2,773 reputation
1518
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location Spain , Basque Country , Bilbao
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visits member for 3 years
seen 20 hours ago

i got a degree on Physics and have several ideas on my own , unfortunately i do not have a physics sponsor :) in order to get an investigation grant :)


Jan
3
comment What is so interesting about the zeroes of the $\zeta$ function
WU-Sprung is the physical realization of the Riemann Hypothesis, the Hilbert-Polya operator is then an Hamiltonian (second order perator) on the half-line $$ [0, \infty) $$ , this potential is connected to a solution of the riemann hypothesis and the Eigenvalues are the square of the Riemann Zeros $$ E_{n}= \gamma ^{2} $$
Jan
3
answered What is so interesting about the zeroes of the $\zeta$ function
Jan
2
comment argument of the Riemann zeta function
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ arg\zeta(1/2+it)/\sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
Jan
2
comment Plot of a Bessel function if possible
it is strange i believed it was real... how about $$J_{ait}(2\pi i)+ J_{-ait}(2\pi i) $$
Jan
2
asked Plot of a Bessel function if possible
Jan
2
accepted stuck with an integral involving an exponential function
Jan
2
asked stuck with an integral involving an exponential function
Jan
2
comment argument of the Riemann zeta function
then stopple what you mean is that $$ \frac{1}{T}\int_{T}^{2T}dt \frac{arg \xi(1/2+it)}{\sqrt{(1/2)loglog(t)}}= \frac{1}{\sqrt{2\pi}}\int_{E}exp(-x^{2}/2) $$ is this true
Jan
2
accepted is this function invertible ??
Jan
1
comment is this function invertible ??
but for big big 'x' the dominant term is $ f(x)=x $ and the inverse of this is just $ g(x) $ , besides any function which can be drawn can be inverted, just reflect each point trough the line $ y=x $ to get the Numerical inverse of the function
Jan
1
asked is this function invertible ??
Dec
31
comment Zeros of $ \frac{1}{B(xi)^{1/2}}((iA)^{ix})(ix)^{ix}+ \frac{1}{B(-xi)^{1/2}}((-iA)^{-ix})(-ix)^{-ix}=H(x)$
yes 'x' is real i search for the real zeros only :)
Dec
31
revised Zeros of $ \frac{1}{B(xi)^{1/2}}((iA)^{ix})(ix)^{ix}+ \frac{1}{B(-xi)^{1/2}}((-iA)^{-ix})(-ix)^{-ix}=H(x)$
error in one term
Dec
31
asked Zeros of $ \frac{1}{B(xi)^{1/2}}((iA)^{ix})(ix)^{ix}+ \frac{1}{B(-xi)^{1/2}}((-iA)^{-ix})(-ix)^{-ix}=H(x)$
Dec
31
accepted argument of the Riemann zeta function
Dec
31
asked argument of the Riemann zeta function
Dec
31
accepted Is a Macdonald function a Bessel function with imaginary argument??
Dec
31
asked Is a Macdonald function a Bessel function with imaginary argument??
Dec
19
accepted Chebyshev function identity
Dec
19
asked Chebyshev function identity