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Answers without the blur. Sign up and see all textbooks for free! Q. 55

Expert-verified Found in: Page 680 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In Exercises 49–56 find the Taylor series for the specified function and the given value of ${x}_{0}$. Note: These are the same functions and values as in Exercises 41–48. $55.\mathrm{cos}2x,\frac{\pi }{4}$

The Taylor series for the function $f\left(x\right)=\mathrm{cos}\left(2x\right)$ at $x=\frac{\mathrm{\pi }}{4}$ is ${P}_{n}\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k+1}{2}^{2k+1}}{\left(2k+1\right)!}{\left(x-\frac{\pi }{4}\right)}^{2k+1}$

See the step by step solution

## Step 1. Given data

We have the function $f\left(x\right)=\mathrm{cos}\left(2x\right)$

## Step 2. Table of the taylor series

Any function $f$ with a derivative of order $n$, the taylor series at $x=\frac{\mathrm{\pi }}{4}$ is given by,

$\begin{array}{rl}{P}_{n}\left(x\right)=& f\left(\frac{\pi }{4}\right)+{f}^{\mathrm{\prime }}\left(\frac{\pi }{4}\right)\left(x-\frac{\pi }{4}\right)+\frac{{f}^{\prime \prime }\left(\frac{\pi }{4}\right)}{2!}{\left(x-\frac{\pi }{4}\right)}^{2}+\frac{{f}^{\prime \prime }\left(\frac{\pi }{4}\right)}{3!}{\left(x-\frac{\pi }{4}\right)}^{3}+\frac{{f}^{\prime \prime \prime \prime }\left(\frac{\pi }{4}\right)}{4!}{\left(x-\frac{\pi }{4}\right)}^{4}+\dots \end{array}$

we can write the general of the Taylor series of the function $f$ is,

${P}_{n}\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{f}^{k}\left({x}_{0}\right)}{k!}{\left(x-{x}_{0}\right)}^{n}$

So, let us first construct the table of the Taylor series for the function $f\left(x\right)=\mathrm{cos}\left(2x\right)$ at $x=\frac{\mathrm{\pi }}{4}$

 $n$ ${f}^{n}\left(x\right)$ ${f}^{n}\left(\frac{\mathrm{\pi }}{4}\right)$ $\frac{{f}^{n}\left(\frac{\mathrm{\pi }}{4}\right)}{n!}$ 0 $\mathrm{cos}2x$ 0 0 1 $-2\mathrm{sin}2x$ -2 -2 2 $-{2}^{2}\mathrm{cos}2x$ 0 0 3 ${2}^{3}\mathrm{sin}2x$ ${2}^{3}$ $\frac{{2}^{3}}{3!}$ ... ... ... ... 2k $\left(-1{\right)}^{k}{2}^{2k}\mathrm{cos}2x$ 0 0 2k+1 $\left(-1{\right)}^{k+1}{2}^{2k+1}\mathrm{sin}2x$ $\left(-1{\right)}^{k+1}{2}^{2k+1}$ $\frac{\left(-1{\right)}^{k+1}{2}^{2k+1}}{\left(2k+1\right)!}$ ... ... ... ...

## Step 3. Taylor series for the f(x)=cos2x

The Taylor series for the function $f\left(x\right)=\mathrm{cos}\left(2x\right)$ at $x=\frac{\mathrm{\pi }}{4}$ is

$\begin{array}{rl}{P}_{n}\left(x\right)=& 0+-2\cdot \left(x-\frac{\pi }{4}\right)+\frac{0}{2!}{\left(x-\frac{\pi }{4}\right)}^{2}+\frac{{2}^{3}}{3!}{\left(x-\frac{\pi }{4}\right)}^{3}+\cdots +\frac{0}{\left(2k\right)!}{\left(x-\frac{\pi }{4}\right)}^{2k}+\frac{\left(-1{\right)}^{k+1}{2}^{2k+1}}{\left(2k+1\right)!}{\left(x-\frac{\pi }{4}\right)}^{2k+1}+\cdots \end{array}$

Or we can write this as ${P}_{n}\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k+1}{2}^{2k+1}}{\left(2k+1\right)!}{\left(x-\frac{\pi }{4}\right)}^{2k+1}$ ### Want to see more solutions like these? 