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location Spain , Basque Country , Bilbao
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visits member for 2 years, 9 months
seen Jul 21 at 20:09

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 :)


Aug
8
comment Evaluating $\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$
anyway thank you all for your answers :D
Aug
8
comment Evaluating $\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$
shouldn't it be $ \frac{\pi }{2} J_{0}(au/2) $ due to the change of variable $ t \rightarrow t/2 $
Aug
8
comment Evaluating $\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$
aja, thanks what bessel function if possible :) thanks again
Aug
8
comment Riemann Siegel formula modification?
yes intx is integer part (or floor function) of 'x' i did not know how to put it
Jul
31
comment evaluation of $ \operatorname{Arg}\zeta (1/2+is) $ ??
umm is there or are there graphics for $ arg\zeta (1/2+it) $ or graphical representaiton of the function $ S(T)= \frac{1}{\pi}arg\zeta (1/2+iT) $ for several values of T , can i see or evaluate the eigenvalue staircase for teh Riemann zeros ??
Jul
31
comment evaluation of $ \operatorname{Arg}\zeta (1/2+is) $ ??
but preciselyt this theta functio n $ \theta (t) $ is just the smooth part of the zeros :S as seen on en.wikipedia.org/wiki/Riemann_hypothesis#Number_of_zeros
Jul
31
comment evaluation of $ \operatorname{Arg}\zeta (1/2+is) $ ??
however since $ Z(t) $ is real then the argument of $ \zeta (1/2+it) $ is just the function $ \theta (t) $ but this theta function is just the same as the smooth part of the zeros ¡¡
Jul
28
comment root-finding methods to invert numerically a function
OK thanks, we could use a bisection method and then a newton method to imporve the convergence :)
Jul
24
comment Eisenstein series solution
er sorry about that $ z=x+iy $ and $ i= \sqrt -1 $
Jul
24
comment how to solve an ODE with boundary conditions $ y(0)=y(\infty) $ by shooting method
another possibility is to make the change of variable $ x=uL $ so the new boundary value problem turns into $ y(0)=0=y(1) $
Jul
24
comment how to solve an ODE with boundary conditions $ y(0)=y(\infty) $ by shooting method
for example $ x= \frac{u}{u-1} $ but now the ode would have a singular point at $ u=1 $ how can i use shooting method to deal with it ?
Jul
23
comment how to solve an ODE with boundary conditions $ y(0)=y(\infty) $ by shooting method
thanks a lot .. so i must find first a number L so $ y''(L)\approx y'(L)\approx0$ by numerical methods and then once i have found it i solve the differential equation by linear shooting method
Jul
13
comment stuck on a differential equation
OK, thanks.. :)
Jul
11
comment inverse function for big 'x'
so your function is $ y=|x| $ the inverse is then $ x=|y| $ and i can draw it numerically
Jul
4
comment Xi function plotter ??
the idea is to see if the Riemann Xi function is a functional determinant in the sense $ det(H+(s-1)s) $ of a certain Hamiltonian oeprator with potential defined by $ f^{-1}(x)= 2 \sqrt \pi \frac{d^{1/2}}{dx^{1/2}} \frac{1}{\pi}arg\xi(1/2+i \sqrt x) $
Jul
4
comment Xi function plotter ??
am not i was interested in the $\xi (s) $ function to compare a method to obtain the Riemann Zeros and Riemann zeta function as a functional determinant with the known values of Riemann Xi function.. thanks anyway
Jun
28
comment Evaluation of these integrals
aha thanks.. :)
Jun
13
comment Operator from the trace
's' here is a real number $ s >0 $
Jun
13
comment Operator from the trace
what is $ I^{-1}(g) $ ?? is the inverse function of $ g(s) $ thanks
Jun
10
comment Fourier integral
OK thanks :) , by the way where i could get some info about the Cosine integral evaluation :) as you have pointed ? thanks.