Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 24


$1-\dfrac{3x}{2}+\dfrac{15 x^2}{8}-\dfrac{35x^3}{16}+...$

Work Step by Step

We have the Maclaurin series for $(1+x)^a$ as follows $(1+x)^a=1+ax+\dfrac{a(a-1)x^2}{2!}x^2+\dfrac{a(a-1)(a-2)}{3!}x^3+...$ Replace $\dfrac{-3}{2}$ for $x$ in the above form through the first four terms. Therefore, we have: $(1+x)^{-3/2}=1+\dfrac{-3x}{2}+\dfrac{\dfrac{-3}{2}(\dfrac{-3}{2}-1)x^2}{2!}x^2+\dfrac{\dfrac{-3}{2}(\dfrac{-3}{2}-1)(\dfrac{-3}{2}-2)}{3!}x^3+...\\=1-\dfrac{3x}{2}+\dfrac{15 x^2}{8}-\dfrac{35x^3}{16}+...$
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