6.1 Power series and functions  (Page 3/8)

From the graph in [link] you see that as N increases, $S_N$ becomes a better approximation for $f\left(x\right)=\frac{1}{1-x}$ for x in the interval $\left(\mathrm{-1},1\right).$

1. You should recognize this function f as the sum of a geometric series, because
   
   $\frac{1}{1+x^3}=\frac{1}{1-\left(\text{−}{x}^3\right)}.$
   
   Using the fact that, for $\left|r\right|<1,\frac{a}{1-r}$ is the sum of the geometric series
   
   ${\displaystyle \sum _{n=0}^{\infty }a}{r}^n=a+ar+a{r}^2+\text{⋯},$
   
   we see that, for $\left|\text{−}{x}^3\right|<1,$
   
   $\begin{array}{cc}\hfill \frac{1}{1+x^3}& =\frac{1}{1-\left(\text{−}{x}^3\right)}\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }{\left(\text{−}{x}^3\right)}^n}\hfill \\ & =1-x^3+x^6-x^9+\text{⋯}.\hfill \end{array}$
   
   Since this series converges if and only if $\left|\text{−}{x}^3\right|<1,$ the interval of convergence is $\left(\mathrm{-1},1\right),$ and we have
   
   $\frac{1}{1+x^3}=1-x^3+x^6-x^9+\text{⋯}\text{for}\left|x\right|<1.$
2. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain
   
   $\begin{array}{cc}\hfill \frac{x^2}{4-x^2}& =\frac{x^2}{4\left(\frac{1-x^2}{4}\right)}\hfill \\ & =\frac{x^2}{4\left(1-{\left(\frac{x}{2}\right)}^2\right)}.\hfill \end{array}$
   
   Therefore, we have
   
   $\begin{array}{cc}\hfill \frac{x^2}{4-x^2}& =\frac{x^2}{4\left(1-{\left(\frac{x}{2}\right)}^2\right)}\hfill \\ & =\frac{\frac{x^2}{4}}{1-{\left(\frac{x}{2}\right)}^2}\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }\frac{x^2}{4}{\left(\frac{x}{2}\right)}^{2n}}.\hfill \end{array}$
   
   The series converges as long as $\left|{\left(\frac{x}{2}\right)}^2\right|<1$ (note that when $\left|{\left(\frac{x}{2}\right)}^2\right|=1$ the series does not converge). Solving this inequality, we conclude that the interval of convergence is $\left(\mathrm{-2},2\right)$ and
   
   $\begin{array}{cc}\hfill \frac{x^2}{4-x^2}& ={\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n+2}}{4^{n+1}}}\hfill \\ & =\frac{x^2}{4}+\frac{x^4}{4^2}+\frac{x^6}{4^3}+\text{⋯}\hfill \end{array}$
   
   for $\left|x\right|<2.$