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

Published by Pearson

# Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 33

#### Answer

$-1-2x-\dfrac{5x^2}{2}+....$

#### Work Step by Step

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

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