University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 18


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

Work Step by Step

Since, we know that the Maclaurin Series for $\cos x$ is defined as: $ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ We also know that $1-\cos 2x=2 \sin^2 x$ Now, $ sin^2 x=\dfrac{1}{2}-\dfrac{\cos 2x}{2}=\dfrac{1}{2}-\dfrac{1}{2}[1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+...]$ or, $=\Sigma_{n=1}^\infty \dfrac{(-1)^{n+1} 2^{2n-1}x^{2n}}{(2n)!}$
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