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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


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