:58
homework first?
19
00:00:58 --> 00:01:03
And maybe open a
conversation about it?
20
00:01:03 --> 00:01:10
So there's really two different
problems that I'm personally
21
00:01:10 --> 00:01:13
quite interested in.
22
00:01:13 --> 00:01:16
Two model, I'll say model
problems because they're
23
00:01:16 --> 00:01:20
there for regular
polyons in a circle.
24
00:01:20 --> 00:01:31
And I'll draw an octagon
again, so M sides.
25
00:01:31 --> 00:01:34
And I'm interested in
if M goes to infinity.
26
00:01:34 --> 00:01:37
And I'm interested in
two different problems.
27
00:01:37 --> 00:01:41
So one of them is our
MATLAB problem, minus
28
00:01:41 --> 00:01:46
is Laplace's equation.
29
00:01:46 --> 00:01:50
What was it, four?
30
00:01:50 --> 00:01:58
With u=0 on the circle.
31
00:01:58 --> 00:02:02
OK, so that's our problem,
totally open for discussion.
32
00:02:02 --> 00:02:05
How many have started on that?
33
00:02:05 --> 00:02:06
Oh, good.
34
00:02:06 --> 00:02:06
OK.
35
00:02:06 --> 00:02:08
Well, then you all know
more about it than I.
36
00:02:08 --> 00:02:10
And that's great.
37
00:02:10 --> 00:02:14
I'd be happy to learn.
38
00:02:14 --> 00:02:15
So have I said
everything there?
39
00:02:15 --> 00:02:18
Yeah, we've got Poisson's
equation inside.
40
00:02:18 --> 00:02:23
We've got u=0 on the circle, so
the problem's well defined and
41
00:02:23 --> 00:02:28
the solution should be
one minus x squared
42
00:02:28 --> 00:02:33
minus y squared.
43
00:02:33 --> 00:02:35
So that's the correct solution.
44
00:02:35 --> 00:02:38
Maybe I can also tell you
about the second problem
45
00:02:38 --> 00:02:40
that I'm interested in.
46
00:02:40 --> 00:02:43
Because it hasn't come
up in class but it's
47
00:02:43 --> 00:02:45
very important, too.
48
00:02:45 --> 00:02:47
It would be the
eigenvalue problem.
49
00:02:47 --> 00:02:52
So this is problem number one,
the steady state problem
50
00:02:52 --> 00:02:57
when you've got a source and
you want to find out the
51
00:02:57 --> 00:02:59
temperature distribution.
52
00:02:59 --> 00:03:01
The problem number two
would be the eigenvalue
53
00:03:01 --> 00:03:03
problem, -u_xx-u_yy.
54
00:03:03 --> 00:03:06
55
00:03:06 --> 00:03:10
I take those minuses so
that the eigenvalue
56
00:03:10 --> 00:03:12
will be positive.
57
00:03:12 --> 00:03:15
So that's what the eigenvalue
problem might look like.
58
00:03:15 --> 00:03:20
And again let me say, with
u=0 on the boundary.
59
00:03:20 --> 00:03:25
On the third one.
60
00:03:25 --> 00:03:33
OK, so a person would say
this is Laplace's eigenvalue
61
00:03:33 --> 00:03:35
problem because we have
Laplace's equation.
62
00:03:35 --> 00:03:38
We've got eigenvalue.
63
00:03:38 --> 00:03:42
As always, it's not linear
because we have two unknowns,
64
00:03:42 --> 00:03:45
lambda's multiplying u.
65
00:03:45 --> 00:03:48
And we have boundary
conditions, and this would
66
00:03:48 --> 00:03:56
describe the normal modes, for
example, of a circular drum.
67
00:03:56 --> 00:04:00
If I had a drum, or
a polygon drum.
68
00:04:00 --> 00:04:08
So maybe I connect to actually
build the drum, I might fold in
69
00:04:08 --> 00:04:11
the sides there and
have a polygon.
70
00:04:11 --> 00:04:17
And again, I hope that the
eigenvalues of the polygon,
71
00:04:17 --> 00:04:20
this equation in the
polygon, which are not
72
00:04:20 --> 00:04:22
known, by the way.
73
00:04:22 --> 00:04:27
To the best to my knowledge, we
know it only for M=3, which
74
00:04:27 --> 00:04:32
would be an equilateral
triangle, and M=4, which
75
00:04:32 --> 00:04:34
would be a square.
76
00:04:34 --> 00:04:39
And those eigenvalues, because
of Fourier or something
77
00:04:39 --> 00:04:42
are humanly doable.
78
00:04:42 --> 00:04:49
But I think five on up is, I
may be wrong about six I'm
79
00:04:49 --> 00:04:52
not sure about M=6,
a hexagon sometimes
80
00:04:52 --> 00:04:53
gives you enough help.
81
00:04:53 --> 00:04:59
But beyond that
you're on your own.
82
00:04:59 --> 00:05:03
With finite elements
to help you.
83
00:05:03 --> 00:05:09
So there's a whole sequence of
eigenfunctions, u, eigenvalues,
84
00:05:09 --> 00:05:15
lambda, just the way there
were in one dimension.
85
00:05:15 --> 00:05:18
And on the circle
they involve Bessel.
86
00:05:18 --> 00:05:21
That's where Bessel showed up.
87
00:05:21 --> 00:05:25
He figured out the functions
and they're not especially
88
00:05:25 --> 00:05:29
nice functions.
89
00:05:29 --> 00:05:32
But they're studied
for centuries.
90
00:05:32 --> 00:05:36
Bessel functions
come into that.
91
00:05:36 --> 00:05:40
But, here I have
the same question.
92
00:05:40 --> 00:05:45
I mean, let me just say, for me
this could be a UROP project if
93
00:05:45 --> 00:05:48
anybody was an undergraduate,
or it could be a project
94
00:05:48 --> 00:05:51
over January or something.
95
00:05:51 --> 00:05:57
I'd like to know something
about what happens as M goes
96
00:05:57 --> 00:06:03
to infinity as the polygon
approaches the circle.
97
00:06:03 --> 00:06:11
So I'm hoping maybe on the
homework that comes, if it's
98
00:06:11 --> 00:06:18
not too difficult, and maybe
it's not, to let m go up a bit.
99
00:06:18 --> 00:06:20
There is one thing.
100
00:06:20 --> 00:06:25
That the code we're working
with is linear elements, right?
101
00:06:25 --> 00:06:27
We're using linear
finite elements.
102
00:06:27 --> 00:06:32
So we're not getting
high accuracy.
103
00:06:32 --> 00:06:38
So I would really like to move
up to quadratic elements, at
104
00:06:38 --> 00:06:42
least, you remember quadratic
elements would be ones where -
105
00:06:42 --> 00:06:52
well, let me draw the one that
we've drawn in class before.
106
00:06:52 --> 00:06:56
We only have to look at one
triangle and then we cut it up
107
00:06:56 --> 00:07:02
into triangular elements by
taking some pieces here, taking
108
00:07:02 --> 00:07:07
the points above, which I hope
are now correct on the website.
109
00:07:07 --> 00:07:11
Connecting those edges, and
then connecting these.
110
00:07:11 --> 00:07:12
Is that right?
111
00:07:12 --> 00:07:18
Is that our mesh?
112
00:07:18 --> 00:07:20
So that mesh is
controlled by N.
113
00:07:20 --> 00:07:23
One, two, N points.
114
00:07:23 --> 00:07:32
Also, N is going to have to get
large too, to give me accuracy.
115
00:07:32 --> 00:07:38
And another way toward more
accuracy is, instead of linear
116
00:07:38 --> 00:07:42
elements, second degree.
117
00:07:42 --> 00:07:44
So do you remember I
wrote those down?
118
00:07:44 --> 00:07:48
Let me take that little
triangle out here as
119
00:07:48 --> 00:07:51
a bigger triangle.
120
00:07:51 --> 00:07:53
It would look something
like that, I guess.
121
00:07:53 --> 00:07:58
The second degree elements
have those six mesh points.
122
00:07:58 --> 00:08:01
You remember I drew those but
we didn't really have time
123
00:08:01 --> 00:08:07
to get further with them.
124
00:08:07 --> 00:08:12
The trial functions phi, which
are one at a typical mesh point
125
00:08:12 --> 00:08:18
and zero at all the others,
they are computable.
126
00:08:18 --> 00:08:22
We're up to second degree, so
it's a little, second degree
127
00:08:22 --> 00:08:26
things, then the first
derivatives, which come into
128
00:08:26 --> 00:08:28
the integrations, are linear.
129
00:08:28 --> 00:08:29
And not constant.
130
00:08:29 --> 00:08:31
So a little bit harder.
131
00:08:31 --> 00:08:37
But finite elements,
linear or quadratic.
132
00:08:37 --> 00:08:38
Or higher.
133
00:08:38 --> 00:08:40
Could be used for this problem.
134
00:08:40 --> 00:08:45
As we know, and
for this problem.
135
00:08:45 --> 00:08:48
What I wanted to add, that I've
not mentioned in class, and I
136
00:08:48 --> 00:08:55
think we may just not get a
chance to do it, is what does
137
00:08:55 --> 00:08:58
the finite element method look
like for an eigenvalue problem?
138
00:08:58 --> 00:09:02
Because eigenvalues
are highly important.
139
00:09:02 --> 00:09:06
That's the different
way to understand.
140
00:09:06 --> 00:09:08
There's the matrix
K and its entries.
141
00:09:08 --> 00:09:11
But then there eigenvalues.
142
00:09:11 --> 00:09:14
And you might think that, what
do you think is the discrete
143
00:09:14 --> 00:09:19
eigenvalue problem
copying this one?
144
00:09:19 --> 00:09:21
Here's my point.
145
00:09:21 --> 00:09:26
Your first guess would be,
well this is like K, right?
146
00:09:26 --> 00:09:34
This is like KU, right? (K2D)U,
I should call it, maybe.
147
00:09:34 --> 00:09:39
Well, I'll call it K, because
K2D I have specifically
148
00:09:39 --> 00:09:47
reserved for the Laplace
stiffness matrix on a square
149
00:09:47 --> 00:09:50
mesh, square mesh with
triangles, the K2D.
150
00:09:51 --> 00:09:57
That was one specific matrix
for one specific mesh, and here
151
00:09:57 --> 00:09:59
we have a different mesh.
152
00:09:59 --> 00:10:01
So I should just call it K.
153
00:10:01 --> 00:10:06
Ok, I think if anybody was
going to make a guess, they
154
00:10:06 --> 00:10:07
would say OK, KU=LAMBDA*U.
155
00:10:09 --> 00:10:13
Maybe I'll use capital LAMBDA,
because I'm using capital U.
156
00:10:13 --> 00:10:28
Is this the finite element
method eigenvalue problem.
157
00:10:28 --> 00:10:33
And if you answered yes, I
would have to say, well
158
00:10:33 --> 00:10:35
that's a reasonable answer.
159
00:10:35 --> 00:10:38
But it's wrong.
160
00:10:38 --> 00:10:43
The eigenvalue problem, when I
take the differential equation
161
00:10:43 --> 00:10:47
for the Laplace, Laplace's
equation, lambda u on the right
162
00:10:47 --> 00:10:55
side, and I go to do finite
elements, it produces K.
163
00:10:55 --> 00:11:00
Out of this stuff, out of the
weak form, all that stuff.
164
00:11:00 --> 00:11:04
But it produces another matrix
on the right-hand side from the
165
00:11:04 --> 00:11:07
constant term, and we have
not really mentioned it,
166
00:11:07 --> 00:11:09
it's the mass matrix.
167
00:11:09 --> 00:11:13
So this, instead of just
the identity here,
168
00:11:13 --> 00:11:16
there's a mass matrix.
169
00:11:16 --> 00:11:21
So that is the problem
that you could do.
170
00:11:21 --> 00:11:28
I could've made a
MATLAB project.
171
00:11:28 --> 00:11:32
I bet I'd do it
next fall. right?
172
00:11:32 --> 00:11:39
You guys did the
first one, this one.
173
00:11:39 --> 00:11:40
Or you are doing it now.
174
00:11:40 --> 00:11:43
And I'm going to pause in a
minute for questions about
175
00:11:43 --> 00:11:46
it, or discussion of it.
176
00:11:46 --> 00:11:49
But this one brings
in something called
177
00:11:49 --> 00:11:50
the mass matrix.
178
00:11:50 --> 00:11:58
So let me just say
what those are.
179
00:11:58 --> 00:12:02
If I write down the entries in
the mass matrix, you'll sort of
180
00:12:02 --> 00:12:04
get an idea of why they are.
181
00:12:04 --> 00:12:07
So what are the entries
in the stiffness matrix?
182
00:12:07 --> 00:12:15
K_ij, you remember, is
the integral of the d
183
00:12:15 --> 00:12:18
phi_i/dx, d phi_j/dx.
184
00:12:20 --> 00:12:33
Plus d phi_i/dy, d phi_j/dy,
dxdy, and that's what's
185
00:12:33 --> 00:12:34
you're computing.
186
00:12:34 --> 00:12:36
And that's what that
code is computing.
187
00:12:36 --> 00:12:42
And when phi is linear,
phi linear, then
188
00:12:42 --> 00:12:47
slopes are constant.
189
00:12:47 --> 00:12:52
So all you have to do, and what
that code in the book is doing,
190
00:12:52 --> 00:12:55
is figuring out what
are the slopes.
191
00:12:55 --> 00:13:01
These things are constant, so
we just need to know the area
192
00:13:01 --> 00:13:06
of the integration where
we're integrating.
193
00:13:06 --> 00:13:09
The area, triangle by triangle.
194
00:13:09 --> 00:13:11
Fine.
195
00:13:11 --> 00:13:12
That's what we're doing.
196
00:13:12 --> 00:13:16
That's what that code
is just set up to do.
197
00:13:16 --> 00:13:20
Now, I have to tell you what
is M_ij, the mass matrix.
198
00:13:20 --> 00:13:25
I just think you don't want to
have - we haven't done too
199
00:13:25 --> 00:13:27
badly with finite that
elements in here.
200
00:13:27 --> 00:13:31
We did it in 1-D, where we
got it kind of straight.
201
00:13:31 --> 00:13:34
And now we're seeing
what it looks like 2-D.
202
00:13:34 --> 00:13:38
But I had not really
mentioned a mass matrix.
203
00:13:38 --> 00:13:42
So here it comes.
204
00:13:42 --> 00:13:45
The mass matrix will
be the integral of
205
00:13:45 --> 00:13:48
phi_i i times phi_j.
206
00:13:49 --> 00:13:50
dxdy.
207
00:13:50 --> 00:13:57
It's the zero order, no
derivatives, just plain zero
208
00:13:57 --> 00:14:07
order, as you'd expect from the
fact that the term in the
209
00:14:07 --> 00:14:09
continuous part is zero order.
210
00:14:09 --> 00:14:12
So it's this mass
matrix that comes in.
211
00:14:12 --> 00:14:20
And maybe we could just look
to see which entries will
212
00:14:20 --> 00:14:23
be zero and which will not.
213
00:14:23 --> 00:14:25
How sparse is it?
214
00:14:25 --> 00:14:28
What does the mass
matrix look like?
215
00:14:28 --> 00:14:33
And can we just, let
me do 1-D first.
216
00:14:33 --> 00:14:36
So there's a phi, right?
217
00:14:36 --> 00:14:38
There's another one.
218
00:14:38 --> 00:14:40
There's another one.
219
00:14:40 --> 00:14:45
So, what do you think about the
mass matrix, one phi multiplied
220
00:14:45 --> 00:14:48
by another phi and integrated?
221
00:14:48 --> 00:14:51
Is it diagonal?
222
00:14:51 --> 00:15:00
No, because each phi
overlaps its two neighbors.
223
00:15:00 --> 00:15:02
So tell me what kind of a
matrix m is going to be?
224
00:15:02 --> 00:15:04
In 1-D.
225
00:15:04 --> 00:15:07
Tri-diagonal.
226
00:15:07 --> 00:15:08
It'll be tri-diagonal.
227
00:15:08 --> 00:15:11
Now, so was K.
228
00:15:11 --> 00:15:15
So K and M actually have
non-zeroes in the same places.
229
00:15:15 --> 00:15:17
the same sparsity pattern.
230
00:15:17 --> 00:15:20
But, of course, not the
same numbers in there.
231
00:15:20 --> 00:15:31
K had minus ones and twos
and fours and minus ones.
232
00:15:31 --> 00:15:36
What can you tell me about
this tri-diagonal matrix?
233
00:15:36 --> 00:15:42
When I integrate that against
this, well, again I would do it
234
00:15:42 --> 00:15:45
element by element because this
against this, they
235
00:15:45 --> 00:15:48
only overlap here.
236
00:15:48 --> 00:15:48
Right?
237
00:15:48 --> 00:15:51
I'll just draw the one
place that they overlap.
238
00:15:51 --> 00:15:54
And what's the point?
239
00:15:54 --> 00:15:56
They're both positive.
240
00:15:56 --> 00:16:02
So the mass matrix is, its
rows don't add to zero.
241
00:16:02 --> 00:16:04
Its rows tend to add to one.
242
00:16:04 --> 00:16:08
But it's not diagonal,
that's the difference.
243
00:16:08 --> 00:16:16
OK, so I just felt I couldn't
feel as though I'd done a
244
00:16:16 --> 00:16:21
decent job in describing
finite elements if I
245
00:16:21 --> 00:16:23
didn't describe this.
246
00:16:23 --> 00:16:26
Didn't mention
this mass matrix.
247
00:16:26 --> 00:16:30
And maybe I'd better say
where it comes from.
248
00:16:30 --> 00:16:35
Because eigenvalue problems, it
may come number two, but that's
249
00:16:35 --> 00:16:37
pretty high up the list.
250
00:16:37 --> 00:16:50
So let me tell you where does
this mass matrix come from.
251
00:16:50 --> 00:16:52
First, let me tell you
about eigenvalues of
252
00:16:52 --> 00:16:56
a, matrix eigenvalues.
253
00:16:56 --> 00:17:00
So the answer was, is
this the finite element
254
00:17:00 --> 00:17:01
eigenvalue problem?
255
00:17:01 --> 00:17:03
Only if there's an M there.
256
00:17:03 --> 00:17:11
And now I want to, OK, first
of all, what MATLAB command
257
00:17:11 --> 00:17:13
solves that problem?
258
00:17:13 --> 00:17:16
Let's just be a little
practical for a moment.
259
00:17:16 --> 00:17:23
What MATLAB command gives me
the matrix of eigenvectors, the
260
00:17:23 --> 00:17:32
matrix of eigenvalues would
come from eig of what?
261
00:17:32 --> 00:17:35
I'd call this the generalized
eigenvalue problem.
262
00:17:35 --> 00:17:38
Generalized because it's
got somebody over here.
263
00:17:38 --> 00:17:39
And it's just K,M.
264
00:17:42 --> 00:17:46
Or of course you get the same
answer, well you get the same
265
00:17:46 --> 00:17:50
eigenvalues, I guess the same
eigenvectors, yeah, if you
266
00:17:50 --> 00:17:57
or, eig(M^-1,K), of course.
267
00:17:57 --> 00:18:00
If you want to do it with
just one matrix then bring
268
00:18:00 --> 00:18:01
M inverse over here.
269
00:18:01 --> 00:18:06
But and M inverse, the
inverse of this tridiagonal
270
00:18:06 --> 00:18:08
matrix, is full.
271
00:18:08 --> 00:18:11
No zeroes in the inverse.
272
00:18:11 --> 00:18:14
So everybody would much
prefer this tridiagonal
273
00:18:14 --> 00:18:16
tridiagonal one.
274
00:18:16 --> 00:18:19
So that's how MATLAB
would do it.
275
00:18:19 --> 00:18:27
And what I want to know is,
back in this problem, how close
276
00:18:27 --> 00:18:39
do the finite element guys
come, on polygons, come to the
277
00:18:39 --> 00:18:41
correct solution on circles.
278
00:18:41 --> 00:18:47
I'm hoping that for problem
one you can maybe keep M
279
00:18:47 --> 00:18:53
and N equal, or maybe four
times M or something.
280
00:18:53 --> 00:18:58
And let them grow and see.
281
00:18:58 --> 00:19:02
Well, for example, at the
center of the circle, or how
282
00:19:02 --> 00:19:05
quickly do you approach the
correct answer one at the
283
00:19:05 --> 00:19:07
center of the circle?
284
00:19:07 --> 00:19:09
I think it's going to
be a good problem.
285
00:19:09 --> 00:19:13
Let me open to, so I started
out just talking there.
286
00:19:13 --> 00:19:17
What about the MATLAB problem.
287
00:19:17 --> 00:19:24
You made a start on
it, is it going?
288
00:19:24 --> 00:19:30
Have you got a graph, maybe, or
what's reasonable to graph,
289
00:19:30 --> 00:19:35
to give Peter to look at?
290
00:19:35 --> 00:19:39
Who's done something on
that MATLAB problem?
291
00:19:39 --> 00:19:43
Yeah, go ahead tell
us all what to do.
292
00:19:43 --> 00:19:45
AUDIENCE: I made the
triangle pi section
293
00:19:45 --> 00:19:48
PROFESSOR STRANG: OK, right
294
00:19:48 --> 00:19:49
AUDIENCE: [INAUDIBLE]
295
00:19:49 --> 00:19:55
and I found that
the [INAUDIBLE]
296
00:19:55 --> 00:19:57
changes to M.
297
00:19:57 --> 00:19:59
PROFESSOR STRANG:
With M more, I see.
298
00:19:59 --> 00:20:07
So if you just fixed M like
eight, and let n get, it
299
00:20:07 --> 00:20:09
didn't change significantly.
300
00:20:09 --> 00:20:12
It wouldn't, of course,
converge to the right answer.
301
00:20:12 --> 00:20:15
It'll converge, if it
does, to some kind of an
302
00:20:15 --> 00:20:18
answer, for the polygon.
303
00:20:18 --> 00:20:19
Right.
304
00:20:19 --> 00:20:19
That's right.
305
00:20:19 --> 00:20:26
So you know, as I wrote the
problem I didn't know whether I
306
00:20:26 --> 00:20:31
dared say let M get increased
too, but of course that's
307
00:20:31 --> 00:20:32
the real question.
308
00:20:32 --> 00:20:34
And what happened then?
309
00:20:34 --> 00:20:37
Did error shrink?
310
00:20:37 --> 00:20:41
OK, and now maybe it's
possible to see how fast or
311
00:20:41 --> 00:20:42
something that's always--
312
00:20:42 --> 00:20:46
AUDIENCE: [INAUDIBLE]
313
00:20:46 --> 00:20:47
PROFESSOR STRANG: Ah.
314
00:20:47 --> 00:20:49
OK, at the center.
315
00:20:49 --> 00:20:53
OK, then I hope
for more comment.
316
00:20:53 --> 00:20:55
Let me say one more thing.
317
00:20:55 --> 00:21:01
My theory is that the error
at the center is quite a
318
00:21:01 --> 00:21:06
bit smaller than the error
closer to the boundary.
319
00:21:06 --> 00:21:13
I would be interested in an
error, is it fairly even?
320
00:21:13 --> 00:21:16
Oh, my theory's wrong.
321
00:21:16 --> 00:21:18
It wouldn't be the first time.
322
00:21:18 --> 00:21:21
And maybe because it's linear.
323
00:21:21 --> 00:21:28
Yeah, my theory is more for
better elements, like these.
324
00:21:28 --> 00:21:30
I'd be interested to know.
325
00:21:30 --> 00:21:37
Why do I think, why do I have
this theory, which you guys
326
00:21:37 --> 00:21:41
are going to prove wrong
anyway, but still.
327
00:21:41 --> 00:21:43
After you've proved it wrong,
you won't listen to me
328
00:21:43 --> 00:21:44
if I tell it to you.
329
00:21:44 --> 00:21:45
So now I'll tell it.
330
00:21:45 --> 00:21:54
My theory is that the error
around the boundary is, there's
331
00:21:54 --> 00:21:57
no error at these vertices, and
then there's sort of a going to
332
00:21:57 --> 00:22:01
be an error because the real
answer is not zero along here.
333
00:22:01 --> 00:22:04
It's sort of near
zero, but not quite.
334
00:22:04 --> 00:22:07
You know, there's an error.
335
00:22:07 --> 00:22:10
So there's errors around
here, from getting
336
00:22:10 --> 00:22:13
the boundary wrong.
337
00:22:13 --> 00:22:16
Squaring it off.
338
00:22:16 --> 00:22:20
But my theory is that errors,
the boundary stuff, drops off
339
00:22:20 --> 00:22:22
quickly as you go inside.
340
00:22:22 --> 00:22:26
That's why I think, from those,
you remember those - well,
341
00:22:26 --> 00:22:31
we'll see them again either
today or Friday, those
342
00:22:31 --> 00:22:36
r^n*cos(nx) type things?
343
00:22:36 --> 00:22:37
That cos(n*theta)?
344
00:22:39 --> 00:22:42
Yeah, you remember those
are the typical solutions
345
00:22:42 --> 00:22:44
to Laplace's equation.
346
00:22:44 --> 00:22:48
And then so that if, and
it has some coefficient,
347
00:22:48 --> 00:22:50
of course, a n.
348
00:22:50 --> 00:22:56
And I look at that, that
might be a piece of error.
349
00:22:56 --> 00:22:59
And it's way bigger
when r is one and way
350
00:22:59 --> 00:23:01
smaller when r is zero.
351
00:23:01 --> 00:23:05
So anyway, that's
sort of my theory.
352
00:23:05 --> 00:23:10
That if you have,
like physically.
353
00:23:10 --> 00:23:18
You have a circular plate
and you're maintaining the
354
00:23:18 --> 00:23:22
boundary temperature at
some sort of oscillation.
355
00:23:22 --> 00:23:27
Like, near one but up
and down from one.
356
00:23:27 --> 00:23:33
Then I think further
inside, it doesn't know.
357
00:23:33 --> 00:23:36
It hardly knows about
that oscillation.
358
00:23:36 --> 00:23:38
This is my theory.
359
00:23:38 --> 00:23:43
That toward the center of the
circle it only sees kind of an
360
00:23:43 --> 00:23:47
average boundary temperature
and not your little
361
00:23:47 --> 00:23:49
ups and downs.
362
00:23:49 --> 00:23:56
So when M is big, I expect that
part of the up and down part to
363
00:23:56 --> 00:23:59
be not so significant
in the center.
364
00:23:59 --> 00:24:01
Anyway, now that's my theory.
365
00:24:01 --> 00:24:04
AUDIENCE: [INAUDIBLE]
366
00:24:04 --> 00:24:11
PROFESSOR STRANG:
Ah, good question.
367
00:24:11 --> 00:24:15
So if we only looked
at the center, would
368
00:24:15 --> 00:24:17
it all be the same?
369
00:24:17 --> 00:24:20
I mean, if we're only looking
at that one point where it
370
00:24:20 --> 00:24:32
should be 1 at the center, but
along the thing, I don't know.
371
00:24:32 --> 00:24:38
If you look at both, and see a
significant difference in the
372
00:24:38 --> 00:24:40
behavior I'd be interested.
373
00:24:40 --> 00:24:41
Yeah, yeah.
374
00:24:41 --> 00:24:43
You know, all these problems
are things that there's
375
00:24:43 --> 00:24:47
no single solution to.
376
00:24:47 --> 00:24:53
AUDIENCE: [INAUDIBLE]
377
00:24:53 --> 00:24:55
PROFESSOR STRANG: The error
between one minus r squared
378
00:24:55 --> 00:25:04
AUDIENCE: [INAUDIBLE]
379
00:25:04 --> 00:25:06
PROFESSOR STRANG: Oh, right,
we've got slope error, too.
380
00:25:06 --> 00:25:10
That's a very
significant point.
381
00:25:10 --> 00:25:13
I see, right.
382
00:25:13 --> 00:25:14
So the slope error's in there.
383
00:25:14 --> 00:25:19
Everybody knows, then,
everybody in working the
384
00:25:19 --> 00:25:26
problem, I mentioned that the
boundary conditions in this
385
00:25:26 --> 00:25:32
piece of pie were zero along
here and normal derivative,
386
00:25:32 --> 00:25:36
somehow it got printed du/dh,
but that was an accident.
387
00:25:36 --> 00:25:42
It should've been
du/dn, dn is zero.
388
00:25:42 --> 00:25:46
So Neumann conditions on this
thing and then I was a little
389
00:25:46 --> 00:25:50
scared about that point,
but I think phooey on it.
390
00:25:50 --> 00:25:56
It's just, don't
worry about it.
391
00:25:56 --> 00:25:59
But what I was going to say.
392
00:25:59 --> 00:26:04
How do you, what do you do to
take into account this du/dn=0?
393
00:26:04 --> 00:26:07
394
00:26:07 --> 00:26:13
This slope condition on
these long boundaries?
395
00:26:13 --> 00:26:15
What should you do in
finite elements to
396
00:26:15 --> 00:26:17
take account for that?
397
00:26:17 --> 00:26:21
And the answer is,
in one nice word?
398
00:26:21 --> 00:26:22
Nothing.
399
00:26:22 --> 00:26:24
Right, nothing.
400
00:26:24 --> 00:26:27
Your finite element method
should not, you don't
401
00:26:27 --> 00:26:30
impose any condition
along these boundaries.
402
00:26:30 --> 00:26:35
Just use the code as it is
with zeroes on this boundary.
403
00:26:35 --> 00:26:39
And it should work, yeah.
404
00:26:39 --> 00:26:39
It should work.
405
00:26:39 --> 00:26:43
Any comments on other people.
406
00:26:43 --> 00:26:48
Did you get reasonable
results, or?
407
00:26:48 --> 00:26:49
Tell me something.
408
00:26:49 --> 00:26:55
Because you guys looked at
those graphs and I have not.
409
00:26:55 --> 00:26:57
Any feedback yet?
410
00:26:57 --> 00:26:58
On those?
411
00:26:58 --> 00:27:01
I'm happy to get
email, too, about.
412
00:27:01 --> 00:27:04
So all the email, first of
all they've corrected the
413
00:27:04 --> 00:27:09
typos in the original
coordinate positions.
414
00:27:09 --> 00:27:14
And now they've pointed
out I'd better look at M
415
00:27:14 --> 00:27:19
is very, very welcome.
416
00:27:19 --> 00:27:22
It doesn't mean that everybody
has to do this, if you've
417
00:27:22 --> 00:27:25
completed that MATLAB
assignment, you never want to
418
00:27:25 --> 00:27:30
see it again, and you've
kept M=8, it's ok.
419
00:27:30 --> 00:27:36
But if you're interested to see
what happens if M goes to 16
420
00:27:36 --> 00:27:39
or 32, I'm interested also.
421
00:27:39 --> 00:27:41
Right, yeah.
422
00:27:41 --> 00:27:45
OK, so anyway that's the
problem we're really
423
00:27:45 --> 00:27:45
thinking about.
424
00:27:45 --> 00:27:51
And that's the problem that is
equally important, but it
425
00:27:51 --> 00:27:55
seemed reasonable just
to do one of the two.
426
00:27:55 --> 00:27:59
And we were set up to do, we
have the code for the stiffness
427
00:27:59 --> 00:28:07
matrix, we would need a new
code to do these integrals.
428
00:28:07 --> 00:28:13
Because this will be linear
times linear, right?
429
00:28:13 --> 00:28:19
I'll have to compute that
one times this one and I
430
00:28:19 --> 00:28:23
would need new formulas
that are not there.
431
00:28:23 --> 00:28:26
I'd need formulas for, this
will be linear times linear
432
00:28:26 --> 00:28:31
so I'll be integrating
x squared type stuff.
433
00:28:31 --> 00:28:36
And xy's, because I'm
2-D, and y squareds.
434
00:28:36 --> 00:28:43
So it would take a little
more code, but not much.
435
00:28:43 --> 00:28:47
I think the math, oh here's
a question for you.
436
00:28:47 --> 00:28:49
Here's a question for you.
437
00:28:49 --> 00:28:52
Suppose I have my trial
functions, phi_i(x).
438
00:28:52 --> 00:28:56
439
00:28:56 --> 00:29:00
What do they add up to?
440
00:29:00 --> 00:29:05
Let me again draw a mesh,
so I've got a mesh.
441
00:29:05 --> 00:29:10
These are you know, I'm
sorry, I want to put in
442
00:29:10 --> 00:29:14
some more triangles here.
443
00:29:14 --> 00:29:18
Lots of triangles, whatever.
444
00:29:18 --> 00:29:23
Let me get some more
vertices, too.
445
00:29:23 --> 00:29:25
I'm getting in trouble.
446
00:29:25 --> 00:29:28
OK, whatever.
447
00:29:28 --> 00:29:33
So phi_i, is the piecewise
linear guy that
448
00:29:33 --> 00:29:35
is one at node i.
449
00:29:35 --> 00:29:38
So I've got all these
different nodes.
450
00:29:38 --> 00:29:41
I need a node there, so I've
got one, two, three, there's
451
00:29:41 --> 00:29:44
a node, there's more nodes.
452
00:29:44 --> 00:29:48
If I add them all up,
this is just like in
453
00:29:48 --> 00:29:51
an insights question.
454
00:29:51 --> 00:29:57
I've got all these, you could
add up these hats in 1-D.
455
00:29:57 --> 00:30:00
What's the sum of the
hats in one dimension?
456
00:30:00 --> 00:30:01
One.
457
00:30:01 --> 00:30:03
Good.
458
00:30:03 --> 00:30:05
The sum is one.
459
00:30:05 --> 00:30:09
It's a nice fact that
these guys add up to one.
460
00:30:09 --> 00:30:14
And now why is it still true
here in 2-D, that these little
461
00:30:14 --> 00:30:18
pyramids will add to one?
462
00:30:18 --> 00:30:22
That's an inside question, but
it's worth thinking about.
463
00:30:22 --> 00:30:25
Why do those pyramids
add to one?
464
00:30:25 --> 00:30:29
Let me leave that question.
465
00:30:29 --> 00:30:32
I'm thinking about, we
haven't imposed any
466
00:30:32 --> 00:30:34
boundary conditions yet.
467
00:30:34 --> 00:30:39
We've got them all. and I claim
that if we add up all the
468
00:30:39 --> 00:30:44
pyramids including the boundary
chopped off pyramids from the
469
00:30:44 --> 00:30:48
boundary, that we'll get
one throughout the whole,
470
00:30:48 --> 00:30:49
now it'll be phi(x,y).
471
00:30:49 --> 00:30:52
472
00:30:52 --> 00:30:56
Because now I'm moving
to 2-D, with pyramids.
473
00:30:56 --> 00:31:00
I think we'll still have one.
474
00:31:00 --> 00:31:02
Let me give you a minute
to think about that one.
475
00:31:02 --> 00:31:08
And then we could turn to
Fourier questions if you
476
00:31:08 --> 00:31:15
would like, we could do some
problems from the text.
477
00:31:15 --> 00:31:17
Any thoughts about this guy?
478
00:31:17 --> 00:31:24
Why should all those
individual pyramids add
479
00:31:24 --> 00:31:30
up to a flat group?
480
00:31:30 --> 00:31:31
Why did it work here?
481
00:31:31 --> 00:31:41
Well, it worked because you
could see it, right, somehow?
482
00:31:41 --> 00:31:47
Does it still work if the
nodes are not equally spaced?
483
00:31:47 --> 00:31:51
So we've got a hat function for
that guy, and a hat function
484
00:31:51 --> 00:31:54
for this guy, and a hat
function for this guy.
485
00:31:54 --> 00:31:57
And these guys are
in there, too.
486
00:31:57 --> 00:32:00
We haven't imposed anything.
487
00:32:00 --> 00:32:08
So those one, two, three, four,
five functions, five phis,
488
00:32:08 --> 00:32:14
they add up to one and y.
489
00:32:14 --> 00:32:18
Well, you're going to say it's
obvious, but that's what
490
00:32:18 --> 00:32:20
professors are allowed to say.
491
00:32:20 --> 00:32:24
Things are obvious, you
have to actually say why.
492
00:32:24 --> 00:32:26
Which is not as easy.
493
00:32:26 --> 00:32:36
So, why do they add to one?
494
00:32:36 --> 00:32:41
Let me look inside one element.
495
00:32:41 --> 00:32:47
Why does the sum of these
two guys add to a flat
496
00:32:47 --> 00:32:51
top inside that interval?
497
00:32:51 --> 00:32:57
AUDIENCE: [INAUDIBLE]
498
00:32:57 --> 00:33:00
PROFESSOR STRANG: At the
end points, you've got it.
499
00:33:00 --> 00:33:04
Because what's happening
at the end points?
500
00:33:04 --> 00:33:10
This guy, one of the guys,
the right guy is one.
501
00:33:10 --> 00:33:14
And all other guys
are zero, right.
502
00:33:14 --> 00:33:17
And this guy is also at one.
503
00:33:17 --> 00:33:20
Because it's the right guy.
504
00:33:20 --> 00:33:23
It has height one and
all others zero.
505
00:33:23 --> 00:33:27
So at the nodes we are at
one, because of one person,
506
00:33:27 --> 00:33:29
really, one element.
507
00:33:29 --> 00:33:30
And then?
508
00:33:30 --> 00:33:32
AUDIENCE: [INAUDIBLE]
509
00:33:32 --> 00:33:38
PROFESSOR STRANG: Right.
510
00:33:38 --> 00:33:42
But the sum of them is, why
is the sum of them always
511
00:33:42 --> 00:33:43
one, why is slope zero?
512
00:33:44 --> 00:33:48
Yeah.
513
00:33:48 --> 00:33:51
The slopes cancel, right.
514
00:33:51 --> 00:33:54
We know that in between it
will be a linear function.
515
00:33:54 --> 00:33:56
That would be one
way to look at it.
516
00:33:56 --> 00:33:59
If I add up a linear function
and a linear function the
517
00:33:59 --> 00:34:01
sum is a linear function.
518
00:34:01 --> 00:34:04
So I'm getting a linear
function, which is one at
519
00:34:04 --> 00:34:08
those points, so what
is that function?
520
00:34:08 --> 00:34:09
One.
521
00:34:09 --> 00:34:11
Right, you know that's
the straight line.
522
00:34:11 --> 00:34:16
So, that idea will
work here too.
523
00:34:16 --> 00:34:20
Look inside some
little triangle here.
524
00:34:20 --> 00:34:25
OK, that's got one, two,
three corners, OK.
525
00:34:25 --> 00:34:31
And if I look at this sum,
what is it at this point?
526
00:34:31 --> 00:34:36
If I look at that sum at this
corner, one guy is one,
527
00:34:36 --> 00:34:37
the one for that pyramid.
528
00:34:37 --> 00:34:40
And all others are?
529
00:34:40 --> 00:34:41
Zero.
530
00:34:41 --> 00:34:44
So the sum is one there, the
sum is one there, the sum is
531
00:34:44 --> 00:34:50
one there, so that blowing up
this little triangle, this is
532
00:34:50 --> 00:34:53
at height one, this is at
height one, this is at height
533
00:34:53 --> 00:34:56
one, so what's the roof?
534
00:34:56 --> 00:34:59
Flat.
535
00:34:59 --> 00:35:04
It's just a nice way to see the
nice property of these phis.
536
00:35:04 --> 00:35:14
That there's a phi for every
node, and they add to one.
537
00:35:14 --> 00:35:17
To that's it.
538
00:35:17 --> 00:35:24
OK, well I was going to say one
more thing and I am, about this
539
00:35:24 --> 00:35:27
eigenvalue problem, just
because I'll never
540
00:35:27 --> 00:35:29
have a chance again.
541
00:35:29 --> 00:35:33
So this is the moment
to say something about
542
00:35:33 --> 00:35:34
the eigenvalues.
543
00:35:34 --> 00:35:35
Lambda.
544
00:35:35 --> 00:35:41
Eigenvalue.
545
00:35:41 --> 00:35:44
I'm answering the question
where does K come from,
546
00:35:44 --> 00:35:45
where does M come from?
547
00:35:45 --> 00:35:56
Well, eigenvalue is, boy we
really got dramatic music here.
548
00:35:56 --> 00:36:00
That's a great Gates of
Kiev, I think might be.
549
00:36:00 --> 00:36:01
Mussorgski.
550
00:36:01 --> 00:36:05
If you like drums and big
noise, it's not music
551
00:36:05 --> 00:36:11
actually, but you got a
lot of noise out of it.
552
00:36:11 --> 00:36:16
Well, of course, he'd know
more than we do, but still.
553
00:36:16 --> 00:36:26
OK, so the eigenvalues in the
matrix case for Kx=lambda*M*x,
554
00:36:26 --> 00:36:31
the eigenvalue problem, lambda,
the lowest eigenvalue, lambda
555
00:36:31 --> 00:36:34
lowest, has a nice property.
556
00:36:34 --> 00:36:46
It's the minimum of sort of our
energy over our other energy.
557
00:36:46 --> 00:36:51
I just think, well this is
something you should see.
558
00:36:51 --> 00:36:54
This is a quotient here.
559
00:36:54 --> 00:36:56
It's got a name called
the Rayleigh quotient.
560
00:36:56 --> 00:36:59
And it would appear
in the book.
561
00:36:59 --> 00:37:02
So really, I guess what
I'm doing is calling your
562
00:37:02 --> 00:37:06
attention to something
that's in the book.
563
00:37:06 --> 00:37:10
That this a ratio of x
transpose K x to x transpose M
564
00:37:10 --> 00:37:19
x, if I look over all vectors
x, the lowest one is
565
00:37:19 --> 00:37:20
the eigenvector.
566
00:37:20 --> 00:37:23
The best x is the eigenvector
and the ratio is
567
00:37:23 --> 00:37:26
the eigenvalue.
568
00:37:26 --> 00:37:29
This is like my point
that I wanted to mention
569
00:37:29 --> 00:37:31
the Rayleigh quotient.
570
00:37:31 --> 00:37:34
Here it is in the matrix case,
and there would be similar
571
00:37:34 --> 00:37:38
Rayleigh quotient in
the continuous case.
572
00:37:38 --> 00:37:41
I'll just leave it at that.
573
00:37:41 --> 00:37:45
That in describing eigenvalues,
we can talk about
574
00:37:45 --> 00:37:49
Kx=lambda*M*x, like this.
575
00:37:49 --> 00:37:51
Or we can get energy into it.
576
00:37:51 --> 00:37:54
And you remember the whole
point about finite elements
577
00:37:54 --> 00:37:57
is, look at the energy.
578
00:37:57 --> 00:37:59
Look at that the quadratics.
579
00:37:59 --> 00:38:04
Multiply things by things.
580
00:38:04 --> 00:38:07
It came from the weak
form, it didn't come
581
00:38:07 --> 00:38:10
from the strong form.
582
00:38:10 --> 00:38:13
In the differential equation
here, we just have
583
00:38:13 --> 00:38:15
single terms.
584
00:38:15 --> 00:38:20
We got to these things through
that process of multiplying
585
00:38:20 --> 00:38:23
by u's and integrating.
586
00:38:23 --> 00:38:25
That's what gave us these
products and it works
587
00:38:25 --> 00:38:29
also in the matrix case.
588
00:38:29 --> 00:38:36
OK, that was a lot of
speechmaking about topics
589
00:38:36 --> 00:38:41
that we simply didn't
have time for in class.
590
00:38:41 --> 00:38:45
I'm ready for any question, or
I'm ready to maybe do a Fourier
591
00:38:45 --> 00:38:48
example, would you like that?
592
00:38:48 --> 00:38:51
Because this is where
we really are.
593
00:38:51 --> 00:38:55
I'll even take one that
will be on the homework.
594
00:38:55 --> 00:39:01
Just so you'll have a start.
595
00:39:01 --> 00:39:09
OK, let me take a square
pulse, yeah this is
596
00:39:09 --> 00:39:16
a good one, I think.
597
00:39:16 --> 00:39:19
In Section 4.1, there's a
question for the Fourier
598
00:39:19 --> 00:39:21
series of a square pulse.
599
00:39:21 --> 00:39:24
OK, so what does the
square pulse look like?
600
00:39:24 --> 00:39:29
Here's minus pi to pi.
601
00:39:29 --> 00:39:30
Here's zero.
602
00:39:30 --> 00:39:35
The square pulse goes along
here, up square pulse and down.
603
00:39:35 --> 00:39:48
Actually, let me go to L/2,
oh I'll just call it h.
604
00:39:48 --> 00:39:56
Let me find the Fourier
series for this function.
605
00:39:56 --> 00:40:02
It goes along at 0, it jumps up
to 1 over a interval of length
606
00:40:02 --> 00:40:06
2 h, going from minus h to h,
and then back down to
607
00:40:06 --> 00:40:08
0 and then repeat.
608
00:40:08 --> 00:40:11
So bip bip bip, square pulse.
609
00:40:11 --> 00:40:14
So that's my function.
610
00:40:14 --> 00:40:18
Is that function odd, or
even, or neither one?
611
00:40:18 --> 00:40:21
It's even, so I can
call that C(x).
612
00:40:22 --> 00:40:25
And figure that I'm going
to use cosines for
613
00:40:25 --> 00:40:27
that one, right?
614
00:40:27 --> 00:40:31
So tell me a formula for the
coefficients, what's the
615
00:40:31 --> 00:40:33
integral that I have to do?
616
00:40:33 --> 00:40:40
So my C(x) o is going to be
some a_0, we have to think
617
00:40:40 --> 00:40:46
what's a_0, then a_1*cos(x),
a_2*cos, and so on.
618
00:40:46 --> 00:40:48
So on. a_k*cos(kx).
619
00:40:48 --> 00:40:52
620
00:40:52 --> 00:41:01
OK, what's the formula for a_k?
621
00:41:01 --> 00:41:03
Before I plug in that
function I would like
622
00:41:03 --> 00:41:04
to get the formula.
623
00:41:04 --> 00:41:07
So I'm looking for the formula.
624
00:41:07 --> 00:41:10
It's a formula to remember.
625
00:41:10 --> 00:41:12
So I'm not wasting your time.
626
00:41:12 --> 00:41:14
Because you're going to see it
on the board and it'll just
627
00:41:14 --> 00:41:16
take a mental photograph of it.
628
00:41:16 --> 00:41:18
What do you think
it's going to be?
629
00:41:18 --> 00:41:20
How am I going to get it?
630
00:41:20 --> 00:41:26
I'll multiply both sides of the
equation by cos(kx), right?
631
00:41:26 --> 00:41:28
And I'll integrate.
632
00:41:28 --> 00:41:32
So and then when I integrate,
the cosines are orthogonal.
633
00:41:32 --> 00:41:34
Just like the sines
this morning.
634
00:41:34 --> 00:41:38
All those terms will go,
except for this term.
635
00:41:38 --> 00:41:40
When I multiply this
by cos(kx), I'll have
636
00:41:40 --> 00:41:42
cos(kx) squared.
637
00:41:42 --> 00:41:46
Here I'll have a cos(kx), and
here I'll have a whole lot of
638
00:41:46 --> 00:41:51
cos(kx)'s but when I
integrate, all this stuff is
639
00:41:51 --> 00:41:55
going to disappear.
640
00:41:55 --> 00:41:57
And this will all disappear.
641
00:41:57 --> 00:41:58
This is it.
642
00:41:58 --> 00:42:04
So a_k is going to be the
integral of my function,
643
00:42:04 --> 00:42:04
times cos(kx)dx.
644
00:42:04 --> 00:42:07
645
00:42:07 --> 00:42:09
Divided by what?
646
00:42:09 --> 00:42:14
Divided by the integral
of cos(kx) squared.
647
00:42:14 --> 00:42:18
Because I haven't
normalized things.
648
00:42:18 --> 00:42:21
So I don't know that that's
one, and in fact it isn't one.
649
00:42:21 --> 00:42:25
So I have to remember
to put that number in.
650
00:42:25 --> 00:42:28
OK, so that's the formula
and that number turns
651
00:42:28 --> 00:42:32
out to be pi, again.
652
00:42:32 --> 00:42:37
If I'm integrating from minus
pi to pi, then the average
653
00:42:37 --> 00:42:41
value of the cosine squared is
a 1/2, it's sort of as much
654
00:42:41 --> 00:42:47
above 1/2 as it is below 1/2,
and so the average of the half,
655
00:42:47 --> 00:42:51
the interval is 2pi, so pi.
656
00:42:51 --> 00:42:54
OK, that's the formula.
657
00:42:54 --> 00:42:59
Please just take a
mental photograph.
658
00:42:59 --> 00:43:00
Catch that one.
659
00:43:00 --> 00:43:07
Alright, now I've got my
particular C(x), my square
660
00:43:07 --> 00:43:10
wave, square pulse.
661
00:43:10 --> 00:43:11
Very, very important.
662
00:43:11 --> 00:43:16
Very important
Fourier series here.
663
00:43:16 --> 00:43:18
Famous one.
664
00:43:18 --> 00:43:20
OK, so what do I have?
665
00:43:20 --> 00:43:24
From minus pi to pi, so
what's my integral?
666
00:43:24 --> 00:43:27
Well, my integral really
doesn't go from minus
667
00:43:27 --> 00:43:31
pi to pi because my
function is mostly zero.
668
00:43:31 --> 00:43:34
Where does my integral go?
669
00:43:34 --> 00:43:36
Negative h to h, right?
670
00:43:36 --> 00:43:39
And in that region,
what is C(x)?
671
00:43:40 --> 00:43:44
One.
672
00:43:44 --> 00:43:47
So you see it's
going to be nice.
673
00:43:47 --> 00:43:53
From negative h to h, where
this is one, I just have to
674
00:43:53 --> 00:44:00
integrate cos(kx), so what do I
get? sin(kx), over k, and the
675
00:44:00 --> 00:44:05
pi so you see again that that k
is showing up in the
676
00:44:05 --> 00:44:09
denominator, and that's going
to give me the typical
677
00:44:09 --> 00:44:20
decay rate of 1/k for
functions with steps.
678
00:44:20 --> 00:44:21
For step functions.
679
00:44:21 --> 00:44:28
And now I have to evaluate
this between minus pi and pi.
680
00:44:28 --> 00:44:30
And no, h.
681
00:44:30 --> 00:44:32
Better be h.
682
00:44:32 --> 00:44:35
I mean, minus h and h.
683
00:44:35 --> 00:44:37
So what do I get for that?
684
00:44:37 --> 00:44:42
I get sine(kh), right?
685
00:44:42 --> 00:44:48
At the top, and what
do I get at minus?
686
00:44:48 --> 00:44:51
So I now I want to subtract,
what is the sin(-kh)?
687
00:44:51 --> 00:44:54
688
00:44:54 --> 00:44:57
It's a negative, right?
689
00:44:57 --> 00:45:04
So as I expect with an even
function like cosine, am
690
00:45:04 --> 00:45:08
I just getting twice?
691
00:45:08 --> 00:45:13
I could take it from 0 to h,
and it would give me one
692
00:45:13 --> 00:45:14
of them and the other one.
693
00:45:14 --> 00:45:21
Yep, I think so, and
divide by k pi.
694
00:45:21 --> 00:45:24
So those are the
Fourier coefficients.
695
00:45:24 --> 00:45:25
Except for a_0.
696
00:45:27 --> 00:45:33
a_0 has a slightly different
formula, because for a_0,
697
00:45:33 --> 00:45:34
why is a_0 different?
698
00:45:34 --> 00:45:39
How do you come up with a_0,
and what's its meaning? a_0
699
00:45:39 --> 00:45:45
has a nice meeting, so this
is worth having come
700
00:45:45 --> 00:45:46
this afternoon for.
701
00:45:46 --> 00:45:50
a_0 will be what?
702
00:45:50 --> 00:45:52
Well, I could get
it the same way.
703
00:45:52 --> 00:45:57
What will I multiply
both sides by?
704
00:45:57 --> 00:45:58
If I want to pick off a_0?
705
00:46:00 --> 00:46:01
Just one.
706
00:46:01 --> 00:46:06
It's not a cosine, it's the
cos(0x), it's the one.
707
00:46:06 --> 00:46:08
And then I integrate.
708
00:46:08 --> 00:46:13
I'm just going to get the
integral from minus pi to pi
709
00:46:13 --> 00:46:19
of C(x) times one, divided
by the integral from minus
710
00:46:19 --> 00:46:30
pi of one times one.
711
00:46:30 --> 00:46:37
Same method. multiply both
sides by one, which was
712
00:46:37 --> 00:46:41
the very first of my
orthogonal functions.
713
00:46:41 --> 00:46:45
Integrate it, all the other
integrals went away, right?
714
00:46:45 --> 00:46:48
The integral of cosine
over a whole interval.
715
00:46:48 --> 00:46:51
Its periodic.
716
00:46:51 --> 00:46:53
You get the same at
the two ends, so the
717
00:46:53 --> 00:46:56
difference is zero.
718
00:46:56 --> 00:47:00
So we just, the only term
left was a constant.
719
00:47:00 --> 00:47:03
And now what is the integral,
row what's the denominator now?
720
00:47:03 --> 00:47:06
That was the little,
slight twist.
721
00:47:06 --> 00:47:06
2pi.
722
00:47:07 --> 00:47:08
The denominator's 2pi.
723
00:47:09 --> 00:47:10
Yeah.
724
00:47:10 --> 00:47:15
That's that's why it's not,
yeah, it's slightly irregular,
725
00:47:15 --> 00:47:16
I have to divide by 2pi.
726
00:47:18 --> 00:47:22
And now, what word would you
use to describe, if I have a
727
00:47:22 --> 00:47:25
function, and integrate
it, and I divide by the
728
00:47:25 --> 00:47:30
length, what am I getting?
729
00:47:30 --> 00:47:35
There's an English word that
would describe what this is.
730
00:47:35 --> 00:47:37
Average.
731
00:47:37 --> 00:47:44
This is the average.
732
00:47:44 --> 00:47:46
And it has to be.
733
00:47:46 --> 00:47:49
This constant term is
always the average.
734
00:47:49 --> 00:47:52
And what will it be for this?
735
00:47:52 --> 00:47:59
So this was a_k, and
what is a_0, then?
736
00:47:59 --> 00:48:03
So you can now tell me, so
everybody's remembering this
737
00:48:03 --> 00:48:06
formula, you integrate the
function and divide by the 2pi.
738
00:48:07 --> 00:48:10
Now we've got a particular
function, so what is the
739
00:48:10 --> 00:48:12
integral of that function?
740
00:48:12 --> 00:48:16
So what does this equal?
741
00:48:16 --> 00:48:17
For this particular C(x)?
742
00:48:18 --> 00:48:20
What's the area under
that function C(x)?
743
00:48:22 --> 00:48:24
2h.
744
00:48:24 --> 00:48:26
Right?
745
00:48:26 --> 00:48:29
So 2h/2pi cancel twos.
746
00:48:29 --> 00:48:40
So there's a constant term, a_0
is h/pi and the and the cosine
747
00:48:40 --> 00:48:43
terms are, yeah, actually we're
going to get something nice.
748
00:48:43 --> 00:48:50
A really nice way to complete
this will be if I put this
749
00:48:50 --> 00:48:55
together, put this
series together.
750
00:48:55 --> 00:49:05
So now I'm saying that this
square pulse is that constant
751
00:49:05 --> 00:49:13
term h/pi plus the next
term a_1, you can see all
752
00:49:13 --> 00:49:14
these terms have 2/pi's.
753
00:49:14 --> 00:49:17
754
00:49:17 --> 00:49:21
I'm a little surprised that h
over, yeah, I guess it's right.
755
00:49:21 --> 00:49:22
2/pi.
756
00:49:22 --> 00:49:29
757
00:49:29 --> 00:49:33
So I've got sin(h), I think.
758
00:49:33 --> 00:49:35
And now I'm just copying this.
759
00:49:35 --> 00:49:47
2/pi*sin(h), sin(h),
is that what I want?
760
00:49:47 --> 00:49:51
Over one.
761
00:49:51 --> 00:49:54
That's the coefficient
version of sine, of cos(x).
762
00:49:54 --> 00:49:58
763
00:49:58 --> 00:50:00
a_1 was the coefficient
of cos(1x).
764
00:50:01 --> 00:50:04
And then a_2 is the
coefficient of cos(2x).
765
00:50:06 --> 00:50:07
So that will be sin(2h).
766
00:50:07 --> 00:50:09
767
00:50:09 --> 00:50:13
k is two, so I have a
two down here, cos(2x).
768
00:50:14 --> 00:50:20
And so on.
769
00:50:20 --> 00:50:23
Yeah, I think that's
the Fourier series.
770
00:50:23 --> 00:50:33
That would be the Fourier
series for the square pulse.
771
00:50:33 --> 00:50:34
Yeah.
772
00:50:34 --> 00:50:38
That would be the Fourier
series for the square pulse.
773
00:50:38 --> 00:50:40
Could I test any
interesting cases?
774
00:50:40 --> 00:50:45
Suppose h is all
the way out to pi.
775
00:50:45 --> 00:50:46
Suppose I take that case.
776
00:50:46 --> 00:50:55
Let h go all the way out to
pi, then what's my function?
777
00:50:55 --> 00:51:01
If h=pi, then what have
I got a graph of?
778
00:51:01 --> 00:51:02
Just one.
779
00:51:02 --> 00:51:03
It's just a one.
780
00:51:03 --> 00:51:07
If h is pi, what happens?
781
00:51:07 --> 00:51:11
That becomes a one, and what
about these other things?
782
00:51:11 --> 00:51:15
What is this thing
when h is pi?
783
00:51:15 --> 00:51:15
Zero.
784
00:51:15 --> 00:51:18
All the other terms go away.
785
00:51:18 --> 00:51:22
It's just a sin(2pi)
that would go away.
786
00:51:22 --> 00:51:27
Yeah, so if h is pi, if I go
out to the place where I don't
787
00:51:27 --> 00:51:33
have any jumps at all because
it's now all the way out there,
788
00:51:33 --> 00:51:37
then these terms all disappear
and I just have this.
789
00:51:37 --> 00:51:40
And I would like to ask you
and it's going to come up on
790
00:51:40 --> 00:51:47
Friday, too, what happens
if h goes to zero?
791
00:51:47 --> 00:51:50
Well, let me just take
h going to zero.
792
00:51:50 --> 00:51:52
What happens to
this whole thing?
793
00:51:52 --> 00:51:55
What happens to my function
if h goes to zero?
794
00:51:55 --> 00:51:58
Goes to zero, right, then
squeezed it to nothing.
795
00:51:58 --> 00:52:05
And if h is zero then sin(h) is
zero, I get 0=0, that's not
796
00:52:05 --> 00:52:10
interesting enough to mention
on Friday But there is one
797
00:52:10 --> 00:52:13
case that is important.
798
00:52:13 --> 00:52:16
Suppose I make the
height, yeah.
799
00:52:16 --> 00:52:17
Make a guess.
800
00:52:17 --> 00:52:24
Suppose I make the
height higher as I
801
00:52:24 --> 00:52:27
make the base smaller.
802
00:52:27 --> 00:52:31
I'm going to keep the area as
one, so if this has a base of
803
00:52:31 --> 00:52:34
2h, I'm going to have
a height of 1/2h.
804
00:52:35 --> 00:52:41
So if I keep the area at one,
so the height now is 1/2h,
805
00:52:41 --> 00:52:45
so now my square pulse
I've divided it by 2h.
806
00:52:46 --> 00:52:50
I have a 1/2h
multiplying everything.
807
00:52:50 --> 00:52:56
And now if I let h go to
zero, something more
808
00:52:56 --> 00:52:58
interesting will happen.
809
00:52:58 --> 00:52:59
And what?
810
00:52:59 --> 00:53:05
Just tell me first, what
would you expect to happen?
811
00:53:05 --> 00:53:05
Delta.
812
00:53:05 --> 00:53:08
Right, delta.
813
00:53:08 --> 00:53:12
So what I'll see show up
will be the Fourier series
814
00:53:12 --> 00:53:15
for the delta function.
815
00:53:15 --> 00:53:23
When I divide by 2h, so I have
sin(h)'s over h's, and of
816
00:53:23 --> 00:53:25
course what's the great
fact about sin(h)/h?
817
00:53:27 --> 00:53:34
As h goes to zero, it
goes to, everybody know,
818
00:53:34 --> 00:53:36
that's the big deal.
819
00:53:36 --> 00:53:36
Yeah.
820
00:53:36 --> 00:53:42
One. sin(h) is the same
size as h for a very small
821
00:53:42 --> 00:53:44
h, and approaches one.
822
00:53:44 --> 00:53:49
Yeah so we'll see the
delta function Friday.
823
00:53:49 --> 00:53:53
OK, so you've got a sort of
mini-lecture instead of a real
824
00:53:53 --> 00:53:56
chance to ask about homework.
825
00:53:56 --> 00:53:59
Next Wednesday should be
different because there will be
826
00:53:59 --> 00:54:04
Fourier series homework, and
I'll be ready to answer
827
00:54:04 --> 00:54:05
questions about it.
828
00:54:05 --> 00:54:07
OK, thanks. |
Materials
All the instructional material for EGEE 102 is presented online. As described above, EGEE 102 consists of 10 online lessons. These lessons include text, graphics, videos, animations, interactive Flash activites, numerical problems, and electronic whiteboard discussions of numerical problems. Quizzes and Home Activities have been developed to test your understanding of the material covered in the 10 lessons. Some examples of the instructional material used in EGEE 102 are provided below. |
Basic Math - Number Patterns Studying number patterns is important for two reasons. First, they help one better understand the concepts of arithmetic and provide a basis for understanding the concepts of more complex mathematics (algebra, trigonometry, calculus). Second, pattern recognition is a useful problem-solving skill, both in mathematics and in real-world situations. Patterns involving odd and even numbers are investigated. Patterns in multiples of certain numbers lead to an understanding of divisibility rules. Seque Author(s): No creator set
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Bioservers This site contains user-friendly tools to launch DNA database searches, statistical analyses, and population modeling from a centralized workspace. Educational databases support investigations of an Alu insertion polymorphism on human chromosome 16 and single nucleotide polymorphisms (SNPs) in the human mitochondrial control region. Author(s): No creator set
Best of the web - November 2010 Read more:
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CSET Science Subtest II: Heat Transfer and Thermodynamics
This module includes the following chemistry topics:
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Statistical Reasoning II Statistical Reasoning in Public Health II provides an introduction to selected important topics in biostatistical concepts and reasoning through lectures, exercises, and bulletin board discussions. Author(s): John McGreadyMethods in Biostatistics II Presents fundamental concepts in applied probability, exploratory data analysis, and statistical inference, focusing on probability and analysis of one and two samples. Author(s): Brian Caffo
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Analyzing Statistics S.S. Europe and Russia Students will gather statistical information on countries in Europe and Russia from almanacs. The information will be recorded in a chart. Students will then take the information and make line or bar graphs. Students will analyze the information by answering higher level thinking questions. Author(s): No creator set
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Statistics Online Computational Resource for Education and Research The goals of the Statistics Online Computational Resource ( are to design, validate and freely disseminate knowledge. Specifically, SOCR provides portable online aids for probability and statistics education, technology based instruction and statistical computing. SOCR tools and resources include a repository of interactive applets, computational and graphing tools, instructional and course materials.
The core SOCR educational and computational components include: Distribution Author(s): No creator set
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Supporting Teachers Intervention in Collaborative Knowledge Building In the context of distributed collaborative learning, the teacher's role is different from traditional teacher-centered environments, they are coordinators/facilitators, guides, and co-learners. They monitor the collaboration activities within a group, detect problems and intervene in the collaboration to give advice and learn alongside students at the same time. We have designed an Assistant to support teachers intervention in collaborative knowledge building. The Assistant monitors the collabo Author(s): Chen Weiqin
Artificial Intelligence: Natural Language Processing This course is designed to introduce students to the fundamental concepts and ideas in natural language processing (NLP), and to get them up to speed with current research in the area. It develops an in-depth understanding of both the algorithms available for the processing of linguistic information and the underlying computational properties of natural languages. Wordlevel, syntactic, and semantic processing from both a linguistic and an algorithmic perspective are considered. The focus is on m Author(s): No creator set
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US History II Upon completion of this course you will: Demonstrate comprehension of a broad body of historical knowledge; Express ideas clearly in writing; Work with classmates to research an historical issue; Interpret and apply data from original documents; Identify underrepresented historical viewpoints; Write to persuade with evidence; Compare and contrast alternate interpretations of an historical figure, event, or trend; Explain how an historical event connects to or causes a larger trend or theme; Deve Author(s): No creator set |
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Overview
Now With A Full-Color Design, The New Fourth Edition Of Zill's Advanced Engineering Mathematics Provides An In-Depth Overview Of The Many Mathematical Topics Necessary For Students Planning A Career In Engineering Or The Sciences. A Key Strength Of This Text Is Zill's Emphasis On Differential Equations As Mathematical Models, Discussing The Constructs And Pitfalls Of Each. The Fourth Edition Is Comprehensive, Yet Flexible, To Meet The Unique Needs Of Various Course Offerings Ranging From Ordinary Differential Equations To Vector Calculus. Numerous New Projects Contributed By Esteemed Mathematicians Have Been Added. New Modern Applications And Projects, Coupled With A New Statistics And Probability CD-ROM Included With Every New Copy Makes Zill's Classic Text A Must-Have Text And Resource For Engineering Math Students |
Here are my online notes for my Algebra course that I teach
here at Lamar University, although I have to admit that it's been years since I
last taught this course. At this point
in my career I mostly teach Calculus and Differential Equations.
Despite the fact that these are my "class notes", they should
be accessible to anyone wanting to learn Algebra or needing a refresher for
algebra. I've tried to make the notes as
self contained as possible and do not reference any book. However, they do assume that you've had some
exposure to the basics of algebra at some point prior to this. While there is some review of exponents,
factoring and graphing it is assumed that not a lot of review will be needed to
remind you how these topics work.
Here are a couple of warnings to my students who may be here
to get a copy of what happened on a day that you missed.
Because
I wanted to make this a fairly complete set of notes for anyone wanting to
learn algebra I have included some material that I do not usually have
time to cover in class and because this changes from semester to semester
it is not noted here. You will need
to find one of your fellow class mates to see if there is something in
these notes that wasn't covered in class.
Because
I want these notes to provide some more examples for you to read through,
I don't always work the same problems in class as those given in the
notes. Likewise, even if I do work
some of the problems in here I may work fewer problems in class than are
presented here.
Sometimes
questions in class will lead down paths that are not covered here. I try to anticipate as many of the
questions as possible in writing these up, but the reality is that I can't
anticipate all the questions.
Sometimes a very good question gets asked in class that leads to
insights that I've not included here.
You should always talk to someone who was in class on the day you
missed and compare these notes to their notes and see what the differences
are.
This
is somewhat related to the previous three items, but is important enough
to merit its own item. THESE NOTES
ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!
Using these notes as a substitute for class is liable to get you in
trouble. As already noted not everything in these notes is covered in
class and often material or insights not in these notes is covered in
class.
Here is a listing of all the material that is currently
available in these notes.
Integer Exponents In this section we will start looking at
exponents and their properties.
Rational Exponents We will define rational exponents in this
section and extend the properties from the previous section to rational
exponents.
Real Exponents This is a short acknowledgment that the
exponent properties from the previous two sections will hold for any real exponent.
Radicals Here we will define radical notation and relate
radicals to rational exponents. We will
also give the properties of radicals.
Polynomials We will introduce the basics of polynomials in
this section including adding, subtracting and multiplying polynomials.
Factoring Polynomials This is the most important section of all the
preliminaries. Factoring polynomials
will appear in pretty much every chapter in this course. Without the ability to factor polynomials you
will be unable to complete this course.
Rational Expressions In this section we will define rational
expressions and discuss adding, subtracting, multiplying and dividing them.
Complex Numbers Here is a very quick primer on complex numbers
and how to manipulate them.
Miscellaneous Functions In this section we will graph a couple of
common functions that don't really take all that much work to so. We'll be looking at the constant function,
square root, absolute value and a simple cubic function.
Transformations We will be looking at shifts and reflections
of graphs in this section. Collectively
these are often called transformations.
Symmetry We will briefly discuss the topic of symmetry
in this section.
Rational Functions In this section we will graph some rational
functions. We will also be taking a look
at vertical and horizontal asymptotes. |
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