Multivariable Calculus Concepts and Contexts, 3rd EditionSet yourlearning in motion with these valuable tools!
Tools for Enriching™ Calculus CD-ROM (TEC) This CD-ROM helps youunderstand and visualize calculus by exploring conceptsthrough
interactive modules and animations. TEC icons in the text directyou to the appropriate module for
exploration. TEC also has homework hints for specially markedexercises in each section.
Interactive Video Skillbuilder CD-ROM This CD-ROM contains videoinstruction for every major concept in the text. It lets youreview
material that you may not have understood the first time you heardit, or if you missed class.
In order to help you evaluate your progress, each section containsa 10-question Web quiz
per section (the results of which can be emailed to the instructor)and a test for each chapter,
with answers.
iLrn™ Student Version Get expert help from your own tutor . . .online. Through iLrn, you have access to live online
tutoring at vMentor™. The tutors at this free service willskillfully guide you through a problem
using unique two-way audio and whiteboard features. The iLrn systemalso provides you with
access to self-directed study via instructor-assigned,text-specific homework problems. Use the
access code packaged with this text to get started today! vMentoris for use by proprietary, college,
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1
ARITHMETIC OPERATIONS
GEOMETRIC FORMULAS
Triangle Circle Sector of Circle
Sphere Cylinder Cone
Point-slope equation of line through with slope m:
Slope-intercept equation of line with slope m and y-interceptb:
CIRCLES
x h2 y k2 r 2
h, k
P1x1, y1
m y2 y1
x1 x2
P2x2, y2P1x1, y1
h
r
r
3 r 3
b
s r in radians C 2r 1 2 ab sin
A 1 2 r 2 A r 2 A 1
2 bh
a 0
a cb ca b
2a ax 2 bx c 0
n
1 2 3 k
kx nkyk nxyn1 yn
x yn x n nx n1y nn 1
2 x n2y2
x y3 x 3 3x 2y 3xy2 y3
x y3 x 3 3x 2y 3xy2 y3
x y2 x 2 2xy y2x y2 x 2 2xy y2
x 3 y3 x yx 2 xy y2
x 3 y3 x yx 2 xy y2
x 2 y2 x yx y
nx
s n xy s
n xs n y
x mn s n x m (sn x )mx 1n s
n x
xn 1
x m
x n x mnx mx n x mn
a
b
c
d
a
b
d
c
ad
bc
2
radians
0
π 2π x
s radians 180
T R I G O N O M E T R Y
FUNDAMENTAL IDENTITIES
1 tan2x
cos 2x cos2x sin2x 2 cos2x 1 1 2 sin2x
sin 2x 2 sin x cos x
tanx y tan x tan y
1 tan x tan y
tanx y tan x tan y
1 tan x tan y
cosx y cos x cos y sin x sin y
cosx y cos x cos y sin x sin y
sinx y sin x cos y cos x sin y
sinx y sin x cos y cos x sin y
c 2 a 2 b 2 2ab cos C
b 2 a 2 c 2 2ac cos B
a 2 b 2 c 2 2bc cos A
A
b
c
a
B
C
sin2 cos2 1cot 1
3
S P E C I A L F U N C T I O N S
POWER FUNCTIONS
(i i i)
INVERSE TRIGONOMETRIC FUNCTIONS
x
1
y
10
y=
x
x
y
0
x
y
0
π
2
_ π
2
y
0
x
4
S P E C I A L F U N C T I O N S
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
1.
2.
3.
1 x y tanh1x &? tanh y x
cosh1x ln(x sx 2 1) y cosh1x &? cosh y x and y 0
sinh1x ln(x sx 2 1) y sinh1x &? sinh y x
coth x cosh x
sinh x tanh x
loga x
y loga x loga ye ln x x lnex x
logaxy loga x loga ya loga x x logaax x
ey x&?ln x y
ln e 1ln x loge x
ay x&?loga x y
y
1
0
ln x
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
d
d
d
d
d
d
d
d
d
d
dx f tx f txtx
d
tx txf x f xtx
tx 2
d
dx f xtx f xtx txf x
d
d
d
d
dx c 0
D I F F E R E N T I A T I O N R U L E S
M U L T I V A R I A B L E
The cover photograph shows the Walt Disney Concert Hall in LosAngeles, designed and built 1992–2003 by Frank Gehry andAssociates. It is a daring building, a layered composition ofcurved surfaces in the form of billowing sails with brushed stain-less steel cladding.
The highly complex structures that Frank Gehry designs would beimpossible to build with- out the computer. The CATIA software thathis architects and engineers use to produce the com- puter modelsis based on principles of calculus— fitting curves by matchingtangent lines, making sure the curvature isn’t too large, andcontrol- ling parametric surfaces. “Consequently,” says Gehry, “wehave a lot of freedom. I can play with shapes.”
The process starts with Gehry’s initial sketches, which aretranslated into a succession of physical models. (Hundreds ofdifferent physical models were constructed during the design of thebuilding, first with basic wooden blocks and then evolving intomore sculptural forms.) Then an engineer uses a digitizer to recordthe coordinates of a series of points on a physical model. Thedigitized points are fed into a computer and the CATIA software isused to link these points with smooth curves. (It joins curves sothat their tangent lines coincide.) The architect has considerablefreedom in creating these curves, guided by displays of the curve,its derivative, and its curvature. Then the
Calculus and the Architecture of Curves
Images not available due to copyright restrictions
Image not available due to copyright restrictions
curves are connected to each other by a parametric surface, andagain the architect can do so in many possible ways with theguidance of displays of the geo- metric characteristics of thesurface.
The CATIA model is then used to produce another physical model,which, in turn, suggests modifications and leads to additionalcomputer and physical models.
The CATIA program was developed in France by Dassault Systèmes,originally for designing airplanes, and was subsequently employedin the automotive industry. Frank Gehry, because of his complexsculptural shapes, is the first to use it in architecture. It helpshim answer his question, “How wiggly can you get and still make abuilding?”
Image not available due to copyright restrictions
Image not available due to copyright restrictions
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Calculus Concepts and Contexts 3E
James Stewart M C M A S T E R U N I V E R S I T Y
M U L T I V A R I A B L E
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Credits continue on page A51.
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8.1 Sequences 557
8.2 Series 567
8.4 Other Convergence Tests 586
8.5 Power Series 594
8.7 Taylor and Maclaurin Series 605
Laboratory Project An Elusive Limit 617
8.8 The Binomial Series 617
Writing Project How Newton Discovered the Binomial Series 621
8.9 Applications of Taylor Polynomials 621
Applied Project Radiation from the Stars 630
Review 631
Vectors and the Geometry of Space 636
9.1 Three-Dimensional Coordinate Systems 637
9.2 Vectors 642
Discovery Project The Geometry of a Tetrahedron 665
9.5 Equations of Lines and Planes 666
Laboratory Project Putting 3D in Perspective 675
9
8
vii
Contents
9.6 Functions and Surfaces 676
9.7 Cylindrical and Spherical Coordinates 685
Laboratory Project Families of Surfaces 690
Review 690
Vector Functions 694
10.2 Derivatives and Integrals of Vector Functions 702
10.3 Arc Length and Curvature 708
10.4 Motion in Space: Velocity and Acceleration 716
Applied Project Kepler’s Laws 727
10.5 Parametric Surfaces 728
Partial Derivatives 738
11.2 Limits and Continuity 750
11.3 Partial Derivatives 756
11.5 The Chain Rule 780
11.6 Directional Derivatives and the Gradient Vector 788
11.7 Maximum and Minimum Values 801
Applied Project Designing a Dumpster 811
Discovery Project Quadratic Approximations and Critical Points812
11.8 Lagrange Multipliers 813
Review 822
Multiple Integrals 828
12.2 Iterated Integrals 837
12.5 Applications of Double Integrals 857
12.6 Surface Area 867
12.7 Triple Integrals 872
12.8 Triple Integrals in Cylindrical and Spherical Coordinates882
Applied Project Roller Derby 888
Discovery Project The Intersection of Three Cylinders 889
12.9 Change of Variables in Multiple Integrals 889
Review 898
Vector Calculus 904
13.4 Green’s Theorem 933
13.5 Curl and Divergence 940
13.6 Surface Integrals 948
13.7 Stokes’ Theorem 959
13.8 The Divergence Theorem 966
13.9 Summary 973
Appendixes A1
E A Few Proofs A3
H Polar Coordinates A6
I Complex Numbers A22
Index A53
x
When the first edition of this book appeared eight years ago, aheated debate about calculus reform was taking place. Such issuesas the use of technology, the relevance of rigor, and the role ofdiscovery versus that of drill were causing deep splits in math-ematics departments. Since then the rhetoric has calmed downsomewhat as reform- ers and traditionalists have realized that theyhave a common goal: to enable students to understand and appreciatecalculus.
The first and second editions were intended to be a synthesis ofreform and tradi- tional approaches to calculus instruction. Inthis third edition I continue to follow that path by emphasizingconceptual understanding through visual, numerical, and alge- braicapproaches.
What’s New in the Third Edition
By way of preparing to write the third edition of this text, Ispent a year teaching cal- culus at the University of Toronto. Ilistened carefully to my students’ questions and my colleagues’suggestions. And as I prepared each lecture I sometimes realizedthat an additional example was needed, or a sentence could beclarified, or a section could use a few more exercises of a certaintype. In addition, I paid attention to the sugges- tions sent to meby many users and to the comments of the reviewers.
Many hundreds of improvements, large and small, have beenincorporated into this edition. Here are some of them.
Many examples have been added or changed.
Extra steps have been provided in some of the existingexamples.
The data in examples and exercises have been updated to be moretimely.
More than 25% of the exercises in each chapter are new. Here are afew of my favorites:
Preface
8.4.36 593 8.6.37–38 605 8.9.22 628
10.1.37–38 701 11.4.38 779 11.5.36 787
Image not available due to copyright restrictions
New phrases and margin notes have been added to clarify theexposition.
A number of pieces of art have been redrawn.
I’ve also added new problems to the Focus on Problem Solvingsections. See, for instance, Problems 14 and 16 on page 635.
Two new projects have been added. The project on page 617 shows howcomputer algebra systems use Taylor series to compute limits, andthe project on page 675 shows how computer graphics programmers useclipping planes and hidden line rendering to portraythree-dimensional objects on a two- dimensional screen.
The CD called Tools for Enriching Calculus (TEC) has beencompletely redesigned and now includes what we call Visuals, briefanimations of various figures in the text. In addition there arenow Visuals, Modules, and Homework Hints for the multivariablechapters. See the description on page xiii.
The symbol has been placed beside examples (an average of three persection) for which there are videos of instructors explaining theexample in more detail. These videos are free to adopters. Thismaterial is also included on an Interactive VideoSkillbuilder CD.See the description of the Interactive Video Skillbuilder on pagexiii.
Conscious of the need to control the size of the book, I’ve put newtopics (as well as expanded coverage of some topics already in thebook) on the revamped web site www.stewartcalculus.com rather thanin the text itself. (See the list of additional topics in thedescription of the web site on page xvii.) As a result, the numberof pages in the text is actually a bit less than in the secondedition.
Features
Conceptual Exercises The most important way to foster conceptualunderstanding is through the problems that we assign. To that end Ihave devised various types of problems. Some exercise sets beginwith requests to explain the meanings of the basic concepts of thesection. (See, for instance, the first couple of exercises inSections 8.2, 11.2, and 11.3. I often use them as a basis forclassroom discussions.) Similarly, review sections begin with aConcept Check and a True-False Quiz. Other exercises testconceptual understand- ing through graphs or tables (see Exercises8.7.2, 10.2.1–2, 10.3.27–33, 11.1.1–2, 11.1.9–14, 11.3.3–8,11.6.1–2, 11.7.3–4, 12.1.5–10, 13.1.11–18, 13.2.15–16, and13.3.1–2).
Graded Exercise Sets Each exercise set is carefully graded,progressing from basic conceptual exercises and skill-developmentproblems to more challenging problems involving applications andproofs.
Real-World Data My assistants and I have spent a great deal of timelooking in libraries, contacting companies and government agencies,and searching the Internet for interesting real- world data tointroduce, motivate, and illustrate the concepts of calculus. As aresult, many of the examples and exercises deal with functionsdefined by such numerical
PREFACE xi
xii PREFACE
data or graphs. For instance, functions of two variables areillustrated by a table of values of wave heights (Example 3 inSection 9.6) and by a table of values of the wind-chill index as afunction of air temperature and wind speed (Example 1 in Sec- tion11.1). Partial derivatives are introduced in Section 11.3 byexamining a column in a table of values of the heat index(perceived air temperature) as a function of the actual temperatureand the relative humidity. This example is pursued further in con-nection with linear approximations (Example 3 in Section 11.4).Directional deriva- tives are introduced in Section 11.6 by using atemperature contour map to estimate the rate of change oftemperature at Reno in the direction of Las Vegas. Double inte-grals are used to estimate the average snowfall in Colorado onDecember 24, 1982 (Example 4 in Section 12.1). Vector fields areintroduced in Section 13.1 by depictions of actual velocity vectorfields showing San Francisco Bay wind patterns.
Projects One way of involving students and making them activelearners is to have them work (perhaps in groups) on extendedprojects that give a feeling of substantial accom- plishment whencompleted. Applied Projects involve applications that are designedto appeal to the imagination of students. The project after Section11.8 uses Lagrange multipliers to determine the masses of the threestages of a rocket so as to minimize the total mass while enablingthe rocket to reach a desired velocity. Discovery Projects exploreaspects of geometry: tetrahedra (after Section 9.4), hyperspheres(after Sec- tion 12.7), and intersections of three cylinders (afterSection 12.8). The Laboratory Project on page 690 uses technologyto discover how interesting the shapes of sur- faces can be and howthese shapes evolve as the parameters change in a family. TheWriting Project on page 965 explores the historical and physicalorigins of Green’s Theorem and Stokes’ Theorem and the interactionsof the three men involved. Many additional projects are provided inthe Instructor’s Guide.
Problem Solving Students usually have difficulties with problemsfor which there is no single well- defined procedure for obtainingthe answer. I think nobody has improved very much on George Polya’sfour-stage problem-solving strategy and, accordingly, I haveincluded a version of his problem-solving principles at the end ofChapter 1. They are applied, both explicitly and implicitly,throughout the book. After the other chapters I have placedsections called Focus on Problem Solving, which feature examples ofhow to tackle challenging calculus problems. In selecting thevaried problems for these sections I kept in mind the followingadvice from David Hilbert: “A mathemati-cal problem should bedifficult in order to entice us, yet not inaccessible lest it mockour efforts.” When I put these challenging problems on assignmentsand tests I grade them in a different way. Here I reward a studentsignificantly for ideas toward a solution and for recognizing whichproblem-solving principles are relevant.
Technology The availability of technology makes it not lessimportant but more important to understand clearly the conceptsthat underlie the images on the screen. But, when properly used,graphing calculators and computers are powerful tools fordiscovering and understanding those concepts. I assume that thestudent has access to either a graphing calculator or a computeralgebra system. The icon ; indicates an exercise that definitelyrequires the use of such technology, but that is not to say that agraph- ing device can’t be used on the other exercises as well. Thesymbol is reserved for problems in which the full resources of acomputer algebra system (like Derive, Maple, Mathematica, or theTI-89/92) are required. But technology doesn’t make pen- cil andpaper obsolete. Hand calculation and sketches are often preferableto technol-
CAS
PREFACE xiii
ogy for illustrating and reinforcing some concepts. Bothinstructors and students need to develop the ability to decidewhere the hand or the machine is appropriate.
Tools for Enriching™ Calculus The CD-ROM called TEC is a companionto the text and is intended to enrich and complement its contents.Developed by Harvey Keynes, Dan Clegg, Hubert Hohn, and myself, TECuses a discovery and exploratory approach. In sections of the bookwhere technology is particularly appropriate, marginal icons directstudents to TEC Visuals and Modules that provide a laboratoryenvironment in which they can explore the topic in different waysand at different levels. Visuals are animations of figures in thetext; Modules are more elaborate activities and include exercises.Instructors can choose to become involved at several differentlevels, ranging from simply encourag- ing students to use theVisuals and Modules for independent exploration, to assigningspecific exercises from those included with each Module, or tocreating additional exercises, labs, and projects that make use ofthe Visuals and Modules.
TEC also includes Homework Hints for representative exercises(usually odd- numbered) in every section of the text, indicated byprinting the exercise number in red. These hints are usuallypresented in the form of questions and try to imitate an effectiveteaching assistant by functioning as a silent tutor. They areconstructed so as not to reveal any more of the actual solutionthan is minimally necessary to make fur- ther progress.
Interactive Video Skilbuilder CD-ROM The Interactive VideoSkillbuilder CD-ROM contains more than eight hours of videoinstruction. The problems worked during each video lesson are shownnext to the viewing screen so that students can try working thembefore watching the solution. To help students evaluate theirprogress, each section contains a ten-question Web quiz (theresults of which can be emailed to the instructor) and each chaptercontains a chap- ter test, with answers to each problem.
Web Site: www.stewartcalculus.com This has been renovated and nowincludes the following.
Algebra Review
Lies My Calculator and Computer Told Me
History of Mathematics, with links to the better historical websites
Additional Topics (complete with exercise sets): TrigonometricIntegrals, Trigonometric Substitution, Strategy for Integration,Volumes by Cylindrical Shells, Strategy for Testing Series, FourierSeries, Formulas for the Remainder Term in Taylor Series, LinearDifferential Equations, Second-Order Linear Differential Equations,Nonhom*ogeneous Linear Equations, Applications of Second-OrderDifferential Equations, Using Series to Solve Differential Equa-tions, Rotation of Axes
Links, for each chapter, to outside Web resources
Archived Problems (drill exercises that appeared in previouseditions, together with their solutions)
Challenge Problems (some from the Focus on Problem Solving sectionsof prior editions)
Downloadable versions of CalcLabs for Derive and TI graphingcalculators
Chapter 8 Infinite Sequences and Series
Tests for the convergence of series are considered briefly, withintuitive rather than formal justifications. Numerical estimates ofsums of series are based on which test was used to proveconvergence. The emphasis is on Taylor series and polynomials andtheir applications to physics. Error estimates include those fromgraphing devices.
Chapter 9 Vectors and the Geometry of Space
The dot product and cross product of vectors are given geometricdefinitions, moti- vated by work and torque, before the algebraicexpressions are deduced. To facilitate the discussion of surfaces,functions of two variables and their graphs are introducedhere.
Chapter 10 Vector Functions
The calculus of vector functions is used to prove Kepler’s FirstLaw of planetary motion, with the proofs of the other laws left asa project. In keeping with the intro- duction of parametric curvesin Chapter 1, parametric surfaces are introduced as soon aspossible, namely, in this chapter. I think an early familiaritywith such surfaces is desirable, especially with the capability ofcomputers to produce their graphs. Then tangent planes and areas ofparametric surfaces can be discussed in Sections 11.4 and12.6.
Chapter 11 Partial Derivatives
Functions of two or more variables are studied from verbal,numerical, visual, and algebraic points of view. In particular, Iintroduce partial derivatives by looking at a specific column in atable of values of the heat index (perceived air temperature) as afunction of the actual temperature and the relative humidity.Directional derivatives are estimated from contour maps oftemperature, pressure, and snowfall.
Chapter 12 Multiple Integrals
Contour maps and the Midpoint Rule are used to estimate the averagesnowfall and average temperature in given regions. Double andtriple integrals are used to compute probabilities, areas ofparametric surfaces, volumes of hyperspheres, and the volume ofintersection of three cylinders.
Chapter 13 Vector Fields
Vector fields are introduced through pictures of velocity fieldsshowing San Francisco Bay wind patterns. The similarities among theFundamental Theorem for line integrals, Green’s Theorem, Stokes’Theorem, and the Divergence Theorem are emphasized.
Ancillaries
Multivariable Calculus: Concepts and Contexts, Third Edition, issupported by a com- plete set of ancillaries developed under mydirection. Each piece has been designed to enhance studentunderstanding and to facilitate creative instruction. The table onpages xv and xvi lists ancillaries available for instructors andstudents.
PREFACE xv
Contains Electronic Instructor’s Guide, Resource Integra- tionGuide, iLrn Testing, Instructions for iLrn Homework, and PowerPoint Lecture notes.
Tools for Enriching™ Calculus CD-ROM by James Stewart and Dan CleggISBN 0-534-40989-X
Completely revised and updated, TEC provides a laboratoryenvironment in which students can explore selected topics. TEC alsoincludes homework hints for representative exercises.
Instructor’s Guide by Douglas Shaw ISBN 0-534-41030-8
Each section of the main text is discussed from several view-points and contains suggested time to allot, points to stress, textdiscussion topics, core materials for lecture, workshop/ discussionsuggestions, group work exercises in a form suitable for handout,with solutions, and suggested home- work problems. An electronicversion is available on the Instructor’s Resource CD-ROM.
Complete Solutions Manual, Multivariable by Dan Clegg ISBN0-534-41012-X
Includes worked-out solutions to all exercises in the text.
Printed Test Bank By William Tomhave & Xueqi Zeng ISBN0-534-41031-6
Contains multiple-choice and short-answer test items that keydirectly to the text.
ILrn Adopter’s Fulfillment Folder Kit ISBN 0-534-41033-2
Featuring full algorithmic generation of problems and free-response mathematics, iLrn allows you to customize exams and trackstudent progress in an accessible, browser-based format, withresults flowing automatically into your grade- book! This kitcontains the generic Instructor’s Guide, Mathematics InstructorsUser’s Guide, Installation CD-ROM (for offline users), atext-specific content CD-ROM, a Quick
Start Guide, and a “How do I” quick introduction to widely usedfunctions in iLrn.
Text-Specific Videos ISBN 0-534-41037-5
Text-specific videotape sets, available at no charge to adopters,consisting of one tape per text chapter. Each tape features a 10-to 20-minute problem-solving lesson for each section of thechapter. Covers both single- and multi- variable calculus.
Transparencies, Multivariable by James Stewart ISBN0-534-41015-4
Full-color, large-scale sheets of reproductions of material fromthe text.
Solutions Builder CD-ROM ISBN 0534410383
This CD is an electronic version of the complete solutions manual.It provides instructors with an efficient method for creatingsolution sets to homework or exams. Instructors can easily view,select, and save solution sets that can then be printed orposted.
Resources for Instructors and Students
Stewart Specialty Web Site: www.stewartcalculus.com
Contents: Algebra Review Additional Topics Drill exer- cisesProblems Plus Web Links History of Mathematics Downloadableversions of CalcLabs for Derive and TI graphing calculators MapleProjects Mathematica Projects
iLrn Homework [http:// iLrn.com] ISBN 0-534-40988-1
iLrn Homework allows instructors to assign machine- gradablehomework problems that help students identify where they needadditional help. That assistance is available through worked-outsolutions that guide students through the steps of problem solving,or via live online tutoring at vMentor. The tutors at this onlineservice will skillfully guide students through a problem, usingunique two-way audio and whiteboard features.
Electronic items Printed items
(cont.)
The Brooks/Cole Mathematics Resource Center Web Sitehttp://mathematics.brookscole.com
When you adopt a Thomson–Brooks/Cole mathematics text, you and yourstudents will have access to a variety of
teaching and learning resources. This Web site features everythingfrom book-specific resources to newsgroups. It’s a great way tomake teaching and learning an interactive and intriguingexperience.
WebTutor Advantage™ on WebCT ISBN 0-534-41028-6
Lecture notes, discussion threads, and quizzes on WebCT.
WebTutor Advantage™ on Blackboard ISBN 0-534-41039-1
Lecture notes, discussion threads, and quizzes on Blackboard.
Student Resources
Tools for Enriching™ Calculus CD-ROM by James Stewart and Dan CleggISBN 0-534-40989-X
TEC provides a laboratory environment in which students can exploreselected topics. TEC also includes homework hints forrepresentative exercises.
Interactive Video SkillBuilder CD-ROM ISBN 0-534-41036-7
Think of it as portable office hours! The Interactive VideoSkillbuilder CD-ROM contains more than eight hours of videoinstruction. The problems worked during each video lesson are shownnext to the viewing screen so that students can try working thembefore watching the solution. To help students evaluate theirprogress, each section contains a ten-question Web quiz (theresults of which can be emailed to the instructor) and each chaptercontains a chapter test, with answers to each problem.
iLrn Student Resource Kit ISBN 0-534-39914-2
This helpful kit provides your students with a CD-ROM that containsthe plug-ins needed to use the iLrn system and a Student Guide thatoffers additional assistance for students using iLrn.
Study Guide, Multivariable
by Robert Burton & Dennis Garity ISBN 0-534-41006-5
Contains key concepts, skills to master, a brief discussion of theideas of the section, and worked-out examples with tips on how tofind the solution.
Student Solutions Manual, Multivariable by Dan Clegg ISBN0-534-41005-7
Provides completely worked-out solutions to all odd- numberedexercises within the text, giving students a way to check theiranswers and ensure that they took the correct steps to arrive at ananswer.
CalcLabs with Maple, Multivariable by Philip Yasskin and ArtBelmonte ISBN 0-534-41010-3
This comprehensive lab manual will help students learn toeffectively use the technology tools available to them. Each labcontains clearly explained exercises and a variety of labs andprojects to accompany the text.
Linear Algebra for Calculus by Konrad J. Heuvers, William P.Francis, John H. Kuisti, Deborah F. Lockhart, Daniel S. Moak, andGene M. Ortner ISBN 0-534-25248-6
This comprehensive book, designed to supplement the calculuscourse, provides an introduction to and review of the basic ideasof linear algebra.
Electronic items Printed items
Acknowledgments
I am grateful to the following reviewers for sharing theirknowledge and judgment with me. I have learned something from eachof them.
William Ardis, Collin County Community College
Judith Broadwin, Jericho High School
Charles Bu, Wellesley University
Robert A. Chaffer, Central Michigan University
Joe W. Fisher, University of Cincinnati
Barry D. Hughes, University of Melbourne
Prem K. Kythe, University of New Orleans
Joyce Riseberg, Montgomery College
Richard Rochberg, Washington University
Denise Taunton Reid, Valdosta State University
Clifton Wingard, Delta State University
Teri E. Woodington, Colorado School of Mines
Second Edition Reviewers
Martina Bode, Northwestern University
Roxanne M. Byrne, University of Colorado at Denver
Larry Cannon, Utah State University
Deborah Troutman Cantrell, Chattanooga State Technical CommunityCollege
Barbara R. Fink, DeAnza College
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PREFACE xvii
xviii PREFACE
I also thank those who have responded to a survey about attitudesto calculus reform:
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PREFACE xix
In addition, I would like to thank George Bergman, Emile LeBlanc,Martin Erick- son, Stuart Goldenberg, Gerald Leibowitz, LarryPeterson, Charles Pugh, Marina Rat- ner, Peter Rosenthal, and AlanWeinstein for their suggestions; Dan Clegg for his
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xx PREFACE
research in libraries and on the Internet; Arnold Good for histreatment of optimiza- tion problems with implicit differentiation;Al Shenk and Dennis Zill for permission to use exercises from theircalculus texts; COMAP for permission to use project mate- rial;George Bergman, David Bleecker, Dan Clegg, John Hagood, VictorKaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz, Lowell Smylie,and Larry Wallen for ideas for exercises; Dan Drucker for theroller derby project; Tom Farmer, Fred Gass, John Ramsay, LarryRiddle, V. K. Srinivasan, and Philip Straffin for ideas forprojects; and Jeff Cole and Dan Clegg for preparing the answermanuscript. I’m grateful to Jeff Cole for suggesting ways toimprove the exercises. Dan Clegg acted as my assistant throughout;he proofread, made suggestions, and contributed some of the newexercises.
In addition, I thank those who have contributed to past editions:Ed Barbeau, Fred Brauer, Andy Bulman-Fleming, Tom DiCiccio, GarretEtgen, Chris Fisher, Gene Hecht, Harvey Keynes, Kevin Kreider, E.L. Koh, Zdislav Kovarik, David Leep, Lothar Redlin, Carl Riehm,Doug Shaw, and Saleem Watson.
I also thank Brian Betsill, Stephanie Kuhns, and Kathi Townes ofTECH-arts for their production services, Tom Bonner for the coverimage, and the following Brooks/ Cole staff: Janet Hill, editorialproduction project manager; Vernon Boes, art director; KarinSandberg, Erin Mitchell, and Bryan Vann, marketing team; EarlPerry, technol- ogy project manager; Stacy Green, assistant editor;Katherine Cook, editorial assistant; Joohee Lee, permissionseditor; Karen Hunt, print/media buyer; and Denise Davidson, coverdesigner. They have all done an outstanding job.
I have been very fortunate to have worked with some of the bestmathematics edi- tors in the business over the past two decades:Ron Munro, Harry Campbell, Craig Barth, Jeremy Hayhurst, GaryOstedt, and now Bob Pirtle. Bob continues in that tra- dition ofeditors who, while offering sound advice and ample assistance,trust my instincts and allow me to write the books that I want towrite.
JAMES STEWART
Calculus Concepts and Contexts 3E
M U L T I V A R I A B L E
Infinite Sequences and Series
Infinite sequences and series were introduced briefly in A Previewof Calculus in connection with Zeno’s paradoxes and the decimalrepresenta- tion of numbers. Their importance in calculus stemsfrom Newton’s idea of representing functions as sums of infi- niteseries. For instance, in finding areas he often integrated afunction by first expressing it as a series and then integratingeach term of the series. We will pursue his idea in Section 8.7 inorder to integrate such functions as
. (Recall that we have previously been unable to do this.) Many ofthe functions that arise in mathematical physics and chemistry,such as Bessel functions, are defined as sums of series, so it isimportant to be familiar with the basic concepts of convergence ofinfinite sequences and series.
Physicists also use series in another way, as we will see inSection 8.9. In studying fields as diverse as optics, specialrelativity, and electromag- netism, they analyze phenomena byreplacing a function with the first few terms in the series thatrepresents it.
ex2
Sequences
A sequence can be thought of as a list of numbers written in adefinite order:
The number is called the first term, is the second term, and ingeneral is the nth term. We will deal exclusively with infinitesequences and so each term will have a successor .
Notice that for every positive integer there is a correspondingnumber and so a sequence can be defined as a function whose domainis the set of positive integers. But we usually write instead ofthe function notation for the value of the func- tion at the number.
NOTATION The sequence { , , , . . .} is also denoted by
EXAMPLE 1 Some sequences can be defined by giving a formula for thenth term. In the following examples we give three descriptions ofthe sequence: one by using the preceding notation, another by usingthe defining formula, and a third by writing out the terms of thesequence. Notice that doesn’t have to start at 1.
(a)
(b)
(c)
(d)
EXAMPLE 2 Find a formula for the general term of the sequence
assuming that the pattern of the first few terms continues.
SOLUTION We are given that
Notice that the numerators of these fractions start with 3 andincrease by 1 whenever we go to the next term. The second term hasnumerator 4, the third term has numer- ator 5; in general, the thterm will have numerator . The denominators are n 2n
a 5 7
3125 a 4
{0, 1, s2, s3, . . . , sn 3, . . .}an sn 3, n 3{sn 3}n3
2
3 ,
3
9 ,
4
27 ,
5
81 , . . . ,
1nn 1 3n1nn 1
3n 1
a3a2a1
8.1
557
the powers of 5, so has denominator . The signs of the terms arealternately positive and negative, so we need to multiply by apower of . In Example 1(b) the factor meant we started with anegative term. Here we want to start with a positive term and so weuse or . Therefore,
EXAMPLE 3 Here are some sequences that don’t have a simple definingequation.
(a) The sequence , where is the population of the world as ofJanuary 1 in the year .
(b) If we let be the digit in the nth decimal place of the number ,then is a well-defined sequence whose first few terms are
(c) The Fibonacci sequence is defined recursively by theconditions
Each term is the sum of the two preceding terms. The first fewterms are
This sequence arose when the 13th-century Italian mathematicianknown as Fibonacci solved a problem concerning the breeding ofrabbits (see Exercise 39).
. . . . . .
From Figure 1 or 2 it appears that the terms of the sequence areapproaching 1 as becomes large. In fact, the difference
can be made as small as we like by taking sufficiently large. Weindicate this by writing
In general, the notation
means that the terms of the sequence approach as becomes large.Notice that the following definition of the limit of a sequence isvery similar to the definition of a limit of a function at infinitygiven in Section 2.5.
nLan
n, an 3, a32, a21, a1
an nn 1
fn
7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . .
an ean
5 n
1 5 nan
0 11
7
Definition A sequence has the limit and we write
if we can make the terms as close to as we like by takingsufficiently large. If exists, we say the sequence converges (or isconvergent). Otherwise, we say the sequence diverges (or isdivergent).
Figure 3 illustrates Definition 1 by showing the graphs of twosequences that have the limit .
If you compare Definition 1 with Definition 2.5.4 you will see thatthe only differ- ence between and is that is required to be aninte- ger. Thus, we have the following theorem, which isillustrated by Figure 4.
Theorem If and when is an integer, then .
In particular, since we know from Section 2.5 that when wehave
if
If becomes large as n becomes large, we use the notation
In this case the sequence is divergent, but in a special way. Wesay that di- verges to .
The Limit Laws given in Section 2.3 also hold for the limits ofsequences and their proofs are similar.
an an
FIGURE 4
y=ƒ
20 x
L
limn l an L nf n anlimx l f x L2
nlimx l f x Llimn l an L
0 n
an
L
FIGURE 3 Graphs of two sequences with lim an= L n `
L
an L
SECTION 8.1 SEQUENCES 559
A more precise definition of the limit of a sequence is given inAppendix D.
If and are convergent sequences and is a constant, then
The Squeeze Theorem can also be adapted for sequences as follows(see Figure 5).
If for and , then .
Another useful fact about limits of sequences is given by thefollowing theorem, which follows from the Squeeze Theorem because.
Theorem If , then .
EXAMPLE 4 Find .
SOLUTION The method is similar to the one we used in Section 2.5:Divide numerator and denominator by the highest power of thatoccurs in the denominator and then use the Limit Laws.
Here we used Equation 3 with .
EXAMPLE 5 Calculate .lim n l
ln n
lim n l
an p [lim
lim n l
an
an lim n l
an lim n l
Limit Laws for Convergent Sequences
Squeeze Theorem for Sequences
This shows that the guess we made earlier from Figures 1 and 2 wascorrect.
FIGURE 5 The sequence bn is squeezed between the sequences an andcn.
0 n
cn
an
bn
SOLUTION Notice that both numerator and denominator approachinfinity as . We can’t apply l’Hospital’s Rule directly because itapplies not to sequences but to functions of a real variable.However, we can apply l’Hospital’s Rule to the related function andobtain
Therefore, by Theorem 2 we have
EXAMPLE 6 Determine whether the sequence is convergent ordivergent.
SOLUTION If we write out the terms of the sequence, we obtain
The graph of this sequence is shown in Figure 6. Since the termsoscillate between 1 and infinitely often, does not approach anynumber. Thus, does not exist; that is, the sequence isdivergent.
EXAMPLE 7 Evaluate if it exists.
SOLUTION
EXAMPLE 8 Discuss the convergence of the sequence , where .
SOLUTION Both numerator and denominator approach infinity as buthere we have no corresponding function for use with l’Hospital’sRule ( is not defined when is not an integer). Let’s write out afew terms to get a feeling for what happens to as gets large:
It appears from these expressions and the graph in Figure 8 thatthe terms are decreasing and perhaps approach 0. To confirm this,observe from Equation 5 that
an 1
a3 1 2 3
3 3 3 a2
lim n l
1, 1, 1, 1, 1, 1, 1, . . .
an 1n
lim n l
n l
FIGURE 6
The graph of the sequence in Example 7 is shown in Figure 7 andsupports the answer.
FIGURE 7
0 n
an
1
1
_1
Notice that the expression in parentheses is at most 1 because thenumerator is less than (or equal to) the denominator. So
We know that as . Therefore, as by the Squeeze Theorem.
EXAMPLE 9 For what values of is the sequence convergent?
SOLUTION We know from Section 2.5 and the graphs of the exponentialfunctions in Section 1.5 that for and for . There- fore, puttingand using Theorem 2, we have
For the cases and we have
and
If , then , so
and therefore by Theorem 4. If , then diverges as in Example 6.Figure 9 shows the graphs for various values of . (The case isshown in Figure 6.)
The results of Example 9 are summarized for future use asfollows.
The sequence is convergent if and divergent for all other values of.
Definition A sequence is called increasing if for all , that is, Itis called decreasing if for all . A sequence is monotonic if it iseither increasing or decreasing.
n 1an an1a1 a2 a3 . n 1an an1an
lim n l
r n 0
r>1
r=1
FIGURE 9 The sequence an=rn
r 1r r n r 1lim n l r n 0
lim n l
r n 0
lim n l
0 0lim n l
1n lim n l
if 0 r 1
a r 0 a 1limx l ax 0a 1limx l ax
r n r
n l an l 0n l 1n l 0
0 an 1
562 CHAPTER 8 INFINITE SEQUENCES AND SERIES
CREATING GRAPHS OF SEQUENCES Some computer algebra systems havespecial commands that enable us to create sequences and graph themdirectly. With most graphing calculators, however, sequences can begraphed by using parametric equations. For instance, the sequencein Example 8 can be graphed by entering the parametricequations
and graphing in dot mode starting with , setting the -step equal to. The
result is shown in Figure 8. 1tt 1
x t y t!t t
FIGURE 8
and so for all .
EXAMPLE 11 Show that the sequence is decreasing.
SOLUTION 1 We must show that , that is,
This inequality is equivalent to the one we get bycross-multiplication:
Since , we know that the inequality is true. Therefore, and so isdecreasing.
SOLUTION 2 Consider the function :
Thus, is decreasing on and so . Therefore, is decreasing.
Definition A sequence is bounded above if there is a number suchthat
It is bounded below if there is a number such that
If it is bounded above and below, then is a bounded sequence.
For instance, the sequence is bounded below but not above. Thesequence is bounded because for all .n0 an 1an nn 1
an 0an n
m
Man
whenever x2 1f x x 2 1 2x 2
x 2 12 1 x 2
x 2 12 0
f x x
x 2 1
1 n2 n&?
n3 n2 n 1 n3 2n2 2n&?
n 1n2 1 nn 12 1 &? n 1
n 12 1
n 5 SECTION 8.1 SEQUENCES 563
The right side is smaller because it has a largerdenominator.
We know that not every bounded sequence is convergent [forinstance, the se- quence satisfies but is divergent from Example 6]and not every monotonic sequence is convergent . But if a sequenceis both bounded and monotonic, then it must be convergent. Thisfact is stated without proof as Theorem 7, but intuitively you canunderstand why it is true by looking at Fig- ure 10. If isincreasing and for all , then the terms are forced to crowdtogether and approach some number .
Monotonic Sequence Theorem Every bounded, monotonic sequence isconvergent.
EXAMPLE 12 Investigate the sequence defined by the recurrencerelation
SOLUTION We begin by computing the first several terms:
These initial terms suggest that the sequence is increasing and theterms are approaching 6. To confirm that the sequence isincreasing, we use mathematical induction to show that for all .This is true for because
. If we assume that it is true for , then we have
so
and
Thus
We have deduced that is true for . Therefore, the inequality istrue for all by induction.
Next we verify that is bounded by showing that for all . (Since thesequence is increasing, we already know that it has a lower bound:for all .) We know that , so the assertion is true for . Suppose itis true for
. Then
so
and
Thus
This shows, by mathematical induction, that for all .nan 6
ak1 6
ak 6 12
an a1 2 nan 6an
n n k 1an1 an
ak2 ak1
ak1 6 ak 6
ak1 ak
a9 5.984375 a8 5.96875 a7 5.9375
a6 5.875 a5 5.75 a4 1 2 5 6 5.5
a3 1 2 4 6 5 a2 1
2 2 6 4 a1 2
for n 1, 2 , 3, . . .an1 1 2 an 6a1 2
an
7
564 CHAPTER 8 INFINITE SEQUENCES AND SERIES
20 n
FIGURE 10
Mathematical induction is often used in dealing with recursivesequences. See page 87 for a discussion of the Principle ofMathematical Induction.
Since the sequence is increasing and bounded, the MonotonicSequence Theorem guarantees that it has a limit. The theoremdoesn’t tell us what the value of the limit is. But now that weknow exists, we can use the given recurrence relation towrite
Since , it follows that too (as , also). So we have
Solving this equation for , we get , as we predicted.L 6L
L 1 2 L 6
n 1 l n l an1 l Lan l L
lim n l
12 an 6 1 2 ( lim
n l an 6) 1
2 L 6
; 29–34 Use a graph of the sequence to decide whether the sequenceis convergent or divergent. If the sequence is conver- gent, guessthe value of the limit from the graph and then prove your guess.(See the margin note on page 562 for advice on graphingsequences.)
29. 30.
31. 32.
33. 34.
35. If $1000 is invested at 6% interest, compounded annually, thenafter years the investment is worth dollars. (a) Find the firstfive terms of the sequence . (b) Is the sequence convergent ordivergent? Explain.
36. Find the first 40 terms of the sequence defined by
and . Do the same if . Make a conjecture about this type ofsequence.
a1 25a1 11
3an 1
an
n3
n!
n 1
an ln n2
n 0, 1, 0, 0, 1, 0, 0, 0, 1, . . . 25.
an sin 2n
2n21. 1. (a) What is a sequence?
(b) What does it mean to say that ? (c) What does it mean to saythat ?
2. (a) What is a convergent sequence? Give two examples. (b) Whatis a divergent sequence? Give two examples.
List the first six terms of the sequence defined by
Does the sequence appear to have a limit? If so, find it.
4. List the first nine terms of the sequence . Does this sequenceappear to have a limit? If so, find it. If not, explain why.
5–8 Find a formula for the general term of the sequence, assumingthat the pattern of the first few terms continues.
6.
7. 8.
9–28 Determine whether the sequence converges or diverges. If itconverges, find the limit.
10.
an cos2n e n e n
e 2n 1 an
5, 1, 5, 1, 5, 1, . . .2, 7, 12, 17, . . .
{ 1 4 , 29 , 3
an
Exercises8.1
is increasing and for all . Deduce that is con- vergent and findits limit.
48. Show that the sequence defined by
satisfies and is decreasing. Deduce that the sequence is convergentand find its limit.
.
estimating the value of to five decimal places.
51. The size of an undisturbed fish population has been modeled bythe formula
where is the fish population after years and and are positiveconstants that depend on the species and its environ- ment. Supposethat the population in year 0 is . (a) Show that if is convergent,then the only possible
values for its limit are 0 and . (b) Show that . (c) Use part (b)to show that if , then ;
in other words, the population dies out. (d) Now assume that . Showthat if , then
is increasing and . Show also that if , then is decreasing and .Deduce that if , then .
52. A sequence is defined recursively by
Find the first eight terms of the sequence . What do you noticeabout the odd terms and the even terms? By consider- ing the oddand even terms separately, show that is convergent and deducethat
This gives the continued fraction expansion
s2 1 1
an1 1 1
1 an a1 1
limn l pn b aa b pn b a pnp 0 b a
0 pn b a pn p 0 b aa b
limn l pn 0a b pn1 bapn
b a pn
f L Llimn l an L fan1 f an
a3 f a2 f f aa2 f aa1 a
0.8n 0.000001 n
0 an 2
an1 1
3 an a1 2
an nan 3(a) Determine whether the sequence defined as follows isconvergent or divergent:
(b) What happens if the first term is ?
38. (a) If , what is the value of ? (b) A sequence is definedby
Find the first ten terms of the sequence correct to five decimalplaces. Does it appear that the sequence is con- vergent? If so,estimate the value of the limit to three decimal places.
(c) Assuming that the sequence in part (b) has a limit, use part(a) to find its exact value. Compare with your esti- mate from part(b).
39. (a) Fibonacci posed the following problem: Suppose that rabbitslive forever and that every month each pair pro- duces a new pairwhich becomes productive at age 2 months. If we start with onenewborn pair, how many pairs of rabbits will we have in the month?Show that the answer is , where is the Fibonacci sequence definedin Example 3(c).
(b) Let and show that . Assuming that is convergent, find itslimit.
40. Find the limit of the sequence
41–44 Determine whether the sequence is increasing, decreasing, ornot monotonic. Is the sequence bounded?
42.
Suppose you know that is a decreasing sequence and all its termslie between the numbers 5 and 8. Explain why the sequence has alimit. What can you say about the value of the limit?
46. A sequence is given by , . (a) By induction or otherwise, showthat is increasing
and bounded above by 3. Apply the Monotonic Sequence Theorem toshow that exists.
(b) Find .
an1 3 1
a1 s2an
an 2n 3
3n 4 an
an an1 1 1an2an fn1fn
fn fn
a1 1 an1 11 an for n 1
an limn l an1limn l an L
a1 2
37.
SECTION 8.2 SERIES 567
Logistic Sequences
A sequence that arises in ecology as a model for population growthis defined by the logis- tic difference equation
where measures the size of the population of the generation of asingle species. To keep the numbers manageable, is a fraction ofthe maximal size of the population, so
. Notice that the form of this equation is similar to the logisticdifferential equa- tion in Section 7.5. The discrete model—withsequences instead of continuous functions—is preferable formodeling insect populations, where mating and death occur in aperiodic fashion.
An ecologist is interested in predicting the size of the populationas time goes on, and asks these questions: Will it stabilize at alimiting value? Will it change in a cyclical fashion? Or will itexhibit random behavior?
Write a program to compute the first terms of this sequencestarting with an initial population . Use this program to do thefollowing.
1. Calculate 20 or 30 terms of the sequence for and for two valuesof such that . Graph the sequences. Do they appear to converge?Repeat for a different
value of between 0 and 1. Does the limit depend on the choice of ?Does it depend on the choice of ?
2. Calculate terms of the sequence for a value of between 3 and 3.4and plot them. What do you notice about the behavior of theterms?
3. Experiment with values of between 3.4 and 3.5. What happens tothe terms?
4. For values of between 3.6 and 4, compute and plot at least 100terms and comment on the behavior of the sequence. What happens ifyou change by 0.001? This type of behavior is called chaotic and isexhibited by insect populations under certain conditions.
p0
k
k
k
2
0 pn 1 pn
CAS
L A B O R A T O R Y P R O J E C T
Series
If we try to add the terms of an infinite sequence we get anexpression of the form
which is called an infinite series (or just a series) and isdenoted, for short, by the symbol
But does it make sense to talk about the sum of infinitely manyterms? It would be impossible to find a finite sum for theseries
because if we start adding the terms we get the cumulative sums 1,3, 6, 10, 15, 21, . . . and, after the term, we get , which becomesvery large as increases.
nnn 12nth
anor
an n1
8.2
However, if we start to add the terms of the series
we get , , , , , , . . . , , . . . . The table shows that as we addmore and more terms, these partial sums become closer and closer to1. (See also Figure 11 in A Preview of Calculus, page 7.) In fact,by adding sufficiently many terms of the series we can make thepartial sums as close as we like to 1. So it seems reasonable tosay that the sum of this infinite series is 1 and to write
We use a similar idea to determine whether or not a general series(1) has a sum. We consider the partial sums
and, in general,
These partial sums form a new sequence , which may or may not havea limit. If exists (as a finite number), then, as in the precedingexample, we call it
the sum of the infinite series .
Definition Given a series , let denote its th partial sum:
If the sequence is convergent and exists as a real number, then theseries is called convergent and we write
The number is called the sum of the series. If the sequence isdivergent, then the series is called divergent.
an
sn n
n sn
an
i1 ai
s3 a1 a2 a3
n Sum of first n terms
1 0.50000000 2 0.75000000 3 0.87500000 4 0.93750000 5 0.96875000 60.98437500 7 0.99218750
10 0.99902344 15 0.99996948 20 0.99999905 25 0.99999997
Compare with the improper integral
To find this integral we integrate from 1 to and then let . For aseries, we sum from 1 to
and then let .n l n t l
t
y
t l y
1 f x dx
EXAMPLE 1 An important example of an infinite series is thegeometric series
Each term is obtained from the preceding one by multiplying it bythe common ratio . (We have already considered the special casewhere and .)
If , then . Since doesn’t exist, the geometric series diverges inthis case.
If , we have
If , we know from (8.1.6) that as , so
Thus, when the geometric series is convergent and its sum is . Ifor , the sequence is divergent by (8.1.6) and so, by Equation3,
does not exist. Therefore, the geometric series diverges in thosecases.
We summarize the results of Example 1 as follows.
The geometric series
If , the geometric series is divergent.
EXAMPLE 2 Find the sum of the geometric series
SOLUTION The first term is and the common ratio is . Since , theseries is convergent by (4) and its sum is
5 10
4
lim n l
rn a
1 r
sn a1 rn
r 1
lim n l snsn a a a na l r 1 r 1
2a 1 2
n1 arn1
SECTION 8.2 SERIES 569
Figure 1 provides a geometric demonstra- tion of the result inExample 1. If the triangles are constructed as shown and is the sumof the series, then, by similar triangles,
s
a
a
aa
a
ara-ar
ar
ar@
ar#
ar@
s
In words: The sum of a convergent geo- metric series is
first term
EXAMPLE 3 Is the series convergent or divergent?
SOLUTION Let’s rewrite the nth term of the series in the form:
We recognize this series as a geometric series with and . Since ,the series diverges by (4).
EXAMPLE 4 Write the number . . . as a ratio of integers.
SOLUTION
After the first term we have a geometric series with and .Therefore
EXAMPLE 5 Find the sum of the series , where .
SOLUTION Notice that this series starts with and so the first termis . (With series, we adopt the convention that even when .)Thus
n0 xn 1 x x 2 x 3 x 4
x 0x 0 1 x 0 1n 0
x 1
2.3171717. . . 2.3 17
n
1 5.000000 2 1.666667 3 3.888889 4 2.407407 5 3.395062 6 2.736626 73.175583 8 2.882945 9 3.078037
10 2.947975
sn
What do we really mean when we say that the sum of the series inExample 2 is ? Of course, we can’t literally add an infinite numberof terms, one by one. But, according to Definition 2, the total sumis the limit of the sequence of partial sums. So, by taking the sumof sufficiently many terms, we can get as close as we like to thenumber . The table shows the first ten partial sums and the graphin Figure 2 shows how the sequence of partial sums approaches.3
sn
3
3
Another way to identify and is to write out the first fewterms:
4 16 3
ra
EXAMPLE 6 Show that the series is convergent, and find itssum.
SOLUTION This is not a geometric series, so we go back to thedefinition of a conver- gent series and compute the partialsums.
We can simplify this expression if we use the partial fractiondecomposition
(see Section 5.7). Thus, we have
and so
is divergent.
SOLUTION For this particular series it’s convenient to consider thepartial sums , , , and show that they become large.
1 1 2
3 1 4 ) ( 1
5 1 8 ) ( 1
3 1 4 ) ( 1
3 1 4 ) 1
1 1
n 1
1 1
2 1
SECTION 8.2 SERIES 571
Notice that the terms cancel in pairs. This is an example of atelescoping sum: Because of all the cancellations, the sumcollapses (like a pirate’s collapsing telescope) into just twoterms.
Figure 3 illustrates Example 6 by show- ing the graphs of thesequence of terms
and the sequence of partial sums. Notice that and
. See Exercises 46 and 47 for two geometric interpretations ofExample 6. sn l 1
an l 0 sn an 1[nn 1]
FIGURE 3
Similarly, , , and in general
This shows that as and so is divergent. Therefore, the harmonicseries diverges.
Theorem If the series is convergent, then .
Proof Let . Then . Since is convergent, the sequence is convergent.Let . Since as , we also have . Therefore
NOTE 1 With any series we associate two sequences: the sequence ofits partial sums and the sequence of its terms. If is convergent,then the limit of the sequence is (the sum of the series) and, asTheorem 6 asserts, the limit of the sequence is 0.
| NOTE 2 The converse of Theorem 6 is not true in general. If , wecannot conclude that is convergent. Observe that for the harmonicseries we have as , but we showed in Example 7 that isdivergent.
The Test for Divergence If does not exist or if , then the
series is divergent.
The Test for Divergence follows from Theorem 6 because, if theseries is not diver- gent, then it is convergent, and so .
EXAMPLE 8 Show that the series diverges.
SOLUTION
So the series diverges by the Test for Divergence.
NOTE 3 If we find that , we know that is divergent. If we find that, we know nothing about the convergence or divergence of . Rememberthe warning in Note 2: If , the series might converge or it mightdiverge.
anlim n l an 0 anlim n l an 0
anlim n l an 0
lim n l
5 0
an7
1nn l an 1n l 0 1n an
lim n l an 0
an ssn
sn lim n l
sn1
lim n l sn1 s n l n 1 l lim n l sn ssn
anan sn sn1sn a1 a2 an
lim n l
s2n 1 n
5 2
572 CHAPTER 8 INFINITE SEQUENCES AND SERIES
The method used in Example 7 for show- ing that the harmonic seriesdiverges is due to the French scholar Nicole Oresme(1323–1382).
Theorem If and are convergent series, then so are the series (whereis a constant), , and , and
(i) (ii)
(iii)
These properties of convergent series follow from the correspondingLimit Laws for Convergent Sequences in Section 8.1. For instance,here is how part (ii) of Theo- rem 8 is proved:
Let
and, using Equation 5.2.9, we have
Therefore, is convergent and its sum is
EXAMPLE 9 Find the sum of the series .
SOLUTION The series is a geometric series with and , so
In Example 6 we found that
So, by Theorem 8, the given series is convergent and
3 1 1 4
n l
SECTION 8.2 SERIES 573
NOTE 4 A finite number of terms doesn’t affect the convergence ordivergence of a series. For instance, suppose that we were able toshow that the series
is convergent. Since
it follows that the entire series is convergent. Similarly, if itis known that the series converges, then the full series
is also convergent.
13. 14.
17–26 Determine whether the series is convergent or diver- gent. Ifit is convergent, find its sum.
17. 18.
19. 20.
21. 22.
23. 24.
27–30 Determine whether the series is convergent or diver- gent byexpressing as a telescoping sum (as in Example 6). If it isconvergent, find its sum.
28.
31–34 Express the number as a ratio of integers.
32.
2 n
3 n
5 n1
n1 5( 2
3 )n11. (a) What is the difference between a sequence and a series?(b) What is a convergent series? What is a divergent series?
2. Explain what it means to say that .
; 3–8 Find at least 10 partial sums of the series. Graph both thesequence of terms and the sequence of partial sums on the samescreen. Does it appear that the series is convergent or divergent?If it is convergent, find the sum. If it is divergent, explainwhy.
4.
(a) Determine whether is convergent. (b) Determine whether isconvergent.
10. (a) Explain the difference between
(b) Explain the difference between
11–16 Determine whether the geometric series is convergent ordivergent. If it is convergent, find its sum.
11.
5 10 3
Exercises8.2
(b) Calculate the total time that the ball travels. (c) Supposethat each time the ball strikes the surface
with velocity it rebounds with velocity , where . How long will ittake for the ball to come
to rest?
What is the value of if ?
; 46. Graph the curves , , for on a common screen. By finding theareas between
successive curves, give a geometric demonstration of the fact,shown in Example 6, that
47. The figure shows two circles and of radius 1 that touch at . isa common tangent line; is the circle that touches , , and ; is thecircle that touches , , and ; is the circle that touches , , and .This procedure can be continued indefinitely and produces aninfinite sequence of circles . Find an expression for the diameterof and thus provide another geometric demonstration of Example6.
48. A right triangle is given with and . is drawn perpendicular to, is drawn perpendicu-
lar to , , and this process is continued indefi- nitely as shown inthe figure. Find the total length of all the perpendiculars
in terms of and .
0 k 1 kvv
35–37 Find the values of for which the series converges. Find thesum of the series for those values of .
36.
37.
38. We have seen that the harmonic series is a divergent serieswhose terms approach 0. Show that
is another series with this property.
39–40 Use the partial fraction command on your CAS to find aconvenient expression for the partial sum, and then use thisexpression to find the sum of the series. Check your answer byusing the CAS to sum the series directly.
39. 40.
find and .
42. If the partial sum of a series is , find and .
43. When money is spent on goods and services, those that receivethe money also spend some of it. The people receiv- ing some of thetwice-spent money will spend some of that, and so on. Economistscall this chain reaction the multiplier effect. In a hypotheticalisolated community, the local gov- ernment begins the process byspending dollars. Suppose that each recipient of spent money spendsand saves
of the money that he or she receives. The values and s are calledthe marginal propensity to consume and the marginal propensity tosave and, of course, . (a) Let be the total spending that has beengenerated after
transactions. Find an equation for . (b) Show that , where . Thenumber
is called the multiplier. What is the multiplier if the marginalpropensity to consume is ?
Note: The federal government uses this principle to justify deficitspending. Banks use this principle to justify lend- ing a largepercentage of the money that they receive in deposits.
44. A certain ball has the property that each time it falls from aheight onto a hard, level surface, it rebounds to a height
, where . Suppose that the ball is dropped from an initial heightof meters. (a) Assuming that the ball continues to bounceindefinitely,
find the total distance that it travels. (Use the fact that theball falls in .)t seconds1
2 tt 2 meters
H 0 r 1rh
Sn
n1 anan
sn n 1
n2 n2
2n
SECTION 8.2 SERIES 575
(b) The Sierpinski carpet is a two-dimensional counterpart of theCantor set. It is constructed by removing the cen- ter one-ninth ofa square of side 1, then removing the centers of the eight smallerremaining squares, and so on. (The figure shows the first threesteps of the construction.) Show that the sum of the areas of theremoved squares is 1. This implies that the Sierpinski carpet hasarea 0.
56. (a) A sequence is defined recursively by the equation for ,where and can be
any real numbers. Experiment with various values of and and useyour calculator to guess the limit of the sequence.
(b) Find in terms of and by expressing in terms of and summing aseries.
57. Consider the series
(a) Find the partial sums and . Do you recognize the denominators?Use the pattern to guess a formula for .
(b) Use mathematical induction to prove your guess. (c) Show thatthe given infinite series is convergent, and
find its sum.
58. In the figure there are infinitely many circles approaching thevertices of an equilateral triangle, each circle touching othercircles and sides of the triangle. If the triangle has sides oflength 1, find the total area occupied by the circles.
sn
an
What is wrong with the following calculation?
(Guido Ubaldus thought that this proved the existence of Godbecause “something has been created out of nothing.”)
50. Suppose that is known to be a convergent series. Prove that isa divergent series.
51. If is convergent and is divergent, show that the series isdivergent. [Hint: Argue by contradiction.]
52. If and are both divergent, is necessar- ily divergent?
Suppose that a series has positive terms and its partial sumssatisfy the inequality for all . Explain why must beconvergent.
54. The Fibonacci sequence was defined in Section 8.1 by theequations
Show that each of the following statements is true.
(a)
(b)
(c)
The Cantor set, named after the German mathematician Georg Cantor(1845–1918), is constructed as follows. We start with the closedinterval and remove the open interval . That leaves the twointervals and and we remove the open middle third of each. Fourintervals remain and again we remove the open middle third of eachof them. We continue this procedure indefinitely, at each stepremoving the open middle third of every interval that remains fromthe preceding step. The Cantor set consists of the numbers thatremain in after all those intervals have been removed. (a) Showthat the total length of all the intervals that are
removed is 1. Despite that, the Cantor set contains infi- nitelymany numbers. Give examples of some numbers in the Cantorset.
[0, 1]
3, 23 ) [0, 1]
1
an
1 1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
0 0 0 0
576 CHAPTER 8 INFINITE SEQUENCES AND SERIES
SECTION 8.3 THE INTEGRAL AND COMPARISON TESTS; EST IMATING SUMS577
The Integral and Comparison Tests; Estimating Sums
In general, it is difficult to find the exact sum of a series. Wewere able to accomplish this for geometric series and the seriesbecause in each of those cases we could find a simple formula forthe partial sum . But usually it isn’t easy to compute . Therefore,in this section and the next we develop tests that enable us todetermine whether a series is convergent or divergent withoutexplicitly finding its sum. In some cases, however, our methodswill enable us to find good esti- mates of the sum.
In this section we deal only with series with positive terms, sothe partial sums are increasing. In view of the Monotonic SequenceTheorem, to decide whether a series is convergent or divergent, weneed to determine whether the partial sums are bounded ornot.
Testing with an Integral
Let’s investigate the series whose terms are the reciprocals of thesquares of the posi- tive integers:
There’s no simple formula for the sum of the first terms, but thecomputer- generated table of values given in the margin suggeststhat the partial sums are ap- proaching a number near 1.64 as andso it looks as if the series is convergent.
We can confirm this impression with a geometric argument. Figure 1shows the curve and rectangles that lie below the curve. The baseof each rectangle is an interval of length 1; the height is equalto the value of the function at the right endpoint of the interval.So the sum of the areas of the rectangles is
If we exclude the first rectangle, the total area of the remainingrectangles is smaller than the area under the curve for , which isthe value of the integral . In Section 5.10 we discovered that thisimproper integral is con- vergent and has value 1. So the pictureshows that all the partial sums are less than
1
1 1x 2 dx x 1y 1x 2
FIGURE 1
y= 1
n l
8.3
n
100 1.6350 500 1.6429
1000 1.6439 5000 1.6447
i 2
Thus, the partial sums are bounded and the series converges. Thesum of the series (the limit of the partial sums) is also less than2:
[The exact sum of this series was found by the Swiss mathematicianLeonhard Euler (1707–1783) to be , but the proof of this fact isbeyond the scope of this book.]
Now let’s look at the series
The table of values of suggests that the partial sums aren’tapproaching a finite num- ber, so we suspect that the given seriesmay be divergent. Again we use a picture for confirmation. Figure 2shows the curve , but this time we use rectangles whose tops lieabove the curve.
The base of each rectangle is an interval of length 1. The heightis equal to the value of the function at the left endpoint of theinterval. So the sum of the areas of all the rectangles is
This total area is greater than the area under the curve for ,which is equal to the integral . But we know from Section 5.10 thatthis improper integral is divergent. In other words, the area underthe curve is infinite. So the sum of the series must be infinite,that is, the series is divergent.
The same sort of geometric reasoning that we used for these twoseries can be used to prove the following test.
The Integral Test Suppose is a continuous, positive, decreasingfunction on and let . Then the series is convergent if and onlyif
the improper integral is convergent. In other words:
(a) If is convergent, then is convergent.
(b) If is divergent, then is divergent.
n1 any
x
1
s1
1
s2
1
s3
1
s4
1
s5
area= 1
n
100 18.5896 500 43.2834
1000 61.8010 5000 139.9681
1
si
NOTE When we use the Integral Test it is not necessary to start theseries or the inte- gral at . For instance, in testing theseries
Also, it is not necessary that be always decreasing. What isimportant is that be ultimately decreasing, that is, decreasing forlarger than some number . Then
is convergent, so is convergent by Note 4 of Section 8.2.
EXAMPLE 1 Determine whether the series converges or diverges.
SOLUTION The function is positive and continuous for because thelogarithm function is continuous. But it is not obvious whether ornot is decreasing, so we compute its derivative:
Thus, when , that is, . It follows that is decreasing when and sowe can apply the Integral Test:
Since this improper integral is divergent, the series is alsodivergent by the Integral Test.
EXAMPLE 2 For what values of is the series convergent?
SOLUTION If , then . If , then . In either case , so the givenseries diverges by the Test for Divergence [see (8.2.7)].
If , then the function is clearly continuous, positive, anddecreasing on . We found in Chapter 5 [see (5.10.2)] that
It follows from the Integral Test that the series converges if anddiverges if . (For , this series is the harmonic series discussedin Example 7 in Section 8.2.)
The series in Example 2 is called the p-series. It is important inthe rest of this chapter, so we summarize the results of Example 2for future reference as follows.
The -series is convergent if and divergent if .p 1p 1
n1
y
1
1
xp dx converges if p 1 and diverges if p 1
1, f x 1xpp 0
2 1
f x x1x ln x
x 2 1 ln x
x 2
n4
n 1
SECTION 8.3 THE INTEGRAL AND COMPARISON TESTS; EST IMATING SUMS579
In order to use the Integral Test we need to be able to evaluateand therefore we have to be able to find an antiderivative of .Frequently this is difficult or impossible, so we need other testsfor convergence too.
f
For instance, the series
is convergent because it is a p-series with . But the series
is divergent because it is a p-series with .
Testing by Comparing
The series
reminds us of the series , which is a geometric series with and andis therefore convergent. Because the series (2) is so similar to aconvergent series, we have the feeling that it too must beconvergent. Indeed, it is. The inequality
shows that our given series (2) has smaller terms than those of thegeometric series and therefore all its partial sums are alsosmaller than 1 (the sum of the geometric series). This means thatit* partial sums form a bounded increasing sequence, which isconvergent. It also follows that the sum of the series is less thanthe sum of the geometric series:
Similar reasoning can be used to prove the following test, whichapplies only to series whose terms are positive. The first partsays that if we have a series whose terms are smaller than those ofa known convergent series, then our series is also convergent. Thesecond part says that if we start with a series whose terms arelarger than those of a known divergent series, then it too isdivergent.
The Comparison Test Suppose that and are series with positiveterms.
(a) If is convergent and for all , then is also convergent.
(b) If is divergent and for all , then is also divergent.
In using the Comparison Test we must, of course, have some knownseries for the purpose of comparison. Most of the time we use oneof these series:
A -series [ converges if and diverges if ; see (1)] A geometricseries [ converges if and diverges if ;
see (8.2.4)] r 1 r 1 arn1
p 1p 1 1npp
bn
Standard Series for Use with the Comparison Test
EXAMPLE 3 Determine whether the series converges or diverges.
SOLUTION For large the dominant term in the denominator is , so wecompare the given series with the series . Observe that
because the left side has a bigger denominator. (In the notation ofthe Comparison Test, is the left side and is the right side.) Weknow that
is convergent ( -series with ). Therefore
is convergent by part (a) of the Comparison Test.
Although the condition or in the Comparison Test is given for all ,we need verify only that it holds for , where is some fixedinteger, because
the convergence of a series is not affected by a finite number ofterms. This is illus- trated in the next example.
EXAMPLE 4 Test the series for convergence or divergence.
SOLUTION We used the Integral Test to test this series in Example1, but we can also test it by comparing it with the harmonicseries. Observe that for and so
We know that is divergent ( -series with ). Thus, the given seriesis divergent by the Comparison Test.
NOTE The terms of the series being tested must be smaller thanthose of a conver- gent series or larger than those of a divergentseries. If the terms are larger than the terms of a convergentseries or smaller than those of a divergent series, then the Com-parison Test doesn’t apply. Consider, for instance, theseries
The inequality
is useless as far as the Comparison Test is concerned because iscon- vergent and . Nonetheless, we have the feeling that ought tobe 12n 1an bn
bn ( 1 2 )n
2n 2 4n 3
SECTION 8.3 THE INTEGRAL AND COMPARISON TESTS; EST IMATING SUMS581
convergent because it is very similar to the convergent geometricseries . In such cases the following test can be used.
The Limit Comparison Test Suppose that and are series with positiveterms. If
where c is a finite number and , then either both series convergeor both diverge.
Although we won’t prove the Limit Comparison Test, it seemsreasonable because for large .
EXAMPLE 5 Test the series for convergence or divergence.
SOLUTION We use the Limit Comparison Test with
and obtain
Since this limit exists and is a convergent geometric series, thegiven series converges by the Limit Comparison Test.
Estimating the Sum of a Series
Suppose we have been able to use the Integral Test to show that aseries is con- vergent and we now want to find an approximation tothe sum of the series. Of course, any partial sum is anapproximation to because . But how good is such an approximation?To find out, we need to estimate the size of the remainder
The remainder is the error made when , the sum of the first terms,is used as an approximation to the total sum.
We use the same notation and ideas as in the Integral Test,assuming that is decreasing on . Comparing the areas of therectangles with the area under
for in Figure 3, we see that
Similarly, we see from Figure 4 that
Rn an1 an2 y
n1 f x dx
Rn an1 an2 y
n f x dx
f
nsnRn
limn l sn sssn
2n
bn 1
2nan 1
2n 1
FIGURE 3
0 x
So we have proved the following error estimate.
Remainder Estimate for the Integral Test Suppose , where is a con-tinuous, positive, decreasing function for and is convergent.If
, then
EXAMPLE 6 (a) Approximate the sum of the series by using the sum ofthe first 10 terms. Estimate the error involved in thisapproximation. (b) How many terms are required to ensure that thesum is accurate to within ?
SOLUTION In both parts (a) and (b) we need to know . With , whichsatisfies the conditions of the Integral Test, we have
(a)
According to the remainder estimate in (3), we have
So the size of the error is at most .
(b) Accuracy to within means that we have to find a va
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