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1

My answer here is effectively an answer to this question as well; I'll leave to other people the decision of whether to close this as a dupe. Details on the proceedings can be read in my other answer. Here, I will merely present results: pts = SortBy[{{0., 0.}, {3., 4.}, {-1., 5.}, {-4., 0.}, {-5., -3.}, {-10., -11.}, {-11., -12.}}, First]; m ...


5

As can be seen from the images presented, the interpolation being done under the hood by VertexColors depends on a prior triangulation of the polygon, thus resulting in visible triangular bands. The approaches presented thus far have all needed to perform a preliminary triangulation; I shall now present a method that avoids this preprocessing step. One ...


4

The procedure for doing a weighted B-spline interpolation is not too different from the unweighted case. I'll use the same point set in the docs, and add a weight vector that gives higher weight to the second and fifth points: pts = {{1., 2.}, {0., 1.}, {2., 0.}, {2., 2.}, {3., 3.}, {5., 2.}}; wts = {1, 4, 1, 1, 4, 1}; Here again is Lee's algorithm, which ...


6

This crash has been fixed in version 10.2. In[1]:= if = NDSolveValue[{D[u[t, x, y, z], t] == 0, u[0, x, y, z] == 0}, u, {x, y, z} \[Element] Ball[], {t, 0, 1}, Method -> "FiniteElement"]; In[2]:= Export["if.wdx", if]; ...


3

As m_goldberg said in the comment A interpolation function must represent a single-vaued function of one variable. A BSpline is not a function but the parametric representation of a curve To achieve the same interpolation result, I must interpolate the data in two directions. The interpolateCurve function gives the interpolation of curves. ...


6

You can use the undocumented ReturnMeshObject method like @Simon Woods used here to get ListSurfacePlot3D to do the smoothing for you. With this option added, it returns a GraphicsComplex ready to be used by DiscretizeGraphics. Graphics`Mesh`MeshInit[]; mc = MeshCoordinates[surface]; extractedmesh = DiscretizeGraphics[ First@ListSurfacePlot3D[mc, ...


0

I am using Mathematica 10.1 on Win7-64 bit. Using Interpolation and the data you posted, the following seems to work fine on my system: interpfun = Interpolation[data, InterpolationOrder -> All] You can then use the interpfun object we obtained to calculate interpolated values: interpfun[15, 4.3, 1.32] (* Out: -102.963 *)



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