Calculus Volume 2

# Chapter 6

### Checkpoint

6.1

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

6.2 6.3

$∑n=0∞xn+32n+1∑n=0∞xn+32n+1$ 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

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

6.7

$1+2x+3x2+4x3+⋯1+2x+3x2+4x3+⋯$

6.8

$∑n=0∞(n+2)(n+1)xn∑n=0∞(n+2)(n+1)xn$

6.9

$∑n=2∞(−1)nxnn(n−1)∑n=2∞(−1)nxnn(n−1)$

6.10

$p0(x)=1;p1(x)=1−2(x−1);p2(x)=1−2(x−1)+3(x−1)2;p3(x)=1−2(x−1)+3(x−1)2−4(x−1)3p0(x)=1;p1(x)=1−2(x−1);p2(x)=1−2(x−1)+3(x−1)2;p3(x)=1−2(x−1)+3(x−1)2−4(x−1)3$

6.11

$p0(x)=1;p1(x)=1−x;p2(x)=1−x+x2;p3(x)=1−x+x2−x3;pn(x)=1−x+x2−x3+⋯+(−1)nxn=∑k=0n(−1)kxkp0(x)=1;p1(x)=1−x;p2(x)=1−x+x2;p3(x)=1−x+x2−x3;pn(x)=1−x+x2−x3+⋯+(−1)nxn=∑k=0n(−1)kxk$

6.12

$p1(x)=2+14(x−4);p2(x)=2+14(x−4)−164(x−4)2;p1(6)=2.5;p2(6)=2.4375;p1(x)=2+14(x−4);p2(x)=2+14(x−4)−164(x−4)2;p1(6)=2.5;p2(6)=2.4375;$

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

6.13

0.96593

6.14

$∑n=0∞(2−x2n+2 )n.∑n=0∞(2−x2n+2 )n.$ 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

$∑n=0∞(−1)n(n+1)xn∑n=0∞(−1)n(n+1)xn$

6.17

$∑n=0∞(−1)nx4n+2(2n+1)!∑n=0∞(−1)nx4n+2(2n+1)!$

6.18

$∑n=1∞(−1)nn!1·3·5⋯(2n−1)2nxn∑n=1∞(−1)nn!1·3·5⋯(2n−1)2nxn$

6.19

$y=5e2xy=5e2x$

6.20

$y=a(1−x43·4+x83·4·7·8−⋯)+b(x−x54·5+x94·5·8·9−⋯)y=a(1−x43·4+x83·4·7·8−⋯)+b(x−x54·5+x94·5·8·9−⋯)$

6.21

$C+∑n=1∞(−1)n+1xnn(2n−2)!C+∑n=1∞(−1)n+1xnn(2n−2)!$ 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 $anxn→0anxn→0$ for $|x| If $an=nn,an=nn,$ then $anxn=(nx)nanxn=(nx)n$ does not tend to zero for any $x≠0.x≠0.$

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 $an+1xan→2x.an+1xan→2x.$ 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 $an+1xan→x2an+1xan→x2$ 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.

$an=10nn!,an+1xan=10xn+1→0<1an=10nn!,an+1xan=10xn+1→0<1$ 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.

$ak=k!1·3·5⋯(2k−1)ak=k!1·3·5⋯(2k−1)$ so $ak+1ak=k+12k+1→12ak+1ak=k+12k+1→12$ 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 $an+1an=(n+1)!n!nn(n+1)n+1=(nn+1)n→1ean+1an=(n+1)!n!nn(n+1)n+1=(nn+1)n→1e$ so $R=eR=e$

33.

$f(x)=∑n=0∞(1−x)nf(x)=∑n=0∞(1−x)n$ on $I=(0,2)I=(0,2)$

35.

$∑n=0∞x2n+1∑n=0∞x2n+1$ on $I=(−1,1)I=(−1,1)$

37.

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

39.

$∑n=0∞2nxn∑n=0∞2nxn$ on $(−12,12)(−12,12)$

41.

$∑n=0∞4nx2n+2∑n=0∞4nx2n+2$ 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 $n→∞n→∞$ and $|x|r<1|x|r<1$ when $|x|<1r.|x|<1r.$ Therefore, $∑n=1∞anxn∑n=1∞anxn$ converges when $|x|<1r|x|<1r$ by the nth root test.

45.

$ak=(k−12k+3)kak=(k−12k+3)k$ so $(ak)1/k→12<1(ak)1/k→12<1$ so $R=2R=2$

47.

$an=(n1/n−1)nan=(n1/n−1)n$ so $(an)1/n→0(an)1/n→0$ so $R=∞R=∞$

49.

We can rewrite $p(x)=∑n=0∞a2n+1x2n+1p(x)=∑n=0∞a2n+1x2n+1$ 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 $y=2x−1∈[−1,1]y=2x−1∈[−1,1]$ so $p(2x−1)=p(y)=∑n=0∞anynp(2x−1)=p(y)=∑n=0∞anyn$ converges.

53.

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

55.

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

57. The approximation is more accurate near $x=−1.x=−1.$ The partial sums follow $11−x11−x$ more closely as N increases but are never accurate near $x=1x=1$ since the series diverges there.

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

61. 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.

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

65.

$4(x−3)(x+1)=1x−3−1x+1=−13(1−x3)−11−(−x)=−13∑n=0∞(x3)n−∑n=0∞(−1)nxn=∑n=0∞((−1)n+1−13n+1)xn4(x−3)(x+1)=1x−3−1x+1=−13(1−x3)−11−(−x)=−13∑n=0∞(x3)n−∑n=0∞(−1)nxn=∑n=0∞((−1)n+1−13n+1)xn$

67.

$5(x2+4)(x2−1)=1x2−1−1411+(x2)2=−∑n=0∞x2n−14∑n=0∞(−1)n(x2)2n=∑n=0∞((−1)+(−1)n+112n+2)x2n5(x2+4)(x2−1)=1x2−1−1411+(x2)2=−∑n=0∞x2n−14∑n=0∞(−1)n(x2)2n=∑n=0∞((−1)+(−1)n+112n+2)x2n$

69.

$1x∑n=0∞1xn=1x11−1x=1x−11x∑n=0∞1xn=1x11−1x=1x−1$

71.

$1x−311−1(x−3)2=x−3(x−3)2−11x−311−1(x−3)2=x−3(x−3)2−1$

73.

$P=P1+⋯+P20P=P1+⋯+P20$ where $Pk=10,0001(1+r)k.Pk=10,0001(1+r)k.$ Then $P=10,000∑k=1201(1+r)k=10,0001−(1+r)−20r.P=10,000∑k=1201(1+r)k=10,0001−(1+r)−20r.$ When $r=0.03,P≈10,000×14.8775=148,775.r=0.03,P≈10,000×14.8775=148,775.$ When $r=0.05,P≈10,000×12.4622=124,622.r=0.05,P≈10,000×12.4622=124,622.$ When $r=0.07,P≈105,940.r=0.07,P≈105,940.$

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 $C≈9439.29C≈9439.29$ 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.

$(x+x2−x3)(1+x3+x6+⋯)=x+x2−x31−x3(x+x2−x3)(1+x3+x6+⋯)=x+x2−x31−x3$

81.

$(x−x2−x3)(1+x3+x6+⋯)=x−x2−x31−x3(x−x2−x3)(1+x3+x6+⋯)=x−x2−x31−x3$

83.

$an=2,bn=nan=2,bn=n$ so $cn=∑k=0nbkan−k=2∑k=0nk=(n)(n+1)cn=∑k=0nbkan−k=2∑k=0nk=(n)(n+1)$ and $f(x)g(x)=∑n=1∞n(n+1)xnf(x)g(x)=∑n=1∞n(n+1)xn$

85.

$an=bn=2−nan=bn=2−n$ so $cn=∑k=1nbkan−k=2−n∑k=1n1=n2ncn=∑k=1nbkan−k=2−n∑k=1n1=n2n$ and $f(x)g(x)=∑n=1∞n(x2)nf(x)g(x)=∑n=1∞n(x2)n$

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.

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

93.

$f(x)=∑n=0∞xn=11−x;f″(12)=∑n=2∞n(n−1)2n−2=d2dx2(1−x)−1|x=1/2=2(1−x)3|x=1/2=16f(x)=∑n=0∞xn=11−x;f″(12)=∑n=2∞n(n−1)2n−2=d2dx2(1−x)−1|x=1/2=2(1−x)3|x=1/2=16$ so $∑n=2∞n(n−1)2n=4.∑n=2∞n(n−1)2n=4.$

95.

$∫∑(1−x)ndx=∫∑(−1)n(x−1)ndx=∑(−1)n(x−1)n+1n+1∫∑(1−x)ndx=∫∑(−1)n(x−1)ndx=∑(−1)n(x−1)n+1n+1$

97.

$−∫t=0x211−tdt=−∑n=0∞∫0x2tndx−∑n=0∞x2(n+1)n+1=−∑n=1∞x2nn−∫t=0x211−tdt=−∑n=0∞∫0x2tndx−∑n=0∞x2(n+1)n+1=−∑n=1∞x2nn$

99.

$∫0x2dt1+t2=∑n=0∞(−1)n∫0x2t2ndt=∑n=0∞(−1)nt2n+12n+1|t=0x2=∑n=0∞(−1)nx4n+22n+1∫0x2dt1+t2=∑n=0∞(−1)n∫0x2t2ndt=∑n=0∞(−1)nt2n+12n+1|t=0x2=∑n=0∞(−1)nx4n+22n+1$

101.

Term-by-term integration gives $∫0xlntdt=∑n=1∞(−1)n−1(x−1)n+1n(n+1)=∑n=1∞(−1)n−1(1n−1n+1)(x−1)n+1=(x−1)lnx+∑n=2∞(−1)n(x−1)nn=xlnx−x.∫0xlntdt=∑n=1∞(−1)n−1(x−1)n+1n(n+1)=∑n=1∞(−1)n−1(1n−1n+1)(x−1)n+1=(x−1)lnx+∑n=2∞(−1)n(x−1)nn=xlnx−x.$

103.

We have $ln(1−x)=−∑n=1∞xnnln(1−x)=−∑n=1∞xnn$ so $ln(1+x)=∑n=1∞(−1)n−1xnn.ln(1+x)=∑n=1∞(−1)n−1xnn.$ Thus, $ln(1+x1−x)=∑n=1∞(1+(−1)n−1)xnn=2∑n=1∞x2n−12n−1.ln(1+x1−x)=∑n=1∞(1+(−1)n−1)xnn=2∑n=1∞x2n−12n−1.$ When $x=13x=13$ we obtain $ln(2)=2∑n=1∞132n−1(2n−1).ln(2)=2∑n=1∞132n−1(2n−1).$ We have $2∑n=13132n−1(2n−1)=0.69300…,2∑n=13132n−1(2n−1)=0.69300…,$ while $2∑n=14132n−1(2n−1)=0.69313…2∑n=14132n−1(2n−1)=0.69313…$ and $ln(2)=0.69314…;ln(2)=0.69314…;$ therefore, $N=4.N=4.$

105.

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

107.

If $y=2−x,y=2−x,$ then $∑k=1∞yk=y1−y=2−x1−2−x=12x−1.∑k=1∞yk=y1−y=2−x1−2−x=12x−1.$ If $ak=2−kx,ak=2−kx,$ then $ak+1ak=2−x<1ak+1ak=2−x<1$ when $x>0.x>0.$ So the series converges for all $x>0.x>0.$

109.

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

113.

When $x=−12,−ln(2)=ln(12)=−∑n=1∞1n2n.x=−12,−ln(2)=ln(12)=−∑n=1∞1n2n.$ Since $∑n=11∞1n2n<∑n=11∞12n=1210,∑n=11∞1n2n<∑n=11∞12n=1210,$ one has $∑n=1101n2n=0.69306…∑n=1101n2n=0.69306…$ whereas $ln(2)=0.69314…;ln(2)=0.69314…;$ therefore, $N=10.N=10.$

115.

$6SN(13)=23∑n=0N(−1)n13n(2n+1).6SN(13)=23∑n=0N(−1)n13n(2n+1).$ One has $π−6S4(13)=0.00101…π−6S4(13)=0.00101…$ and $π−6S5(13)=0.00028…π−6S5(13)=0.00028…$ so $N=5N=5$ is the smallest partial sum with accuracy to within 0.001. Also, $π−6S7(13)=0.00002…π−6S7(13)=0.00002…$ while $π−6S8(13)=−0.000007…π−6S8(13)=−0.000007…$ 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.

$f′(x)=2cos(2x);f″(x)=−4sin(2x);p2(x)=−2(x−π2)f′(x)=2cos(2x);f″(x)=−4sin(2x);p2(x)=−2(x−π2)$

121.

$f′(x)=1x;f″(x)=−1x2;p2(x)=0+(x−1)−12(x−1)2f′(x)=1x;f″(x)=−1x2;p2(x)=0+(x−1)−12(x−1)2$

123.

$p2(x)=e+e(x−1)+e2(x−1)2p2(x)=e+e(x−1)+e2(x−1)2$

125.

$d2dx2x1/3=−29x5/3≥−0.00092…d2dx2x1/3=−29x5/3≥−0.00092…$ when $x≥28x≥28$ so the remainder estimate applies to the linear approximation $x1/3≈p1(27)=3+x−2727,x1/3≈p1(27)=3+x−2727,$ which gives $(28)1/3≈3+127=3.037¯,(28)1/3≈3+127=3.037¯,$ while $(28)1/3≈3.03658.(28)1/3≈3.03658.$

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 $e2≈p9(2)=1+2+222+236+⋯+299!=7.3887…e2≈p9(2)=1+2+222+236+⋯+299!=7.3887…$ whereas $e2≈7.3891.e2≈7.3891.$

129.

Since $dndxn(lnx)=(−1)n−1(n−1)!xn,R1000≈11001.dndxn(lnx)=(−1)n−1(n−1)!xn,R1000≈11001.$ One has $p1000(1)=∑n=11000(−1)n−1n≈0.6936p1000(1)=∑n=11000(−1)n−1n≈0.6936$ whereas $ln(2)≈0.6931⋯.ln(2)≈0.6931⋯.$

131.

$∫01(1−x2+x42−x66+x824−x10120+x12720)dx∫01(1−x2+x42−x66+x824−x10120+x12720)dx$

$=1−133+1510−1742+199·24−111120·11+113720·13≈0.74683=1−133+1510−1742+199·24−111120·11+113720·13≈0.74683$ whereas $∫01e−x2dx≈0.74682.∫01e−x2dx≈0.74682.$

133.

Since $f(n+1)(z)f(n+1)(z)$ is $sinzsinz$ or $cosz,cosz,$ we have $M=1.M=1.$ Since $|x−0|≤π2,|x−0|≤π2,$ 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 $R7≤0.00092.R7≤0.00092.$

135.

Since $f(n+1)(z)=±e−zf(n+1)(z)=±e−z$ one has $M=e3.M=e3.$ Since $|x−0|≤3,|x−0|≤3,$ 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 $R14≤0.000220.R14≤0.000220.$

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

139. Since the second derivative of $cosxcosx$ is $−cosx−cosx$ and since $cosxcosx$ is decreasing away from $x=0,x=0,$ the estimate applies when $R2cosR≤0.2R2cosR≤0.2$ or $R≤0.447.R≤0.447.$

141.

$(x+1)3−2(x+1)2+2(x+1)(x+1)3−2(x+1)2+2(x+1)$

143.

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

145.

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

147.

The derivatives are $f(n)(1)=ef(n)(1)=e$ so $ex=e∑n=0∞(x−1)nn!.ex=e∑n=0∞(x−1)nn!.$

149.

$1(x−1)3=−(12)d2dx211−x=−∑n=0∞((n+2)(n+1)xn2)1(x−1)3=−(12)d2dx211−x=−∑n=0∞((n+2)(n+1)xn2)$

151.

$2−x=1−(x−1)2−x=1−(x−1)$

153.

$((x−1)−1)2=(x−1)2−2(x−1)+1((x−1)−1)2=(x−1)2−2(x−1)+1$

155.

$11−(1−x)=∑n=0∞(−1)n(x−1)n11−(1−x)=∑n=0∞(−1)n(x−1)n$

157.

$x∑n=0∞2n(1−x)2n=∑n=0∞2n(x−1)2n+1+∑n=0∞2n(x−1)2nx∑n=0∞2n(1−x)2n=∑n=0∞2n(x−1)2n+1+∑n=0∞2n(x−1)2n$

159.

$e2x=e2(x−1)+2=e2∑n=0∞2n(x−1)nn!e2x=e2(x−1)+2=e2∑n=0∞2n(x−1)nn!$

161.

$x=e2;S10=34,9134725≈7.3889947x=e2;S10=34,9134725≈7.3889947$

163.

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

165. The difference is small on the interior of the interval but approaches $11$ near the endpoints. The remainder estimate is $|R4|=π5120≈2.552.|R4|=π5120≈2.552.$

167. 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.

$ln(1−x2)x2→−1ln(1−x2)x2→−1$

173.

$cos(x)−12x≈(1−x2+x24!−⋯)−12x→−14cos(x)−12x≈(1−x2+x24!−⋯)−12x→−14$

### Section 6.4 Exercises

175.

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

177.

$(1−2x)2/3=∑n=0∞(−1)n2n(23n)xn(1−2x)2/3=∑n=0∞(−1)n2n(23n)xn$

179.

$2+x2=∑n=0∞2(1/2)−n(12n)x2n;(|x2|<2)2+x2=∑n=0∞2(1/2)−n(12n)x2n;(|x2|<2)$

181.

$2x−x2=1−(x−1)22x−x2=1−(x−1)2$ so $2x−x2=∑n=0∞(−1)n(12n)(x−1)2n2x−x2=∑n=0∞(−1)n(12n)(x−1)2n$

183.

$x=21+x−44x=21+x−44$ so $x=∑n=0∞21−2n(12n)(x−4)nx=∑n=0∞21−2n(12n)(x−4)n$

185.

$x=∑n=0∞31−3n(12n)(x−9)nx=∑n=0∞31−3n(12n)(x−9)n$

187.

$10(1+x1000)1/3=∑n=0∞101−3n(13n)xn.10(1+x1000)1/3=∑n=0∞101−3n(13n)xn.$ Using, for example, a fourth-degree estimate at $x=1x=1$ gives $(1001)1/3≈10(1+(131)10−3+(132)10−6+(133)10−9+(134)10−12)=10(1+13.103−19.106+581.109−10243.1012)=10.00333222...(1001)1/3≈10(1+(131)10−3+(132)10−6+(133)10−9+(134)10−12)=10(1+13.103−19.106+581.109−10243.1012)=10.00333222...$ 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.

$1−x2=1−x22−x48−x616−5x8128+⋯.1−x2=1−x22−x48−x616−5x8128+⋯.$ Thus

$∫−111−x2dx=x−x36−x540−x77·16−5x99·128+⋯|−11≈2−13−120−156−109·128+error=1.590...∫−111−x2dx=x−x36−x540−x77·16−5x99·128+⋯|−11≈2−13−120−156−109·128+error=1.590...$ whereas $π2=1.570...π2=1.570...$

195.

$(1+x)4/3=(1+x)(1+13x−19x2+581x3−10243x4+⋯)=1+4x3+2x29−4x381+5x4243+⋯(1+x)4/3=(1+x)(1+13x−19x2+581x3−10243x4+⋯)=1+4x3+2x29−4x381+5x4243+⋯$

197.

$(1+(x+3)2)1/3=1+13(x+3)2−19(x+3)4+581(x+3)6−10243(x+3)8+⋯(1+(x+3)2)1/3=1+13(x+3)2−19(x+3)4+581(x+3)6−10243(x+3)8+⋯$

199.

Twice the approximation is $1.260…1.260…$ 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.

$ex3=∑n=0∞x3nn!ex3=∑n=0∞x3nn!$

209.

$sin2x=−∑k=1∞(−1)k22k−1x2k(2k)!sin2x=−∑k=1∞(−1)k22k−1x2k(2k)!$

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.

$1+x−x33−x46+⋯1+x−x33−x46+⋯$

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.

$ln(1+x2)=∑n=1∞(−1)n−1nx2nln(1+x2)=∑n=1∞(−1)n−1nx2n$ so $R=1R=1$ by the ratio test.

231.

Add series of $exex$ and $e−xe−x$ term by term. Odd terms cancel and $coshx=∑n=0∞x2n(2n)!.coshx=∑n=0∞x2n(2n)!.$

233. 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 $N≥3.N≥3.$ The dashed curves are $SnCn−tanSnCn−tan$ 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 $p7−tanx.p7−tanx.$

235.

By the term-by-term differentiation theorem, $y′=∑n=1∞nanxn−1y′=∑n=1∞nanxn−1$ so $y′=∑n=1∞nanxn−1xy′=∑n=1∞nanxn,y′=∑n=1∞nanxn−1xy′=∑n=1∞nanxn,$ whereas $y′=∑n=2∞n(n−1)anxn−2y′=∑n=2∞n(n−1)anxn−2$ so $xy″=∑n=2∞n(n−1)anxn.xy″=∑n=2∞n(n−1)anxn.$

237.

The probability is $p=12π∫(a−μ)/σ(b−μ)/σe−x2/2dxp=12π∫(a−μ)/σ(b−μ)/σe−x2/2dx$ where $a=90a=90$ and $b=100,b=100,$ that is, $p=12π∫−11e−x2/2dx=12π∫−11∑n=05(−1)nx2n2nn!dx=22π∑n=05(−1)n1(2n+1)2nn!≈0.6827.p=12π∫−11e−x2/2dx=12π∫−11∑n=05(−1)nx2n2nn!dx=22π∑n=05(−1)n1(2n+1)2nn!≈0.6827.$

239. 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 $y″−y′+y=0y″−y′+y=0$ 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 $an=an−1n−an−2n(n−1)an=an−1n−an−2n(n−1)$ 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 $Rs≤0.1(9)!π9≈0.0082<0.01.Rs≤0.1(9)!π9≈0.0082<0.01.$ We have $∫0π(1−x23!+x45!−x67!+x89!)dx=π−π33·3!+π55·5!−π77·7!+π99·9!=1.852...,∫0π(1−x23!+x45!−x67!+x89!)dx=π−π33·3!+π55·5!−π77·7!+π99·9!=1.852...,$ whereas $∫0πsinttdt=1.85194...,∫0πsinttdt=1.85194...,$ so the actual error is approximately $0.00006.0.00006.$

245. 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.

$∫01/4x(1−x2−x28−x316−5x4128−7x5256)dx∫01/4x(1−x2−x28−x316−5x4128−7x5256)dx$

$=232−3−12252−5−18272−7−116292−9−51282112−11−72562132−13=0.0767732...=232−3−12252−5−18272−7−116292−9−51282112−11−72562132−13=0.0767732...$

whereas $∫01/4x−x2dx=0.076773.∫01/4x−x2dx=0.076773.$

249.

$T≈2π109.8(1+sin2(θ/12)4)≈6.453T≈2π109.8(1+sin2(θ/12)4)≈6.453$ seconds. The small angle estimate is $T≈2π109.8≈6.347.T≈2π109.8≈6.347.$ The relative error is around $22$ percent.

251.

$∫0π/2sin4θdθ=3π16.∫0π/2sin4θdθ=3π16.$ Hence $T≈2πLg(1+k24+9256k4).T≈2πLg(1+k24+9256k4).$

### 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: $∑n=0∞(−1)n2n+1(2x)2n+1∑n=0∞(−1)n2n+1(2x)2n+1$

265.

$p4(x)=(x+3)3−11(x+3)2+39(x+3)−41;p4(x)=(x+3)3−11(x+3)2+39(x+3)−41;$ exact

267.

$∑n=0∞(−1)n(3x)2n2n!∑n=0∞(−1)n(3x)2n2n!$

269.

$∑n=0∞(−1)n(2n)!(x−π2)2n∑n=0∞(−1)n(2n)!(x−π2)2n$

271.

$∑n=1∞(−1)nn!x2n∑n=1∞(−1)nn!x2n$

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.

277.

$2.5%2.5%$