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Calculus / Calculus: Early Transcendentals 1 / Chapter 9.2 / Problem 34E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Answer: Differentiating and integrating power series Find

Chapter 8, Problem 34E

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QUESTION:

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.

\(g(x)=\frac{1}{(1-x)^{3}}\) using \(f(x)=\frac{1}{1-x}\)

Questions & Answers

QUESTION:

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.

\(g(x)=\frac{1}{(1-x)^{3}}\) using \(f(x)=\frac{1}{1-x}\)

ANSWER:

Solution 34E

Step 1:

In this problem we need to find the the power series representation for $g(x)=\frac{1}{(1\ -\ x{)}^{3}}$ centered at 0 by differentiating or integrating the power series for $f(x)=\frac{1}{1\ -\ x\ }$

We already know a power series for $f(x)=\frac{1}{1\ -\ x}$ ,namely $\frac{1}{1\ -\ x}=\sum\limits_{n\ =\ 0}^{\infty{}}{x}^{n}$ for $-1<x<1$

$\frac{1}{1\ -\ x}=\sum\limits_{n\ =\ 0}^{\infty{}}{x}^{n}$

$\Rightarrow{}\frac{d}{dx}(\frac{1}{1\ -\ x})=\frac{d}{dx}(\sum\limits_{n\ =\ 0}^{\infty{}}{x}^{n})$

$\Rightarrow{}\frac{2}{(1\ -\ x{)}^{2}}=\sum\limits_{n\ =\ 0}^{\infty{}}n{x}^{n-1}$

$\Rightarrow{}\frac{d}{dx}(\frac{2}{(1\ -\ x{)}^{2}})=\frac{d}{dx}(\sum\limits_{n\ =\ 0}^{\infty{}}n{x}^{n-1})$

$\Rightarrow{}\frac{2}{(1\ -\ x{)}^{3}}=\sum\limits_{n=0}^{\infty{}}n(n-1){x}^{n-2}$

$\Rightarrow{}\frac{1}{(1\ -\ x{)}^{3}}=\frac{1}{2}\sum\limits_{n\ =\ 0}^{\infty{}}n(n-1){x}^{n-2}$

Therefore power series representation for $g(x)=\frac{1}{(1\ -\ x{)}^{3}}$ is $\frac{1}{2}\sum\limits_{n\ =\ 0}^{\infty{}}n(n-1){x}^{n-2}$

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