Re: Re: Complex Plot

*To*: mathgroup at smc.vnet.net*Subject*: [mg88311] Re: [mg88288] Re: Complex Plot*From*: "W_Craig Carter" <ccarter at mit.edu>*Date*: Thu, 1 May 2008 03:22:01 -0400 (EDT)*References*: <fv9a83$mjo$1@smc.vnet.net> <200804301103.HAA04939@smc.vnet.net>

Hello Jean-Marc, In 6.0, we can use ContourPlot for implicit equations, but the timing seems much slower that what you are getting: Timing[cp1 = ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, 23}];] (*about 1 second my machine) cp1 Timing[cp2 = ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, 23},PlotPoints-> 25];] (*about 3 seconds my machine*) cp2 Timing[cp3 = ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, 23},PlotPoints-> 50];] (*about 15 seconds my machine*) Timing[cp3 = ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, 23}, PlotPoints -> 75];] (*about 36 seconds my machine*) The estimate is about 1/2 hour on my machine (if no swapping)... So, unless your machine is about 6 times faster than mine, the implicit version is slower... However, PlotPoints->50, seems to be sufficient to see the lovely structure. Craig On Wed, Apr 30, 2008 at 7:03 AM, Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote: > Chris Degnen wrote: > > > Can anyone tell me how I can re-express this > > equation in terms of x for plotting? > > > > Sin[x^2*y]==Log[x/y] > > ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, > ImageSize -> 500] > > ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, > ImageSize -> 500, ColorFunction -> (White &), ContourStyle -> Black, > MaxRecursion -> 0, PlotPoints -> 500] > Hope this helps, > -- Jean-Marc > > -- W. Craig Carter