Chapter 8: Q 24 (page 704)

Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.
$\sqrt{x}, x_{0} = 2$

Short Answer

Expert verified

The radius of convergence for the series is $P_{n} (x) = 1 + \frac{1}{2 \sqrt{2}} (x - 2) + \sum_{k = 2}^{\infty} (- 1)^{k + 1} \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k} k!} 2^{- (\frac{2 k - 1}{2})} (x - 2)^{k}$

Step by step solution

Given information

The function is $f (x) = \sqrt{x}$

Find the general of the Taylor series of the function f

The Taylor series at $x = 2$ for any function $f$ with a derivative of order $n$ is given by

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

As a result, first determine the function's value as well as $f^{'} (x), f^{''} (x), f^{'''} (x)$ and $f^{''''} (x)$ at $x = 2$

Furthermore, the function's general Taylor series is $P_{n} (x) = \sum_{k = 0}^{\infty} \frac{f^{k} (x_{0})}{k!} {(x - x_{0})}^{n}$

Make a table of the Taylor series for the function f(x)=x at x=2

Let us begin by constructing the Taylor series table for the function $f (x) = \sqrt{x}$ at $x = 2$

$n$	$f^{''} (x)$	$f^{''} (2)$	$\frac{f^{''} (2)}{n!}$
0	$\sqrt{x}$	$\sqrt{2}$	$\sqrt{2}$
$1$	$\frac{1}{2 \sqrt{x}}$	$\frac{1}{2 \sqrt{2}}$	$\frac{1}{2 \sqrt{2}}$
$2$	$- \frac{1}{2^{2}} {(x)}^{- \frac{3}{2}}$	$- \frac{1}{2^{2}} 2^{- \frac{3}{2}}$	$\frac{- \frac{1}{2^{2}} 2^{- \frac{3}{2}}}{2!}$
$3$	$\frac{1 \cdot 3}{2^{3}} {(x)}^{- \frac{5}{2}}$	$\frac{1 \cdot 3}{2^{3}} 2^{- \frac{5}{2}}$	$\frac{\frac{1 \cdot 3}{2^{3}} 2^{- \frac{5}{2}}}{3!}$
$4$	$- \frac{1 \cdot 3 \cdot 5}{2^{4}} {(x)}^{- \frac{7}{2}}$	$- \frac{1 \cdot 3 \cdot 5}{2^{4}} 2^{- \frac{7}{2}}$	$\frac{- \frac{1 \cdot 3 \cdot 5}{2^{4}} 2^{- \frac{7}{2}}}{4!}$
$. . .$	$. . .$	$. . .$	$. . .$
$k$	${(- 1)}^{k + 1} \frac{1 \cdot 3 \cdot 5 . . . . (2 k - 3)}{2}$ ^k(x)-(2k-12)	${(- 1)}^{k + 1} \frac{1 \cdot 3 \cdot 5 . . . (2 k - 3)}{2}$ ^k2-(2k-12)	${(- 1)}^{k + 1} \frac{1 \cdot 3 \cdot 5 . . . . (2 k - 3)}{2}$ ^k2^-(2k-12)k!

Find the Taylor series for the function f(x)=x at x=2

The Taylor series for the function $f (x) = \sqrt{x}$ at $x = 2$ is

$\sqrt{2} + \frac{1}{2 \sqrt{2}} \cdot (x - 2) + \frac{- \frac{1}{2^{2}} 2^{- \frac{3}{2}}}{2!} {(x - 2)}^{2} + \frac{1 \cdot 3}{2}$ ³2^-523!(x-2)3+-1·3·52⁴2^-724!(x-2)4+.....+(-1)k+11·3·5.....(2k-3)2^k2-(2k-12)k!(x-2)k

Or,

$P_{n} (x) = 1 + \frac{1}{2 \sqrt{2}} (x - 2) + \sum_{k = 2}^{\infty} (- 1)^{k + 1} \frac{1 \cdot 3 \cdot 5 \dots (2 k - 3)}{2^{k} k!} 2^{- (\frac{2 k - 1}{2})} (x - 2)^{k}$

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Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.
$\sqrt{x}, x_{0} = 2$

Short Answer

Step by step solution

Given information

Find the general of the Taylor series of the function f

Make a table of the Taylor series for the function f(x)=x at x=2

Find the Taylor series for the function f(x)=x at x=2

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Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.x,x0=2

Short Answer

Step by step solution

Given information

Find the general of the Taylor series of the function f

Make a table of the Taylor series for the function f(x)=x at x=2

Find the Taylor series for the function f(x)=x at x=2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Applied Mathematics

Calculus

Statistics

Discrete Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.
$\sqrt{x}, x_{0} = 2$