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 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 618: 36

Answer

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

Work Step by Step

The Maclaurin series for $\sin x$ can be defined as: $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......=\Sigma_{n=0}^{\infty} \dfrac{ (-1)^n x^{2n+1}}{(2n+1)!}$ Now, $x \times \sin^2 x=x [\Sigma_{n=0}^{\infty}\dfrac{ (-1)^n x^{2n+1}}{(2n+1)!}]^2$ $\implies x \times \sin^2 x=x[x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......]^2$ Thus, we have the result: $x \times \sin^2 x =x^3-\dfrac{x^5}{3}+\dfrac{2 x^7}{45}+....$
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