# Tag Info

6

You can do this with GeoLabels: GeoRegionValuePlot[Counts[temp], GeoLabels -> (Tooltip[#1, #2] &)]

5

Update Alternatively, a much much easier way: p1 = Plot3D[BetheSalpeter[a, b] , {a, 0.0001, 5}, {b, 0.0001, 5} , MeshFunctions -> {Function[x[#1, #2]]} , Mesh -> {Range[0, 1, 0.25]} , PlotPoints -> 30] Plot3D[Log@BetheSalpeter[a, b] , {a, 0.0001, 5}, {b, 0.0001, 5} , MeshFunctions -> {Function[x[#1, #2]]} , ...

5

Based on the now-deleted answer by belisarius is forth, I suggest ContourPlot[(1/2) u^(2) + 1 - Cos[Theta], {Theta, -4 Pi, 4 Pi}, {u, -5, 5}, PlotLegends -> LineLegend[{Blue, Black, Red}, {"Closed", "Separatrix", "Open"}], PlotLabel -> "Contours of E", ContourShading -> False, Contours -> Range[0, 15, .5], ContourStyle -> ...

5

This solves your specific problem. It does not generalize in the sense that you need to know the values of the contours beforehand and also know which ones correspond to closed and open. Anyway vals = {0.5, 1, 2, 5, 6.5, 8, 10}; s = ContourPlot[(1/2) u^(2) + 1 - Cos[Theta] , {Theta, -4 Pi, 4 Pi}, {u, -5, 5} , ContourShading -> None , Contours -> ...

2

I will answer the question posed in the title: How does ListPlot3D interpolate data? It uses a Delaunay triangulation of the projection of the points on the $xy$ plane. This is essentially the same as using first-order Interpolation (which at present is the only interpolation scheme available for unstructured data). Observe: SeedRandom[0]; points = ...

2

I think you are understanding things correctly and I hope I am understanding your question correctly. The default is to assume a negative feedback but you may set this as an option using FeedbackType which takes values of "Positive", "Negative" or None. With none you are just looking at the poles of your G(s). num = s^2 a1 + s a2 + a3; denom = s^3 b1 + s^2 ...

2

s = ParametricNDSolveValue[{th'[t] == u[t], u'[t] == -Sin[th[t]], th[0] == i, u[0] == 1.5}, {th, u}, {t, -4 Pi, 4 Pi}, {i}] Show[ ContourPlot[(1/2) u^(2) + 1 - Cos[th], {th, -4 Pi, 4 Pi}, {u, -5, 5}, ColorFunction -> "Pastel"], ParametricPlot[Through[s[#][t]], {t, -4 Pi, 4 Pi}] & /@ Pi Range[-4, ...

1

Needs["ErrorBarPlots"] minimal = {{0.0626294, 9.1, 10.4523, 7.9, 0.52, 9.75, 12.73, 1.59, 9.21, 8.49, -19.7381}, {0.154463, 10.98, 2.43, 0.9158, 0.25, 12.33, 13.73, 2.05, 7.59, 8.28, -12.479}, {0.0772834, 3.89214, 5.45804, 1.1667, 0.09, 6.51, 11.78, 1.22, 8.47, 8, -18.8248}, {0.084189, 3.7676, 0.714178, 0.9589, 0.64, 7.43, 9.85, 1.92, ...

1

Does this look right? Graphics[ Scale[First@minimalplot, {1, -1}, {0, 0}], Frame -> True, PlotRange -> {1, -1} PlotRange[minimalplot], FrameTicks -> {{ChartingScaledTicks[{-# &, -# &}], Automatic}, {Automatic, Automatic}}, CoordinatesToolOptions -> {"DisplayFunction" -> ({1, -1} # &), "CopiedValueFunction" -> ({1, -1} ...

1

Here's a stupidly simple thing I came up with for my own question: just divide the domain into smaller blocks, make separate StreamPlots for each of them, and stitch them together. f[x_, y_] := {-1 - x^2 + y, 1 + x - y^2}; xrange = {-3, 3}; yrange = {-3, 3}; xdivs = 3; ydivs = 3; xranges = Partition[Rescale[Range[0, xdivs], {0, xdivs}, xrange], 2, 1]; ...

1

Another way to do it. ListLinePlot[Table[{i, RandomReal[]}, {j, 2}, {i, 10}], PlotLegends -> LineLegend[{"first entry", "second entry"}, LabelStyle -> Small]]

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