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.

$\sqrt{x},x_0=2$

Short answer

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

Step-by-step solution

Given information

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

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

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

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

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

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 $f\left(x\right)=\sqrt{x}$at $x=2$

$n$	$f^{\text{'}\text{'}}\left(x\right)$	$f^{\text{'}\text{'}}\left(2\right)$	$\frac{f^{\text{'}\text{'}}\left(2\right)}{n!}$
0	$\sqrt{x}$	$\sqrt{2}$	$\sqrt{2}$
$1$	$\frac{1}{2\sqrt{x}}$	$\frac{1}{2\sqrt{2}}$	$\frac{1}{2\sqrt{2}}$
$2$	$-\frac{1}{2^2}{\left(x\right)}^{-\frac{3}{2}}$	$-\frac{1}{2^2}{2}^{-\frac{3}{2}}$	$\frac{-{\displaystyle \frac{1}{2^2}}{2}^{-{\displaystyle \frac{3}{2}}}}{2!}$
$3$	$\frac{1·3}{2^3}{\left(x\right)}^{-\frac{5}{2}}$	$\frac{1·3}{2^3}{2}^{-\frac{5}{2}}$	$\frac{{\displaystyle \frac{1·3}{2^3}}{2}^{-{\displaystyle \frac{5}{2}}}}{3!}$
$4$	$-\frac{1·3·5}{2^4}{\left(x\right)}^{-\frac{7}{2}}$	$-\frac{1·3·5}{2^4}{2}^{-\frac{7}{2}}$	$\frac{-{\displaystyle \frac{1·3·5}{2^4}}{2}^{-{\displaystyle \frac{7}{2}}}}{4!}$
$$\begin{gathered} . \\ . \\ . \end{gathered}$$	$$\begin{gathered} . \\ . \\ . \end{gathered}$$	$$\begin{gathered} . \\ . \\ . \end{gathered}$$	$$\begin{gathered} . \\ . \\ . \end{gathered}$$
$k$	${(-1)}^{k+1}\frac{1·3·5....(2k-3)}{2k}{\left(x\right)}^{-\left(\frac{2k-1}{2}\right)}$	${(-1)}^{k+1}\frac{1·3·5...(2k-3)}{2k}{2}^{-\left(\frac{2k-1}{2}\right)}$	${(-1)}^{k+1}\frac{\frac{1·3·5....(2k-3)}{2k}2-\left(\frac{2k-1}{2}\right)}{k!}$

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

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

$\sqrt{2}+\frac{1}{2\sqrt{2}}·(x-2)+\frac{-{\displaystyle \frac{1}{2^2}}{2}^{-{\displaystyle \frac{3}{2}}}}{2!}{(x-2)}^2+\frac{\frac{1·3}{23}2-\frac{5}{2}}{3!}{(x-2)}^3+\frac{-\frac{1·3·5}{24}2-\frac{7}{2}}{4!}{(x-2)}^4+.....+\frac{{(-1)}^{k+1}\frac{1·3·5.....(2k-3)}{2k}{2}^{-\left(\frac{2k-1}{2}\right)}}{k!}{(x-2)}^k$

Or,

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