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insert two criteria function to lsqcurvefit

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Mani Ahmadian
Mani Ahmadian 2014-11-26
评论： Star Strider 2014-12-29
Hi, I have some data points as (X,Y) and I want to use lsqcurvefit function to find the best fit by least square method. My guidance function is a non-linear function Gamma2(h) as bellow:
As above equation shows, it's a two criteria function and because the value of a0 is not clearly known, I have to use lsqcurvefit function simultaneously to reach the a0, but I don't know how to do it.
Please help me to solve problem.
Thanks, Mani

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回答（2 个）

Star Strider
Star Strider 2014-11-26
This works with simulated data:
% PARAMETERS: b(1) = C0, b(2) = a0
gamma2 = @(b,h) b(1).*[(1.5*(h/b(2)) - 0.5*(h./b(2)).^3).*(h<=b(2)) + (h>b(2))];
% SIMULATE FUNCTION
b = [10; 5]; % Created Parameters
h = linspace(0,7,25); % ‘h’ (Independent Variable)
gam2data = gamma2(b,h)+0.5*randn(1,length(h)); % Created Data (Dependent Variable)
B0 = [1; 1];
B = lsqcurvefit(gamma2, B0, h, gam2data); % Estimate Parameters
figure(1)
plot(h, gam2data,'bp')
hold on
plot(h, gamma2(B,h), '-r')
hold off
grid
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Mani Ahmadian
Mani Ahmadian 2014-12-29
编辑：Mani Ahmadian 2014-12-29
Hi
I used above code on another data points and plot the result as bellow:
It's not a good fit based on my personal experiences. I think I have to change my view point to solve the problem.
Can I use regression functions to find the curve and use gamma2 as a limit for it?
gamma2 = @(b,h) b(1).*[(1.5*(h/b(2)) - 0.5*(h./b(2)).^3).*(h<=b(2)) + (h>b(2))];
If it is possible, please help me to do.
Thanks so much
Mani
Star Strider
Star Strider 2014-12-29
Your equation does not appear to accurately describe your data. I tried several options, including setting lower bounds on ‘b(1)’=C0=max(gam2data), and could not get a good fit.
I suggest you consider a different model. Your current model does not appear to describe the process that is generating the data you are fitting to it.

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Matt J
Matt J 2014-11-26
编辑：Matt J 2014-11-26
Beware differentiability issues. Since your curve is not differentiable with respect to a0, the least squares cost function might not be either. It may be more trustworthy to use a derivative-free method like fminsearch instead,
gamma2 = @(b,h) b(1).*[(1.5*(h/b(2)) - 0.5*(h./b(2)).^3).*(h<=b(2)) + (h>b(2))];
A0=fminsearch( @(a0) norm(gamma2(a0,X)-Y) , initial_a0 )

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