New answers tagged visualization
7
Following this question you can define:
invmollweide[{x_, y_}] :=
With[{theta = ArcSin[y]}, {Pi (x)/(2 Cos[theta]),
ArcSin[(2 theta + Sin[2 theta])/Pi]}];
fc[phi_] := Block[{theta}, If[Abs[phi] == Pi/2, phi, theta /.
FindRoot[2 theta + Sin[2 theta] == Pi Sin[phi], {theta, phi}]]];
cart[{lambda_, phi_}] :=
With[{theta = fc[phi]}, {2/Pi*lambda ...
16
Edit: I added more explanations below, because this visualization method is quite different from conventional vector plots
For just this purpose I had at some point invented the following visualization technique. I'll reproduce your definition first. It defines a complex vector field on the surface of a unit sphere.
Clear[\[Epsilon]];(*Polarization ...
2
You can use reciprocal to avoid $\infty$ points:
ParametricPlot[Evaluate[
ComplexExpand[
Through[{Re, Im}[(x + y I + I)/(x + y I - I) /. {x -> x^#1, y -> y^#2}]]] & @@@
Flatten[Outer[List, {1, -1}, {1, -1}], 1]
], {x, 0, 1}, {y, 0, 1}, PlotRange -> {{-3, 3}, {-.5, 4}}] // Quiet
1
Since a complex function is involved, it seems natural to use plotting functions that treat complex objects directly, without having to overtly separate them into real and imaginary parts. David Park's Presentations add-on (http://home.comcast.net/~djmpark/DrawGraphicsPage.html) allows this:
<< Presentations`
With[{f = Function[z, (z + I)/(z - I)],
...
4
I like all the existing answers because they reveal the nature of this map--but they are all deceiving. The problem is that they miss obvious and important parts of the region that cannot easily be drawn parametrically because they are associated with extremely large values of $x$ and $y$.
Here is a simple solution that is natural in the sense that it ...
9
Take it in steps:
Extract the coefficients and locations into an appropriate data structure.
Use that data structure to create the graphics.
By examining the FullForm of the original expression, we can cobble a rule to find the key data: the coefficients $c$, $d$, and $-1$ and the offsets to the indexes. First, the expression itself:
s = Subscript;
exp ...
6
Although you haven't exactly asked this, you might like to generate your graphic automatically by applying pattern matching on your difference operator. The basic idea is as below:
Clear[i, j, k];
op = Plus @@
MapThread[Subscript[u, i - #1, j - #2, k - #3] &,
RotateRight[{0, -1, 1, 0, 0, 0, 0}, #] & /@ {0, 2, 4}]
Giving $op = ...
4
Not sure if this is what you're after. Anyway:
r = .1;
Graphics3D[{Specularity[White, 10],
Red, Sphere[{0, 0, 0}, r],
Blue, Sphere[{0, 0, 1}, r], Sphere[{1, 0, 0}, r], Sphere[{-1, 0, 0}, r],
Sphere[{0, -1, 0}, r], Sphere[{0, 0, -1}, r],
Green, Sphere[{1, 1, 0}, r], Sphere[{1, ...
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