Problem Set: Working with Taylor Series

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

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

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

Show Solution

${\left(1+{x}^{2}\right)}^{\frac{-1}{3}}=\displaystyle\sum _{n=0}^{\infty }\left(\begin{array}{c}-\frac{1}{3}\hfill \\ \hfill n\hfill \end{array}\right){x}^{2n}$

${\left(1-x\right)}^{1.01}$

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

Show Solution

${\left(1 - 2x\right)}^{\frac{2}{3}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{2}^{n}\left(\begin{array}{c}\frac{2}{3}\hfill \\ n\hfill \end{array}\right){x}^{n}$

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

$\sqrt{x+2}$ at

$a=0$

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

Show Solution

$\sqrt{2+{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{2}^{\left(\frac{1}{2}\right)-n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){x}^{2n};\left(|{x}^{2}|<2\right)$

$\sqrt{x+2}$ at

$a=1$

8. $\sqrt{2x-{x}^{2}}$ at $a=1$ (Hint: $2x-{x}^{2}=1-{\left(x - 1\right)}^{2}$ )

Show Solution

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

$\sqrt{2x-{x}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x - 1\right)}^{2n}$

${\left(x - 8\right)}^{\frac{1}{3}}$ at

$a=9$

10. $\sqrt{x}$ at $a=4$

Show Solution

$\sqrt{x}=2\sqrt{1+\frac{x - 4}{4}}$ so

$\sqrt{x}=\displaystyle\sum _{n=0}^{\infty }{2}^{1 - 2n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x - 4\right)}^{n}$

11.

${x}^{\frac{1}{3}}$ at

$a=27$

12. $\sqrt{x}$ at $x=9$

Show Solution

$\sqrt{x}=\displaystyle\sum _{n=0}^{\infty }{3}^{1 - 3n}\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right){\left(x - 9\right)}^{n}$

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 $\frac{1}{1000}$ .

13. [T]

${\left(15\right)}^{\frac{1}{4}}$ using

${\left(16-x\right)}^{\frac{1}{4}}$

14. [T] ${\left(1001\right)}^{\frac{1}{3}}$ using ${\left(1000+x\right)}^{\frac{1}{3}}$

Show Solution

$10{\left(1+\frac{x}{1000}\right)}^{\frac{1}{3}}=\displaystyle\sum _{n=0}^{\infty }{10}^{1 - 3n}\left(\begin{array}{c}\frac{1}{3}\hfill \\ n\hfill \end{array}\right){x}^{n}$ . Using, for example, a fourth-degree estimate at

$x=1$ gives

$\begin{array}{cc}\hfill {\left(1001\right)}^{\frac{1}{3}}& \approx 10\left(1+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 1\hfill \end{array}\right){10}^{-3}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 2\hfill \end{array}\right){10}^{-6}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 3\hfill \end{array}\right){10}^{-9}+\left(\begin{array}{c}\frac{1}{3}\hfill \\ 4\hfill \end{array}\right){10}^{-12}\right)\hfill \\ & =10\left(1+\frac{1}{{3.10}^{3}}-\frac{1}{{9.10}^{6}}+\frac{5}{{81.10}^{9}}-\frac{10}{{243.10}^{12}}\right)=10.00333222...\hfill \end{array}$ whereas

${\left(1001\right)}^{\frac{1}{3}}=10.00332222839093...$ . Two terms would suffice for three-digit accuracy.

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

15. [T]

$\frac{1}{\sqrt{2}}$ using

$x=\frac{1}{2}$ in

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

16. [T] $\sqrt{5}=5\times \frac{1}{\sqrt{5}}$ using $x=\frac{4}{5}$ in ${\left(1-x\right)}^{\frac{1}{2}}$

Show Solution

The approximation is

$2.3152$ ; the CAS value is

$2.23\ldots$ .

17. [T]

$\sqrt{3}=\frac{3}{\sqrt{3}}$ using

$x=\frac{2}{3}$ in

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

18. [T] $\sqrt{6}$ using $x=\frac{5}{6}$ in ${\left(1-x\right)}^{\frac{1}{2}}$

Show Solution

The approximation is

$2.583\ldots$ ; the CAS value is

$2.449\ldots$ .

19. Integrate the binomial approximation of

$\sqrt{1-x}$ to find an approximation of

${\displaystyle\int }_{0}^{x}\sqrt{1-t}dt$ .

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

Show Solution

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

${\displaystyle\int }_{-1}^{1}\sqrt{1-{x}^{2}}dx=x-\frac{{x}^{3}}{6}-\frac{{x}^{5}}{40}-\frac{{x}^{7}}{7\cdot 16}-\frac{5{x}^{9}}{9\cdot 128}+\cdots{|}_{-1}^{1}\approx 2-\frac{1}{3}-\frac{1}{20}-\frac{1}{56}-\frac{10}{9\cdot 128}+\text{error}=1.590..$ . whereas

$\frac{\pi }{2}=1.570..$ .

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

21.

${\left(1+4x\right)}^{\frac{1}{3}};a=0$

22. ${\left(1+4x\right)}^{\frac{4}{3}};a=0$

Show Solution

${\left(1+4x\right)}^{\frac{4}{3}}=\left(1+4x\right)\left(1+4x\right)^{\frac{1}{3}}=\left(1+4x\right)\left(1+\frac{4x}{3}-\frac{16x^{3}}{9}+\frac{320x^{3}}{81}-\frac{2560x^{4}}{243}\right)=1+\frac{16}{3}x+\frac{32}{9}x^{2}-\frac{256}{81}x^{3}+\frac{1280}{243}x^{4}-\frac{10240}{243}x^{5}$

23.

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

24. ${\left({x}^{2}+6x+10\right)}^{\frac{1}{3}};a=-3$

Show Solution

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

25. Use

${\left(1+x\right)}^{\frac{1}{3}}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}+\cdots$ with

$x=1$ to approximate

${2}^{\frac{1}{3}}$ .

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

Show Solution

Twice the approximation is

$1.260\ldots$ whereas

${2}^{\frac{1}{3}}=1.2599...$ .

27. Find the

$25\text{th}$ derivative of

$f\left(x\right)={\left(1+{x}^{2}\right)}^{13}$ at

$x=0$ .

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

Show Solution

${f}^{\left(99\right)}\left(0\right)=0$

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

29.

$f\left(x\right)=x{e}^{2x}$

30. $f\left(x\right)={2}^{x}$

Show Solution

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

31.

$f\left(x\right)=\frac{\sin{x}}{x}$

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

Show Solution

For

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

33.

$f\left(x\right)=\sin\left({x}^{2}\right)$

34. $f\left(x\right)={e}^{{x}^{3}}$

Show Solution

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

35.

$f\left(x\right)={\cos}^{2}x$ using the identity

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

36. $f\left(x\right)={\sin}^{2}x$ using the identity ${\sin}^{2}x=\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)$

Show Solution

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

In the following exercises, find the Maclaurin series of $F\left(x\right)={\displaystyle\int }_{0}^{x}f\left(t\right)dt$ by integrating the Maclaurin series of $f$ term by term. If $f$ is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

37.

$F\left(x\right)={\displaystyle\int }_{0}^{x}{e}^{\text{-}{t}^{2}}dt;f\left(t\right)={e}^{\text{-}{t}^{2}}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{2n}}{n\text{!}}$

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

Show Solution

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

39.

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

40. $F\left(x\right)={\sin}^{-1}x;f\left(t\right)=\frac{1}{\sqrt{1-{t}^{2}}}=\displaystyle\sum _{k=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\hfill \\ k\hfill \end{array}\right)\frac{{t}^{2k}}{k\text{!}}$

Show Solution

${\sin}^{-1}x=\displaystyle\sum _{n=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\hfill \\ n\hfill \end{array}\right)\frac{{x}^{2n+1}}{\left(2n+1\right)n\text{!}}$

41.

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

42. $F\left(x\right)={\displaystyle\int }_{0}^{x}\cos\left(\sqrt{t}\right)dt;f\left(t\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\left(2n\right)\text{!}}$

Show Solution

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

43.

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

44. $F\left(x\right)={\displaystyle\int }_{0}^{x}\frac{\text{ln}\left(1+t\right)}{t}dt;f\left(t\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{n}}{n+1}$

Show Solution

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

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

45.

$f\left(x\right)=\sin\left(x+\frac{\pi }{4}\right)=\sin{x}\cos\left(\frac{\pi }{4}\right)+\cos{x}\sin\left(\frac{\pi }{4}\right)$

46. $f\left(x\right)=\tan{x}$

Show Solution

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

47.

$f\left(x\right)=\text{ln}\left(\cos{x}\right)$

48. $f\left(x\right)={e}^{x}\cos{x}$

Show Solution

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

49.

$f\left(x\right)={e}^{\sin{x}}$

50. $f\left(x\right)={\sec}^{2}x$

Show Solution

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

51.

$f\left(x\right)=\text{tanh}x$

52. $f\left(x\right)=\frac{\tan\sqrt{x}}{\sqrt{x}}$ (see expansion for $\tan{x}$ )

Show Solution

Using the expansion for

$\tan{x}$ gives

$1+\frac{x}{3}+\frac{2{x}^{2}}{15}$ .

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

53.

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

54. $\frac{1}{1+{x}^{2}}$

Show Solution

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

$R=1$ by the ratio test.

55.

${\tan}^{-1}x$

56. $\text{ln}\left(1+{x}^{2}\right)$

Show Solution

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

$R=1$ by the ratio test.

57. Find the Maclaurin series of

$\text{sinh}x=\frac{{e}^{x}-{e}^{\text{-}x}}{2}$ .

58. Find the Maclaurin series of $\text{cosh}x=\frac{{e}^{x}+{e}^{\text{-}x}}{2}$ .

Show Solution

Add series of

${e}^{x}$ and

${e}^{\text{-}x}$ term by term. Odd terms cancel and

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

59. Differentiate term by term the Maclaurin series of

$\text{sinh}x$ and compare the result with the Maclaurin series of

$\text{cosh}x$ .

60. [T] Let ${S}_{n}\left(x\right)=\displaystyle\sum _{k=0}^{n}{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{\left(2k+1\right)\text{!}}$ and ${C}_{n}\left(x\right)=\displaystyle\sum _{n=0}^{n}{\left(-1\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)\text{!}}$ denote the respective Maclaurin polynomials of degree $2n+1$ of $\sin{x}$ and degree $2n$ of $\cos{x}$ . Plot the errors $\frac{{S}_{n}\left(x\right)}{{C}_{n}\left(x\right)}-\tan{x}$ for $n=1,..,5$ and compare them to $x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}-\tan{x}$ on $\left(-\frac{\pi }{4},\frac{\pi }{4}\right)$ .

Show Solution

This graph has two curves. The first one is a decreasing function passing through the origin. The second is a broken line which is an increasing function passing through the origin. The two curves are very close around the origin.

The ratio $\frac{{S}_{n}\left(x\right)}{{C}_{n}\left(x\right)}$ approximates $\tan{x}$ better than does ${p}_{7}\left(x\right)=x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}+\frac{17{x}^{7}}{315}$ for $N\ge 3$ . The dashed curves are $\frac{{S}_{n}}{{C}_{n}}-\tan$ for $n=1,2$ . The dotted curve corresponds to $n=3$ , and the dash-dotted curve corresponds to $n=4$ . The solid curve is ${p}_{7}-\tan{x}$ .

61. Use the identity

$2\sin{x}\cos{x}=\sin\left(2x\right)$ to find the power series expansion of

${\sin}^{2}x$ at

$x=0$ . (Hint: Integrate the Maclaurin series of

$\sin\left(2x\right)$ term by term.)

62. If $y=\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ , find the power series expansions of $x{y}^{\prime }$ and ${x}^{2}y\text{''}$ .

Show Solution

By the term-by-term differentiation theorem,

${y}^{\prime }=\displaystyle\sum _{n=1}^{\infty }n{a}_{n}{x}^{n - 1}$ so

${y}^{\prime }=\displaystyle\sum _{n=1}^{\infty }n{a}_{n}{x}^{n - 1}x{y}^{\prime }=\displaystyle\sum _{n=1}^{\infty }n{a}_{n}{x}^{n}$ , whereas

${y}^{\prime }=\displaystyle\sum _{n=2}^{\infty }n\left(n - 1\right){a}_{n}{x}^{n - 2}$ so

$xy\text{''}=\displaystyle\sum _{n=2}^{\infty }n\left(n - 1\right){a}_{n}{x}^{n}$ .

63. [T] Suppose that

$y=\displaystyle\sum _{k=0}^{\infty }{a}_{k}{x}^{k}$ satisfies

${y}^{\prime }=-2xy$ and

$y\left(0\right)=0$ . Show that

${a}_{2k+1}=0$ for all

$k$ and that

${a}_{2k+2}=\frac{\text{-}{a}_{2k}}{k+1}$ . Plot the partial sum

${S}_{20}$ of

$y$ on the interval

$\left[-4,4\right]$ .

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

Show Solution

The probability is

$p=\frac{1}{\sqrt{2\pi}}{\displaystyle\int }_{\frac{(a-\mu}{\sigma }}^\frac{(b-\mu)}{\sigma}{e}^{\frac{-{x}^{2}}{2}}dx$ where

$a=90$ and

$b=100$ , that is,

$p=\frac{1}{\sqrt{2\pi }}{\displaystyle\int }_{-1}^{1}{e}^{\frac{\text{-}{x}^{2}}{2}}dx=\frac{1}{\sqrt{2\pi }}{\displaystyle\int }_{-1}^{1}\displaystyle\sum _{n=0}^{5}{\left(-1\right)}^{n}\frac{{x}^{2n}}{{2}^{n}n\text{!}}dx=\frac{2}{\sqrt{2\pi }}\displaystyle\sum _{n=0}^{5}{\left(-1\right)}^{n}\frac{1}{\left(2n+1\right){2}^{n}n\text{!}}\approx 0.6827$ .

65. [T] Suppose that a set of standardized test scores is normally distributed with mean

$\mu =100$ and standard deviation

$\sigma =10$ . Set up an integral that represents the probability that a test score will be between

$70$ and

$130$ and use the integral of the degree

$50$ Maclaurin polynomial of

$\frac{1}{\sqrt{2\pi }}{e}^{\frac{\text{-}{x}^{2}}{2}}$ to estimate this probability.

66. [T] Suppose that $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ converges to a function $f\left(x\right)$ such that $f\left(0\right)=1,{f}^{\prime }\left(0\right)=0$ , and $f\text{''}\left(x\right)=\text{-}f\left(x\right)$ . Find a formula for ${a}_{n}$ and plot the partial sum ${S}_{N}$ for $N=20$ on $\left[-5,5\right]$ .

Show Solution

This graph is a wave curve symmetrical about the origin. It has a peak at y = 1 above the origin. It has lowest points at -3 and 3.

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

67. [T] Suppose that

$\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ converges to a function

$f\left(x\right)$ such that

$f\left(0\right)=0,{f}^{\prime }\left(0\right)=1$ , and

$f\text{''}\left(x\right)=\text{-}f\left(x\right)$ . Find a formula for

${a}_{n}$ and plot the partial sum

${S}_{N}$ for

$N=10$ on

$\left[-5,5\right]$ .

68. Suppose that $\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ converges to a function $y$ such that $y\text{''}-{y}^{\prime }+y=0$ where $y\left(0\right)=1$ and $y^{\prime} \left(0\right)=0$ . Find a formula that relates ${a}_{n+2},{a}_{n+1}$ , and ${a}_{n}$ and compute ${a}_{0},...,{a}_{5}$ .

Show Solution

$y\text{''}=\displaystyle\sum _{n=0}^{\infty }\left(n+2\right)\left(n+1\right){a}_{n+2}{x}^{n}$ and

${y}^{\prime }=\displaystyle\sum _{n=0}^{\infty }\left(n+1\right){a}_{n+1}{x}^{n}$ so

$y\text{''}-{y}^{\prime }+y=0$ implies that

$\left(n+2\right)\left(n+1\right){a}_{n+2}-\left(n+1\right){a}_{n+1}+{a}_{n}=0$ or

${a}_{n}=\frac{{a}_{n - 1}}{n}-\frac{{a}_{n - 2}}{n\left(n - 1\right)}$ for all

$n\cdot y\left(0\right)={a}_{0}=1$ and

${y}^{\prime }\left(0\right)={a}_{1}=0$ , so

${a}_{2}=\frac{1}{2},{a}_{3}=\frac{1}{6},{a}_{4}=0$ , and

${a}_{5}=-\frac{1}{120}$ .

69. Suppose that

$\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ converges to a function

$y$ such that

$y\text{''}-{y}^{\prime }+y=0$ where

$y\left(0\right)=0$ and

${y}^{\prime }\left(0\right)=1$ . Find a formula that relates

${a}_{n+2},{a}_{n+1}$ , and

${a}_{n}$ and compute

${a}_{1},...,{a}_{5}$ .

The error in approximating the integral ${\displaystyle\int }_{a}^{b}f\left(t\right)dt$ by that of a Taylor approximation ${\displaystyle\int }_{a}^{b}{P}_{n}\left(t\right)dt$ is at most ${\displaystyle\int }_{a}^{b}{R}_{n}\left(t\right)dt$ . In the following exercises, the Taylor remainder estimate ${R}_{n}\le \frac{M}{\left(n+1\right)\text{!}}{|x-a|}^{n+1}$ guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $f$ with an error less than $\frac{1}{10}$ .

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

70. [T] ${\displaystyle\int }_{0}^{\pi }\frac{\sin{t}}{t}dt;{P}_{s}=1-\frac{{x}^{2}}{3\text{!}}+\frac{{x}^{4}}{5\text{!}}-\frac{{x}^{6}}{7\text{!}}+\frac{{x}^{8}}{9\text{!}}$ (You may assume that the absolute value of the ninth derivative of $\frac{\sin{t}}{t}$ is bounded by $0.1.$ )

Show Solution

a. (Proof) b. We have

${R}_{s}\le \frac{0.1}{\left(9\right)\text{!}}{\pi }^{9}\approx 0.0082<0.01$ . We have

${\displaystyle\int }_{0}^{\pi }\left(1-\frac{{x}^{2}}{3\text{!}}+\frac{{x}^{4}}{5\text{!}}-\frac{{x}^{6}}{7\text{!}}+\frac{{x}^{8}}{9\text{!}}\right)dx=\pi -\frac{{\pi }^{3}}{3\cdot 3\text{!}}+\frac{{\pi }^{5}}{5\cdot 5\text{!}}-\frac{{\pi }^{7}}{7\cdot 7\text{!}}+\frac{{\pi }^{9}}{9\cdot 9\text{!}}=1.852...$ , whereas

${\displaystyle\int }_{0}^{\pi }\frac{\sin{t}}{t}dt=1.85194...$ , so the actual error is approximately

$0.00006$ .

71. [T]

${\displaystyle\int }_{0}^{2}{e}^{\text{-}{x}^{2}}dx;{p}_{11}=1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{3\text{!}}+\cdots-\frac{{x}^{22}}{11\text{!}}$ (You may assume that the absolute value of the

$23\text{rd}$ derivative of

${e}^{\text{-}{x}^{2}}$ is less than

$2\times {10}^{14}.$ )

The following exercises deal with Fresnel integrals.

72. The Fresnel integrals are defined by $C\left(x\right)={\displaystyle\int }_{0}^{x}\cos\left({t}^{2}\right)dt$ and $S\left(x\right)={\displaystyle\int }_{0}^{x}\sin\left({t}^{2}\right)dt$ . Compute the power series of $C\left(x\right)$ and $S\left(x\right)$ and plot the sums ${C}_{N}\left(x\right)$ and ${S}_{N}\left(x\right)$ of the first $N=50$ nonzero terms on $\left[0,2\pi \right]$ .

Show Solution

This graph has two curves. The first one is a solid curve labeled Csub50(x). It begins at the origin and is a wave that gradually decreases in amplitude. The highest it reaches is y = 1. The second curve is labeled Ssub50(x). It is a wave that gradually decreases in amplitude. The highest it reaches is 0.9. It is very close to the pattern of the first curve with a slight shift to the right.

Since $\cos\left({t}^{2}\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{4n}}{\left(2n\right)\text{!}}$ and $\sin\left({t}^{2}\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{t}^{4n+2}}{\left(2n+1\right)\text{!}}$ , one has $S\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{4n+3}}{\left(4n+3\right)\left(2n+1\right)\text{!}}$ and $C\left(x\right)=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{4n+1}}{\left(4n+1\right)\left(2n\right)\text{!}}$ . The sums of the first $50$ nonzero terms are plotted below with ${C}_{50}\left(x\right)$ the solid curve and ${S}_{50}\left(x\right)$ 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

$\left(C\left(t\right),S\left(t\right)\right)$ . Plot the curve

$\left({C}_{50},{S}_{50}\right)$ for

$0\le t\le 2\pi$ , the coordinates of which were computed in the previous exercise.

74. Estimate ${\displaystyle\int }_{0}^{\frac{1}{4}}\sqrt{x-{x}^{2}}dx$ by approximating $\sqrt{1-x}$ using the binomial approximation $1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{2128}-\frac{7{x}^{5}}{256}$ .

Show Solution

${\displaystyle\int }_{0}^{\frac{1}{4}}\sqrt{x}\left(1-\frac{x}{2}-\frac{{x}^{2}}{8}-\frac{{x}^{3}}{16}-\frac{5{x}^{4}}{128}-\frac{7{x}^{5}}{256}\right)dx$

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

whereas ${\displaystyle\int }_{0}^{\frac{1}{4}}\sqrt{x-{x}^{2}}dx=0.076773$ .

75. [T] Use Newton’s approximation of the binomial

$\sqrt{1-{x}^{2}}$ to approximate

$\pi$ as follows. The circle centered at

$\left(\frac{1}{2},0\right)$ with radius

$\frac{1}{2}$ has upper semicircle

$y=\sqrt{x}\sqrt{1-x}$ . The sector of this circle bounded by the

$x$ -axis between

$x=0$ and

$x=\frac{1}{2}$ and by the line joining

$\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right)$ corresponds to

$\frac{1}{6}$ of the circle and has area

$\frac{\pi }{24}$ . This sector is the union of a right triangle with height

$\frac{\sqrt{3}}{4}$ and base

$\frac{1}{4}$ and the region below the graph between

$x=0$ and

$x=\frac{1}{4}$ . To find the area of this region you can write

$y=\sqrt{x}\sqrt{1-x}=\sqrt{x}\times \left(\text{binomial expansion of}\sqrt{1-x}\right)$ and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate

$\pi$ .

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

Show Solution

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

$T\approx 2\pi \sqrt{\frac{10}{9.8}\approx 6.347}$ . The relative error is around

$2$ percent.

77. Suppose that a pendulum is to have a period of

$2$ seconds and a maximum angle of

${\theta }_{\text{max}}=\frac{\pi }{6}$ . Use

$T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate

$T\approx 2\pi \sqrt{\frac{L}{g}}?$

78. Evaluate ${\displaystyle\int }_{0}^{\frac{\pi}{2}}{\sin}^{4}\theta d\theta$ in the approximation $T=4\sqrt{\frac{L}{g}}{\displaystyle\int }_{0}^{\frac{\pi}{2}}\left(1+\frac{1}{2}{k}^{2}{\sin}^{2}\theta +\frac{3}{8}{k}^{4}{\sin}^{4}\theta +\cdots\right)d\theta$ to obtain an improved estimate for $T$ .

Show Solution

${\displaystyle\int }_{0}^{\frac{\pi}{2}}{\sin}^{4}\theta d\theta =\frac{3\pi }{16}$ . Hence

$T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}+\frac{9}{256}{k}^{4}\right)$ .

79. [T] An equivalent formula for the period of a pendulum with amplitude

${\theta }_{\text{max}}$ is

$T\left({\theta }_{\text{max}}\right)=2\sqrt{2}\sqrt{\frac{L}{g}}{\displaystyle\int }_{0}^{{\theta }_{\text{max}}}\frac{d\theta }{\sqrt{\cos\theta }-\cos\left({\theta }_{\text{max}}\right)}$ where

$L$ is the pendulum length and

$g$ is the gravitational acceleration constant. When

${\theta }_{\text{max}}=\frac{\pi }{3}$ we get

$\frac{1}{\sqrt{\cos{t} - \frac{1}{2}}}\approx \sqrt{2}\left(1+\frac{{t}^{2}}{2}+\frac{{t}^{4}}{3}+\frac{181{t}^{6}}{720}\right)$ . Integrate this approximation to estimate

$T\left(\frac{\pi }{3}\right)$ in terms of

$L$ and

$g$ . Assuming

$g=9.806$ meters per second squared, find an approximate length

$L$ such that

$T\left(\frac{\pi }{3}\right)=2$ seconds.

Power Series: Problem Sets

Problem Set: Working with Taylor Series

Candela Citations