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: 32


$ x-\dfrac{7}{6} x^3+\dfrac{61}{120} x^5-\dfrac{1247}{5040} x^7+....$

Work Step by Step

The Taylor series for $\cos x $ can be defined as: $\cos x=1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....$ and the Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$ We have $\cos^2 x \sin x =[1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....] \times [x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....]$ or, $=x-\dfrac{7}{6} x^3+\dfrac{61}{120} x^5-\dfrac{1247}{5040} x^7+....$
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