Problem Set: Properties of Power Series

1. If $f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$ and $g\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{n\text{!}}$ , find the power series of $\frac{1}{2}\left(f\left(x\right)+g\left(x\right)\right)$ and of $\frac{1}{2}\left(f\left(x\right)-g\left(x\right)\right)$ .

Show Solution

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

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

2. If

$C\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{2n}}{\left(2n\right)\text{!}}$ and

$S\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}$ , find the power series of

$C\left(x\right)+S\left(x\right)$ and of

$C\left(x\right)-S\left(x\right)$ .

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

3. $\frac{4}{\left(x - 3\right)\left(x+1\right)}$

Show Solution

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

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

5. $\frac{5}{\left({x}^{2}+4\right)\left({x}^{2}-1\right)}$

Show Solution

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

$\frac{30}{\left({x}^{2}+1\right)\left({x}^{2}-9\right)}$

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

7. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{x}^{n}}$

Show Solution

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

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{x}^{2n}}$

9. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{\left(x - 3\right)}^{2n - 1}}$

Show Solution

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

10.

$\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{{\left(x - 3\right)}^{2n - 1}}-\frac{1}{{\left(x - 2\right)}^{2n - 1}}\right)$

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 $r=0.03,r=0.05$ , and $r=0.07$ .

Show Solution

$P={P}_{1}+\cdots+{P}_{20}$ where

${P}_{k}=10,000\frac{1}{{\left(1+r\right)}^{k}}$ . Then

$P=10,000\displaystyle\sum _{k=1}^{20}\frac{1}{{\left(1+r\right)}^{k}}=10,000\frac{1-{\left(1+r\right)}^{-20}}{r}$ . 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

$r=0.07,P\approx 105,940$ .

12. Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of

$r=0.03,r=0.05$ and

$r=0.07$ .

13. Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of $r=0.03,r=0.05$ , and $r=0.07$ .

Show Solution

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

$r=0.07$ .

14. Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of

$r=0.03,r=0.05$ , and

$r=0.07$ .

15. Suppose that an annuity has a present value $P=1\text{million dollars}$ . What interest rate r would allow for perpetual annual payouts of $50,000?

Show Solution

In general,

$P=\frac{C}{r}$ . Thus,

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

16. Suppose that an annuity has a present value

$P=10\text{million dollars}\text{.}$ 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. $x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\cdots$ (Hint: Group powers x^3k, ${x}^{3k - 1}$ , and ${x}^{3k - 2}.$ )

Show Solution

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

18.

$x+{x}^{2}-{x}^{3}-{x}^{4}+{x}^{5}+{x}^{6}-{x}^{7}-{x}^{8}+\cdots$ (Hint: Group powers x^4k,

${x}^{4k - 1}$ , etc.)

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

Show Solution

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

20.

$\frac{x}{2}+\frac{{x}^{2}}{4}-\frac{{x}^{3}}{8}+\frac{{x}^{4}}{16}+\frac{{x}^{5}}{32}-\frac{{x}^{6}}{64}+\cdots$ (Hint: Group powers

${\left(\frac{x}{2}\right)}^{3k},{\left(\frac{x}{2}\right)}^{3k - 1}$ , and

${\left(\frac{x}{2}\right)}^{3k - 2}.$ )

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

21. $f\left(x\right)=2\displaystyle\sum _{n=0}^{\infty }{x}^{n},g\left(x\right)=\displaystyle\sum _{n=0}^{\infty }n{x}^{n}$

Show Solution

${a}_{n}=2,{b}_{n}=n$ so

${c}_{n}=\displaystyle\sum _{k=0}^{n}{b}_{k}{a}_{n-k}=2\displaystyle\sum _{k=0}^{n}k=\left(n\right)\left(n+1\right)$ and

$f\left(x\right)g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }n\left(n+1\right){x}^{n}$

22.

$f\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{x}^{n},g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}{x}^{n}$ . Express the coefficients of

$f\left(x\right)g\left(x\right)$ in terms of

${H}_{n}=\displaystyle\sum _{k=1}^{n}\frac{1}{k}$ .

23. $f\left(x\right)=g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(\frac{x}{2}\right)}^{n}$

Show Solution

${a}_{n}={b}_{n}={2}^{\text{-}n}$ so

${c}_{n}=\displaystyle\sum _{k=1}^{n}{b}_{k}{a}_{n-k}={2}^{\text{-}n}\displaystyle\sum _{k=1}^{n}1=\frac{n}{{2}^{n}}$ and

$f\left(x\right)g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }n{\left(\frac{x}{2}\right)}^{n}$

24.

$f\left(x\right)=g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }n{x}^{n}$

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. $f\left(x\right)=\frac{1}{1+x}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{n}$

Show Solution

The derivative of

$f$ is

$-\frac{1}{{\left(1+x\right)}^{2}}=\text{-}\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\left(n+1\right){x}^{n}$ .

26.

$f\left(x\right)=\frac{1}{1-{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{x}^{2n}$

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

27. $f\left(x\right)=\frac{2x}{{\left(1+{x}^{2}\right)}^{2}}=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\left(2n\right){x}^{2n - 1}$

Show Solution

The indefinite integral of

$f$ is

$\frac{1}{1+{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n}$ .

28.

$f\left(x\right)=\frac{2x}{1+{x}^{2}}=2\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n+1}$

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

29. Evaluate $\displaystyle\sum _{n=1}^{\infty }\frac{n}{{2}^{n}}$ as ${f}^{\prime }\left(\frac{1}{2}\right)$ where $f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{x}^{n}$ .

Show Solution

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

$\displaystyle\sum _{n=1}^{\infty }\frac{n}{{2}^{n}}=2$ .

30. Evaluate

$\displaystyle\sum _{n=1}^{\infty }\frac{n}{{3}^{n}}$ as

${f}^{\prime }\left(\frac{1}{3}\right)$ where

$f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{x}^{n}$ .

31. Evaluate $\displaystyle\sum _{n=2}^{\infty }\frac{n\left(n - 1\right)}{{2}^{n}}$ as $f''\left(\frac{1}{2}\right)$ where $f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{x}^{n}$ .

Show Solution

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

$\displaystyle\sum _{n=2}^{\infty }\frac{n\left(n - 1\right)}{{2}^{n}}=4$ .

32. Evaluate

$\displaystyle\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{n+1}$ as

${\displaystyle\int }_{0}^{1}f\left(t\right)dt$ where

$f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n}=\frac{1}{1+{x}^{2}}$ .

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

33. $f\left(x\right)=\text{ln}x$ centered at $x=1$ (Hint: $x=1-\left(1-x\right)$ )

Show Solution

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

34.

$\text{ln}\left(1-x\right)$ at

$x=0$

35. $\text{ln}\left(1-{x}^{2}\right)$ at $x=0$

Show Solution

$\text{-}{\displaystyle\int }_{t=0}^{{x}^{2}}\frac{1}{1-t}dt=\text{-}\displaystyle\sum _{n=0}^{\infty }{\displaystyle\int }_{0}^{{x}^{2}}{t}^{n}dx-\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{2\left(n+1\right)}}{n+1}=\text{-}\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{2n}}{n}$

36.

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

$x=0$

37. $f\left(x\right)={\tan}^{-1}\left({x}^{2}\right)$ at $x=0$

Show Solution

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

38.

$f\left(x\right)=\text{ln}\left(1+{x}^{2}\right)$ at

$x=0$

39. $f\left(x\right)={\displaystyle\int }_{0}^{x}\text{ln}tdt$ where $\text{ln}\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{\left(x - 1\right)}^{n}}{n}$

Show Solution

Term-by-term integration gives

${\displaystyle\int }_{0}^{x}\text{ln}tdt=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{\left(x - 1\right)}^{n+1}}{n\left(n+1\right)}=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\left(\frac{1}{n}-\frac{1}{n+1}\right){\left(x - 1\right)}^{n+1}=\left(x - 1\right)\text{ln}x+\displaystyle\sum _{n=2}^{\infty }{\left(-1\right)}^{n}\frac{{\left(x - 1\right)}^{n}}{n}= x\text{ln}x-x$ .

40. [T] Evaluate the power series expansion

$\text{ln}\left(1+x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{x}^{n}}{n}$ at

$x=1$ to show that

$\text{ln}\left(2\right)$ 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

$\text{ln}\left(2\right)$ accurate to within 0.001, and find such an approximation.

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

Show Solution

We have

$\text{ln}\left(1-x\right)=\text{-}\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{n}$ so

$\text{ln}\left(1+x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{x}^{n}}{n}$ . Thus,

$\text{ln}\left(\frac{1+x}{1-x}\right)=\displaystyle\sum _{n=1}^{\infty }\left(1+{\left(-1\right)}^{n - 1}\right)\frac{{x}^{n}}{n}=2\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{2n - 1}}{2n - 1}$ . When

$x=\frac{1}{3}$ we obtain

$\text{ln}\left(2\right)=2\displaystyle\sum _{n=1}^{\infty }\frac{1}{{3}^{2n - 1}\left(2n - 1\right)}$ . We have

$2\displaystyle\sum _{n=1}^{3}\frac{1}{{3}^{2n - 1}\left(2n - 1\right)}=0.69300\ldots$ , while

$2\displaystyle\sum _{n=1}^{4}\frac{1}{{3}^{2n - 1}\left(2n - 1\right)}=0.69313\ldots$ and

$\text{ln}\left(2\right)=0.69314\ldots$ ; therefore,

$N=4$ .

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.

$\displaystyle\sum _{k=0}^{\infty }\left({x}^{k}-{x}^{2k+1}\right)$

43. $\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{3k}}{6k}$

Show Solution

$\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{k}}{k}=\text{-}\text{ln}\left(1-x\right)$ so

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

44.

$\displaystyle\sum _{k=1}^{\infty }{\left(1+{x}^{2}\right)}^{\text{-}k}$ using

$y=\frac{1}{1+{x}^{2}}$

45. $\displaystyle\sum _{k=1}^{\infty }{2}^{\text{-}kx}$ using $y={2}^{\text{-}x}$

Show Solution

$y={2}^{\text{-}x}$ , then

$\displaystyle\sum _{k=1}^{\infty }{y}^{k}=\frac{y}{1-y}=\frac{{2}^{\text{-}x}}{1-{2}^{\text{-}x}}=\frac{1}{{2}^{x}-1}$ . 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

$x>0$ .

46. Show that, up to powers x³ and y³,

$E\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$ satisfies

$E\left(x+y\right)=E\left(x\right)E\left(y\right)$ .

47. Differentiate the series $E\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$ term-by-term to show that $E\left(x\right)$ is equal to its derivative.

Show Solution

48. Show that if

$f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ is a sum of even powers, that is,

${a}_{n}=0$ if n is odd, then

$F={\displaystyle\int }_{0}^{x}f\left(t\right)dt$ is a sum of odd powers, while if f is a sum of odd powers, then F is a sum of even powers.

49. [T] Suppose that the coefficients a_n of the series $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ are defined by the recurrence relation ${a}_{n}=\frac{{a}_{n - 1}}{n}+\frac{{a}_{n - 2}}{n\left(n - 1\right)}$ . For ${a}_{0}=0$ and ${a}_{1}=1$ , compute and plot the sums ${S}_{N}=\displaystyle\sum _{n=0}^{N}{a}_{n}{x}^{n}$ for $N=2,3,4,5$ on $\left[-1,1\right]$ .

Show Solution

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

The solid curve is ${S}_{5}$ . The dashed curve is ${S}_{2}$ , dotted is ${S}_{3}$ , and dash-dotted is ${S}_{4}$

50. [T] Suppose that the coefficients a_n of the series

$\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ are defined by the recurrence relation

${a}_{n}=\frac{{a}_{n - 1}}{\sqrt{n}}-\frac{{a}_{n - 2}}{\sqrt{n\left(n - 1\right)}}$ . For

${a}_{0}=1$ and

${a}_{1}=0$ , compute and plot the sums

${S}_{N}=\displaystyle\sum _{n=0}^{N}{a}_{n}{x}^{n}$ for

$N=2,3,4,5$ on

$\left[-1,1\right]$ .

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

Show Solution

When

$x=-\frac{1}{2},\text{-}\text{ln}\left(2\right)=\text{ln}\left(\frac{1}{2}\right)=\text{-}\displaystyle\sum _{n=1}^{\infty }\frac{1}{n{2}^{n}}$ . Since

$\displaystyle\sum _{n=11}^{\infty }\frac{1}{n{2}^{n}}<\displaystyle\sum _{n=11}^{\infty }\frac{1}{{2}^{n}}=\frac{1}{{2}^{10}}$ , one has

$\displaystyle\sum _{n=1}^{10}\frac{1}{n{2}^{n}}=0.69306\ldots$ whereas

$\text{ln}\left(2\right)=0.69314\ldots$ ; therefore,

$N=10$ .

52. [T] Given the power series expansion

${\tan}^{-1}\left(x\right)=\displaystyle\sum _{k=0}^{\infty }{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{2k+1}$ , use the alternating series test to determine how many terms N of the sum evaluated at

$x=1$ are needed to approximate

${\tan}^{-1}\left(1\right)=\frac{\pi }{4}$ accurate to within

$\frac{1}{1000}$ . Evaluate the corresponding partial sum

$\displaystyle\sum _{k=0}^{N}{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{2k+1}$ .

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

Show Solution

$6{S}_{N}\left(\frac{1}{\sqrt{3}}\right)=2\sqrt{3}\displaystyle\sum _{n=0}^{N}{\left(-1\right)}^{n}\frac{1}{{3}^{n}\left(2n+1\right)}$ . One has

$\pi -6{S}_{4}\left(\frac{1}{\sqrt{3}}\right)=0.00101\ldots$ and

$\pi -6{S}_{5}\left(\frac{1}{\sqrt{3}}\right)=0.00028\ldots$ so

$N=5$ is the smallest partial sum with accuracy to within 0.001. Also,

$\pi -6{S}_{7}\left(\frac{1}{\sqrt{3}}\right)=0.00002\ldots$ while

$\pi -6{S}_{8}\left(\frac{1}{\sqrt{3}}\right)=-0.000007\ldots$ so

$N=8$ is the smallest N to give accuracy to within 0.00001.

Power Series: Problem Sets

Problem Set: Properties of Power Series

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