Re: ContourPlot non rectangular evaluation?
- To: mathgroup at smc.vnet.net
- Subject: [mg127839] Re: ContourPlot non rectangular evaluation?
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Sun, 26 Aug 2012 02:50:22 -0400 (EDT)
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- References: <20120825082641.60C04689F@smc.vnet.net>
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 >
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- Re: ContourPlot non rectangular evaluation?
- References:
- ContourPlot non rectangular evaluation?
- From: Sam McDermott <samwell187@gmail.com>
- ContourPlot non rectangular evaluation?