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: 68

Answer

$1+\frac{x^{2}}{2}+\frac{5x^{4}}{24}+...$

Work Step by Step

$y=secx=\frac{1}{cosx}$ $cosx=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-....$ $secx=\frac{1}{cosx}=\dfrac{1}{1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-....}$ $\approx 1+\frac{x^{2}}{2}+\frac{5x^{4}}{24}+...$
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