Chapter 8.6, Problem 11E

a.

To determine

To find: A power series representation for $f(x)=\frac{1}{{(1+x)}^2}$ using differentiation and to find the radios of convergence.

Answer to Problem 11E

  $\frac{1}{{(1+x)}^2}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)x^n}$ and the radius of convergence of this series is $R=1$ .

Explanation of Solution

Given Information: Given power series is $f(x)=\frac{1}{{(1+x)}^2}$ .

Calculation:

Replacing $t$ with $-x$ in $\frac{1}{1-t}={\displaystyle \sum _{n=0}^{\infty }{t}^n}$ ,

  $\begin{array}{l}\frac{1}{1-(-x)}={\displaystyle \sum _{n=0}^{\infty }{(-x)}^n}\\ \Rightarrow \frac{1}{1+x}={\displaystyle \sum _{n=0}^{\infty }{(-x)}^n}\end{array}$

Differentiating both sides with respect to $x$ ,

  $\begin{array}{l}\frac{d}{dx}\left(\frac{1}{1+x}\right)={\displaystyle \sum _{n=1}^{\infty }\frac{d({(-x)}^n)}{dx}}\\ \Rightarrow \frac{-1}{{(1+x)}^2}={\displaystyle \sum _{n=1}^{\infty }n{(-x)}^{n-1}{(-1)}^n}\\ \Rightarrow \frac{1}{{(1+x)}^2}={\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}n{(-x)}^{n-1}}\\ \Rightarrow \frac{1}{{(1+x)}^2}={\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}n{x}^{n-1}}\\ \Rightarrow \frac{1}{{(1+x)}^2}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)x^n}\end{array}$

Now,

  $\begin{array}{l}$ a_n={(-1)}^n(n+1)x^n,a_{n+1}={(-1)}^{n+1}(n+2)x^{n+1}$\therefore \frac{a_{n+1}}{a_n}=\frac{{(-1)}^{n+1}(n+2)x^{n+1}}{{(-1)}^n(n+1)x^n}=\frac{-(n+2)x}{n+1}\\ \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{-(n+2)x}{n+1}\right|\\ =\underset{n\to \infty }{\lim }\frac{n+2}{n+1}\left|x\right|\\ =\underset{n\to \infty }{\lim }\left(1+\frac{1}{n+1}\right)\left|x\right|\\ =(1+0)\left|x\right|\\ =\left|x\right|\end{array}$

By Ratio Test, this series is convergent if $\left|x\right|<1$ .

Thus, $\frac{1}{{(1+x)}^2}={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)x^n}$ and the radius of convergence of this series is $R=1$ .

b.

To determine

To find: A power series for $f(x)=\frac{1}{{(1+x)}^3}$ .

Answer to Problem 11E

  $\frac{1}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n-1}}$ and the radius of convergence of this series is $R=1$ .

Explanation of Solution

Given Information: Power series from part (a).

Calculation:

From part (a),

  $\frac{1}{{(1+x)}^2}={\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}n{x}^{n-1}}$

Differentiating both sides with respect to $x$ ,

  $\begin{array}{l}\frac{d}{dx}\left(\frac{1}{{(1+x)}^2}\right)={\displaystyle \sum _{n=1}^{\infty }\frac{d[{(-1)}^{n-1}n{x}^{n-1})}{dx}}\\ \Rightarrow \frac{-2}{{(1+x)}^3}={\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}n(n-1)x^{n-2}}\\ \Rightarrow \frac{1}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}n(n-1)x^{n-2}}\\ \Rightarrow \frac{1}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n-1}}\end{array}$

  $\begin{array}{l}$ a_n={(-1)}^{n}n(n+1)x^{n-1},a_{n+1}={(-1)}^{n+1}(n+1)(n+2)x^n$\therefore \frac{a_{n+1}}{a_n}=\frac{{(-1)}^{n+1}(n+1)(n+2)x^n}{{(-1)}^{n}n(n+1)x^{n-1}}=\frac{-(n+2)x}{n}\\ \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{-(n+2)x}{n}\right|\\ =\underset{n\to \infty }{\lim }\frac{n+2}{n}\left|x\right|\\ =\underset{n\to \infty }{\lim }\left(1+\frac{2}{n}\right)\left|x\right|\\ =(1+0)\left|x\right|\\ =\left|x\right|\end{array}$

By Ratio Test, this series is convergent if $\left|x\right|<1$ .

Thus, $\frac{1}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n-1}}$ and the radius of convergence of this series is $R=1$ .

c.

To determine

To find: A power series for $f(x)=\frac{x^2}{{(1+x)}^3}$ .

Answer to Problem 11E

  $\frac{x^2}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n+1}}$ and the radius of convergence of this series is $R=1$ .

Explanation of Solution

Given Information: Power series from part (b).

Calculation:

From part (b),

  $\begin{array}{l}\frac{1}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n-1}}\\ \Rightarrow \frac{x^2}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n+1}}\end{array}$

  $\begin{array}{l}$ a_n=-\frac{1}{2}{(-1)}^n(n+1)n{x}^{n+1},a_{n+1}=-\frac{1}{2}{(-1)}^{n+1}(n+2)(n+1)x^{n+2}$& \therefore \frac{a_{n+1}}{a_n}=\frac{{(-1)}^{n+1}(n+2)(n+1)x^{n+2}}{{(-1)}^n(n+1)n{x}^{n+1}}=\frac{-(n+2)x}{n}\\ \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{-(n+2)x}{n}\right|\\ =\underset{n\to \infty }{\lim }\frac{n+2}{n}\left|x\right|\\ =\underset{n\to \infty }{\lim }\left(1+\frac{2}{n}\right)\left|x\right|\\ =(1+0)\left|x\right|\\ =\left|x\right|\end{array}$

By Ratio Test, this series is convergent if $\left|x\right|<1$ .

Thus, $\frac{x^2}{{(1+x)}^3}=-\frac{1}{2}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n(n+1)n{x}^{n+1}}$ and the radius of convergence of this series is $R=1$ .