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Description: These roughly 100 lesson plans are related to mathematics technology The lessons incorporate calculators and graphing calculators, non-electronic tools and manipulatives, geometry software, spreadsheets, and computer algebra systems. This is an extensive and thorough resource for first-cycle college mathematics teachers and teachers of advanced high school math.
Calculus is more than just about applying formulae and rules to solve problems. It is, in my view, one of the most diverse topics in mathematics. The concept in itself is beautiful and, of course, useful. The course aims at making the reader understand actual concept of calculus, introduce him/her to the concept of limits, functions, and practice problems that would open up his/her mind to more diverse applications of calculus, like trigonometry and co-ordinate geometry. Prior knowledge of linear equations will be helpful, though not compusory.
What jobs use college algebra and what concepts of college algebra do they use? I have to write a paper for my college algebra class on jobs that use college algebra and the concepts that the jobs use. I've researched and found jobs, just not the concepts. Someone please help me out! Asked By: - 6/29/2010 Best Answer - Chosen by Asker All engineering jobs use college algebra either directly or indirectly. If the engineers don't use the algebra in their designs, maintenance or repairs, they will use meters or computers and programs that do. Engineers need to use vectors, scalar products, vector products and linear equations. In stress analysis, the... More Answered By: Pisces ♥ Math - 7/2/2010 Additional Answers (1) Great answer. Also in business college algebra principles will be used to calculate prices, break even points, and advertising budgets versus costs and expenses. Another is to calculate the advertisement to editorial ratio of a newspaper or website. Where are the jobs? Is productivity and globalization creating a permanent "recession" of jobs? My main issue is I am doing a paper for school and have no idea where to begin. I was hoping suggestions...
This is a collection of free math resources. This material was written by the founders of Pass Math Class to not be like a textbook. We have seen many math textbooks that are great at rigorous proof and complex notation but terrible at actually explaining the material. We have written these hoping that we can get straight to the important points, and help people actually learn. We are constantly adding new lessons and improving the existing ones, in an effort to make a truly complete resource for all students of mathematics. If you want to request a specific lesson, just let us know. We have left out the rigorous proofs and confusing vocabulary – we have kept the most simple essence of each concept.
ncertsolutions of class11mathsncertsolutions for class11th maths chapter 2 relations and functions national council educational research training book syllabus classes at the elementary level 72 i ii counts number objects in a collection makes corresponding to – (ncert) 06Math (I-V).pdfRead Now Guidelines for CBSE Proficiency Test in Mathematics – 2011 ncertsolutions of class11maths cbse guide ncertsolutionsncert cbse class 8 mathematics a complete with chapter wise in addition to notes questions and mcqs their answers on various subjects classes section i: multiple choice questions this section contains for 1 only one the four options is correct you have indicate your notes,
Note: Citations are based on reference standards. However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied. "2. Systems of Equations on a Finite Interval. 3. Equations on an Infinite Interval. 4. Systems of Equations on an Infinite Interval. 5. Equations and Systems in the Complex Plane. 6. Turning Points. 7. A Problem on Scattering, Adiabatic Invariants and a Problem on Eigenvalues. 8. Examples." "Ch. 1. The Analytic Theory of Differential Equations. 1. Analyticity of the Solutions of a System of Ordinary Differential Equations. 2. Regular Singular Points. 3. Irregular Singular Points -- Ch. 2. Second-Order Equations on the Real Line. 1. Transformations of Second-Order Equations. 2. WKB-Bounds. 3. Asymptotic Behaviour of Solutions of a Second-Order Equation for Large Values of the Parameter. 4. Systems of Two Equations Containing a Large Parameter. 5. Systems of Equations Close to Diagonal Form. 6. Asymptotic Behaviour of the Solutions for Large Values of the Argument. 7. Dual Asymptotic Behaviour. 8. Counterexamples. 9. Roots of Constant Multiplicity. 10. Problems on Eigenvalues. 11. A Problem on Scattering -- Ch. 3. Second-Order Equations in the Complex Plane. 1. Stokes Lines and the Domains Bounded by them. 2. WKB-Bounds in the Complex Plane. 3. Equations with Polynomial Coefficients. Asymptotic Behaviour of a Solution in the Large."
Course in Modern Geometries - 2nd edition Summary: A Course in Modern Geometries, Second Edition, is designed for a junior-senior level course for mathematics majors, including those who plan to teach in secondary school. Chapter 1 presents several finite geometries in an axiomatic framework. Chapter 2 continues the synthetic approach as it introduces Euclid's geometry and ideas of non-Euclidean geometry. In Chapter 3, a new introduction to symmetry and hands-on explorations of isometries precede the extensive analyt...show moreic treatment of isometries, similarities and affinities. A new concluding section explores isometries of space. Chapter 4 presents plane projective geometry both synthetically and analytically. The extensive use of matrix representations of groups of transformations in Chapters 3-4 reinforces ideas from linear algebra and serves as excellent preparation for a course in abstract algebra. The new Chapter 5 uses a descriptive and exploratory approach to introduce chaos theory and fractal geometry, stressing the self-similarity of fractals and their generation by transformations from Chapter 3. Each chapter includes a list of suggested resources for applications or related topics in areas such as art and history. The second edition also includes pointers to the web location of author-developed guides for dynamic software explorations of the Poincaré model, isometries, projectivities, conics and fractals. Parallel versions of these explorations are available for Cabri Geometry and Geometer's Sketchpad. ...show less Judith N. Cederberg is an associate professor of mathematics at St. Olaf College in Minnesota where she often teaches a modern geometry course. In the last decade, she has been one of the Principal Investigators for two St. Olaf-based NSF Teacher Enhancement projects designed to revitalize the teaching of geometry in secondary school. 2001-01-01 Hardcover 2nd Fair There is no writing or highlighting on the pages. About six pages have this corner folded down. There is black marker on the page edges that you can't really see when t...show morehe book is open. The covers have some rubbing and soiling. The first 10 pages are little ripple but no stains. 20. ...show less $52.70 +$3.99 s/h New indoo Avenel, NJ BRAND NEW $57.0260.86 +$3.99 s/h New PaperbackshopUS Secaucus, NJ New Book. Shipped from US within 4 to 14 business days. Established seller since 2000 2000$103
Resources Sunday, October 30, 2011 I am a student self-studying mathematics from Singapore's textbook at the 10th to 11th (Secondary 5) grade levels. When I finish these, I would like to continue following Singapore's curriculum. Which books (series) do you recommend. I plan to study advanced topics in depth, like statistics and calculus? I would like to follow the newest books, ones followed currently in Singapore by JC students. Student in USA In JC, students do not use a textbook because the lecturers provide their own lecture notes, like in the university. If you have not done Additional Mathematics, you may work on this - it is a book that includes trigonometry, calculus and other advanced topics. This is used in Grade 9 - 10 by advanced students. It is available from the same place you got your other books. Wednesday, October 19, 2011 I hope you still remember me. Last time I contacted you regarding congruence and similarity. Now, my son is studying in Grade 8. If you don't mind can you clear my confusion regarding one problem of percentage. I have difference of opinion with the mathematics teacher and I want to clear my concept. The problem is as follows: When we are converting 2/5 to percentage we write as follows: 2/5 x 100% = 40%. ---------- (1) The teacher says it is not necassary to write symbol(%) with 100. 2/5 x 100 = 40%. ------------(2) You will notice that % is missing from (2). Can you please explain is it also the correct practice not to write the symbol % with 100. My view is that 2/5 x 100 will result in 40 and not 40%. I am afraid that this mistake will results in marks being deducted in the IGCSE. A father in Saudi Arabia You are right 2/5 is equal to 2/ 5 x 1 and 1 = 100/100 which is written as 100%. 2/5 x 100 = 40 as you said. 40% is equal to 0.4 (not 40). Perhaps the teacher knows that, in a lenient way or marking, candidates get full credit whether they did (1) or (2). But I am sure you rather your son learn what is mathematically correct rather what is minimally acceptable to earn a credit in the examination. Please advise your son to write the mathematically correct sentence. I also trust that you will help him understand why (2) is not correct (although it may still earn a credit in the examinations, according to the teacher). Wednesday, October 5, 2011 I'm a homeschooling mother from the Netherlands and I have used (and still use) Singapore Math for my children in all the years of their primary education. I can't rave enough about this method of math education: the books are great, the bar diagrams are marvellous - I wish I had learned mathematics this way in my time. Now my oldest child is starting Secondary Education the upcoming year. In the Netherlands secondary education consists of 6 years and the level of math at the final exams is relatively high (higher than say, American High School level - to give comparison). However, instead of preparing my children for their final exams with Dutch secondary math books, I prefer to keep on working with Singapore Math. Even though I can't really compare Dutch and Singapore High School Math, I very much like the way Singapore Math has build a strong math education in my children in such a thorough and painless way. But in my search for information I got a little confused about all the available series. There seem to be: 1) New Mathematics Counts Series 2) New Elementary Maths Series and 3) New Syllabus Mathematics Would you be able to explain to me the difference between these series or can you give me an advice on what to use for the High School years of my children? I'm aware this is not the type of problem question you usually receive on your blog, but I very much hope you can answer me just the same. For the primary years we're using the My Pals are Here Maths Series (Grades 1-6), and I'd like use the best available SM high school sequence. I prefer to use the series that is used in most Singapore and/or International High Schools. For example: I understand that for the primary years My Pals are Here Maths and Shaping Maths are the most common used series in Singapore and International schools: MPAH in 80% of the schools, Shaping Maths in 20% - roughly estimated. Do you happen to know the ratio of the three above mentioned series for the secondary years (New Mathematics Counts Series / New Elementary Maths Series / New Syllabus Mathematics)? Or do you know any distinctive features? It seems that all Singapore Math distributors, either in the US, UK or Singapore are all pushing the series that offers them the most profit, so I don't know who to believe in that area. Buying all the books for four children is quite an investment, so it would be a pity if I invested in the wrong series. I was even on the verge of abandoning Singapore Math for a curriculum called The Art of Problem Solving ( because I didn't have any clue of the right series. But then I found your blog, I hope you can find the time to answer me. Thanks in advance for your effort. A Parent in The Netherlands In Singapore, some students study 4 years and others 5 years for their secondary education. They then move to another two years in junior college (some opt to go to a polytechnic instead). The first four years leads to GCE O Levels (Grades 7-10) and the last two years leads to GCE A Levels (Grades 11-12). You are right - there are many textbook series for secondary levels. In fact, more than the three you listed. Essentially they are all the same. Internationally, most schools / parents use New Syllabus Mathematics (NSM) and New Elementary Mathematics (NEM) simply because these has been around for many years. NSM is still sued in Singapore schools. NEM is no longer used - the publisher sometimes did not submit them for review or update it to make it a 100% fit with topics in the revised curriculum. All our textbooks must be reviewed by Ministry of Education. In my opinion, NEM is of good quality too. New Mathematics Counts (NMC) is designed for academically weaker students - the program is to be done over five, instead of four years. NSM is designed for students who have a strong foundation in mathematics. That is why other than the four books, there is a fifth book that advanced students opt for in Grade 9/10 (they use this book over two year to supplement the main text). This is Additional Mathematics. Many kids in Singapore study this subject. By the time they complete Additional Mathematics, they would have done basic calculus. Generally, if your child do well using the Singapore textbooks, he/she should be ready for ay kind of test. In brief, NMC if the child is struggling with math. NEM or NSM if the student is quite good in math. NSM has the option for advanced topics. See for textbooks used in Singapore. Titles with Sec 5N are designed for students who tend to struggle somewhat with mathematics - the topics coverage is the same but done over five instead of four years. Books listed as NT (Nornal Technical) are for students who are moving to vocational course after Grade 10. This morning I taught Primary 4 students word problems using bar models. I expected the students to get a clear picture by it, but actually not. They were so impatient to compute directly, even without reading the question :( They just depended on my instruction whether to multiply or to divide. So, I'm thinking of an activity at the beginning to introduce the bar model to them - perhaps by making the bars using color papers. Teacher in Indonesia This is not unexpected if the students are already used to a computational approach. All they want to do is to compute. Without a clear understanding of the problem, they will not be able to identify the correct operations. Bar modelling begins in kindergarten when a teacher models the story using the real things and later pictures of the real thing. Later, 5 unifix cubes / snap cubes are used to represent, say 5 sweets. By Primary 2, they begin to use a bar to represent quantities. Your idea of using paper strips is excellent. I do it all the time. You can also give problems without numbers. Johan has more sweets than Siti. How many sweets does Siti have? After students show the correct bars for number of sweets Johan has (longer) and number of sweets Siti has (shorter), tell them Johan has 6 sweets more than Siti. Ask them to put in the new information on the bar models. Finally tell them that Johan has 14 sweets. Get them to solve the problem. Welcome to askyeapbanhar.blogspot.com This blog is for me to post the answers to many questions that parents (and some teachers) ask about mathematics teaching and learning. I hope other teachers will add to my answers by clicking on comments to write their views relevant to the question posed. Questions are most welcome - email me at yeapbanhar@gmail.com I will post the answers on this blog as soon as time permits. Have a good day! About Me E-mail yeapbanhar@gmail.com This is the blog of Dr Yeap Ban Har who taught at National Institute of Education, Nanyang Technological University in Singapore for more than ten years. Presently, he holds two concurrent positions as the director of curriculum and professional development at Pathlight School, an autism-oriented K-10 school in Singapore, and the principal of Marshall Cavendish Institute, a global teacher professional development institute.
Schaum's Outline of Probability - 2nd edition Summary: Many students in disciplines outside mathematics, including the humanities and science, must take courses in statistics, of which probability forms an integral part, demanding that the information be simplified and updated. This book will give students, particularly in applied statistics courses, an easy-to-understand methodology for manipulating and solving simple and complex problems involving probability. Since the book does not employ calculus, it assumes familia...show morerity only with arithmetic and algebra, making it ideal for the liberal arts student. However, even in beginning statistics courses for mathematics students, 90% of the problems are non-calculus-related, making the Outline suitable as well for understanding the probability elements of those courses00713520312.99 +$3.99 s/h Acceptable Dream Books Company, LLC Englewood, CO 2000 Paperback
Homeschool Connections has created a brand new 6-week course to focus solely on trigonometry. It is designed to help Algebra II students prepare for Advanced Mathematics, give Advanced Mathematics students a review for Calculus, or give a refresher for students preparing for their ACT or SAT test. Introduction to Trigonometry Dates: Mondays, May 5 to June 9, 2014 Time: 4:00 pm Eastern (3:00 Central; 2:00 Mountain; 1:00 Pacific) Number of classes: 6 Prerequisites: Algebra 2 Suggested high school credit: 1/2 semester Math Instructor: Jean Hoeft, MS Course description: This course will polish students' Algebra 2 skills and prepare them for Pre-calculus / Advanced Mathematics. We will study the trigonometric functions and their inverses, polar and rectangular coordinates, the calculations of a triangle, vectors and more. Students are expected to do homework sets which are provided free by the instructor. There is no text book necessary, reference tables and notes will be provided. Students will have a working knowledge of trigonometric functions and their properties at the conclusion of the course. Course materials: None, all course materials provided Free by the instructor. Topics covered: Trigonometric functions, changing degrees to radians, recriprocal identities, arc lengths, writing the equations of sine and cosine functions of sinusoid graphs, graphs of secant and cosecant, simplifying trig identities, solving equations involving trigonometric identities, reference triangles, solving triangles for their angles and side lengths, and complex numbers with the complex coordinate plane. Homework: Students should expect to spend 3-4 hours on homework a week to complete the assigned work. Fee: $75 for all 6 classes It's time again for Homeschool Connections Annual Refresh Conference. This is an online conference for Catholic homeschooling parents. It is designed to help you get through the winter months refreshed and ready for all that homeschooling has in store for you and your family. Attendance is limited. If it fills up, you will be placed on a wait list. All webinars are recorded and made available for free viewing within 24 hours. Meet Our Presenters: Andrew Pudewa Andrew Pudewa is the director of the Institute for Excellence in Writing and a homeschooling father of seven. Presenting throughout North America, he addresses issues relating to teaching, writing, thinking, spelling, and music with clarity, insight, practical experience, and humor. His seminars for parents, students and teachers have helped transform many a reluctant writer and have equipped educators with powerful tools to dramatically improve students' skills. Although he is a graduate of the Talent Education Institute in Japan, and holds a Certificate of Child Brain Development from the Institutes for the Achievement of Human Potential in Philadelphia, Pennsylvania, his best endorsement is from a young Alaskan boy who called him "the funny man with the wonderful words." He and his beautiful, heroic wife Robin currently teach their three youngest at home in Atascadero, California. Ginny Seuffert Mrs. Virginia (Ginny) Seuffert, a native New Yorker and mother of 12, has been homeschooling for over 20 years! While in New York, Mrs. Seuffert lectured, debated, and wrote a number of articles for the Pro-Life movement. After moving to Illinois, she became a founding member of the Network of Illinois Catholic Home Educators, helped establish the "Round Table" (a Catholic homeschool leadership discussion group), and became a founder and officer of the Catholic Home School Network of America. In addition to appearing on EWTN, she has been a guest on numerous radio shows, lectured at Catholic family conferences all over the United States and Canada, and has authored several articles on such topics as home education, teaching children the virtues, and household management. The Seton Magazine regularly features her columns, and recently Seton Press published her first two books of a series, Ginny's Gems: Home Management Essentials and Ginny's Gems: 10 Essentials for Teaching Your Preschooler at Home. Mary Ellen Barrett Mary Ellen is a homeschooling mother of eight children. She is a longtime columnist for The Long Island Catholic (Life in Our Domestic Church) and speaks on a variety of topics at Catholic conferences and parishes around the United States. Mary Ellen is currently working on two books, on about mothering a large family and the other is an Advent Books of Days. She can be found blogging at Tales from the Bonny Blue Houseand during the Advent season at O Night Divine. Besides writing and blogging, Mary Ellen enjoys cooking, sewing, photography, and studying theology. Katie Moran Dr. Moran and her husband reside in Niles, OH with their five children, including two who were adopted from the Ukraine. Dr. Moran has her Bachelor of Science in General Studies from Kent State University, her Master of Science in Education from Breyer State University, and her Doctor of Philosophy in Education from University St. John of the Cross –Universidad San Juan del la Cruz. Dr. Moran is host of the Homeschool Lifeline talk show on Radio Maria, She brings to our conference 23 years of homeschooling experience, in addition to her experience as a private tutor and homeschool consultant. She has been a guest speaker on the History Channel, served on the board of Northern Ohio Adoption Services, and is Secretary/Treasurer of the Blue Army of Our Lady of Fatima (Byzantine Chapter). Additionally, she is the president of CHSNA (Catholic Home School Network of America), member of the Round Table, a national organization of Catholic Home Schooling Leaders, founder and past leader of the Ohio Educators' Catholic Home Schooling Network , member of CHSNA delegation to Pope John Paul II, various congregations and curias, 1995, 1997, and again in 2006 to Pope Benedict XVI. Dr. Katie Moran is the author of Doorway to Heaven, The Unique Learner – Homeschooling Children with Learning Disabilities, Philip's Fast – 40 Days of Advent Meditations According to the Byzantine Rite, The Story of the Church Workbook with Answer Key and Study Guide, based on the textbook, The Story of Church, and a DVD on Fatima and the Holy Angels. Her latest book still in the editing and writing stage: Home Schooling for Heaven, not Harvard. Maria Rioux Maria began her undergraduate studies at Thomas Aquinas College. There she met her husband, Jean, Homeschool Connections' philosophy instructor and chair of the philosophy department at Benedictine College (where Maria is a theology major slowly completing her degree). Together they have been homeschooled their nine children for more than 20 years. In those early days of homeschooling there were not many resources available. As a consequence, they developed their own curriculum which reflects their love for classical education as well as their affection for Charlotte Mason. Maria is owner and co-moderator of the yahoo group The History Place. She loves to write and is a regular contributor to mater et magistra magazine. She is also a reviewer for Love2Learn Literature Alive! Rachel Watkins Rachel Watkins is wife to Matthew, homeschooling mom to 11 great kids, creator/writer of the Little Flowers Girls Club (ages 5 and up) and Honor Guard (ages 12 and up) Contributor to Ave Maria Radio's More 2 Life with Dr. Greg and Lisa Popcak and their companion blog Exceptional Marriages. In the midst of life, she has found some time to be published in a number of Catholic publications and websites. Her life could be defined as a daily attempt to fulfill the words of Jesus who assures us that He came so that our joy would be full! She doesn't always succeed but the efforts have been surprising. NOTE: All talks are recorded and available within 24 hours. They can be found, along with talks from past conferences, here: Homeschool Connections Free Webinars. You'll need to "Login as a guest". Homeschool Connections writing program, Aquinas Writing Advantage, is a complete online program for you and your student. It is designed to help students become skilled writers and be prepared for their futures. Aquinas Writing Advantage graduates are ready for college and beyond. Parents often asked us, "Where do I start?" To answer that question, we offer the following scope and sequence, based on your student's grade level in the fall. Whether your child is starting with Homeschool Connections in 7th grade or 12th grade, we can help you. Our live, interactive classes provide grading and feedback, giving you ease of mind and freeing your time. We also offer recorded, independent-learning classes, providing you with yet another homeschool option.To learn more, please visit our website at or email us at homeschoolconnections@gmail.com. SUGGESTED SCOPE AND SEQUENCE For the Student Beginning in the 12th Grade 12th GRADE Fall How to Be an Excellent Student (short course) Elements of Writing for High School: Punctuation and Grammar / Simplified Writing for High School Vocabulary and Writing I Spring Advance Writing and Rhetoric Advanced Research Writing Vocabulary and Writing II SUGGESTED SCOPE AND SEQUENCE For the Student Beginning in the 11th Grade 11th GRADE Fall Elements of Writing for High School: Punctuation and Grammar / Simplified Writing for High School Vocabulary and Writing I Spring How to Be an Excellent Student (short course) High School Writing Essentials: Excellent Paragraph and Essay/Test Writing Vocabulary and Writing II 12th GRADE Fall Advanced Writing and Rhetoric The Hero's Journey and Mythic Structure for Writers 1: Archetypes Spring Advanced Research Writing The Hero's Journey and Mythic Structure for Writers 2: Form SUGGESTED SCOPE AND SEQUENCE For the Student Beginning in the 10th Grade 10th GRADE Fall How to Be an Excellent Student (short course) Elements of Writing for High School: Punctuation and Grammar / Simplified Writing for High School Vocabulary and Writing I Spring Vocabulary and Writing II Fiction Writing Series 9th Grade 9th GRADE Fall How to Be an Excellent Student (or in the spring) Fiction Writing Series Spring Fiction Writing Series 10th GRADE Fall Elements of Writing for High School: Punctuation and Grammar/Simplified Writing for High School Vocabulary and Writing I Spring Vocabulary and Writing II 8th Grade How to Be an Excellent Student Middle School Writing II Spring Fiction Writing Series 10th GRADE Fall Elements of Writing for High School: Punctuation and Grammar / Simplified Writing for High School Voc 7th Grade Elements of Writing for High School: Punctuation and Grammar / Simplified Writing for High School 10th GRADE Fall VocHave you ever wondered what it is like to be a subscriber with Homeschool Connections? To get a free look click on this link! After clicking on the link, please login as a Guest. If you would like to have access to all 120+ courses then simply click on the Subscribe button in the right margin. For only $1 for the first 7 days and $30 a month thereafter you have access to the best instruction money can buy. Please let us know if you have any questions by emailing us at homeschoolconnections@gmail.com. Build Your Teen's College Skill Set Are you and your high school student(s) planning for college? If so, there are certain skill sets that are particularly important to acquire: Study Skills: Students need to know how to manage their time and meet deadlines. The brightest student can still flounder if these skills are not learned. The successful college student also needs good note taking and basic study skills so that they can get the most out of their classes and homework. After completing HSC's Study Skills and Note Taking course, students will put these skills into practice through their high school years and will therefore be better prepared for college. Communication Skills: Strong communication skills will greatly benefit your student in any college major or career field. HSC offers a course to help students learn and practice good communications. In the Leadership and Communications Skills course, students learn speaking skills, listening skills, conflict management, and more. Leadership Skills: The most successful students are often the ones who are also leaders. As Catholics, it is important that our students become people who are a positive influence at school and in the world. HSC's Leadership and Communications Skills course will encourage them to be people of service, show them how to be a faith-filled leader, and more. Writing Skills: It's not enough to learn lessons taught in school, students need to be able to communicate the lessons learned in writing. Strong writing skills are vital for college success. HSC offers a strong writing program (Aquinas Writing Advantage) that will take your student from the basics (grammar, punctuation, vocabulary) to the advanced (rhetoric, research, academic papers). Your student will be ready for college writing after successfully completing these writing courses. Critical Thinking Skills: Education should not be about cramming facts into children's heads. It should be about giving them a love for learning and the ability to think. We highly recommend formal logic and philosophy to help your student think critically and therefore succeed in all their school subjects. Logic and Philosophy are not electives — they are vital to a core curriculum. HSC offers a variety of courses that teach your student critical thinking skills, while at the same time raising their hearts to God and finding the beauty of their Catholic faith. ACT/SAT Test Skills: To help your student get into the college of his choice, and get the best scholarship possible, we offer courses on preparing for the ACT and SAT tests. Your student will learn how to prepare for the test, what to expect, manage time, and more for success. Latin studies should also be considered, for a variety of reasons including the evidence that Latin studies increase ACT and SAT scores. Most Importantly — How to Evaluate Ideas through a Catholic Lens: In college your student will encounter many new ideas and assumptions. Some of them will be potentially damaging. We want to give your student the necessary tools to recognize and understand the worldviews they encounter and know how to articulate their own beliefs effectively and convincingly. All of HSC instructors are Catholic and teach their courses through a Catholic lens, thus demonstrating to your student how God is evident in everything. Our theology courses will specifically prepare your student to defend his faith when he goes out into the world, as well as help him build a solid foundation of faith for his life. How to Use Technology in Education: In HSC's online courses students become familiar with the same, or similar, technology they'll encounter in college. They learn how to be engaged participants in a live, interactive webinar and gain experience using online tools to collaborate with their instructor and fellow students from all across the country and the world. This is a skill set that will help them advance in higher education as well as the business place. Recommended Homeschool Connections College Skill Set Courses Note: We offer a wide variety of courses and this recommend scope and sequence can easily be adjusted to fit your student's needs. Of course, you'll also want to include history, science, and moth. 9th Grade How to be an Excellent Student: Note Taking, Test Taking, and How to Get an A (4 weeks) We are very happy to announce, thrilled in fact, that Professor Carol will be teaching music and art appreciation for Homeschool Connections in the fall semester. Yes, THE Professor Carol! We hope you'll join us in the adventure of art and music through history … Class dates: Fridays, September 13 to December 13, 2013. No class October 11 and November 29 Total classes: 12 classes plus recorded lectures Starting time: 2:30 PM Eastern (1:30 PM Central) Duration: 1 hour Prerequisite: None. No musical background is necessary. Suggested grade level: 9th to 12th grade Suggested high school credit: 1 full semester Music/Art Appreciation Fee: $175 if you register on or before August 1, 2013. $195 after Aug. 1st for all 12 classes Instructor: Carol Reynolds, Ph.D. (Professor Carol) Course description: Journey with Professor Carol through Western History, using music as the focal point but weaving in visual art (painting, sculpture), dance, theater, architecture, and literature. The study of music and the Fine Arts supports the understanding of history, geography, and culture. Elements of science, technology, and language are included in the course as well. Sessions will focus on the years between 1600 and World War One, but will present an overview of Medieval/Renaissance Sacred Music. Course materials: 1. Discovering Music online curriculum by Professor Carol will be made available to students for half of the regular price ($30 for four months subscription). 2. Music selections assigned by the instructor. These can be accessed in one of four ways. Choose the one that best suits your family: a) Free by searching your public library or YouTube; b) Classical Archives ($8 per month); c) Naxos ($20 per year); OR d) purchase 3-CD set from the instructor (HSC discounted price $34.95). Homework: This is not a course for the faint of heart. We'll have a lot of fun as we discover music together, but students should expect a good amount of work outside of the classroom in that discovery. Homework will entail: 1. Viewing recorded classes in advance to the live classes. 2. Viewing assigned artwork and listening to music. 3. Interactive quizzes. 4. A midterm and a final exam (fill-in-blank, short essays, long essays, with answers/suggested answers). 5. Unit projects to be determined. Due to the nature of the medium, we encourage students more than ever to share their learning experience and the resources used in this course with the rest of their family. There are a lot of different ways you can use Homeschool Connections' recorded classes (aka Unlimited Access) to keep learning alive and fun over the summer. Here are ten ideas to get you started. 10. Take school with you. All you need for recorded classes is a power source, internet, and a computer. You should add ear buds or a headset to the list if you need privacy. We've had students take classes from hotel rooms, Grandma's house, the library, and even the beach. Though we don't recommend taking your laptop anywhere near sand! 9. Plug the computer into the television. This is a really fun way to learn together as a family. Pick a subject that everyone is interested in learning. It may be Catholic Apologetics or Civil War or Lord of the Rings or something completely different. Make some popcorn and watch together. You may need an HDMI cable and a newer TV (Mac users will need a converter). Do what I do and have a teen set it up for you. 8. Pick a time that works best for you. Recorded classes are available 24/7. You could watch classes first thing in the morning, getting them done early so the rest of the day can be spent outdoors. Perhaps, you could would prefer to watch classes during lunch or just before bed in the evening. Pick the time that is going to help you keep up on your work throughout the whole of summer. 7. Audit a course. Watch a lecture each day and forgo the homework. For example, instead of taking 12 weeks for World History: 12 Inventions that Changed the World, watch the lectures over 12 days. When auditing, pick a subject that is easy for you. 6. Buckle down on tough subjects. Really need help with algebra? Struggled with science last year? If so, buckle down and get to work. Set aside time each and every day (Sundays off!) and stick to the schedule. Complete all of the homework before moving to the next recorded lecture. If you want extra help, sign up for the optional grading support (Instructor Access). 5. Catch up on subjects for September. Planning on taking Latin II next year but not quite ready? Perhaps illness or something else kept you from finishing Latin I this year. Whether you simply need a refresher or need to make up for lost time, there are a number of "Bootcamps" available in recording (math, Latin, and more). 4. Ask yourself, "What do I love?" Perhaps you love to read. If so, choose a literature course on a book you love. Reread The Hobbit as you watch Dr. Russell'sHobbit lectures over a couple of weeks. Or Screwtape Letters, or Space Trilogy, or The Man Who Was Thursday. You can choose from over 20 literature courses. 3. Summer is a great time to hone your writing skills Writing is a key skill for success in all other school subjects. Focusing on writing skills over summer will help you do better in history, literature, and more when fall arrives. Other courses that help you succeed in core subjects include: Note Taking Skillsand How to Use Microsoft Word. 2. Keep a schedule and stick to it. How many times have we all laid out grand plans, only to forget about them as the excitement wore off? Write out a reasonable schedule on a white board or print it and post it. Program your computer or smart phone to remind you each day. Do something tangible to keep you on schedule. 1. Keep it simple. You don't need a complicated schedule to be effective. Pick just one or two subjects. For example, maybe you weren't able to make time for philosophy in the fall and spring, but you know it would help you a lot to learn it and it sounds interesting. Focus just on philosophy courses for summer. Bonus. Unlimited Access means just that! You have unlimited access to over 100 courses for your entire family. Yes, it's true! You can't beat the price ($30 per month!!!) and you can't beat the convenience. Middle school, high school, and adult students can easily learn year round with this independent learning program. It can be as easy or as complicated as you want to make it. It's YOUR program.
Rent Book Buy New Book Used Book We're Sorry Sold Out eBook We're Sorry Not Available More New and Used from Private Sellers Starting at $4Kaplan SAT Math Workbook Kaplan SAT Math Workbook 3rd edition Summary The complete test preparation tool that contains tips, strategies, and practice for students who want to score higher on the math section of the SAT!As colleges across the country grow increasingly selective in their admissions standards, students seek out SAT prep in an effort to get a top score and stand out in the college applicant pool. No other products on the market can match the quality and experience behind Kaplanrs"s SAT guides.Kaplan SAT Math Workbookprovides everything students need to conquer the Quantitative Reasoning section of the exam. This targeted guide includes in-depth coverage of all pertinent math skills and information, as well as effective score-raising strategies for building speed and accuracy from math experts. Not only does this tool contain everything a student needs to conquer all the math included in the test, it also provides key information about the SAT in general, such as Kaplanrs"s methods for answering multiple-choice questions and more.Kaplan SAT Math Workbookcontains many essential and unique features to help improve test scores, including:Two realistic math tests with detailed answer explanations covering all parts of the SAT math section The top 100 SAT math concepts Effective score-raising methods and strategies for building speed and accuracy Proven strategies for avoiding common math errorsKaplan SAT Math Workbookprovides students with everything they need to improve their scores-guaranteed. Kaplanrs"sKaplan SAT Math Workbookis the must-have preparation tool for every student looking to score higher! Author Biography The Best Choice for SAT Prep With nearly 70 years of experience, Kaplan's
Link Details Have you ever wondered how your GPS can find the fastest way to your destination, selecting one route from seemingly countless possibilities in mere seconds? How your credit card account number is protected when you make a purchase over the Internet? The answer is algorithms. And how do these mathematical formulations translate themselves into your GPS, your laptop, or your smart phone? This book offers an engagingly written guide to the basics of computer algorithms. In Algorithms Unlocked, Thomas Cormen—coauthor of the leading college textbook on the subject—provides a general explanation, with limited mathematics, of how algorithms enable computers to solve problems.
Introduction to Real Analysis Book Description: This text provides the fundamental concepts and techniques of real analysis for students in all of these areas. It helps one develop the ability to think deductively, analyse mathematical situations and extend ideas to a new context. Like the first three editions, this edition maintains the same spirit and user-friendly approach with addition examples and expansion on Logical Operations and Set Theory. There is also content revision in the following areas: introducing point-set topology before discussing continuity, including a more thorough discussion of limsup and limimf, covering series directly following sequences, adding coverage of Lebesgue Integral and the construction of the reals, and drawing student attention to possible applications wherever possible
Algebra textbook is a college-level, introductory textbook that covers the important subject of Algebra -- one of the basic building blocks of studies in higher mathematics. Boundless works with subject matter experts to select the best open educational resources available on the web, review the content for quality, and create introductory, college-level textbooks designed to meet the study needs of university students.This textbook covers:The Building Blocks of Algebra -- Real Numbers, Exponents, Scientific Notation, Order of Operations, Working with Polynomials, Factoring, Rational Expressions, Radical Notation and Exponents, Basics of Equation SolvingGraphs, Functions, and Models -- Graphing, Functions: An Introduction, Modeling Equations of Lines, Functions Revisited, Algebra of Functions, TransformationsFunctions, Equations, and Inequalities -- Linear Equations and Functions, Complex Numbers, Quadratic Equations, Functions, and Applications, Graphs of Quadratic Functions, Further Equation Solving, Working with Linear InequalitiesPolynomial and Rational Functions -- Polynomial Functions and Models, Graphing Polynomial Functions, Polynomial Division; The Remainder and Factor Theorems, Zeroes of Polynomial Functions and Their Theorems, Rational Functions, Inequalities, Variation and Problem SolvingExponents and Logarithms -- Inverse Functions, Graphing Exponential Functions, Graphing Logarithmic Functions, Properties of Logarithmic Functions, Growth and Decay; Compound InterestSystems of Equations and Matrices -- Systems of Equations in Two Variables, Systems of Equations in Three Variables, Matrices, Matrix Operations, Inverses of Matrices, Determinants and Cramer's Rule, Systems of Inequalities and Linear Programming, Partial FractionsConic Sections -- The Parabola, The Circle and the Ellipse, The Hyperbola, Nonlinear Systems of Equations and InequalitiesSequences, Series and Combinatorics -- Sequences and Series, Arithmetic Sequences and Series, Geometric Sequences and Series, Mathematical Inductions, Combinatorics, The Binomial Theorem, Probability Calculus textbook is a college-level, introductory textbook that covers the fascinating subject of Calculus. Boundless works with subject matter experts to select the best open educational resources available on the web, review the content for quality, and create introductory, college-level textbooks designed to meet the study needs of university students.This textbook covers:Building Blocks of Calculus -- Precalculus Review, Functions and Models, LimitsDerivatives and Integrals -- Derivatives, Applications of Differentiation, Integrals, Applications of IntegrationInverse Functions and Advanced Integration -- Inverse Functions: Exponential, Logarithmic, and Trigonometric Functions, Techniques of Integration, Further Applications of IntegrationDifferential Equations, Parametric Equations, and Sequences and Series -- Differential Equations, Parametric Equations and Polar Coordinates, Infinite Sequences and SeriesAdvanced Topics in Single-Variable Calculus and an Multivariable Calculus -- Vectors and the Geometry of Space, Vector Functions, Partial Derivatives, Multiple Integrals, Vector Calculus, Second-Order Linear EquationsA Closer Look at Tests of Significance -- Which Test?, A Closer Look at Tests of Significance Algebra I Second Edition is a clear presentation of algebra for the high school student. Volume 1 includes the first 6 chapters and covers the following topics: Equations and Functions, Real Numbers, Equations of Lines, Graphs of Equations and Functions, Writing Linear Equations, and Linear Inequalities.'
Carnegie Learning develops textbooks that support a collaborative, student-centered classroom. Our classroom activities address both mathematical content and process standards. Students develop skills to work cooperatively to solve problems and improve their reasoning and sense-making skills. Program Components Click icons for details » Program Components Click icons for details » Supplemental & Intervention Solutions Some students will need additional support and intervention to meet the high expectations of state standards. Carnegie Learning can help you implement tiered interventions in mathematics. In addition to the core instruction we provide in our textbooks, we provide interactive math instruction in our Cognitive Tutor software. Our Algebra Readiness curriculum is a one-year course designed to remediate students who have completed a middle school math sequence of instruction but still exhibit gaps in their math knowledge and skills. The course covers the five major NCTM strands: Number and Operations, Algebra, Geometry, Measurement, Data Analysis and Probability. Whitepapers I'm very happy that Carnegie Learning also provides 24/7 Math Help for students where in the middle of the night or wee hours of the morning can get online and get their questions answered so they're not stuck and frustrated, and can move on in their learning. – Suzanne Etheridge, Mathematics Instructor Pellissippi State Community College
Annual High School Mathematics Competition Every fall, the Mathematics Department at Clarion University hosts a high school mathematics competition. Teams of area students compete for individual and school recognition. Top scores receive prizes but every student gets to enjoy a day on campus, a college lunch, and a T-shirt. For more information, contact Dr. Carey Childers at 393-2599 or by email at cchilders@clarion.edu. Facilities Facilities for the Mathematics department include the Computational Science Laboratory that contains 32 computers and all the software a mathematics major needs such as Mathematica, MatLab, Geometer's Sketchpad, etc. Tutoring Tutoring for mathematics classes is free to all university students and coordinated through the Department of Academic Enrichment. More information can be found here.
Mathematical Applications in Agriculture Summary Invaluable in any area of agriculture or as a hands-on learning tool in introductory math courses, the 2nd Edition of MATHEMATICAL APPLICATIONS IN AGRICULTURE demonstrates industry-specific methods for solving real-world problems using applied math and logic skills students already have.
The student workbook is 669 pages with 119 lessons, while the four CD-ROMs provide step-by-step audiovisual solutions to every homework and test problem. The CD-ROM's digital gradebook grades answers as soon as they are entered and calculates percentages for each assignment; a softcover answer booklet is also provided. Windows 2000. Teaching Textbooks Grade 4.
Excursions in Modern Mathematics - 7th edition Summary: Excursions in Modern Mathematics, Seventh Edition, shows readers that math is a lively, interesting, useful, and surprisingly rich subject. With a new chapter on financial math and an improved supplements package, this book helps students appreciate that math is more than just a set of classroom theories: math can enrich the life of any one who appreciates and knows how to use it
This video was created by a student for a graduate class in digital storytelling. The author decided to use this format to... see more This video was created by a student for a graduate class in digital storytelling. The author decided to use this format to illustrate functional relationships in math. The student claimed that "The idea of functions is very difficult for most of my students to understand because it's usually taught very abstractly." This is a humorous way to get the point across to which students can relate.
eter-Hutchison Series in Mathematics: Basic Mathematical Skills with Geometry The "Streeter-Hutchison Series in Mathematics: Basic Mathematical Skills with Geometry, 7/e" by Baratto/Bergman is designed for a one-semester basic ...Show synopsisThe "Streeter-Hutchison Series in Mathematics: Basic Mathematical Skills with Geometry, 7/e" by Baratto/Bergman is designed for a one-semester basic math course. This successful worktext series is appropriate for lecture, learning center, laboratory, or self-paced courses. "Basic Mathematical Skills with Geometry" continues with it's hallmark approach of encouraging the learning mathematics by focusing its coverage on mastering math through practice. The "Streeter-Hutchison" series worktexts seek to provide carefully detailed explanations and accessible pedagogy to introduce basic mathematical skills and put the content in context. With repeated exposure and consistent structure of Streeter's hallmark three-pronged approach to the introduction of basic mathematical skills, students are able to advance quickly in grasping the concepts of the mathematical skill at hand
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
Fundamentals of Math 2nd Ed. Home School Kit This kit includes one of each of the following Fundamentals of Math 2nd Edition items: Teacher's Edition bju244228 Student Text bju218933 Student Activities Book bju262626 Student Activities Answer Key bju262634 Testpack bju218958 Tests Answer Key bju218941 Grade 7 List $214.44 Sale Price $184.00 Fundamentals of Math 2nd Ed. Teacher's Edition with CD Publisher: BJU Press Item #bju244228 ISBN-13: 9781591667216 Grade 7 List $83.33 Sale Price $71.50 Fundamentals of Math 2nd Ed. Student Text Publisher: BJU Press Item #bju218933 ISBN-13: 9781591663751 The new Fundamentals of Math Student Text is a larger 8.5" by 11" paperback book. The larger size is a good change visually because offers a more relaxed layout with larger font and figures. Features include: Chapter Reviews Chapters about whole numbers, decimals, number theory, fractions, rational numbers, percents, measurement, intro to geometry, area and volume, probability and statistics, integers, intro to algebra, relations and functions, and logic and set theory. Each lesson has problems that focus on the lesson. Most lessons have cumulative review problems at the end of the lesson exercises. Fundamentals of Math 2nd Ed. Tests Answer Key Pre-Algebra Home School Kit (2nd Ed.) Publisher: BJU Press Item #bju271270 The new 2nd Edition of Pre-Algebra helps your student make the transition from arithmetic to algebra. Topics include Integers, Expressions, Basic Equations and Inequalities, Number Theory, Rational Numbers, Operations on Rational Numbers, Percents, Applying Equations and Inequalities, Relations and Functions, Statistics and Probabilities, Radicals, Geometry, Area and Volume, and Polynomials. The BJU Press Pre-Algebra Subject Kit includes one of each of the following: With a larger size (8.5" by 11"), the new Pre-Algebra Student Text is improved because it has a larger font, larger figures to show problems, and more roomy layout. Each chapter has a review. Most lessons have cumulative review problems at the end of the exercises. The exercise problems for each lesson focus on problems relating to the lesson. For the student who needs to work a lot of the same type of problems to master a lesson, this textbooks is a good choice. With the review problems grouped at the end of the lesson and the chapter, your student will have enough review to keep previously learned pre-algebra concepts fresh. It has a good mix of arithmetic review and new pre-algebra topics that will help prepare your student for algebra.
Welcome! Happy Spring! The countdown has begun! With just weeks left of school, let's finish strong!! Summer School Summer school is highly recommended for any student who will not pass Algebra 1. In order to graduate with a high school diploma, students are required to have 4 credits of math. This means 4 full years, including Algebra 1. Please pick up an application/registration for summer school in the front office of the high school if this pertains to you. Thanks! OGT Blitz Week 2/28 - 3/11 For 2 weeks, all math classes with students in grades 9 and 10 are working on specific goals to improve the understsanding for material from the Ohio Graduation Test. Sophomores and Juniors/Seniors who have not yet passed, will be taking the OGT the week of 3/14 - 3/21. Freshman need to arrive at 10 am - 2 pm. Enjoy sleeping in! Here you will find helpful tools for SUCCESS in Algebra 1. Just a few key items include: Worksheets (did you miss a day of class? or misplace an assignment? not to worry!) Vocabulary Lists (finish your homework in math class and looking for additional math to brush up on? find it here!) Links to websites that I think are cool/helpful for your math learning or somehow relevant to what we're learning.
Mathematical Modelling [NOOK Book] ... More About This Book modelling at work through published papers; Modelling in mechanics. Written in the lively interactive style of the Modular Mathematics Series, this text will encourage the reader to take part in the modelling process. Audience: First year mathematics undergraduates and other students taking a first course in mathematical modelling
Algebra I Essentials For Dummies by Mary Jane Sterling Just the critical concepts you need for cramming, homework help, and reference Whether you're cramming, you're studying new material, or you just need a refresher, this compact guide gives you a concise, easy-to-follow review of the most important concepts covered in a typical Algebra I course. Free of review and ramp-up materials, it lets you skip right to the parts where you need the most help. It's that easy! Set the scene — get the lowdown on everything you'll encounter in algebra, from words andsymbols to decimals and fractions
GeoGebra is a dynamic mathematics software that joins geometry, algebra, and calculus. Two views are characteristic of GeoGebra: an expression in the algebra window corresponds to an object in the geometry window and vice versa
Get connected with like-minded individuals in the movement dedicated to innovative thinking and action to improve and advance STEM education. Sign up below for our updates on STEM events in USA and around the globe. Students will locate points in the plane and work to solve problems using the coordinate plane. The module also addresses the solution of an equation in two variables, focusing on functions and relations, the vocabulary of functions, and linear functions. Lessons on equations of lines include graphing a line, slope of a line, parallel and perpendicular lines, slope-intercept form of a line, and point-slope form of a line. When solving systems of linear equations students relate algebraic and geometric interpretations of simultaneous solutions. Data and Probability Data and Probability Students will discuss appropriate sample sizes, organize and interpret data, and use measures of central tendency and range. Students will then explore probability and use their understanding of probability to form conclusions on events. Equations and Formulas Equations and Formulas This module emphasizes the meaning of equality and the function of the equal sign in an equation. Students will understand and use the Properties of Equality to solve one and two step equations. They will then apply those strategies to transform formulas and solve formulas for a specified variable. Exponents Exponents Students will develop and apply the meaning of positive integer exponents in contextual situations, and extend patterns to conceptualize the meaning of a zero exponent and negative exponents. They will generalize patterns within context to develop the properties of exponents while applying exponents to both numbers and variables. Inequalities Inequalities This module introduces the notation and vocabulary of inequalities, explores the properties of inequalities and how to solve them, and has students illustrate what the solutions look like on a coordinate plane. Number Theory Number Theory This module reinforces number sense and the language of numbers, factors, multiples, and divisors. Students will work with the set of Natural Numbers then extend their understanding to algebraic expressions. Patterns Patterns Students will examine and analyze numerical and algebraic patterns. They will connect patterns to physical models and make generalizations. To describe patterns, students will use recursive and explicit rules. They will develop their understanding of functions as they connect the physical models of patterns to the numerical relationships in input/output tables. This module will improve logic, mathematical reasoning, and problem-solving skills. Students will be able to connect the patterns they study in this module to graphic representations of those patterns. Proportional Reasoning Proportional Reasoning Students will apply their understanding of rations to reason about proportional relationships. They will use conceptually-based strategies to gain a deeper understanding of percents and unit conversions. Ratio and Proportion Ratio and Proportion Students make comparisons and distinguish between additive (absolute) and multiplicative (relative) thinking. They learn to use ratios and rates to make statements about one measurement in relation to another. They are able to determine equivalent ratios and compare ratios. The concept of scale and similar figures is applied in various contexts. Several strategies are developed to solve proportions: ratio tables, equivalent fractions, and solving proportions algebraically. Rational Numbers Rational Numbers This module reinforces the concepts involved in working with rational numbers. Students use models to conceptualize the meaning of rational numbers and operations with rational numbers. They will also extend their understanding to rational expressions. Signed Number Operations Signed Number Operations This module reinforces the concepts involved in working with signed numbers. Students use models to conceptualize operations with signed numbers. They will also work with absolute values and solve number puzzles. Variables and Expressions Variables and Expressions Students represent quantities with variables to write, interpret, and evaluate expressions. They translate expressions into words and words into algebraic expressions. Students simplify algebraic expressions using the order of operations and the commutative, associative, and distributive properties.
Deep Dive into Mathematica's Numerics: Applications and Tips Andrew Moylan In this course from the Wolfram Mathematica Virtual Conference 2011, you'll learn how to best use Mathematica's numerics functions in advanced settings. Topics include techniques and best practices for using multiple numerics functions together, advanced numeric features, and understanding precision and accuracy
The book is intended as the primary text for an introductory course in proving theorems, as well as for self-study or as a reference. Throughout the text, some pieces (usually proofs) are left as exercises. Part V gives hints to help students find good approaches to the exercises. Part I introduces the language of mathematics and the methods of proof. The mathematical content of Parts II through IV were chosen so as not to seriously overlap the standard mathematics major. In Part II, students study sets, functions, equivalence and order relations, and cardinality. Part III concerns algebra. The goal is to prove that the real numbers form the unique, up to isomorphism, ordered field with the least upper bound; in the process, we construct the real numbers starting with the natural numbers. Students will be prepared for an abstract linear algebra or modern algebra course. Part IV studies analysis. Continuity and differentiation are considered in the context of time scales (nonempty closed subsets of the real numbers). Students will be prepared for advanced calculus and general topology courses. There is a lot of room for instructors to skip and choose topics from among those that are presented. Table of Contents Some Notes on Notation To the Students To Those Beginning the Journey into Proof Writing How to Use This Text Do the Exercises! Acknowledgments For the Professors To Those Leading the Development of Proof Writing For Students in a Broad Range of Disciplines I. The Axiomatic Method 1. Introduction 2. Statements in Mathematics 3. Proofs in Mathematics II. Set Theory 4. Basic Set 5. Functions 6. Relations on a Set 7. Cardinality III. Number Systems 8. Algebra of Number Systems 9. The Natural Numbers 10. The Integers 11. The Rational Numbers 12. The Real Numbers 13. Cantor's Reals 14. The Complex Numbers IV. Time Scales 15. Time Scales 16. The Delta Derivative V. Hints 17. Hints for (and Comments on) the Exercises Bibliography Index About the Authors Excerpt: Ch. 2 Statements in Mathematics (p. 9) Whatever mathematics may be as a mental activity, it is communicated as a language. Therefore, it has its specific syntax, its own technical terms, and its own conventions. Mathematics is also an exact science, which means that we are obliged to express our mathematical thoughts with high precision. A deviation from the norm may lead to a complete distortion of the intended meaning. The purpose of this chapter is to make explicit a number of principles that pertain to mathematical logic. This branch of mathematics formalizes the principles of mathematical reasoning, principles that permeate mathematical thinking, be it consciously or subconsciously. In mathematics, we assert the truth of some statements. Other statements are to be proved or disproved. This is a fundamental dichotomy: No mathematical statement is both true and false. Viewed axiomatically, you can take true and false to be undefined terms; the above dichotomy can be taken as an axiom. We should clarify what is meant by a mathematical statement. About the Authors Ralph W. Oberste-Vorth was born in Brooklyn, New York and attended New York City public schools. His interests in mathematics began in junior high school and developed at Stuyvesant High School and Hunter College of the City University of New York. He met Aristides Mouzakitis at Hunter, where they became good friends while earning BA andMA degrees in mathematics. Ralph continued his studies at Cornell University where he earned his PhD in dynamical systems under the direction of John Hamal Hubbard. After oneyear positions at Yale University and the Institute for Advanced Study in Princeton, New Jersey, he moved to the University of South Florida in 1989. In 2000, Ralph approached Aristides with the idea of writing a "proofs text." During an intense 19-day session at his home in Corfu, Greece, they wrote the first draft of this book. Several instructors at South Florida used it. Ralph moved to Marshall University as the Chairman of the Department of Mathematics in 2002. In 2009, he invited his Marshall colleague, Bonita Lawrence, to help them put the book into a publishable form. The book had been used several times at Marshall. This project was started during the summer of 2009 and completed in the summer of 2011, with input from the MAA. In August 2011, Ralph became the Chairman of the Department of Mathematics and Computer Science at Indiana State University. Ralph lives in Terre Haute, Indiana with his wife and three children. Aristides Mouzakitis was born in the village of Avliotes on the Greek island Corfu in the Ionian Sea. He attended primary school in Avliotes and moved to Kerkyra, the main town of Corfu, to attend high school. In 1980, Aristides moved to New York City to attend Hunter College of the City University of New York. There, he earned his BA in the Special Honors Curriculum and his MA in mathematics. Aristides met Ralph Oberste-Vorth while at Hunter, where they laid the foundations for an enduring friendship. He began doctoral studies in mathematics, but Greece beckoned. In Greece, he has worked as a teacher in secondary education and as an English - Greek translator of popular mathematics books and articles. Eventually, he took on further formal studies and in 2009 he earned his doctorate in mathematics education fromthe University of Exeter in England under the direction of Paul Ernest. Aristides stays active in the Astronomical Society of Corfu and the Corfu branch of the Hellenic Mathematical Society. He lives in Kerkyra with his wife and his daughter, and enjoys reading and swimming, especially in winter time. Bonita Lawrence is currently a Professor of Mathematics at Marshall University in Huntington, West Virginia. She was born to a military family when her father was stationed with the U.S. Army in Stuttgart, Germany. Her father retired at Ft. Sill near Lawton, Oklahoma when she was in junior high school. She received her baccalaureate degree in Mathematics Education from Cameron University in Lawton in 1979. After ten years of teaching, she returned to school to study for a Master's degree in Mathematics at Auburn University. Upon completion of her Master's degree in 1990, she continued her academic training at the University of Texas at Arlington, earning a Ph.D. in Mathematics in 1994. In her first teaching position after completing her Ph.D., at North Carolina Wesleyan College, she was the 1996 Professor of the Year. After a few years at small institutions, North Carolina Wesleyan College and the Beaufort Campus of the University of South Carolina, she made the move to Marshall University to expand her teaching opportunities and to work with graduate students at the Master's level. She served as either Associate Chairman or Assistant Chairman for Graduate Studies for 10 of the 11 years under the leadership of Dr. Ralph Oberste-Vorth. During her time at Marshall University, she has received the following research and teaching awards: Marshall University Distinguished Artists and Scholars Award—Junior Recipient for Excellence in All Fields (Spring 2002); Shirley and Marshall Reynolds Outstanding Teaching Award (Spring 2005); Marshall University Distinguished Artists and Scholars Award—Team Award for Distinguished Scholarly Activity, with one of my coauthors, Dr. Ralph Oberste-Vorth (Spring 2007); Charles E. Hedrick Outstanding Faculty Award (April 2009); and the West Virginia Professor of the Year (March 2010). Dr. Lawrence currently is the lead researcher for the Marshall Differential Analyzer Lab, a mathematics lab that houses the Marshall Differential Analyzers. These machines, built by students of replicated Meccano components, are models of the machines that were first built in the late 1920's to solve differential equations. The largest of the machines, a four integrator model that can run up to fourth order equations, is the only publicly accessible machine of its size and type in the country. The lab offers the opportunity for the investigation of new research ideas as well as educational experiences for students of mathematics at many levels. This is her first book as a coauthor. She served as a reviewer for a linear algebra textbook and the solutions manual, The Keys to Linear Algebra, by Daniel Solow. Dr. Lawrence shares her life with her husband of 15 years, Dr. Clayton Brooks, a colleague in the Mathematics Department.
Pearls in Graph Theory: A Comprehensive Introduction "Innovative introductory text . . . clear exposition of unusual and more advanced topics . . . Develops material to substantial level."--American Mathematical Monthly "Refreshingly different . . . an ideal training ground for the mathematical process of investigation, generalization, and conjecture leading to the discovery of proofs and counterexamples."--American Mathematical Monthly " . . . An excellent textbook for an undergraduate course."--Australian Computer Journal A stimulating view of mathematics that appeals to students as well as teachers, this undergraduate-level text is written in an informal style that does not sacrifice depth or challenge. Based on 20 years of teaching by the leading researcher in graph theory, it offers a solid foundation on the subject. This revised and augmented edition features new exercises, simplifications, and other improvements suggested by classroom users and reviewers. Topics include basic graph theory, colorings of graphs, circuits and cycles, labeling graphs, drawings of graphs, measurements of closeness to planarity, graphs on surfaces, and applications and algorithms. 1994 ed. Review: Pearls in Graph Theory: A Comprehensive Introduction User Review - Goodreads Skimmed through most of this; it looked good, and there were several wonderful examples I'd never seen before (generally, results I'd seen achieved with trivial topological results, but not through a "pure" graph theory methodology). Notes for the Research Colloquium Notes for the Research Colloquium. A graph G is a finite non-empty set of objects called vertices together with a (possibly empty) finite set of unordered ... csc.colstate.edu/ Bosworth/ GraphTheory/ ResearchNotes.htm Gerhard Ringel —— 维客(wiki) Gerhard Ringel (born 10/28/1919, Kolnbrunn, Austria) is a German mathematician who earned his Ph.D. from the University of Bonn in 1951. ... wiki/ Gerhard_Ringel
263003 / ISBN-13: 9781904263005 Useful Mathematical and Physical Formulae A compact volume of mathematical and physical formulae presented in a concise manner for general use. Collected in this book are commonly used ...Show synopsisA compact volume of mathematical and physical formulae presented in a concise manner for general use. Collected in this book are commonly used formulae for studies such as quadratics, calculus and trigonometry; in addition are simplified explanations of Newton's Laws of Gravity and Snell's Laws of Refraction. A glossary, a table of mathematical and physical constants, and a listing of Imperial and Metric conversions is also included
P R E F A C ESee also and Structure of the BookThis book provides a comprehensive, thorough, and up-to-date treatment of engineeringmathematics. It is intended to introduce students of engineering, physics, mathematics,computer science, and related fields to those areas of applied mathematics that are mostrelevant for solving practical problems. A course in elementary calculus is the soleprerequisite. (However, a concise refresher of basic calculus for the student is includedon the inside cover and in Appendix 3.)The subject matter is arranged into seven parts as follows:A. Ordinary Differential Equations (ODEs) in Chapters 1–6B. Linear Algebra. Vector Calculus. See Chapters 7–10C. Fourier Analysis. Partial Differential Equations (PDEs). See Chapters 11 and 12D. Complex Analysis in Chapters 13–18E. Numeric Analysis in Chapters 19–21F. Optimization, Graphs in Chapters 22 and 23G. Probability, Statistics in Chapters 24 and 25.These are followed by five appendices: 1. References, 2. Answers to Odd-NumberedProblems, 3. Auxiliary Materials (see also inside covers of book), 4. Additional Proofs,5. Table of Functions. This is shown in a block diagram on the next page.The parts of the book are kept independent. In addition, individual chapters are kept asindependent as possible. (If so needed, any prerequisites—to the level of individualsections of prior chapters—are clearly stated at the opening of each chapter.) We give theinstructor maximum flexibility in selecting the material and tailoring it to his or herneed. The book has helped to pave the way for the present development of engineeringmathematics. This new edition will prepare the student for the current tasks and the futureby a modern approach to the areas listed above. We provide the material and learningtools for the students to get a good foundation of engineering mathematics that will helpthem in their careers and in further studies.General Features of the Book Include:• Simplicity of examples to make the book teachable—why choose complicatedexamples when simple ones are as instructive or even better?• Independence of parts and blocks of chapters to provide flexibility in tailoringcourses to specific needs.• Self-contained presentation, except for a few clearly marked places where a proofwould exceed the level of the book and a reference is given instead.• Gradual increase in difficulty of material with no jumps or gaps to ensure anenjoyable teaching and learning experience.• Modern standard notation to help students with other courses, modern books, andjournals in mathematics, engineering, statistics, physics, computer science, and others.Furthermore, we designed the book to be a single, self-contained, authoritative, andconvenient source for studying and teaching applied mathematics, eliminating the needfor time-consuming searches on the Internet or time-consuming trips to the library to geta particular reference book.viifpref.qxd 11/8/10 3:16 PM Page vii Four Underlying Themes of the BookThe driving force in engineering mathematics is the rapid growth of technology and thesciences. New areas—often drawing from several disciplines—come into existence.Electric cars, solar energy, wind energy, green manufacturing, nanotechnology, riskmanagement, biotechnology, biomedical engineering, computer vision, robotics, spacetravel, communication systems, green logistics, transportation systems, financialengineering, economics, and many other areas are advancing rapidly. What does this meanfor engineering mathematics? The engineer has to take a problem from any diverse areaand be able to model it. This leads to the first of four underlying themes of the book.1. Modeling is the process in engineering, physics, computer science, biology,chemistry, environmental science, economics, and other fields whereby a physical situationor some other observation is translated into a mathematical model. This mathematicalmodel could be a system of differential equations, such as in population control (Sec. 4.5),a probabilistic model (Chap. 24), such as in risk management, a linear programmingproblem (Secs. 22.2–22.4) in minimizing environmental damage due to pollutants, afinancial problem of valuing a bond leading to an algebraic equation that has to be solvedby Newton's method (Sec. 19.2), and many others.The next step is solving the mathematical problem obtained by one of the manytechniques covered in Advanced Engineering Mathematics.The third step is interpreting the mathematical result in physical or other terms tosee what it means in practice and any implications.Finally, we may have to make a decision that may be of an industrial nature orrecommend a public policy. For example, the population control model may implythe policy to stop fishing for 3 years. Or the valuation of the bond may lead to arecommendation to buy. The variety is endless, but the underlying mathematics issurprisingly powerful and able to provide advice leading to the achievement of goalstoward the betterment of society, for example, by recommending wise policiesconcerning global warming, better allocation of resources in a manufacturing process,or making statistical decisions (such as in Sec. 25.4 whether a drug is effective in treatinga disease).While we cannot predict what the future holds, we do know that the student has topractice modeling by being given problems from many different applications as is donein this book. We teach modeling from scratch, right in Sec. 1.1, and give many examplesin Sec. 1.3, and continue to reinforce the modeling process throughout the book.2. Judicious use of powerful software for numerics (listed in the beginning of Part E)and statistics (Part G) is of growing importance. Projects in engineering and industrialcompanies may involve large problems of modeling very complex systems with hundredsof thousands of equations or even more. They require the use of such software. However,our policy has always been to leave it up to the instructor to determine the degree of use ofcomputers, from none or little use to extensive use. More on this below.3. The beauty of engineering mathematics. Engineering mathematics relies onrelatively few basic concepts and involves powerful unifying principles. We point themout whenever they are clearly visible, such as in Sec. 4.1 where we "grow" a mixingproblem from one tank to two tanks and a circuit problem from one circuit to two circuits,thereby also increasing the number of ODEs from one ODE to two ODEs. This is anexample of an attractive mathematical model because the "growth" in the problem isreflected by an "increase" in ODEs.Preface ixfpref.qxd 11/8/10 3:16 PM Page ix 4. To clearly identify the conceptual structure of subject matters. For example,complex analysis (in Part D) is a field that is not monolithic in structure but was formedby three distinct schools of mathematics. Each gave a different approach, which we clearlymark. The first approach is solving complex integrals by Cauchy's integral formula (Chaps.13 and 14), the second approach is to use the Laurent series and solve complex integralsby residue integration (Chaps. 15 and 16), and finally we use a geometric approach ofconformal mapping to solve boundary value problems (Chaps. 17 and 18). Learning theconceptual structure and terminology of the different areas of engineering mathematics isvery important for three reasons:a. It allows the student to identify a new problem and put it into the right group ofproblems. The areas of engineering mathematics are growing but most often retain theirconceptual structure.b. The student can absorb new information more rapidly by being able to fit it into theconceptual structure.c. Knowledge of the conceptual structure and terminology is also important when usingthe Internet to search for mathematical information. Since the search proceeds by puttingin key words (i.e., terms) into the search engine, the student has to remember the importantconcepts (or be able to look them up in the book) that identify the application and areaof engineering mathematics.Big Changes in This EditionProblem Sets ChangedThe problem sets have been revised and rebalanced with some problem sets having moreproblems and some less, reflecting changes in engineering mathematics. There is a greateremphasis on modeling. Now there are also problems on the discrete Fourier transform(in Sec. 11.9).Series Solutions of ODEs, Special Functions and Fourier Analysis ReorganizedChap. 5, on series solutions of ODEs and special functions, has been shortened. Chap. 11on Fourier Analysis now contains Sturm–Liouville problems, orthogonal functions, andorthogonal eigenfunction expansions (Secs. 11.5, 11.6), where they fit better conceptually(rather than in Chap. 5), being extensions of Fourier's idea of using orthogonal functions.Openings of Parts and Chapters Rewritten As Well As Parts of SectionsIn order to give the student a better idea of the structure of the material (see UnderlyingTheme 4 above), we have entirely rewritten the openings of parts and chapters.Furthermore, large parts or individual paragraphs of sections have been rewritten or newsentences inserted into the text. This should give the students a better intuitiveunderstanding of the material (see Theme 3 above), let them draw conclusions on theirown, and be able to tackle more advanced material. Overall, we feel that the book hasbecome more detailed and leisurely written.Student Solutions Manual and Study Guide EnlargedUpon the explicit request of the users, the answers provided are more detailed andcomplete. More explanations are given on how to learn the material effectively by pointingout what is most important.More Historical Footnotes, Some EnlargedHistorical footnotes are there to show the student that many people from different countriesworking in different professions, such as surveyors, researchers in industry, etc., contributed54321x Prefacefpref.qxd 11/8/10 3:16 PM Page x to the field of engineering mathematics. It should encourage the students to be creative intheir own interests and careers and perhaps also to make contributions to engineeringmathematics.Further Changes and New Features• Parts of Chap. 1 on first-order ODEs are rewritten. More emphasis on modeling, alsonew block diagram explaining this concept in Sec. 1.1. Early introduction of Euler'smethod in Sec. 1.2 to familiarize student with basic numerics. More examples ofseparable ODEs in Sec. 1.3.• For Chap. 2, on second-order ODEs, note the following changes: For ease of reading,the first part of Sec. 2.4, which deals with setting up the mass-spring system, hasbeen rewritten; also some rewriting in Sec. 2.5 on the Euler–Cauchy equation.• Substantially shortened Chap. 5, Series Solutions of ODEs. Special Functions:combined Secs. 5.1 and 5.2 into one section called "Power Series Method," shortenedmaterial in Sec. 5.4 Bessel's Equation (of the first kind), removed Sec. 5.7(Sturm–Liouville Problems) and Sec. 5.8 (Orthogonal Eigenfunction Expansions) andmoved material into Chap. 11 (see "Major Changes" above).• New equivalent definition of basis (Sec. 7.4).• In Sec. 7.9, completely new part on composition of linear transformations withtwo new examples. Also, more detailed explanation of the role of axioms, inconnection with the definition of vector space.• New table of orientation (opening of Chap. 8 "Linear Algebra: Matrix EigenvalueProblems") where eigenvalue problems occur in the book. More intuitive explanationof what an eigenvalue is at the begining of Sec. 8.1.• Better definition of cross product (in vector differential calculus) by properlyidentifying the degenerate case (in Sec. 9.3).• Chap. 11 on Fourier Analysis extensively rearranged: Secs. 11.2 and 11.3combined into one section (Sec. 11.2), old Sec. 11.4 on complex Fourier Seriesremoved and new Secs. 11.5 (Sturm–Liouville Problems) and 11.6 (OrthogonalSeries) put in (see "Major Changes" above). New problems (new!) in problem set11.9 on discrete Fourier transform.• New section 12.5 on modeling heat flow from a body in space by setting up the heatequation. Modeling PDEs is more difficult so we separated the modeling processfrom the solving process (in Sec. 12.6).• Introduction to Numerics rewritten for greater clarity and better presentation; newExample 1 on how to round a number. Sec. 19.3 on interpolation shortened byremoving the less important central difference formula and giving a reference instead.• Large new footnote with historical details in Sec. 22.3, honoring George Dantzig,the inventor of the simplex method.• Traveling salesman problem now described better as a "difficult" problem, typicalof combinatorial optimization (in Sec. 23.2). More careful explanation on how tocompute the capacity of a cut set in Sec. 23.6 (Flows on Networks).• In Chap. 24, material on data representation and characterization restructured interms of five examples and enlarged to include empirical rule on distribution ofPreface xifpref.qxd 11/8/10 3:16 PM Page xi xii Prefacedata, outliers, and the z-score (Sec. 24.1). Furthermore, new example on encription(Sec. 24.4).• Lists of software for numerics (Part E) and statistics (Part G) updated.• References in Appendix 1 updated to include new editions and some references towebsites.Use of ComputersThe presentation in this book is adaptable to various degrees of use of software,Computer Algebra Systems (CAS's), or programmable graphic calculators, rangingfrom no use, very little use, medium use, to intensive use of such technology. The choiceof how much computer content the course should have is left up to the instructor, therebyexhibiting our philosophy of maximum flexibility and adaptability. And, no matter whatthe instructor decides, there will be no gaps or jumps in the text or problem set. Someproblems are clearly designed as routine and drill exercises and should be solved byhand (paper and pencil, or typing on your computer). Other problems require morethinking and can also be solved without computers. Then there are problems where thecomputer can give the student a hand. And finally, the book has CAS projects, CASproblems and CAS experiments, which do require a computer, and show its power insolving problems that are difficult or impossible to access otherwise. Here our goal isto combine intelligent computer use with high-quality mathematics. The computerinvites visualization, experimentation, and independent discovery work. In summary,the high degree of flexibility of computer use for the book is possible since there areplenty of problems to choose from and the CAS problems can be omitted if desired.Note that information on software (what is available and where to order it) is at thebeginning of Part E on Numeric Analysis and Part G on Probability and Statistics. SinceMaple and Mathematica are popular Computer Algebra Systems, there are two computerguides available that are specifically tailored to Advanced Engineering Mathematics:E. Kreyszig and E.J. Norminton, Maple Computer Guide, 10th Edition and MathematicaComputer Guide, 10th Edition. Their use is completely optional as the text in the book iswritten without the guides in mind.Suggestions for Courses: A Four-Semester SequenceThe material, when taken in sequence, is suitable for four consecutive semester courses,meeting 3 to 4 hours a week:1st Semester ODEs (Chaps. 1–5 or 1–6)2nd Semester Linear Algebra. Vector Analysis (Chaps. 7–10)3rd Semester Complex Analysis (Chaps. 13–18)4th Semester Numeric Methods (Chaps. 19–21)Suggestions for Independent One-Semester CoursesThe book is also suitable for various independent one-semester courses meeting 3 hoursa week. For instance,Introduction to ODEs (Chaps. 1–2, 21.1)Laplace Transforms (Chap. 6)Matrices and Linear Systems (Chaps. 7–8)fpref.qxd 11/8/10 3:16 PM Page xii CHAPTER 1 First-Order ODEsCHAPTER 2 Second-Order Linear ODEsCHAPTER 3 Higher Order Linear ODEsCHAPTER 4 Systems of ODEs. Phase Plane. Qualitative MethodsCHAPTER 5 Series Solutions of ODEs. Special FunctionsCHAPTER 6 Laplace TransformsMany physical laws and relations can be expressed mathematically in the form of differentialequations. Thus it is natural that this book opens with the study of differential equations andtheir solutions. Indeed, many engineering problems appear as differential equations.The main objectives of Part A are twofold: the study of ordinary differential equationsand their most important methods for solving them and the study of modeling.Ordinary differential equations (ODEs) are differential equations that depend on a singlevariable. The more difficult study of partial differential equations (PDEs), that is,differential equations that depend on several variables, is covered in Part C.Modeling is a crucial general process in engineering, physics, computer science, biology,medicine, environmental science, chemistry, economics, and other fields that translates aphysical situation or some other observations into a "mathematical model." Numerousexamples from engineering (e.g., mixing problem), physics (e.g., Newton's law of cooling),biology (e.g., Gompertz model), chemistry (e.g., radiocarbon dating), environmental science(e.g., population control), etc. shall be given, whereby this process is explained in detail,that is, how to set up the problems correctly in terms of differential equations.For those interested in solving ODEs numerically on the computer, look at Secs. 21.1–21.3of Chapter 21 of Part F, that is, numeric methods for ODEs. These sections are keptindependent by design of the other sections on numerics. This allows for the study ofnumerics for ODEs directly after Chap. 1 or 2.1P A R T AOrdinaryDifferentialEquations (ODEs)c01.qxd 7/30/10 8:14 PM Page 1 2C H A P T E R 1First-Order ODEsChapter 1 begins the study of ordinary differential equations (ODEs) by deriving them fromphysical or other problems (modeling), solving them by standard mathematical methods,and interpreting solutions and their graphs in terms of a given problem. The simplest ODEsto be discussed are ODEs of the first order because they involve only the first derivativeof the unknown function and no higher derivatives. These unknown functions will usuallybe denoted by or when the independent variable denotes time t. The chapter endswith a study of the existence and uniqueness of solutions of ODEs in Sec. 1.7.Understanding the basics of ODEs requires solving problems by hand (paper and pencil,or typing on your computer, but first without the aid of a CAS). In doing so, you willgain an important conceptual understanding and feel for the basic terms, such as ODEs,direction field, and initial value problem. If you wish, you can use your Computer AlgebraSystem (CAS) for checking solutions.COMMENT. Numerics for first-order ODEs can be studied immediately after thischapter. See Secs. 21.1–21.2, which are independent of other sections on numerics.Prerequisite: Integral calculus.Sections that may be omitted in a shorter course: 1.6, 1.7.References and Answers to Problems: App. 1 Part A, and App. 2.1.1 Basic Concepts. ModelingIf we want to solve an engineering problem (usually of a physical nature), we firsthave to formulate the problem as a mathematical expression in terms of variables,functions, and equations. Such an expression is known as a mathematical model of thegiven problem. The process of setting up a model, solving it mathematically, andinterpreting the result in physical or other terms is called mathematical modeling or,briefly, modeling.Modeling needs experience, which we shall gain by discussing various examples andproblems. (Your computer may often help you in solving but rarely in setting up models.)Now many physical concepts, such as velocity and acceleration, are derivatives. Hencea model is very often an equation containing derivatives of an unknown function. Sucha model is called a differential equation. Of course, we then want to find a solution (afunction that satisfies the equation), explore its properties, graph it, find values of it, andinterpret it in physical terms so that we can understand the behavior of the physical systemin our given problem. However, before we can turn to methods of solution, we must firstdefine some basic concepts needed throughout this chapter.y1t2y1x2PhysicalSystemPhysicalInterpretationMathematicalModelMathematicalSolutionFig. 1. Modeling,solving, interpretingc01.qxd 7/30/10 8:14 PM Page 2 are ordinary differential equations (ODEs). Here, as in calculus, denotes ,etc. The term ordinary distinguishes them from partial differentialequations (PDEs), which involve partial derivatives of an unknown function of twoor more variables. For instance, a PDE with unknown function u of two variables xand y isPDEs have important engineering applications, but they are more complicated than ODEs;they will be considered in Chap. 12.An ODE is said to be of order n if the nth derivative of the unknown function y is thehighest derivative of y in the equation. The concept of order gives a useful classificationinto ODEs of first order, second order, and so on. Thus, (1) is of first order, (2) of secondorder, and (3) of third order.In this chapter we shall consider first-order ODEs. Such equations contain only thefirst derivative and may contain y and any given functions of x. Hence we can writethem as(4)or often in the formThis is called the explicit form, in contrast to the implicit form (4). For instance, the implicitODE (where ) can be written explicitly asConcept of SolutionA functionis called a solution of a given ODE (4) on some open interval if isdefined and differentiable throughout the interval and is such that the equation becomesan identity if y and are replaced with h and , respectively. The curve (the graph) ofh is called a solution curve.Here, open interval means that the endpoints a and b are not regarded aspoints belonging to the interval. Also, includes infinite intervals(the real line) as special cases.E X A M P L E 1 Verification of SolutionVerify that (c an arbitrary constant) is a solution of the ODE for all Indeed, differentiateto get Multiply this by x, obtaining thus, the given ODE. ᭿xyr ϭ Ϫy,xyr ϭ Ϫc>x;yr ϭ Ϫc>x2.y ϭ c>xx 0.xyr ϭ Ϫyy ϭ c>xa Ͻ x Ͻ ϱ, Ϫϱ Ͻ x Ͻ ϱϪϱ Ͻ x Ͻ b,a Ͻ x Ͻ ba Ͻ x Ͻ bhryrh(x)a Ͻ x Ͻ by ϭ h(x)yr ϭ 4x3y2.x 0x؊3yr Ϫ 4y2ϭ 0yr ϭ f(x, y).F(x, y, yr) ϭ 0yr02u0x2ϩ02u0y2ϭ 0.ys ϭ d2y>dx2,dy>dxyr4 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:14 PM Page 4 We see that each ODE in these examples has a solution that contains an arbitraryconstant c. Such a solution containing an arbitrary constant c is called a general solutionof the ODE.(We shall see that c is sometimes not completely arbitrary but must be restricted to someinterval to avoid complex expressions in the solution.)We shall develop methods that will give general solutions uniquely (perhaps except fornotation). Hence we shall say the general solution of a given ODE (instead of a generalsolution).Geometrically, the general solution of an ODE is a family of infinitely many solutioncurves, one for each value of the constant c. If we choose a specific c (e.g., or 0or ) we obtain what is called a particular solution of the ODE. A particular solutiondoes not contain any arbitrary constants.In most cases, general solutions exist, and every solution not containing an arbitraryconstant is obtained as a particular solution by assigning a suitable value to c. Exceptionsto these rules occur but are of minor interest in applications; see Prob. 16 in ProblemSet 1.1.Initial Value ProblemIn most cases the unique solution of a given problem, hence a particular solution, isobtained from a general solution by an initial condition with given valuesand , that is used to determine a value of the arbitrary constant c. Geometricallythis condition means that the solution curve should pass through the pointin the xy-plane. An ODE, together with an initial condition, is called an initial valueproblem. Thus, if the ODE is explicit, the initial value problem is ofthe form(5)E X A M P L E 4 Initial Value ProblemSolve the initial value problemSolution. The general solution is ; see Example 3. From this solution and the initial conditionwe obtain Hence the initial value problem has the solution . This is aparticular solution.More on ModelingThe general importance of modeling to the engineer and physicist was emphasized at thebeginning of this section. We shall now consider a basic physical problem that will showthe details of the typical steps of modeling. Step 1: the transition from the physical situation(the physical system) to its mathematical formulation (its mathematical model); Step 2:the solution by a mathematical method; and Step 3: the physical interpretation of the result.This may be the easiest way to obtain a first idea of the nature and purpose of differentialequations and their applications. Realize at the outset that your computer (your CAS)may perhaps give you a hand in Step 2, but Steps 1 and 3 are basically your work.᭿y(x) ϭ 5.7e3xy(0) ϭ ce0ϭ c ϭ 5.7.y(x) ϭ ce3xy(0) ϭ 5.7.yr ϭdydxϭ 3y,y(x0) ϭ y0.yr ϭ f(x, y),yr ϭ f(x, y),(x0, y0)y0x0y(x0) ϭ y0,Ϫ2.01c ϭ 6.456 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:14 PM Page 6 And Step 2 requires a solid knowledge and good understanding of solution methodsavailable to you—you have to choose the method for your work by hand or by thecomputer. Keep this in mind, and always check computer results for errors (which mayarise, for instance, from false inputs).E X A M P L E 5 Radioactivity. Exponential DecayGiven an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.Physical Information. Experiments show that at each instant a radioactive substance decomposes—and is thusdecaying in time—proportional to the amount of substance present.Step 1. Setting up a mathematical model of the physical process. Denote by the amount of substance stillpresent at any time t. By the physical law, the time rate of change is proportional to . Thisgives the first-order ODE(6)where the constant k is positive, so that, because of the minus, we do get decay (as in [B] of Example 3).The value of k is known from experiments for various radioactive substances (e.g.,approximately, for radium ).Now the given initial amount is 0.5 g, and we can call the corresponding instant Then we have theinitial condition This is the instant at which our observation of the process begins. It motivatesthe term initial condition (which, however, is also used when the independent variable is not time or whenwe choose a t other than ). Hence the mathematical model of the physical process is the initial valueproblem(7)Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decayand has the general solution (with arbitrary constant c but definite given k)(8)We now determine c by using the initial condition. Since from (8), this gives Hencethe particular solution governing our process is (cf. Fig. 5)(9)Always check your result—it may involve human or computer errors! Verify by differentiation (chain rule!)that your solution (9) satisfies (7) as well asStep 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts fromthe correct initial amount and decreases with time because k is positive. The limit of y as is zero. ᭿t : ϱdydtϭ Ϫ0.5ke؊ktϭ Ϫk ؒ 0.5e؊ktϭ Ϫky, y(0) ϭ 0.5e0ϭ 0.5.y(0) ϭ 0.5:(k Ͼ 0).y(t) ϭ 0.5e؊kty(0) ϭ c ϭ 0.5.y(0) ϭ cy(t) ϭ ce؊kt.dydtϭ Ϫky, y(0) ϭ 0.5.t ϭ 0y(0) ϭ 0.5.t ϭ 0.22688Rak ϭ 1.4 ؒ 10؊11sec؊1,dydtϭ Ϫkyy(t)yr(t) ϭ dy>dty(t)SEC. 1.1 Basic Concepts. Modeling 70.10.20.30.40.50y0 0.5 1.5 2 2.5 31 tFig. 5. Radioactivity (Exponential decay,with as an example)k ϭ 1.5y ϭ 0.5eϪkt,c01.qxd 7/30/10 8:14 PM Page 7 1.2 Geometric Meaning ofDirection Fields, Euler's MethodA first-order ODE(1)has a simple geometric interpretation. From calculus you know that the derivative ofis the slope of . Hence a solution curve of (1) that passes through a pointmust have, at that point, the slope equal to the value of f at that point; that is,Using this fact, we can develop graphic or numeric methods for obtaining approximatesolutions of ODEs (1). This will lead to a better conceptual understanding of an ODE (1).Moreover, such methods are of practical importance since many ODEs have complicatedsolution formulas or no solution formulas at all, whereby numeric methods are needed.Graphic Method of Direction Fields. Practical Example Illustrated in Fig. 7. Wecan show directions of solution curves of a given ODE (1) by drawing short straight-linesegments (lineal elements) in the xy-plane. This gives a direction field (or slope field)into which you can then fit (approximate) solution curves. This may reveal typicalproperties of the whole family of solutions.Figure 7 shows a direction field for the ODE(2)obtained by a CAS (Computer Algebra System) and some approximate solution curvesfitted in.yr ϭ y ϩ xyr(x0) ϭ f(x0, y0).yr(x0)(x0, y0)y(x)y(x)yr(x)yr ϭ f(x, y)yr ϭ f(x, y).SEC. 1.2 Geometric Meaning of yЈ ϭ ƒ(x, y). Direction Fields, Euler's Method 9120.5 1–0.5–1–1.5–2–1–2yxFig. 7. Direction field of with three approximate solutioncurves passing through (0, 1), (0, 0), (0, ), respectivelyϪ1yr ϭ y ϩ x,c01.qxd 7/30/10 8:15 PM Page 9 If you have no CAS, first draw a few level curves const of , then parallellineal elements along each such curve (which is also called an isocline, meaning a curveof equal inclination), and finally draw approximation curves fit to the lineal elements.We shall now illustrate how numeric methods work by applying the simplest numericmethod, that is Euler's method, to an initial value problem involving ODE (2). First wegive a brief description of Euler's method.Numeric Method by EulerGiven an ODE (1) and an initial value Euler's method yields approximatesolution values at equidistant x-values namely,(Fig. 8), etc.In general,where the step h equals, e.g., 0.1 or 0.2 (as in Table 1.1) or a smaller value for greateraccuracy.yn ϭ ynϪ1 ϩ hf(xnϪ1, ynϪ1)y2 ϭ y1 ϩ hf(x1, y1)y1 ϭ y0 ϩ hf(x0, y0)x0, x1 ϭ x0 ϩ h, x2 ϭ x0 ϩ 2h, Á ,y(x0) ϭ y0,f(x, y)f(x, y) ϭ10 CHAP. 1 First-Order ODEsyxx0x1y0y1y(x1)Solution curveError of y1hf(x0, y0)hFig. 8. First Euler step, showing a solution curve, its tangent at ( ),step h and increment in the formula for y1hf(x0, y0)x0, y0Table 1.1 shows the computation of steps with step for the ODE (2) andinitial condition corresponding to the middle curve in the direction field. Weshall solve the ODE exactly in Sec. 1.5. For the time being, verify that the initial valueproblem has the solution . The solution curve and the values in Table 1.1are shown in Fig. 9. These values are rather inaccurate. The errors are shownin Table 1.1 as well as in Fig. 9. Decreasing h would improve the values, but would soonrequire an impractical amount of computation. Much better methods of a similar naturewill be discussed in Sec. 21.1.y(xn) Ϫ yny ϭ exϪ x Ϫ 1y(0) ϭ 0,h ϭ 0.2n ϭ 5c01.qxd 7/30/10 8:15 PM Page 10 1.3 Separable ODEs. ModelingMany practically useful ODEs can be reduced to the form(1)by purely algebraic manipulations. Then we can integrate on both sides with respect to x,obtaining(2)On the left we can switch to y as the variable of integration. By calculus, , so that(3)If f and g are continuous functions, the integrals in (3) exist, and by evaluating them weobtain a general solution of (1). This method of solving ODEs is called the method ofseparating variables, and (1) is called a separable equation, because in (3) the variablesare now separated: x appears only on the right and y only on the left.E X A M P L E 1 Separable ODEThe ODE is separable because it can be writtenBy integration, or .It is very important to introduce the constant of integration immediately when the integration is performed.If we wrote then and then introduced c, we would have obtained whichis not a solution (when ). Verify this. ᭿c 0y ϭ tan x ϩ c,y ϭ tan x,arctan y ϭ x,y ϭ tan (x ϩ c)arctan y ϭ x ϩ cdy1 ϩ y2ϭ dx.yr ϭ 1 ϩ y2Ύg(y) dy ϭ Ύf(x) dx ϩ c.yrdx ϭ dyΎg(y) yrdx ϭ Ύf(x) dx ϩ c.g(y) yr ϭ f(x)12 CHAP. 1 First-Order ODEs16. CAS PROJECT. Direction Fields. Discuss directionfields as follows.(a) Graph portions of the direction field of the ODE (2)(see Fig. 7), for instance,Explain what you have gained by this enlargement ofthe portion of the field.(b) Using implicit differentiation, find an ODE withthe general solution Graph itsdirection field. Does the field give the impressionthat the solution curves may be semi-ellipses? Can youdo similar work for circles? Hyperbolas? Parabolas?Other curves?(c) Make a conjecture about the solutions offrom the direction field.(d) Graph the direction field of and somesolutions of your choice. How do they behave? Whydo they decrease for ?y Ͼ 0yr ϭ Ϫ12 yyr ϭ Ϫx>yx2ϩ 9y2ϭ c (y Ͼ 0).Ϫ5 Ϲ x Ϲ 2, Ϫ1 Ϲ y Ϲ 5.17–20 EULER'S METHODThis is the simplest method to explain numerically solvingan ODE, more precisely, an initial value problem (IVP).(More accurate methods based on the same principle areexplained in Sec. 21.1.) Using the method, to get a feel fornumerics as well as for the nature of IVPs, solve the IVPnumerically with a PC or a calculator, 10 steps. Graph thecomputed values and the solution curve on the samecoordinate axes.17.18.19.Sol.20.Sol. y ϭ 1>(1 ϩ x)5yr ϭ Ϫ5x4y2, y(0) ϭ 1, h ϭ 0.2y ϭ x Ϫ tanh xyr ϭ (y Ϫ x)2, y(0) ϭ 0, h ϭ 0.1yr ϭ y, y(0) ϭ 1, h ϭ 0.01yr ϭ y, y(0) ϭ 1, h ϭ 0.1c01.qxd 7/30/10 8:15 PM Page 12 E X A M P L E 2 Separable ODEThe ODE is separable; we obtainE X A M P L E 3 Initial Value Problem (IVP). Bell-Shaped CurveSolveSolution. By separation and integration,This is the general solution. From it and the initial condition, Hence the IVP has thesolution This is a particular solution, representing a bell-shaped curve (Fig. 10). ᭿y ϭ 1.8e؊x2.y(0) ϭ ce0ϭ c ϭ 1.8.dyyϭ Ϫ2x dx, ln y ϭ Ϫx2ϩ cෂ, y ϭ ce؊x2.yr ϭ Ϫ2xy, y(0) ϭ 1.8.᭿By integration, Ϫy؊1ϭ Ϫ(x ϩ 2)e؊xϩ c, y ϭ1(x ϩ 2)eϪxϪ c.y؊2dy ϭ (x ϩ 1)e؊xdx.yr ϭ (x ϩ 1)e؊xy2SEC. 1.3 Separable ODEs. Modeling 13110–1–2 2 xyFig. 10. Solution in Example 3 (bell-shaped curve)ModelingThe importance of modeling was emphasized in Sec. 1.1, and separable equations yieldvarious useful models. Let us discuss this in terms of some typical examples.E X A M P L E 4 Radiocarbon Dating2In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found inthe ice of the Oetztal Alps (hence the name "Oetzi") in Southern Tyrolia near the Austrian–Italian border, causeda scientific sensation. When did Oetzi approximately live and die if the ratio of carbon to carbon inthis mummy is 52.5% of that of a living organism?Physical Information. In the atmosphere and in living organisms, the ratio of radioactive carbon (maderadioactive by cosmic rays) to ordinary carbon is constant. When an organism dies, its absorption ofby breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactivecarbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of , whichis 5715 years (CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002, page 11–52,line 9).Solution. Modeling. Radioactive decay is governed by the ODE (see Sec. 1.1, Example 5). Byseparation and integration (where t is time and is the initial ratio of to )(y0 ϭ ec).y ϭ y0 ektln ƒ yƒ ϭ kt ϩ c,dyyϭ k dt,126C146Cy0yr ϭ ky146C146C126C146C126C146C2Method by WILLARD FRANK LIBBY (1908–1980), American chemist, who was awarded for this workthe 1960 Nobel Prize in chemistry.c01.qxd 7/30/10 8:15 PM Page 13 Next we use the half-life to determine k. When , half of the original substance is still present. Thus,Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed),Answer: About 5300 years ago.Other methods show that radiocarbon dating values are usually too small. According to recent research, this isdue to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing.E X A M P L E 5 Mixing ProblemMixing problems occur quite frequently in chemical industry. We explain here how to solve the basic modelinvolving a single tank. The tank in Fig. 11 contains 1000 gal of water in which initially 100 lb of salt is dissolved.Brine runs in at a rate of 10 gal min, and each gallon contains 5 lb of dissoved salt. The mixture in the tank iskept uniform by stirring. Brine runs out at 10 gal min. Find the amount of salt in the tank at any time t.Solution. Step 1. Setting up a model. Let denote the amount of salt in the tank at time t. Its time rateof change isBalance law.5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This isof the total brine content in the tank, hence 0.01 of the salt content , that is, 0.01 . Thus themodel is the ODE(4)Step 2. Solution of the model. The ODE (4) is separable. Separation, integration, and taking exponents on bothsides givesInitially the tank contains 100 lb of salt. Hence is the initial condition that will give the uniquesolution. Substituting and in the last equation gives HenceHence the amount of salt in the tank at time t is(5)This function shows an exponential approach to the limit 5000 lb; see Fig. 11. Can you explain physically thatshould increase with time? That its limit is 5000 lb? Can you see the limit directly from the ODE?The model discussed becomes more realistic in problems on pollutants in lakes (see Problem Set 1.5, Prob. 35)or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flowrates (in and out) may be different and known only very roughly. ᭿y(t)y(t) ϭ 5000 Ϫ 4900e؊0.01t.c ϭ Ϫ4900.100 Ϫ 5000 ϭ ce0ϭ c.t ϭ 0y ϭ 100y(0) ϭ 100y Ϫ 5000 ϭ ce؊0.01t.ln ƒy Ϫ 5000ƒ ϭ Ϫ0.01t ϩ c,dyy Ϫ 5000ϭ Ϫ0.01 dt,yr ϭ 50 Ϫ 0.01y ϭ Ϫ0.01(y Ϫ 5000).y(t)y(t)(ϭ 1%)10>1000 ϭ 0.01yr ϭ Salt inflow rate Ϫ Salt outflow ratey(t)>>᭿t ϭln 0.525Ϫ0.0001213ϭ 5312.ektϭ e؊0.0001213tϭ 0.525,k ϭln 0.5Hϭ Ϫ0.6935715ϭ Ϫ0.0001213.ekHϭ 0.5,y0ekHϭ 0.5y0,t ϭ HH ϭ 571514 CHAP. 1 First-Order ODEs100200030001000500040001000 300200 400 500Salt content y(t)tTankTankyFig. 11. Mixing problem in Example 5c01.qxd 7/30/10 8:15 PM Page 14 E X A M P L E 6 Heating an Office Building (Newton's Law of Cooling3)Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The heatingis shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M.was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 A.M. Whatwas the temperature inside the building when the heat was turned on at 6 A.M.?Physical information. Experiments show that the time rate of change of the temperature T of a body B (whichconducts heat well, for example, as a copper ball does) is proportional to the difference between T and thetemperature of the surrounding medium (Newton's law of cooling).Solution. Step 1. Setting up a model. Let be the temperature inside the building and TA the outsidetemperature (assumed to be constant in Newton's law). Then by Newton's law,(6)Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if amodel seems to fit the reality only poorly (as in the present case), it may still give valuable qualitative information.To see how good a model is, the engineer will collect experimental data and compare them with calculationsfrom the model.Step 2. General solution. We cannot solve (6) because we do not know TA, just that it varied between 50°Fand 40°F, so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one. Wesolve (6) with the unknown function TA replaced with the average of the two known values, or 45°F. For physicalreasons we may expect that this will give us a reasonable approximate value of T in the building at 6 A.M.For constant (or any other constant value) the ODE (6) is separable. Separation, integration, andtaking exponents gives the general solutionStep 3. Particular solution. We choose 10 P.M. to be Then the given initial condition is andyields a particular solution, call it . By substitution,Step 4. Determination of k. We use where is 2 A.M. Solving algebraically for k and insertingk into gives (Fig. 12)Tp(t) ϭ 45 ϩ 25e؊0.056t.k ϭ 14 ln 0.8 ϭ Ϫ0.056,e4kϭ 0.8,Tp(4) ϭ 45 ϩ 25e4kϭ 65,Tp(t)t ϭ 4T(4) ϭ 65,Tp(t) ϭ 45 ϩ 25ekt.c ϭ 70 Ϫ 45 ϭ 25,T(0) ϭ 45 ϩ ce0ϭ 70,TpT(0) ϭ 70t ϭ 0.(c ϭ ec).T(t) ϭ 45 ϩ cektln ƒ T Ϫ 45 ƒ ϭ kt ϩ c,dTT Ϫ 45ϭ k dt,TA ϭ 45dTdtϭ k(T Ϫ TA).T(t)SEC. 1.3 Separable ODEs. Modeling 156264687060y2 4 6 80 t666165Fig. 12. Particular solution (temperature) in Example 63Sir ISAAC NEWTON (1642–1727), great English physicist and mathematician, became a professor atCambridge in 1669 and Master of the Mint in 1699. He and the German mathematician and philosopherGOTTFRIED WILHELM LEIBNIZ (1646–1716) invented (independently) the differential and integral calculus.Newton discovered many basic physical laws and created the method of investigating physical problems bymeans of calculus. His Philosophiae naturalis principia mathematica (Mathematical Principles of NaturalPhilosophy, 1687) contains the development of classical mechanics. His work is of greatest importance to bothmathematics and physics.c01.qxd 7/30/10 8:15 PM Page 15 Step 5. Answer and interpretation. 6 A.M. is (namely, 8 hours after 10 P.M.), andHence the temperature in the building dropped 9°F, a result that looks reasonable.E X A M P L E 7 Leaking Tank. Outflow of Water Through a Hole (Torricelli's Law)This is another prototype engineering problem that leads to an ODE. It concerns the outflow of water from acylindrical tank with a hole at the bottom (Fig. 13). You are asked to find the height of the water in the tank atany time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when thehole is opened is 2.25 m. When will the tank be empty?Physical information. Under the influence of gravity the outflowing water has velocity(7) (Torricelli's law4),where is the height of the water above the hole at time t, and is theacceleration of gravity at the surface of the earth.Solution. Step 1. Setting up the model. To get an equation, we relate the decrease in water level to theoutflow. The volume of the outflow during a short time is(A Area of hole).must equal the change of the volume of the water in the tank. Now(B Cross-sectional area of tank)where is the decrease of the height of the water. The minus sign appears because the volume ofthe water in the tank decreases. Equating and givesWe now express v according to Torricelli's law and then let (the length of the time interval considered)approach 0—this is a standard way of obtaining an ODE as a model. That is, we haveand by letting we obtain the ODE,where This is our model, a first-order ODE.Step 2. General solution. Our ODE is separable. is constant. Separation and integration givesandDividing by 2 and squaring gives . Insertingyields the general solutionh(t) ϭ (c Ϫ 0.000332t)2.13.28A>B ϭ 13.28 ؒ 0.52p>1002p ϭ 0.000332h ϭ (c Ϫ 13.28At>B)221h ϭ c Ϫ 26.56ABt.dh1hϭ Ϫ26.56ABdtA>B26.56 ϭ 0.60022 ؒ 980.dhdtϭ Ϫ26.56AB1h¢t : 0¢h¢tϭ ϪABv ϭ ϪAB0.60012gh(t)¢tϪB ¢h ϭ Av ¢t.¢V¢Vh(t)¢h (Ͼ 0)ϭ¢V ϭ ϪB ¢h¢V¢Vϭ¢V ϭ Av ¢t¢t¢Vh(t)g ϭ 980 cm>sec2ϭ 32.17 ft>sec2h(t)v(t) ϭ 0.60022gh(t)᭿Tp(8) ϭ 45 ϩ 25e؊0.056 ؒ 8ϭ 613°F4.t ϭ 816 CHAP. 1 First-Order ODEs4EVANGELISTA TORRICELLI (1608–1647), Italian physicist, pupil and successor of GALILEO GALILEI(1564–1642) at Florence. The "contraction factor" 0.600 was introduced by J. C. BORDA in 1766 because thestream has a smaller cross section than the area of the hole.c01.qxd 7/30/10 8:15 PM Page 16 Step 3. Particular solution. The initial height (the initial condition) is cm. Substitution ofand gives from the general solution and thus the particular solution (Fig. 13)Step 4. Tank empty. if [hours].Here you see distinctly the importance of the choice of units—we have been working with the cgs system,in which time is measured in seconds! We usedStep 5. Checking. Check the result. ᭿g ϭ 980 cm>sec2.t ϭ 15.00>0.000332 ϭ 45,181 c sec d ϭ 12.6hp(t) ϭ 0hp(t) ϭ (15.00 Ϫ 0.000332t)2.c2ϭ 225, c ϭ 15.00h ϭ 225t ϭ 0h(0) ϭ 225SEC. 1.3 Separable ODEs. Modeling 172.25 m2.00 mh(t)OutflowingwaterWater levelat time tht250200150100500100000 30000 50000Tank Water level h(t) in tankFig. 13. Example 7. Outflow from a cylindrical tank ("leaking tank").Torricelli's lawExtended Method: Reduction to Separable FormCertain nonseparable ODEs can be made separable by transformations that introduce fory a new unknown function. We discuss this technique for a class of ODEs of practicalimportance, namely, for equations(8)Here, f is any (differentiable) function of , such as sin , , and so on. (Suchan ODE is sometimes called a homogeneous ODE, a term we shall not use but reservefor a more important purpose in Sec. 1.5.)The form of such an ODE suggests that we set ; thus,(9) and by product differentiationSubstitution into then gives or . We see thatif , this can be separated:(10)duf(u) Ϫ uϭdxx.f(u) Ϫ u 0urx ϭ f(u) Ϫ uurx ϩ u ϭ f(u)yr ϭ f(y>x)yr ϭ urx ϩ u.y ϭ uxy>x ϭ u(y>x)4(y>x)y>xyr ϭ f ayxb.c01.qxd 7/30/10 8:15 PM Page 17 19–36 MODELING, APPLICATIONS19. Exponential growth. If the growth rate of the numberof bacteria at any time t is proportional to the numberpresent at t and doubles in 1 week, how many bacteriacan be expected after 2 weeks? After 4 weeks?20. Another population model.(a) If the birth rate and death rate of the number ofbacteria are proportional to the number of bacteriapresent, what is the population as a function of time.(b) What is the limiting situation for increasing time?Interpret it.21. Radiocarbon dating. What should be the content(in percent of ) of a fossilized tree that is claimed tobe 3000 years old? (See Example 4.)22. Linear accelerators are used in physics foraccelerating charged particles. Suppose that an alphaparticle enters an accelerator and undergoes a constantacceleration that increases the speed of the particlefrom to sec. Find theacceleration a and the distance traveled during thatperiod of sec.23. Boyle–Mariotte's law for ideal gases.5Experimentsshow for a gas at low pressure p (and constanttemperature) the rate of change of the volumeequals . Solve the model.24. Mixing problem. A tank contains 400 gal of brinein which 100 lb of salt are dissolved. Fresh water runsinto the tank at a rate of The mixture, keptpractically uniform by stirring, runs out at the samerate. How much salt will there be in the tank at theend of 1 hour?25. Newton's law of cooling. A thermometer, reading5°C, is brought into a room whose temperature is 22°C.One minute later the thermometer reading is 12°C.How long does it take until the reading is practically22°C, say, 21.9°C?26. Gompertz growth in tumors. The Gompertz modelis , where is the mass oftumor cells at time t. The model agrees well withclinical observations. The declining growth rate withincreasing corresponds to the fact that cells inthe interior of a tumor may die because of insufficientoxygen and nutrients. Use the ODE to discuss thegrowth and decline of solutions (tumors) and to findconstant solutions. Then solve the ODE.27. Dryer. If a wet sheet in a dryer loses its moisture ata rate proportional to its moisture content, and if itloses half of its moisture during the first 10 min ofy Ͼ 1y(t)yr ϭ ϪAy ln y (A Ͼ 0)2 gal>min.ϪV>pV(p)10؊3104m>sec in 10؊3103m>secy0146CSEC. 1.3 Separable ODEs. Modeling 19drying, when will it be practically dry, say, when willit have lost 99% of its moisture? First guess, thencalculate.28. Estimation. Could you see, practically without calcu-lation, that the answer in Prob. 27 must lie between60 and 70 min? Explain.29. Alibi? Jack, arrested when leaving a bar, claims thathe has been inside for at least half an hour (whichwould provide him with an alibi). The police checkthe water temperature of his car (parked near theentrance of the bar) at the instant of arrest and again30 min later, obtaining the values 190°F and 110°F,respectively. Do these results give Jack an alibi?(Solve by inspection.)30. Rocket. A rocket is shot straight up from the earth,with a net acceleration ( acceleration by the rocketengine minus gravitational pullback) ofduring the initial stage of flight until the engine cut outat sec. How high will it go, air resistanceneglected?31. Solution curves of Show that any(nonvertical) straight line through the origin of thexy-plane intersects all these curves of a given ODE atthe same angle.32. Friction. If a body slides on a surface, it experiencesfriction F (a force against the direction of motion).Experiments show that (Coulomb's6law ofkinetic friction without lubrication), where N is thenormal force (force that holds the two surfaces together;see Fig. 15) and the constant of proportionality iscalled the coefficient of kinetic friction. In Fig. 15assume that the body weighs 45 nt (about 10 lb; seefront cover for conversion). (correspondingto steel on steel), the slide is 10 m long, theinitial velocity is zero, and air resistance isnegligible. Find the velocity of the body at the endof the slide.a ϭ 30°,␮ ϭ 0.20␮ƒFƒ ϭ ␮ƒNƒyr ‫؍‬ g1y>x2.t ϭ 107t m>sec2ϭ5ROBERT BOYLE (1627–1691), English physicist and chemist, one of the founders of the Royal Society. EDME MARIOTTE (about1620–1684), French physicist and prior of a monastry near Dijon. They found the law experimentally in 1662 and 1676, respectively.6CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer.v(t)WNBodyαs(t)Fig. 15. Problem 32c01.qxd 7/30/10 8:15 PM Page 19 33. Rope. To tie a boat in a harbor, how many timesmust a rope be wound around a bollard (a verticalrough cylindrical post fixed on the ground) so that aman holding one end of the rope can resist a forceexerted by the boat 1000 times greater than the mancan exert? First guess. Experiments show that thechange of the force S in a small portion of therope is proportional to S and to the small anglein Fig. 16. Take the proportionality constant 0.15.The result should surprise you!¢␾¢S20 CHAP. 1 First-Order ODEsthis as the condition for the two families to beorthogonal (i.e., to intersect at right angles)? Do yourgraphs confirm this?(e) Sketch families of curves of your own choice andfind their ODEs. Can every family of curves be givenby an ODE?35. CAS PROJECT. Graphing Solutions. A CAS canusually graph solutions, even if they are integrals thatcannot be evaluated by the usual analytical methods ofcalculus.(a) Show this for the five initial value problems, , graphing all five curveson the same axes.(b) Graph approximate solution curves, using the firstfew terms of the Maclaurin series (obtained by term-wise integration of that of ) and compare with theexact curves.(c) Repeat the work in (a) for another ODE and initialconditions of your own choice, leading to an integralthat cannot be evaluated as indicated.36. TEAM PROJECT. Torricelli's Law. Suppose thatthe tank in Example 7 is hemispherical, of radius R,initially full of water, and has an outlet of 5 cm2cross-sectional area at the bottom. (Make a sketch.) Setup the model for outflow. Indicate what portion ofyour work in Example 7 you can use (so that it canbecome part of the general method independent of theshape of the tank). Find the time t to empty the tank(a) for any R, (b) for Plot t as function ofR. Find the time when (a) for any R, (b) forR ϭ 1 m.h ϭ R>2R ϭ 1 m.yry(0) ϭ 0, Ϯ1, Ϯ2yr ϭ e؊x2S + ΔSΔ␾SSmallportionof ropeFig. 16. Problem 3334. TEAM PROJECT. Family of Curves. A family ofcurves can often be characterized as the generalsolution of(a) Show that for the circles with center at the originwe get(b) Graph some of the hyperbolas Find anODE for them.(c) Find an ODE for the straight lines through theorigin.(d) You will see that the product of the right sides ofthe ODEs in (a) and (c) equals Do you recognizeϪ1.xy ϭ c.yr ϭ Ϫx>y.yr ϭ f(x, y).1.4 Exact ODEs. Integrating FactorsWe recall from calculus that if a function has continuous partial derivatives, itsdifferential (also called its total differential) isFrom this it follows that if thenFor example, if , thenoryr ϭdydxϭ Ϫ1 ϩ 2xy33x2y2,du ϭ (1 ϩ 2xy3) dx ϩ 3x2y2dy ϭ 0u ϭ x ϩ x2y3ϭ cdu ϭ 0.u(x, y) ϭ c ϭ const,du ϭ0u0xdx ϩ0u0ydy.u(x, y)c01.qxd 7/30/10 8:15 PM Page 20 an ODE that we can solve by going backward. This idea leads to a powerful solutionmethod as follows.A first-order ODE written as (use as in Sec. 1.3)(1)is called an exact differential equation if the differential formis exact, that is, this form is the differential(2)of some function . Then (1) can be writtenBy integration we immediately obtain the general solution of (1) in the form(3)This is called an implicit solution, in contrast to a solution as defined in Sec.1.1, which is also called an explicit solution, for distinction. Sometimes an implicit solutioncan be converted to explicit form. (Do this for ) If this is not possible, yourCAS may graph a figure of the contour lines (3) of the function and help you inunderstanding the solution.Comparing (1) and (2), we see that (1) is an exact differential equation if there is somefunction such that(4) (a) (b)From this we can derive a formula for checking whether (1) is exact or not, as follows.Let M and N be continuous and have continuous first partial derivatives in a region inthe xy-plane whose boundary is a closed curve without self-intersections. Then by partialdifferentiation of (4) (see App. 3.2 for notation),By the assumption of continuity the two second partial derivaties are equal. Thus(5)0M0yϭ0N0x.0N0xϭ02u0x 0y.0M0yϭ02u0y 0x,0u0yϭ N.0u0xϭ M,u(x, y)u(x, y)x2ϩ y2ϭ 1.y ϭ h(x)u(x, y) ϭ c.du ϭ 0.u(x, y)du ϭ0u0xdx ϩ0u0ydyM(x, y) dx ϩ N(x, y) dyM(x, y) dx ϩ N(x, y) dy ϭ 0dy ϭ yrdxM(x, y) ϩ N(x, y)yr ϭ 0,SEC. 1.4 Exact ODEs. Integrating Factors 21c01.qxd 7/30/10 8:15 PM Page 21 This condition is not only necessary but also sufficient for (1) to be an exact differentialequation. (We shall prove this in Sec. 10.2 in another context. Some calculus books, forinstance, [GenRef 12], also contain a proof.)If (1) is exact, the function can be found by inspection or in the followingsystematic way. From (4a) we have by integration with respect to x(6)in this integration, y is to be regarded as a constant, and plays the role of a "constant"of integration. To determine , we derive from (6), use (4b) to get , andintegrate to get k. (See Example 1, below.)Formula (6) was obtained from (4a). Instead of (4a) we may equally well use (4b).Then, instead of (6), we first have by integration with respect to y(6)To determine , we derive from (6), use (4a) to get , and integrate. Weillustrate all this by the following typical examples.E X A M P L E 1 An Exact ODESolve(7)Solution. Step 1. Test for exactness. Our equation is of the form (1) withThusFrom this and (5) we see that (7) is exact.Step 2. Implicit general solution. From (6) we obtain by integration(8)To find , we differentiate this formula with respect to y and use formula (4b), obtainingHence By integration, Inserting this result into (8) and observing (3),we obtain the answeru(x, y) ϭ sin (x ϩ y) ϩ y3ϩ y2ϭ c.k ϭ y3ϩ y2ϩ c.dk>dy ϭ 3y2ϩ 2y.0u0yϭ cos (x ϩ y) ϩdkdyϭ N ϭ 3y2ϩ 2y ϩ cos (x ϩ y).k(y)u ϭ ΎM dx ϩ k(y) ϭ Ύcos (x ϩ y) dx ϩ k(y) ϭ sin (x ϩ y) ϩ k(y).0N0xϭ Ϫsin (x ϩ y).0M0yϭ Ϫsin (x ϩ y),N ϭ 3y2ϩ 2y ϩ cos (x ϩ y).M ϭ cos (x ϩ y),cos (x ϩ y) dx ϩ (3y2ϩ 2y ϩ cos (x ϩ y)) dy ϭ 0.dl>dx0u>0xl(x)u ϭ ΎN dy ϩ l(x).dk>dydk>dy0u>0yk(y)k(y)u ϭ ΎM dx ϩ k(y);u(x, y)22 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:15 PM Page 22 This example gives the idea. All we did was to multiply a given nonexact equation, say,(12)by a function F that, in general, will be a function of both x and y. The result was an equation(13)that is exact, so we can solve it as just discussed. Such a function is then calledan integrating factor of (12).E X A M P L E 4 Integrating FactorThe integrating factor in (11) is Hence in this case the exact equation (13) isSolutionThese are straight lines through the origin. (Note that is also a solution of )It is remarkable that we can readily find other integrating factors for the equation namely,and because(14)How to Find Integrating FactorsIn simpler cases we may find integrating factors by inspection or perhaps after some trials,keeping (14) in mind. In the general case, the idea is the following.For the exactness condition (5) is Hence for (13),the exactness condition is(15)By the product rule, with subscripts denoting partial derivatives, this givesIn the general case, this would be complicated and useless. So we follow the Golden Rule:If you cannot solve your problem, try to solve a simpler one—the result may be useful(and may also help you later on). Hence we look for an integrating factor depending onlyon one variable: fortunately, in many practical cases, there are such factors, as we shallsee. Thus, let Then and so that (15) becomesDividing by FQ and reshuffling terms, we have(16) where R ϭ1Qa0P0yϪ0Q0xb.1FdFdxϭ R,FPy ϭ FrQ ϩ FQx.Fx ϭ Fr ϭ dF>dx,Fy ϭ 0,F ϭ F(x).FyP ϩ FPy ϭ FxQ ϩ FQx.00y(FP) ϭ00x(FQ).FP dx ϩ FQ dy ϭ 0,0M>0y ϭ 0N>0x.M dx ϩ N dy ϭ 0᭿Ϫy dx ϩ x dyx2ϩ y2ϭ d aarctanyxb.Ϫy dx ϩ x dyxyϭ Ϫd alnxyb,Ϫy dx ϩ x dyy2ϭ d axyb,1>(x2ϩ y2),1>y2, 1>(xy),Ϫy dx ϩ x dy ϭ 0,Ϫy dx ϩ x dy ϭ 0.x ϭ 0y ϭ cxyxϭ c.FP dx ϩ FQ dy ϭϪy dx ϩ x dyx2ϭ d ayxb ϭ 0.F ϭ 1>x2.F(x, y)FP dx ϩ FQ dy ϭ 0P(x, y) dx ϩ Q(x, y) dy ϭ 0,24 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:15 PM Page 24 16. TEAM PROJECT. Solution by Several Methods.Show this as indicated. Compare the amount of work.(a) as an exact ODEand by separation.(b) by Theorem 2and by separation.(c) by Theorem 1 or 2 andby separation with(d) by Theorems 1 and 2 andby separation.(e) Search the text and the problems for further ODEsthat can be solved by more than one of the methodsdiscussed so far. Make a list of these ODEs. Findfurther cases of your own.17. WRITING PROJECT. Working Backward.Working backward from the solution to the problemis useful in many areas. Euler, Lagrange, and othergreat masters did it. To get additional insight intothe idea of integrating factors, start from a ofyour choice, find destroy exactness bydivision by some and see what ODE'ssolvable by integrating factors you can get. Can youproceed systematically, beginning with the simplestF(x, y)?F(x, y),du ϭ 0,u(x, y)3x2y dx ϩ 4x3dy ϭ 0v ϭ y>x.(x2ϩ y2) dx Ϫ 2xy dy ϭ 0(1 ϩ 2x) cos y dx ϩ dy>cos y ϭ 0ey(sinh x dx ϩ cosh x dy) ϭ 0SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 27yx04π2π–1–2–31233ππParticular solutions in CAS Project 1818. CAS PROJECT. Graphing Particular Solutions.Graph particular solutions of the following ODE,proceeding as explained.(21)(a) Show that (21) is not exact. Find an integratingfactor using either Theorem 1 or 2. Solve (21).(b) Solve (21) by separating variables. Is this simplerthan (a)?(c) Graph the seven particular solutions satisfying thefollowing initial conditions(see figure below).(d) Which solution of (21) do we not get in (a) or (b)?Ϯ23, Ϯ1y(p>2) ϭ Ϯ12,y(0) ϭ 1,dy Ϫ y2sin x dx ϭ 0.1.5 Linear ODEs. Bernoulli Equation.Population DynamicsLinear ODEs or ODEs that can be transformed to linear form are models of variousphenomena, for instance, in physics, biology, population dynamics, and ecology, as weshall see. A first-order ODE is said to be linear if it can be brought into the form(1)by algebra, and nonlinear if it cannot be brought into this form.The defining feature of the linear ODE (1) is that it is linear in both the unknownfunction y and its derivative whereas p and r may be any given functions ofx. If in an application the independent variable is time, we write t instead of x.If the first term is (instead of ), divide the equation by to get the standardform (1), with as the first term, which is practical.For instance, is a linear ODE, and its standard form isThe function on the right may be a force, and the solution a displacement ina motion or an electrical current or some other physical quantity. In engineering, isfrequently called the input, and is called the output or the response to the input (and,if given, to the initial condition).y(x)r(x)y(x)r(x)yr ϩ y tan x ϭ x sec x.yr cos x ϩ y sin x ϭ xyrf(x)yrf(x)yryr ϭ dy>dx,yr ϩ p(x)y ϭ r(x),c01.qxd 7/30/10 8:15 PM Page 27 28 CHAP. 1 First-Order ODEsHomogeneous Linear ODE. We want to solve (1) in some interval callit J, and we begin with the simpler special case that is zero for all x in J. (This issometimes written ) Then the ODE (1) becomes(2)and is called homogeneous. By separating variables and integrating we then obtainthusTaking exponents on both sides, we obtain the general solution of the homogeneousODE (2),(3)here we may also choose and obtain the trivial solution for all x in thatinterval.Nonhomogeneous Linear ODE. We now solve (1) in the case that in (1) is noteverywhere zero in the interval J considered. Then the ODE (1) is called nonhomogeneous.It turns out that in this case, (1) has a pleasant property; namely, it has an integrating factordepending only on x. We can find this factor by Theorem 1 in the previous sectionor we can proceed directly, as follows. We multiply (1) by obtainingF(x),F(x)r(x)y(x) ϭ 0c ϭ 0(c ϭ Ϯecwhen y ѥ 0);y(x) ϭ ce؊͐p(x)dxln ƒy ƒ ϭ ϪΎp(x)dx ϩ c.dyyϭ Ϫp(x)dx,yr ϩ p(x)y ϭ 0r(x) ϵ 0.r(x)a Ͻ x Ͻ b,(1)The left side is the derivative of the product Fy ifBy separating variables, By integration, writingWith this F and Eq. (1) becomesBy integration,Dividing by we obtain the desired solution formula(4) y(x) ϭ e؊ha Ύehr dx ϩ cb, h ϭ Ύp(x) dx.eh,ehy ϭ Ύehr dx ϩ c.ehyr ϩ hrehy ϭ ehyr ϩ (eh)ry ϭ (ehy)r ϭ reh.hr ϭ p,ln ƒFƒ ϭ h ϭ Ύp dx, thus F ϭ eh.h ϭ ͐p dx,dF>F ϭ p dx.pFy ϭ Fry, thus pF ϭ Fr.(Fy)r ϭ Fry ϩ FyrFyr ϩ pFy ϭ rF.c01.qxd 7/30/10 8:15 PM Page 28 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 29This reduces solving (1) to the generally simpler task of evaluating integrals. For ODEsfor which this is still difficult, you may have to use a numeric method for integrals fromSec. 19.5 or for the ODE itself from Sec. 21.1. We mention that h has nothing to do within Sec. 1.1 and that the constant of integration in h does not matter; see Prob. 2.The structure of (4) is interesting. The only quantity depending on a given initialcondition is c. Accordingly, writing (4) as a sum of two terms,(4)we see the following:(5)E X A M P L E 1 First-Order ODE, General Solution, Initial Value ProblemSolve the initial value problemSolution. Here andFrom this we see that in (4),and the general solution of our equation isFrom this and the initial condition, thus and the solution of our initial value problemis Here 3 cos x is the response to the initial data, and is the response to theinput sin 2x.E X A M P L E 2 Electric CircuitModel the RL-circuit in Fig. 19 and solve the resulting ODE for the current A (amperes), where t istime. Assume that the circuit contains as an EMF (electromotive force) a battery of V (volts), whichis constant, a resistor of (ohms), and an inductor of H (henrys), and that the current is initiallyzero.Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm's law) anda voltage drop across the conductor, and the sum of these two voltage drops equals the EMF(Kirchhoff's Voltage Law, KVL).Remark. In general, KVL states that "The voltage (the electromotive force EMF) impressed on a closedloop is equal to the sum of the voltage drops across all the other elements of the loop." For Kirchoff's CurrentLaw (KCL) and historical information, see footnote 7 in Sec. 2.9.Solution. According to these laws the model of the RL-circuit is in standard form(6) Ir ϩRLI ϭE(t)L.LIr ϩ RI ϭ E(t),LIr ϭ L dI>dtL ϭ 0.1R ϭ 11 ⍀E ϭ 48E(t)I(t)᭿Ϫ2 cos2xy ϭ 3 cos x Ϫ 2 cos2x.c ϭ 31 ϭ c # 1 Ϫ 2 # 12;y(x) ϭ cos x a2 Ύsin x dx ϩ cb ϭ c cos x Ϫ 2 cos2x.ehr ϭ (sec x)(2 sin x cos x) ϭ 2 sin x,e؊hϭ cos x,ehϭ sec x,h ϭ Ύp dx ϭ Ύtan x dx ϭ ln ƒ sec xƒ.p ϭ tan x, r ϭ sin 2x ϭ 2 sin x cos x,y(0) ϭ 1.yr ϩ y tan x ϭ sin 2x,Total Output ϭ Response to the Input r ϩ Response to the Initial Data.y(x) ϭ e؊hΎehr dx ϩ ce؊h,h(x)c01.qxd 7/30/10 8:15 PM Page 29 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 31Fig. 20. Particular solution in Example 3010152025100 20005tyThe last term decreases to 0 as t increases, practically after a short time and regardless of c (that is, of the initialcondition). The other part of is called the steady-state solution because it consists of constant and periodicterms. The entire solution is called the transient-state solution because it models the transition from rest to thesteady state. These terms are used quite generally for physical and other systems whose behavior depends on time.Step 3. Particular solution. Setting in and choosing we havethusInserting this result into we obtain the particular solutionwith the steady-state part as before. To plot we must specify values for the constants, say,and Figure 20 shows this solution. Notice that the transition period is relatively short (althoughK is small), and the curve soon looks sinusoidal; this is the response to the input᭿1 ϩ cos ( 112 pt).A ϩ B cos ( 112 pt) ϭK ϭ 0.05.A ϭ B ϭ 1ypartypart(t) ϭAKϩBK2ϩ (p>12)2aK cospt12ϩp12sinpt12b Ϫ aAKϩKBK2ϩ (p>12)2b e؊Ky(t),c ϭ ϪAKϪKBK2ϩ (p>12)2.y(0) ϭAKϩBK2ϩ (p>12)2upK ϩ c ϭ 0,y0 ϭ 0,y(t)t ϭ 0y(t)ϭAKϩBK 2ϩ (p>12)2aK cospt12ϩp12sinpt12b ϩ ce؊Kt.ϭ e؊KteKtcAKϩBK 2ϩ v2aK cos vt ϩ v sin vtbd ϩ ce؊Kty(t) ϭ e؊KtΎeKtaA ϩ B cos vtb dt ϩ ce؊KtReduction to Linear Form. Bernoulli EquationNumerous applications can be modeled by ODEs that are nonlinear but can be transformedto linear ODEs. One of the most useful ones of these is the Bernoulli equation7(9) (a any real number).yr ϩ p(x)y ϭ g(x)ya7JAKOB BERNOULLI (1654–1705), Swiss mathematician, professor at Basel, also known for his contributionto elasticity theory and mathematical probability. The method for solving Bernoulli's equation was discovered byLeibniz in 1696. Jakob Bernoulli's students included his nephew NIKLAUS BERNOULLI (1687–1759), whocontributed to probability theory and infinite series, and his youngest brother JOHANN BERNOULLI (1667–1748),who had profound influence on the development of calculus, became Jakob's successor at Basel, and had amonghis students GABRIEL CRAMER (see Sec. 7.7) and LEONHARD EULER (see Sec. 2.5). His son DANIELBERNOULLI (1700–1782) is known for his basic work in fluid flow and the kinetic theory of gases.c01.qxd 7/30/10 8:15 PM Page 31 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 33Fig. 21. Logistic population model. Curves (9) in Example 4 with A>B ϭ 41 2 3 4Population yTime t20= 468ABPopulation DynamicsThe logistic equation (11) plays an important role in population dynamics, a fieldthat models the evolution of populations of plants, animals, or humans over time t.If then (11) is In this case its solution (12) isand gives exponential growth, as for a small population in a large country (theUnited States in early times!). This is called Malthus's law. (See also Example 3 inSec. 1.1.)The term in (11) is a "braking term" that prevents the population from growingwithout bound. Indeed, if we write we see that if thenso that an initially small population keeps growing as long as But ifthen and the population is decreasing as long as The limitis the same in both cases, namely, See Fig. 21.We see that in the logistic equation (11) the independent variable t does not occurexplicitly. An ODE in which t does not occur explicitly is of the form(13)and is called an autonomous ODE. Thus the logistic equation (11) is autonomous.Equation (13) has constant solutions, called equilibrium solutions or equilibriumpoints. These are determined by the zeros of because gives by(13); hence These zeros are known as critical points of (13). Anequilibrium solution is called stable if solutions close to it for some t remain closeto it for all further t. It is called unstable if solutions initially close to it do not remainclose to it as t increases. For instance, in Fig. 21 is an unstable equilibriumsolution, and is a stable one. Note that (11) has the critical points andE X A M P L E 5 Stable and Unstable Equilibrium Solutions. "Phase Line Plot"The ODE has the stable equilibrium solution and the unstable as the directionfield in Fig. 22 suggests. The values and are the zeros of the parabola in the figure.Now, since the ODE is autonomous, we can "condense" the direction field to a "phase line plot" giving andand the direction (upward or downward) of the arrows in the field, and thus giving information about thestability or instability of the equilibrium solutions. ᭿y2,y1f(y) ϭ (y Ϫ 1)(y Ϫ 2)y2y1y2 ϭ 2,y1 ϭ 1yr ϭ (y Ϫ 1)(y Ϫ 2)y ϭ A>B.y ϭ 0y ϭ 4y ϭ 0y ϭ const.yr ϭ 0f(y) ϭ 0f(y),yr ϭ f(y)yr ϭ f(t, y)A>B.y Ͼ A>B.yr Ͻ 0y Ͼ A>B,y Ͻ A>B.yr Ͼ 0,y Ͻ A>B,yr ϭ Ay31 Ϫ (B>A)y4,ϪBy2y ϭ (1>c)eAtyr ϭ dy>dt ϭ Ay.B ϭ 0,c01.qxd 7/30/10 8:15 PM Page 33 SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 3516. (that is, for all x, also written )is a solution of (2) [not of (1) if !], called thetrivial solution.17. The sum of a solution of (1) and a solution of (2) is asolution of (1).18. The difference of two solutions of (1) is a solution of (2).19. If is a solution of (1), what can you say about20. If and are solutions of andrespectively (with the same p!), whatcan you say about the sum21. Variation of parameter. Another method of obtaining(4) results from the following idea. Write (3) aswhere is the exponential function, which is a solutionof the homogeneous linear ODEReplace the arbitrary constant c in (3) with a functionu to be determined so that the resulting functionis a solution of the nonhomogeneous linear ODE22–28 NONLINEAR ODEsUsing a method of this section or separating variables, findthe general solution. If an initial condition is given, findalso the particular solution and sketch or graph it.22.23.24.25.26.27.28.29. REPORT PROJECT. Transformation of ODEs.We have transformed ODEs to separable form, to exactform, and to linear form. The purpose of suchtransformations is an extension of solution methods tolarger classes of ODEs. Describe the key idea of eachof these transformations and give three typical exam-ples of your choice for each transformation. Show eachstep (not just the transformed ODE).30. TEAM PROJECT. Riccati Equation. ClairautEquation. Singular Solution.A Riccati equation is of the form(14)A Clairaut equation is of the form(15)(a) Apply the transformation to theRiccati equation (14), where Y is a solution of (14), andobtain for u the linear ODEExplain the effect of the transformation by writing itas y ϭ Y ϩ v, v ϭ 1>u.ur ϩ (2Yg Ϫ p)u ϭ Ϫg.y ϭ Y ϩ 1>uy ϭ xyr ϩ g(yr).yr ϩ p(x)y ϭ g(x)y2ϩ h(x).2xyyr ϩ (x Ϫ 1)y2ϭ x2ex(Set y2ϭ z)yr ϭ 1>(6eyϪ 2x)y(0) ϭ 12 pyr ϭ (tan y)>(x Ϫ 1),yr ϭ 3.2y Ϫ 10y2yr ϩ y ϭ Ϫx>yyr ϩ xy ϭ xy؊1, y(0) ϭ 3yr ϩ y ϭ y2, y(0) ϭ Ϫ13yr ϩ py ϭ r.y ϭ uyyr ϩ py ϭ 0.ycy,y1 ϩ y2?y2r ϩ py2 ϭ r2,y1r ϩ py1 ϭ r1y2y1cy1?y1r(x) 0y(x) ϵ 0y(x) ϭ 0y ϭ 0 (b) Show that is a solution of the ODEand solve thisRiccati equation, showing the details.(c) Solve the Clairaut equation asfollows. Differentiate it with respect to x, obtainingThen solve (A) and (B)separately and substitute the two solutions(a) and (b) of (A) and (B) into the given ODE. Thusobtain (a) a general solution (straight lines) and (b) aparabola for which those lines (a) are tangents (Fig. 6in Prob. Set 1.1); so (b) is the envelope of (a). Such asolution (b) that cannot be obtained from a generalsolution is called a singular solution.(d) Show that the Clairaut equation (15) has assolutions a family of straight lines anda singular solution determined by wherethat forms the envelope of that family.31–40 MODELING. FURTHER APPLICATIONS31. Newton's law of cooling. If the temperature of a cakeis when it leaves the oven and is tenminutes later, when will it be practically equal to theroom temperature of say, when will it be32. Heating and cooling of a building. Heating andcooling of a building can be modeled by the ODEwhere is the temperature in the building attime t, the outside temperature, the temperaturewanted in the building, and P the rate of increase of Tdue to machines and people in the building, and andare (negative) constants. Solve this ODE, assumingand varying sinusoidallyover 24 hours, say, Discussthe effect of each term of the equation on the solution.33. Drug injection. Find and solve the model for druginjection into the bloodstream if, beginning at aconstant amount A g min is injected and the drug issimultaneously removed at a rate proportional to theamount of the drug present at time t.34. Epidemics. A model for the spread of contagiousdiseases is obtained by assuming that the rate of spreadis proportional to the number of contacts betweeninfected and noninfected persons, who are assumed tomove freely among each other. Set up the model. Findthe equilibrium solutions and indicate their stability orinstability. Solve the ODE. Find the limit of theproportion of infected persons as and explainwhat it means.35. Lake Erie. Lake Erie has a water volume of aboutand a flow rate (in and out) of about 175 km2450 km3t : ϱ>t ϭ 0,Ta ϭ A Ϫ C cos(2p>24)t.TaP ϭ const, Tw ϭ const,k2k1TwTaT ϭ T(t)Tr ϭ k1(T Ϫ Ta) ϩ k2(T Ϫ Tv) ϩ P,61°F?60°F,200°F300°Fs ϭ yr,gr(s) ϭ Ϫx,y ϭ cx ϩ g(c)2yr Ϫ x ϭ 0ys ϭ 0ys(2yr Ϫ x) ϭ 0.yr2Ϫ xyr ϩ y ϭ 0y ϭ Ϫx2y2Ϫ x4Ϫ x ϩ1(2x3ϩ 1)yr Ϫy ϭ Y ϭ xc01.qxd 7/30/10 10:01 PM Page 35 36 CHAP. 1 First-Order ODEsper year. If at some instant the lake has pollutionconcentration how long, approximately,will it take to decrease it to p 2, assuming that theinflow is much cleaner, say, it has pollutionconcentration p 4, and the mixture is uniform (anassumption that is only imperfectly true)? First guess.36. Harvesting renewable resources. Fishing. Supposethat the population of a certain kind of fish is givenby the logistic equation (11), and fish are caught at arate Hy proportional to y. Solve this so-called Schaefermodel. Find the equilibrium solutions andwhen The expression is calledthe equilibrium harvest or sustainable yield corre-sponding to H. Why?37. Harvesting. In Prob. 36 find and graph the solutionsatisfying when (for simplicity)and What is the limit? What does it mean?What if there were no fishing?38. Intermittent harvesting. In Prob. 36 assume that youfish for 3 years, then fishing is banned for the next3 years. Thereafter you start again. And so on. This iscalled intermittent harvesting. Describe qualitativelyhow the population will develop if intermitting iscontinued periodically. Find and graph the solution forthe first 9 years, assuming thatand y(0) ϭ 2.A ϭ B ϭ 1, H ϭ 0.2,H ϭ 0.2.A ϭ B ϭ 1y(0) ϭ 2Y ϭ Hy2H Ͻ A.y2 (Ͼ 0)y1y(t)>>p ϭ 0.04%,39. Extinction vs. unlimited growth. If in a populationthe death rate is proportional to the population, andthe birth rate is proportional to the chance encountersof meeting mates for reproduction, what will the modelbe? Without solving, find out what will eventuallyhappen to a small initial population. To a large one.Then solve the model.40. Air circulation. In a room containing of air,of fresh air flows in per minute, and the mixture(made practically uniform by circulating fans) isexhausted at a rate of 600 cubic feet per minute (cfm).What is the amount of fresh air at any time ifAfter what time will 90% of the air be fresh?y(0) ϭ 0?y(t)600 ft320,000 ft3y(t)Fig. 23. Fish population in Problem 380.811.21.41.61.822 4 6 80 ty1.6 Orthogonal Trajectories. OptionalAn important type of problem in physics or geometry is to find a family of curves thatintersects a given family of curves at right angles. The new curves are called orthogonaltrajectories of the given curves (and conversely). Examples are curves of equaltemperature (isotherms) and curves of heat flow, curves of equal altitude (contour lines)on a map and curves of steepest descent on that map, curves of equal potential(equipotential curves, curves of equal voltage—the ellipses in Fig. 24) and curves ofelectric force (the parabolas in Fig. 24).Here the angle of intersection between two curves is defined to be the angle betweenthe tangents of the curves at the intersection point. Orthogonal is another word forperpendicular.In many cases orthogonal trajectories can be found using ODEs. In general, if weconsider to be a given family of curves in the xy-plane, then each value ofc gives a particular curve. Since c is one parameter, such a family is called a one-parameter family of curves.In detail, let us explain this method by a family of ellipses(1) (c Ͼ 0)12 x2ϩ y2ϭ cG(x, y, c) ϭ 0c01.qxd 7/30/10 8:15 PM Page 36 Step 2. Find an ODE for the orthogonal trajectories This ODE is(3)with the same f as in (2). Why? Well, a given curve passing through a point hasslope at that point, by (2). The trajectory through has slopeby (3). The product of these slopes is , as we see. From calculus it is known that thisis the condition for orthogonality (perpendicularity) of two straight lines (the tangents at), hence of the curve and its orthogonal trajectory at .Step 3. Solve (3) by separating variables, integrating, and taking exponents:This is the family of orthogonal trajectories, the quadratic parabolas along which electronsor other charged particles (of very small mass) would move in the electric field betweenthe black ellipses (elliptic cylinders).yෂϭ cx2.ln ƒyෂƒ ϭ 2 ln x ϩ c,dyෂyෂ ϭ 2dxx,(x0, y0)(x0, y0)Ϫ1Ϫ1>f(x0, y0)(x0, y0)f(x0, y0)(x0, y0)yෂr ϭ Ϫ1f(x, yෂ)ϭ ϩ2yෂxyෂϭ yෂ(x).SEC. 1.6 Orthogonal Trajectories. Optional 37–6 6yx4–4Fig. 24. Electrostatic field between two ellipses (elliptic cylinders in space):Elliptic equipotential curves (equipotential surfaces) and orthogonaltrajectories (parabolas)and illustrated in Fig. 24. We assume that this family of ellipses represents electricequipotential curves between the two black ellipses (equipotential surfaces between twoelliptic cylinders in space, of which Fig. 24 shows a cross-section). We seek theorthogonal trajectories, the curves of electric force. Equation (1) is a one-parameter familywith parameter c. Each value of c corresponds to one of these ellipses.Step 1. Find an ODE for which the given family is a general solution. Of course, thisODE must no longer contain the parameter c. Differentiating (1), we haveHence the ODE of the given curves is(2) yr ϭ f(x, y) ϭ Ϫx2y.x ϩ 2yyr ϭ 0.(Ͼ 0)c01.qxd 7/30/10 8:15 PM Page 37 38 CHAP. 1 First-Order ODEs1–3 FAMILIES OF CURVESRepresent the given family of curves in the formand sketch some of the curves.1. All ellipses with foci and 3 on the x-axis.2. All circles with centers on the cubic parabolaand passing through the origin3. The catenaries obtained by translating the catenaryin the direction of the straight line .4–10 ORTHOGONAL TRAJECTORIES (OTs)Sketch or graph some of the given curves. Guess what theirOTs may look like. Find these OTs.4. 5.6. 7.8. 9.10.11–16 APPLICATIONS, EXTENSIONS11. Electric field. Let the electric equipotential lines(curves of constant potential) between two concentriccylinders with the z-axis in space be given by(these are circular cylinders inthe xyz-space). Using the method in the text, find theirorthogonal trajectories (the curves of electric force).12. Electric field. The lines of electric force of two oppositecharges of the same strength at and arethe circles through and . Show that thesecircles are given by . Showthat the equipotential lines (which are orthogonaltrajectories of those circles) are the circles given by(dashed in Fig. 25).(x ϩ c)2ϩ yෂ2ϭ c2Ϫ 1x2ϩ (y Ϫ c)2ϭ 1 ϩ c2(1, 0)(Ϫ1, 0)(1, 0)(Ϫ1, 0)u(x, y) ϭ x2ϩ y2ϭ cx2ϩ (y Ϫ c)2ϭ c2y ϭ ce؊x2y ϭ 2x ϩ cy ϭ c>x2xy ϭ cy ϭ cxy ϭ x2ϩ cy ϭ xy ϭ cosh x(0, 0).y ϭ x3Ϫ3G(x, y; c) ϭ 0P R O B L E M S E T 1 . 6Fig. 25. Electric field in Problem 1213. Temperature field. Let the isotherms (curves ofconstant temperature) in a body in the upper half-planebe given by . Find the ortho-gonal trajectories (the curves along which heat willflow in regions filled with heat-conducting material andfree of heat sources or heat sinks).14. Conic sections. Find the conditions under whichthe orthogonal trajectories of families of ellipsesare again conic sections. Illustrateyour result graphically by sketches or by using yourCAS. What happens if If15. Cauchy–Riemann equations. Show that for a familyconst the orthogonal trajectoriesconst can be obtained from the followingCauchy–Riemann equations (which are basic incomplex analysis in Chap. 13) and use them to find theorthogonal trajectories of const. (Here, sub-scripts denote partial derivatives.)16. Congruent OTs. If with f independent of y,show that the curves of the corresponding family arecongruent, and so are their OTs.yr ϭ f(x)uy ϭ Ϫvxux ϭ vy,exsin y ϭc ϭv(x, y) ϭu(x, y) ϭ c ϭb : 0?a : 0?x2>a2ϩ y2>b2ϭ c4x2ϩ 9y2ϭ cy Ͼ 01.7 Existence and Uniqueness of Solutionsfor Initial Value ProblemsThe initial value problemhas no solution because (that is, for all x) is the only solution of the ODE.The initial value problemy(0) ϭ 1yr ϭ 2x,y(x) ϭ 0y ϭ 0y(0) ϭ 1ƒyr ƒ ϩ ƒyƒ ϭ 0,c01.qxd 7/30/10 8:15 PM Page 38 SEC. 1.7 Existence and Uniqueness of Solutions 39Theorems that state such conditions are called existence theorems and uniquenesstheorems, respectively.Of course, for our simple examples, we need no theorems because we can solve theseexamples by inspection; however, for complicated ODEs such theorems may be ofconsiderable practical importance. Even when you are sure that your physical or othersystem behaves uniquely, occasionally your model may be oversimplified and may notgive a faithful picture of reality.T H E O R E M 1 Existence TheoremLet the right side of the ODE in the initial value problem(1)be continuous at all points in some rectangle(Fig. 26)and bounded in R; that is, there is a number K such that(2) for all in R.Then the initial value problem (1) has at least one solution . This solution existsat least for all x in the subinterval of the intervalhere, is the smaller of the two numbers a and b K.>aƒx Ϫ x0 ƒ Ͻ a;ƒx Ϫ x0 ƒ Ͻ ay(x)(x, y)ƒf(x, y) ƒ Ϲ Kƒy Ϫ y0 ƒ Ͻ bR: ƒx Ϫ x0 ƒ Ͻ a,(x, y)y(x0) ϭ y0yr ϭ f(x, y),f(x, y)has precisely one solution, namely, The initial value problemhas infinitely many solutions, namely, where c is an arbitrary constant becausefor all c.From these examples we see that an initial value problem(1)may have no solution, precisely one solution, or more than one solution. This fact leadsto the following two fundamental questions.Problem of ExistenceUnder what conditions does an initial value problem of the form (1) have at leastone solution (hence one or several solutions)?Problem of UniquenessUnder what conditions does that problem have at most one solution (hence excludingthe case that is has more than one solution)?y(x0) ϭ y0yr ϭ f(x, y),y(0) ϭ 1y ϭ 1 ϩ cx,y(0) ϭ 1xyr ϭ y Ϫ 1,y ϭ x2ϩ 1.c01.qxd 7/30/10 8:15 PM Page 39 40 CHAP. 1 First-Order ODEsyxy0+ bx0+ ax0– a x0y0y0– bRFig. 26. Rectangle R in the existence and uniqueness theorems(Example of Boundedness. The function is bounded (with ) in thesquare . The function is not bounded for. Explain!)T H E O R E M 2 Uniqueness TheoremLet f and its partial derivative be continuous for all in the rectangleR (Fig. 26) and bounded, say,(3) (a) (b) for all in R.Then the initial value problem (1) has at most one solution . Thus, by Theorem 1,the problem has precisely one solution. This solution exists at least for all x in thatsubinterval ƒx Ϫ x0 ƒ Ͻ a.y(x)(x, y)ƒ fy(x, y) ƒ Ϲ Mƒ f(x, y) ƒ Ϲ K,(x, y)fy ϭ 0f>0yƒx ϩ yƒ Ͻ p>2f(x, y) ϭ tan (x ϩ y)ƒxƒ Ͻ 1, ƒy ƒ Ͻ 1K ϭ 2f(x, y) ϭ x2ϩ y2Understanding These TheoremsThese two theorems take care of almost all practical cases. Theorem 1 says that ifis continuous in some region in the xy-plane containing the point , then the initialvalue problem (1) has at least one solution.Theorem 2 says that if, moreover, the partial derivative of f with respect to yexists and is continuous in that region, then (1) can have at most one solution; hence, byTheorem 1, it has precisely one solution.Read again what you have just read—these are entirely new ideas in our discussion.Proofs of these theorems are beyond the level of this book (see Ref. [A11] in App. 1);however, the following remarks and examples may help you to a good understanding ofthe theorems.Since , the condition (2) implies that that is, the slope of anysolution curve in R is at least and at most K. Hence a solution curve that passesthrough the point must lie in the colored region in Fig. 27 bounded by the linesand whose slopes are and K, respectively. Depending on the form of R, twodifferent cases may arise. In the first case, shown in Fig. 27a, we have andtherefore in the existence theorem, which then asserts that the solution exists for allx between and . In the second case, shown in Fig. 27b, we have .Therefore, and all we can conclude from the theorems is that the solutiona ϭ b>K Ͻ a,b>K Ͻ ax0 ϩ ax0 Ϫ aa ϭ ab>K м aϪKl2l1(x0, y0)ϪKy(x)ƒyr ƒ Ϲ K;yr ϭ f(x, y)0f>0y(x0, y0)f(x, y)c01.qxd 7/30/10 8:15 PM Page 40 and take the rectangle Then , andIndeed, the solution of the problem is (see Sec. 1.3, Example 1). This solution is discontinuous at, and there is no continuous solution valid in the entire interval from which we started.The conditions in the two theorems are sufficient conditions rather than necessary ones,and can be lessened. In particular, by the mean value theorem of differential calculus wehavewhere and are assumed to be in R, and is a suitable value betweenand . From this and (3b) it follows that(4) ƒ f(x, y2) Ϫ f(x, y1)ƒ Ϲ Mƒy2 Ϫ y1 ƒ.y2y1yෂ(x, y2)(x, y1)f(x, y2) Ϫ f(x, y1) ϭ (y2 Ϫ y1)0f0y`yϭyෂ᭿ƒ xƒ Ͻ 5Ϯp>2y ϭ tan xa ϭbKϭ 0.3 Ͻ a.`0f0y` ϭ 2ƒ yƒ Ϲ M ϭ 6,ƒ f(x, y)ƒ ϭ ƒ1 ϩ y2ƒ Ϲ K ϭ 10,a ϭ 5, b ϭ 3R; ƒ xƒ Ͻ 5, ƒyƒ Ͻ 3.SEC. 1.7 Existence and Uniqueness of Solutions 41y yxy0+ bl1l2x0(a)y0y0– bRxy0+ bl1l2x0(b)y0y0– bRa a= a = aα αα αLet us illustrate our discussion with a simple example. We shall see that our choice ofa rectangle R with a large base (a long x-interval) will lead to the case in Fig. 27b.E X A M P L E 1 Choice of a RectangleConsider the initial value problemy(0) ϭ 0yr ϭ 1 ϩ y2,exists for all x between and . For larger or smaller x's the solutioncurve may leave the rectangle R, and since nothing is assumed about f outside R, nothingcan be concluded about the solution for those larger or amaller x's; that is, for such x'sthe solution may or may not exist—we don't know.x0 ϩ b>Kx0 Ϫ b>KFig. 27. The condition (2) of the existence theorem. (a) First case. (b) Second casec01.qxd 7/30/10 8:15 PM Page 41 42 CHAP. 1 First-Order ODEs9RUDOLF LIPSCHITZ (1832–1903), German mathematician. Lipschitz and similar conditions are importantin modern theories, for instance, in partial differential equations.10EMILE PICARD (1856–1941). French mathematician, also known for his important contributions tocomplex analysis (see Sec. 16.2 for his famous theorem). Picard used his method to prove Theorems 1 and 2as well as the convergence of the sequence (7) to the solution of (1). In precomputer times, the iteration was oflittle practical value because of the integrations.It can be shown that (3b) may be replaced by the weaker condition (4), which is knownas a Lipschitz condition.9However, continuity of is not enough to guarantee theuniqueness of the solution. This may be illustrated by the following example.E X A M P L E 2 NonuniquenessThe initial value problemhas the two solutionsandalthough is continuous for all y. The Lipschitz condition (4) is violated in any region that includesthe line , because for and positive we have(5)and this can be made as large as we please by choosing sufficiently small, whereas (4) requires that thequotient on the left side of (5) should not exceed a fixed constant M. ᭿y2(2y2 Ͼ 0)ƒ f(x, y2) Ϫ f(x, y1)ƒƒ y2 Ϫ y1 ƒϭ2y2y2ϭ12y2,y2y1 ϭ 0y ϭ 0f(x, y) ϭ 2 ƒyƒy* ϭ ex2>4 if x м 0Ϫx2>4 if x Ͻ 0y ϭ 0y(0) ϭ 0yr ϭ 2 ƒ yƒ.f(x, y)1. Linear ODE. If p and r in arecontinuous for all x in an interval showthat in this ODE satisfies the conditions of ourpresent theorems, so that a corresponding initial valueproblem has a unique solution. Do you actually needthese theorems for this ODE?2. Existence? Does the initial value problemhave a solution? Does yourresult contradict our present theorems?3. Vertical strip. If the assumptions of Theorems 1 and2 are satisfied not merely in a rectangle but in a verticalinfinite strip in what interval will thesolution of (1) exist?4. Change of initial condition. What happens in Prob.2 if you replace with5. Length of x-interval. In most cases the solution of aninitial value problem (1) exists in an x-interval larger thanthat guaranteed by the present theorems. Show this factfor by finding the best possible ayr ϭ 2y2, y(1) ϭ 1y(2) ϭ k?y(2) ϭ 1ƒx Ϫ x0 ƒ Ͻ a,(x Ϫ 2)yr ϭ y, y(2) ϭ 1f(x, y)ƒx Ϫ x0 ƒ Յ a,yr ϩ p(x)y ϭ r(x) (choosing b optimally) and comparing the result with theactual solution.6. CAS PROJECT. Picard Iteration. (a) Show that byintegrating the ODE in (1) and observing the initialcondition you obtain(6)This form (6) of (1) suggests Picard's Iteration Method10which is defined by(7)It gives approximations of the unknownsolution y of (1). Indeed, you obtain by substitutingon the right and integrating—this is the firststep—then by substituting on the right andintegrating—this is the second step—and so on. Writey ϭ y1y2y ϭ y0y1y1, y2, y3, . . .yn(x) ϭ y0 ϩ Ύxx0f(t, yn؊1(t) dt, n ϭ 1, 2, Á .y(x) ϭ y0 ϩ Ύxx0f(t, y(t))dt.P R O B L E M S E T 1 . 7c01.qxd 7/30/10 8:15 PM Page 42 Chapter 1 Review Questions and Problems 43a program of the iteration that gives a printout of thefirst approximations as well as theirgraphs on common axes. Try your program on twoinitial value problems of your own choice.(b) Apply the iteration to Alsosolve the problem exactly.(c) Apply the iteration to Alsosolve the problem exactly.(d) Find all solutions of Whichof them does Picard's iteration approximate?(e) Experiment with the conjecture that Picard'siteration converges to the solution of the problem forany initial choice of y in the integrand in (7) (leavingoutside the integral as it is). Begin with a simple ODEand see what happens. When you are reasonably sure,take a slightly more complicated ODE and give it a try.y0yr ϭ 2 1y, y(1) ϭ 0.yr ϭ 2y2, y(0) ϭ 1.yr ϭ x ϩ y, y(0) ϭ 0.y0, y1, . . . , yN7. Maximum . What is the largest possible inExample 1 in the text?8. Lipschitz condition. Show that for a linear ODEwith continuous p and r ina Lipschitz condition holds. This isremarkable because it means that for a linear ODE thecontinuity of guarantees not only the existencebut also the uniqueness of the solution of an initialvalue problem. (Of course, this also follows directlyfrom (4) in Sec. 1.5.)9. Common points. Can two solution curves of the sameODE have a common point in a rectangle in which theassumptions of the present theorems are satisfied?10. Three possible cases. Find all initial conditions suchthat has no solution, preciselyone solution, and more than one solution.(x2Ϫ x)yr ϭ (2x Ϫ 1)yf(x, y)ƒx Ϫ x0 ƒ Ϲ ayr ϩ p(x)y ϭ r(x)aA14.15.16. Solve by Euler's method(10 steps, ). Solve exactly and compute the error.17–21 GENERAL SOLUTIONFind the general solution. Indicate which method in thischapter you are using. Show the details of your work.17.18.19.20.21.22–26 INITIAL VALUE PROBLEM (IVP)Solve the IVP. Indicate the method used. Show the detailsof your work.22.23.24.25.26.27–30 MODELING, APPLICATIONS27. Exponential growth. If the growth rate of a cultureof bacteria is proportional to the number of bacteriapresent and after 1 day is 1.25 times the originalnumber, within what interval of time will the numberof bacteria (a) double, (b) triple?x sinh y dy ϭ cosh y dx, y(3) ϭ 03 sec y dx ϩ 13 sec x dy ϭ 0, y(0) ϭ 0yr ϩ 12 y ϭ y3, y(0) ϭ 13yr ϭ 21 Ϫ y2, y(0) ϭ 1>12yr ϩ 4xy ϭ eϪ2x2, y(0) ϭ Ϫ4.3(3xeyϩ 2y) dx ϩ (x2eyϩ x) dy ϭ 0yr ϭ ay ϩ by2(a 0)25yyr Ϫ 4x ϭ 0yr Ϫ 0.4y ϭ 29 sin xyr ϩ 2.5y ϭ 1.6xh ϭ 0.1yr ϭ y Ϫ y2, y(0) ϭ 0.2yr ϩ y ϭ 1.01 cos 10xxyr ϭ y ϩ x21. Explain the basic concepts ordinary and partialdifferential equations (ODEs, PDEs), order, generaland particular solutions, initial value problems (IVPs).Give examples.2. What is a linear ODE? Why is it easier to solve thana nonlinear ODE?3. Does every first-order ODE have a solution? A solutionformula? Give examples.4. What is a direction field? A numeric method for first-order ODEs?5. What is an exact ODE? Isalways exact?6. Explain the idea of an integrating factor. Give twoexamples.7. What other solution methods did we consider in thischapter?8. Can an ODE sometimes be solved by several methods?Give three examples.9. What does modeling mean? Can a CAS solve a modelgiven by a first-order ODE? Can a CAS set up a model?10. Give problems from mechanics, heat conduction, andpopulation dynamics that can be modeled by first-orderODEs.11–16 DIRECTION FIELD: NUMERIC SOLUTIONGraph a direction field (by a CAS or by hand) and sketchsome solution curves. Solve the ODE exactly and compare.In Prob. 16 use Euler's method.11.12.13. yr ϭ y Ϫ 4y2yr ϭ 1 Ϫ y2yr ϩ 2y ϭ 0f(x) dx ϩ g(y) dy ϭ 0C H A P T E R 1 R E V I E W Q U E S T I O N S A N D P R O B L E M Sc01.qxd 7/30/10 8:15 PM Page 43 44 CHAP. 1 First-Order ODEs28. Mixing problem. The tank in Fig. 28 contains 80 lbof salt dissolved in 500 gal of water. The inflow perminute is 20 lb of salt dissolved in 20 gal of water. Theoutflow is 20 gal min of the uniform mixture. Find thetime when the salt content in the tank reaches 95%of its limiting value (as ).t : ϱy(t)>Fig. 28. Tank in Problem 2829. Half-life. If in a reactor, uranium loses 10% ofits weight within one day, what is its half-life? Howlong would it take for 99% of the original amount todisappear?30. Newton's law of cooling. A metal bar whosetemperature is is placed in boiling water. Howlong does it take to heat the bar to practicallysay, to , if the temperature of the bar after 1 minof heating is First guess, then calculate.51.5°C?99.9°C100°C,20°C23797UThis chapter concerns ordinary differential equations (ODEs) of first order andtheir applications. These are equations of the form(1) or in explicit forminvolving the derivative of an unknown function y, given functions ofx, and, perhaps, y itself. If the independent variable x is time, we denote it by t.In Sec. 1.1 we explained the basic concepts and the process of modeling, that is,of expressing a physical or other problem in some mathematical form and solvingit. Then we discussed the method of direction fields (Sec. 1.2), solution methodsand models (Secs. 1.3–1.6), and, finally, ideas on existence and uniqueness ofsolutions (Sec. 1.7).A first-order ODE usually has a general solution, that is, a solution involving anarbitrary constant, which we denote by c. In applications we usually have to find aunique solution by determining a value of c from an initial condition .Together with the ODE this is called an initial value problem(2)and its solution is a particular solution of the ODE. Geometrically, a generalsolution represents a family of curves, which can be graphed by using directionfields (Sec. 1.2). And each particular solution corresponds to one of these curves.A separable ODE is one that we can put into the form(3) (Sec. 1.3)by algebraic manipulations (possibly combined with transformations, such as) and solve by integrating on both sides.y>x ϭ ug(y) dy ϭ f(x) dx(x0, y0 given numbers)y(x0) ϭ y0yr ϭ f(x, y),y(x0) ϭ y0yr ϭ dy>dxyr ϭ f(x, y)F(x, y, yr) ϭ 0SUMMARY OF CHAPTER 1First-Order ODEsc01.qxd 7/30/10 8:15 PM Page 44 An exact ODE is of the form(4) (Sec. 1.4)where is the differentialof a function so that from we immediately get the implicit generalsolution This method extends to nonexact ODEs that can be made exactby multiplying them by some function called an integrating factor (Sec. 1.4).Linear ODEs(5)are very important. Their solutions are given by the integral formula (4), Sec. 1.5.Certain nonlinear ODEs can be transformed to linear form in terms of new variables.This holds for the Bernoulli equation(Sec. 1.5).Applications and modeling are discussed throughout the chapter, in particular inSecs. 1.1, 1.3, 1.5 (population dynamics, etc.), and 1.6 (trajectories).Picard's existence and uniqueness theorems are explained in Sec. 1.7 (andPicard's iteration in Problem Set 1.7).Numeric methods for first-order ODEs can be studied in Secs. 21.1 and 21.2immediately after this chapter, as indicated in the chapter opening.yr ϩ p(x)y ϭ g(x)yayr ϩ p(x)y ϭ r(x)F(x, y,),u(x, y) ϭ c.du ϭ 0u(x, y),du ϭ ux dx ϩ uy dyM dx ϩ N dyM(x, y) dx ϩ N(x, y) dy ϭ 0Summary of Chapter 1 45c01.qxd 7/30/10 8:15 PM Page 45 46C H A P T E R 2Second-Order Linear ODEsMany important applications in mechanical and electrical engineering, as shown in Secs.2.4, 2.8, and 2.9, are modeled by linear ordinary differential equations (linear ODEs) of thesecond order. Their theory is representative of all linear ODEs as is seen when comparedto linear ODEs of third and higher order, respectively. However, the solution formulas forsecond-order linear ODEs are simpler than those of higher order, so it is a natural progressionto study ODEs of second order first in this chapter and then of higher order in Chap. 3.Although ordinary differential equations (ODEs) can be grouped into linear and nonlinearODEs, nonlinear ODEs are difficult to solve in contrast to linear ODEs for which manybeautiful standard methods exist.Chapter 2 includes the derivation of general and particular solutions, the latter inconnection with initial value problems.For those interested in solution methods for Legendre's, Bessel's, and the hypergeometricequations consult Chap. 5 and for Sturm–Liouville problems Chap. 11.COMMENT. Numerics for second-order ODEs can be studied immediately after thischapter. See Sec. 21.3, which is independent of other sections in Chaps. 19–21.Prerequisite: Chap. 1, in particular, Sec. 1.5.Sections that may be omitted in a shorter course: 2.3, 2.9, 2.10.References and Answers to Problems: App. 1 Part A, and App. 2.2.1 Homogeneous Linear ODEs of Second OrderWe have already considered first-order linear ODEs (Sec. 1.5) and shall now define anddiscuss linear ODEs of second order. These equations have important engineeringapplications, especially in connection with mechanical and electrical vibrations (Secs. 2.4,2.8, 2.9) as well as in wave motion, heat conduction, and other parts of physics, as weshall see in Chap. 12.A second-order ODE is called linear if it can be written(1)and nonlinear if it cannot be written in this form.The distinctive feature of this equation is that it is linear in y and its derivatives, whereasthe functions p, q, and r on the right may be any given functions of x. If the equationbegins with, say, then divide by to have the standard form (1) with as thefirst term.ysf(x)f(x)ys,ys ϩ p(x)yr ϩ q(x)y ϭ r(x)c02.qxd 10/27/10 6:06 PM Page 46 The definitions of homogeneous and nonhomogenous second-order linear ODEs arevery similar to those of first-order ODEs discussed in Sec. 1.5. Indeed, if (thatis, for all x considered; read " is identically zero"), then (1) reduces to(2)and is called homogeneous. If then (1) is called nonhomogeneous. This issimilar to Sec. 1.5.An example of a nonhomogeneous linear ODE isand a homogeneous linear ODE iswritten in standard form .Finally, an example of a nonlinear ODE is.The functions p and q in (1) and (2) are called the coefficients of the ODEs.Solutions are defined similarly as for first-order ODEs in Chap. 1. A functionis called a solution of a (linear or nonlinear) second-order ODE on some open interval Iif h is defined and twice differentiable throughout that interval and is such that the ODEbecomes an identity if we replace the unknown y by h, the derivative by , and thesecond derivative by . Examples are given below.Homogeneous Linear ODEs: Superposition PrincipleSections 2.1–2.6 will be devoted to homogeneous linear ODEs (2) and the remainingsections of the chapter to nonhomogeneous linear ODEs.Linear ODEs have a rich solution structure. For the homogeneous equation the backboneof this structure is the superposition principle or linearity principle, which says that wecan obtain further solutions from given ones by adding them or by multiplying them withany constants. Of course, this is a great advantage of homogeneous linear ODEs. Let usfirst discuss an example.E X A M P L E 1 Homogeneous Linear ODEs: Superposition of SolutionsThe functions and are solutions of the homogeneous linear ODEfor all x. We verify this by differentiation and substitution. We obtain ; henceys ϩ y ϭ (cos x)s ϩ cos x ϭ Ϫcos x ϩ cos x ϭ 0.(cos x)s ϭ Ϫcos xys ϩ y ϭ 0y ϭ sin xy ϭ cos xhsyshryry ϭ h(x)ysy ϩ yr2ϭ 0ys ϩ1x yr ϩ y ϭ 0xys ϩ yr ϩ xy ϭ 0,ys ϩ 25y ϭ e؊xcos x,r(x) [ 0,ys ϩ p(x)yr ϩ q(x)y ϭ 0r(x)r(x) ϭ 0r(x) ϵ 0SEC. 2.1 Homogeneous Linear ODEs of Second Order 47c02.qxd 10/27/10 6:06 PM Page 47 Initial Value Problem. Basis. General SolutionRecall from Chap. 1 that for a first-order ODE, an initial value problem consists of theODE and one initial condition . The initial condition is used to determine thearbitrary constant c in the general solution of the ODE. This results in a unique solution,as we need it in most applications. That solution is called a particular solution of theODE. These ideas extend to second-order ODEs as follows.For a second-order homogeneous linear ODE (2) an initial value problem consists of(2) and two initial conditions(4)These conditions prescribe given values and of the solution and its first derivative(the slope of its curve) at the same given in the open interval considered.The conditions (4) are used to determine the two arbitrary constants and in ageneral solution(5)of the ODE; here, and are suitable solutions of the ODE, with "suitable" to beexplained after the next example. This results in a unique solution, passing through thepoint with as the tangent direction (the slope) at that point. That solution iscalled a particular solution of the ODE (2).E X A M P L E 4 Initial Value ProblemSolve the initial value problemSolution. Step 1. General solution. The functions and are solutions of the ODE (by Example 1),and we takeThis will turn out to be a general solution as defined below.Step 2. Particular solution. We need the derivative . From this and theinitial values we obtain, since and ,This gives as the solution of our initial value problem the particular solutionFigure 29 shows that at it has the value 3.0 and the slope , so that its tangent intersectsthe x-axis at . (The scales on the axes differ!)Observation. Our choice of and was general enough to satisfy both initialconditions. Now let us take instead two proportional solutions andso that . Then we can write in the form.y ϭ c1 cos x ϩ c2(k cos x) ϭ C cos x where C ϭ c1 ϩ c2ky ϭ c1 y1 ϩ c2 y2y1/y2 ϭ 1/k ϭ consty2 ϭ k cos x,y1 ϭ cos xy2y1᭿x ϭ 3.0>0.5 ϭ 6.0Ϫ0.5x ϭ 0y ϭ 3.0 cos x Ϫ 0.5 sin x.y(0) ϭ c1 ϭ 3.0 and yr(0) ϭ c2 ϭ Ϫ0.5.sin 0 ϭ 0cos 0 ϭ 1yr ϭ Ϫc1 sin x ϩ c2 cos xy ϭ c1 cos x ϩ c2 sin x.sin xcos xys ϩ y ϭ 0, y(0) ϭ 3.0, yr(0) ϭ Ϫ0.5.K1(x0, K0)y2y1y ϭ c1 y1 ϩ c2 y2c2c1x ϭ x0K1K0y(x0) ϭ K0, yr(x0) ϭ K1.y(x0) ϭ y0SEC. 2.1 Homogeneous Linear ODEs of Second Order 492 4 6 108 x–3–2–10123yFig. 29. Particular solutionand initial tangent inExample 4c02.qxd 10/27/10 6:06 PM Page 49 Hence we are no longer able to satisfy two initial conditions with only one arbitraryconstant C. Consequently, in defining the concept of a general solution, we must excludeproportionality. And we see at the same time why the concept of a general solution is ofimportance in connection with initial value problems.D E F I N I T I O N General Solution, Basis, Particular SolutionA general solution of an ODE (2) on an open interval I is a solution (5) in whichand are solutions of (2) on I that are not proportional, and and are arbitraryconstants. These , are called a basis (or a fundamental system) of solutionsof (2) on I.A particular solution of (2) on I is obtained if we assign specific values toand in (5).For the definition of an interval see Sec. 1.1. Furthermore, as usual, and are calledproportional on I if for all x on I,(6) (a) or (b)where k and l are numbers, zero or not. (Note that (a) implies (b) if and only if ).Actually, we can reformulate our definition of a basis by using a concept of generalimportance. Namely, two functions and are called linearly independent on aninterval I where they are defined if(7) everywhere on I implies .And and are called linearly dependent on I if (7) also holds for some constants ,not both zero. Then, if , we can divide and see that and areproportional,orIn contrast, in the case of linear independence these functions are not proportional becausethen we cannot divide in (7). This gives the followingD E F I N I T I O N Basis (Reformulated)A basis of solutions of (2) on an open interval I is a pair of linearly independentsolutions of (2) on I.If the coefficients p and q of (2) are continuous on some open interval I, then (2) has ageneral solution. It yields the unique solution of any initial value problem (2), (4). Itincludes all solutions of (2) on I; hence (2) has no singular solutions (solutions notobtainable from of a general solution; see also Problem Set 1.1). All this will be shownin Sec. 2.6.y2 ϭ Ϫk1k2y1.y1 ϭ Ϫk2k1y2y2y1k1 0 or k2 0k2k1y2y1k1 ϭ 0 and k2 ϭ 0k1y1(x) ϩ k2y2(x) ϭ 0y2y1k 0y2 ϭ ly1y1 ϭ ky2y2y1c2c1y2y1c2c1y2y150 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 50 E X A M P L E 5 Basis, General Solution, Particular Solutionand in Example 4 form a basis of solutions of the ODE for all x because theirquotient is (or ). Hence is a general solution. The solutionof the initial value problem is a particular solution.E X A M P L E 6 Basis, General Solution, Particular SolutionVerify by substitution that and are solutions of the ODE . Then solve the initialvalue problem.Solution. and show that and are solutions. They are notproportional, . Hence , form a basis for all x. We now write down the correspondinggeneral solution and its derivative and equate their values at 0 to the given initial conditions,.By addition and subtraction, , so that the answer is . This is the particular solutionsatisfying the two initial conditions.Find a Basis if One Solution Is Known.Reduction of OrderIt happens quite often that one solution can be found by inspection or in some other way.Then a second linearly independent solution can be obtained by solving a first-order ODE.This is called the method of reduction of order.1We first show how this method worksin an example and then in general.E X A M P L E 7 Reduction of Order if a Solution Is Known. BasisFind a basis of solutions of the ODE.Solution. Inspection shows that is a solution because and , so that the first termvanishes identically and the second and third terms cancel. The idea of the method is to substituteinto the ODE. This givesux and –xu cancel and we are left with the following ODE, which we divide by x, order, and simplify,,This ODE is of first order in , namely, . Separation of variables and integrationgives, .ln ƒv ƒ ϭ ln ƒ x Ϫ 1 ƒ Ϫ 2 ln ƒ xƒ ϭ lnƒ x Ϫ 1ƒx2dvvϭ Ϫx Ϫ 2x2Ϫ xdx ϭ a1x Ϫ 1Ϫ2xb dx(x2Ϫ x)vr ϩ (x Ϫ 2)v ϭ 0v ϭ ur(x2Ϫ x)us ϩ (x Ϫ 2)ur ϭ 0.(x2Ϫ x)(usx ϩ 2ur) Ϫ x2ur ϭ 0(x2Ϫ x)(usx ϩ 2ur) Ϫ x(urx ϩ u) ϩ ux ϭ 0.y ϭ uy1 ϭ ux, yr ϭ urx ϩ u, ys ϭ usx ϩ 2urys1 ϭ 0yr1 ϭ 1y1 ϭ x(x2Ϫ x)ys Ϫ xyr ϩ y ϭ 0᭿y ϭ 2exϩ 4e؊xc1 ϭ 2, c2 ϭ 4y ϭ c1exϩ c2e؊x, yr ϭ c1exϪ c2e؊x, y(0) ϭ c1 ϩ c2 ϭ 6, yr(0) ϭ c1 Ϫ c2 ϭ Ϫ2e؊xexex/e؊xϭ e2xconste؊xex(e؊x)s Ϫ e؊xϭ 0(ex)s Ϫ exϭ 0ys Ϫ y ϭ 0, y(0) ϭ 6, yr(0) ϭ Ϫ2ys Ϫ y ϭ 0y2 ϭ e؊xy1 ϭ ex᭿y ϭ 3.0 cos x Ϫ 0.5 sin xy ϭ c1 cos x ϩ c2 sin xtan x constcot x constys ϩ y ϭ 0sin xcos xSEC. 2.1 Homogeneous Linear ODEs of Second Order 511Credited to the great mathematician JOSEPH LOUIS LAGRANGE (1736–1813), who was born in Turin,of French extraction, got his first professorship when he was 19 (at the Military Academy of Turin), becamedirector of the mathematical section of the Berlin Academy in 1766, and moved to Paris in 1787. His importantmajor work was in the calculus of variations, celestial mechanics, general mechanics (Mécanique analytique,Paris, 1788), differential equations, approximation theory, algebra, and number theory.c02.qxd 10/27/10 6:06 PM Page 51 13. Linear operator. Illustrate the linearity of L in (2) bytaking , and .Prove that L is linear.14. Double root. If has distinct rootsand , show that a particular solution is. Obtain from this a solutionby letting and applying l'Hôpital's rule.␮ : lxelxy ϭ (e␮xϪ elx)>(␮ Ϫ l)l␮D2ϩ aD ϩ bIw ϭ cos 2xc ϭ 4, k ϭ Ϫ6, y ϭ e2x62 CHAP. 2 Second-Order Linear ODEs15. Definition of linearity. Show that the definition oflinearity in the text is equivalent to the following. Ifand exist, then exists andand exist for all constants c and k, andas well asand .L[kw] ϭ kL[w]L[cy] ϭ cL[y]L[y ϩ w] ϭ L[y] ϩ L[w]L[kw]L[cy]L[y ϩ w]L[w]L[y]2.4 Modeling of Free Oscillationsof a Mass–Spring SystemLinear ODEs with constant coefficients have important applications in mechanics, as weshow in this section as well as in Sec. 2.8, and in electrical circuits as we show in Sec. 2.9.In this section we model and solve a basic mechanical system consisting of a mass on anelastic spring (a so-called "mass–spring system," Fig. 33), which moves up and down.Setting Up the ModelWe take an ordinary coil spring that resists extension as well as compression. We suspendit vertically from a fixed support and attach a body at its lower end, for instance, an ironball, as shown in Fig. 33. We let denote the position of the ball when the systemis at rest (Fig. 33b). Furthermore, we choose the downward direction as positive, thusregarding downward forces as positive and upward forces as negative.y ϭ 02ROBERT HOOKE (1635–1703), English physicist, a forerunner of Newton with respect to the law ofgravitation.UnstretchedspringSystem atrestSystem inmotion(a) (b) (c)s0y(y = 0)Fig. 33. Mechanical mass–spring systemWe now let the ball move, as follows. We pull it down by an amount (Fig. 33c).This causes a spring force(1) (Hooke's law2)proportional to the stretch y, with called the spring constant. The minus signindicates that points upward, against the displacement. It is a restoring force: It wantsto restore the system, that is, to pull it back to . Stiff springs have large k.y ϭ 0F1k (Ͼ0)F1 ϭ Ϫkyy Ͼ 0c02.qxd 10/27/10 6:06 PM Page 62 Note that an additional force is present in the spring, caused by stretching it infastening the ball, but has no effect on the motion because it is in equilibrium withthe weight W of the ball, , whereis the constant of gravity at the Earth's surface (not to be confused withthe universal gravitational constant , which weshall not need; here and are the Earth's radius andmass, respectively).The motion of our mass–spring system is determined by Newton's second law(2)where and "Force" is the resultant of all the forces acting on the ball. (Forsystems of units, see the inside of the front cover.)ODE of the Undamped SystemEvery system has damping. Otherwise it would keep moving forever. But if the dampingis small and the motion of the system is considered over a relatively short time, wemay disregard damping. Then Newton's law with gives the modelthus(3) .This is a homogeneous linear ODE with constant coefficients. A general solution isobtained as in Sec. 2.2, namely (see Example 6 in Sec. 2.2)(4)This motion is called a harmonic oscillation (Fig. 34). Its frequency is Hertz3because and in (4) have the period . The frequency f is calledthe natural frequency of the system. (We write to reserve for Sec. 2.8.)vv02p>v0sincos(ϭ cycles>sec)f ϭ v0>2pv0 ϭBkm.y(t) ϭ A cos v0t ϩ B sin v0tmys ϩ ky ϭ 0mys ϭ ϪF1 ϭ Ϫky;F ϭ ϪF1ys ϭ d2y>dt2Mass ϫ Acceleration ϭ mys ϭ ForceM ϭ 5.98 # 1024kgR ϭ 6.37 # 106mG ϭ gR2>M ϭ 6.67 # 10؊11nt m2>kg232.17 ft>sec2g ϭ 980 cm>sec2ϭ 9.8 m>sec2ϭϪF0 ϭ W ϭ mgF0ϪF0SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 63yt123123PositiveZeroNegativeInitial velocityFig. 34. Typical harmonic oscillations (4) and with the same anddifferent initial velocities , positive 1 , zero 2 , negative 3yr(0) ϭ v0By(0) ϭ A(4)3HEINRICH HERTZ (1857–1894), German physicist, who discovered electromagnetic waves, as the basisof wireless communication developed by GUGLIELMO MARCONI (1874–1937), Italian physicist (Nobel prizein 1909).c02.qxd 10/27/10 6:06 PM Page 63 An alternative representation of (4), which shows the physical characteristics of amplitudeand phase shift of (4), is(4)with and phase angle , where . This follows from theaddition formula (6) in App. 3.1.E X A M P L E 1 Harmonic Oscillation of an Undamped Mass–Spring SystemIf a mass–spring system with an iron ball of weight nt (about 22 lb) can be regarded as undamped, andthe spring is such that the ball stretches it 1.09 m (about 43 in.), how many cycles per minute will the systemexecute? What will its motion be if we pull the ball down from rest by 16 cm (about 6 in.) and let it start withzero initial velocity?Solution. Hooke's law (1) with W as the force and 1.09 meter as the stretch gives ; thus. The mass is . Thisgives the frequency .From (4) and the initial conditions, . Hence the motion is(Fig. 35).If you have a chance of experimenting with a mass–spring system, don't miss it. You will be surprised aboutthe good agreement between theory and experiment, usually within a fraction of one percent if you measurecarefully. ᭿y(t) ϭ 0.16 cos 3t [meter] or 0.52 cos 3t [ft]y(0) ϭ A ϭ 0.16 [meter] and yr(0) ϭ v0B ϭ 0v0>(2p) ϭ 2k>m>(2p) ϭ 3>(2p) ϭ 0.48 [Hz] ϭ 29 [cycles>min]m ϭ W>g ϭ 98>9.8 ϭ 10 [kg]98>1.09 ϭ 90 [kg>sec2] ϭ 90 [nt>meter]k ϭ W>1.09 ϭW ϭ 1.09kW ϭ 98tan d ϭ B>AdC ϭ 2A2ϩ B2y(t) ϭ C cos (v0t Ϫ d)64 CHAP. 2 Second-Order Linear ODEs102 4 6 8 t–0.1–0.200.10.2yFig. 35. Harmonic oscillation in Example 1ODE of the Damped SystemTo our model we now add a damping forceobtaining ; thus the ODE of the damped mass–spring system is(5) (Fig. 36)Physically this can be done by connecting the ball to a dashpot; see Fig. 36. We assumethis damping force to be proportional to the velocity . This is generally a goodapproximation for small velocities.yr ϭ dy>dtmys ϩ cyr ϩ ky ϭ 0.mys ϭ Ϫky Ϫ cyrF2 ϭ Ϫcyr,mys ϭ ϪkyFig. 36.Damped systemDashpotBallSpringkmcc02.qxd 10/27/10 6:06 PM Page 64 SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 65Case I. . Distinct real roots . (Overdamping)Case II. . A real double root. (Critical damping)Case III. . Complex conjugate roots. (Underdamping)c2Ͻ 4mkc2ϭ 4mkl1, l2c2Ͼ 4mkThey correspond to the three Cases I, II, III in Sec. 2.2.Discussion of the Three CasesCase I. OverdampingIf the damping constant c is so large that , then are distinct real roots.In this case the corresponding general solution of (5) is(7) .We see that in this case, damping takes out energy so quickly that the body does notoscillate. For both exponents in (7) are negative because , and. Hence both terms in (7) approach zero as . Practicallyspeaking, after a sufficiently long time the mass will be at rest at the static equilibriumposition . Figure 37 shows (7) for some typical initial conditions.(y ϭ 0)t : ϱb2ϭ a2Ϫ k>m Ͻ a2a Ͼ 0, b Ͼ 0t Ͼ 0y(t) ϭ c1e؊(a؊b)tϩ c2e؊(a؉b)tl1 and l2c2Ͼ 4mkThe constant c is called the damping constant. Let us show that c is positive. Indeed,the damping force acts against the motion; hence for a downward motion wehave which for positive c makes F negative (an upward force), as it should be.Similarly, for an upward motion we have which, for makes positive (adownward force).The ODE (5) is homogeneous linear and has constant coefficients. Hence we can solveit by the method in Sec. 2.2. The characteristic equation is (divide (5) by m).By the usual formula for the roots of a quadratic equation we obtain, as in Sec. 2.2,(6) , where and .It is now interesting that depending on the amount of damping present—whether a lot ofdamping, a medium amount of damping or little damping—three types of motions occur,respectively:b ϭ12m2c2Ϫ 4mka ϭc2ml1 ϭ Ϫa ϩ b, l2 ϭ Ϫa Ϫ bl2ϩcm l ϩkm ϭ 0F2c Ͼ 0yr Ͻ 0yr Ͼ 0F2 ϭ Ϫcyrc02.qxd 10/27/10 6:06 PM Page 65 66 CHAP. 2 Second-Order Linear ODEsty123(a)yt11232PositiveZeroNegativeInitial velocity3(b)Fig. 37. Typical motions (7) in the overdamped case(a) Positive initial displacement(b) Negative initial displacementCase II. Critical DampingCritical damping is the border case between nonoscillatory motions (Case I) and oscillations(Case III). It occurs if the characteristic equation has a double root, that is, if ,so that . Then the corresponding general solution of (5) is(8) .This solution can pass through the equilibrium position at most once becauseis never zero and can have at most one positive zero. If both are positive(or both negative), it has no positive zero, so that y does not pass through 0 at all. Figure 38shows typical forms of (8). Note that they look almost like those in the previous figure.c1 and c2c1 ϩ c2te؊aty ϭ 0y(t) ϭ (c1 ϩ c2t)e؊atb ϭ 0, l1 ϭ l2 ϭ Ϫac2ϭ 4mkyt123123PositiveZeroNegativeInitial velocityFig. 38. Critical damping [see (8)]c02.qxd 10/27/10 6:06 PM Page 66 Case III. UnderdampingThis is the most interesting case. It occurs if the damping constant c is so small that. Then in (6) is no longer real but pure imaginary, say,(9) where .(We now write to reserve for driving and electromotive forces in Secs. 2.8 and 2.9.)The roots of the characteristic equation are now complex conjugates,with , as given in (6). Hence the corresponding general solution is(10)where , as in .This represents damped oscillations. Their curve lies between the dashed curvesin Fig. 39, touching them when is an integer multipleof because these are the points at which equals 1 or .The frequency is Hz (hertz, cycles/sec). From (9) we see that the smalleris, the larger is and the more rapid the oscillations become. If c approaches 0,then approaches , giving the harmonic oscillation (4), whose frequencyis the natural frequency of the system.v0>(2p)v0 ϭ 2k>mvvc (Ͼ0)v>(2p)Ϫ1cos (vt Ϫ d)pvt Ϫ dy ϭ Ce؊atand y ϭ ϪCe؊at(4)C2ϭ A2ϩ B2and tan d ϭ B>Ay(t) ϭ e؊at(A cos vt ϩ B sin vt) ϭ Ce؊atcos (vt Ϫ d)a ϭ c>(2m)l1 ϭ Ϫa ϩ iv, l2 ϭ Ϫa Ϫ ivvv(Ͼ0)v* ϭ12m24mk Ϫ c2ϭBkmϪc24m2b ϭ ivbc2Ͻ 4mkSEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 67Fig. 39. Damped oscillation in Case III [see (10)]tyCe– tα–Ce– tαE X A M P L E 2 The Three Cases of Damped MotionHow does the motion in Example 1 change if we change the damping constant c from one to another of thefollowing three values, with as before?(I) , (II) , (III) .Solution. It is interesting to see how the behavior of the system changes due to the effect of the damping,which takes energy from the system, so that the oscillations decrease in amplitude (Case III) or even disappear(Cases II and I).(I) With , as in Example 1, the model is the initial value problem.10ys ϩ 100yr ϩ 90y ϭ 0, y(0) ϭ 0.16 [meter], yr(0) ϭ 0m ϭ 10 and k ϭ 90c ϭ 10 kg>secc ϭ 60 kg>secc ϭ 100 kg>secy(0) ϭ 0.16 and yr(0) ϭ 0c02.qxd 10/27/10 6:06 PM Page 67 SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 691–10 HARMONIC OSCILLATIONS(UNDAMPED MOTION)1. Initial value problem. Find the harmonic motion (4)that starts from with initial velocity . Graph orsketch the solutions for , and variousof your choice on common axes. At what t-valuesdo all these curves intersect? Why?2. Frequency. If a weight of 20 nt (about 4.5 lb) stretchesa certain spring by 2 cm, what will the frequency of thecorresponding harmonic oscillation be? The period?3. Frequency. How does the frequency of the harmonicoscillation change if we (i) double the mass, (ii) takea spring of twice the modulus? First find qualitativeanswers by physics, then look at formulas.4. Initial velocity. Could you make a harmonic oscillationmove faster by giving the body a greater initial push?5. Springs in parallel. What are the frequencies ofvibration of a body of mass kg (i) on a springof modulus , (ii) on a spring of modulus, (iii) on the two springs in parallel? SeeFig. 41.k2 ϭ 45 nt>mk1 ϭ 20 nt>mm ϭ 5v0v0 ϭ p, y0 ϭ 1v0y0P R O B L E M S E T 2 . 4The cylindrical buoy of diameter 60 cm in Fig. 43 isfloating in water with its axis vertical. When depresseddownward in the water and released, it vibrates withperiod 2 sec. What is its weight?Fig. 41. Parallel springs (Problem 5)Fig. 42. Pendulum (Problem 7)6. Spring in series. If a body hangs on a spring ofmodulus , which in turn hangs on a springof modulus , what is the modulus k of thiscombination of springs?7. Pendulum. Find the frequency of oscillation of apendulum of length L (Fig. 42), neglecting airresistance and the weight of the rod, and assumingto be so small that practically equals .usin uuk2 ϭ 12s2k1 ϭ 8s110. TEAM PROJECT. Harmonic Motions of SimilarModels. The unifying power of mathematical meth-ods results to a large extent from the fact that differentphysical (or other) systems may have the same or verysimilar models. Illustrate this for the following threesystems(a) Pendulum clock. A clock has a 1-meter pendulum.The clock ticks once for each time the pendulumcompletes a full swing, returning to its original position.How many times a minute does the clock tick?(b) Flat spring (Fig. 45). The harmonic oscillationsof a flat spring with a body attached at one end andhorizontally clamped at the other are also governed by(3). Find its motions, assuming that the body weighs8 nt (about 1.8 lb), the system has its static equilibrium1 cm below the horizontal line, and we let it start fromthis position with initial velocity 10 cm/sec.8. Archimedian principle. This principle states that thebuoyancy force equals the weight of the waterdisplaced by the body (partly or totally submerged).Fig. 44. Tube (Problem 9)9. Vibration of water in a tube. If 1 liter of water (about1.06 US quart) is vibrating up and down under theinfluence of gravitation in a U-shaped tube of diameter2 cm (Fig. 44), what is the frequency? Neglect friction.First guess.Fig. 43. Buoy (Problem 8)LθBody ofmass mWaterlevel( y = 0)yyFig. 45. Flat springc02.qxd 10/27/10 6:06 PM Page 69 (c) Torsional vibrations (Fig. 46). Undampedtorsional vibrations (rotations back and forth) of awheel attached to an elastic thin rod or wire aregoverned by the equation , whereis the angle measured from the state of equilibrium.Solve this equation for , initialangle and initial angular velocity.20° sec؊1(ϭ 0.349 rad # sec؊1)30°(ϭ 0.5235 rad)K>I0 ϭ 13.69 sec؊2uI0us ϩ Ku ϭ 070 CHAP. 2 Second-Order Linear ODEs11–20 DAMPED MOTION11. Overdamping. Show that for (7) to satisfy initial condi-tions and we must haveand.12. Overdamping. Show that in the overdamped case, thebody can pass through at most once (Fig. 37).13. Initial value problem. Find the critical motion (8)that starts from with initial velocity . Graphsolution curves for and several suchthat (i) the curve does not intersect the t-axis, (ii) itintersects it at respectively.14. Shock absorber. What is the smallest value of thedamping constant of a shock absorber in the suspen-sion of a wheel of a car (consisting of a spring and anabsorber) that will provide (theoretically) an oscillation-free ride if the mass of the car is 2000 kg and the springconstant equals ?15. Frequency. Find an approximation formula for interms of by applying the binomial theorem in (9)and retaining only the first two terms. How good is theapproximation in Example 2, III?16. Maxima. Show that the maxima of an underdampedmotion occur at equidistant t-values and find thedistance.17. Underdamping. Determine the values of t corre-sponding to the maxima and minima of the oscillation. Check your result by graphing .18. Logarithmic decrement. Show that the ratio oftwo consecutive maximum amplitudes of a dampedoscillation (10) is constant, and the natural logarithmof this ratio called the logarithmic decrement,y(t)y(t) ϭ e؊tsin tv0v4500 kg>sec2t ϭ 1, 2, . . . , 5,v0a ϭ 1, y0 ϭ 1v0y0y ϭ 0v0>b]>2c2 ϭ [(1 Ϫ a>b)y0 Ϫ[(1 ϩ a>b)y0 ϩ v0>b]>2c1 ϭv(0) ϭ v0y(0) ϭ y0equals . Find for the solutions of.19. Damping constant. Consider an underdamped motionof a body of mass . If the time between twoconsecutive maxima is 3 sec and the maximumamplitude decreases to its initial value after 10 cycles,what is the damping constant of the system?20. CAS PROJECT. Transition Between Cases I, II,III. Study this transition in terms of graphs of typicalsolutions. (Cf. Fig. 47.)(a) Avoiding unnecessary generality is part of goodmodeling. Show that the initial value problems (A)and (B),(A)(B) the same with different c and (insteadof 0), will give practically as much information as aproblem with other m, k, .(b) Consider (A). Choose suitable values of c,perhaps better ones than in Fig. 47, for the transitionfrom Case III to II and I. Guess c for the curves in thefigure.(c) Time to go to rest. Theoretically, this time isinfinite (why?). Practically, the system is at rest whenits motion has become very small, say, less than 0.1%of the initial displacement (this choice being up to us),that is in our case,(11) for all t greater than some .In engineering constructions, damping can often bevaried without too much trouble. Experimenting withyour graphs, find empirically a relation betweenand c.(d) Solve (A) analytically. Give a reason why thesolution c of , with the solution of, will give you the best possible c satisfying(11).(e) Consider (B) empirically as in (a) and (b). Whatis the main difference between (B) and (A)?yr(t) ϭ 0t2y(t2) ϭ Ϫ0.001t1t1ƒy(t)ƒ Ͻ 0.001y(0), yr(0)yr(0) ϭ Ϫ2ys ϩ cyr ϩ y ϭ 0, y(0) ϭ 1, yr(0) ϭ 012m ϭ 0.5 kgys ϩ 2yr ϩ 5y ϭ 0¢¢ ϭ 2pa>vFig. 47. CAS Project 20Fig. 46. Torsional vibrationsθ10.5–0.5–16 1084yt2c02.qxd 10/27/10 6:06 PM Page 70 2.5 Euler–Cauchy EquationsEuler–Cauchy equations4are ODEs of the form(1)with given constants a and b and unknown function . We substituteinto (1). This givesand we now see that was a rather natural choice because we have obtained a com-mon factor . Dropping it, we have the auxiliary equation or(2) . (Note: , not a.)Hence is a solution of (1) if and only if m is a root of (2). The roots of (2) are(3) .Case I. Real different roots give two real solutionsand .These are linearly independent since their quotient is not constant. Hence they constitutea basis of solutions of (1) for all x for which they are real. The corresponding generalsolution for all these x is(4) (c1, c2 arbitrary).E X A M P L E 1 General Solution in the Case of Different Real RootsThe Euler–Cauchy equation has the auxiliary equation . Theroots are 0.5 and . Hence a basis of solutions for all positive x is and and gives the generalsolution. ᭿(x Ͼ 0)y ϭ c1 1x ϩc2xy2 ϭ 1>xy1 ϭ x0.5Ϫ1m2ϩ 0.5m Ϫ 0.5 ϭ 0x2ys ϩ 1.5xyr Ϫ 0.5y ϭ 0y ϭ c1xm1ϩ c2xm2y2(x) ϭ xm2y1(x) ϭ xm1m1 and m2m1 ϭ 12 (1 Ϫ a) ϩ 214 (1 Ϫ a)2Ϫ b, m2 ϭ 12 (1 Ϫ a) Ϫ 214 (1 Ϫ a)2Ϫ by ϭ xma Ϫ 1m2ϩ (a Ϫ 1)m ϩ b ϭ 0m(m Ϫ 1) ϩ am ϩ b ϭ 0xmy ϭ xmx2m(m Ϫ 1)xmϪ2ϩ axmxmϪ1ϩ bxmϭ 0y ϭ xm, yr ϭ mxm؊1, ys ϭ m(m Ϫ 1)xm؊2y(x)x2ys ϩ axyr ϩ by ϭ 0SEC. 2.5 Euler–Cauchy Equations 714LEONHARD EULER (1707–1783) was an enormously creative Swiss mathematician. He madefundamental contributions to almost all branches of mathematics and its application to physics. His importantbooks on algebra and calculus contain numerous basic results of his own research. The great Frenchmathematician AUGUSTIN LOUIS CAUCHY (1789–1857) is the father of modern analysis. He is the creatorof complex analysis and had great influence on ODEs, PDEs, infinite series, elasticity theory, and optics.c02.qxd 10/27/10 6:06 PM Page 71 The proof of existence uses the same prerequisites as the existence proof in Sec. 1.7and will not be presented here; it can be found in Ref. [A11] listed in App. 1. Uniquenessproofs are usually simpler than existence proofs. But for Theorem 1, even the uniquenessproof is long, and we give it as an additional proof in App. 4.Linear Independence of SolutionsRemember from Sec. 2.1 that a general solution on an open interval I is made up from abasis on I, that is, from a pair of linearly independent solutions on I. Here we calllinearly independent on I if the equation(4) .We call linearly dependent on I if this equation also holds for constantsnot both 0. In this case, and only in this case, are proportional on I, that is (seeSec. 2.1),(5) (a) or (b) for all on I.For our discussion the following criterion of linear independence and dependence ofsolutions will be helpful.T H E O R E M 2 Linear Dependence and Independence of SolutionsLet the ODE (1) have continuous coefficients and on an open interval I.Then two solutions of (1) on I are linearly dependent on I if and only iftheir "Wronskian"(6)is 0 at some in I. Furthermore, if at an in I, then on I;hence, if there is an in I at which W is not 0, then are linearly independenton I.P R O O F (a) Let be linearly dependent on I. Then (5a) or (5b) holds on I. If (5a) holds,thenSimilarly if (5b) holds.(b) Conversely, we let for some and show that this implies lineardependence of on I. We consider the linear system of equations in the unknowns(7)k1 y1r(x0) ϩ k2 y2r(x0) ϭ 0.k1 y1(x0) ϩ k2 y2(x0) ϭ 0k1, k2y1 and y2x ϭ x0W(y1, y2) ϭ 0W(y1, y2) ϭ y1y2r Ϫ y2y1r ϭ ky2y2r Ϫ y2ky2r ϭ 0.y1 and y2y1, y2x1W ϭ 0x ϭ x0W ϭ 0x0W(y1, y2) ϭ y1y2r Ϫ y2y1ry1 and y2q(x)p(x)y2 ϭ ly1y1 ϭ ky2y1 and y2k1, k2y1, y2k1y1(x) ϩ k2y2(x) ϭ 0 on I implies k1 ϭ 0, k2 ϭ 0y1, y2y1, y2SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian 75c02.qxd 10/27/10 6:06 PM Page 75 To eliminate , multiply the first equation by and the second by and add theresulting equations. This gives.Similarly, to eliminate , multiply the first equation by and the second by andadd the resulting equations. This gives.If W were not 0 at , we could divide by W and conclude that . Since W is0, division is not possible, and the system has a solution for which are not both0. Using these numbers , we introduce the function.Since (1) is homogeneous linear, Fundamental Theorem 1 in Sec. 2.1 (the superpositionprinciple) implies that this function is a solution of (1) on I. From (7) we see that it satisfiesthe initial conditions . Now another solution of (1) satisfying thesame initial conditions is . Since the coefficients p and q of (1) are continuous,Theorem 1 applies and gives uniqueness, that is, , written outon I.Now since and are not both zero, this means linear dependence of , on I.(c) We prove the last statement of the theorem. If at an in I, we havelinear dependence of on I by part (b), hence by part (a) of this proof. Hencein the case of linear dependence it cannot happen that at an in I. If it doeshappen, it thus implies linear independence as claimed.For calculations, the following formulas are often simpler than (6).(6) or (b)These formulas follow from the quotient rule of differentiation.Remark. Determinants. Students familiar with second-order determinants may havenoticed that.This determinant is called the Wronski determinant5or, briefly, the Wronskian, of twosolutions and of (1), as has already been mentioned in (6). Note that its four entriesoccupy the same positions as in the linear system (7).y2y1W(y1, y2) ϭ `y1 y2yr1 yr2` ϭ y1yr2 Ϫ y2yr1Ϫay1y2bry22 (y2 0).W(y1, y2) ϭ (a) ay2y1bry21 (y1 0)᭿x1W(x1) 0W ϵ 0y1, y2x0W(x0) ϭ 0y2y1k2k1k1y1 ϩ k2y2 ϵ 0y ϵ yy ϵ 0y(x0) ϭ 0, yr(x0) ϭ 0y(x) ϭ k1y1(x) ϩ k2y2(x)k1, k2k1 and k2k1 ϭ k2 ϭ 0x0k2W(y1(x0), y2(x0)) ϭ 0y1Ϫy1rk1k1y1(x0)y2r(x0) Ϫ k1y1r(x0)y2(x0) ϭ k1W(y1(x0), y2(x0)) ϭ 0Ϫy2yr2k276 CHAP. 2 Second-Order Linear ODEs5Introduced by WRONSKI (JOSEF MARIA HÖNE, 1776–1853), Polish mathematician.c02.qxd 10/27/10 6:06 PM Page 76 E X A M P L E 1 Illustration of Theorem 2The functions and are solutions of . Their Wronskian is.Theorem 2 shows that these solutions are linearly independent if and only if . Of course, we can seethis directly from the quotient . For we have , which implies linear dependence(why?).E X A M P L E 2 Illustration of Theorem 2 for a Double RootA general solution of on any interval is . (Verify!). The correspondingWronskian is not 0, which shows linear independence of and on any interval. Namely,.A General Solution of (1) Includes All SolutionsThis will be our second main result, as announced at the beginning. Let us start withexistence.T H E O R E M 3 Existence of a General SolutionIf p(x) and q(x) are continuous on an open interval I, then (1) has a general solutionon I.P R O O F By Theorem 1, the ODE (1) has a solution on I satisfying the initial conditionsand a solution on I satisfying the initial conditionsThe Wronskian of these two solutions has at the valueHence, by Theorem 2, these solutions are linearly independent on I. They form a basis ofsolutions of (1) on I, and with arbitrary is a general solution of (1)on I, whose existence we wanted to prove. ᭿c1, c2y ϭ c1y1 ϩ c2˛y2W(y1(0), y2(0)) ϭ y1(x0)y2r(x0) Ϫ y2(x0)y1r(x0) ϭ 1.x ϭ x0y2r(x0) ϭ 1.y2(x0) ϭ 0,y2(x)y1r(x0) ϭ 0y1(x0) ϭ 1,y1(x)᭿W(x, xex) ϭ `exxexex(x ϩ 1)ex` ϭ (x ϩ 1)e2xϪ xe2xϭ e2x0xexexy ϭ (c1 ϩ c2x)exys Ϫ 2yr ϩ y ϭ 0᭿y2 ϭ 0v ϭ 0y2>y1 ϭ tan vxv 0W(cos vx, sin vx) ϭ `cos vx sin vxϪv sin vx v cos vx` ϭ y1y2r Ϫ y2y1r ϭ v cos2vx ϩ v sin2vx ϭ vys ϩ v2y ϭ 0y2 ϭ sin vxy1 ϭ cos vxSEC. 2.6 Existence and Uniqueness of Solutions. Wronskian 77c02.qxd 10/27/10 6:06 PM Page 77 We finally show that a general solution is as general as it can possibly be.T H E O R E M 4 A General Solution Includes All SolutionsIf the ODE (1) has continuous coefficients p(x) and q(x) on some open interval I,then every solution of (1) on I is of the form(8)where is any basis of solutions of (1) on I and are suitable constants.Hence (1) does not have singular solutions (that is, solutions not obtainable froma general solution).P R O O F Let be any solution of (1) on I. Now, by Theorem 3 the ODE (1) has a generalsolution(9)on I. We have to find suitable values of such that on I. We choose anyin I and show first that we can find values of such that we reach agreement atthat is, and . Written out in terms of (9), this becomes(10)(a)(b)We determine the unknowns and . To eliminate we multiply (10a) by and(10b) by and add the resulting equations. This gives an equation for Then wemultiply (10a) by and (10b) by and add the resulting equations. This givesan equation for These new equations are as follows, where we take the values ofatSince is a basis, the Wronskian W in these equations is not 0, and we can solve forand We call the (unique) solution By substituting it into (9) weobtain from (9) the particular solutionNow since is a solution of (10), we see from (10) thatFrom the uniqueness stated in Theorem 1 this implies that y* and Y must be equaleverywhere on I, and the proof is complete. ᭿yr(x0) ϭ Yr(x0).y(x0) ϭ Y(x0),C1, C2y(x) ϭ C1y1(x) ϩ C2y2(x).c1 ϭ C1, c2 ϭ C2.c2.c1y1, y2c2(y1y2r Ϫ y2y1r) ϭ c2W(y1, y2) ϭ y1Yr Ϫ Yy1r.c1(y1y2r Ϫ y2y1r) ϭ c1W(y1, y2) ϭ Yy2r Ϫ y2Yrx0.y1, y1r, y2, y2r, Y, Yrc2.y1(x0)Ϫy1r(x0)c1.Ϫy2(x0)y2r(x0)c2,c2c1c1y1r(x0) ϩ c2y2r(x0) ϭ Yr(x0).c1y1(x0) ϩ c2y2(x0) ϭ Y(x0)yr(x0) ϭ Yr(x0)y(x0) ϭ Y(x0)x0,c1, c2x0y(x) ϭ Y(x)c1, c2y(x) ϭ c1y1(x) ϩ c2y2(x)y ϭ Y(x)C1, C2y1, y2Y(x) ϭ C1y1(x) ϩ C2y2(x)y ϭ Y(x)78 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 78 (2)and a "particular solution" of (1). These two new terms "general solution of (1)" and"particular solution of (1)" are defined as follows.D E F I N I T I O N General Solution, Particular SolutionA general solution of the nonhomogeneous ODE (1) on an open interval I is asolution of the form(3)here, is a general solution of the homogeneous ODE (2) on I andis any solution of (1) on I containing no arbitrary constants.A particular solution of (1) on I is a solution obtained from (3) by assigningspecific values to the arbitrary constants and in .Our task is now twofold, first to justify these definitions and then to develop a methodfor finding a solution of (1).Accordingly, we first show that a general solution as just defined satisfies (1) and thatthe solutions of (1) and (2) are related in a very simple way.T H E O R E M 1 Relations of Solutions of (1) to Those of (2)(a) The sum of a solution y of (1) on some open interval I and a solution of(2) on I is a solution of (1) on I. In particular, (3) is a solution of (1) on I.(b) The difference of two solutions of (1) on I is a solution of (2) on I.P R O O F (a) Let denote the left side of (1). Then for any solutions y of (1) and of (2) on I,(b) For any solutions y and y of (1) on I we haveNow for homogeneous ODEs (2) we know that general solutions include all solutions.We show that the same is true for nonhomogeneous ODEs (1).T H E O R E M 2 A General Solution of a Nonhomogeneous ODE Includes All SolutionsIf the coefficients p(x), q(x), and the function r(x) in (1) are continuous on someopen interval I, then every solution of (1) on I is obtained by assigning suitablevalues to the arbitrary constants and in a general solution (3) of (1) on I.P R O O F Let be any solution of (1) on I and any x in I. Let (3) be any general solution of(1) on I. This solution exists. Indeed, exists by Theorem 3 in Sec. 2.6yh ϭ c1y1 ϩ c2y2x0yc2c1᭿r Ϫ r ϭ 0.L[y Ϫ y] ϭ L[y] Ϫ L[y] ϭL[y ϩ y~] ϭ L[y] ϩ L[y~] ϭ r ϩ 0 ϭ r.y~L[y]y~ypyhc2c1ypyh ϭ c1y1 ϩ c2y2y(x) ϭ yh(x) ϩ yp1x2;ys ϩ p(x)yr ϩ q(x)y ϭ 080 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 80 because of the continuity assumption, and exists according to a construction to beshown in Sec. 2.10. Now, by Theorem 1(b) just proved, the difference is asolution of (2) on I. At we haveTheorem 1 in Sec. 2.6 implies that for these conditions, as for any other initial conditionsin I, there exists a unique particular solution of (2) obtained by assigning suitable valuesto in . From this and the statement follows.Method of Undetermined CoefficientsOur discussion suggests the following. To solve the nonhomogeneous ODE (1) or an initialvalue problem for (1), we have to solve the homogeneous ODE (2) and find any solutionof (1), so that we obtain a general solution (3) of (1).How can we find a solution of (1)? One method is the so-called method ofundetermined coefficients. It is much simpler than another, more general, method (givenin Sec. 2.10). Since it applies to models of vibrational systems and electric circuits to beshown in the next two sections, it is frequently used in engineering.More precisely, the method of undetermined coefficients is suitable for linear ODEswith constant coefficients a and b(4)when is an exponential function, a power of x, a cosine or sine, or sums or productsof such functions. These functions have derivatives similar to itself. This gives theidea. We choose a form for similar to , but with unknown coefficients to bedetermined by substituting that and its derivatives into the ODE. Table 2.1 on p. 82shows the choice of for practically important forms of . Corresponding rules areas follows.Choice Rules for the Method of Undetermined Coefficients(a) Basic Rule. If in (4) is one of the functions in the first column inTable 2.1, choose in the same line and determine its undeterminedcoefficients by substituting and its derivatives into (4).(b) Modification Rule. If a term in your choice for happens to be asolution of the homogeneous ODE corresponding to (4), multiply this termby x (or by if this solution corresponds to a double root of thecharacteristic equation of the homogeneous ODE).(c) Sum Rule. If is a sum of functions in the first column of Table 2.1,choose for the sum of the functions in the corresponding lines of thesecond column.The Basic Rule applies when is a single term. The Modification Rule helps in theindicated case, and to recognize such a case, we have to solve the homogeneous ODEfirst. The Sum Rule follows by noting that the sum of two solutions of (1) withand (and the same left side!) is a solution of (1) with . (Verify!)r ϭ r1 ϩ r2r ϭ r2r ϭ r1r(x)ypr(x)x2ypypypr(x)r(x)ypypr(x)ypr(x)r(x)ys ϩ ayr ϩ by ϭ r(x)ypyp᭿y ϭ Y ϩ ypyhc1, c2Yr1x02 ϭ yr1x02 Ϫ yrp1x02.Y1x02 ϭ y1x02 Ϫ yp(x0).x0Y ϭ y* Ϫ ypypSEC. 2.7 Nonhomogeneous ODEs 81c02.qxd 10/27/10 6:06 PM Page 81 We now extend our model by including an additional force, that is, the external forceon the right. Then we have(2)Mechanically this means that at each instant t the resultant of the internal forces is inequilibrium with The resulting motion is called a forced motion with forcing functionwhich is also known as input or driving force, and the solution to be obtainedis called the output or the response of the system to the driving force.Of special interest are periodic external forces, and we shall consider a driving forceof the formThen we have the nonhomogeneous ODE(2)Its solution will reveal facts that are fundamental in engineering mathematics and allowus to model resonance.Solving the Nonhomogeneous ODE (2)From Sec. 2.7 we know that a general solution of (2) is the sum of a general solutionof the homogeneous ODE (1) plus any solution of (2). To find we use the methodof undetermined coefficients (Sec. 2.7), starting from(3)By differentiating this function (chain rule!) we obtainSubstituting and into (2) and collecting the cosine and the sine terms, we getThe cosine terms on both sides must be equal, and the coefficient of the sine termon the left must be zero since there is no sine term on the right. This gives the twoequations(4)(k Ϫ mv2)b ϭ 0ϩϪvcaϭ F0vcb(k Ϫ mv2)a ϩ[(k Ϫ mv2)a ϩ vcb] cos vt ϩ [Ϫvca ϩ (k Ϫ mv2)b] sin vt ϭ F0 cos vt.yspyp, yrp,ysp ϭ Ϫv2a cos vt Ϫ v2b sin vt.yrp ϭ Ϫva sin vt ϩ vb cos vt,yp(t) ϭ a cos vt ϩ b sin vt.yp,ypyhmys ϩ cyr ϩ ky ϭ F0 cos vt.(F0 Ͼ 0, v Ͼ 0).r(t) ϭ F0 cos vty(t)r(t),r(t).mys ϩ cyr ϩ ky ϭ r(t).r(t),86 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 86 for determining the unknown coefficients a and b. This is a linear system. We can solveit by elimination. To eliminate b, multiply the first equation by and the secondby and add the results, obtainingSimilarly, to eliminate a, multiply (the first equation by and the second byand add to getIf the factor is not zero, we can divide by this factor and solve for aand b,If we set as in Sec. 2.4, then and we obtain(5)We thus obtain the general solution of the nonhomogeneous ODE (2) in the form(6)Here is a general solution of the homogeneous ODE (1) and is given by (3) withcoefficients (5).We shall now discuss the behavior of the mechanical system, distinguishing betweenthe two cases (no damping) and (damping). These cases will correspond totwo basically different types of output.Case 1. Undamped Forced Oscillations. ResonanceIf the damping of the physical system is so small that its effect can be neglected over thetime interval considered, we can set Then (5) reduces toand Hence (3) becomes (use )(7)Here we must assume that ; physically, the frequency ofthe driving force is different from the natural frequency of the system, which isthe frequency of the free undamped motion [see (4) in Sec. 2.4]. From (7) and from (4)in Sec. 2.4 we have the general solution of the "undamped system"(8)We see that this output is a superposition of two harmonic oscillations of the frequenciesjust mentioned.y(t) ϭ C cos (v0t Ϫ d) ϩF0m(v02Ϫ v2)cos vt.v0>(2p)v>(2p) [cycles>sec]v2v02yp(t) ϭF0m(v02Ϫ v2)cos vt ϭF0k[1 Ϫ (v>v0)2]cos vt.v02ϭ k>mb ϭ 0.a ϭ F0>[m(v02Ϫ v2)]c ϭ 0.c Ͼ 0c ϭ 0ypyhy(t) ϭ yh(t) ϩ yp(t).b ϭ F0vcm2(v02Ϫ v2)2ϩ v2c2.a ϭ F0m(v02Ϫ v2)m2(v02Ϫ v2)2ϩ v2c2,k ϭ mv022k>m ϭ v0 (Ͼ 0)b ϭ F0vc(k Ϫ mv2)2ϩ v2c2.a ϭ F0k Ϫ mv2(k Ϫ mv2)2ϩ v2c2,(k Ϫ mv2)2ϩ v2c2v2c2b ϩ (k Ϫ mv2)2b ϭ F0vc.k Ϫ mv2vc(k Ϫ mv2)2a ϩ v2c2a ϭ F0(k Ϫ mv2).Ϫvck Ϫ mv2SEC. 2.8 Modeling: Forced Oscillations. Resonance 87c02.qxd 10/27/10 6:06 PM Page 87 Resonance. We discuss (7). We see that the maximum amplitude of is (put(9) wheredepends on and If , then and tend to infinity. This excitation of largeoscillations by matching input and natural frequencies is called resonance. iscalled the resonance factor (Fig. 54), and from (9) we see that is the ratioof the amplitudes of the particular solution and of the input We shall seelater in this section that resonance is of basic importance in the study of vibrating systems.In the case of resonance the nonhomogeneous ODE (2) becomes(10)Then (7) is no longer valid, and, from the Modification Rule in Sec. 2.7, we conclude thata particular solution of (10) is of the formyp(t) ϭ t(a cos v0t ϩ b sin v0t).ys ϩ v02y ϭF0m cos v0t.F0 cos vt.ypr>k ϭ a0>F0r(v ϭ v0)a0rv : v0v0.va0r ϭ11 Ϫ (v>v0)2.a0 ϭF0krcos vt ϭ 1)yp88 CHAP. 2 Second-Order Linear ODEsωρω0ω1Fig. 54. Resonance factor r(v)By substituting this into (10) we find and . Hence (Fig. 55)(11) yp(t) ϭF02mv0t sin v0t.b ϭ F0>(2mv0)a ϭ 0yptFig. 55. Particular solution in the case of resonanceWe see that, because of the factor t, the amplitude of the vibration becomes larger andlarger. Practically speaking, systems with very little damping may undergo large vibrationsc02.qxd 10/27/10 6:06 PM Page 88 that can destroy the system. We shall return to this practical aspect of resonance later inthis section.Beats. Another interesting and highly important type of oscillation is obtained if isclose to . Take, for example, the particular solution [see (8)](12)Using (12) in App. 3.1, we may write this asSince is close to , the difference is small. Hence the period of the last sinefunction is large, and we obtain an oscillation of the type shown in Fig. 56, the dashedcurve resulting from the first sine factor. This is what musicians are listening to whenthey tune their instruments.v0 Ϫ vv0vy(t) ϭ2F0m(v02Ϫ v2)sin av0 ϩ v2tb sin av0 Ϫ v2tb.(v v0).y(t) ϭF0m(v02Ϫ v2)(cos vt Ϫ cos v0t)v0vSEC. 2.8 Modeling: Forced Oscillations. Resonance 89ytFig. 56. Forced undamped oscillation when the difference of the inputand natural frequencies is small ("beats")Case 2. Damped Forced OscillationsIf the damping of the mass–spring system is not negligibly small, we have anda damping term in (1) and (2). Then the general solution of the homogeneousODE (1) approaches zero as t goes to infinity, as we know from Sec. 2.4. Practically,it is zero after a sufficiently long time. Hence the "transient solution" (6) of (2),given by approaches the "steady-state solution" . This proves thefollowing.T H E O R E M 1 Steady-State SolutionAfter a sufficiently long time the output of a damped vibrating system under a purelysinusoidal driving force [see (2)] will practically be a harmonic oscillation whosefrequency is that of the input.ypy ϭ yh ϩ yp,yhcyrc Ͼ 0c02.qxd 10/27/10 6:06 PM Page 89 Amplitude of the Steady-State Solution. Practical ResonanceWhereas in the undamped case the amplitude of approaches infinity as approaches, this will not happen in the damped case. In this case the amplitude will always befinite. But it may have a maximum for some depending on the damping constant c.This may be called practical resonance. It is of great importance because if is not toolarge, then some input may excite oscillations large enough to damage or even destroythe system. Such cases happened, in particular in earlier times when less was known aboutresonance. Machines, cars, ships, airplanes, bridges, and high-rising buildings are vibratingmechanical systems, and it is sometimes rather difficult to find constructions that arecompletely free of undesired resonance effects, caused, for instance, by an engine or bystrong winds.To study the amplitude of as a function of , we write (3) in the form(13)C* is called the amplitude of and the phase angle or phase lag because it measuresthe lag of the output behind the input. According to (5), these quantities are(14)Let us see whether has a maximum and, if so, find its location and then its size.We denote the radicand in the second root in C* by R. Equating the derivative of C* tozero, we obtainThe expression in the brackets [. . .] is zero if(15)By reshuffling terms we haveThe right side of this equation becomes negative if so that then (15) has noreal solution and C* decreases monotone as increases, as the lowest curve in Fig. 57shows. If c is smaller, then (15) has a real solution where(15)From (15) we see that this solution increases as c decreases and approaches as capproaches zero. See also Fig. 57.v0vmax2ϭ v02Ϫc22m2.v ϭ vmax,c2Ͻ 2mk,vc2Ͼ 2mk,2m2v2ϭ 2m2v02Ϫ c2ϭ 2mk Ϫ c2.(v02ϭ k>m).c2ϭ 2m2(v02Ϫ v2)dCdvϭ F0 aϪ12RϪ3>2b [2m2(v02Ϫ v2)(Ϫ2v) ϩ 2vc2].C(v)tan h(v) ϭbaϭvcm(v02Ϫ v2).C(v) ϭ 2a2ϩ b2ϭF02m2(v02Ϫ v2)2ϩ v2c2,hypyp(t) ϭ C cos (vt Ϫ h).vypcvv0vyp90 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 90 The size of is obtained from (14), with given by (15). For thiswe obtain in the second radicand in (14) from (15)andThe sum of the right sides of these two formulas isSubstitution into (14) gives(16)We see that is always finite when Furthermore, since the expressionin the denominator of (16) decreases monotone to zero as goes to zero, the maximumamplitude (16) increases monotone to infinity, in agreement with our result in Case 1. Figure 57shows the amplification (ratio of the amplitudes of output and input) as a function offor hence and various values of the damping constant c.Figure 58 shows the phase angle (the lag of the output behind the input), which is lessthan when and greater than for v Ͼ v0.p>2v Ͻ v0,p>2v0 ϭ 1,m ϭ 1, k ϭ 1,vC>F0c2(Ͻ2mk)c24m2v02Ϫ c4ϭ c2(4mk Ϫ c2)c Ͼ 0.C(vmax)C(vmax) ϭ2mF0c24m2v02Ϫ c2.(c4ϩ 4m2v02c2Ϫ 2c4)>(4m2) ϭ c2(4m2v02Ϫ c2)>(4m2).vmax2c2ϭ av02Ϫc22m2b c2.m2(v02Ϫ vmax2)2ϭc44m2v2v2ϭ vmax2C(vmax)SEC. 2.8 Modeling: Forced Oscillations. Resonance 9143200 1 2c = 1c = 2c = 1_4c = 1_2CF01ωFig. 57. Amplification as a function offor and various values of thedamping constant cm ϭ 1, k ϭ 1,vC>F0ηωc = 1/2__2c = 0c = 1c = 2ππ001 2Fig. 58. Phase lag as a function of forthus and various valuesof the damping constant cv0 ϭ 1,m ϭ 1, k ϭ 1,vh1. WRITING REPORT. Free and Forced Vibrations.Write a condensed report of 2–3 pages on the mostimportant similarities and differences of free and forcedvibrations, with examples of your own. No proofs.2. Which of Probs. 1–18 in Sec. 2.7 (with time t)can be models of mass–spring systems with a harmonicoscillation as steady-state solution?x ϭ3–7 STEADY-STATE SOLUTIONSFind the steady-state motion of the mass–spring systemmodeled by the ODE. Show the details of your work.3.4.5. (D2ϩ D ϩ 4.25I)y ϭ 22.1 cos 4.5tys ϩ 2.5yr ϩ 10y ϭ Ϫ13.6 sin 4tys ϩ 6yr ϩ 8y ϭ 42.5 cos 2tP R O B L E M S E T 2 . 8c02.qxd 10/27/10 6:06 PM Page 91 2.9 Modeling: Electric CircuitsDesigning good models is a task the computer cannot do. Hence setting up models hasbecome an important task in modern applied mathematics. The best way to gain experiencein successful modeling is to carefully examine the modeling process in various fields andapplications. Accordingly, modeling electric circuits will be profitable for all students,not just for electrical engineers and computer scientists.Figure 61 shows an RLC-circuit, as it occurs as a basic building block of large electricnetworks in computers and elsewhere. An RLC-circuit is obtained from an RL-circuit byadding a capacitor. Recall Example 2 on the RL-circuit in Sec. 1.5: The model of theRL-circuit is It was obtained by KVL (Kirchhoff's Voltage Law)7byequating the voltage drops across the resistor and the inductor to the EMF (electromotiveforce). Hence we obtain the model of the RLC-circuit simply by adding the voltage dropQ C across the capacitor. Here, C F (farads) is the capacitance of the capacitor. Q coulombsis the charge on the capacitor, related to the current bySee also Fig. 62. Assuming a sinusoidal EMF as in Fig. 61, we thus have the model ofthe RLC-circuitI(t) ϭdQdt, equivalently Q(t) ϭ ΎI(t) dt.>LIr ϩ RI ϭ E(t).SEC. 2.9 Modeling: Electric Circuits 937GUSTAV ROBERT KIRCHHOFF (1824–1887), German physicist. Later we shall also need Kirchhoff'sCurrent Law (KCL):At any point of a circuit, the sum of the inflowing currents is equal to the sum of the outflowing currents.The units of measurement of electrical quantities are named after ANDRÉ MARIE AMPÈRE (1775–1836),French physicist, CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer,MICHAEL FARADAY (1791–1867), English physicist, JOSEPH HENRY (1797–1878), American physicist,GEORG SIMON OHM (1789–1854), German physicist, and ALESSANDRO VOLTA (1745–1827), Italianphysicist.R LCE(t) = E0sin ωtωFig. 61. RLC-circuitFig. 62. Elements in an RLC-circuitNameOhm's ResistorInductorCapacitorSymbol NotationR Ohm's ResistanceL InductanceC CapacitanceUnitohms (⍀)henrys (H)farads (F)Voltage DropRILQ/CdIdtc02.qxd 10/27/10 6:06 PM Page 93 This is an "integro-differential equation." To get rid of the integral, we differentiatewith respect to t, obtaining(1)This shows that the current in an RLC-circuit is obtained as the solution of thisnonhomogeneous second-order ODE (1) with constant coefficients.In connection with initial value problems, we shall occasionally useobtained from andSolving the ODE (1) for the Current in an RLC-CircuitA general solution of (1) is the sum where is a general solution of thehomogeneous ODE corresponding to (1) and is a particular solution of (1). We firstdetermine by the method of undetermined coefficients, proceeding as in the previoussection. We substitute(2)into (1). Then we collect the cosine terms and equate them to on the right,and we equate the sine terms to zero because there is no sine term on the right,(Cosine terms)(Sine terms).Before solving this system for a and b, we first introduce a combination of L and C, calledthe reactance(3)Dividing the previous two equations by ordering them, and substituting S givesϪRa Ϫ Sb ϭ 0.ϪSa ϩ Rb ϭ E0v,S ϭ vL Ϫ1vC.Lv2(Ϫb) ϩ Rv(Ϫa) ϩ b>C ϭ 0Lv2(Ϫa) ϩ Rvb ϩ a>C ϭ E0vE0v cos vtIps ϭ v2(Ϫa cos vt Ϫ b sin vt)Ipr ϭ v(Ϫa sin vt ϩ b cos vt)Ip ϭ a cos vt ϩ b sin vtIpIpIhI ϭ Ih ϩ Ip,I ϭ Qr.(1r)LQs ϩ RQs ϩ1CQ ϭ E(t),(1s)LIs ϩ RIr ϩ1CI ϭ Er(t) ϭ E0v cos vt.(1r)LIr ϩ RI ϩ1C ΎI dt ϭ E(t) ϭ E0 sin vt.(1r)94 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 94 We now eliminate b by multiplying the first equation by S and the second by R, andadding. Then we eliminate a by multiplying the first equation by R and the second byand adding. This givesWe can solve for a and b,(4)Equation (2) with coefficients a and b given by (4) is the desired particular solution ofthe nonhomogeneous ODE (1) governing the current I in an RLC-circuit with sinusoidalelectromotive force.Using (4), we can write in terms of "physically visible" quantities, namely, amplitudeand phase lag of the current behind the EMF, that is,(5)where [see (14) in App. A3.1]The quantity is called the impedance. Our formula shows that the impedanceequals the ratio This is somewhat analogous to (Ohm's law) and, becauseof this analogy, the impedance is also known as the apparent resistance.A general solution of the homogeneous equation corresponding to (1) iswhere and are the roots of the characteristic equationWe can write these roots in the form and whereNow in an actual circuit, R is never zero (hence ). From this it follows thatapproaches zero, theoretically as but practically after a relatively short time. Hencethe transient current tends to the steady-state current and after some timethe output will practically be a harmonic oscillation, which is given by (5) and whosefrequency is that of the input (of the electromotive force).Ip,I ϭ Ih ϩ Ipt : ϱ,IhR Ͼ 0b ϭBR24L2Ϫ1LCϭ12L BR2Ϫ4LC.a ϭR2L,l2 ϭ Ϫa Ϫ b,l1 ϭ Ϫa ϩ bl2ϩRLl ϩ1LCϭ 0.l2l1Ih ϭ c1el1tϩ c2el2tE>I ϭ RE0>I0.2R2ϩ S2tan u ϭ ϪabϭSR.I0 ϭ 2a2ϩ b2ϭE02R2ϩ S2,Ip(t) ϭ I0 sin (vt Ϫ u)uI0IpIpb ϭE0RR2ϩ S2.a ϭϪE0SR2ϩ S2,(R2ϩ S2)b ϭ E0R.Ϫ(S2ϩ R2)a ϭ E0S,ϪS,SEC. 2.9 Modeling: Electric Circuits 95c02.qxd 10/27/10 6:06 PM Page 95 Analogy of Electrical and Mechanical QuantitiesEntirely different physical or other systems may have the same mathematical model.For instance, we have seen this from the various applications of the ODE inChap. 1. Another impressive demonstration of this unifying power of mathematics isgiven by the ODE (1) for an electric RLC-circuit and the ODE (2) in the last section fora mass–spring system. Both equationsandare of the same form. Table 2.2 shows the analogy between the various quantities involved.The inductance L corresponds to the mass m and, indeed, an inductor opposes a changein current, having an "inertia effect" similar to that of a mass. The resistance R correspondsto the damping constant c, and a resistor causes loss of energy, just as a damping dashpotdoes. And so on.This analogy is strictly quantitative in the sense that to a given mechanical system wecan construct an electric circuit whose current will give the exact values of the displacementin the mechanical system when suitable scale factors are introduced.The practical importance of this analogy is almost obvious. The analogy may be usedfor constructing an "electrical model" of a given mechanical model, resulting in substantialsavings of time and money because electric circuits are easy to assemble, and electricquantities can be measured much more quickly and accurately than mechanical ones.mys ϩ cyr ϩ ky ϭ F0 cos vtLIs ϩ RIr ϩ1CI ϭ E0v cos vtyr ϭ kySEC. 2.9 Modeling: Electric Circuits 97yt0 0.02 0.03 0.04 0.050.012–2–31–13I(t)Fig. 63. Transient (upper curve) and steady-state currents in Example 1Table 2.2 Analogy of Electrical and Mechanical QuantitiesElectrical System Mechanical SystemInductance L Mass mResistance R Damping constant cReciprocal 1 C of capacitance Spring modulus kDerivative of} Driving forceelectromotive forceCurrent Displacement y(t)I(t)F0 cos vtE0v cos vt>c02.qxd 10/27/10 6:06 PM Page 97 11.12.13.14. Prove the claim in the text that if (hencethen the transient current approaches as15. Cases of damping. What are the conditions for anRLC-circuit to be (I) overdamped, (II) critically damped,(III) underdamped? What is the critical resistance(the analog of the critical damping constant )?16–18 Solve the initial value problem for the RLC-circuit in Fig. 61 with the given data, assuming zero initialcurrent and charge. Graph or sketch the solution. Show thedetails of your work.21mkRcritt : ϱ.IpR Ͼ 0),R 0E ϭ 12,000 sin 25t VR ϭ 12, L ϭ 1.2 H, C ϭ 203# 10؊3F,R ϭ 0.2 ⍀, L ϭ 0.1 H, C ϭ 2 F, E ϭ 220 sin 314t VE ϭ 220 sin 10t VR ϭ 12 ⍀, L ϭ 0.4 H, C ϭ 180 F,SEC. 2.10 Solution by Variation of Parameters 9916.17.18.19. WRITING REPORT. Mechanic-Electric Analogy.Explain Table 2.2 in a 1–2 page report with examples,e.g., the analog (with ) of a mass–spring systemof mass 5 kg, damping constant 10 kg sec, spring constant, and driving force20. Complex Solution Method. Solveby substituting(K unknown) and its derivatives and taking the realpart of the solution . Show agreement with (2), (4).Hint: Use (11) cf. Sec. 2.2,and i2ϭ Ϫ1.eivtϭ cos vt ϩ i sin vt;I~pIpIp ϭ Keivti ϭ 1Ϫ1,I~>C ϭ E0eivt,LI~s ϩ RI~r ϩ220 cos 10t kg>sec.60 kg>sec2>L ϭ 1 HE ϭ 820 cos 10t VR ϭ 18 ⍀, L ϭ 1 H, C ϭ 12.5 # 10؊3F,E ϭ 600 (cos t ϩ 4 sin t) VR ϭ 6 ⍀, L ϭ 1 H, C ϭ 0.04 F,E ϭ 100 sin 10t VR ϭ 8 ⍀, L ϭ 0.2 H, C ϭ 12.5 # 10؊3F,2.10 Solution by Variation of ParametersWe continue our discussion of nonhomogeneous linear ODEs, that is(1)In Sec. 2.6 we have seen that a general solution of (1) is the sum of a general solutionof the corresponding homogeneous ODE and any particular solution of (1). To obtainwhen is not too complicated, we can often use the method of undetermined coefficients,as we have shown in Sec. 2.7 and applied to basic engineering models in Secs. 2.8 and 2.9.However, since this method is restricted to functions whose derivatives are of a formsimilar to itself (powers, exponential functions, etc.), it is desirable to have a method validfor more general ODEs (1), which we shall now develop. It is called the method of variationof parameters and is credited to Lagrange (Sec. 2.1). Here p, q, r in (1) may be variable(given functions of x), but we assume that they are continuous on some open interval I.Lagrange's method gives a particular solution of (1) on I in the form(2)where form a basis of solutions of the corresponding homogeneous ODE(3)on I, and W is the Wronskian of(4) (see Sec. 2.6).CAUTION! The solution formula (2) is obtained under the assumption that the ODEis written in standard form, with as the first term as shown in (1). If it starts withdivide first by f(x).f(x)ys,ysW ϭ y1y2r Ϫ y2y1ry1, y2,ys ϩ p(x)yr ϩ q(x)y ϭ 0y1, y2yp(x) ϭ Ϫy1 Ύy2rWdx ϩ y2 Ύy1rWdxypr(x)r(x)r(x)ypypyhys ϩ p(x)yr ϩ q(x)y ϭ r(x).c02.qxd 10/27/10 6:06 PM Page 99 The integration in (2) may often cause difficulties, and so may the determination ofif (1) has variable coefficients. If you have a choice, use the previous method. It issimpler. Before deriving (2) let us work an example for which you do need the newmethod. (Try otherwise.)E X A M P L E 1 Method of Variation of ParametersSolve the nonhomogeneous ODESolution. A basis of solutions of the homogeneous ODE on any interval is . This givesthe WronskianFrom (2), choosing zero constants of integration, we get the particular solution of the given ODE(Fig. 70)Figure 70 shows and its first term, which is small, so that essentially determines the shape of the curveof . (Recall from Sec. 2.8 that we have seen in connection with resonance, except for notation.) Fromand the general solution of the homogeneous ODE we obtain the answerHad we included integration constants in (2), then (2) would have given the additionalthat is, a general solution of the given ODE directly from (2). This willalways be the case. ᭿c1 cos x ϩ c2 sin x ϭ c1y1 ϩ c2y2,Ϫc1, c2y ϭ yh ϩ yp ϭ (c1 ϩ ln ƒ cos x ƒ) cos x ϩ (c2 ϩ x) sin x.yh ϭ c1y1 ϩ c2y2ypx sin xypx sin xypϭ cos x ln ƒ cos x ƒ ϩ x sin xyp ϭ Ϫcos xΎsin x sec x dx ϩ sin x Ύcos x sec x dxW(y1, y2) ϭ cos x cos x Ϫ sin x (Ϫsin x) ϭ 1.y1 ϭ cos x, y2 ϭ sin xys ϩ y ϭ sec x ϭ1cos x.y1, y2100 CHAP. 2 Second-Order Linear ODEsyx04 82510–5–106 10 12Fig. 70. Particular solution yp and its first term in Example 1Idea of the Method. Derivation of (2)What idea did Lagrange have? What gave the method the name? Where do we use thecontinuity assumptions?The idea is to start from a general solutionyh(x) ϭ c1y1(x) ϩ c2y2(x)c02.qxd 10/27/10 6:06 PM Page 100 of the homogeneous ODE (3) on an open interval I and to replace the constants ("theparameters") and by functions and this suggests the name of the method.We shall determine u and v so that the resulting function(5)is a particular solution of the nonhomogeneous ODE (1). Note that exists by Theorem3 in Sec. 2.6 because of the continuity of p and q on I. (The continuity of r will be usedlater.)We determine u and v by substituting (5) and its derivatives into (1). Differentiating (5),we obtainNow must satisfy (1). This is one condition for two functions u and v. It seems plausiblethat we may impose a second condition. Indeed, our calculation will show that we candetermine u and v such that satisfies (1) and u and v satisfy as a second condition theequation(6)This reduces the first derivative to the simpler form(7)Differentiating (7), we obtain(8)We now substitute and its derivatives according to (5), (7), (8) into (1). Collectingterms in u and terms in v, we obtainSince and are solutions of the homogeneous ODE (3), this reduces to(9a)Equation (6) is(9b)This is a linear system of two algebraic equations for the unknown functions andWe can solve it by elimination as follows (or by Cramer's rule in Sec. 7.6). To eliminatewe multiply (9a) by and (9b) by and add, obtainingHere, W is the Wronskian (4) of To eliminate we multiply (9a) by and (9b)by and add, obtainingϪy1ry1,ury1, y2.ur(y1y2r Ϫ y2y1r) ϭ Ϫy2r, thus urW ϭ Ϫy2r.y2rϪy2vr,vr.urury1 ϩ vry2 ϭ 0.ury1r ϩ vry2r ϭ r.y2y1u(y1s ϩ py1r ϩ qy1) ϩ v(y2s ϩ py2r ϩ qy2) ϩ ury1r ϩ vry2r ϭ r.ypyps ϭ ury1r ϩ uy1s ϩ vry2r ϩ vy2s.ypr ϭ uy1r ϩ vy2r.yprury1 ϩ vry2 ϭ 0.ypypypr ϭ ury1 ϩ uy1r ϩ vry2 ϩ vy2r.yhyp(x) ϭ u(x)y1(x) ϩ v(x)y2(x)v(x);u(x)c2c1SEC. 2.10 Solution by Variation of Parameters 101c02.qxd 10/27/10 6:06 PM Page 101 For the homogeneous ODE (2) we have the important superposition principle (Sec.2.1) that a linear combination of two solutions is again a solution.Two linearly independent solutions of (2) on an open interval I form a basis(or fundamental system) of solutions on I. and with arbitraryconstants a general solution of (2) on I. From it we obtain a particularsolution if we specify numeric values (numbers) for and usually by prescribingtwo initial conditions(3) given numbers; Sec. 2.1).(2) and (3) together form an initial value problem. Similarly for (1) and (3).For a nonhomogeneous ODE (1) a general solution is of the form(4) (Sec. 2.7).Here is a general solution of (2) and is a particular solution of (1). Such acan be determined by a general method (variation of parameters, Sec. 2.10) or inmany practical cases by the method of undetermined coefficients. The latter applieswhen (1) has constant coefficients p and q, and is a power of x, sine, cosine,etc. (Sec. 2.7). Then we write (1) as(5) (Sec. 2.7).The corresponding homogeneous ODE has solutionswhere is a root of(6)Hence there are three cases (Sec. 2.2):l2ϩ al ϩ b ϭ 0.ly ϭ elx,yr ϩ ayr ϩ by ϭ 0ys ϩ ayr ϩ by ϭ r(x)r(x)ypypyhy ϭ yh ϩ yp(x0, K0, K1yr(x0) ϭ K1y(x0) ϭ K0,c2,c1c1, c2y ϭ c1y1 ϩ c2y2y1, y2y1, y2y ϭ ky1 ϩ ly2104 CHAP. 2 Second-Order Linear ODEsCase Type of Roots General SolutionI Distinct realII DoubleIII Complex y ϭ e؊ax>2(A cos vx ϩ B sin vx)Ϫ12 a Ϯ ivy ϭ (c1 ϩ c2x)eϪax>2Ϫ12 ay ϭ c1el1xϩ c2el2xl1, l2Here is used since is needed in driving forces.Important applications of (5) in mechanical and electrical engineering in connectionwith vibrations and resonance are discussed in Secs. 2.4, 2.7, and 2.8.Another large class of ODEs solvable "algebraically" consists of the Euler–Cauchyequations(7) (Sec. 2.5).These have solutions of the form where m is a solution of the auxiliary equation(8)Existence and uniqueness of solutions of (1) and (2) is discussed in Secs. 2.6and 2.7, and reduction of order in Sec. 2.1.m2ϩ (a Ϫ 1)m ϩ b ϭ 0.y ϭ xm,x2ys ϩ axyr ϩ by ϭ 0vvc02.qxd 10/27/10 6:06 PM Page 104 105C H A P T E R 3Higher Order Linear ODEsThe concepts and methods of solving linear ODEs of order extend nicely to linearODEs of higher order n, that is, etc. This shows that the theory explained inChap. 2 for second-order linear ODEs is attractive, since it can be extended in astraightforward way to arbitrary n. We do so in this chapter and notice that the formulasbecome more involved, the variety of roots of the characteristic equation (in Sec. 3.2)becomes much larger with increasing n, and the Wronskian plays a more prominent role.The concepts and methods of solving second-order linear ODEs extend readily to linearODEs of higher order.This chapter follows Chap. 2 naturally, since the results of Chap. 2 can be readilyextended to that of Chap. 3.Prerequisite: Secs. 2.1, 2.2, 2.6, 2.7, 2.10.References and Answers to Problems: App. 1 Part A, and App. 2.3.1 Homogeneous Linear ODEsRecall from Sec. 1.1 that an ODE is of nth order if the nth derivative ofthe unknown function is the highest occurring derivative. Thus the ODE is of the formwhere lower order derivatives and y itself may or may not occur. Such an ODE is calledlinear if it can be written(1)(For this is (1) in Sec. 2.1 with and .) The coefficientsand the function r on the right are any given functions of x, and y is unknown. hascoefficient 1. We call this the standard form. (If you have divide byto get this form.) An nth-order ODE that cannot be written in the form (1) is callednonlinear.If is identically zero, (zero for all x considered, usually in some openinterval I), then (1) becomes(2) y(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ 0r(x) ϵ 0r(x)pn(x)pn(x)y(n),y(n)p0, Á , pn؊1p0 ϭ qp1 ϭ pn ϭ 2y(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ r(x).F(x, y, yr, Á , y(n)) ϭ 0y(x)y(n)ϭ dny>dxnn ϭ 3, 4,n ϭ 2c03.qxd 10/27/10 6:20 PM Page 105 and is called homogeneous. If is not identically zero, then the ODE is callednonhomogeneous. This is as in Sec. 2.1.A solution of an nth-order (linear or nonlinear) ODE on some open interval I is afunction that is defined and n times differentiable on I and is such that the ODEbecomes an identity if we replace the unknown function y and its derivatives by h and itscorresponding derivatives.Sections 3.1–3.2 will be devoted to homogeneous linear ODEs and Section 3.3 tononhomogeneous linear ODEs.Homogeneous Linear ODE: Superposition Principle,General SolutionThe basic superposition or linearity principle of Sec. 2.1 extends to nth orderhomogeneous linear ODEs as follows.T H E O R E M 1 Fundamental Theorem for the Homogeneous Linear ODE (2)For a homogeneous linear ODE (2), sums and constant multiples of solutions onsome open interval I are again solutions on I. (This does not hold for anonhomogeneous or nonlinear ODE!)The proof is a simple generalization of that in Sec. 2.1 and we leave it to the student.Our further discussion parallels and extends that for second-order ODEs in Sec. 2.1.So we next define a general solution of (2), which will require an extension of linearindependence from 2 to n functions.D E F I N I T I O N General Solution, Basis, Particular SolutionA general solution of (2) on an open interval I is a solution of (2) on I of the form(3)where is a basis (or fundamental system) of solutions of (2) on I; thatis, these solutions are linearly independent on I, as defined below.A particular solution of (2) on I is obtained if we assign specific values to then constants in (3).D E F I N I T I O N Linear Independence and DependenceConsider n functions defined on some interval I.These functions are called linearly independent on I if the equation(4)implies that all are zero. These functions are called linearly dependenton I if this equation also holds on I for some not all zero.k1, Á , knk1, Á , knk1y1(x) ϩ Á ϩ knyn(x) ϭ 0 on Iy1(x), Á , yn(x)c1, Á , cny1, Á , yn(c1, Á , cn arbitrary)y(x) ϭ c1y1(x) ϩ Á ϩ cnyn(x)y ϭ h(x)r(x)106 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 106 In extension of the existence and uniqueness theorem in Sec. 2.6 we now have thefollowing.T H E O R E M 2 Existence and Uniqueness Theorem for Initial Value ProblemsIf the coefficients of (2) are continuous on some open interval Iand is in I, then the initial value problem (2), (5) has a unique solution on I.Existence is proved in Ref. [A11] in App. 1. Uniqueness can be proved by a slightgeneralization of the uniqueness proof at the beginning of App. 4.E X A M P L E 4 Initial Value Problem for a Third-Order Euler–Cauchy EquationSolve the following initial value problem on any open interval I on the positive x-axis containingSolution. Step 1. General solution. As in Sec. 2.5 we try By differentiation and substitution,Dropping and ordering gives If we can guess the root We can divideby and find the other roots 2 and 3, thus obtaining the solutions which are linearly independenton I (see Example 2). [In general one shall need a root-finding method, such as Newton's (Sec. 19.2), alsoavailable in a CAS (Computer Algebra System).] Hence a general solution isvalid on any interval I, even when it includes where the coefficients of the ODE divided by (to havethe standard form) are not continuous.Step 2. Particular solution. The derivatives are and From this, andy and the initial conditions, we get by setting(a)(b)(c)This is solved by Cramer's rule (Sec. 7.6), or by elimination, which is simple, as follows. gives(d) Then (c) (d) gives Then (c) gives Finally from (a).Answer:Linear Independence of Solutions. WronskianLinear independence of solutions is crucial for obtaining general solutions. Although it canoften be seen by inspection, it would be good to have a criterion for it. Now Theorem 2in Sec. 2.6 extends from order to any n. This extended criterion uses the WronskianW of n solutions defined as the nth-order determinant(6) W(y1, Á , yn) ϭ 5y1 y2Á yny1r y2r Á ynr# # Á #y1(n؊1)y2(n؊1) Á yn(n؊1)5 .y1, Á , ynn ϭ 2᭿y ϭ 2x ϩ x2Ϫ x3.c1 ϭ 2c2 ϭ 1.c3 ϭ Ϫ1.Ϫ 2c2 ϩ 2c3 ϭ Ϫ1.(b) Ϫ (a)ys(1) ϭ 2c2 ϩ 6c3 ϭ Ϫ4.yr(1) ϭ c1 ϩ 2c2 ϩ 3c3 ϭ 1y(1) ϭ c1 ϩ c2 ϩ c3 ϭ 2x ϭ 1ys ϭ 2c2 ϩ 6c3x.yr ϭ c1 ϩ 2c2x ϩ 3c3x2x3x ϭ 0y ϭ c1x ϩ c2x2ϩ c3x3x, x2, x3,m Ϫ 1m ϭ 1.m3Ϫ 6m2ϩ 11m Ϫ 6 ϭ 0.xmm(m Ϫ 1)(m Ϫ 2)xmϪ 3m(m Ϫ 1)xmϩ 6mxmϪ 6xmϭ 0.y ϭ xm.ys(1) ϭ Ϫ4.yr(1) ϭ 1,y(1) ϭ 2,x3yt Ϫ 3x2ys ϩ 6xyr Ϫ 6y ϭ 0,x ϭ 1.y(x)x0p0(x), Á , pn؊1(x)108 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 108 Note that W depends on x since do. The criterion states that these solutionsform a basis if and only if W is not zero; more precisely:T H E O R E M 3 Linear Dependence and Independence of SolutionsLet the ODE (2) have continuous coefficients on an open intervalI. Then n solutions of (2) on I are linearly dependent on I if and only if theirWronskian is zero for some in I. Furthermore, if W is zero for then Wis identically zero on I. Hence if there is an in I at which W is not zero, thenare linearly independent on I, so that they form a basis of solutions of (2) on I.P R O O F (a) Let be linearly dependent solutions of (2) on I. Then, by definition, thereare constants not all zero, such that for all x in I,(7)By differentiations of (7) we obtain for all x in I(8)(7), (8) is a homogeneous linear system of algebraic equations with a nontrivial solutionHence its coefficient determinant must be zero for every x on I, by Cramer'stheorem (Sec. 7.7). But that determinant is the Wronskian W, as we see from (6). HenceW is zero for every x on I.(b) Conversely, if W is zero at an in I, then the system (7), (8) with has asolution not all zero, by the same theorem. With these constants we definethe solution of (2) on I. By (7), (8) this solution satisfies theinitial conditions But another solution satisfying thesame conditions is Hence by Theorem 2, which applies since the coefficientsof (2) are continuous. Together, on I. This means lineardependence of on I.(c) If W is zero at an in I, we have linear dependence by (b) and then by (a).Hence if W is not zero at an in I, the solutions must be linearly independenton I.E X A M P L E 5 Basis, WronskianWe can now prove that in Example 3 we do have a basis. In evaluating W, pull out the exponential functionscolumnwise. In the result, subtract Column 1 from Columns 2, 3, 4 (without changing Column 1). Then expand byRow 1. In the resulting third-order determinant, subtract Column 1 from Column 2 and expand the result by Row 2:᭿W ϭ 6e؊2xe؊xexe2xϪ2e؊2xϪe؊xex2e2x4e؊2xe؊xex4e2xϪ8e؊2xϪe؊xex8e2x6 ϭ 61 1 1 1Ϫ2 Ϫ1 1 24 1 1 4Ϫ8 Ϫ1 1 86 ϭ 31 3 4Ϫ3 Ϫ3 07 9 163 ϭ 72.᭿y1, Á , ynx1W ϵ 0x0y1, Á , yny* ϭ k1y1 ϩ Á ϩ knyn ϵ 0y* ϵ yy ϵ 0.y(x0) ϭ 0, Á , y(n؊1)(x0) ϭ 0.y* ϭ k1y1 ϩ Á ϩ knynk1, Á , kn,x ϭ x0x0k1, Á , kn.k1y1(n؊1)ϩ Á ϩ knyn(n؊1)ϭ 0....k1y1r ϩ Á ϩ knynr ϭ 0n Ϫ 1k1y1 ϩ Á ϩ knyn ϭ 0.k1, Á , kny1, Á , yny1, Á , ynx1x ϭ x0,x ϭ x0y1, Á , ynp0(x), Á , pn؊1(x)y1, Á , ynSEC. 3.1 Homogeneous Linear ODEs 109c03.qxd 10/27/10 6:20 PM Page 109 A General Solution of (2) Includes All SolutionsLet us first show that general solutions always exist. Indeed, Theorem 3 in Sec. 2.6 extendsas follows.T H E O R E M 4 Existence of a General SolutionIf the coefficients of (2) are continuous on some open interval I,then (2) has a general solution on I.P R O O F We choose any fixed in I. By Theorem 2 the ODE (2) has n solutions wheresatisfies initial conditions (5) with and all other K's equal to zero. TheirWronskian at equals 1. For instance, when thenand the other initial values are zero. Thus, as claimed,Hence for any n those solutions are linearly independent on I, by Theorem 3.They form a basis on I, and is a general solution of (2) on I.We can now prove the basic property that, from a general solution of (2), every solutionof (2) can be obtained by choosing suitable values of the arbitrary constants. Hence annth-order linear ODE has no singular solutions, that is, solutions that cannot be obtainedfrom a general solution.T H E O R E M 5 General Solution Includes All SolutionsIf the ODE (2) has continuous coefficients on some open intervalI, then every solution of (2) on I is of the form(9)where is a basis of solutions of (2) on I and are suitable constants.P R O O F Let Y be a given solution and a general solution of (2) on I. Wechoose any fixed in I and show that we can find constants for which y andits first derivatives agree with Y and its corresponding derivatives at That is,we should have at(10)But this is a linear system of equations in the unknowns Its coefficientdeterminant is the Wronskian W of at Since form a basis, theyy1, Á , ynx0.y1, Á , ync1, Á , cn.c1y1(n؊1)ϩ Á ϩ cnyn(n؊1)ϭ Y(n؊1)....c1y1r ϩ Á ϩ cnynr ϭ Y rc1y1 ϩ Á ϩ cnyn ϭ Yx ϭ x0x0.n Ϫ 1c1, Á , cnx0y ϭ c1y1 ϩ Á ϩ cnynC1, Á , Cny1, Á , ynY(x) ϭ C1y1(x) ϩ Á ϩ Cnyn(x)y ϭ Y(x)p0(x), Á , pn؊1(x)᭿y ϭ c1y1 ϩ Á ϩ cnyny1, Á , ynW(y1(x0), y2(x0), y3(x0)) ϭ 4y1(x0) y2(x0) y3(x0)y1r(x0) y2r(x0) y3r(x0)y1s(x0) y2s(x0) y3s(x0)4 ϭ 41 0 00 1 00 0 14 ϭ 1.y3s(x0) ϭ 1,y1(x0) ϭ 1, y2r(x0) ϭ 1,n ϭ 3,x0Kj؊1 ϭ 1yjy1, Á , yn,x0p0(x), Á , pn؊1(x)110 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 110 are linearly independent, so that W is not zero by Theorem 3. Hence (10) has a uniquesolution (by Cramer's theorem in Sec. 7.7). With these values weobtain the particular solutionon I. Equation (10) shows that and its first derivatives agree at with Y andits corresponding derivatives. That is, and Y satisfy, at , the same initial conditions.The uniqueness theorem (Theorem 2) now implies that on I. This proves thetheorem.This completes our theory of the homogeneous linear ODE (2). Note that for it isidentical with that in Sec. 2.6. This had to be expected.n ϭ 2᭿y* ϵ Yx0yx0n Ϫ 1yy(x) ϭ C1y1(x) ϩ Á ϩ Cnyn(x)c1 ϭ C1, Á , cn ϭ CnSEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 1111–6 BASES: TYPICAL EXAMPLESTo get a feel for higher order ODEs, show that the givenfunctions are solutions and form a basis on any interval.Use Wronskians. In Prob. 6,1.2.3.4.5.6.7. TEAM PROJECT. General Properties of Solutionsof Linear ODEs. These properties are important inobtaining new solutions from given ones. Thereforeextend Team Project 38 in Sec. 2.2 to nth-order ODEs.Explore statements on sums and multiples of solutionsof (1) and (2) systematically and with proofs.Recognize clearly that no new ideas are needed in thisextension from to general n.8–15 LINEAR INDEPENDENCEAre the given functions linearly independent or dependenton the half-axis Give reason.8. 9. tan x, cot x, 1x2, 1>x2, 0x Ն 0?n ϭ 21, x2, x4, x2yt Ϫ 3xys ϩ 3yr ϭ 01, e؊xcos 2x, e؊xsin 2x, yt ϩ 2ys ϩ 5yr ϭ 0e؊4x, xe؊4x, x2e؊4x, ytϩ 12ysϩ 48yrϩ 64y ϭ 0cos x, sin x, x cos x, x sin x, yivϩ 2ys ϩ y ϭ 0ex, e؊x, e2x, yt Ϫ 2ys Ϫ yr ϩ 2y ϭ 01, x, x2, x3, yivϭ 0x Ͼ 0,P R O B L E M S E T 3 . 110. 11.12. 13.14. 15.16. TEAM PROJECT. Linear Independence andDependence. (a) Investigate the given question abouta set S of functions on an interval I. Give an example.Prove your answer.(1) If S contains the zero function, can S be linearlyindependent?(2) If S is linearly independent on a subinterval J of I,is it linearly independent on I?(3) If S is linearly dependent on a subinterval J of I,is it linearly dependent on I?(4) If S is linearly independent on I, is it linearlyindependent on a subinterval J?(5) If S is linearly dependent on I, is it linearlyindependent on a subinterval J?(6) If S is linearly dependent on I, and if T contains S,is T linearly dependent on I?(b) In what cases can you use the Wronskian fortesting linear independence? By what other means canyou perform such a test?cosh 2x, sinh 2x, e2xcos2x, sin2x, 2psin x, cos x, sin 2xsin2x, cos2x, cos 2xexcos x, exsin x, exe2x, xe2x, x2e2x3.2 Homogeneous Linear ODEswith Constant CoefficientsWe proceed along the lines of Sec. 2.2, and generalize the results from to arbitrary n.We want to solve an nth-order homogeneous linear ODE with constant coefficients,written as(1) y(n)ϩ an؊1y(n؊1)ϩ Á ϩ a1yr ϩ a0y ϭ 0n ϭ 2c03.qxd 10/27/10 6:20 PM Page 111 where etc. As in Sec. 2.2, we substitute to obtain the characteristicequation(2)of (1). If is a root of (2), then is a solution of (1). To find these roots, you mayneed a numeric method, such as Newton's in Sec. 19.2, also available on the usual CASs.For general n there are more cases than for We can have distinct real roots, simplecomplex roots, multiple roots, and multiple complex roots, respectively. This will be shownnext and illustrated by examples.Distinct Real RootsIf all the n roots of (2) are real and different, then the n solutions(3)constitute a basis for all x. The corresponding general solution of (1) is(4)Indeed, the solutions in (3) are linearly independent, as we shall see after the example.E X A M P L E 1 Distinct Real RootsSolve the ODESolution. The characteristic equation is It has the roots if you find oneof them by inspection, you can obtain the other two roots by solving a quadratic equation (explain!). Thecorresponding general solution (4) isLinear Independence of (3). Students familiar with nth-order determinants may verifythat, by pulling out all exponential functions from the columns and denoting their productby the Wronskian of the solutions in (3) becomes(5)ϭ E 71 1 Á 1l1 l2Á lnl12l22 Á ln2# # Á #l1n؊1l2n؊1 Á lnn؊17.W ϭ 7el1xel2x Á elnxl1el1xl2el2x Á lnelnxl12el1xl22el2x Á ln2elnx# # Á #l1n؊1el1xl2n؊1el2x Á lnn؊1elnx7E ϭ exp [l1 ϩ Á ϩ ln)x],᭿y ϭ c1e؊xϩ c2exϩ c3e2x.Ϫ1, 1, 2;l3Ϫ 2l2Ϫ l ϩ 2 ϭ 0.yt Ϫ 2ys Ϫ yr ϩ 2y ϭ 0.y ϭ c1el1xϩ Á ϩ cnelnx.y1 ϭ el1x, Á , yn ϭ elnx.l1, Á , lnn ϭ 2.y ϭ elxll(n)ϩ an؊1l(n؊1)ϩ Á ϩ a1l ϩ a0y ϭ 0y ϭ elxy(n)ϭ dny>dxn,112 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 112 The exponential function E is never zero. Hence if and only if the determinant onthe right is zero. This is a so-called Vandermonde or Cauchy determinant.1It can beshown that it equals(6)where V is the product of all factors with for instance, whenwe get This shows that the Wronskian is not zeroif and only if all the n roots of (2) are different and thus gives the following.T H E O R E M 1 BasisSolutions of (1) (with any real or complex 's) form abasis of solutions of (1) on any open interval if and only if all n roots of (2) aredifferent.Actually, Theorem 1 is an important special case of our more general result obtainedfrom (5) and (6):T H E O R E M 2 Linear IndependenceAny number of solutions of (1) of the form are linearly independent on an openinterval I if and only if the corresponding are all different.Simple Complex RootsIf complex roots occur, they must occur in conjugate pairs since the coefficients of (1)are real. Thus, if is a simple root of (2), so is the conjugate andtwo corresponding linearly independent solutions are (as in Sec. 2.2, except for notation)E X A M P L E 2 Simple Complex Roots. Initial Value ProblemSolve the initial value problemSolution. The characteristic equation is It has the root 1, as can perhaps beseen by inspection. Then division by shows that the other roots are Hence a general solution andits derivatives (obtained by differentiation) areys ϭ c1exϪ 100A cos 10x Ϫ 100B sin 10x.yr ϭ c1exϪ 10A sin 10x ϩ 10B cos 10x,y ϭ c1exϩ A cos 10x ϩ B sin 10x,Ϯ10i.l Ϫ 1l3Ϫ l2ϩ 100l Ϫ 100 ϭ 0.ys(0) ϭ Ϫ299.yr(0) ϭ 11,y(0) ϭ 4,yt Ϫ ys ϩ 100yr Ϫ 100y ϭ 0,y2 ϭ egxsin vx.y1 ϭ egxcos vx,l ϭ g Ϫ iv,l ϭ g ϩ ivlelxljy1 ϭ el1x, Á , yn ϭ elnxϪV ϭ Ϫ(l1 Ϫ l2)(l1 Ϫ l3)(l2 Ϫ l3).n ϭ 3j Ͻ k (Ϲ n);lj Ϫ lk(Ϫ1)n(n؊1)>2VW ϭ 0SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 1131ALEXANDRE THÉOPHILE VANDERMONDE (1735–1796), French mathematician, who worked onsolution of equations by determinants. For CAUCHY see footnote 4, in Sec. 2.5.c03.qxd 10/27/10 6:20 PM Page 113 Now let be a root of mth order of the polynomial on the right, where Forlet be the other roots, all different from Writing the polynomial inproduct form, we then havewith if and if Now comes thekey idea: We differentiate on both sides with respect to(9)The differentiations with respect to x and are independent and the resulting derivativesare continuous, so that we can interchange their order on the left:(10)The right side of (9) is zero for because of the factors (and sincewe have a multiple root!). Hence by (9) and (10). This proves that isa solution of (1).We can repeat this step and produce by another suchdifferentiations with respect to Going one step further would no longer give zero on theright because the lowest power of would then be multiplied byand because has no factors so we get precisely the solutions in (7).We finally show that the solutions (7) are linearly independent. For a specific n thiscan be seen by calculating their Wronskian, which turns out to be nonzero. For arbitrarym we can pull out the exponential functions from the Wronskian. This givestimes a determinant which by "row operations" can be reduced to the Wronskian of 1,The latter is constant and different from zero (equal toThese functions are solutions of the ODE so that linear independence followsfrom Theroem 3 in Sec. 3.1.Multiple Complex RootsIn this case, real solutions are obtained as for complex simple roots above. Consequently,if is a complex double root, so is the conjugate Correspondinglinearly independent solutions are(11)The first two of these result from and as before, and the second two fromand in the same fashion. Obviously, the corresponding general solution is(12)For complex triple roots (which hardly ever occur in applications), one would obtaintwo more solutions and so on.x2egxcos vx, x2egxsin vx,y ϭ egx[(A1 ϩ A2x) cos vx ϩ (B1 ϩ B2x) sin vx].xelxxelxelxelxegxcos vx, egxsin vx, xegxcos vx, xegxsin vx.l ϭ g Ϫ iv.l ϭ g ϩ ivy(m)ϭ 0,1!2! Á (m Ϫ 1)!).x, Á , xm؊1.(elx)mϭ elmxl Ϫ l1;h(l)h(l1) 0m!h(l)(l Ϫ l1)0,l Ϫ l1l.m Ϫ 2x2el1x, Á , xm؊1el1xxel1xL[xel1x] ϭ 0m м 2l Ϫ l1l ϭ l100lL[elx] ϭ Lc00lelxd ϭ L[xelx].l00lL[elx] ϭ m(l Ϫ l1)m؊1h(l)elxϩ (l Ϫ l1)m 00l[h(l)elx].l,m Ͻ n.h(l) ϭ (l Ϫ lmϩ1) Á (l Ϫ ln)m ϭ n,h(l) ϭ 1L[elx] ϭ (l Ϫ l1)mh(l)elxl1.lm؉1, Á , lnm Ͻ nm Ϲ n.l1SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 115c03.qxd 10/27/10 6:20 PM Page 115 An initial value problem for (1) consists of (1) and n initial conditions(4)with in I. Under those continuity assumptions it has a unique solution. The ideas ofproof are the same as those for in Sec. 2.7.Method of Undetermined CoefficientsEquation (2) shows that for solving (1) we have to determine a particular solution of (1).For a constant-coefficient equation(5)( constant) and special as in Sec. 2.7, such a can be determined bythe method of undetermined coefficients, as in Sec. 2.7, using the following rules.(A) Basic Rule as in Sec. 2.7.(B) Modification Rule. If a term in your choice for is a solution of thehomogeneous equation (3), then multiply this term by where k is the smallestpositive integer such that this term times is not a solution of (3).(C) Sum Rule as in Sec. 2.7.The practical application of the method is the same as that in Sec. 2.7. It suffices toillustrate the typical steps of solving an initial value problem and, in particular, the newModification Rule, which includes the old Modification Rule as a particular case (withor 2). We shall see that the technicalities are the same as for except perhapsfor the more involved determination of the constants.E X A M P L E 1 Initial Value Problem. Modification RuleSolve the initial value problem(6)Solution. Step 1. The characteristic equation is It has the triple rootHence a general solution of the homogeneous ODE isStep 2. If we try we get which has no solution. Try andThe Modification Rule calls forThenypt ϭ C(6 Ϫ 18x ϩ 9x2Ϫ x3)e؊x.yps ϭ C(6x Ϫ 6x2ϩ x3)e؊x,ypr ϭ C(3x2Ϫ x3)e؊x,yp ϭ Cx3e؊x.Cx2e؊x.Cxe؊xϪC ϩ 3C Ϫ 3C ϩ C ϭ 30,yp ϭ Ce؊x,ϭ (c1 ϩ c2x ϩ c3x2)e؊x.yh ϭ c1e؊xϩ c2xe؊xϩ c3x2e؊xl ϭ Ϫ1.l3ϩ 3l2ϩ 3l ϩ 1 ϭ (l ϩ 1)3ϭ 0.yt ϩ 3ys ϩ 3yr ϩ y ϭ 30e؊x, y(0) ϭ 3, yr(0) ϭ Ϫ3, ys(0) ϭ Ϫ47.n ϭ 2,k ϭ 1xkxk,yp(x)yp(x)r(x)a0, Á , an؊1y(n)ϩ an؊1y(n؊1)ϩ Á ϩ a1yr ϩ a0y ϭ r(x)n ϭ 2x0y(x0) ϭ K0, yr(x0) ϭ K1, Á , y(n؊1)(x0) ϭ Kn؊1SEC. 3.3 Nonhomogeneous Linear ODEs 117c03.qxd 10/27/10 6:20 PM Page 117 Application: Elastic BeamsWhereas second-order ODEs have various applications, of which we have discussed someof the more important ones, higher order ODEs have much fewer engineering applications.An important fourth-order ODE governs the bending of elastic beams, such as wooden oriron girders in a building or a bridge.A related application of vibration of beams does not fit in here since it leads to PDEsand will therefore be discussed in Sec. 12.3.E X A M P L E 3 Bending of an Elastic Beam under a LoadWe consider a beam B of length L and constant (e.g., rectangular) cross section and homogeneous elasticmaterial (e.g., steel); see Fig. 76. We assume that under its own weight the beam is bent so little that it ispractically straight. If we apply a load to B in a vertical plane through the axis of symmetry (the x-axis inFig. 76), B is bent. Its axis is curved into the so-called elastic curve C (or deflection curve). It is shown inelasticity theory that the bending moment is proportional to the curvature of C. We assume the bendingto be small, so that the deflection and its derivative (determining the tangent direction of C) are small.Then, by calculus, HenceEI is the constant of proportionality. E is Young's modulus of elasticity of the material of the beam. I is themoment of inertia of the cross section about the (horizontal) z-axis in Fig. 76.Elasticity theory shows further that where is the load per unit length. Together,(8) EIyivϭ f(x).f(x)Ms(x) ϭ f(x),M(x) ϭ EIys(x).k ϭ ys>(1 ϩ yr2)3>2Ϸ ys.yr(x)y(x)k(x)M(x)120 CHAP. 3 Higher Order Linear ODEs–20205 xy3010–10100Fig. 75. Particular solution of the nonhomogeneousEuler–Cauchy equation in Example 2ypLUndeformed beamDeformed beamunder uniform load(simply supported)xzyxzyFig. 76. Elastic beamc03.qxd 10/27/10 6:20 PM Page 120 Summary of Chapter 3 123Compare with the similar Summary of Chap. 2 (the case ).Chapter 3 extends Chap. 2 from order to arbitrary order n. An nth-orderlinear ODE is an ODE that can be written(1)with as the first term; we again call this the standard form. Equation(1) is called homogeneous if on a given open interval I considered,nonhomogeneous if on I. For the homogeneous ODE(2)the superposition principle (Sec. 3.1) holds, just as in the case A basis orfundamental system of solutions of (2) on I consists of n linearly independentsolutions of (2) on I. A general solution of (2) on I is a linear combinationof these,(3) ( arbitrary constants).A general solution of the nonhomogeneous ODE (1) on I is of the form(4) (Sec. 3.3).Here, is a particular solution of (1) and is obtained by two methods (undeterminedcoefficients or variation of parameters) explained in Sec. 3.3.An initial value problem for (1) or (2) consists of one of these ODEs and ninitial conditions (Secs. 3.1, 3.3)(5)with given in I and given If are continuous on I,then general solutions of (1) and (2) on I exist, and initial value problems (1), (5)or (2), (5) have a unique solution.p0, Á , pn؊1, rK0, Á , Kn؊1.x0y(x0) ϭ K0, yr(x0) ϭ K1, Á , y(n؊1)(x0) ϭ Kn؊1ypy ϭ yh ϩ ypc1, Á , cny ϭ c1y1 ϩ Á ϩ cnyny1, Á , ynn ϭ 2.y(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ 0r(x) [ 0r(x) ϵ 0y(n)ϭ dny>dxny(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ r(x)n ϭ 2n ‫؍‬ 2SUMMARY OF CHAPTER 3Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 123 124C H A P T E R 4Systems of ODEs. Phase Plane.Qualitative MethodsTying in with Chap. 3, we present another method of solving higher order ODEs inSec. 4.1. This converts any nth-order ODE into a system of n first-order ODEs. We alsoshow some applications. Moreover, in the same section we solve systems of first-orderODEs that occur directly in applications, that is, not derived from an nth-order ODE butdictated by the application such as two tanks in mixing problems and two circuits inelectrical networks. (The elementary aspects of vectors and matrices needed in this chapterare reviewed in Sec. 4.0 and are probably familiar to most students.)In Sec. 4.3 we introduce a totally different way of looking at systems of ODEs. Themethod consists of examining the general behavior of whole families of solutions of ODEsin the phase plane, and aptly is called the phase plane method. It gives information on thestability of solutions. (Stability of a physical system is desirable and means roughly that asmall change at some instant causes only a small change in the behavior of the system atlater times.) This approach to systems of ODEs is a qualitative method because it dependsonly on the nature of the ODEs and does not require the actual solutions. This can be veryuseful because it is often difficult or even impossible to solve systems of ODEs. In contrast,the approach of actually solving a system is known as a quantitative method.The phase plane method has many applications in control theory, circuit theory,population dynamics and so on. Its use in linear systems is discussed in Secs. 4.3, 4.4,and 4.6 and its even more important use in nonlinear systems is discussed in Sec. 4.5 withapplications to the pendulum equation and the Lokta–Volterra population model. Thechapter closes with a discussion of nonhomogeneous linear systems of ODEs.NOTATION. We continue to denote unknown functions by y; thus, —analogous to Chaps. 1–3. (Note that some authors use x for functions, whendealing with systems of ODEs.)Prerequisite: Chap. 2.References and Answers to Problems: App. 1 Part A, and App. 2.4.0 For Reference:Basics of Matrices and VectorsFor clarity and simplicity of notation, we use matrices and vectors in our discussionof linear systems of ODEs. We need only a few elementary facts (and not the bulk ofthe material of Chaps. 7 and 8). Most students will very likely be already familiarx1(t), x2(t)y1(t), y2(t)c04.qxd 10/27/10 9:32 PM Page 124 Using matrix multiplication and differentiation, we can now write (1) as(7) .Similarly for (2) by means of an n n matrix A and a column vector y with n components,namely, . The vector equation (7) is equivalent to two equations for thecomponents, and these are precisely the two ODEs in (1).Some Further Operations and TermsTransposition is the operation of writing columns as rows and conversely and is indicatedby T. Thus the transpose of the 2 2 matrixis .The transpose of a column vector, say,, is a row vector, ,and conversely.Inverse of a Matrix. The n n unit matrix I is the n n matrix with main diagonaland all other entries zero. If, for a given n n matrix A, there is an n nmatrix B such that , then A is called nonsingular and B is called the inverseof A and is denoted by ; thus(8) .The inverse exists if the determinant det A of A is not zero.If A has no inverse, it is called singular. For ,(9)where the determinant of A is(10) .(For general n, see Sec. 7.7, but this will not be needed in this chapter.)Linear Independence. r given vectors with n components are called alinearly independent set or, more briefly, linearly independent, if(11) c1v(1)ϩ Á ϩ crv(r)ϭ 0v(1), Á , v(r)det A ϭ 2a11 a12a21 a222 ϭ a11a22 Ϫ a12a21A؊1ϭ1det Aca22 Ϫa12Ϫa21 a11d,n ϭ 2AA؊1ϭ A؊1A ϭ IA؊1AB ϭ BA ϭ Iϫϫ1, 1, Á , 1ϫϫvTϭ [v1 v2]v ϭ cv1v2dATϭ ca11 a21a12 a22d ϭ cϪ5 132 12dA ϭ ca11 a12a21 a22d ϭ cϪ5 213 12dϫATyr ϭ Ayϫyr ϭ cyr1yr2d ϭ Ay ϭ ca11 a12a21 a22d cy1y2d, e.g., yr ϭ cϪ5 213 12d cy1y2d128 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 128 implies that all scalars must be zero; here, 0 denotes the zero vector, whose ncomponents are all zero. If (11) also holds for scalars not all zero (so that at least one ofthese scalars is not zero), then these vectors are called a linearly dependent set or, briefly,linearly dependent, because then at least one of them can be expressed as a linearcombination of the others; that is, if, for instance, in (11), then we can obtainEigenvalues, EigenvectorsEigenvalues and eigenvectors will be very important in this chapter (and, as a matter offact, throughout mathematics).Let be an n n matrix. Consider the equation(12)where is a scalar (a real or complex number) to be determined and x is a vector to bedetermined. Now, for every , a solution is . A scalar such that (12) holds forsome vector is called an eigenvalue of A, and this vector is called an eigenvectorof A corresponding to this eigenvalue .We can write (12) as or(13) .These are n linear algebraic equations in the n unknowns (the componentsof x). For these equations to have a solution , the determinant of the coefficientmatrix must be zero. This is proved as a basic fact in linear algebra (Theorem 4in Sec. 7.7). In this chapter we need this only for . Then (13) is(14) ;in components,Now is singular if and only if its determinant , called the characteristicdeterminant of A (also for general n), is zero. This gives(15)ϭ l2Ϫ (a11 ϩ a22)l ϩ a11a22 Ϫ a12a21 ϭ 0.ϭ (a11 Ϫ l)(a22 Ϫ l) Ϫ a12a21det (A Ϫ lI) ϭ 2a11 Ϫ l a12a21 a22 Ϫ l2det (A Ϫ lI)A Ϫ lIa21 x1 ϩ (a22 Ϫ l)x2 ϭ 0.(a11 Ϫ l)x1 ϩ a12 x2 ϭ 0(14)ca11 Ϫ l a12a21 a22 Ϫ ld cx1x2d ϭ c00dn ϭ 2A Ϫ lIx 0x1, Á , xn(A Ϫ lI)x ϭ 0Ax Ϫ lx ϭ 0lx 0lx ϭ 0llAx ϭ lxϫA ϭ [ajk]v(1)ϭ Ϫ1c1(c2v(2)ϩ Á ϩ crv(r)).c1 0c1, Á , crSEC. 4.0 For Reference: Basics of Matrices and Vectors 129c04.qxd 10/27/10 9:32 PM Page 129 This quadratic equation in is called the characteristic equation of A. Its solutions arethe eigenvalues of A. First determine these. Then use with todetermine an eigenvector of A corresponding to . Finally use withto find an eigenvector of A corresponding to . Note that if x is an eigenvector ofA, so is kx with any .E X A M P L E 1 Eigenvalue ProblemFind the eigenvalues and eigenvectors of the matrix(16)Solution. The characteristic equation is the quadratic equation.It has the solutions . These are the eigenvalues of A.Eigenvectors are obtained from . For we have fromA solution of the first equation is . This also satisfies the second equation. (Why?) Hence aneigenvector of A corresponding to is(17) . Similarly,is an eigenvector of A corresponding to , as obtained from with . Verify this.4.1 Systems of ODEs as Modelsin Engineering ApplicationsWe show how systems of ODEs are of practical importance as follows. We first illustratehow systems of ODEs can serve as models in various applications. Then we show how ahigher order ODE (with the highest derivative standing alone on one side) can be reducedto a first-order system.E X A M P L E 1 Mixing Problem Involving Two TanksA mixing problem involving a single tank is modeled by a single ODE, and you may first review thecorresponding Example 3 in Sec. 1.3 because the principle of modeling will be the same for two tanks. Themodel will be a system of two first-order ODEs.Tank and in Fig. 78 contain initially 100 gal of water each. In the water is pure, whereas 150 lb offertilizer are dissolved in . By circulating liquid at a rate of and stirring (to keep the mixture uniform)the amounts of fertilizer in and in change with time t. How long should we let the liquid circulateso that will contain at least half as much fertilizer as there will be left in ?T2T1T2y2(t)T1y1(t)2 gal>minT2T1T2T1᭿l ϭ l2(14)l2 ϭ Ϫ0.8x(2)ϭ c10.8dx(1)ϭ c21dl1 ϭ Ϫ2.0x1 ϭ 2, x2 ϭ 1Ϫ1.6x1 ϩ (1.2 ϩ 2.0)x2 ϭ 0.(Ϫ4.0 ϩ 2.0)x1 ϩ 4.0x2 ϭ 0(14)l ϭ l1 ϭ Ϫ2(14)l1 ϭ Ϫ2 and l2 ϭ Ϫ0.8det ƒ A Ϫ lIƒ ϭ 2Ϫ4 Ϫ l 4Ϫ1.6 1.2 Ϫ l2 ϭ l2ϩ 2.8l ϩ 1.6 ϭ 0A ϭ cϪ4.0 4.0Ϫ1.6 1.2dk 0l2x(2)l ϭ l2(14)l1x(1)l ϭ l1(14)l1 and l2l130 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 130 Solution. Step 1. Setting up the model. As for a single tank, the time rate of change of equalsinflow minus outflow. Similarly for tank . From Fig. 78 we see that(Tank )(Tank ).Hence the mathematical model of our mixture problem is the system of first-order ODEs(Tank )(Tank ).As a vector equation with column vector and matrix A this becomes.Step 2. General solution. As for a single equation, we try an exponential function of t,(1) .Dividing the last equation by and interchanging the left and right sides, we obtain.We need nontrivial solutions (solutions that are not identically zero). Hence we have to look for eigenvaluesand eigenvectors of A. The eigenvalues are the solutions of the characteristic equation(2) .We see that (which can very well happen—don't get mixed up—it is eigenvectors that must not be zero)and . Eigenvectors are obtained from in Sec. 4.0 with and . For our presentA this gives [we need only the first equation in ]and ,(Ϫ0.02 ϩ 0.04)x1 ϩ 0.02x2 ϭ 0Ϫ0.02x1 ϩ 0.02x2 ϭ 0(14)l ϭ Ϫ0.04l ϭ 0(14)l2 ϭ Ϫ0.04l1 ϭ 0det (A Ϫ lI) ϭ 2Ϫ0.02 Ϫ l 0.020.02 Ϫ0.02 Ϫ l2 ϭ (Ϫ0.02 Ϫ l)2Ϫ 0.022ϭ l(l ϩ 0.04) ϭ 0Ax ϭ lxeltlxeltϭ Axelty ϭ xelt. Then yr ϭ lxeltϭ Axeltyr ϭ Ay, where A ϭ cϪ0.02 0.020.02 Ϫ0.02dy ϭ cy1y2dT2yr2 ϭ 0.02y1 Ϫ 0.02y2T1yr1 ϭ Ϫ0.02y1 ϩ 0.02y2T2yr2 ϭ Inflow>min Ϫ Outflow>min ϭ2100y1 Ϫ2100y2T1yr1 ϭ Inflow>min Ϫ Outflow>min ϭ2100y2 Ϫ2100y1T2y1(t)yr1(t)SEC. 4.1 Systems of ODEs as Models in Engineering Applications 131T15001005027.50System of tanks ty(t)y1(t)y2(t)10075150T22 gal/min2 gal/minFig. 78. Fertilizer content in Tanks (lower curve) and T2T1c04.qxd 10/27/10 9:32 PM Page 131 The initial conditions giveHence and . As the solution of our problem we thus obtain(7)In components (Fig. 80b),Now comes an important idea, on which we shall elaborate further, beginning in Sec. 4.3. Figure 80a showsand as two separate curves. Figure 80b shows these two currents as a single curve in the-plane. This is a parametric representation with time t as the parameter. It is often important to know inwhich sense such a curve is traced. This can be indicated by an arrow in the sense of increasing t, as is shown.The -plane is called the phase plane of our system (5), and the curve in Fig. 80b is called a trajectory. Weshall see that such "phase plane representations" are far more important than graphs as in Fig. 80a becausethey will give a much better qualitative overall impression of the general behavior of whole families of solutions,not merely of one solution as in the present case. ᭿I1I2I1I2[I1(t), I2(t)]I2(t)I1(t)I2 ϭ Ϫ4e؊2tϩ 4e؊0.8t.I1 ϭ Ϫ8e؊2tϩ 5e؊0.8tϩ 3J ϭ Ϫ4x(1)e؊2tϩ 5x(2)e؊0.8tϩ a.c2 ϭ 5c1 ϭ Ϫ4I2(0) ϭ c1 ϩ 0.8c2 ϭ 0.I1(0) ϭ 2c1 ϩ c2 ϩ 3 ϭ 0134 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods0.5015432101.5I1I210254321034tI(t)(a) Currents I1(upper curve)and I2(b) Trajectory [I1(t), I2(t)]in the I1I2-plane(the "phase plane")I1(t)I2(t)Fig. 80. Currents in Example 2Remark. In both examples, by growing the dimension of the problem (from one tank totwo tanks or one circuit to two circuits) we also increased the number of ODEs (from oneODE to two ODEs). This "growth" in the problem being reflected by an "increase" in themathematical model is attractive and affirms the quality of our mathematical modeling andtheory.Conversion of an nth-Order ODE to a SystemWe show that an nth-order ODE of the general form (8) (see Theorem 1) can be convertedto a system of n first-order ODEs. This is practically and theoretically important—practically because it permits the study and solution of single ODEs by methods forsystems, and theoretically because it opens a way of including the theory of higher orderODEs into that of first-order systems. This conversion is another reason for the importanceof systems, in addition to their use as models in various basic applications. The idea ofthe conversion is simple and straightforward, as follows.c04.qxd 10/27/10 9:32 PM Page 134 136 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods15. CAS EXPERIMENT. Electrical Network. (a) InExample 2 choose a sequence of values of C thatincreases beyond bound, and compare the correspondingsequences of eigenvalues of A. What limits of thesesequences do your numeric values (approximately)suggest?(b) Find these limits analytically.(c) Explain your result physically.(d) Below what value (approximately) must you decreaseC to get vibrations?k1= 3k2= 2 (Net change inspring length= y2– y1)System inmotionSystem instaticequilibriumm1= 1(y1= 0)(y2= 0) m2= 1y1y2y2y1Fig. 81. Mechanical system in Team Project1–6 MIXING PROBLEMS1. Find out, without calculation, whether doubling theflow rate in Example 1 has the same effect as halfingthe tank sizes. (Give a reason.)2. What happens in Example 1 if we replace by a tankcontaining 200 gal of water and 150 lb of fertilizerdissolved in it?3. Derive the eigenvectors in Example 1 without consultingthis book.4. In Example 1 find a "general solution" for any ratio, tank sizes being equal.Comment on the result.5. If you extend Example 1 by a tank of the same sizeas the others and connected to by two tubes withflow rates as between and , what system of ODEswill you get?6. Find a "general solution" of the system in Prob. 5.7–9 ELECTRICAL NETWORKIn Example 2 find the currents:7. If the initial currents are 0 A and A (minus meaningthat flows against the direction of the arrow).8. If the capacitance is changed to . (Generalsolution only.)9. If the initial currents in Example 2 are 28 A and 14 A.10–13 CONVERSION TO SYSTEMSFind a general solution of the given ODE (a) by first convertingit to a system, (b), as given. Show the details of your work.10. 11.12.13. ys ϩ 2yr Ϫ 24y ϭ 0yt ϩ 2ys Ϫ yr Ϫ 2y ϭ 04ys Ϫ 15yr Ϫ 4y ϭ 0ys ϩ 3yr ϩ 2y ϭ 0C ϭ 5>27 FI2(0)Ϫ3T2T1T2T3a ϭ (flow rate)>(tank size)T114. TEAM PROJECT. Two Masses on Springs. (a) Setup the model for the (undamped) system in Fig. 81.(b) Solve the system of ODEs obtained. Hint. Tryand set . Proceed as in Example 1 or2. (c) Describe the influence of initial conditions on thepossible kind of motions.v2ϭ ly ϭ xevtP R O B L E M S E T 4 . 1It agrees with that in Sec. 2.4. For an illustrative computation, let , and . ThenThis gives the eigenvalues and . Eigenvectors follow from the first equation inwhich is . For this gives , say, , . For it gives, say, , . These eigenvectorsgiveThis vector solution has the first componentwhich is the expected solution. The second component is its derivative᭿y2 ϭ yr1 ϭ yr ϭ Ϫc1e؊0.5tϪ 1.5c2e؊1.5t.y ϭ y1 ϭ 2c1e؊0.5tϩ c2e؊1.5ty ϭ c1 c2Ϫ1d e؊0.5tϩ c2 c1Ϫ1.5d e؊1.5t.x(1)ϭ c2Ϫ1d, x(2)ϭ c1Ϫ1.5dx2 ϭ Ϫ1.5x1 ϭ 11.5x1 ϩ x2 ϭ 0l2 ϭ Ϫ1.5x2 ϭ Ϫ1x1 ϭ 20.5x1 ϩ x2 ϭ 0l1Ϫlx1 ϩ x2 ϭ 0A Ϫ lI ϭ 0,l2 ϭ Ϫ1.5l1 ϭ Ϫ0.5l2ϩ 2l ϩ 0.75 ϭ (l ϩ 0.5)(l ϩ 1.5) ϭ 0.k ϭ 0.75m ϭ 1,c ϭ 2c04.qxd 10/27/10 9:32 PM Page 136 4.2 Basic Theory of Systems of ODEs.WronskianIn this section we discuss some basic concepts and facts about system of ODEs that arequite similar to those for single ODEs.The first-order systems in the last section were special cases of the more general system(1)We can write the system (1) as a vector equation by introducing the column vectorsand (where means transposition and saves usthe space that would be needed for writing y and f as columns). This gives(1)This system (1) includes almost all cases of practical interest. For it becomesor, simply, , well known to us from Chap. 1.A solution of (1) on some interval is a set of n differentiable functionson that satisfy (1) throughout this interval. In vector from, introducing the"solution vector" (a column vector!) we can writeAn initial value problem for (1) consists of (1) and n given initial conditions(2)in vector form, , where is a specified value of t in the interval considered andthe components of are given numbers. Sufficient conditions for theexistence and uniqueness of a solution of an initial value problem (1), (2) are stated inthe following theorem, which extends the theorems in Sec. 1.7 for a single equation. (Fora proof, see Ref. [A7].)T H E O R E M 1 Existence and Uniqueness TheoremLet in (1) be continuous functions having continuous partial derivativesin some domain R of -spacecontaining the point . Then (1) has a solution on some intervalsatisfying (2), and this solution is unique.t0 Ϫ a Ͻ t Ͻ t0 ϩ a(t0, K1, Á , Kn)ty1 y2Á yn0f1>0y1, Á , 0f1>0yn, Á , 0fn>0ynf1, Á , fnK ϭ [K1Á Kn]Tt0y(t0) ϭ Ky1(t0) ϭ K1, y2(t0) ϭ K2, Á , yn(t0) ϭ Kn,y ϭ h(t).h ϭ [h1Á hn]Ta Ͻ t Ͻ by1 ϭ h1(t), Á , yn ϭ hn(t)a Ͻ t Ͻ byr ϭ f(t, y)yr1 ϭ f1(t, y1)n ϭ 1yr ϭ f(t, y).Tf ϭ [ f1Á fn]Ty ϭ [y1Á yn]Tyr1 ϭ f1(t, y1, Á , yn)yr2 ϭ f2(t, y1, Á , yn)Áyrn ϭ fn(t, y1, Á , yn).SEC. 4.2 Basic Theory of Systems of ODEs. Wronskian 137c04.qxd 10/27/10 9:32 PM Page 137 Linear SystemsExtending the notion of a linear ODE, we call (1) a linear system if it is linear inthat is, if it can be written(3)As a vector equation this becomes(3)whereThis system is called homogeneous if so that it is(4)If then (3) is called nonhomogeneous. For example, the systems in Examples 1 and3 of Sec. 4.1 are homogeneous. The system in Example 2 of that section is nonhomogeneous.For a linear system (3) we have in Theorem 1.Hence for a linear system we simply obtain the following.T H E O R E M 2 Existence and Uniqueness in the Linear CaseLet the 's and 's in (3) be continuous functions of t on an open intervalcontaining the point Then (3) has a solution y(t) on this intervalsatisfying (2), and this solution is unique.As for a single homogeneous linear ODE we haveT H E O R E M 3 Superposition Principle or Linearity PrincipleIf and are solutions of the homogeneous linear system (4) on some interval,so is any linear combination .P R O O F Differentiating and using (4), we obtain᭿ϭ A(c1 y(1)ϩ c2 y(2)) ϭ Ay.ϭ c1Ay(1)ϩ c2Ay(2)ϭ c1y(1)r ϩ c2y(2)ryr ϭ [c1 y(1)ϩ c1 y(2)]ry ϭ c1 y(1)ϩ c1 y(2)y(2)y(1)t ϭ t0.a Ͻ t Ͻ bgjajk0f1 >0y1 ϭ a11(t), Á , 0fn >0yn ϭ ann(t)g 0,yr ϭ Ay.g ϭ 0,A ϭ Da11Á a1n. Á .an1Á annT, y ϭ Dy1oynT, g ϭ Dg1ognT.yr ϭ Ay ϩ gyr1 ϭ a11(t)y1 ϩ Á ϩ a1n(t)yn ϩ g1(t)oyrn ϭ an1(t)y1 ϩ Á ϩ ann(t)yn ϩ gn(t).y1, Á , yn;138 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 138 The general theory of linear systems of ODEs is quite similar to that of a single linearODE in Secs. 2.6 and 2.7. To see this, we explain the most basic concepts and facts. Forproofs we refer to more advanced texts, such as [A7].Basis. General Solution. WronskianBy a basis or a fundamental system of solutions of the homogeneous system (4) on someinterval J we mean a linearly independent set of n solutions of (4) on thatinterval. (We write J because we need I to denote the unit matrix.) We call a correspondinglinear combination(5)a general solution of (4) on J. It can be shown that if the (t) in (4) are continuous onJ, then (4) has a basis of solutions on J, hence a general solution, which includes everysolution of (4) on J.We can write n solutions of (4) on some interval J as columns of anmatrix(6)The determinant of Y is called the Wronskian of , written(7)The columns are these solutions, each in terms of components. These solutions form abasis on J if and only if W is not zero at any in this interval. W is either identicallyzero or nowhere zero in J. (This is similar to Secs. 2.6 and 3.1.)If the solutions in (5) form a basis (a fundamental system), then (6) isoften called a fundamental matrix. Introducing a column vectorwe can now write (5) simply as(8)Furthermore, we can relate (7) to Sec. 2.6, as follows. If y and z are solutions of asecond-order homogeneous linear ODE, their Wronskian isTo write this ODE as a system, we have to set and similarly for z(see Sec. 4.1). But then becomes (7), except for notation.W(y, z)y ϭ y1, yr ϭ y1r ϭ y2W(y, z) ϭ 2y zyr zr2.y ϭ Yc.c ϭ [c1 c2Á cn]T,y(1), Á , y(n)t1W(y(1), Á , y(n)) ϭ 5y1(1)y1(2) Á y1(n)y2(1)y2(2) Á y2(n)# # Á #yn(1)yn(2) Á yn(n)5.y(1), Á , y(n)Y ϭ [y(1) Á y(n)].n ϫ ny(1), Á , y(n)ajk(c1, Á , cn arbitrary)y ϭ c1y(1) Á ϩ cny(n)y(1), Á , y(n)SEC. 4.2 Basic Theory of Systems of ODEs. Wronskian 139c04.qxd 10/27/10 9:32 PM Page 139 4.3 Constant-Coefficient Systems.Phase Plane MethodContinuing, we now assume that our homogeneous linear system(1)under discussion has constant coefficients, so that the matrix has entriesnot depending on t. We want to solve (1). Now a single ODE has the solution. So let us try(2)Substitution into (1) gives . Dividing by , we obtain theeigenvalue problem(3)Thus the nontrivial solutions of (1) (solutions that are not zero vectors) are of the form(2), where is an eigenvalue of A and x is a corresponding eigenvector.We assume that A has a linearly independent set of n eigenvectors. This holds in mostapplications, in particular if A is symmetric or skew-symmetricor has n different eigenvalues.Let those eigenvectors be and let them correspond to eigenvalues(which may be all different, or some––or even all––may be equal). Then thecorresponding solutions (2) are(4)Their Wronskian [(7) in Sec. 4.2] is given byOn the right, the exponential function is never zero, and the determinant is not zero eitherbecause its columns are the n linearly independent eigenvectors. This proves the followingtheorem, whose assumption is true if the matrix A is symmetric or skew-symmetric, or ifthe n eigenvalues of A are all different.W ϭ (y(1), Á , y(n)) ϭ 5x1(1)el1t Á x1(n)elntx2(1)el1t Á x2(n)elnt# Á #xn(1)el1t Á xn(n)elnt5 ϭ el1tϩ Á ϩlnt5x1(1) Á x1(n)x2(1) Á x2(n)# Á #xn(1) Á xn(n)5.W ϭ W(y(1), Á , y(n))y(4)ϭ x(1)el1t, Á , y(n)ϭ x(n)elnt.l1, Á , lnx(1), Á , x(n)(akj ϭ Ϫajk)(akj ϭ ajk)lAx ϭ lx.eltyr ϭ lxeltϭ Ay ϭ Axelty ϭ xelt.y ϭ Cektyr ϭ kyA ϭ [ajk]n ϫ ny؅ ϭ Ay140 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 140 T H E O R E M 1 General SolutionIf the constant matrix A in the system (1) has a linearly independent set of neigenvectors, then the corresponding solutions in (4) form a basis ofsolutions of (1), and the corresponding general solution is(5)How to Graph Solutions in the Phase PlaneWe shall now concentrate on systems (1) with constant coefficients consisting of twoODEs(6) in components,Of course, we can graph solutions of (6),(7)as two curves over the t-axis, one for each component of y(t). (Figure 80a in Sec. 4.1 showsan example.) But we can also graph (7) as a single curve in the -plane. This is a parametricrepresentation (parametric equation) with parameter t. (See Fig. 80b for an example. Manymore follow. Parametric equations also occur in calculus.) Such a curve is called a trajectory(or sometimes an orbit or path) of (6). The -plane is called the phase plane.1If we fillthe phase plane with trajectories of (6), we obtain the so-called phase portrait of (6).Studies of solutions in the phase plane have become quite important, along withadvances in computer graphics, because a phase portrait gives a good general qualitativeimpression of the entire family of solutions. Consider the following example, in whichwe develop such a phase portrait.E X A M P L E 1 Trajectories in the Phase Plane (Phase Portrait)Find and graph solutions of the system.In order to see what is going on, let us find and graph solutions of the system(8) thusy1r ϭ Ϫ3y1 ϩ y2y2r ϭ y1 Ϫ 3y2.yr ϭ Ay ϭ cϪ3 11 Ϫ3d y,y1 y2y1 y2y(t) ϭ cy1(t)y2(t)d,y1r ϭ a11 y1 ϩ a12 y2y2r ϭ a21 y1 ϩ a22 y2.y؅ ϭ Ay;y ϭ c1x(1)el1tϩ Á ϩ cnx(n)elnt.y(1), Á , y(n)SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 1411A name that comes from physics, where it is the y-(mv)-plane, used to plot a motion in terms of position yand velocity yЈ ϭ v (m ϭ mass); but the name is now used quite generally for the y1 y2-plane.The use of the phase plane is a qualitative method, a method of obtaining general qualitative informationon solutions without actually solving an ODE or a system. This method was created by HENRI POINCARÉ(1854–1912), a great French mathematician, whose work was also fundamental in complex analysis, divergentseries, topology, and astronomy.c04.qxd 10/27/10 9:32 PM Page 141 Solution. By substituting and and dropping the exponential function we getThe characteristic equation isThis gives the eigenvalues and . Eigenvectors are then obtained fromFor this is . Hence we can take . For this becomesand an eigenvector is . This gives the general solutionFigure 82 shows a phase portrait of some of the trajectories (to which more trajectories could be added if sodesired). The two straight trajectories correspond to and and the others to other choices ofThe method of the phase plane is particularly valuable in the frequent cases when solvingan ODE or a system is inconvenient of impossible.Critical Points of the System (6)The point in Fig. 82 seems to be a common point of all trajectories, and we wantto explore the reason for this remarkable observation. The answer will follow by calculus.Indeed, from (6) we obtain(9)This associates with every point a unique tangent direction of thetrajectory passing through P, except for the point , where the right side of (9)becomes . This point , at which becomes undetermined, is called a criticalpoint of (6).Five Types of Critical PointsThere are five types of critical points depending on the geometric shape of the trajectoriesnear them. They are called improper nodes, proper nodes, saddle points, centers, andspiral points. We define and illustrate them in Examples 1–5.E X A M P L E 1 (Continued ) Improper Node (Fig. 82)An improper node is a critical point at which all the trajectories, except for two of them, have the samelimiting direction of the tangent. The two exceptional trajectories also have a limiting direction of the tangentat which, however, is different.The system (8) has an improper node at 0, as its phase portrait Fig. 82 shows. The common limiting directionat 0 is that of the eigenvector because goes to zero faster than as t increases. The twoexceptional limiting tangent directions are those of and . ᭿Ϫx(2)ϭ [Ϫ1 1]Tx(2)ϭ [1 Ϫ1]Te؊2te؊4tx(1)ϭ [1 1]TP0P0dy2>dy1P00>0P ϭ P0 :(0, 0)dy2>dy1P: (y1, y2)dy2dy1ϭy2r dty1r dtϭy2ry1rϭa21 y1 ϩ a22 y2a11 y1 ϩ a12 y2.y ϭ 0᭿c1, c2.c2 ϭ 0c1 ϭ 0y ϭ cy1y2d ϭ c1 y(1)ϩ c2 y(2)ϭ c1 c11d e؊2tϩ c2 c1Ϫ1d e؊4t.x(2)ϭ [1 Ϫ1]Tx1 ϩ x2 ϭ 0,l2 ϭ Ϫ4x(1)ϭ [1 1]TϪx1 ϩ x2 ϭ 0l1 ϭ Ϫ2(Ϫ3 Ϫ l)x1 ϩ x2 ϭ 0.l2 ϭ Ϫ4l1 ϭ Ϫ2ϭ l2ϩ 6l ϩ 8 ϭ 0.det (A Ϫ lI) ϭ 2Ϫ3 Ϫ l 11 Ϫ3 Ϫ l2Ax ϭ lx.yr ϭ lxelty ϭ xelt142 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 142 E X A M P L E 2 Proper Node (Fig. 83)A proper node is a critical point at which every trajectory has a definite limiting direction and for any givendirection d at there is a trajectory having d as its limiting direction.The system(10)has a proper node at the origin (see Fig. 83). Indeed, the matrix is the unit matrix. Its characteristic equationhas the root . Any is an eigenvector, and we can take and . Hencea general solution is᭿y ϭ c1 c10d etϩ c2 c01d etory1 ϭ c1ety2 ϭ c2etor c1 y2 ϭ c2 y1.[0 1]T[1 0]Tx 0l ϭ 1(1 Ϫ l)2ϭ 0yr ϭ c1 00 1d y, thusy1r ϭ y1y2r ϭ y2P0P0SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 143y2y1y(1)(t)y(2)(t)Fig. 82. Trajectories of the system (8)(Improper node)y2y1Fig. 83. Trajectories of the system (10)(Proper node)E X A M P L E 3 Saddle Point (Fig. 84)A saddle point is a critical point at which there are two incoming trajectories, two outgoing trajectories, andall the other trajectories in a neighborhood of bypass .The system(11)has a saddle point at the origin. Its characteristic equation has the roots and. For an eigenvector is obtained from the second row of that is,. For the first row gives . Hence a general solution isThis is a family of hyperbolas (and the coordinate axes); see Fig. 84. ᭿y ϭ c1 c10d etϩ c2 c01d e؊tory1 ϭ c1ety2 ϭ c2e؊tor y1 y2 ϭ const.[0 1]Tl2 ϭ Ϫ10x1 ϩ (Ϫ1 Ϫ 1)x2 ϭ 0(A Ϫ lI)x ϭ 0,[1 0]Tl ϭ 1l2 ϭ Ϫ1l1 ϭ 1(1 Ϫ l)(Ϫ1 Ϫ l) ϭ 0yr ϭ c1 00 Ϫ1d y, thusy1r ϭ y1y1r ϭ Ϫy2P0P0P0c04.qxd 10/27/10 9:32 PM Page 143 E X A M P L E 4 Center (Fig. 85)A center is a critical point that is enclosed by infinitely many closed trajectories.The system(12)has a center at the origin. The characteristic equation gives the eigenvalues 2i and . For 2i aneigenvector follows from the first equation of , say, . For thatequation is and gives, say, . Hence a complex general solution is(12 )A real solution is obtained from (12 ) by the Euler formula or directly from (12) by a trick. (Remember thetrick and call it a method when you apply it again.) Namely, the left side of (a) times the right side of (b) is. This must equal the left side of (b) times the right side of (a). Thus,. By integration, .This is a family of ellipses (see Fig. 85) enclosing the center at the origin. ᭿2y12ϩ 12 y22ϭ constϪ4y1y1r ϭ y2y2rϪ4y1y1ry ϭ c1 c12id e2itϩ c2 c1Ϫ2id e؊2it, thusy1 ϭ c1e2itϩ c2e؊2ity2 ϭ 2ic1e2itϪ 2ic2e؊2it.[1 Ϫ2i]TϪ(Ϫ2i)x1 ϩ x2 ϭ 0l ϭ Ϫ2i[1 2i]T(A Ϫ lI)x ϭ 0Ϫ2ix1 ϩ x2 ϭ 0Ϫ2il2ϩ 4 ϭ 0yr ϭ c0 1Ϫ4 0d y, thus(a)(b)y1r ϭ y2y2r ϭ Ϫ4y1144 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsy2y1Fig. 84. Trajectories of the system (11)(Saddle point)y2y1Fig. 85. Trajectories of the system (12)(Center)E X A M P L E 5 Spiral Point (Fig. 86)A spiral point is a critical point about which the trajectories spiral, approaching as (or tracing thesespirals in the opposite sense, away from ).The system(13)has a spiral point at the origin, as we shall see. The characteristic equation is . It gives theeigenvalues and . Corresponding eigenvectors are obtained from . For(Ϫ1 Ϫ l)x1 ϩ x2 ϭ 0Ϫ1 Ϫ iϪ1 ϩ il2ϩ 2l ϩ 2 ϭ 0yr ϭ cϪ1 1Ϫ1 Ϫ1d y, thusy1r ϭ Ϫy1 ϩ y2y2r ϭ Ϫy1 Ϫ y2P0t : ϱP0P0c04.qxd 10/27/10 9:32 PM Page 144 this becomes and we can take as an eigenvector. Similarly, an eigenvectorcorresponding to is . This gives the complex general solutionThe next step would be the transformation of this complex solution to a real general solution by the Eulerformula. But, as in the last example, we just wanted to see what eigenvalues to expect in the case of a spiralpoint. Accordingly, we start again from the beginning and instead of that rather lengthy systematic calculationwe use a shortcut. We multiply the first equation in (13) by , the second by , and add, obtaining.We now introduce polar coordinates r, t, where . Differentiating this with respect to t gives. Hence the previous equation can be written, Thus, , , .For each real c this is a spiral, as claimed (see Fig. 86). ᭿r ϭ ce؊tln ƒ rƒ ϭ Ϫt ϩ c,dr>r ϭ Ϫdtrr ϭ Ϫrrrr ϭ Ϫr22rrr ϭ 2y1 yr1 ϩ 2y2 yr2r 2ϭ y12ϩ y22y1 yr1 ϩ y2 yr2 ϭ Ϫ(y12ϩ y22)y2y1y ϭ c1 c1id e(؊1؉i)tϩ c2 c1Ϫid e(؊1؊i)t.[1 Ϫi]TϪ1 Ϫ i[1 i]TϪix1 ϩ x2 ϭ 0l ϭ Ϫ1 ϩ iSEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 145y2y1Fig. 86. Trajectories of the system (13) (Spiral point)E X A M P L E 6 No Basis of Eigenvectors Available. Degenerate Node (Fig. 87)This cannot happen if A in (1) is symmetric , as in Examples 1–3) or skew-symmetricthus . And it does not happen in many other cases (see Examples 4 and 5). Hence it suffices to explainthe method to be used by an example.Find and graph a general solution of(14)Solution. A is not skew-symmetric! Its characteristic equation is.det (A Ϫ lI) ϭ 24 Ϫ l 1Ϫ1 2 Ϫ l2 ϭ l2Ϫ 6l ϩ 9 ϭ (l Ϫ 3)2ϭ 0yr ϭ Ay ϭ c4 1Ϫ1 2d y.ajj ϭ 0)(akj ϭ Ϫajk,(akj ϭ ajkc04.qxd 10/27/10 9:32 PM Page 145 It has a double root . Hence eigenvectors are obtained from , thus fromsay, and nonzero multiples of it (which do not help). The method now is to substitutewith constant into (14). (The xt-term alone, the analog of what we did in Sec. 2.2 in the caseof a double root, would not be enough. Try it.) This gives.On the right, . Hence the terms cancel, and then division by gives, thus .Here and , so that, thusA solution, linearly independent of , is . This yields the answer (Fig. 87)The critical point at the origin is often called a degenerate node. gives the heavy straight line, withthe lower part and the upper part of it. gives the right part of the heavy curve from 0 throughthe second, first, and—finally—fourth quadrants. gives the other part of that curve. ᭿Ϫy(2)y(2)c1 Ͻ 0c1 Ͼ 0c1y(1)y ϭ c1y(1)ϩ c2y(2)ϭ c1 c1Ϫ1d e3tϩ c2 £c1Ϫ1d t ϩ c01d≥ e3t.u ϭ [0 1]Tx ϭ [1 Ϫ1]Tu1 ϩ u2 ϭ 1Ϫu1 Ϫ u2 ϭ Ϫ1.(A Ϫ 3I)u ϭ c4 Ϫ 3 1Ϫ1 2 Ϫ 3d u ϭ c1Ϫ1dx ϭ [1 Ϫ1]Tl ϭ 3(A Ϫ lI)u ϭ xx ϩ lu ϭ AueltlxteltAx ϭ lxy(2)r ϭ xeltϩ lxteltϩ lueltϭ Ay(2)ϭ Axteltϩ Aueltu ϭ [u1 u2]Ty(2)ϭ xteltϩ ueltx(1)ϭ [1 Ϫ1]Tx1 ϩ x2 ϭ 0,(4 Ϫ l)x1 ϩ x2 ϭ 0l ϭ 3146 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsy2y1y(1)y(2)Fig. 87. Degenerate node in Example 6We mention that for a system (1) with three or more equations and a triple eigenvaluewith only one linearly independent eigenvector, one will get two solutions, as justdiscussed, and a third linearly independent one fromwith v from u ϩ lv ϭ Av.y(3)ϭ 12 xt2eltϩ uteltϩ veltc04.qxd 10/27/10 9:32 PM Page 146 148 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods4.4 Criteria for Critical Points. StabilityWe continue our discussion of homogeneous linear systems with constant coefficients (1).Let us review where we are. From Sec. 4.3 we have(1) in components,From the examples in the last section, we have seen that we can obtain an overview offamilies of solution curves if we represent them parametrically asand graph them as curves in the -plane, called the phase plane. Such a curve is calleda trajectory of (1), and their totality is known as the phase portrait of (1).Now we have seen that solutions are of the form. Substitution into (1) gives .Dropping the common factor , we have(2)Hence is a (nonzero) solution of (1) if is an eigenvalue of A and x a correspondingeigenvector.Our examples in the last section show that the general form of the phase portrait isdetermined to a large extent by the type of critical point of the system (1) defined as apoint at which becomes undetermined, ; here [see (9) in Sec. 4.3](3)We also recall from Sec. 4.3 that there are various types of critical points.What is now new, is that we shall see how these types of critical points are relatedto the eigenvalues. The latter are solutions and of the characteristic equation(4) .This is a quadratic equation with coefficients p, q and discriminantgiven by(5) , , .From algebra we know that the solutions of this equation are(6) , .l2 ϭ 12 (p Ϫ 1¢)l1 ϭ 12 (p ϩ 1¢)¢ ϭ p2Ϫ 4qq ϭ det A ϭ a11a22 Ϫ a12a21p ϭ a11 ϩ a22¢l2Ϫ pl ϩ q ϭ 0det (A Ϫ lI) ϭ 2a11 Ϫ l a12a21 a22 Ϫ l2 ϭ l 2Ϫ (a11 ϩ a22)l ϩ det A ϭ 0l2l ϭ l1dy2dy1ϭyr2 dtyr1 dtϭa21 y1 ϩ a22 y2a11 y1 ϩ a12 y2.0>0dy2>dy1ly(t)Ax ϭ lx.eltyr(t) ϭ lxeltϭ Ay ϭ Axelty(t) ϭ xelty1 y2y(t) ϭ [y1(t) y2(t)]Tyr1 ϭ a11 y1 ϩ a12 y2yr2 ϭ a21 y1 ϩ a22 y2.yr ϭ Ay ϭ ca11 a12a21 a22d y,c04.qxd 10/27/10 9:32 PM Page 148 Furthermore, the product representation of the equation gives.Hence p is the sum and q the product of the eigenvalues. Also from (6).Together,(7) , , .This gives the criteria in Table 4.1 for classifying critical points. A derivation will beindicated later in this section.¢ ϭ (l1 Ϫ l2)2q ϭ l1l2p ϭ l1 ϩ l2l1 Ϫ l2 ϭ 1¢l2Ϫ pl ϩ q ϭ (l Ϫ l1)(l Ϫ l2) ϭ l2Ϫ (l1 ϩ l2)l ϩ l1l2SEC. 4.4 Criteria for Critical Points. Stability 149Table 4.1 Eigenvalue Criteria for Critical Points(Derivation after Table 4.2)Name Comments on(a) Node Real, same sign(b) Saddle point Real, opposite signs(c) Center Pure imaginary(d) Spiral point Complex, not pureimaginary¢ Ͻ 0p 0q Ͼ 0p ϭ 0q Ͻ 0¢ м 0q Ͼ 0l1, l2¢ ϭ (l1 Ϫ l2)2q ϭ l1l2p ϭ l1 ϩ l2StabilityCritical points may also be classified in terms of their stability. Stability concepts are basicin engineering and other applications. They are suggested by physics, where stabilitymeans, roughly speaking, that a small change (a small disturbance) of a physical systemat some instant changes the behavior of the system only slightly at all future times t. Forcritical points, the following concepts are appropriate.D E F I N I T I O N S Stable, Unstable, Stable and AttractiveA critical point of (1) is called stable2if, roughly, all trajectories of (1) that atsome instant are close to remain close to at all future times; precisely: if forevery disk of radius with center there is a disk of radius withcenter such that every trajectory of (1) that has a point (corresponding tosay) in has all its points corresponding to in . See Fig. 90.is called unstable if is not stable.is called stable and attractive (or asymptotically stable) if is stable andevery trajectory that has a point in approaches as . See Fig. 91.Classification criteria for critical points in terms of stability are given in Table 4.2. Bothtables are summarized in the stability chart in Fig. 92. In this chart region of instabilityis dark blue.t : ϱP0DdP0P0P0P0DPt м t1Ddt ϭ t1,P1P0d Ͼ 0DdP0P Ͼ 0DPP0P0P02In the sense of the Russian mathematician ALEXANDER MICHAILOVICH LJAPUNOV (1857–1918),whose work was fundamental in stability theory for ODEs. This is perhaps the most appropriate definition ofstability (and the only we shall use), but there are others, too.c04.qxd 10/27/10 9:32 PM Page 149
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Algebra and Trig.: Graphs and Models - Text - 5th edition Summary: The Graphs and Models series by Bittinger, Beecher, Ellenbogen, and Penna is known for helping students ''see the math'' through its focus on visualization and technology. These books continue to maintain the features that have helped students succeed for years: focus on functions, visual emphasis, side-by-side algebraic and graphical solutions, and real-data applications. This package contains134.85177.45202.28 +$3.99 s/h New PaperbackshopUS Secaucus, NJ New Book. Shipped from US within 4 to 14 business days. Established seller since 2000 $247.05 +$3.99 s/h New Supreme Bookstore San Jose, CA 1-19-12 Hardback 5
Math 2 Tuesday 6-21-11Presentation Transcript Monday June 21, 2011 Math 2 Objectives and Agenda Math 2 Course Objectives Number sense:  Statistics and probability Represent and use numbers in Collect, organize and represent data equivalent forms Read and interpret data representations Understand meanings of operations and Describe data using numerical descriptions, how they relate to one another statistics and trend terminology Compute fluently and make reasonable Make and evaluate arguments or statements estimates by applying knowledge of data analysis Know and apply basic probability concepts Patterns, functions and algebra  Geometry and measurements Explore, identify, analyze, and extend Use and apply geometric properties and patterns in mathematical and adult relationships to describe the physical world contextual situations and identify and analyze the characteristics of Articulate and represent number and geometric figures data relationships using words, tables, Use transformations and symmetry to and graphs analyze mathematical situations Recognize and use algebraic symbols Specify locations and describe spatial to model mathematical and contextual relationships using coordinate geometry and situations other representational systems Analyze change in various contexts Understand measurable attributes of objects and the units, systems, and processes of measurement and apply appropriate techniques, tools and formulas to determine measurements What are we doing today? Today's ObjectivesWe will set goals for this class and our education in general.We will check our current understanding of the math concepts covered in Math 2.We will brainstorm places we see and use math in the real world. Welcome to Math 2!Here are some things you should know… Class schedule Materials to bring with you What will we be studying? Class expectations Class website and Andrea's contact info: Math survey How do you feel about math? Answer honestly - this will help me to customize our class to our interests Goal Setting What is a goal? Why is it important to set goals? Hopes and dreams for this class Hopes and dreams for education in general Math 2 -Show what you already know •Used to customize the class to our needs. •We can go more quickly over the things people know, and take more time on the areas we need more help. Math in the Real World  How can we apply the math skills we learn in class everyday?  Think about the world around us…  Each day, we will have a Math in the Real World problem to get us started with class.  Students are welcome to present a Math in the Real World problem to the class, as well!  Answer garden  Exit CardOne thing I think will be interesting about this class is ________.One question I have about this class is _________________?One thing I want Andrea to know is___________. Thanks for a great day! See you all back again tomorrow… Same time, same place!
Complex Analysis 9780387950693 ISBN: 0387950699 Pub Date: 2001 Publisher: Springer Verlag Summary: The book provides an introduction to complex analysis for students with some familiarity with complex numbers from high school. The first part comprises the basic core of a course in complex analysis for junior and senior undergraduates. The second part includes various more specialized topics as the argument principle the Poisson integral, and the Riemann mapping theorem. The third part consists of a selection of to...pics designed to complete the coverage of all background necessary for passing Ph.D. qualifying exams in complex analysis. Gamelin, Theodore W. is the author of Complex Analysis, published 2001 under ISBN 9780387950693 and 0387950699. Nine hundred fifty Complex Analysis textbooks are available for sale on ValoreBooks.com, two hundred thirteen used from the cheapest price of $25.41, or buy new starting at $42.80
Teaching Textbooks: Math books designed for homeschoolers Plain language, friendly fonts, highlighted phrases, constant review and flexibility make Teaching Textbooks a very popular math program for homeschool students. The Teaching Textbooks program is designed to make learning math in a homeschool setting the best possible experience. Since it was designed specifically for homeschoolers, the text is self-explanatory and the CD-ROM teaching allows students to work through problems with a tutor in the comfort of their own homes. Having a solution worked out for each problem is invaluable, especially in the upper levels. Write a Review: Have you used the Teaching Textbooks program to teach your children? Share your opinion of the program with other homeschoolers. What level did you use? How did it work? Do you recommend it to others? Read what others have to say.
Mathematics eBooks Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. eBookMall offers math eBooks on the subjects of in calculus, geometry, research, reference, and more. Humans have been aware of mathematics ever since they began to count objects but the ancient Greeks were the first to start a systematic study of mathematics. Applied mathematics, the branch of mathematics concerned with application of mathematical knowledge to other fields, inspires and makes use of new mathematical discoveries and sometimes leads to the development of entirely new mathematical disciplines, such as statistics and game theory. There are over 200 eBooks in the category Mathematics. Use our eBook search to find a specific book or author.
The following pages provide information on useful resources for mathematics students at BHSU. Click on a link to learn more. Math Assistance Center Located in Jonas 154, the Math Assistance Center is available to help math students in courses at all levels. Visit the MAC's webpage for current hours and other information. CAAP/Rising Juniors Exam mathematics review information As explained in the academic catalog, all schools in the South Dakota Regental System require students to complete the Collegiate Assessment of Academic Proficiency (CAAP) exam at or near the end of their second year of college. The Math Assistance Center and Mathematics Program provide students with review materials and occasional review sessions for the mathematics portion of the exam.
Elementary Algebra for College Students (8th Edition) 9780321620934 ISBN: 0321620933 Edition: 8 Pub Date: 2010 Publisher: Prentice Hall Summary: Angel, Allen R. is the author of Elementary Algebra for College Students (8th Edition), published 2010 under ISBN 9780321620934 and 0321620933. Four hundred twenty seven Elementary Algebra for College Students (8th Edition) textbooks are available for sale on ValoreBooks.com, one hundred seventy seven used from the cheapest price of $30.58, or buy new starting at $150.46the class that required me to use this book was math 101 at Rockland community college. the class was very effective especially with the professor who taught us each topic. it was a very cooperative class. there is nothing i would change about this book. it offered problems to do and even showed exactly how to do them with examples provided.
The new ebook "Engineering Mathematics: YouTube Workbook" takes learning to a new level by combining free written lessons with free online video tutorials. About the book Description "Free ebooks + free videos = better education" is the equation that describes this book's commitment to free and open education across the globe. Download the book and discover free video lessons on the Author's YouTube channel. " . Author, Dr Chris Tisdell, is a mathematician at UNSW, Sydney and a YouTube Partner in Education. He is passionate about free educational resources. Chris' YouTube mathematics videos have enjoyed a truly global reach, being seen by learners in every country on earth. Preface How to use this workbook This workbook is designed to be used in conjunction with the author's free online video tutorials. Inside this workbook each chapter is divided into learning modules (subsections), each having its own dedicated video tutorial. View the online video via the hyperlink located at the top of the page of each learning module, with workbook and paper / tablet at the ready. Or click on the Engineering Mathematics YouTube Workbook playlist where all the videos for the workbook are located in chronological order: The delivery method for each learning module in the workbook is as follows: Briefly motivate the topic under consideration; Carefully discuss a concrete example; Mention how the ideas generalize; Provide a few exercises (with answers) for the reader to try. Incorporating YouTube as an educational tool means enhanced eLearning benefits, for example, the student can easily control the delivery of learning by pausing, rewinding (or fast-forwarding) the video as needed. The subject material is based on the author's lectures to engineering students at UNSW, Sydney. The style is informal. It is anticipated that most readers will use this workbook as a revision tool and have their own set of problems to solve -- this is one reason why the number of exercises herein are limited. Two semesters of calculus is an essential prerequisite for anyone using this workbook. Content How to use this workbook About the author Acknowledgments Partial derivatives & applications Partial derivatives & partial differential equations Partial derivatives & chain rule Taylor polynomial approximations: two variables Error estimation Differentiate under integral signs: Leibniz rule Some max/min problems for multivariable functions How to determine & classify critical points More on determining & classifying critical points The method of Lagrange multipliers Another example on Lagrange multipliers More on Lagrange multipliers: 2 constraints A glimpse at vector calculus Vector functions of one variable The gradient field of a function The divergence of a vector field The curl of a vector field Introduction to line integrals More on line integrals Fundamental theorem of line integrals Flux in the plane + line integrals Double integrals and applications How to integrate over rectangles Double integrals over general regions How to reverse the order of integration How to determine area of 2D shapes Double integrals in polar co-ordinates More on integration & polar co-ordinates Ordinary differential equations Separable differential equations Linear, first–order differential equations Homogeneous, first–order ODEs 2nd–order linear ordinary differential equations Nonhomogeneous differential equations Variation of constants / parameters Laplace transforms and applications Introduction to the Laplace transform Laplace transforms + the first shifting theorem Laplace transforms + the 2nd shifting theorem Laplace transforms + differential equations Fourier series Introduction to Fourier series Odd + even functions + Fourier series More on Fourier series Applications of Fourier series to ODEs PDEs & separation of variables Deriving the heat equation Heat equation & separation of variables Heat equation & Fourier series Wave equation and Fourier series Bibliography About the Author "With more than a million YouTube hits, Dr Chris Tisdell is the equivalent of a best-selling author or chart-topping musician. And the unlikely subject of this mass popularity? University mathematics." [Sydney Morning Herald, 14/6/2012 Chris Tisdell has been inspiring, motivating and engaging large mathematics classes at UNSW, Sydney for over a decade. His lectures are performance-like, with emphasis on contextualisation, clarity in presentation and a strong connection between student and teacher. He blends the live experience with out-of-class learning, underpinned by flexibility, sharing and openness. Enabling this has been his creation, freely sharing and management of future-oriented online learning resources, known as Open Educational Resources (OER). They are designed to empower learners by granting them unlimited access to knowledge at a time, location and pace that suits their needs. This includes: hundreds of YouTube educational videos of his lectures and tutorials; an etextbook with each section strategically linked with his online videos; and live interactive classes streamed over the internet. His approach has changed the way students learn mathematics, moving from a traditional closed classroom environment to an open, flexible and forward-looking learning model. Indicators of esteem include: a prestigious educational partnership with Google; an etextbook with over 500,000 unique downloads; mathematics videos enjoying millions of hits from over 200 countries; a UNSW Vice-Chancellor's Award for Teaching Excellence; and 100% student satisfaction rating in teaching surveys across 15 different courses at UNSW over eight years. Chris has been an educational consultant to The Australian Broadcasting Corporation and has advised the Chief Scientist of Australia on educational policyChiva ★★★★★ 1 January 0001 I think it is a really good book for me as a student when I research. Also, it is really useful for every student. DAVID ★★★★★ 1 January 0001 Best of its kind but should contain simpler examples Adam ★★★★☆ 1 January 0001 very good when it comes to tackling technical problems Mohamed Anas ★★★★★ 1 January 0001 AWESOME, it covers the sufficient amount of subject areas than I expected. And yeah YOUTUBE VIDEOS, <<--- PURE AWESOME Aidan O'Brien ★★★★★ 1 January 0001 It was an amazingly useful textbook. I was doing Engineering Maths at the time and the way that Dr Tisdell explained the concepts was a breath of fresh air from the hour long lecture format. It was a great way to understand the concepts. Definitely passed the subject with a lot less effort because of this textbook than I would have otherwise.
More About This Textbook Overview After an introduction to the geometry of polynomials and a discussion of refinements of the Fundamental Theorem of Algebra, the book turns to a consideration of various special polynomials. Chebyshev and Descartes systems are then introduced, and Müntz systems and rational systems are examined in detail. Subsequent chapters discuss denseness questions and the inequalities satisfied by polynomials and rational functions. Appendices on algorithms and computational concerns, on the interpolation theorem, and on orthogonality and irrationality round off the text. The book is self-contained and assumes at most a senior-undergraduate familiarity with real and complex analysis. Editorial Reviews From the Publisher "This is a wonderful book which is strongly recommended for use in a class with students who are willing to work on the proofs, rather than to digest fully prepared and worked out proofs and examples." Jnl of Approximation
Everyone could use some help improving their math skills at one time or another! Refresh your memory on a host of various mathematic functions, including fractions, decimals, percents, algebra, equations, graphs, probability, geometry and trigonometry. Step by Step instructions and practical application problems help keep information easy to understand and relevant to everyday life. Multiple choice tests help to monitor progress and a final test allows you to see where you still need improvement. 280 pages with glossary and index.
ALGEBRA II LECTURES - PARABOLA GRAPHING Summary: These webpage videotapes show how to solve and graph step-by-step second-degree algebra II functions. The videos demonstrate easy techniques to solve and graph parabola equations by finding a few points of the functions. Links Example links This project is on a class website meant to be supplemental to an Algebra II Class. In this course page, students of the class may have an opportunity to watch videos of second-degree algebra problems being solved. At the end of the videotaped lectures, students undertake similar problems to solve and graph. Then students post the solutions for further feedback. Additionally, students can download course assignments, upload their work, and review the class schedule online. Also, students can interact among themselves by posting messages in the student zone. Students will have access to valuable educational websites. PROJECT PURPOSE The purpose of this class website is to assist students who miss classes and support English Language Learners (ELL) who cannot keep up with the lecture pace. Another reason for creating this website is to help students have an opportunity to review previously covered materials. Lastly, the rationale of the page is to facilitate students to have online-access course assignments and submit their work online, interact in the students' zone, and follow links of other useful educational websites. ALGEBRA II CLASS WEBSITE It is appropriate for the students of this class to have course-supplemental materials online so that they can download assignments or project documents, upload their class work, view the class schedule online, collaborate online (Kim, 2002), and review previously covered lectures (harasim, 2000). Additionally, students will have links to other educational pages. ASYNCHRONOUS COMMUNICATION LECTURES Conveniently, students can access archives any time they like, and it enables them to collaborate among themselves posting messages in the student zone (Taylor, 2006; Petrakou, 2010). Online course content is beneficial to students who miss classes and ELLs. They can review the material as many times as they want and prepare for exams (Harasim, 2000; Merryfield, 2003). REFERCES: Harasim, L. (2000). Shift happens: online education as a new paradigm in learning. The Internet and Higher Education, 3(1-2), 41-61
In this applet, the user applies Euler's Method to modeling population growth using the Malthus exponential model and the... see more In this applet, the user applies Euler's Method to modeling population growth using the Malthus exponential model and the Verhulst constrained growth model. After finding the Euler solution, the user can "check" the solution with the Adaptive Euler Approximation or with a slope field. Also, the user can enter an exact solution obtained from separating variables (or whatever) and again check the Euler solution graphically. This site has has interactive explanations and simulations of math from alegrbra to trigonometry. Just click the... see more This site has has interactive explanations and simulations of math from alegrbra to trigonometry. Just click the "interactive" tab on the top left menu and you can choose different simulations. It includes, the complete definition of parabolas, reaching beyond the ability to graph into the realm of why the graph appears as it does. It also has vivid descriptions of angles including circle angles for geometry. It also has calculators for principal nth roots, gdc, matrices, and prime factorization. It's definitely worth checking out. Quote from site: "A parabola is actually a locus of a point and a line. The point is called the focus and the line the directrix. That means that all points on a parabola are equidistant from the focus and the directrix. To change the equation and the graph of the interactive parabola below just click and drag either the point A, which is the focus, or point B, which controls the directrix." This is an interactive site that allows people to change the graph to understand why directrix and focus dictate parabolic graphs. To calculate the value of an apple and an orange from 2 purchases.About mental arithmetic, with a pre-algebra tool... see more To calculate the value of an apple and an orange from 2 purchases.About mental arithmetic, with a pre-algebra tool introducing the Gaussian elimination.In the mirror site, there's the Android 2.2 (and up) version of this program. This application is a basic tool for teaching algebra.Trying to solve a system of linear equations with two equations and two... see more This application is a basic tool for teaching algebra.Trying to solve a system of linear equations with two equations and two unknown variables. (Two purchases)Based on the same problem seen at the free "Adding Apples and Oranges." ( where only used integers.Now, this new program works externally with a decimal.It contains:. A button that generates new questions (״two new purchases of fruits״).. Two drop-downs to select a solution and a button to verify: The verification button checks the solution, and adds it to the list of the session.You can edit the statements. The program checks if the determinant of the proposed statement is nonzero.. The 5 tools to help resolve the problem:* Tool 1: Gauss-Jordan elimination:The two equations are represented as two purchases of fruits.Contains tools to add one column to another, multiply, divide and change the sign for to get to have one apple on a side and and an orange at the other.The result of the two additions are the values of apple and orange.* Tool 2: Graphic:Contains a graphic frame and 2 drop-downs to find the solution, using the method essay / error.The problem posed is a linear system of equations, so the results must be ever into two lines that intersect.To get the answer, one should find this point.You can also click directly on the chart.The graph is divided by the diagonal.Below the diagonal, apple values are entered, for to get the values of the orange.Above the diagonal, one enters the price of the orange.The meeting point of the two lines is represented by a yellow dot.* Tool 3: Two weighing scales:Shows two weighing balances. On the left, due to its drop-down, which determines the value proposed for apples: results gives the price of oranges in terms of the two equations.The right hand scale gives the results from the proposed price for oranges.When a scale is balanced, gives the results of both the price of the apples and oranges.* Tool 4: Cramer's Rule:This tool solves the problem at once.Contrary to what happens in other tools, here we have to manually enter the data of the problem. The two equations (two purchases) are represented in horizontal rows.If you have not entered data, the tool warns that you can not solve the problem.It can also warn when the determinant is zero:The determinant is zero when the two lines that we saw in the graphical tool, are not crossing. This happens when they are parallel. And the lines are parallel when a purchase (equation) is linear function of the other. (eg 1 A + 2 O =n1 and 2 A + 4 O = n2)The program automatically never generates such problems.* Tool 5: Inverse matrix:The tool represents the terms of the equation automatically and in landscape format (each purchase is a row)The resolution tool there are the terms of the two equations on the left and the identity matrix at right.The objective is to calculate the inverse matrix.This is achieved by operating the rows, until the data from the equations (on the left) become the identity matrix.At this time to the right we have the inverse matrix.The multiplication of the results vector by the inverse matrix gives the solution. Application on addition of fractions.Working with italian pizzas, improper fractions and mixed fractions.The program raises... see more Application on addition of fractions.Working with italian pizzas, improper fractions and mixed fractions.The program raises the problem from a hypothetical party where the guests ask for a specific pizza fraction. Guests can only choose two types of pizza for each party.The program calls for the total amount you end up ordering pizza.In order to solve the most difficult problems is a tool to help solve, finding the common denominator.There is a video as support material: soundtrack: 'O surdato nnammurato' (written by: Aniello Califano Composer: Enrico Cannio) Voice by: Beniamino Gigli) (In Neapolitan language)This program has a castilian (spanish) version and catalan version in the nummolt site:״Cuanta pizza encargamos?״ & "Quanta pizza encarreguem?״There's a free Android (V2.2 and up) version in the mirror link: (multilingual version: en, ca, es, fr, it)
Graph theory applications This text offers an introduction to the theory of graphs and its application in engineering and science. The first part covers the main graph theoretic topics: connectivity, trees, traversability, planarity, coloring, covering, matching, digraphs, networks, matrices of a graph, graph theoretic algorithms, and matroids. In the second part, these
East Elmhurst ACT Math hand...Vector Spaces can be studied in a more general sense in abstract algebra as well. This area of math is concerned with abstract mathematical structures such as groups, rings, and fields. After all, that's all a vector space is, a set of vectors that satisfy a short list of axioms.(These being th...
0387941029 9780387941028 Topology of Surfaces:This book aims to provide undergraduates with an understanding of geometric topology. Topics covered include a sampling from pointset, geometric, and algebraic topology. The presentation is pragmatic, and includes many line drawings to help the student visualize the concepts, and tries to avoid "that famous pedagogical method whereby one begins with the general and proceeds to the particular only after the student is too confused to understand even that anymore." Exercises are an integral part of the text. Students taking the course are assumed to have some exposure to algebraic ideas, as would be acquired in a linear algebra course. All of the group theory necessary is developed in the appendix. Back to top Rent Topology of Surfaces 1st edition today, or search our site for L. Christine textbooks. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Springer.
Mathematics with Applications As in previous editions, the focus in ESSENTIAL MATHEMATICS with APPLICATIONS remains on the Aufmann Interactive Method (AIM). Students are ...Show synopsisAs in previous editions, the focus in ESSENTIAL MATHEMATICS with APPLICATIONS remains on the Aufmann Interactive Method (AIM). Students are encouraged to be active participants in the classroom and in their own studies as they work through the How To examples and the paired Examples and You Try It problems. The role of "active participant" is crucial to success. Presenting students with worked examples, and then providing them with the opportunity to immediately work similar problems, helps them build their confidence and eventually master the concepts. To this point, simplicity plays a key factor in the organization of this edition, as in all other editions. All lessons, exercise sets, tests, and supplements are organized around a carefully-constructed hierarchy of objectives. This "objective-based" approach not only serves the needs of students, in terms of helping them to clearly organize their thoughts around the content, but instructors as well, as they work to design syllabi, lesson plans, and other administrative documents. The Eighth Edition features a new design, enhancing the Aufmann Interactive Method and the organization of the text around objectives, making the pages easier for both students and instructors to follow
Frequently Asked Questions You will attend one required weekly class meeting, work in a computer lab called the Math Emporium for an additional three hours each week, and complete one unit of lessons each week. Faculty and tutors are available to provide personalized assistance. The lessons are delivered by specialized software and consist of videos, homework, and quizzes. The goal was to redesign the courses to increase student learning through the effective use of technology. The content of the courses remains the same. The delivery method changed to a combination of a teacher-directed classroom meeting and a student-directed learning lab that requires much more active participation from students than passive lecture-only formats of the past. In the redesigned courses, each student works whenever it best fits his schedule as long as he attends class and meets weekly deadlines for completing homework and quizzes. Math 100 and Math 110 students must also work in the Math Emporium during scheduled hours. Once a student has mastered the content of one lesson, he can immediately start the next. The instructor will present new material for the upcoming week, follow-up on weaknesses identified by quizzes and tests, provide tips on how to learn in the emporium environment, and check on students' progress. Attendance is required. You will not be permitted to enter the Emporium without your pass. You will swipe in at one of the card scanners at the front desk, and the attendant will let you know if you have been successfully swiped in. The Math Emporium is only available to students enrolled in the residential sections of MATH 100, 110, 115, 121, and 201. Online students may not use the Math Emporium facilities. To check out of the Emporium, make sure to log out of the computer that you were using, and then return to the front desk and swipe out. Again, the attendant will let you know if you have been successfully swiped out, and you must ALWAYS swipe out. If you must leave to grab a snack, answer a phone call, or use the restroom, you do not have to log off of your computer; however, be advised that the Math Emporium staff cannot be held responsible for items left at your computer station or any work being completed on your computer. While you are in the Emporium, you are expected to work on your math assignments. After swiping in, you will sit in an area designated for your course and sign onto a computer. Each week, you will be given a set of assignments to complete.. Every 4 to 5 weeks, you will be tested over the material covered in the previous weeks. All assignments are delivered by specialized software called MyLabsPlus. The web address is Each lesson has a short video with concepts and examples followed by homework exercises for practice. The software provides immediate feedback and tutorial help. The homework may be attempted an unlimited number of times on any computer until mastery has been achieved. Instructors and tutors give immediate, personalized instruction as needed. Tests and the final exam must be worked on in the Math Emporium. The Math Emporium maintains a "lost and found" at the back of the room. Items such as phones, personal calculators, and car keys will be kept at the front desk. If you believe you have left an item at the Emporium you may come in during our scheduled hours or call us at (434)-592-5956 and inquire about the item. All items left after the last day of classes will be thrown away. A student with a documented disability may contact the Office of Disability Academic Support (ODAS) in DeMoss 2016 to arrange for academic accommodations. ODAS will notify your professor of the appropriate accommodations, which may be extended time on tests or a quiet environment. You will not be assigned to a specific computer; however, you must sit in a designated area. Each row will be labeled with a course number and an arrow. By sitting in this row, it will be easier for tutors to help you with your questions. For MATH 100 and 110 students, your class will meet in a specific row each week. It is possible that all available tutors and faculty are working with other students or unavailable to assist you. If you have been waiting for an excessive amount of time, notify the attendant at the front desk and he/she will assist you if possible. Unlike homework and test reviews, quizzes and practice tests are used to gauge your understanding of the material that you should already have learned while doing homework. Additionally, each quiz can be taken up to three times and the practice test can be taken an unlimited number of times. In each of these assignments, your highest grade will be kept. Optional lectures and test reviews are held in the Math Emporium classroom. If you are struggling with your work, please contact your professor immediately and seek additional help. Some students may benefit from one-on-one tutoring sessions. Each computer is connected to a power strip below the desk. If your computer suddenly shuts off, check to see that it is securely plugged in. Any work that you have completed will be saved automatically. This includes work completed during a test. If your computer is securely plugged in and still will not turn on or receives a blue screen error, place your red cup on top of the computer and someone will assist you Many times, people will accidentally hit the caps lock button. Since all passwords are case sensitive, check to see that caps lock is not enabled. If your password does not work for the math website, simply close the browser and reopen it. If the problem persists, place your red cup on top of the computer and someone will assist you. During peak hours it is possible that the servers will be overloaded. This may cause your computer to run slower and load webpages/ modules slowly. If you have already completed your hours for the week, you may complete the remaining assignments in your dorm or apartment. If you fall behind due to an illness or emergency, please contact your instructor as soon as you can. You are more than welcome to stay in the Emporium and work on math as long as you need to catch up. Tutors will always be happy to assist you. Please understand though that excessive use of the tutors' help hinders their ability to assist other students.
What's New in Mathematica 9 Jon McLoone Mathematica 9 adds major computational areas and introduces a new interface paradigm—further expanding Mathematica's unrivaled base of algorithmic, knowledge, and interface capabilities. Get an overview of what's new in Mathematica 9 in this video Includes Spanish audio. Using interactive manipulations created with Mathematica, the Illinois State Water Survey studies groundwater recharge and discharge, as deemed critical by the National Research Council. Hydrogeologist Yu-Feng Lin explains the advantages gained from using Mathematica in this video. Brian Frezza describes how Emerald Therapeutics uses Mathematica for all tasks in the company's antiviral research workflow, from developing functions to processing and storing data, designing and managing experiments, presenting findings, and even controlling lab instruments Chinese Russian Spanish audio. Mathematica gives students the power to manipulate interactive graphics and develop complex data models. High-school teacher Abby Brown shares the success she experiences by using Mathematica in her classroom. Mathematica gives students the power to manipulate interactive graphics and develop complex data models. High-school teacher Abby Brown shares the success she experiences by using Mathematica in her classroom. Includes Spanish audio Includes Spanish audio. In this talk from the Wolfram Technology Conference 2011, Yves Klett from the Institute of Aircraft Design & University of Stuttgart, Germany, shares a few mechanical engineering examples that make good use of Mathematica.
11-NewPreAlgLessonBOct7Order of Operations lesson plan linda bolin lesson title simplifying expressions using the order of operations course pre-algebra date october 11-NewPreAlgLessonBOct7Order of Operations MATHEMATICS PROGRAMS AND RESOURCES A DESCRIPTION 3 mcdougal littell larson learning larsons algebra 1 larsonlearning-com is a multimedia mathematics program that can be MATHEMATICS PROGRAMS AND RESOURCES A DESCRIPTION
.... More. It also has vivid descriptions of angles including circle angles for geometry. It also has calculators for principal nth roots, gdc, matrices, and prime factorization. It's definitely worth checking out. Quote from site: "A parabola is actually a locus of a point and a line. The point is called the focus and the line the directrix. That means that all points on a parabola are equidistant from the focus and the directrix. To change the equation and the graph of the interactive parabola below just click and drag either the point A, which is the focus, or point B, which controls the directrix." This is an interactive site that allows people to change the graph to understand why directrix and focus dictate parabolic graphs. Discussion for Math Warehouse Anhtai Huynh (Student) I only had a chance to look through it briefly, going through the main menu (the links), probably for about 10 minutes or so, and so far it looks pretty useful for a student taking an Algebra course. I looked at the links provided, and it's pretty useful for both student and teachers who want to review the algebraic concepts. The content is pretty simple to use. Under the main menu, there are links that takes you to the different concepts, which provides an brief explanation about what it is, and then continues on to show examples of how it is used in the math. It takes you from angles and circles all the way to vectors, and although it is not as thorough and detailed like it would be in a textbook, it does provide to be a good reference. They even have math games and riddles for the student to ponder and try out as well. It's effective as a reference, and maybe a brief study on the concepts before taking it head on in a math course or text. A student can prepare him/herself using this site. The layout of the site isn't as intuitive as other sights, so it might be a little harder on the eyes to navigate around. But once you've learn to navigate around, it makes it a lot easier for you to bookmark and figure out where to go.
Just the math skills you need to excel in the study or practice of engineering Good math skills are indispensable for all engineers regardless of their specialty, yet only a relatively small portion of the math that engineering students study in college mathematics courses is used on a frequent basis in the study or practice of engineering. That's... more... Trigonometry has always been the black sheep of mathematics. It has a reputation as a dry and difficult subject, a glorified form of geometry complicated by tedious computation. In this book, Eli Maor draws on his remarkable talents as a guide to the world of numbers to dispel that view. Rejecting the usual arid descriptions of sine, cosine, and theirPraise for the Second Edition "This book is an excellent introduction to the wide field of boundary value problems."—Journal of Engineering Mathematics "No doubt this textbook will be useful for both students and research workers."—Mathematical Reviews A new edition of the highly-acclaimed guide to boundary value problems, now featuring
History of Mathematics An Introduction 9780072885231 ISBN: 0072885238 Publisher: McGraw-Hill Higher Education Summary: This text is designed for the junior/senior mathematics major who intends to teach mathematics in high school or college. It concentrates on the history of those topics typically covered in an undergraduate curriculum or in elementary schools or high schools. At least one year of calculus is a prerequisite for this course. This book contains enough material for a 2 semester course but it is flexible enough to be used... in the more common 1 semester course. Burton, David M. is the author of History of Mathematics An Introduction, published under ISBN 9780072885231 and 0072885238. Eleven History of Mathematics An Introduction textbooks are available for sale on ValoreBooks.com, six used from the cheapest price of $28.55, or buy new starting at $94
Supplemental Workshops in Math help students master math concepts and promote collaborative learning, including concept enhancement, problem-solving and mock exams. Workshops are held twice a week and earn one unit of academic credit. Because the workshops are interactive, the more students are involved, the more they learn. The workshops are guided by trained facilitators who are upper-division students Math workshops at Cal Poly have been successful in providing unique environments for study and learning since 1988. Workshops are designed to operate directly with specific sections of math courses so that all participants are working on similar material from the same instructor at each workshop session. Supplemental Workshops in Math students focus on learning more, achieving more in their math courses and excelling in their math courses; as a result, students in math workshops achieve high percentages of A, B and C grades in their math courses
Algebra for College Students _ Text Only - 6th edition ISBN13:978-0073384344 ISBN10: 0073384348 This edition has also been released as: ISBN13: 978-0077441371 ISBN10: 0077441370 Summary: The Dugopolski series in developmental mathematics has helped thousands of students succeed in their developmental math courses.Algebra for College Students,6eis part of the latest offerings in the successful Dugopolski series in mathematics. In his books, students and faculty will find short, precise explanations of terms and concepts written in clear, understandable language that is mathematically accurate. Dugopolski also includes a double cross-referencing system be...show moretween the examples and exercise sets, so no matter where the students start, they will see the connection between the two. Finally, the author finds it important to not only provide quality but also a wide variety and quantity of exercises and applications26.2527.6200 +$3.99 s/h Good txtbroker Murfreesboro, TN 6th Edition. Used - Good. Used books do not include online codes or other supplements unless noted. Choose EXPEDITED shipping for faster delivery! n $4047 +$3.99 s/h New Lyric Vibes Geneva, IL Hardcover New 0073384348 New Condition ~~~ Right off the Shelf-BUY NOW & INCREASE IN KNOWLEDGE... $132
Schaum's Outlines present all the essential course information in an easy-to-follow, topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills. Suitable for an introductory combinatorics course lasting one or two semesters, this book includes an extensive list of problems, ranging from routine exercises to research questions. It walks the reader
Find an East Rutherford CalCalculus is used extensively in numerous fields: business, physics, biology, medicine, engineering. Its ability to deal with change makes it a useful tool for describing the constantly changing world. Without calculus, modern science and technology would not exist.
The Hampshire College Summer Studies in Mathematics is a rigorous math program that is both demanding and expanding. Participants are expected to spend a major portion of each day actively engaged in learning, doing, and sharing mathematics
Using History to Teach Mathematics : An International Perspective - 00 edition Marketplace Sellers for Using History to Teach Mathematics : An International Perspective This book is a collection of articles by international specialists in the history of mathematics and its use in teaching, based on presentations g...show moreiven at an international conference in 1996. Although the articles vary in technical or educational level and in the level of generality, they show how and why an understanding of the history of mathematics is necessary for informed teaching of various subjects in the mathematics curriculum, both at secondary and at university levels. Many of the articles can serve teachers directly as the basis of classroom lessons, while others will give teachers plenty to think about in designing courses or entire curricula. For example, there are articles dealing with the teaching of geometry and quadratic equations to high school students, of the notion of pi at various levels, and of linear algebra, combinatorics, and geometry to university students. But there is also an article showing how to use historical problems in various courses and one dealing with mathematical anomalies and their classroom use.Although the primary aim of the book is the teaching of mathematics through its history, some of the articles deal more directly with topics in the history of mathematics not usually found in textbooks. These articles will give teachers valuable background. They include one on the background of Mesopotamian mathematics by one of the world's experts in this field, one on the development of mathematics in Latin America by a mathematician who has done much primary research in this little known field, and another on the development of mathematics in Portugal, a country whose mathematical accomplishments are little known. Finally, an article on the reasons for studying mathematics in Medieval Islam will give all teachers food for thought when they discuss similar questions, while a short play dealing with the work of Augustus DeMorgan will help teachers provide an outlet for their class thespians. ...show less $41
algebra, 2e, is a book for the student. The author's goal is to help build students' confidence, their understanding and appreciation of math author has framed three goals for Prealgebra: 1) reduce math anxiety, 2) teach for understanding, and 3) foster critical thinking and enthusiasm. The author's writing style is extremely student-friendly. He talks to students in their own language and walks them through the concepts, explaining not only how to do the math, but also why it works and where it comes from, rather than using the "monkey-see, monkey-do" approach that some books take. The second edition has been revised to include an increase in mid-level examples and exercises. In addition, the explanations and annotations are now more concise, yet keep the ever-important thoroughness that is the Carson style. Whole Numbers; Integers; Expressions and Polynomials; Equations; Fractions and Rational Expressions; Decimals; Ratios, Proportions, and Measurement; Percents; Statistics and Graphs For all readers interested in prealgebra.
Arithmetic and Algebra Again: Leaving Math Anxiety Behind Forever Book Description: The bestselling guide updated and expanded for today's mathphobes Written by two pioneers of the concept of math anxiety and how to overcome it, Arithmetic and Algebra Again has helped tens of thousands of people conquer their irrational fear of math. This revised and expanded second edition of the perennial bestseller: Features the latest techniques for breaking through common anxieties about numbers Takes a real-world approach that lets mathphobes learn the math they need as they need it Covers all key math areas--from whole numbers and fractions to basic algebra Features a section on practical math for banking, mortgages, interest, and statistics and probability Includes a new section on the graphing calculator, a chapter on the metric system, a section on word problems, and all updated exercises
More About This Textbook Overview This third edition of Arem's CONQUERING MATH ANXIETY workbook presents a comprehensive, multifaceted treatment approach to reduce math anxiety and math avoidance. The author offers tips on specific strategies, as well as relaxation exercises. The book's major focus is to encourage students to take action. Hands-on activities help readers explore both the underlying causes of their problem and viable solutions. Many activities are followed by illustrated examples completed by other students. The free accompanying CD contains recordings of powerful relaxation and visualization exercises for reducing math anxiety. Related Subjects Meet the Author Cynthia A. Arem is Chair of the Social Sciences department at Pima Community College. She received her Ph.D. in Educational Psychology from the University of Arizona, and was Math and Sciences Counselor at Pima Community College from 1982 to 1998. She is President of National League of American Pen Women, Tucson Branch
The Common Core State Standards for Mathematics are notable for raising the rigor of student language demands during math instruction. Students are expected to understand complex problems, engage in constructive classroom conversations about math, and clearly support their reasoning with evidence. In this course teachers will be provided with a range of practical tools for gathering and analyzing language samples that show how students learn and what supports they need in elementary math classrooms. This course on Division and Multiplication of Whole Numbers introduces a learning trajectory approach to students' multiplicative reasoning, exploring a stronger conceptual basis for multiplicative reasoning, so that, eventually, multiplication and division of fractions is an extension of multiplication and division of whole numbers, instead of a new and mystifying monster of its own. This course offers participants an opportunity to engage in a community of learners using an inquiry cycle focusing on math formative assessments as a strategy for implementing CCSS in math. It focuses on the implementation of a Classroom Challenge: a 1 – 2 day lesson developed by the Mathematics Assessment Project (MAP) based on formative assessment and the CCSSM. M2O2C2 provides a first taste of multivariable differential calculus. By introducing the machinery of linear algebra, this course provides helpful tools for understanding the derivative of a function of many variables.
Linear Algebra Syllabus Course Description: Linear algebra is a fundamental branch of mathematics dealing with the solution of linear equations. We will consider systems of linear equations, properties of matrices and determinants, vector spaces, linear transformations, inner products, orthogonality, eigenvectors and eigenvalues, and the canonical representation of linear transformations. Office Hours: 10am MTWF, 2pm MTW. Other times by appointment. My schedule and office hours are also listed on the webpage. Practice Problems: Once material has been covered in class it is expected that you will work through problems in the relevant section of the text. This daily homework is the absolute minimum work required to succeed in the course. Homework: Regular assignments and their due date will be announced in class. These assignments will be graded, and are due at the beginning of the relevant class. Exams: Exams will be held in class on Monday February 2nd, Monday March 2nd, and Monday April 6th. The Final Exam will be held on Monday May 4th, 9:00am - 11:00am. Attendance: Students are expected to be present at every class. Success in this course requires regular attendance. Grading: Grades will be based on 3 one-hour exams (a total of 300 points), homework assignments (equally weighted and scaled to a total of 100 points), a final project (100 points) and a final comprehensive exam (150 points). A total of 650 points are available, and the cut-o s for the final letter grade are as follows: A 85% B 70% C 60% D 50% B+ 80% C+ 67% D+ 57% F Below 50% Makeup Exams: Departmental policy dictates that make-up exams are to be given under extenuating circumstances only. No make-up quizzes will be given.
Website Detail Page published by the Concord Consortium supported by the National Science Foundation This is the portal to SmartGraphs, a project developed to promote learner understanding of graph functions through use of interactive digital graphing tools. The project goals are threefold: 1) Development of free digital graphing activities which are scaffolded to allow learner inputs, 2) Classroom testing of the graphing tools to document their effectiveness, and 3) Dissemination of free authoring tools for teachers to allow creation and sharing of new activities as open education resources. Each activity includes the interactive activity, lesson plan, and assessment with answer key. This item is part of the Concord Consortium, a nonprofit research and development organization dedicated to transforming education through technology. The Concord Consortium develops deeply digital learning innovations for science, mathematics, and engineering. Please note that this resource requires Java. Editor's Note:SmartGraphs are appropriate for use in physical science courses from Grade 6-10, but can also be adapted for preparatory physics courses to assist in student remediation. Users must register to access full functionality of all the tools available with SmartGraphs, which include graph sketching, acquiring and sharing real-time data, creating databases for classroom record-keeping and assessment, and access to authoring tools. Standards (20) AAAS Benchmark Alignments (2008 Version) 2. The Nature of Mathematics 2B. Mathematics, Science, and Technology 6-8: 2B/M1. Mathematics is helpful in almost every kind of human endeavor—from laying bricks to prescribing medicine or drawing a face. 9-12: 2B/H3. Mathematics provides a precise language to describe objects and events and the relationships among them. In addition, mathematics provides tools for solving problems, analyzing data, and making logical arguments. 9. The Mathematical World 9B. Symbolic Relationships 6-8: 9B/M3. Graphs can show a variety of possible relationships between two variables. As one variable increases uniformly, the other may do one of the following: increase or decrease steadily, increase or decrease faster and faster, get closer and closer to some limiting value, reach some intermediate maximum or minimum, alternately increase and decrease, increase or decrease in steps, or do something different from any of these. 9C. Shapes 6-8: 9C/M4. The graphic display of numbers may help to show patterns such as trends, varying rates of change, gaps, or clusters that are useful when making predictions about the phenomena being graphed. 6-8: 9C/M6. The scale chosen for a graph or drawing makes a big difference in how useful it is. 9-12: 9C/H3b. The position of any point on a surface can be specified by two numbers. 11. Common Themes 11C. Constancy and Change 9-12: 11C/H4. Graphs and equations are useful (and often equivalent) ways for depicting and analyzing patterns of change. Next Generation Science Standards Crosscutting Concepts (K-12) Patterns (K-12) Graphs and charts can be used to identify patterns in data. (6-8) Patterns can be used to identify cause and effect relationships. (6-8) Science and Engineering Practices (K-12) Analyzing and Interpreting Data (K-12) Analyzing data in 6–8 builds on K–5 and progresses to extending quantitative analysis to investigations, distinguishing between correlation and causation, and basic statistical techniques of data and error analysis. (6-8) Mathematical and computational thinking at the 6–8 level builds on K–5 and progresses to identifying patterns in large data sets and using mathematical concepts to support explanations and arguments. (6-8) Use mathematical representations to describe and/or support scientific conclusions and design solutions. (6-8) Mathematical and computational thinking at the 9–12 level builds on K–8 and progresses to using algebraic thinking and analysis, a range of linear and nonlinear functions including trigonometric functions, exponentials and logarithms, and computational tools for statistical analysis to analyze, represent, and model data. Simple computational simulations are created and used based on mathematical models of basic assumptions. (9-12) Use mathematical representations of phenomena to describe explanations. (9-12) Common Core State Standards for Mathematics Alignments Standards for Mathematical Practice (K-12) MP.2 Reason abstractly and quantitatively. Geometry (K-8) Graph points on the coordinate plane to solve real-world and mathematical problems. (5) 5.G5.G.2 Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation. Ratios and Proportional Relationships (6-7) Analyze proportional relationships and use them to solve real-world and mathematical problems. (7) 7.RP.2.a Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. 7.RP.2.d Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. Functions (8) Define, evaluate, and compare functions. (8) 8.F.1 Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. 8.F.3 Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. Use functions to model relationships between quantities. (8) 8.F.5 Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally. Units (1) This resource is part of a Physics Front Topical Unit. Topic: Kinematics: The Physics of Motion Unit Title: Special Collections This free collection will impress teachers in the way it promotes depth of understanding about graphs. Learners use interactive digital tools to predict how a motion graph will look, then they watch as the computer simulates process in real time. Next, they place inputs on the graphs and use language to explain what is happening. Finally, they compare their own predictions with the simulated process to analyze why the graphs appear as they do. As with all Concord Consortium materials, the resources are subjected to rigorous classroom testing to ensure their effectiveness. Disclaimer: ComPADRE offers citation styles as a guide only. We cannot offer interpretations about citations as this is an automated procedure. Please refer to the style manuals in the Citation Source Information area for clarifications.
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Reviews This book is a welcome addition to the set of textbooks available to beginning graduate students and advanced undergraduates in mechanical engineering. I have previously taught the subject of elasticity from textbooks written by Barber, Salughter, and Shames with frequent references to the classics by Timoshenko, Love, Sokolnikoff, and Green and Zerna. However, students have found these books either too difficult to understand or too dated in their notation. When I received the new textbook by Professor Sadd, I read it briefly and then handed it over to one of my graduate students who was preparing for his qualifying examinations. His response was unqualified admiration of the ease with which he was able to navigate the book and grasp its contents. - Biswajit Banerjee - Department of Mechanical Engineering, University of Utah Contents Part I: Foundations and Elementary Applications (Part I - the first half of the text, provides a standard first semester course in beginning elasticity theory and applications. Particular topics from Part II could also be used to supplement such a first course.)1. Mathematical Preliminaries (A self-contained review is provided of mathematical principles and notation needed in the text. This material can be covered at the beginning of the course to bring the entire class to a common point or various sections can be referred to as later course material is presented. Vector and index notation is introduced and Cartesian tensor notation will become the primary notational scheme for the formulation part of the course. MATLAB is first introduced here and is used to conduct rotational transformations and solve eigenvalue problems.)1.1 Introduction1.2 Scalar, Vector and Matrix Notation1.3 Index Notational Properties1.4 Kronecker Delta and Alternating Symbol1.5 Determinants1.6 Coordinate Rotation Transformations (This section is the first MATLAB application, and a specific code is to be introduced for coordinate rotations. This will reinforce the theoretical material presented and allow students to explicitly calculate particular transformations and evaluate the nature of this concept.)1.7 Cartesian Tensors1.8 Principal Values, Axes and Invariants of Symmetric Second Order Tensors(Another MATLAB application is covered, which will solve the general eigenvalue problem. This will again reinforce the theoretical material covered in this section.)1.9 Algebra of Cartesian Tensors1.10 Tensor Fields, Derivatives and Integral Theorems1.11 Orthogonal Curvilinear Coordinates2. Deformation: Displacements and Strains (Starting with a brief presentation of general deformation theory, the chapter moves into small deformation kinematics, and develops basic definitions and relations for linear elasticity.) 2.1 Introduction2.2 General Deformations2.3 Small Deformation Theory - Strain-Displacement Relations2.4 Principal Strains2.5 Deviatoric Strains2.6 Strain Compatibility2.7 Strain Transformation (MATLAB applications from sections 1.6 and 1.8 are applied to specific strain transformation problems. Applications related to strain measurement systems will be included.)3. Stress and Equilibrium (This chapter will cover the appropriate theory for force and stress distribution in solids undergoing small deformation.) 3.1 Introduction3.2 Body and Surface Forces3.3 The Traction Vector3.4 The Stress Tensor3.5 Principal Stresses3.6 Deviatoric Stresses3.7 Stress Transformation (Similar to section 2.7, MATLAB applications will be introduced for specific stress transformation problems)3.8 Equilibrium Equations4. Material Behavior-Constitutive Equations (The chapter introduces linear elastic material behavior as formalized by Hooke's Law. The formulation starts with the general case, pointing out the complexities for heterogeneous anisotropic materials. Further more detailed study of anisotropic elasticity is made in chapter 11. Reduction for homogeneous and isotropic materials is then made. Thermoelastic relations are initially presented, and further applications of this case are made in chapter 12. Discussion of bounds on elastic constants is saved for chapter 6 where strain energy is introduced.)4.1 Introduction4.2 Generalized Hooke's Law4.3 Isotropic Case4.4 Thermoelastic Relations4.5 Material Characterization 4.6 Physical Data (In order to provide students with some experience of real materials, this section will present tables of elastic constants for common materials used in engineering applications.)5. Formulation and Solution Strategies (This chapter is very important for student understanding of theory and application for problem solution. I feel that I have provided more educational innovation than found in competitor texts. Collecting and understanding all of the previous theory and equations is not an easy task for students. Developing the final formulations will be done concisely and organized around solution strategies. Material in section 5.7 will provide useful material that will link with later chapters in the text.) 5.1 Introduction5.2 Boundary Conditions and Fundamental Problem Classifications (Some additional material is presented here to provide students with more experience on proper formulation of boundary conditions, since this has been a perceived area of student difficulty)5.3 Stress Formulation (Stress function solution strategy is discussed)5.4 Displacement Formulation (Displacement potential solution strategy is mentioned)5.5 Principal of Superposition5.6 Saint-Venant's Principle5.7 General Solution Strategies (This particular section will contain special extended discussions on solution strategies in order to provide students with a more broad and general understanding of the variety of such techniques. Many of these solution methods will then be explicitly demonstrated in appropriate problems in later chapters. Students wishing to apply elasticity theory to their own future work will need such information to make a successful solution method choice.)- Direct MethodExample 5.1 Stretching of Prismatic Bar Under Its Own Weight- Inverse MethodExample 5.2 Pure Beam Bending - Semi_Inverse MethodExample 5.3 Torsion of Prismatic Bars - Analytical Solution ProceduresPower Series, Fourier Methods, Integral Transform and Complex VariableTechniques - Approximate Solution ProceduresRitz Technique - Numerical Solution ProceduresFinite Difference, Finite and Boundary Element Techniques6. Strain Energy and Related Principles (Final elasticity formulation is completed in this chapter through the introduction of strain energy and related topics. Standard uniqueness and bounds on elastic moduli are discussed. The reciprocal theorem is presented and this will lead into the integral formulation that is later used in developing the numerical boundary element method in chapter 15.)6.1 Introduction6.2 Strain Energy6.3 Uniqueness of the Elasticity Boundary-Value Problem6.4 Bounds on the Elastic Constants6.5 Reciprocal Theorem6.6 Integral Formulation of Elasticity: Somigliana's Identity6.7 Principle of Virtual Work 6.8 Principles of Minimum Potential and Complimentary Energy6.9 Ritz Method7. Two-Dimensional Formulations (This chapter establishes the standard two-dimensional theories of plane strain, plane stress and generalized plane stress. The Airy stress function solution methodology is covered, and is latter used to develop several analytical solutions in chapter 8.)7.1 Introduction7.2 Plane Strain7.3 Plane Stress7.4 Generalized Plane Stress7.5 Airy Stress Function7.6 Polar Coordinate Formulation8. Two-Dimensional Problem Solution (Numerous analytical solutions are developed in this chapter. In order to provide comparisons with student's previous knowledge and experience, particular stress and displacement solutions are compared with elementary mechanics of materials solutions. Continued use of MATLAB software is made in many examples to evaluate and plot the analytical stress and displacement solution fields. This is accomplished through the use of X-Y, polar and contour plots as illustrated in Figures 8.4, 8.5, 8.13, 8.22, and 8.36. A unique feature of this activity is the plotting of maximum shear stress contours and the subsequent comparison with the corresponding photoelastic fringe patterns. This is done in Figure 8.36 and in exercises 8-18 and 8-20. It is felt that students will appreciate and understand more clearly such solutions when they are able to view such graphical representations. This concept also fits nicely with the numerical solution procedures presented in chapter 15.)8.1 Introduction8.2 Cartesian Coordinate Solutions Using Polynomials Example 8.1 Uniaxial Tension of a Beam Example 8.2 Pure Bending of a Beam Example 8.3 Bending of a Beam by Uniform Transverse Loading8.3 Cartesian Coordinate Solutions Using Fourier Methods Example 8.4 Beam Subjected to Transverse Sinusoidal Loading Applications Involving Fourier Series Example 8.5 Rectangular Domain with Arbitrary Boundary Loading8.4 General Solutions in Polar Coordinates - Axisymmetric Solution - Mitchell Solution8.5 Polar Coordinate Solutions Example 8.6 Thick-Walled Cylinder Under Uniform Boundary Pressure Pressurized Hole in an Infinite Medium Stress Free Hole in an Infinite Medium Under Uniform Tension at Infinity Example 8.7 Infinite Medium with a Stress Free Hole Under Uniform Far Field Loading Biaxial and Shear Loading Cases Example 8.8 Wedge and Semi-Infinite Domain Problems Quarter Plane Example Half Space Examples Half Space Under Uniform Normal Stress Over x [0 Half Space Under Concentrated Force at the Origin (Flamant Solution) Half Space Under a Surface Concentrated Moment Half Space Under Uniform Normal Loading Over -a / x / a Notch and Crack Problems Example 8.9 Curved Beam Problems Pure Bending Example Curved Cantilever Under End Loading Example 8.10 Disk Under Diametrical Compression Example 8.11 Rotating Disk Problem9. Saint - Venant Extension, Torsion and Flexure (Some basic three dimensional analytical solutions are presented in this chapter. Similar to the previous chapter, MATLAB is again used to evaluate particular stress and displacement solutions of examples 9.1, 9.2 and 9.3.9.1 Introduction9.2 Extension Formulation Example 9.1 Longitudinal Loading Example9.3 Torsion Formulation Stress Function Formulation Displacement Formulation Membrane Analogy9.4 Torsion Solutions Derived from Boundary Equation Example 9.2 Elliptical Section Example 9.3 Equilateral Triangular Section9.5 Torsion Solution Using Fourier Methods Example 9.4 Rectangular Section9.6 Torsion of Multiply Connected Cross Sections Example 9.5 Hollow Elliptical Section9.7 Torsion of Circular Shafts of Variable Diameter Example 9.6 Conical Shaft9.8 Flexure Formulation9.9 Flexure Problems without Twist Example 9.7 Circular Section Example 9.8 Rectangular SectionPart II: Advanced Applications (This second part of the text includes more advanced topics which could form the basis of a second elasticity course or selected topics could be used to supplement a first course.) 10. Complex Variable Methods (This chapter provides an overview of complex variable methods for solution of two-dimensional elasticity problems. A brief mathematical review of complex variable theory is provided within the chapter. Formulation and solution schemes are based on using Kolosov-Muskhelishvili complex potentials. Applications include examples of stress concentration and fracture mechanics (previously introduced in example 8.8). MATLAB will again be used to evaluate and plot particular solutions in the chapter examples.)10.1 Introduction10.2 Review of Complex Variable Theory10.3 Complex Formulation of the Plane Problem Example 10.1 Constant Stress State Example10.4 Resultant Boundary Conditions10.5 General Structure of the Complex Potentials- Finite Simply-Connected Domains- Finite Multiply-Connected Domains- Infinite Domains10.6 Circular Domain Examples Example 10.2 Disk Under Uniform Compression Example 10.3 Circular Plate with Concentrated Edge Loading10.7 Plane and Half-Plane Problems Example 10.4 Concentrated Force-Moment System in an Infinite Plane Example 10.5 Concentrated Force System on the Surface of a Half-Plane Example 10.6 Stressed Infinite Plate with a Circular Hole10.8 Applications Using the Method of Conformal Mapping Example 10.7 Stressed Infinite Plate with an Elliptical Hole 10.9 Applications to Fracture Mechanics Example 10.8 Infinite Plate with a Central Crack 10.10 Westergaard Method for Crack Analysis11. Anisotropic Elasticity (This chapter picks up from the material in section 4.2 and presents a concise development of elasticity applied to anisotropic materials. After a brief description of a torsion example, the chapter focuses primarily on two-dimensional applications using the complex potentials as introduced in the preceding chapter. Similar to chapter 10, applications to stress concentration and fracture mechanics are developed.)11.1 Introduction11.2 Material Symmetry- Plane of Symmetry (Monoclinic Material)- Three Perpendicular Planes of Symmetry (Orthotropic Material)- Axis of Symmetry (Transversely Isotropic Material)- Complete Symmetry (Isotropic Material) Example 11.1 Hydrostatic Compression of a Monoclinic Cube11.3 Restrictions on the Elastic Moduli11.4 Torsion of a Solid Possessing a Plane of Material Symmetry- Stress Formulation- Displacement Formulation- General Solution to Governing Equation Example 11.2 Torsion of an Elliptical Orthotropic Bar11.5 Plane Deformation Problems Example 11.3 Simple Tension of an Anisotropic Sheet Example 11.4 Concentrated Force in an Infinite Plane Example 11.5 Concentrated Force on the Surface of a Half-Plane Example 11.6 Infinite Plate with an Elliptical Hole Example 11.7 Stressed Infinite Plate with an Elliptical Hole11.6 Applications to Fracture Mechanics12. Thermoelasticity (Starting from the introductory material in section 4.4, this chapter provides a brief formulation of thermoelasticity and solution of problems of engineering interest. Most of the application problems are to be two-dimensional, and examples will be solved using methods previously discussed in chapters 8, 10 and 11. As in previous chapters, examples of stress concentration and fracture mechanics will be presented.)12.1 Introduction12.2 Heat Conduction and the Energy Equation12.3 General Uncoupled Formulation12.4 Two-Dimensional Formulation- Plane Strain- Plane Stress12.5 Displacement Potential Solution12.6 Stress Function Formulation Example 12.1 Thermal Stresses in an Elastic Strip12.7 Polar Coordinate Formulation12.8 Radially Symmetric Problems Example 12.2 Circular Plate Problems12.9 Complex Variable Methods for Plane Problems Example 12.3 Annular Plate Problem Example 12.4 Stresses Around a Circular Hole in an Infinite Plane Under Uniform Heat Flow Example 12.5 Stresses Around an Elliptical Hole in an Infinite Plane Under Uniform Heat Flow13. Displacement Potentials and Stress Functions (This chapter will present the use of displacement potentials and stress functions for the solution of three-dimensional problems. These important techniques provide convenient solution methods for a variety of classical problems. Further applications of this technique (the Papkovich method) appear in chapter 14 for development of singular stress states.)13.1 Introduction13.2 Helmoholtz Displacement Vector Representation13.3 LamJ's Strain Potential13.4 The Galerkin Vector Representation Example 13.1 Kelvin's Problem: A Concentrated Force Acting in the Interior of an Infinite Solid Example 13.2 Boussinesq's Problem: A Concentrated Force Acting Normal to the Free Surface of a Semi-Infinite Solid Example 13.3 Cerruti's Problem: A Concentrated Force Acting Parallel to the Free Surface of a Semi-Infinite Solid13.5 The Papkovich-Neuber Representation Example 13.4 Boussinesq's Problem13.6 Spherical Coordinate Formulations Example 13.5 Spherical Cavity in an Infinite Medium Subject to Uniform Far-Field Tension13.7 Stress Functions - Maxwell Stress Function Representation- Morera Stress Function Representation14. Micromechanics Applications (This unique chapter provides an introduction to the use of elasticity theory in micromechanical modeling of materials. A series of topics are covered to provide a background on several of the more common and popular theories that have been proposed in the literature. No other elasticity book provides such a presentation, and this material should prove to be useful in connecting the course with current material modeling.) 14.1 Introduction14.2 Dislocation ModelingExample 14.1 Edge Dislocation in x-DirectionExample 14.2 Screw Dislocation in z-Direction14.3 Singular Stress States Example 14.3 Concentrated Force in an Infinite Medium (Kelvin Problem) Example 14.4 Kelvin State with Unit Loads in Coordinate Directions Example 14.5 Force Doublet Example 14.6 Force Doublet with a Moment (About (-Axis) Example 14.7 Center of Compression/Dilatation Example 14.8 Center of Rotation Example 14.9 Half-Line of Dilatation14.4 Elasticity Theory with Distributed Cracks Example 14.10 Isotropic Dilute Crack Distribution Example 14.11 Planar Transverse Isotropy with Dilute Crack Distribution Example 14.12 Isotropic Self-Consistent Crack Distribution14.5 Micropolar/Couple-Stress Elasticity - Two-Dimensional Couple-Stress TheoryExample 14.13 Stress Concentration Around a Circular Hole: Micropolar Elasticity14.5 Elasticity Theory with Voids Example 14.14 Stress Concentration Around a Circular Hole: Elasticity with Voids15. Numerical Methods: Finite and Boundary Element Methods (A brief introduction is given to the important numerical finite and boundary elements methods. Two-dimensional formulations are developed for each method, and some example applications are provided. Use of the MATLAB PDE Toolbox allows easy finite element solutions to be developed, and comparisons are made with analytical solutions from chapter 8.) 15.1 Introduction15.2 Virtual Work Formulation15.3 Weighted Residual Formulation15.4 Displacement Interpolation and Element Development15.5 Model Assembly and Boundary Conditions Example 15.1 Rectangular Plate15.6 MATLAB FEM CODE: PDE Toolbox Example 15.2 Circular Hole in a Stressed Plate Example 15.3 Circular Disk Under Diametrical Compression15.7 Boundary Integral Formulation15.8 Boundary Element Formulation15.9 Calculation of Internal Stresses and Displacements15.10 Boundary Element Example Example 15.4 Circular Hole in a Stressed PlateAppendix A: Basic Field Equations in Cartesian, Cylindrical and Spherical CoordinatesAppendix B: Stress Transformation Relations Between Cartesian, Cylindrical and SphericalAppendix C: Theory of PhotoelasticityAppendix D: MATLAB Primer
Harvard_University_-_Linear_AlDocument Transcript USE OF LINEAR ALGEBRA I Math 21b, O. Knill This is not a list of topics covered in the course. It is rather a lose selection of subjects, in which linear algebra is useful or relevant. The aim is to convince you that it is worth learning this subject. Most likely, some of this handout does not make much sense yet to you. Look at this page at the end of the course again, some of the content will become more interesting then. GRAPHS, NETWORKS. Linear al- gebra can be used to understand networks which is a collection of nodes connected by edges. Networks 4 are also called graphs. The adja- Application: An is the number of n- ij step walks in the graph which start cency matrix of a graph is defined by Aij = 1 if there is an edge from at the vertex i and end at the vertex 1 2 3 node i to node j in the graph. Oth- j. erwise the entry is zero. A prob- lem using such matrices appeared on a blackboard at MIT in the movie "Good will hunting". CHEMISTRY, MECHANICS Complicated objects like a The solution x(t) = exp(At) of the bridge (the picture shows Storrow differential equation x = Ax can be ˙ Drive connection bridge which is understood and computed by find- part of the "big dig"), or a molecule ing the eigenvalues of the matrix A. (i.e. a protein) can be modeled by Knowing these frequencies is impor- finitely many parts (bridge elements tant for the design of a mechani- or atoms) coupled with attractive cal object because the engineer can and repelling forces. The vibrations damp dangerous frequencies. In of the system are described by a chemistry or medicine, the knowl- differential equation x = Ax, where ˙ edge of the vibration resonances al- x(t) is a vector which depends on lows to determine the shape of a time. Differential equations are an molecule. important part of this course. QUANTUM COMPUTING A quantum computer is a quantum Theoretically, quantum computa- mechanical system which is used to tions could speed up conventional perform computations. The state x computations significantly. They of a machine is no more a sequence could be used for example for cryp- of bits like in a classical computer tological purposes. Freely available but a sequence of qubits, where quantum computer language (QCL) each qubit is a vector. The memory interpreters can simulate quantum of the computer can be represented computers with an arbitrary num- as a vector. Each computation step ber of qubits. is a multiplication x → Ax with a suitable matrix A. CHAOS THEORY. Dynamical systems theory deals with the iteration of maps or the analysis of Examples of dynamical systems are solutions of differential equations. our solar system or the stars in a At each time t, one has a map galaxy, electrons in a plasma or par- T (t) on the vector space. The ticles in a fluid. The theoretical linear approximation DT (t) is study is intrinsically linked to linear called Jacobean is a matrix. If the algebra because stability properties largest eigenvalue of DT (t) grows often depends on linear approxima- exponentially in t, then the system tions. shows "sensitive dependence on initial conditions" which is also called "chaos". USE OF LINEAR ALGEBRA II Math 21b, O. Knill CODING, ERROR CORRECTION Coding theory is used for encryp- tion or error correction. In the first case, the data x are maped by a map T into code y=Tx. For a good Linear algebra enters in different code, T is a "trapdoor function" in ways, often directly because the ob- the sense that it is hard to get x jects are vectors but also indirectly back when y is known. In the sec- like for example in algorithms which ond case, a code is a linear subspace aim at cracking encryption schemes. X of a vector space and T is a map describing the transmission with er- rors. The projection of T x onto the subspace X corrects the error. DATA COMPRESSION Image- Typically, a picture, a sound or (i.e. JPG), video- (MPG4) and a movie is cut into smaller junks. sound compression algorithms These parts are represented by vec- (i.e. MP3) make use of linear tors. If U denotes the Fourier trans- transformations like the Fourier form and P is a cutoff function, then transform. In all cases, the com- y = P U x is transferred or stored on pression makes use of the fact that a CD or DVD. The receiver obtains in the Fourier space, information back U T y which is close to x in the can be cut away without disturbing sense that the human eye or ear does the main information. not notice a big difference. SOLVING SYSTEMS OR EQUA- Solving systems of nonlinear equa- TIONS When extremizing a func- tions can be tricky. Already for sys- tion f on data which satisfy a con- tems of polynomial equations, one straint g(x) = 0, the method of has to work with linear spaces of Lagrange multipliers asks to solve polynomials. Even if the Lagrange a nonlinear system of equations system is a linear system, the task f (x) = λ g(x), g(x) = 0 for the of solving it can be done more ef- (n + 1) unknowns (x, l), where f ficiently using a solid foundation of is the gradient of f . linear algebra. Rotations are represented by or- thogonal matrices. For example, GAMES Moving around in a world if an object located at (0, 0, 0), described in a computer game re- turning around the y-axes by an quires rotations and translations to angle φ, every point in the ob- be implemented efficiently. Hard- ject gets transformed by matrix the ware acceleration can help to handle  cos(φ) 0 sin(φ) this.  0 1 0  − sin(φ) 0 cos(φ) USE OF LINEAR ALGEBRA (III) Math 21b, Oliver Knill STATISTICS When analyzing data statistically, one often is interested in the correlation matrix Aij = For example, if the random variables E[Yi Yj ] of a random vector X = have a Gaussian (=Bell shaped) dis- (X1 , . . . , Xn ) with Yi = Xi − E[Xi ]. tribution, the correlation matrix to- This matrix is derived from the data gether with the expectation E[Xi ] and determines often the random determines the random variables. variables when the type of the dis- tribution is fixed. A famous example is the prisoner dilemma. Each player has the choice to corporate or to cheat.. The game is described by a 2x2 matrix 3 0 like for example . If a 5 1 GAME THEORY Abstract Games player cooperates and his partner are often represented by pay-off ma- also, both get 3 points. If his partner trices. These matrices tell the out- cheats and he cooperates, he gets 5 come when the decisions of each points. If both cheat, both get 1 player are known. point. More generally, in a game with two players where each player can chose from n strategies, the pay- off matrix is a n times n matrix A. A Nash equilibrium is a vector p ∈ S = { i pi = 1, pi ≥ 0 } for which qAp ≤ pAp for all q ∈ S. NEURAL NETWORK In part of neural network theory, for exam- ple Hopfield networks, the state space is a 2n-dimensional vector space. Every state of the network is given by a vector x, where each For example, if Wij = xi yj , then x component takes the values −1 or is a fixed point of the learning map. 1. If W is a symmetric nxn matrix, one can define a "learning map" T : x → signW x, where the sign is taken component wise. The en- ergy of the state is the dot prod- uct −(x, W x)/2. One is interested in fixed points of the map. USE OF LINEAR ALGEBRA (IV) Math 21b, Oliver Knill MARKOV. Suppose we have three bags with 10 balls each. Every time The problem defines a Markov we throw a dice and a 5 shows up, chain described   by a matrix we move a ball from bag 1 to bag 2, 5/6 1/6 0 if the dice shows 1 or 2, we move a  0 2/3 1/3 . ball from bag 2 to bag 3, if 3 or 4 1/6 1/6 2/3 turns up, we move a ball from bag 3 From this matrix, the equilibrium to bag 1 and a ball from bag 3 to bag distribution can be read off as an 2. What distribution of balls will we eigenvector of a matrix. see in average? SPLINES In computer aided de- sign (CAD) used for example to If we write down the conditions, we construct cars, one wants to inter- will have to solve a system of 4 equa- polate points with smooth curves. tions for four unknowns. Graphic One example: assume you want to artists (i.e. at the company "Pixar") find a curve connecting two points need to have linear algebra skills also P and Q and the direction is given at many other topics in computer at each point. Find a cubic function graphics. f (x, y) = ax3 + bx2 y + cxy 2 + dy 3 which interpolates. SYMBOLIC DYNAMICS Assume that a system can be in three dif- The dynamics of the system is ferent states a, b, c and that tran- coded with a symbolic dynamical sitions a → b, b → a, b → c, system. The transition matrix is   c → c, c → a are allowed. A 0 1 0 a c possible evolution of the system is then a, b, a, b, a, c, c, c, a, b, c, a... One  1 0 1 . 1 0 1 calls this a description of the system Information theoretical quantities with symbolic dynamics. This like the "entropy" can be read off language is used in information the- b ory or in dynamical systems theory. from this matrix. INVERSE PROBLEMS The recon- struction of a density function from Toy problem: We have 4 containers projections along lines reduces to with density a, b, c, d arranged in a the solution of the Radon trans- square. We are able and measure q r form. Studied first in 1917, it is to- the light absorption by by sending day a basic tool in applications like light through it. Like this, we get a b medical diagnosis, tokamak moni- o = a + b,p = c + d,q = a + c and o toring, in plasma physics or for as- r = b + d. The problem is to re- trophysical applications. The re- cover a, b, c, d. The system of equa- c d construction is also called tomogra- p tions is equivalent to Ax = b, with phy. Mathematical tools developed x = (a, b, c, d) and b = (o, p, q, r) and   for the solution of this problem lead 1 1 0 0 to the construction of sophisticated  0 0 1 1  A=  1 0 1 0 .  scanners. It is important that the inversion h = R(f ) → f is fast, ac- 0 1 0 1 curate, robust and requires as few datas as possible. ICE: A BLACKBOARD PROBLEM Math 21b, O. Knill In the movie "Good Will Hunting", the main character Will Hunting (Matt Damon) solves a black- board challenge problem, which is given as a challenge to a linear algebra class. THE "WILL HUNTING" PROBLEM. 4 G is the graph 1 2 3 Find. 1) the adjacency matrix A. 2) the matrix giving the number of 3 step walks. 3) the generating function for walks from point i → j. 4) the generating function for walks from points 1 → 3. This problem belongs to linear algebra and calculus evenso the problem origins from graph the- ory or combinatorics. For a calculus student who has never seen the connection between graph theory, calculus and linear algebra, the assignment is actually hard – probably too hard - as the movie correctly indicates. The problem was posed in the last part of a linear algebra course. An explanation of some terms: THE ADJACENCY MATRIX. The structure of the graph can be encoded with a 4 × 4 array which encodes how many paths of length 1, one can take in the graph from one node to an other:   no one no one which can more conve- 0 1 0 1  1 0 2 1  one none two one niently be written as L=    no two no no an array of numbers  0 2 0 0  one one no no called a matrix: 1 1 0 0 Problem 2 asks to find the matrix which encodes all possible paths of length 3. GENERATING FUNCTION. To any graph one can assign for every pair of nodes i, j a series f (z) = ∞ a(ij) z n , where a(ij) is the number of possible walks from node i to node j with n n=0 n n steps. Problem 3) asks for an explicit expression of f (z) and problem 4) asks for an explicit expression in the special case i = 1, j = 3. Linear algebra has many relations to other fields in mathematics. It is not true that linear algebra is just about solving systems of linear equations. LINEAR EQUATIONS Math 21b, O. Knill SYSTEM OF LINEAR EQUATIONS. A collection of linear equations is called a system of linear equations. An example is 3x − y − z = 0 −x + 2y − z = 0 . −x − y + 3z = 9 It consists of three equations for three unknowns x, y, z. Linear means that no nonlinear terms like x2 , x3 , xy, yz 3 , sin(x) etc. appear. A formal definition of linearity will be given later. LINEAR EQUATION. More precicely, ax + by = c is the general linear equation in two variables. ax + by + cz = d is the general linear equation in three variables. The general linear equation in n variables is a1 x1 + a2 x2 + ... + an xn = 0 Finitely many such equations form a system of linear equations. SOLVING BY ELIMINATION. Eliminate variables. In the example the first equation gives z = 3x − y. Putting this into the second and third equation gives −x + 2y − (3x − y) =0 −x − y + 3(3x − y) =9 or −4x + 3y = 0 . 8x − 4y = 9 The first equation gives y = 4/3x and plugging this into the other equation gives 8x − 16/3x = 9 or 8x = 27 which means x = 27/8. The other values y = 9/2, z = 45/8 can now be obtained. SOLVE BY SUITABLE SUBTRACTION. Addition of equations. If we subtract the third equation from the second, we get 3y − 4z = −9 and add three times the second equation to the first, we get 5y − 4z = 0. Subtracting this equation to the previous one gives −2y = −9 or y = 2/9. SOLVE BY COMPUTER. Use the computer. In Mathematica: Solve[{3x − y − z == 0, −x + 2y − z == 0, −x − y + 3z == 9}, {x, y, z}] . But what did Mathematica do to solve this equation? We will look in this course at some efficient algorithms. GEOMETRIC SOLUTION. Each of the three equations represents a plane in three-dimensional space. Points on the first plane satisfy the first equation. The second plane is the solution set to the second equation. To satisfy the first two equations means to be on the intersection of these two planes which is here a line. To satisfy all three equations, we have to intersect the line with the plane representing the third equation which is a point. LINES, PLANES, HYPERPLANES. The set of points in the plane satisfying ax + by = c form a line. The set of points in space satisfying ax + by + cd = d form a plane. The set of points in n-dimensional space satisfying a 1 x1 + ... + an xn = a0 define a set called a hyperplane. RIDDLES: "15 kids have bicycles or tricycles. Together they Solution. With x bicycles and y tricycles, then x + count 37 wheels. How many have bicycles?" y = 15, 2x + 3y = 37. The solution is x = 8, y = 7. Solution If there are x brothers and y systers, then "Tom, the brother of Carry has twice as many sisters Tom has y sisters and x − 1 brothers while Carry has as brothers while Carry has equal number of systers x brothers and y − 1 sisters. We know y = 2(x − and brothers. How many kids is there in total in this 1), x = y − 1 so that x + 1 = 2(x − 1) and so x = family?" 3, y = 4. INTERPOLATION. Solution. Assume the parabola is Find the equation of the ax2 + bx + c = 0. So, c = −1, a + b + c = parabola which passes through 4, 4a+2b+c = 13. Elimination of c gives the points P = (0, −1), a + b = 5, 4a + 2b = 14 so that 2b = 6 Q = (1, 4) and R = (2, 13). and b = 3, a = 2. The parabola has the equation 2x2 + 3x − 1 = 0 TOMOGRAPHY Here is a toy example of a problem one has to solve for magnetic resonance imaging (MRI), which makes use of the absorption and emission of energy in the radio frequency range of the electromag- netic spectrum. q r Assume we have 4 hydrogen atoms whose nuclei are excited with intensity a, b, c, d. We measure the spin echo intensity in 4 different a b directions. 3 = a + b,7 = c + d,5 = a + c and 5 = b + d. What o is a, b, c, d? Solution: a = 2, b = 1, c = 3, d = 4. However, also a = 0, b = 3, c = 5, d = 2 solves the problem. This system has not a unique solution even so there are 4 equations and 4 c d unknowns. A good introduction into MRI can be found online at p ( INCONSISTENT. x − y = 4, y + z = 5, x + z = 6 is a system with no solutions. It is called inconsistent. EQUILIBRIUM. As an example of a system with many variables, consider a drum modeled by a fine net. The heights at each interior node needs the average the a11 a12 a13 a14 heights of the 4 neighboring nodes. The height at the boundary is fixed. With n2 nodes in the interior, we a21 x11 x12 a24 have to solve a system of n2 equations. For exam- ple, for n = 2 (see left), the n2 = 4 equations are x11 = a21 + a12 + x21 + x12 , x12 = x11 + x13 + x22 + x22 , a31 x21 x22 a34 x21 = x31 +x11 +x22 +a43 , x22 = x12 +x21 +a43 +a34 . To the right, we see the solution to a problem with n = 300, a41 a42 a43 a44 where the computer had to solve a system with 90 000 variables. LINEAR OR NONLINEAR? a) The ideal gas law P V = nKT for the P, V, T , the pressure, volume and temperature of the gas. b) The Hook law F = k(x − x0 ) relates the force F pulling a string at position x which is relaxed at x 0 . MATRICES AND GAUSS-JORDAN Math 21b, O. Knill MATRIX FORMULATION. Consider the sys- The system can be written as Ax = b, where A is a matrix tem of linear equations (called coefficient matrix) and and x and b are vectors.       3x − y − z = 0 3 −1 −1 x 0 −x + 2y − z = 0 A =  −1 2 −1  , x =  y  , b =  0  . −x − y + 3z = 9 −1 −1 3 z 9 ((Ax)i is the dot product  of the i'th row with x).  3 −1 −1 \| 0 We also look at the augmented matrix B =  −1 2 −1 \| 0  . where one puts separators for clarity reasons. −1 −1 3 \| 9 MATRIX "JARGON". A rectangular array of numbers is called a matrix. If the matrix has m rows and n columns, it is called a m × n matrix. A matrix with one column only is called a column m vector, a matrix with one row a row vector. The entries of a matrix are denoted by aij , where i is the row and j is the column. In the case of the linear equation above, the matrix A is a square matrix and the augmented matrix B above is a 3 × 4 matrix. n GAUSS-JORDAN ELIMINATION. Gauss-Jordan Elimination is a process, where successive subtraction of multiples of other rows or scaling brings the matrix into reduced row echelon form. The elimination process consists of three possible steps which are called elementary row operations: • Swap two rows. • Divide a row by a scalar • Subtract a multiple of a row from an other row. The process transfers a given matrix A into a new matrix rref(A) REDUCED ECHELON FORM. A matrix is called in reduced row echelon form 1) if a row has nonzero entries, then the first nonzero entry is 1. (leading one) 2) if a column contains a leading 1, then the other column entries are 0. 3) if a row has a leading 1, then every row above has leading 1's to the left. Pro memoriam: Leaders like to be number one, are lonely and want other leaders above to their left. RANK. The number of leading 1 in rref(A) is called the rank of A. SOLUTIONS OF LINEAR EQUATIONS. If Ax = b is a linear system of equations with m equations and n unknowns, then A is a m × n matrix. We have the following three possibilities: • Exactly one solution. There is a leading 1 in each row but not in the last row. • Zero solutions. There is a leading 1 in the last row. • Infinitely many solutions. There are rows without leading 1 and no leading 1 is in the last row. JIUZHANG SUANSHU. The technique of successively eliminating variables from systems of linear equations is called Gauss elimination or Gauss Jordan elimination and appeared already in the Chinese manuscript "Jiuzhang Suanshu" ('Nine Chapters on the Mathematical art'). The manuscript appeared around 200 BC in the Han dynasty and was probably used as a textbook. For more history of Chinese Mathematics, see ˜ djoyce/mathhist/china.html. (exactly one solution) (no solution) (infinitely many solutions) MURPHYS LAW. "If anything can go wrong, it will go wrong". "If you are feeling good, don't worry, you will get over it!" "Whenever you do Gauss-Jordan elimination, you screw up during the first couple of steps." MURPHYS LAW IS TRUE. Two equations could contradict each other. Geometrically this means that the two planes do not intersect. This is possible if they are parallel. Even without two planes be- ing parallel, it is possible that there is no intersection between all three of them. Also possible is that we don't have enough equations (for example because two equations are the same) and that there are many solutions. Furthermore, we can have too many equations and the four planes would not intersect. RELEVANCE OF EXCEPTIONAL CASES. There are important applications, where "unusual" situations happen: For example in medical tomography, systems of equations appear which are "ill posed". In this case one has to be careful with the method. The linear equations are then obtained from a method called the Radon transform. The task for finding a good method had led to a Nobel prize in Medicis 1979 for Allan Cormack. Cormack had sabbaticals at Harvard and probably has done part of his work on tomography here. Tomography helps today for example for cancer treatment. MATRIX ALGEBRA. Matrices can be added, subtracted if they have the same size:       a11 a12 · · · a1n b11 b12 · · · b1n a11 + b11 a12 + b12 · · · a1n + b1n  a21 a22 · · · a2n   b21 b22 · · · b2n   a21 + b21 a22 + b22 · · · a2n + b2n  A+B =   ··· + =  ··· ··· ···   ··· ··· ··· ···   ··· ··· ··· ···  am1 am2 · · · amn bm1 bm2 · · · bmn am1 + bm2 am2 + bm2 · · · amn + bmn They can also be scaled by a scalar λ:     a11 a12 · · · a1n λa11 λa12 · · · λa1n  a21 a22 · · · a2n   λa21 λa22 · · · λa2n  λA = λ   ··· =  ··· ··· ···   ··· ··· ··· ···  am1 am2 · · · amn λam1 λam2 · · · λamn LINEAR TRANSFORMATIONS Math 21b, O. Knill TRANSFORMATIONS. A transformation T from a set X to a set Y is a rule, which assigns to every element in X an element y = T (x) in Y . One calls X the domain and Y the codomain. A transformation is also called a map. LINEAR TRANSFORMATION. A map T from Rn to Rm is called a linear transformation if there is a m × n matrix A such that T (x) = Ax . EXAMPLES. 3 4 • To the linear transformation T (x, y) = (3x+4y, x+5y) belongs the matrix . This transformation 1 5 maps the plane onto itself. • T (x) = −3x is a linear transformation from the real line onto itself. The matrix is A = [−3]. • To T (x) = y · x from R3 to R belongs the matrix A = y = y1 y2 y3 . This 1 × 3 matrix is also called a row vector. If the codomain is the real axes, one calls the map also a function. function defined on space.   y1 • T (x) = xy from R to R3 . A = y =  y2  is a 3 × 1 matrix which is also called a column vector. The y3 map defines a line in space.   1 0 • T (x, y, z) = (x, y) from R3 to R2 , A is the 2 × 3 matrix A =  0 1 . The map projects space onto a 0 0 plane. 1 1 2 • To the map T (x, y) = (x + y, x − y, 2x − 3y) belongs the 3 × 2 matrix A = . The image 1 −1 −3 of the map is a plane in three dimensional space. • If T (x) = x, then T is called the identity transformation. PROPERTIES OF LINEAR TRANSFORMATIONS. T (0) = 0 T (x + y) = T (x) + T (y) T (λx) = λT (x) In words: Linear transformations are compatible with addition and scalar multiplication. It does not matter, whether we add two vectors before the transformation or add the transformed vectors. ON LINEAR TRANSFORMATIONS. Linear transformations generalize the scaling transformation x → ax in one dimensions. They are important in • geometry (i.e. rotations, dilations, projections or reflections) • art (i.e. perspective, coordinate transformations), • CAD applications (i.e. projections), • physics (i.e. Lorentz transformations), • dynamics (linearizations of general maps are linear maps), • compression (i.e. using Fourier transform or Wavelet trans- form), • coding (many codes are linear codes), • probability (i.e. Markov processes). • linear equations (inversion is solving the equation) LINEAR TRANSFORMATION OR NOT? (The square to the right is the image of the square to the left):   \| \| ··· \| COLUMN VECTORS. A linear transformation T (x) = Ax with A =  v1 v2 · · · vn  has the property \| \| ·· \|  ·    1 0 0  ·   ·   ·        that the column vector v1 , vi , vn are the images of the standard vectors e1 =  ·   . e i =  1  . e n =  · .       ·   ·   ·  0 0 1 In order to find the matrix of a linear transformation, look at the image of the standard vectors and use those to build the columns of the matrix. QUIZ. a) Find the matrix belonging to the linear transformation, which rotates a cube around the diagonal (1, 1, 1) by 120 degrees (2π/3). b) Find the linear transformation, which reflects a vector at the line containing the vector (1, 1, 1). INVERSE OF A TRANSFORMATION. If S is a second transformation such that S(T x) = x, for every x, then S is called the inverse of T . We will discuss this more later. SOLVING A LINEAR SYSTEM OF EQUATIONS. Ax = b means to invert the linear transformation x → Ax. If the linear system has exactly one solution, then an inverse exists. We will write x = A −1 b and see that the inverse of a linear transformation is again a linear transformation. THE BRETSCHER CODE. Otto Bretschers book contains as a motivation a "code", where the encryption happens with the linear map T (x, y) = (x + 3y, 2x + 5y). The map has the inverse T −1 (x, y) = (−5x + 3y, 2x − y). Cryptologists use often the following approach to crack a encryption. If one knows the input and output of some data, one often can decode the key. Assume we know, the enemy uses a Bretscher code and we know that a b T (1, 1) = (3, 5) and T (2, 1) = (7, 5). How do we get the code? The problem is to find the matrix A = . c d 2x2 MATRIX. It is useful to decode the Bretscher code in general If ax + by = X and cx + dy = Y , then a b x = (dX − bY )/(ad − bc), y = (cX − aY )/(ad − bc). This is a linear transformation with matrix A = c d −1 d −b and the corresponding matrix is A = /(ad − bc). −c a "Switch diagonally, negate the wings and scale with a cross". ROTATION: −1 0 cos(α) − sin(α) A= A= 0 −1 sin(α) cos(α) Any rotation has the form of the matrix to the right. ROTATION-DILATION: 2 −3 a −b A= A= 3 2 b a A rotation dilation is a composition of a rotation by angle arctan(y/x) and a dilation by a factor x2 + y 2 . If z = x + iy and w = a + ib and T (x, y) = (X, Y ), then X + iY = zw. So a rotation dilation is tied to the process of the multiplication with a complex number. BOOST: The boost is a basic Lorentz transformation cosh(α) sinh(α) in special relativity. It acts on vectors (x, ct), A= sinh(α) cosh(α) where t is time, c is the speed of light and x is space. Unlike in Galileo transformation (x, t) → (x + vt, t) (which is a shear), time t also changes during the transformation. The transformation has the effect that it changes length (Lorentz contraction). The angle α is related to v by tanh(α) = v/c. One can write also A(x, ct) = ((x + vt)/γ, t + (v/c 2 )/γx), with γ = 1 − v 2 /c2 . ROTATION IN SPACE. Rotations in space are defined by an axes of rotation and an angle. A rotation by 120◦ around a line containing (0, 0, 0) and (1, 1, 1)   0 0 1 blongs to A =  1 0 0  which permutes e1 → e2 → e3 . 0 1 0 REFLECTION AT PLANE. To a reflection at the xy-plane belongs the matrix   1 0 0 A =  0 1 0  as can be seen by looking at the images of ei . The picture 0 0 −1 to the right shows the textbook and reflections of it at two different mirrors. PROJECTION ONTO SPACE. To project a 4d-object into xyz-space, use   1 0 0 0  0 1 0 0  for example the matrix A =    0 0 1 0 . The picture shows the pro- 0 0 0 0 jection of the four dimensional cube (tesseract, hypercube) with 16 edges (±1, ±1, ±1, ±1). The tesseract is the theme of the horror movie "hypercube". FRACTALS. Closely related to linear maps are affine maps x → Ax + b. They are compositions of a linear map with a translation. It is not a linear map if B(0) = 0. Affine maps can be disguised as linear maps x A b Ax + b in the following way: let y = and defne the (n+1)∗(n+1) matrix B = . Then By = . 1 0 1 1 Fractals can be constructed by taking for example 3 affine maps R, S, T which contract area. For a given object Y0 define Y1 = R(Y0 ) ∪ S(Y0 ) ∪ T (Y0 ) and recursively Yk = R(Yk−1 ) ∪ S(Yk−1 ) ∪ T (Yk−1 ). The above picture shows Yk after some iterations. In the limit, for example if R(Y 0 ), S(Y0 ) and T (Y0 ) are disjoint, the sets Yk converge to a fractal, an object with dimension strictly between 1 and 2. x 2x + 2 sin(x) − y CHAOS. Consider a map in the plane like T : → We apply this map again and y x again and follow the points (x1 , y1 ) = T (x, y), (x2 , y2 ) = T (T (x, y)), etc. One writes T n for the n-th iteration of the map and (xn , yn ) for the image of (x, y) under the map T n . The linear approximation of the map at a 2 + 2 cos(x) − 1 x f (x, y) point (x, y) is the matrix DT (x, y) = . (If T = , then the row vectors of 1 y g(x, y) DT (x, y) are just the gradients of f and g). T is called chaotic at (x, y), if the entries of D(T n )(x, y) grow exponentially fast with n. By the chain rule, D(T n ) is the product of matrices DT (xi , yi ). For example, T is chaotic at (0, 0). If there is a positive probability to hit a chaotic point, then T is called chaotic. FALSE COLORS. Any color can be represented as a vector (r, g, b), where r ∈ [0, 1] is the red g ∈ [0, 1] is the green and b ∈ [0, 1] is the blue component. Changing colors in a picture means applying a transformation on the cube. Let T : (r, g, b) → (g, b, r) and S : (r, g, b) → (r, g, 0). What is the composition of these two linear maps? OPTICS. Matrices help to calculate the motion of light rays through lenses. A light ray y(s) = x + ms in the plane is described by a vector (x, m). Following the light ray over a distance of length L corresponds to the map (x, m) → (x + mL, m). In the lens, the ray is bent depending on the height x. The transformation in the lens is (x, m) → (x, m − kx), where k is the strength of the lense. x x 1 L x x x 1 0 x → AL = , → Bk = . m m 0 1 m m m −k 1 m Examples: 1) Eye looking far: AR Bk . 2) Eye looking at distance L: AR Bk AL . 3) Telescope: Bk2 AL Bk1 . (More about it in problem 80 in section 2.4). IMAGE AND KERNEL Math 21b, O. Knill IMAGE. If T : Rn → Rm is a linear transformation, then {T (x) \| x ∈ Rn } is called the image of T . If T (x) = Ax, then the image of T is also called the image of A. We write im(A) or im(T ). EXAMPLES.      x 1 0 0 x 1) If T (x, y, z) = (x, y, 0), then T (x) = A  y  =  0 1 0   y . The image of T is the x − y plane. z 0 0 0 z 2) If T (x, y)(cos(φ)x − sin(φ)y, sin(φ)x + cos(φ)y) is a rotation in the plane, then the image of T is the whole plane. 3) If T (x, y, z) = x + y + z, then the image of T is R. SPAN. The span of vectors v1 , . . . , vk in Rn is the set of all combinations c1 v1 + . . . ck vk , where ci are real numbers. PROPERTIES. The image of a linear transformation x → Ax is the span of the column vectors of A. The image of a linear transformation contains 0 and is closed under addition and scalar multiplication. KERNEL. If T : Rn → Rm is a linear transformation, then the set {x \| T (x) = 0 } is called the kernel of T . If T (x) = Ax, then the kernel of T is also called the kernel of A. We write ker(A) or ker(T ). EXAMPLES. (The same examples as above) 1) The kernel is the z-axes. Every vector (0, 0, z) is mapped to 0. 2) The kernel consists only of the point (0, 0, 0). 3) The kernel consists of all vector (x, y, z) for which x + y + z = 0. The kernel is a plane. PROPERTIES. The kernel of a linear transformation contains 0 and is closed under addition and scalar multiplication. IMAGE AND KERNEL OF INVERTIBLE MAPS. A linear map x → Ax, Rn → Rn is invertible if and only if ker(A) = {0} if and only if im(A) = Rn . HOW DO WE COMPUTE THE IMAGE? The rank of rref(A) is the dimension of the image. The column vectors of A span the image. (Dimension will be discussed later in detail). EXAMPLES. (The     same examples as above) 1 0 cos(φ) sin(φ) 1)  0  and  1  2) and 3) The 1D vector 1 spans the − sin(φ) cos(φ) 0 0 image. span the image. span the image. HOW DO WE COMPUTE THE KERNEL? Just solve Ax = 0. Form rref(A). For every column without leading 1 we can introduce a free variable si . If x is the solution to Axi = 0, where all sj are zero except si = 1, then x = j sj xj is a general vector in the kernel.   1 3 0  2 6 5  EXAMPLE. Find the kernel of the linear map R3 → R4 , x → Ax with A =   3 . Gauss-Jordan 9 1  −2 −6 0   1 3 0  0 0 1  elimination gives: B = rref(A) =    0 0 0 . We see one column without leading 1 (the second one). The 0 0 0 equation Bx = 0 is equivalent to the system x + 3y = 0, z = 0. After fixing z = 0,  chose y = t freely and can  −3 obtain from the first equation x = −3t. Therefore, the kernel consists of vectors t  1 . In the book, you 0 have a detailed calculation, in a case, where the kernel is 2 dimensional. kernel image domain codomain WHY DO WE LOOK AT THE KERNEL? WHY DO WE LOOK AT THE IMAGE? • It is useful to understand linear maps. To which • A solution Ax = b can be solved if and only if b degree are they non-invertible? is in the image of A. • Helpful to understand quantitatively how many • Knowing about the kernel and the image is use- solutions a linear equation Ax = b has. If x is ful in the similar way that it is useful to know a solution and y is in the kernel of A, then also about the domain and range of a general map A(x + y) = b, so that x + y solves the system and to understand the graph of the map. also. In general, the abstraction helps to understand topics like error correcing codes (Problem 53/54 in Bretschers book), where two matrices H, M with the property that ker(H) = im(M ) appear. The encoding x → M x is robust in the sense that adding an error e to the result M x → M x + e can be corrected: H(M x + e) = He allows to find e and so M x. This allows to recover x = P M x with a projection P . PROBLEM. Find ker(A) and im(A) for the 1 × 3 matrix A = [5, 1, 4], a row vector. ANSWER. A · x = Ax = 5x + y + 4z = 0 shows that the kernel is a plane with normal vector [5, 1, 4] through the origin. The image is the codomain, which is R. PROBLEM. Find ker(A) and im(A) of the linear map x → v × x, (the cross product with v. ANSWER. The kernel consists of the line spanned by v, the image is the plane orthogonal to v. PROBLEM. Fix a vector w in space. Find ker(A) and image im(A) of the linear map from R 6 to R3 given by x, y → [x, v, y] = (x × y) · w. ANSWER. The kernel consist of all (x, y) such that their cross product orthogonal to w. This means that the plane spanned by x, y contains w. PROBLEM Find ker(T ) and im(T ) if T is a composition of a rotation R by 90 degrees around the z-axes with with a projection onto the x-z plane. ANSWER. The kernel of the projection is the y axes. The x axes is rotated into the y axes and therefore the kernel of T . The image is the x-z plane. PROBLEM. Can the kernel of a square matrix A be trivial if A 2 = 0, where 0 is the matrix containing only 0? ANSWER. No: if the kernel were trivial, then A were invertible and A 2 were invertible and be different from 0. PROBLEM. Is it possible that a 3 × 3 matrix A satisfies ker(A) = R 3 without A = 0? ANSWER. No, if A = 0, then A contains a nonzero entry and therefore, a column vector which is nonzero. PROBLEM. What is the kernel and image of a projection onto the plane Σ : x − y + 2z = 0? ANSWER. The kernel consists of all vectors orthogonal to Σ, the image is the plane Σ. PROBLEM. Given two square matrices A, B and assume AB = BA. You know ker(A) and ker(B). What can you say about ker(AB)? ANSWER. ker(A) is contained in ker(BA). Similar ker(B) is contained in ker(AB). Because AB = BA, the 0 1 kernel of AB contains both ker(A) and ker(B). (It can be bigger: A = B = .) 0 0 A 0 PROBLEM. What is the kernel of the partitioned matrix if ker(A) and ker(B) are known? 0 B ANSWER. The kernel consists of all vectors (x, y), where x in ker(A) and y ∈ ker(B). BASIS Math 21b, O. Knill LINEAR SUBSPACE. A subset X of Rn which is closed under addition and scalar multiplication is called a linear subspace of Rn . WHICH OF THE FOLLOWING SETS ARE LINEAR SPACES? a) The kernel of a linear map. d) the line x + y = 0. b) The image of a linear map. e) The plane x + y + z = 1. c) The upper half plane. f) The unit circle. BASIS. A set of vectors v1 , . . . , vm is a basis of a linear subspace X of Rn if they are linear independent and if they span the space X. Linear independent means that there are no nontrivial linear relations ai v1 + . . . + am vm = 0. Spanning the space means that very vector v can be written as a linear combination v = a 1 v1 +. . .+am vm of basis vectors. A linear subspace is a set containing {0} which is closed under addition and scaling.       1 0 1 EXAMPLE 1) The vectors v1 =  1  , v2 =  1  , v3 =  0  form a basis in the three dimensional space. 0 1 1   4 If v =  3 , then v = v1 + 2v2 + 3v3 and this representation is unique. We can find the coefficients by solving 5      1 0 1 x 4 Ax = v, where A has the vi as column vectors. In our case, A =  1 1 0   y  =  3  had the unique 0 1 1 z 5 solution x = 1, y = 2, z = 3 leading to v = v1 + 2v2 + 3v3 . EXAMPLE 2) Two nonzero vectors in the plane which are not parallel form a basis. EXAMPLE 3) Three vectors in R3 which are in a plane form not a basis. EXAMPLE 4) Two vectors in R3 do not form a basis. EXAMPLE 5) Three nonzero vectors in R3 which are not contained in a single plane form a basis in R 3 . EXAMPLE 6) The columns of an invertible n × n matrix form a basis in R n as we will see. REASON. There is at least one representation because the vectors FACT. If v1 , ..., vn is a basis, then every vi span the space. If there were two different representations v = vector v can be represented uniquely a1 v1 +. . .+an vn and v = b1 v1 +. . .+bn vn , then subtraction would as a linear combination of the vi . lead to 0 = (a1 − b1 )v1 + ... + (an − bn )vn . This nontrivial linear v = a 1 v1 + . . . + a n vn . relation of the vi is forbidden by assumption. FACT. If n vectors v1 , ..., vn span a space and w1 , ..., wm are linear independent, then m ≤ n. REASON. This is intuitively clear in dimensions up to 3. You can not have more then 4 vectors in space which are linearly independent. We will give a precise reason later.   \| \| \| A BASIS DEFINES AN INVERTIBLE MATRIX. The n × n matrix A =  v1 v2 . . . vn  is invertible if \| \| \| and only if v1 , . . . , vn define a basis in Rn . EXAMPLE. In the example 1), the 3 × 3 matrix A is invertible. FINDING A BASIS FOR THE KERNEL. To solve Ax = 0, we bring the matrix A into the reduced row echelon form rref(A). For every non-leading entry in rref(A), we will get a free variable t i . Writing the system Ax = 0 with these free variables gives us an equation x = i ti vi . The vectors vi form a basis of the kernel of A. REMARK. The problem to find a basis for all vectors wi which are orthogonal to a given set of vectors, is equivalent to the problem to find a basis for the kernel of the matrix which has the vectors w i in its rows. FINDING A BASIS FOR THE IMAGE. Bring the m × n matrix A into the form rref(A). Call a column a pivot column, if it contains a leading 1. The corresponding set of column vectors of the original matrix A form a basis for the image because they are linearly independent and are in the image. Assume there are k of them. They span the image because there are (k − n) non-leading entries in the matrix. REMARK. The problem to find a basis of the subspace generated by v 1 , . . . , vn , is the problem to find a basis for the image of the matrix A with column vectors v 1 , ..., vn . EXAMPLES. 1) Two vectors on a line are linear dependent. One is a multiple of the other. 2) Three vectors in the plane are linear dependent. One can find a relation av 1 + bv2 = v3 by changing the size of the lengths of the vectors v1 , v2 until v3 becomes the diagonal of the parallelogram spanned by v 1 , v2 . 3) Four vectors in three dimensional space are linearly dependent. As in the plane one can change the length of the vectors to make v4 a diagonal of the parallelepiped spanned by v 1 , v2 , v3 . 0 0 1 1 1 0 EXAMPLE. Let A be the matrix A = 1 1 0 . In reduced row echelon form is B = rref(A) = 0 0 1 . 1 1 1 0 0 0 To determine a basis of the kernel we write Bx = 0 as a system of linear equations: x + y = 0, z = 0. The   variable. With y  t, x = −t is fixed. The linear system rref(A)x = 0 is solved by variable yis the free   = x −1 −1 x =  y  = t  1 . So, v =  1  is a basis of the kernel. z 0 0 EXAMPLE. Because the first and third vectors in rref(A) are columns with leading 1's, the first and third     0 1 columns v1 =  1  , v2 =  0  of A form a basis of the image of A. 1 1 WHY DO WE INTRODUCE BASIS VECTORS? Wouldn't it be just easier to look at the standard basis vectors e1 , . . . , en only? The rea- son for more general basis vectors is that they allow a more flexible adaptation to the situation. A person in Paris prefers a different set of basis vectors than a person in Boston. We will also see that in many applications, problems can be solved easier with the right basis. For example, to describe the reflection of a ray at a plane or at a curve, it is preferable to use basis vec- tors which are tangent or orthogonal. When looking at a rotation, it is good to have one basis vector in the axis of rotation, the other two orthogonal to the axis. Choosing the right basis will be especially im- portant when studying differential equations.   1 2 3 A PROBLEM. Let A =  1 1 1 . Find a basis for ker(A) and im(A). 0 1 2   1 0 −1 1 SOLUTION. From rref(A) = 0 1 2 we see that = v =  −2  is in the kernel. The two column vectors 0 0 0 1   1  1  , 2, 1, 1 of A form a basis of the image because the first and third column are pivot columns. 0 DIMENSION Math 21b, O. Knill REVIEW LINEAR SPACE. X linear space: 0 ∈ X REVIEW BASIS. B = {v1 , . . . , vn } ⊂ X closed under addition and scalar multiplication. B linear independent: c1 v1 +...+cn vn = 0 implies c1 = ... = cn = 0. Examples: Rn , X = ker(A), X = im(A) are linear B span X: v ∈ X then v = a1 v1 + ... + an vn . spaces. B basis: both linear independent and span. BASIS: ENOUGH BUT NOT TOO MUCH. The spanning condition for a basis assures that there are enough vectors to represent any other vector, the linear independence condi- tion assures that there are not too many vectors. A basis is, where J.Lo meets A.Hi: Left: J.Lopez in "Enough", right "The man who new too much" by A.Hitchcock AN UNUSUAL EXAMPLE. Let X be the space of polynomials up to degree 4. For example p(x) = 3x4 + 2x3 + x + 5 is an element in this space. It is straightforward to check that X is a linear space. The "zero vector" is the function f (x) = 0 which is zero everywhere. We claim that e1 (x) = 1, e2 (x) = x, e3 (x) = x2 , e4 (x) = x3 and e5 (x) = x4 form a basis in X. PROOF. The vectors span the space: every polynomial f (x) = c 0 + c1 x + c2 x2 + c3 x3 + c4 x4 is a sum f = c0 e1 + c1 e2 + c2 e3 + c3 e4 + c4 e5 of basis elements. The vectors are linearly independent: a nontrivial relation 0 = c 0 e1 + c1 e2 + c2 e3 + c3 e4 + c4 e5 would mean that c0 + c1 x + c2 x2 + c3 x3 + c4 x4 = 0 for all x which is not possible unless all cj are zero. DIMENSION. The number of elements in a basis of UNIQUE REPRESENTATION. v1 , . . . , vn ∈ X ba- X is independent of the choice of the basis. It is sis ⇒ every v ∈ X can be written uniquely as a sum called the dimension of X. v = a 1 v1 + . . . + a n vn . EXAMPLES. The dimension of {0} is zero. The dimension of any line 1. The dimension of a plane is 2, the dimension of three dimensional space is 3. The dimension is independent on where the space is embedded in. For example: a line in the plane and a line in space have dimension 1. IN THE UNUSUAL EXAMPLE. The set of polynomials of degree ≤ 4 form a linear space of dimension 5. REVIEW: KERNEL AND IMAGE. We can construct a basis of the kernel and image of a linear transformation T (x) = Ax by forming B = rrefA. The set of Pivot columns in A form a basis of the image of T , a basis for the kernel is obtained by solving Bx = 0 and introducing free variables for each non-pivot column. EXAMPLE. Let X the linear space from above. Define the linear transformation T (f )(x) = f (x). For example: T (x3 + 2x4 ) = 3x2 + 8x3 . Find a basis for the kernel and image of this transformation. SOLUTION.  Because T (e1 ) = 0, T (e2 ) = e1 , T (e3 ) = 2e2 , T (e4 ) = 3e3 , T (e5 ) = 4e4 , the matrix of T is  0 1 0 0 0  0 0 2 0 0  which is almost in row reduced echelon form. You see that the last four columns   are pivot columns. The kernel is spanned by e1 which corresponds to the con- A= 0 0  0 3 0   stant function f (x) = 1. The image is the 4 dimensional space of polynomials  0 0 0 0 4  of degree ≤ 3. 0 0 0 0 0 Mathematicians call a fact a "lemma" if it is used to prove a theorem and if does not deserve the be honored by the name "theorem": LEMMA. If n vectors v1 , ..., vn span a space and w1 , ..., wm are linearly independent, then m ≤ n. REASON. Assume m > n. Because vi span, each vector wi can be written as j aij vj = wi . After doing a11 . . . a1n \| w1 Gauss-Jordan elimination of the augmented (m × (n + 1))-matrix . . . . . . . . . \| . . . which belong to am1 . . . amn \| wm these relations, we end up with a matrix, which has in the last line 0 ... 0 \| b1 w1 + ... + bm wm . But this means that b1 w1 + ... + bm wm = 0 and this is a nontrivial relation between the w i . Contradiction. The assumption m > n was absurd. THEOREM. Given a basis A = {v1 , ..., vn } and a basis B = {w1 , ..., .wm } of X, then m = n. PROOF. Because A spans X and B is linearly independent, we know that n ≤ m. Because B spans X and A is linearly independent also m ≤ n holds. Together, n ≤ m and m ≤ n implies n = m. DIMENSION OF THE KERNEL. The number of columns in rref(A) without leading 1's is the dimension of the kernel 1 * * dim(ker(A)): we can introduce a parameter for each such col- 1 * umn when solving Ax = 0 using Gauss-Jordan elimination. 1 DIMENSION OF THE IMAGE. The number of leading 1 * in rref(A), the rank of A is the dimension of the image 1 dim(im(A)) because every such leading 1 produces a differ- ent column vector (called pivot column vectors) and these column vectors are linearly independent. DIMENSION FORMULA: (A : Rn → Rm ) EXAMPLE: A invertible is equivalant that the dimension of the image is n and that the dim(ker(A)) + dim(im(A)) = n dim(ker)(A) = 0. PROOF. There are n columns. dim(ker(A)) is the number of columns without leading 1, dim(im(A)) is the number of columns with leading 1. EXAMPLE. In the space X of polynomials of degree 4 define T (f )(x) = f (x). The kernel consists of linear polynomials spanned by e1 , e2 , the image consists of all polynomials of degree ≤ 2. It is spanned by e 3 , e4 , e5 . Indeed dim(ker(T )) + dim(im(T )) = 2 + 3 = 5 = n. FRACTAL DIMENSION. Mathematicians study objects with non-integer dimension since the early 20'th cen- tury. The topic became fashion in the 80'ies, when people started to generate fractals on computers. To define fractals, the notion of dimension is extended: define a s-volume of accuracy r of a bounded set X in R n as the infimum of all hs,r (X) = Uj \|Uj \|s , where Uj are cubes of length ≤ r covering X and \|Uj \| is the length of Uj . The s-volume is then defined as the limit hs (X) of hs (X) = hs,r (X) when r → 0. The dimension is the limiting value s, where hs (X) jumps from 0 to ∞. Examples: 1) A smooth curve X of length 1 in the plane can be covered with n squares U j of length \|Uj \| = 1/n and n hs,1/n (X) = j=1 (1/n)s = n(1/n)s . If s < 1, this converges, if s > 1 it diverges for n → ∞. So dim(X) = 1. 2) A square X in space of area 1 can be covered with n 2 cubes Uj of length \|Uj \| = 1/n and hs,1/n (X) = n2 s j=1 (1/n) = n2 (1/n)s which converges to 0 for s < 2 and diverges for s > 2 so that dim(X) = 2. 3) The Shirpinski carpet is constructed recur- sively by dividing a square in 9 equal squares and cutting away the middle one, repeating this proce- dure with each of the squares etc. At the k'th step, we need 8k squares of length 1/3k to cover the car- pet. The s−volume hs,1/3k (X) of accuracy 1/3k is 8k (1/3k )s = 8k /3ks , which goes to 0 for k → ∞ if 3ks < 8k or s < d = log(8)/ log(3) and diverges if s > d. The dimension is d = log(8)/ log(3) = 1.893.. INFINITE DIMENSIONS. Linear spaces also can have infinite dimensions. An example is the set X of all continuous maps from the real R to R. It contains all polynomials and because Xn the space of polynomials of degree n with dimension n + 1 is contained in X, the space X is infinite dimensional. By the ∞ way, there are functions like g(x) = n=0 sin(2n x)/2n in X which have graphs of fractal dimension > 1 and which are not differentiable at any point x. COORDINATES Math 21b, O. Knill   \| ... \| B-COORDINATES. Given a basis v1 , . . . vn , define the matrix S =  v1 . . . vn . It is invertible. If \| . . .  \|   c1 x1 x = i ci vi , then ci are called the B-coordinates of v. We write [x]B =  . . . . If x =  . . . , we have cn xn x = S([x]B ). B-coordinates of x are obtained by applying S −1 to the coordinates of the standard basis: [x]B = S −1 (x) . 1 3 1 3 6 EXAMPLE. If v1 = and v2 = , then S = . A vector v = has the coordinates 2 5 2 5 9 −5 3 6 −3 S −1 v = = 2 −1 9 3 Indeed, as we can check, −3v1 + 3v2 = v. EXAMPLE. Let V be the plane x + y − z = 1. Find a basis, in which every vector in the plane has the form   a  b . SOLUTION. Find a basis, such that two vectors v1 , v2 are in the plane and such that a third vector 0 v3 is linearly independent to the first two. Since (1, 0, 1), (0, 1, are points in the plane and (0, 0, 0) is in the      1) 1 0 1 plane, we can choose v1 =  0  v2 =  1  and v3 =  1  which is perpendicular to the plane. 1 1 −1 2 1 1 EXAMPLE. Find the coordinates of v = with respect to the basis B = {v1 = , v2 = }. 3 0 1 1 1 1 −1 −1 We have S = and S −1 = . Therefore [v]B = S −1 v = . Indeed −1v1 + 3v2 = v. 0 1 0 1 3   B-MATRIX. If B = {v1 , . . . , vn } is a basis in \| ... \| Rn and T is a linear transformation on Rn , B =  [T (v1 )]B . . . [T (vn )]B  then the B-matrix of T is defined as \| ... \| COORDINATES HISTORY. Cartesian geometry was introduced by Fermat and Descartes (1596-1650) around 1636. It had a large influence on mathematics. Algebraic methods were introduced into geometry. The beginning of the vector concept came only later at the beginning of the 19'th Century with the work of Bolzano (1781-1848). The full power of coordinates becomes possible if we allow to chose our coordinate system adapted to the situation. Descartes biography shows how far dedication to the teaching of mathematics can go ...: (...) In 1649 Queen Christina of Sweden persuaded Descartes to go to Stockholm. However the Queen wanted to draw tangents at 5 a.m. in the morning and Descartes broke the habit of his lifetime of getting up at 11 o'clock. After only a few months in the cold northern climate, walking to the palace at 5 o'clock every morning, he died of pneumonia. Fermat Descartes Christina Bolzano CREATIVITY THROUGH LAZINESS? Legend tells that Descartes (1596-1650) introduced coordinates while lying on the bed, watching a fly (around 1630), that Archimedes (285-212 BC) discovered a method to find the volume of bodies while relaxing in the bath and that Newton (1643-1727) discovered New- ton's law while lying under an apple tree. Other examples are August Kekul´'s analysis of the Benzene e molecular structure in a dream (a snake biting in its tail revieled the ring structure) or Steven Hawkings discovery that black holes can radiate (while shaving). While unclear which of this is actually true, there is a pattern: According David Perkins (at Harvard school of education): "The Eureka effect", many creative breakthroughs have in common: a long search without apparent progress, a prevailing moment and break through, and finally, a transformation and realization. A breakthrough in a lazy moment is typical - but only after long struggle and hard work. EXAMPLE. Let T be the reflection at the plane x + 2y + 3z = 0. Find the transformation matrix B in the       2 1 0 basis v1 =  −1  v2 =  2  v3 =  3 . Because T (v1 ) = v1 = [e1 ]B , T (v2 ) = v2 = [e2 ]B , T (v3 ) = −v3 = 0 3 −2   1 0 0 −[e3 ]B , the solution is B =  0 −1 0 . 0 0 1   \| ... \| SIMILARITY. The B matrix of A is B = S −1 AS, where S =  v1 . . . vn . One says B is similar to A. \| ... \| EXAMPLE. If A is similar to B, then A2 +A+1 is similar to B 2 +B +1. B = S −1 AS, B 2 = S −1 B 2 S, S −1 S = 1, S −1 (A2 + A + 1)S = B 2 + B + 1. PROPERTIES OF SIMILARITY. A, B similar and B, C similar, then A, C are similar. If A is similar to B, then B is similar to A. 0 1 QUIZZ: If A is a 2 × 2 matrix and let S = , What is S −1 AS? 1 0 MAIN IDEA OF CONJUGATION S. The transformation S −1 maps the coordinates from the standard basis into the coordinates of the new basis. In order to see what a transformation A does in the new coordinates, we map it back to the old coordinates, apply A and then map it back again to the new coordinates: B = S −1 AS. S v ← w = [v]B The transformation in A↓ ↓B The transformation in standard coordinates. S −1 B-coordinates. Av → Bw QUESTION. Can the matrix A which belongs to a projection from R 3 to a plane x + y + 6z = 0 be similar to a matrix which is a rotation by 20 degrees around the z axis? No: a non-invertible A can not be similar to an invertible B: if it were, the inverse A = SBS −1 would exist: A−1 = SB −1 S −1 . 1 −2 PROBLEM. Find a clever basis for the reflection of a light ray at the line x + 2y = 0. v 1 = , v2 = . 2 1 1 0 1 −2 SOLUTION. You can achieve B = with S = . 0 −1 2 1 1 a PROBLEM. Are all shears A = with a = 0 similar? Yes, use a basis v1 = ae1 and v2 = e2 . 0 1 3 0 1 0 0 −1 PROBLEM. You know A = is similar to B = with S = . Find eA = −1 2 0 −1 1 1 1/e 0 1+A+A2 +A3 /3!+.... SOLUTION. Because B k = S −1 Ak S for every k we have eA = SeB S −1 = . e + 1/e e ON THE RELEVANCE OF ORTHOGONALITY. 1) During the pyramid age in Egypt (from -2800 til -2300 BC), the Egyptians used ropes divided into length ratios 3 : 4 : 5 to build triangles. This allowed them to triangulate areas quite pre- cisely: for example to build irrigation needed because the Nile was reshaping the land constantly or to build the pyramids: for the great pyramid at Giza with a base length of 230 meters, the average error on each side is less then 20cm, an error of less then 1/1000. A key to achieve this was orthogonality. 2) During one of Thales (-624 til -548 BC) journeys to Egypt, he used a geometrical trick to measure the height of the great pyramid. He measured the size of the shadow of the pyramid. Using a stick, he found the relation between the length of the stick and the length of its shadow. The same length ratio applies to the pyramid (orthogonal triangles). Thales found also that triangles inscribed into a circle and having as the base as the diameter must have a right angle. 3) The Pythagoreans (-572 until -507) were interested in the dis- covery that the squares of a lengths of a triangle with two or- thogonal sides would add up as a2 + b2 = c2 . They were puzzled in √ assigning a length to the diagonal of the unit square, which √ is 2. This number is irrational because 2 = p/q would imply that q 2 = 2p2 . While the prime factorization of q 2 contains an even power of 2, the prime factorization of 2p2 contains an odd power of 2. 4) Eratosthenes (-274 until 194) realized that while the sun rays were orthogonal to the ground in the town of Scene, this did no more do so at the town of Alexandria, where they would hit the ground at 7.2 degrees). Because the distance was about 500 miles and 7.2 is 1/50 of 360 degrees, he measured the circumference of the earth as 25'000 miles - pretty close to the actual value 24'874 miles. 5) Closely related to orthogonality is parallelism. For a long time mathematicians tried to prove Euclid's parallel axiom using other postulates of Euclid (-325 until -265). These attempts had to fail because there are geometries in which parallel lines always meet (like on the sphere) or geometries, where parallel lines never meet (the Poincar´ e half plane). Also these geometries can be studied using linear algebra. The geometry on the sphere with rotations, the geometry on the half plane uses M¨bius transformations, o 2 × 2 matrices with determinant one. 6) The question whether the angles of a right triangle are in reality always add up to 180 degrees became an issue when geometries where discovered, in which the measure- ment depends on the position in space. Riemannian geometry, founded 150 years ago, is the foundation of general relativity a theory which describes gravity geometrically: the presence of mass bends space-time, where the dot product can depend on space. Orthogonality becomes relative too. 7) In probability theory the notion of independence or decorrelation is used. For example, when throwing a dice, the number shown by the first dice is independent and decorrelated from the number shown by the second dice. Decorrelation is identical to orthogonality, when vectors are associated to the random variables. The correlation coefficient between two vectors v, w is defined as v · w/(\|v\|w\|). It is the cosine of the angle between these vectors. 8) In quantum mechanics, states of atoms are described by functions in a linear space of functions. The states with energy −EB /n2 (where EB = 13.6eV is the Bohr energy) in a hydrogen atom. States in an atom are orthogonal. Two states of two different atoms which don't interact are orthogonal. One of the challenges in quantum computing, where the computation deals with qubits (=vectors) is that orthogonality is not preserved during the computation. Different states can interact. This coupling is called decoherence.   2 1 1 BACK TO THE EXAMPLE. The matrix with the vectors v1 , v2 , v3 is A =  0 3 2 . 0 0 5     v1 = \|\|v1 \|\|w1 1 0 0 2 1 1 v2 = (w1 · v2 )w1 + \|\|u2 \|\|w2 so that Q =  0 1 0  and R =  0 3 2 . v3 = (w1 · v3 )w1 + (w2 · v3 )w2 + \|\|u3 \|\|w3 , 0 0 1 0 0 5 PRO MEMORIA. While building the matrix R we keep track of the vectors u i during the Gram-Schmidt procedure. At the end you have vectors ui , vi , wi and the matrix R has \|\|ui \|\| in the diagonal as well as the dot products wi · vj in the upper right triangle. 0 −1 0 −1 0 −1 PROBLEM. Make the QR decomposition of A = . w1 = . u2 = − = . 1 1 −1 1 1 0 0 −1 1 1 w2 = u 2 . A = = QR. 1 0 0 1 WHY do we care to have an orthonormal basis? • An orthonormal basis looks like the standard basis v 1 = (1, 0, ..., 0), . . . , vn = (0, 0, ..., 1). Actually, we will see that an orthonormal basis into a standard basis or a mirror of the standard basis. • The Gram-Schmidt process is tied to the factorization A = QR. The later helps to solve linear equations. In physical problems like in astrophysics, the numerical methods to simulate the problems one needs to invert huge matrices in every time step of the evolution. The reason why this is necessary sometimes is to assure the numerical method is stable implicit methods. Inverting A −1 = R−1 Q−1 is easy because R and Q are easy to invert. • For many physical problems like in quantum mechanics or dynamical systems, matrices are symmetric A∗ = A, where A∗ = Aji . For such matrices, there will a natural orthonormal basis. ij • The formula for the projection onto a linear subspace V simplifies with an orthonormal basis v j in V : projV (x) = (v1 · x)w1 + · · · + (wn · x)wn . • An orthonormal basis simplifies compuations due to the presence of many zeros w j · wi = 0. This is especially the case for problems with symmetry. • The Gram Schmidt process can be used to define and construct classes of classical polynomials, which are important in physics. Examples are Chebyshev polynomials, Laguerre polynomials or Hermite polynomi- als. • QR factorization allows fast computation of the determinant, least square solutions R −1 Q−1 b of overde- termined systems Ax = b or finding eigenvalues - all topics which will appear later. SOME HISTORY. The recursive formulae of the process were stated by Erhard Schmidt (1876-1959) in 1907. The essence of the formulae were already in a 1883 paper of J.P.Gram in 1883 which Schmidt mentions in a footnote. The process seems already have been used by Laplace (1749-1827) and was also used by Cauchy (1789-1857) in 1836. Gram Schmidt Laplace Cauchy ORTHOGONAL MATRICES 10/27/2002 Math 21b, O. Knill TRANSPOSE The transpose of a matrix A is the matrix (AT )ij = Aji . If A is a n × m matrix, then AT is a m × n matrix. For square matrices, the transposed matrix is obtained by reflecting the matrix at the diagonal.   1 EXAMPLES The transpose of a vector A =  2  is the row vector AT = 1 2 3 . 3 1 2 1 3 The transpose of the matrix is the matrix . 3 4 2 4 PROOFS. A PROPERTY OF THE TRANSPOSE. a) Because x · Ay = j i xi Aij yj and a) x · Ay = AT x · y. AT x · y = j i Aji xi yj the two expressions are the same by renaming i and j. b) (AB)T = B T AT . b) (AB)kl = i Aki Bil . (AB)T = kl i Ali Bik = T T A B . c) (AT )T = A. c) ((AT )T )ij = (AT )ji = Aij . ORTHOGONAL MATRIX. A n × n matrix A is called orthogonal if AT A = 1. The corresponding linear transformation is called orthogonal. INVERSE. It is easy to invert an orthogonal matrix: A−1 = AT . cos(φ) sin(φ) EXAMPLES. The rotation matrix A = is orthogonal because its column vectors have − sin(φ) cos(φ) cos(φ) sin(φ) cos(φ) − sin(φ) length 1 and are orthogonal to each other. Indeed: A T A = · = − sin(φ) cos(φ) sin(φ) cos(φ) 1 0 . A reflection at a line is an orthogonal transformation because the columns of the matrix A have 0 1 cos(2φ) sin(2φ) cos(2φ) sin(2φ) 1 0 length 1 and are orthogonal. Indeed: AT A = · = . sin(2φ) − cos(2φ) sin(2φ) − cos(2φ) 0 1 FACTS. An orthogonal transformation preserves the dot product: Ax · Ay = x · y Proof: this is a homework assignment: Hint: just look at the properties of the transpose. Orthogonal transformations preserve the length of vectors as well as the angles between them. Proof. We have \|\|Ax\|\|2 = Ax · Ax = x · x \|\|x\|\|2 . Let α be the angle between x and y and let β denote the angle between Ax and Ay and α the angle between x and y. Using Ax·Ay = x·y we get \|\|Ax\|\|\|\|Ay\|\| cos(β) = Ax·Ay = x · y = \|\|x\|\|\|\|y\|\| cos(α). Because \|\|Ax\|\| = \|\|x\|\|, \|\|Ay\|\| = \|\|y\|\|, this means cos(α) = cos(β). Because this property holds for all vectors we can rotate x in plane V spanned by x and y by an angle φ to get cos(α + φ) = cos(β + φ) for all φ. Differentiation with respect to φ at φ = 0 shows also sin(α) = sin(β) so that α = β. ORTHOGONAL MATRICES AND BASIS. A linear transformation A is orthogonal if and only if the column vectors of A form an orthonormal basis. (That is what A T A = 1n means.) COMPOSITION OF ORTHOGONAL TRANSFORMATIONS. The composition of two orthogonal transformations is orthogonal. The inverse of an orthogonal transformation is orthogonal. Proof. The properties of the transpose give (AB)T AB = B T AT AB = B T B = 1 and (A−1 )T A−1 = (AT )−1 A−1 = (AAT )−1 = 1. EXAMPLES. The composition of two reflections at a line is a rotation. The composition of two rotations is a rotation. The composition of a reflections at a plane with a reflection at an other plane is a rotation (the axis of rotation is the intersection of the planes). ORTHOGONAL PROJECTIONS. The orthogonal projection P onto a linear space with orthonormal basis v1 , . . . , vn is the matrix AAT , where A is the matrix with column vectors vi . To see this just translate the formula P x = (v1 · x)v1 + . . . + (vn · x)vn into the language of matrices: AT x is a vector with components bi = (vi · x) and Ab is the sum of the bi vi , where vi are the column vectors of A. Orthogonal projections are no orthogonal transformations in general!   0 EXAMPLE. Find the orthogonal projection P from R3 to the linear space spanned by v1 =  3  1 and 5       4 1 0 1 1 0 0 0 3/5 4/5 v 2 =  0 . Solution: AAT =  3/5 0  =  0 9/25 12/25 . 1 0 0 0 4/5 0 0 12/25 16/25 WHY ARE ORTHOGONAL TRANSFORMATIONS USEFUL? • In Physics, Galileo transformations are compositions of translations with orthogonal transformations. The laws of classical mechanics are invariant under such transformations. This is a symmetry. • Many coordinate transformations are orthogonal transformations. We will see examples when dealing with differential equations. • In the QR decomposition of a matrix A, the matrix Q is orthogonal. Because Q −1 = Qt , this allows to invert A easier. • Fourier transformations are orthogonal transformations. We will see this transformation later in the course. In application, it is useful in computer graphics (i.e. JPG) and sound compression (i.e. MP3). • Quantum mechanical evolutions (when written as real matrices) are orthogonal transformations. WHICH OF THE FOLLOWING MAPS ARE ORTHOGONAL TRANSFORMATIONS?: Yes No Shear in the plane. Yes No Projection in three dimensions onto a plane. Yes No Reflection in two dimensions at the origin. Yes No Reflection in three dimensions at a plane. Yes No Dilation with factor 2. cosh(α) sinh(α) Yes No The Lorenz boost x → Ax in the plane with A = sinh(α) cosh(α) Yes No A translation. CHANGING COORDINATES ON THE EARTH. Problem: what is the matrix which rotates a point on earth with (latitude,longitude)=(a1 , b1 ) to a point with (latitude,longitude)=(a2 , b2 )? Solution: The ma- trix which rotate the point (0, 0) to (a, b) a composition of two rotations. The first rotation brings  point into the right latitude, the second brings the point into the right longitude. the   R a,b = cos(b) − sin(b) 0 cos(a) 0 − sin(a)  sin(b) cos(b) 0   0 1 0 . To bring a point (a1 , b1 ) to a point (a2 , b2 ), we form 0 0 1 sin(a) 0 cos(a) −1 A = Ra2 ,b2 Ra1 ,b1 . EXAMPLE: With Cambridge (USA): (a1 , b1 ) = (42.366944, 288.893889)π/180 and Z¨rich u (Switzerland): (a2 , b2 ) = (47.377778, 8.551111)π/180, we get the matrix   0.178313 −0.980176 −0.0863732 A =  0.983567 0.180074 −0.0129873 . 0.028284 −0.082638 0.996178 LEAST SQUARES AND DATA Math 21b, O. Knill GOAL. The best possible "solution" of an inconsistent linear systems Ax = b will be called the least square solution. It is the orthogonal projection of b onto the image im(A) of A. The theory of the kernel and the image of linear transformations helps to understand this situation and leads to an explicit formula for the least square fit. Why do we care about non-consistent systems? Often we have to solve linear systems of equations with more constraints than variables. An example is when we try to find the best polynomial which passes through a set of points. This problem is called data fitting. If we wanted to accommodate all data, the degree of the polynomial would become too large, the fit would look too wiggly. Taking a smaller degree polynomial will not only be more convenient but also give a better picture. Especially important is regression, the fitting of data with lines. The above pictures show 30 data points which are fitted best with polynomials of degree 1, 6, 11 and 16. The first linear fit maybe tells most about the trend of the data. THE ORTHOGONAL COMPLEMENT OF im(A). Because a vector is in the kernel of A T if and only if it is orthogonal to the rows of AT and so to the columns of A, the kernel of AT is the orthogonal complement of im(A): (im(A))⊥ = ker(AT ) EXAMPLES.   a 1) A =  b . The kernel V of AT = a b c consists of all vectors satisfying ax + by + cz = 0. V is a c   a plane. The orthogonal complement is the image of A which is spanned by the normal vector  b  to the plane. c 1 1 1 0 2) A = . The image of A is spanned by the kernel of AT is spanned by . 0 0 0 1 ORTHOGONAL PROJECTION. If b is a vector and V is a linear subspace, then projV (b) is the vector closest to b on V : given any other vector v on V , one can form the triangle b, v, projV (b) which has a right angle at projV (b) and invoke Pythagoras. THE KERNEL OF AT A. For any m × n matrix ker(A) = ker(AT A) Proof. ⊂ is clear. On the other T T hand A Av = 0 means that Av is in the kernel of A . But since the image of A is orthogonal to the kernel of AT , we have Av = 0, which means v is in the kernel of A. LEAST SQUARE SOLUTION. The least square so- lution of Ax = b is the vector x∗ such that Ax∗ is closest to b from all other vectors Ax. In other words, b T T Ax∗ = projV (b), where V = im(V ). Because b − Ax∗ ker(A )= im(A) is in V ⊥ = im(A)⊥ = ker(AT ), we have AT (b − Ax∗ ) = T 0. The last equation means that x∗ is a solution of A (b-Ax)=0 T -1 Ax=A(A A) AT b * T T A Ax = A b, the normal . If the kernel of A is triv- equation of Ax = b V=im(A) Ax * ial, then the kernel of AT A is trivial and AT A can be inverted. Therefore x∗ = (AT A)−1 AT b is the least square solution. WHY LEAST SQUARES? If x∗ is the least square solution of Ax = b then \|\|Ax∗ − b\|\| ≤ \|\|Ax − b\|\| for all x. Proof. AT (Ax∗ − b) = 0 means that Ax∗ − b is in the kernel of AT which is orthogonal to V = im(A). That is projV (b) = Ax∗ which is the closest point to b on V . THE DIMENSION OF THE MASS COAST LINE Math 21b, O.Knill To measure the dimension of an object, one can count the number f (n) of boxes of length 1/n needed to cover the object and see how f (n) grows. If f (n) grows like n 2 , then the dimension is 2, if n(k) grows like n, the dimension is 1. For fractal objects, like coast lines, the number of boxes grows like n s for a number s between 1 and 2. The dimension is obtained by correlating y k = log2 f (k) with xk = log2 (k). We measure: The Massachusetts coast line is a fractal of dimension 1.3. We measure the data: f (1) = 5, f (2) = 12, f (4) = 32 and f (8) = 72. A plot of the data xk yk (xk , yk ) = (log2 (k), log2 f (k)) together with a 0 log2 (5) least square fit can be seen to the right. The 1 log2 (12) slope of the line is the dimension of the coast. 2 log2 (32) It is about 1.295. In order to measure the di- 3 log2 (72) mension better, one would need better maps. DETERMINANTS I Math 21b, O. Knill PERMUTATIONS. A permutation of n elements {1, 2, . . . , n} is a rearrangement of {1, 2, . . . , n}. There are n! = n · (n − 1)... · 1 different permutations of {1, 2, . . . , n}: fixing the position of first element leaves (n − 1)! possibilities to permute the rest. EXAMPLE. There are 6 permutations of {1, 2, 3}: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). PATTERNS AND SIGN. The matrix A with zeros everywhere except Ai,π(i) = 1 is called a permutation matrix or the pattern of π. An inversion is a pair k < l such that σ(k) < σ(l). The sign of a permutation π, denoted by (−1)σ(π) is (−1) if there are an odd number of inversions in the pattern. Otherwise, the sign is defined to be +1. EXAMPLES. σ(1, 2) = 0, (−1)σ = 1, σ(2, 1) = 1, (−1)σ = −1. The permutations(1, 2, 3), (3, 2, 1), (2, 3, 1) have sign 1, the permutations (1, 3, 2), (3, 2, 1), (2, 1, 3) have sign −1. DETERMINANT The determinant of a n × n matrix A is the sum π (−1)σ(π) A1π(1) A2π(2) · · · Anπ(n) , where π runs over all permutations of {1, 2, . . . , n} and σ(π) is the sign of the permutation. a b 2 × 2 CASE. The determinant of A = is ad − bc. There are two permutations of (1, 2). The identity c d permutation (1, 2) gives A11 A12 , the permutation (2, 1) gives A21 A22 . If you have seen some multi-variable calculus, you know that det(A) is the area of the parallelogram spanned by the column vectors of A. The two vectors form a basis if and only if det(A) = 0. a b c 3 × 3 CASE. The determinant of A = d e f is aei + bf g + cdh − ceg − f ha − bdi corresponding to the g h i 6 permutations of (1, 2, 3). Geometrically, det(A) is the volume of the parallelepiped spanned by the column vectors of A. The three vectors form a basis if and only if det(A) = 0. EXAMPLE DIAGONAL AND TRIANGULAR MATRICES. The determinant of a diagonal or triangular matrix is the product of the diagonal elements. EXAMPLE PERMUTATION MATRICES. The determinant of a matrix which has everywhere zeros except Aiπ(j) = 1 is just the sign (−1)σ(π) of the permutation. HOW FAST CAN WE COMPUTE THE DETERMINANT?. The cost to find the determinant is the same as for the Gauss- Jordan elimination as we will see below The graph to the left shows some measurements of the time needed for a CAS to calculate the determinant in dependence on the size of the n × n matrix. The matrix size ranges from n=1 to n=300. We also see a best cubic fit of these data using the least square method from the last lesson. It is the cubic p(x) = a + bx + cx2 + dx3 which fits best through the 300 data points. WHY DO WE CARE ABOUT DETERMINANTS? • check invertibility of matrices • allow to define orientation in any dimensions • have geometric interpretation as volume • appear in change of variable formulas in higher dimensional integration. • explicit algebraic expressions for inverting a ma- trix • proposed alternative concepts are unnatural, hard to teach and and harder to understand • as a natural functional on matrices it appears in formulas in particle or statistical physics • determinants are fun PARTITIONED MATRICES. TRIANGULAR AND DIAGONAL MATRICES. The determinant of a partitioned matrix The determinant of a diagonal or triangular ma- A 0 trix is the product of its diagonal elements.   is the product det(A)det(B). 0 B 1 0 0 0    4 5 0 0  3 4 0 0 Example: det(   2 3 4 0 ) = 20.  1 2 0 0  Example det(   0 0 4 −2 ) = 2 · 12 = 24. 1 1 2 1 0 0 2 2 LINEARITY OF THE DETERMINANT. If the columns of A and B are the same except for the i'th column, det([v1 , ..., v, ...vn ]] + det([v1 , ..., w, ...vn ]] = det([v1 , ..., v + w, ...vn ]] . In general, one has det([v1 , ..., kv, ...vn ]] = k det([v1 , ..., v, ...vn ]] . The same holds for rows. These identities follow directly from the original definition of the determimant. det(AB) = det(A)det(B) . det(SAS −1 ) = det(A) . de PROPERTIES OF DETERMINANTS. det(AT ) = det(A) det(− det(A−1 ) = det(A)−1 If B is obtained from A by switching two rows, then det(B) = −det(A). If B is obtained by adding an other row to a given row, then this does not change the value of the determinant. PROOF OF det(AB) = det(A)det(B), one brings the n × n matrix [A\|AB] into row reduced echelon form. Similar than the augmented matrix [A\|b] was brought into the form [1\|A −1 b], we end up with [1\|A−1 AB] = [1\|B]. By looking at the n × n matrix to the left during Gauss-Jordan elimination, the determinant has changed by a factor det(A). We end up with a matrix B which has determinant det(B). Therefore, det(AB) = det(A)det(B). PROOF OF det(AT ) = det(A). The transpose of a pattern is a pattern with the same signature.  PROBLEM.  Find the determi-  1 2 4 5 0 0 0 2  0 7 2 9   1 2 4 5  SOLUTION. Three row transpositions give B =    0 0 6 4  a ma- nant of A =   0 7 2 . 9  0 0 0 2 0 0 6 4 trix which has determinant 84. Therefore det(A) = (−1) 3 det(B) = −84. 1 2 PROBLEM. Determine det(A100 ), where A is the matrix . 3 16 SOLUTION. det(A) = 10, det(A100 ) = (det(A))100 = 10100 = 1 · gogool. This name as well as the gogoolplex = 100 51 1010 are official. They are huge numbers: the mass of the universe for example is 10 52 kg and 1/1010 is the chance to find yourself on Mars by quantum fluctuations. (R.E. Crandall, Scient. Amer., Feb. 1997). ROW REDUCED ECHELON FORM. Determining rref(A) also determines det(A). If A is a matrix and αi are the factors which are used to scale different rows and s is the number of times, two rows are switched, then det(A) = (−1)s α1 · · · αn det(rref(A)) . INVERTIBILITY. Because of the last formula: A n × n matrix A is invertible if and only if det(A) = 0. THE LAPLACE EXPANSION. In order to calculate by hand the determinant of n × n matrices A = a ij for n > 3, the following expansion is useful. Choose a column i. For each entry aji in that column, take the (n − 1) × (n − 1) matrix Aij called minor which does not contain the i'th column and j'th row. Then n det(A) = (−1)i+1 ai1 det(Ai1 ) + · · · + (−1)i+n ain det(Ain ) = j=1 (−1) i+j aij det(Aij ) ORTHOGONAL MATRICES. Because QT Q = 1, we have det(Q)2 = 1 and so \|det(Q)\| = 1 Rotations have determinant 1, reflections have determinant −1. QR DECOMPOSITION. If A = QR, then det(A) = det(Q)det(R). The determinant of Q is ±1, the determinant of R is the product of the diagonal elements of R. DETERMINANTS II, Math 21b, O.Knill DETERMINANT AND VOLUME. If A is a n × n matrix, then \|det(A)\| is the volume of the n-dimensional parallelepiped En spanned by the n column vectors vj of A. Proof. Use the QR decomposition A = QR, where Q is orthogonal and R is upper triangular. From QQT = 1, we get 1 = det(Q)det(QT ) = det(Q)2 see that \|det(Q)\| = 1. Therefore, det(A) = ±det(R). The determinant of R is the product of the \|\|ui \|\| = \|\|vi − projVj−1 vi \|\| which was the distance from vi to Vj−1 . The volume vol(Ej ) of a j-dimensional parallelepiped Ej with base Ej−1 in Vj−1 and height \|\|ui \|\| is vol(Ej−1 )\|\|uj \|\|. Inductively vol(Ej ) = \|\|uj \|\|vol(Ej−1 ) and therefore n vol(En ) = j=1 \|\|uj \|\| = det(R). The volume of a k dimensional parallelepiped defined by the vectors v 1 , . . . , vk is det(AT A). k Proof. QT Q = In gives AT A = (QR)T (QR) = RT QT QR = RT R. So, det(RT R) = det(R)2 = ( j=1 \|\|uj \|\|)2 . (Note that A is a n × k matrix and that AT A = RT R and R are k × k matrices.) ORIENTATION. Determinants allow to define the orientation of n vectors in n-dimensional space. This is "handy" because there is no "right hand rule" in hyperspace... To do so, define the matrix A with column vectors vj and define the orientation as the sign of det(A). In three dimensions, this agrees with the right hand rule: if v1 is the thumb, v2 is the pointing finger and v3 is the middle finger, then their orientation is positive. x i det(A) = CRAMER'S RULE. This is an explicit formula for the solution of Ax = b. If Ai denotes the matrix, where the column vi of A is replaced by b, then det b xi = det(Ai )/det(A) Proof. det(Ai ) = det([v1 , . . . , b, . . . , vn ] = det([v1 , . . . , (Ax), . . . , vn ] = det([v1 , . . . , i xi vi , . . . , vn ] = xi det([v1 , . . . , vi , . . . , vn ]) = xi det(A) i 5 3 EXAMPLE. Solve the system 5x+3y = 8, 8x+5y = 2 using Cramers rule. This linear system with A = 8 5 8 8 3 5 8 and b = . We get x = det = 34y = det = −54. 2 2 5 8 2 GABRIEL CRAMER. (1704-1752), born in Geneva, Switzerland, he worked on geometry and analysis. Cramer used the rule named after him in a book "Introduction a l'analyse ` des lignes courbes alg´braique", where he solved like this a system of equations with 5 e unknowns. According to a short biography of Cramer by J.J O'Connor and E F Robertson, the rule had however been used already before by other mathematicians. Solving systems with Cramers formulas is slower than by Gaussian elimination. The rule is still important. For example, if A or b depends on a parameter t, and we want to see how x depends on the parameter t one can find explicit formulas for (d/dt)x i (t). THE INVERSE OF A MATRIX. Because the columns of A−1 are solutions of Ax = ei , where ej are basis vectors, Cramers rule together with the Laplace expansion gives the formula: [A−1 ]ij = (−1)i+j det(Aji )/det(A) Bij = (−1)i+j det(Aji ) is called the classical adjoint of A. Note the change ij → ji. Don't confuse the classical adjoint with the transpose AT which is sometimes also called the adjoint. COMPUTING EIGENVALUES Math 21b, O.Knill THE TRACE. The trace of a matrix A is the sum of its diagonal elements.   1 2 3 EXAMPLES. The trace of A =  3 4 5  is 1 + 4 + 8 = 13. The trace of a skew symmetric matrix A is zero 6 7 8 because there are zeros in the diagonal. The trace of I n is n. CHARACTERISTIC POLYNOMIAL. The polynomial fA (λ) = det(λIn − A) is called the characteristic polynomial of A. EXAMPLE. The characteristic polynomial of A above is x 3 − 13x2 + 15x. The eigenvalues of A are the roots of the characteristic polynomial f A (λ). Proof. If λ is an eigenvalue of A with eigenfunction v, then A − λ has v in the kernel and A − λ is not invertible so that fA (λ) = det(λ − A) = 0. The polynomial has the form fA (λ) = λn − tr(A)λn−1 + · · · + (−1)n det(A) a b THE 2x2 CASE. is the trace and D is the determinant. In order that this is real, we must have (T /2)2 ≥ D. Away from that parabola, there are two different eigenvalues. The map A contracts volume for \|D\| < 1. 1 2 EXAMPLE. The characteristic polynomial of A = is λ2 − 3λ + 2 which has the roots 1, 2: pA (λ) = 0 2 (λ − 1)(λ − 2). THE FIBONNACCI RABBITS. The Fibonacci's recursion un+1 = un + un−1 defines the growth of the rabbit un+1 un 1 1 population. We have seen that it can be rewritten as =A with A = . The roots un un−1 1 0 √ √ of the characteristic polynomial pA (x) = λ2 − λ − 1. are ( 5 + 1)/2, ( 5 − 1)/2. ALGEBRAIC MULTIPLICITY. If fA (λ) = (λ − λ0 )k g(λ), where g(λ0 ) = 0 then λ is said to be an eigenvalue of algebraic multiplicity k.   1 1 1 EXAMPLE:  0 1 1  has the eigenvalue λ = 1 with algebraic multiplicity 2 and the eigenvalue λ = 2 with 0 0 2 algebraic multiplicity 1. HOW TO COMPUTE EIGENVECTORS? Because (λ − A)v = 0, the vector v is in the kernel of λ − A. We know how to compute the kernel. λ± − 1 −1 √ EXAMPLE FIBONNACCI. The kernel of λI2 − A = is spanned by v+ = (1 + 5)/2, 1 −1 λ± √ and v− = (1 − 5)/2, 1 . They form a basis B. 1 1 √ SOLUTION OF FIBONNACCI. To obtain a formula for An v with v = , we form [v]B = / 5. 0 −1 un+1 √ √ √ √ √ √ Now, = An v = An (v+ / 5 − v− / 5) = An v+ / 5 − An (v− / 5 = λn v+ / 5 + λn v+ / 5. We see that + − un √ √ √ un = [( 1+2 5 )n + ( 1−2 5 )n ]/ 5. ROOTS OF POLYNOMIALS. For polynomials of degree 3 and 4 there exist explicit formulas in terms of radicals. As Galois (1811-1832) and Abel (1802-1829) have shown, it is not possible for equations of degree 5 or higher. Still, one can compute the roots numerically. REAL SOLUTIONS. A (2n + 1) × (2n + 1) matrix A always has a real eigenvalue because the characteristic polynomial p(x) = x 5 + ... + det(A) has the property that p(x) goes to ±∞ for x → ±∞. Because there exist values a, b for which p(a) < 0 and p(b) > 0, by the intermediate value theorem, there exists a real x with p(x) = 0. Application: A rotation in 11 dimensional space has all eigenvalues \|λ\| = 1. The real eigenvalue must have an eigenvalue 1 or −1. EIGENVALUES OF TRANSPOSE. We know (see homework) that the characteristic polynomials of A and A T agree. Therefore A and AT have the same eigenvalues. APPLICATION: MARKOV MATRICES. A matrix A for which  each column sums up to 1 is called a Markov  1  1  matrix. The transpose of a Markov matrix has the eigenvector    . . .  with eigenvalue 1. Therefore: 1 A Markov matrix has an eigenvector v to the eigenvalue 1. This vector v defines an equilibrium point of the Markov process. 1/3 1/2 EXAMPLE. If A = . Then [3/7, 4/7] is the equilibrium eigenvector to the eigenvalue 1. 2/3 1/2 BRETSCHERS HOMETOWN. Problem 28 in the book deals with a Markov problem in Andelfingen the hometown of Bretscher, where people shop in two shops. (Andelfingen is a beautiful village at the Thur river in the middle of a "wine country"). Initially all shop in shop W . After a new shop opens, every week 20 percent switch to the other shop M . Missing something at the new place, every week, 10 percent switch back. This leads to a Markov matrix A = 8/10 1/10 . After some time, things will settle 2/10 9/10 down and we will have certain percentage shopping in W and other percentage shopping in M . This is the equilibrium. . MARKOV PROCESS IN PROBABILITY. Assume we have a graph like a network and at each node i, the probability to go from i to j in the next step is [A]ij , where Aij is a Markov matrix. We know from the above result that there is an eigenvector p which satisfies Ap = p. It can be normalized that i pi = 1. The interpretation is that p i is the probability that the walker is on A C the node p. For example, on a triangle, we can have the probabilities: P (A → B) = 1/2, P (A → C) = 1/4, P (A → A) = 1/4, P (B → A) = 1/3, P (B → B) = 1/6, P (B → C) = 1/2, P (C → A) = 1/2, P (C → B) = 1/3, P (C → C) = 1/6. The corresponding matrix is   B 1/4 1/3 1/2 A =  1/2 1/6 1/3  . 1/4 1/2 1/6 In this case, the eigenvector to the eigenvalue 1 is p = [38/107, 36/107, 33/107]T . CALCULATING EIGENVECTORS Math 21b, O.Knill NOTATION. We often just write 1 instead of the identity matrix 1 n . COMPUTING EIGENVALUES. Recall: because λ−A has v in the kernel if λ is an eigenvalue the characteristic polynomial fA (λ) = det(λ − A) = 0 has eigenvalues as roots. a b 2 × 2 CASE. Recall: = a + d is the trace and D = ad − bc is the determinant of A. If (T /2)2 ≥ D, then the eigenvalues are real. Away from that parabola in the (T, D) space, there are two different eigenvalues. The map A contracts volume for \|D\| < 1. NUMBER OF ROOTS. Recall: There are examples with no real eigenvalue (i.e. rotations). By inspecting the graphs of the polynomials, one can deduce that n × n matrices with odd n always have a real eigenvalue. Also n × n matrixes with even n and a negative determinant always have a real eigenvalue. IF ALL ROOTS ARE REAL. fA (λ) = λn − tr(A)λn−1 + ... + (−1)n det(A) = (λ − λ1 )...(λ − λn ), we see that i λi = trace(A) and i λi = det(A). HOW TO COMPUTE EIGENVECTORS? Because (λ − A)v = 0, the vector v is in the kernel of λ − A. a b 1 0 EIGENVECTORS of are v± with eigen- If c = 0 and d = 0, then and are eigen- c d 0 1 value λ± . vectors. If c = 0, then the eigenvectors to λ± are v± = If c = 0 and a = d, then the eigenvectors to a and d λ± − d 1 b/(d − a) . are and . c 0 1 ALGEBRAIC MULTIPLICITY. If fA (λ) = (λ−λ0 )k g(λ), where g(λ0 ) = 0, then f has algebraic multiplicity k. The algebraic multiplicity counts the number of times, an eigenvector occurs.  1 1 1 EXAMPLE:  0 1 1  has the eigenvalue λ = 1 with algebraic multiplicity 2. 0 0 2 GEOMETRIC MULTIPLICITY. The dimension of the eigenspace E λ of an eigenvalue λ is called the geometric multiplicity of λ. 1 1 EXAMPLE: the matrix of a shear is . It has the eigenvalue 1 with algebraic multiplicity 2. The kernel 0 1 0 1 1 of A − 1 = is spanned by and the geometric multiplicity is 1. It is different from the algebraic 0 0 0 multiplicity.   1 1 1 EXAMPLE: The matrix  0 0 1  has eigenvalue 1 with algebraic multiplicity 2 and the eigenvalue 0 0 0 1 with multiplicity 1. Eigenvectors    to  eigenvalue λ = 1 are in the kernel of A − 1 which is the kernel of the 0 1 1 1  0 −1 1  and spanned by  0 . The geometric multiplicity is 1. 0 0 0 0 RELATION BETWEEN ALGEBRAIC AND GEOMETRIC MULTIPLICIY. (Proof later in the course). The geometric multiplicity is smaller or equal than the algebraic multiplicity. √ PRO MEMORIA. Remember that the geometric mean ab of two numbers is smaller or equal to the alge- braic mean (a + b)/2? (This fact is totally∗ unrelated to the above fact and a mere coincidence of expressions, but it helps to remember it). Quite deeply buried there is a connection in terms of convexity. But this is rather philosophical. . DIAGONALIZATION Math 21b, O.Knill SUMMARY. A n × n matrix, Av = λv with eigenvalue λ and eigenvector v. The eigenvalues are the roots of the characteristic polynomial fA (λ) = det(λ − A) = λn − tr(A)λn−1 + ... + (−1)n det(A). The eigenvectors to the eigenvalue λ are in ker(λ − A). The number of times, an eigenvalue λ occurs in the full list of n roots of fA (λ) is called algebraic multiplicity. It is bigger or equal than the geometric multiplicity: dim(ker(λ − A). a b EXAMPLE. The eigenvalues of are λ± = T /2 + T 2 /4 − D, where T = a + d is the trace and c d λ± − d D = ad − bc is the determinant of A. If c = 0, the eigenvectors are v ± = if c = 0, then a, d are c a −b eigenvalues to the eigenvectors and . If a = d, then the second eigenvector is parallel to the 0 a−d first and the geometric multiplicity of the eigenvalue a = d is 1. EIGENBASIS. If A has n different eigenvalues, then A has an eigenbasis, consisting of eigenvectors of A. DIAGONALIZATION. How does the matrix A look in an eigenbasis? If S is the matrix with the eigenvectors as columns, then we know B = S −1 AS . We have Sei = vi and ASei = λi vi we know S −1 ASei = λi ei . Therefore, B is diagonal with diagonal entries λi . 2 3 √ √ EXAMPLE. A = 3 with eigenvector v1 = [ 3, 1] and the eigenvalues has the eigenvalues λ1 = 2 + 1 2 √ √ √ √ 3 − 3 λ2 = 2 − 3 with eigenvector v2 = [− 3, 1]. Form S = and check S −1 AS = D is diagonal. 1 1 APPLICATION: FUNCTIONAL CALCULUS. Let A be the matrix in the above example. What is A 100 + A37 − 1? The trick is to diagonalize A: B = S −1 AS, then B k = S −1 Ak S and We can compute A100 + A37 − 1 = S(B 100 + B 37 − 1)S −1 . APPLICATION: SOLVING LINEAR SYSTEMS. x(t+1) = Ax(t) has the solution x(n) = A n x(0). To compute An , we diagonalize A and get x(n) = SB n S −1 x(0). This is an explicit formula. SIMILAR MATRICES HAVE THE SAME EIGENVALUES. One can see this in two ways: 1) If B = S −1 AS and v is an eigenvector of B to the eigenvalue λ, then Sv is an eigenvector of A to the eigenvalue λ. 2) From det(S −1 AS) = det(A), we know that the characteristic polynomials f B (λ) = det(λ − B) = det(λ − S −1 AS) = det(S −1 (λ − AS) = det((λ − A) = fA (λ) are the same. CONSEQUENCES. 1) Because the characteristic polynomials of similar matrices agree, the trace tr(A) of similar matrices agrees. 2) The trace is the sum of the eigenvalues of A. (Compare the trace of A with the trace of the diagonalize matrix.) THE CAYLEY HAMILTON THEOREM. If A is diagonalizable, then fA (A) = 0. PROOF. The DIAGONALIZATION B = S −1 AS has the eigenvalues in the diagonal. So fA (B), which contains fA (λi ) in the diagonal is zero. From fA (B) = 0 we get SfA (B)S −1 = fA (A) = 0. By the way: the theorem holds for all matrices: the coefficients of a general matrix can be changed a tiny bit so that all eigenvalues are different. For any such perturbations one has fA (A) = 0. Because the coefficients of fA (A) depend continuously on A, they are zero 0 in general. CRITERIA FOR SIMILARITY. • If A and B have the same characteristic polynomial and diagonalizable, then they are similar. • If A and B have a different determinant or trace, they are not similar. • If A has an eigenvalue which is not an eigenvalue of B, then they are not similar. COMPLEX EIGENVALUES Math 21b, O. Knill NOTATION. Complex numbers are written as z = x + iy = r exp(iφ) = r cos(φ) + ir sin(φ). The real number r = \|z\| is called the absolute value of z, the value φ is the argument and denoted by arg(z). Complex numbers contain the real numbers z = x+i0 as a subset. One writes Re(z) = x and Im(z) = y if z = x + iy. ARITHMETIC. Complex numbers are added like vectors: x + iy + u + iv = (x + u) + i(y + v) and multiplied as z ∗ w = (x + iy)(u + iv) = xu − yv + i(yu − xv). If z = 0, one can divide 1/z = 1/(x + iy) = (x − iy)/(x 2 + y 2 ). ABSOLUTE VALUE AND ARGUMENT. The absolute value \|z\| = x2 + y 2 satisfies \|zw\| = \|z\| \|w\|. The argument satisfies arg(zw) = arg(z) + arg(w). These are direct consequences of the polar representation z = r exp(iφ), w = s exp(iψ), zw = rs exp(i(φ + ψ)). x GEOMETRIC INTERPRETATION. If z = x + iy is written as a vector , then multiplication with an y other complex number w is a dilation-rotation: a scaling by \|w\| and a rotation by arg(w). THE DE MOIVRE FORMULA. z n = exp(inφ) = cos(nφ) + i sin(nφ) = (cos(φ) + i sin(φ))n follows directly from z = exp(iφ) but it is magic: it leads for example to formulas like cos(3φ) = cos(φ) 3 − 3 cos(φ) sin2 (φ) which would be more difficult to come by using geometrical or power series arguments. This formula is useful for example in integration problems like cos(x)3 dx, which can be solved by using the above deMoivre formula. THE UNIT CIRCLE. Complex numbers of length 1 have the form z = exp(iφ) and are located on the unit circle. The characteristic polynomial f A (λ) =   0 1 0 0 0  0 0 1 0 0    λ − 1 of the matrix  0 0 0 1 0  has all roots on the unit circle. The 5    0 0 0 0 1  1 0 0 0 0 roots exp(2πki/5), for k = 0, . . . , 4 lye on the unit circle. THE LOGARITHM. log(z) is defined for z = 0 as log \|z\| + iarg(z). For example, log(2i) = log(2) + iπ/2. Riddle: what is ii ? (ii = ei log(i) = eiiπ/2 = e−π/2 ). The logarithm is not defined at 0 and the imaginary part is define only up to 2π. For example, both iπ/2 and 5iπ/2 are equal to log(i). √ HISTORY. The struggle with −1 is historically quite interesting. Nagging questions appeared for example when trying to find closed solutions for roots of polynomials. Cardano (1501-1576) was one of the mathemati- cians who at least considered complex numbers but called them arithmetic subtleties which were "as refined as useless". With Bombelli (1526-1573), complex numbers found some practical use. Descartes (1596-1650) called roots of negative numbers "imaginary". Although the fundamental theorem of algebra (below) was still not proved in the 18th century, and complex √ numbers were not fully understood, the square root of minus one −1 was used more and more. Euler (1707- 1783) made the observation that exp(ix) = cos x + i sin x which has as a special case the magic formula eiπ + 1 = 0 which relate the constants 0, 1, π, e in one equation. For decades, many mathematicians still thought complex numbers were a waste of time. Others used complex numbers extensively in their work. In 1620, Girard suggested that an equation may have as many roots as its degree in 1620. Leibniz (1646-1716) spent quite a bit of time trying to apply the laws of algebra to complex numbers. He and Johann Bernoulli used imaginary numbers as integration aids. Lambert used complex numbers for map projections, d'Alembert used them in hydrodynamics, while Euler, D'Alembert and Lagrange used them √ in their incorrect proofs of the fundamental theorem of algebra. Euler write first the symbol i for −1. Gauss published the first correct proof of the fundamental theorem of algebra in his doctoral thesis, but still claimed in 1825 that the true metaphysics of the square root of −1 is elusive as late as 1825. By 1831 Gauss overcame his uncertainty about complex numbers and published his work on the geometric representation of complex numbers as points in the plane. In 1797, a Norwegian Caspar Wessel (1745-1818) and in 1806 a Swiss clerk named Jean Robert Argand (1768-1822) (who stated the theorem the first time for polynomials with complex coefficients) did similar work. But these efforts went unnoticed. William Rowan Hamilton (1805-1865) (who would also discover the quaternions while walking over a bridge) expressed in 1833 complex numbers as vectors. Complex numbers continued to develop to complex function theory or chaos theory, a branch of dynamical systems theory. Complex numbers are helpful in geometry in number theory or in quantum mechanics. Once believed fictitious they are now most "natural numbers" and the "natural numbers" themselves are in fact the √ most "complex". A philospher who asks "does −1 really exist?" might be shown the representation of x + iy x −y as . When adding or multiplying such dilation-rotation matrices, they behave like complex numbers: y x 0 −1 for example plays the role of i. 1 0 FUNDAMENTAL THEOREM OF ALGEBRA. (Gauss 1799) A polynomial of degree n has exactly n roots. CONSEQUENCE: A n × n MATRIX HAS n EIGENVALUES. The characteristic polynomial f A (λ) = λn + an−1 λn−1 + . . . + a1 λ + a0 satisfies fA (λ) = (λ − λ1 ) . . . (λ − λn ), where λi are the roots of f . TRACE AND DETERMINANT. Comparing fA (λ) = (λ − λn ) . . . (λ − λn ) with λn − tr(A) + .. + (−1)n det(A) gives tr(A) = λ1 + · · · + λn , det(A) = λ1 · · · λn . COMPLEX FUNCTIONS. The characteristic polynomial is an example of a function f from C to C. The graph of this function would live in C × C which corresponds to a l l four dimensional real space. One can visualize the function however with the real-valued function z → \|f (z)\|. The figure to the left shows the contour lines of such a function z → \|f (z)\|, where f is a polynomial. ITERATION OF POLYNOMIALS. A topic which is off this course (it would be a course by itself) is the iteration of polynomials like fc (z) = z 2 + c. The set of parameter values c for which the iterates 2 n fc (0), fc (0) = fc (fc (0)), . . . , fc (0) stay bounded is called the Mandelbrot set. It is the fractal black region in the picture to the left. The now already dusty object appears everywhere, from photoshop plugins to decorations. In Mathematica, you can compute the set very quickly (see COMPLEX √ NUMBERS IN MATHEMATICA OR MAPLE. In both computer algebra systems, the letter I is used for i = −1. In Maple, you can ask log(1 + I), in Mathematica, this would be Log[1 + I]. Eigenvalues or eigenvectors of a matrix will in general involve complex numbers. For example, in Mathematica, Eigenvalues[A] gives the eigenvalues of a matrix A and Eigensystem[A] gives the eigenvalues and the corresponding eigenvectors. cos(φ) sin(φ) EIGENVALUES AND EIGENVECTORS OF A ROTATION. The rotation matrix A = − sin(φ) cos(φ) has the characteristic polynomial λ2 − 2 cos(φ) + 1. The eigenvalues are cos(φ) ± cos2 (φ) − 1 = cos(φ) ± −i i sin(φ) = exp(±iφ). The eigenvector to λ1 = exp(iφ) is v1 = and the eigenvector to the eigenvector 1 i λ2 = exp(−iφ) is v2 = . 1 STABILITY Math 21b, O. Knill LINEAR DYNAMICAL SYSTEM. A linear map x → Ax defines a dynamical system. Iterating the map produces an orbit x0 , x1 = Ax, x2 = A2 = AAx, .... The vector xn = An x0 describes the situation of the system at time n. Where does xn go when time evolves? Can one describe what happens asymptotically when time n goes to infinity? In the case of the Fibonacci sequence xn which gives the number of rabbits in a rabbit population at time n, the population grows essentially exponentially. Such a behavior would be called unstable. On the other hand, if A is a rotation, then An v stays bounded which is a type of stability. If A is a dilation with a dilation factor < 1, then An v → 0 for all v, a thing which we will call asymptotic stability. The next pictures show experiments with some orbits An v with different matrices. 0.99 −1 0.54 −1 0.99 −1 1 0 0.95 0 0.99 0 stable (not asymptotic asymptotic asymptotic) stable stable 0.54 −1 2.5 −1 1 0.1 1.01 0 1 0 0 1 unstable unstable unstable ASYMPTOTIC STABILITY. The origin 0 is invariant under a linear map T (x) = Ax. It is called asymptot- ically stable if An (x) → 0 for all x ∈ I n . R p −q EXAMPLE. Let A = be a dilation rotation matrix. Because multiplication wich such a matrix is q p analogue to the multiplication with a complex number z = p + iq, the matrix A n corresponds to a multiplication with (p + iq)n . Since \|(p + iq)\|n = \|p + iq\|n , the origin is asymptotically stable if and only if \|p + iq\| < 1. Because det(A) = \|p+iq\|2 = \|z\|2 , rotation-dilation matrices A have an asymptotic stable origin if and only if \|det(A)\| < 1. p −q Dilation-rotation matrices have eigenvalues p ± iq and can be diagonalized in the complex. q p EXAMPLE. If a matrix A has an eigenvalue \|λ\| ≥ 1 to an eigenvector v, then A n v = λn v, whose length is \|λn \| times the length of v. So, we have no asymptotic stability if an eigenvalue satisfies \|λ\| ≥ 1. STABILITY. The book also writes "stable" for "asymptotically stable". This is ok to abbreviate. Note however that the commonly used term "stable" also includes linear maps like rotations, reflections or the identity. It is therefore preferable to leave the attribute "asymptotic" in front of "stable". cos(φ) − sin(φ) ROTATIONS. Rotations have the eigenvalue exp(±iφ) = cos(φ) + i sin(φ) and are not sin(φ) cos(φ) asymptotically stable. r 0 DILATIONS. Dilations have the eigenvalue r with algebraic and geometric multiplicity 2. Dilations 0 r are asymptotically stable if \|r\| < 1. A linear dynamical system x → Ax has an asymptotically stable origin if and CRITERION. only if all its eigenvalues have an absolute value < 1. PROOF. We have already seen in Example 3, that if one eigenvalue satisfies \|λ\| > 1, then the origin is not asymptotically stable. If \|λi \| < 1 for all i and all eigenvalues are different, there is an eigenbasis v 1 , . . . , vn . n n n Every x can be written as x = j=1 xj vj . Then, An x = An ( j=1 xj vj ) = j=1 xj λn vj and because \|λj \|n → 0, j there is stability. The proof of the general (nondiagonalizable) case will be accessible later. a b THE 2-DIMENSIONAL CASE. The characteristic polynomial of a 2 × 2 matrix A = is c d fA (λ) = λ2 − tr(A)λ + det(A). If c = 0, the eigenvalues are λ± = tr(A)/2 ± (tr(A)/2)2 − det(A). If the discriminant (tr(A)/2)2 − det(A) is nonnegative, then the eigenvalues are real. This happens below the parabola, where the discriminant is zero. CRITERION. In two dimensions we have asymptotic stability if and only if (tr(A), det(A)) is contained in the stability triangle bounded by the lines det(A) = 1, det(A) = tr(A)−1 and det(A) = −tr(A)−1. PROOF. Write T = tr(A)/2, D = det(A). √ \|D\| ≥ 1, If there is no asymptotic stability. If λ = T + T 2 − D = ±1, then T 2 − D = (±1 − T )2 and D = 1 ± 2T . For D ≤ −1 + \|2T \| we have a real eigenvalue ≥ 1. The conditions for stability is therefore D > \|2T \| − 1. It implies automatically D > −1 so that the triangle can be described shortly as \|tr(A)\| − 1 < det(A) < 1 . EXAMPLES. 1 1/2 1) The matrix A = has determinant 5/4 and trace 2 and the origin is unstable. It is a dilation- −1/2 1 rotation matrix which corresponds to the complex number 1 + i/2 which has an absolute value > 1. 2) A rotation A is never asymptotically stable: det(A)1 and tr(A) = 2 cos(φ). Rotations are the upper side of the stability triangle. 3) A dilation is asymptotically stable if and only if the scaling factor has norm < 1. 4) If det(A) = 1 and tr(A) < 2 then the eigenvalues are on the unit circle and there is no asymptotic stability. 5) If det(A) = −1 (like for example Fibonacci) there is no asymptotic stability. For tr(A) = 0, we are a corner of the stability triangle and the map is a reflection, which is not asymptotically stable neither. SOME PROBLEMS. 1) If A is a matrix with asymptotically stable origin, what is the stability of 0 with respect to A T ? 2) If A is a matrix which has an asymptotically stable origin, what is the stability with respect to to A −1 ? 3) If A is a matrix which has an asymptotically stable origin, what is the stability with respect to to A 100 ? ON THE STABILITY QUESTION. For general dynamical systems, the question of stability can be very difficult. We deal here only with linear dynamical systems, where the eigenvalues determine everything. For nonlinear systems, the story is not so simple even for simple maps like the Henon map. The questions go deeper: it is for example not known, whether our solar system is stable. We don't know whether in some future, one of the planets could get expelled from the solar system (this is a mathematical question because the escape time would be larger than the life time of the sun). For other dynamical systems like the atmosphere of the earth or the stock market, we would really like to know what happens in the near future ... A pioneer in stability theory was Alek- sandr Lyapunov (1857-1918). For nonlin- ear systems like xn+1 = gxn − x3 − xn−1 n the stability of the origin is nontrivial. As with Fibonacci, this can be written as (xn+1 , xn ) = (gxn − x2 − xn−1 , xn ) = n A(xn , xn−1 ) called cubic Henon map in the plane. To the right are orbits in the cases g = 1.5, g = 2.5. The first case is stable (but proving this requires a fancy theory called KAM theory), the second case is unstable (in this case actually the linearization at 0 determines the picture). SYMMETRIC MATRICES Math 21b, O. Knill SYMMETRIC MATRICES. A matrix A with real entries is symmetric, if AT = A. 1 2 1 1 EXAMPLES. A = is symmetric, A = is not symmetric. 2 3 0 3 EIGENVALUES OF SYMMETRIC MATRICES. Symmetric matrices A have real eigenvalues. PROOF. The dot product is extend to complex vectors as (v, w) = i v i wi . For real vectors it satisfies (v, w) = v · w and has the property (Av, w) = (v, AT w) for real matrices A and (λv, w) = λ(v, w) as well as (v, λw) = λ(v, w). Now λ(v, v) = (λv, v) = (Av, v) = (v, AT v) = (v, Av) = (v, λv) = λ(v, v) shows that λ = λ because (v, v) = 0 for v = 0. p −q EXAMPLE. A = has eigenvalues p + iq which are real if and only if q = 0. q p EIGENVECTORS OF SYMMETRIC MATRICES. Symmetric matrices have an orthonormal eigenbasis PROOF. If Av = λv and Aw = µw. The relation λ(v, w) = (λv, w) = (Av, w) = (v, A T w) = (v, Aw) = (v, µw) = µ(v, w) is only possible if (v, w) = 0 if λ = µ. WHY ARE SYMMETRIC MATRICES IMPORTANT? In applications, matrices are often symmetric. For ex- ample in geometry as generalized dot products v·Av, or in statistics as correlation matrices Cov[X k , Xl ] or in quantum mechanics as observables or in neural networks as learning maps x → sign(W x) or in graph theory as adjacency matrices etc. etc. Symmetric matrices play the same role as real numbers do among the complex numbers. Their eigenvalues often have physical or geometrical interpretations. One can also calculate with symmetric matrices like with numbers: for example, we can solve B 2 = A for B if A is symmetric matrix 0 1 and B is square root of A.) This is not possible in general: try to find a matrix B such that B 2 = ... 0 0 RECALL. We have seen when an eigenbasis exists, a matrix A can be transformed to a diagonal matrix B = S −1 AS, where S = [v1 , ..., vn ]. The matrices A and B are similar. B is called the diagonalization of A. Similar matrices have the same characteristic polynomial det(B − λ) = det(S −1 (A − λ)S) = det(A − λ) and have therefore the same determinant, trace and eigenvalues. Physicists call the set of eigenvalues also the spectrum. They say that these matrices are isospectral. The spectrum is what you "see" (etymologically the name origins from the fact that in quantum mechanics the spectrum of radiation can be associated with eigenvalues of matrices.) SPECTRAL THEOREM. Symmetric matrices A can be diagonalized B = S −1 AS with an orthogonal S. PROOF. If all eigenvalues are different, there is an eigenbasis and diagonalization is possible. The eigenvectors are all orthogonal and B = S −1 AS is diagonal containing the eigenvalues. In general, we can change the matrix A to A = A + (C − A)t where C is a matrix with pairwise different eigenvalues. Then the eigenvalues are different for all except finitely many t. The orthogonal matrices St converges for t → 0 to an orthogonal matrix S and S diagonalizes A. WAIT A SECOND ... Why could we not perturb a general matrix At to have disjoint eigenvalues and At could −1 be diagonalized: St At St = Bt ? The problem is that St might become singular for t → 0. See problem 5) first practice exam. a b 1 √ EXAMPLE 1. The matrix A = has the eigenvalues a + b, a − b and the eigenvectors v1 = / 2 b a 1 −1 √ and v2 = / 2. They are orthogonal. The orthogonal matrix S = v1 v2 diagonalized A. 1   1 1 1 EXAMPLE 2. The 3 × 3 matrix A =  1 1 1  has 2 eigenvalues 0 to the eigenvectors 1 −1 0 , 1 1 1 1 0 −1 and one eigenvalue 3 to the eigenvector 1 1 1 . All these vectors can be made orthogonal and a diagonalization is possible even so the eigenvalues have multiplicities. SQUARE ROOT OF A MATRIX. How do we find a square root of a given symmetric matrix? Because S −1 AS = B is diagonal and we know how to take a square root of the diagonal matrix B, we can form √ √ √ C = S BS −1 which satisfies C 2 = S BS −1 S BS −1 = SBS −1 = A. RAYLEIGH FORMULA. We write also (v, w) = v · w. If v(t) is an eigenvector of length 1 to the eigenvalue λ(t) of a symmetric matrix A(t) which depends on t, differentiation of (A(t) − λ(t))v(t) = 0 with respect to t gives (A −λ )v +(A−λ)v = 0. The symmetry of A−λ implies 0 = (v, (A −λ )v)+(v, (A−λ)v ) = (v, (A −λ )v). We see that the Rayleigh quotient λ = (A v, v) is a polynomial in t if A(t) only involves terms t, t2 , . . . , tm . The 1 t2 formula shows how λ(t) changes, when t varies. For example, A(t) = has for t = 2 the eigenvector t2 1 √ 0 4 v = [1, 1]/ 2 to the eigenvalue λ = 5. The formula tells that λ (2) = (A (2)v, v) = ( v, v) = 4. Indeed, 4 0 λ(t) = 1 + t2 has at t = 2 the derivative 2t = 4. EXHIBITION. "Where do symmetric matrices occur?" Some informal motivation: I) PHYSICS: In quantum mechanics a system is described with a vector v(t) which depends on time t. The evolution is given by the Schroedinger equation v = i¯ Lv, where L is a symmetric matrix and h is a small ˙ h ¯ number called the Planck constant. As for any linear differential equation, one has v(t) = e i¯ Lt v(0). If v(0) is h an eigenvector to the eigenvalue λ, then v(t) = eit¯ λ v(0). Physical observables are given by symmetric matrices h too. The matrix L represents the energy. Given v(t), the value of the observable A(t) is v(t) · Av(t). For example, if v is an eigenvector to an eigenvalue λ of the energy matrix L, then the energy of v(t) is λ. This is called the Heisenberg picture. In order that v · A(t)v = v(t) · Av(t) = S(t)v · AS(t)v we have A(t) = S(T )∗ AS(t), where S ∗ = S T is the correct generalization of the adjoint to complex matrices. S(t) satisfies S(t)∗ S(t) = 1 which is called unitary and the complex analogue of orthogonal. The matrix A(t) = S(t)∗ AS(t) has the same eigenvalues as A and is similar to A. II) CHEMISTRY. The adjacency matrix A of a graph with n vertices determines the graph: one has A ij = 1 if the two vertices i, j are connected and zero otherwise. The matrix A is symmetric. The eigenvalues λ j are real and can be used to analyze the graph. One interesting question is to what extent the eigenvalues determine the graph. In chemistry, one is interested in such problems because it allows to make rough computations of the electron density distribution of molecules. In this so called H¨ ckel theory, the molecule is represented as a graph. The u eigenvalues λj of that graph approximate the energies an electron on the molecule. The eigenvectors describe the electron density distribution.   This matrix A has the eigenvalue 0 with 0 1 1 1 1 multiplicity 3 (ker(A) is obtained im- The Freon molecule for  1 0 0 0 0    mediately from the fact that 4 rows are example has 5 atoms. The  1 0 0 0 0 . the same) and the eigenvalues 2, −2.   adjacency matrix is  1 0 0 0 0  The eigenvector to the eigenvalue ±2 is 1 0 0 0 0 T ±2 1 1 1 1 . III) STATISTICS. If we have a random vector X = [X1 , · · · , Xn ] and E[Xk ] denotes the expected value of Xk , then [A]kl = E[(Xk − E[Xk ])(Xl − E[Xl ])] = E[Xk Xl ] − E[Xk ]E[Xl ] is called the covariance matrix of the random vector X. It is a symmetric n × n matrix. Diagonalizing this matrix B = S −1 AS produces new random variables which are uncorrelated. For example, if X is is the sum of two dice and Y is the value of the second dice then E[X] = [(1 + 1) + (1 + 2) + ... + (6 + 6)]/36 = 7, you throw in average a sum of 7 and E[Y ] = (1 + 2 + ... + 6)/6 = 7/2. The matrix entry A11 = E[X 2 ] − E[X]2 = [(1 + 1) + (1 + 2) + ... + (6 + 6)]/36 − 72 = 35/6 known as the variance of X, and A22 = E[Y 2 ] − E[Y ]2 = (12 + 22 + ... + 62 )/6 − (7/2)2 = 35/12 known as the variance of Y and 35/6 35/12 A12 = E[XY ] − E[X]E[Y ] = 35/12. The covariance matrix is the symmetric matrix A = . 35/12 35/12 DIFFERENTIAL EQUATIONS, Math 21b, O. Knill LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS. Df = T f = f is a linear map on the space of smooth functions C ∞ . If p(x) = a0 + a1 x + ... + an xn is a polynomial, then p(D) = a0 + a1 D + ... + an Dn is a linear map on C ∞ (R) too. We will see here how to find the general solution of p(D)f = g. EXAMPLE. For p(x) = x2 − x + 6 and g(x) = cos(x) the problem p(D)f = g is the differential equation f (x) − f (x) − 6f (x) = cos(x). It has the solution c1 e−2x + c2 e3x − (sin(x) + 7 cos(x))/50, where c1 , c2 are arbitrary constants. How do we find these solutions? THE IDEA. In general, a differential equation p(D)f = g has many solution. For example, for p(D) = D 3 , the equation D 3 f = 0 has solutions (c0 + c1 x + c2 x2 ). The constants come from integrating three times. Integrating means applying D −1 but since D has as a kernel the constant functions, integration gives a one dimensional space of anti-derivatives (we can add a constant to the result and still have an anti-derivative). In order to solve D 3 f = g, we integrate g three times. We will generalize this idea by writing T = p(D) as a product of simpler transformations which we can invert. These simpler transformations have the form (D − λ)f = g. FINDING THE KERNEL OF A POLYNOMIAL IN D. How do we find a basis for the kernel of T f = f +2f +f ? The linear map T can be written as a polynomial in D which means T = D 2 − D − 2 = (D + 1)(D − 2). The kernel of T contains the kernel of D − 2 which is one-dimensional and spanned by f 1 = e2x . The kernel of T = (D − 2)(D + 1) also contains the kernel of D + 1 which is spanned by f 2 = e−x . The kernel of T is therefore two dimensional and spanned by e2x and e−x . THEOREM: If T = p(D) = D n + an−1 Dn−1 + ... + a1 D + a0 on C ∞ then dim(ker(T )) = n. PROOF. T = p(D) = (D − λj ), where λj are the roots of the polynomial p. The kernel of T contains the kernel of D − λj which is spanned by fj (t) = eλj t . In the case when we have a factor (D − λj )k of T , then we have to consider the kernel of (D − λj )k which is q(t)eλt , where q is a polynomial of degree k − 1. For example, the kernel of (D − 1)3 consists of all functions (a + bt + ct2 )et . T SECOND PROOF. Write this as Ag = 0, where A is a n × n matrix and g = ˙ f, f˙, · · · , f (n−1) , where (k) k f = D f is the k'th derivative. The linear map T = AD acts on vectors of functions. If all eigenvalues λ j of A are different (they are the same λj as before), then A can be diagonalized. Solving the diagonal case BD = 0 is easy. It has a n dimensional kernel of vectors F = [f 1 , . . . , fn ]T , where fi (t) = t. If B = SAS −1 , and F is in the kernel of BD, then SF is in the kernel of AD. REMARK. The result can be generalized to the case, when a j are functions of x. Especially, T f = g has a solution, when T is of the above form. It is important that the function in front of the highest power D n is bounded away from 0 for all t. For example xDf (x) = e x has no solution in C ∞ , because we can not integrate ex /x. An example of a ODE with variable coefficients is the Sturm-Liouville eigenvalue problem T (f )(x) = a(x)f (x) + a (x)f (x) + q(x)f (x) = λf (x) like for example the Legendre differential equation (1 − x2 )f (x) − 2xf (x) + n(n + 1)f (x) = 0. BACKUP • Equations T f = 0, where T = p(D) form linear differential equations with constant coefficients for which we want to understand the solution space. Such equations are called homogeneous. Solving differential equations often means finding a basis of the kernel of T . In the above example, a general solution of f + 2f + f = 0 can be written as f (t) = a1 f1 (t) + a2 f2 (t). If we fix two values like f (0), f (0) or f (0), f (1), the solution is unique. • If we want to solve T f = g, an inhomogeneous equation then T −1 is not unique because we have a kernel. If g is in the image of T there is at least one solution f . The general solution is then f + ker(T ). For example, for T = D 2 , which has C ∞ as its image, we can find a solution to D 2 f = t3 by integrating twice: f (t) = t5 /20. The kernel of T consists of all linear functions at + b. The general solution to D 2 = t3 is at + b + t5 /20. The integration constants parameterize actually the kernel of a linear map. A GENERAL REGION. For a general region, one uses numerical methods. One pos- sibility is by conformal transportation: to map the region into the the disc using a complex map F , which maps the region Ω into the disc. Solve then the problem in the disc with boundary value T (F −1 (x, y)) on the disc. If S(x, y) is the solution there, then T (x, y) = S(F −1 (x, y)) is the solution in the region Ω. The picture shows the example of the Joukowski map z → (z + 1/z)/2 which has been used in fluid dynamics (N.J. Joukowski (1847-1921) was a Russian aerodynamics researcher.) A tourists view on related PDE topics POISSON EQUATION. The Poisson equation ∆f = g in a square [0, π]2 can be solved via Fourier theory: If f (x, y) = k,m an,m cos(nx) cos(my) and g(x, y) = k,m bn,m cos(nx) cos(my), then fxx + fyy = n,m −(n2 + m2 )an,m cos(nx) cos(my) = k,m bn,m cos(nx) cos(my), then an,m = −bn,m /(n2 − m2 ). The poisson equation is important in electrostatics, for example to determine the electromagnetic field when the charge and current distribution is known: ∆U (x) = −(1/ 0 )ρ, then E = U is the electric field to the charge distribution ρ(x). EIGENVALUES. The functions sin(nx) sin(ny) are eigenfunctions for the Laplace operator fxx + fyy = n2 f on the disc. For a general bounded region Ω, we can look at all smooth functions which are zero on the boundary of Ω. The possi- ble eigenvalues of ∆f = λf are the possible ener- gies of a particle in Ω. COMPLEX DYNAMICS. Also in complex dy- namics, harmonic functions appear. Finding properties of complicated sets like the Mandel- brot set is done by mapping the exterior to the outside of the unit circle. If the Mandelbrot set is charged then the contour lines of equal poten- tial can be obtained as the corresponding contour lines in the disc case (where the lines are circles). SCHRODINGER EQUATION. If H is the energy operator, then i¯ f˙ = Hf is called the Schr¨dinger equa- ¨ h o tion. If H = −¯ 2 ∆/(2m), this looks very much like the heat equation, if there were not the i. If f is an h eigenvalue of H, then i¯ f˙ = λf and f (t) = ei¯ t f (0). In the heat equation, we would get f (t) = e−µ(t) f (0). The h h evolution of the Schr¨dinger equation is very smilar to the wave equation. o QUANTUM CHAOS studies the eigenvalues and eigenvectors of the Laplacian in a bounded region. If the billiard in the region is chaotic, the study of the eigensystem is called quantum chaos. The eigenvalue problem ∆f = λf and the billiard problem in the region are related. A famous open problem is whether two smooth convex regions in the plane for which the eigenvalues λj are the same are equivalent up to rotation and translation. 2 π In a square of size L we know all the eigenvalues L2 (n2 + m2 ). The eigenvectors are sin( pi nx) sin( pi mx). We have explicit formulas for the eigenfunctions and eigenvalues. L L On the classical level, for the billiard, we have also a complete picture, how a billiard path will look like. The square is an example, where we have no quantum chaos.
Algebraic Equations Algebraic Equations Algebra!! Oh my god! How I will solve algebraic problems, god please help me. Many students are afraid of this branch of mathematics. Now, I am here to tell you that how you can practice algebra in an easy way. Algebra is the basic or we can say ... More Algebraic Equations Algebra!! Oh my god! How I will solve algebraic problems, god please help me. Many students are afraid of this branch of mathematics. Now, I am here to tell you that how you can practice algebra in an easy way. Algebra is the basic or we can say fundamental branch of mathematics, in which we learn how to use operators, numbers variables in a well defined manner. Algebra covers many topics which are commonly used in daily life and has great real time applications. Main parts of Algebra starts with equations and moves on and on. Do you know what are Algebraic Equations ? 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Tutorvista is an online math tutoring website which helps students in grasping different math concepts. Tutorvista covers ... More Solving Fractions Hey! Friends, if you want to become master of mathematics then you must know all the fundamental concept of math and its different topics. Tutorvista is an online math tutoring website which helps students in grasping different math concepts. Tutorvista covers all the math topics whether it is of initial mathematics or higher studies, which are explained by expert online math tutors. Today, we will discuss about fractions, an initial topic of algebra. Fraction means a part of whole number which is represented as numerator and denominator, 2/ 3 is a fractional form in which 2 is numerator and 3 is denominator. There are 3 possible forms in which a fraction can exist: Common form of fraction =2/ 3 Decimal form = 0. 22 Percentage form = 22 % = 22/ 100 Now move on the general rules that are required to remember while performing normal arithmetic operation to solve fractions: First is the standard property for performing addition or subtraction o Less Graphing Calculator Graphing is the useful procedure in mathematics for explaining complex equations, functions and relations and solving them. Graphing of any equation means its corresponding 2D paper representation of any linear shape in respect of Cartesian axes coordinates and tangent at which it is angled with X- axis. In... More Graphing Calculator Graphing is the useful procedure in mathematics for explaining complex equations, functions and relations and solving them. Graphing of any equation means its corresponding 2D paper representation of any linear shape in respect of Cartesian axes coordinates and tangent at which it is angled with X- axis. In present time online math tutoring is a dynamic and advance way of math learning for students which also offers the use of some online math tools to solve math queries easily. These online tools are coded JavaScript programs which runs on compatible Internet browser. Graphing calculator is also an online tools uses for graphing of mathematical equations. As we all know every program have its fixed process for solving queries which is implemented in its algorithm and for graphing calculator this algorithm are graphing functions which are distinct for various kind of mathematical equations. Learn More about Graphing Functions Less 5th Grade Math Get 5th grade math help from expert online tutors of tutorvista who are highly qualified and well trained as tutors. Below is a representative list of topics covered in our 5th Grade Math Program - however all programs will be customized for the individual student. Our online tutors for 5th grade math help improve... More 5th Grade Math Get 5th grade math help from expert online tutors of tutorvista who are highly qualified and well trained as tutors. Below is a representative list of topics covered in our 5th Grade Math Program - however all programs will be customized for the individual student. Our online tutors for 5th grade math help improve your knowledge over the topics below. Important Topics Covered in 5th Grade Math * Place value and number sense * Fractions and mixed numbers * Geometry * Add and subtract fractions * Data, charts and graphs * Patterns * Algebra Word Problems"> Algebra Word Problems Note :Please maintain Math worksheets,a review page, and maintain a list of important topics covered under each of the main topics. Learn More about 4th Grade Math Less Online Help In algebra we have got a lot of different types of equations like linear equation, quadratic equation, and polynomial equation. In mathematics, an equation is a statement which has two expressions separated by = ... More Online Help In algebra we have got a lot of different types of equations like linear equation, quadratic equation, and polynomial equation. In mathematics, an equation is a statement which has two expressions separated by = sign having same value or in other words we can say it shows the balance situation where the value of one side is exactly same as the opposite side. While we go through these all different equations. We got a lot of problems and we need guidance to solve them. On internet, Online Help is available to solve our problems related to all different types of equations. Today we will discuss about linear equations. A linear equation is an algebraic equation where all the terms can be a constant or product of a constant and a single variable. It can have multiple variables also. For example : 7h + c = 18 4h + c = 9 To find out the solution either we can add or subtract both equations. For given example we will use elimination method. Less How To Factor Polynomials Factoring Polynomials refers to factoring a polynomial into irreducible polynomials over a given field. It gives out the factors that together form a polynomial function. A polynomial function is of the form xn + xn -1 + xn - 2 + . . . . + k = 0, where k is a constant and n is a power. Polynomials are... More How To Factor Polynomials Factoring Polynomials refers to factoring a polynomial into irreducible polynomials over a given field. It gives out the factors that together form a polynomial function. A polynomial function is of the form xn + xn -1 + xn - 2 + . . . . + k = 0, where k is a constant and n is a power. Polynomials are expressions that are formed by adding or subtracting several variables called monomials. Monomials are variables that are formed with a constant and a variable of some degree. Examples of monomials are 5x3, 6a2. Monomials having different exponents such as 5x3 and 3x4 cannot be added or subtracted but can be multiplied or divided by them. Any polynomial of the form F(a) can also be written as F(a) = Q(a) x D (a) + R (a) using Dividend = Quotient x Divisor + Remainder. If the polynomial F(a) is divisible by Q(a), then the remainder is zero. Thus, F(a) = Q(a) x D(a). That is, the polynomial F(a) is a product of two other polynomials Q(a) and D(a). Fo Less Simplifying Rational Expressions Simplifying Rational Expressions Friends in this session we will talk all about rational expressions, sometimes an rational expression can have fractions of complex form. This complexity is caused because of polynomials functions which are included in ... More Simplifying Rational Expressions Friends in this session we will talk all about rational expressions, sometimes an rational expression can have fractions of complex form. This complexity is caused because of polynomials functions which are included in numerator and denominator of the Rational Expression. So the solution of these complex fractions or rational expressions is not so easy it requires an appropriate way. Lets have some brief discussion on the way of solving any rational expression, while solving complex fraction the first thing is to reduce the polynomials functions of both numerator and denominator to the lowest form Reducing the polynomials to the lowest form is done by using factorization process for polynomials. After converting polynomial into their factors the common factor is taken out so that the numerator and denominator is divided by their common factor. Once all the common factor are eliminated than the value of vari Less Linear Equations Learn More about Graphing Linear Equations What is a linear equation: An equation is a condition on a variable. A variable takes on different values; its value is not fixed . Variables are denoted usually by letter of alphabets, such as x, y , z , l , m , n , p etc. From variables we form expression. Linear... More Linear Equations Learn More about Graphing Linear Equations What is a linear equation: An equation is a condition on a variable. A variable takes on different values; its value is not fixed . Variables are denoted usually by letter of alphabets, such as x, y , z , l , m , n , p etc. From variables we form expression. Linear equation in one variable :- These are the type of equation which have unique (i. e, only one and one ) solution. For example :- 2 x + 5 = 0 is a linear equation in one variable. Root of the equation is −52 Example 1 :- Convert the following equation in statement form. x – 5 = 9 Less 4th Grade Math Get 4th grade math help from expert tutorvista online tutors and make your math easy. It is always necessary to have a clear basic knowledge about everything, sign up for 4th grade Math help from our online tutors and improve your knowledge. We also offer 4th grade Math help on getting homework help as well. This is a... More 4th Grade Math Get 4th grade math help from expert tutorvista online tutors and make your math easy. It is always necessary to have a clear basic knowledge about everything, sign up for 4th grade Math help from our online tutors and improve your knowledge. We also offer 4th grade Math help on getting homework help as well. This is a representative list of topics covered in our Grade 4 Math program - however all programs will be customized for the individual student. You can also avail to our other grade math programs. Grade 3 Math Grade 5 Math Grade 6 Math Grade 7 Math Grade 8 Math Grade 9 Math 4th Grade Math Topics :Find the list of topics given below, each of these topics have sub topics, have sample examples to explain the topic & sample test papers. We have a list of important topics and some tips that can make learning easy and effective. Important topics covered under Grade 4 :* Fraction Learn More about 5th Grade Math Less
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CHAPTER 3 Linear Programming: Geometric Approach OUTLINE 3.1 Systems of Linear Inequalities 3.2 A Geometric Approach to Linear Programming Problems 3.3 Applications • Chapter Review • Chapter Project • Mathematical Questions from Professional Exams I t's the weekend after midterms, and a hiking trip to Pinnacles National good. But in what proportions should they be mixed? And what about Monument in California is on the meeting some minimum calorie agenda. Hiking in Pinnacles will requirements? What about carbohy- require some advance planning. drates and protein? And, of course, Some trails go through caves, so a fat should be minimized! Fortunately, flashlight is needed. Others go to this chapter was covered before the top of low mountain peaks. No midterms, so these questions can be matter where the hiking ends up, answered. The Chapter Project at the some food, such as a trail mix, will be end of the chapter will guide you. required. Peanuts and raisins sound 158 Systems of Linear Inequalities 159 A L O O K B AC K , A L O O K F O RWA R D In Chapter 1, we discussed linear equations and some appli- Historically, linear programming problems evolved out cations that involve linear equations. In Chapter 2, we stud- of the need to solve problems involving resource allocation ied systems of linear equations. In this chapter, we discuss by the U.S. Army during World War II. Among those who systems of linear inequalities and an extremely important worked on such problems was George Dantzig, who later application involving linear equations and systems of linear gave a general formulation of the linear programming inequalities: linear programming. problem and offered a method for solving it, called the Whenever the analysis of a problem leads to minimiz- simplex method. This method is discussed in Chapter 4. ing or maximizing a linear expression in which the In this chapter we study ways to solve linear program- variables must obey a collection of linear inequalities, a ming problems that involve only two variables. As a result, solution may be obtained using linear programming we can use a geometric approach utilizing the graph of a techniques. system of linear inequalities to solve the problem. 3.1 Systems of Linear Inequalities PREPARING FOR THIS SECTION Before getting started, review the following: > Inequalities (Appendix A, Section A.2, pp. xx) > Pairs of Lines (Section 1.2, pp. 19– 23) > Lines (Section 1.1, pp. 2– 14) OBJECTIVES 1 Graph linear inequalities 2 Graph systems of linear inequalities In Section 1.1, we discussed linear equations (linear equalities) in two variables x and y. Recall that these are equations of the form Ax By C (1) where A, B, C are real numbers and A and B are not both zero. If in Equation (1) we replace the equal sign by an inequality symbol, namely, one of the symbols , , , or , we obtain a linear inequality in two variables x and y. For example, the expressions 3x 2y 4 2x 3y 0 3x 5y 8 are each linear inequalities in two variables. The first of these is called a nonstrict inequal- ity since the expression is satisfied when 3x 2y 4, as well as when 3x 2y 4. The remaining two linear inequalities are strict. The Graph of a Linear Inequality Graph linear inequalities 1 The graph of a linear inequality in two variables x and y is the set of all points (x, y) for which the inequality is satisfied. Let's look at an example. EXAMPLE 1 Graphing a Linear Inequality Graph the inequality: 2x 3y 6 160 Chapter 3 Linear Programming: Geometric Approach SOLUTION The inequality 2x 3y 6 is equivalent to 2x 3y 6 or 2x 3y 6. So we begin by graphing the line L: 2x 3y 6, noting that any point on the line must satisfy the inequality 2x 3y 6. See Figure 1(a). FIGURE 1 y y 6 (5, 5) 6 (5, 5) 4 4 (0, 2) (0, 2) (4, 0) x (4, 0) x (–4, 0) –2 (3, 0) 6 (–4, 0) –2 (3, 0) 6 (–1, –1) –2 (–1, –1) –2 2x + 3y = 6 2x + 3y = 6 (a) (b) Now let's test a few points, such as ( 1, 1), (5, 5), (4, 0), ( 4, 0), to see if they satisfy the inequality. We do this by substituting the coordinates of each point into the inequality and determining whether the result is 6 or 6. 2x 3y Conclusion ( 1, 1): 2( 1) 3( 1) 2 3 5 6 Not part of graph (5, 5): 2(5) 3(5) 25 6 Part of graph (4, 0): 2(4) 3(0) 8 6 Part of graph ( 4, 0): 2( 4) 3(0) 8 6 Not part of graph Notice that the two points (4, 0) and (5, 5) that are part of the graph both lie on one side of L, while the points ( 4, 0) and ( 1, 1) (not part of the graph) lie on the oth- er side of L. This is not an accident. The graph of the inequality consists of all points on the same side of L as (4, 0) and (5, 5). The shaded region of Figure 1(b) illustrates the graph of the inequality. ◗ The inequality in Example 1 was nonstrict, so the corresponding line was part of the graph of the inequality. If the inequality is strict, the corresponding line is not part of the graph of the inequality. We will indicate a strict inequality by using dashes to graph the line. Let's outline the procedure for graphing a linear inequality: L Steps for Graphing a Linear Inequality STEP 1 Graph the corresponding linear equation, a line L. If the inequality is non- strict, graph L using a solid line; if the inequality is strict, graph L using dashes. STEP 2 Select a test point P not on the line L. STEP 3 Substitute the coordinates of the test point P into the given inequality. If the coordinates of this point P satisfy the linear inequality, then all points on the same side of L as the point P satisfy the inequality. If the coordinates of the point P do not satisfy the linear inequality, then all points on the opposite side of L from P satisfy the inequality. Systems of Linear Inequalities 161 EXAMPLE 2 Graphing a Linear Inequality Graph the linear inequality: 2x y 4 SOLUTION The corresponding linear equation is the line L: 2x y 4 Since the inequality is strict, points on L are not part of the graph of the linear inequal- ity. When we graph L, we use a dashed line to indicate this fact. See Figure 2(a). We select a point not on the line L to be tested, for example, (0, 0): 2x y 2(0) 0 0 2x y 4 Since 0 is not less then 4, the point (0, 0) does not satisfy the inequality. As a result, all points on the opposite side of L from (0, 0) satisfy the inequality. The graph of 2x y 4 is the shaded region of Figure 2(b). ◗ FIGURE 2 y y L L 6 6 (0, 4) (0, 4) 2 2 (–2, 0) x (–2, 0) x –4 2 4 –4 2 4 –2 –2 (0, 0) 2x – y = –4 2x – y = –4 (a) (b) NOW WORK PROBLEM 7. EXAMPLE 3 Graphing Linear Inequalities Graph: (a) x 3 (b) 2x y SOLUTION (a) The corresponding linear equation is x 3, a vertical line. If we choose (0, 0) as the test point, we find that it satisfies the inequality [0 3], so all points to the left of, and on, the vertical line also satisfy the inequality. See Figure 3(a). (b) The corresponding linear equation is 2x y. Since its graph passes through (0, 0), we choose the point (0, 2) as the test point. The inequality is satisfied by the point (0, 2)[2(0) 2], so all points on the same side of the line as (0, 2) also satisfy the inequality. See Figure 3(b). FIGURE 3 y y 2 (0, 2) (1, 2) (0, 0) x (0, 0) x –4 –2 2 4 –4 –2 2 4 –2 –2 x=3 2x = y (a) (b) ◗ 162 Chapter 3 Linear Programming: Geometric Approach The set of points belonging to the graph of a linear inequality [for example, the shaded region in Figure 3(b)] is called a half-plane. NOW WORK PROBLEM 5. COMMENT: A graphing utility can be used to obtain the graph of a linear inequality. See "Using a graphing utility to graph inequalities" in Appendix C, Section C.4, for a discussion. ◗ Systems of Linear Inequalities Graph systems of linear 2 A system of linear inequalities is a collection of two or more linear inequalities. To inequalities graph a system of two inequalities we locate all points whose coordinates satisfy each of the linear inequalities of the system. Determining Whether a Point Belongs to the Graph of a System of EXAMPLE 4 Two Linear Inequalities Determine which of the following points are part of the graph of the system of linear inequalities: 2x y 6 (1) x y 3 (2) (a) P1 (6, 0) (b) P2 (3, 5) (c) P3 (0, 0) (d) P4 (3, 2) SOLUTION We check to see if the given point satisfies each of the inequalities of the system. (a) P1 (6, 0) 2x y 2(6) 0 12 x y 6 0 6 2x y 6 x y 3 P1 satisfies inequality (2) but not inequality (1), so P1 is not part of the graph of the system. (b) P2 (3, 5) 2x y 2(3) 5 11 x y 3 5 2 2x y 6 x y 3 P2 satisfies neither inequality (1) nor inequality (2), so P2 is not part of the graph of the system. (c) P3 (0, 0) 2x y 2(0) 0 0 x y 0 0 0 2x y 6 x y 3 P3 satisfies inequality (1) but not inequality (2), so P3 is not part of the graph of the system. (d) P4 (3, 2) 2x y 2(3) ( 2) 4 x y 3 ( 2) 5 2x y 6 x y 3 P4 satisfies both inequality (1) and inequality (2), so P4 is part of the graph of the system. ◗ NOW WORK PROBLEM 13. Systems of Linear Inequalities 163 Let's graph the information from Example 4. Figure 4(a) shows the graph of each of the lines 2x y 6 and x y 3 and the four points P1, P2, P3, and P4. Notice that because the two lines of the system intersect, the plane is divided into four regions. Since the graph of each linear inequality of the system is a half-plane, the graph of the system of linear inequalities is the intersection of these two half-planes. As a result, the region containing P4 is the graph of the system. See Figure 4(b). We use the method described above in the next example. FIGURE 4 y y (0, 6) (0, 6) P2 x–y = 3 P2 x–y = 3 4 4 (3, 0) (3, 0) x x –4 P3 4 P1 –4 P3 4 P1 P4 P4 (0, –3) (0, –3) –4 –4 2x + y = 6 2x + y = 6 (a) (b) EXAMPLE 5 Graphing a System of Two Linear Inequalities 2x y 4 Graph the system: x y 1 SOLUTION First we graph each inequality separately. See Figures 5(a) and 5(b). The solution of the system consists of all points common to these two half-planes. The dark blue shaded region in Figure 6 represents the solution of the system. FIGURE 5 FIGURE 6 y y y 6 6 6 2x – y ≤ –4 x + y ≥ –1 4 (0, 4) (0, 4) 2 2 2 (–2, 0) x x (–2, 0) (–1, 0) x –4 2 –4 (–1, 0) 2 –4 2 (0, –1) –2 (0, –1) –2 –4 –4 2x – y = –4 x + y = –1 2x – y = –4 x + y = –1 (a) (b) ◗ NOW WORK PROBLEM 17. The lines in the system of linear inequalities given in Example 5 intersect. If the two lines of a system of two linear inequalities are parallel, the system of linear inequalities may or may not have a solution. Examples of such situations follow. 164 Chapter 3 Linear Programming: Geometric Approach EXAMPLE 6 Graphing a System of Two Linear Inequalities 2x y 4 Graph the system: 2x y 2 SOLUTION First we graph each inequality separately. See Figures 7(a) and 7(b). The grey shaded region in Figure 8 represents the solution of the system. FIGURE 7 FIGURE 8 y y y 6 6 6 2x – y ≤ –4 2x – y ≤ –2 (0, 4) 4 (0, 4) 2 (0, 2) (–2, 0) x (–1, 0) x (0, 2) (–2, 0) x –4 2 –4 –2 2 –4 (–1, 0) 2 –2 –2 –2 –4 –4 –4 2x – y = –4 2x – y = –2 (a) (b) ◗ Notice that the solution of this system is the same as that of the single linear inequality 2x y 4. EXAMPLE 7 Graphing a System of Two Linear Inequalities The solution of the system 2x y 4 2x y 2 is the grey shaded region in Figure 9. FIGURE 9 y 6 (0, 4) (0, 2) (–2, 0) x –4 (–1, 0) 2 –2 –4 2x – y = – 4 2x – y = –2 ◗ EXAMPLE 8 Graphing a System of Two Linear Inequalities The system 2x y 4 2x y 2 Systems of Linear Inequalities 165 has no solution, as Figure 10 indicates, because the two half-planes have no points in common. FIGURE 10 y 6 (0, 4) (0, 2) (–2, 0) x –4 (–1, 0) 2 –2 –4 2x – y = –4 2x – y = –2 ◗ Until now, we have considered systems of only two linear inequalities. The next example is of a system of four linear inequalities. As we shall see, the technique for graphing such systems is the same as that used for graphing systems of two linear inequalities in two variables. EXAMPLE 9 Graphing a System of Four Linear Inequalities x y 2 2x y 3 Graph the system: x 0 y 0 SOLUTION Again we first graph the four lines: L1: x y 2 L2: 2x y 3 L3: x 0 (the y-axis) L4: y 0 (the x-axis) The lines L1 and L2 intersect at the point (1, 1). (Do you see why?) The inequalities x 0 and y 0 indicate that the graph of the system lies in quadrant I. The graph of the system consists of that part of the graphs of the inequalities x y 2 and 2x y 3 that lies in quadrant I. See Figure 11. FIGURE 11 y L2 L1 6 (0, 3) (0, 2) (1, 1) (2, 0) x –2 4 6 –2 (3 , 0) 2 2x + y = 3 x + y = 2 ◗ 166 Chapter 3 Linear Programming: Geometric Approach EXAMPLE 10 Graphing a System of Four Linear Inequalities x y 2 2x y 3 Graph the system: x 0 y 0 SOLUTION The lines associated with these linear inequalities are the same as those of the previous example. Again the inequalities x 0, y 0 indicate that the graph of the system lies in quadrant I. The graph of the system consists of that part of the graphs of the inequalities x y 2 and 2x y 3 that lie in quadrant I. See Figure 12. FIGURE 12 y L2 L1 6 (0, 3) (0, 2) (1, 1) (0, 0) (2, 0) x 4 –2 (3 , 0) 2 2x+ y = 3 x+y = 2 ◗ Some Terminology Compare the graphs of the systems of linear inequalities given in Figures 11 and 12. The graph in Figure 11 is said to be unbounded in the sense that it extends infinitely far in some direction. The graph in Figure 12 is bounded in the sense that it can be enclosed by some circle of sufficiently large radius. See Figure 13. FIGURE 13 y y x x (a) Bounded graph (b) Unbounded graph The boundary of each of the graphs in Figures 11 and 12 consists of line segments. In fact, the graph of any system of linear inequalities will have line segments as boundaries. The point of intersection of two line segments that form the boundary is called a corner point of the graph. For example, the graph of the system given in Example 9 has the corner points (0, 3), (1, 1), and (2, 0). See Figure 11. The graph of the system given in Example 10 has the corner points (0, 2), (0, 0), (3 , 0), (1, 1). See 2 Figure 12. Systems of Linear Inequalities 167 We shall soon see that the corner points of the graph of a system of linear inequali- ties play a major role in the procedure for solving linear programming problems. NOW WORK PROBLEMS 25 AND 29. Application EXAMPLE 11 Analyzing a Mixture Problem(a) Use x to denote the number of packages of the low-grade mixture and use y to denote the number of packages of the high-grade mixture to be made and write a system of linear inequalities that describes the possible number of each kind of package. (b) Graph the system and list its corner points. SOLUTION (a) We begin by naming the variables: x Number of packages of low-grade mixture y Number of packages of high-grade mixture First, we note that the only meaningful values for x and y are nonnegative values. We restrict x and y so that x 0 and y 0 Next, we note that there is a limit to the number of pounds of cashews and peanuts available. That is, the total number of pounds of cashews cannot exceed 75 pounds (1200 ounces), and the number of pounds of peanuts cannot exceed 120 pounds (1920 ounces). This means that Ounces of Ounces of Number of Number of cashews cashews packages packages of cannot required required for of high- 1200 low-grade exceed for low-grade high-grade grade mixture mixture mixture mixture Ounces of Ounces of Number of Number of peanuts peanuts packages packages of cannot required for high- of high- 1920 low-grade exceed for low-grade grade grade mixture mixture mixture mixture In terms of the data given and the variables introduced, we can write these state- ments compactly as 4x 8y 1200 12x 8y 1920 168 Chapter 3 Linear Programming: Geometric Approach The system of linear inequalities that gives the possible values x and y can take on is 4x 8y 1200 12x 8y 1920 x 0 y 0 (b) The system of linear inequalities given above can be simplified to the equivalent form x 2y 300 (1) 3x 2y 480 (2) x 0 (3) y 0 (4) The graph of the system is given in Figure 14. Notice that the corner points of the graph are labeled. Three are easy to identify by inspection: (0, 0), (0, 150), and (160, 0). We found the remaining one (90, 105) by solving the system of equations x 2y 300 3x 2y 480 By subtracting the first equation from the second, we find 2x 180 or x 90. Back-substituting in the first equation, we find y 105. FIGURE 14 ◗ EXERCISE 3.1 Answers to Odd-Numbered Problems Begin on Page AN-00. In Problems 1–12, graph each inequality. 1. x 0 2. y 0 3. x 4 4. y 6 5. y 1 6. x 2 7. 2x 3y 6 8. 3x 2y 6 9. 5x y 10 10. x 2y 4 11. x 5y 5 12. 3x y 3 13. Without graphing, determine which of the points 14. Without graphing, determine which of the points P1 (3, 8), P2 (12, 9), P3 (5, 1) are part of the graph of P1 (9, 5), P2 (12, 4), P3 (4, 1) are part of the the following system: graph of the following system: x 3y 0 x 4y 0 3x 2y 0 5x 2y 0 Systems of Linear Inequalities 169 15. Without graphing, determine which of the points 16. Without graphing, determine which of the points P1 (2, 3), P2 (10, 10), P3 (5, 1) are part of the graph of P1 (2, 6), P2 (12, 4), P3 (4, 2) are part of the graph of the following system: the following system: 3x 2y 0 2x 5y 0 x y 15 x 3y 15 In Problems 17–24, determine which region a, b, c, or d represents the graph of the given system of lin- ear inequalities. The regions a, b, c, and d are nonoverlapping regions bounded by the indicated lines. 5x 4y 8 4x 5y 0 2x 5y 0 17. 18. 19. 2x 5y 23 4x 2y 28 x 3y 15 y y y d c a 8 8 8 d b 4 a 4 c b d x x x –8 –4 4 8 –8 –4 4 –8 4 8 –4 –4 a –8 –8 –8 c b 6x 5y 5 5x 3y 3 5x 5y 10 20. 21. 22. 2x 4y 30 2x 6y 30 6x 4y 48 y y y a a b 8 8 d d 4 4 4 c b a x b x x –8 –4 4 8 –8 –4 4 8 –8 –4 4 8 –4 –4 –8 c –8 –8 d c 5x 4y 0 2x 5y 5 23. 24. 2x 4y 28 3x 5y 30 y y d a 8 a 4 4 b c x d x –8 –4 4 8 –8 4 8 –4 –4 c –8 b –8 170 Chapter 3 Linear Programming: Geometric Approach In Problems 25 – 36, graph each system of linear inequalities. Tell whether the graph is bounded or unbounded and list each corner point of the graph. x y 2 x y 2 x y 2 2x 3y 6 2x 3y 12 2x 3y 6 25. x 0 26. x 0 27. 28. 3x 2y 12 x 0 y 0 y 0 x 0 y 0 y 0 x y 2 x y 2 x y 2 x y 2 x y 8 x y 8 2x 3y 12 2x y 3 29. 2x y 10 30. x 2y 1 31. 3x y 12 32. x 0 x 0 x 0 x 0 y 0 y 0 y 0 y 0 x 2y 1 x 2y 2 2x y 2 x 2y 1 x 2y 10 x y 4 3x 2y 6 y 4 x y 2 33. 34. 35. 3x y 3 36. x y 2 x 0 x y 8 x 0 x 0 y 0 x 0 y 0 y 0 y 0 37. Rework Example 11 if 60 pounds of cashews and 90 pounds (a) Using x to denote the amount of money invested in of peanuts are available. Treasury bills and y to denote the amount invested in corporate bonds, write a system of inequalities that 38. Rework Example 11 if the high-grade mixture contains describes this situation. 10 ounces of cashews and 6 ounces of peanuts. (b) Graph the system and list its corner points. 39. Manufacturing Mike's Famous Toy Trucks company manu- (c) Interpret the meaning of each corner point in relation to factures two kinds of toy trucks — a dumpster and a tanker. the investments it represents. In the manufacturing process, each dumpster requires 3 hours 42. Financial Planning Use the information supplied in Problem of grinding and 4 hours of finishing, while each tanker 41, along with the fact that the couple will invest at least requires 2 hours of grinding and 3 hours of finishing. The $20,000, to answer parts (a), (b), and (c). company has two grinders and three finishers, each of whom works at most 40 hours per week. 43. Nutrition A farmer prepares feed for livestock by combining two types of grain. Each unit of the first grain contains (a) Using x to denote the number of dumpsters and y to 1 unit of protein and 5 units of iron while each unit of the denote the number of tankers, write a system of linear second grain contains 2 units of protein and 1 unit of iron. inequalities that describes the possible numbers of each Each animal must receive at least 5 units of protein and truck that can be manufactured. 16 units of iron each day. (b) Graph the system and list its corner points. (a) Write a system of linear inequalities that describes the 40. Manufacturing Repeat Problem 39 if one grinder and two possible amounts of each grain the farmer needs to pre- finishers, each of whom works at most 40 hours per week, pare. are available. (b) Graph the system and list the corner points. 41. Financial Planning A retired couple have up to $25,000 to 44. Investment Strategy Kathleen wishes to invest up to a total invest. As their financial adviser, you recommend they place of $40,000 in class AA bonds and stocks. Furthermore, she at least $15,000 in Treasury bills yielding 6% and at most believes that the amount invested in class AA bonds should $10,000 in corporate bonds yielding 9%. be at most one-third of the amount invested in stocks. A Geometric Approach to Linear Programming Problems 171 (a) Write a system of linear inequalities that describes the microwaves from the eastern plant to its central warehouse possible amount of investments in each security. and 20 hours from the Midwest plant to its central ware- (b) Graph the system and list the corner points. house. It costs $80 to transport an order from the eastern plant to the central warehouse and $40 from the midwestern 45. Nutrition To maintain an adequate daily diet, nutritionists plant to its central warehouse. There are 1000 work-hours recommend the following: at least 85 g of carbohydrate, 70 g of available for packing, transportation, and so on, and $3000 fat, and 50 g of protein. An ounce of food A contains 5 g of car- for transportation cost. bohydrate, 3 g of fat, and 2 g of protein, while an ounce of food B contains 4 g of carbohydrate, 3 g of fat, and 3 g of protein. (a) Write a system of linear inequalities that describes the transportation system. (a) Write a system of linear inequalities that describes the (b) Graph the system and list the corner points. possible quantities of each food. (b) Graph the system and list the corner points. 47. Make up a system of linear inequalities that has no solution. 46. Transportation A microwave company has two plants, one 48. Make up a system of linear inequalities that has a single on the East Coast and one in the Midwest. It takes 25 hours point as solution. (packing, transportation, and so on) to transport an order of 3.2 A Geometric Approach to Linear Programming Problems OBJECTIVES 1 Identify a linear programming problem 2 Solve a linear programming problem Identify a linear program- 1 To help see the characteristics of a linear programming problem, we look again at ming problem Example 11 of the previous section.Suppose that in addition to the information given above, we also know what the profit will be on each type of mixture. For example, suppose the profit is $0.25 on each package of the low-grade mixture and is $0.45 on each package of the high-grade mix- ture. The question of importance to the manager is "How many packages of each type of mixture should be prepared to maximize the profit?" If P symbolizes the profit, x the number of packages of low-grade mixture, and y the number of high-grade packages, then the question can be restated as "What are the val- ues of x and y so that the expression P $0.25x $0.45y is a maximum?" This problem is typical of a linear programming problem. It requires that a cer- tain linear expression, in this case the profit, be maximized. This linear expression is called the objective function. Furthermore, the problem requires that the maxi- mum profit be achieved under certain restrictions or constraints, each of which are linear inequalities involving the variables. The linear programming problem may be restated as Maximize P $0.25x $0.45y Objective function 172 Chapter 3 Linear Programming: Geometric Approach subject to the conditions that x 2y 300 Cashew constraint 3x 2y 480 Peanut constraint x 0 Nonnegativity constraint y 0 Nonnegativity constraint In general, every linear programming problem has two components: 1. A linear objective function to be maximized or minimized. 2. A collection of linear inequalities that must be satisfied simultaneously. L Linear Programming Problem A linear programming problem in two variables, x and y, consists of maximiz- ing or minimizing an objective function z Ax By where A and B are given real numbers, not both zero, subject to certain conditions or constraints expressible as a system of linear inequalities in x and y. Let's look at this definition more closely. To maximize (or minimize) the quantity z Ax By means to locate the points (x, y) that result in the largest (or smallest) val- ue of z. But not all points (x, y) are eligible. Only the points that obey all the con- straints are potential solutions. We refer to such points as feasible points. Solve a linear program- 2 In a linear programming problem we want to find the feasible point that maximizes ming problem (or minimizes) the objective function. By a solution to a linear programming problem we mean a feasible point (x, y), together with the value of the objective function at that point, which maximizes (or minimizes) the objective function. If none of the feasible points maximizes (or mini- mizes) the objective function, or if there are no feasible points, then the linear pro- gramming problem has no solution. EXAMPLE 1 Solving a Linear Programming Problem Minimize the quantity z x 2y subject to the constraints x y 1 x 0 y 0 A Geometric Approach to Linear Programming Problems 173 SOLUTION The objective function to be minimized is z x 2y. The constraints are the linear inequalities x y 1 x 0 y 0 FIGURE 15 The shaded portion of Figure 15 illustrates the set of feasible points. To see if there is a smallest z, we graph z x 2y for some choice of z, say, z 3. y See Figure 16. By moving the line x 2y 3 parallel to itself, we can observe what happens for different values of z. Since we want a minimum value for z, we try to move z x 2y down as far as possible while keeping some part of the line within the set of (0, 1) (1, 0) x feasible points. The "best" solution is obtained when the line just touches a corner x+y = 1 point of the set of feasible points. If you refer to Figure 16, you will see that the best solution is x 1, y 0, which yields z 1. There is no other feasible point for which z is smaller. FIGURE 16 y x + 2y = z = 3 x + 2y = z = 2 x + 2y = z = 1 x (1, 0) z=3 z=2 z=1 x+y = 1 ◗ The next example illustrates a linear programming problem that has no solution. EXAMPLE 2 A Linear Programming Problem without a Solution Maximize the quantity z x 2y subject to the constraints x y 1 x 0 y 0 SOLUTION First, we graph the constraints. The shaded portion of Figure 17 illustrates the set of feasible points. The graphs of the objective function z x 2y for z 2, z 8, and z 12 are also shown in Figure 17. Observe that we continue to get larger values for z by moving the graph of the objective function upward. But there is no feasible point that will make z largest. No matter how large a value is assigned to z, there is a feasible point that 174 Chapter 3 Linear Programming: Geometric Approach FIGURE 17 will give a larger value. Since there is no feasible point that makes z largest, we conclude y that this linear programming problem has no solution. ◗ 6 Examples 1 and 2 demonstrate that sometimes a linear programming problem has 5 a solution and sometimes it does not. The next result gives conditions on the set of 4 x + 2y = z = 12 feasible points that determine when a solution to a linear programming problem x+y = 1 3 exists. 2 x + 2y = z = 8 (0, 1) 1 x L Existence of a Solution (1, 0) x + 2y = z = 2 Consider a linear programming problem with the set R of feasible points and x+y = 1 objective function z Ax By. 1. If R is bounded, then z has both a maximum and a minimum value on R. 2. If R is unbounded and A 0, B 0, and the constraints include x 0 and y 0, then z has a minimum value on R but not a maximum value (see Example 2). 3. If R is the empty set, then the linear programming problem has no solution and z has neither a maximum nor a minimum value. In Example 1 we found that the feasible point that minimizes z occurs at a corner point. This is not an unusual situation. If there are feasible points minimizing (or max- imizing) the objective function, at least one will be at a corner point of the set of feasi- ble points. L Fundamental Theorem of Linear Programming If a linear programming problem has a solution, it is located at a corner point of the set of feasible points; if a linear programming problem has multiple solutions, at least one of them is located at a corner point of the set of feasible points. In either case the corresponding value of the objective function is unique. The result just stated indicates that it is possible for a feasible point that is not a corner point to minimize (or maximize) the objective function. For example, if the slope of the objective function is the same as the slope of one of the boundaries of the set of feasible points and if the two adjacent corner points are solutions, then so are all the points on the line segment joining them. The following example illustrates this situation. EXAMPLE 3 A Linear Programming Problem with Multiple Solutions Minimize the quantity z x 2y subject to the constraints A Geometric Approach to Linear Programming Problems 175 x y 1 2x 4y 3 x 0 y 0 SOLUTION Again we first graph the constraints. The shaded portion of Figure 18 illustrates the set of feasible points. FIGURE 18 y 2 (0, 1) (1 , 1) 2 2 x 1 2 z = x + 2y (3 , 0)x + y = 1 2 z = x + 2y = 3 2 (2x + 4y = 3) If we graph the objective equation z x 2y for some choice of z and move it down, we find that a minimum is reached when z 3. In fact, any point on the line 2 2x 4y 3 between the adjacent corner points (2, 1) and (3, 0) and including these cor- 1 2 2 ner points will minimize the objective function. Of course, the reason any feasible point on 2x 4y 3 minimizes the objective equation z x 2y is that these two lines each have slope 1. This linear programming problem has infinitely many solutions. 2 ◗ NOW WORK PROBLEM 1. Since the objective function attains its maximum or minimum value at the corner points of the set of feasible points, we can outline a procedure for solving a linear pro- gramming problem provided that it has a solution. L Steps for Solving a Linear Programming Problem If a linear programming problem has a solution, follow these steps to find it: STEP 1 Write an expression for the quantity that is to be maximized or minimized (the objective function). STEP 2 Determine all the constraints and graph the set of feasible points. STEP 3 List the corner points of the set of feasible points. STEP 4 Determine the value of the objective function at each corner point. STEP 5 Select the maximum or minimum value of the objective function. Let's look at some examples. 176 Chapter 3 Linear Programming: Geometric Approach EXAMPLE 4 Solving a Linear Programming Problem Maximize and minimize the objective function z x 5y subject to the constraints x 4y 12 (1) x 8 (2) x y 2 (3) x 0 (4) y 0 (5) SOLUTION The objective function is z x 5y and the constraints consist of a system of five lin- ear inequalities. We proceed to graph the system of five linear inequalities. The shaded portion of Figure 19 illustrates the graph, the set of feasible points. Since this set is bounded, we know a solution to the linear programming problem exists. Notice in Figure 19 that we have labeled each line from the system of linear inequalities. We have also labeled the corner points. FIGURE 19 x=0 x=8 y 4 (0, 3) (0, 2) (8, 1) x y=0 (2, 0) 4 (8, 0) 12 x + 4y = 12 x+y = 2 The corner points of the set of feasible points are (0, 3) (8, 1) (8, 0) (2, 0) (0, 2) To find the maximum and minimum value of the objective function z x 5y, we construct Table 1: TABLE 1 Corner Point Value of Objective Function (x, y) z x 5y (0, 3) z 0 5(3) 15 (8, 1) z 8 5(1) 13 (8, 0) z 8 5(0) 8 (2, 0) z 2 5(0) 2 (0, 2) z 0 5(2) 10 The maximum value of z is 15, and it occurs at the point (0, 3). The minimum value of z is 2, and it occurs at the point (2, 0). ◗ NOW WORK PROBLEMS 17 AND 29. Now let's solve the problem of the cashews and peanuts that we discussed at the start of this section. A Geometric Approach to Linear Programming Problems 177 EXAMPLE 5 Maximizing Profit Maximize P 0.25x 0.45y subject to the constraints x 2y 300 (1) 3x 2y 480 (2) x 0 (3) y 0 (4) SOLUTION Before applying the method of this chapter to solve this problem, let's discuss a solu- tion that might be suggested by intuition. Namely, since the profit is higher for the high-grade mixture, you might think that Nutt's Nuts should prepare as many packages of the high-grade mixture as possible. If this were done, then there would be a total of 150 packages (8 ounces divides into 75 pounds of cashews exactly 150 times) and the total profit would be 150(0.45) $67.50 As we shall see, this is not the best solution to the problem. To obtain the maximum profit, we use linear programming. We reproduce here in Figure 20 the graph of the set of feasible points we obtained earlier (see Figure 14, page 168) FIGURE 20 Notice that this set is bounded. The corner points of the set of feasible points are (0, 0) (0, 150) (160, 0) (90, 105) It remains only to evaluate the objective function at each corner point: see Table 2. TABLE 2 Corner Point Value of Objective Function (x, y) P ($0.25)x ($0.45)y (0, 0) P (0.25)(0) (0.45)(0) 0 (0, 150) P (0.25)(0) (0.45)(150) $67.50 (160, 0) P (0.25)(160) (0.45)(0) $40.00 (90, 105) P (0.25)(90) (0.45)(105) $69.75 178 Chapter 3 Linear Programming: Geometric Approach A maximum profit is obtained if 90 packages of low-grade mixture and 105 packages of high-grade mixture are made. The maximum profit obtainable under the conditions described is $69.75. ◗ NOW WORK PROBLEM 49. EXERCISE 3.2 Answers to Odd-Numbered Problems Begin on Page AN-00. In Problems 1– 10, the given figure illustrates the graph of the set of feasible points of a linear programming problem. Find the maximum and minimum values of each objective function. y 1. z 2x 3y 2. z 3x 2y (7, 8) 8 (2, 7) 3. z x y 4. z 3x 3y 6 5. z x 6y 6. z 6x y 4 7. z 3x 4y 8. z 4x 3y 2 (2, 2) (8, 1) 9. z 10x y 10. z x 10y x 2 4 6 8 10 In Problems 11–16, list the corner points for each collection of constraints of a linear programming problem. 11. x 13 12. x 8 13. y 10 4x 3y 12 2x 3y 6 x y 15 x 0 x 0 x 0 y 0 y 0 y 0 14. y 8 15. x 10 16. x 9 2x y 10 y 8 y 12 x 0 4x 3y 12 2x 3y 24 y 0 x 0 x 0 y 0 y 0 In Problems 17–24, maximize (if possible) the quantity z 5x 7y subject to the given constraints. 17. x y 2 18. 2x 3y 6 19. x y 2 20. x y 2 y 1 x 2 2x 3y 6 2x 3y 12 x 0 x 0 x 0 3x 2y 12 y 0 y 0 y 0 x 0 y 0 21. x y 2 22. x y 2 23. x y 10 24. x y 8 x y 8 x y 8 x 6 y 2 2x y 10 x 2y 1 x 0 x 0 x 0 x 2y 10 y 0 y 0 y 0 x 0 y 0 Applications 179 In Problems 25–32, minimize (if possible) the quantity z 2x 3y subject to the given constraints. 25. x y 2 26. 3x y 3 27. x y 2 28. x y 8 y x y x x 3y 12 2x 3y 6 x 0 x 0 3x y 12 x y 2 y 0 y 0 x 0 x 0 y 0 y 0 29. x y 2 30. 2y x 31. x 2y 1 32. 2x y 2 x y 10 x 2y 10 x 2y 10 x y 6 2x 3y 6 x 2y 4 y 2x 2x y x 0 x 0 x y 8 x 0 y 0 y 0 x 0 y 0 y 0 In Problems 33–40, find the maximum and minimum values (if possible) of the given objective function subject to the constraints x y 10 2x y 10 x 2y 10 x 0 y 0 33. z x y 34. z 2x 3y 35. z 5x 2y 36. z x 2y 37. z 3x 4y 38. z 3x 6y 39. z 10x y 40. z x 10y 41. Find the maximum and minimum values of z 18x 30y 46. Maximize z 10x 10y subject to the constraints subject to the constraints 3x 3y 9, x 4y 12, and 0 x 15, 0 y 10, 6x y 6, and 3x y 7, 4x y 12, where x 0 and y 0. where x 0 and y 0. 42. Find the maximum and minimum values of z 20x 16y subject to the constraints 4x 3y 12, 2x 4y 16, 47. Maximize z 12x 24y subject to the constraints and 6x y 18, where x 0 and y 0. 0 x 15, 0 y 10, 3x 3y 9, and 3x 2y 14, where x 0 and y 0. 43. Find the maximum and minimum values of z 7x 6y subject to the constraints 2x 3y 6, 3x 4y 8, and 48. Maximize z 20x 10y subject to the constraints 5x y 15, where x 0 and y 0. 0 x 15, 0 y 10, 4x 3y 12, and 3x y 7, 44. Find the maximum and minimum values of z 6x 3y where x 0 and y 0. subject to the constraints 2x 2y 4, x 5y 10, and 3x 3y 6, where x 0 and y 0. 49. In Example 5, if the profit on the low-grade mixture is $0.30 45. Maximize z 20x 30y subject to the constraints per package and the profit on the high-grade mixture is 0 x 15, 0 y 10, 5x 3y 15, and 3x 3y 21, $0.40 per package, how many packages of each mixture where x 0 and y 0. should be made for a maximum profit? 3.3 Applications OBJECTIVES 1 Solve applied problems Solve applied problems 1 In this section, several situations that lead to linear programming problems are presented. 180 Chapter 3 Linear Programming: Geometric Approach EXAMPLE 1 Maximizing Profit Mike's Famous Toy Trucks manufactures two kinds of toy trucks — a standard model and a deluxe model. In the manufacturing process each standard model requires 2 hours of grinding and 2 hours of finishing, and each deluxe model needs 2 hours of grinding and 4 hours of finishing. The company has two grinders and three finishers, each of whom works at most 40 hours per week. Each standard model toy truck brings a profit of $3 and each deluxe model a profit of $4. Assuming that every truck made will be sold, how many of each should be made to maximize profits? SOLUTION First, we name the variables: x Number of standard models made y Number of deluxe models made The quantity to be maximized is the profit, which we denote by P: P $3x $4y This is the objective function. To manufacture one standard model requires 2 grinding hours and to make one deluxe model requires 2 grinding hours. The number of grind- ing hours needed to manufacture x standard and y deluxe models is 2x 2y But the total amount of grinding time available is only 80 hours per week. This means we have the constraint 2x 2y 80 Grinding time constraint Similarly, for the finishing time we have the constraint 2x 4y 120 Finishing time constraint By simplifying each of these constraints and adding the nonnegativity constraints x 0 and y 0, we may list all the constraints for this problem: x y 40 (1) x 2y 60 (2) x 0 (3) y 0 (4) Figure 21 illustrates the set of feasible points, which is bounded. FIGURE 21 x=0 y (0, 40) (0, 30) (20, 20) 20 10 (60, 0) x y=0 (0, 0) 10 20 30 50 70 80 (40, 0) x + y = 40 x + 2y = 60 The corner points of the set of feasible points are (0, 0) (0, 30) (40, 0) (20, 20) Applications 181 Table 3 lists the corresponding values of the objective equation: TABLE 3 Corner Point Value of Objective Function (x, y) P $3x $4y (0, 0) P 0 (0, 30) P $120 (40, 0) P $120 (20, 20) P 3(20) 4(20) $140 A maximum profit is obtained if 20 standard trucks and 20 deluxe trucks are manufac- tured. The maximum profit is $140. ◗ NOW WORK PROBLEM 1. EXAMPLE 2 Financial Planning A retired couple have up to $30,000 to invest in fixed-income securities. Their broker recommends investing in two bonds: one a AAA bond yielding 8%; the other a B bond paying 12%. After some consideration, the couple decide to invest at most $12,000 in the B -rated bond and at least $6000 in the AAA bond. They also want the amount invested in the AAA bond to exceed or equal the amount invested in the B bond. What should the broker recommend if the couple (quite naturally) want to max- imize the return on their investment? SOLUTION First, we name the variables: x Amount invested in the AAA bond y Amount invested in the B bond The quantity to be maximized, the couple's return on investment, which we denote by P, is P 0.08x 0.12y This is the objective function. The conditions specified by the problem are Up to $30,000 available to invest x y 30,000 Invest at most $12,000 in the B bond y 12,000 Invest at least $6000 in the AAA bond x 6000 Amount in the AAA bond must exceed x y or equal the amount in the B bond In addition, we must have the conditions x 0 and y 0. The total list of con- straints is x y 30,000 (1) y 12,000 (2) x 6000 (3) x y (4) x 0 (5) y 0 (6) 182 Chapter 3 Linear Programming: Geometric Approach Figure 22 illustrates the set of feasible points, which is bounded. The corner points of the set of feasible points are (6000, 0) (6000, 6000) (12000, 12000) (18000, 12000) (30000, 0) FIGURE 22 x=0 y x = 6,000 30,000 x=y 24,000 18,000 (18,000, 12,000) y = 12,000 12,000 (12,000, 12,000) 6,000 (6,000, 6,000) (30,000, 0) x y=0 6,000 12,000 18,000 24,000 30,000 36,000 42,000 x + y = 30,000 The corresponding return on investment at each corner point is P 0.08(6000) 0.12(0) $480 P 0.08(6000) 0.12(6000) 480 720 $1200 P 0.08(12,000) 0.12(12,000) 960 1440 $2400 P 0.08(18,000) 0.12(12,000) 1440 1440 $2880 P 0.08(30,000) 0.12(0) $2400 The maximum return on investment is $2880, obtained by placing $18,000 in the AAA bond and $12,000 in the B bond. ◗ NOW WORK PROBLEM 3. EXAMPLE 3 Manufacturing Vitamin Pills — Maximizing Profit A pharmaceutical company makes two types of vitamins at its New Jersey plant — a high-potency, antioxidant vitamin and a vitamin enriched with added calcium. Each high-potency vitamin contains, among other things, 500 mg of vitamin C and 40 mg of calcium and generates a profit of $0.10 per tablet. A calcium-enriched vitamin tablet contains 100 mg of vitamin C and 400 mg of calcium and generates a profit of $0.05 per tablet. Each day the company has available 235 kg of vitamin C and 156 kg of calcium for use. Assuming all vitamins made are sold, how many of each type of vitamin should be manufactured to maximize profit? Source: Centrum Vitamin Supplements. Applications 183 SOLUTION First we name the variables: x Number (in thousands) of high-potency vitamins to be produced y Number (in thousands) of calcium-enriched vitamins to be produced We want to maximize the profit, P, which is given by: P 0.10x 0.05y x and y in thousands Since 1 kg 1,000,000 mg, the constraints, in mg, take the form 500x 100y 235,000 vitamin C constraint (in thousands of mg) 40x 400y 156,000 calcium constraint (in thousands of mg) x 0 non-negativity constraints (in thousands) y 0 Figure 23 illustrates the set of feasible points, which is bounded. The corner points of the set of feasible points are (0, 0) (0, 390) (400, 350) (470, 0) FIGURE 23 x=0 y (0, 2,350) (0, 390) (400, 350) 40x + 400y = 156,000 x y=0 (0, 0) (470, 0) (3,900, 0) 500x + 100y = 235,000 184 Chapter 3 Linear Programming: Geometric Approach Since x and y are in thousands, the profit corresponding to each corner point is P 0.10(0) 0.05(0) 0 $0 P 0.10(0) 0.05(390) 19.5 thousand $19,500 P 0.10(400) 0.05(350) 57.5 thousand $57,500 P 0.10(470) 0.05(0) 47 thousand $47,000 The maximum profit is $57,500, obtained when 400,000 high-potency vitamins are produced (x 400 thousand units) and 350,000 calcium-enriched vitamins are pro- duced (y 350 thousand units). ◗ NOW WORK PROBLEM 5. EXERCISE 3.3 Answers to Odd-Numbered Problems Begin on Page AN-00. 1. Optimal Land Use A farmer has 70 acres of land available type B bond yielding a 15% return on the amount invested. on which to grow some soybeans and some corn. The cost of She wants to invest at least as much in the type A bond as in cultivation per acre, the workdays needed per acre, and the the type B bond. She will also invest at least $5000 in the profit per acre are indicated in the table: type A bond and no more than $8000 in the type B bond. How much should she invest in each type of bond to maxi- mize her return? Total Soybeans Corn Available 4. Diet A diet is to contain at least 400 units of vitamins, 500 units of minerals, and 1400 calories. Two foods are Cultivation Cost available: F1, which costs $0.05 per unit, and F2, which per Acre $60 $30 $1800 costs $0.03 per unit. A unit of food F1 contains 2 units Days of Work of vitamins, 1 unit of minerals, and 4 calories; a unit of per Acre 3 days 4 days 120 days food F2 contains 1 unit of vitamins, 2 units of minerals, and 4 calories. Find the minimum cost for a diet that consists of Profit per Acre $300 $150 a mixture of these two foods and also meets the minimal nutrition requirements. As indicated in the last column, the acreage to be cultivated 5. Manufacturing Vitamin Pills After changing suppliers, the is limited by the amount of money available for cultivation pharmaceutical company in Example 3 has 300 kg of costs and by the number of working days that can be put vitamin C and 220 kg of calcium available each day for the into this part of the business. Find the number of acres of manufacture of the high-potency, antioxidant vitamins and each crop that should be planted in order to maximize the vitamins enriched with added calcium. If each high-potency profit. vitamin contains, among other things, 500 mg of vitamin C and 40 mg of calcium and generates a profit of $0.10 per 2. Manufacturing A factory manufactures two products, each tablet, and each calcium-enriched vitamin tablet contains requiring the use of three machines. The first machine can 100 mg of vitamin C and 400 mg of calcium and generates a be used at most 70 hours; the second machine at most profit of $0.05 per tablet, how many of each type of vitamin 40 hours; and the third machine at most 90 hours. The first should be manufactured to maximize profit? product requires 2 hours on machine 1, 1 hour on machine 2, and 1 hour on machine 3; the second product requires Source: Centrum Vitamin Supplements. 1 hour each on machines 1 and 2, and 3 hours on machine 3. 6. Investment Strategy A financial consultant wishes to If the profit is $40 per unit for the first product and $60 per invest up to a total of $30,000 in two types of securities, one unit for the second product, how many units of each prod- that yields 10% per year and another that yields 8% per year. uct should be manufactured to maximize profit? Furthermore, she believes that the amount invested in the 3. Investment Strategy An investment broker wants to invest first security should be at most one-third of the amount up to $20,000. She can purchase a type A bond yielding a invested in the second security. What investment program 10% return on the amount invested, and she can purchase a should the consultant pursue in order to maximize income? Applications 185 7. Scheduling Blink Appliances has a sale on microwaves and tains 60 calories, 13 g of carbohydrates, and 100% of the stoves. Each microwave requires 2 hours to unpack and set up, recommended daily allowance of vitamin C. Determine and each stove requires 1 hour. The storeroom space is limited how many servings of each of the above foods would be to 50 items. The budget of the store allows only 80 hours of needed to provide a child at least 160 calories, 40 g of car- employee time for unpacking and setup. Microwaves sell for bohydrates, and 70% of the recommended daily allowance $300 each, and stoves sell for $200 each. How many of each of vitamin C at minimum cost. Fractions of servings are should the store order to maximize revenue? permitted. 8. Transportation An appliance company has a warehouse and Source: Gerber website and Safeway Stores, Inc. two terminals. To minimize shipping costs, the manager must decide how many appliances should be shipped to each 12. Production Scheduling A company produces two types of terminal. There is a total supply of 1200 units in the ware- steel. Type 1 requires 2 hours of melting, 4 hours of cutting, house and a demand for 400 units in terminal A and 500 and 10 hours of rolling per ton. Type 2 requires 5 hours of units in terminal B. It costs $12 to ship each unit to terminal melting, 1 hour of cutting, and 5 hours of rolling per ton. A and $16 to ship to terminal B. How many units should be Forty hours are available for melting, 20 for cutting, and 60 shipped to each terminal in order to minimize cost? for rolling. Each ton of Type 1 produces $240 profit, and each ton of Type 2 yields $80 profit. Find the maximum 9. Pension Fund Investments A pension fund has decided to profit and the production schedule that will produce this invest $45,000 in the two high-yield stocks listed in the table profit. below. 13. Website Ads Nielson's Net Ratings for the month of Price per Share December 2002 indicated that in the United States, AOL had 2/21/03 Yield a unique audience of about 76.4 million people and Yahoo! Duke Energy Corp. $14 8% had a unique audience of about 66.2 million people. An advertising company wants to purchase website ads to pro- Eastman Kodak $30 6% mote a new product. Suppose that the monthly cost of an ad on the AOL website is $1200 and the monthly cost of an ad This pension fund has decided to invest at least 25% of the on the Yahoo! website is $1100. Determine how many $45,000 in each of the two stocks. Further, it has been decid- months an ad should run on each website to maximize the ed that at most 63% of the $45,000 can be invested in either number of people who would be exposed to it. Assume that, one of the stocks. How many shares of each stock should be for future months, the monthly website audience remains purchased in order to maximize the annual yield, while the same as given for December 2002. Also assume that the meeting the stipulated requirements? What is the annual advertising budget is $35,000 and that it has been decided to yield in dollars for the optimal investment plan? advertise on Yahoo! for at least ten months. Source: Yahoo! Finance. Prices and yields have been rounded. Source: Nielson's Net Ratings. 10. Pollution Control A chemical plant produces two items A 14. Diet Danny's Chicken Farm is a producer of frying chick- and B. For each item A produced, 2 cubic feet of carbon ens. In order to produce the best fryers possible, the regular monoxide and 6 cubic feet of sulfur dioxide are emitted chicken feed is supplemented by four vitamins. The mini- into the atmosphere; to produce item B, 4 cubic feet of car- mum amount of each vitamin required per 100 ounces of bon monoxide and 3 cubic feet of sulfur dioxide are emit- feed is: vitamin 1, 50 units; vitamin 2, 100 units; vitamin 3, ted into the atmosphere. Government pollution standards 60 units; vitamin 4, 180 units. Two supplements are avail- permit the manufacturer to emit a maximum of 3000 cubic able: supplement I costs $0.03 per ounce and contains 5 feet of carbon monoxide and 5400 cubic feet of sulfur units of vitamin 1 per ounce, 25 units of vitamin 2 per dioxide per week. The manufacturer can sell all of the items ounce, 10 units of vitamin 3 per ounce, and 35 units of vita- that it produces and makes a profit of $1.50 per unit for min 4 per ounce. Supplement II costs $0.04 per ounce and item A and $1.00 per unit for item B. Determine the num- contains 25 units of vitamin 1 per ounce, 10 units of vitamin ber of units of each item to be produced each week to max- 2 per ounce, 10 units of vitamin 3 per ounce, and 20 units of imize profit without exceeding government standards. vitamin 4 per ounce. How much of each supplement should Danny buy to add to each 100 ounces of feed in order to 11. Baby Food Servings Gerber Banana Plum Granola minimize his cost, but still have the desired vitamin amounts costs $0.89 per 5.5-oz serving; each serving contains 140 present? calories, 31 g of carbohydrates, and 0% of the recommend- ed daily allowance of vitamin C. Gerber Mixed Fruit 15. Maximizing Income J. B. Rug Manufacturers has available Carrot Juice costs $0.79 per 4-oz serving; each serving con- 1200 square yards of wool and 1000 square yards of nylon for 186 Chapter 3 Linear Programming: Geometric Approach the manufacture of two grades of carpeting: high-grade, square yard. How many rolls of each type of carpet should be which sells for $500 per roll, and low-grade, which sells for manufactured to maximize income? [Hint: Income $300 per roll. Twenty square yards of wool and 40 square Revenue from sale (Production cost for material labor)] yards of nylon are used in a roll of high-grade carpet, and 40 square yards of nylon are used in a roll of low-grade carpet. 16. The rug manufacturer in Problem 15 finds that maximum Forty work-hours are required to manufacture each roll of the income occurs when no high-grade carpet is produced. If the high-grade carpet, and 20 work-hours are required for each price of the low-grade carpet is kept at $300 per roll, in what roll of the low-grade carpet, at an average cost of $6 per work- price range should the high-grade carpet be sold so that hour. A maximum of 800 work-hours are available. The cost income is maximized by selling some rolls of each type of of wool is $5 per square yard and the cost of nylon is $2 per carpet? Assume all other data remain the same. Chapter 3 Review OBJECTIVES Section You should be able to Review Exercises 3.1 1 Graph linear inequalities 1–4 2 Graph systems of linear inequalities 9–14 3.2 1 Identify a linear programming problem 33 – 38 2 Solve a linear programming problem 15 – 32 3.3 1 Solve applied problems 33 – 38 THINGS TO KNOW Graphs of Inequalities (p. 160) Linear Programming (p. 172) The graph of a strict inequality is represented by a dashed Maximize (or minimize) a linear objective function, line and the half-plane satisfying the inequality. z Ax By, subject to certain conditions, or constraints, The graph of a nonstrict inequality is represented by a sol- expressible as linear inequalities in x and y. A feasible point id line and the half-plane satisfying the inequality. (x, y) is a point that satisfies the constraints of a linear pro- gramming problem. Graphs of Systems of Linear Inequalities (p. 162) The graph of a system of linear inequalities is the set of all Solution to a Linear Programming Problem (p. 172) points that satisfy each inequality in the system. A solution to a linear programming problem is a feasible point that maximizes (or minimizes) the objective function Bounded (p. 166) together with the value of the objective function at that point. The graph is called bounded if some circle can be drawn around it. Location of Solution (p. 174) The graph is called unbounded if it extends infinitely far If a linear programming problem has a solution, it is locat- in at least one direction. ed at a corner point of the graph of the feasible points. If a linear programming problem has multiple solutions, Corner Point (p. 166) at least one of them is located at a corner point of the A corner point is the intersection of two line segments graph of the feasible points. that form the boundary of the graph of a system of linear In either case, the corresponding value of the objective inequalities. function is unique. Chapter Review 187 TRUE–FALSE ITEMS Answers are on page AN-00. T F 1. The graph of a system of linear inequalities may T F 4. In a linear programming problem, there may be be bounded or unbounded. more than one point that maximizes or mini- T F 2. The graph of the set of constraints of a linear mizes the objective function. programming problem, under certain conditions, T F 5. Some linear programming problems will have no could have a circle for a boundary. solution. T F 3. The objective function of a linear programming T F 6. If a linear programming problem has a solution, problem is always a linear equation involving the it is located at the center of the set of feasible variables. points. FILL IN THE BLANKS Answers are on page AN-00. 1. The graph of a linear inequality in two variables is 4. A linear programming problem will always have a solution called a . if the set of feasible points is . 2. In a linear programming problem the quantity to be maxi- 5. If a linear programming problem has a solution, it is mized or minimized is referred to as the located at a of the set of feasible points. function. 3. The points that obey the collection of constraints of a linear programming problem are called points. REVIEW EXERCISES Answers to odd-numbered problems begin on page AN-00. Blue problem numbers indicate the author's suggestions for use in a practice test. In Problems 1–4, graph each linear inequality. 1. x 3y 0 2. 4x y 0 3. 5x y 10 4. 2x 3y 6 5. Without graphing, determine which of the points 6. Without graphing, determine which of the points P1 (4, 3), P2 (2, 6), P3 (8, 3) are part of the P1 (8, 6), P2 (2, 5), P3 (4, 1) are part of the graph of the following system: graph of the following system: x 2y 8 5x y 2 2x y 4 x 4y 2 In Problems 7–8, determine which region — a, b, c, or d — represents the graph of the given system of linear inequalities. 6x 4y 12 y 6x 5y 5 y 7. b 8. b 3x 2y 18 6x 6y 60 8 8 c 4 c 4 a a x x –8 –4 4 8 –8 –4 4 8 –4 –8 d –8 d In Problems 9–14, graph each system of linear inequalities. Locate the corner points and tell whether the graph is bounded or unbounded. 3x 2y 12 x y 8 x 2y 4 x y 4 2x y 4 3x y 6 9. 10. 11. x 0 x 0 x 0 y 0 y 0 y 0 188 Chapter 3 Linear Programming: Geometric Approach 2x y 4 3x 2y 6 x y 8 3x 2y 6 3x 2y 12 x 2y 10 12. x 0 13. x 2y 8 14. y 8 y 0 x 0 x 0 y 0 y 0 In Problems 15 – 22, use the constraints below to solve each linear programming problem. x 2y 40 2x y 40 x y 10 x 0 y 0 15. Maximize z x y 16. Maximize z 2x 3y 17. Minimize z 5x 2y 18. Minimize z 3x 2y 19. Maximize z 2x y 20. Maximize z x 2y 21. Minimize z 2x 5y 22. Minimize z x y In Problems 23 – 26, maximize and minimize (if possible) the quantity z 15x 20y subject to the given constraints. 23. x 5 24. x 2y 4 25. 2x 3y 22 26. x 2y 20 y 8 3x 2y 6 x 5 x 10y 36 3x 4y 12 x 0 y 6 5x 2y 36 x 0 y 0 x 0 x 0 y 0 y 0 y 0 In Problems 27 – 32, solve each linear programming problem. 27. Maximize 28. Maximize 29. Maximize z 2x 3y z 4x y z x 2y subject to the constraints subject to the constraints subject to the constraints x 9 x 7 x y 1 y 8 y 8 y 2x x y 3 x y 2 x 8 x 0 x 0 y 8 y 0 y 0 x 0 y 0 30. Maximize 31. Minimize 32. Minimize z 3x 4y z 3x 2y z 2x 5y subject to the constraints subject to the constraints subject to the constraints x 2y 2 x 2y 8 x 10 3x 2y 12 3x y 6 y 12 y 5 x 8 2x y 10 x 0 x 0 x y 0 y 0 y 0 x 0 y 0 Chapter Review 189 33. Maximizing Profit A ski manufacturer makes two types contains 60 calories, 15 g of carbohydrates, and 100% of skis: downhill and cross-country. Using the informa- of the recommended daily allowance of vitamin C. tion given in the table below, how many of each type of Determine how many servings of each of the above ski should be made for a maximum profit to be achieved? foods would be needed to provide a child with at least What is the maximum profit? 130 calories, 30 g of carbohydrates, and 60% of the rec- ommended daily allowance of vitamin C at minimum cost. Fractions of servings are permitted. Source: Gerber website and Safeway Stores, Inc. Maximum Cross- Time 38. Maximizing Profit A company makes two explosives: Downhill Country Available type I and type II. Due to storage problems, a Manufacturing maximum of 100 pounds of type I and 150 pounds of type II can be mixed and packaged each week. One Time per Ski 2 hours 1 hour 40 hours pound of type I takes 60 hours to mix and 70 hours to Finishing Time package; 1 pound of type II takes 40 hours to mix and per Ski 1 hour 1 hour 32 hours 40 hours to package. The mixing department has at most 7200 work-hours available each week, and pack- Profit per Ski $70 $50 aging has at most 7800 work-hours available. If the profit for 1 pound of type I is $60 and for 1 pound of type II is $40, what is the maximum profit possible each week? 34. Rework Problem 33 if the manufacturing unit has a maxi- mum of 48 hours available. 39. A pharmaceutical company makes two types of vitamins at its New Jersey plant — a high-potency, antioxidant vitamin 35. Nutrition Katy needs at least 60 units of carbohydrates, and a vitamin enriched with added calcium. Each high- 45 units of protein, and 30 units of fat each month. potency vitamin contains, among other things, 500 mg From each pound of food A, she receives 5 units of of vitamin C, 40 mg of calcium, and 100 mg of magne- carbohydrates, 3 of protein, and 4 of fat. Food B con- sium and generates a profit of $0.10 per tablet. A calci- tains 2 units of carbohydrates, 2 units of protein, and um-enriched vitamin tablet contains 100 mg of vitamin 1 unit of fat per pound. If food A costs $1.30 per pound C, 400 mg of calcium, and 40 mg of magnesium and and food B costs $0.80 per pound, how many pounds of generates a profit of $0.05 per tablet. Each day the com- each food should Katy buy each month to keep costs at pany has available 300 kg of vitamin C, 122 kg of calci- a minimum? um, and 65 kg of magnesium for use in the production of the vitamins. How many of each type of vitamin should 36. Production Scheduling A company sells two types of be manufactured to maximize profit? shoes. The first uses 2 units of leather and 2 units of synthetic material and yields a profit of $8 per pair. The Source: Centrum Vitamin Supplements. second type requires 5 units of leather and 1 unit of syn- thetic material and gives a profit of $10 per pair. If there 40. Mixture A company makes two kinds of animal food, A are 40 units of leather and 16 units of synthetic material and B, which contain two food supplements. It takes available, how many pairs of each type of shoe should be 2 pounds of the first supplement and one pound of the manufactured to maximize profit? What is the maximum second to make a dozen cans of food A, and 4 pounds of profit? the first supplement and 5 pounds of the second to make a dozen cans of food B. On a certain day 80 pounds of the 37. Baby Food Servings Gerber Banana Oatmeal and Peach first supplement and 70 pounds of the second are avail- costs $0.79 per 4-oz serving; each serving contains 90 able. How many cans of food A and food B should be calories, 19 g of carbohydrates, and 45% of the recom- made to maximize company profits, if the profit on a mended daily allowance of vitamin C. Gerber Mixed dozen cans of food A is $3.00 and the profit on a dozen Fruit Juice costs $0.65 per 4-oz serving; each serving cans of food B is $10.00. 190 Chapter 3 Linear Programming: Geometric Approach Chapter 3 Project BALANCING NUTRIENTS In preparing a recipe you must decide what ingredients and least 1 of the amount of peanuts and the amount of peanuts 2 how much of each ingredient you will use. In these health- to be at least 1 of the amount of raisins. You want the 2 conscious days, you may also want to consider the amount of entire amount of trail mix you make to have fewer than 4000 certain nutrients in your recipe. You may even be interested calories, and you want to maximize the amount of carbohy- in minimizing some quantities (like calories or fat) or maxi- drates in the mix. mizing others (like carbohydrates or protein). Linear pro- gramming techniques can help to do this. 1. Let x be the number of cups of peanuts you will use, let y For example, consider making a very simple trail mix from be the number of cups of raisins you will use, and let c be dry-roasted, unsalted peanuts and seedless raisins. Table 1 lists the amount of carbohydrates in the mix. Find the objec- the amounts of various dietary quantitites for these ingredi- tive function. ents. The amounts are given per serving of the ingredient. 2. What constraints must be placed on the objective function? 3. Graph the set of feasible points for this problem. TABLE 1 NUTRIENTS IN PEANUTS AND RAISINS 4. Find the number of cups of peanuts and raisins that Peanuts Raisins maximize the amount of carbohydrates in the mix. Serving Serving 5. How many grams of carbohydrates are in a cup of the Nutrient Size 1 Cup Size 1 Cup final mix? How many calories? Calories (kcal) 850 440 6. Under all of the constraints given above, what recipe for Protein (g) 34.57 4.67 trail mix will maximize the amount of protein in the mix? How many grams of protein are in a cup of this mix? Fat(g) 72.50 .67 How many calories? Carbohydrates (g) 31.40 114.74 7. Suppose you decide to eat at least 3 cups of the trail mix. Source: USDA National Nutrient Database for Standard Keeping the constraints given above, what recipe for trail Reference. mix will minimize the amount of fat in the mix? 8. How many grams of carbohydrates are in this mix? Suppose that you want to make at most 6 cups of trail 9. How many grams of protein are in this mix? mix for a day hike. You don't want either ingredient to domi- nate the mixture, so you want the amount of raisins to be at 10. Which of the three trail mixes would you use? Why? Chapter Project 191 MATHEMATICAL QUESTIONS FROM PROFESSIONAL EXAMS Use the following information to do Problems 1–3: The Random Company manufactures two products, Zeta and Beta. Each product must pass through two processing operations. All materials are introduced at the start of process 1. There are no work-in-process inventories. Random may produce either one product exclusively or var- ious combinations of both products subject to the following constraints: Contribution Process Process Margin No. 1 No. 2 per Unit Hours required to produce one unit of Zeta 1 hour 1 hour $4.00 Beta 2 hours 3 hours 5.25 Total capacity in hours per day 1000 hours 1275 hours A shortage of technical labor has limited Beta production to 400 units per day. There are no constraints on the production of Zeta other than the hour constraints in the above schedule. Assume that all relationships between capacity and production are linear, and that all of the above data and relationships are deterministic rather than probabilistic. 1. CPA Exam Given the objective to maximize total contri- 6 hours of time on machine 1 and 12 hours of time on bution margin, what is the production constraint for machine 2. Product Y requires 4 hours of time on process 1? machine 1 and no time on machine 2. Both machines are available for 24 hours. Assuming that the objective (a) Zeta Beta 1000 function of the total contribution margin is $2X (b) Zeta 2 Beta 1000 $1Y, what product mix will produce the maximum (c) Zeta Beta 1000 profit? (d) Zeta 2 Beta 1000 (a) No units of product X and 6 units of product Y. 2. CPA Exam Given the objective to maximize total contri- (b) 1 unit of product X and 4 units of product Y. bution margin, what is the labor constraint for produc- (c) 2 units of product X and 3 units of product Y. tion of Beta? (d) 4 units of product X and no units of product Y. (a) Beta 400 (b) Beta 400 5. CPA Exam Quepea Company manufactures two prod- (c) Beta 425 (d) Beta 425 ucts, Q and P, in a small building with limited capacity. 3. CPA Exam What is the objective function of the data The selling price, cost data, and production time are presented? given below: (a) Zeta 2 Beta $9.25 (b) ($4.00)Zeta 3($5.25)Beta Total contribution margin Product Q Product P (c) ($4.00)Zeta ($5.25)Beta Total contribution margin Selling price per unit $20 $17 (d) 2($4.00)Zeta 3($5.25)Beta Total contribution Variable costs of producing margin and selling a unit $12 $13 4. CPA Exam Williamson Manufacturing intends to Hours to produce a unit 3 1 produce two products, X and Y. Product X requires 192 Chapter 3 Linear Programming: Geometric Approach Based on this information, the profit maximization number of units of each product that may be processed in objective function for a linear programming solution the two departments. may be stated as y (a) Maximize $20Q $17P. Units of product B 40 Polishing (b) Maximize $12Q $13P. constraint 30 (restriction) (c) Maximize $3Q $1P. Grinding (d) Maximize $8Q $4P. 20 constraint (restriction) 10 6. CPA Exam Patsy, Inc., manufactures two products, X and x Y. Each product must be processed in each of three depart- 0 10 20 30 40 ments: machining, assembling, and finishing. The hours Units of product A needed to produce one unit of product per department and the maximum possible hours per department follow: Considering the constraints (restrictions) on processing, which combination of products A and B maximizes the Production total contribution margin? Hours per Unit Maximum Capacity (a) 0 units of A and 20 units of B. Department X Y in Hours (b) 20 units of A and 10 units of B. (c) 30 units of A and 0 units of B. Machining 2 1 420 (d) 40 units of A and 0 units of B. Assembling 2 2 500 9. CPA Exam Johnson, Inc., manufactures product X and Finishing 2 3 600 product Y, which are processed as follows: Other restrictions follow: Machine A Machine B X 50 Y 50 Product X 6 hours 4 hours The objective function is to maximize profits where prof- Product Y 9 hours 5 hours it $4X $2Y. Given the objective and constraints, what is the most profitable number of units of X and Y, respectively, to manufacture? The contribution margin is $12 for product X and $7 for product Y. The available time daily for processing the two (a) 150 and 100 (b) 165 and 90 products is 120 hours for machine A and 80 hours for (c) 170 and 80 (d) 200 and 50 machine B. How would the restriction (constraint) for machine B be expressed? 7. CPA Exam Milford Company manufactures two models, medium and large. The contribution margin expected is (a) 4X 5Y (b) 4X 5Y 80 $12 for the medium model and $20 for the large model. (c) 6X 9Y 120 (d) 12X 7Y The medium model is processed 2 hours in the machin- 10. CPA Exam A small company makes only two products, ing department and 4 hours in the polishing department. with the following two production constraints represent- The large model is processed 3 hours in the machining ing two machines and their maximum availability: department and 6 hours in the polishing department. How would the formula for determining the maximiza- 2X 3Y 18 tion of total contribution margin be expressed? 2X Y 10 (a) 5X 10Y (b) 6X 9Y where X Units of the first product (c) 12X 20Y (d) 12X(2 4) 20Y(3 6) Y Units of the second product 8. CPA Exam Hale Company manufactures products A and If the profit equation is Z $4X $2Y, the maximum B, each of which requires two processes, polishing and possible profit is grinding. The contribution margin is $3 for product A and (a) $20 (b) $21 (c) $18 (d) $24 $4 for product B. The illustration shows the maximum (e) Some profit other than those given above Chapter Project 193 Questions 11– 13 are based on the Jarten Company, which manufactures and sells two prod- ucts. Demand for the two products has grown to such a level that Jarten can no longer meet the demand with its facilities. The company can work a total of 600,000 direct labor-hours annual- ly using three shifts. A total of 200,000 hours of machine time is available annually. The com- pany plans to use linear programming to determine a production schedule that will maximize its net return. The company spends $2,000,000 in advertising and promotion and incurs $1,000,000 for general and administrative costs. The unit sale price for model A is $27.50; model B sells for $75.00 each. The unit manufacturing requirements and unit cost data are as shown below. Overhead is assigned on a machine-hour (MH) basis. Model A Model B Raw material $3 $ 7 Direct labor 1 DLH @ $8 8 1.5 DLH @ $8 12 Variable overhead 0.5 MH @ $12 6 2.0 MH @ $12 24 Fixed overhead 0.5 MH @ $4 2 2.0 MH @ $4 8 $19 $ 51 11. CMA Exam The objective function that would maximize (c) 0.5A 2B 200,000 Jarten's net income is (d) (0.5 0.5)A (2 2)B 200,000 (e) (0.5 1) (1.5 2.00) (200,000 600,000) (a) 10.50A 32.00B (b) 8.50A 24.00B (c) 27.50A 75.00B (d) 19.00A 51.00B 14. CPA Exam Boaz Company manufactures two models, (e) 17.00A 43.00B medium (X) and large (Y). The contribution margin 12. CMA Exam The constraint function for the direct labor expected is $24 for the medium model and $40 for the is large model. The medium model is processed 2 hours in the machining department and 4 hours in the polishing (a) 1A 1.5B 200,000 (b) 8A 12B 600,000 department. The large model is processed 3 hours in (c) 8A 12B 200,000 (d) 1A 1.5B 4,800,000 the machining department and 6 hours in the polishing (e) 1A 1.5B 600,000 department. If total contribution margin is to be maxi- 13. CMA Exam The constraint function for the machine mized, using linear programming, how would the objec- capacity is tive function be expressed? (a) 6A 24B 200,000 (a) 24X(2 4) 40Y(3 6) (b) 24X 40Y (b) (1/0.5)A (1.5/2.0)B 800,000 (c) 6X 9Y (d) 5X 10
How to Be Happy as a self-help book is not unique in the advice it gives, however, this book is far more than that and from page 1, reading it is like a breath of fresh air. The subject matter in itself is intriguing, as happiness is not easy to define unlike, wealth, health or material success. Most people when asked what happiness is have no idea and it is a subject that has exercised the minds of philosophers over the ages. The answer is not all that clear but has a lot to do with having a balanced approach to life. The author reflects and considers just how people's emotional well-being is affected on a daily basis by outside factors such as work, relationships,… Read more This Edexcel A-level textbook for the Core 1 course is very clearly set out. There is a CD of exam questions which you can try and which are also fully worked out for you to check that you are doing things right (or where you have gone wrong).It is worth remember that the working out for each answer also goes towards the marks. Excellent book with clear working out - a must for all maths students.
MS Math 3.0 Rolls Out This Month By Michelle Rutledge 05/17/07 ##AUTHORSPLIT##<---> Microsoft recently announced the launch of Math 3.0, a math and science educational tool for students in grade levels 6-12, as well as entry-level college students. The software is designed for use at home, to assist students with math and science concepts and homework, or for visual examples in the classroom. "Microsoft Math provides a space for nurturing student learning in mathematics with dynamic visualizations. The program provides essential ingredients for classroom environments designed to challenge all students to engage in visual thinking," said Margaret L. Neiss, mathematics education professor at Oregon State University in a prepared statement. According to Microsoft, the software includes study material for six different math and science subjects. Features in the program include: Graphing calculator; Step-by-step math solutions; Formulas and equation library; Unit conversion tool; Triangle solver; and Handwriting support. Math 3.0 runs $19.95 for a single license. Volume licensing is available for
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Topology is a branch of mathematics packed with intriguing concepts, fascinating geometrical objects, and ingenious methods for studying them. The...show more authors have written this textbook to make this material accessible to undergraduate students without requiring extensive prerequisites in upper-level mathematics. The approach is to cultivate the intuitive ideas of continuity, convergence, and connectedness so students can quickly delve into knot theory, the topology of surfaces and three-dimensional manifolds, fixed points, and elementary homotopy theory. The fundamental concepts of point-set topology appear at the end of the book when students can see how this level of abstraction provides a sound logical basis for the geometrical ideas that have come before. This organization exposes students to the exciting geometrical ideas of topology now(!) rather than later.Students using this textbook should have some exposure to the geometry of objects in higher-dimensional Euclidean spaces together with an appreciation of precise mathematical definitions and proofs. Multivariable calculus, linear algebra, and one further proof-oriented mathematics course are suitable preparation. ...show less 2006 Hardcover New Book New and in stock. 5/11
More About This Textbook Overview This lively introduction to measure theory and Lebesgue integration is motivated by the historical questions that led to its development. The author stresses the original purpose of the definitions and theorems, highlighting the difficulties mathematicians encountered as these ideas were refined. The story begins with Riemann's definition of the integral, and then follows the efforts of those who wrestled with the difficulties inherent in it, until Lebesgue finally broke with Riemann's definition. With his new way of understanding integration, Lebesgue opened the door to fresh and productive approaches to the previously intractable problems of analysis. Editorial Reviews From the Publisher "Bressoud is an insightful writer, and he presents this material in an enchanting fashion. The writing is scholarly but inviting, rigorous but readable. There are heaps of exercises, and they are quite accessible. I know of no other source with such a wealth of information about the genesis of the modern integral concept. This book will be valuable for mathematicians, for scholars of mathematical history, and certainly for students." Steven G. Krantz, American Institute of Mathematics for The UMAP Journal "A new and noteworthy title from Cambridge University Press! An outstanding book meant to advance undergraduate and beginning graduate students in mathematics." B. Crstici, Mathematical Reviews "I find it difficult to think of a better introduction to this corner store of modern mathematics and highly recommend the book to a very broad readership of students and researchers alike." Paul Embrechts, ETH Zurich for the
their maths studies. 1 TYPES OF BEAMS A beam is a structure, which is loaded transversely (sideways). ... programmes for solving beamproblems. The Archon Engineering web site has many such programmes. WORKED EXAMPLE No.1 Teacher guide Solving Geometry Problems: Floodlights T-1 Solving Geometry Problems: Floodlights MATHEMATICAL GOALS This lesson unit is intended to help you assess how well students are able to identify and use Finite deformation beam models and triality theory in dynamical post-buckling analysis1 ... In statical frictional contact problems, if the beam is supported on a rigid obstacle, and the shape of the obstacle is described by a strictly concave /(x) ... A-level Maths Tutor [email protected] Statics : Rigid Bodies The Moment of a Force The ... Parallel forces acting on a beam When attempting problems concerning a balance points or fulcrum, remember that there is Solutions of Mechanical (or) Electrical problems involving discontinuous force functions (R.H.S function ) (or) Periodic functions other than and are obtained easily. The Laplace Transformation is a very powerful technique, that it replaces operations of calculus by operations of ... Solving Structural Dynamic Problems Using DCALC p.7-1 Chapter 7: Solving Structural Dynamic Problems Using DCALC By Karl Hanson, S ... The above frame shows a modeled representation of the beam in the previous problem. As before, we will prepare this analysis running the following order of DCALC ... BeamMaths years 0–2 Storybook Maths years 1–2 MiniMaths ... Big Book of Word Problems All years Maths all Week Each Big Book of Word Problems set includes a fully illustrated big book to share with the class, a teachers' book of notes and Maths has applications to many problems that are vital to human health and happiness. In this article we are going to describe how the mathematics of tomography has ... If a light beam is shone through several bottles, then this absorption adds up. BEAM Education (1999) Maths together, London: Walker Books, available from BEAM Education ... understanding maths for a wide range of magazines and book publishers. ... measuring and solving problems are all needed for simple tasks like measuring out washing Problem solving (Beam) Problem solving (nrich) You tube ... When I talk through problems, I understand numbers better. Pupil ... pupils with the opportunity to talk about their maths more frequently, and in Mental Maths & Problem Solving Rebeca Muñoz San Millán ... Listen to the problems and complete these sets according to the operations: Set 1 Set 4 Set 2 Set 3 Set 4 ... 2.-Add standard weights to the other pan until the beam is balance 2.3 The Loaded Beam ... endpoint problems and we will also exemplify the need for such generalisations to ... Calculus of Variation An Introduction To Isoperimetric Problems, 2013. http: // ... challenging Olympiad problems, several surprising applications, ... build a simple machine which applies a gradually increasing force to the centre of the beam while it is ... runs a maths competition for schools in the Cape Town area, and directs a nationwide Mathematical Talent Search, ... I can use my maths to solve real life problems I can solve problems and puzzles in my head by counting, adding, ... "Beam" materials Paper resources such as problem solving spinners and scaffolded frames ~ see Maths Subject Leader Use geometric reasoning to solve problems A ... 45° Ground Narrow Beam Ladder Ladder It is built on sloping ground. The beam, AB, is parallel to the ground. The support posts, AE and BD, are vertical. The support post, AE, makes an angle of 80º with the ground. radiographers sometimes use X-ray beam filters and need to be able to quantify the impact of filtration on the intensity and quality of the resulting images, as well as on patient exposure. Once the images are processed, radiographers' next e-mail: sandrab@maths.usyd.edu.au ABSTRACT ... example, has studied the problems students of physics and engineering have with math-ematics. Jackman et al ... ponential decay of the number of photons in a photon beam, a microbiology problem to the beam's curvature which is assumed to equal the second derivative of the out-of-plane displacement w with respect to x. However, ... In statics problems it is often more convenient to formulate the governing equations using Two distinct problems are considered: the first is where the stress is assumed continuous across the boundary ... There, the solution is simplified by assumingthat the turning angle α throughwhich the beam is bent is specified, mathematical problems their ability to work at some mathematics harmoniously in a group their ability to listen and ... resource ideas for pupils learning maths outdoors. BEAM Education publishes two books full of ideas for Early Years outdoor maths activities: Carole ... magnuso@maths.lth.se, kalle@maths.lth.se, nco@maths.lth.se Reflector Angle meter a b ... motion problems have been used to solve structure and mo- ... A laser beam generated by a ver-tical laser in the scanner, ... which can be applied to problems they will encounter in their Engineering and Science subjects. ... to the Maths I textbook, Engineering Mathematics by KA Stroud (page indicated in brackets) ... bending in a beam. Use matrices to do calculations in survey problems : 1 Star Problems ADD TO ONE-THOUSAND Problem There are exactly three different pairs of positive integers that add to make six. 1 + 5 = 6 ... If a beam of light is fired through the top left corner of a 2 by 50 rectangle, which corner will it emerge from? Pupil switches attention between objects on a table that are highlighted in turn using a torch beam, or are tapped to make distinct sounds. NUMBER . P LEVEL 2(i) ... Pupil begins to use their developing mathematical understanding of counting to solve simple problems Maths Outside and In Published by BEAM Education, a division of Nelson Thornes Maze Workshops 72a Southgate Road London N1 3JT ... This book is packed with illustrated mathsproblems designed to push the child who always finishes first.
How to Teach Mathematics - 2nd edition Summary: This expanded edition of the original bestseller, How to Teach Mathematics, offers hands-on guidance for teaching mathematics in the modern classroom setting. Twelve appendices have been added that are written by experts who have a wide range of opinions and viewpoints on the major teaching issues. Eschewing generalities, the award-winning author and teacher, Steven Krantz, addresses issues such as preparation, presentation, discipline, and grading. ...show moreHe also emphasizes specifics--from how to deal with students who beg for extra points on an exam to mastering blackboard technique to how to use applications effectively. No other contemporary book addresses the principles of good teaching in such a comprehensive and cogent manner. 85 +$3.99 s/h VeryGood SellBackYourBook Aurora, IL 0821813986 Item in very good condition and at a great price! Textbooks may not include supplemental items i.e. CDs, access codes etc... All day low prices, buy from us sell to us we do it all!! $20.5121New Textbook Charlie Nashville, TN Brand New! Ships same day or next business day. Free USPS Tracking Number. Excellent Customer Service. Ships from TN $26.99 +$3.99 s/h Good One Stop Text Books Store Sherman Oaks, CA 1999-01-01 Paperback Good Expedited shipping is available for this item! $29.00 +$3.99 s/h LikeNew Gian Luigi Fine Books, Inc. NY Albany, NY Providence, RI 2000 PAPERBACK SECOND EDITION FINE 0821813986. $30.00 +$3.99 s/h New Sequitur Books Boonsboro, MD Brand new. We distribute directly for the publisher. This expanded edition of the original bestseller, How to Teach Mathematics, offers hands-on guidance for teaching mathematics in the modern classr...show moreoom setting. Twelve appendices have been added that are written by experts who have a wide range of opinions and viewpoints on the major teaching issues.Eschewing generalities, the award-winning author and teacher, Steven Krantz, addresses issues such as preparation, presentation, discipline, and grading. He also emphasizes specifics--from how to deal with students who beg for extra points on an exam to mastering blackboard technique to how to use applications effectively. No other contemporary book addresses the principles of good teaching in such a comprehensive and cogent manner. ...show less $32.11 +$3.99 s/h Good Big Planet Books Burbank, CA 1999
Intermediate Algebra (Preview Edition) - 12 edition Summary: INTERMEDIATE ALGEBRA: CONNECTING CONCEPTS THROUGH APPLICATIONS shows students how to apply traditional mathematical skills in real-world contexts. The emphasis on skill building and applications engages students as they master concepts, problem solving, and communication skills. It modifies the rule of four, integrating algebraic techniques, graphing, the use of data in tables, and writing sentences to communicate solutions to application problems. The authors have developed several ...show morekey ideas to make concepts real and vivid for students. First, the authors integrate applications, drawing on real-world data to show students why they need to know and how to apply math. The applications help students develop the skills needed to explain the meaning of answers in the context of the application. Second, they emphasize strong algebra skills. These skills support the applications and enhance student comprehension. Third, the authors use an eyeball best-fit approach to modeling. Doing models by hand helps students focus on the characteristics of each function type. Fourth, the text underscores the importance of graphs and graphing. Students learn graphing by hand, while the graphing calculator is used to display real-life data problems. In short, INTERMEDIATE ALGEBRA: CONNECTING CONCEPTS THROUGH APPLICATIONS takes an application-driven approach to algebra, using appropriate calculator technology as students master algebraic concepts and skillsIMPORTANT! Please read before buying. AIE - Instructor's Edition with same identical content as regular student edition but may include instructor's notes and ALL answers. $24.30 +$3.99 s/h LikeNew textbooknook Knoxville, TN INSTRUCTORS EDITION!!! EDGES BARELY WORN. BARELY SHELFWEAR. SAME AS THE STUDENT EDITION, BUT MAY CONTAIN ADDITIONAL TEACHING NOTES & ADDITIONAL/ALL ANSWERS IN THE MARGINS!!! THIS IS THE TEXT ONLY;...show more NO CD's, DVD's, ACCESS CODES, OR ANY OTHER SUPPLEMENTS!!! ...show less $29.81 +$3.99 s/h VeryGood SellBackYourBook Aurora, IL 0534496369 Item in very good condition and at a great price! Textbooks may not include supplemental items i.e. CDs, access codes etc... All day low prices, buy from us sell to us we do it all!! $30.97 +$3.99 s/h Acceptable SellBackYourBook Aurora, IL 053449636951.99 +$3.99 s/h New BookStore101 SUNNY ISLES BEACH, FL SHIPS FAST!!! SAME DAY OR W/N 24 HOURS.EXPEDITED SHIPPING AVAILABLE TOO!!! $95.7190153.19 +$3.99 s/h New Textbookcenter.com Columbia, MO Ships same day or next business day! UPS(AK/HI Priority Mail)/ NEW book $153
If you do a search on "from arithmetic to algebra" as a verbatim phrase, you'll get about 600 hits, with the ones from Google Books reaching back into the nineteenth century. About three out of every four will be about helping students make the transition from arithmetic to algebra -- it has been known for a very long time that that's where we lose many people who are never able to advance much further in math. As I noted before, 25 years ago RAND surveyed the then-nascent field of educational software and found many effective arithmetic teaching programs and practically nothing that taught any of the important aspects of algebra (abstract relations, strategy, fundamental concepts and so on). Even back in 1988, it was clearly understood that arithmetic training programs should not be the model for developing algebra educational software because arithmetic is taught as procedural training. When you teach the quadratic formula, polynomial factoring or Cramer's Rule as if they were mere complex recipes like long division, you miss the fundamental concepts that are the whole point of studying algebra. Depressingly, my survey of algebra-teaching software revealed that 25 years later the situation remains the same. Plenty of programs will drill a student on algebraic procedures, but most do not even attempt to teach any sort of insight, strategy or deeper understanding. Even the best merely offer supporting text or "guess the next step," the same wrong-headed approach that the pioneering math educator Mary Everest Boole identified about the arithmetic-to-algebra transition in 1909. Proceduralism is about performing tasks (now write this number here and do this ...), but conceptualism is the heart of mathematics (how are these numbers connected or related?). In the leap from "what do I do?" to "what is it?" we're losing many students who might otherwise have gained not only the higher incomes, but the much better understanding of the world, to which mathematics is the gateway. Now, there may be an educational software design company out there right now about to fix this problem, but probably there's not. And parents and teachers who need to get a seventh-to-ninth grader across the gap right now can't very well wait until he or she is halfway through college, or longer, for algebra teaching software that actually teaches algebra. And yet there is a piece of instructional software right on your computer that can be used to teach all levels of algebra to all levels of student, in a fully conceptual way. It's the spreadsheet. You can find plenty of discussion and advice about how to do this from excellent teachers like Tom Button, Sue Johnston-Wilder and David Pimm, Teresa Rojano and Ros Sutherland, but let's just quickly hit the highlights of how exploring spreadsheets, and then exploring with spreadsheets, can provide a conceptual doorway into algebra. If you're interested, you'll find all but limitless resources for this. Consider, to begin with, that variables and parameters in spreadsheets are very similar to what they are in ordinary algebra. For that matter, Microsoft Excel notation (and most of the Open Office software notation) is either algebra notation outright, or so close that only simple explanations need to be given ("in algebra the multiplication asterisk is understood, in Excel you have to put it in," "what we call a function in algebra is what a formula is in Excel," etc.). A basic insight of algebra is that a function can be thought of as a rule OR a table OR a graph. (I'm capitalizing because it's the Boolean logical OR rather than ordinary English "or.") In fact, they are three different ways of looking at the same thing. Similarly, a spreadsheet formula can be used to generate a table of data, and the spreadsheet's graphing features can be used to turn it into a graph. I found in teaching elementary algebra to disadvantaged adult learners that the progression from table to graph and then to formula/function/equation might have been the most powerful tool for "selling" algebra I ever encountered. Anyone can see intuitively that for many problems, if you just make a table big enough, trying out all the possible values will lead to a solution. From there, it's a short step to, but what if there are millions of values? Then they're ready to graph it and the answers are right there at the intersections. From there it's just the step to, but what if you need an answer more exact than the line you can draw, or more dimensions than two? Well, by then, they're used to the idea of formula/function/equation as description, and if a point satisfies more than one description, it's a solution. And they've crossed over to doing algebra. The spreadsheet provides a natural bridge from arithmetic procedure to formula/function. At first, students will just input numbers, as if the spreadsheet were a calculator. Then they see that if they input variables, they don't have to type nearly so much, and then that this means having not just this answer this time, but all the answers to all problems of this type, all the time. It's a wide and easy-to-cross bridge from the specific to the general, and from procedure to concept. One of the big changes in mathematics in the last 30 years has been the idea of experimental math, i.e., of exploring how numbers work by setting up numerical processes and looking at the results; it's at the heart of chaos research, for example. Just as the computer has become the equivalent of the telescope or microscope for mathematicians, Excel can be used as an amateur "scope" for exploring numbers, in a way very much analogous to the way countless students have gotten a handle on science by finding a planet in the sky or exploring the ecology of pond water. Among other things, I've used Excel to teach how every fraction is a division, and division is equivalent to multiplying by the inverse. It could easily be used for many other projects beginning even from a very early age in arithmetic. Spreadsheet algebra is such an effective and intuitive idea that it has been re-invented several times in the last 20 years, and some Googling around will turn up immense amounts about it. (Caution: "spreadsheet algebra" is also a term used in advanced mathematics research for a kind of non-linear matrix algebra, so your Googling may very well turn up an article or two that's a bit beyond you. Don't worry, just keep looking!) Although there still needs to be a human being there to guide the student in exploring and using the spreadsheet, as a teaching device for actual algebra (as opposed to a drilling device for standardized tests) the spreadsheet still beats out thousands of purpose-designed
02012518 Solving Experiences in General Mathematics Designed to reflect the sequence of most standard general math, algebra, and geometry texts, this series builds on students' skills, and challenges them with new problem-solving approaches. Each book provides 28 sets of three field-tested problems--academic skill, process, and real world. These sets can be taught as a unit or used as problems of the day. An appendix on using computers in problem solving, with references to appropriate problems throughout the program, is included. Teacher Sourcebooks include teaching strategies, extension problems, evaluation devices, and blackline masters of the student problem sets. General Math/Grades 7-12
Indispensable for students of modern physics, this text provides the necessary background in mathematics to study the concepts of electromagnetic theory and quantum mechanics. Topics include vector algebra, matrix and tensor algebra, vector calculus, functions of a complex variable, integral transforms, linear differential equations, and partial differential equations. 1967 edition. Description: Selected Mathematical Methods in Theoretical Physics shows how a scientist, knowing the answer to a problem intuitively or through experiment, can develop a mathematical method to prove that answer. The author first discusses the formulation of differential or integral ...
Print Book Key Features @bul:* Includes a thematic presentation of linear algebra * Provides a systematic integration of Mathematica * Encourages students to appreciate the benefits of mathematical rigor * All exercises can be solved with Mathematica Description Linear Algebra: An Introduction With Mathematica uses a matrix-based presentation and covers the standard topics any mathematician will need to understand linear algebra while using Mathematica. Development of analytical and computational skills is emphasized, and worked examples provide step-by-step methods for solving basic problems using Mathematica. The subject's rich pertinence to problem solving across disciplines is illustrated with applications in engineering, the natural sciences, computer animation, and statistics. Readership For researchers, librarians, professionals, and the general public who want to enhance their knowledge of linear algebra and Mathematica.
The fun and easy way to understand and solve complex equations Many of the fundamental laws of physics, chemistry, biology, and economics can be formulated as differential equations. This plain-English guide explores the many applications of this mathematical tool and shows how differential equations can help us understand the world around us. Differential... more... Malliavin calculus provides an infinite-dimensional differential calculus in the context of continuous paths stochastic processes. This book, demonstrating the relevance of Malliavin calculus for Mathematical Finance, starts with an exposition from scratch of this theory. Greeks (price sensitivities) are interpreted in terms of Malliavin calculus. more... Offers a systematic presentation of Lipschitzian-type mappings in metric and Banach spaces. This book covers some basic properties of metric and Banach spaces. It also provides background in terms of convexity, smoothness and geometric coefficients of Banach spaces including duality mappings and metric projection mappings. more... Master pre-calculus from the comfort of home! Want to "know it ALL" when it comes to pre-calculus? This book gives you the expert, one-on-one instruction you need, whether you're new to pre-calculus or you're looking to ramp up your skills. Providing easy-to-understand concepts and thoroughly explained exercises, math whiz Stan Gibilisco... more...
Elementary Number Theory 9780073051888 ISBN: 0073051888 Edition: 6 Pub Date: 2005 Publisher: McGraw-Hill College Summary: Elementary Number Theory, Sixth Edition, is written for the one-semester undergraduate number theory course taken by math majors, secondary education majors, and computer science students. This contemporary text provides a simple account of classical number theory, set against a historical background that shows the subject's evolution from antiquity to recent research. Written in David Burton's engaging style, Elemen...tary Number Theory reveals the attraction that has drawn leading mathematicians and amateurs alike to number theory over the course of history. Burton, David M. is the author of Elementary Number Theory, published 2005 under ISBN 9780073051888 and 0073051888. Two hundred seventeen Elementary Number Theory textbooks are available for sale on ValoreBooks.com, eleven used from the cheapest price of $42.23, or buy new starting at $161.87 policy. Contact Customer Service for questions.[less] Highlighting, underlining, marks & notes in pen & pencil, covers scuffed/scratched, minimal shelf wear, stickers on back cover & spine
Algebra: Word Problems Help and Practice Problems Find study help on linear applications for algebra. Use the links below to select the specific area of linear applications you're looking for help with. Each guide comes complete with an explanation, example problems, and practice problems with solutions to help you learn linear applications for algebra. Introduction to Simplifying Expressions and Solving Equation Word Problems Mathematics may be defined as the economy of counting. There is no problem in the whole of mathematics which cannot be solved by direct counting. Math Word Problems Practice Quiz This test has 30 multiple-choice questions and is in the same format as the introductory test (Introductory Math Word Problems Practice Quiz). Each question corresponds ...
books.google.com - This volume offers a new, "hands-on" approach to teaching Discrete Mathematics. A simple functional language is used to allow students to experiment with mathematical notations which are traditionally difficult to pick up. This practical approach provides students with instant feedback and also allows... Mathematics Using a Computer
9Marge Lial Marge Lial was always MJohn Hornsby When ten years of writing mathematics textbooks, both of his goals have been realized. His love for teaching and for mathematics is evident in his passion for working with students and fellow teachers as well. His specific professional interests are recreational mathematics, mathematics history, and incorporating graphing calculators into the curriculum. Johncollecting. He loves the music of the 1960s and has an extensive collection of the recorded works of Frankie Valli and the Four Seasons. Terry McGinnis - A native Midwesterner, Terry received her Bachelor's of Science in Elementary Education (Mathematics concentration) from Iowa State University. She has taught elementary and middle school mathematics, and developed and implemented the curriculum used with her students. Terry has been involved in college mathematics publishing for over 15 years, working with a variety of authors on textbooks in developmental and precalculus mathematics. After working behind the scenes on many of the Lial/Hornsby textbooks and supplements for the past ten years, Terry most recently joined Margaret Lial and John Hornsby as co-author of their developmental mathematics series that includes Introductory Algebra, Intermediate Algebra, and Introductory and Intermediate Algebra, all published by Addison-Wesley. When not working, Terry enjoys spinning at a local health club and walking. She is a devoted parent of two sons, Andrew and Tyler. Table of Contents (Note: Each chapter ends with a Group Activity, Summary, Review Exercises, a Chapter Test and, with the exception of Ch. 1, a Cumulative Review). List of Applications. Preface. Feature Walkthrough. 1. The Real Number System. Fractions. Exponents, Order of Operations, and Inequality. Variables, Expressions, and Equations. Real Numbers and the Number Line. Adding and Subtracting Real Numbers. Multiplying and Dividing Real Numbers. Summary Exercises on Operations with Real Numbers. Properties of Real Numbers. Simplifying Expressions. 2. Linear Equations and Inequalities in One Variable. The Addition Property of Equality. The Multiplication Property of Equality. More on Solving Linear Equations. Summary Exercises on Solving Linear Equations . An Introduction to Applications of Linear Equations. Formulas and Applications from Geometry. Ratios and Proportions. More About Problem Solving. Solving Linear Inequalities. 3. Linear Equations and Inequalities in Two Variables; Functions. Reading Graphs; Linear Equations in Two Variables. Graphing Linear Equations in Two Variables. The Slope of a Line. Equations of a Line. Graphing Linear Inequalities in Two Variables. Introduction to Functions. The Product Rule and Power Rules for Exponents. Integer Exponents and the Quotient Rule. Summary Exercises on the Rules for Exponents. An Application of Exponents: Scientific Notation. Adding and Subtracting Polynomials; Graphing Simple Polynomials. Multiplying Polynomials. Special Products. Dividing Polynomials. 6. Factoring and Applications. The Greatest Common Factor; Factoring by Grouping. Factoring Trinomials. More on Factoring Trinomials. Special Factoring Rules. Summary Exercises on Factoring. Solving Quadratic Equations by Factoring. Applications of Quadratic May 12, 2007 A reviewer Lial does a magnificent job in this introductory algebra textbook. It makes algebra (to those who dislike math) fun and very interesting. This is THE textbook every high school or community college should adopt. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged.
Discrete Mathematics with Applications, 3rd Edition Susanna Epp's DISCRETE MATHEMATICS, THIRD concepts experience. Cengage Learning reserves the right to remove content from eBooks at any time if subsequent rights restrictions require it.
About the Author Gary Haggard is Professor of Computer Science at Bucknell University. His research in data structures focuses on the implementation of effective algorithms for computing invariants for large combinatorial structures such as graphs. Dr. Haggard¿s current work is directed towards finding chromatic polynomials of large graphs. John Schlipf is a Professor of Computer Science in the Department of Electrical and Computer Engineering and Computer Science at the University of Cincinnati. His research interests include logic programming and deductive databases, algorithms for satisfiability, computability and complexity, formal verification, and model theory. Sue Whitesides is Professor of Computer Science at McGill University. She holds a Ph.D. from University of Wisconsin and a Masters from Stanford University. Her research interests lie within combinatorial mathematics and theoretical computer science required textbook for a course at my university. My professor pulled all the homework from the ends of each chapter. This part of the book is one of my biggest gripes. The reading sections of this book pack a large amount of material in a brief page or two for each section followed by homework exercises. The exercise sections have are about as long as the actual information sections, meaning they are packed with questions. This would be a positive for this book except the questions aren't similar, so the included CD with the odd problems solved will often be of little help because question 3 will be a completely different sort of problem than question 4. Since each problem is so unique, you'll often be left dealing with problems that are considerably more complex than anything found in the reading sections of the text. If you are using the questions of this book for homework, be prepared to use google extensively. As an example, the book may explain how to perform an operation on 2 sets of numbers. Then in the homework, it will ask you to perform the same operation on 5 sets abstract sets without ever explaining how to go about doing that. I ended up receiving an A in the course, but that was after spending ~8 hours for each 10-14 question homework. Most of that time was spent on the internet trying to learn the material from whatever sites I could find. The reading sections of this text are an excersize in frustration. In one of the explanations for a concept in the book, the author literally uses the phrase "from [problem], it is obvious that the answer is [answer]." That was the entire explanation on the topic. A textbook should never say the phrase "from X, it is obvious that Y" if the whole section is supposed to be telling you how to find Y from X in the first place. This is an introductory text into formal logic, proofs, and set mathematics. Yet, you'll often find that the author skips steps in his solutions which may be obvious to someone familiar with the material but that is obviously not the target of this text. There is an occasional table for reference which doesn't explain what the relationship between anything on the table is (I'm looking at you, Table of Commonly Used Tautologies....). This book covers a great number of topics in a fairly small book, for a textbook that is. However, this book suffers from a lack of depth necessary to reach its potential. If you have a choice, skip this text. If, like me, you are required to use this text.... Google everything and god help you. 1.0 out of 5 starsExtremely poor organization.Jan. 13 2014 By Dan G. - Published on Amazon.com Format:Hardcover\|Amazon Verified Purchase This book has an extremely poor organization of information. It's like the authors just threw a bunch of information at the book without thinking about how a student has to go through learning the mathematical concepts. The only reason I have to use this book is because a professor from my university was one of the authors. Get another book on discrete mathematics if you want to really learn the material. 1 of 8 people found the following review helpful 5.0 out of 5 starsGreat TextbookSept. 7 2011 By mfox - Published on Amazon.com Format:Hardcover\|Amazon Verified Purchase This textbook was the exact same one I needed for class and was MUCH cheaper than buying from the school store. It was even in better condition than what was advertised! I would definitely recommend this book.
Math 1350 Foundations of Mathematics I Information LSC-CyFair Math Department Catalog Description This is designed specifically for students who seek elementary and middle school teacher certification. Topics include set theory, functions, numerations systems, number theory, emphasis on problem solving and critical thinking. Course Learning Outcomes The student will: • Use models and manipulatives to demonstrate the four basic operations of the rational numbers. • Demonstrate an understanding of place value through multiple representations including the use of grouping manipulatives, place value manipulatives and abstract representations such as with exponents and different number bases. • Demonstrate an understanding of the attributes of numeration systems. • Analyze mathematical situations and solve problems using mathematical heuristics
More About This Textbook Overview This text is an unbound, binder-ready edition. The 10th edition of Elementary Differential Equations, like its predecessors, is written from the viewpoint of the applied mathematician, whose interest in differential equations may sometimes be quite theoretical, sometimes intensely practical, and often somewhere in between. The authors have sought to combine a sound and accurate (but not abstract) exposition of the elementary theory of differential equations with considerable material on methods of solution, analysis, and approximation that have proved useful in a wide variety of applications. While the general structure of the book remains unchanged, some notable changes have been made to improve the clarity and readability of basic material about differential equations and their applications. In addition to expanded explanations, the 10th edition includes new problems, updated figures and examples to help motivate students ] or three ]semester course sequence or its equivalent. Some familiarity with matrices will also be helpful in the chapters on systems of differential equations. Editorial Reviews Booknews To accommodate the changing learning environment, this edition takes a more visual, quantitative, project, and example-oriented approach for undergraduate students in mathematics, science, or engineering whose interest in differential equations ranges from the totally theoretical to the diehard practical. Other changes include an introduction to mathematical modeling, direction fields, the basic ideas of stability and instability, and Euler's method of numerical approximation. Coverage includes first-, second-, and higher-order linear and nonlinear equations; the Laplace transform; and numerical methods. Includes chapter problems and appended answers. Boyce is professor emeritus, and DiPrima is a professor, respectively, in the department of mathematical sciences at Rensselaer Polytechnic Institute in Troy, New York. Annotation c. Book News, Inc., Portland, OR (booknews.com) Related Subjects Meet the Author Dr for. In 1991 he received the William H.Wiley Distinguished Faculty Award given by Rensselaer 2001 Good exercises, but a little wordy for an intro course I used this course in an introductory differential equations course targeted at engineers, physicists, chemists, and mathematicians. Coupled with a good instructor, it can be a valuable tool. My favorite feature is the inclusion of solutions (in the back of the book) to ALL exercises that do not require plotting. This helped me to learn much more as I always worked out my solutions until I had the right answer, knowing that the process is the most important. It is a little long to read everything in a section, so it needs to be used by a good instructor, but not too bad overall. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged.
Educational material for courses in algebra, trigonometry, calculus I, and calculus II can be found at this unique site. Users can navigate between topics within each of these four subjects by using a visual index,... Louise Huttner of Burlington County College designed this activity to help students learn and practice their calculus skills. Students will collect data on the monthly temperature of a city. They will then graph and... Back in the late 1990s, the Scout Report first reported on this site, which was a collection of useful mathematics resources intended to serve both students interested in college-level mathematics and their teachers.... The thought of learning calculus has struck fear into the heart of many students for several centuries, but this most intriguing subject need do so no longer. Developed with assistance from the National Science... Working in tandem with the Harvard-Smithsonian Center for Astrophysics, the team of experts at the Annenberg Media Foundation has created this excellent instructional series. In this eight-part program, educators can...
This is a booklet containing 31 problem sets that involve a variety of math skills, including scientific notation, simple algebra, and calculus. Each set of problems is contained on one page. Learners will use mathematics to explore varied space...(View More) science topics including black holes, ice on Mercury, a mathematical model of the Sun's interior, sunspots, the heliopause, and coronal mass ejections, among many others
Modelling With Circular Motion The School Mathematics Project 9780521408899 ISBN: 052140889X Pub Date: 1993 Publisher: Cambridge University Press Summary: The aim of 16-19 Mathematics has been to produce a course which, while challenging, is accessible and enjoyable to all students. The course develops ability and confidence in mathematics and its applications, together with an appreciation of how mathematical ideas help in the understanding of the world and society in which we live. This unit: - develops an understanding of work and energy through the modelling of rea...l situations involving circular motion; - provides insight into the potential of mathematics for modelling physical phenomena; - helps foster an appreciation of the links between mathematics and the real world; - develops a basis for further study in engineering and science; - fosters an ability to model both in familiar and unfamiliar contexts within the field of mechanics. School Mathematics Project Staff is the author of Modelling With Circular Motion The School Mathematics Project, published 1993 under ISBN 9780521408899 and 052140889X. Fifty six Modelling With Circular Motion The School Mathematics Project textbooks are available for sale on ValoreBooks.com, twenty two used from the cheapest price of $1.53, or buy new starting at $11

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