Chapter 8.5, Problem 32E

To determine

To find: the interval of convergence of the series and a formula for f(x).

Answer to Problem 32E

The values of x for which f(x) converges:

  $x\in (-1,1)$ .

  $f(x)=\frac{c_0+x{c}_1+x^2{c}_2+x^3{c}_3}{1-x^4}$ .

Explanation of Solution

Given information: Given function $f(x)={\displaystyle \sum _{n=0}^{\infty }{c}_n}{x}^n,\text{Where}{c}_{n+4}=c_n\text{for}\text{all}n\ge 0.$

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{c}_n}{x}^n=c_1{x}^1+c_2{x}^2+\mathrm{.....}{c}_n{x}^n\\ f(x)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_0{x}^4+c_1{x}^5+c_2{x}^6+c_3{x}^7+\mathrm{.....}\end{array}$

  $f(x)=c_0[1+x+x^2+x^3+\mathrm{...}{x}^n]+c_1[x+x^5+x^9+\mathrm{..}{x}^{4n+1}]+c_2[x^2+x^6+x^{10}+\mathrm{..}{x}^{4n+2}]+c_3[x^3+x^7+x^{11}+\mathrm{..}{x}^{4n+3}]$

  $f(x)={\displaystyle \sum _{n=0}^{\infty }[c_0}{x}^{4n}+c_1{x}^{4n+1}+c_2{x}^{4n+2}+c_3{x}^{4n+3}]$

  $f(x)={\displaystyle \sum _{n=0}^{\infty }[c_0}{x}^{4n}]+{\displaystyle \sum _{n=0}^{\infty }[}{c}_1{x}^{4n+1}]+{\displaystyle \sum _{n=0}^{\infty }[}{c}_2{x}^{4n+2}]+{\displaystyle \sum _{n=0}^{\infty }[}{c}_3{x}^{4n+3}].$

  $f(x)={\displaystyle \sum _{n=0}^{\infty }[c_0}{x}^{4n}]+{\displaystyle \sum _{n=0}^{\infty }[x{c}_1}{x}^{4n}]+{\displaystyle \sum _{n=0}^{\infty }[x^2{c}_2}{x}^{4n}]+{\displaystyle \sum _{n=0}^{\infty }[x^3{c}_3}{x}^{4n}]$

  $\begin{array}{l}\text{Geomatric}\text{series}\text{.}\\ {\displaystyle \sum _0^{\infty }a{r}^n}=\frac{1}{1-r}.\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }[c_0}{x}^{4n}]=\frac{c_0}{1-x^4}.\\ {\displaystyle \sum _{n=0}^{\infty }[x{c}_1}{x}^{4n}]=\frac{x{c}_1}{1-x^4}.\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }[x^2{c}_2}{x}^{4n}]=\frac{x^2{c}_2}{1-x^4}.\\ {\displaystyle \sum _{n=0}^{\infty }[x^3{c}_2}{x}^{4n}]=\frac{x^3{c}_3}{1-x^4}.\end{array}$

  $f(x)=\frac{c_0}{1-x^4}+\frac{x{c}_1}{1-x^4}+\frac{x^2{c}_2}{1-x^4}+\frac{x^3{c}_3}{1-x^4}.$

  $f(x)=\frac{c_0+x{c}_1+x^2{c}_2+x^3{c}_3}{1-x^4}.$

As for the interval of convergence.

  ${\displaystyle \sum _{n=0}^{\infty }a{r}^n}=\frac{1}{1-r}.$

The series converges is $\begin{array}{l}\left|r\right|>1,\text{so}\text{let}r=x:\\ \left|x\right|<1\end{array}$

Then let $r=x^2,$ and for all 4 series, the same exact result:

  $\left|x\right|<1$ .

So, the values of x for which f(x) converges:

  $x\in (-1,1)$