Finding a function from a Maclaurin expansion Given the following Maclaurin expansion:
$$1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots + \frac{x^r}{(2r)!}+ \ldots$$
How would you go about finding the function? All I can see to do is use forms of trial and error.
 A: So you're looking for a closed form for the series:
$$\sum_{r=0}^{\infty} \frac{x^r}{(2r)!} = 1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots $$
You probably know that:
$$e^x = \sum_{r=0}^{\infty} \frac{x^r}{r!} = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$
And evaluating this at $\sqrt{x}$ makes the powers agree with the factorials:
$$e^\sqrt{x} = \sum_{r=0}^{\infty} \frac{\sqrt{x}^r}{r!} = 1\color{red}{+\sqrt{x}}+\frac{x}{2!} \color{red}{+ \frac{x\sqrt{x}}{3!}} + \ldots \tag{1}$$
You don't want the odd powers (in red), so add the following series to $(1)$:
$$e^{-\sqrt{x}} = \sum_{r=0}^{\infty} \frac{\left(-\sqrt{x}\right)^r}{r!} = 1\color{red}{-\sqrt{x}}+\frac{x}{2!} \color{red}{- \frac{x\sqrt{x}}{3!}} + \ldots$$
And divide by $2$ to get:
$$\frac{e^{\sqrt{x}}+e^{-\sqrt{x}}}{2}=1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots $$
Notice that this is also $\cosh(\sqrt{x})$.
A: Another method consists in trying to find a differential equation verified by this series.
$\displaystyle y=\sum\limits_{r=0}^{\infty}\frac{x^r}{(2r)!}$
$\displaystyle 2y'=\sum\limits_{r=1}^{\infty}2r\frac{x^{r-1}}{(2r)!}=\sum\limits_{r=1}^{\infty}\frac{x^{r-1}}{(2r-1)!}=\sum\limits_{r=0}^{\infty}\frac{x^r}{(2r+1)!}=\sum\limits_{r=0}^{\infty}\frac{x^r}{(2r)!}\bigg(\frac 1{2r+1}\bigg)$
$\displaystyle 4y''=\sum\limits_{r=2}^{\infty} 4r(r-1)\frac{x^{r-2}}{(2r)!}=\sum\limits_{r=1}^{\infty}4r(r+1)\frac{x^{r-1}}{(2r+2)!}=\sum\limits_{r=1}^{\infty}\frac{x^{r-1}}{(2r)!}\bigg(\frac{2r}{2r+1}\bigg)$
Here we have the choice of shifting once again the index, but it's also a matter of guessing. 
In fact $\displaystyle \bigg(\frac{2r}{2r+1}\bigg)=1-\frac 1{2r+1}$ the coefficients in $y$ and $y'$ should instead incite us to consider multiplying the series for $y''$ by $x$.
$4xy''=\sum\limits_{r=1}^{\infty}\frac{x^r}{(2r)!}\bigg(1-\frac{1}{2r+1}\bigg)$ the formula is also true for $r=0$ because the term is zero.

$4xy''=y-2y'$

Let's have 
$\begin{cases}
u(x)=y(x^2)\\
u'(x)=2xy'(x^2)\\
u''(x)=4x^2y''(x^2)+2y'(x^2)=\bigg[y(x^2)-2y'(x^2)\bigg]+2y'(x^2)=y(x^2)=u(x)
\end{cases}$
$u$ verifies $u''=u$ 
so $u(x)=A\cosh(x)+B\sinh(x)$ but since $u$ is even $B=0$.
Finally $y=A\cosh(\pm\sqrt{x})=A\cosh(\sqrt{x})$ and the series gives $y(0)=A=1$.

$y(x)=\cosh(\sqrt{x})$

And we verify a posteriori that, it corresponds to the series given, that convergence is fine, and derivable two times...
I admit, there is a lot of guessing and tricks in this method, but if there is a closed form we can hope to find it. If the series does not verifies a linear ODE then it is almost hopeless to guess it, but in general shifting indexes or/and multiply $y',y'',y''',...$ by powers $x,x^2,x^3,...$ often lead to elimination of terms and finding the ODE.
Then it's another matter to solve the ODE... Here I was knowing what I was searching and cheated a little with my variable change with $u$, or you may use a CAS to have clues.
