Square root in $\mathbb C$ Denote $\sqrt{}$ the square root defined on $\mathcal D:=\mathbb C\setminus]-\infty,0]$ thanks to the principal logarithm by the formula:
$$\forall z\in\mathcal D,\ \sqrt z=e^{\frac12\log(z)}$$
Can one explicit $\sum_{n\ge0}\frac{z^n}{(2n)!}$ in function of $\cosh$ and $\sqrt{}$. If $z\in\mathcal D$, it is obvious. But what about if $z\in]-\infty,0]$?
Thanks in advance for any hints or solution.
 A: Use the following facts:
$$\cosh(z) = \frac{e^z + e^{-z}}{2}$$
$$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$
Then, we have:
$$\cosh(z) = \frac{1}{2}\left(\sum_{n=0}^{\infty} \frac{z^n}{n!} + \sum_{n=0}^{\infty} \frac{z^n (-1)^n}{n!}\right)$$
Expand out a few terms and you can see the pattern:
$$\cosh(z) = \frac{1}{2} \left[ (1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdot\cdot\cdot) + (1-z+\frac{z^2}{2!}-\frac{z^3}{3!}+\cdot\cdot\cdot)\right]$$
The odd exponents of $z$ cancel, and the even terms get multiplied by $2$:
$$\cosh(z) = \frac{1}{2}\left( 2\cdot(1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdot\cdot\cdot)\right) = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}$$
In order to get the power of $z$ to be $n$ instead of $2n$, perform the transformation $z\mapsto\sqrt{z}$:
$$\cosh(\sqrt{z}) = \sum_{n=0}^{\infty} \frac{z^n}{(2n)!}$$
This is the sum that you want. Note that whether $z$ is in your domain $\mathcal{D}$ is irrelevant. Because remember we derived it from the Laurent/Taylor series for $e^z$, which is valid for the whole complex plane. If $z\in [-\infty,0)$, just substitute it into $\cosh(z)$, and evaluate the principle square root. Here's if $z = -1:$
$$\cosh(\sqrt{-1}) = \cosh(i)\underset{(*)}{=} \cos(1) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}$$
$(*)$ I have used the fact that $\cosh(iz)=\cos(z)$ for all $z$ here.
Long story short, this series for $\cosh(\sqrt{z})$ is perfectly well-defined, but if $z\in (-\infty,0)$, you just have to evaluate the square root, which will have an imaginary part, and use an identity to turn it into cosine. Nothing wrong with $\cosh$ having a complex number inside it. Feel free to accept the answer if it helped, and let me know if there's any questions.
