6.2 Properties of power series  (Page 2/10)

1. Since the interval $\left(\mathrm{-1},1\right)$ is a common interval of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ and ${\displaystyle \sum _{n=0}^{\infty }{b}_n{x}^n},$ the interval of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\left(a_n{x}^n+b_n{x}^n\right)}$ is $\left(\mathrm{-1},1\right).$
2. Since ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ is a power series centered at zero with radius of convergence 1, it converges for all x in the interval $\left(\mathrm{-1},1\right).$ By [link] , the series
   
   ${\displaystyle \sum _{n=0}^{\infty }{a}_n{3}^n{x}^n}={\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(3x\right)}^n}$
   
   converges if 3 x is in the interval $\left(\mathrm{-1},1\right).$ Therefore, the series converges for all x in the interval $\left(-\frac{1}{3},\frac{1}{3}\right).$

1. First write $f\left(x\right)$ as
   
   $f\left(x\right)=3x\left(\frac{1}{1-\left(\text{−}{x}^2\right)}\right).$
   
   Using the power series representation for $f\left(x\right)=\frac{1}{1-x}$ and parts ii. and iii. of [link] , we find that a power series representation for f is given by
   
   ${\displaystyle \sum _{n=0}^{\infty }3x{\left(\text{−}{x}^2\right)}^n}={\displaystyle \sum _{n=0}^{\infty }3{\left(\mathrm{-1}\right)}^n{x}^{2n+1}}.$
   
   Since the interval of convergence of the series for $\frac{1}{1-x}$ is $\left(\mathrm{-1},1\right),$ the interval of convergence for this new series is the set of real numbers x such that $\left|x^2\right|<1.$ Therefore, the interval of convergence is $\left(\mathrm{-1},1\right).$
2. To find the power series representation, use partial fractions to write $f\left(x\right)=\frac{1}{\left(1-x\right)\left(x-3\right)}$ as the sum of two fractions. We have
   
   $\begin{array}{cc}\hfill \frac{1}{\left(x-1\right)\left(x-3\right)}& =\frac{\text{−}1\text{/}2}{x-1}+\frac{1\text{/}2}{x-3}\hfill \\ & =\frac{1\text{/}2}{1-x}-\frac{1\text{/}2}{3-x}\hfill \\ & =\frac{1\text{/}2}{1-x}-\frac{1\text{/}6}{1-\frac{x}{3}}.\hfill \end{array}$
   
   First, using part ii. of [link] , we obtain
   
   $\frac{1\text{/}2}{1-x}={\displaystyle \sum _{n=0}^{\infty }\frac{1}{2}{x}^n}\text{for}\left|x\right|<1.$
   
   Then, using parts ii. and iii. of [link] , we have
   
   $\frac{1\text{/}6}{1-x\text{/}3}={\displaystyle \sum _{n=0}^{\infty }\frac{1}{6}{\left(\frac{x}{3}\right)}^n}\text{for}\left|x\right|<3.$
   
   Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of [link] , we have
   
   $\frac{1}{\left(x-1\right)\left(x-3\right)}={\displaystyle \sum _{n=0}^{\infty }\left(\frac{1}{2}-\frac{1}{6·3^n}\right)x^n}$
   
   where the interval of convergence is $\left(\mathrm{-1},1\right).$

Writing the given series as

we can recognize this series as the power series for

Since this is a geometric series, the series converges if and only if $\left|2x\right|<1.$ Therefore, the interval of convergence is $\left(-\frac{1}{2},\frac{1}{2}\right).$