Multivariable Calculus, 7th Edition

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: 37

Answer

$\Sigma_{n=1}^{\infty}(-1)^{(n+1)}\frac{{(2)}^{2n-1}x^{2n}}{(2n)!}$ $R=\infty$

Work Step by Step

$sin^{2}x=\frac{1}{2}(1-cos(2x))$ $=\dfrac{1}{2}(1-\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(2x)}^{2n}}{(2n)!})$ $=\dfrac{1}{2}(1-(1+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{{(2)}^{2n}x^{2n}}{(2n)!})$ $=\Sigma_{n=1}^{\infty}(-1)^{(n+1)}\frac{{(2)}^{2n-1}x^{2n}}{(2n)!}$ $R=\infty$
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