Calculus Concepts and Contexts 4th Edition Stewart Solutions Manual

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COMPLETE SOLUTIONS MANUAL for Stewart’s MULTIVARIABLE CALCULUS: CONCEPTS AND CONTEXTS FOURTH EDITION

DAN CLEGG Palomar College

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NOT FOR SALE PREFACE This Complete Solutions Manual contains detailed solutions to all exercises in the text Multivariable Calculus: Concepts and Contexts, Fourth Edition (Chapters 8–13 of Calculus: Concepts and Contexts, Fourth Edition) by James Stewart. A Student Solutions Manual is also available, which contains solutions to the odd-numbered exercises in each chapter section, review section, True-False Quiz, and Focus on Problem Solving section as well as all solutions to the Concept Check questions. (It does not, however, include solutions to any of the projects.)

While I have extended every effort to ensure the accuracy of the solutions presented, I would appreciate correspondence regarding any errors that may exist. Other suggestions or comments are also welcome, and can be sent to me at the email address or mailing address below.

I would like to thank James Stewart for entrusting me with the writing of this manual and offering suggestions, Kathi Townes, Stephanie Kuhns, and Rebekah Steele of TECH-arts for typesetting and producing this manual, and Brian Betsill of TECH-arts for creating the illustrations.

Brian Karasek prepared solutions for comparison of accuracy and style in addition to proofreading manuscript; his assistance and suggestions were very helpful and much appreciated. Finally, I would like to thank Richard Stratton and Elizabeth Neustaetter of Brooks/Cole, Cengage Learning for their trust, assistance, and patience.

DA N C L E G G

dclegg@palomar.edu Palomar College Department of Mathematics 1140 West Mission Road San Marcos, CA 92069

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iii

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NOT FOR SALE CONTENTS ■

8

8.1

Sequences

1

Laboratory Project

■

Logistic Sequences

Series

8.3

The Integral and Comparison Tests; Estimating Sums

8.4

Other Convergence Tests

8.5

Power Series

8.6

Representations of Functions as Power Series

8.7

Taylor and Maclaurin Series

8.8

13

32

■

46

55

An Elusive Limit

Applications of Taylor Polynomials

Review

69

70

Radiation from the Stars

■

81

83

Focus on Problem Solving

95

VECTORS AND THE GEOMETRY OF SPACE 9.1

Three-Dimensional Coordinate Systems

9.2

Vectors

9.3

The Dot Product

9.4

The Cross Product

101

115 122 The Geometry of a Tetrahedron

■

Equations of Lines and Planes Laboratory Project

■

Putting 3D in Perspective

Functions and Surfaces

9.7

Cylindrical and Spherical Coordinates Laboratory Project

130

132

9.6

Review

101

108

Discovery Project

9.5

26

39

Applied Project

■

9

8.2

Laboratory Project

9

1

INFINITE SEQUENCES AND SERIES

141

143

■

151

Families of Surfaces

156

158

Focus on Problem Solving

169

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v

vi

■

NOT FOR SALE

CONTENTS

10 ■

VECTOR FUNCTIONS

175

10.1

Vector Functions and Space Curves

10.2

Derivatives and Integrals of Vector Functions

10.3

Arc Length and Curvature

10.4

Motion in Space: Velocity and Acceleration Applied Project

10.5

218

225

231

PARTIAL DERIVATIVES

239

11.1

Functions of Several Variables

11.2

Limits and Continuity

11.3

Partial Derivatives

11.4

Tangent Planes and Linear Approximations

11.5

The Chain Rule

11.6

Directional Derivatives and the Gradient Vector

11.7

Maximum and Minimum Values

249

272

280

■

■

Lagrange Multipliers

290

302

Designing a Dumpster

Discovery Project

11.8

239

256

Applied Project

318

Quadratic Approximations and Critical Points

320

323

Applied Project

■

Rocket Science

Applied Project

■

Hydro-Turbine Optimization

Review

333 335

338

Focus on Problem Solving

12 ■

208

219

Focus on Problem Solving

11 ■

185

195

Kepler’s Laws

■

Parametric Surfaces Review

175

MULTIPLE INTEGRALS

351

357

12.1

Double Integrals over Rectangles

12.2

Iterated Integrals

12.3

Double Integrals over General Regions

12.4

Double Integrals in Polar Coordinates

12.5

Applications of Double Integrals

12.6

Surface Area

357

362 368 380

386

395

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NOT FOR SALE 12.7

Triple Integrals

Volumes of Hyperspheres

■

Roller Derby

■

Discovery Project

The Intersection of Three Cylinders

■

429

VECTOR CALCULUS

447

453

13.1

Vector Fields

453

13.2

Line Integrals

458

13.3

The Fundamental Theorem for Line Integrals

13.4

Green’s Theorem

13.5

Curl and Divergence

13.6

Surface Integrals

486

13.7

Stokes’ Theorem

497

13.8

The Divergence Theorem

APPENDIXES

478

501

505

515

519

D

Precise Definitions of Limits

H

Polar Coordinates Complex Numbers

519

519

Discovery Project

I

466

471

Focus on Problem Solving

■

427

435

Focus on Problem Solving

Review

417

425

Change of Variables in Multiple Integrals Review

13 ■

416

Triple Integrals in Cylindrical and Spherical Coordinates Applied Project

12.9

■

400

Discovery Project

12.8

CONTENTS

■

Conic Sections in Polar Coordinates

543

544

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NOT FOR SALE 8

INFINITE SEQUENCES AND SERIES

8.1 Sequences 1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers.

(b) The terms dq approach 8 as q becomes large. In fact, we can make dq as close to 8 as we like by taking q sufficiently large. (c) The terms dq become large as q becomes large. In fact, we can make dq as large as we like by taking q sufficiently large.

2. (a) From Definition 1, a convergent sequence is a sequence for which lim dq exists. Examples: {1@q}, {1@2q } q

(b) A divergent sequence is a sequence for which lim dq does not exist. Examples: {q}, {sin q} q

3. The first six terms of dq =

1 2 3 4 5 6 1 q are , , , , , . It appears that the sequence is approaching .

2q + 1 3 5 7 9 11 13 2

q 1 1 = lim = q 2q + 1 q 2 + 1@q 2 lim

4. {cos(q @3)}9q=1 =

1 2

> 12 > 1> 12 > 12 > 1> 12 > 12 > 1 . The sequence does not appear to have a limit. The values will cycle

through the first six numbers in the sequence—never approaching a particular number.

5.

1 1 1 1

1> 3 > 5 > 7 > 9 > = = = . The denominator of the nth term is the nth positive odd integer, so dq =

1 .

2q 1

6.

1 1 1> 3 > 9 >

3q 1

. The denominator of the nth term is the (q 1)st power of 3, so dq =

1 1 27 > 81 > = = =

7. {2> 7> 12> 17> = = =}.

8.

.

Each term is larger than the preceding one by 5, so dq = d1 + g(q 1) = 2 + 5(q 1) = 5q 3.

1 2

3 4 4 > 9 > 16 > 25 > = = = . The numerator of the qth term is q and its denominator is (q + 1)2 . Including the alternating signs, we get dq = ( 1)q

9.

1

q . (q + 1)2

q 1 8 1> 23 > 49 > 27 > = = = . Each term is 23 times the preceding one, so dq = 23 .

10. {5> 1> 5> 1> 5> 1> = = =}.

The average of 5 and 1 is 3, so we can think of the sequence as alternately adding 2 and 2 to 3.

Thus, dq = 3 + ( 1)q+1 · 2.

11. dq =

3 + 5q2 (3 + 5q2 )@q2 5 + 3@q2 5+0 , so dq = 5 as q . Converges = = q + q2 (q + q2 )@q2 1 + 1@q 1+0

12. dq =

q3@q3 q3 1 1 = = 1 as q . Converges = , so dq q3 + 1 (q3 + 1)@q3 1 + 1@q3 1+0

13. dq = 1 (0=2)q , so lim dq = 1 0 = 1 by (7). q

Converges

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CHAPTER 8

14. dq =

NOT FOR SALE

INFINITE SEQUENCES EQUENCES AND AN SERIES

q3@q q2 q3 = = , so dq as q since lim q2 = and lim (1 + 1@q2 ) = 1. Diverges q q q+1 (q + 1)@q 1 + 1@q2

15. Because the natural exponential function is continuous at 0, Theorem 5 enables us to write

lim dq = lim h1@q = hlimq (1@q) = h0 = 1= Converges

q

16. dq =

q

q q 3q+2 32 3q = q = 9 35 , so lim dq = 9 lim 35 = 9 · 0 = 0 by (7) with u = 35 . Converges q q q 5 5

(2q )@q 2q 2 2

, then lim eq = lim = lim = = . Since tan is continuous at q q (1 + 8q)@q q 1@q + 8 1 + 8q 8 4

2q

2q Theorem 5, lim tan = tan lim = tan = 1. Converges q q 1 + 8q 1 + 8q 4

17. If eq =

, 4

by

18. Using the last limit law for sequences and the continuity of the square root function,

u lim dq = lim

q

19. dq =

q

q+1 = 9q + 1

u

q+1 = lim q 9q + 1

v 1 + 1@q = lim q 9 + 1@q

u

1 1 = . Converges 9 3

( 1)q 1 q 1 ( 1)q 1 1 = , so 0 |dq | = 0 as q , so dq 0 by the Squeeze Theorem and 2 q +1 q + 1@q q + 1@q q

Theorem 4. Converges 20. dq =

1 ( 1)q q3 q3 . Now |dq | = 3 = 2 + 2q + 1 q + 2q2 + 1 1 + q2 +

q3

1 q3

1 as q , but the terms of the sequence {dq }

alternate in sign, so the sequence d1 > d3 > d5> = = = converges to 1 and the sequence d2 > d4 > d6 > = = = converges to +1.

This shows that the given sequence diverges since its terms don’t approach a single real number.

21. dq =

hq + h q h q 1 + h 2q = 0 as q because 1 + h 2q 1 and hq h q . Converges · h2q 1 h q hq h q

22. dq = cos(2@q).

23. dq = q2 h q =

As q , 2@q 0, so cos(2@q) cos 0 = 1 because cos is continuous. Converges {2 H 2{ H 2 q2 . Since lim { = lim { = lim { = 0, it follows from Theorem 2 that lim dq = 0. Converges q { h { h { h q h

24. 2q as q , so since lim arctan { = {

25. 0

cos2 q 1 q 2q 2

, 2

we have lim arctan 2q = q

1 = 0, q 2q

[since 0 cos2 q 1], so since lim

26. dq = q cos q = q( 1)q .

cos2 q 2q

.

2

Converges

converges to 0 by the Squeeze Theorem.

Since |dq | = q as q , the given sequence diverges.

{

2 27. | = 1 + {

2 , so ln | = { ln 1 + {

1 2 2 1 + 2@{ { ln(1 + 2@{) H 2 lim ln | = lim =2 = lim = lim { { { { 1 + 2@{ 1@{ 1@{2

{ q 2 2 lim 1 + = lim hln | = h2 , so by Theorem 2, lim 1 + = h2 . Convergent { { q { q

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SECTION 8.1 SEC

28. dq =

SEQUENCES

¤

q 1+3q 2 = (21+3q )1@q = (21 23q )1@q = 21@q 23 = 8 · 21@q , so

lim dq = 8 lim 21@q = 8 · 2limq (1@q) = 8 · 20 = 8 by Theorem 5, since the function i ({) = 2{ is continuous at 0.

q

q

Convergent 29. dq =

(2q 1)!

1 (2q 1)! = = 0 as q . Converges (2q + 1)!

(2q + 1)(2q)(2q 1)!

(2q + 1)(2q)

30. dq =

1 sin 2q 1 1 1 . |dq | and lim = 0, so dq

q 1 + 1+ q 1+ q q 1+ q 1+ q

lim dq = 0 by the

q

Squeeze Theorem. Converges 31. {0> 1> 0> 0> 1> 0> 0> 0> 1> = = =} diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to

either one (or any other value) for q sufficiently large.

32. lim

{

(ln {)2 H 2(ln {)(1@{) ln { H 1@{ (ln q)2 = 2 lim = 0, so by Theorem 3, lim = 0. Convergent = lim = 2 lim { { { { 1 q { 1 q

33. dq = ln(2q2 + 1) ln(q2 + 1) = ln

34. 0 ? |dq | =

2q2 + 1 q2 + 1

= ln

2 + 1@q2 1 + 1@q2

3 3 3 3 3 3 3 3 3q = · · · ··· · · · · q!

1 2 3 (q 1) q 1 2 q

ln 2 as q . Convergent

[for q A 2] =

27 0 as q , so by the Squeeze 2q

Theorem and Theorem 4, {( 3)q@q!} converges to 0.

35.

From the graph, it appears that the sequence converges to 1. {( 2@h)q } converges to 0 by (7), and hence {1 + ( 2@h)q } converges to 1 + 0 = 1.

36.

From the graph, it appears that the sequence converges to a number greater than 3.

sin @ q

lim dq = lim q sin = lim · q q q q @ q k l sin { · { = @ q = 1 · = = = lim { { 0+

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CHAPTER 8

NOT FOR SALE

INFINITE SEQUENCES EQUENCES AND AN SERIES

From the graph, it appears that the sequence converges to 12 .

37.

As q , v u 3 + 2q2 3@q2 + 2 = dq = 2 8q + q 8 + 1@q

u

0+2 = 8+0

u

1 1 = , 4 2

so lim dq = 12 . q

From the graph, it appears that the sequence converges to 5=

q q q q 5= 5 q 3q + 5q q 5q + 5q = 2 5q k l

q = 2 · 5 5 as q lim 21@q = 20 = 1

38.

q

Hence, dq 5 by the Squeeze Theorem.

Alternate solution: Let | = (3{ + 5{ )1@{ . Then ln (3{ + 5{ ) H 3{ ln 3 + 5{ ln 5 = lim = lim lim ln | = lim { { { { { 3{ + 5{ so lim | = hln 5 = 5, and so {

3 { ln 3 + ln 5 5 3 { = ln 5, +1 5

q q 3 + 5q converges to 5.

From the graph, it appears that the sequence {dq } =

39.

q2 cos q 1 + q2

is

divergent, since it oscillates between 1 and 1 (approximately). To prove this, suppose that {dq } converges to O. If eq = {eq } converges to 1, and lim

q

lim

q

q2 , then 1 + q2

dq dq O = O. But = = cos q, so eq 1 eq

dq does not exist. This contradiction shows that {dq } diverges. eq

From the graph, it appears that the sequence approaches 0.

40.

0 ? dq =

3 5 2q 1 1 · 3 · 5 · · · · · (2q 1) 1 · · · ··· · = (2q)q 2q 2q 2q 2q

1 1 · (1) · (1) · · · · · (1) = 0 as q 2q 2q

1 · 3 · 5 · · · · · (2q 1) converges to 0.

So by the Squeeze Theorem, (2q)q

41. (a) dq = 1000(1=06)q

d1 = 1060, d2 = 1123=60, d3 = 1191=02, d4 = 1262=48, and d5 = 1338=23.

(b) lim dq = 1000 lim (1=06)q , so the sequence diverges by (7) with u = 1=06 A 1.

INSTRUCTOR USE ONLY q q

q q

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SECTION 8.1 SEC

42. (a) Substitute 1 to 6 for q in Lq = 100

SEQUENCES

¤

5

1=0025q 1 q to get L1 = $0, L2 = $0=25, L3 = $0=75, L4 = $1=50, 0=0025

L5 = $2=51, and L6 = $3=76. (b) For two years, use 2 · 12 = 24 for q to get $70=28.

43. (a) We are given that the initial population is 5000, so S0 = 5000. The number of catfish increases by 8% per month and is

decreased by 300 per month, so S1 = S0 + 8%S0 300 = 1=08S0 300, S2 = 1=08S1 300, and so on. Thus, Sq = 1=08Sq 1 300. (b) Using the recursive formula with S0 = 5000, we get S1 = 5100, S2 = 5208, S3 = 5325 (rounding any portion of a catfish), S4 = 5451, S5 = 5587, and S6 = 5734, which is the number of catfish in the pond after six months. +1

2 dq

44. dq+1 =

if dq is an even number

3dq + 1 if dq is an odd number

When d1 = 11, the first 40 terms are 11, 34, 17, 52, 26, 13, 40, 20, 10, 5,

16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4. When d1 = 25, the first 40 terms are 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4.

The famous Collatz conjecture is that this sequence always reaches 1, regardless of the starting point d1 .

45. (a) d1 = 1, dq+1 = 4 dq for q 1.

d1 = 1, d2 = 4 d1 = 4 1 = 3, d3 = 4 d2 = 4 3 = 1,

d4 = 4 d3 = 4 1 = 3, d5 = 4 d4 = 4 3 = 1. Since the terms of the sequence alternate between 1 and 3, the sequence is divergent. (b) d1 = 2, d2 = 4 d1 = 4 2 = 2, d3 = 4 d2 = 4 2 = 2. Since all of the terms are 2, lim dq = 2 and hence, the q

sequence is convergent.

46. (a) Since lim dq = O, the terms dq approach O as q becomes large. Because we can make dq as close to O as we wish, q

dq+1 will also be close, and so lim dq+1 = O. q

(b) d1 = 1, d2 =

1 1 1 1 1 = = 0=5, d3 = = = 1 + d1 1+1 2 1 + d2 1+

1 2

=

2 0=66667, 3

d4 =

1 1 = 1 + d3 1+

2 3

d6 =

1 1 = 1 + d5 1+

5 8

d8 =

1 1 21 1 1 34 = = = 0=61765, d9 = 0=61818, 21 = 1 + d7 34 1 + d 55 1 + 13 1 + 8 21 34

d10 =

=

3 1 1 = 0=6, d5 = = 5 1 + d4 1+

=

8 1 1 13 = 0=61538, d7 = 0=61905, 8 = 13 1 + d6 21 1 + 13

3 5

=

5 = 0=625, 8

1 1 55 = 0=61800. It appears that lim dq 0=618; hence, the sequence is convergent.

34 = q 1 + d9 89 1 + 55

(c) If O = lim dq then lim dq+1 = O also, so O must satisfy q

O = 1@(1 + O)

q

O2 + O 1 = 0

O=

1 + 2

5

0=618 (since O has to be non-negative if it exists).

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CHAPTER 8

NOT FOR SALE

INFINITE SEQUENCES EQUENCES AND AN SERIES

47. (a) Let dq be the number of rabbit pairs in the nth month. Clearly d1 = 1 = d2 . In the nth month, each pair that is

2 or more months old (that is, dq 2 pairs) will produce a new pair to add to the dq 1 pairs already present. Thus, dq = dq 1 + dq 2 , so that {dq } = {iq }, the Fibonacci sequence. iq+1 iq

(b) dq =

dq 1 =

iq iq 1 + iq 2 iq 2 1 1 = =1+ =1+ =1+ . If O = lim dq , q iq 1 iq 1 iq 1 iq 1 /iq 2 dq 2

then O = lim dq 1 and O = lim dq 2 , so O must satisfy O = 1 + q

q

1 O

O2 O 1 = 0 O =

1+ 5 2

[since O must be positive].

48. For

t s s

q q q 2, 2 2, 2 2 2, = = = , d1 = 21@2 , d2 = 23@4 , d3 = 27@8 , = = =, so dq = 2(2 1)@2 = 21 (1@2 ) . q)

lim dq = lim 21 (1@2

q

q

= 21 = 2.

Alternate solution: Let O = lim dq . (We could show the limit exists by showing that {dq } is bounded and increasing.) q

Then O must satisfy O = 49. dq =

1 1 1 1 is decreasing since dq+1 = = ? = dq for each q 1. The sequence is 2q + 3 2(q + 1) + 3 2q + 5 2q + 3

bounded since 0 ? dq 50. dq =

2 · O O2 = 2O O(O 2) = 0. O 6= 0 since the sequence increases, so O = 2.

1 5

for all q 1. Note that d1 = 15 .

2{ 3 2q 3 defines an increasing sequence since for i ({) = , 3q + 4 3{ + 4

i 0 ({) =

(3{ + 4)(2) (2{ 3)(3) 17 = A 0. The sequence is bounded since dq d1 = 17 for q 1, 2 (3{ + 4) (3{ + 4)2

and dq ?

2q 2 2q 3 ? = for q 1.

3q 3q 3

51. The terms of dq = q( 1)q alternate in sign, so the sequence is not monotonic. The first five terms are 1, 2, 3, 4, and 5.

Since lim |dq | = lim q = , the sequence is not bounded. q

52. dq = q +

q

1 1 defines an increasing sequence since the function j({) = { + is increasing for { A 1. [j 0 ({) = 1 1@{2 A 0 q {

for { A 1.] The sequence is unbounded since dq as q . (It is, however, bounded below by d1 = 2.)

53. Since {dq } is a decreasing sequence, dq A dq+1 for all q 1. Because all of its terms lie between 5 and 8, {dq } is a

bounded sequence. By the Monotonic Sequence Theorem, {dq } is convergent; that is, {dq } has a limit O. O must be less than 8 since {dq } is decreasing, so 5 O ? 8.

54. (a) Let Sq be the statement that dq+1 dq and dq 3. S1 is obviously true. We will assume that Sq is true and

then show that as a consequence Sq+1 must also be true. dq+2 dq+1

2 + dq+1 2 + dq

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SECTION 8.1 SEC

2 + dq+1 2 + dq

dq+1 dq , which is the induction hypothesis. dq+1 3

¤

SEQUENCES

7

2 + dq 3

2 + dq 9 dq 7, which is certainly true because we are assuming that dq 3. So Sq is true for all q, and so d1 dq 3 (showing that the sequence is bounded), and hence by the Monotonic Sequence Theorem, lim dq exists. q

(b) If O = lim dq , then lim dq+1 = O also, so O = q

q

2 + O O2 = 2 + O O2 O 2 = 0

(O + 1)(O 2) = 0 O = 2 [since O can’t be negative].

55. d1 = 1, dq+1 = 3

1 . We show by induction that {dq } is increasing and bounded above by 3. Let Sq be the proposition dq

that dq+1 A dq and 0 ? dq ? 3. Clearly S1 is true. Assume that Sq is true. Then dq+1 A dq

1 1 1 1 A . Now dq+2 = 3 A3 = dq+1 dq+1 dq dq+1 dq

1 dq+1

?

1 dq

Sq+1 . This proves that {dq } is increasing and bounded

above by 3, so 1 = d1 ? dq ? 3, that is, {dq } is bounded, and hence convergent by the Monotonic Sequence Theorem.

If O = lim dq , then lim dq+1 = O also, so O must satisfy O = 3 1@O O2 3O + 1 = 0 O = q

q

But O A 1, so O = 56. d1 = 2, dq+1 =

3+ 5 .

2

1 . We use induction. Let Sq be the statement that 0 ? dq+1 dq 2. Clearly S1 is true, since 3 dq

d2 = 1@(3 2) = 1. Now assume that Sq is true. Then dq+1 dq dq+2 =

3± 5 .

2

dq+1 dq

3 dq+1 3 dq

1 1

= dq+1 . Also dq+2 A 0 [since 3 dq+1 is positive] and dq+1 2 by the induction 3 dq+1 3 dq

hypothesis, so Sq+1 is true. To find the limit, we use the fact that lim dq = lim dq+1 q

O2 3O + 1 = 0 O = q

57. (0=8) ? 0=000001

3± 5 .

2

But O 2, so we must have O =

ln(0=8)q ? ln(0=000001)

q

O=

1 3 O

3 5 .

2

q ln(0=8) ? ln(0=000001) q A

ln(0=000001) ln(0=8)

q A 61=9, so q must be at least 62 to satisfy the given inequality.

58. (a) If i is continuous, then i(O) = i

lim dq = lim i (dq ) = lim dq+1 = lim dq = O by Exercise 46(a).

q

q

q

q

(b) By repeatedly pressing the cosine key on the calculator (that is, taking cosine of the previous answer) until the displayed value stabilizes, we see that O 0=73909.

59. (a) Suppose {sq } converges to s. Then sq+1 =

s2 + ds = es

(b) sq+1

s(s + d e) = 0

esq d + sq

lim sq+1 =

q

e lim sq q

d + lim sq

q

s=

es d+s

s = 0 or s = e d.

e

sq d esq sq e = = ? sq since 1 + A 1. sq d + sq d d 1+ d

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8

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NOT FOR SALE

CHAPTER 8

INFINITE SEQUENCES EQUENCES AND AN SERIES

2

3 q e e e e e e (c) By part (b), s1 ? s0 , s3 ? s0 , etc. In general, sq ? s0 , s0 , s2 ? s1 ? s2 ? d d d d d d

q

e e · s0 = 0 since e ? d. By (7)> lim uq = 0 if 1 ? u ? 1. Here u = (0> 1) . so lim sq lim q q q d d (d) Let d ? e. We first show, by induction, that if s0 ? e d, then sq ? e d and sq+1 A sq .

For q = 0, we have s1 s0 =

es0 s0 (e d s0 ) s0 = A 0 since s0 ? e d. So s1 A s0 . d + s0 d + s0

Now we suppose the assertion is true for q = n, that is, sn ? e d and sn+1 A sn . Then d(e d) + esn dsn esn d(e d sn ) esn = = A 0 because sn ? e d. So d + sn d + sn d + sn

e d sn+1 = e d

sn+1 ? e d. And sn+2 sn+1 =

esn+1 sn+1 (e d sn+1 ) sn+1 = A 0 since sn+1 ? e d. Therefore, d + sn+1 d + sn+1

sn+2 A sn+1 . Thus, the assertion is true for q = n + 1. It is therefore true for all q by mathematical induction.

A similar proof by induction shows that if s0 A e d, then sq A e d and {sq } is decreasing.

In either case the sequence {sq } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem.

It then follows from part (a) that lim sq = e d. q

1 1+1

60. d1 = 1, d2 = 1 +

d5 = 1 +

1 29@12

d8 = 1 +

1 408@169

=

41 29

=

3 2

=

= 1=5, d3 = 1 +

1=413793, d6 = 1 +

577 408

1 5@2

=

1 70@29

7 5

=

= 1=4, d4 = 1 + 99 70

1 12@5

=

17 12

1=414286, d7 = 1 +

= 1=416,

1 169@70

=

239 169

1=414201,

1=414216. Notice that d1 ? d3 ? d5 ? d7 and d2 A d4 A d6 A d8 . It appears that the odd terms

are increasing and the even terms are decreasing. Let’s prove that d2q 2 A d2q and d2q 1 ? d2q+1 by mathematical induction. Suppose that d2n 2 A d2n . Then 1 + d2n 2 A 1 + d2n 1 1 ?

1 + d2n 2 1 + d2n

1+

1 + d2n 1 ? 1 + d2n+1

1 1 ?1+ 1 + d2n 2 1 + d2n

1 1 A 1 + d2n 1 1 + d2n+1

d2n 1 ? d2n+1

1+

1 1 A1+ 1 + d2n 1 1 + d2n+1

d2n A d2n+2 .

We have thus shown, by induction, that the odd terms are increasing and the even terms are decreasing. Also all terms lie between 1 and 2, so both {dq } and {eq } are bounded monotonic sequences and therefore convergent by the Monotonic Sequence Theorem. Let lim d2q = O. Then lim d2q+2 = O also. We have q

dq+2 = 1 +

1 4 + 3d2q 1 4 + 3dq , so d2q+2 = . Taking limits of both =1+ = 1 + 1 + 1@ (1 + dq ) (3 + 2dq ) @ (1 + dq ) 3 + 2dq 3 + 2d2q

sides, we get O = lim d2q =

q

q

4 + 3O 3 + 2O

3O + 2O2 = 4 + 3O O2 = 2 O =

2 [since O A 0]. Thus,

2.

Similarly, we find that lim d2q+1 = q

2. Since the even terms approach 2 and the odd terms also approach 2, it

follows that the sequence as a whole approaches

2, that is, lim dq = 2. q

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NOT FOR SALE

LABORATORY PROJECT LOGISTIC SEQUENCES

¤

LABORATORY PROJECT Logistic Sequences 1. To write such a program in Maple it is best to calculate all the points first and then graph them. One possible sequence of

commands [taking s0 =

1 2

and n = 1=5 for the difference equation] is

t:=’t’;p(0):=1/2;k:=1.5; for j from 1 to 20

do p(j):=k*p(j-1)*(1-p(j-1)) od;

plot([seq([t,p(t)] t=0..20)],t=0..20,p=0..0.5,style=point); In Mathematica, we can use the following program: p[0]=1/2 k=1.5 p[j_]:=k*p[j-1]*(1-p[j-1]) P=Table[p[t],{t,20}] ListPlot[P]

With s0 =

1 2

q

and n = 1=5:

sq

q

sq

q

sq

0

0=5

7

0=3338465076

14

0=3333373303

1

0=375

8

0=3335895255

15

0=3333353318

2

0=3515625

9

0=3334613309

16

0=3333343326

3

0=3419494629

10

0=3333973076

17

0=3333338329

4

0=3375300416

11

0=3333653143

18

0=3333335831

5

0=3354052689

12

0=3333493223

19

0=3333334582

6

0=3343628617

13

0=3333413274

20

0=3333333958

q

sq

q

sq

With s0 =

1 2

q

and n = 2=5:

sq

0

0=5

7

0=6004164790

14

0=5999967417

1

0=625

8

0=5997913269

15

0=6000016291

2

0=5859375

9

0=6001042277

16

0=5999991854

3

0=6065368651

10

0=5999478590

17

0=6000004073

4

0=5966247409

11

0=6000260637

18

0=5999997964

5

0=6016591486

12

0=5999869664

19

0=6000001018

6

0=5991635437

13

0=6000065164

20

0=5999999491

Both of these sequences seem to converge the first first to about 13 , thee second to about 0.60 .

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9

10

¤

NOT FOR SALE

CHAPTER 8 INFINITE SEQUENCES EQUENCES AND AN SERIES

With s0 =

7 8

q

and n = 1=5:

sq

q

sq

q

sq

0

0=875

7

0=3239166554

14

0=3332554829

1

0=1640625

8

0=3284919837

15

0=3332943990

2

0=2057189941

9

0=3308775005

16

0=3333138639

3

0=2450980344

10

0=3320963702

17

0=3333235980

4

0=2775374819

11

0=3327125567

18

0=3333284655

5

0=3007656421

12

0=3330223670

19

0=3333308994

6

0=3154585059

13

0=3331777051

20

0=3333321164

q

sq

q

sq

With s0 =

7 8

q

and n = 2=5:

sq

0

0=875

7

0=6016572368

14

0=5999869815

1

0=2734375

8

0=5991645155

15

0=6000065088

2

0=4966735840

9

0=6004159972

16

0=5999967455

3

0=6249723374

10

0=5997915688

17

0=6000016272

4

0=5859547872

11

0=6001041070

18

0=5999991864

5

0=6065294364

12

0=5999479194

19

0=6000004068

6

0=5966286980

13

0=6000260335

20

0=5999997966

The limit of the sequence seems to depend on n, but not on s0 .

2. With s0 =

7 8

q

and n = 3=2:

sq

q

sq

q

sq

0

0=875

7

0=5830728495

14

0=7990633827

1

0=35

8

0=7779164854

15

0=5137954979

2

0=728

9

0=5528397669

16

0=7993909896

3

0=6336512

10

0=7910654689

17

0=5131681132

4

0=7428395416

11

0=5288988570

18

0=7994451225

5

0=6112926626

12

0=7973275394

19

0=5130643795

6

0=7603646184

13

0=5171082698

20

0=7994538304

It seems that eventually the terms fluctuate between two values (about 0=5 and 0=8 in this case).

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NOT FOR SALE

LABORATORY PROJECT LOGISTIC SEQUENCES

3. With s0 =

7 8

q

q

sq

q

sq

0

0=875

7

0=4523028596

14

0=8442074951

1

0=3740625

8

0=8472194412

15

0=4498025048

2

0=8007579316

9

0=4426802161

16

0=8463823232

3

0=5456427596

10

0=8437633929

17

0=4446659586

4

0=8478752457

11

0=4508474156

18

0=8445284520

5

0=4411212220

12

0=8467373602

19

0=4490464985

6

0=8431438501

13

0=4438243545

20

0=8461207931

q

sq

q

sq

7 8

q

11

and n = 3=42:

sq

With s0 =

¤

and n = 3=45:

sq

0

0=875

7

0=4670259170

14

0=8403376122

1

0=37734375

8

0=8587488490

15

0=4628875685

2

0=8105962830

9

0=4184824586

16

0=8577482026

3

0=5296783241

10

0=8395743720

17

0=4209559716

4

0=8594612299

11

0=4646778983

18

0=8409445432

5

0=4167173034

12

0=8581956045

19

0=4614610237

6

0=8385707740

13

0=4198508858

20

0=8573758782

From the graphs above, it seems that for n between 3=4 and 3=5, the terms eventually fluctuate between four values. In the graph below, the pattern followed by the terms is 0=395> 0=832> 0=487> 0=869> 0=395> = = =. Note that even for n = 3=42 (as in the first graph), there are four distinct “branches”; even after 1000 terms, the first and third terms in the pattern differ by about 2 × 10 9 , while the first and fifth terms differ by only 2 × 10 10 . With s0 =

7 8

and n = 3=48:

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¤

CHAPTER 8 INFINITE SEQUENCES EQUENCES AND AN SERIES

4.

s0 = 0=5, n = 3=7

s0 = 0=501, n = 3=7

s0 = 0=75, n = 3=9

s0 = 0=749, n = 3=9

s0 = 0=5, n = 3=999 From the graphs, it seems that if s0 is changed by 0=001, the whole graph changes completely. (Note, however, that this might be partially due to accumulated round-off error in the CAS. These graphs were generated by Maple with 100-digit accuracy, and different degrees of accuracy give different graphs.) There seem to be some some fleeting patterns in these graphs, but on the whole they are certainly very chaotic. As n increases, the graph spreads out vertically, with more extreme values close to 0 or 1.

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