Answer
$\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi ^{(2n+1)}x^{(2n+1)}}{(2n+1)!}$
Work Step by Step
$sin\pi x=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(\pi x)^{(2n+1)}}{(2n+1)!}$
$=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi ^{(2n+1)}x^{(2n+1)}}{(2n+1)!}$
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Chapter 11 - Infinite Sequences and Series - 11.10 Exercises - Page 789: 29
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