jose
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• Champaign, IL, United States

Dataset was restructured in the 12.1 release in order to support expanded formatting options and interactivity such as hiding and sorting. As a result, some Dataset outputs showed a slowdown due to ...

The definition used (motivated by exterior calculus) is as follows: Given a rectangular array $a$ of depth $n$, with dimensions $\{d, ..., d\}$ (so there are $n$ $d$'s) and a list $x = \{x_1, ..., ... View answer 16 votes What about some 2D Geo functionality for this? points = {{-0.9207, -0.3896, 0.0091}, {-0.8272, 0.5077, -0.2399}, {0.2544, -0.3511, 0.9010}, {0.3510, 0.6527, 0.6712}, {0.5436, -0.6326, -0.5513}, {0.... View answer 16 votes I think you need to use a group-theoretical construction. In this way you will have full freedom in specifying any group of permutations you need. In your case the group is G = PermutationGroup[{... View answer Accepted answer 15 votes I take a screenshot of your image and assign it to the image variable. In[2]:= ImageDimensions[image] Out[2]= {1326, 1150} This computes the positions of the lines in your image: In[3]:= lines = ... View answer Accepted answer 14 votes This is not a bug. When dealing with a map there are two different coordinate systems you need to handle, related by the cartographic projection you are using. First you have the {lat, lon} ... View answer 12 votes This is code to compute Easter Sunday for each year (the "Computus"), from Gauss, in the proleptic Gregorian calendar: computusGauss[year_Integer] := Module[{a, b, c, d, e, f, g, h, i, j, k, month,... View answer Accepted answer 10 votes You can use something like this: p = Entity["City", {"NewYork", "NewYork", "UnitedStates"}]; rOut = Quantity[2000, "Miles"]; rIn = Quantity[1000, "Miles"]; GeoGraphics[FilledCurve[{{GeoCircle[p, ... View answer Accepted answer 9 votes There are a few things that can be improved here: Don't call {"Latitude", "Longitude"} separately in EntityValue. It's better to use the "Position" property, that ... View answer Accepted answer 9 votes Due to the non-sphericity of the Earth, there is no exact way of rotating a polygon, but the following is a good approximation. Suppose we want to rotate the UK polygon so that London is moved to ... View answer Accepted answer 9 votes Yes, it is possible to use the ellipsoidal Mercator projection by specifying an ellipsoidal "ReferenceModel" in the projection. To compare, let me define a spherical Mercator projection: In[1]:= ... View answer Accepted answer 8 votes I think some of the rotations must be corrected: rot1 = Cycles[{{1, 2, 4, 3}, {5, 24, 9, 7}, {6, 23, 10, 8}}]; rot2 = Cycles[{{21, 22, 24, 23}, {1, 11, 20, 10}, {2, 5, 19, 16}}]; rot3 = Cycles[{{11, ... View answer Accepted answer 8 votes The problem here is that the geo circle is being resolved into a line as if you were drawing the full primitive, with insufficient resolution for very low scales. To alleviate that, use segments of ... View answer 7 votes A possible way to rotate a map is to use the freedom provided by an oblique projection. Obliqueness adds the three degrees of freedom of a general 3D rotation, namely the {lat, lon} coordinates of the ... View answer 7 votes Let me propose an alternative, very similar in spirit to Chip's idea, but avoiding the use of the internal GeoGraphics`GeoEvaluate. I'll base the construction on a L-type displacement that moves a ... View answer Accepted answer 7 votes GeoElevationData has elevation data for the whole world. If you know the position (I hope I interpreted correctly the Wikipedia data from your link): In[]:= p = GeoPosition[{FromDMS[{42, 19, 32.}], ... View answer Accepted answer 7 votes Take the points of the geo path, and thread the GeoPosition head to have a list of individual geo positions: points = Thread[gpath[[1]]]; Take them in pairs and compute the direction from the first ... View answer 7 votes My approximation to that projection is proj = {"Stereographic", "ReferenceModel" -> "Bessel1841", "GridOrigin" -> {155000, 463000}, "Centering" -> {52.1562, 5.38764}, "CentralScaleFactor" -&... View answer Accepted answer 7 votes The multiplication table is itself a list of permutations of a representation of the group so you can do In[1]:= m = {{1, 2, 3, 4}, {2, 3, 4, 1}, {3, 4, 1, 2}, {4, 1, 2, 3}}; In[2]:= G = ... View answer Accepted answer 6 votes Along the same lines of the answer by Daniel Huber, I think what you need here is this combination: RandomSymmetrizedArray[dims_, sym_, dist_] := Normal@ SymmetrizedArray[_ :> RandomVariate[dist], ... View answer Accepted answer 6 votes Tensor simplification can be an expensive computation, so it is not performed automatically. You need to use TensorReduce on symbolic tensor expressions, like you would use Simplify in more general ... View answer Accepted answer 6 votes The area-preserving projections are listed in GeoProjectionData["EqualArea"]. Most of these projections have formulas for the sphere only, but some of them, like "Albers" or "LambertAzimuthal" in WL ... View answer Accepted answer 6 votes Symmetric[{1,2,3}] means that all six permutations of {1,2,3} are symmetries of the tensor: Equal[tensor, Transpose[tensor, #]] & /@ Permutations[{1, 2, 3}] (* {True, True, True, True, True, True}... View answer 6 votes I think the simplest way to handle the general case is to use TensorProduct and TensorContract, as follows: Take a rank 3 array for example, in dimension 100: In[1]:= A = RandomReal[{-1, 1}, {100, ... View answer 6 votes I think you first need to start by generating the graphics with ListContourPlot. For example, take this arbitrary data as a 31x31 matrix: data = Table[Exp[-(90 - lat) Degree] Cos[lon Degree] + ... View answer 6 votes Suppose we work with objects in some symbolic dimension dim: In[1]:=$Assumptions = (e[1] | e[2] | e[3] | e[4]) \[Element] Vectors[dim] && (a | b) \[Element] Matrices[{dim, dim}, ...

Define your scalar product of the h polynomials: p[h[n_], h[m_]] := 16 Pi S Sum[Binomial[8 S + 2, k] (8 S + 1 - n - m - k), {k, 0, 8 S - n - m}] / (2^(8 S + 1) (n + m + 2) (n + m + 1) Binomial[8 S + ...

You could also do \$Assumptions = (a | b | c | d) \[Element] Matrices[{k, k}] KroneckerProduct[a, b] . KroneckerProduct[c, d] // TensorExpand which returns KroneckerProduct[a.c, b.d]