Find the Maclaurin expansion for f(x)=coshx.
Solution Since coshx=ex+e−x2
its Maclaurin expansion can be found by adding corresponding terms of the Maclaurin expansions for ex and e−x and then dividing by 2. The result is coshx=ex+e−x2=(1+x+x22!+x33!+⋯+xnn!+⋯)+(1−x+x22!−x33!+⋯+(−1)nxnn!+⋯)2=1+12+x−x2+x2+x22⋅2!+x3−x32⋅3!+⋯+xn+(−1)nxn2⋅n!+⋯
\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.