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: 20



Work Step by Step

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