## Calculus (3rd Edition)

Published by W. H. Freeman

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

#### Answer

$$f(x)=\tan^{-1}(x^2)= x^2-\frac{x^6}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+...$$ Convergent for $|x|\leq 1$

#### Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=\tan^{-1}x$ as follows $$f(x)=\tan^{-1}= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$$ Now by the comparison with the function $f(x)=\tan^{-1}(x^2)$, we have the Maclaurin series as follows $$f(x)=\tan^{-1}(x^2)= x^2-\frac{x^6}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+...$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=\tan^{-1}x$ is convergent for the values $|x|\leq 1$. Hence the Maclaurin series for $f(x)=\tan^{-1}(x^2)$ is convergent for the values $|x^2|=x^2\leq 1$ or $|x|\leq 1$.

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