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.

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 $\cosh(\sqrt{x})$.
