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Q. 1

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Calculus
Found in: Page 703
Calculus

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

The Calculus of Power Series: Let \(\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}\) be a power series in \(x-x_{0}\) that converges to a function \(f(x)\) on an interval \(I\).

The derivative of \(f\) is given by \(f^{\prime}\left( x \right )=\)____.

The interval of convergence for this new series is ______, with the possible exception that _____.

The derivative of \(f\) is given by \(f^{\prime}\left( x \right )=\sum_{k=1}^{\infty }ka_{k}\left ( x-x_{0} \right )^{k-1}\).

The interval of convergence for this new series is \(I\), with the possible exception that if \(I\) is a finite interval, the convergence at the endpoints of \(I\) may be different than it was for the initial series.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1. Derivative of \(f\)

The given function is \(\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}\).

So, the derivative of the function is

\(f^{\prime}\left ( x \right )=\frac{d}{dx}\left ( \sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k} \right )\)

\(f^{\prime}\left ( x \right )=\sum_{k=0}^{\infty }a_{k}\frac{d}{dx}\left ( \left ( x-x_{0} \right )^{k} \right)\)

\(f^{\prime}\left ( x \right )=\sum_{k=1}^{\infty }ka_{k}\left ( x-x_{0} \right )^{k-1}\)

Step 2. Derivative of \(f\) 

So, the derivative of \(f\) is given by \(f^{\prime}\left( x \right )=\sum_{k=1}^{\infty }ka_{k}\left ( x-x_{0} \right )^{k-1}\).

The interval of convergence for this new series is \(I\), with the possible exception that if \(I\) is a finite interval, the convergence at the endpoints of \(I\) may be different than it was for the initial series.

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