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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Checkpoint

6.1

The interval of convergence is [−1,1).[−1,1). The radius of convergence is R=1.R=1.

6.2


This figure is the graph of y = 1/(1-x^2), which is a curve concave up, symmetrical about the y axis. Also on this graph are three partial sums of the function, S sub 6, S sub 4, and S sub 2. These curves, in order, gradually become flatter.
6.3

n=0xn+32n+1n=0xn+32n+1 with interval of convergence (−2,2)(−2,2)

6.4

Interval of convergence is (−2,2).(−2,2).

6.5

n=0(−1+12n+1)xn.n=0(−1+12n+1)xn. The interval of convergence is (−1,1).(−1,1).

6.6

f(x)=33x.f(x)=33x. The interval of convergence is (−3,3).(−3,3).

6.7

1+2x+3x2+4x3+1+2x+3x2+4x3+

6.8

n=0(n+2)(n+1)xnn=0(n+2)(n+1)xn

6.9

n=2(−1)nxnn(n1)n=2(−1)nxnn(n1)

6.10

p0(x)=1;p1(x)=12(x1);p2(x)=12(x1)+3(x1)2;p3(x)=12(x1)+3(x1)24(x1)3p0(x)=1;p1(x)=12(x1);p2(x)=12(x1)+3(x1)2;p3(x)=12(x1)+3(x1)24(x1)3

6.11

p0(x)=1;p1(x)=1x;p2(x)=1x+x2;p3(x)=1x+x2x3;pn(x)=1x+x2x3++(−1)nxn=k=0n(−1)kxkp0(x)=1;p1(x)=1x;p2(x)=1x+x2;p3(x)=1x+x2x3;pn(x)=1x+x2x3++(−1)nxn=k=0n(−1)kxk

6.12


p1(x)=2+14(x4);p2(x)=2+14(x4)164(x4)2;p1(6)=2.5;p2(6)=2.4375;p1(x)=2+14(x4);p2(x)=2+14(x4)164(x4)2;p1(6)=2.5;p2(6)=2.4375;

|R1(6)|0.0625;|R2(6)|0.015625|R1(6)|0.0625;|R2(6)|0.015625

6.13

0.96593

6.14

n=0(2x2n+2 )n.n=0(2x2n+2 )n. The interval of convergence is (0,4).(0,4).

6.15

n=0(−1)nx2n(2n)!n=0(−1)nx2n(2n)!

By the ratio test, the interval of convergence is (,).(,). Since |Rn(x)||x|n+1(n+1)!,|Rn(x)||x|n+1(n+1)!, the series converges to cosxcosx for all real x.

6.16

n=0(−1)n(n+1)xnn=0(−1)n(n+1)xn

6.17

n=0(−1)nx4n+2(2n+1)!n=0(−1)nx4n+2(2n+1)!

6.18

n=1(−1)nn!1·3·5(2n1)2nxnn=1(−1)nn!1·3·5(2n1)2nxn

6.19

y=5e2xy=5e2x

6.20

y=a(1x43·4+x83·4·7·8)+b(xx54·5+x94·5·8·9)y=a(1x43·4+x83·4·7·8)+b(xx54·5+x94·5·8·9)

6.21

C+n=1(−1)n+1xnn(2n2)!C+n=1(−1)n+1xnn(2n2)! The definite integral is approximately 0.5140.514 to within an error of 0.01.0.01.

6.22

The estimate is approximately 0.3414.0.3414. This estimate is accurate to within 0.0000094.0.0000094.

Section 6.1 Exercises

1.

True. If a series converges then its terms tend to zero.

3.

False. It would imply that anxn0anxn0 for |x|<R.|x|<R. If an=nn,an=nn, then anxn=(nx)nanxn=(nx)n does not tend to zero for any x0.x0.

5.

It must converge on (0,6](0,6] and hence at: a. x=1;x=1; b. x=2;x=2; c. x=3;x=3; d. x=0;x=0; e. x=5.99;x=5.99; and f. x=0.000001.x=0.000001.

7.

|an+12n+1xn+1an2nxn|=2|x||an+1an|2|x||an+12n+1xn+1an2nxn|=2|x||an+1an|2|x| so R=12R=12

9.

|an+1(πe)n+1xn+1an(πe)nxn|=π|x|e|an+1an|π|x|e|an+1(πe)n+1xn+1an(πe)nxn|=π|x|e|an+1an|π|x|e so R=eπR=eπ

11.

|an+1(−1)n+1x2n+2an(−1)nx2n|=|x2||an+1an||x2||an+1(−1)n+1x2n+2an(−1)nx2n|=|x2||an+1an||x2| so R=1R=1

13.

an=2nnan=2nn so an+1xan2x.an+1xan2x. so R=12.R=12. When x=12x=12 the series is harmonic and diverges. When x=12x=12 the series is alternating harmonic and converges. The interval of convergence is I=[12,12).I=[12,12).

15.

an=n2nan=n2n so an+1xanx2an+1xanx2 so R=2.R=2. When x=±2x=±2 the series diverges by the divergence test. The interval of convergence is I=(−2,2).I=(−2,2).

17.

an=n22nan=n22n so R=2.R=2. When x=±2x=±2 the series diverges by the divergence test. The interval of convergence is I=(−2,2).I=(−2,2).

19.

ak=πkkπak=πkkπ so R=1π.R=1π. When x=±1πx=±1π the series is an absolutely convergent p-series. The interval of convergence is I=[1π,1π].I=[1π,1π].

21.

an=10nn!,an+1xan=10xn+10<1an=10nn!,an+1xan=10xn+10<1 so the series converges for all x by the ratio test and I=(,).I=(,).

23.

ak=(k!)2(2k)!ak=(k!)2(2k)! so ak+1ak=(k+1)2(2k+2)(2k+1)14ak+1ak=(k+1)2(2k+2)(2k+1)14 so R=4R=4

25.

ak=k!1·3·5(2k1)ak=k!1·3·5(2k1) so ak+1ak=k+12k+112ak+1ak=k+12k+112 so R=2R=2

27.

an=1(2nn)an=1(2nn) so an+1an=((n+1)!)2(2n+2)!2n!(n!)2=(n+1)2(2n+2)(2n+1)14an+1an=((n+1)!)2(2n+2)!2n!(n!)2=(n+1)2(2n+2)(2n+1)14 so R=4R=4

29.

an+1an=(n+1)3(3n+3)(3n+2)(3n+1)127an+1an=(n+1)3(3n+3)(3n+2)(3n+1)127 so R=27R=27

31.

an=n!nnan=n!nn so an+1an=(n+1)!n!nn(n+1)n+1=(nn+1)n1ean+1an=(n+1)!n!nn(n+1)n+1=(nn+1)n1e so R=eR=e

33.

f(x)=n=0(1x)nf(x)=n=0(1x)n on I=(0,2)I=(0,2)

35.

n=0x2n+1n=0x2n+1 on I=(−1,1)I=(−1,1)

37.

n=0(−1)nx2n+2n=0(−1)nx2n+2 on I=(−1,1)I=(−1,1)

39.

n=02nxnn=02nxn on (12,12)(12,12)

41.

n=04nx2n+2n=04nx2n+2 on (12,12)(12,12)

43.

|anxn|1/n=|an|1/n|x||x|r|anxn|1/n=|an|1/n|x||x|r as nn and |x|r<1|x|r<1 when |x|<1r.|x|<1r. Therefore, n=1anxnn=1anxn converges when |x|<1r|x|<1r by the nth root test.

45.

ak=(k12k+3)kak=(k12k+3)k so (ak)1/k12<1(ak)1/k12<1 so R=2R=2

47.

an=(n1/n1)nan=(n1/n1)n so (an)1/n0(an)1/n0 so R=R=

49.

We can rewrite p(x)=n=0a2n+1x2n+1p(x)=n=0a2n+1x2n+1 and p(x)=p(x)p(x)=p(x) since x2n+1=(x)2n+1.x2n+1=(x)2n+1.

51.

If x[0,1],x[0,1], then y=2x1[−1,1]y=2x1[−1,1] so p(2x1)=p(y)=n=0anynp(2x1)=p(y)=n=0anyn converges.

53.

Converges on (−1,1)(−1,1) by the ratio test

55.

Consider the series bkxkbkxk where bk=akbk=ak if k=n2k=n2 and bk=0bk=0 otherwise. Then bkakbkak and so the series converges on (−1,1)(−1,1) by the comparison test.

57.


This figure is the graph of y = 1/(1-x), which is an increasing curve with vertical asymptote at 1.


The approximation is more accurate near x=−1.x=−1. The partial sums follow 11x11x more closely as N increases but are never accurate near x=1x=1 since the series diverges there.

59.


This figure is the graph of y = -ln(1-x) which is an increasing curve passing through the origin.


The approximation appears to stabilize quickly near both x=±1.x=±1.

61.


This figure is the graph of the partial sums of (-1)^n times x^(2n+1) divided by (2n+1)! For n=3,5,10. The curves approximate the sine curve close to the origin and then separate as the curves move away from the origin.


The polynomial curves have roots close to those of sinxsinx up to their degree and then the polynomials diverge from sinx.sinx.

Section 6.2 Exercises

63.

12(f(x)+g(x))=n=0x2n(2n)!12(f(x)+g(x))=n=0x2n(2n)! and 12(f(x)g(x))=n=0x2n+1(2n+1)!.12(f(x)g(x))=n=0x2n+1(2n+1)!.

65.

4(x3)(x+1)=1x31x+1=13(1x3)11(x)=13n=0(x3)nn=0(−1)nxn=n=0((−1)n+113n+1)xn4(x3)(x+1)=1x31x+1=13(1x3)11(x)=13n=0(x3)nn=0(−1)nxn=n=0((−1)n+113n+1)xn

67.

5(x2+4)(x21)=1x211411+(x2)2=n=0x2n14n=0(−1)n(x2)2n=n=0((−1)+(−1)n+112n+2)x2n5(x2+4)(x21)=1x211411+(x2)2=n=0x2n14n=0(−1)n(x2)2n=n=0((−1)+(−1)n+112n+2)x2n

69.

1xn=01xn=1x111x=1x11xn=01xn=1x111x=1x1

71.

1x3111(x3)2=x3(x3)211x3111(x3)2=x3(x3)21

73.

P=P1++P20P=P1++P20 where Pk=10,0001(1+r)k.Pk=10,0001(1+r)k. Then P=10,000k=1201(1+r)k=10,0001(1+r)−20r.P=10,000k=1201(1+r)k=10,0001(1+r)−20r. When r=0.03,P10,000×14.8775=148,775.r=0.03,P10,000×14.8775=148,775. When r=0.05,P10,000×12.4622=124,622.r=0.05,P10,000×12.4622=124,622. When r=0.07,P105,940.r=0.07,P105,940.

75.

In general, P=C(1(1+r)N)rP=C(1(1+r)N)r for N years of payouts, or C=Pr1(1+r)N.C=Pr1(1+r)N. For N=20N=20 and P=100,000,P=100,000, one has C=6721.57C=6721.57 when r=0.03;C=8024.26r=0.03;C=8024.26 when r=0.05;r=0.05; and C9439.29C9439.29 when r=0.07.r=0.07.

77.

In general, P=Cr.P=Cr. Thus, r=CP=5×104106=0.05.r=CP=5×104106=0.05.

79.

(x+x2x3)(1+x3+x6+)=x+x2x31x3(x+x2x3)(1+x3+x6+)=x+x2x31x3

81.

(xx2x3)(1+x3+x6+)=xx2x31x3(xx2x3)(1+x3+x6+)=xx2x31x3

83.

an=2,bn=nan=2,bn=n so cn=k=0nbkank=2k=0nk=(n)(n+1)cn=k=0nbkank=2k=0nk=(n)(n+1) and f(x)g(x)=n=1n(n+1)xnf(x)g(x)=n=1n(n+1)xn

85.

an=bn=2nan=bn=2n so cn=k=1nbkank=2nk=1n1=n2ncn=k=1nbkank=2nk=1n1=n2n and f(x)g(x)=n=1n(x2)nf(x)g(x)=n=1n(x2)n

87.

The derivative of ff is 1(1+x)2=n=0(−1)n(n+1)xn.1(1+x)2=n=0(−1)n(n+1)xn.

89.

The indefinite integral of ff is 11+x2=n=0(−1)nx2n.11+x2=n=0(−1)nx2n.

91.

f(x)=n=0xn=11x;f(12)=n=1n2n1=ddx(1x)−1|x=1/2=1(1x)2|x=1/2=4f(x)=n=0xn=11x;f(12)=n=1n2n1=ddx(1x)−1|x=1/2=1(1x)2|x=1/2=4 so n=1n2n=2.n=1n2n=2.

93.

f(x)=n=0xn=11x;f(12)=n=2n(n1)2n2=d2dx2(1x)−1|x=1/2=2(1x)3|x=1/2=16f(x)=n=0xn=11x;f(12)=n=2n(n1)2n2=d2dx2(1x)−1|x=1/2=2(1x)3|x=1/2=16 so n=2n(n1)2n=4.n=2n(n1)2n=4.

95.

(1x)ndx=(−1)n(x1)ndx=(−1)n(x1)n+1n+1(1x)ndx=(−1)n(x1)ndx=(−1)n(x1)n+1n+1

97.

t=0x211tdt=n=00x2tndxn=0x2(n+1)n+1=n=1x2nnt=0x211tdt=n=00x2tndxn=0x2(n+1)n+1=n=1x2nn

99.

0x2dt1+t2=n=0(−1)n0x2t2ndt=n=0(−1)nt2n+12n+1|t=0x2=n=0(−1)nx4n+22n+10x2dt1+t2=n=0(−1)n0x2t2ndt=n=0(−1)nt2n+12n+1|t=0x2=n=0(−1)nx4n+22n+1

101.

Term-by-term integration gives 0xlntdt=n=1(−1)n1(x1)n+1n(n+1)=n=1(−1)n1(1n1n+1)(x1)n+1=(x1)lnx+n=2(−1)n(x1)nn=xlnxx.0xlntdt=n=1(−1)n1(x1)n+1n(n+1)=n=1(−1)n1(1n1n+1)(x1)n+1=(x1)lnx+n=2(−1)n(x1)nn=xlnxx.

103.

We have ln(1x)=n=1xnnln(1x)=n=1xnn so ln(1+x)=n=1(−1)n1xnn.ln(1+x)=n=1(−1)n1xnn. Thus, ln(1+x1x)=n=1(1+(−1)n1)xnn=2n=1x2n12n1.ln(1+x1x)=n=1(1+(−1)n1)xnn=2n=1x2n12n1. When x=13x=13 we obtain ln(2)=2n=1132n1(2n1).ln(2)=2n=1132n1(2n1). We have 2n=13132n1(2n1)=0.69300,2n=13132n1(2n1)=0.69300, while 2n=14132n1(2n1)=0.693132n=14132n1(2n1)=0.69313 and ln(2)=0.69314;ln(2)=0.69314; therefore, N=4.N=4.

105.

k=1xkk=ln(1x)k=1xkk=ln(1x) so k=1x3k6k=16ln(1x3).k=1x3k6k=16ln(1x3). The radius of convergence is equal to 1 by the ratio test.

107.

If y=2x,y=2x, then k=1yk=y1y=2x12x=12x1.k=1yk=y1y=2x12x=12x1. If ak=2kx,ak=2kx, then ak+1ak=2x<1ak+1ak=2x<1 when x>0.x>0. So the series converges for all x>0.x>0.

109.

Answers will vary.

111.


This is a graph of three curves. They are all increasing and become very close as the curves approach x = 0. Then they separate as x moves away from 0.


The solid curve is S5. The dashed curve is S2, dotted is S3, and dash-dotted is S4

113.

When x=12,ln(2)=ln(12)=n=11n2n.x=12,ln(2)=ln(12)=n=11n2n. Since n=111n2n<n=1112n=1210,n=111n2n<n=1112n=1210, one has n=1101n2n=0.69306n=1101n2n=0.69306 whereas ln(2)=0.69314;ln(2)=0.69314; therefore, N=10.N=10.

115.

6SN(13)=23n=0N(−1)n13n(2n+1).6SN(13)=23n=0N(−1)n13n(2n+1). One has π6S4(13)=0.00101π6S4(13)=0.00101 and π6S5(13)=0.00028π6S5(13)=0.00028 so N=5N=5 is the smallest partial sum with accuracy to within 0.001. Also, π6S7(13)=0.00002π6S7(13)=0.00002 while π6S8(13)=−0.000007π6S8(13)=−0.000007 so N=8N=8 is the smallest N to give accuracy to within 0.00001.

Section 6.3 Exercises

117.

f(−1)=1;f(−1)=−1;f(−1)=2;f(x)=1(x+1)+(x+1)2f(−1)=1;f(−1)=−1;f(−1)=2;f(x)=1(x+1)+(x+1)2

119.

f(x)=2cos(2x);f(x)=−4sin(2x);p2(x)=−2(xπ2)f(x)=2cos(2x);f(x)=−4sin(2x);p2(x)=−2(xπ2)

121.

f(x)=1x;f(x)=1x2;p2(x)=0+(x1)12(x1)2f(x)=1x;f(x)=1x2;p2(x)=0+(x1)12(x1)2

123.

p2(x)=e+e(x1)+e2(x1)2p2(x)=e+e(x1)+e2(x1)2

125.

d2dx2x1/3=29x5/3−0.00092d2dx2x1/3=29x5/3−0.00092 when x28x28 so the remainder estimate applies to the linear approximation x1/3p1(27)=3+x2727,x1/3p1(27)=3+x2727, which gives (28)1/33+127=3.037¯,(28)1/33+127=3.037¯, while (28)1/33.03658.(28)1/33.03658.

127.

Using the estimate 21010!<0.00028321010!<0.000283 we can use the Taylor expansion of order 9 to estimate ex at x=2.x=2. as e2p9(2)=1+2+222+236++299!=7.3887e2p9(2)=1+2+222+236++299!=7.3887 whereas e27.3891.e27.3891.

129.

Since dndxn(lnx)=(−1)n1(n1)!xn,R100011001.dndxn(lnx)=(−1)n1(n1)!xn,R100011001. One has p1000(1)=n=11000(−1)n1n0.6936p1000(1)=n=11000(−1)n1n0.6936 whereas ln(2)0.6931.ln(2)0.6931.

131.

01(1x2+x42x66+x824x10120+x12720)dx01(1x2+x42x66+x824x10120+x12720)dx

=1133+15101742+199·24111120·11+113720·130.74683=1133+15101742+199·24111120·11+113720·130.74683 whereas 01ex2dx0.74682.01ex2dx0.74682.

133.

Since f(n+1)(z)f(n+1)(z) is sinzsinz or cosz,cosz, we have M=1.M=1. Since |x0|π2,|x0|π2, we seek the smallest n such that πn+12n+1(n+1)!0.001.πn+12n+1(n+1)!0.001. The smallest such value is n=7.n=7. The remainder estimate is R70.00092.R70.00092.

135.

Since f(n+1)(z)=±ezf(n+1)(z)=±ez one has M=e3.M=e3. Since |x0|3,|x0|3, one seeks the smallest n such that 3n+1e3(n+1)!0.001.3n+1e3(n+1)!0.001. The smallest such value is n=14.n=14. The remainder estimate is R140.000220.R140.000220.

137.


This graph has a horizontal line at y=0.2. It also has a curve starting at the origin and concave up. The curve and the line intersect at the ordered pair (0.5966, 0.2).


Since sinxsinx is increasing for small x and since sinx=sinx,sinx=sinx, the estimate applies whenever R2sin(R)0.2,R2sin(R)0.2, which applies up to R=0.596.R=0.596.

139.


This graph has a horizontal line at y=0.2. It also has a curve starting at the origin and concave up. The curve and the line intersect at the ordered pair (0.44720, 0.2).


Since the second derivative of cosxcosx is cosxcosx and since cosxcosx is decreasing away from x=0,x=0, the estimate applies when R2cosR0.2R2cosR0.2 or R0.447.R0.447.

141.

(x+1)32(x+1)2+2(x+1)(x+1)32(x+1)2+2(x+1)

143.

Values of derivatives are the same as for x=0x=0 so cosx=n=0(−1)n(x2π)2n(2n)!cosx=n=0(−1)n(x2π)2n(2n)!

145.

cos(π2)=0,sin(π2)=−1cos(π2)=0,sin(π2)=−1 so cosx=n=0(−1)n+1(xπ2)2n+1(2n+1)!,cosx=n=0(−1)n+1(xπ2)2n+1(2n+1)!, which is also cos(xπ2).cos(xπ2).

147.

The derivatives are f(n)(1)=ef(n)(1)=e so ex=en=0(x1)nn!.ex=en=0(x1)nn!.

149.

1(x1)3=(12)d2dx211x=n=0((n+2)(n+1)xn2)1(x1)3=(12)d2dx211x=n=0((n+2)(n+1)xn2)

151.

2x=1(x1)2x=1(x1)

153.

((x1)1)2=(x1)22(x1)+1((x1)1)2=(x1)22(x1)+1

155.

11(1x)=n=0(−1)n(x1)n11(1x)=n=0(−1)n(x1)n

157.

xn=02n(1x)2n=n=02n(x1)2n+1+n=02n(x1)2nxn=02n(1x)2n=n=02n(x1)2n+1+n=02n(x1)2n

159.

e2x=e2(x1)+2=e2n=02n(x1)nn!e2x=e2(x1)+2=e2n=02n(x1)nn!

161.

x=e2;S10=34,91347257.3889947x=e2;S10=34,91347257.3889947

163.

sin(2π)=0;S10=8.27×10−5sin(2π)=0;S10=8.27×10−5

165.


This graph has a concave up curve that is symmetrical about the y axis. The lowest point of the graph is the origin with the rest of the curve above the x-axis.


The difference is small on the interior of the interval but approaches 11 near the endpoints. The remainder estimate is |R4|=π51202.552.|R4|=π51202.552.

167.


This graph has two curves. The solid curve is very flat and close to the x-axis. It passes through the origin. The second curve, a broken line, is concave down and symmetrical about the y-axis. It is very close to the x-axis between -3 and 3.


The difference is on the order of 10−410−4 on [−1,1][−1,1] while the Taylor approximation error is around 0.10.1 near ±1.±1. The top curve is a plot of tan2x(S5(x)C4(x))2tan2x(S5(x)C4(x))2 and the lower dashed plot shows t2(S5C4)2.t2(S5C4)2.

169.

a. Answers will vary. b. The following are the xnxn values after 1010 iterations of Newton’s method to approximation a root of pN(x)2=0:pN(x)2=0: for N=4,x=0.6939...;N=4,x=0.6939...; for N=5,x=0.6932...;N=5,x=0.6932...; for N=6,x=0.69315...;.N=6,x=0.69315...;. (Note: ln(2)=0.69314...)ln(2)=0.69314...) c. Answers will vary.

171.

ln(1x2)x21ln(1x2)x21

173.

cos(x)12x(1x2+x24!)12x14cos(x)12x(1x2+x24!)12x14

Section 6.4 Exercises

175.

(1+x2)−1/3=n=0(13n)x2n(1+x2)−1/3=n=0(13n)x2n

177.

(12x)2/3=n=0(−1)n2n(23n)xn(12x)2/3=n=0(−1)n2n(23n)xn

179.

2+x2=n=02(1/2)n(12n)x2n;(|x2|<2)2+x2=n=02(1/2)n(12n)x2n;(|x2|<2)

181.

2xx2=1(x1)22xx2=1(x1)2 so 2xx2=n=0(−1)n(12n)(x1)2n2xx2=n=0(−1)n(12n)(x1)2n

183.

x=21+x44x=21+x44 so x=n=0212n(12n)(x4)nx=n=0212n(12n)(x4)n

185.

x=n=0313n(12n)(x9)nx=n=0313n(12n)(x9)n

187.

10(1+x1000)1/3=n=01013n(13n)xn.10(1+x1000)1/3=n=01013n(13n)xn. Using, for example, a fourth-degree estimate at x=1x=1 gives (1001)1/310(1+(131)10−3+(132)10−6+(133)10−9+(134)10−12)=10(1+13.10319.106+581.10910243.1012)=10.00333222...(1001)1/310(1+(131)10−3+(132)10−6+(133)10−9+(134)10−12)=10(1+13.10319.106+581.10910243.1012)=10.00333222... whereas (1001)1/3=10.00332222839093....(1001)1/3=10.00332222839093.... Two terms would suffice for three-digit accuracy.

189.

The approximation is 2.3152;2.3152; the CAS value is 2.23.2.23.

191.

The approximation is 2.583;2.583; the CAS value is 2.449.2.449.

193.


1x2=1x22x48x6165x8128+.1x2=1x22x48x6165x8128+. Thus

−111x2dx=xx36x540x77·165x99·128+|−11213120156109·128+error=1.590...−111x2dx=xx36x540x77·165x99·128+|−11213120156109·128+error=1.590... whereas π2=1.570...π2=1.570...

195.

(1+x)4/3=(1+x)(1+13x19x2+581x310243x4+)=1+4x3+2x294x381+5x4243+(1+x)4/3=(1+x)(1+13x19x2+581x310243x4+)=1+4x3+2x294x381+5x4243+

197.

(1+(x+3)2)1/3=1+13(x+3)219(x+3)4+581(x+3)610243(x+3)8+(1+(x+3)2)1/3=1+13(x+3)219(x+3)4+581(x+3)610243(x+3)8+

199.

Twice the approximation is 1.2601.260 whereas 21/3=1.2599....21/3=1.2599....

201.

f(99)(0)=0f(99)(0)=0

203.

n=0(ln(2)x)nn!n=0(ln(2)x)nn!

205.

For x>0,sin(x)=n=0(−1)nx(2n+1)/2x(2n+1)!=n=0(−1)nxn(2n+1)!.x>0,sin(x)=n=0(−1)nx(2n+1)/2x(2n+1)!=n=0(−1)nxn(2n+1)!.

207.

ex3=n=0x3nn!ex3=n=0x3nn!

209.

sin2x=k=1(−1)k22k1x2k(2k)!sin2x=k=1(−1)k22k1x2k(2k)!

211.

tan−1x=k=0(−1)kx2k+12k+1tan−1x=k=0(−1)kx2k+12k+1

213.

sin−1x=n=0(12n)x2n+1(2n+1)n!sin−1x=n=0(12n)x2n+1(2n+1)n!

215.

F(x)=n=0(−1)nxn+1(n+1)(2n)!F(x)=n=0(−1)nxn+1(n+1)(2n)!

217.

F(x)=n=1(−1)n+1xnn2F(x)=n=1(−1)n+1xnn2

219.

x+x33+2x515+x+x33+2x515+

221.

1+xx33x46+1+xx33x46+

223.

1+x2+2x43+17x645+1+x2+2x43+17x645+

225.

Using the expansion for tanxtanx gives 1+x3+2x215.1+x3+2x215.

227.

11+x2=n=0(−1)nx2n11+x2=n=0(−1)nx2n so R=1R=1 by the ratio test.

229.

ln(1+x2)=n=1(−1)n1nx2nln(1+x2)=n=1(−1)n1nx2n so R=1R=1 by the ratio test.

231.

Add series of exex and exex term by term. Odd terms cancel and coshx=n=0x2n(2n)!.coshx=n=0x2n(2n)!.

233.


This graph has two curves. The first one is a decreasing function passing through the origin. The second is a broken line which is an increasing function passing through the origin. The two curves are very close around the origin.


The ratio Sn(x)Cn(x)Sn(x)Cn(x) approximates tanxtanx better than does p7(x)=x+x33+2x515+17x7315p7(x)=x+x33+2x515+17x7315 for N3.N3. The dashed curves are SnCntanSnCntan for n=1,2.n=1,2. The dotted curve corresponds to n=3,n=3, and the dash-dotted curve corresponds to n=4.n=4. The solid curve is p7tanx.p7tanx.

235.

By the term-by-term differentiation theorem, y=n=1nanxn1y=n=1nanxn1 so y=n=1nanxn1xy=n=1nanxn,y=n=1nanxn1xy=n=1nanxn, whereas y=n=2n(n1)anxn2y=n=2n(n1)anxn2 so xy=n=2n(n1)anxn.xy=n=2n(n1)anxn.

237.

The probability is p=12π(aμ)/σ(bμ)/σex2/2dxp=12π(aμ)/σ(bμ)/σex2/2dx where a=90a=90 and b=100,b=100, that is, p=12π−11ex2/2dx=12π−11n=05(−1)nx2n2nn!dx=22πn=05(−1)n1(2n+1)2nn!0.6827.p=12π−11ex2/2dx=12π−11n=05(−1)nx2n2nn!dx=22πn=05(−1)n1(2n+1)2nn!0.6827.

239.


This graph is a wave curve symmetrical about the origin. It has a peak at y = 1 above the origin. It has lowest points at -3 and 3.


As in the previous problem one obtains an=0an=0 if nn is odd and an=(n+2)(n+1)an+2an=(n+2)(n+1)an+2 if nn is even, so a0=1a0=1 leads to a2n=(−1)n(2n)!.a2n=(−1)n(2n)!.

241.

y=n=0(n+2)(n+1)an+2xny=n=0(n+2)(n+1)an+2xn and y=n=0(n+1)an+1xny=n=0(n+1)an+1xn so yy+y=0yy+y=0 implies that (n+2)(n+1)an+2(n+1)an+1+an=0(n+2)(n+1)an+2(n+1)an+1+an=0 or an=an1nan2n(n1)an=an1nan2n(n1) for all n·y(0)=a0=1n·y(0)=a0=1 and y(0)=a1=0,y(0)=a1=0, so a2=12,a3=16,a4=0,a2=12,a3=16,a4=0, and a5=1120.a5=1120.

243.

a. (Proof) b. We have Rs0.1(9)!π90.0082<0.01.Rs0.1(9)!π90.0082<0.01. We have 0π(1x23!+x45!x67!+x89!)dx=ππ33·3!+π55·5!π77·7!+π99·9!=1.852...,0π(1x23!+x45!x67!+x89!)dx=ππ33·3!+π55·5!π77·7!+π99·9!=1.852..., whereas 0πsinttdt=1.85194...,0πsinttdt=1.85194..., so the actual error is approximately 0.00006.0.00006.

245.


This graph has two curves. The first one is a solid curve labeled Csub50(x). It begins at the origin and is a wave that gradually decreases in amplitude. The highest it reaches is y = 1. The second curve is labeled Ssub50(x). It is a wave that gradually decreases in amplitude. The highest it reaches is 0.9. It is very close to the pattern of the first curve with a slight shift to the right.


Since cos(t2)=n=0(−1)nt4n(2n)!cos(t2)=n=0(−1)nt4n(2n)! and sin(t2)=n=0(−1)nt4n+2(2n+1)!,sin(t2)=n=0(−1)nt4n+2(2n+1)!, one has S(x)=n=0(−1)nx4n+3(4n+3)(2n+1)!S(x)=n=0(−1)nx4n+3(4n+3)(2n+1)! and C(x)=n=0(−1)nx4n+1(4n+1)(2n)!.C(x)=n=0(−1)nx4n+1(4n+1)(2n)!. The sums of the first 5050 nonzero terms are plotted below with C50(x)C50(x) the solid curve and S50(x)S50(x) the dashed curve.

247.

01/4x(1x2x28x3165x41287x5256)dx01/4x(1x2x28x3165x41287x5256)dx

=232−312252−518272−7116292−951282112−1172562132−13=0.0767732...=232−312252−518272−7116292−951282112−1172562132−13=0.0767732...

whereas 01/4xx2dx=0.076773.01/4xx2dx=0.076773.

249.

T2π109.8(1+sin2(θ/12)4)6.453T2π109.8(1+sin2(θ/12)4)6.453 seconds. The small angle estimate is T2π109.86.347.T2π109.86.347. The relative error is around 22 percent.

251.

0π/2sin4θdθ=3π16.0π/2sin4θdθ=3π16. Hence T2πLg(1+k24+9256k4).T2πLg(1+k24+9256k4).

Chapter Review Exercises

253.

True

255.

True

257.

ROC: 1;1; IOC: (0,2)(0,2)

259.

ROC: 12;12; IOC: (−16,8)(−16,8)

261.

n=0(−1)n3n+1xn;n=0(−1)n3n+1xn; ROC: 3;3; IOC: (−3,3)(−3,3)

263.

integration: n=0(−1)n2n+1(2x)2n+1n=0(−1)n2n+1(2x)2n+1

265.

p4(x)=(x+3)311(x+3)2+39(x+3)41;p4(x)=(x+3)311(x+3)2+39(x+3)41; exact

267.

n=0(−1)n(3x)2n2n!n=0(−1)n(3x)2n2n!

269.

n=0(−1)n(2n)!(xπ2)2nn=0(−1)n(2n)!(xπ2)2n

271.

n=1(−1)nn!x2nn=1(−1)nn!x2n

273.

F(x)=n=0(−1)n(2n+1)(2n+1)!x2n+1F(x)=n=0(−1)n(2n+1)(2n+1)!x2n+1

275.

Answers may vary.

277.

2.5%2.5%

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