10.3E: Exercises for Section 10.3

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Gilbert Strang & Edwin “Jed” Herman
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Taylor Polynomials

In exercises 1 - 8, find the Taylor polynomials of degree two approximating the given function centered at the given point.

1) $ f(x)=1+x+x^2$ at $ a=1$

2) $ f(x)=1+x+x^2$ at $ a=−1$

Answer: $ f(−1)=1;\;f′(−1)=−1;\;f''(−1)=2;\quad p_2(x)=1−(x+1)+(x+1)^2$

3) $ f(x)=\cos(2x)$ at $ a=π$

4) $ f(x)=\sin(2x)$ at $ a=\frac{π}{2}$

Answer: $ f′(x)=2\cos(2x);\;f''(x)=−4\sin(2x);\quad p_2(x)=−2(x−\frac{π}{2})$

5) $ f(x)=\sqrt{x}$ at $ a=4$

6) $ f(x)=\ln x$ at $ a=1$

Answer: $ f′(x)=\dfrac{1}{x};\; f''(x)=−\dfrac{1}{x^2};\quad p_2(x)=0+(x−1)−\frac{1}{2}(x−1)^2$

7) $ f(x)=\dfrac{1}{x}$ at $ a=1$

8) $ f(x)=e^x$ at $ a=1$

Answer: $ p_2(x)=e+e(x−1)+\dfrac{e}{2}(x−1)^2$

Taylor Remainder Theorem

In exercises 9 - 14, verify that the given choice of $n$ in the remainder estimate $ |R_n|≤\dfrac{M}{(n+1)!}(x−a)^{n+1}$, where $M$ is the maximum value of $ ∣f^{(n+1)}(z)∣$ on the interval between $a$ and the indicated point, yields $ |R_n|≤\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$

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

Answer: $ \dfrac{d^2}{dx^2}x^{1/3}=−\dfrac{2}{9x^{5/3}}≥−0.00092…$ when $ x≥28$ so the remainder estimate applies to the linear approximation $ x^{1/3}≈p_1(27)=3+\dfrac{x−27}{27}$, which gives $ (28)^{1/3}≈3+\frac{1}{27}=3.\bar{037}$, while $ (28)^{1/3}≈3.03658.$

11) [T] $ \sin(6);\; a=2π,\; n=5$

12) [T] $ e^2; \; a=0,\; n=9$

Answer: Using the estimate $ \dfrac{2^{10}}{10!}<0.000283$ we can use the Taylor expansion of order 9 to estimate $ e^x$ at $ x=2$. as $ e^2≈p_9(2)=1+2+\frac{2^2}{2}+\frac{2^3}{6}+⋯+\frac{2^9}{9!}=7.3887$… whereas $ e^2≈7.3891.$

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

14) [T] $ \ln(2);\; a=1,\; n=1000$

Answer: Since $ \dfrac{d^n}{dx^n}(\ln x)=(−1)^{n−1}\dfrac{(n−1)!}{x^n},R_{1000}≈\frac{1}{1001}$. One has $\displaystyle p_{1000}(1)=\sum_{n=1}^{1000}\dfrac{(−1)^{n−1}}{n}≈0.6936$ whereas $ \ln(2)≈0.6931⋯.$

Approximating Definite Integrals Using Taylor Series

15) Integrate the approximation $\sin t≈t−\dfrac{t^3}{6}+\dfrac{t^5}{120}−\dfrac{t^7}{5040}$ evaluated at $ π$t to approximate $\displaystyle ∫^1_0\frac{\sin πt}{πt}\,dt$.

16) Integrate the approximation $ e^x≈1+x+\dfrac{x^2}{2}+⋯+\dfrac{x^6}{720}$ evaluated at $ −x^2$ to approximate $\displaystyle ∫^1_0e^{−x^2}\,dx.$

Answer: $\displaystyle ∫^1_0\left(1−x^2+\frac{x^4}{2}−\frac{x^6}{6}+\frac{x^8}{24}−\frac{x^{10}}{120}+\frac{x^{12}}{720}\right)\,dx =1−\frac{1^3}{3}+\frac{1^5}{10}−\frac{1^7}{42}+\frac{1^9}{9⋅24}−\frac{1^{11}}{120⋅11}+\frac{1^{13}}{720⋅13}≈0.74683$ whereas $\displaystyle ∫^1_0e^{−x^2}dx≈0.74682.$

Taylor Series

In exercises 25 - 35, find the Taylor series of the given function centered at the indicated point.

25) $f(x) = x^4$ at $ a=−1$

26) $f(x) = 1+x+x^2+x^3$ at $ a=−1$

Answer: $ (x+1)^3−2(x+1)^2+2(x+1)$

27) $f(x) = \sin x$ at $ a=π$

28) $f(x) = \cos x$ at $ a=2π$

Answer: Values of derivatives are the same as for $ x=0$ so $\displaystyle \cos x=\sum_{n=0}^∞(−1)^n\frac{(x−2π)^{2n}}{(2n)!}$

29) $f(x) = \sin x$ at $ x=\frac{π}{2}$

30) $f(x) = \cos x$ at $ x=\frac{π}{2}$

Answer: $ \cos(\frac{π}{2})=0,\;−\sin(\frac{π}{2})=−1$ so $\displaystyle \cos x=\sum_{n=0}^∞(−1)^{n+1}\frac{(x−\frac{π}{2})^{2n+1}}{(2n+1)!}$, which is also $ −\cos(x−\frac{π}{2})$.

31) $f(x) = e^x$ at $ a=−1$

32) $f(x) = e^x$ at $ a=1$

Answer: The derivatives are $ f^{(n)}(1)=e,$ so $\displaystyle e^x=e\sum_{n=0}^∞\frac{(x−1)^n}{n!}.$

33) $f(x) = \dfrac{1}{(x−1)^2}$ at $ a=0$ (Hint: Differentiate the Taylor Series for$ \dfrac{1}{1−x}$.)

34) $f(x) = \dfrac{1}{(x−1)^3}$ at $ a=0$

Answer: $\displaystyle \frac{1}{(x−1)^3}=−\frac{1}{2}\frac{d^2}{dx^2}\left(\frac{1}{1−x}\right)=−\sum_{n=0}^∞\left(\frac{(n+2)(n+1)x^n}{2}\right)$

35) $\displaystyle F(x)=∫^x_0\cos(\sqrt{t})\,dt;\quad \text{where}\; f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{(2n)!}$ at a=0 (Note: $ f$ is the Taylor series of $\cos(\sqrt{t}).)$

In exercises 36 - 44, compute the Taylor series of each function around $ x=1$.

36) $ f(x)=2−x$

Answer: $ 2−x=1−(x−1)$

37) $ f(x)=x^3$

38) $ f(x)=(x−2)^2$

Answer: $ ((x−1)−1)^2=(x−1)^2−2(x−1)+1$

39) $ f(x)=\ln x$

40) $ f(x)=\dfrac{1}{x}$

Answer: $\displaystyle \frac{1}{1−(1−x)}=\sum_{n=0}^∞(−1)^n(x−1)^n$

41) $ f(x)=\dfrac{1}{2x−x^2}$

42) $ f(x)=\dfrac{x}{4x−2x^2−1}$

Answer: $\displaystyle x\sum_{n=0}^∞2^n(1−x)^{2n}=\sum_{n=0}^∞2^n(x−1)^{2n+1}+\sum_{n=0}^∞2^n(x−1)^{2n}$

43) $ f(x)=e^{−x}$

44) $ f(x)=e^{2x}$

Answer: $\displaystyle e^{2x}=e^{2(x−1)+2}=e^2\sum_{n=0}^∞\frac{2^n(x−1)^n}{n!}$

Maclaurin Series

[T] In exercises 45 - 48, identify the value of $x$ such that the given series $\displaystyle \sum_{n=0}^∞a_n$ is the value of the Maclaurin series of $ f(x)$ at $ x$. Approximate the value of $ f(x)$ using $\displaystyle S_{10}=\sum_{n=0}^{10}a_n$.

45) $\displaystyle \sum_{n=0}^∞\frac{1}{n!}$

46) $\displaystyle \sum_{n=0}^∞\frac{2^n}{n!}$

Answer: $ x=e^2;\quad S_{10}=\dfrac{34,913}{4725}≈7.3889947$

47) $\displaystyle \sum_{n=0}^∞\frac{(−1)^n(2π)^{2n}}{(2n)!}$

48) $\displaystyle \sum_{n=0}^∞\frac{(−1)^n(2π)^{2n+1}}{(2n+1)!}$

Answer: $\sin(2π)=0;\quad S_{10}=8.27×10^{−5}$

In exercises 49 - 52 use the functions $ S_5(x)=x−\dfrac{x^3}{6}+\dfrac{x^5}{120}$ and $ C_4(x)=1−\dfrac{x^2}{2}+\dfrac{x^4}{24}$ on $ [−π,π]$.

49) [T] Plot $\sin^2x−(S_5(x))^2$ on $ [−π,π]$. Compare the maximum difference with the square of the Taylor remainder estimate for $ \sin x.$

50) [T] Plot $\cos^2x−(C_4(x))^2$ on $ [−π,π]$. Compare the maximum difference with the square of the Taylor remainder estimate for $ \cos x$.

Answer

The difference is small on the interior of the interval but approaches $ 1$ near the endpoints. The remainder estimate is $ |R_4|=\frac{π^5}{120}≈2.552.$

$CNX_Calc_Figure_10_03_206.jpeg$

51) [T] Plot $ |2S_5(x)C_4(x)−\sin(2x)|$ on $ [−π,π]$.

52) [T] Compare $ \dfrac{S_5(x)}{C_4(x)}$ on $ [−1,1]$ to $ \tan x$. Compare this with the Taylor remainder estimate for the approximation of $ \tan x$ by $ x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}$.

Answer

The difference is on the order of $ 10^{−4}$ on $ [−1,1]$ while the Taylor approximation error is around $ 0.1$ near $ ±1$. The top curve is a plot of $\tan^2x−\left(\dfrac{S_5(x)}{C_4(x)}\right)^2$ and the lower dashed plot shows $ t^2−\left(\dfrac{S_5}{C_4}\right)^2$.

$CNX_Calc_Figure_10_03_208.jpeg$

53) [T] Plot $ e^x−e_4(x)$ where $ e_4(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}$ on $ [0,2]$. 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−\dfrac{f(x_n)}{f'(x_n)}$ approximates solutions of $ f(x)=0$ near the input $ x_0$.

a. If $ f$ and $ g$ are inverse functions, explain why a solution of $ g(x)=a$ is the value $ f(a)$ of $ f$.

b. Let $ p_N(x)$ be the $ N^{\text{th}}$ degree Maclaurin polynomial of $ e^x$. Use Newton’s method to approximate solutions of $ p_N(x)−2=0$ for $ N=4,5,6.$

c. Explain why the approximate roots of $ p_N(x)−2=0$ are approximate values of $\ln(2).$

Answer: 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(x)−2=0$: for $ N=4,x=0.6939...;$ for $ N=5,x=0.6932...;$ for $ N=6,x=0.69315...;.$ (Note: $ \ln(2)=0.69314...$)
c. Answers will vary.

Evaluating Limits using Taylor Series

In exercises 55 - 58, use the fact that if $\displaystyle q(x)=\sum_{n=1}^∞a_n(x−c)^n$ converges in an interval containing $ c$, then $\displaystyle \lim_{x→c}q(x)=a_0$ to evaluate each limit using Taylor series.

55) $\displaystyle \lim_{x→0}\frac{\cos x−1}{x^2}$

56) $\displaystyle \lim_{x→0}\frac{\ln(1−x^2)}{x^2}$

Answer: $ \dfrac{\ln(1−x^2)}{x^2}→−1$

57) $\displaystyle \lim_{x→0}\frac{e^{x^2}−x^2−1}{x^4}$

58) $\displaystyle \lim_{x→0^+}\frac{\cos(\sqrt{x})−1}{2x}$

Answer: $\displaystyle \frac{\cos(\sqrt{x})−1}{2x}≈\frac{(1−\frac{x}{2}+\frac{x^2}{4!}−⋯)−1}{2x}→−\frac{1}{4}$