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: 24


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

Work Step by Step

We know that $\sin x \cos x=\dfrac{\sin 2x}{2}$ We know that the Maclaurin Series for $\sin x$ is defined as: $ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ Then $\sin x \cos x=\dfrac{\sin 2x}{2}=\dfrac{2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-\dfrac{(2x)^7}{7!}+...}{2}$ or, $=x-\dfrac{4x^3}{3!}+\dfrac{16x^5}{5!}-\dfrac{64x^7}{7!}+...$ or, $=\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{2n}x^{2n+1}}{(2n+1)!}$
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