Chapter 8.6, Problem 5E

To determine

To find: The power series representation of $f(x)=\frac{2}{3-x}$ .

Answer to Problem 5E

Interval of convergence $(-3,3)$

Explanation of Solution

Given information: We know that $\frac{1}{1-x}$is a geometric series and its expression is $\frac{1}{1-x}=1+x+x^2+x^3+\mathrm{...}={\displaystyle \sum _{n=0}^{\infty }{x}^n}\left|x\right|<1.\mathrm{.....}\text{(1)}$

Solution: $f(x)=\frac{2}{3-x}$

  $=2·\frac{1}{1-x}$

  $=\frac{2}{3[1-(\frac{x}{3})]}$ [ by help of (1)]

  $=\frac{2}{3}·\frac{1}{[1-(\frac{x}{3})]}$

The power series expression is

  $f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{2}{3}}·{\left(\frac{x}{3}\right)}^n\text{ , }\left|\frac{x}{3}\right|<1$

[ by help of (1)]

= ${\displaystyle \sum _{n=0}^{\infty }\frac{2}{3}·\frac{x^n}{3^n}}$ , $\left|\frac{x}{3}\right|<1$

  $={\displaystyle \sum _{n=0}^{\infty }\frac{2}{3^{n+1}}·}{x}^n$ , $\left|\frac{x}{3}\right|<1$

Interval of convergence $\left|\frac{x}{3}\right|<1$

i.e $\left|x\right|<3$

i.e $-3<x<3$

And interval of convergence $(-3,3)$