Re: ContourPlot non rectangular evaluation?

*To*: mathgroup at smc.vnet.net*Subject*: [mg127840] Re: ContourPlot non rectangular evaluation?*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Sun, 26 Aug 2012 02:50:43 -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>

Since you say you can find where the denominator g[x, y] == 0 and imply that tis is a "curve", I presume you already have, at least numerically, a function k[x] given implicitly by g[x, y] == 0, that is, satisfying g[x, k[x]] == 0. The condition for a point {x, y} to be above the curve g[x, y] == 0 is therefore that y > k[x]. Then use that condition in the RegionFunction option. For example: f[x_, y_] := x^3y - 4 Exp[y -x] g[x_, y_] := x^2 - y^3 h[x_, y_] := f[x, y]/g[x ,y] k[x_] := Abs[x]^(2/3) ContourPlot[h[x, y], {x, -2, 2}, {y, 0, 3}, Exclusions -> {g[x, y] == 0}, RegionFunction -> Function[{x, y}, y > k[x]], Contours -> 10] On Aug 25, 2012, at 4:26 AM, Sam McDermott <samwell187 at gmail.com> wrote: > > 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 really uglifies the plot. It is easy for me to identify where the = function becomes singular (i.e., when the denominator becomes 0!), but = I'm having trouble telling Mathematica to stop evaluating before then, = because this singularity 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 = evaluating an "If" conditional inside of ContourPlot but can't find a = good way of doing it. Any help would be much appreciated! --- 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>