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1619
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location Spain , Basque Country , Bilbao
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visits member for 3 years, 2 months
seen Nov 26 at 13:24

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 :)


Sep
30
comment Solving an inverse spectral problem
assume all the spectrum is discrete for example the potential $ V (x) \to \infty $ whenever $ |x| \to \infty$
Sep
19
comment Demystify integration of $\int \frac{1}{x} \mathrm dx$
you get the expression $ x^{0}$ by direct integration howver if you 'regularize ' this divergence by substraction of the term $ 1/0 $ you get $ \frac{x^{0}-1}{0}=ln(x)$ by expanding the expression by a taylor series
Sep
17
comment Unitary matrices and Riemann zeros is there an error?
but the term $ arg \zeta (1/2+it) $ is also important because you can have thousands of operators with real or imaginary eigenvalues so the mean density of zeros is about $ N(T)= \frac{T}{2\pi}log \frac{T}{2\pi e} $ this does not prove all the zeros are real.
Sep
12
comment Is there a link between divergent series and discontinuities in Fourier integrals?
www3.nd.edu/~lnicolae/Enyart.pdf
Sep
12
comment Scale invariance and the Mellin transform?
only for scale invariant functions $ f(kx)=f(x) $ for the rest this is not true
Sep
12
comment find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$
a=b=1 is the correct solution and the minimum is 0
Sep
12
comment Definition of distribution, pseudofunction, and tempered distribution
umm why the downvote ? :(
Sep
10
comment Convergence of integral $\int_{0}^{\infty}{\frac{1}{x^{3}-1}}dx$
Ramanujan master theorem should give $ PV \int_{0}^{\infty} \frac{x^{s-1}}{x-a}= a^{s-1}\pi cotg( \pi s) $
Sep
10
comment self adjoint linear operator and integration
consider an opeartor that includes only $ x $ and $ \partial _{x} $
Sep
9
comment Can the system $\partial_x f(x,y) = \dot{y}$, $\partial_y f(x,y) = \dot{x}$ be related to some Hamiltonian system?
NO, see en.wikipedia.org/wiki/… one of your equation must have a minus sign
Sep
8
comment argument principle and evaluation of a complex integral
ok Michael, thanks for the edit :)
Sep
6
comment How to find asymptotics?
should a taylor series near $ x=0 $ work ??
Aug
29
comment Laurent expansion of $ \log\zeta(s) $
perhaps if we could prove that (approximately) $ log\zeta (s) \sim -log(1-1/s) $ then we would have the expansion $log\zeta (s) \approx \sum_{n=1}^{\infty} \frac{1}{ns^{n}}$
Aug
19
comment Laurent expansion of $ \log\zeta(s) $
mistake corrected, the idea is to expand the logarithm of riemann zeta $ log\zeta (s) $ as a Z-transform , see en.wikipedia.org/wiki/Z_transform
Aug
7
comment What are applications of number theory in physics?
zeta regularization $ 1+2^{k}+3^{k}+....$ understood as $ \zeta (-k) $ appears whenever trying to regularize series and integrals :) that are divergents
Aug
7
comment Integrating the Gram Series
how did Gramm got his Gramm series ??
Jul
29
comment abel summation and Harmonic series
no, you are wrong is $ log(1-e^{-\epsilon}) 4 not $ log(1+\epsilon) $
Jul
18
comment Laguerre transform function
for non integer 'm' we can use fractional derivative to define $ L_{m}(x)= \frac{e^{x}}{\Gamma (m+1)}D_{x}^{n}(x^{n}e^{-x}) $
Jul
11
comment zero raised to infinity
$ o^{0}=1 $ since $ x^{x}=e^{xln(x)} $ and $ x \to 0 $ $ xln(x)=1 $
Jul
6
comment Computing the value of logarithmic series: $Q(s,n) = \ln(1)^s + \ln(2)^s + \ln(3)^s + \cdots+ \ln(n)^s $
the sum is divergent , regularizatio is needed so $ (-1)^{s}\zeta ^{(s)} (0) $