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'Nother Inverse Function Question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg126607] 'Nother Inverse Function Question
  • From: Bill Freed <billfreed at shaw.ca>
  • Date: Fri, 25 May 2012 04:54:46 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

Thanks for the previous hints on using InverseFunction. Will be helpful 
for me.
I am working on parameterizing  regions bounded by 4 curves in the plane 
or 6 surfaces in 3D.
Below is an example for the region bounded by
      x=2+1.2(2-y)^2, x=-3.2-1/3(y-1/3)^2
      y=(-x/3)^3+x/2+2.5, y=(x/4)^3+1

Table[FindRoot[{x == s (2 + 1.2 (2 - y)^2) + (1 - s) (-3.2 - 1/3 (y - 
1.3)^2),
y == t (-(x/3)^3 + x/2 + 2.5) + (1 - t) ((x/4)^3 + 1)}, {x, 0}, {y, 2}], 
{s, 0, 1, .1}, {t, 0, 1, .1}];
x = ListInterpolation[x /. %, {{0, 1}, {0, 1}}];
y = ListInterpolation[y /. %%, {{0, 1}, {0, 1}}];
ParametricPlot3D[{x[s, t], y[s, t], 0}, {s, 0, 1}, {t, 0, 1}, ViewPoint 
-> {0, 0, +Infinity},
Boxed -> False, Axes -> None,  PlotPoints -> {25, 11}]

The problem with this homemade inverter is frequent error messages 
concerning accuracy and convergence and also problems in choosing the 
starting points, here {x,0}, {y,2}.
Is there a way of using InverseFunction or other Mathematica command 
that is more robust?

Thanks
Bll Freed



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