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.

$\frac{1}{1-x},x_0=2$

Short answer

The Taylor series for the function is$P_n\left(x\right)=\sum _{k=0}^{\infty }(-1{)}^{k+1}(x-2{)}^k$

Step-by-step solution

Given information

The function is$f\left(x\right)=\frac{1}{1-x}$

 Find the general of the Taylor series of the function 

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

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

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 $x=2$

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)=11-x at x=2 

$n$	$f^n\left(x\right)$	$f^n\left(2\right)$	$\frac{f^n\left(2\right)}{n!}$
$0$	$\frac{1}{1-x}$	$-1$	$-1$
$1$	$\frac{1}{(1-x{)}^2}$	$1$	$1$
$2$	$\frac{2}{(1-x{)}^3}$	$-2$	$-1$
$$\begin{gathered} . \\ . \\ . \end{gathered}$$	$$\begin{gathered} . \\ . \\ . \end{gathered}$$	$$\begin{gathered} . \\ . \\ . \end{gathered}$$	$$\begin{gathered} . \\ . \\ . \end{gathered}$$
$k$	$\frac{(-1{)}^{k+1}k!}{(1-x{)}^{2k+1}}$	$(-1{)}^{k+1}k!$	$(-1{)}^{k+1}$

 Find the Taylor series for the function f(x)=11-x at x=2 

The Taylor series for the function $f\left(x\right)=\frac{1}{1-x}$at $x=2$is:

$-1+1(x-2)+(-1)(x-2{)}^2+\cdots +(-1{)}^{k+1}(x-2{)}^k+\dots$

Or, we can writte as:

$P_n\left(x\right)=\sum _{k=0}^{\infty }(-1{)}^{k+1}(x-2{)}^k$