Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 67

Answer

$1-\frac{3x^{2}}{2}+\frac{25x^{4}}{24}+...$

Work Step by Step

$y=e^{-x^{2}}cosx$ $cosx=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-....$ and $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$ Thus, $e^{-x^{2}}=1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+...$ $e^{-x^{2}}cosx=(1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+...)(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-....)$ $\approx 1-\frac{3x^{2}}{2}+\frac{25x^{4}}{24}+...$
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