Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 633: 8

Answer

$1-\dfrac{1}{3}x^3+\dfrac{2}{9}x^4-\dfrac{14}{81}x^6$

Work Step by Step

Apply Binomial series formula to find the first four terms. $(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$ and $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$ Thus, we have $(1+x^2)^{-1/3}=1-\dfrac{1}{3}(x^2)+\dfrac{(\dfrac{-1}{3})(-\dfrac{4}{3})(x^2)^2}{2!}+\dfrac{(-\dfrac{1}{3})(-\dfrac{4}{3})(-\dfrac{7}{3})(x^2)^3}{3!}+...=1-\dfrac{1}{3}x^3+\dfrac{2}{9}x^4-\dfrac{14}{81}x^6+...$ Hence, we have the first four terms: $1-\dfrac{1}{3}x^3+\dfrac{2}{9}x^4-\dfrac{14}{81}x^6$
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