How to i find a range of parameters that admit positive solutions ?

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msh 2013-7-6

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采纳的回答： Matt J

在 MATLAB Online 中打开

Hi,

I have this non linear equation.

x+6-(beta/(1+beta))*(1-alpha)*x^(alpha)+(exp(sigma^2*(1/2 - x/((beta/(1+beta))*(1-alpha)*x^(alpha)))))*((beta/(1+beta))*alpha*x^(2*alpha-1)-(alpha)*x^(alpha))=0

my alpha's, beta's and sigma, are parameters. I am wondering how someone can find in MATLAB the range or value/values of those parameters that admit positive solution/s for my x.

Thanks

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msh 2013-7-6

by "fit" I mean to get the unknown x necessary positive, by "fitting" or finding alphas and betas that can produce such solution. X is unknown but WE KNOW that MUST be positive ALL THE TIME, and this solution depends on those parameters. This is the basic point.

So, if

x=sqrt(a) and we want x>=0, then any a>=0 is fine, in a lot of cases is not that clear or obvious.

Matt J 2013-7-6

编辑：Matt J 2013-7-6

by "fit" I mean to get the unknown x necessary positive, by "fitting" or finding alphas and betas that can produce such solution.

So, if I present you with one particular triple x>=0, alpha, beta, that satisfies your equation, you will be done? Mission accomplished?

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Matt J 2013-7-6

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编辑：Matt J 2013-7-6

在 MATLAB Online 中打开

Here is a combination of parameters that seems to satisfy your equation, with x>=0,

alpha =
     -0.0786
beta =
      1.8827
x =
      0.2446
sigma =
      4.2678

I obtained it by minimizing abs(LHS) over all 4 parameters via the code below.

    LHS=@(alpha,beta,x,sigma) x+6-(beta/(1+beta))*(1-alpha)*x^(alpha)+(exp(sigma^2*(1/2 - x/((beta/(1+beta))*(1-alpha)*x^(alpha)))))*((beta/(1+beta))*alpha*x^(2*alpha-1)-(alpha)*x^(alpha));
      fun=@(p) abs(LHS(p(1),p(2),abs(p(3)),p(4)));
      options=optimset('TolFun',1e-6);
      [p,fval]=fminsearch(fun,[1,1,1,3],options);
      alpha=p(1),
      beta=p(2),
      x=abs(p(3)),
      sigma=abs(p(4)),

So does this complete your mission?

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Matt J 2013-7-6

编辑：Matt J 2013-7-6

And just for future reference, are those solutions the only ones that this code gives?

No, you have 1 equation in 4 unknowns, so there will inevitably be an infinite space of solutions. You can get different solutions by feeding different starting guesses to FMINSEARCH.

I am asking this because in general, we might not wish to have, let's beta negative, or alpha above 10 etc etc.. so ruling out some solutions.

If you have the Optimization Toolbox, you can use LSQNONLIN instead of FMINSEARCH. LSQNONLIN let's you impose whatever upper/lower bounds you want, and handles them more gracefully.

msh 2013-7-6