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EXAMPLE 5Finding the Maclaurin Expansion for f(x)=coshx

Find the Maclaurin expansion for f(x)=coshx.

Solution Since coshx=ex+ex2

its Maclaurin expansion can be found by adding corresponding terms of the Maclaurin expansions for ex and ex and then dividing by 2. The result is coshx=ex+ex2=(1+x+x22!+x33!++xnn!+)+(1x+x22!x33!++(1)nxnn!+)2=1+12+xx2+x2+x222!+x3x323!++xn+(1)nxn2n!+

\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.