Hi,
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.
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.
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,
x]
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.
The expression is very complicated, and Mathematica gives an oultine of the structure. Click on “show more”, “show all” to explore.
