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Oct
4
revised How can I illustrate the solution space of these equations in 3D?
added 1152 characters in body
Oct
4
answered How can I illustrate the solution space of these equations in 3D?
Sep
30
awarded  Notable Question
Sep
26
comment Diagonals of a regular octagon
In Mathematica 10.1 I believe you can just do pts = CirclePoints[8].
Sep
21
comment Challenge: deblurring images
Also previously: How to enhance a fuzzy image
Sep
20
revised Plotting precise intersections involving singular functions
added 34 characters in body
Sep
5
comment Is there a way to change PlotRange without redrawing?
@belisarius: We don't know whether the OP is using Plot or some other drawing function. But still, suppose you used Plot but the PlotRange was too small in the $y$-axis, might the Show trick still work, or does Plot throw those points away? (I'd try it myself but I don't have Mathematica on this machine.)
Aug
28
comment Creating a certain image (3D shadows, highlighted intersections)
For the right half with the 3D curves with white outlines, see my answer here.
Aug
25
revised How can I superimpose a set of points on a ContourPlot?
fixed spelling of my name, and improved the grammar while I was at it
Aug
1
comment Redefining a built-in operator
If you follow @ciao's suggestion, you can also use it in infix notation via a ~p~ b.
Jul
29
awarded  Nice Question
Jul
28
comment Gauss Fit to List of 2D Points
I didn't know about EstimatedDistribution, thanks for that! By the way, given that this is a probability distribution, the scaling factor ought to just be the total number of observations times the spacing between bins, i.e. c = 0.1 Total@data[[All,2]]. That gives $264.9$ compared to $266.675$ from your method; I don't have a good explanation for the difference.
Jul
27
comment Gauss Fit to List of 2D Points
@infinitezero: By the way, that is known as the Rayleigh distribution.
Jul
25
revised InverseFunction of [E, 1, 1][c/Sqrt[K[1]]] isn't it just Log[c/Sqrt[K[1]]?
added 24 characters in body
Jul
23
comment How do I remove grid from this photo?
Well, you have all their code; why don't you try it out yourself on the high-resolution image?
Jul
22
comment How do I remove grid from this photo?
I think the reason the grid is still left is because RidgeFilter looks for thin bright ridges, so in the regions where the grid is darker than the background, RidgeFilter is selecting parts not in the grid to be part of the mask. For illustration, take a look at ImageMultiply[img, mask] and zoom in on the white window frame behind the cat. Not sure what would be a better way of doing it though.
Jul
16
answered How can you get the area of a region from an image?
Jul
16
comment How can you get the area of a region from an image?
@Mr.Wizard: Because I wasn't sure whether I was answering the intended question. Going by the title it's conceivable the OP wants to count the area of the shaded region in the actual raster image rather than in the geometric shape it depicts. Fernando: Can you clarify? Also, what do you mean by "modify something out imaginary parts"?
Jul
16
comment How can you get the area of a region from an image?
Area[RegionIntersection[Disk[{l, 0}, l], Disk[{0, l}, l], Disk[{l/2, l/2}, l/2]]]
Jul
14
comment Surface Area of a Region inside a Region
In principle, you could just do Area[RegionIntersection[RegionBoundary[myRegion], container]], but I'm not near a Mathematica instance right now to try it out and I recall it having trouble with intersections of regions of different dimensionalities.