Chapter 8.5, Problem 31E

To determine

To calculate: The interval of convergence of the series and an explicit formula for $f\left(x\right)$ .

Answer to Problem 31E

The interval of convergence of the series is $\left(-1,1\right)$ and explicit formula for $f\left(x\right)=\frac{1+2x}{1-x^2}$ .

Explanation of Solution

Given information:

  $f\left(x\right)=1+2x+x^2+2x^3+x^4+\mathrm{.......................}$

There is, its coefficients are $c_{2n}=1$ and $c_{2n+1}=2$ for all $n\ge 0$ .

Concept Used:

The infinite sum of the geometric progression is

  $s_n=a+ar+a{r}^2+\mathrm{................}$

  $s_n=\frac{a}{1-r}$

Calculation:

The series given is

  $f\left(x\right)=1+2x+x^2+2x^3+x^4+\mathrm{.......................}$

Then function can also be written as

  $\begin{array}{l}f\left(x\right)={\displaystyle \sum _{k=0}^{\infty }\left(x^{2k}+2x^{2k+1}\right)}\\ ={\displaystyle \sum _{k=0}^{\infty }\left(1+2x\right)}{x}^{2k}\end{array}$

This function is a geometric series with first term $a=\left(1+2x\right)x^0=1+2x$ and the common ratio is $r=x^2$ .

Therefore

  $\begin{array}{l}f\left(x\right)=\frac{a}{1-r}\\ =\frac{1+2x}{1-x^2}\end{array}$

This geometric series converges when the value of common ratio

  $\begin{array}{l}\left|r\right|=\left|x^2\right|<1\\ \Rightarrow x^2<1\end{array}$

Therefore, the interval of convergence is $\left(-1,1\right)$

Conclusion:

The interval of convergence of the series is $\left(-1,1\right)$ and explicit formula for $f\left(x\right)=\frac{1+2x}{1-x^2}$ .