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Jan
3
comment Convexity Coefficient Calculation
Ah, it doesn't seem to like MeshRegions. If you keep them as graphics primitives it works: RegionDimension[RegionDifference[lines[[1]], Polygon[{{0, 0}, {1, 0}, {0.1, 0.5}, {1, 1}, {0, 1}}]]] gives 1, showing that the line is not entirely contained in the polygon.
Jan
3
comment Convexity Coefficient Calculation
Related: How to check if a line segment intersects with a polygon? Just replacing RegionIntersection with RegionDifference in some of those answers should work.
Jan
3
comment Help with finding the point(s) inside a closed shape with the highest average ray length?
Please enter the statements into Mathematica one at a time and show me what happens. By the way the easier way to upload an image is to add it to your question and then just copy the URL without submitting the edit.
Jan
3
comment Interpolation of 4D data on a triangular grid
Your data points lie entirely in a 2D plane, so it is not possible to perform interpolation across 3D space. Maybe discard one of the three coordinates and perform 2D interpolation?
Jan
2
comment Help with finding the point(s) inside a closed shape with the highest average ray length?
Sorry, I don't know what could be the problem. I assume you must've tried restarting Mathematica at some point. When you enter the curve = ... line, do you get the first image (the blue polygon)? At what point do you start seeing errors?
Jan
1
revised Help with finding the point(s) inside a closed shape with the highest average ray length?
added 336 characters in body
Jan
1
comment Help with finding the point(s) inside a closed shape with the highest average ray length?
Sorry, I forgot to include the definition of the curve.
Jan
1
revised Help with finding the point(s) inside a closed shape with the highest average ray length?
added 27 characters in body
Dec
31
revised Help with finding the point(s) inside a closed shape with the highest average ray length?
added 355 characters in body
Dec
31
answered Help with finding the point(s) inside a closed shape with the highest average ray length?
Dec
29
comment Is the real spherical harmonic (l = 1, m = 0) really 'bigger' than (l = 1, m = 1)?
From the documentation: "The spherical harmonics are orthonormal with respect to integration over the surface of the unit sphere." The issue you're running into is that the $m\ne0$ harmonic is complex-valued, with half its energy in the real component and half in the imaginary one. So if you just plot the real component, its amplitude turns out to be only $1/\sqrt2$ of the amplitude of the purely real $m=0$ harmonic. If you want to plot atomic orbitals, you should probably plot both Sqrt@2 Abs@Re[...] and Sqrt@2 Abs@Im[...] for all the $m>0$ harmonics.
Dec
28
comment How to restructure a list of {x,y,f[x,y]}-triples for use with Interpolation?
Apparently an identical question was closed(?!) two months ago.
Dec
25
awarded  Nice Question
Dec
25
revised Revolution of Koch Snowflake
added 2392 characters in body
Dec
25
answered Prevent Plus from threading
Dec
25
comment Prevent Plus from threading
Thanks for all the answers, everyone! I guess I should have been more forthcoming about my ultimate goal, namely getting a correct PiecewiseExpand[f[t, 0, 2]] for my Koch snowflake parametrization, because I couldn't figure out make the answers here work for it. But seeing all the different approaches inspired a solution of my own that did work out (though still not as elegant as I would like).
Dec
24
asked Prevent Plus from threading
Dec
24
answered Revolution of Koch Snowflake
Dec
23
comment Eigenvectors choose intuitive ordering/sorting
But you said you want the eigenvectors of a diagonal matrix to form the identity matrix.
Dec
23
comment Eigenvectors choose intuitive ordering/sorting
In general what you are asking for is not possible. For example, the eigenvectors for $\begin{bmatrix}1&\epsilon\\\epsilon&1\end{bmatrix}$ are $(1,1)^T$ and $(1,-1)^T$ for any $\epsilon>0$, so there is no way for the eigenvector matrix to approach the identity as $\epsilon\to0$. You may have to write your own eigenvalue algorithm for your specific problem.