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equation solving - Is there a way to symbolically invert a piecewise function


I have a piecewise function that is continuous and strictly monotonic, like this:


f[t_] = Piecewise[{{t/4, t < 0}, {t/2, t < 3}, {3/2 + (t - 3)*3, True}}]


Now, I would like to get the inverse of that function. I would like an expression like this:


fInverse[x_] = Piecewise[{{4*x, x < 0}, {2*x, x < 1.5}, {5/2 + x/3, True}}]

(I later want to convert that expression or something derived from it of it to C code that will run on a microprocessor. So I need symbolic result, not a numerical one.)


I have tried InverseFunction[f], but I think it the result is a numerical inverse. I also tried Solve[f[t] == x, t], but it returns no result at all.


Simplify[Reduce[f[t] == x, t]] with proper assumptions ($Assumptions = {t ∈ Reals, x ∈ Reals}) is promising:


(t == 2 x && 0 <= x < 3/2) || (t == 4 x && x < 0) || (6 t == 15 + 2 x && 2 x >= 3)

but it's not a piecewise expression.


My next step would be to write a function piecewiseInvert that iterates through the alternatives in the Piecewise expression, solves each one for t, changes the conditions appropriately, and creates a new Piecewise, but I'm hoping that there's a simpler way to do this.




Answer



I think you're on the right track with Reduce. If the domain Reals is specified, Reduce will return results that can be converted to a Piecewise expression straightforwardly.


Clear[f];
f[t_] = Piecewise[{{t/4, t < 0}, {t/2, t < 3}, {3/2 + (t - 3)*3, True}}];

invPW[f_] := Evaluate @ Module[{t}, 
  Piecewise[
   List @@ Reduce[# == f[t], t, Reals] /. {cond_ && t == expr_ :> {expr, cond}}]] &;

fInverse = invPW[f]

fInverse[x]

Mathematica graphics


One can remove the extraneous True case by replacing the last condition by True:


invPWC[f_] := Evaluate @ Module[{t}, 
     Piecewise[
      ReplacePart[
       List @@ Reduce[# == f[t], t, Reals] /. {cond_ && t == expr_ :> {expr, cond}},
       {-1, -1} -> True]]] &;


fInverse = invPWC[f]
fInverse[x]

Mathematica graphics



Note: Solve almost does it, also yielding a result that can be converted to Piecewise easily. But it does not include the end points of the intervals when appropriate.


Solve[x == f[t], t, Reals]

(* {{t -> ConditionalExpression[2 x, 0 < x < 3/2]},
    {t -> ConditionalExpression[4 x, x < 0]}, 

    {t -> ConditionalExpression[1/6 (15 + 2 x), x > 3/2]}}  *)

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