       Re: 'Nother Inverse Function Question

• To: mathgroup at smc.vnet.net
• Subject: [mg126639] Re: 'Nother Inverse Function Question
• From: Ingolf Dahl <ingolf.dahl at telia.com>
• Date: Mon, 28 May 2012 05:09:37 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201205250854.EAA26865@smc.vnet.net>

```Bill,
We might first note that it is possible to solve for s and t as functions of
x and y

In:= Simplify[Solve[
x == s (2 + 1.2 (2 - y)^2) + (1 - s) (-3.2 - 1/3 (y - 1.3)^2), s]]
Out= {{s -> (3.76333\[VeryThinSpace]+ 1. x - 0.866667 y + 0.333333 y^2)/(
10.5633\[VeryThinSpace]- 5.66667 y + 1.53333 y^2)}}
In:= Simplify[Solve[y == t (-(x/3)^3 + x/2 + 2.5) + (1 - t) ((x/4)^3 +
1),
t]]
Out= {{t -> (1.\[VeryThinSpace]+ 0.015625 x^3 - 1. y)/(-1.5 - 0.5 x +
0.052662 x^3)}}

Now put

In:= s[x_, y_] := (3.7633333333333336`\[VeryThinSpace]+ 1.` x -
0.8666666666666667` y +
0.3333333333333333` y^2)/(10.563333333333333`\[VeryThinSpace]-
5.666666666666666` y + 1.5333333333333332` y^2)
In:= t[x_, y_] := (
1.`\[VeryThinSpace]+ 0.015625` x^3 - 1.` y)/(-1.5` - 0.5` x +
0.052662037037037035` x^3)

We might now plot s amd t as functions of x and y. Here is one method which
give a little jagged edges.

In:= ContourPlot[s1 = s[x, y]; t1 = t[x, y];
If[Or[s1 < 0, s1 > 1, t1 < 0, t1 > 1], -0.1, s1], {x, -4, 4}, {y,
0.0, 3.6}, Contours -> Table[i, {i, 0.0, 1.0, 0.1}]]
In:= ContourPlot[s1 = s[x, y]; t1 = t[x, y];
If[Or[s1 < 0, s1 > 1, t1 < 0, t1 > 1], -0.1, t1], {x, -4, 4}, {y,
0.0, 3.6}, Contours -> Table[i, {i, 0.0, 1.0, 0.1}]]

The transformation seems smooth and invertible in the region. I was a little
bit curious to check if we could use reverse interpolation, using my
http://www.familydahl.se/mathematica/index.html). Ignore the warnings when

In:= Needs["Obtuse`"];
During evaluation of In:= General::compat :  "Combinatorica Graph and
Permutations \
functionality has been superseded by preloaded functionaliy. The \
package now being loaded may conflict with this. Please see the \
Compatibility Guide for details."

Create a table of x and y values as function of s and t values.

In:= tabxy3 = Flatten[
Table[{{s[x, y], t[x, y]}, {x, y}}, {x, -4, 3.5, 0.3}, {y, 0.0,
3.6, 0.3}], {1, 2}];

Something fishy happens outside the region of interest, so remove all points
outside. An alternative is to allow points slightly outside the region.

In:= tabxy3 = tabxy3 /. {{{a_, b_}, {c_, d_}} /; (Or[a < 0, a > 1, b <
0,
b > 1]) :> Sequence[]};

RBF interpolation should work very well in this case. Create an
interpolation function, which gives x and y as function of s and t. This
type of interpolation can be used in 3D also, with 3 parameters.

In:= ip = Interpolation[tabxy3, Method -> "RBF"];

Test in one point (where s and t are both 0.5). If a more exact value is
needed, this should at least provide a good start value for FindRoot. One
could also fiddle with the options of the RBF interpolation or the number of
points in the table.

ip[{0.5, 0.5}]
{-0.529058, 1.61451}

Make a parametric plot. We see that we get a quite good, but not perfect,
approximation.

In:=  ParametricPlot[ip[{s, t}], {s, 0., 1.}, {t, 0., 1.},
PlotPoints -> {25, 11}, PlotRange -> {{-4, 3.5}, {0.0, 3.5}},
AspectRatio -> 3.5/7.5]

Have fun!

Ingolf Dahl
Sweden

> -----Original Message-----
> From: Bill Freed [mailto:billfreed at shaw.ca]
> Sent: den 25 maj 2012 10:55
> To: mathgroup at smc.vnet.net
> Subject: 'Nother Inverse Function Question
>
> 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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