Finding the Maclaurin Expansion for \(f(x) =\cosh x\)

Find the Maclaurin expansion for \(f(x) =\cosh x.\)

Solution Since \[ \cosh x=\frac{e^{x}+e^{-x}}{2} \]

its Maclaurin expansion can be found by adding corresponding terms of the Maclaurin expansions for \(e^{x}\) and \(e^{-x}\) and then dividing by 2. The result is \[ \begin{eqnarray*} \hspace{-10pc} \cosh x &=&\frac{e^{x}+e^{-x}}{2}=\frac{\left( 1+x+\frac{x^{2}}{2!}+ \frac{x^{3}}{3!}+\cdots +\frac{x^{n}}{n!}+\cdots \right) +\left( 1-x+ \frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\cdots +(-1)^{n}\frac{x^{n}}{n!}+\cdots \right) }{2} \\[10pt] \hspace{-10pc} &=&\frac{1+1}{2}+\frac{x-x}{2}+\frac{x^{2}+x^{2}}{2\cdot 2!}+\frac{ x^{3}-x^{3}}{2\cdot 3!}+\cdots+\frac{x^{n}+(-1) ^{n}x^{n}}{ 2\cdot n!}+\cdots \end{eqnarray*} \]

\[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \cosh x=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}+\cdots + \frac{x^{2n}}{(2n)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,\frac{x^{2k}}{(2k) !}}} \]

for all \(x.\)