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Apr
1
comment How to integrate Mathematica with other software
If you're not using WebMathematica and try to integrate a "regular" copy with a web interface, you're probably violating the license agreement. But to decide that, you'd (a) need a specific example and (b) ultimately still may need to ask Wolfram support.
Mar
31
comment Poisson PDE in a rectangular domain
@SjoerdC.deVries I would definitely agree, but in an earlier question it seemed that the OP has a version of Mathematica in which FEM was still buggy, so he needs a finite-difference type solution.
Mar
31
comment Poisson PDE in a rectangular domain
Then I would say (again) that this is a duplicate of Poisson solver using Mathematica, because your problem is a special case of that one.
Mar
31
comment Poisson PDE in a rectangular domain
I'm sorry, but I don't understand what your actual Mathematica question is.
Mar
30
comment Improving a jagged Tube[ ] edge
To avoid the atrocious look I see on version 10 Mac, I would get rid of the replacement Line -> Tube and simply add Tube[0.2] to the PlotStyle options instead, combined with Exclusions -> None.
Mar
30
comment Improving a jagged Tube[ ] edge
In my MMA version 8 on Mac, it looks exactly like what you show in the question. But in version 10, it looks like someone left it in the bathtub too long: all wrinkly... so in other words, on version 10 (Mac) I don't think I can even begin to help in any way because the problem is deeper than antialiasing.
Mar
30
comment Laplace PDE in a polar coordinate system
A previous version of this question was closed. To avoid the same fate with this version of the question, you should show what your specific issue in Mathematica is and how you have so far tried to address it.
Mar
30
comment Expansion of a hypergeometric function takes ages with Mathematica 9 and 10 (regression?)
To make a general answer, this shouldn't have the order of expansion hard-coded. As it is, the result could also be obtained with Limit. I understood the comment to mean that he needs higher orders, and that was the whole point of my answer.
Mar
30
comment Expansion of a hypergeometric function takes ages with Mathematica 9 and 10 (regression?)
Oh, actually, this doesn't really work for higher orders, so it's as limited as the approach of taking the Limit. Yes, it appears to spit out a result if you do the series with 1 instead of 0, but it's in terms of derivatives of the hypergeometric function, which when I try to evaluate numerically doesn't return a result. So this doesn't get us back to the behavior in version 8, as my answer does.
Mar
30
comment Expansion of a hypergeometric function takes ages with Mathematica 9 and 10 (regression?)
I wonder why this wasn't the accepted answer - it seems the most straightforward (+1). Maybe the reason this works is the same as why my answer works: the series expansions for the hypergeometric function for undefined arguments always seem to work best.
Mar
29
comment Issue with FindRoot and initial value
Yes, infinities probably get counted as crossings of the axis by mistake. So infinities have to be removed, either by choosing the brackets properly, or by other tricks such as searching for a root of the ArcTan of your function instead.
Mar
29
comment Issue with FindRoot and initial value
If you want to be guaranteed that you'll always get a root between two given brackets, I would definitely recommend the function findAllRoots in my answer here. Then you can choose the brackets by extrapolation from the previous root as a function of the parameter. With the built-in FindRoot, you're not guaranteed that the two initial values will bracket the root that is returned (as far as I recall).
Mar
29
comment Expansion of a hypergeometric function takes ages with Mathematica 9 and 10 (regression?)
@vsht Yes, that's why I also kept version 8... but in principle it would be possible to automate the method I used here. You "just" have to wrap it in a rule that picks out all hypergeometric functions. The implementation will of course depend on how different the various terms look. So I agree, it's in many cases easier to not upgrade from version 8 if you want to avoid having to fix stuff that will eventually change again anyway.
Mar
29
comment Issue with FindRoot and initial value
It works if you bracket the root by specifying two values: FindRoot[c[k, 210*10^-3], {k, 500, 700}] yields {k->686.816}. You could also use this
Mar
28
comment Exporting a Mathematica Document as LaTeX
This is not an issue with Mathematica, but with the other platforms you mention. Apparently they don't accept the $\LaTeX$ code you want.
Mar
28
comment Poisson PDE over a irregular region with FDM
@wlkyr If you liked this answer, you should accept it by clicking on the check mark.
Mar
28
comment Poisson PDE over a irregular region with FDM
This post is closely related: Poisson solver using Mathematica I would have marked your post as a duplicate of the linked question, but didn't see this post until now.
Mar
28
comment Proving positive definiteness or semi-definiteness of a matrix
The dimensions of your matrix are $3\times 4$ because you omitted a comma. Also, just a warning: You shouldn't use capital letters as variable names. K is a built-in symbol that's reserved for other uses.
Mar
28
comment Perturbation theory with Mathematica: Definite integral of polynomial times exponential times hypergeometric function of imaginary argument
You have an error in the math post linked from your question here: the hydrogen ground state doesn't have a factor of $r$ in front of the exponential. That would be $\ell=1$. The case $n=0$ in your question never actually arises. With your notation, you need $n=2$ for the ground state.
Mar
27
comment Perturbation theory with Mathematica: Definite integral of polynomial times exponential times hypergeometric function of imaginary argument
I don't think you can easily use the analogous approach for the $k$ integral because the integrands there involve absolute squares. But in any case, that goes beyond what the original question contains. Since you're doing perturbation theory, it seems reasonable to approximate the $k$ integral first before trying anything with the exact form. Maybe I'll have an idea later...