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solving a complex equation

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

0 Comments

cm

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2 Answers

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.

5 Comments

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.

cm on 20 Sep 2013

Yes You are right! the power of 2*p makes it so difficult to reach the closed form. I have to look for another way to solve this equation, but it seems a bit confusing! thanks anyway

Matt J
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]

2 Comments

Roger Stafford on 20 Sep 2013

It also has a solution at p = 0 and q = 0.

cm on 21 Sep 2013

Yes that's true!

I also added another equation to the first one, so that it may seem easier to find better answers for these equations. Thank you

Walter Roberson

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