University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 1



Work Step by Step

Since, we know that the Maclaurin Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Now, $e^{-5x}=\Sigma_{n=0}^\infty \dfrac{(-5x)^n}{n!}=1-5x+\dfrac{(-5x)^2}{2!}+\dfrac{(-5x)^3}{3!}-\dfrac{(-5x)^4}{4!}...$ or, $=1-5x+\dfrac{25}{2}x^2-\dfrac{125}{6}x^3+\dfrac{625}{24}x^4-...$
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