Chapter 8.6, Problem 7E

To determine

To find: The power series representation of $f(x)={\displaystyle \sum _{n=0}^{\infty }\frac{x}{9+x^2}}$ .

Answer to Problem 7E

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: $f(x)=\frac{1}{x^2+9}$

  $=x·\frac{1}{9\left(1+{(\frac{x}{3})}^2\right)}$

  $=x·\frac{1}{9[1-(-\frac{x^2}{9})]}$ [ by help of (1)]

  $=x·\frac{1}{9[1-\{-{(\frac{x}{3})}^2\}]}$

The power series expression is

  $f(x)=\frac{x}{9}{\displaystyle \sum _{n=0}^{\infty }\frac{1}{10}}·{\left({(\frac{x}{3})}^2\right)}^n\text{ , }\left|-{(\frac{x}{3})}^2\right|<1$

[ by help of (1)]

= $\frac{x}{9}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n}}{9^n}}$ , $\left|-1\right|·\left|-{(\frac{x}{3})}^2\right|<1$

  $={\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n}{9^{n+1}}·}{x}^{2n+1}$ , $\left|-{(\frac{x}{3})}^2\right|<1$

And the Interval of convergence is $\left|-{(\frac{x}{3})}^2\right|<1$

i.e, $\left|\frac{x}{3}\right|<1$

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

i.e $-3<x<3$

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