
#1
Oct610, 04:07 AM

PF Gold
P: 2,884

I notice that the trick to define Dirichlet Eta Function can be repeated for each prime number, let p a prime, and then
[tex]\eta_p= (1  p^{s}) \zeta(s)  p^{s} \zeta(s) = (1  2 p^s) \zeta(s)[/tex] So each prime p defines a function [tex]\eta_p[/tex] that adds a family of zeros at [itex]s= (\log 2 + 2 n i \pi) / \log p} [/itex]. Particularly, for p=2 it kills the only pole of zeta in s = 1. Repeating the trick for each prime, we seem to obtain a function [tex]\eta_\infty(s)[/tex] having the same zeros that the Riemann zeta plus families of zeros of density 1/log p placed at the lines r=log(2)/log(p) Or, if we dont like to crowd the critical strip, we can use only the left part of the eta, the [tex] \eta^F_p(s)= (1  p^{s}) \zeta(s) [/tex], and then for [tex]\eta^F_\infty(s)[/tex] we get all the families cummulated in the r=0 line. But on other hand it can be argued that [tex]\eta^F_\infty(s)=1[/tex], as we have removed all the factors in Euler product. So there should be some relation between the n/log(p) zeros in the imaginary line and the other zeros in the Riemann function 


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