5
$\begingroup$

Let $\rho_n$ be the $n$-th non-trivial zero of $\zeta(s)$ and $z_n = \Im(\rho_n)$ with $z_n > 0$ and $z_{n+1} \ge z_n$.

A well known method to establish that all $\rho$s reside on the critical line ($\Re(\rho)=\frac12)$ up to a certain height $T$, is to make use of the regularly distributed and easy to compute Gram points ($g_n$).

The process boils down to assuring that each Gram point interval $[g_n,g_{n+1})$ contains only a single $z$. Heuristics show this is already the case for most intervals, however it is also known to fail infinitely often resulting in empty intervals or intervals containing multiple $z$ (i.e. violations of Gram's law and Rosser's rule).

Failures will typically be induced when larger swings in the $Z(t)$-function occur, where $Z(t) = e^{i\Theta(t)} \zeta\left(\frac12 + it\right)$ and $\Theta(t)$ is the Riemann-Siegel Theta function. Some accounting is then required to ensure there are as many $z$ as intervals in a Gram block with 'bad' Gram points. Morally this process is equivalent to mapping each 'wrongly' located $z$ to its unique interval.

So, when $Z(t)$ oscillates rather 'calmly', a $z$ will typically reside not too far from the middle of its Gram point interval. However, when oscillations of $Z(t)$ become more 'fierce' in amplitude, the $z$ will be pushed more towards the edges of their intervals or even hop into the neighbouring intervals.

With this in mind, we could compute how far ($=d_n$) each $z_n$ has been 'pushed' away from the mid point of the interval it 'belongs to'. This could be done as follows (note: the index of a Gram point $g_n$ starts at $-1$ and for $z$ at $1$):

$$d_n = z_n - \frac{g_{n-2}+g_{n-1}}{2}$$

The picture below aims to visualise the process and how $d_n$ is computed for real data around the first Lehmer pair. (note: the legend of the red line should be: $\Im\left( e^{(i\sin(\Theta(t))}\right)$)

enter image description here

The next graphs show the normalised histograms of the differences $d_n$ for increasing ranges of $n$.

enter image description here

Question:

The shape of the distribution seems to become increasingly 'normal' for higher $n$. Could this indeed be the case and if so, is there a logical explanation for this phenomenon?

Here is the $d_n$-data for $n=1..999999$.

ADDED 1:

Since, on average, the densities of $z_n$ and $g_n$ increase when we go up the critical line, the variances of the distributions shown above will become smaller and their peaks higher. This can be normalised as follows:

$$d_n^* = z_n \log(z_n) - \frac{g_{n-2}+g_{n-1}}{2}\log\left(\frac{g_{n-2}+g_{n-1}}{2}\right)$$

Here is the $d_n^*$-data for $n=1..999999$.

A Normal distribution with $\mu=0$ and $\sigma=2.65$ then already provides a pretty decent visual fit (red line).

enter image description here

ADDED 2:

Here is the distribution of the first 100 mln $d_n^*$ (blue dots). The red line is a Normal distribution with $\mu=0$ and $\sigma=2.71$.

enter image description here

The real data have a $\mu=-5.301\cdot 10^{-8}$ and $\sigma=2.642$. The fit with the Normal distribution clearly gets weaker in the tails. I do expect this to improve for larger $n$, since the "turmoil" in $Z(t)$ is expected to increase thereby inducing a relatively higher proportion of larger deviations $d_n^*$.

A Gram point is defined as the $t$ where $\Theta(t)=k\pi$ (i.e. where $\Im\left(\zeta(\frac12+it)\right)=0, t\ne z$). I realised that the mid point of the Gram interval (calculated as a simple average above), quickly converges to the $t$ where $\Theta(t)=\frac{(2k+1)\pi}{2}$ (i.e. where $\Re\left(\zeta(\frac12+it)\right)=0, t\ne z$). Let's label these points as $\hat{g}_n$.

My question is then equivalent to: "Is $\hat{d}_n=z_n -\hat{g}_{n-2}$ uniformly distributed?" or in other words: "Is the difference between 'paired' zeros of $\Re\zeta(\frac12+it)$ uniformly distributed?".

$\endgroup$
11
  • $\begingroup$ Maybe you can find relevant results about some kind of "normal distribution of critical zeros related quantities" in Selberg's collected papers. $\endgroup$ Jul 22, 2020 at 6:05
  • $\begingroup$ Not what you ask for, but somewhat related and maybe interesting for you (if you don't know it already): Arias de Reyna, J.: On the distribution (mod 1) of the normalized zeros of the Riemann zeta-function. J. Number Theory 153 (2015), 37–53. That paper is not about the distance between the zero and the Gram point it "belongs to", but about the relative location of the zero with respect to the Gram points between which it is actually located. $\endgroup$ Jul 22, 2020 at 6:55
  • 2
    $\begingroup$ @Kurisuto Asutora Thanks for the useful link! Arias de Reyna indeed follows a related but different approach. The uniformity of the resulting distributions (for $n=1..2mln$) also appears to be a bit less strong, but this is based on inspecting the visuals only. An intriguing sentence is his paper I find: "In a non published report, Odlyzko [14, p. 60] conjectured that the ordinates of the zeros $z_n$ are not related to the Gram points for $z_n$ large". This conjecture seems to contradict my observations above. $\endgroup$
    – Agno
    Jul 22, 2020 at 14:22
  • 2
    $\begingroup$ Odlyzko's unpublished paper is at dtc.umn.edu/~odlyzko/unpublished/zeta.10to20.1992.pdf $\endgroup$ Jul 22, 2020 at 15:20
  • $\begingroup$ Is there a rationale for $\sigma=2.65$? I observe that the $10^7$-th zero occurs at height $T=4.99238\cdot 10^6$, and $\log \log T=2.73589$. $\endgroup$
    – Stopple
    Jul 24, 2020 at 21:13

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.