The Riemann Hypothesis

by -Job-
Tags: hypothesis, riemann
zetafunction is offline
Jan17-12, 04:19 AM
P: 399

the Riemann Xi function(s) [tex] \xi(1/2+z) [/tex] and [tex] \xi(1/2+iz) [/tex] can be expressed as a functional determinant of a Hamiltonian operator, functional determinants may be evaluated by zeta regularization, using in both cases the Theta functions , semiclassical and spectral ones :)
zetafunction is offline
Jan25-12, 07:33 AM
P: 399
[tex] \xi (s) = \xi(1-s) [/tex] with [tex] \frac{\xi(s)}{\xi(0)}= \frac{det(H+1/4-s(1-s))}{det(H+1/4)} [/tex]

with [tex] H= - \partial _{x}^{2}+ f(x) [/tex] and

[tex] f^{-1}(x)= \frac{2}{\sqrt \pi }\frac{d^{1/2}{dx^{1/2}}Arg (1/2+i \sqrt x ) [/tex]

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