Mathematica Asked on March 2, 2021
Consider the following sample code
pts = Table[{i, i^2}, {i, -10, 10}];
foo = Interpolation[pts];
Plot[InverseFunction[foo][y], {y, 0, 10}]
From the plot, we see that InverseFunction[foo][y]
only finds the positive x
corresponding to the interpolation function foo
, but ideally, I would like the inverse function to be something like ifoo[y]={x1,x2,...}
, where x1,x2,...
are the possible solutions for $y=x^2$.
In the end, I would like to generalize this to a multi-dimensional function. More specifically, let’s say I have a set of data {{{x1,y1},{f1,g1}},...}
and apply interpolation so that we get the function foo[x,y]={f[x,y],g[x,y]}
. Now I would like to invert this function so that ifoo[z,w]={{x1,y1},{x2,y2},...}
where {x1,y1},{x2,y2},...
are the possible solutions towards foo[x,y]={z,w}
.
How would one be able to do this?
Define the inverse with FindRoot
Clear["Global`*"]
pts = Table[{i, i^2}, {i, -10, 10}];
foo = Interpolation[pts];
inv[y_?NumericQ, branch_ : Automatic] :=
Module[{x, init = If[branch === Automatic, -10, 10]},
x /. FindRoot[foo[x] == y, {x, init}]]
Augmenting InverseFunction
Plot[{InverseFunction[foo][y], inv[y]},
{y, 0, 100}, PlotLegends -> Placed["Expressions", {.85, .25}]]
Replacing InverseFunction
Plot[{inv[y, "+"], inv[y]}, {y, 0, 100},
PlotLegends -> Placed["Expressions", {.85, .25}]]
Answered by Bob Hanlon on March 2, 2021
You can find the two inverses by generating a differential equation of foo and solve with NDSolve applied to two initial conditions. Even derivatives of inverse function can be produced.
pts = Table[{i, i^2}, {i, -10, 10}];
foo = Interpolation[pts];
dinv = D[foo[x[y]] == y, y]
xsol[y_] =
x[y] /. NDSolve[{dinv, #}, x, {y, 100, 0}] & /@ {x[100] == -10,
x[100] == 10} // Quiet
{Plot[xsol[y], {y, 0, 100}, ImageSize -> 300, AspectRatio -> 1],
Plot[xsol'[y], {y, 0, 100}, ImageSize -> 300, AspectRatio -> 1]
}
Answered by Akku14 on March 2, 2021
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