# Evaluating a trigonometric product $\prod_{n=1}^{\infty}\cos^2\left(\frac{1}{n^2}\right)$

I'm interested in finding a closed form for $$\prod_{n=1}^{\infty}\cos^2\left(\frac{1}{n^2}\right)$$

Wolfram Alpha confirms that it converges, but I can't find any plausible closed forms. I've made some efforts to rewrite it in terms of stuff of the form $re^{i\theta}$ and make it a geometric series, but I think a more high powered solution may be needed. Any ideas?

• Do you have a reason to believe that a simple expression for this exists? – DanielV Oct 12 '14 at 3:37

By the Weierstrass product for the cosine function we know that: $$\cos z=\prod_{n=0}^{+\infty}\left(1-\frac{4z^2}{(2n+1)^2 \pi^2}\right)$$ hence: $$\log\cos z=-\sum_{n=0}^{+\infty}\sum_{m=1}^{+\infty}\frac{4^m z^{2m}}{m(2n+1)^{2m}\pi^{2m}}=-\sum_{m=1}^{+\infty}\frac{(4^m-1)\zeta(2m)z^{2m}}{m\pi^{2m}}$$ and: $$\sum_{n=1}^{+\infty}\log\cos\frac{1}{n^2}=-\sum_{m=1}^{+\infty}\frac{(4^m-1)\zeta(2m)\zeta(4m)}{m\pi^{2m}}$$ so: $$\prod_{n=1}^{+\infty}\cos^2\left(\frac{1}{n^2}\right)=\exp\left(-2\sum_{m=1}^{+\infty}\frac{(4^m-1)\zeta(2m)\zeta(4m)}{m\pi^{2m}}\right).\tag{1}$$ By approximating $2\log\cos x$ as $-x^2-\frac{x^4}{6}$ we get, for instance: $$\prod_{n=1}^{+\infty}\cos^2\left(\frac{1}{n^2}\right)\leq \cos^2(1)\exp\left(-\zeta(4)-\frac{\zeta(8)}{6}+\frac{7}{6}\right).$$
• I would not bet that a closed form exist, but another possible approach may be given by exploiting the identity: $$\prod_{n=1}^{+\infty}\left(1-\frac{4}{(2m+1)^2 \pi^2 n^4}\right)=\frac{2m+1}{2\pi}\sin\sqrt{\frac{2\pi}{2m+1}}\sinh\sqrt{\frac{2\pi}{2m+1}}.$$ – Jack D'Aurizio Oct 12 '14 at 3:29