Exercises: Taylor Series

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Exercises relating to Taylor series and their computation.

Use the Maclaurin series for $\sin x$ to give a complete formula for the Maclaurin series for $x \sin x^2$ : $x \sin (x^2) = \sum _{m=0}^\infty \answer {\frac {(-1)^m}{(2m+1)!}} x^{\answer {4m+3}}$

Use the Maclaurin series for $e^x$ and $e^{-x}$ to give a complete formula for the Maclaurin series for $\cosh x$ : $\cosh x = \sum _{n=0}^\infty \answer {\frac {1}{(2n)!}} x^{\answer {2n}}.$

Your formula should account for the fact that there are only even powers of $x$ .

Compute the Taylor series of the function $f(x) = \ln x$ centered at the point $x_0 = 2$ . $\answer { \ln 2} + \answer {\frac {1}{2}} (x-2) + \answer {- \frac {1}{8}}(x-2)^2 + \answer {\frac {1}{24}}(x-2)^3 + \cdots = \ln 2 + \sum _{n=1}^\infty \answer {\frac {(-1)^{n-1}2^{-n}}{n}} (x-2)^n$

Compute the Maclaurin series of the function identified below. $\int _0^x \frac {\sin t}{t} dt = \sum _{n=0}^\infty \answer {\frac {(-1)^n}{(2n+1)! (2n+1)}} x^{\answer {2n+1}}$

Sample Quiz Questions

Compute the first $4$ nonzero terms in the Taylor series at $x=0$ of the function $\frac {d}{dx} \left [ x e^{x^2} \right ].$

$\displaystyle 1 + 3x^{2} + \frac {5}{2}x^{4} + \frac {7}{6}x^{6}$ $\displaystyle - 1 - 3x^{2} + \frac {5}{2}x^{4} + \frac {7}{6}x^{6}$ $\displaystyle 1 - 6x^{2} - \frac {5}{4}x^{4} + \frac {7}{6}x^{6}$ $\displaystyle - 1 - 6x^{2} + \frac {5}{4}x^{4} + \frac {7}{6}x^{6}$ $\displaystyle 2 + 3x^{2} + \frac {5}{4}x^{4} + \frac {7}{6}x^{6}$ $\displaystyle 2 - 3x^{2} - \frac {5}{4}x^{4} + \frac {7}{6}x^{6}$

Compute the series in stages beginning with substitution into known series: $e^{x^2} = 1 + x^{2} + \frac {1}{2}x^{4} + \frac {1}{6}x^{6} + \cdots$ $x e^{x^2} = x + x^{3} + \frac {1}{2}x^{5} + \frac {1}{6}x^{7} + \cdots$ $\frac {d}{dx} \left [ x e^{x^2} \right ] = 1 + 3x^{2} + \frac {5}{2}x^{4} + \frac {7}{6}x^{6} + \cdots$

Compute the first $4$ nonzero terms in the Taylor series at $x=0$ of the function $\int _0^x \left ( x \ln (1-x) \right ) ~ dx.$

$\displaystyle - \frac {1}{6}x^{3} - \frac {1}{8}x^{4} - \frac {2}{15}x^{5} - \frac {1}{24}x^{6}$ $\displaystyle \frac {1}{6}x^{3} + \frac {1}{8}x^{4} - \frac {2}{15}x^{5} - \frac {1}{24}x^{6}$ $\displaystyle - \frac {1}{6}x^{3} + \frac {1}{4}x^{4} + \frac {1}{15}x^{5} - \frac {1}{24}x^{6}$ $\displaystyle \frac {1}{6}x^{3} + \frac {1}{4}x^{4} - \frac {1}{15}x^{5} - \frac {1}{24}x^{6}$ $\displaystyle - \frac {1}{3}x^{3} - \frac {1}{8}x^{4} - \frac {1}{15}x^{5} - \frac {1}{24}x^{6}$ $\displaystyle - \frac {1}{3}x^{3} + \frac {1}{8}x^{4} + \frac {1}{15}x^{5} - \frac {1}{24}x^{6}$

Compute the series in stages beginning with substitution into known series: $\ln (1-x) = - x - \frac {1}{2}x^{2} - \frac {1}{3}x^{3} - \frac {1}{4}x^{4} + \cdots$ $x \ln (1-x) = - x^{2} - \frac {1}{2}x^{3} - \frac {1}{3}x^{4} - \frac {1}{4}x^{5} + \cdots$ $\int _0^x \left ( x \ln (1-x) \right ) ~ dx = - \frac {1}{3}x^{3} - \frac {1}{8}x^{4} - \frac {1}{15}x^{5} - \frac {1}{24}x^{6} + \cdots$

Sample Exam Questions

The first few nonzero terms of the Maclaurin series for $f(x) = \ln ( 1 + \sin x)$ are:

$\displaystyle 1 + \frac {1}{2} x - \frac {1}{8} x^2 + \frac {1}{24} x^3 + \cdots$ $\displaystyle 1 + \frac {1}{2} x - \frac {1}{8} x^2 - \frac {1}{48} x^3 + \cdots$ $\displaystyle x - \frac {1}{2} x^2 + \frac {1}{8} x^3 - \frac {1}{24} x^4 + \cdots$ $\displaystyle 1 + x + \frac {1}{2} x^2 + \frac {1}{3} x^3 + \frac {1}{6} x^4 \cdots$ $\displaystyle x - \frac {1}{2} x^2 + \frac {1}{6} x^3 - \frac {1}{12} x^4 + \cdots$ $\displaystyle 1 + x + \frac {1}{2} x^2 + \frac {1}{3} x^3 - \frac {1}{12} x^4 + \cdots$

(Hints will not be revealed until you choose your response.)

The first few derivatives of $f(x)$ are: $\begin {aligned} f(x) & = \ln (1 + \sin (x)) \\ f'(x) & = \frac { \cos x}{1 + \sin x} \\ f''(x) & = - \frac {\sin x}{1 + \sin x} - \frac {\cos ^2 x}{(1 + \sin x)^2} \\ f'''(x) & = - \frac {\cos x}{1 + \sin x} + \frac {\sin x \cos x}{(1 + \sin x)^2} - \frac {2 \sin x \cos x}{(1 + \sin x)^2} + 2 \frac {\cos ^3 x}{(1 + \sin x)^3} \end {aligned}$

Evaluating at $x = 0$ gives $f(0) = \ln 1 = 0$ , $f'(0) = 1$ , $f''(0) = -1$ , and $f'''(0) = 1$ . Therefore the series starts with the terms $x - \frac {1}{2} x^2 + \frac {1}{6} x^3 + \cdots$ .

Find the Taylor polynomial of degree 2 for $f(x) = \sqrt {x+16}$ centered at $x=9$ .

$\displaystyle 5 + \frac {4}{5} x + \frac {9}{250} x^2$ $\displaystyle 5 - \frac {3}{5} (x-5) + \frac {1}{125} (x-5)^2$ $\displaystyle 5 + \frac {1}{10} (x-9) - \frac {1}{1000} (x-9)^2$ $\displaystyle 5 + \frac {3}{5} (x-5) + \frac {8}{125} (x-5)^2$ $\displaystyle 5 + \frac {1}{5} (x-9) + \frac {16}{125} (x-9)^2$ none of these

Use the Taylor polynomial of degree $3$ for $f(x) = \ln (1+x)$ centered at $x_0 = 0$ to approximate the value of $\displaystyle \ln \left ( \frac {3}{2} \right )$ .

$\displaystyle \frac {2}{3}$ $\displaystyle \frac {3}{2}$ $\displaystyle \frac {15}{4}$ $\displaystyle \frac {5}{12}$ $\displaystyle \frac {9}{24}$ $\displaystyle \frac {11}{24}$

Let $F(x)$ be the unique function that satisfies $F(0) = 0$ , $F'(0) = 0$ , and $F'(x) = \frac {1}{x} \sin x^3$ for all $x \neq 0$ . Find the Taylor Series of $F(x)$ centered at $x_0 = 0$ .

$\displaystyle \sum _{n=0}^\infty \frac {(-1)^n x^{6n+3}}{(2n+1)!}$ $\displaystyle \sum _{n=0}^\infty \frac {(-1)^n (6n+3) x^{6n+2}}{(2n+1)!}$ $\displaystyle \sum _{n=0}^\infty \frac {(-1)^n x^{6n+3}}{(6n+3)(2n+1)!}$ $\displaystyle \sum _{n=0}^\infty \frac {(-1)^n x^{6n+2}}{(2n+1)!}$ $\displaystyle \sum _{n=0}^\infty \frac {(-1)^n (6n+2) x^{6n+2}}{(2n+1)!}$ $\displaystyle \sum _{n=0}^\infty \frac {(-1)^n x^{2n+3}}{(6n+3)(2n+1)!}$