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Give the first three nonzero terms of the Maclaurin series of the given functions.
1) \((1−x)^{1/3}\)
- Answer
- \(\displaystyle 1 - \frac 1 3 x - \frac 1 9 x^2 + \cdots \)
2) \( \sqrt{1+2x }\)
- Answer
- \(\displaystyle 1+x- \frac 1 2 x^2 + \cdots \)
3) \( \displaystyle \sqrt[4]{1-x^2} \)
- Answer
- \(\displaystyle 1- \frac 1 4 x^2 - \frac{3}{32} x^4 + \cdots\)
4) \( \displaystyle \frac{x^2}{\sqrt[3]{1+2x^3}} \)
- Answer
- \(\displaystyle x^2 - \frac{2}{3}x^5 + \frac 8 9 x^8 + \cdots \)
In the following exercises, use series method to calculate the given functions.
5) \(\displaystyle f(x)=e^{−x^2}\)
- Answer
- \(\displaystyle e^{−x^2}=\sum_{n=0}^∞(−1)^n\frac{x^{2n}}{n!} \)
6) \(\displaystyle F(x)=\int e^{−x^2}\,dx \)
- Answer
- Hint: Use result from problem (5) to get \(\displaystyle F(x) = C + \sum_{n=0}^∞(−1)^n\frac{x^{2n+1}}{n! \, (2n+1)} \)
7) \(\displaystyle f(x)=\tan^{−1}x \)
- Answer
- Show that the Maclaurin series of the integrand \(\displaystyle \tan^{−1}x = \sum_{n=0}^∞(−1)^n x^{2n}\). Answer is \( \displaystyle f(x)=\sum_{k=0}^∞\frac{(−1)^kx^{2k+1}}{2k+1}\).
8) \(\displaystyle F(x)=∫\frac{\sin x}{x}\,dx \)
- Answer
- \(\displaystyle F(x) = C+\sum_{n=0}^∞\frac{(−1)^n x^{2n+1}}{(2n+1)(2n+1)!}\).
9) \(\displaystyle F(x)=∫\cos\left(\sqrt{x}\right)\,dx \)
- Answer
- \(\displaystyle F(x)=C+\sum_{n=0}^∞(−1)^n\frac{x^{n+1}}{(n+1)(2n)!}\)
10) \(\displaystyle F(x)=∫ \frac{\cos x}{x} \,dx \)
- Answer
- \(\displaystyle F(x)=C+ \ln |x| + \sum_{n=1}^∞(−1)^n \frac{x^{2n}}{2n \cdot (2n)!}\)
11) \(\displaystyle f(x)=\ln (4+9x^2) \)
- Answer
- \(\displaystyle f(x)=\ln 4 +\sum_{n=1}^∞(−1)^{n+1} \left( \frac 9 4 \right)^n \frac{x^{2n}}{n} \)
In the following exercises, compute the first three nonzero terms of the Maclaurin series of the given functions.
12) \(\displaystyle f(x)=\frac{e^x}{1-x^2} \)
- Answer
- \(\displaystyle f(x)= 1+x +\frac{3}{2} x^{2}+ \cdots \)
13) \(f(x)=\tan x\)
- Answer
- \(f(x) = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+⋯\)
14) \(f(x)=e^x\cos x\)
- Answer
- \(f(x) = 1+x−\dfrac{x^3}{3} +⋯\)
15) \(f(x)=e^{\sin x}\)
- Answer
- \(1+x +\frac{1}{2} x^{2}+⋯\)
16) \(f(x)=\sec^2x\)
- Answer
- \(1+x^2+\dfrac{2x^4}{3} +⋯\)
17) \(\displaystyle f(x)=\frac{e^x}{1+\sin x} \)
- Answer
- \( f(x) = 1+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}+⋯\)
Taylor Polynomials
In exercises 18 - 25, find the Taylor polynomials of degree two approximating the given function centered at the given point.
18) \( f(x)=1+3x+2x^2+5x^3\) at \( a=1\)
- Answer
- \( p_2(x)=11+22(x-1)+17(x-1)^2\)
19) \( f(x)=x-x^2+x^3-x^4\) at \( a=−1\)
- Answer
- \( p_2(x)=-4+10(x+1)-10(x+1)^2\)
20) \( f(x)=\cos(2x)\) at \( a=π\)
- Answer
- \( p_2(x)=1−2(x-\pi)^2\)
21) \( f(x)=2\sin^2 x\) at \( a=\frac{π}{3}\)
- Answer
- \( p_2(x)= \frac 3 2 + \sqrt{3} \left( x - \frac{\pi}{3} \right) - \left( x - \frac{\pi}{3} \right)^2 \)
22) \( f(x)=\sqrt{x}\) at \( a=9\)
- Answer
- \( p_2(x)= 3 + \frac 1 6 (x-9) - \frac{1}{216} (x-9)^2 \)
23) \( f(x)=\ln x\) at \( a=1\)
- Answer
- \( p_2(x)=(x−1)−\frac{1}{2}(x−1)^2\)
24) \( f(x)=\dfrac{1}{x}\) at \( a= - \frac 1 2 \)
- Answer
- \( p_2(x)=-2-4 \left( x + \frac{1}{2} \right) - 8\left( x + \frac{1}{2} \right)^2 \)
25) \( f(x)=e^x\) at \( a=1\)
- Answer
- \( p_2(x)=e+e(x−1)+\dfrac{e}{2}(x−1)^2\)
Approximating Definite Integrals Using Taylor Series
26) 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 27 - 32, find the Taylor series of the given function centered at the indicated point.
27) \(f(x) = -3+2x-x^4\) at \( a=0\)
- Answer
- \( -3+2(x-0)-(x-0)^4 = -3+2x-x^4 \)
28) \(f(x) = 1+x+x^2+x^3\) at \( a=−1\)
- Answer
- \( (x+1)^3−2(x+1)^2+2(x+1)\)
29) \(f(x) = \cos x\) at \( a=2π\)
- Answer
- \(\displaystyle \cos x=\sum_{n=0}^∞(−1)^n\frac{(x−2π)^{2n}}{(2n)!}\)
30) \(f(x) = \cos x\) at \( x=\frac{π}{2}\)
- Answer
- \(\displaystyle \cos x=\sum_{n=0}^∞(−1)^{n+1}\frac{(x−\frac{π}{2})^{2n+1}}{(2n+1)!}\), which is also \( −\cos(x−\frac{π}{2})\).
31) \(f(x) = e^x\) at \( a=1\)
- Answer
- \(\displaystyle e^x=e\sum_{n=0}^∞\frac{(x−1)^n}{n!}.\)
32) \(f(x) = \dfrac{1}{(x−1)^3}\) at \( a=0\)
- Answer
- \(\displaystyle \frac{1}{(x−1)^3}=−\frac{1}{2}\frac{d^2}{dx^2}\left(\frac{1}{1−x}\right)=−\sum_{n=0}^∞\left(\frac{(n+2)(n+1)x^n}{2}\right)\)
In exercises 33 - 36, compute the Taylor series of each function around \( x=1\).
33) \( f(x)=2−x\)
- Answer
- \( 2−x=1−(x−1)\)
34) \( f(x)=x^3\)
- Answer
- \( x^3 = 1+3(x-1)+3(x-1)^2+(x-1)^3 \)
35) \( f(x)=(x−2)^2\)
- Answer
- \( ((x−1)−1)^2=(x−1)^2−2(x−1)+1\)
36) \( f(x)=\dfrac{1}{x}\)
- Answer
- \(\displaystyle \frac{1}{1−(1−x)}=\sum_{n=0}^∞(−1)^n(x−1)^n\)
Evaluating Limits using Taylor Series
In the following exercises, compute each limit using Taylor series.
37) \(\displaystyle \lim_{x→0}\frac{\cos x−1}{x^2}\)
- Answer
- \(\displaystyle -\frac{1}{2} \)
38) \(\displaystyle \lim_{x→0}\frac{e^{2x}−2x^2-2x−1}{x^3}\)
- Answer
- \(\displaystyle \frac{4}{3} \)
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
This page titled 11.10E: Exercises for Section 11.10 was last modified on Wed, 16 Jul 2025 02:30:55 GMT and is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.
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