Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 21


$x+x^2+\dfrac{ 5x^3}{6}+\dfrac{5x^4}{6}+...$

Work Step by Step

We have the Maclaurin series for $ \sin x$ and $e^x$ as follows $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$ and $$\dfrac{1}{ (1-x)}=1+x+x^2+x^3+...$$ Now, we have the Maclaurin series of $\dfrac{\sin x}{1-x}$ as: $$\dfrac{\sin x}{1-x}=\\ =\dfrac{(\displaystyle x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...)}{1+x+x^2+x^3+...}\\ =x+x^2+x^3-\dfrac{x^3}{3!}+x^4-\dfrac{x^4}{3!}+...\\=x+x^2+\dfrac{ 5x^3}{6}+\dfrac{5x^4}{6}+...$$
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