What is the function representative of $\sum \frac{x^n}{(2n)!}$ I've tried relating it to $\cosh x$ but couldn't do anything. It definitely has something to do with $\cosh x$ though, probably.. any hints?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 0}^{\infty}{x^{n} \over \pars{2n}!} & =
\sum_{n = 0}^{\infty}{\pars{\root{x}}^{2n} \over \pars{2n}!} =
\sum_{n = 0}^{\infty}{\pars{\root{x}}^{n} \over n!}
\,{1 + \pars{-1}^{n} \over 2}
\\[5mm] & =
{1 \over 2}\sum_{n = 0}^{\infty}{\pars{\root{x}}^{n} \over n!} +
{1 \over 2}\sum_{n = 0}^{\infty}{\pars{-\root{x}}^{n} \over n!} =
{1 \over 2}\,\expo{\root{x}} + {1 \over 2}\,\expo{-\root{x}} =
\bbx{\ds{\cosh\pars{\!\!\root{x}\!}}}
\end{align}
A: Let $y=\sqrt{x}$.  Then
\begin{align}
f(y) &= \sum_{n=0}^\infty \frac{y^{2n}}{(2n)!}
\\
f'(y) &= \sum_{n=0}^\infty \frac{2n y^{2n-1}}{(2n)!}
= \sum_{n=1}^\infty \frac{ y^{2n-1}}{(2n-1)!}
\\
f''(y) &= \sum_{n=1}^\infty \frac{(2n-1) y^{2n-2}}{(2n-1)!}
= \sum_{n=1}^\infty \frac{ y^{2n-2}}{(2n-2)!}
=\sum_{k=0}^\infty \frac{ y^{2k}}{(2k)!} = f(y)
\end{align}
Note: differentiation term-by-term is valid inside the
radius of convergence of a power series.  So (before you begin)
compute the radius of convergence using the ratio test.  
Solve the differential equation $f''(y)=f(y)$.  Use the
initial conditions $f(0)=1, f'(0)=0$ to determine the constants in the general solution.
