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: 7



Work Step by Step

Since, we know that the Maclaurin Series for $\ln(1+x)$ is defined as: $ \ln (1+x)=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{n+1}}{(n+1)!}=x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+...$ Now, $ \ln (1+x^2)=x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-\dfrac{(x^2)^4}{4!}+...$ or, $=x^2-\dfrac{x^4}{2}+\dfrac{x^{6}}{3}-\dfrac{x^{8}}{4}$
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