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: 18


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

Work Step by Step

Consider the Maclaurin Series for $\cos x$ is defined as: $ \cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ Use formula such as: $1-\cos 2x=2 \sin^2 x$ Then, we have $\sin^2 x=\dfrac{1}{2}-\dfrac{\cos 2x}{2}=\dfrac{1}{2}-\dfrac{1}{2}[1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+...]$ $ \implies \Sigma_{n=1}^\infty (-1)^{(n+1)} \dfrac{2^{(2n-1)}x^{(2n)}}{(2n)!}$
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