How can I deduce $\cos\pi z=\prod_{n=0}^{\infty}(1-4z^2/(2n+1)^2)$?

Question

Using the infinite product of $\sin(\pi z)$, one can find the Hadamard product for $e^z-1$:

$$e^z-1 =2ie^{z/2}\sin(-iz/2)= 2i e^{z/2} (-iz/2) \prod_n \left(1+\frac{z^2}{4\pi n^2}\right)\\= e^{z/2} z \prod_n \left(1+\frac{z^2}{4\pi n^2}\right).$$

I don't see a way to find the product for $\cos\pi z$. A naive attempt is letting $\{a_n\}\subset{\Bbb C}$ be all the zeros of $\cos(\pi z)$ and showing the possible convergence of $$ \prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right) $$

Is there an alternative way to find the Hadamard product in the title for $\cos\pi z$?

Answer 1

Hint: Use $\sin(2z)=2\sin(z)\cos(z)$ so that $$\cos(z)=\frac{\sin(2z)}{2\sin(z)}.$$ If you're careful about how you write it, you will see that all of the 'even terms' cancel nicely. I do not have time right now, but if you haven't been able to solve it within a few hours, I will return and post my solution.

Answer 2

Well, you can perform a logarithmic differentiation and get a series that may be summed using the residue theorem.

Let $p(z)$ be the product in question; we intend to prove that $p(z)=\cos{\pi z}$.

$$\log{p} = \sum_{n=0}^{\infty} \log{\left ( 1-\frac{4 z^2}{(2 n+1)^2}\right)}$$

$$\frac{d}{dz} \log{p} = -z \sum_{n=-\infty}^{\infty} \frac{1}{(n+(1/2))^2-z^2}$$

Note that we were able to use the symmetry of the sum to change the lower limit to $-\infty$. This sum is in a form that may be evaluated using the residue theorem:

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{Res}_{s=s_k} [\pi \cot{\pi s} \, f(s)]$$

where the $s_k$ are the non-integral poles of $f$. In this case, $f(s) = 1/((s+(1/2))^2-z^2)$, so that the poles of $f$ are at $s_{\pm} = -1/2 \pm z$. The residues of these poles are

$$\frac{\pi \cot{(-\pi/2 + \pi z)}}{2 z} - \frac{\pi \cot{(-\pi/2 - \pi z)}}{2 z} = -\frac{\pi \tan{\pi z}}{z}$$

Therefore

$$\frac{d}{dz} \log{p} = -\pi \tan{\pi z} \implies \log{p} = \log{\cos{\pi z}} + C$$

where $C$ is a constant of integration, which using $p(0)=1$ implies that $C=0$. Then

$$p(z) = \cos{\pi z}$$

as was to be shown.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\prod_{n = 0}^{m} \bracks{1 - 4z^{2}/\pars{2n + 1}^{2}} \right\vert_{\,m\ \in\ \mathbb{N}_{\,\geq 1}}} = \prod_{n = 0}^{m} {\pars{n + 1/2}^{2} - z^{2} \over \pars{n + 1/2}^{2}} \\[5mm] = &\ \on{f}_{m}\pars{z}\on{f}_{m}\pars{-z} \end{align} where \begin{align} \on{f}_{m}\pars{z} & \equiv \prod_{n = 0}^{m} {n + 1/2 - z \over n + 1/2} = {\pars{1/2 - z}^{\overline{m + 1}} \over \pars{1/2}^{\overline{m + 1}}} \\[5mm] & = {\pars{1/2 - z + m}!\,/\,\Gamma\pars{1/2 - z} \over \pars{1/2 + m}!\,/\,\Gamma\pars{1/2}} \\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\, {\root{\pi} \over \Gamma\pars{1/2 - z}}\ \times \\[2mm] &\ {\root{2\pi}\pars{1/2 - z + m}^{1 - z + m}\,\,\, \expo{-1/2 + z - m} \over \root{2\pi}\pars{1/2 + m}^{1 + m}\,\, \expo{-1/2 - m}} \\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\, {\root{\pi} \over \Gamma\pars{1/2 - z}}\ \times \\[2mm] &\ {m^{1 - z + m}\,\,\,\,\bracks{1 + \pars{1/2 - z}/m}^{\,m} \over m^{m + 1}\,\,\,\bracks{1 + \pars{1/2}/m}^{\,m}}\expo{z} \\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\, {\root{\pi} \over \Gamma\pars{1/2 - z}}\,m^{-z} \end{align} Then, \begin{align} &\bbox[5px,#ffd]{\left.\prod_{n = 0}^{m} \bracks{1 - 4z^{2}/\pars{2n + 1}^{2}} \right\vert_{\,m\ \in\ \mathbb{N}_{\,\geq 1}}} \\[5mm] = &\ \lim_{m \to \infty}\braces{\bracks{{\root{\pi} \over \Gamma\pars{1/2 - z}}\,m^{-z}}\bracks{{\root{\pi} \over \Gamma\pars{1/2 + z}}\,m^{z}}} \\[5mm] = &\ {\pi \over \pi/\sin\pars{\pi\bracks{1/2 + z}}} = \bbx{\cos\pars{\pi z}} \\ & \end{align}