Convergence of $\prod_{i=1}^{\infty}\cos(a^{-i}x)$

Question

If $a>1$, $x$ is real, then how to see that $\prod_{i=1}^{\infty}\cos(a^{-i}x)$ converges except for countably many $x$.

Answer 1

Some ideas.

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:

$$\sum_{n=1}^{+\infty}\log\cos\frac{x}{a^n}=-\sum_{m=1}^{+\infty}\frac{(4^m-1)\zeta(2m)x^{2m}\zeta(a^{2m})}{m\pi^{2m}}$$ so:

$$\prod_{n=1}^{+\infty}\cos\frac{x}{a^n} = \exp\left(-\sum_{m=1}^{+\infty}\frac{(4^m-1)\zeta(2m)x^{2m}\zeta(a^{2m})}{m\pi^{2m}}\right)$$

The first terms of the series inside the $\exp$ are, for example

$$\frac{x^6 \zeta \left(a^6\right)}{45}+\frac{x^4 \zeta \left(a^4\right)}{12}+\frac{x^2 \zeta \left(a^2\right)}{2}$$ Due to the minus sign in front of all, and because of $\exp$ we can approximate it for example as

$$1-\frac{x^2 \zeta \left(a^2\right)}{2}+O\left(x^4\right)$$

Further ideas are left to be developed.