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

Solution 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.

Solution 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.