Chapter 10.1, Problem 64E

To determine

To find: The series for the given function and also the interval of convergence of series.

Answer to Problem 64E

The series $f(x)$ is converge when $\left|x\right|<1,$ so the interval of convergence is (−1,1)

Explanation of Solution

Given information:

The function is:

  $f(x)=\frac{2}{{(1-x)}^3}$

Calculation:

Let's take the function $g(x)$ such that $g(x)=\int f(x)dx$

Since, $f(x)=\frac{2}{{(1-x)}^3}\int f(x)dx=\int \frac{2}{{(1-x)}^3}dx={(1-x)}^2+C$

Therefore, $g(x)=\frac{1}{{(x-1)}^2}$ (ignoring the constant C)

  $g(x)=\frac{1}{{(x-1)}^2}=1+2x+3x^2+4x^3+\dots \mathrm{..}+n{x}^{n-1}+\dots \dots .$

To find the power series, let's differentiate on both the sides with respect to $x$

  $\frac{d[g(x)]}{dx}=\frac{d}{dx}\left(1+2x+3x^2+4x^3+\dots \mathrm{..}+n{x}^{n-1}+\dots \dots \right)$

  $\frac{2}{-{(x-1)}^3}=0+2+6x+12x^2+\dots \dots +n(n-1)x^{n-2}+\dots \dots \dots$

  $\frac{2}{{(1-x)}^3}=2+6x+12x^2+\dots \dots +n(n-1)x^{n-2}+\dots \dots \dots$

Therefore, power series of $f(x)$ is,

  $\frac{2}{{(1-x)}^3}=2+6x+12x^2+\dots \dots +n(n-1)x^{n-2}+\dots \dots \mathrm{..}$

  $f(x)=2+6x+12x^2+\dots \dots +n(n-1)x^{n-2}+\dots \dots .$

The series $f(x)$ is converge when $\left|x\right|<1,$ so the interval of convergence is (−1,1)