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I'm trying to solve the following polynomial equation for $x$:

$$ (qx)^\alpha (1-qx)^{1-\alpha} = [(1-q)x]^\beta [1-(1-q)x]^{1-\beta} $$

where $\alpha, \beta, q, x$ are all strictly between 0 and 1.

Sadly both Solve and Reduce refuse to help. I've also tried log-transforming, to no luck. What makes this polynomial so hard to solve?

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    $\begingroup$ Since alpfa and beta are between 0 and 1, this equation is is nonlinear but not a polynomial one. This makes it impossible to solve. $\endgroup$
    – Alexei Boulbitch
    Commented Nov 8, 2023 at 18:07
  • $\begingroup$ That's very interesting. In what sense are non-linear, non-polynomial equations impossible to solve? What should I call this equation instead? Do you have a reference for these kinds of problem? How would one go about solving them? $\endgroup$
    – raving-bandit
    Commented Nov 8, 2023 at 18:10
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    $\begingroup$ If you assign the constants reasonable rational values, then can use poly=GrobenerBasis[eqn,x] to reduce eqn to a polynomial, then poly[[1]] is a polynomial representation of eqn, then can solve for the roots then subset of roots are solutions to original equation. $\endgroup$
    – josh
    Commented Nov 8, 2023 at 18:29
  • $\begingroup$ Thank you! I'm not sure what "reasonable" means in this case. Is there any resource I could read on this kind of computation? $\endgroup$
    – raving-bandit
    Commented Nov 8, 2023 at 18:45
  • $\begingroup$ I mean low order exponents like 1/2 or 1/3 or 1/4, 1/5. Generally users here like to see some attempt at solving problem although some don't mind. Can you format an error-free expression for your equation in Mathematica code as myEqn=leftSide==rightSide with $\alpha=1/2$, $\beta=1/3$ and $q=1/4$ and update your post above? $\endgroup$
    – josh
    Commented Nov 8, 2023 at 19:03

1 Answer 1

Reset to default
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$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global`*"]

Manipulate[
 {a, b} = Rationalize[{av, bv}];
 {eqn = (q x)^a (1 - q x)^(1 - a) == ((1 - q) x)^
       b (1 - (1 - q) x)^(1 - b), "",
   StringForm["solution:  ``",
    sol = NSolve[{eqn /. q -> Rationalize[qv], 0 < x < 1},
       x, Reals, WorkingPrecision -> 15] // N], "",
   ContourPlot[Evaluate@eqn,
    {x, 10^-6, 1 - 10^-6}, {q, 10^-6, 1 - 10^-6},
    GridLines -> {x /. sol, {qv}},
    GridLinesStyle ->
     Directive[Red, Dashed, AbsoluteThickness[1]],
    FrameLabel -> (Style[#, 14] & /@ {x, q}),
    ImageSize -> 360]} //
  Column,
 {{av, 0.2, "α"}, 0.005, 0.995, 0.005, Appearance -> "Labeled"},
 {{bv, 0.3, "β"}, 0.005, 0.995, 0.005, 
  Appearance -> "Labeled"},
 {{qv, 0.15, "q"}, 0.005, 0.995, 0.005, Appearance -> "Labeled"},
 SynchronousUpdating -> False,
 TrackedSymbols :> {av, bv, qv}]

enter image description here

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  • $\begingroup$ Thank you! It will take me some time to parse through the code and understand it all. Is there some precise reason why Mathematica can't offer a closed form solution? $\endgroup$
    – raving-bandit
    Commented Nov 9, 2023 at 16:32
  • $\begingroup$ Either there isn’t one or Mathematica doesn’t how to get there. $\endgroup$
    – Bob Hanlon
    Commented Nov 9, 2023 at 18:46

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