Skip to main content

plotting - Visualizing a Complex Vector Field near Poles


I've been playing around with a visualization technique for complex functions where one views the function $f: \mathbb{C} \rightarrow \mathbb{C}$ as the vector field $f: \mathbb{R^2} \rightarrow \mathbb{R^2}$. These vector fields have some nice properties as a consequence of the Cauchy-Riemann equations, and usually look pretty neat. I'm surprised I haven't heard of this until recently (they're known as Pólya plots). Here's an example:


f[z_] := Exp[-z^2]
VectorPlot[{Re[f[x + I*y]], Im[f[x + I*y]]}, {x, -1.5, 1.5}, {y, -1, 1},
VectorPoints -> Fine]

Plot of the vector field f(x,y)=(Re(exp(x+iy)),Im(x+iy))


The problem I'm having is trying to do this near the poles of functions. This is understandable, however Mathematica usually has no trouble plotting functions with singularities. Here's an attempt to plot $z^{-1}$:



Attempt at 1/z


I tried upping MaxRecursion and a couple of other things, but I figured you guys might know what to do immediately.




Now that the pole issue has been taken care of (thanks to everyone who contributed), here are some very intriguing plots:


Poles of $\Gamma(z)$ at -4, -3, and -2:


PolyaPlot[g, {-4.5, -1.5}, {-1, 1}, 50]

Poles of $\Gamma(z)$ at -4, -3, and -2


$\sin(z)$:


PolyaPlot[F, {-3 Pi/2, 3 Pi/2}, {-4, 4}, 45]


sin(z)


Now, here is a function that has poles over a subset of the Gaussian integers. The plot immediately reveals the symmetry of the zeros of the nontrivial polynomial


$35900-(72768-72768 i) z-128304 i z^2+(64392+64392 i) z^3-40305 z^4+(8064-8064 i) z^5+2016 i z^6-(144+144 i) z^7+9 z^8$


$\displaystyle \sum_{m=1}^{3} \sum_{n=1}^{3} \frac{1}{z-(m+in)}$:


PolyaPlot[G, {.7,3.3},{.7,3.3},60]

interesting rational function


where the function PolyaPlot is given by:


PolyaPlot[f_,ReBounds_,ImBounds_,vPoints_]:=Module[{reMin=ReBounds[[1]],reMax=ReBounds[[2]],imMin=ImBounds[[1]],imMax=ImBounds[[2]]},

Return[VectorPlot[{Re[f[x+I*y]],Im[f[x+I*y]]},
{x,reMin,reMax},{y,imMin,imMax},
VectorPoints->vPoints,
VectorScale->{Automatic,Automatic,None},
VectorColorFunction -> (Hue[2 ArcTan[#5]/Pi]&),
VectorColorFunctionScaling->False]];
]

Answer



Here are two suggestions for the function


f[z_] := 1/z;


First, instead of defining a region to omit from your plot, you should base the omission criterion on the length of the vectors (so that you don't have to adjust the criterion manually when switching to a function with different pole locations). That can be achieved like this:


With[{maximumModulus = 10},
VectorPlot[{Re[f[x + I*y]], Im[f[x + I*y]]}, {x, -1.5, 1.5}, {y, -1,
1}, VectorPoints -> Fine,
VectorScale -> {Automatic, Automatic,
If[#5 > maximumModulus, 0, #5] &}]
]

vector plot 1



The main thing here is that as the third element of the VectorScale option I provided a function that takes the 5th argument (which is the norm of the vector field) and outputs a nonzero vector scale only when the field is smaller than the cutoff value maximumModulus.


Another possibility is to encode the modulus not in the vector length at all, but in the color of the arrows:


VectorPlot[{Re[f[x + I*y]], Im[f[x + I*y]]}, {x, -1.5, 1.5}, {y, -1, 
1}, VectorPoints -> Fine,
VectorScale -> {Automatic, Automatic, None},
VectorColorFunction -> (Hue[2 ArcTan[#5]/Pi] &),
VectorColorFunctionScaling -> False]

vector color plot


What I did here is to suppress the automatic re-scaling colors in VectorColorFunction and provided my own scaling that can easily deal with infinite values. It's based on the ArcTan function.



As a mix between these two approaches, you could also use the ArcTan to rescale vector length.


Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....