Show $\cosh(z) - \cos(z) = z^2\prod_{n = 1}^{\infty}\left(1 + \frac{z^4}{4\pi^4n^4} \right)$ I would like to show that $\cosh(z) - \cos(z) = z^2\prod_{n = 1}^{\infty}\left(1 + \frac{z^4}{4\pi^4n^4} \right)$.
Firstly, I have found that solutions to the equation $\cosh(z) - \cos(z) = 0$ are of the form $z = (1 \pm i)n\pi$ ; $n \in \mathbb{Z}$.
I then would like to appeal to the Weierstrass factorization theorem.
By theorem, it will follow that:
$$\cosh(z) - \cos(z) = ze^{g(z)}\prod_{n = 1}^{\infty}\left(\frac{z}{(1 \pm i)n\pi}\right)E_{p_n} $$
where the $E_{p_n}$ are the Weierstrass elementary factors and $g$ is some entire function.
At this point, I am pretty stuck as to how to transform this into the desired infinite product shown above. Any tips?
 A: Let's start on the RHS.
It is
$$\left[z\prod_{n=1}^\infty\left(1+\frac{iz^2}{2\pi^2 n^2}\right)\right]
\left[z\prod_{n=1}^\infty\left(1-\frac{iz^2}{2\pi^2 n^2}\right)\right].$$
The first bracket here is
$$z\prod_{n=1}^\infty\left(1-\frac{((1-i)z)^2}{4\pi^2 n^2}\right)
=(1+i)\sin\frac{(1-i)z}2$$
and the second is
$$(1-i)\sin\frac{(1+i)z}2.$$
The product is
$$2\sin\frac{(1-i)z}2\sin\frac{(1+i)z}2
=\cos iz-\cos z=\cosh z-\cos z.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{N \in \mathbb{N}_{\geq 1}}$:

\begin{align}
&\bbox[10px,#ffd]{z^{2}\prod_{n = 1}^{N}
\pars{1 + {z^{4} \over 4\pi^{4}n^{4}}}} =
z^{2}\,{\prod_{n = 1}^{N}\bracks{n^{4} + z^{4}/\pars{4\pi^{4}}} \over
\pars{N!}^{4}}
\\[8mm] = &\
z^{2}\,
\verts{\mrm{f}_{N}\pars{{z \over
\root{2}\pi}\,\expo{\ic\pi/4}}}^{2}\
\verts{\mrm{f}_{N}\pars{{z \over
\root{2}\pi}\,\expo{3\ic\pi/4}}}^{2}\label{1}\tag{1}
\end{align}
where
\begin{align}
\mrm{f}_{N}\pars{\alpha} & \equiv{\prod_{n = 1}^{N}\pars{n - \alpha} \over N!} =
{\pars{1 - \alpha}^{\overline{N}} \over N!} =
{\Gamma\pars{1 - \alpha + N}/\Gamma\pars{1 - \alpha} \over N!}
\\[5mm] & =
{1 \over \Gamma\pars{1 - \alpha}}\,{\pars{N - \alpha}! \over N!}
\\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{1 \over \Gamma\pars{1 - \alpha}}\,
{\root{2\pi}\pars{N - \alpha}^{N - \alpha + 1/2}\expo{-N + \alpha} \over
\root{2\pi}N^{N + 1/2}\expo{-N}}
\\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{1 \over \Gamma\pars{1 - \alpha}}\,
{N^{N - \alpha + 1/2}\pars{1 - \alpha/N}^{N} \over
N^{N + 1/2}}\,\expo{\alpha}
\\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{N^{-\alpha} \over \Gamma\pars{1 - \alpha}}
\end{align}

and $\ds{\verts{\mrm{f}_{N}\pars{\alpha}}^{2}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{1 \over \Gamma\pars{1 - \alpha}\Gamma\pars{1 + \alpha}} =
\mrm{sinc}\pars{\pi\alpha}}$


Then,
\begin{align}
&\bbox[10px,#ffd]{z^{2}\prod_{n = 1}^{N}
\pars{1 + {z^{4} \over 4\pi^{4}n^{4}}}} =
z^{2}\,\mrm{sinc}\pars{{z \over \root{2}}\,\expo{\ic\pi/4}}
\,\mrm{sinc}\pars{{z \over \root{2}}\,\expo{3\ic\pi/4}}
\\[5mm] = &\
z^{2}\,{\sin\pars{\bracks{1 + \ic}z/2} \over \pars{1 + \ic}z/2}\,
{\sin\pars{\bracks{-1 + \ic}z/2} \over \pars{-1 + \ic}z/2} =
2\,\verts{\sin\pars{{1 + \ic \over 2}\,z}}^{2}
\\[5mm] = &\
2\,\verts{\sin\pars{z \over 2}\cosh\pars{z \over 2} +
\ic\cos\pars{z \over 2}\sinh\pars{z \over 2}}^{\, 2}
\\[5mm] = &\
2\,\bracks{\sin^{2}\pars{z \over 2}\cosh^{2}\pars{z \over 2} +
\cos^{2}\pars{z \over 2}\sinh^{2}\pars{z \over 2}}
\\[5mm] = &\
\bbx{\cosh\pars{z} - \cos\pars{z}}
\end{align}
