1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Can you explain how the factor of $(1+i)$ appears when you simplify the first bracket? Looking at the product formula for $\sin(z)$, it is not immediately clear. $\endgroup$ – Nicholas Roberts Apr 12 at 19:59
0
$\begingroup$

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

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.