Hello.
I have this function ## v(x) = -\sum_{i=1} x^i \sqrt{2}^{i-2} \int_{-\infty}^{\infty} m^{i-1} \cosh(m)^{-4} dm## which I can not seem to figure out how to simplify.I tried looking at some partial integration but repeated integration of ## \cosh ## gives polylogarithms which seemed to...
I can not see why this changes a whole lot. I just assumed some distribution of when a sensor will fail.
What sort of distribution will they follow since it is not Gaussian( ignoring the fact that it extends to ##-\infty##)? You obviously know a lot more about it than me, so please enlighten...
Sorry for the late reply. I have been traveling.
Ahh of cause I get it now. Thanks for the help!
What they give is a number of how long they say the sensor should at least work( for example 2000 hours.). If the sensor company does not want to many RMA's this number will be some low quantile...
Thank you both for your answers. A few comments.
Yes t is the same for all sensors. It is time since installation. I do not understand why the distribution of failed sensors will be binomial and do you mean the distribution of the amount of sensors failed at a given time? Can you elaborate?
It...
(Sorry for the terrible title. If anybody have a better idea, post and I will edit. Also I have no idea of the level so now I just put undergraduate since the problem is fairly easy to state.)
Suppose I buy ## N## sensors which the manufacturer tells me will fail at some point and the failure...
If your equation is given by ( please please please learn to tex )
$$ \frac{r^2}{f(r)} \frac{d^2 f(r)}{dr^2} + \frac{2mr^2}{h^2}\left(E + \frac{zt^2}{kr}\right) -a^2 = 0 $$
Then you can rewrite to
$$ \left( \frac{d^2}{dr^2} + \frac{2mE}{h^2} + \frac{2mzt^2}{h^2 kr} - \frac{a^2}{r^2} \right)...
It is in order to fulfill the boundary conditions. Using this trick the slope "out" of the lattice becomes zero.
Other methods of solution could be using Chebyshev polynomials.
Is p_{i,j} the fourier transform? You should take care that the method does not produce some unwanted complex phase which might ruin the solution.
The kx=0 entry is the mean value for that "row" so you can not just set it equal to zero. I would imagine it depends a bit on your RHS how to best...
Why would you think there is a problem with non-linear terms? I really can not see how this should pose a problem as long as he uses some explicit time integration scheme ( and even if he used an implicit I can not see why the difficulties would even be related to the finite difference method ).
I used a simple 5 or 9 point finite difference stensil.
You result in #28 is to be expected since you are differentiating a function given by ## f(x,y) = e^{-(x^2 + y^2)} + 1 ##, in the interior and ## f(x,u) = 0 ## on the boundary. That will generate problems. This however, is not the...
@Johnny
A max error on the laplacian of around 0.005 for even large dx for this function seems pretty good to me. On the other hand in your 2d plots it looks weird that the error does not go towards zero away from the middle( assuming red is not = 0. You should put in a colorbar on those plots...
It looks like matlab colormap. In matlab depending on your implementation you do not need a matrix formulation. You can directly vectorize the input. It is a bit hard to see the lattice size but already 512^3 needs some memory.
For validation, perhaps you can look at the velocity statistics of...
How important is system size? If not so much how about a super simple exponential time integration scheme in fourier space with some simple(or no) anti-aliasing on the non-linear term? I seem to recall it worked pretty well back when I worked with it. Have you read The world of the complex...
Lets just do it. Since ## A[h] ## is only dependent on the gradient of ## h ## the variational derivative reduces to
## \frac{\delta A}{\delta h} =- \nabla \cdot \frac{\partial \sqrt{1 + (\nabla h)^{2} } }{\partial \nabla h} =- \nabla \cdot \left( \frac{1}{\sqrt{1 + (\nabla h)^{2} } } \nabla h...
Shouldn't you tell us what ## V ## is? At any rate the area of a curved surface can be calculated as ## A = \int dx (a) = \int dx \sqrt{1 + (\nabla h(x))^{2}} ## with ## h(x) ## being the heigth of the surface. Your equations looks like a ficks type of law with "diffusion" coefficient given by...
I suppose you mean ## \lim_{x=\pm \infty} y(x) = 0 ##? And no the function doesn't have to be ## 0 ## everywhere. An example is ## y(x) = \tanh(x)^{2}(1-\tanh(x)^{2}) ##. (You will have to work out ## V(x) ## yourself.)
I do not know much about this stuff but according to the wiki article on Landau poles (https://en.wikipedia.org/wiki/Landau_pole), some russian guy recently showed that the beta function does not blow up in QED. Isn't this in contrast with the calculations discussed in the text?
Also somewhat...
As far as I could figure out by reading some old old document Whittaker shewed that solutions to the 3d laplace equation can be put on the form ## \int_{0}^{2\pi} f(x + iy\cos(v) + iz\sin(v)) dv ##. By pure guessing I found that
## \int_{0}^{2\pi} f(x,y,z,v) dv = \frac{2\pi x }{\left(x^2 + y^2...
Hi all.
Sometimes I have a few questions in my head I like to post at some point but which at the moment aren't completely ready. It would be nice if we had some personal draft section where I could write down my question and do updates and so on before releasing it for the general public...