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Q-18E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 434

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Question 18: In Problems, find a power series expansion for $f (X)$ , given the expansion for f(x).
role="math" localid="1664283848012" $f (x) = \sin x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k + 1}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: differentiation of given power series

The differentiation for the series

$f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} f (x) = \sum_{n = 0}^{\infty} (a_{n} x^{n = 1})$

The given series is

$f (x) = \sin x = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k + 1}$

The differentiation of the above will be

$f (x) = \sum_{k = 0}^{\infty} (2 k + 1) (\frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k})$

Step 2: Final proof

$f (x) = \sum_{k = 0}^{\infty} (2 k + 1) (\frac{{(- 1)}^{k}}{(2 k + 1)!} x^{2 k})$

Most popular questions for Math Textbooks

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