$p_{0} (x) = 1; p_{1} (x) = 1 - 2 (x - 1); p_{2} (x) = 1 - 2 (x - 1) + 3 {(x - 1)}^{2}; p_{3} (x) = 1 - 2 (x - 1) + 3 {(x - 1)}^{2} - 4 {(x - 1)}^{3}$

6.11

$p_{0} (x) = 1; p_{1} (x) = 1 - x; p_{2} (x) = 1 - x + x^{2}; p_{3} (x) = 1 - x + x^{2} - x^{3}; p_{n} (x) = 1 - x + x^{2} - x^{3} + ⋯ + {(−1)}^{n} x^{n} = \sum_{k = 0}^{n} {(−1)}^{k} x^{k}$

6.12

$p_{1} (x) = 2 + \frac{1}{4} (x - 4); p_{2} (x) = 2 + \frac{1}{4} (x - 4) - \frac{1}{64} {(x - 4)}^{2}; p_{1} (6) = 2.5; p_{2} (6) = 2.4375;$

$| R_{1} (6) | \leq 0.0625; | R_{2} (6) | \leq 0.015625$

6.13

0.96593

6.14

$\sum_{n = 0}^{\infty} \frac{{(2 - x)}^{n}}{2^{n + 1}} .$ The interval of convergence is $(0, 4) .$

6.15

$\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} x^{2 n}}{(2 n)!}$

By the ratio test, the interval of convergence is $(− \infty, \infty) .$ Since $| R_{n} (x) | \leq \frac{{| x |}^{n + 1}}{(n + 1)!},$ the series converges to $cos x$ for all real x.

6.16

$\sum_{n = 0}^{\infty} {(−1)}^{n} (n + 1) x^{n}$

6.17

$\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} x^{4 n + 2}}{(2 n + 1)!}$

6.18

$\sum_{n = 1}^{\infty} \frac{{(−1)}^{n}}{n!} \frac{1 \cdot 3 \cdot 5 ⋯ (2 n - 1)}{2^{n}} x^{n}$

6.19

$y = 5 e^{2 x}$

6.20

$y = a (1 - \frac{x^{4}}{3 \cdot 4} + \frac{x^{8}}{3 \cdot 4 \cdot 7 \cdot 8} - ⋯) + b (x - \frac{x^{5}}{4 \cdot 5} + \frac{x^{9}}{4 \cdot 5 \cdot 8 \cdot 9} - ⋯)$

6.21

$C + \sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n (2 n - 2)!}$ The definite integral is approximately $0.514$ to within an error of $0.01 .$

6.22

The estimate is approximately $0.3414 .$ This estimate is accurate to within $0.0000094 .$

Section 6.1 Exercises

1.

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

3.

False. It would imply that $a_{n} x^{n} \to 0$ for $| x | < R .$ If $a_{n} = n^{n},$ then $a_{n} x^{n} = {(n x)}^{n}$ does not tend to zero for any $x \neq 0 .$

5.

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

7.

$| \frac{a_{n + 1} 2^{n + 1} x^{n + 1}}{a_{n} 2^{n} x^{n}} | = 2 | x | | \frac{a_{n + 1}}{a_{n}} | \to 2 | x |$ so $R = \frac{1}{2}$

9.

$| \frac{a_{n + 1} {(\frac{π}{e})}^{n + 1} x^{n + 1}}{a_{n} {(\frac{π}{e})}^{n} x^{n}} | = \frac{π | x |}{e} | \frac{a_{n + 1}}{a_{n}} | \to \frac{π | x |}{e}$ so $R = \frac{e}{π}$

11.

$| \frac{a_{n + 1} {(−1)}^{n + 1} x^{2 n + 2}}{a_{n} {(−1)}^{n} x^{2 n}} | = | x^{2} | | \frac{a_{n + 1}}{a_{n}} | \to | x^{2} |$ so $R = 1$

13.

$a_{n} = \frac{2^{n}}{n}$ so $\frac{a_{n + 1} x}{a_{n}} \to 2 x .$ so $R = \frac{1}{2} .$ When $x = \frac{1}{2}$ the series is harmonic and diverges. When $x = - \frac{1}{2}$ the series is alternating harmonic and converges. The interval of convergence is $I = [- \frac{1}{2}, \frac{1}{2}) .$

15.

$a_{n} = \frac{n}{2^{n}}$ so $\frac{a_{n + 1} x}{a_{n}} \to \frac{x}{2}$ so $R = 2 .$ When $x = ± 2$ the series diverges by the divergence test. The interval of convergence is $I = (−2, 2) .$

17.

$a_{n} = \frac{n^{2}}{2^{n}}$ so $R = 2 .$ When $x = ± 2$ the series diverges by the divergence test. The interval of convergence is $I = (−2, 2) .$

19.

$a_{k} = \frac{π^{k}}{k^{π}}$ so $R = \frac{1}{π} .$ When $x = ± \frac{1}{π}$ the series is an absolutely convergent p-series. The interval of convergence is $I = [- \frac{1}{π}, \frac{1}{π}] .$

21.

$a_{n} = \frac{10^{n}}{n!}, \frac{a_{n + 1} x}{a_{n}} = \frac{10 x}{n + 1} \to 0 < 1$ so the series converges for all x by the ratio test and $I = (− \infty, \infty) .$

23.

$a_{k} = \frac{{(k!)}^{2}}{(2 k)!}$ so $\frac{a_{k + 1}}{a_{k}} = \frac{{(k + 1)}^{2}}{(2 k + 2) (2 k + 1)} \to \frac{1}{4}$ so $R = 4$

25.

$a_{k} = \frac{k!}{1 \cdot 3 \cdot 5 ⋯ (2 k - 1)}$ so $\frac{a_{k + 1}}{a_{k}} = \frac{k + 1}{2 k + 1} \to \frac{1}{2}$ so $R = 2$

27.

$a_{n} = \frac{1}{(\begin{matrix} 2 n \\ n \end{matrix})}$ so $\frac{a_{n + 1}}{a_{n}} = \frac{{((n + 1)!)}^{2}}{(2 n + 2)!} \frac{2 n!}{{(n!)}^{2}} = \frac{{(n + 1)}^{2}}{(2 n + 2) (2 n + 1)} \to \frac{1}{4}$ so $R = 4$

29.

$\frac{a_{n + 1}}{a_{n}} = \frac{{(n + 1)}^{3}}{(3 n + 3) (3 n + 2) (3 n + 1)} \to \frac{1}{27}$ so $R = 27$

31.

$a_{n} = \frac{n!}{n^{n}}$ so $\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1)!}{n!} \frac{n^{n}}{{(n + 1)}^{n + 1}} = {(\frac{n}{n + 1})}^{n} \to \frac{1}{e}$ so $R = e$

33.

$f (x) = \sum_{n = 0}^{\infty} {(1 - x)}^{n}$ on $I = (0, 2)$

35.

$\sum_{n = 0}^{\infty} x^{2 n + 1}$ on $I = (−1, 1)$

37.

$\sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n + 2}$ on $I = (−1, 1)$

39.

$\sum_{n = 0}^{\infty} 2^{n} x^{n}$ on $(- \frac{1}{2}, \frac{1}{2})$

41.

$\sum_{n = 0}^{\infty} 4^{n} x^{2 n + 2}$ on $(- \frac{1}{2}, \frac{1}{2})$

43.

${| a_{n} x^{n} |}^{1 / n} = {| a_{n} |}^{1 / n} | x | \to | x | r$ as $n \to \infty$ and $| x | r < 1$ when $| x | < \frac{1}{r} .$ Therefore, $\sum_{n = 1}^{\infty} a_{n} x^{n}$ converges when $| x | < \frac{1}{r}$ by the nth root test.

45.

$a_{k} = {(\frac{k - 1}{2 k + 3})}^{k}$ so ${(a_{k})}^{1 / k} \to \frac{1}{2} < 1$ so $R = 2$

47.

$a_{n} = {(n^{1 / n} - 1)}^{n}$ so ${(a_{n})}^{1 / n} \to 0$ so $R = \infty$

49.

We can rewrite $p (x) = \sum_{n = 0}^{\infty} a_{2 n + 1} x^{2 n + 1}$ and $p (x) = p (− x)$ since only even powers of $x$ remain, $p (x)$ is an even function, for which, by definition $p (x) = p (–x)$ .

51.

If $x \in [0, 1],$ then $y = 2 x - 1 \in [−1, 1]$ so $p (2 x - 1) = p (y) = \sum_{n = 0}^{\infty} a_{n} y^{n}$ converges.

53.

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

55.

Consider the series $\sum b_{k} x^{k}$ where $b_{k} = a_{k}$ if $k = n^{2}$ and $b_{k} = 0$ otherwise. Then $b_{k} \leq a_{k}$ and so the series converges on $(−1, 1)$ by the comparison test.

57.

The approximation is more accurate near $x = −1 .$ The partial sums follow $\frac{1}{1 - x}$ more closely as N increases but are never accurate near $x = 1$ since the series diverges there.

59.

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

61.

The polynomial curves have roots close to those of $sin x$ up to their degree and then the polynomials diverge from $sin x .$

Section 6.2 Exercises

63.

$\frac{1}{2} (f (x) + g (x)) = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!}$ and $\frac{1}{2} (f (x) - g (x)) = \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)!} .$

65.

$\frac{4}{(x - 3) (x + 1)} = \frac{1}{x - 3} - \frac{1}{x + 1} = - \frac{1}{3 (1 - \frac{x}{3})} - \frac{1}{1 - (− x)} = - \frac{1}{3} \sum_{n = 0}^{\infty} {(\frac{x}{3})}^{n} - \sum_{n = 0}^{\infty} {(−1)}^{n} x^{n} = \sum_{n = 0}^{\infty} ({(−1)}^{n + 1} - \frac{1}{3^{n + 1}}) x^{n}$

67.

$\frac{5}{(x^{2} + 4) (x^{2} - 1)} = \frac{1}{x^{2} - 1} - \frac{1}{4} \frac{1}{1 + {(\frac{x}{2})}^{2}} = − \sum_{n = 0}^{\infty} x^{2 n} - \frac{1}{4} \sum_{n = 0}^{\infty} {(−1)}^{n} {(\frac{x}{2})}^{2 n} = \sum_{n = 0}^{\infty} (- 1 + (−1) \cdot^{n + 1} \frac{1}{2^{n + 2}}) x^{2 n}$

69.

$\frac{1}{x} \sum_{n = 0}^{\infty} \frac{1}{x^{n}} = \frac{1}{x} \frac{1}{1 - \frac{1}{x}} = \frac{1}{x - 1}$

71.

$\frac{1}{x - 3} \frac{1}{1 - \frac{1}{{(x - 3)}^{2}}} = \frac{x - 3}{{(x - 3)}^{2} - 1}$

73.

$P = P_{1} + ⋯ + P_{20}$ where $P_{k} = 10,000 \frac{1}{{(1 + r)}^{k}} .$ Then $P = 10,000 \sum_{k = 1}^{20} \frac{1}{{(1 + r)}^{k}} = 10,000 \frac{1 - {(1 + r)}^{−20}}{r} .$ When $r = 0.03, P \approx 10,000 \times 14.8775 = 148,775 .$ When $r = 0.05, P \approx 10,000 \times 12.4622 = 124,622 .$ When $r = 0.07, P \approx 105,940 .$

75.

In general, $P = \frac{C (1 - {(1 + r)}^{− N})}{r}$ for N years of payouts, or $C = \frac{P r}{1 - {(1 + r)}^{− N}} .$ For $N = 20$ and $P = 100,000,$ one has $C = 6721.57$ when $r = 0.03; C = 8024.26$ when $r = 0.05;$ and $C \approx 9439.29$ when $r = 0.07 .$

77.

In general, $P = \frac{C}{r} .$ Thus, $r = \frac{C}{P} = 5 \times \frac{10^{4}}{10^{6}} = 0.05 .$

79.

$(x + x^{2} - x^{3}) (1 + x^{3} + x^{6} + ⋯) = \frac{x + x^{2} - x^{3}}{1 - x^{3}}$

81.

$(x - x^{2} - x^{3}) (1 + x^{3} + x^{6} + ⋯) = \frac{x - x^{2} - x^{3}}{1 - x^{3}}$

83.

$a_{n} = 2, b_{n} = n$ so $c_{n} = \sum_{k = 0}^{n} b_{k} a_{n - k} = 2 \sum_{k = 0}^{n} k = (n) (n + 1)$ and $f (x) g (x) = \sum_{n = 1}^{\infty} n (n + 1) x^{n}$

85.

$a_{n} = b_{n} = 2^{− n}$ so $c_{n} = \sum_{k = 1}^{n - 1} b_{k} a_{n - k} = 2^{− n} \sum_{k = 1}^{n - 1} 1 = \frac{n - 1}{2^{n}}$ and $f (x) g (x) = \sum_{n = 2}^{\infty} (n - 1) {(\frac{x}{2})}^{n}$

87.

The derivative of $f$ is $- \frac{1}{{(1 + x)}^{2}} = − \sum_{n = 0}^{\infty} {(−1)}^{n} (n + 1) x^{n} .$

89.

The indefinite integral of $f$ is $- \frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n} .$

91.

$f (x) = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}; f^{'} (\frac{1}{2}) = \sum_{n = 1}^{\infty} \frac{n}{2^{n - 1}} = {\frac{d}{d x} {(1 - x)}^{−1} |}_{x = 1 / 2} = {\frac{1}{{(1 - x)}^{2}} |}_{x = 1 / 2} = 4$ so $\sum_{n = 1}^{\infty} \frac{n}{2^{n}} = 2 .$

93.

$f (x) = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}; f^{″} (\frac{1}{2}) = \sum_{n = 2}^{\infty} \frac{n (n - 1)}{2^{n - 2}} = {\frac{d^{2}}{d x^{2}} {(1 - x)}^{−1} |}_{x = 1 / 2} = {\frac{2}{{(1 - x)}^{3}} |}_{x = 1 / 2} = 16$ so $\sum_{n = 2}^{\infty} \frac{n (n - 1)}{2^{n}} = 4 .$

95.

$\int \sum {(1 - x)}^{n} d x = \int \sum {(−1)}^{n} {(x - 1)}^{n} d x = \sum \frac{{(−1)}^{n} {(x - 1)}^{n + 1}}{n + 1}$

97.

$− \int_{t = 0}^{x^{2}} \frac{1}{1 - t} d t = − \sum_{n = 0}^{\infty} \int_{0}^{x^{2}} t^{n} d x - \sum_{n = 0}^{\infty} \frac{x^{2 (n + 1)}}{n + 1} = − \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n}$

99.

$\int_{0}^{x^{2}} \frac{d t}{1 + t^{2}} = \sum_{n = 0}^{\infty} {(−1)}^{n} \int_{0}^{x^{2}} t^{2 n} d t = {\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{2 n + 1}}{2 n + 1} |}_{t = 0}^{x^{2}} = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 2}}{2 n + 1}$

101.

Term-by-term integration gives $\int_{0}^{x} ln t d t = \sum_{n = 1}^{\infty} {(−1)}^{n - 1} \frac{{(x - 1)}^{n + 1}}{n (n + 1)} = \sum_{n = 1}^{\infty} {(−1)}^{n - 1} (\frac{1}{n} - \frac{1}{n + 1}) {(x - 1)}^{n + 1} = (x - 1) ln x + \sum_{n = 2}^{\infty} {(−1)}^{n} \frac{{(x - 1)}^{n}}{n} = x ln x - x .$

103.

We have $ln (1 - x) = − \sum_{n = 1}^{\infty} \frac{x^{n}}{n}$ so $ln (1 + x) = \sum_{n = 1}^{\infty} {(−1)}^{n - 1} \frac{x^{n}}{n} .$ Thus, $ln (\frac{1 + x}{1 - x}) = \sum_{n = 1}^{\infty} (1 + {(−1)}^{n - 1}) \frac{x^{n}}{n} = 2 \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n - 1} .$ When $x = \frac{1}{3}$ we obtain $ln (2) = 2 \sum_{n = 1}^{\infty} \frac{1}{3^{2 n - 1} (2 n - 1)} .$ We have $2 \sum_{n = 1}^{3} \frac{1}{3^{2 n - 1} (2 n - 1)} = 0.69300 …,$ while $2 \sum_{n = 1}^{4} \frac{1}{3^{2 n - 1} (2 n - 1)} = 0.69313 …$ and $ln (2) = 0.69314 …;$ therefore, $N = 4 .$

105.

$\sum_{k = 1}^{\infty} \frac{x^{k}}{k} = − ln (1 - x)$ so $\sum_{k = 1}^{\infty} \frac{x^{3 k}}{6 k} = - \frac{1}{6} ln (1 - x^{3}) .$ The radius of convergence is equal to 1 by the ratio test.

107.

If $y = 2^{− x},$ then $\sum_{k = 1}^{\infty} y^{k} = \frac{y}{1 - y} = \frac{2^{− x}}{1 - 2^{− x}} = \frac{1}{2^{x} - 1} .$ If $a_{k} = 2^{− k x},$ then $\frac{a_{k + 1}}{a_{k}} = 2^{− x} < 1$ when $x > 0 .$ So the series converges for all $x > 0 .$

109.

Answers will vary.

111.

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

113.

When $x = - \frac{1}{2}, − ln (2) = ln (\frac{1}{2}) = − \sum_{n = 1}^{\infty} \frac{1}{n 2^{n}} .$ Since $\sum_{n = 11}^{\infty} \frac{1}{n 2^{n}} < \sum_{n = 11}^{\infty} \frac{1}{2^{n}} = \frac{1}{2^{10}},$ one has $\sum_{n = 1}^{10} \frac{1}{n 2^{n}} = 0.69306 …$ whereas $ln (2) = 0.69314 …;$ therefore, $N = 10 .$

115.

$6 S_{N} (\frac{1}{\sqrt{3}}) = 2 \sqrt{3} \sum_{n = 0}^{N} {(−1)}^{n} \frac{1}{3^{n} (2 n + 1)} .$ One has $π - 6 S_{4} (\frac{1}{\sqrt{3}}) = 0.00101 …$ and $π - 6 S_{5} (\frac{1}{\sqrt{3}}) = 0.00028 …$ so $N = 5$ is the smallest partial sum with accuracy to within 0.001. Also, $π - 6 S_{7} (\frac{1}{\sqrt{3}}) = 0.00002 …$ while $π - 6 S_{8} (\frac{1}{\sqrt{3}}) = −0.000007 …$ so $N = 8$ is the smallest N to give accuracy to within 0.00001.

Section 6.3 Exercises

117.

$f (−1) = 1; f^{'} (−1) = −1; f^{″} (−1) = 2; f (x) = 1 - (x + 1) + {(x + 1)}^{2}$

119.

$f^{'} (x) = 2 cos (2 x); f^{″} (x) = −4 sin (2 x); p_{2} (x) = −2 (x - \frac{π}{2})$

121.

$f^{'} (x) = \frac{1}{x}; f^{″} (x) = - \frac{1}{x^{2}}; p_{2} (x) = 0 + (x - 1) - \frac{1}{2} {(x - 1)}^{2}$

123.

$p_{2} (x) = e + e (x - 1) + \frac{e}{2} {(x - 1)}^{2}$

125.

$\frac{d^{2}}{d x^{2}} x^{1 / 3} = - \frac{2}{9 x^{5 / 3}} \geq −0.00092 …$ when $x \geq 28$ so the remainder estimate applies to the linear approximation $x^{1 / 3} \approx p_{1} (27) = 3 + \frac{x - 27}{27},$ which gives ${(28)}^{1 / 3} \approx 3 + \frac{1}{27} = 3. \bar{037},$ while ${(28)}^{1 / 3} \approx 3.03658 .$

127.

Using the estimate $\frac{2^{10}}{10!} < 0.000283$ we can use the Taylor expansion of order 9 to estimate e^x at $x = 2 .$ as $e^{2} \approx p_{9} (2) = 1 + 2 + \frac{2^{2}}{2} + \frac{2^{3}}{6} + ⋯ + \frac{2^{9}}{9!} = 7.3887 …$ whereas $e^{2} \approx 7.3891 .$

129.

Since $\frac{d^{n}}{d x^{n}} (ln x) = {(−1)}^{n - 1} \frac{(n - 1)!}{x^{n}}, R_{1000} \approx \frac{1}{1001} .$ One has $p_{1000} (1) = \sum_{n = 1}^{1000} \frac{{(−1)}^{n - 1}}{n} \approx 0.6936$ whereas $ln (2) \approx 0.6931 ⋯ .$

131.

$\int_{0}^{1} (1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \frac{x^{8}}{24} - \frac{x^{10}}{120} + \frac{x^{12}}{720}) d x$

$= 1 - \frac{1^{3}}{3} + \frac{1^{5}}{10} - \frac{1^{7}}{42} + \frac{1^{9}}{9 \cdot 24} - \frac{1^{11}}{120 \cdot 11} + \frac{1^{13}}{720 \cdot 13} \approx 0.74683$ whereas $\int_{0}^{1} e^{− x^{2}} d x \approx 0.74682 .$

133.

Since $f^{(n + 1)} (z)$ is $sin z$ or $cos z,$ we have $M = 1 .$ Since $| x - 0 | \leq \frac{π}{2},$ we seek the smallest n such that $\frac{π^{n + 1}}{2^{n + 1} (n + 1)!} \leq 0.001 .$ The smallest such value is $n = 7 .$ The remainder estimate is $R_{7} \leq 0.00092 .$

135.

Since $f^{(n + 1)} (z) = ± e^{− z}$ one has $M = e^{3} .$ Since $| x - 0 | \leq 3,$ one seeks the smallest n such that $\frac{3^{n + 1} e^{3}}{(n + 1)!} \leq 0.001 .$ The smallest such value is $n = 14 .$ The remainder estimate is $R_{14} \leq 0.000220 .$

137.

Since $sin x$ is increasing for small x and since $si n^{″} x = − sin x,$ the estimate applies whenever $R^{2} sin (R) \leq 0.2,$ which applies up to $R = 0.596 .$

139.

Since the second derivative of $cos x$ is $− cos x$ and since $cos x$ is decreasing away from $x = 0,$ the estimate applies when $R^{2} cos R \leq 0.2$ or $R \leq 0.447 .$

141.

${(x + 1)}^{3} - 2 {(x + 1)}^{2} + 2 (x + 1)$

143.

Values of derivatives are the same as for $x = 0$ so $cos x = {\sum_{n = 0}^{\infty} (−1)}^{n} \frac{{(x - 2 π)}^{2 n}}{(2 n)!}$

145.

$cos (\frac{π}{2}) = 0, − sin (\frac{π}{2}) = −1$ so $cos x = \sum_{n = 0}^{\infty} {(−1)}^{n + 1} \frac{{(x - \frac{π}{2})}^{2 n + 1}}{(2 n + 1)!},$ which is also $− cos (x - \frac{π}{2}) .$

147.

The derivatives are $f^{(n)} (1) = e$ so $e^{x} = e \sum_{n = 0}^{\infty} \frac{{(x - 1)}^{n}}{n!} .$

149.

$\frac{1}{{(x - 1)}^{3}} = − (\frac{1}{2}) \frac{d^{2}}{d x^{2}} \frac{1}{1 - x} = − \sum_{n = 0}^{\infty} (\frac{(n + 2) (n + 1) x^{n}}{2})$

151.

$2 - x = 1 - (x - 1)$

153.

${((x - 1) - 1)}^{2} = {(x - 1)}^{2} - 2 (x - 1) + 1$

155.

$\frac{1}{1 - (1 - x)} = \sum_{n = 0}^{\infty} {(−1)}^{n} {(x - 1)}^{n}$

157.

$x \sum_{n = 0}^{\infty} 2^{n} {(1 - x)}^{2 n} = \sum_{n = 0}^{\infty} 2^{n} {(x - 1)}^{2 n + 1} + \sum_{n = 0}^{\infty} 2^{n} {(x - 1)}^{2 n}$

159.

$e^{2 x} = e^{2 (x - 1) + 2} = e^{2} \sum_{n = 0}^{\infty} \frac{2^{n} {(x - 1)}^{n}}{n!}$

161.

$x = e^{2}; S_{10} = \frac{34,913}{4725} \approx 7.3889947$

163.

$sin (2 π) = 0; S_{10} = 8.27 \times 10^{−5}$

165.

The difference is small on the interior of the interval but approaches $1$ near the endpoints. The remainder estimate is $| R_{4} | = \frac{π^{5}}{120} \approx 2.552 .$

167.

The difference is on the order of $10^{−4}$ on $[−1, 1]$ while the Taylor approximation error is around $0.1$ near $\pm 1 .$ The top curve is a plot of ${tan}^{2} x - {(\frac{S_{5} (x)}{C_{4} (x)})}^{2}$ and the lower dashed plot shows $t^{2} - {(\frac{S_{5}}{C_{4}})}^{2} .$

169.

a. Answers will vary. b. The following are the $x_{n}$ values after $10$ iterations of Newton’s method to approximation a root of $p_{N} (x) - 2 = 0 :$ for $N = 4, x = 0.6939...;$ for $N = 5, x = 0.6932...;$ for $N = 6, x = 0.69315...; .$ (Note: $ln (2) = 0.69314 ...)$ c. Answers will vary.

171.

$\frac{ln (1 - x^{2})}{x^{2}} \to − 1$

173.

$\frac{cos (\sqrt{x}) - 1}{2 x} \approx \frac{(1 - \frac{x}{2} + \frac{x^{2}}{4!} - ⋯) - 1}{2 x} \to - \frac{1}{4}$

Section 6.4 Exercises

175.

${(1 + x^{2})}^{−1 / 3} = \sum_{n = 0}^{\infty} (\begin{matrix} - \frac{1}{3} \\ n \end{matrix}) x^{2 n}$

177.

${(1 - 2 x)}^{2 / 3} = \sum_{n = 0}^{\infty} {(−1)}^{n} 2^{n} (\begin{matrix} \frac{2}{3} \\ n \end{matrix}) x^{n}$

179.

$\sqrt{2 + x^{2}} = \sum_{n = 0}^{\infty} 2^{(1 / 2) - n} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) x^{2 n}; (| x^{2} | < 2)$

181.

$\sqrt{2 x - x^{2}} = \sqrt{1 - {(x - 1)}^{2}}$ so $\sqrt{2 x - x^{2}} = \sum_{n = 0}^{\infty} {(−1)}^{n} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) {(x - 1)}^{2 n}$

183.

$\sqrt{x} = 2 \sqrt{1 + \frac{x - 4}{4}}$ so $\sqrt{x} = \sum_{n = 0}^{\infty} 2^{1 - 2 n} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) {(x - 4)}^{n}$

185.

$\sqrt{x} = \sum_{n = 0}^{\infty} 3^{1 - 2 n} (\frac{1}{2} n) {(x - 9)}^{n}$

187.

$10 {(1 + \frac{x}{1000})}^{1 / 3} = \sum_{n = 0}^{\infty} 10^{1 - 3 n} (\begin{matrix} \frac{1}{3} \\ n \end{matrix}) x^{n} .$ Using, for example, a fourth-degree estimate at $x = 1$ gives $\begin{matrix} {(1001)}^{1 / 3} & \approx 10 (1 + (\begin{matrix} \frac{1}{3} \\ 1 \end{matrix}) 10^{−3} + (\begin{matrix} \frac{1}{3} \\ 2 \end{matrix}) 10^{−6} + (\begin{matrix} \frac{1}{3} \\ 3 \end{matrix}) 10^{−9} + (\begin{matrix} \frac{1}{3} \\ 4 \end{matrix}) 10^{−12}) \\ = 10 (1 + \frac{1}{{3.10}^{3}} - \frac{1}{{9.10}^{6}} + \frac{5}{{81.10}^{9}} - \frac{10}{{243.10}^{12}}) = 10.00333222... \end{matrix}$ whereas ${(1001)}^{1 / 3} = 10.00332222839093 ....$ Two terms would suffice for three-digit accuracy.

189.

The approximation is $2.3152;$ the CAS value is $2.23 … .$

191.

The approximation is $2.583 …;$ the CAS value is $2.449 … .$

193.

$\sqrt{1 - x^{2}} = 1 - \frac{x^{2}}{2} - \frac{x^{4}}{8} - \frac{x^{6}}{16} - \frac{5 x^{8}}{128} + ⋯ .$ Thus

$\int_{−1}^{1} \sqrt{1 - x^{2}} d x = x - \frac{x^{3}}{6} - \frac{x^{5}}{40} - \frac{x^{7}}{7 \cdot 16} - \frac{5 x^{9}}{9 \cdot 128} + ⋯ |_{−1}^{1} \approx 2 - \frac{1}{3} - \frac{1}{20} - \frac{1}{56} - \frac{10}{9 \cdot 128} + error = 1.590 ...$ whereas $\frac{π}{2} = 1.570 ...$

195.

$\begin{array}{l} {(1 + 4 x)}^{4 / 3} = (1 + 4 x) {(1 + 4 x)}^{1 / 3} \\ = (1 + 4 x) (1 + \frac{4 x}{3} - \frac{16 x^{3}}{9} + \frac{320 x^{3}}{81} - \frac{2560 x^{4}}{243}) \\ = 1 + \frac{16}{3} x + \frac{32}{9} x^{2} - \frac{256}{81} x^{3} + \frac{1280}{243} x^{4} - \frac{10240}{243} x^{5} \end{array}$

197.

${(1 + {(x + 3)}^{2})}^{1 / 3} = 1 + \frac{1}{3} {(x + 3)}^{2} - \frac{1}{9} {(x + 3)}^{4} + \frac{5}{81} {(x + 3)}^{6} - \frac{10}{243} {(x + 3)}^{8} + ⋯$

199.

Twice the approximation is $1.260 …$ whereas $2^{1 / 3} = 1.2599. ...$

201.

$f^{(99)} (0) = 0$

203.

$\sum_{n = 0}^{\infty} \frac{{(ln (2) x)}^{n}}{n!}$

205.

For $x > 0, sin (\sqrt{x}) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{(2 n + 1) / 2}}{\sqrt{x} (2 n + 1)!} = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n}}{(2 n + 1)!} .$

207.

$e^{x^{3}} = \sum_{n = 0}^{\infty} \frac{x^{3 n}}{n!}$

209.

${sin}^{2} x = − \sum_{k = 1}^{\infty} \frac{{(−1)}^{k} 2^{2 k - 1} x^{2 k}}{(2 k)!}$

211.

${tan}^{−1} x = \sum_{k = 0}^{\infty} \frac{{(−1)}^{k} x^{2 k + 1}}{2 k + 1}$

213.

${sin}^{−1} x = \sum_{n = 0}^{\infty} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) \frac{x^{2 n + 1}}{(2 n + 1) n!}$

215.

$F (x) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n + 1}}{(n + 1) (2 n)!}$

217.

$F (x) = \sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n^{2}}$

219.

$x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + ⋯$

221.

$1 + x - \frac{x^{3}}{3} - \frac{x^{4}}{6} + ⋯$

223.

$1 + x^{2} + \frac{2 x^{4}}{3} + \frac{17 x^{6}}{45} + ⋯$

225.

Using the expansion for $tan x$ gives $1 + \frac{x}{3} + \frac{2 x^{2}}{15} .$

227.

$\frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n}$ so $R = 1$ by the ratio test.

229.

$ln (1 + x^{2}) = \sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n} x^{2 n}$ so $R = 1$ by the ratio test.

231.

Add series of $e^{x}$ and $e^{− x}$ term by term. Odd terms cancel and $cosh x = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!} .$

233.

The ratio $\frac{S_{n} (x)}{C_{n} (x)}$ approximates $tan x$ better than does $p_{7} (x) = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \frac{17 x^{7}}{315}$ for $N \geq 3 .$ The dashed curves are $\frac{S_{n}}{C_{n}} - tan$ for $n = 1, 2 .$ The dotted curve corresponds to $n = 3,$ and the dash-dotted curve corresponds to $n = 4 .$ The solid curve is $p_{7} - tan x .$

235.

By the term-by-term differentiation theorem, $y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ so $y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} x y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n},$ whereas $y^{'} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ so $x y^{″} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} .$

237.

The probability is $p = \frac{1}{\sqrt{2 π}} \int_{(a - μ) / σ}^{(b - μ) / σ} e^{− x^{2} / 2} d x$ where $a = 90$ and $b = 100,$ that is, $p = \frac{1}{\sqrt{2 π}} \int_{−1}^{1} e^{− x^{2} / 2} d x = \frac{1}{\sqrt{2 π}} \int_{−1}^{1} \sum_{n = 0}^{5} {(−1)}^{n} \frac{x^{2 n}}{2^{n} n!} d x = \frac{2}{\sqrt{2 π}} \sum_{n = 0}^{5} {(−1)}^{n} \frac{1}{(2 n + 1) 2^{n} n!} \approx 0.6827 .$

239.

As in the previous problem one obtains $a_{n} = 0$ if $n$ is odd and $a_{n} = − (n + 2) (n + 1) a_{n + 2}$ if $n$ is even, so $a_{0} = 1$ leads to $a_{2 n} = \frac{{(−1)}^{n}}{(2 n)!} .$

241.

$y^{″} = \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n}$ and $y^{'} = \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n}$ so $y^{″} - y^{'} + y = 0$ implies that $(n + 2) (n + 1) a_{n + 2} - (n + 1) a_{n + 1} + a_{n} = 0$ or $a_{n} = \frac{a_{n - 1}}{n} - \frac{a_{n - 2}}{n (n - 1)}$ for all $n \cdot y (0) = a_{0} = 1$ and $y^{'} (0) = a_{1} = 0,$ so $a_{2} = \frac{1}{2}, a_{3} = \frac{1}{6}, a_{4} = 0,$ and $a_{5} = - \frac{1}{120} .$

243.

a. (Proof) b. We have $R_{s} \leq \frac{0.1}{(9)!} π^{9} \approx 0.0082 < 0.01 .$ We have $\int_{0}^{π} (1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \frac{x^{8}}{9!}) d x = π - \frac{π^{3}}{3 \cdot 3!} + \frac{π^{5}}{5 \cdot 5!} - \frac{π^{7}}{7 \cdot 7!} + \frac{π^{9}}{9 \cdot 9!} = 1.852...,$ whereas $\int_{0}^{π} \frac{sin t}{t} d t = 1.85194...,$ so the actual error is approximately $0.00006 .$

245.

Since $cos (t^{2}) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{4 n}}{(2 n)!}$ and $sin (t^{2}) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{4 n + 2}}{(2 n + 1)!},$ one has $S (x) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 3}}{(4 n + 3) (2 n + 1)!}$ and $C (x) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 1}}{(4 n + 1) (2 n)!} .$ The sums of the first $50$ nonzero terms are plotted below with $C_{50} (x)$ the solid curve and $S_{50} (x)$ the dashed curve.

247.

$\int_{0}^{1 / 4} \sqrt{x} (1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \frac{5 x^{4}}{128} - \frac{7 x^{5}}{256}) d x$

$= \frac{2}{3} 2^{−3} - \frac{1}{2} \frac{2}{5} 2^{−5} - \frac{1}{8} \frac{2}{7} 2^{−7} - \frac{1}{16} \frac{2}{9} 2^{−9} - \frac{5}{128} \frac{2}{11} 2^{−11} - \frac{7}{256} \frac{2}{13} 2^{−13} = 0.0767732 ...$

whereas $\int_{0}^{1 / 4} \sqrt{x - x^{2}} d x = 0.076773 .$

249.

$T \approx 2 π \sqrt{\frac{10}{9.8}} (1 + \frac{{sin}^{2} (θ / 12)}{4}) \approx 6.453$ seconds. The small angle estimate is $T \approx 2 π \sqrt{\frac{10}{9.8} \approx 6.347} .$ The relative error is around $2$ percent.