Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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.
$55 . \cos 2 x, \frac{π}{4}$

The Taylor series for the function $f (x) = \cos (2 x)$ at $x = \frac{π}{4}$ is $P_{n} (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1} 2^{2 k + 1}}{(2 k + 1)!} {(x - \frac{π}{4})}^{2 k + 1}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given data

We have the function $f (x) = \cos (2 x)$

Step 2. Table of the taylor series

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

$\begin{aligned} P_{n} (x) = & f (\frac{π}{4}) + f^{'} (\frac{π}{4}) (x - \frac{π}{4}) + \frac{f^{''} (\frac{π}{4})}{2!} {(x - \frac{π}{4})}^{2} + \frac{f^{''} (\frac{π}{4})}{3!} {(x - \frac{π}{4})}^{3} + \frac{f^{''''} (\frac{π}{4})}{4!} {(x - \frac{π}{4})}^{4} + \dots \end{aligned}$

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

$P_{n} (x) = \sum_{k = 0}^{\infty} \frac{f^{k} (x_{0})}{k!} {(x - x_{0})}^{n}$

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

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

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

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

$\begin{aligned} P_{n} (x) = & 0 + - 2 \cdot (x - \frac{π}{4}) + \frac{0}{2!} {(x - \frac{π}{4})}^{2} + \frac{2^{3}}{3!} {(x - \frac{π}{4})}^{3} + \dots + \frac{0}{(2 k)!} {(x - \frac{π}{4})}^{2 k} + \frac{(- 1)^{k + 1} 2^{2 k + 1}}{(2 k + 1)!} {(x - \frac{π}{4})}^{2 k + 1} + \dots \end{aligned}$

Or we can write this as $P_{n} (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k + 1} 2^{2 k + 1}}{(2 k + 1)!} {(x - \frac{π}{4})}^{2 k + 1}$

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Calculus

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Short Answer

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.
$55 . \cos 2 x, \frac{π}{4}$

Step by Step Solution

Step 1. Given data

Step 2. Table of the taylor series

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

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Calculus

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Short Answer

In Exercises 49–56 find the Taylor series for the specified function and the given value of x0. Note: These are the same functions and values as in Exercises 41–48. 55. cos2x,π4

Step by Step Solution

Step 1. Given data

Step 2. Table of the taylor series

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

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Pure Maths

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.
$55 . \cos 2 x, \frac{π}{4}$