1. If [latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex] and [latex]g\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{n\text{!}}[/latex], find the power series of [latex]\frac{1}{2}\left(f\left(x\right)+g\left(x\right)\right)[/latex] and of [latex]\frac{1}{2}\left(f\left(x\right)-g\left(x\right)\right)[/latex].
Show Solution
[latex]\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{!}}[/latex] and [latex]\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{!}}[/latex].
2. If [latex]C\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{2n}}{\left(2n\right)\text{!}}[/latex] and [latex]S\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}[/latex], find the power series of [latex]C\left(x\right)+S\left(x\right)[/latex] and of [latex]C\left(x\right)-S\left(x\right)[/latex].
In the following exercises, use partial fractions to find the power series of each function.
3. [latex]\frac{4}{\left(x - 3\right)\left(x+1\right)}[/latex]
Show Solution
[latex]\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}[/latex]
4. [latex]\frac{3}{\left(x+2\right)\left(x - 1\right)}[/latex]
5. [latex]\frac{5}{\left({x}^{2}+4\right)\left({x}^{2}-1\right)}[/latex]
Show Solution
[latex]\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}[/latex]
6. [latex]\frac{30}{\left({x}^{2}+1\right)\left({x}^{2}-9\right)}[/latex]
In the following exercises, express each series as a rational function.
7. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{x}^{n}}[/latex]
Show Solution
[latex]\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}[/latex]
8. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{x}^{2n}}[/latex]
9. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{\left(x - 3\right)}^{2n - 1}}[/latex]
Show Solution
[latex]\frac{1}{x - 3}\frac{1}{1-\frac{1}{{\left(x - 3\right)}^{2}}}=\frac{x - 3}{{\left(x - 3\right)}^{2}-1}[/latex]
10. [latex]\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{{\left(x - 3\right)}^{2n - 1}}-\frac{1}{{\left(x - 2\right)}^{2n - 1}}\right)[/latex]
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 [latex]r=0.03,r=0.05[/latex], and [latex]r=0.07[/latex].
Show Solution
[latex]P={P}_{1}+\cdots+{P}_{20}[/latex] where [latex]{P}_{k}=10,000\frac{1}{{\left(1+r\right)}^{k}}[/latex]. Then [latex]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}[/latex]. When [latex]r=0.03,P\approx 10,000\times 14.8775=148,775[/latex]. When [latex]r=0.05,P\approx 10,000\times 12.4622=124,622[/latex]. When [latex]r=0.07,P\approx 105,940[/latex].
12. Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of [latex]r=0.03,r=0.05[/latex] and [latex]r=0.07[/latex].
13. Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of [latex]r=0.03,r=0.05[/latex], and [latex]r=0.07[/latex].
Show Solution
In general, [latex]P=\frac{C\left(1-{\left(1+r\right)}^{\text{-}N}\right)}{r}[/latex] for N years of payouts, or [latex]C=\frac{Pr}{1-{\left(1+r\right)}^{\text{-}N}}[/latex]. For [latex]N=20[/latex] and [latex]P=100,000[/latex], one has [latex]C=6721.57[/latex] when [latex]r=0.03;C=8024.26[/latex] when [latex]r=0.05[/latex]; and [latex]C\approx 9439.29[/latex] when [latex]r=0.07[/latex].
14. Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of [latex]r=0.03,r=0.05[/latex], and [latex]r=0.07[/latex].
15. Suppose that an annuity has a present value [latex]P=1\text{million dollars}[/latex]. What interest rate r would allow for perpetual annual payouts of $50,000?
Show Solution
In general, [latex]P=\frac{C}{r}[/latex]. Thus, [latex]r=\frac{C}{P}=5\times \frac{{10}^{4}}{{10}^{6}}=0.05[/latex].
16. Suppose that an annuity has a present value [latex]P=10\text{million dollars}\text{.}[/latex] 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. [latex]x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\cdots[/latex] (Hint: Group powers x3k, [latex]{x}^{3k - 1}[/latex], and [latex]{x}^{3k - 2}.[/latex])
Show Solution
[latex]\left(x+{x}^{2}-{x}^{3}\right)\left(1+{x}^{3}+{x}^{6}+\cdots\right)=\frac{x+{x}^{2}-{x}^{3}}{1-{x}^{3}}[/latex]
18. [latex]x+{x}^{2}-{x}^{3}-{x}^{4}+{x}^{5}+{x}^{6}-{x}^{7}-{x}^{8}+\cdots[/latex] (Hint: Group powers x4k, [latex]{x}^{4k - 1}[/latex], etc.)
19. [latex]x-{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}-{x}^{6}+{x}^{7}-\cdots[/latex] (Hint: Group powers x3k, [latex]{x}^{3k - 1}[/latex], and [latex]{x}^{3k - 2}.[/latex])
Show Solution
[latex]\left(x-{x}^{2}-{x}^{3}\right)\left(1+{x}^{3}+{x}^{6}+\cdots\right)=\frac{x-{x}^{2}-{x}^{3}}{1-{x}^{3}}[/latex]
20. [latex]\frac{x}{2}+\frac{{x}^{2}}{4}-\frac{{x}^{3}}{8}+\frac{{x}^{4}}{16}+\frac{{x}^{5}}{32}-\frac{{x}^{6}}{64}+\cdots[/latex] (Hint: Group powers [latex]{\left(\frac{x}{2}\right)}^{3k},{\left(\frac{x}{2}\right)}^{3k - 1}[/latex], and [latex]{\left(\frac{x}{2}\right)}^{3k - 2}.[/latex])
In the following exercises, find the power series of [latex]f\left(x\right)g\left(x\right)[/latex] given f and g as defined.
21. [latex]f\left(x\right)=2\displaystyle\sum _{n=0}^{\infty }{x}^{n},g\left(x\right)=\displaystyle\sum _{n=0}^{\infty }n{x}^{n}[/latex]
Show Solution
[latex]{a}_{n}=2,{b}_{n}=n[/latex] so [latex]{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)[/latex] and [latex]f\left(x\right)g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }n\left(n+1\right){x}^{n}[/latex]
22. [latex]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}[/latex]. Express the coefficients of [latex]f\left(x\right)g\left(x\right)[/latex] in terms of [latex]{H}_{n}=\displaystyle\sum _{k=1}^{n}\frac{1}{k}[/latex].
23. [latex]f\left(x\right)=g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(\frac{x}{2}\right)}^{n}[/latex]
Show Solution
[latex]{a}_{n}={b}_{n}={2}^{\text{-}n}[/latex] so [latex]{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}}[/latex] and [latex]f\left(x\right)g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }n{\left(\frac{x}{2}\right)}^{n}[/latex]
24. [latex]f\left(x\right)=g\left(x\right)=\displaystyle\sum _{n=1}^{\infty }n{x}^{n}[/latex]
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. [latex]f\left(x\right)=\frac{1}{1+x}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{n}[/latex]
Show Solution
The derivative of [latex]f[/latex] is [latex]-\frac{1}{{\left(1+x\right)}^{2}}=\text{-}\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\left(n+1\right){x}^{n}[/latex].
26. [latex]f\left(x\right)=\frac{1}{1-{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{x}^{2n}[/latex]
In the following exercises, integrate the given series expansion of [latex]f[/latex] term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of [latex]f[/latex].
27. [latex]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}[/latex]
Show Solution
The indefinite integral of [latex]f[/latex] is [latex]\frac{1}{1+{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n}[/latex].
28. [latex]f\left(x\right)=\frac{2x}{1+{x}^{2}}=2\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n+1}[/latex]
In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.
29. Evaluate [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n}{{2}^{n}}[/latex] as [latex]{f}^{\prime }\left(\frac{1}{2}\right)[/latex] where [latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex].
Show Solution
[latex]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[/latex] so [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n}{{2}^{n}}=2[/latex].
30. Evaluate [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n}{{3}^{n}}[/latex] as [latex]{f}^{\prime }\left(\frac{1}{3}\right)[/latex] where [latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex].
31. Evaluate [latex]\displaystyle\sum _{n=2}^{\infty }\frac{n\left(n - 1\right)}{{2}^{n}}[/latex] as [latex]f''\left(\frac{1}{2}\right)[/latex] where [latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex].
Show Solution
[latex]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[/latex] so [latex]\displaystyle\sum _{n=2}^{\infty }\frac{n\left(n - 1\right)}{{2}^{n}}=4[/latex].
32. Evaluate [latex]\displaystyle\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{n+1}[/latex] as [latex]{\displaystyle\int }_{0}^{1}f\left(t\right)dt[/latex] where [latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n}=\frac{1}{1+{x}^{2}}[/latex].
In the following exercises, given that [latex]\frac{1}{1-x}=\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex], use term-by-term differentiation or integration to find power series for each function centered at the given point.
33. [latex]f\left(x\right)=\text{ln}x[/latex] centered at [latex]x=1[/latex] (Hint: [latex]x=1-\left(1-x\right)[/latex])
Show Solution
[latex]\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}[/latex]
34. [latex]\text{ln}\left(1-x\right)[/latex] at [latex]x=0[/latex]
35. [latex]\text{ln}\left(1-{x}^{2}\right)[/latex] at [latex]x=0[/latex]
Show Solution
[latex]\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}[/latex]
36. [latex]f\left(x\right)=\frac{2x}{{\left(1-{x}^{2}\right)}^{2}}[/latex] at [latex]x=0[/latex]
37. [latex]f\left(x\right)={\tan}^{-1}\left({x}^{2}\right)[/latex] at [latex]x=0[/latex]
Show Solution
[latex]{\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}[/latex]
38. [latex]f\left(x\right)=\text{ln}\left(1+{x}^{2}\right)[/latex] at [latex]x=0[/latex]
39. [latex]f\left(x\right)={\displaystyle\int }_{0}^{x}\text{ln}tdt[/latex] where [latex]\text{ln}\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{\left(x - 1\right)}^{n}}{n}[/latex]
Show Solution
Term-by-term integration gives [latex]{\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[/latex].
40. [T] Evaluate the power series expansion [latex]\text{ln}\left(1+x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{x}^{n}}{n}[/latex] at [latex]x=1[/latex] to show that [latex]\text{ln}\left(2\right)[/latex] 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 [latex]\text{ln}\left(2\right)[/latex] accurate to within 0.001, and find such an approximation.
41. [T] Subtract the infinite series of [latex]\text{ln}\left(1-x\right)[/latex] from [latex]\text{ln}\left(1+x\right)[/latex] to get a power series for [latex]\text{ln}\left(\frac{1+x}{1-x}\right)[/latex]. Evaluate at [latex]x=\frac{1}{3}[/latex]. What is the smallest N such that the Nth partial sum of this series approximates [latex]\text{ln}\left(2\right)[/latex] with an error less than 0.001?
Show Solution
We have [latex]\text{ln}\left(1-x\right)=\text{-}\displaystyle\sum _{n=1}^{\infty }\frac{{x}^{n}}{n}[/latex] so [latex]\text{ln}\left(1+x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{x}^{n}}{n}[/latex]. Thus, [latex]\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}[/latex]. When [latex]x=\frac{1}{3}[/latex] we obtain [latex]\text{ln}\left(2\right)=2\displaystyle\sum _{n=1}^{\infty }\frac{1}{{3}^{2n - 1}\left(2n - 1\right)}[/latex]. We have [latex]2\displaystyle\sum _{n=1}^{3}\frac{1}{{3}^{2n - 1}\left(2n - 1\right)}=0.69300\ldots[/latex], while [latex]2\displaystyle\sum _{n=1}^{4}\frac{1}{{3}^{2n - 1}\left(2n - 1\right)}=0.69313\ldots[/latex] and [latex]\text{ln}\left(2\right)=0.69314\ldots[/latex]; therefore, [latex]N=4[/latex].
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. [latex]\displaystyle\sum _{k=0}^{\infty }\left({x}^{k}-{x}^{2k+1}\right)[/latex]
43. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{3k}}{6k}[/latex]
Show Solution
[latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{k}}{k}=\text{-}\text{ln}\left(1-x\right)[/latex] so [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{3k}}{6k}=-\frac{1}{6}\text{ln}\left(1-{x}^{3}\right)[/latex]. The radius of convergence is equal to 1 by the ratio test.
44. [latex]\displaystyle\sum _{k=1}^{\infty }{\left(1+{x}^{2}\right)}^{\text{-}k}[/latex] using [latex]y=\frac{1}{1+{x}^{2}}[/latex]
45. [latex]\displaystyle\sum _{k=1}^{\infty }{2}^{\text{-}kx}[/latex] using [latex]y={2}^{\text{-}x}[/latex]
Show Solution
If [latex]y={2}^{\text{-}x}[/latex], then [latex]\displaystyle\sum _{k=1}^{\infty }{y}^{k}=\frac{y}{1-y}=\frac{{2}^{\text{-}x}}{1-{2}^{\text{-}x}}=\frac{1}{{2}^{x}-1}[/latex]. If [latex]{a}_{k}={2}^{\text{-}kx}[/latex], then [latex]\frac{{a}_{k+1}}{{a}_{k}}={2}^{\text{-}x}<1[/latex] when [latex]x>0[/latex]. So the series converges for all [latex]x>0[/latex].
46. Show that, up to powers x3 and y3, [latex]E\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex] satisfies [latex]E\left(x+y\right)=E\left(x\right)E\left(y\right)[/latex].
47. Differentiate the series [latex]E\left(x\right)=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex] term-by-term to show that [latex]E\left(x\right)[/latex] is equal to its derivative.
Show Solution
Answers will vary.
48. Show that if [latex]f\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] is a sum of even powers, that is, [latex]{a}_{n}=0[/latex] if n is odd, then [latex]F={\displaystyle\int }_{0}^{x}f\left(t\right)dt[/latex] 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 an of the series [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] are defined by the recurrence relation [latex]{a}_{n}=\frac{{a}_{n - 1}}{n}+\frac{{a}_{n - 2}}{n\left(n - 1\right)}[/latex]. For [latex]{a}_{0}=0[/latex] and [latex]{a}_{1}=1[/latex], compute and plot the sums [latex]{S}_{N}=\displaystyle\sum _{n=0}^{N}{a}_{n}{x}^{n}[/latex] for [latex]N=2,3,4,5[/latex] on [latex]\left[-1,1\right][/latex].
Show Solution
The solid curve is [latex]{S}_{5}[/latex]. The dashed curve is [latex]{S}_{2}[/latex], dotted is [latex]{S}_{3}[/latex], and dash-dotted is [latex]{S}_{4}[/latex]
50. [T] Suppose that the coefficients an of the series [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] are defined by the recurrence relation [latex]{a}_{n}=\frac{{a}_{n - 1}}{\sqrt{n}}-\frac{{a}_{n - 2}}{\sqrt{n\left(n - 1\right)}}[/latex]. For [latex]{a}_{0}=1[/latex] and [latex]{a}_{1}=0[/latex], compute and plot the sums [latex]{S}_{N}=\displaystyle\sum _{n=0}^{N}{a}_{n}{x}^{n}[/latex] for [latex]N=2,3,4,5[/latex] on [latex]\left[-1,1\right][/latex].
51. [T] Given the power series expansion [latex]\text{ln}\left(1+x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n - 1}\frac{{x}^{n}}{n}[/latex], determine how many terms N of the sum evaluated at [latex]x=-\frac{1}{2}[/latex] are needed to approximate [latex]\text{ln}\left(2\right)[/latex] accurate to within [latex]\frac{1}{1000}[/latex]. Evaluate the corresponding partial sum [latex]\displaystyle\sum _{n=1}^{N}{\left(-1\right)}^{n - 1}\frac{{x}^{n}}{n}[/latex].
Show Solution
When [latex]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}}[/latex]. Since [latex]\displaystyle\sum _{n=11}^{\infty }\frac{1}{n{2}^{n}}<\displaystyle\sum _{n=11}^{\infty }\frac{1}{{2}^{n}}=\frac{1}{{2}^{10}}[/latex], one has [latex]\displaystyle\sum _{n=1}^{10}\frac{1}{n{2}^{n}}=0.69306\ldots[/latex] whereas [latex]\text{ln}\left(2\right)=0.69314\ldots[/latex]; therefore, [latex]N=10[/latex].
52. [T] Given the power series expansion [latex]{\tan}^{-1}\left(x\right)=\displaystyle\sum _{k=0}^{\infty }{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{2k+1}[/latex], use the alternating series test to determine how many terms N of the sum evaluated at [latex]x=1[/latex] are needed to approximate [latex]{\tan}^{-1}\left(1\right)=\frac{\pi }{4}[/latex] accurate to within [latex]\frac{1}{1000}[/latex]. Evaluate the corresponding partial sum [latex]\displaystyle\sum _{k=0}^{N}{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{2k+1}[/latex].
53. [T] Recall that [latex]{\tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}[/latex]. Assuming an exact value of [latex]\left(\frac{1}{\sqrt{3}}\right)[/latex], estimate [latex]\frac{\pi }{6}[/latex] by evaluating partial sums [latex]{S}_{N}\left(\frac{1}{\sqrt{3}}\right)[/latex] of the power series expansion [latex]{\tan}^{-1}\left(x\right)=\displaystyle\sum _{k=0}^{\infty }{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{2k+1}[/latex] at [latex]x=\frac{1}{\sqrt{3}}[/latex]. What is the smallest number N such that [latex]6{S}_{N}\left(\frac{1}{\sqrt{3}}\right)[/latex] approximates π accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?
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[latex]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)}[/latex]. One has [latex]\pi -6{S}_{4}\left(\frac{1}{\sqrt{3}}\right)=0.00101\ldots[/latex] and [latex]\pi -6{S}_{5}\left(\frac{1}{\sqrt{3}}\right)=0.00028\ldots[/latex] so [latex]N=5[/latex] is the smallest partial sum with accuracy to within 0.001. Also, [latex]\pi -6{S}_{7}\left(\frac{1}{\sqrt{3}}\right)=0.00002\ldots[/latex] while [latex]\pi -6{S}_{8}\left(\frac{1}{\sqrt{3}}\right)=-0.000007\ldots[/latex] so [latex]N=8[/latex] is the smallest N to give accuracy to within 0.00001.
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