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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.

Egor Maximenko
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2 Answers2

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$\mathrm{h}_a$ 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

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Reference

http://demonstrations.wolfram.com/ApproximateSolutionsOfAFunctionalDifferentialEquation/

Oleg Kravchenko
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    Thanks, this is a nice generalization. However, I don't think you are actually adressing OP's question with your answer. @Egor asked for a closed form expression of the corresponding function in terms of a hypergeometric function. – kram1032 Dec 14 '14 at 19:27
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    This is a great contribution to this topic, thank you very much! In fact, the (inverse) Fourier transform of $f_q$ is more important for me than $f_q$ itself, but formally speaking, the original question is open. – Egor Maximenko Dec 20 '14 at 04:44
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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 $$R(x) = \prod\limits_{n=0}^{m-1}\mathrm{sinc}\dfrac{\pi x}{q^n}\cdot\dfrac{\left(x^2, \dfrac1q\right)_\infty}{\left(x^2, \dfrac1q\right)_m}\cdot\exp\left(\sum\limits_{k=1}^\infty\ \dfrac{c_k}{1-q^{-mk}} \left(\dfrac{x}{q^m}\right)^{2k}\right), \quad |x| < q^{m+1}.$$

Oleg Kravchenko
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