Non-linear curvefitting in MATLAB
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Hey guys! I'm some given some huge set of data. I am trying to fit a set of data into a model of functional form as described below:
z(x, y) = c0. * x^0 * y^2 + c1. * x^1 * y^1 + c2. * x^2 *y^1
where c0, c1, c2 are the coefficients to be found.
My attempt is to use the nlinfit function to solve it.
So far I have tried:
% i have just added a small portion of my data
a= [ 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011, 0.011];
x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
y = x.* a;
z = [ -.304860225, .170315374, .343019354, .370114906, .373180536, .36719579, .363397853, .363417755, .366962504, .379710865, -.304860225, .170315374, .343019354, .370114906, .373180536, .36719579, .363397853, .363417755, .366962504, .379710865];
model= c0.* (x(:).^0).* (y(:).^2) + c1.* (x(:).^1).* (y(:).^1) + c2.* (x(:).^2).* (y(:).^0)
[c0 c1 c2] = [0.001 0.007 0.788]
C= nlinfit( [x,y], z, 'model', [0.001 0.007 0.788])
% Here x,y are independent variables and z is dependent variable.
How can one set these initial values for the coefficients? I'm not getting how to pass the arguments. I'm getting this error "??? Undefined function or variable 'c0' ". Please help!!!
Thanks in advance, Syeda
Accepted Answer
More Answers (1)
Matt J
on 8 Oct 2013
1 vote
Using a nonlinear solver for a linear fitting problem seems like the wrong way to go. A better option might be
10 Comments
Syeda
on 8 Oct 2013
Matt J
on 8 Oct 2013
I've never used the tool myself. Perhaps you need to columnize z,
p = polyfitn([x,y], z(:), 'model');
Syeda
on 8 Oct 2013
Matt J
on 8 Oct 2013
Maybe this
p = polyfitn([x(:),y(:)], z(:), {'y', 'x*y', 'x^2*y'});
Matt J
on 8 Oct 2013
I can't see why the syntax
p = polyfitn([x(:),y(:)], z(:), 'model')
would work. And it is obviously not working, since it gives you the wrong number of coefficients.
Syeda
on 8 Oct 2013
Matt J
on 8 Oct 2013
Dunno. Instead of 'x^2*y' maybe you should try either 'x*x*y' or 'y*x^2'.
Syeda
on 9 Oct 2013
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