Computing a power series I came a across this power series during the resolution of a problem,
$$\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n)!}x^n$$
but I haven't been able to crack it, it seems to be some sort of exponential function or trigonometric, but so far nothing. I'm also puzzled with its counterpart
$$\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^n$$
But I think knowing how to solve one will deliver the other one with the same machinery. I would appreciate any hint or ideas on how to tackle this series.
 A: In the comments of your question you were given half the correct answer, but we can do a better work!
First, let's notice that the given series converge $\forall x \in \mathbb{R}$, which can be easily proven with the ratio test. Then you can show that the first series converges to $\cos{\sqrt{x}}$ for $x>0$... but this function it's not differentiable at $x_0=0$, the center of the series, a necessary condition for the Taylor series to exist. So this far, the series
$$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}x^n$$ converges to a sum function $S(x)$ such that it's equal to $\cos{\sqrt{x}}$ for $x>0$. You can easily show that at $x=0$, $S(0)=1$. Let's find out the rest of the expression by inspecting what other Taylor series could match the previous series. The candidate now is:
$$\cosh{u}=\sum_{n=0}^{+\infty} \frac{u^{2n}}{(2n)!}, \; \forall u \in \mathbb{R} $$
now we will substitute in $u=\sqrt{-x}=(-x)^{1/2}$, giving:
$$\cosh{\sqrt{-x}}=\sum_{n=0}^{+\infty} \frac{((-x)^{1/2})^{2n}}{(2n)!} =\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!} x^{n}, \; x<0 $$
Finally,
$$\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}x^n=S(x)= \left\{
\begin{array} 
 .\cos{\sqrt{x}}& \text{if} \; x>0 \\
1& \text{if} \; x=0 \\ 
\cosh{\sqrt{-x}}& \text{if} \; x<0
 \end{array} \right.$$
Where you can prove that $S(x)$ admits infinite order derivatives evaluated everywhere.
As for the "counterpart", unfortunetly the series doesn't converge to $\sin{\sqrt{x}}$ when $x>0$, because $(x^{1/2})^{2n+1} \neq x^n$. But you can try with
$$\frac{\sin{\sqrt{x}}}{\sqrt{x}}$$
and I'll leave you the rest as an excercise ;) Hope you find this useful!
