University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 36


$x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$

Work Step by Step

The Maclaurin's series for $\sin x$ is as follows: $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}$ Therefore, $f(x)=x \sin^2 x$ $x \sin^2 x=x [\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}]^2$ or, $=x[x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......]^2$ or, $=x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$
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