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Asked by cm on 20 Sep 2013

I want to solve this equation to find a formula that computes "p" in terms of "q". Said differently, the variable is "p" and the parameter is "q".

the equation is " p.^2 .*(1-q).*(2.*q.*(1-p.^2)).^(2.*p) -2.*(1-p).*(1-q.*(1-p)).*(p+q.*(1-p))"

and there is a constraint for "p" and "q" since they are probabilities between 0 to 1

I will be so pleased if some one kindly let me know how to do it?

Thanks in advance

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Answer by Matt J on 20 Sep 2013

Edited by Matt J on 20 Sep 2013

Accepted answer

I doubt a closed form expression exists. You could solve numerically, using fzero

fun=@(p) p.^2 .*(1-q).*(2.*q.*(1-p.^2)).^(2.*p) -2.*(1-p).*(1-q.*(1-p)).*(p+q.*(1-p));

p=fzero(fun,[0,1]);

However, it is unclear to me whether multiple solutions for p might exist for a given q, even when both are constrained to [0,1]. If so, fzero will find only one of them.

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cm on 20 Sep 2013

Ok! So you mean it's not possible to reach a closed form for this question?

anyway, thanks for your help!

Matt J on 20 Sep 2013

No, I don't think a closed form expression is possible. If you didn't have the '^2*p' in there, it would be a 4th order polynomial in p. Closed form expressions do exist in that case for all the roots. They would be fairly ugly expressions and you would have to analyze which root(s) lay in [0,1].

With the power of 2*p, though, it looks rather unlikely you'd find something closed form. But, if you have the Symbolic Math Toolbox, you could try the SOLVE command, to see what it gives you.

Answer by Walter Roberson on 20 Sep 2013

If the overall expression is to equal 0 (rather than it being of the form y = .... and needing to solve for given y and p), then the expression given has two solution families:

p = 0, q = 1 p = 1, q = anything in [0 to 1]

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