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.

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

Short answer

The radius of convergence for the series is$P_n\left(x\right)=\sum _{k=0}^{\infty }(-1{)}^k(x-1{)}^k$

Step-by-step solution

Given information

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

 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$P_n\left(x\right)=\sum _{k=0}^{\infty }\frac{f^k\left(x_0\right)}{k!}{\left(x-x_0\right)}^n$

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

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

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

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

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

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