Q. 52 In Exercises 49鈥�56 find the Ta... [FREE SOLUTION]

Chapter 8: Q. 52 (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.
$52 . \sqrt{x}, 1$

Short Answer

Expert verified

The Taylor series for the function $f (x) = \sqrt{x}$ at $x = 1$ is $P_{n} (x) = 1 + \frac{1}{2} (x 鈭� 1) + {鈭�}_{k = 2}^{鈭�} (鈭� 1)^{k + 1} \frac{1 鈰� 3 鈰� 5 鈰� (2 k 鈭� 3)}{2^{k} k!} (x 鈭� 1)^{k}$

Step by step solution

Step 1. Given data

We have the function $f (x) = \sqrt{x}$ .

Step 2. Table of the taylor series

Any function $f$ with a derivative of order $n$ , the taylor series at $x = 1$ is given by,

$P_{n} (x) = f (1) + f^{鈥�} (1) (x 鈭� 1) + \frac{f^{鈥测赌�} (1)}{2!} (x 鈭� 1)^{2} + \frac{f^{鈥测赌� 鈥�} (1)}{3!} (x 鈭� 1)^{3} + \frac{f^{鈥测赌� 鈥测赌�} (1)}{4!} (x 鈭� 1)^{4} + 鈥�$

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 $f (x) = \sqrt{x}$ at $x = 1$

n	$f^{n} (x)$	$f^{n} (0)$	$\frac{f^{n} (0)}{n!}$
0	$\sqrt{x}$	1	1
1	$\frac{1}{2 \sqrt{x}}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$鈭� \frac{1}{2^{2}} (x)^{鈭� \frac{3}{2}}$	$鈭� \frac{1}{2^{2}}$	$\frac{鈭� \frac{1}{2^{2}}}{2!}$
3	$\frac{1 鈰� 3}{2^{3}} (x)^{鈭� \frac{5}{2}}$	$\frac{1 鈰� 3}{2^{3}}$	$\frac{\frac{1 鈰� 3}{2^{3}}}{3!}$
4	$鈭� \frac{1 鈰� 3 鈰� 5}{2^{4}} (x)^{鈭� \frac{7}{2}}$	$鈭� \frac{1 鈰� 3 鈰� 5}{2^{4}}$	$\frac{鈭� \frac{1 鈰� 3 鈰� 5}{2^{4}}}{4!}$
. . .	. . .	. . .	. . .
k	$(鈭� 1)^{k + 1} \frac{1 鈰� 3 鈰� 5 鈰� (2 k 鈭� 3)}{2^{4}} (1 鈭� x)^{鈭� (\frac{2 k 鈭� 1}{2})}$	$(鈭� 1)^{k + 1} \frac{1 鈰� 3 鈰� 5 鈰� (2 k 鈭� 3)}{2^{4}}$	$(鈭� 1)^{k + 1} \frac{\frac{1 鈰� 3 鈰� 5 鈰� (2 k 鈭� 3)}{2^{4}}}{k!}$

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

The Taylor series for the function $f (x) = \sqrt{x}$ at $x = 1$ is $P_{n} (x) = \begin{aligned} 1 & + \frac{1}{2} 鈰� (x 鈭� 1) + \frac{鈭� \frac{1}{2^{2}}}{2!} (x 鈭� 1)^{2} + \frac{\frac{1 鈰� 3}{2^{3}}}{3!} (x 鈭� 1)^{3} + \frac{鈭� \frac{1 鈰� 3 鈰� 5}{2^{4}}}{4!} (x 鈭� 1)^{4} + . . . . 鈥� + \frac{(鈭� 1)^{k + 1} \frac{1 鈰� 3 鈰� 5 鈰� (2 k 鈭� 3)}{2^{k}}}{k!} (x 鈭� 1)^{k} \end{aligned}$

Or we can write this as $P_{n} (x) = 1 + \frac{1}{2} (x 鈭� 1) + {鈭�}_{k = 2}^{鈭�} (鈭� 1)^{k + 1} \frac{1 鈰� 3 鈰� 5 鈰� (2 k 鈭� 3)}{2^{k} k!} (x 鈭� 1)^{k}$

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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.
$52 . \sqrt{x}, 1$

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)=x

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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. 52.x,1

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)=x

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Calculus

Pure Maths

Geometry

Probability and Statistics

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.
$52 . \sqrt{x}, 1$