- #1
eljose
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Heilbronn proved that the Epstein Zeta function did not satisfy RH...but the Zeta function [tex]\zeta(s)[/tex] can be put in a form of an Epstein function but a factor k..let be the functional equation for Epstein functions:
[tex]\pi^{-s}\Gamma(s)Z_{Q^{-1}}(s)=|Q|^{1/2}\pi^{s-n/2}\Gamma(n/2-s)Z_{Q}(n/2-s) [/tex] (1)
then if we set the Quadratic form with associated Matrix Q so [tex]Q=Q^{-1}=R[/tex] with DetQ=1 if we set n=2 (bidimensional) and make the change of variable s=p/2 the functional equation (1) is the functional equation for the Riemann Zeta function or what is the same:
[tex]\frac{Z_{R}(p/2)}{Z_{R}(1-p/2)}=\frac{\zeta(p)}{\zeta(1-p)}[/tex]
or what is the same the riemann Zeta function is (but a factor) an epstein zeta function with Q^2=I being Q the matrix of a bidimensional quadratic form...but if Heilbronn proved that RH is falso for any Epstein function then RH is false for Riemann zeta function...
[tex]\pi^{-s}\Gamma(s)Z_{Q^{-1}}(s)=|Q|^{1/2}\pi^{s-n/2}\Gamma(n/2-s)Z_{Q}(n/2-s) [/tex] (1)
then if we set the Quadratic form with associated Matrix Q so [tex]Q=Q^{-1}=R[/tex] with DetQ=1 if we set n=2 (bidimensional) and make the change of variable s=p/2 the functional equation (1) is the functional equation for the Riemann Zeta function or what is the same:
[tex]\frac{Z_{R}(p/2)}{Z_{R}(1-p/2)}=\frac{\zeta(p)}{\zeta(1-p)}[/tex]
or what is the same the riemann Zeta function is (but a factor) an epstein zeta function with Q^2=I being Q the matrix of a bidimensional quadratic form...but if Heilbronn proved that RH is falso for any Epstein function then RH is false for Riemann zeta function...