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zetafunction
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HERE http://vixra.org/pdf/1007.0005v1.pdf
is my proposed proof of an operator whose Eigenvalues would be the Imaginary part of the zeros for the Riemann Hypothesis
the ideas are the following* for semiclassical WKB evaluation of energies the number of levels N(E) is related to the integral of pdq along the classical orbits
* for one dimensional systems, in the WKB approach the inverse of the potential V(x) is proportional to the Half-derivative of N(E)
* For the case of Riemann Zeros the N(E) is given by the argument of Riemann Xi function
Xi(s)= s(s-1)G(s)Z(s) G=gamma function Z= zeta function
evaluated on the critical line [tex] Arg\xi(1/2+iE) [/tex]
If we combine these points then the inverse of the potential V(x) inside the Hamiltonian H=p^2 +V(x) would be proportional to the HALF INTEGRAL of the logarithmic derivative of
[tex] \xi(1/2+iE) [/tex] as i express in the attached .PDF see formulae 3 4 and 5 inside the paper
Also as a final NUMERICAL test, i think that the functional determinant of this operator [tex] H=p^2 +V(x) [/tex] is related to the Xi function, since the Hadamard product applied to Xi function can be used to express the Xi function as a product of the eigenvalues of some operator
in brief, the INVERSE function of the potential V(x) is given by the half-integral of the logarithmic derivative of the Xi function evaluated on the critical line 1/2+is
also the FUNCTIONAL DETERMINANT of the operator [tex] H+m^2 [/tex]is the Riemann Xi-function
is my proposed proof of an operator whose Eigenvalues would be the Imaginary part of the zeros for the Riemann Hypothesis
the ideas are the following* for semiclassical WKB evaluation of energies the number of levels N(E) is related to the integral of pdq along the classical orbits
* for one dimensional systems, in the WKB approach the inverse of the potential V(x) is proportional to the Half-derivative of N(E)
* For the case of Riemann Zeros the N(E) is given by the argument of Riemann Xi function
Xi(s)= s(s-1)G(s)Z(s) G=gamma function Z= zeta function
evaluated on the critical line [tex] Arg\xi(1/2+iE) [/tex]
If we combine these points then the inverse of the potential V(x) inside the Hamiltonian H=p^2 +V(x) would be proportional to the HALF INTEGRAL of the logarithmic derivative of
[tex] \xi(1/2+iE) [/tex] as i express in the attached .PDF see formulae 3 4 and 5 inside the paper
Also as a final NUMERICAL test, i think that the functional determinant of this operator [tex] H=p^2 +V(x) [/tex] is related to the Xi function, since the Hadamard product applied to Xi function can be used to express the Xi function as a product of the eigenvalues of some operator
in brief, the INVERSE function of the potential V(x) is given by the half-integral of the logarithmic derivative of the Xi function evaluated on the critical line 1/2+is
also the FUNCTIONAL DETERMINANT of the operator [tex] H+m^2 [/tex]is the Riemann Xi-function