Infinite product of sinc functions Calculate the infinite product
$f_q(x):=\prod_{n=0}^\infty\frac{\sin(q^n x)}{q^n x}$,
where $x$ is real and $0<q<1$.
In other words, $f_q$ must satisfy the functional equation $f_q(x)=f_q(qx)\operatorname{sinc}(x)$ with an initial condition $f_q(0)=1$,
where $\operatorname{sinc}(x):=\frac{\sin(x)}{x}$.
Using the Euler's factorization of sinc one can rewrite $f_q(x)$ as a certain infinite product of $q$-Pochhammer symbols. Therefore I hope that $f_q$ can be expressed in terms of some hypergeometric functions. The Fourier transform of $f_q$ is a smooth function with compact support.
 A: $\mathrm{h}_a$ atomic function
I suppose, that inverse Fourier transform of $f_q(x)$ leads finite function, which known as $\mathrm{h}_a(x)$.
It could be approximated by Fourier cosine series as follows:
$$
\begin{cases}
\mathrm{h}_a(x,a,M,N)=(a-1)\biggl(\dfrac{1}{2}+\sum\limits_{k=1}^{N}\prod\limits_{m=1}^{M}\mathrm{sinc}(m(a-1)\pi)\,\cos(k(a-1)\pi x)\biggr) ~~~\text{if}~~~\\ \hspace{11cm} x\,\in\,[-\frac{1}{a-1},\frac{1}{a-1}],\\  
0 \quad elsewhere.
\end{cases}
$$
Code
Wolfram Mathematica:
$$
FTha[t_, a_, N_] := Product[Sinc[t a^-k], {k, 1, N}];
ha[x_, a_, M_, N_] := 
 If[-1/(a - 1) <= x && x <= 1/(a - 1), (a - 1) (1/2 + Sum[FTha[(a - 1) \[Pi] k, a, N] Cos[(a - 1) \[Pi] k x], {k, 1, 
       M}]), 0];
$$
Plots





Reference
http://demonstrations.wolfram.com/ApproximateSolutionsOfAFunctionalDifferentialEquation/
A: I guess, expression $(10)$ from here is the most fresh suggestion on the original question up to this moment for the case, when $q=2.$ Substitute $2\rightarrow q$ and obtain the following expression
$$R(x) = \prod\limits_{n=0}^{m-1}\mathrm{sinc}\dfrac{\pi x}{q^n}\cdot\dfrac{\left(x^2, \dfrac1q\right)_\infty}{\left(x^2, \dfrac1q\right)_m}\cdot\exp\left(\sum\limits_{k=1}^\infty\ \dfrac{c_k}{1-q^{-mk}} \left(\dfrac{x}{q^m}\right)^{2k}\right), \quad |x| < q^{m+1}.$$
