Mathematica Optimizing Resolution in RegionPlot for Complex Conditions in Mathematica 6

[dukebdx12]

can someone help me out with a mathematica problem ( i use mathematica 6)? I have never used this program before and really don't know what to do or where to start. So if someone could give me a little light that would be great.

http://i29.tinypic.com/25rkn5f.jpg

-last graph is g(x,y)=

 

[EnumaElish]

These must have been produced by Plot3D in Math'ca.

For example,

Plot3D[Cos[x y], {x, -3 \[Pi], 3 \[Pi]}, {y, -3 \[Pi], 3 \[Pi]}]

then shift+enter.

First thing I'd do is to hold & rotate the graph to change the ViewPoint.

To change it explicitly, you can type for example:

Plot3D[Cos[x y], {x, -3 \[Pi], 3 \[Pi]}, {y, -3 \[Pi], 3 \[Pi]}, ViewPoint -> {-1, 0, -3}]

Another thing to try is to change the plot limits from {x, -3 \[Pi], 3 \[Pi]} to (say) {x, -\[Pi], \[Pi]} (same for y).

 

[dukebdx12]

ok thanks for the reply. my question is, it says make surface and contour plots for suitable intervals, viewpoints, and resolutions. For example f(x,y) = cos(xy) for -3pi <= x,y <= 3pi , that is 1 of the equations for the question. So to make surface and contour plots(many of them) do I just plug other numbers in for -3pi and 3pi? Confused on what I need to do.

 

[EnumaElish]

You should make a surface plot and a contour plot for:

a. suitable intervals,

b. suitable viewpoints, and

c. suitable resolutions

for each function f, g, etc.

"Plugging other numbers in for -3pi and 3pi" addresses item a.

 

[dukebdx12]

i've got it all down now except the resolutions. I have tried searching for it on google and mathematica and i can't find anything on how to do it. It says. to draw surface and contour plots for resolutions. I don't know what that means? can someone explain?

 

[mrandersdk]

try searching in the master index in mathematica it is very good (help -> master index)

 

[EnumaElish]

My guess is, "resolution" means "density" of the plot. For example in a ContourPlot, the number of contours you specify manually, as an option.

Alternatively, resolution can mean sampling frequency. Mathematica has a default sampling frequency when plotting graphs. You can change it by specifying a step size as part of the domain specification. For example:

{x, -\[Pi], \[Pi]} produces default sampling

{x, -\[Pi], \[Pi], \[Pi]/180} produces sampling along the x-axis with step size = 1 degree.

 

[dingpud]

I noticed that a version of convert-pdf was used. If that free trial runs out, try searching for "cute pdf"...

sorry i didn't have anything of much value to add other than that...

 

[EnumaElish]

Re my last post above:

If {x, -\[Pi], \[Pi], \[Pi]/180} doesn't work try putting in the option WorkingPrecision -> value in the Plot statement where the default value is MachinePrecision but you can specify a numeric value to increase plot precision (e.g. WorkingPrecision -> 20).

You can also experiment with Mesh and MeshFunctions options. For example:
Plot[Tan[x], {x, 0, Pi/2}, Mesh -> {5, 10}, MeshFunctions -> {#1 &, #2 &}]

 

[big-geek]

I have struggled with bad resolution and quality in RegionPlot for a long time. In particular this is a problem for complicated multiple conditions. The way to solve it is to add PerformanceGoal and MaxRecursion:

RegionPlot[ f(x,y)>0 && g(x,y)>0, {x,xmin,xmax},{y,ymin,ymax},
PerformanceGoal -> "Quality", MaxRecursion -> 8]

I suspect that the default values are PerformanceGoal -> "Speed", MaxRecursion -> 1. The speed and quality seems to be very sensitive to MaxRecursion, so experiment a bit with different values.

( google keywords: resolution RegionPlot Mathematica )