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: 31


$ x^2-\dfrac{2}{3}x^4+\dfrac{23}{45}x^6-\dfrac{44}{105}x^8+.....$

Work Step by Step

The Taylor series for $\tan^{-1} x $ at $ x=0$ can be defined as: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....$ Consider the given series: $(\tan^{-1} x)^2= (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) \times (x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....) $ or, $=x^2+(-\dfrac{1}{3}-\dfrac{1}{3}) x^4+(\dfrac{1}{5}+\dfrac{1}{(3)(3)}+\dfrac{1}{5}) x^6+......$ or, $=x^2-\dfrac{2}{3}x^4+\dfrac{23}{45}x^6-\dfrac{44}{105}x^8+.....$
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