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: 35

Answer

$x^2-\dfrac{x^3}{2}+\dfrac{x^4}{6}-....$

Work Step by Step

The Maclaurin's series for $\sin x$ is as follows: $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}$ Now, we have the Maclaurin's series for $\ln (1+x)$ as follows: $x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-......=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{n+1}}{(n+1)}$ Therefore, $f(x)=\sin x \ln (1+x)$ $\sin x \ln (1+x)=[\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}]-[\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{n+1}}{(n+1)}]$ or, $=x^2-\dfrac{x^3}{2}+\dfrac{x^4}{6}-....$
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