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: 13

Answer

$\dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\dfrac{1}{8!}x^8-\dfrac{1}{10!}x^{10}+...$

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!}+...$ Then $\dfrac{x^2}{2}-1+\cos x=\dfrac{x^2}{2}-1+1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ or, $\implies \dfrac{1}{4!}x^4-\dfrac{1}{6!}x^6+\dfrac{1}{8!}x^8-\dfrac{1}{10!}x^{10}+...$
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