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Calculus / Calculus: Early Transcendentals 1 / Chapter 9.2 / Problem 5E

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

Do the interval and radius of convergence of a power

Chapter 8, Problem 5E

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QUESTION:

Do the interval and radius of convergence of a power series change when the series is differentiated or integrated? Explain.

Questions & Answers

QUESTION:

Do the interval and radius of convergence of a power series change when the series is differentiated or integrated? Explain.

ANSWER:

Solution 5E

Step 1:

Consider the function $f(x)=\frac{4}{(2-x{)}^{2}}$ ,

which is the derivative of the function $g(x)=\frac{2x}{2-x}$

This $g(x)$ has the power series representation of the form $g(x)=\frac{2x}{2-x}=\sum\limits_{n=1}^{\infty{}}\frac{1}{{2}^{n-1}}{x}^{n}$

This series converges when $-2<x<2$ .

The power series representation of $f(x)$ is obtained by differentiating the series term by term.thus we get

$f(x)=\frac{4}{(2-x{)}^{2}}=\sum\limits_{n=1}^{\infty{}}\frac{1}{{2}^{n-1}}n{x}^{n-1}=\sum\limits_{n=0}^{\infty{}}\frac{(n+1)}{{2}^{n}}{x}^{n}$

This also converges when $-2<x<2$ .

Thus the interval and radius of convergence of a power series does not change on differentiating.

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Do the interval and radius of convergence of a power

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Solution Found!

Do the interval and radius of convergence of a power

Questions & Answers

Study Tools You Might Need

Chapter: 8 Problem: 25P

Chapter: 16 Problem: 93

Chapter: 4 Problem: 22

Chapter: 15 Problem: 3

Chapter: 7 Problem: 15

Chapter: 4 Problem: 7

Chapter: 20 Problem: 30

Chapter: 17 Problem: 34

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