LSQ of exponential function without linearizing

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Hi. I have a problem solving the inverse problem of this form: y = a*exp(b*x1)*exp(c*x2)

I need to solve for a,b,c given data: y, x1, x2. I believe this requires an iterative LSQ approach using damping by the Levenberg-Marquardt algorithm. However, I find that the solution is highly dependent on starting point. y is in the range 10^10 while x1 and x2 are in the range [0:1]. I have an expectation that b and c are in the range [-2:2] meaning that "a" is much bigger in the range [10^9:10^10].

DO you guys have any good advices for how to make the inverse method stable? Standardization / normalization?

Best regards, Commat

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Matt J 2014-3-4

编辑：Matt J 2014-3-4

I have an expectation that b and c are in the range [-2:2] meaning that "a" is much bigger in the range [10^9:10^10].

I don't think that follows. The magnitude of a will depend on the magnitudes of x1 and x2 as well as on b and c.

commat 2014-3-4

编辑：commat 2014-3-4

在 MATLAB Online 中打开

Hi Matt. Thanks for your reply.

cond([x1(:), x2(:)]) = 2.044

and

cond([x1(:), x2(:), ones(size(x1(:)))]) = 27.0614

What is this telling me?

I agree with you on the magnitude but getting an approximate solution by linearinzing the expression and then solve with traditional Gauss-Newton yields b=1.68 and c=-1.70.

I believe the problem with my LSQ approach is that the Jacobian is very indifferent in values being that. J(:,1) = small while J(:,2) and J(:,3) are very big. I think that this would disturb how the solution is iteratively directed?

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Matt J 2014-3-4

编辑：Matt J 2014-3-5

在 MATLAB Online 中打开

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I don't know exactly what you mean by "linearizing". If you take the log of both sides of your model equation, you obtain linear equations in b,c, and log(a)

log(y(:))=x1(:)*b +x2(:)*c +log(a)

Your result for cond([x1(:), x2(:), ones(size(x1(:)))]) says you should be able to get a pretty stable solution using mldivide,

params=[x1(:), x2(:), ones(size(x1(:)))]\log(y(:));
   a=exp(params(3));
   b=params(1);
   c=params(2);

These kinds of transformations don't always play well when you have additive measurement noise, but the above should at least be a good initial guess [a0,b0,c0] for an iterative method.

Without actual data to run, I can only guess what might be wrong with the iterative method you're using. However, I would recommend normalizing your data to smaller, more manageable, numbers y=y/10^10. This should have no effect other than to re-express a in different units. You might also try using FMINSPLEAS , which can take advantage of the fact that y is linear w.r.t a.

    flist={@(bc,x) exp(x*bc(:))};
    [bc,a] = fminspleas(flist,[b0,c0],[x1(:),x2(:)],y(:)/1e10);
    a=a*1e10;
    b=bc(1);
    c=bc(2);

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commat 2014-3-6

Hi Matt. Yes, I understand the fact that the cost function can have a local minima but if it actually returns the least misfit to data should the least square then stay in that minima? A

Matt J 2014-3-6

Yes, it should, but I still don't understand what you say you're witnessing. You think it's starting to converge to the global solution, then jumping to another sub-optimal one? What is the evidence of that?

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LSQ of exponential function without linearizing

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