(related to my previous question)

Consider the function $$f(x)=\prod_{n=0}^\infty\operatorname{sinc}\left(\frac{\pi \, x}{2^n}\right),\quad\color{gray}{x\ge0},\tag1$$ where $\operatorname{sinc}(z)$ denotes the sinc function. The function $f(x)$ has zeros at positive integers, and oscillates with a quickly decaying amplitude. Its signs on the intervals between consecutive zeros follow the same pattern as the Thue–Morse sequence. graph of f(x)

Can we find a real-valued function $g(x)$ (elementary, if possible), such that it is analytic, positive and monotone decreasing (and having monotone derivatives of any order, if possible) and satisfies $$\lim_{t \to \infty} \frac 1 {t-t_0} \int_{t_0}^t \frac{|f(x)|}{g(x)} \, dx=1\tag2$$ for large enough $t_0$? Or, at least, $$\color{gray}{\exists A > 0, \, \exists B > A, \, \forall t > t_0,} \, A < \frac 1 {t-t_0} \int_{t_0}^t \frac{|f(x)|}{g(x)} \, dx < B\tag3$$ for large enough $t_0$?

The "for large enough" provision means that $g(x)$ is permitted to be not monotone, or have zeros or discontinuities, or be not defined at all for small $t,$ and we only care about its "eventual" behavior.

Empirically, it looks like $g(x)$ should decay faster than any negative power of $x$, but slower than exponentially. I'm thinking something close to $\exp(-\log^2x)$, but perhaps not exactly that.

Update: Maybe this expansion can be useful: $$\prod _{n=0}^m \operatorname{sinc}\left(\frac{\pi\,x}{2^n}\right) = \frac{2^{\binom m2}}{(\pi\,x)^{m+1}} \sum_{n=0}^{2^m-1} t_n\,\sin\left(\!\frac {\pi\,m}2+\frac{2n+1}{2^m}\,\pi\,x\!\right),\tag4$$ where $t_0=1,\,t_n=(-1)^n\,t_{\lfloor n/2\rfloor}$ (the signed Thue–Morse sequence).

  • $\begingroup$ Don't you have the limit in $(2)$ equal to $0$ for $t_0$ an integer? $\endgroup$ – zhw. May 1 '18 at 0:33
  • $\begingroup$ @zhw. A choice of $t_0$ does not affect the limit (if it exists), because we are averaging the ratio of functions over an infinite ray. Changing something in a finite initial segment of it does not affect the result. $\endgroup$ – Vladimir Reshetnikov May 1 '18 at 22:58
  • $\begingroup$ Ah sorry, my bad. Didn't see $t\to \infty$ $\endgroup$ – zhw. May 2 '18 at 0:08

$\mathbf{\text Definition}$

Let $$R(x) = \prod_{n=0}^\infty\mathrm{sinc}\dfrac{\pi x}{2^n}.\tag1$$

$\mathbf{\text{Formulas for } {sinc(\pi z)}}$

Using symmetry formulas for the gamma function $$\Gamma(z)\Gamma(1-z)=\dfrac\pi{\sin(\pi z)}\tag2$$ easy to obtain $$\dfrac{\sin(\pi z)}{\pi z} = \dfrac1{z\Gamma(z)\Gamma(1-z)},$$ $$\mathrm{sinc(\pi z)} = \dfrac1{\Gamma(1+z)\Gamma(1-z)},\tag3$$ $$\ln\mathrm{sinc}(\pi z) = -\left(\ln\Gamma(1+z)+\ln\Gamma(1-z)\right).\tag4$$

Using the Maclaurin series in the form of $$\ln\Gamma(1+z)=-\ln(1+z)+(1-\gamma)z + \sum\limits_{j=2}^\infty(-1)^j\dfrac{\zeta(j)-1}{j}z^j,\quad |z|<2,\tag5$$ where $\gamma = 0.57721...,$ one can get $$\ln\Gamma(1-z)=-\ln(1-z)-(1-\gamma)z + \sum\limits_{j=2}^\infty\dfrac{\zeta(j)-1}{j}z^n,\quad$$ $$\ln\mathrm{sinc}(\pi z)=\ln(1-z^2) - \sum\limits_{k=1}^\infty\dfrac{\zeta(2k)-1}{k}z^{2k}.$$ Zeta function of the even argument is $$\zeta(2k) = \dfrac{(2\pi)^{2k}}{2(2k)!}|B_{2k}|,\tag6$$ where $B_{2k}$ are the Bernoulli numbers. Then $$\ln\mathrm{sinc}(\pi z)=\ln(1-z^2) - \sum\limits_{k=1}^\infty c_k z^{2k},\tag7$$ where $$c_k = \dfrac1{2k}\left(\dfrac{(2\pi)^{2k}}{(2k)!}|B_{2k}|-2\right).\tag8$$

$\mathbf{\text{Formulas for R(x)}}$

Let us find the sum \begin{align} &\ln R_m(x) = \sum_{n=m}^\infty \ln\mathrm{sinc}\dfrac{\pi x}{2^n},\\ \end{align}

\begin{align} &\ln R_m(x) = \sum_{n=m}^\infty \left(\ln\left(1-\left(\dfrac{x}{2^n}\right)^2\right) - \sum\limits_{k=1}^\infty c_k\left(\dfrac{x}{2^n}\right)^{2k}\right) =\\ & \sum_{n=m}^\infty \ln\left(1-\left(\dfrac{x}{2^n}\right)^2\right) - \sum\limits_{k=1}^\infty\left(c_k x^{2k}\sum\limits_{n=m}^{\infty}2^{-2nk}\right) = \\ &\sum_{n=m}^\infty \ln\left(1-\left(\dfrac{x}{2^n}\right)^2\right) - \sum\limits_{k=1}^\infty\dfrac{c_k}{1-2^{-2k}}\left(\dfrac{x}{2^m}\right)^{2k}. \end{align}

The first term can be presented over the q-Pochhammer Symbol in the form of $$\sum_{n=0}^m \ln\left(1-\left(\dfrac{x}{2^n}\right)^2\right) = \ln \left(x^2, \dfrac14\right)_m.\tag{9}$$

This allows to obtain the formulas $$R(x) = \prod\limits_{n=0}^{m-1}\mathrm{sinc}\dfrac{\pi x}{2^n}\cdot\dfrac{\left(x^2, \dfrac14\right)_\infty}{\left(x^2, \dfrac14\right)_m}\cdot\exp\left(\sum\limits_{k=1}^\infty\ \dfrac{c_k}{1-2^{-mk}} \left(\dfrac{x}{2^m}\right)^{2k}\right), \quad |x| < 2^{m+1}.\tag{10}$$

Radius of convergence in the formula $(10)$ shows that it gives $R(x)=0$ outside this radius, such as the Stirling asymptotics. Easy to see that parameter $m$ allows to manage this radius. At the same time, it allows to provide the reqired length of sum.

If the bounds of the required interval are predefined, then paraneter $m$ can be considered as constant. This allows to use the common function for this interval.

For example, for the predefined interval $x=(10,100)$ the value $m=6$ can be chosen. Then formula $(10)$ presents the function $|R(x)|$ as the production of the three factors:

partial sinc production,

partial sinc production

q-Pochhammer ratio

q-Pochhammer ratio

and the reminder series exponent

The reminder series exponent

In the general case, the most convenient way looks using of formula $(10)$ for a piecewise approximation on the required intervals .

Anyway, this allows to take in account the multiplicity of preceeding zeros.

  • $\begingroup$ Anyway, the function $f(x)$ is known to be continuous and analytic, its Taylor–Maclaurin series and Fourier transform are studied pretty well, see, for example, the references in my previous question. $\endgroup$ – Vladimir Reshetnikov Apr 30 '18 at 3:56
  • $\begingroup$ @VladimirReshetnikov New version. Waiting for comments. $\endgroup$ – Yuri Negometyanov May 1 '18 at 5:06
  • $\begingroup$ In formula $(10)$, there is a free variable $m$ on the right, but not on the left -- is it a typo? Also, the summation over $k$ in $(10)$ starts from $k=0$, but the summand contains $c_k$. Looking at the definition $(8)$, we can see that $c_0$ is undefined because it involves a division by zero. Also, there is another division by zero because of the denominator $1-2^{-mk}$. $\endgroup$ – Vladimir Reshetnikov May 1 '18 at 23:10
  • $\begingroup$ @VladimirReshetnikov Of course, this was a typo. The earlier formulas contains $k=1$. Thank you for accuracy. $\endgroup$ – Yuri Negometyanov May 1 '18 at 23:20
  • $\begingroup$ @VladimirReshetnikov By the way. Is known the inequaliity $\dfrac{2(2n)!}{(2\pi)^{2n}}\left(\dfrac1{1-2^{1-2n}}\right) > (-1)^{n+1}B_{2n} > \dfrac{2(2n)!}{(2\pi)^{2n}}, \quad n=1,2,\dots.$ $\endgroup$ – Yuri Negometyanov May 1 '18 at 23:36

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