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

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

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

$+\frac{f^{\text{'}\text{'}\text{'}\text{'}}(-1)}{4!}(x+1{)}^4+\dots$

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 

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

 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

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

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