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: 16


$x^2-\dfrac{x^6}{2!}+\dfrac{ x^{10}}{4!}-\dfrac{x^{14}}{6!}+..$

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!}+...$ Now, $ x^2 \cos (x^2)=x^2(1-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^4}{4!}-\dfrac{(x^2)^6}{6!}+...)=x^2-\dfrac{x^6}{2!}+\dfrac{ x^{10}}{4!}-\dfrac{x^{14}}{6!}+..$
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