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# 10.E: Power Series (Exercises)

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## 10.1: Power Series and Functions

In the following exercises, state whether each statement is true, or give an example to show that it is false.

1) If $$\displaystyle \sum_{n=1}^∞a_nx^n$$ converges, then $$\displaystyle a_nx^n→0$$ as $$\displaystyle n→∞.$$

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

2) $$\displaystyle \sum_{n=1}^∞a_nx^n$$ converges at $$\displaystyle x=0$$ for any real numbers $$\displaystyle a_n$$.

3) Given any sequence $$\displaystyle a_n$$, there is always some $$\displaystyle R>0$$, possibly very small, such that $$\displaystyle \sum_{n=1}^∞a_nx^n$$ converges on $$\displaystyle (−R,R)$$.

Solution: False. It would imply that $$\displaystyle a_nx^n→0$$ for $$\displaystyle |x|<R$$. If $$\displaystyle a_n=n^n$$, then $$\displaystyle a_nx^n=(nx)^n$$ does not tend to zero for any $$\displaystyle x≠0$$.

4) If $$\displaystyle \sum_{n=1}^∞a_nx^n$$ has radius of convergence $$\displaystyle R>0$$ and if $$\displaystyle |b_n|≤|a_n|$$ for all $$\displaystyle n$$, then the radius of convergence of $$\displaystyle \sum_{n=1}^∞b_nx^n$$ is greater than or equal to $$\displaystyle R$$.

5) Suppose that $$\displaystyle \sum_{n=0}^∞a_n(x−3)^n$$ converges at $$\displaystyle x=6$$. At which of the following points must the series also converge? Use the fact that if $$\displaystyle \sum a_n(x−c)^n$$ converges at $$\displaystyle x$$, then it converges at any point closer to $$\displaystyle c$$ than $$\displaystyle x$$.

a. $$\displaystyle x=1$$

b. $$\displaystyle x=2$$

c. $$\displaystyle x=3$$

d. $$\displaystyle x=0$$

e. $$\displaystyle x=5.99$$

f. $$\displaystyle x=0.000001$$

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

6) Suppose that $$\displaystyle \sum_{n=0}^∞a_n(x+1)^n$$ converges at $$\displaystyle x=−2$$. At which of the following points must the series also converge? Use the fact that if $$\displaystyle \sum a_n(x−c)^n$$ converges at $$\displaystyle x$$, then it converges at any point closer to $$\displaystyle c$$ than $$\displaystyle x$$.

a. $$\displaystyle x=2$$

b. $$\displaystyle x=−1$$

c. $$\displaystyle x=−3$$

d. $$\displaystyle x=0$$

e. $$\displaystyle x=0.99$$

f. $$\displaystyle x=0.000001$$

In the following exercises, suppose that $$\displaystyle ∣\frac{a_{n+1}}{a_n}∣→1$$ as $$\displaystyle n→∞.$$ Find the radius of convergence for each series.

7) $$\displaystyle \sum_{n=0}^∞a_n2^nx^n$$

Solution: $$\displaystyle ∣\frac{a_{n+1}2^{n+1}x^{n+1}}{a_n2^nx^n}∣ =2|x|∣\frac{a_{n+1}}{a_n}∣→2|x|$$ so $$\displaystyle R=\frac{1}{2}$$

8) $$\displaystyle \sum_{n=0}^∞\frac{a_nx^n}{2^n}$$

9) $$\displaystyle \sum_{n=0}^∞\frac{a_nπ^nx^n}{e^n}$$

Solution: $$\displaystyle ∣\frac{a_{n+1}(\frac{π}{e})^{n+1}x^{n+1}}{a_n(\frac{π}{e})^nx^n}∣ =\frac{π|x|}{e}∣\frac{a_{n+1}}{a_n}∣→\frac{π|x|}{e}$$ so $$\displaystyle R=\frac{e}{π}$$

10) $$\displaystyle sum_{n=0}^∞\frac{a_n(−1)^nx^n}{10^n}$$

11) $$\displaystyle \sum_{n=0}^∞a_n(−1)^nx^{2n}$$

Solution: $$\displaystyle ∣\frac{a_{n+1}(−1)^{n+1}x^{2n+2}}{a_n(−1)^nx^{2n}}∣ =∣x^2∣∣\frac{a_{n+1}}{a_n}∣→∣x^2∣$$ so $$\displaystyle R=1$$

12) $$\displaystyle \sum_{n=0}^∞a_n(−4)^nx^{2n}$$

In the following exercises, find the radius of convergence $$\displaystyle R$$ and interval of convergence for $$\displaystyle \sum a_nx^n$$ with the given coefficients $$\displaystyle a_n$$.

13) $$\displaystyle \sum_{n=1}^∞\frac{(2x)^n}{n}$$

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

14) $$\displaystyle \sum_{n=1}^∞(−1)^n\frac{x^n}{\sqrt{n}}$$

15) $$\displaystyle \sum_{n=1}^∞\frac{nx^n}{2^n}$$

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

16) $$\displaystyle \sum_{n=1}^∞\frac{nx^n}{e^n}$$

17) $$\displaystyle \sum_{n=1}^∞\frac{n^2x^n}{2^n}$$

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

18) $$\displaystyle \sum_{k=1}^∞\frac{k^ex^k}{e^k}$$

19) $$\displaystyle \sum_{k=1}^∞\frac{π^kx^k}{k^π}$$

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

20) $$\displaystyle \sum_{n=1}^∞\frac{x^n}{n!}$$

21) $$\displaystyle \sum_{n=1}^∞\frac{10^nx^n}{n!}$$

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

22) $$\displaystyle \sum_{n=1}^∞(−1)^n\frac{x^n}{ln(2n)}$$

In the following exercises, find the radius of convergence of each series.

23) $$\displaystyle \sum_{k=1}^∞\frac{(k!)^2x^k}{(2k)!}$$

Solution: $$\displaystyle a_k=\frac{(k!)^2}{(2k)!}$$ so $$\displaystyle \frac{a_{k+1}}{a_k}=\frac{(k+1)^2}{(2k+2)(2k+1)}→\frac{1}{4}$$ so $$\displaystyle R=4$$

24) $$\displaystyle \sum_{n=1}^∞\frac{(2n)!x^n}{n^{2n}}$$

25) $$\displaystyle \sum_{k=1}^∞\frac{k!}{1⋅3⋅5⋯(2k−1)}x^k$$

Solution: $$\displaystyle a_k=\frac{k!}{1⋅3⋅5⋯(2k−1)}$$ so $$\displaystyle \frac{a_{k+1}}{a_k}=\frac{k+1}{2k+1}→\frac{1}{2}$$ so $$\displaystyle R=2$$

26) $$\displaystyle \sum_{k=1}^∞\frac{2⋅4⋅6⋯2k}{(2k)!}x^k$$

27) $$\displaystyle \sum_{n=1}^∞\frac{x^n}{(^{2n}_n)}$$ where $$\displaystyle (^n_k)=\frac{n!}{k!(n−k)!}$$

Solution: $$\displaystyle a_n=\frac{1}{(^{2n}_n)}$$ so $$\displaystyle \frac{a_{n+1}}{a_n}=\frac{((n+1)!)^2}{(2n+2)!}\frac{2n!}{(n!)^2}=\frac{(n+1)^2}{(2n+2)(2n+1)}→\frac{1}{4}$$ so $$\displaystyle R=4$$

28) $$\displaystyle \sum_{n=1}^∞sin^2nx^n$$

In the following exercises, use the ratio test to determine the radius of convergence of each series.

29) $$\displaystyle \sum_{n=1}^∞\frac{(n!)^3}{(3n)!}x^n$$

Solution: $$\displaystyle \frac{a_{n+1}}{a_n}=\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}→\frac{1}{27}$$ so $$\displaystyle R=27$$

30) $$\displaystyle \sum_{n=1}^∞\frac{2^{3n}(n!)^3}{(3n)!}x^n$$

31) $$\displaystyle \sum_{n=1}^∞\frac{n!}{n^n}x^n$$

Solution: $$\displaystyle a_n=\frac{n!}{n^n}$$ so $$\displaystyle \frac{a_{n+1}}{a_n}=\frac{(n+1)!}{n!}\frac{n^n}{(n+1)^{n+1}}=(\frac{n}{n+1})^n→\frac{1}{e}$$ so $$\displaystyle R=e$$

32) $$\displaystyle \sum_{n=1}^∞\frac{(2n)!}{n^{2n}}x^n$$

In the following exercises, given that $$\displaystyle \frac{1}{1−x}=\sum_{n=0}^∞x^n$$ with convergence in $$\displaystyle (−1,1)$$, find the power series for each function with the given center a, and identify its interval of convergence.

33) $$\displaystyle f(x)=\frac{1}{x};a=1$$ (Hint: $$\displaystyle \frac{1}{x}=\frac{1}{1−(1−x)})$$

Solution: $$\displaystyle f(x)=\sum_{n=0}^∞(1−x)^n$$ on $$\displaystyle I=(0,2)$$

34) $$\displaystyle f(x)=\frac{1}{1−x^2};a=0$$

35) $$\displaystyle f(x)=\frac{x}{1−x^2};a=0$$

Solution: $$\displaystyle \sum_{n=0}^∞x^{2n+1}$$ on $$\displaystyle I=(−1,1)$$

36) $$\displaystyle f(x)=\frac{1}{1+x^2};a=0$$

37) $$\displaystyle f(x)=\frac{x^2}{1+x^2};a=0$$

Solution: $$\displaystyle \sum_{n=0}^∞(−1)^nx^{2n+2}$$ on $$\displaystyle I=(−1,1)$$

38) $$\displaystyle f(x)=\frac{1}{2−x};a=1$$

39) $$\displaystyle f(x)=\frac{1}{1−2x};a=0.$$

Solution: $$\displaystyle \sum_{n=0}^∞2^nx^n$$ on $$\displaystyle (−\frac{1}{2},\frac{1}{2})$$

40) $$\displaystyle f(x)=\frac{1}{1−4x^2};a=0$$

41) $$\displaystyle f(x)=\frac{x^2}{1−4x^2};a=0$$

Solution: $$\displaystyle \sum_{n=0}^∞4^nx^{2n+2}$$ on $$\displaystyle (−\frac{1}{2},\frac{1}{2})$$

42) $$\displaystyle f(x)=\frac{x^2}{5−4x+x^2};a=2$$

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

43) Explain why, if $$\displaystyle |a_n|^{1/n}→r>0,$$ then $$\displaystyle |a_nx^n|^{1/n}→|x|r<1$$ whenever $$\displaystyle |x|<\frac{1}{r}$$ and, therefore, the radius of convergence of $$\displaystyle \sum_{n=1}^∞a_nx^n$$ is $$\displaystyle R=\frac{1}{r}$$.

Solution: $$\displaystyle |a_nx^n|^{1/n}=|a_n|^{1/n}|x|→|x|r$$ as $$\displaystyle n→∞$$ and $$\displaystyle |x|r<1$$ when $$\displaystyle |x|<\frac{1}{r}$$. Therefore, $$\displaystyle \sum_{n=1}^∞a_nx^n$$ converges when $$\displaystyle |x|<\frac{1}{r}$$ by the nth root test.

44) $$\displaystyle \sum_{n=1}^∞\frac{x^n}{n^n}$$

45) $$\displaystyle \sum_{k=1}^∞(\frac{k−1}{2k+3})^kx^k$$

Solution: $$\displaystyle a_k=(\frac{k−1}{2k+3})^k$$ so $$\displaystyle (a_k)^{1/k}→\frac{1}{2}<1$$ so $$\displaystyle R=2$$

46) $$\displaystyle \sum_{k=1}^∞(\frac{2k^2−1}{k^2+3})^kx^k$$

47) $$\displaystyle \sum_{n=1}^∞a_n=(n^{1/n}−1)^nx^n$$

Solution: $$\displaystyle a_n=(n^{1/n}−1)^n$$ so $$\displaystyle (a_n)^{1/n}→0$$ so $$\displaystyle R=∞$$

48) Suppose that $$\displaystyle p(x)=\sum_{n=0}^∞a_nx^n$$ such that $$\displaystyle a_n=0$$ if $$\displaystyle n$$ is even. Explain why $$\displaystyle p(x)=p(−x).$$

49) Suppose that $$\displaystyle p(x)=\sum_{n=0}^∞a_nx^n$$ such that $$\displaystyle a_n=0$$ if $$\displaystyle n$$ is odd. Explain why $$\displaystyle p(x)=−p(−x).$$

Solution: We can rewrite $$\displaystyle p(x)=\sum_{n=0}^∞a_{2n+1}x^{2n+1}$$ and $$\displaystyle p(x)=p(−x)$$ since $$\displaystyle x^{2n+1}=−(−x)^{2n+1}$$.

50) Suppose that $$\displaystyle p(x)=\sum_{n=0}^∞a_nx^n$$ converges on $$\displaystyle (−1,1]$$. Find the interval of convergence of $$\displaystyle p(Ax)$$.

51) Suppose that $$\displaystyle p(x)=\sum_{n=0}^∞a_nx^n$$ converges on $$\displaystyle (−1,1]$$. Find the interval of convergence of $$\displaystyle p(2x−1)$$.

Solution: If $$\displaystyle x∈[0,1],$$ then $$\displaystyle y=2x−1∈[−1,1]$$ so $$\displaystyle p(2x−1)=p(y)=\sum_{n=0}^∞a_ny^n$$ converges.

In the following exercises, suppose that $$\displaystyle p(x)=\sum_{n=0}^∞a_nx^n$$ satisfies $$\displaystyle \lim_{n→∞}\frac{a_{n+1}}{a_n}=1$$ where $$\displaystyle a_n≥0$$ for each $$\displaystyle n$$. State whether each series converges on the full interval $$\displaystyle (−1,1)$$, or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

52) $$\displaystyle \sum_{n=0}^∞a_nx^{2n}$$

53) $$\displaystyle \sum_{n=0}^∞a_{2n}x^{2n}$$

Solution: Converges on $$\displaystyle (−1,1)$$ by the ratio test

54) $$\displaystyle \sum_{n=0}^∞a_{2n}x^n$$ (Hint:$$\displaystyle x=±\sqrt{x^2}$$)

55) $$\displaystyle \sum_{n=0}^∞a_{n^2}x^{n^2}$$ (Hint: Let $$\displaystyle b_k=a_k$$ if $$\displaystyle k=n^2$$ for some $$\displaystyle n$$, otherwise $$\displaystyle b_k=0$$.)

Solution: Consider the series $$\displaystyle \sum b_kx^k$$ where $$\displaystyle b_k=a_k$$ if $$\displaystyle k=n^2$$ and $$\displaystyle b_k=0$$ otherwise. Then $$\displaystyle b_k≤a_k$$ and so the series converges on $$\displaystyle (−1,1)$$ by the comparison test.

56) Suppose that $$\displaystyle p(x)$$ is a polynomial of degree $$\displaystyle N$$. Find the radius and interval of convergence of $$\displaystyle \sum_{n=1}^∞p(n)x^n$$.

57) [T] Plot the graphs of $$\displaystyle \frac{1}{1−x}$$ and of the partial sums $$\displaystyle S_N=\sum_{n=0}^Nx^n$$ for $$\displaystyle n=10,20,30$$ on the interval $$\displaystyle [−0.99,0.99]$$. Comment on the approximation of $$\displaystyle \frac{1}{1−x}$$ by $$\displaystyle S_N$$ near $$\displaystyle x=−1$$ and near $$\displaystyle x=1$$ as $$\displaystyle N$$ increases.

Solution: The approximation is more accurate near $$\displaystyle x=−1$$. The partial sums follow $$\displaystyle \frac{1}{1−x}$$ more closely as $$\displaystyle N$$ increases but are never accurate near $$\displaystyle x=1$$ since the series diverges there. 58) [T] Plot the graphs of $$\displaystyle −ln(1−x)$$ and of the partial sums $$\displaystyle S_N=\sum_{n=1}^N\frac{x^n}{n}$$ for $$\displaystyle n=10,50,100$$ on the interval $$\displaystyle [−0.99,0.99]$$. Comment on the behavior of the sums near $$\displaystyle x=−1$$ and near $$\displaystyle x=1$$ as $$\displaystyle N$$ increases.

59) [T] Plot the graphs of the partial sums $$\displaystyle S_n=\sum_{n=1}^N\frac{x^n}{n^2}$$ for $$\displaystyle n=10,50,100$$ on the interval $$\displaystyle [−0.99,0.99]$$. Comment on the behavior of the sums near $$\displaystyle x=−1$$ and near $$\displaystyle x=1$$ as $$\displaystyle N$$ increases.

Solution: The approximation appears to stabilize quickly near both $$\displaystyle x=±1$$. 60) [T] Plot the graphs of the partial sums $$\displaystyle S_N=\sum_{n=1}^Nsinnx^n$$ for $$\displaystyle n=10,50,100$$ on the interval $$\displaystyle [−0.99,0.99]$$. Comment on the behavior of the sums near $$\displaystyle x=−1$$ and near $$\displaystyle x=1$$ as $$\displaystyle N$$ increases.

61) [T] Plot the graphs of the partial sums $$\displaystyle S_N=\sum_{n=0}^N(−1)^n\frac{x^{2n+1}}{(2n+1)!}$$ for $$\displaystyle n=3,5,10$$ on the interval $$\displaystyle [−2π,2π]$$. Comment on how these plots approximate $$\displaystyle sinx$$ as $$\displaystyle N$$ increases.

Solution: The polynomial curves have roots close to those of $$\displaystyle sinx$$ up to their degree and then the polynomials diverge from $$\displaystyle sinx$$. 62) [T] Plot the graphs of the partial sums $$\displaystyle S_N=\sum_{n=0}^N(−1)^n\frac{x^{2n}}{(2n)!}$$ for $$\displaystyle n=3,5,10$$ on the interval $$\displaystyle [−2π,2π]$$. Comment on how these plots approximate $$\displaystyle cosx$$ as $$\displaystyle N$$ increases.

## 10.2: Properties of Power Series

1) If $$\displaystyle f(x)=\sum_{n=0}^∞\frac{x^n}{n!}$$ and $$\displaystyle g(x)=\sum_{n=0}^∞(−1)^n\frac{x^n}{n!}$$, find the power series of $$\displaystyle \frac{1}{2}(f(x)+g(x))$$ and of $$\displaystyle \frac{1}{2}(f(x)−g(x))$$.

Solution: $$\displaystyle \frac{1}{2}(f(x)+g(x))=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$$ and $$\displaystyle \frac{1}{2}(f(x)−g(x))=\sum_{n=0}^∞\frac{x^{2n+1}}{(2n+1)!}$$.

2) If $$\displaystyle C(x)=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$$ and $$\displaystyle S(x)=\sum_{n=0}^∞\frac{x^{2n+1}}{(2n+1)!}$$, find the power series of $$\displaystyle C(x)+S(x)$$ and of $$\displaystyle C(x)−S(x)$$.

In the following exercises, use partial fractions to find the power series of each function.

3) $$\displaystyle \frac{4}{(x−3)(x+1)}$$

Solution: $$\displaystyle \frac{4}{(x−3)(x+1)}=\frac{1}{x−3}−\frac{1}{x+1}=−\frac{1}{3(1−\frac{x}{3})}−\frac{1}{1−(−x)}=−\frac{1}{3}\sum_{n=0}^∞(\frac{x}{3})^n−\sum_{n=0}^∞(−1)^nx^n=\sum_{n=0}^∞((−1)^{n+1}−\frac{1}{3n+1})x^n$$

4) $$\displaystyle \frac{3}{(x+2)(x−1)}$$

5) $$\displaystyle \frac{5}{(x^2+4)(x^2−1)}$$

Soution: $$\displaystyle \frac{5}{(x^2+4)(x^2−1)}=\frac{1}{x^2−1}−\frac{1}{4}\frac{1}{1+(\frac{x}{2})^2}=−\sum_{n=0}^∞x^{2n}−\frac{1}{4}\sum_{n=0}^∞(−1)^n(\frac{x}{2})^n=\sum_{n=0}^∞((−1)+(−1)^{n+1}\frac{1}{2^{n+2}})x^{2n}$$

6) $$\displaystyle \frac{30}{(x^2+1)(x^2−9)}$$

In the following exercises, express each series as a rational function.

7) $$\displaystyle \sum_{n=1}^∞\frac{1}{x^n}$$

Solution: $$\displaystyle \frac{1}{x}\sum_{n=0}^∞\frac{1}{x^n}=\frac{1}{x}\frac{1}{1−\frac{1}{x}}=\frac{1}{x−1}$$

8) $$\displaystyle \sum_{n=1}^∞\frac{1}{x^{2n}}$$

9) $$\displaystyle \sum_{n=1}^∞\frac{1}{(x−3)^{2n−1}}$$

Solution: $$\displaystyle \frac{1}{x−3}\frac{1}{1−\frac{1}{(x−3)^2}}=\frac{x−3}{(x−3)^2−1}$$

10) $$\displaystyle \sum_{n=1}^∞(\frac{1}{(x−3)^{2n−1}}−\frac{1}{(x−2)^{2n−1}})$$

The following exercises explore applications of annuities.

11) Calculate the present values P of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of $$\displaystyle r=0.03,r=0.05$$, and $$\displaystyle r=0.07$$. Solution: $$\displaystyle P=P_1+⋯+P_{20}$$ where $$\displaystyle P_k=10,000\frac{1}{(1+r)^k}$$. Then $$\displaystyle P=10,000\sum_{k=1}^{20}\frac{1}{(1+r)^k}=10,000\frac{1−(1+r)^{−20}}{r}$$. When $$\displaystyle r=0.03,P≈10,000×14.8775=148,775.$$ When $$\displaystyle r=0.05,P≈10,000×12.4622=124,622.$$ When $$\displaystyle r=0.07,P≈105,940$$. 12) Calculate the present values P of annuities in which$9,000 is to be paid out annually perpetually, assuming interest rates of $$\displaystyle r=0.03,r=0.05$$ and $$\displaystyle r=0.07$$.

13) Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of $$\displaystyle r=0.03,r=0.05,$$ and $$\displaystyle r=0.07.$$ Solution: In general, $$\displaystyle P=\frac{C(1−(1+r)^{−N})}{r}$$ for N years of payouts, or $$\displaystyle C=\frac{Pr}{1−(1+r)^{−N}}$$. For $$\displaystyle N=20$$ and $$\displaystyle P=100,000$$, one has $$\displaystyle C=6721.57$$ when $$\displaystyle r=0.03;C=8024.26$$ when $$\displaystyle r=0.05$$; and $$\displaystyle C≈9439.29$$ when $$\displaystyle r=0.07$$. 14) Calculate the annual payouts C to be given perpetually on annuities having present value$100,000 assuming respective interest rates of $$\displaystyle r=0.03,r=0.05,$$ and $$\displaystyle r=0.07$$.

15) Suppose that an annuity has a present value $$\displaystyle P=1$$ million dollars. What interest rate r would allow for perpetual annual payouts of $50,000? Solution: In general, $$\displaystyle P=\frac{C}{r}.$$ Thus, $$\displaystyle r=\frac{C}{P}=5×\frac{10^4}{10^6}=0.05.$$ 16) Suppose that an annuity has a present value $$\displaystyle P=10$$ million dollars. What interest rate r would allow for perpetual annual payouts of$100,000?

In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.

17) $$\displaystyle x+x^2−x^3+x^4+x^5−x^6+⋯$$ (Hint: Group powers $$\displaystyle x^{3k}, x^{3k−1},$$ and $$\displaystyle x^{3k−2}$$.)

Solution: $$\displaystyle (x+x^2−x^3)(1+x^3+x^6+⋯)=\frac{x+x^2−x^3}{1−x^3}$$

18) $$\displaystyle x+x^2−x^3−x^4+x^5+x^6−x^7−x^8+⋯$$ (Hint: Group powers $$\displaystyle x^{4k}, x^{4k−1},$$ etc.)

19) $$\displaystyle x−x^2−x^3+x^4−x^5−x^6+x^7−⋯$$ (Hint: Group powers $$\displaystyle x^{3k}, x^{3k−1}$$, and $$\displaystyle x^{3k−2}$$.)

Solution: $$\displaystyle (x−x^2−x^3)(1+x^3+x^6+⋯)=\frac{x−x^2−x^3}{1−x^3}$$

20) $$\displaystyle \frac{x}{2}+\frac{x^2}{4}−\frac{x^3}{8}+\frac{x^4}{16}+\frac{x^5}{32}−\frac{x^6}{64}+⋯$$ (Hint: Group powers $$\displaystyle \frac{x}{2})^{3k},(\frac{x}{2})^{3k−1},$$ and $$\displaystyle \frac{x}{2})^{3k−2}$$.)

In the following exercises, find the power series of $$\displaystyle f(x)g(x)$$ given f and g as defined.

21) $$\displaystyle f(x)=2\sum_{n=0}^∞x^n,g(x)=\sum_{n=0}^∞nx^n$$

Solution: $$\displaystyle a_n=2,b_n=n$$ so $$\displaystyle c_n=\sum_{k=0}^nb_ka_{n−k}=2\sum_{k=0}^nk=(n)(n+1)$$ and $$\displaystyle f(x)g(x)=\sum_{n=1}^∞n(n+1)x^n$$

22) $$\displaystyle f(x)=\sum_{n=1}^∞x^n,g(x)=\sum_{n=1}^∞\frac{1}{n}x^n$$. Express the coefficients of $$\displaystyle f(x)g(x)$$ in terms of $$\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}$$.

23) $$\displaystyle f(x)=g(x)=\sum_{n=1}^∞(\frac{x}{2})^n$$

Solution: $$\displaystyle a_n=b_n=2^{−n}$$ so $$\displaystyle c_n=\sum_{k=1}^nb_ka_{n−k}=2^{−n}\sum_{k=1}^n1=\frac{n}{2^n}$$ and $$\displaystyle f(x)g(x)=\sum_{n=1}^∞n(\frac{x}{2})^n$$

24) $$\displaystyle f(x)=g(x)=\sum_{n=1}^∞nx^n$$

In the following exercises, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f.

25) $$\displaystyle f(x)=\frac{1}{1+x}=\sum_{n=0}^∞(−1)^nx^n$$

Solution: The derivative of $$\displaystyle f$$ is $$\displaystyle −\frac{1}{(1+x)^2}=−\sum_{n=0}^∞(−1)^n(n+1)x^n$$.

26) $$\displaystyle f(x)=\frac{1}{1−x^2}=\sum_{n=0}^∞x^{2n}$$

In the following exercises, integrate the given series expansion of $$\displaystyle f$$ term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of $$\displaystyle f$$.

27) $$\displaystyle f(x)=\frac{2x}{(1+x^2)^2}=\sum_{n=1}^∞(−1)^n(2n)x^{2n−1}$$

Solution: The indefinite integral of $$\displaystyle f$$ is $$\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^∞(−1)^nx^{2n}$$.

28) $$\displaystyle f(x)=\frac{2x}{1+x^2}=2\sum_{n=0}^∞(−1)^nx^{2n+1}$$

In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.

29) Evaluate $$\displaystyle \sum_{n=1}^∞\frac{n}{2^n}$$ as $$\displaystyle f′(\frac{1}{2})$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x^n$$.

Solution: $$\displaystyle f(x)=\sum_{n=0}^∞x^n=\frac{1}{1−x};f′(\frac{1}{2})=\sum_{n=1}^∞\frac{n}{2^{n−1}}=\frac{d}{dx}(1−x)^{−1}∣_{x=1/2}=\frac{1}{(1−x)^2}∣_{x=1/2}=4$$ so $$\displaystyle \sum_{n=1}^∞\frac{n}{2^n}=2.$$

30) Evaluate $$\displaystyle \sum_{n=1}^∞\frac{n}{3^n}$$ as $$\displaystyle f′(\frac{1}{3})$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x6n$$.

31) Evaluate $$\displaystyle \sum_{n=2}^∞\frac{n(n−1)}{2^n}$$ as $$\displaystyle f''(\frac{1}{2})$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x^n$$.

Solution: $$\displaystyle f(x)=\sum_{n=0}^∞x^n=\frac{1}{1−x};f''(\frac{1}{2})=\sum_{n=2}^∞\frac{n(n−1)}{2^{n−2}}=\frac{d^2}{dx^2}(1−x)^{−1}∣_{x=1/2}=\frac{2}{(1−x)^3}∣_{x=1/2}=16$$ so $$\displaystyle \sum_{n=2}^∞n\frac{(n−1)}{2^n}=4.$$

32) Evaluate $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n}{n+1}$$ as $$\displaystyle ∫^1_0f(t)dt$$ where $$\displaystyle f(x)=\sum_{n=0}^∞(−1)^nx^{2n}=\frac{1}{1+x^2}$$.

In the following exercises, given that $$\displaystyle \frac{1}{1−x}=\sum_{n=0}^∞x^n$$, use term-by-term differentiation or integration to find power series for each function centered at the given point.

33) $$\displaystyle f(x)=lnx$$ centered at $$\displaystyle x=1$$ (Hint: $$\displaystyle x=1−(1−x)$$)

Solution: $$\displaystyle ∫\sum(1−x)^ndx=∫\sum(−1)^n(x−1)^ndx=\sum \frac{(−1)^n(x−1)^{n+1}}{n+1}$$

34) $$\displaystyle ln(1−x)$$ at $$\displaystyle x=0$$

35) $$\displaystyle ln(1−x^2)$$ at $$\displaystyle x=0$$

Solution: $$\displaystyle −∫^{x^2}_{t=0}\frac{1}{1−t}dt=−\sum_{n=0}^∞∫^{x^2}_0t^ndx−\sum_{n=0}^∞\frac{x^{2(n+1)}}{n+1}=−\sum_{n=1}^∞\frac{x^{2n}}{n}$$

36) $$\displaystyle f(x)=\frac{2x}{(1−x^2)^2}$$ at $$\displaystyle x=0$$

37) $$\displaystyle f(x)=tan^{−1}(x^2)$$ at $$\displaystyle x=0$$

Solution: $$\displaystyle ∫^{x^2}_0\frac{dt}{1+t^2}=\sum_{n=0}^∞(−1)^n∫^{x^2}_0t^{2n}dt=\sum_{n=0}^∞(−1)^n\frac{t^{2n+1}}{2n+1}∣^{x^2}_{t=0}=\sum_{n=0}^∞(−1)^n\frac{x^{4n+2}}{2n+1}$$

38) $$\displaystyle f(x)=ln(1+x^2)$$ at $$\displaystyle x=0$$

39) $$\displaystyle f(x)=∫^x_0lntdt$$ where $$\displaystyle ln(x)=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^n}{n}$$

Solution: Term-by-term integration gives $$\displaystyle ∫^x_0lntdt=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^{n+1}}{n(n+1)}=\sum_{n=1}^∞(−1)^{n−1}(\frac{1}{n}−\frac{1}{n+1})(x−1)^{n+1}=(x−1)lnx+\sum_{n=2}^∞(−1)^n\frac{(x−1)^n}{n}=xlnx−x.$$

40) [T] Evaluate the power series expansion $$\displaystyle ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$$ at $$\displaystyle x=1$$ to show that $$\displaystyle ln(2)$$ is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate $$\displaystyle ln(2)$$

accurate to within 0.001, and find such an approximation.

41) [T] Subtract the infinite series of $$\displaystyle ln(1−x)$$ from $$\displaystyle ln(1+x)$$ to get a power series for $$\displaystyle ln(\frac{1+x}{1−x})$$. Evaluate at $$\displaystyle x=\frac{1}{3}$$. What is the smallest N such that the Nth partial sum of this series approximates $$\displaystyle ln(2)$$ with an error less than 0.001?

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

In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.

42) $$\displaystyle \sum_{k=0}^∞(x^k−x^{2k+1})$$

43) $$\displaystyle \sum_{k=1}^∞\frac{x^{3k}}{6k}$$

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

44) $$\displaystyle \sum_{k=1}^∞(1+x^2)^{−k}$$ using $$\displaystyle y=\frac{1}{1+x^2}$$

45) $$\displaystyle \sum_{k=1}^∞2^{−kx}$$ using $$\displaystyle y=2^{−x}$$

Solution: If $$\displaystyle y=2^{−x}$$, then $$\displaystyle \sum_{k=1}^∞y^k=\frac{y}{1−y}=\frac{2^{−x}}{1−2^{−x}}=\frac{1}{2^x−1}$$. If $$\displaystyle a_k=2^{−kx}$$, then $$\displaystyle \frac{a_{k+1}}{a_k}=2^{−x}<1$$ when $$\displaystyle x>0$$. So the series converges for all $$\displaystyle x>0$$.

46) Show that, up to powers $$\displaystyle x^3$$ and $$\displaystyle y^3$$, $$\displaystyle E(x)=\sum_{n=0}^∞\frac{x^n}{n!}$$ satisfies $$\displaystyle E(x+y)=E(x)E(y)$$.

47) Differentiate the series $$\displaystyle E(x)=\sum_{n=0}^∞\frac{x^n}{n!}$$ term-by-term to show that $$\displaystyle E(x)$$ is equal to its derivative.

48) Show that if $$\displaystyle f(x)=\sum_{n=0}^∞a_nx^n$$ is a sum of even powers, that is, $$\displaystyle a_n=0$$ if $$\displaystyle n$$ is odd, then $$\displaystyle F=∫^x_0f(t)dt$$ is a sum of odd powers, while if I is a sum of odd powers, then F is a sum of even powers.

49) [T] Suppose that the coefficients an of the series $$\displaystyle \sum_{n=0}^∞a_nx^n$$ are defined by the recurrence relation $$\displaystyle a_n=\frac{a_{n−1}}{n}+\frac{a_{n−2}}{n(n−1)}$$. For $$\displaystyle a_0=0$$ and $$\displaystyle a_1=1$$, compute and plot the sums $$\displaystyle S_N=\sum_{n=0}^Na_nx^n$$ for $$\displaystyle N=2,3,4,5$$ on $$\displaystyle [−1,1].$$

Solution: The solid curve is $$\displaystyle S_5$$. The dashed curve is $$\displaystyle S_2$$, dotted is $$\displaystyle S_3$$, and dash-dotted is $$\displaystyle S_4$$ 50) [T] Suppose that the coefficients an of the series $$\displaystyle \sum_{n=0}^∞a_nx^n$$ are defined by the recurrence relation $$\displaystyle a_n=\frac{a_{n−1}}{\sqrt{n}}−\frac{a_{n−2}}{\sqrt{n(n−1)}}$$. For $$\displaystyle a_0=1$$ and $$\displaystyle a_1=0$$, compute and plot the sums $$\displaystyle S_N=\sum_{n=0}^Na_nx^n$$ for $$\displaystyle N=2,3,4,5$$ on $$\displaystyle [−1,1]$$.

51) [T] Given the power series expansion $$\displaystyle ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$$, determine how many terms N of the sum evaluated at $$\displaystyle x=−1/2$$ are needed to approximate $$\displaystyle ln(2)$$ accurate to within 1/1000. Evaluate the corresponding partial sum $$\displaystyle \sum_{n=1}^N(−1)^{n−1}\frac{x^n}{n}$$.

Solution: When $$\displaystyle x=−\frac{1}{2},−ln(2)=ln(\frac{1}{2})=−\sum_{n=1}^∞\frac{1}{n2^n}$$. Since $$\displaystyle \sum^∞_{n=11}\frac{1}{n2^n}<\sum_{n=11}^∞\frac{1}{2^n}=\frac{1}{2^{10}},$$ one has $$\displaystyle \sum_{n=1}^{10}\frac{1}{n2^n}=0.69306…$$ whereas $$\displaystyle ln(2)=0.69314…;$$ therefore, $$\displaystyle N=10.$$

52) [T] Given the power series expansion $$\displaystyle tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1}$$, use the alternating series test to determine how many terms N of the sum evaluated at $$\displaystyle x=1$$ are needed to approximate $$\displaystyle tan^{−1}(1)=\frac{π}{4}$$ accurate to within 1/1000. Evaluate the corresponding partial sum $$\displaystyle \sum_{k=0}^N(−1)^k\frac{x^{2k+1}}{2k+1}$$.

53) [T] Recall that $$\displaystyle tan^{−1}(\frac{1}{\sqrt{3}})=\frac{π}{6}.$$ Assuming an exact value of $$\displaystyle \frac{1}{\sqrt{3}})$$, estimate $$\displaystyle \frac{π}{6}$$ by evaluating partial sums $$\displaystyle S_N(\frac{1}{\sqrt{3}})$$ of the power series expansion $$\displaystyle tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1}$$ at $$\displaystyle x=\frac{1}{\sqrt{3}}$$. What is the smallest number $$\displaystyle N$$ such that $$\displaystyle 6S_N(\frac{1}{\sqrt{3}})$$ approximates $$\displaystyle π$$ accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?

Solution: $$\displaystyle 6S_N(\frac{1}{\sqrt{3}})=2\sqrt{3}\sum_{n=0}^N(−1)^n\frac{1}{3^n(2n+1).}$$ One has $$\displaystyle π−6S_4(\frac{1}{\sqrt{3}})=0.00101…$$ and $$\displaystyle π−6S_5(\frac{1}{\sqrt{3}})=0.00028…$$ so $$\displaystyle N=5$$ is the smallest partial sum with accuracy to within 0.001. Also, $$\displaystyle π−6S_7(\frac{1}{\sqrt{3}})=0.00002…$$ while $$\displaystyle π−6S_8(\frac{1}{\sqrt{3}})=−0.000007…$$ so $$\displaystyle N=8$$ is the smallest N to give accuracy to within 0.00001.

## 10.3: Taylor and Maclaurin Series

### Taylor Polynomials

In exercises 1 - 8, find the Taylor polynomials of degree two approximating the given function centered at the given point.

1) $$f(x)=1+x+x^2$$ at $$a=1$$

2) $$f(x)=1+x+x^2$$ at $$a=−1$$

$$f(−1)=1;\;f′(−1)=−1;\;f''(−1)=2;\quad p_2(x)=1−(x+1)+(x+1)^2$$

3) $$f(x)=\cos(2x)$$ at $$a=π$$

4) $$f(x)=\sin(2x)$$ at $$a=\frac{π}{2}$$

$$f′(x)=2\cos(2x);\;f''(x)=−4\sin(2x);\quad p_2(x)=−2(x−\frac{π}{2})$$

5) $$f(x)=\sqrt{x}$$ at $$a=4$$

6) $$f(x)=\ln x$$ at $$a=1$$

$$f′(x)=\dfrac{1}{x};\; f''(x)=−\dfrac{1}{x^2};\quad p_2(x)=0+(x−1)−\frac{1}{2}(x−1)^2$$

7) $$f(x)=\dfrac{1}{x}$$ at $$a=1$$

8) $$f(x)=e^x$$ at $$a=1$$

$$p_2(x)=e+e(x−1)+\dfrac{e}{2}(x−1)^2$$

### Taylor Remainder Theorem

In exercises 9 - 14, verify that the given choice of $$n$$ in the remainder estimate $$|R_n|≤\dfrac{M}{(n+1)!}(x−a)^{n+1}$$, where $$M$$ is the maximum value of $$∣f^{(n+1)}(z)∣$$ on the interval between $$a$$ and the indicated point, yields $$|R_n|≤\frac{1}{1000}$$. Find the value of the Taylor polynomial $$p_n$$ of $$f$$ at the indicated point.

9) [T] $$\sqrt{10};\; a=9,\; n=3$$

10) [T] $$(28)^{1/3};\; a=27,\; n=1$$

$$\dfrac{d^2}{dx^2}x^{1/3}=−\dfrac{2}{9x^{5/3}}≥−0.00092…$$ when $$x≥28$$ so the remainder estimate applies to the linear approximation $$x^{1/3}≈p_1(27)=3+\dfrac{x−27}{27}$$, which gives $$(28)^{1/3}≈3+\frac{1}{27}=3.\bar{037}$$, while $$(28)^{1/3}≈3.03658.$$

11) [T] $$\sin(6);\; a=2π,\; n=5$$

12) [T] $$e^2; \; a=0,\; n=9$$

Using the estimate $$\dfrac{2^{10}}{10!}<0.000283$$ we can use the Taylor expansion of order 9 to estimate $$e^x$$ at $$x=2$$. as $$e^2≈p_9(2)=1+2+\frac{2^2}{2}+\frac{2^3}{6}+⋯+\frac{2^9}{9!}=7.3887$$… whereas $$e^2≈7.3891.$$

13) [T] $$\cos(\frac{π}{5});\; a=0,\; n=4$$

14) [T] $$\ln(2);\; a=1,\; n=1000$$

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

### Approximating Definite Integrals Using Taylor Series

15) Integrate the approximation $$\sin t≈t−\dfrac{t^3}{6}+\dfrac{t^5}{120}−\dfrac{t^7}{5040}$$ evaluated at $$π$$t to approximate $$\displaystyle ∫^1_0\frac{\sin πt}{πt}\,dt$$.

16) Integrate the approximation $$e^x≈1+x+\dfrac{x^2}{2}+⋯+\dfrac{x^6}{720}$$ evaluated at $$−x^2$$ to approximate $$\displaystyle ∫^1_0e^{−x^2}\,dx.$$

$$\displaystyle ∫^1_0\left(1−x^2+\frac{x^4}{2}−\frac{x^6}{6}+\frac{x^8}{24}−\frac{x^{10}}{120}+\frac{x^{12}}{720}\right)\,dx =1−\frac{1^3}{3}+\frac{1^5}{10}−\frac{1^7}{42}+\frac{1^9}{9⋅24}−\frac{1^{11}}{120⋅11}+\frac{1^{13}}{720⋅13}≈0.74683$$ whereas $$\displaystyle ∫^1_0e^{−x^2}dx≈0.74682.$$

### More Taylor Remainder Theorem Problems

In exercises 17 - 20, find the smallest value of $$n$$ such that the remainder estimate $$|R_n|≤\dfrac{M}{(n+1)!}(x−a)^{n+1}$$, where $$M$$ is the maximum value of $$∣f^{(n+1)}(z)∣$$ on the interval between $$a$$ and the indicated point, yields $$|R_n|≤\frac{1}{1000}$$ on the indicated interval.

17) $$f(x)=\sin x$$ on $$[−π,π],\; a=0$$

18) $$f(x)=\cos x$$ on $$[−\frac{π}{2},\frac{π}{2}],\; a=0$$

Since $$f^{(n+1)}(z)$$ is $$\sin z$$ or $$\cos z$$, we have $$M=1$$. Since $$|x−0|≤\frac{π}{2}$$, we seek the smallest $$n$$ such that $$\dfrac{π^{n+1}}{2^{n+1}(n+1)!}≤0.001$$. The smallest such value is $$n=7$$. The remainder estimate is $$R_7≤0.00092.$$

19) $$f(x)=e^{−2x}$$ on $$[−1,1],a=0$$

20) $$f(x)=e^{−x}$$ on $$[−3,3],a=0$$

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

In exercises 21 - 24, the maximum of the right-hand side of the remainder estimate $$|R_1|≤\dfrac{max|f''(z)|}{2}R^2$$ on $$[a−R,a+R]$$ occurs at $$a$$ or $$a±R$$. Estimate the maximum value of $$R$$ such that $$\dfrac{max|f''(z)|}{2}R^2≤0.1$$ on $$[a−R,a+R]$$ by plotting this maximum as a function of $$R$$.

21) [T] $$e^x$$ approximated by $$1+x,\; a=0$$

22) [T] $$\sin x$$ approximated by $$x,\; a=0$$

Since $$\sin x$$ is increasing for small $$x$$ and since $$\frac{d^2}{dx^2}\left(\sin x\right)=−\sin x$$, the estimate applies whenever $$R^2\sin(R)≤0.2$$, which applies up to $$R=0.596.$$ 23) [T] $$\ln x$$ approximated by $$x−1,\; a=1$$

24) [T] $$\cos x$$ approximated by $$1,\; a=0$$

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

In exercises 25 - 35, find the Taylor series of the given function centered at the indicated point.

25) $$f(x) = x^4$$ at $$a=−1$$

26) $$f(x) = 1+x+x^2+x^3$$ at $$a=−1$$

$$(x+1)^3−2(x+1)^2+2(x+1)$$

27) $$f(x) = \sin x$$ at $$a=π$$

28) $$f(x) = \cos x$$ at $$a=2π$$

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

29) $$f(x) = \sin x$$ at $$x=\frac{π}{2}$$

30) $$f(x) = \cos x$$ at $$x=\frac{π}{2}$$

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

31) $$f(x) = e^x$$ at $$a=−1$$

32) $$f(x) = e^x$$ at $$a=1$$

The derivatives are $$f^{(n)}(1)=e,$$ so $$\displaystyle e^x=e\sum_{n=0}^∞\frac{(x−1)^n}{n!}.$$

33) $$f(x) = \dfrac{1}{(x−1)^2}$$ at $$a=0$$ (Hint: Differentiate the Taylor Series for$$\dfrac{1}{1−x}$$.)

34) $$f(x) = \dfrac{1}{(x−1)^3}$$ at $$a=0$$

$$\displaystyle \frac{1}{(x−1)^3}=−\frac{1}{2}\frac{d^2}{dx^2}\left(\frac{1}{1−x}\right)=−\sum_{n=0}^∞\left(\frac{(n+2)(n+1)x^n}{2}\right)$$

35) $$\displaystyle F(x)=∫^x_0\cos(\sqrt{t})\,dt;\quad \text{where}\; f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{(2n)!}$$ at a=0 (Note: $$f$$ is the Taylor series of $$\cos(\sqrt{t}).)$$

In exercises 36 - 44, compute the Taylor series of each function around $$x=1$$.

36) $$f(x)=2−x$$

$$2−x=1−(x−1)$$

37) $$f(x)=x^3$$

38) $$f(x)=(x−2)^2$$

$$((x−1)−1)^2=(x−1)^2−2(x−1)+1$$

39) $$f(x)=\ln x$$

40) $$f(x)=\dfrac{1}{x}$$

$$\displaystyle \frac{1}{1−(1−x)}=\sum_{n=0}^∞(−1)^n(x−1)^n$$

41) $$f(x)=\dfrac{1}{2x−x^2}$$

42) $$f(x)=\dfrac{x}{4x−2x^2−1}$$

$$\displaystyle x\sum_{n=0}^∞2^n(1−x)^{2n}=\sum_{n=0}^∞2^n(x−1)^{2n+1}+\sum_{n=0}^∞2^n(x−1)^{2n}$$

43) $$f(x)=e^{−x}$$

44) $$f(x)=e^{2x}$$

$$\displaystyle e^{2x}=e^{2(x−1)+2}=e^2\sum_{n=0}^∞\frac{2^n(x−1)^n}{n!}$$

### Maclaurin Series

[T] In exercises 45 - 48, identify the value of $$x$$ such that the given series $$\displaystyle \sum_{n=0}^∞a_n$$ is the value of the Maclaurin series of $$f(x)$$ at $$x$$. Approximate the value of $$f(x)$$ using $$\displaystyle S_{10}=\sum_{n=0}^{10}a_n$$.

45) $$\displaystyle \sum_{n=0}^∞\frac{1}{n!}$$

46) $$\displaystyle \sum_{n=0}^∞\frac{2^n}{n!}$$

$$x=e^2;\quad S_{10}=\dfrac{34,913}{4725}≈7.3889947$$

47) $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n(2π)^{2n}}{(2n)!}$$

48) $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n(2π)^{2n+1}}{(2n+1)!}$$

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

In exercises 49 - 52 use the functions $$S_5(x)=x−\dfrac{x^3}{6}+\dfrac{x^5}{120}$$ and $$C_4(x)=1−\dfrac{x^2}{2}+\dfrac{x^4}{24}$$ on $$[−π,π]$$.

49) [T] Plot $$\sin^2x−(S_5(x))^2$$ on $$[−π,π]$$. Compare the maximum difference with the square of the Taylor remainder estimate for $$\sin x.$$

50) [T] Plot $$\cos^2x−(C_4(x))^2$$ on $$[−π,π]$$. Compare the maximum difference with the square of the Taylor remainder estimate for $$\cos x$$.

The difference is small on the interior of the interval but approaches $$1$$ near the endpoints. The remainder estimate is $$|R_4|=\frac{π^5}{120}≈2.552.$$ 51) [T] Plot $$|2S_5(x)C_4(x)−\sin(2x)|$$ on $$[−π,π]$$.

52) [T] Compare $$\dfrac{S_5(x)}{C_4(x)}$$ on $$[−1,1]$$ to $$\tan x$$. Compare this with the Taylor remainder estimate for the approximation of $$\tan x$$ by $$x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}$$.

The difference is on the order of $$10^{−4}$$ on $$[−1,1]$$ while the Taylor approximation error is around $$0.1$$ near $$±1$$. The top curve is a plot of $$\tan^2x−\left(\dfrac{S_5(x)}{C_4(x)}\right)^2$$ and the lower dashed plot shows $$t^2−\left(\dfrac{S_5}{C_4}\right)^2$$. 53) [T] Plot $$e^x−e_4(x)$$ where $$e_4(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}$$ on $$[0,2]$$. Compare the maximum error with the Taylor remainder estimate.

54) (Taylor approximations and root finding.) Recall that Newton’s method $$x_{n+1}=x_n−\dfrac{f(x_n)}{f'(x_n)}$$ approximates solutions of $$f(x)=0$$ near the input $$x_0$$.

a. If $$f$$ and $$g$$ are inverse functions, explain why a solution of $$g(x)=a$$ is the value $$f(a)$$ of $$f$$.

b. Let $$p_N(x)$$ be the $$N^{\text{th}}$$ degree Maclaurin polynomial of $$e^x$$. Use Newton’s method to approximate solutions of $$p_N(x)−2=0$$ for $$N=4,5,6.$$

c. Explain why the approximate roots of $$p_N(x)−2=0$$ are approximate values of $$\ln(2).$$

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: $$\ln(2)=0.69314...$$)

### Evaluating Limits using Taylor Series

In exercises 55 - 58, use the fact that if $$\displaystyle q(x)=\sum_{n=1}^∞a_n(x−c)^n$$ converges in an interval containing $$c$$, then $$\displaystyle \lim_{x→c}q(x)=a_0$$ to evaluate each limit using Taylor series.

55) $$\displaystyle \lim_{x→0}\frac{\cos x−1}{x^2}$$

56) $$\displaystyle \lim_{x→0}\frac{\ln(1−x^2)}{x^2}$$

$$\dfrac{\ln(1−x^2)}{x^2}→−1$$

57) $$\displaystyle \lim_{x→0}\frac{e^{x^2}−x^2−1}{x^4}$$

58) $$\displaystyle \lim_{x→0^+}\frac{\cos(\sqrt{x})−1}{2x}$$

$$\displaystyle \frac{\cos(\sqrt{x})−1}{2x}≈\frac{(1−\frac{x}{2}+\frac{x^2}{4!}−⋯)−1}{2x}→−\frac{1}{4}$$

## 10.4: Working with Taylor Series

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

1) $$\displaystyle (1−x)^{1/3}$$

2) $$\displaystyle (1+x^2)^{−1/3}$$

Solution: $$\displaystyle (1+x^2)^{−1/3}=\sum_{n=0}^∞(^{−\frac{1}{3}}_n)x^{2n}$$

3) $$\displaystyle (1−x)^{1.01}$$

4) $$\displaystyle (1−2x)^{2/3}$$

Solution: $$\displaystyle (1−2x)^{2/3}=\sum_{n=0}^∞(−1)^n2^n(^{\frac{2}{3}}_n)x^n$$

In the following exercises, use the substitution $$\displaystyle (b+x)^r=(b+a)^r(1+\frac{x−a}{b+a})^r$$ in the binomial expansion to find the Taylor series of each function with the given center.

5) (\sqrt{x+2}\) at $$\displaystyle a=0$$

6) $$\displaystyle \sqrt{x^2+2}$$ at $$\displaystyle a=0$$

Solution: $$\displaystyle \sqrt{2+x^2}=\sum_{n=0}^∞2^{(1/2)−n}(^{\frac{1}{2}}_n)x^{2n};(∣x^2∣<2)$$

7) $$\displaystyle \sqrt{x+2}$$ at $$\displaystyle a=1$$

8) $$\displaystyle \sqrt{2x−x^2}$$ at $$\displaystyle a=1$$ (Hint: $$\displaystyle 2x−x^2=1−(x−1)^2$$)

Solution: $$\displaystyle \sqrt{2x−x^2}=\sqrt{1−(x−1)^2}$$ so $$\displaystyle \sqrt{2x−x^2}=\sum_{n=0}^∞(−1)^n(^{\frac{1}{2}}_n)(x−1)^{2n}$$

9) $$\displaystyle (x−8)^{1/3}$$ at $$\displaystyle a=9$$

10) $$\displaystyle \sqrt{x}$$ at $$\displaystyle a=4$$

Solution: $$\displaystyle \sqrt{x}=2\sqrt{1+\frac{x−4}{4}}$$ so $$\displaystyle \sqrt{x}=\sum_{n=0}^∞2^{1−2n}(^{\frac{1}{2}}_n)(x−4)^n$$

11) $$\displaystyle x^{1/3}$$ at $$\displaystyle a=27$$

12) $$\displaystyle \sqrt{x}$$ at $$\displaystyle x=9$$

Solution: $$\displaystyle \sqrt{x}=\sum_{n=0}^∞3^{1−3n}(^{\frac{1}{2}}_n)(x−9)^n$$

In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most $$\displaystyle 1/1000.$$

13) [T] $$\displaystyle (15)^{1/4}$$ using $$\displaystyle (16−x)^{1/4}$$

14) [T] $$\displaystyle (1001)^{1/3}$$ using $$\displaystyle (1000+x)^{1/3}$$

Solution: $$\displaystyle 10(1+\frac{x}{1000})^{1/3}=\sum_{n=0}^∞10^{1−3n}(^{\frac{1}{3}}_n)x^n$$. Using, for example, a fourth-degree estimate at $$\displaystyle x=1$$ gives $$\displaystyle (1001)^{1/3}≈10(1+(^{\frac{1}{3}}_1)10^{−3}+(^{\frac{1}{3}}_2)10^{−6}+(^{\frac{1}{3}}_3)10^{−9}+(^{\frac{1}{3}}_4)10^{−12})=10(1+\frac{1}{3.10^3}−\frac{1}{9.10^6}+\frac{5}{81.10^9}−\frac{10}{243.10^{12}})=10.00333222...$$ whereas $$\displaystyle (1001)^{1/3}=10.00332222839093....$$ Two terms would suffice for three-digit accuracy.

In the following exercises, use the binomial approximation $$\displaystyle \sqrt{1−x}≈1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{128}−\frac{7x^5}{256}$$ for $$\displaystyle |x|<1$$ to approximate each number. Compare this value to the value given by a scientific calculator.

15) [T] $$\displaystyle \frac{1}{\sqrt{2}}$$ using $$\displaystyle x=\frac{1}{2}$$ in $$\displaystyle (1−x)^{1/2}$$

16) [T] $$\displaystyle \sqrt{5}=5×\frac{1}{\sqrt{5}}$$ using $$\displaystyle x=\frac{4}{5}$$ in $$\displaystyle (1−x)^{1/2}$$

Solution: The approximation is $$\displaystyle 2.3152$$; the CAS value is $$\displaystyle 2.23….$$

17) [T] $$\displaystyle \sqrt{3}=\frac{3}{\sqrt{3}}$$ using $$\displaystyle x=\frac{2}{3}$$ in $$\displaystyle (1−x)^{1/2}$$

18) [T] $$\displaystyle \sqrt{6}$$ using $$\displaystyle x=\frac{5}{6}$$ in $$\displaystyle (1−x)^{1/2}$$

Solution: The approximation is $$\displaystyle 2.583…$$; the CAS value is $$\displaystyle 2.449….$$

19) Integrate the binomial approximation of $$\displaystyle \sqrt{1−x}$$ to find an approximation of $$\displaystyle ∫^x_0\sqrt{1−t}dt$$.

20) [T] Recall that the graph of $$\displaystyle \sqrt{1−x^2}$$ is an upper semicircle of radius $$\displaystyle 1$$. Integrate the binomial approximation of $$\displaystyle \sqrt{1−x^2}$$ up to order $$\displaystyle 8$$ from $$\displaystyle x=−1$$ to $$\displaystyle x=1$$ to estimate $$\displaystyle \frac{π}{2}$$.

Solution: $$\displaystyle \sqrt{1−x^2}=1−\frac{x^2}{2}−\frac{x^4}{8}−\frac{x^6}{16}−\frac{5x^8}{128}+⋯.$$ Thus $$\displaystyle ∫^1_{−1}\sqrt{1−x^2}dx=x−\frac{x^3}{6}−\frac{x^5}{40}−\frac{x^7}{7⋅16}−\frac{5x^9}{9⋅128}+⋯∣^1_{−1}≈2−\frac{1}{3}−\frac{1}{20}−\frac{1}{56}−\frac{10}{9⋅128}+error=1.590...$$ whereas $$\displaystyle \frac{π}{2}=1.570...$$

In the following exercises, use the expansion $$\displaystyle (1+x)^{1/3}=1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯$$ to write the first five terms (not necessarily a quartic polynomial) of each expression.

21) $$\displaystyle (1+4x)^{1/3};a=0$$

22) $$\displaystyle (1+4x)^{4/3};a=0$$

Solution: $$\displaystyle (1+x)^{4/3}=(1+x)(1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯)=1+\frac{4x}{3}+\frac{2x^2}{9}−\frac{4x^3}{81}+\frac{5x^4}{243}+⋯$$

23) $$\displaystyle (3+2x)^{1/3};a=−1$$

24) $$\displaystyle (x^2+6x+10)^{1/3};a=−3$$

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

25) Use $$\displaystyle (1+x)^{1/3}=1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯$$ with $$\displaystyle x=1$$ to approximate $$\displaystyle 2^{1/3}$$.

26) Use the approximation $$\displaystyle (1−x)^{2/3}=1−\frac{2x}{3}−\frac{x^2}{9}−\frac{4x^3}{81}−\frac{7x^4}{243}−\frac{14x^5}{729}+⋯$$ for $$\displaystyle |x|<1$$ to approximate $$\displaystyle 2^{1/3}=2.2^{−2/3}$$.

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

27) Find the $$\displaystyle 25th$$ derivative of $$\displaystyle f(x)=(1+x^2)^{13}$$ at $$\displaystyle x=0$$.

28) Find the $$\displaystyle 99$$ th derivative of $$\displaystyle f(x)=(1+x^4)^{25}$$.

Solution: $$\displaystyle f^{(99)}(0)=0$$

In the following exercises, find the Maclaurin series of each function.

29) $$\displaystyle f(x)=xe^{2x}$$

30) $$\displaystyle f(x)=2^x$$

Solution: $$\displaystyle \sum_{n=0}^∞\frac{(ln(2)x)^n}{n!}$$

31) $$\displaystyle f(x)=\frac{sinx}{x}$$

32) $$\displaystyle f(x)=\frac{sin(\sqrt{x})}{\sqrt{x}},(x>0),$$

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

33) $$\displaystyle f(x)=sin(x^2)$$

34) $$\displaystyle f(x)=e^{x^3}$$

Solution: $$\displaystyle e^{x^3}=\sum_{n=0}^∞\frac{x^{3n}}{n!}$$

35) $$\displaystyle f(x)=cos^2x$$ using the identity $$\displaystyle cos^2x=\frac{1}{2}+\frac{1}{2}cos(2x)$$

36) $$\displaystyle f(x)=sin^2x$$ using the identity $$\displaystyle sin^2x=\frac{1}{2}−\frac{1}{2}cos(2x)$$

Solution: $$\displaystyle sin^2x=−\sum_{k=1}^∞\frac{(−1)^k2^{2k−1}x^{2k}}{(2k)!}$$

In the following exercises, find the Maclaurin series of $$\displaystyle F(x)=∫^x_0f(t)dt$$ by integrating the Maclaurin series of $$\displaystyle f$$ term by term. If $$\displaystyle f$$ is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

37) $$\displaystyle F(x)=∫^x_0e^{−t^2}dt;f(t)=e^{−t^2}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{n!}$$

38) $$\displaystyle F(x)=tan^{−1}x;f(t)=\frac{1}{1+t^2}=\sum_{n=0}^∞(−1)^nt^{2n}$$

Solution: $$\displaystyle tan^{−1}x=\sum_{k=0}^∞\frac{(−1)^kx^{2k+1}}{2k+1}$$

39) $$\displaystyle F(x)=tanh^{−1}x;f(t)=\frac{1}{1−t^2}=\sum_{n=0}^∞t^{2n}$$

40) $$\displaystyle F(x)=sin^{−1}x;f(t)=\frac{1}{\sqrt{1−t^2}}=\sum_{k=0}^∞(^{\frac{1}{2}}_k)\frac{t^{2k}}{k!}$$

Solution: $$\displaystyle sin^{−1}x=\sum_{n=0}^∞(^{\frac{1}{2}}_n)\frac{x^{2n+1}}{(2n+1)n!}$$

41) $$\displaystyle F(x)=∫^x_0\frac{sint}{t}dt;f(t)=\frac{sint}{t}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{(2n+1)!}$$

42) $$\displaystyle F(x)=∫^x_0cos(\sqrt{t})dt;f(t)=\sum_{n=0}^∞(−1)^n\frac{x^n}{(2n)!}$$

Solution: $$\displaystyle F(x)=\sum_{n=0}^∞(−1)^n\frac{x^{n+1}}{(n+1)(2n)!}$$

43) $$\displaystyle F(x)=∫^x_0\frac{1−cost}{t^2}dt;f(t)=\frac{1−cost}{t^2}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{(2n+2)!}$$

44) $$\displaystyle F(x)=∫^x_0\frac{ln(1+t)}{t}dt;f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{n+1}$$

Solution: $$\displaystyle F(x)=\sum_{n=1}^∞(−1)^{n+1}\frac{x^n}{n^2}$$

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of $$\displaystyle f$$.

45) $$\displaystyle f(x)=sin(x+\frac{π}{4})=sinxcos(\frac{π}{4})+cosxsin(\frac{π}{4})$$

46) $$\displaystyle f(x)=tanx$$

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

47) $$\displaystyle f(x)=ln(cosx)$$

48) $$\displaystyle f(x)=e^xcosx$$

Solution: $$\displaystyle 1+x−\frac{x^3}{3}−\frac{x^4}{6}+⋯$$

49) $$\displaystyle f(x)=e^{sinx}$$

50) $$\displaystyle f(x)=sec^2x$$

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

51) $$\displaystyle f(x)=tanhx$$

52) $$\displaystyle f(x)=\frac{tan\sqrt{x}}{\sqrt{x}}$$ (see expansion for $$\displaystyle tanx$$)

Solution: Using the expansion for $$\displaystyle tanx$$ gives $$\displaystyle 1+\frac{x}{3}+\frac{2x^2}{15}$$.

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

53) $$\displaystyle ln(1+x)$$

54) $$\displaystyle \frac{1}{1+x^2}$$

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

55) $$\displaystyle tan^{−1}x$$

56) $$\displaystyle ln(1+x^2)$$

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

57) Find the Maclaurin series of $$\displaystyle sinhx=\frac{e^x−e^{−x}}{2}$$.

58) Find the Maclaurin series of $$\displaystyle coshx=\frac{e^x+e^{−x}}{2}$$.

Solution: Add series of $$\displaystyle e^x$$ and $$\displaystyle e^{−x}$$ term by term. Odd terms cancel and $$\displaystyle coshx=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$$.

59) Differentiate term by term the Maclaurin series of $$\displaystyle sinhx$$ and compare the result with the Maclaurin series of $$\displaystyle coshx$$.

60) [T] Let $$\displaystyle S_n(x)=\sum_{k=0}^n(−1)^k\frac{x^{2k+1}}{(2k+1)!}$$ and $$\displaystyle C_n(x)=\sum_{n=0}^n(−1)^k\frac{x^{2k}}{(2k)!}$$ denote the respective Maclaurin polynomials of degree $$\displaystyle 2n+1$$ of $$\displaystyle sinx$$ and degree $$\displaystyle 2n$$ of $$\displaystyle cosx$$. Plot the errors $$\displaystyle \frac{S_n(x)}{C_n(x)}−tanx$$ for $$\displaystyle n=1,..,5$$ and compare them to $$\displaystyle x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}−tanx$$ on $$\displaystyle (−\frac{π}{4},\frac{π}{4})$$.

Solution: The ratio $$\displaystyle \frac{S_n(x)}{C_n(x)}$$ approximates $$\displaystyle tanx$$ better than does $$\displaystyle p_7(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}$$ for $$\displaystyle N≥3$$. The dashed curves are $$\displaystyle \frac{S_n}{C_n}−tan$$ for $$\displaystyle n=1,2$$. The dotted curve corresponds to $$\displaystyle n=3$$, and the dash-dotted curve corresponds to $$\displaystyle n=4$$. The solid curve is $$\displaystyle p_7−tanx$$. 61) Use the identity $$\displaystyle 2sinxcosx=sin(2x)$$ to find the power series expansion of $$\displaystyle sin^2x$$ at $$\displaystyle x=0$$. (Hint: Integrate the Maclaurin series of $$\displaystyle sin(2x)$$ term by term.)

62) If $$\displaystyle y=\sum_{n=0}^∞a_nx^n$$, find the power series expansions of $$\displaystyle xy′$$ and $$\displaystyle x^2y''$$.

Solution: By the term-by-term differentiation theorem, $$\displaystyle y′=\sum_{n=1}^∞na_nx^{n−1}$$ so $$\displaystyle y′=\sum_{n=1}^∞na_nx^{n−1}xy′=\sum_{n=1}^∞na_nx^n$$, whereas $$\displaystyle y′=\sum_{n=2}^∞n(n−1)a_nx^{n−2}$$ so $$\displaystyle xy''=\sum_{n=2}^∞n(n−1)a_nx^n$$.

63) [T] Suppose that $$\displaystyle y=\sum_{k=0}^∞a^kx^k$$ satisfies $$\displaystyle y′=−2xy$$ and $$\displaystyle y(0)=0$$. Show that $$\displaystyle a_{2k+1}=0$$ for all $$\displaystyle k$$ and that $$\displaystyle a_{2k+2}=\frac{−a_{2k}}{k+1}$$. Plot the partial sum $$\displaystyle S_{20}$$ of $$\displaystyle y$$ on the interval $$\displaystyle [−4,4]$$.

64) [T] Suppose that a set of standardized test scores is normally distributed with mean $$\displaystyle μ=100$$ and standard deviation $$\displaystyle σ=10$$. Set up an integral that represents the probability that a test score will be between $$\displaystyle 90$$ and $$\displaystyle 110$$ and use the integral of the degree $$\displaystyle 10$$ Maclaurin polynomial of $$\displaystyle \frac{1}{\sqrt{2π}}e^{−x^2/2}$$ to estimate this probability.

Solution: The probability is $$\displaystyle p=\frac{1}{\sqrt{2π}}∫^{(b−μ)/σ}_{(a−μ)/σ}e^{−x^2/2}dx$$ where $$\displaystyle a=90$$ and $$\displaystyle b=100$$, that is, $$\displaystyle p=\frac{1}{\sqrt{2π}}∫^1_{−1}e^{−x^2/2}dx=\frac{1}{\sqrt{2π}}∫^1_{−1}\sum_{n=0}^5(−1)^n\frac{x^{2n}}{2^nn!}dx=\frac{2}{\sqrt{2π}}\sum_{n=0}^5(−1)^n\frac{1}{(2n+1)2^nn!}≈0.6827.$$

65) [T] Suppose that a set of standardized test scores is normally distributed with mean $$\displaystyle μ=100$$ and standard deviation $$\displaystyle σ=10$$. Set up an integral that represents the probability that a test score will be between $$\displaystyle 70$$ and $$\displaystyle 130$$ and use the integral of the degree $$\displaystyle 50$$ Maclaurin polynomial of $$\displaystyle \frac{1}{\sqrt{2π}}e^{−x^2/2}$$ to estimate this probability.

66) [T] Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle f(x)$$ such that $$\displaystyle f(0)=1,f′(0)=0$$, and $$\displaystyle f''(x)=−f(x)$$. Find a formula for $$\displaystyle a_n$$ and plot the partial sum $$\displaystyle S_N$$ for $$\displaystyle N=20$$ on $$\displaystyle [−5,5].$$

Solution: As in the previous problem one obtains $$\displaystyle a_n=0$$ if $$\displaystyle n$$ is odd and $$\displaystyle a_n=−(n+2)(n+1)a_{n+2}$$ if $$\displaystyle n$$ is even, so $$\displaystyle a_0=1$$ leads to $$\displaystyle a_{2n}=\frac{(−1)^n}{(2n)!}$$. 67) [T] Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle f(x)$$ such that $$\displaystyle f(0)=0,f′(0)=1$$, and $$\displaystyle f''(x)=−f(x)$$. Find a formula for an and plot the partial sum $$\displaystyle S_N$$ for $$\displaystyle N=10$$ on $$\displaystyle [−5,5]$$.

68) Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle y$$ such that $$\displaystyle y''−y′+y=0$$ where $$\displaystyle y(0)=1$$ and $$\displaystyle y'(0)=0.$$ Find a formula that relates $$\displaystyle a_{n+2},a_{n+1},$$ and an and compute $$\displaystyle a_0,...,a_5$$.

Solution: $$\displaystyle y''=\sum_{n=0}^∞(n+2)(n+1)a_{n+2}x^n$$ and $$\displaystyle y′=\sum_{n=0}^∞(n+1)a_{n+1}x^n$$ so $$\displaystyle y''−y′+y=0$$ implies that $$\displaystyle (n+2)(n+1)a_{n+2}−(n+1)a_{n+1}+a_n=0$$ or $$\displaystyle a_n=\frac{a_{n−1}}{n}−\frac{a_{n−2}}{n(n−1)}$$ for all $$\displaystyle n⋅y(0)=a_0=1$$ and $$\displaystyle y′(0)=a_1=0,$$ so $$\displaystyle a_2=\frac{1}{2},a_3=\frac{1}{6},a_4=0$$, and $$\displaystyle a_5=−\frac{1}{120}$$.

69) Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle y$$ such that $$\displaystyle y''−y′+y=0$$ where $$\displaystyle y(0)=0$$ and $$\displaystyle y′(0)=1$$. Find a formula that relates $$\displaystyle a_{n+2},a_{n+1}$$, and an and compute $$\displaystyle a_1,...,a_5$$.

The error in approximating the integral $$\displaystyle ∫^b_af(t)dt$$ by that of a Taylor approximation $$\displaystyle ∫^b_aPn(t)dt$$ is at most $$\displaystyle ∫^b_aR_n(t)dt$$. In the following exercises, the Taylor remainder estimate $$\displaystyle R_n≤\frac{M}{(n+1)!}|x−a|^{n+1}$$ guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $$\displaystyle f$$ with an error less than $$\displaystyle \frac{1}{10}$$.

a. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than $$\displaystyle \frac{1}{100}$$.

b. Compare the accuracy of the polynomial integral estimate with the remainder estimate.

70) [T] $$\displaystyle ∫^π_0\frac{sint}{t}dt;P_s=1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!}$$ (You may assume that the absolute value of the ninth derivative of $$\displaystyle \frac{sint}{t}$$ is bounded by $$\displaystyle 0.1$$.)

Solution: a. (Proof) b. We have $$\displaystyle R_s≤\frac{0.1}{(9)!}π^9≈0.0082<0.01.$$ We have $$\displaystyle ∫^π_0(1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!})dx=π−\frac{π^3}{3⋅3!}+\frac{π^5}{5⋅5!}−\frac{π^7}{7⋅7!}+\frac{π^9}{9⋅9!}=1.852...,$$ whereas $$\displaystyle ∫^π_0\frac{sint}{t}dt=1.85194...$$, so the actual error is approximately $$\displaystyle 0.00006.$$

71) [T] $$\displaystyle ∫^2_0e^{−x^2}dx;p_{11}=1−x^2+\frac{x^4}{2}−\frac{x^6}{3!}+⋯−\frac{x^{22}}{11!}$$ (You may assume that the absolute value of the $$\displaystyle 23rd$$ derivative of $$\displaystyle e^{−x^2}$$ is less than $$\displaystyle 2×10^{14}$$.)

The following exercises deal with Fresnel integrals.

72) The Fresnel integrals are defined by $$\displaystyle C(x)=∫^x_0cos(t^2)dt$$ and $$\displaystyle S(x)=∫^x_0sin(t^2)dt$$. Compute the power series of $$\displaystyle C(x)$$ and $$\displaystyle S(x)$$ and plot the sums $$\displaystyle C_N(x)$$ and $$\displaystyle S_N(x)$$ of the first $$\displaystyle N=50$$ nonzero terms on $$\displaystyle [0,2π]$$.

Solution: Since $$\displaystyle cos(t^2)=\sum_{n=0}^∞(−1)^n\frac{t^{4n}}{(2n)!}$$ and $$\displaystyle sin(t^2)=\sum_{n=0}^∞(−1)^n\frac{t^{4n+2}}{(2n+1)!}$$, one has $$\displaystyle S(x)=_sum_{n=0}^∞(−1)^n\frac{x^{4n+3}}{(4n+3)(2n+1)!}$$ and $$\displaystyle C(x)=\sum_{n=0}^∞(−1)^n\frac{x^{4n+1}}{(4n+1)(2n)!}$$. The sums of the first $$\displaystyle 50$$ nonzero terms are plotted below with $$\displaystyle C_{50}(x)$$ the solid curve and $$\displaystyle S_{50}(x)$$ the dashed curve. 73) [T] The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates $$\displaystyle (C(t),S(t))$$. Plot the curve $$\displaystyle (C_{50},S_{50})$$ for $$\displaystyle 0≤t≤2π$$, the coordinates of which were computed in the previous exercise.

74) Estimate $$\displaystyle ∫^{1/4}_0\sqrt{x−x^2}dx$$ by approximating $$\displaystyle \sqrt{1−x}$$ using the binomial approximation $$\displaystyle 1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{2128}−\frac{7x^5}{256}$$.

Solution: $$\displaystyle ∫^{1/4}_0\sqrt{x}(1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{128}−\frac{7x^5}{256})dx =\frac{2}{3}2^{−3}−\frac{1}{2}\frac{2}{5}2^{−5}−\frac{1}{8}\frac{2}{7}2^{−7}−\frac{1}{16}\frac{2}{9}2^{−9}−\frac{5}{128}\frac{2}{11}2^{−11}−\frac{7}{256}\frac{2}{13}2^{−13}=0.0767732...$$ whereas $$\displaystyle ∫^{1/4}_0\sqrt{x−x^2}dx=0.076773.$$

75) [T] Use Newton’s approximation of the binomial $$\displaystyle \sqrt{1−x^2}$$ to approximate $$\displaystyle π$$ as follows. The circle centered at $$\displaystyle (\frac{1}{2},0)$$ with radius $$\displaystyle \frac{1}{2}$$ has upper semicircle $$\displaystyle y=\sqrt{x}\sqrt{1−x}$$. The sector of this circle bounded by the $$\displaystyle x$$-axis between $$\displaystyle x=0$$ and $$\displaystyle x=\frac{1}{2}$$ and by the line joining $$\displaystyle (\frac{1}{4},\frac{\sqrt{3}}{4})$$ corresponds to $$\displaystyle \frac{1}{6}$$ of the circle and has area $$\displaystyle \frac{π}{24}$$. This sector is the union of a right triangle with height $$\displaystyle \frac{\sqrt{3}}{4}$$ and base $$\displaystyle \frac{1}{4}$$ and the region below the graph between $$\displaystyle x=0$$ and $$\displaystyle x=\frac{1}{4}$$. To find the area of this region you can write $$\displaystyle y=\sqrt{x}\sqrt{1−x}=\sqrt{x}×(\text{binomial expansion of} \sqrt{1−x})$$ and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate $$\displaystyle π$$.

76) Use the approximation $$\displaystyle T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4})$$ to approximate the period of a pendulum having length $$\displaystyle 10$$ meters and maximum angle $$\displaystyle θ_{max}=\frac{π}{6}$$ where $$\displaystyle k=sin(\frac{θ_{max}}{2})$$. Compare this with the small angle estimate $$\displaystyle T≈2π\sqrt{\frac{L}{g}}$$.

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

77) Suppose that a pendulum is to have a period of $$\displaystyle 2$$ seconds and a maximum angle of $$\displaystyle θ_{max}=\frac{π}{6}$$. Use $$\displaystyle T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4})$$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $$\displaystyle T≈2π\sqrt{\frac{L}{g}}$$?

78) Evaluate $$\displaystyle ∫^{π/2}_0sin^4θdθ$$ in the approximation $$\displaystyle T=4\sqrt{\frac{L}{g}}∫^{π/2}_0(1+\frac{1}{2}k^2sin^2θ+\frac{3}{8}k^4sin^4θ+⋯)dθ$$ to obtain an improved estimate for $$\displaystyle T$$.

Solution: $$\displaystyle ∫^{π/2}_0sin^4θdθ=\frac{3π}{16}.$$ Hence $$\displaystyle T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4}+\frac{9}{256}k^4).$$

79) [T] An equivalent formula for the period of a pendulum with amplitude $$\displaystyle θ_max$$ is $$\displaystyle T(θ_{max})=2\sqrt{2}\sqrt{\frac{L}{g}}∫^{θ_{max}}_0\frac{dθ}{\sqrt{cosθ}−cos(θ_{max})}$$ where $$\displaystyle L$$ is the pendulum length and $$\displaystyle g$$ is the gravitational acceleration constant. When $$\displaystyle θ_{max}=\frac{π}{3}$$ we get $$\displaystyle \frac{1}{\sqrt{cost−1/2}}≈\sqrt{2}(1+\frac{t^2}{2}+\frac{t^4}{3}+\frac{181t^6}{720})$$. Integrate this approximation to estimate $$\displaystyle T(\frac{π}{3})$$ in terms of $$\displaystyle L$$ and $$\displaystyle g$$. Assuming $$\displaystyle g=9.806$$ meters per second squared, find an approximate length $$\displaystyle L$$ such that $$\displaystyle T(\frac{π}{3})=2$$ seconds.

## Chapter Review Exercise

True or False? In the following exercises, justify your answer with a proof or a counterexample.

1) If the radius of convergence for a power series $$\displaystyle \sum_{n=0}^∞a_nx^n$$ is $$\displaystyle 5$$, then the radius of convergence for the series $$\displaystyle \sum_{n=1}^∞na_nx^{n−1}$$ is also $$\displaystyle 5$$.

Solution: True

2) Power series can be used to show that the derivative of $$\displaystyle e^x$$ is $$\displaystyle e^x$$. (Hint: Recall that $$\displaystyle e^x=\sum_{n=0}^∞\frac{1}{n!}x^n.$$)

3) For small values of $$\displaystyle x,sinx≈x.$$

Solution: True

4) The radius of convergence for the Maclaurin series of $$\displaystyle f(x)=3^x$$ is $$\displaystyle 3$$.

In the following exercises, find the radius of convergence and the interval of convergence for the given series.

5) $$\displaystyle \sum_{n=0}^∞n^2(x−1)^n$$

Solution: ROC: $$\displaystyle 1$$; IOC: $$\displaystyle (0,2)$$

6) $$\displaystyle \sum_{n=0}^∞\frac{x^n}{n^n}$$

7) $$\displaystyle \sum_{n=0}^∞\frac{3nx^n}{12^n}$$

Solution: ROC: $$\displaystyle 12;$$ IOC: $$\displaystyle (−16,8)$$

8) $$\displaystyle \sum_{n=0}^∞\frac{2^n}{e^n}(x−e)^n$$

In the following exercises, find the power series representation for the given function. Determine the radius of convergence and the interval of convergence for that series.

9) $$\displaystyle f(x)=\frac{x^2}{x+3}$$

Solution: $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n}{3^{n+1}}x^n;$$ ROC: $$\displaystyle 3$$; IOC: $$\displaystyle (−3,3)$$

10) $$\displaystyle f(x)=\frac{8x+2}{2x^2−3x+1}$$

In the following exercises, find the power series for the given function using term-by-term differentiation or integration.

11) $$\displaystyle f(x)=tan^{−1}(2x)$$

Solution: integration: $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n}{2n+1}(2x)^{2n+1}$$

12) $$\displaystyle f(x)=\frac{x}{(2+x^2)^2}$$

In the following exercises, evaluate the Taylor series expansion of degree four for the given function at the specified point. What is the error in the approximation?

13) $$\displaystyle f(x)=x^3−2x^2+4,a=−3$$

Solution: $$\displaystyle p_4(x)=(x+3)^3−11(x+3)^2+39(x+3)−41;$$ exact

14) $$\displaystyle f(x)=e^{1/(4x)},a=4$$

In the following exercises, find the Maclaurin series for the given function.

15) $$\displaystyle f(x)=cos(3x)$$

Solution: $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n(3x)^{2n}}{2n!}$$

16) $$\displaystyle f(x)=ln(x+1)$$

In the following exercises, find the Taylor series at the given value.

17) $$\displaystyle f(x)=sinx,a=\frac{π}{2}$$

Solution: $$\displaystyle \sum_{n=0}^∞\frac{(−1)^n}{(2n)!}(x−\frac{π}{2})^{2n}$$

18) $$\displaystyle f(x)=\frac{3}{x},a=1$$

In the following exercises, find the Maclaurin series for the given function.

19) $$\displaystyle f(x)=e^{−x^2}−1$$

Solution: $$\displaystyle \sum_{n=1}^∞\frac{(−1)^n}{n!}x^{2n}$$

20) $$\displaystyle f(x)=cosx−xsinx$$

In the following exercises, find the Maclaurin series for $$\displaystyle F(x)=∫^x_0f(t)dt$$ by integrating the Maclaurin series of $$\displaystyle f(x)$$ term by term.

21) $$\displaystyle f(x)=\frac{sinx}{x}$$

Solution: $$\displaystyle F(x)=\sum_{n=0}^∞\frac{(−1)^n}{(2n+1)(2n+1)!}x^{2n+1}$$

22) $$\displaystyle f(x)=1−e^x$$

23) Use power series to prove Euler’s formula: $$\displaystyle e^{ix}=cosx+isinx$$

The following exercises consider problems of annuity payments.

24) For annuities with a present value of $$\displaystyle 1$$ million, calculate the annual payouts given over $$\displaystyle 25$$ years assuming interest rates of $$\displaystyle 1%,5%$$, and $$\displaystyle 10%.$$

25) A lottery winner has an annuity that has a present value of $$\displaystyle 10$$ million. What interest rate would they need to live on perpetual annual payments of $$\displaystyle 250,000$$?

Solution: $$\displaystyle 2.5%$$

26) Calculate the necessary present value of an annuity in order to support annual payouts of $$\displaystyle 15,000$$ given over $$\displaystyle 25$$ years assuming interest rates of $$\displaystyle 1%,5%$$,and $$\displaystyle 10%.$$