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8

That's the best I could go with Mathematica 10.0.2, with PlotPoints->50 and MaxRecursion->4. ContourPlot[ E^(Sin[x] + Cos[y]) == Sin[E^(x + y)], {x, -10, 10}, {y, -10, 10}, Axes -> True, ImageSize -> Large, PlotPoints -> 50, MaxRecursion -> 4] The rendering took about 1 hour with Mathematica eating all my 16Gb Ram. (I'll never try ...


6

As far as I know there is no documented list of Method options for ContourPlot and DensityPlot. If you want to experiment there is a large list of strings in Charting`CommonDump`$VisualizationMethodOptions to have a look at. Some of these are option settings, some are option values. Most seem to have no effect on a simple ContourPlot and probably apply to ...


6

Since any 2D transformation that maps the sine function onto a circle will distort the functional form, I would suggest drawing the sine function perpendicularly to the 2D plane as follows: frames = With[{m = 10, r = 1, h = .3}, Table[ Show[ ParametricPlot3D[ {r Cos[ϕ], r Sin[ϕ], h Sin[m ϕ - t]}, {ϕ, 0, 2 Pi}, PlotStyle -> ...


6

The quick workaround: You have to add the option PlotRangePadding -> 0 (or its equivalent PlotRangePadding -> None) to Show. Show[SmoothHistogram[RandomVariate[ExponentialDistribution[1/2], 500], Filling -> Bottom], PlotRangeClipping -> True, PlotRange -> {{0, 10}, Automatic}] ...


4

p = Plot3D[x^2 + y^2, {x, -1, 1}, {y, -1, 1}]; pv = VectorPlot3D[{{2 x, 2 y, 0}, {-2 y, 2 x, 0}}, {x, -1, 1}, {y, -1, 1}, {z, -0.01, 0.01}, PlotRange -> {Automatic, Automatic, {-1, 1}}]; Show[pv, p]


3

data=Compile[{},With[{y=Range[-10,10,0.006]}, Table[UnitStep[(E^(Sin[x]+Cos[y])-Sin[E^(x+y)])],{x,y}]]][];//AbsoluteTiming ArrayPlot[data,DataReversed->True]


2

Supposing you import the file so that it produces the following structure.. data = { {"2014-12-11T16:13:13.038337Z", 3.90092}, {"2014-12-11T16:14:13.456558Z", 3.89734}, {"2014-12-11T16:15:12.444318Z", 3.90092}} You would then be best off converting it to a TimeSeries object which is fairly clever about parsing dates automatically. ts = ...


2

With your parameters param1 = {1.56894, 0.548273, 0.548273}; param2 = {5.04353, 0.974998, 0.974998}; param3 = {9.79371, 1.28185, 1.28185}; param4 = {1.96894, 0.748273, 0.348273}; param5 = {5.34353, 0.174998, 0.774998}; param6 = {9.39371, 1.48185, 1.98185}; and a txt-file with the same structure as Export["parameter.txt", {param1, param2, param3, param4, ...


2

It appears that much of what you've posted doesn't work in its present form but it can be simplified greatly in a working form. I've highlighted the key changes that need to be made (with some explanation) and then have put it all together at the end. Highlights I'm not sure what you were trying to do using Table to generate the c values, but it seems that ...


1

If I understand you correctly, here is what I will do: I will the take doc example: g = {Graphics3D[Cylinder[{{0, 0, 0}, {1, 0, 0}}, 0.5]], Graphics3D[Cone[{{-0.5, 0, 0}, {1.5, 0, 0}}, 0.5]]}; v = VectorPlot3D[{y, -x, z}, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, VectorStyle -> {g}, VectorScale -> {Automatic, Scaled[0.3]}, VectorPoints -> ...


1

One should not be confused with method or option. A method in the sense of Mathematica (See: Method) Options in the sense of Mathematica (See: Options) For DensityPlot or ContourPlot you can query for Options with ??DensityPlot or with ??ContourPlot: The Link provided by @Karsten 7. (answer by @Nasser), is a really fabulous strategy to "unearth" ...


1

Manipulate[ PolarPlot[{1, a + Sin[3 θ]}, {θ, 0, 2 π}, PlotRange -> {{-3.5, 3.5}, {-4, 3.2}}], {a, -3, 3}]



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