In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.
1. [latex]{\left(1-x\right)}^{\frac{1}{3}}[/latex]
2. [latex]{\left(1+{x}^{2}\right)}^{\frac{-1}{3}}[/latex]
Show Solution
[latex]{\left(1+{x}^{2}\right)}^{\frac{-1}{3}}=\displaystyle\sum _{n=0}^{\infty }\left(\begin{array}{c}-\frac{1}{3}\hfill \\ \hfill n\hfill \end{array}\right){x}^{2n}[/latex]
3. [latex]{\left(1-x\right)}^{1.01}[/latex]
4. [latex]{\left(1 - 2x\right)}^{\frac{2}{3}}[/latex]
Show Solution
[latex]{\left(1 - 2x\right)}^{\frac{2}{3}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{2}^{n}\left(\begin{array}{c}\frac{2}{3}\hfill \\ n\hfill \end{array}\right){x}^{n}[/latex]
In the following exercises, use the substitution [latex]{\left(b+x\right)}^{r}={\left(b+a\right)}^{r}{\left(1+\frac{x-a}{b+a}\right)}^{r}[/latex] in the binomial expansion to find the Taylor series of each function with the given center.
5. [latex]\sqrt{x+2}[/latex] at [latex]a=0[/latex]
6. [latex]\sqrt{{x}^{2}+2}[/latex] at [latex]a=0[/latex]
Show Solution
[latex]\sqrt{2+{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{2}^{\left(\frac{1}{2}\right)-n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){x}^{2n};\left(|{x}^{2}|<2\right)[/latex]
7. [latex]\sqrt{x+2}[/latex] at [latex]a=1[/latex]
8. [latex]\sqrt{2x-{x}^{2}}[/latex] at [latex]a=1[/latex] (Hint: [latex]2x-{x}^{2}=1-{\left(x - 1\right)}^{2}[/latex])
Show Solution
[latex]\sqrt{2x-{x}^{2}}=\sqrt{1-{\left(x - 1\right)}^{2}}[/latex] so [latex]\sqrt{2x-{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x - 1\right)}^{2n}[/latex]
9. [latex]{\left(x - 8\right)}^{\frac{1}{3}}[/latex] at [latex]a=9[/latex]
10. [latex]\sqrt{x}[/latex] at [latex]a=4[/latex]
Show Solution
[latex]\sqrt{x}=2\sqrt{1+\frac{x - 4}{4}}[/latex] so [latex]\sqrt{x}=\displaystyle\sum _{n=0}^{\infty }{2}^{1 - 2n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x - 4\right)}^{n}[/latex]
11. [latex]{x}^{\frac{1}{3}}[/latex] at [latex]a=27[/latex]
12. [latex]\sqrt{x}[/latex] at [latex]x=9[/latex]
Show Solution
[latex]\sqrt{x}=\displaystyle\sum _{n=0}^{\infty }{3}^{1 - 3n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x - 9\right)}^{n}[/latex]
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 [latex]\frac{1}{1000}[/latex].
13. [T] [latex]{\left(15\right)}^{\frac{1}{4}}[/latex] using [latex]{\left(16-x\right)}^{\frac{1}{4}}[/latex]
14. [T] [latex]{\left(1001\right)}^{\frac{1}{3}}[/latex] using [latex]{\left(1000+x\right)}^{\frac{1}{3}}[/latex]
Show Solution
[latex]10{\left(1+\frac{x}{1000}\right)}^{\frac{1}{3}}=\displaystyle\sum _{n=0}^{\infty }{10}^{1 - 3n}\left(\begin{array}{c}\frac{1}{3}\hfill \\ n\hfill \end{array}\right){x}^{n}[/latex]. Using, for example, a fourth-degree estimate at [latex]x=1[/latex] gives [latex]\begin{array}{cc}\hfill {\left(1001\right)}^{\frac{1}{3}}& \approx 10\left(1+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 1\hfill \end{array}\right){10}^{-3}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 2\hfill \end{array}\right){10}^{-6}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 3\hfill \end{array}\right){10}^{-9}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 4\hfill \end{array}\right){10}^{-12}\right)\hfill \\ & =10\left(1+\frac{1}{{3.10}^{3}}-\frac{1}{{9.10}^{6}}+\frac{5}{{81.10}^{9}}-\frac{10}{{243.10}^{12}}\right)=10.00333222…\hfill \end{array}[/latex] whereas [latex]{\left(1001\right)}^{\frac{1}{3}}=10.00332222839093…[/latex]. Two terms would suffice for three-digit accuracy.
In the following exercises, use the binomial approximation [latex]\sqrt{1-x}\approx 1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{128}-\frac{7{x}^{5}}{256}[/latex] for [latex]|x|<1[/latex] to approximate each number. Compare this value to the value given by a scientific calculator.
15. [T] [latex]\frac{1}{\sqrt{2}}[/latex] using [latex]x=\frac{1}{2}[/latex] in [latex]{\left(1-x\right)}^{\frac{1}{2}}[/latex]
16. [T] [latex]\sqrt{5}=5\times \frac{1}{\sqrt{5}}[/latex] using [latex]x=\frac{4}{5}[/latex] in [latex]{\left(1-x\right)}^{\frac{1}{2}}[/latex]
Show Solution
The approximation is [latex]2.3152[/latex]; the CAS value is [latex]2.23\ldots[/latex].
17. [T] [latex]\sqrt{3}=\frac{3}{\sqrt{3}}[/latex] using [latex]x=\frac{2}{3}[/latex] in [latex]{\left(1-x\right)}^{\frac{1}{2}}[/latex]
18. [T] [latex]\sqrt{6}[/latex] using [latex]x=\frac{5}{6}[/latex] in [latex]{\left(1-x\right)}^{\frac{1}{2}}[/latex]
Show Solution
The approximation is [latex]2.583\ldots[/latex]; the CAS value is [latex]2.449\ldots[/latex].
19. Integrate the binomial approximation of [latex]\sqrt{1-x}[/latex] to find an approximation of [latex]{\displaystyle\int }_{0}^{x}\sqrt{1-t}dt[/latex].
20. [T] Recall that the graph of [latex]\sqrt{1-{x}^{2}}[/latex] is an upper semicircle of radius [latex]1[/latex]. Integrate the binomial approximation of [latex]\sqrt{1-{x}^{2}}[/latex] up to order [latex]8[/latex] from [latex]x=-1[/latex] to [latex]x=1[/latex] to estimate [latex]\frac{\pi }{2}[/latex].
Show Solution
[latex]\sqrt{1-{x}^{2}}=1-\frac{{x}^{2}}{2}-\frac{{x}^{4}}{8}-\frac{{x}^{6}}{16}-\frac{5{x}^{8}}{128}+\cdots[/latex]. Thus [latex]{\displaystyle\int }_{-1}^{1}\sqrt{1-{x}^{2}}dx=x-\frac{{x}^{3}}{6}-\frac{{x}^{5}}{40}-\frac{{x}^{7}}{7\cdot 16}-\frac{5{x}^{9}}{9\cdot 128}+\cdots{|}_{-1}^{1}\approx 2-\frac{1}{3}-\frac{1}{20}-\frac{1}{56}-\frac{10}{9\cdot 128}+\text{error}=1.590..[/latex]. whereas [latex]\frac{\pi }{2}=1.570..[/latex].
In the following exercises, use the expansion [latex]{\left(1+x\right)}^{\frac{1}{3}}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}+\cdots[/latex] to write the first five terms (not necessarily a quartic polynomial) of each expression.
21. [latex]{\left(1+4x\right)}^{\frac{1}{3}};a=0[/latex]
22. [latex]{\left(1+4x\right)}^{\frac{4}{3}};a=0[/latex]
Show Solution
[latex]{\left(1+4x\right)}^{\frac{4}{3}}=\left(1+4x\right)\left(1+4x\right)^{\frac{1}{3}}=\left(1+4x\right)\left(1+\frac{4x}{3}-\frac{16x^{3}}{9}+\frac{320x^{3}}{81}-\frac{2560x^{4}}{243}\right)=1+\frac{16}{3}x+\frac{32}{9}x^{2}-\frac{256}{81}x^{3}+\frac{1280}{243}x^{4}-\frac{10240}{243}x^{5}[/latex]
23. [latex]{\left(3+2x\right)}^{\frac{1}{3}};a=-1[/latex]
24. [latex]{\left({x}^{2}+6x+10\right)}^{\frac{1}{3}};a=-3[/latex]
Show Solution
[latex]\left(1+{\left(x+3\right)}^{2}\right)^{\frac{1}{3}} =1+\frac{1}{3}{\left(x+3\right)}^{2}-\frac{1}{9}{\left(x+3\right)}^{4}+\frac{5}{81}{\left(x+3\right)}^{6}-\frac{10}{243}{\left(x+3\right)}^{8}+\cdots [/latex]
25. Use [latex]{\left(1+x\right)}^{\frac{1}{3}}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}+\cdots[/latex] with [latex]x=1[/latex] to approximate [latex]{2}^{\frac{1}{3}}[/latex].
26. Use the approximation [latex]{\left(1-x\right)}^{\frac{2}{3}}=1-\frac{2x}{3}-\frac{{x}^{2}}{9}-\frac{4{x}^{3}}{81}-\frac{7{x}^{4}}{243}-\frac{14{x}^{5}}{729}+\cdots [/latex] for [latex]|x|<1[/latex] to approximate [latex]{2}^{\frac{1}{3}}={2.2}^{\frac{-2}{3}}[/latex].
Show Solution
Twice the approximation is [latex]1.260\ldots[/latex] whereas [latex]{2}^{\frac{1}{3}}=1.2599…[/latex].
27. Find the [latex]25\text{th}[/latex] derivative of [latex]f\left(x\right)={\left(1+{x}^{2}\right)}^{13}[/latex] at [latex]x=0[/latex].
28. Find the [latex]99[/latex] th derivative of [latex]f\left(x\right)={\left(1+{x}^{4}\right)}^{25}[/latex].
Show Solution
[latex]{f}^{\left(99\right)}\left(0\right)=0[/latex]
In the following exercises, find the Maclaurin series of each function.
29. [latex]f\left(x\right)=x{e}^{2x}[/latex]
30. [latex]f\left(x\right)={2}^{x}[/latex]
Show Solution
[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{\left(\text{ln}\left(2\right)x\right)}^{n}}{n\text{!}}[/latex]
31. [latex]f\left(x\right)=\frac{\sin{x}}{x}[/latex]
32. [latex]f\left(x\right)=\frac{\sin\left(\sqrt{x}\right)}{\sqrt{x}},\left(x>0\right)[/latex],
Show Solution
For [latex]x>0,\sin\left(\sqrt{x}\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{\frac{\left(2n+1\right)}{2}}}{\sqrt{x}\left(2n+1\right)\text{!}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\left(2n+1\right)\text{!}}[/latex].
33. [latex]f\left(x\right)=\sin\left({x}^{2}\right)[/latex]
34. [latex]f\left(x\right)={e}^{{x}^{3}}[/latex]
Show Solution
[latex]{e}^{{x}^{3}}=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{3n}}{n\text{!}}[/latex]
35. [latex]f\left(x\right)={\cos}^{2}x[/latex] using the identity [latex]{\cos}^{2}x=\frac{1}{2}+\frac{1}{2}\cos\left(2x\right)[/latex]
36. [latex]f\left(x\right)={\sin}^{2}x[/latex] using the identity [latex]{\sin}^{2}x=\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)[/latex]
Show Solution
[latex]{\sin}^{2}x=\text{-}\displaystyle\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k}{2}^{2k - 1}{x}^{2k}}{\left(2k\right)\text{!}}[/latex]
In the following exercises, find the Maclaurin series of [latex]F\left(x\right)={\displaystyle\int }_{0}^{x}f\left(t\right)dt[/latex] by integrating the Maclaurin series of [latex]f[/latex] term by term. If [latex]f[/latex] is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.
37. [latex]F\left(x\right)={\displaystyle\int }_{0}^{x}{e}^{\text{-}{t}^{2}}dt;f\left(t\right)={e}^{\text{-}{t}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{2n}}{n\text{!}}[/latex]
38. [latex]F\left(x\right)={\tan}^{-1}x;f\left(t\right)=\frac{1}{1+{t}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{t}^{2n}[/latex]
Show Solution
[latex]{\tan}^{-1}x=\displaystyle\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{x}^{2k+1}}{2k+1}[/latex]
39. [latex]F\left(x\right)={\text{tanh}}^{-1}x;f\left(t\right)=\frac{1}{1-{t}^{2}}=\displaystyle\sum _{n=0}^{\infty }{t}^{2n}[/latex]
40. [latex]F\left(x\right)={\sin}^{-1}x;f\left(t\right)=\frac{1}{\sqrt{1-{t}^{2}}}=\displaystyle\sum _{k=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\hfill \\ k\hfill \end{array}\right)\frac{{t}^{2k}}{k\text{!}}[/latex]
Show Solution
[latex]{\sin}^{-1}x=\displaystyle\sum _{n=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right)\frac{{x}^{2n+1}}{\left(2n+1\right)n\text{!}}[/latex]
41. [latex]F\left(x\right)={\displaystyle\int }_{0}^{x}\frac{\sin{t}}{t}dt;f\left(t\right)=\frac{\sin{t}}{t}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{2n}}{\left(2n+1\right)\text{!}}[/latex]
42. [latex]F\left(x\right)={\displaystyle\int }_{0}^{x}\cos\left(\sqrt{t}\right)dt;f\left(t\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\left(2n\right)\text{!}}[/latex]
Show Solution
[latex]F\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n+1}}{\left(n+1\right)\left(2n\right)\text{!}}[/latex]
43. [latex]F\left(x\right)={\displaystyle\int }_{0}^{x}\frac{1-\cos{t}}{{t}^{2}}dt;f\left(t\right)=\frac{1-\cos{t}}{{t}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{2n}}{\left(2n+2\right)\text{!}}[/latex]
44. [latex]F\left(x\right)={\displaystyle\int }_{0}^{x}\frac{\text{ln}\left(1+t\right)}{t}dt;f\left(t\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{n}}{n+1}[/latex]
Show Solution
[latex]F\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{x}^{n}}{{n}^{2}}[/latex]
In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of [latex]f[/latex].
45. [latex]f\left(x\right)=\sin\left(x+\frac{\pi }{4}\right)=\sin{x}\cos\left(\frac{\pi }{4}\right)+\cos{x}\sin\left(\frac{\pi }{4}\right)[/latex]
46. [latex]f\left(x\right)=\tan{x}[/latex]
Show Solution
[latex]x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\cdots[/latex]
47. [latex]f\left(x\right)=\text{ln}\left(\cos{x}\right)[/latex]
48. [latex]f\left(x\right)={e}^{x}\cos{x}[/latex]
Show Solution
[latex]1+x-\frac{{x}^{3}}{3}-\frac{{x}^{4}}{6}+\cdots[/latex]
49. [latex]f\left(x\right)={e}^{\sin{x}}[/latex]
50. [latex]f\left(x\right)={\sec}^{2}x[/latex]
Show Solution
[latex]1+{x}^{2}+\frac{2{x}^{4}}{3}+\frac{17{x}^{6}}{45}+\cdots[/latex]
51. [latex]f\left(x\right)=\text{tanh}x[/latex]
52. [latex]f\left(x\right)=\frac{\tan\sqrt{x}}{\sqrt{x}}[/latex] (see expansion for [latex]\tan{x}[/latex])
Show Solution
Using the expansion for [latex]\tan{x}[/latex] gives [latex]1+\frac{x}{3}+\frac{2{x}^{2}}{15}[/latex].
In the following exercises, find the radius of convergence of the Maclaurin series of each function.
53. [latex]\text{ln}\left(1+x\right)[/latex]
54. [latex]\frac{1}{1+{x}^{2}}[/latex]
Show Solution
[latex]\frac{1}{1+{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{x}^{2n}[/latex] so [latex]R=1[/latex] by the ratio test.
55. [latex]{\tan}^{-1}x[/latex]
56. [latex]\text{ln}\left(1+{x}^{2}\right)[/latex]
Show Solution
[latex]\text{ln}\left(1+{x}^{2}\right)=\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n - 1}}{n}{x}^{2n}[/latex] so [latex]R=1[/latex] by the ratio test.
57. Find the Maclaurin series of [latex]\text{sinh}x=\frac{{e}^{x}-{e}^{\text{-}x}}{2}[/latex].
58. Find the Maclaurin series of [latex]\text{cosh}x=\frac{{e}^{x}+{e}^{\text{-}x}}{2}[/latex].
Show Solution
Add series of [latex]{e}^{x}[/latex] and [latex]{e}^{\text{-}x}[/latex] term by term. Odd terms cancel and [latex]\text{cosh}x=\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{2n}}{\left(2n\right)\text{!}}[/latex].
59. Differentiate term by term the Maclaurin series of [latex]\text{sinh}x[/latex] and compare the result with the Maclaurin series of [latex]\text{cosh}x[/latex].
60. [T] Let [latex]{S}_{n}\left(x\right)=\displaystyle\sum _{k=0}^{n}{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{\left(2k+1\right)\text{!}}[/latex] and [latex]{C}_{n}\left(x\right)=\displaystyle\sum _{n=0}^{n}{\left(-1\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)\text{!}}[/latex] denote the respective Maclaurin polynomials of degree [latex]2n+1[/latex] of [latex]\sin{x}[/latex] and degree [latex]2n[/latex] of [latex]\cos{x}[/latex]. Plot the errors [latex]\frac{{S}_{n}\left(x\right)}{{C}_{n}\left(x\right)}-\tan{x}[/latex] for [latex]n=1,..,5[/latex] and compare them to [latex]x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}-\tan{x}[/latex] on [latex]\left(-\frac{\pi }{4},\frac{\pi }{4}\right)[/latex].
Show Solution
The ratio [latex]\frac{{S}_{n}\left(x\right)}{{C}_{n}\left(x\right)}[/latex] approximates [latex]\tan{x}[/latex] better than does [latex]{p}_{7}\left(x\right)=x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}[/latex] for [latex]N\ge 3[/latex]. The dashed curves are [latex]\frac{{S}_{n}}{{C}_{n}}-\tan[/latex] for [latex]n=1,2[/latex]. The dotted curve corresponds to [latex]n=3[/latex], and the dash-dotted curve corresponds to [latex]n=4[/latex]. The solid curve is [latex]{p}_{7}-\tan{x}[/latex].
61. Use the identity [latex]2\sin{x}\cos{x}=\sin\left(2x\right)[/latex] to find the power series expansion of [latex]{\sin}^{2}x[/latex] at [latex]x=0[/latex]. (Hint: Integrate the Maclaurin series of [latex]\sin\left(2x\right)[/latex] term by term.)
62. If [latex]y=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex], find the power series expansions of [latex]x{y}^{\prime }[/latex] and [latex]{x}^{2}y\text{''}[/latex].
Show Solution
By the term-by-term differentiation theorem, [latex]{y}^{\prime }=\displaystyle\sum _{n=1}^{\infty }n{a}_{n}{x}^{n - 1}[/latex] so [latex]{y}^{\prime }=\displaystyle\sum _{n=1}^{\infty }n{a}_{n}{x}^{n - 1}x{y}^{\prime }=\displaystyle\sum _{n=1}^{\infty }n{a}_{n}{x}^{n}[/latex], whereas [latex]{y}^{\prime }=\displaystyle\sum _{n=2}^{\infty }n\left(n - 1\right){a}_{n}{x}^{n - 2}[/latex] so [latex]xy\text{''}=\displaystyle\sum _{n=2}^{\infty }n\left(n - 1\right){a}_{n}{x}^{n}[/latex].
63. [T] Suppose that [latex]y=\displaystyle\sum _{k=0}^{\infty }{a}_{k}{x}^{k}[/latex] satisfies [latex]{y}^{\prime }=-2xy[/latex] and [latex]y\left(0\right)=0[/latex]. Show that [latex]{a}_{2k+1}=0[/latex] for all [latex]k[/latex] and that [latex]{a}_{2k+2}=\frac{\text{-}{a}_{2k}}{k+1}[/latex]. Plot the partial sum [latex]{S}_{20}[/latex] of [latex]y[/latex] on the interval [latex]\left[-4,4\right][/latex].
64. [T] Suppose that a set of standardized test scores is normally distributed with mean [latex]\mu =100[/latex] and standard deviation [latex]\sigma =10[/latex]. Set up an integral that represents the probability that a test score will be between [latex]90[/latex] and [latex]110[/latex] and use the integral of the degree [latex]10[/latex] Maclaurin polynomial of [latex]\frac{1}{\sqrt{2\pi }}{e}^{\frac{\text{-}{x}^{2}}{2}}[/latex] to estimate this probability.
Show Solution
The probability is [latex]p=\frac{1}{\sqrt{2\pi}}{\displaystyle\int }_{\frac{(a-\mu}{\sigma }}^\frac{(b-\mu)}{\sigma}{e}^{\frac{-{x}^{2}}{2}}dx[/latex] where [latex]a=90[/latex] and [latex]b=100[/latex], that is, [latex]p=\frac{1}{\sqrt{2\pi }}{\displaystyle\int }_{-1}^{1}{e}^{\frac{\text{-}{x}^{2}}{2}}dx=\frac{1}{\sqrt{2\pi }}{\displaystyle\int }_{-1}^{1}\displaystyle\sum _{n=0}^{5}{\left(-1\right)}^{n}\frac{{x}^{2n}}{{2}^{n}n\text{!}}dx=\frac{2}{\sqrt{2\pi }}\displaystyle\sum _{n=0}^{5}{\left(-1\right)}^{n}\frac{1}{\left(2n+1\right){2}^{n}n\text{!}}\approx 0.6827[/latex].
65. [T] Suppose that a set of standardized test scores is normally distributed with mean [latex]\mu =100[/latex] and standard deviation [latex]\sigma =10[/latex]. Set up an integral that represents the probability that a test score will be between [latex]70[/latex] and [latex]130[/latex] and use the integral of the degree [latex]50[/latex] Maclaurin polynomial of [latex]\frac{1}{\sqrt{2\pi }}{e}^{\frac{\text{-}{x}^{2}}{2}}[/latex] to estimate this probability.
66. [T] Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] converges to a function [latex]f\left(x\right)[/latex] such that [latex]f\left(0\right)=1,{f}^{\prime }\left(0\right)=0[/latex], and [latex]f\text{''}\left(x\right)=\text{-}f\left(x\right)[/latex]. Find a formula for [latex]{a}_{n}[/latex] and plot the partial sum [latex]{S}_{N}[/latex] for [latex]N=20[/latex] on [latex]\left[-5,5\right][/latex].
Show Solution
As in the previous problem one obtains [latex]{a}_{n}=0[/latex] if [latex]n[/latex] is odd and [latex]{a}_{n}=\text{-}\left(n+2\right)\left(n+1\right){a}_{n+2}[/latex] if [latex]n[/latex] is even, so [latex]{a}_{0}=1[/latex] leads to [latex]{a}_{2n}=\frac{{\left(-1\right)}^{n}}{\left(2n\right)\text{!}}[/latex].
67. [T] Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] converges to a function [latex]f\left(x\right)[/latex] such that [latex]f\left(0\right)=0,{f}^{\prime }\left(0\right)=1[/latex], and [latex]f\text{''}\left(x\right)=\text{-}f\left(x\right)[/latex]. Find a formula for [latex]{a}_{n}[/latex] and plot the partial sum [latex]{S}_{N}[/latex] for [latex]N=10[/latex] on [latex]\left[-5,5\right][/latex].
68. Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] converges to a function [latex]y[/latex] such that [latex]y\text{''}-{y}^{\prime }+y=0[/latex] where [latex]y\left(0\right)=1[/latex] and [latex]y^{\prime} \left(0\right)=0[/latex]. Find a formula that relates [latex]{a}_{n+2},{a}_{n+1}[/latex], and [latex]{a}_{n}[/latex] and compute [latex]{a}_{0},…,{a}_{5}[/latex].
Show Solution
[latex]y\text{''}=\displaystyle\sum _{n=0}^{\infty }\left(n+2\right)\left(n+1\right){a}_{n+2}{x}^{n}[/latex] and [latex]{y}^{\prime }=\displaystyle\sum _{n=0}^{\infty }\left(n+1\right){a}_{n+1}{x}^{n}[/latex] so [latex]y\text{''}-{y}^{\prime }+y=0[/latex] implies that [latex]\left(n+2\right)\left(n+1\right){a}_{n+2}-\left(n+1\right){a}_{n+1}+{a}_{n}=0[/latex] or [latex]{a}_{n}=\frac{{a}_{n - 1}}{n}-\frac{{a}_{n - 2}}{n\left(n - 1\right)}[/latex] for all [latex]n\cdot y\left(0\right)={a}_{0}=1[/latex] and [latex]{y}^{\prime }\left(0\right)={a}_{1}=0[/latex], so [latex]{a}_{2}=\frac{1}{2},{a}_{3}=\frac{1}{6},{a}_{4}=0[/latex], and [latex]{a}_{5}=-\frac{1}{120}[/latex].
69. Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] converges to a function [latex]y[/latex] such that [latex]y\text{''}-{y}^{\prime }+y=0[/latex] where [latex]y\left(0\right)=0[/latex] and [latex]{y}^{\prime }\left(0\right)=1[/latex]. Find a formula that relates [latex]{a}_{n+2},{a}_{n+1}[/latex], and [latex]{a}_{n}[/latex] and compute [latex]{a}_{1},…,{a}_{5}[/latex].
The error in approximating the integral [latex]{\displaystyle\int }_{a}^{b}f\left(t\right)dt[/latex] by that of a Taylor approximation [latex]{\displaystyle\int }_{a}^{b}{P}_{n}\left(t\right)dt[/latex] is at most [latex]{\displaystyle\int }_{a}^{b}{R}_{n}\left(t\right)dt[/latex]. In the following exercises, the Taylor remainder estimate [latex]{R}_{n}\le \frac{M}{\left(n+1\right)\text{!}}{|x-a|}^{n+1}[/latex] guarantees that the integral of the Taylor polynomial of the given order approximates the integral of [latex]f[/latex] with an error less than [latex]\frac{1}{10}[/latex].
- Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than [latex]\frac{1}{100}[/latex].
- Compare the accuracy of the polynomial integral estimate with the remainder estimate.
70. [T] [latex]{\displaystyle\int }_{0}^{\pi }\frac{\sin{t}}{t}dt;{P}_{s}=1-\frac{{x}^{2}}{3\text{!}}+\frac{{x}^{4}}{5\text{!}}-\frac{{x}^{6}}{7\text{!}}+\frac{{x}^{8}}{9\text{!}}[/latex] (You may assume that the absolute value of the ninth derivative of [latex]\frac{\sin{t}}{t}[/latex] is bounded by [latex]0.1.[/latex])
Show Solution
a. (Proof) b. We have [latex]{R}_{s}\le \frac{0.1}{\left(9\right)\text{!}}{\pi }^{9}\approx 0.0082<0.01[/latex]. We have [latex]{\displaystyle\int }_{0}^{\pi }\left(1-\frac{{x}^{2}}{3\text{!}}+\frac{{x}^{4}}{5\text{!}}-\frac{{x}^{6}}{7\text{!}}+\frac{{x}^{8}}{9\text{!}}\right)dx=\pi -\frac{{\pi }^{3}}{3\cdot 3\text{!}}+\frac{{\pi }^{5}}{5\cdot 5\text{!}}-\frac{{\pi }^{7}}{7\cdot 7\text{!}}+\frac{{\pi }^{9}}{9\cdot 9\text{!}}=1.852…[/latex], whereas [latex]{\displaystyle\int }_{0}^{\pi }\frac{\sin{t}}{t}dt=1.85194…[/latex], so the actual error is approximately [latex]0.00006[/latex].
71. [T] [latex]{\displaystyle\int }_{0}^{2}{e}^{\text{-}{x}^{2}}dx;{p}_{11}=1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{3\text{!}}+\cdots-\frac{{x}^{22}}{11\text{!}}[/latex] (You may assume that the absolute value of the [latex]23\text{rd}[/latex] derivative of [latex]{e}^{\text{-}{x}^{2}}[/latex] is less than [latex]2\times {10}^{14}.[/latex])
The following exercises deal with Fresnel integrals.
72. The Fresnel integrals are defined by [latex]C\left(x\right)={\displaystyle\int }_{0}^{x}\cos\left({t}^{2}\right)dt[/latex] and [latex]S\left(x\right)={\displaystyle\int }_{0}^{x}\sin\left({t}^{2}\right)dt[/latex]. Compute the power series of [latex]C\left(x\right)[/latex] and [latex]S\left(x\right)[/latex] and plot the sums [latex]{C}_{N}\left(x\right)[/latex] and [latex]{S}_{N}\left(x\right)[/latex] of the first [latex]N=50[/latex] nonzero terms on [latex]\left[0,2\pi \right][/latex].
Show Solution
Since [latex]\cos\left({t}^{2}\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{4n}}{\left(2n\right)\text{!}}[/latex] and [latex]\sin\left({t}^{2}\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{4n+2}}{\left(2n+1\right)\text{!}}[/latex], one has [latex]S\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{4n+3}}{\left(4n+3\right)\left(2n+1\right)\text{!}}[/latex] and [latex]C\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{4n+1}}{\left(4n+1\right)\left(2n\right)\text{!}}[/latex]. The sums of the first [latex]50[/latex] nonzero terms are plotted below with [latex]{C}_{50}\left(x\right)[/latex] the solid curve and [latex]{S}_{50}\left(x\right)[/latex] 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 [latex]\left(C\left(t\right),S\left(t\right)\right)[/latex]. Plot the curve [latex]\left({C}_{50},{S}_{50}\right)[/latex] for [latex]0\le t\le 2\pi [/latex], the coordinates of which were computed in the previous exercise.
74. Estimate [latex]{\displaystyle\int }_{0}^{\frac{1}{4}}\sqrt{x-{x}^{2}}dx[/latex] by approximating [latex]\sqrt{1-x}[/latex] using the binomial approximation [latex]1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{2128}-\frac{7{x}^{5}}{256}[/latex].
Show Solution
[latex]{\displaystyle\int }_{0}^{\frac{1}{4}}\sqrt{x}\left(1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{128}-\frac{7{x}^{5}}{256}\right)dx[/latex]
[latex]=\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..[/latex].
whereas [latex]{\displaystyle\int }_{0}^{\frac{1}{4}}\sqrt{x-{x}^{2}}dx=0.076773[/latex].
75. [T] Use Newton’s approximation of the binomial [latex]\sqrt{1-{x}^{2}}[/latex] to approximate [latex]\pi [/latex] as follows. The circle centered at [latex]\left(\frac{1}{2},0\right)[/latex] with radius [latex]\frac{1}{2}[/latex] has upper semicircle [latex]y=\sqrt{x}\sqrt{1-x}[/latex]. The sector of this circle bounded by the [latex]x[/latex] -axis between [latex]x=0[/latex] and [latex]x=\frac{1}{2}[/latex] and by the line joining [latex]\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right)[/latex] corresponds to [latex]\frac{1}{6}[/latex] of the circle and has area [latex]\frac{\pi }{24}[/latex]. This sector is the union of a right triangle with height [latex]\frac{\sqrt{3}}{4}[/latex] and base [latex]\frac{1}{4}[/latex] and the region below the graph between [latex]x=0[/latex] and [latex]x=\frac{1}{4}[/latex]. To find the area of this region you can write [latex]y=\sqrt{x}\sqrt{1-x}=\sqrt{x}\times \left(\text{binomial expansion of}\sqrt{1-x}\right)[/latex] and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate [latex]\pi [/latex].
76. Use the approximation [latex]T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)[/latex] to approximate the period of a pendulum having length [latex]10[/latex] meters and maximum angle [latex]{\theta }_{\text{max}}=\frac{\pi }{6}[/latex] where [latex]k=\sin\left(\frac{{\theta }_{\text{max}}}{2}\right)[/latex]. Compare this with the small angle estimate [latex]T\approx 2\pi \sqrt{\frac{L}{g}}[/latex].
Show Solution
[latex]T\approx 2\pi \sqrt{\frac{10}{9.8}}\left(1+\frac{{\sin}^{2}\left(\frac{\theta}{12}\right)}{4}\right)\approx 6.453[/latex] seconds. The small angle estimate is [latex]T\approx 2\pi \sqrt{\frac{10}{9.8}\approx 6.347}[/latex]. The relative error is around [latex]2[/latex] percent.
77. Suppose that a pendulum is to have a period of [latex]2[/latex] seconds and a maximum angle of [latex]{\theta }_{\text{max}}=\frac{\pi }{6}[/latex]. Use [latex]T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)[/latex] to approximate the desired length of the pendulum. What length is predicted by the small angle estimate [latex]T\approx 2\pi \sqrt{\frac{L}{g}}?[/latex]
78. Evaluate [latex]{\displaystyle\int }_{0}^{\frac{\pi}{2}}{\sin}^{4}\theta d\theta [/latex] in the approximation [latex]T=4\sqrt{\frac{L}{g}}{\displaystyle\int }_{0}^{\frac{\pi}{2}}\left(1+\frac{1}{2}{k}^{2}{\sin}^{2}\theta +\frac{3}{8}{k}^{4}{\sin}^{4}\theta +\cdots\right)d\theta [/latex] to obtain an improved estimate for [latex]T[/latex].
Show Solution
[latex]{\displaystyle\int }_{0}^{\frac{\pi}{2}}{\sin}^{4}\theta d\theta =\frac{3\pi }{16}[/latex]. Hence [latex]T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}+\frac{9}{256}{k}^{4}\right)[/latex].
79. [T] An equivalent formula for the period of a pendulum with amplitude [latex]{\theta }_{\text{max}}[/latex] is [latex]T\left({\theta }_{\text{max}}\right)=2\sqrt{2}\sqrt{\frac{L}{g}}{\displaystyle\int }_{0}^{{\theta }_{\text{max}}}\frac{d\theta }{\sqrt{\cos\theta }-\cos\left({\theta }_{\text{max}}\right)}[/latex] where [latex]L[/latex] is the pendulum length and [latex]g[/latex] is the gravitational acceleration constant. When [latex]{\theta }_{\text{max}}=\frac{\pi }{3}[/latex] we get [latex]\frac{1}{\sqrt{\cos{t} - \frac{1}{2}}}\approx \sqrt{2}\left(1+\frac{{t}^{2}}{2}+\frac{{t}^{4}}{3}+\frac{181{t}^{6}}{720}\right)[/latex]. Integrate this approximation to estimate [latex]T\left(\frac{\pi }{3}\right)[/latex] in terms of [latex]L[/latex] and [latex]g[/latex]. Assuming [latex]g=9.806[/latex] meters per second squared, find an approximate length [latex]L[/latex] such that [latex]T\left(\frac{\pi }{3}\right)=2[/latex] seconds.
