Q 23 - Calculus

Question

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

$\sqrt{x},x_0=1$

Step-by-Step Solution

Answer

 The radius of convergence for the series is$P_n\left(x\right)=1+\frac{1}{2}(x-1)+\sum _{k=2}^{\infty }(-1{)}^{k+1}\frac{1·3·5\cdots (2k-3)}{2^{k}k!}(x-1{)}^k$

Step 1: Given information

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

Step 2: Find the general of the Taylor series of the function

The Taylor series at $x=1$ for any function $f$ with a derivative of order$n$ is given by

$P_n\left(x\right)=f\left(1\right)+f^{\text{'}}\left(1\right)(x-1)+\frac{f^{\text{'}\text{'}}\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(1\right)}{3!}(x-1{)}^3+\frac{f^{\text{'}\text{'}\text{'}\text{'}}\left(1\right)}{4!}(x-1{)}^4+\dots$

As a result, first determine the function's value as well as  $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$at $x=1$

Furthermore, the function's general Taylor series is $P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{f^k\left(x_0\right)}{k!}{\left(x-x_0\right)}^n$

Step 3: Make a table of the Taylor series for the function f ( x ) = x   at x = 1  

$0$	$\sqrt{x}$	$1$	$1$
$1$	$\frac{1}{2\sqrt{x}}$	$\frac{1}{2}$	$\frac{1}{2}$
$2$	$-\frac{1}{2^2}(x{)}^{-\frac{3}{2}}$	$-\frac{1}{2^2}$	$\frac{-\frac{1}{2^2}}{2!}$
$3$	$\frac{1·3}{2^3}(x{)}^{\frac{5}{2}}$	$\frac{1·3}{2^3}$	$\frac{1·3}{2^3}$
$4$	$-\frac{1·3·5}{2^4}(x{)}^{-\frac{7}{2}}$	$-\frac{1·3·5}{2^4}$	$-\frac{1·3·5}{2^4}$
.	$.$	$.$	$.$
$k$	$(-1{)}^{k+1}\frac{1·3·5\cdots (2k-3)}{2^4}(1-x{)}^{-\left(\frac{2k-1}{2}\right)}$	$(-1{)}^{k+1}\frac{1·3·5\cdots (2k-3)}{2^k}$	$(-1{)}^{k+1}\frac{1·3·5\cdots (2k-3)}{2^k}$

Step 4: Find the Taylor series for the function f ( x ) = x   at x = 1  

The Taylor series for the function $f\left(x\right)=\sqrt{x}$at $x=1$is

$1+\frac{1}{2}·(x-1)+\frac{-{\displaystyle \frac{1}{2^2}}}{2!}{(x-1)}^2+\frac{\frac{1·3}{23}}{3!}{(x-1)}^3+\frac{-\frac{1·3·5}{24}}{4!}{(x-1)}^4+.....+\frac{{(-1)}^{k+1}\frac{1·3·5.....(2k-3)}{2k}}{k!}{(x-1)}^k$

Or, 

$P_n\left(x\right)=1+\frac{1}{2}(x-1)+\sum _{k=2}^{\infty }(-1{)}^{k+1}\frac{1·3·5\cdots (2k-3)}{2^{k}k!}(x-1{)}^k$