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I have a cubic equations with 2 unknowns and constraints on both variables. I am unsure whether or not a closed form solution does exist. How should I go about solving this?

${ -1 + 2b - 2b^2 - \frac{5}{6}b^3 + 2e - e^2 + \frac{e^3}{6} - 2eb + \frac{e^2b}{2} = 0 }$

or alternatively (not developed)

${ \frac{1}{3} - b^2(1 + b + \frac{e}{2}) + \frac{1}{6}(b^3-(-2 -b -e)^3) = 0 }$

given that ${ e \in [0,1]}$ and ${ b \in [-1,0] }$

quasi
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unisateur
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  • There is a cubic formula (Cardano's formula), but it's usually not very useful. What's the application? – quasi May 10 '18 at 14:59
  • Solving for an Incentive Compatibility constraint with continuous payoffs functions, in a game theoretical model. – unisateur May 10 '18 at 15:00
  • Are you trying to maximize or minimize some function of $e$ and $b$, subject to the above constraints? – quasi May 10 '18 at 15:01
  • Kinda: ${b}$ varies over the range defined and represents one player bias', and this player is looking for the largest possible ${e}$ such that this hold. The original IC is just the sum of two integrals with squared terms, if that helps. – unisateur May 10 '18 at 15:03
  • If by solution you mean numerical values for $e$ and $b$, then you need at least one more equation relating them. As it stands, if there is a solution set, then it contains uncountably infinitely many elements. – Allawonder May 10 '18 at 15:14
  • Yes @Allawonder I know that, the question is given that I do not have another equation (or I do it just introduce another variable), and given that I have these constraints on the two variables, is there any way I can characterise ${e}$ and ${b}$, for instance ${e}$ as a function of ${b}$ – unisateur May 10 '18 at 15:17
  • If you're solving a real problem there's always another equation if you look enough. And the constraints are insufficient to determine a unique solution -- that's the point. You need another relationship between the variables to solve this uniquely. Look more, analyse the problem more. You should get one. Otherwise, no unique solution. – Allawonder May 10 '18 at 15:23

1 Answers1

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Based on the comments, it appears you are trying to maximize $e$, subject to the given constraints.

Trying $e=1$, the cubic equation reduces to $$5b^3+12b^2-3b-1=0$$ which has $3$ real solutions for $b$, one of which, $b\approx -.1944051955$, satisfies the given constraints.

Therefore the maximum value of $e$, subject to the given constraints, is $e=1$.

quasi
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  • thank you, that makes sense! but is there anyway to have the value of ${e}$ as a function of ${b}$ in general ? – unisateur May 10 '18 at 15:25
  • Not in any useful way. Rather, treat it as an implicit equation, and analyze it that way. For example, you can use a graphing utility to graph the equation in the rectangle,$-1\le b\le 0,;0\le e \le 1$, and the graph will yield lots of qualitative information. – quasi May 10 '18 at 15:29