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I am using the bellow code. With opA or opD works but not with opB or opD. Perhaps because in the output B and C is used 10^-6! But why?. Also, in OpB and optC the menu reset to OptA . ![enter image description here][1] ![enter image description...
Transpose is used in tree different ways. It spents much time, except when it is added AbsoluteTiming. Why? Thanks. Guillermo Very slow Transpose[{Range[10^5], Range[10^5]}] slow Evaluate[Transpose[{Range[10^5], Range[10^5]}]] ...
Mike, I know why qMultiple2 is faster than qMultiple1, and why uc[t3_] is defined with = (much better than :=).The code works fine, I try to refine it. The question is: How can be modified qMultiple1 (including Integrate inside it) in order to...
I wish to square every term of each element of a list. Let me show the pattern (e.g.): Map[(#^2) &, {rf[a], rf[b], rf[a] + rf[b], rf[a] + rf[b] + rf[c]}, {2}] I obtain as output: Out:= {rf(a^2), rf(b^2), rf(a)^2+rf(b)^2,...
Dear friends I have a code to make a simulation. Here it is a toy example represented by only two functions (the real code has more than 100 line an spend in my computer 10 minute each simulation) data = RandomVariate[NormalDistribution[1,...
Why: AirTemperatureData["LESA", DateObject[{2020, 1}], Mean] and AirTemperatureData["LESA", {DateObject[{2020, 1, 1}], DateObject[{2020, 1, 31}]}, Mean] give different outputs? Thank you Guillermo
Thank you all. My preferred solution is fun[list_,n_]:= list]|>//Values
You are right. The code in the English version is (a light but important difference: - 10^-6): ufun = NDSolveValue[{-Laplacian[u[x, y], {x, y}] == 1, PeriodicBoundaryCondition[u[x, y], (x - 2)^2 + y^2 == 1, ...
Thank you Hans. How did you obtain the solution? g1 = E^(((a - Sqrt[a^2 + 4 b + 4 b k]) z)/(2 b)) C[1] + E^(((a + Sqrt[a^2 + 4 b + 4 b k]) z)/(2 b)) C[2]; h1 = E^((k t)/c) K[1]; I found the solution using Laplace Transform, But...
The insert menu "Inline Free-form Input" in Spanish say: "Entrada de forma libre en linea" (Crt+=) but in the Spanish Keyboard Crt+= doesn´t work. Thank you