Q. 55 In Exercises 49鈥�56 find the Ta... [FREE SOLUTION]

Chapter 8: Q. 55 (page 680)

In Exercises 49鈥�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鈥�48.
$55 . \cos 2 x, \frac{蟺}{4}$

Short Answer

Expert verified

The Taylor series for the function $f (x) = \cos (2 x)$ at $x = \frac{蟺}{4}$ is $P_{n} (x) = {鈭�}_{k = 0}^{鈭�} \frac{(鈭� 1)^{k + 1} 2^{2 k + 1}}{(2 k + 1)!} {(x 鈭� \frac{蟺}{4})}^{2 k + 1}$

Step by step solution

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 $x = \frac{蟺}{4}$ is given by,

$\begin{aligned} P_{n} (x) = & f (\frac{蟺}{4}) + f^{鈥�} (\frac{蟺}{4}) (x 鈭� \frac{蟺}{4}) + \frac{f^{鈥测赌�} (\frac{蟺}{4})}{2!} {(x 鈭� \frac{蟺}{4})}^{2} + \frac{f^{鈥测赌�} (\frac{蟺}{4})}{3!} {(x 鈭� \frac{蟺}{4})}^{3} + \frac{f^{鈥测赌� 鈥测赌�} (\frac{蟺}{4})}{4!} {(x 鈭� \frac{蟺}{4})}^{4} + 鈥� \end{aligned}$

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

$P_{n} (x) = {鈭�}_{k = 0}^{鈭�} \frac{f^{k} (x_{0})}{k!} {(x 鈭� x_{0})}^{n}$

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

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

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

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

$\begin{aligned} P_{n} (x) = & 0 + 鈭� 2 鈰� (x 鈭� \frac{蟺}{4}) + \frac{0}{2!} {(x 鈭� \frac{蟺}{4})}^{2} + \frac{2^{3}}{3!} {(x 鈭� \frac{蟺}{4})}^{3} + 鈰� + \frac{0}{(2 k)!} {(x 鈭� \frac{蟺}{4})}^{2 k} + \frac{(鈭� 1)^{k + 1} 2^{2 k + 1}}{(2 k + 1)!} {(x 鈭� \frac{蟺}{4})}^{2 k + 1} + 鈰� \end{aligned}$

Or we can write this as $P_{n} (x) = {鈭�}_{k = 0}^{鈭�} \frac{(鈭� 1)^{k + 1} 2^{2 k + 1}}{(2 k + 1)!} {(x 鈭� \frac{蟺}{4})}^{2 k + 1}$

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In Exercises 49鈥�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鈥�48.
$55 . \cos 2 x, \frac{蟺}{4}$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Table of the taylor series

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

One App. One Place for Learning.

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91影视

In Exercises 49鈥�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鈥�48. 55.cos2x,蟺4

Short Answer

Step by step solution

Step 1. Given data

Step 2. Table of the taylor series

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Pure Maths

Geometry

Logic and Functions

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Exercises 49鈥�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鈥�48.
$55 . \cos 2 x, \frac{蟺}{4}$