## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 14

#### Answer

$$f(x)=\frac{1-\cos x}{x} = \frac{x}{2!}-\frac{x^3}{4!}+\frac{x^5}{6!}+..$$ Convergent for $x\ne 0$.

#### Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=\cos x$ as follows $$f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$ Now by the comparison with the function $f(x)=\frac{1-\cos x}{x}$, we have the Maclaurin series as follows $$f(x)=\frac{1-\cos x}{x}\\ =\frac{1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)}{x}\\ = \frac{x}{2!}-\frac{x^3}{4!}+\frac{x^5}{6!}+..$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=\cos x$ is convergent for any value of $x$. In our case, the series does not converge for $x=0$ (division by 0). Hence, the Maclaurin series for $f(x)=\frac{1-\cos x}{x}$ is convergent for $x\ne 0$.

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