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I am trying to plot this

which is a numerical simulation of the Montgomery-Odlyzko law for the nontrivial 1st $10^5$ zeros of the Riemann zeta function $ζ(s)$. The solid line is given by `1-(Sin[π x]/(πx))^2`

.

The following is what I can't seem to interpret:

The blue symbols indicate the pair correlation function of normalized spacings $δn=(γ_{n+1}-γ_n)\log(γ_n/2π)/2π$ between two consecutive nontrivial zeros $1/2+iγ_n$ and $1/2+iγ_{n+1},\ n=1\dots10^5)$ of the Riemann zeta function $ζ(s)$.

from this page.
I have looked at various ways of calculating the density, using `Histogram`

, `NearestFunction`

, `FindClusters`

, etc. but am getting nothing like the image above. I realise that this is largely due to a conceptual misunderstanding, but I was hoping someone could point me in the right direction.

# Update

plotted with

```
DeleteCases[Flatten[Table[Abs[z1 - z2] Log[z2/(2 π)]/(2 π),
{z1, Take[zz,1000]}, {z2, Take[zz,1000]}]], 0];
Show[ListPlot[HistogramList[%, {0, 3, 0.1}][[2]]/200,
DataRange -> 3], Plot[1 - (Sin[π u]/(π u))^2, {u, 0, 3}]]
```

using First $10^5$ zeros as `zz`

, thanks to Rahul Narain's comments below. Unfortunately, my computer won't calculate for $10^5$ zeros, but link left for those that will.

... I am now beginning to appreciate the computational power that would have to be used to generate an image like this or this!

I like your question (+1) because of interesting topic even though I think you are asking about a trivial

Mathematicaproblem. Nevertheless I have no time to answer. – Artes – 2014-06-05T10:31:05.397Thank you. I realise it is trivial in terms of

Mathematicacomputation - sorry! I would really appreciate your assistance if you manage to get time at some point :) – martin – 2014-06-05T10:33:41.217I assume you've read

Odlyzko's paper - all the (mathematical) details of how such a graph is created are in it. – ciao – 2014-06-05T10:46:17.453@rasher - do you have a link? – eldo – 2014-06-05T10:53:20.993

@eldo: No, I have it on

JSTOR, read it long ago. If you haveJSTOR(I'm sure it's elsewhere too), worth a look. – ciao – 2014-06-05T10:55:30.527In the paper, it refers to the integral of the pair correlation function. Would one then have to find the derivative for each pair? I am sorry, but I am at a bit of a loss :( – martin – 2014-06-05T11:21:53.283

Using your above link and having a look at the image (you show it without its text) it can be seen that 10^5 zeroes were used to compute it. You overlooked the exponentiation operator. – eldo – 2014-06-05T11:32:29.020

Ah - many thanks! – martin – 2014-06-05T11:37:00.463

Would using $10^3$ yield much? – martin – 2014-06-05T11:39:03.790

@martin - better than just 100. Another question is the somehow vague meaning of "normalized spacings". If you could produce a y-axis like in the image you probably would be on the right track. – eldo – 2014-06-05T11:44:41.757

As I understand it, normalised spacings are given by $δn=(γ_{n+1}-γ_n)\log(γ_n/2π)/2π$, no? – martin – 2014-06-05T11:46:40.420

@martin - yes, they are. I overlooked this definition in the text. Sorry. – eldo – 2014-06-05T12:07:10.913

as well studied as the topic is i'd strongly suspect you can readily find tabulated values for the zeros to work with. – george2079 – 2014-06-05T12:34:20.703

@ george2079, That would be great - it would certainly make calculations faster, but I don't know what to do with them once I have them :/ – martin – 2014-06-05T13:09:15.930

3

I suggest you look up the actual definition of the pair correlation function. Wikipedia: "The radial distribution function (or pair correlation function) is usually determined by calculating the distance between

– None – 2014-06-05T19:26:26.767allparticle pairs and binning them into ahistogram" (emphasis mine). For example, just taking the first 1000 points (because I didn't want to wait a long time):`zzd = Flatten@Table[Abs[zz[[i]] - zz[[j]]], {i, 2, 1000}, {j, 1, i - 1}]; Histogram[zzd, {0, 3, 0.2}]`

http://i.stack.imgur.com/Pc3Tw.pngThank you :) ... I admit it has had me rather confused! – martin – 2014-06-05T19:28:23.080

Please feel free to add answer & I will accept (though if you wouldn't mind skipping the first sentence, I would be eternally grateful ;)). – martin – 2014-06-05T19:30:45.023

1The histogram is supposed to then be "normalized with respect to an ideal gas, where particle histograms are completely uncorrelated" so that the horizontal asymptote is at 1 instead of at about 140 in my plot, but I didn't get around to computing that. – None – 2014-06-05T19:31:15.947

Yes, I figured that :) – martin – 2014-06-05T19:31:49.077

@martin: You can get a text table of the first 100K zeros here, while there, go to home page, follow publications link, you can get a copy of the original paper if you've not already.

– ciao – 2014-06-06T08:27:03.140Yes, thank you. - The limitations are with my computer now - but I understand the concept. Can't plot with 10^5 on my machine - :/ – martin – 2014-06-06T08:29:31.573