Closed form for $\prod\limits_{l=1}^\infty \cos\frac{x}{3^l}$ Is there any closed form for the infinite product $\prod_{l=1}^\infty \cos\dfrac{x}{3^l}$? I think it is convergent for any $x\in\mathbb{R}$.
I think there might be one because there is a closed form for $\prod_{l=1}^\infty\cos\dfrac{x}{2^l}$ if I'm not wrong.
 A: If we start from
$$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$
we have:
$$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m} \pi^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)\,x^{2m}}{m\pi^{2m}}\tag{2} $$
hence:
$$ \sum_{l\geq 1}\log\cos\left(\frac{x}{3^l}\right)=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{(9^m-1)m\pi^{2m}}x^{2m}\tag{3} $$
and 
$$ \prod_{l\geq 1}\cos\left(\frac{x}{3^l}\right) = \color{red}{\exp\left(-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{(9^m-1)m\pi^{2m}}x^{2m}\right)}\tag{4}$$
does not simplify much further.
A: We follow the method of this paper, using the Fourier transform. 
Following his normalizations,
$$
\hat f (\omega) = \frac{1}{2\pi} \int_{\mathbb R} f(x) e^{-i \omega x} \, dx, \qquad f(x) = \int_{\mathbb R} \hat f(\omega) e^{i \omega x} \, d\omega,
$$
the key facts to be used are:


*

*The identity $\cos(bx) = \frac{e^{i b x} + e^{-i b x}}{2}$

*If $f(x) = e^{ibx}$, then $\hat f(\omega) = \delta_b(\omega)$, the Dirac delta centered at $\omega = b$.

*The convolution formula $\widehat{fg} = \hat f * \hat g$.

*The identity $\delta_a * \delta_b = \delta_{a+b}$.


Applying these to your problem,
\begin{align*}
\prod_{k=1}^n \cos\left(\frac{x}{3^k}\right) &= \prod_{k=1}^n \frac{e^{i \frac{x}{3^k}} + e^{-i \frac{x}{3^k}}}{2} \\
\implies \widehat{\prod_{k=1}^n \cos\left(\frac{x}{3^k}\right)} &= \frac{1}{2^n} \left(\delta_{\frac{1}{3^1}} + \delta_{-\frac{1}{3^1}} \right) * \cdots * \left(\delta_{\frac{1}{3^n}} + \delta_{-\frac{1}{3^n}} \right)(\omega) \\
&= \frac{1}{2^n} \sum_{p \in P_n} \delta_p(\omega)
\end{align*}
where $P_n$ is the set of $2^n$ points given by $P_n = \left\{\sum_{k=1}^n \frac{b_k}{3^k} \;\middle|\; b_k \in \{-1,1\} \; \forall k \right\}$. Note that the endpoints of $\bigcup_{n=1}^\infty P_n$ are bounded tightly by $\pm \sum_{k=1}^\infty \frac{1}{3^k} = \pm \frac{1/3}{1 - 1/3} = \pm \frac{1}{2}$, and every point in $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$ is the limit of some sequence $(p_n)_{n=1}^\infty$, where $p_n \in P_n$. Hence $\tfrac{1}{2^n} \sum_{p \in P_n} \delta_p(\omega)$ tends to the uniform density of total mass $1$ over $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$ as $n\to\infty$, which is given by the indicator function $\chi_{\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]}(\omega)$:
\begin{align*}
\widehat{\prod_{k=1}^\infty \cos\left(\frac{x}{3^k}\right)} &= \chi_{\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]}(\omega).
\end{align*}
Taking inverse Fourier transforms, we get
$$
\prod_{k=1}^\infty \cos\left(\frac{x}{3^k}\right) = \int_{\mathbb R} \chi_{\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]}(\omega) e^{i \omega x} \, d\omega = \int_{-1/2}^{1/2} e^{i \omega x} \, d\omega = \frac{\sin\left(\tfrac{x}{2}\right)}{\tfrac{x}{2}}.
$$

UPDATE: here's why the above is wrong. The statement that "every point in $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$ is the limit of some sequence $(p_n)_{n=1}^\infty$, where $p_n \in P_n$" is incorrect; in fact, the points belonging to $P = \bigcup_n P_n$ are those which have balanced ternary expansion consisting of the digits $1$ and $-1$ but not $0$. I'm not sure how to better describe $P$, but we'd proceed as before by integrating over $P$ (rather than all of $\left[ - \tfrac{1}{2}, \tfrac{1}{2} \right]$) to compute the inverse Fourier transform yielding the correct answer. 
A: Let $\displaystyle P=\prod_{i=1}^n\cos\left(\frac x{2^i}\right)$.
$\displaystyle P=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^n}\right)\sin\left(\frac x{2^n}\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac12\,\cos\left(\frac x{2}\right)\cos\left(\frac x{4}\right)\cos\left(\frac x{8}\right)\cdots\cos\left(\frac x{2^{n-1}}\right)\sin\left(\frac x{2^{n-1}}\right)$
...
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac1{2^{n-1}}\,\cos\left(\frac x{2}\right)\sin\left(\frac x2\right)$
$\displaystyle P\sin\left(\frac x{2^n}\right)=\frac1{2^{n}}\,\sin(x)$
$\displaystyle P=\frac{\sin(x)}{2^{n}\sin\left(\frac x{2^n}\right)}$
$\displaystyle \lim_{n\to\infty}P=\lim_{n\to\infty}\frac{\sin(x)}x\frac{2^{-n}x}{\sin\left( 2^{-n}x \right)}=\frac{\sin(x)}x$
However, I am not able to solve the one with $3$ in the denominator.
