## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 10

#### Answer

$$f(x)=(1-x)^{-1/2}=1+\frac{1}{2}x+\frac{3}{2^2}\frac{x^2}{2!}+\frac{3*5}{2^3}\frac{x^3}{3!}+...$$ Convergent for $|x|\lt 1$

#### Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=(1+x)^a$ as follows $$f(x)=(1+x)^a=1+ax+\frac{a(a-1)x^2}{2!}+\frac{a(a-1)(a-2)x^3}{3!}+...$$ Now by the comparison with the function $f(x)=(1-x)^{-1/2}$, we have the Maclaurin series as follows $$f(x)=(1-x)^{-1/2}=1+\frac{1}{2}x+\frac{3}{2^2}\frac{x^2}{2!}+\frac{3*5}{2^3}\frac{x^3}{3!}+...$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=(1+x)^a$ is convergent for the values $|x|\lt 1$. Hence, the Maclaurin series for $f(x)=(1-x)^{-1/2}$ is convergent for the values $|-x^2|=x^2\lt 1$ or $|x|\lt 1$.

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