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1h
comment Discrete time system with Z transform
This site is for the software Mathematica. Do you have a Mathematica related question?
7h
comment How to insert a Matrix into a larger matrix
Your myM has many repeated identical elements. Is that always the case, or are they supposed to be all different? I mean, is there a pattern we can assume for the repetitions? Also, do you only need zeros in the large matrix to start with? This would affect the degree to which one could simplify things.
8h
comment How to insert a Matrix into a larger matrix
@MrZ The positioning is in the PadLeft command. You could use ArrayPad there, too. That would allow arbitrary positioning.
8h
comment How to insert a Matrix into a larger matrix
@MrZ If you want to have a more general approach, then you may want to look at @kguler's answer here. But then your question should have made it clearer that you want a more general outcome.
19h
comment Sum over Binomials and Gammas
Confirmed in version 8 on Mac OS X.
23h
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.
1d
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.
1d
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.
1d
comment Poisson PDE in a rectangular domain
I'm sorry, but I don't understand what your actual Mathematica question is.
2d
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.
2d
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.
2d
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.
2d
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.
2d
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.
2d
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
29
comment It is possible to obtain an equation of motion by using Hamilton's principle?
Since it's a broad question, there won't be an exact duplictae on this site. But there are examples of what you're asking about here: Solution to a T+U = E equation, and Animation of double pendulum