Chapter 10.1, Problem 27E

To determine

To calculate: The power series for $f^{\prime }\left(x\right)$ by the series and the function.

Answer to Problem 27E

The power series is $f^{\prime }\left(x\right)=2+8x+24x^2+64x^3+\cdots$ , where $-\frac{1}{2}<x<\frac{1}{2}$ .

Explanation of Solution

Given Information:

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

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

Formula Used:

Power Rule:

Let the function be $f\left(x\right)=x^n$ .

The derivative is $f^{\prime }\left(x\right)=n{x}^{n-1}$ .

Calculation:

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

Rewrite the function as:

  $f\left(x\right)=\frac{1}{1-2x}={\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}$

Simplify the series.

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{2}^n{x}^n}\\ =1+2x+4x^2+8x^3+\cdots \end{array}$ , where $-\frac{1}{2}<x<\frac{1}{2}$

Now, the function becomes,

  $f\left(x\right)=1+2x+4x^2+8x^3+\cdots$

Find the power series.

  $\begin{array}{c}\frac{d\left[f\left(x\right)\right]}{dx}=\frac{d}{dx}\left[1+2x+4x^2+8x^3+\cdots \right]\\ f^{\prime }\left(x\right)=2+8x+24x^2+64x^3+\cdots \end{array}$

Hence, the power series is $f^{\prime }\left(x\right)=2+8x+24x^2+64x^3+\cdots$ , where $-\frac{1}{2}<x<\frac{1}{2}$ .