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1d
comment Inverse of function including hyperbolic cotangent
Is this a Mathematica or a math question? If Mathematica, what have you tried? (If math, should migrate it to math.stackexchange.com)
1d
comment How to speed up execution of this function?
@Yves Klett That's "Dr. Legolas", to you.
1d
comment How to define a two-variable sequence?
(1) @QuantumDot gives a very good answer (I upvoted the comment, if that's the correct term. I'll also upvote if it becomes a bona fide answer.) (2) Possibly this question should be closed as one that "simply" required checking documentation (e.g. for Sum). But... (3) Sum can evaluate symbolically and that might not always be what is wanted e.g. when there is a "procedural" construct in the summands. So a variant that might sometimes be preferred would be Inactive[Sum][2^(n+m-2j),{j,0,m-1}].
1d
comment How to speed up execution of this function?
@Yves Klett When I was younger I kept long hair to hide my pointy ears (they receded with age).
1d
comment How to speed up execution of this function?
@belisarius (1) `harmonicNumber is almost certainly a better way to go. I failed to see that simplification. (2) I also had that impression of the hat. What can I say? I get sent them from SE Central and I just put them on for a few days.
1d
answered How to speed up execution of this function?
2d
comment Discrete Fourier Transform with x axis data?
Is this any improvement? Fourier[data1, FourierParameters -> {0, 2*Pi/1000}]
2d
comment Matrix form of complex numbers
What would be an example of two different matrices that represent the same complex number?
2d
comment How to find the period of an arbitrary mathematical function?
@alancalvitti Not sure what might be a reliable method offhand. This might work (have not tested for reliability though). Requires several repeats of the periodic part. periodicity[seq_] := Module[ {len = Length[seq], dft = Rest[Abs[Fourier[seq]]], pos}, pos = Position[dft, Max[dft]]; Round[len/pos[[1, 1]]] ] This gives the period length. Now work from that. (Probably you should pose this as a separate question so you can get different approaches.)
Dec
17
comment plot both parts of a two-sheeted hyperboloid
@belisarius Yeah, but only because you read the question correctly. Which probably is cheating.
Dec
17
comment Design matrix rank does not match
It means the 9x6199 submatrix of data positions does not have full rank. Contrast behavior of LinearModelFit[{{0, 0, 1}, {0, 1, 2}, {0, 2, 3}}, {x, y}, {x, y}] with LinearModelFit[{{0, 0, 1}, {0, 1, 2}, {1, 2, 3}}, {x, y}, {x, y}]. The submatrices in question are {{0,0},{0,1},{0,2}} which is rank deficient, and {{0,0},{0,1},{,2}}` which has full rank of 2.
Dec
17
comment plot both parts of a two-sheeted hyperboloid
@belisarius Cute (though at heart that's two parametrizations).
Dec
17
comment plot both parts of a two-sheeted hyperboloid
Cannot do it in one parametrization. A parametrization maps a planar region to its range continuously so it won't be able to jump sheets.
Dec
17
comment Optimizing alphametrics code
A very different method that works tolerably well is to set up an explicit integer linear programming (ILP) branch-and-prune loop. Weight the higher digits first in terms of which to branch on and it goes reasonably well. (Failure to weight means you'll have a long wait, as I have learned from personal experience.) There are several approaches here and the ILP-based method, which does not do this weighting, is indeed slow.
Dec
17
comment Why does this function inside FindRoot fail to evaluate?
@Mr.Wizard I have no idea (that font stuff is all Klingon to me..)
Dec
17
comment Listing all cubic order derivatives of a function of 4 variables
@belisarius I thought it was kind of cuke myself. Actally I learned it from MathGroup back in the 90's. I believe the first to come out with this approach was Andre Deprit, in the context of multivariate series.
Dec
17
answered Listing all cubic order derivatives of a function of 4 variables
Dec
16
answered Why does this function inside FindRoot fail to evaluate?
Dec
16
comment Why does this function inside FindRoot fail to evaluate?
You might want to have a close look at that expression for mu. Maybe check the FullForm or something.
Dec
13
comment How do I understand this big difference in speed:
@YvesKlett Giving it a go, but it's far from a complete answer.