I have a question on prime zeta function approximations

Have you encountered in the math literature one that resembles the following

Thanks,

Catalin

I’d like to thank you for always correcting the mistakes that I may find. I’ve witnessed that occasionally, people react very reluctantly to corrections of all sorts, but I found that those who are the best in their subject will include corrections, because no-one can hurt their scientific reputation.

]]>H(n) = + n/(ß^2+0.25) – sum_(k=1)^n (2*sum_(j=k+1)^n (k^(-1/2)*j^(-1/2)*cos(ß*ln(k/j))))

for those of you that like WolframAlfa, you can try this approximation (for a very low n=20 as WolframAlfa does not want to work for higher values on n):

1) for ß=14.134725, n=20

****** 20/(ß^2+0.25)-H(20) = 20/(ß^2+0.25)-sum_(k-1)^n (1/k) = -3.49776

****** sum_(k=1)^n (2*sum_(j=k+1)^n (k^(-1/2)*j^(-1/2)*cos(ß*ln(k/j)))) = -3.491098

1) for ß=37.586178, n=20

***** 20/(ß^2+0.25)-H(20) = 20/(ß^2+0.25)-sum_(k-1)^n (1/k) = -3.58359

***** sum_(k=1)^n (2*sum_(j=k+1)^n (k^(-1/2)*j^(-1/2)*cos(ß*ln(k/j)))) = -3.57838

with n>500, the error is less than 0.0001.

You can see more results at:

https://drive.google.com/open?id=1t2k690rHr0zes5YrS1dvF7ZWFP1PxEEx

]]>X(n) = n/(ß^2+(1-α)^2)) for α=1/2

O(n) = ∑(k=1)^n ∑(j≠k)^n (k^(-1/2)*j^(-1/2)*cos(ß*(ln(k/j))

H(n) = ∑(k=1)^n (1/k)

Then:

H(n) = X(n) – O(n) for all z*=1/2+ßi non-trivial zero of zeta

This is easily to see graphically as H(n)+O(n) = X(n) is a straight line as n->infinity, and actually it happens very quickly after n>100. You can see that at:

https://drive.google.com/open?id=1c–u6CJR2Rj4p8YmSteaYbWdNTaztdvS

For values of α 1/2, H(n)+O(n) is always a curve, some examples:

https://drive.google.com/open?id=1zHy-C3UM89IRK5Q4SQiCszx_-VeD3T1M

https://drive.google.com/open?id=1x1OlpuDBzvKGeV_CiDgWIZEyZYm1ARoI

It seems like the “music of the primes shuts down at the non-trivial zeros”, the waves become a straight line.

As a corollary, all non-trivial zeros of zeta can be linked algebraically by the following expression.

if z1=1/2+i*ß1 , z2=1/2+i*ß2 are non-trivial zeros, then:

X(n,z) = n/(ß^2+(1-α)^2)

O(n,z) = ∑(k=1)^n ∑(j≠k)^n (k^(-1/2)*j^(-1/2)*cos(ß*(ln(k/j))

then X(n,z1)+O(n,z1) = X(n,z2)+O(n,z2)

I have developed a Python code to find all non-trivial zeros knowing that z1=1/2+14.13472514173469379*I and it works well.

]]>Not quite: if one only has a single pair of zeroes then the two terms in the sum will end up being only and thus negligible compared to the main term (or the existing error term).

However, the Erdos-Selberg proof of the prime number theorem can be interpreted somewhat along these lines, but working with rather than . I discuss this at https://mathoverflow.net/questions/259698/ideas-in-the-elementary-proof-of-the-prime-number-theorem-selberg-erd%C5%91s/259719#259719 .

]]>The natural analogue of a Dirichlet series of an arithmetic function over a p-adic place would be , but this usually diverges for every , so one probably has to work with a hybrid of p-adic places and the place at infinity, e.g.

This generalises the usual Dirichlet series which corresponds to the case . One could also work with multiple primes rather than a single prime, but I think this is already illustrative. For instance, the Dirichlet series of is

One still has good behaviour with respect to Dirichlet convolution:

Presumably, continuous counterparts come from functions on using a two-variable Mellin type transform

but I haven’t computed this carefully to see what it gives as far as heuristic predictions are concerned (presumably it makes predictions for sums like ).

]]>Begin by replacing with where are the usual -adics and defined to the the characteristic function on this set. Define multiplicative convolution by

where denotes the -adic norm with normalization so that and is the multiplicative Haar measure on normalized so that

with denoting the Haar measure on normalized so that has measure 1.

Claim: where for .

This follows from the computation that

for each such that .

Remark: Initially it seemed to me that the condition in the definition of convolution is wrong since it reverses the Archimedean one . However, it makes sense since on the norm is at most one instead of growing on . This also accounts for the minus sign in the definition of .

Carrying on, we define the Mellin type transform as above:

I don’t have the time to follow through at the moment, but this feels like the way forward to me…

]]>- ,
- ,
- and
- the inclusion map

with

- ,
- ,
- and
- the diagonal map

and then define Dirichlet series and Mellin transforms as in the post. One could then take inverse limits (e.g., over powers of ) and work -arithmetically or adelically, but I’m not sure what that would add.

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