Show $\cosh(z) - \cos(z) = z^2\prod_{n = 1}^{\infty}\left(1 + \frac{z^4}{4\pi^4n^4} \right)$

Question

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?

Answer 1

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 $$2\sin\frac{(1-i)z}2\sin\frac{(1+i)z}2 =\cos iz-\cos z=\cosh z-\cos z.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

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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}