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: 23


$x^3-\dfrac{x^7}{3}+\dfrac{x^{11}}{5}-....$ or, $\Sigma_{n=1}^\infty (-1)^{n}\dfrac{x^{4n-1}}{2n-1}$

Work Step by Step

We know that the Taylor Series for $\tan^{-1} x$ is defined as: $\tan^{-1} x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-....$ Now, $ x\tan^{-1} x^2=x[x^2-\dfrac{x^6}{3}+\dfrac{x^{10}}{5}-....]$ or, $=x^3-\dfrac{x^7}{3}+\dfrac{x^{11}}{5}-....$ or, $\Sigma_{n=1}^\infty (-1)^{n}\dfrac{x^{4n-1}}{2n-1}$
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