University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 549: 1

Answer

$1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+...$

Work Step by Step

Formula to find the binomial series is: $(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$ Here, $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$ Now, $(1+x)^{1/2}=1+\dfrac{1}{2}x+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})x^2}{2!}+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-\dfrac{3}{2})x^3}{3!}+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})x^4}{4!}+...$ Thus, the first four terms are: $1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+...$
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