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[?] Solve cubic equation w.r.t X ?

Posted 1 year ago
5 Replies
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I have partially differentiated an equation w.r.t. x, see picture below, and want to isolate x in the equation to solve for a local maximum. This equation is part of my master's thesis and quite crucial. I know that, if rewritten, it will be cubic in x, thus, there can be at least three different solutions. However, at this moment, I am not able to obtain any solutions using Mathematica. Can you help me?

Attached there is a picture of the equation as it looks in the paper and also a picture of how I've written it into Mathematica.

Ideally, I would like to simplify it and find values of x that solves the equation.

I hope you can help.

Equation to be solved for x. Mathematica Code

5 Replies

Hi Mads,

Please post code, not images.

One obvious problem is that solve should be Solve. Functions in WL are case sensitive and all built-in functions start with a capital letter.

Posted 1 year ago

Sorry for the inconvenience. Please find my code below.

Ideally, I would like the expression on a cubic form, if Wolfram can provide it, and then isolate x in the expression. x is the only variable, the others are constants.

Solve[[Alpha](ph - pl)(R1 + R2[Gamma][Lambda]*( { {x(ph)^2 + (1 - x)(pl)^2}, {xph + (1 - x)pl} } - { {x(1 - ph)ph + (1 - x)(1 - pl)pl}, {x(1 - ph) + (1 - x)(1 - pl)} }) + R2*( { {(1 - [Gamma]*[Lambda])}, {1 + r} } )) - 2b[Phi]*x == 0, x]

Did you mean this?

Solve[\[Alpha] (ph - pl) (R1 + 
      R2\[Gamma]*\[Lambda]*(((x (ph)^2 + (1 - x) (pl)^2)/(x*
              ph + (1 - x) pl)) - ((x (1 - ph) ph + (1 - x) (1 - 
                pl) pl)/(x (1 - ph) + (1 - x) (1 - pl)))) + 
      R2*((((1 - \[Gamma]*\[Lambda]))/(1 + r)))) - 2 b*\[Phi]*x == 0,
Posted 1 year ago

Thank you Gianluca!

I am new to Mathematica. The code you provided gives me three different solutions, as expected. However, the solutions are not that meaningful. Is there any way I can simplify these in some way?

and what do the grey rounded boxes with numbers mean? i.e. (--- 1 ---) etc. enter image description here

The expression is very complicated, and Mathematica gives an oultine of the structure. Click on "show more", "show all" to explore.

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