$[-/0\mathrm{.}11$ Points$]$

SCALCET$911\mathrm{.}10\mathrm{.}024.$MI$.0/100$ Submissions Used $.$

ASK YOUR TEACH

Find the Taylor series for $f(x)$ centered at the given value of $a.[$Assume that $f$ has a power series expansion. Do not show that $R_n(x)0.]$

$f(x)=\frac{4}{x},a=-2$

$f(x)={\displaystyle \underset{n=0}{\overset{}{}}}$

Use the binomial series to expand the function as a power series.

$\sqrt[5]{1-x}$

$1-\frac{1}{5}x-{\displaystyle \underset{n=2}{\overset{}{}}}\frac{4*9\text{c}\text{d}\text{o}\text{t}\text{s}(5n-6)}{n!}{x}^n$

$1+{\displaystyle \underset{n=1}{\overset{}{}}}(-1{)}^{n+1}\frac{4*9\text{c}\text{d}\text{o}\text{t}\text{s}(5n-6)}{5^n*n!}{x}^n$

$1+\frac{1}{5}x-{\displaystyle \underset{n=2}{\overset{}{}}}\frac{4*9\text{c}\text{d}\text{o}\text{t}\text{s}(5n-6)}{5^n*n!}{x}^n$

${\displaystyle \underset{n=0}{\overset{}{}}}\frac{(-1{)}^{n+1}(5n-6{)}^n}{{\displaystyle \underset{n=2}{\overset{n}{}}}}{x}^n$

State the radius of convergence, $R.$

$R=$