1.3E: Exercises

Exercise \(\PageIndex{1}\)

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

1. \(\displaystyle f(x)=1+x+x^2\) at \(\displaystyle a=1\)

2. \(\displaystyle f(x)=1+x+x^2\) at \(\displaystyle a=−1\)

Answer

\(\displaystyle f(−1)=1;f′(−1)=−1;f''(−1)=2;f(x)=1−(x+1)+(x+1)^2\)

3. \(\displaystyle f(x)=cos(2x)\) at \(\displaystyle a=π\)

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

Answer

\(\displaystyle f′(x)=2cos(2x);f''(x)=−4sin(2x);p_2(x)=−2(x−\frac{π}{2})\)

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

6. \(\displaystyle f(x)=lnx\) at \(\displaystyle a=1\)

Answer

\(\displaystyle f′(x)=\frac{1}{x};f''(x)=−\frac{1}{x^2};p_2(x)=0+(x−1)−\frac{1}{2}(x−1)^2\)

7. \(\displaystyle f(x)=\frac{1}{x}\) at \(\displaystyle a=1\)

8. \(\displaystyle f(x)=e^x\) at \(\displaystyle a=1\)

Answer

\(\displaystyle p_2(x)=e+e(x−1)+\frac{e}{2}(x−1)^2\)

Exercise \(\PageIndex{2}\)

In the following exercises, verify that the given choice of n in the remainder estimate \(\displaystyle |R_n|≤\frac{M}{(n+1)!}(x−a)^{n+1}\), where M is the maximum value of \(\displaystyle ∣f^{(n+1)}(z)∣\) on the interval between a and the indicated point, yields \(\displaystyle |R_n|≤\frac{1}{1000}\). Find the value of the Taylor polynomial \(\displaystyle p_n\) of \(\displaystyle f\) at the indicated point.

1. (\displaystyle \sqrt{10};a=9,n=3\)

2. \(\displaystyle (28)^{1/3};a=27,n=1\)

Answer

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

3. \(\displaystyle sin(6);a=2π,n=5\)

4. \(\displaystyle e^2; a=0,n=9\)

Answer

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

5. \(\displaystyle cos(\frac{π}{5});a=0,n=4\)

6. \(\displaystyle ln(2);a=1,n=1000\)

Answer

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

Exercise \(\PageIndex{3}\)

1. Integrate the approximation \(\displaystyle sint≈t−\frac{t^3}{6}+\frac{t^5}{120}−\frac{t^7}{5040}\) evaluated at \(\displaystyle π\)t to approximate \(\displaystyle ∫^1_0\frac{sinπt}{πt}dt\).

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

Answer

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

Exercise \(\PageIndex{4}\)

In the following exercises, find the smallest value of n such that the remainder estimate \(\displaystyle |R_n|≤\frac{M}{(n+1)!}(x−a)^{n+1}\), where M is the maximum value of \(\displaystyle ∣f^{(n+1)}(z)∣\) on the interval between a and the indicated point, yields \(\displaystyle |R_n|≤\frac{1}{1000}\) on the indicated interval.

1. \(\displaystyle f(x)=sinx\) on \(\displaystyle [−π,π],a=0\)

2. \(\displaystyle f(x)=cosx\) on \(\displaystyle [−\frac{π}{2},\frac{π}{2}],a=0\)

Answer

Since \(\displaystyle f^{(n+1)}(z)\) is \(\displaystyle sinz\) or \(\displaystyle cosz\), we have \(\displaystyle M=1\). Since \(\displaystyle |x−0|≤\frac{π}{2}\), we seek the smallest n such that \(\displaystyle \frac{π^{n+1}}{2^{n+1}(n+1)!}≤0.001\). The smallest such value is \(\displaystyle n=7\). The remainder estimate is \(\displaystyle R_7≤0.00092.\)

3. \(\displaystyle f(x)=e^{−2x}\) on \(\displaystyle [−1,1],a=0\)

4. \(\displaystyle f(x)=e^{−x}\) on \(\displaystyle [−3,3],a=0\)

Answer

Since \(\displaystyle f^{(n+1)}(z)=±e^{−z}\) one has \(\displaystyle M=e^3\). Since \(\displaystyle |x−0|≤3\), one seeks the smallest n such that \(\displaystyle \frac{3^{n+1}e^3}{(n+1)!}≤0.001\). The smallest such value is \(\displaystyle n=14\). The remainder estimate is \(\displaystyle R_{14}≤0.000220.\)

Exercise \(\PageIndex{5}\)

In the following exercises, the maximum of the right-hand side of the remainder estimate \(\displaystyle |R_1|≤\frac{max|f''(z)|}{2}R^2\) on \(\displaystyle [a−R,a+R]\) occurs at a or \(\displaystyle a±R\). Estimate the maximum value of R such that \(\displaystyle \frac{max|f''(z)|}{2}R^2≤0.1\) on \(\displaystyle [a−R,a+R]\) by plotting this maximum as a function of I.

1. \(\displaystyle e^x\) approximated by \(\displaystyle 1+x,a=0\)

2. \(\displaystyle sinx\) approximated by \(\displaystyle x, a=0\)

Answer

Since \(\displaystyle sinx\) is increasing for small \(\displaystyle x\) and since \(\displaystyle sin''x=−sinx\), the estimate applies whenever \(\displaystyle R^2sin(R)≤0.2\), which applies up to \(\displaystyle R=0.596.\)

3. \(\displaystyle lnx\) approximated by \(\displaystyle x−1,a=1\)

4. \(\displaystyle cosx\) approximated by \(\displaystyle 1,a=0\)

Answer

Since the second derivative of \(\displaystyle cosx\) is \(\displaystyle −cosx\) and since \(\displaystyle cosx\) is decreasing away from \(\displaystyle x=0\), the estimate applies when \(\displaystyle R^2cosR≤0.2\) or \(\displaystyle R≤0.447\).

Exercise \(\PageIndex{6}\)

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

1. \(\displaystyle x^4\) at \(\displaystyle a=−1\)

2. \(\displaystyle 1+x+x^2+x^3\) at \(\displaystyle a=−1\)

Answer

\(\displaystyle (x+1)^3−2(x+1)^2+2(x+1)\)

3. \(\displaystyle sinx\) at \(\displaystyle a=π\)

4. \(\displaystyle cosx\) at \(\displaystyle a=2π\)

Answer

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

5. \(\displaystyle sinx\) at \(\displaystyle x=\frac{π}{2}\)

6. \(\displaystyle cosx\) at \(\displaystyle x=\frac{π}{2}\)

Answer

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

7. \(\displaystyle e^x\) at \(\displaystyle a=−1\)

8. \(\displaystyle e^x\) at \(\displaystyle a=1\)

Answer

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

9. \(\displaystyle \frac{1}{(x−1)^2}\) at \(\displaystyle a=0\) (Hint: Differentiate \(\displaystyle \frac{1}{1−x}\).)

10. \(\displaystyle \frac{1}{(x−1)^3}\) at \(\displaystyle a=0\)

Answer

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

11. \(\displaystyle F(x)=∫^x_0cos(\sqrt{t})dt;f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{(2n)!}\) at a=0 (Note: \(\displaystyle f\) is the Taylor series of \(\displaystyle cos(\sqrt{t}).)\)

Exercise \(\PageIndex{7}\)

In the following exercises, compute the Taylor series of each function around \(\displaystyle x=1\).

1. \(\displaystyle f(x)=2−x\)

Answer

\(\displaystyle 2−x=1−(x−1)\)

2. \(\displaystyle f(x)=x^3\)

3. \(\displaystyle f(x)=(x−2)^2\)

Answer

\(\displaystyle ((x−1)−1)^2=(x−1)^2−2(x−1)+1\)

4. \(\displaystyle f(x)=lnx\)

5. \(\displaystyle f(x)=\frac{1}{x}\)

Answer

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

6. \(\displaystyle f(x)=\frac{1}{2x−x^2}\)

7. \(\displaystyle f(x)=\frac{x}{4x−2x^2−1}\)

Answer

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

8. \(\displaystyle f(x)=e^{−x}\)

9. \(\displaystyle f(x)=e^{2x}\)

Answer

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

Exercise \(\PageIndex{8}\)

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

1. \(\displaystyle \sum_{n=0}^∞\frac{1}{n!}\)

2. \(\displaystyle sum_{n=0}^∞\frac{2^n}{n!}\)

Answer

\(\displaystyle x=e^2;S_{10}=\frac{34,913}{4725}≈7.3889947\)

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

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

Answer

\(\displaystyle sin(2π)=0;S_{10}=8.27×10^{−5}\)

Exercise \(\PageIndex{9}\)

The following exercises make use of the functions \(\displaystyle S_5(x)=x−\frac{x^3}{6}+\frac{x^5}{120}\) and \(\displaystyle C_4(x)=1−\frac{x^2}{2}+\frac{x^4}{24}\) on \(\displaystyle [−π,π]\).

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

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

Answer

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

3. Plot \(\displaystyle |2S_5(x)C_4(x)−sin(2x)|\) on \(\displaystyle [−π,π]\).

4. Compare \(\displaystyle \frac{S_5(x)}{C_4(x)}\) on \(\displaystyle [−1,1]\) to \(\displaystyle tanx\). Compare this with the Taylor remainder estimate for the approximation of \(\displaystyle tanx\) by \(\displaystyle x+\frac{x^3}{3}+\frac{2x^5}{15}\).

Answer

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

5. Plot \(\displaystyle e^x−e_4(x)\) where \(\displaystyle e_4(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\) on \(\displaystyle [0,2]\). Compare the maximum error with the Taylor remainder estimate.

Exercise \(\PageIndex{10}\)

1. (Taylor approximations and root finding.) Recall that Newton’s method \(\displaystyle x_{n+1}=x_n−\frac{f(x_n)}{f'(x_n)}\) approximates solutions of \(\displaystyle f(x)=0\) near the input \(\displaystyle x_0\).

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

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

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

Answer

a. Answers will vary.

b. The following are the \(\displaystyle x_n\) values after \(\displaystyle 10\) iterations of Newton’s method to approximation a root of \(\displaystyle p_N(x)−2=0\): for \(\displaystyle N=4,x=0.6939...;\) for \(\displaystyle N=5,x=0.6932...;\) for \(\displaystyle N=6,x=0.69315...;.\) (Note: \(\displaystyle ln(2)=0.69314...\))

c. Answers will vary.

Exercise \(\PageIndex{11}\)

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

1. \(\displaystyle \lim_{x→0}\frac{cosx−1}{x^2}\)

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

Answer

\(\displaystyle \frac{ln(1−x^2)}{x^2}→−1\)

3. \(\displaystyle \lim_{x→0}\frac{e^{x^2}−x^2−1}{x^4}\)

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

Answer

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