Re: ContourPlot non rectangular evaluation?
- To: mathgroup at smc.vnet.net
- Subject: [mg127851] Re: ContourPlot non rectangular evaluation?
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Sun, 26 Aug 2012 23:35:19 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
- Delivered-to: mathgroup-newsend@smc.vnet.net
- References: <20120825082641.60C04689F@smc.vnet.net> <20120826065022.F1C976866@smc.vnet.net>
The O.P. wanted to exclude the region *below* the curve g([x, y] == 0.
On Aug 26, 2012, at 2:50 AM, Bob Hanlon <hanlonr357 at gmail.com> wrote:
> Try the option Exclusions or using Boole
>
> f[x_, y_] = Sin[x + y];
> g[x_, y_] = x - y;
> h[x_, y_] = f[x, y]/g[x, y];
>
> ContourPlot[h[x, y],
> {x, -Pi, Pi}, {y, -Pi, Pi}] //
> Quiet
>
> ContourPlot[h[x, y],
> {x, -Pi, Pi}, {y, -Pi, Pi},
> Exclusions -> {g[x, y] == 0}] //
> Quiet
>
> ContourPlot[h[x, y]*
> Boole[g[x, y] != 0],
> {x, -Pi, Pi}, {y, -Pi, Pi}] //
> Quiet
>
>
> Bob Hanlon
>
>
> On Sat, Aug 25, 2012 at 4:26 AM, Sam McDermott <samwell187 at gmail.com> wrote:
>> Hi,
>>
>> I'd like to make a plot of level curves of a function, but this =
function =
> becomes singular and passes from infinity to negative infinity, which =
reall=
> y uglifies the plot. It is easy for me to identify where the function =
becom=
> es singular (i.e., when the denominator becomes 0!), but I'm having =
trouble=
> telling Mathematica to stop evaluating before then, because this =
singulari=
> ty is sensitive to both of the variables on the axes of my =
ContourPlot. Is =
> there a simple way of telling Mathematica where to stop?
>>
>> In other words, I have some function
>>
>> h[x_,y_]:=f[x,y]/g[x,y]
>>
>> and I can numerically find the zeroes of g[x,y]. Can I make a =
ContourPlot=
> such that
>>
>> ContourPlot[h[x,y],{x,xmin,xmax},{y,ymin,ymax}]
>>
>> does not evaluate below the curve
>>
>> g[x,y]=0
>>
>> ?
>>
>> I think I'm basically looking for some way of setting assumptions of =
eval=
> uating an "If" conditional inside of ContourPlot but can't find a good =
way =
> of doing it. Any help would be much appreciated!
>>
>> Thanks very much for your time!
>>
>> -Sam
>>
---
Murray Eisenberg =
murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
- References:
- ContourPlot non rectangular evaluation?
- From: Sam McDermott <samwell187@gmail.com>
- Re: ContourPlot non rectangular evaluation?
- From: Bob Hanlon <hanlonr357@gmail.com>
- ContourPlot non rectangular evaluation?