Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 24


$x-\dfrac{4}{3!}x^3+\dfrac{16}{5!}x^5-\dfrac{64}{7!}x^7+...=\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{(2n)}x^{(2n+1)}}{(2n+1)!}$

Work Step by Step

Consider 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!}+...$ Use formula such as: $\sin x \cos x=\dfrac{\sin 2x}{2}$ Then, we have $\sin x \cos x=\dfrac{\sin 2x}{2}=\dfrac{2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-\dfrac{(2x)^7}{7!}+...}{2}$ Thus, $x-\dfrac{4}{3!}x^3+\dfrac{16}{5!}x^5-\dfrac{64}{7!}x^7+...=\Sigma_{n=0}^\infty (-1)^n\dfrac{2^{(2n)}x^{(2n+1)}}{(2n+1)!}$
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