## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 12

#### Answer

$x^3-\dfrac{x^5}{3!}+\dfrac{x^7}{5!}-\dfrac{x^9}{7!}+...$

#### Work Step by Step

Since, we know that the Maclaurin Series for $\sin x$ is defined as: $\sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ Now, $x^2 \sin x=x^2 (x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)=x^3-\dfrac{x^5}{3!}+\dfrac{x^7}{5!}-\dfrac{x^9}{7!}+...$

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