## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 18

#### Answer

$$f(x)=\cosh x=\frac{e^{x}+e^{-x}}{2}\\ =1+\frac{x^2}{2!}+\frac{x^4}{4!}+...$$ Convergent for any value of $x$.

#### Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows $$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ Now by the comparison with the function $f(x)=\cosh x=\frac{e^{x}+e^{-x}}{2}$, we have the Maclaurin series as follows $$f(x)=\cosh x=\frac{e^{x}+e^{-x}}{2}\\ =\frac{1}{2}(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\ +\frac{1}{2}(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+...)\\ =1+\frac{x^2}{2!}+\frac{x^4}{4!}+...$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence, the Maclaurin series for $f(x)=\cosh x$ is also convergent for any value of $x$.

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