Re: ContourPlot non rectangular evaluation?
- To: mathgroup at smc.vnet.net
- Subject: [mg127862] Re: ContourPlot non rectangular evaluation?
- From: Sam McDermott <samwell187 at gmail.com>
- Date: Tue, 28 Aug 2012 04:50:42 -0400 (EDT)
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Hi,
Thanks everybody for the responses - I appreciate your time but suddenly have had to be out of the office and can't find any time of my own to test these solutions. I'll respond in more depth when I can.
Thanks again,
Best,
Sam
On Sunday, August 26, 2012 11:35:23 PM UTC-4, Murray Eisenberg wrote:
> 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 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 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 =
>
>
>
> 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?