# Tag Info

4

Mathematica's special technique for handling complex values in a plot is to simply ignore them -- it plots nothing for at values of the domain for which the range is complex. Experiment with the following code. Manipulate[ PolarPlot[r[t], {t, min °, max °}, PlotRange -> {{0, 3.5}, {-1.5, 1.5}}, PlotRangePadding -> Scaled[.05], Epilog ...

3

In order to study that region where you get the defect try f[t_] := {20 Cos[ t] - ((837 + 800 Cos[2 t] - 35 Cos[12 t] + 40 Sqrt[2] Cos[t] Sqrt[801 + Cos[12 t]]) Cos[300 t] + 480 Sin[t] Sin[6 t] Sin[300 t])/(-1637 + 35 Cos[12 t] - 40 Sqrt[2] Cos[t] Sqrt[801 + Cos[12 t]]), 20 Sin[t] + (4 (40 Cos[t] + Sqrt[2] Sqrt[801 + ...

5

vals = Table[Norm[myfield[x, y]], {x, -3, 3, 6/100}, {y, -3, 3, 6/100}]; m = Mean@Log@Flatten@vals; st = StandardDeviation@Log@Flatten@vals; (* For some cases you may use these instead m = NIntegrate[Log@Norm[myfield[x, y]], {x, -3, 3}, {y, -3, 3}]/36; st = NIntegrate[(m - Log@Norm[myfield[x, y]])^2, {x, -3, 3}, {y, -3, 3}]/36 // Sqrt; *) Manipulate[ ...

6

In version 10.2 there is a new suite of functions that may be helpful: f[x_, y_, z_] := Sin[x/4]*Sin[y/4]*Sin[z/4]; xyzw = Flatten[Table[{x, y, z, f[x, y, z]}, {x, 1, 10}, {y, 1, 10}, {z, 1, 10}] // N, 2]; ListSliceContourPlot3D[xyzw, "CenterPlanes"]

4

There are several options you can try: ContourPlot3D is one way. You can use multiple transparent contours. DensityPlot3D was introduced in version 10.2, but the same effect can be achieved manually using Image3D or Raster3D. The latter has the ClipRange option for cutting away a part (visualizing a cross-section).

1

It is just ListPlot[Flatten[data], Joined -> True, PlotStyle -> {Black, Dashed}, PlotRange -> All, DataRange -> {0, 1000}, Frame -> True, FrameLabel -> {"Zone", "Force"}] EDIT: Oh like it was mentioned in the comments. :)

6

Here is some code I used, which may partially answer your question. The key is to calculate efficiently points near the border of the Mandelbrot set. These points have large iteration counts which, when iterated, produce the Buddhabrot form. The algorithm linked to in the question uses an adaptive mesh of squares to locate border points. The alternate code ...

5

As this is a special-functions question, I feel justified in using a bit of heavy artillery. Here goes nothing... In effect, what the OP seems to want to do is to evaluate $$\sum_{n=1}^\infty \frac{(q^{n+1};q)_\infty}{(q^n;q)_\infty} q^{n-1}$$ (where $(a;q)_n$ is the $q$-Pochhammer symbol) by approximating it with its partial sums. However, there is a ...

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