Chapter 6

> 来源: OpenStax《Calculus Volume 2》| 原页: https://openstax.org/books/calculus-volume-2/pages/chapter-6

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Chapter 6

Calculus Volume 2Chapter 6

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Chapter 6

Checkpoint

6.1

The interval of convergence is $\left\lbrack {-1,1} \right).$ The radius of convergence is $R = 1.$

6.2

6.3

$\sum\limits_{n = 0}^{\infty}\frac{x^{n + 3}}{2^{n + 1}}$ with interval of convergence $\left( {-2,2} \right)$

6.4

Interval of convergence is $\left( {-2,2} \right).$

6.5

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

6.6

$f(x) = \frac{3}{3 - x}.$ The interval of convergence is $\left( {-3,3} \right).$

6.7

$1 + 2x + 3x^{2} + 4x^{3} + \text{⋯}$

6.8

$\sum\limits_{n = 0}^{\infty}{\left( {n + 2} \right)\left( {n + 1} \right)x^{n}}$

6.9

$\sum\limits_{n = 2}^{\infty}\frac{(-1)^{n}x^{n}}{n\left( {n - 1} \right)}$

6.10

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

6.11

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

6.12

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

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

6.13

0.96593

6.14

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

6.15

$\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}x^{2n}}{\left( {2n} \right)\text{!}}$

By the ratio test, the interval of convergence is $\left( {\text{−}\infty,\infty} \right).$ Since $\left| {R_{n}(x)} \right| \leq \frac{|x|^{n + 1}}{\left( {n + 1} \right)\text{!}},$ the series converges to $\text{cos}\mspace{2mu} x$ for all real *x*.

6.16

$\sum\limits_{n = 0}^{\infty}{(-1)^{n}\left( {n + 1} \right)x^{n}}$

6.17

$\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}x^{4n + 2}}{\left( {2n + 1} \right)\text{!}}$

6.18

$\sum\limits_{n = 1}^{\infty}{\frac{(-1)^{n}}{n\text{!}}\ \frac{1 \cdot 3 \cdot 5\text{⋯}\left( {2n - 1} \right)}{2^{n}}x^{n}}$

6.19

$y = 5e^{2x}$

6.20

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

6.21

$C + {\sum\limits_{n = 1}^{\infty}{(-1)^{n + 1}\frac{x^{n}}{n\left( {2n - 2} \right)\text{!}}}}$ 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 $a_{n}x^{n}\rightarrow 0$ for $|x| < R.$ If $a_{n} = n^{n},$ then $a_{n}x^{n} = \left( {nx} \right)^{n}$ does not tend to zero for any $x \neq 0.$

5.

It must converge on $\left( {0,6} \right\rbrack$ 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.

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

9.

$\left| \frac{a_{n + 1}\left( \frac{\pi}{e} \right)^{n + 1}x^{n + 1}}{a_{n}\left( \frac{\pi}{e} \right)^{n}x^{n}} \right| = \frac{\pi|x|}{e}\left| \frac{a_{n + 1}}{a_{n}} \right|\rightarrow\frac{\pi|x|}{e}$ so $R = \frac{e}{\pi}$

11.

$\left| \frac{a_{n + 1}(-1)^{n + 1}x^{2n + 2}}{a_{n}(-1)^{n}x^{2n}} \right| = \left| x^{2} \right|\left| \frac{a_{n + 1}}{a_{n}} \right|\rightarrow\left| x^{2} \right|$ so $R = 1$

13.

$a_{n} = \frac{2^{n}}{n}$ so $\frac{a_{n + 1}x}{a_{n}}\rightarrow 2x.$ 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 $I = \left\lbrack {- \frac{1}{2},\frac{1}{2}} \right).$

15.

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

17.

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

19.

$a_{k} = \frac{\pi^{k}}{k^{\pi}}$ so $R = \frac{1}{\pi}.$ When $x = \text{±}\frac{1}{\pi}$ the series is an absolutely convergent *p*-series. The interval of convergence is $I = \left\lbrack {- \frac{1}{\pi},\frac{1}{\pi}} \right\rbrack.$

21.

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

23.

$a_{k} = \frac{\left( {k\text{!}} \right)^{2}}{\left( {2k} \right)\text{!}}$ so $\frac{a_{k + 1}}{a_{k}} = \frac{\left( {k + 1} \right)^{2}}{\left( {2k + 2} \right)\left( {2k + 1} \right)}\rightarrow\frac{1}{4}$ so $R = 4$

25.

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

27.

$a_{n} = \frac{1}{\begin{pmatrix}

{2n} \\

n

\end{pmatrix}}$ so $\frac{a_{n + 1}}{a_{n}} = \frac{\left( {\left( {n + 1} \right)\text{!}} \right)^{2}}{\left( {2n + 2} \right)\text{!}}\ \frac{2n\text{!}}{\left( {n\text{!}} \right)^{2}} = \frac{\left( {n + 1} \right)^{2}}{\left( {2n + 2} \right)\left( {2n + 1} \right)}\rightarrow\frac{1}{4}$ so $R = 4$

29.

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

31.

$a_{n} = \frac{n\text{!}}{n^{n}}$ so $\frac{a_{n + 1}}{a_{n}} = \frac{\left( {n + 1} \right)\text{!}}{n\text{!}}\ \frac{n^{n}}{\left( {n + 1} \right)^{n + 1}} = \left( \frac{n}{n + 1} \right)^{n}\rightarrow\frac{1}{e}$ so $R = e$

33.

$f(x) = {\sum\limits_{n = 0}^{\infty}\left( {1 - x} \right)^{n}}$ on $I = \left( {0,2} \right)$

35.

$\sum\limits_{n = 0}^{\infty}x^{2n + 1}$ on $I = \left( {-1,1} \right)$

37.

$\sum\limits_{n = 0}^{\infty}{(-1)^{n}x^{2n + 2}}$ on $I = \left( {-1,1} \right)$

39.

$\sum\limits_{n = 0}^{\infty}{2^{n}x^{n}}$ on $\left( {- \frac{1}{2},\frac{1}{2}} \right)$

41.

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

43.

$\left| {a_{n}x^{n}} \right|^{1\text{/}n} = \left| a_{n} \right|^{1\text{/}n}|x|\rightarrow|x|r$ as $n\rightarrow\infty$ and $|x|r < 1$ when $|x| < \frac{1}{r}.$ Therefore, $\sum\limits_{n = 1}^{\infty}{a_{n}x^{n}}$ converges when $|x| < \frac{1}{r}$ by the *n*th root test.

45.

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

47.

$a_{n} = \left( {n^{1\text{/}n} - 1} \right)^{n}$ so $\left( a_{n} \right)^{1\text{/}n}\rightarrow 0$ so $R = \infty$

49.

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

51.

If $x \in \left\lbrack {0,1} \right\rbrack,$ then $y = 2x - 1 \in \left\lbrack {-1,1} \right\rbrack$ so $p\left( {2x - 1} \right) = p(y) = {\sum\limits_{n = 0}^{\infty}{a_{n}y^{n}}}$ converges.

53.

Converges on $\left( {-1,1} \right)$ 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 $\left( {-1,1} \right)$ 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 $x = \text{±}1.$

61.

The polynomial curves have roots close to those of $\text{sin}\mspace{2mu} x$ up to their degree and then the polynomials diverge from $\text{sin}\mspace{2mu} x.$

Section 6.2 Exercises

63.

$\frac{1}{2}\left( {f(x) + g(x)} \right) = {\sum\limits_{n = 0}^{\infty}\frac{x^{2n}}{\left( {2n} \right)\text{!}}}$ and $\frac{1}{2}\left( {f(x) - g(x)} \right) = {\sum\limits_{n = 0}^{\infty}\frac{x^{2n + 1}}{\left( {2n + 1} \right)\text{!}}}.$

65.

$\frac{4}{\left( {x - 3} \right)\left( {x + 1} \right)} = \frac{1}{x - 3} - \frac{1}{x + 1} = - \frac{1}{3\left( {1 - \frac{x}{3}} \right)} - \frac{1}{1 - \left( {\text{−}x} \right)} = - \frac{1}{3}{\sum\limits_{n = 0}^{\infty}\left( \frac{x}{3} \right)^{n}} - {\sum\limits_{n = 0}^{\infty}{(-1)^{n}x^{n}}} = {\sum\limits_{n = 0}^{\infty}{\left( {(-1)^{n + 1} - \frac{1}{3^{n + 1}}} \right)x^{n}}}$

67.

$\frac{5}{\left( x^{2} + 4 \right)\left( x^{2} - 1 \right)} = \frac{1}{x^{2} - 1} - \frac{1}{4}\ \frac{1}{1 + \left( \frac{x}{2} \right)^{2}} = \text{−}\sum\limits_{n = 0}^{\infty}x^{2n} - \frac{1}{4}\sum\limits_{n = 0}^{\infty}(-1)^{n}\left( \frac{x}{2} \right)^{2n} = \sum\limits_{n = 0}^{\infty}\left( - 1 + (-1) \cdot^{n + 1}\frac{1}{2^{n + 2}} \right)x^{2n}$

69.

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

71.

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

73.

$P = P_{1} + \text{⋯} + P_{20}$ where $P_{k} = 10,000\frac{1}{\left( {1 + r} \right)^{k}}.$ Then $P = 10,000{\sum\limits_{k = 1}^{20}\frac{1}{\left( {1 + r} \right)^{k}}} = 10,000\frac{1 - \left( {1 + r} \right)^{-20}}{r}.$ 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, $P = \frac{C\left( {1 - \left( {1 + r} \right)^{\text{−}N}} \right)}{r}$ for *N* years of payouts, or $C = \frac{Pr}{1 - \left( {1 + r} \right)^{\text{−}N}}.$ 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.

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

81.

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

83.

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

85.

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

87.

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

89.

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

91.

$f(x) = {\sum\limits_{n = 0}^{\infty}x^{n}} = \frac{1}{1 - x};f^{\prime}\left( \frac{1}{2} \right) = {\sum\limits_{n = 1}^{\infty}\frac{n}{2^{n - 1}}} = \left. {\frac{d}{dx}\left( {1 - x} \right)^{-1}} \right|_{x = 1\text{/}2} = \left. \frac{1}{\left( {1 - x} \right)^{2}} \right|_{x = 1\text{/}2} = 4$ so ${\sum\limits_{n = 1}^{\infty}\frac{n}{2^{n}}} = 2.$

93.

$f(x) = {\sum\limits_{n = 0}^{\infty}x^{n}} = \frac{1}{1 - x};f^{''}\left( \frac{1}{2} \right) = {\sum\limits_{n = 2}^{\infty}\frac{n\left( {n - 1} \right)}{2^{n - 2}}} = \left. {\frac{d^{2}}{dx^{2}}\left( {1 - x} \right)^{-1}} \right|_{x = 1\text{/}2} = \left. \frac{2}{\left( {1 - x} \right)^{3}} \right|_{x = 1\text{/}2} = 16$ so ${\sum\limits_{n = 2}^{\infty}\frac{n\left( {n - 1} \right)}{2^{n}}} = 4.$

95.

${\int{{\sum\left( {1 - x} \right)^{n}}dx}} = {\int{{\sum(-1)^{n}}\left( {x - 1} \right)^{n}dx}} = {\sum\frac{(-1)^{n}\left( {x - 1} \right)^{n + 1}}{n + 1}}$

97.

$\text{−}{\int_{t = 0}^{x^{2}}{\frac{1}{1 - t}dt}} = \text{−}{\sum\limits_{n = 0}^{\infty}{{\int_{0}^{x^{2}}{t^{n}dx}}\operatorname{}}} - {\sum\limits_{n = 0}^{\infty}\frac{x^{2{({n + 1})}}}{n + 1}} = \text{−}{\sum\limits_{n = 1}^{\infty}\frac{x^{2n}}{n}}$

99.

${\int_{0}^{x^{2}}\frac{dt}{1 + t^{2}}} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}{\int_{0}^{x^{2}}{t^{2n}dt}}}} = {\sum\limits_{n = 0}^{\infty}\left. {(-1)^{n}\frac{t^{2n + 1}}{2n + 1}} \right|}_{t = 0}^{x^{2}} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{4n + 2}}{2n + 1}}}$

101.

Term-by-term integration gives ${\int_{0}^{x}{\text{ln}\mspace{2mu} tdt}} = {\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}\frac{\left( {x - 1} \right)^{n + 1}}{n\left( {n + 1} \right)}}} = {\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}\left( {\frac{1}{n} - \frac{1}{n + 1}} \right)\left( {x - 1} \right)^{n + 1}}} = \left( {x - 1} \right)\text{ln}\mspace{2mu} x + {\sum\limits_{n = 2}^{\infty}{(-1)^{n}\frac{\left( {x - 1} \right)^{n}}{n}}} = x\mspace{2mu}\text{ln}\mspace{2mu} x - x.$

103.

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

105.

${\sum\limits_{k = 1}^{\infty}\frac{x^{k}}{k}} = \text{−}\text{ln}\mspace{2mu}\left( {1 - x} \right)$ so ${\sum\limits_{k = 1}^{\infty}\frac{x^{3k}}{6k}} = - \frac{1}{6}\mspace{2mu}\text{ln}\mspace{2mu}\left( {1 - x^{3}} \right).$ The radius of convergence is equal to 1 by the ratio test.

107.

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

109.

Answers will vary.

111.

The solid curve is *S*₅. The dashed curve is *S*₂, dotted is *S*₃, and dash-dotted is *S*₄

113.

When $x = - \frac{1}{2},\text{−}\text{ln}\mspace{2mu}(2) = \text{ln}\mspace{2mu}\left( \frac{1}{2} \right) = \text{−}{\sum\limits_{n = 1}^{\infty}\frac{1}{n2^{n}}}.$ Since $\sum\limits_{n = 11}^{\infty}\frac{1}{n2^{n}} < {\sum\limits_{n = 11}^{\infty}\frac{1}{2^{n}}} = \frac{1}{2^{10}},$ one has ${\sum\limits_{n = 1}^{10}\frac{1}{n2^{n}}} = 0.69306\text{…}$ whereas $\text{ln}\mspace{2mu}(2) = 0.69314\text{…};$ therefore, $N = 10.$

115.

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

Section 6.3 Exercises

117.

$f(-1) = 1;f^{\prime}(-1) = -1;f^{''}(-1) = 2;f(x) = 1 - \left( {x + 1} \right) + \left( {x + 1} \right)^{2}$

119.

$f^{\prime}(x) = 2\mspace{2mu}\text{cos}\left( {2x} \right);f^{''}(x) = -4\mspace{2mu}\text{sin}\left( {2x} \right);p_{2}(x) = -2\left( {x - \frac{\pi}{2}} \right)$

121.

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

123.

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

125.

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

127.

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

129.

Since $\frac{d^{n}}{dx^{n}}\left( {\text{ln}\mspace{2mu} x} \right) = (-1)^{n - 1}\frac{\left( {n - 1} \right)\text{!}}{x^{n}},R_{1000} \approx \frac{1}{1001}.$ One has $p_{1000}(1) = {\sum\limits_{n = 1}^{1000}\frac{(-1)^{n - 1}}{n}} \approx 0.6936$ whereas $\text{ln}\mspace{2mu}(2) \approx 0.6931\text{⋯}.$

131.

$\int_{0}^{1}{\left( {1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \frac{x^{8}}{24} - \frac{x^{10}}{120} + \frac{x^{12}}{720}} \right)\mspace{2mu} dx}$

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

133.

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

135.

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

137.

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

139.

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

141.

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

143.

Values of derivatives are the same as for $x = 0$ so $\text{cos}\mspace{2mu} x = {\sum\limits_{n = 0}^{\infty}(-1)}^{n}\frac{\left( {x - 2\pi} \right)^{2n}}{\left( {2n} \right)\text{!}}$

145.

$\text{cos}\left( \frac{\pi}{2} \right) = 0,\text{−}\text{sin}\left( \frac{\pi}{2} \right) = -1$ so $\text{cos}\mspace{2mu} x = {\sum\limits_{n = 0}^{\infty}(-1)^{n + 1}}\frac{\left( {x - \frac{\pi}{2}} \right)^{2n + 1}}{\left( {2n + 1} \right)\text{!}},$ which is also $\text{−}\text{cos}\left( {x - \frac{\pi}{2}} \right).$

147.

The derivatives are $f^{(n)}(1) = e$ so $e^{x} = e{\sum\limits_{n = 0}^{\infty}\frac{\left( {x - 1} \right)^{n}}{n\text{!}}}.$

149.

$\frac{1}{\left( {x - 1} \right)^{3}} = \text{−}\left( \frac{1}{2} \right)\frac{d^{2}}{dx^{2}}\ \frac{1}{1 - x} = \text{−}{\sum\limits_{n = 0}^{\infty}\left( \frac{\left( {n + 2} \right)\left( {n + 1} \right)x^{n}}{2} \right)}$

151.

$2 - x = 1 - \left( {x - 1} \right)$

153.

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

155.

$\frac{1}{1 - \left( {1 - x} \right)} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\left( {x - 1} \right)^{n}}}$

157.

$x{\sum\limits_{n = 0}^{\infty}{2^{n}\left( {1 - x} \right)^{2n} =}}{\sum\limits_{n = 0}^{\infty}{2^{n}\left( {x - 1} \right)^{2n + 1} +}}{\sum\limits_{n = 0}^{\infty}{2^{n}\left( {x - 1} \right)^{2n}}}$

159.

$e^{2x} = e^{2{({x - 1})} + 2} = e^{2}{\sum\limits_{n = 0}^{\infty}\frac{2^{n}\left( {x - 1} \right)^{n}}{n\text{!}}}$

161.

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

163.

$\text{sin}\left( {2\pi} \right) = 0;S_{10} = 8.27\ \times \ 10^{-5}$

165.

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

167.

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

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 $p_{N}(x) - 2 = 0\text{:}$ for $N = 4,x = 0.6939...;$ for $N = 5,x = 0.6932...;$ for $N = 6,x = 0.69315...;.$ (*Note:* $\text{ln}\mspace{2mu}(2) = 0.69314...)$ c. Answers will vary.

171.

$\frac{\text{ln}\mspace{2mu}\left( {1 - x^{2}} \right)}{x^{2}}\rightarrow\text{−}1$

173.

$\frac{\text{cos}\left( \sqrt{x} \right) - 1}{2x} \approx \frac{\left( {1 - \frac{x}{2} + \frac{x^{2}}{4\text{!}} - \text{⋯}} \right) - 1}{2x}\rightarrow - \frac{1}{4}$

Section 6.4 Exercises

175.

$\left( {1 + x^{2}} \right)^{-1\text{/}3} = {\sum\limits_{n = 0}^{\infty}{\begin{pmatrix}

{- \frac{1}{3}} \\

n

\end{pmatrix}x^{2n}}}$

177.

$\left( {1 - 2x} \right)^{2\text{/}3} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}2^{n}}}\left( \begin{array}{l}

\frac{2}{3} \\

n

\end{array} \right)x^{n}$

179.

$\sqrt{2 + x^{2}} = {\sum\limits_{n = 0}^{\infty}2^{{({1\text{/}2})} - n}}\left( \begin{array}{l}

\frac{1}{2} \\

n

\end{array} \right)x^{2n};\left( {\left| x^{2} \right| < 2} \right)$

181.

$\sqrt{2x - x^{2}} = \sqrt{1 - \left( {x - 1} \right)^{2}}$ so $\sqrt{2x - x^{2}} = {\sum\limits_{n = 0}^{\infty}(-1)^{n}}\left( \begin{array}{l}

\frac{1}{2} \\

n

\end{array} \right)\left( {x - 1} \right)^{2n}$

183.

$\sqrt{x} = 2\sqrt{1 + \frac{x - 4}{4}}$ so $\sqrt{x} = {\sum\limits_{n = 0}^{\infty}{2^{1 - 2n}\left( \begin{array}{l}

\frac{1}{2} \\

n

\end{array} \right)\left( {x - 4} \right)^{n}}}$

185.

$\sqrt{x} = {\sum\limits_{n = 0}^{\infty}{3^{1 - 2n}\left( \frac{1}{2}n \right)\left( {x - 9} \right)^{n}}}$

187.

$10\left( {1 + \frac{x}{1000}} \right)^{1\text{/}3} = {\sum\limits_{n = 0}^{\infty}10^{1 - 3n}}\left( \begin{array}{l}

\frac{1}{3} \\

n

\end{array} \right)x^{n}.$ Using, for example, a fourth-degree estimate at $x = 1$ gives $\begin{array}{cl}

(1001)^{1\text{/}3} & {\approx 10\left( {1 + \left( \begin{array}{l}

\frac{1}{3} \\

1

\end{array} \right)10^{-3} + \left( \begin{array}{l}

\frac{1}{3} \\

2

\end{array} \right)10^{-6} + \left( \begin{array}{l}

\frac{1}{3} \\

3

\end{array} \right)10^{-9} + \left( \begin{array}{l}

\frac{1}{3} \\

4

\end{array} \right)10^{-12}} \right)} \\

& {= 10\left( {1 + \frac{1}{3.10^{3}} - \frac{1}{9.10^{6}} + \frac{5}{81.10^{9}} - \frac{10}{243.10^{12}}} \right) = 10.00333222...}

\end{array}$ whereas $(1001)^{1\text{/}3} = 10.00332222839093\operatorname{....}$ Two terms would suffice for three-digit accuracy.

189.

The approximation is $2.3152;$ the CAS value is $2.23\text{…}.$

191.

The approximation is $2.583\text{…};$ the CAS value is $2.449\text{…}.$

193.

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

${\int_{-1}^{1}\sqrt{1 - x^{2}}}dx = x - \frac{x^{3}}{6} - \frac{x^{5}}{40} - \frac{x^{7}}{7 \cdot 16} - \frac{5x^{9}}{9 \cdot 128} + \text{⋯}\left. {} \middle| \right._{-1}^{1} \approx 2 - \frac{1}{3} - \frac{1}{20} - \frac{1}{56} - \frac{10}{9 \cdot 128} + \text{error} = 1.590...$ whereas $\frac{\pi}{2} = 1.570...$

195.

$\begin{array}{l}

{(1 + 4x)^{4/3} = (1 + 4x)(1 + 4x)^{1/3}} \\

{= (1 + 4x)\left( 1 + \frac{4x}{3} - \frac{16x^{3}}{9} + \frac{320x^{3}}{81} - \frac{2560x^{4}}{243} \right)} \\

{= 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.

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

199.

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

201.

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

203.

$\sum\limits_{n = 0}^{\infty}\frac{\left( {\text{ln}\mspace{2mu}(2)x} \right)^{n}}{n\text{!}}$

205.

For $x > 0,\text{sin}\left( \sqrt{x} \right) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{{({2n + 1})}\text{/}2}}{\sqrt{x}\left( {2n + 1} \right)\text{!}}}} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{n}}{\left( {2n + 1} \right)\text{!}}}}.$

207.

$e^{x^{3}} = {\sum\limits_{n = 0}^{\infty}\frac{x^{3n}}{n\text{!}}}$

209.

$\text{sin}^{2}x = \text{−}{\sum\limits_{k = 1}^{\infty}\frac{(-1)^{k}2^{2k - 1}x^{2k}}{\left( {2k} \right)\text{!}}}$

211.

$\text{tan}^{-1}x = {\sum\limits_{k = 0}^{\infty}\frac{(-1)^{k}x^{2k + 1}}{2k + 1}}$

213.

$\text{sin}^{-1}x = {\sum\limits_{n = 0}^{\infty}{\left( \begin{array}{l}

\frac{1}{2} \\

n

\end{array} \right)\frac{x^{2n + 1}}{\left( {2n + 1} \right)n\text{!}}}}$

215.

$F(x) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{n + 1}}{\left( {n + 1} \right)\left( {2n} \right)\text{!}}}}$

217.

$F(x) = {\sum\limits_{n = 1}^{\infty}{(-1)^{n + 1}\frac{x^{n}}{n^{2}}}}$

219.

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

221.

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

223.

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

225.

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

227.

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

229.

$\text{ln}\mspace{2mu}\left( {1 + x^{2}} \right) = {\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n}}x^{2n}$ so $R = 1$ by the ratio test.

231.

Add series of $e^{x}$ and $e^{\text{−}x}$ term by term. Odd terms cancel and $\text{cosh}\mspace{2mu} x = {\sum\limits_{n = 0}^{\infty}\frac{x^{2n}}{\left( {2n} \right)\text{!}}}.$

233.

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

235.

By the term-by-term differentiation theorem, $y^{\prime} = {\sum\limits_{n = 1}^{\infty}{na_{n}x^{n - 1}}}$ so $y^{\prime} = {\sum\limits_{n = 1}^{\infty}{na_{n}x^{n - 1}}}\ xy^{\prime} = {\sum\limits_{n = 1}^{\infty}{na_{n}x^{n}}},$ whereas $y^{\prime} = {\sum\limits_{n = 2}^{\infty}{n\left( {n - 1} \right)a_{n}x^{n - 2}}}$ so $xy^{''} = {\sum\limits_{n = 2}^{\infty}{n\left( {n - 1} \right)a_{n}x^{n}}}.$

237.

The probability is $p = \frac{1}{\sqrt{2\pi}}{\int_{{({a - \mu})}\text{/}\sigma}^{{({b - \mu})}\text{/}\sigma}{e^{\text{−}x^{2}\text{/}2}dx}}$ where $a = 90$ and $b = 100,$ that is, $p = \frac{1}{\sqrt{2\pi}}{\int_{-1}^{1}{e^{\text{−}x^{2}\text{/}2}dx}} = \frac{1}{\sqrt{2\pi}}{\int_{-1}^{1}{{\sum\limits_{n = 0}^{5}(-1)^{n}}\frac{x^{2n}}{2^{n}n\text{!}}dx}} = \frac{2}{\sqrt{2\pi}}{\sum\limits_{n = 0}^{5}(-1)^{n}}\frac{1}{\left( {2n + 1} \right)2^{n}n\text{!}} \approx 0.6827.$

239.

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

241.

$y^{''} = {\sum\limits_{n = 0}^{\infty}{\left( {n + 2} \right)\left( {n + 1} \right)a_{n + 2}x^{n}}}$ and $y^{\prime} = {\sum\limits_{n = 0}^{\infty}{\left( {n + 1} \right)a_{n + 1}x^{n}}}$ so $y^{''} - y^{\prime} + y = 0$ implies that $\left( {n + 2} \right)\left( {n + 1} \right)a_{n + 2} - \left( {n + 1} \right)a_{n + 1} + a_{n} = 0$ or $a_{n} = \frac{a_{n - 1}}{n} - \frac{a_{n - 2}}{n\left( {n - 1} \right)}$ for all $n \cdot {y(0) = a_{0} = 1}$ and $y^{\prime}(0) = a_{1} = 0,$ 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 $R_{s} \leq \frac{0.1}{(9)\text{!}}\pi^{9} \approx 0.0082 < 0.01.$ We have ${\int_{0}^{\pi}{\left( {1 - \frac{x^{2}}{3\text{!}} + \frac{x^{4}}{5\text{!}} - \frac{x^{6}}{7\text{!}} + \frac{x^{8}}{9\text{!}}} \right)\mspace{2mu} dx}} = \pi - \frac{\pi^{3}}{3 \cdot 3\text{!}} + \frac{\pi^{5}}{5 \cdot 5\text{!}} - \frac{\pi^{7}}{7 \cdot 7\text{!}} + \frac{\pi^{9}}{9 \cdot 9\text{!}} = 1.852...,$ whereas ${\int_{0}^{\pi}{\frac{\text{sin}\mspace{2mu} t}{t}dt}} = 1.85194...,$ so the actual error is approximately $0.00006.$

245.

Since $\text{cos}\left( t^{2} \right) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{t^{4n}}{\left( {2n} \right)\text{!}}}}$ and $\text{sin}\left( t^{2} \right) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{t^{4n + 2}}{\left( {2n + 1} \right)\text{!}}}},$ one has $S(x) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{4n + 3}}{\left( {4n + 3} \right)\left( {2n + 1} \right)\text{!}}}}$ and $C(x) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{4n + 1}}{\left( {4n + 1} \right)\left( {2n} \right)\text{!}}}}.$ 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.

${\int_{0}^{1\text{/}4}\sqrt{x}}\left( {1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \frac{5x^{4}}{128} - \frac{7x^{5}}{256}} \right)\mspace{2mu} dx$

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

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

249.

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

251.

${\int_{0}^{\pi\text{/}2}{\text{sin}^{4}\theta d\theta}} = \frac{3\pi}{16}.$ Hence $T \approx 2\pi\sqrt{\frac{L}{g}}\left( {1 + \frac{k^{2}}{4} + \frac{9}{256}k^{4}} \right).$

Review Exercises

253.

True

255.

True

257.

ROC: $1;$ IOC: $\left( {0,2} \right)$

259.

ROC: $12;$ IOC: $\left( {-12,12} \right)$

261.

$\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{3^{n + 1}}x^{n + 2};$ ROC: $3;$ IOC: $\left( {-3,3} \right)$

263.

integration: ${\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}}\left( {2x} \right)^{2n + 1}$

265.

$p_{4}(x) = \left( {x + 3} \right)^{3} - 11\left( {x + 3} \right)^{2} + 39\left( {x + 3} \right) - 41;$ exact

267.

$\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}\left( {3x} \right)^{2n}}{2n\text{!}}$

269.

${\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{\left( {2n} \right)\text{!}}}\left( {x - \frac{\pi}{2}} \right)^{2n}$

271.

${\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n}}{n\text{!}}}x^{2n}$

273.

$F(x) = {\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{\left( {2n + 1} \right)\left( {2n + 1} \right)\text{!}}}x^{2n + 1}$

275.

Answers may vary.

277.

$2.5\text{\%}$

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• Book title: Calculus Volume 2
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-2/pages/chapter-6

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