Problem Set: Taylor and Maclaurin Series

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

f (x) = 1 + x + x^{2}

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

a = 1

$a=1$

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

Show Solution

f (- 1) = 1; f^{'} (- 1) = - 1; f'' (- 1) = 2; f (x) = 1 - (x + 1) + {(x + 1)}^{2}

$f\left(-1\right)=1;{f}^{\prime }\left(-1\right)=-1;f\text{''}\left(-1\right)=2;f\left(x\right)=1-\left(x+1\right)+{\left(x+1\right)}^{2}$

f (x) = cos (2 x)

$f\left(x\right)=\cos\left(2x\right)$ at

a = π

$a=\pi$

4. $f\left(x\right)=\sin\left(2x\right)$ at $a=\frac{\pi }{2}$

Show Solution

f^{'} (x) = 2 cos (2 x); f'' (x) = - 4 sin (2 x); p_{2} (x) = - 2 (x - \frac{π}{2})

${f}^{\prime }\left(x\right)=2\cos\left(2x\right);f\text{''}\left(x\right)=-4\sin\left(2x\right);{p}_{2}\left(x\right)=-2\left(x-\frac{\pi }{2}\right)$

f (x) = \sqrt{x}

$f\left(x\right)=\sqrt{x}$ at

a = 4

$a=4$

6. $f\left(x\right)=\text{ln}x$ at $a=1$

Show Solution

f^{'} (x) = \frac{1}{x}; f'' (x) = - \frac{1}{x^{2}}; p_{2} (x) = 0 + (x - 1) - \frac{1}{2} {(x - 1)}^{2}

${f}^{\prime }\left(x\right)=\frac{1}{x};f\text{''}\left(x\right)=-\frac{1}{{x}^{2}};{p}_{2}\left(x\right)=0+\left(x - 1\right)-\frac{1}{2}{\left(x - 1\right)}^{2}$

f (x) = \frac{1}{x}

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

a = 1

$a=1$

8. $f\left(x\right)={e}^{x}$ at $a=1$

Show Solution

p_{2} (x) = e + e (x - 1) + \frac{e}{2} {(x - 1)}^{2}

${p}_{2}\left(x\right)=e+e\left(x - 1\right)+\frac{e}{2}{\left(x - 1\right)}^{2}$

In the following exercises, verify that the given choice of n in the remainder estimate $|{R}_{n}|\le \frac{M}{\left(n+1\right)\text{!}}{\left(x-a\right)}^{n+1}$ , where M is the maximum value of $|{f}^{\left(n+1\right)}\left(z\right)|$ on the interval between a and the indicated point, yields $|{R}_{n}|\le \frac{1}{1000}$ . Find the value of the Taylor polynomial p_n of $f$ at the indicated point.

9. [T]

\sqrt{10}; a = 9, n = 3

$\sqrt{10};a=9,n=3$

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

Show Solution

\frac{d^{2}}{d x^{2}} x^{\frac{1}{3}} = - \frac{2}{9 x^{\frac{5}{3}}} \geq - 0.00092 \dots

$\frac{{d}^{2}}{d{x}^{2}}{x}^{\frac{1}{3}}=-\frac{2}{9{x}^{\frac{5}{3}}}\ge -0.00092\ldots$ when

x \geq 28

$x\ge 28$ so the remainder estimate applies to the linear approximation

x^{\frac{1}{3}} \approx p_{1} (27) = 3 + \frac{x - 27}{27}

${x}^{\frac{1}{3}}\approx {p}_{1}\left(27\right)=3+\frac{x - 27}{27}$ , which gives

{(28)}^{\frac{1}{3}} \approx 3 + \frac{1}{27} = 3. ¯ ¯¯¯¯¯¯ ¯ 037

${\left(28\right)}^{\frac{1}{3}}\approx 3+\frac{1}{27}=3.\overline{037}$ , while

{(28)}^{\frac{1}{3}} \approx 3.03658

${\left(28\right)}^{\frac{1}{3}}\approx 3.03658$ .

11. [T]

sin (6); a = 2 π, n = 5

$\sin\left(6\right);a=2\pi ,n=5$

12. [T] e²; $a=0,n=9$

Show Solution

Using the estimate

\frac{2^{10}}{10!} < 0.000283

$\frac{{2}^{10}}{10\text{!}}<0.000283$ we can use the Taylor expansion of order 9 to estimate

e^{x}

$e^{x}$ at

x = 2

$x=2$ . as

e^{2} \approx p_{9} (2) = 1 + 2 + \frac{2^{2}}{2} + \frac{2^{3}}{6} + \dots + \frac{2^{9}}{9!} = 7.3887 \dots

${e}^{2}\approx {p}_{9}\left(2\right)=1+2+\frac{{2}^{2}}{2}+\frac{{2}^{3}}{6}+\cdots+\frac{{2}^{9}}{9\text{!}}=7.3887\ldots$ whereas

e^{2} \approx 7.3891

${e}^{2}\approx 7.3891$ .

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

14. [T] $\text{ln}\left(2\right);a=1,n=1000$

Show Solution

Since

\frac{d^{n}}{d x^{n}} (ln x) = {(- 1)}^{n - 1} \frac{(n - 1)!}{x^{n}}, R_{1000} \approx \frac{1}{1001}

$\frac{{d}^{n}}{d{x}^{n}}\left(\text{ln}x\right)={\left(-1\right)}^{n - 1}\frac{\left(n - 1\right)\text{!}}{{x}^{n}},{R}_{1000}\approx \frac{1}{1001}$ . One has

p_{1000} (1) = 1000 \sum n = 1 \frac{{(- 1)}^{n - 1}}{n} \approx 0.6936

${p}_{1000}\left(1\right)=\displaystyle\sum _{n=1}^{1000}\frac{{\left(-1\right)}^{n - 1}}{n}\approx 0.6936$ whereas

ln (2) \approx 0.6931 \dots

$\text{ln}\left(2\right)\approx 0.6931\cdots$ .

15. Integrate the approximation

sin t \approx t - \frac{t^{3}}{6} + \frac{t^{5}}{120} - \frac{t^{7}}{5040}

$\sin{t}\approx t-\frac{{t}^{3}}{6}+\frac{{t}^{5}}{120}-\frac{{t}^{7}}{5040}$ evaluated at πt to approximate

\int_{0}^{1} \frac{sin π t}{π t} d t

${\displaystyle\int }_{0}^{1}\frac{\sin\pi t}{\pi t}dt$ .

16. Integrate the approximation ${e}^{x}\approx 1+x+\frac{{x}^{2}}{2}+\cdots+\frac{{x}^{6}}{720}$ evaluated at −x² to approximate ${\displaystyle\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx$ .

Show Solution

\int_{0}^{1} (1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \frac{x^{8}}{24} - \frac{x^{10}}{120} + \frac{x^{12}}{720}) d x

${\displaystyle\int }_{0}^{1}\left(1-{x}^{2}+\frac{{x}^{4}}{2}-\frac{{x}^{6}}{6}+\frac{{x}^{8}}{24}-\frac{{x}^{10}}{120}+\frac{{x}^{12}}{720}\right)dx$

$=1-\frac{{1}^{3}}{3}+\frac{{1}^{5}}{10}-\frac{{1}^{7}}{42}+\frac{{1}^{9}}{9\cdot 24}-\frac{{1}^{11}}{120\cdot 11}+\frac{{1}^{13}}{720\cdot 13}\approx 0.74683$ whereas ${\displaystyle\int }_{0}^{1}{e}^{\text{-}{x}^{2}}dx\approx 0.74682$ .

In the following exercises, find the smallest value of n such that the remainder estimate $|{R}_{n}|\le \frac{M}{\left(n+1\right)\text{!}}{\left(x-a\right)}^{n+1}$ , where M is the maximum value of $|{f}^{\left(n+1\right)}\left(z\right)|$ on the interval between a and the indicated point, yields $|{R}_{n}|\le \frac{1}{1000}$ on the indicated interval.

17.

f (x) = sin x

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

[- π, π], a = 0

$\left[\text{-}\pi ,\pi \right],a=0$

18. $f\left(x\right)=\cos{x}$ on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right],a=0$

Show Solution

Since

f^{(n + 1)} (z)

${f}^{\left(n+1\right)}\left(z\right)$ is

sin z

$\sin{z}$ or

cos z

$\cos{z}$ , we have

M = 1

$M=1$ . Since

| x - 0 | \leq \frac{π}{2}

$|x - 0|\le \frac{\pi }{2}$ , we seek the smallest n such that

\frac{π^{n + 1}}{2^{n + 1} (n + 1)!} \leq 0.001

$\frac{{\pi }^{n+1}}{{2}^{n+1}\left(n+1\right)\text{!}}\le 0.001$ . The smallest such value is

n = 7

$n=7$ . The remainder estimate is

R_{7} \leq 0.00092

${R}_{7}\le 0.00092$ .

19.

f (x) = e^{- 2 x}

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

[- 1, 1], a = 0

$\left[-1,1\right],a=0$

20. $f\left(x\right)={e}^{\text{-}x}$ on $\left[-3,3\right],a=0$

Show Solution

Since

f^{(n + 1)} (z) = \pm e^{- z}

${f}^{\left(n+1\right)}\left(z\right)=\pm{e}^{\text{-}z}$ one has

M = e^{3}

$M={e}^{3}$ . Since

| x - 0 | \leq 3

$|x - 0|\le 3$ , one seeks the smallest n such that

\frac{3^{n + 1} e^{3}}{(n + 1)!} \leq 0.001

$\frac{{3}^{n+1}{e}^{3}}{\left(n+1\right)\text{!}}\le 0.001$ . The smallest such value is

n = 14

$n=14$ . The remainder estimate is

R_{14} \leq 0.000220

${R}_{14}\le 0.000220$ .

In the following exercises, the maximum of the right-hand side of the remainder estimate $|{R}_{1}|\le \frac{\text{max}|f\text{''}\left(z\right)|}{2}{R}^{2}$ on $\left[a-R,a+R\right]$ occurs at a or $a\pm R$ . Estimate the maximum value of R such that $\frac{\text{max}|f\text{''}\left(z\right)|}{2}{R}^{2}\le 0.1$ on $\left[a-R,a+R\right]$ by plotting this maximum as a function of R.

21. [T] e^x approximated by

1 + x, a = 0

$1+x,a=0$

22. [T] $\sin{x}$ approximated by x, $a=0$

Show Solution

This graph has a horizontal line at y=0.2. It also has a curve starting at the origin and concave up. The curve and the line intersect at the ordered pair (0.5966, 0.2).

Since $\sin{x}$ is increasing for small x and since $\mathrm{si}n\text{''}x=\text{-}\sin{x}$ , the estimate applies whenever ${R}^{2}\sin\left(R\right)\le 0.2$ , which applies up to $R=0.596$ .

23. [T]

ln x

$\text{ln}x$ approximated by

x - 1, a = 1

$x - 1,a=1$

24. [T] $\cos{x}$ approximated by $1,a=0$

Show Solution

Since the second derivative of $\cos{x}$ is $\text{-}\cos{x}$ and since $\cos{x}$ is decreasing away from $x=0$ , the estimate applies when ${R}^{2}\cos{R}\le 0.2$ or $R\le 0.447$ .

In the following exercises, find the Taylor series of the given function centered at the indicated point.

25.

x^{4}

${x}^{4}$ at

a = - 1

$a=-1$

26. $1+x+{x}^{2}+{x}^{3}$ at $a=-1$

Show Solution

{(x + 1)}^{3} - 2 {(x + 1)}^{2} + 2 (x + 1)

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

27.

sin x

$\sin{x}$ at

a = π

$a=\pi$

28. $\cos{x}$ at $a=2\pi$

Show Solution

Values of derivatives are the same as for

x = 0

$x=0$ so

cos x = {\infty \sum n = 0 (- 1)}^{n} \frac{{(x - 2 π)}^{2 n}}{(2 n)!}

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

29.

sin x

$\sin{x}$ at

x = \frac{π}{2}

$x=\frac{\pi }{2}$

30. $\cos{x}$ at $x=\frac{\pi }{2}$

Show Solution

cos (\frac{π}{2}) = 0, - sin (\frac{π}{2}) = - 1

$\cos\left(\frac{\pi }{2}\right)=0,\text{-}\sin\left(\frac{\pi }{2}\right)=-1$ so

cos x = \infty \sum n = 0 {(- 1)}^{n + 1} \frac{{(x - \frac{π}{2})}^{2 n + 1}}{(2 n + 1)!}

$\cos{x}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n+1}\frac{{\left(x-\frac{\pi }{2}\right)}^{2n+1}}{\left(2n+1\right)\text{!}}$ , which is also

- cos (x - \frac{π}{2})

$\text{-}\cos\left(x-\frac{\pi }{2}\right)$ .

31.

e^{x}

${e}^{x}$ at

a = - 1

$a=-1$

32. ${e}^{x}$ at $a=1$

Show Solution

The derivatives are

f^{(n)} (1) = e

${f}^{\left(n\right)}\left(1\right)=e$ so

e^{x} = e \infty \sum n = 0 \frac{{(x - 1)}^{n}}{n!}

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

33.

\frac{1}{{(x - 1)}^{2}}

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

a = 0

$a=0$ (Hint: Differentiate

\frac{1}{1 - x} .

$\frac{1}{1-x}.$ )

34. $\frac{1}{{\left(x - 1\right)}^{3}}$ at $a=0$

Show Solution

\frac{1}{{(x - 1)}^{3}} = - (\frac{1}{2}) \frac{d^{2}}{d x^{2}} \frac{1}{1 - x} = - \infty \sum n = 0 (\frac{(n + 2) (n + 1) x^{n}}{2})

$\frac{1}{{\left(x - 1\right)}^{3}}=\text{-}\left(\frac{1}{2}\right)\frac{{d}^{2}}{d{x}^{2}}\frac{1}{1-x}=\text{-}\displaystyle\sum _{n=0}^{\infty }\left(\frac{\left(n+2\right)\left(n+1\right){x}^{n}}{2}\right)$

35.

F (x) = \int_{0}^{x} cos (\sqrt{t}) d t; f (t) = \infty \sum n = 0 {(- 1)}^{n} \frac{t^{n}}{(2 n)!}

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

a = 0

$a=0$ (Note:

f

$f$ is the Taylor series of

cos (\sqrt{t}) .

$\cos\left(\sqrt{t}\right).$ )

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

36. $f\left(x\right)=2-x$

Show Solution

2 - x = 1 - (x - 1)

$2-x=1-\left(x - 1\right)$

37.

f (x) = x^{3}

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

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

Show Solution

{((x - 1) - 1)}^{2} = {(x - 1)}^{2} - 2 (x - 1) + 1

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

39.

f (x) = ln x

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

40. $f\left(x\right)=\frac{1}{x}$

Show Solution

\frac{1}{1 - (1 - x)} = \infty \sum n = 0 {(- 1)}^{n} {(x - 1)}^{n}

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

41.

f (x) = \frac{1}{2 x - x^{2}}

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

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

Show Solution

x \infty \sum n = 0 2^{n} {(1 - x)}^{2 n} = \infty \sum n = 0 2^{n} {(x - 1)}^{2 n + 1} + \infty \sum n = 0 2^{n} {(x - 1)}^{2 n}

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

43.

f (x) = e^{- x}

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

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

Show Solution

e^{2 x} = e^{2 (x - 1) + 2} = e^{2} \infty \sum n = 0 \frac{2^{n} {(x - 1)}^{n}}{n!}

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

[T] In the following exercises, identify the value of x such that the given series $\displaystyle\sum _{n=0}^{\infty }{a}_{n}$ is the value of the Maclaurin series of $f\left(x\right)$ at $x$ . Approximate the value of $f\left(x\right)$ using ${S}_{10}=\displaystyle\sum _{n=0}^{10}{a}_{n}$ .

45.

\infty \sum n = 0 \frac{1}{n!}

$\displaystyle\sum _{n=0}^{\infty }\frac{1}{n\text{!}}$

46. $\displaystyle\sum _{n=0}^{\infty }\frac{{2}^{n}}{n\text{!}}$

Show Solution

$x={e}^{2};{S}_{10}=\frac{34,913}{4725}\approx 7.3889947$

47.

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

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

Show Solution

$\sin\left(2\pi \right)=0;{S}_{10}=8.27\times {10}^{-5}$

The following exercises make use of the functions ${S}_{5}\left(x\right)=x-\frac{{x}^{3}}{6}+\frac{{x}^{5}}{120}$ and ${C}_{4}\left(x\right)=1-\frac{{x}^{2}}{2}+\frac{{x}^{4}}{24}$ on $\left[\text{-}\pi ,\pi \right]$ .

49. [T] Plot

${\sin}^{2}x-{\left({S}_{5}\left(x\right)\right)}^{2}$ on

$\left[\text{-}\pi ,\pi \right]$ . Compare the maximum difference with the square of the Taylor remainder estimate for

$\sin{x}$ .

50. [T] Plot ${\cos}^{2}x-{\left({C}_{4}\left(x\right)\right)}^{2}$ on $\left[\text{-}\pi ,\pi \right]$ . Compare the maximum difference with the square of the Taylor remainder estimate for $\cos{x}$ .

Show Solution

This graph has a concave up curve that is symmetrical about the y axis. The lowest point of the graph is the origin with the rest of the curve above the x-axis.

The difference is small on the interior of the interval but approaches $1$ near the endpoints. The remainder estimate is $|{R}_{4}|=\frac{{\pi }^{5}}{120}\approx 2.552$ .

51. [T] Plot

$|2{S}_{5}\left(x\right){C}_{4}\left(x\right)-\sin\left(2x\right)|$ on

$\left[\text{-}\pi ,\pi \right]$ .

52. [T] Compare $\frac{{S}_{5}\left(x\right)}{{C}_{4}\left(x\right)}$ on $\left[-1,1\right]$ to $\tan{x}$ . Compare this with the Taylor remainder estimate for the approximation of $\tan{x}$ by $x+\frac{{x}^{3}}{3}+\frac{2{x}^{5}}{15}$ .

Show Solution

This graph has two curves. The solid curve is very flat and close to the x-axis. It passes through the origin. The second curve, a broken line, is concave down and symmetrical about the y-axis. It is very close to the x-axis between -3 and 3.

The difference is on the order of ${10}^{-4}$ on $\left[-1,1\right]$ while the Taylor approximation error is around $0.1$ near $\pm 1$ . The top curve is a plot of ${\tan}^{2}x-{\left(\frac{{S}_{5}\left(x\right)}{{C}_{4}\left(x\right)}\right)}^{2}$ and the lower dashed plot shows ${t}^{2}-{\left(\frac{{S}_{5}}{{C}_{4}}\right)}^{2}$ .

53. [T] Plot

${e}^{x}-{e}_{4}\left(x\right)$ where

${e}_{4}\left(x\right)=1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}+\frac{{x}^{4}}{24}$ on

$\left[0,2\right]$ . Compare the maximum error with the Taylor remainder estimate.

54. (Taylor approximations and root finding.) Recall that Newton’s method ${x}_{n+1}={x}_{n}-\frac{f\left({x}_{n}\right)}{f\prime \left({x}_{n}\right)}$ approximates solutions of $f\left(x\right)=0$ near the input ${x}_{0}$ .

If $f$ and $g$ are inverse functions, explain why a solution of $g\left(x\right)=a$ is the value $f\left(a\right)\text{of}f$ .
Let ${p}_{N}\left(x\right)$ be the $N\text{th}$ degree Maclaurin polynomial of ${e}^{x}$ . Use Newton’s method to approximate solutions of ${p}_{N}\left(x\right)-2=0$ for $N=4,5,6$ .
Explain why the approximate roots of ${p}_{N}\left(x\right)-2=0$ are approximate values of $\text{ln}\left(2\right)$ .

Show Solution

a. Answers will vary. b. The following are the

${x}_{n}$ values after

$10$ iterations of Newton’s method to approximation a root of

${p}_{N}\left(x\right)-2=0\text{:}$ for

$N=4,x=0.6939...$ ; for

$N=5,x=0.6932...$ ; for

$N=6,x=0.69315...$ ;. (Note:

$\text{ln}\left(2\right)=0.69314...$ ) c. Answers will vary.

In the following exercises, use the fact that if $q\left(x\right)=\displaystyle\sum _{n=1}^{\infty }{a}_{n}{\left(x-c\right)}^{n}$ converges in an interval containing $c$ , then $\underset{x\to c}{\text{lim}}q\left(x\right)={a}_{0}^{}$ to evaluate each limit using Taylor series.

55.

$\underset{x\to 0}{\text{lim}}\frac{\cos{x} - 1}{{x}^{2}}$

56. $\underset{x\to 0}{\text{lim}}\frac{\text{ln}\left(1-{x}^{2}\right)}{{x}^{2}}$

Show Solution

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

57.

$\underset{x\to 0}{\text{lim}}\frac{{e}^{{x}^{2}}-{x}^{2}-1}{{x}^{4}}$

58. $\underset{x\to {0}^{+}}{\text{lim}}\frac{\cos\left(\sqrt{x}\right)-1}{2x}$

Show Solution

$\frac{\cos\left(\sqrt{x}\right)-1}{2x}\approx \frac{\left(1-\frac{x}{2}+\frac{{x}^{2}}{4\text{!}}-\cdots\right)-1}{2x}\to -\frac{1}{4}$

Power Series: Problem Sets

Problem Set: Taylor and Maclaurin Series

Candela Citations