Chapter 10.1, Problem 21E

To determine

To calculate: The interval of convergence and write the function in terms of $x$ .

Answer to Problem 21E

The required function is $f\left(x\right)=\frac{1}{1-2x}$ .

Explanation of Solution

Given Information:

The series is ${\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}$ .

Formula Used:

The function of the geometric series is $f\left(x\right)=\frac{a}{1-r}$ , where $a$ is initial term and $r$ is the common ratio.

Interval of Convergence:

The geometric series converges then, the interval of convergence is $\left|r\right|<1$ .

Calculation:

The series is ${\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}$ .

Rewrite the series as:

  ${\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}={\displaystyle \sum _{n=0}^{\infty }{\left(2x\right)}^n}$

In the series ${\displaystyle \sum _{n=0}^{\infty }{\left(2x\right)}^n}$ , $a=1$ and $r=2x$ .

Now, find the interval of convergence.

  $\begin{array}{c}\left|2x\right|<1\\ -1<2x<1\\ -\frac{1}{2}<x<\frac{1}{2}\end{array}$

Find the function in terms of $x$ .

Substitute 1 for $a$ and $2x$ for $r$ in the formula $\frac{a}{1-r}$ .

  $f\left(x\right)=\frac{1}{1-2x}$ , where $-\frac{1}{2}<x<\frac{1}{2}$

Hence, the requiredfunction is $f\left(x\right)=\frac{1}{1-2x}$ .