Non-linear curvefitting in MATLAB

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Syeda
Syeda el 8 de Oct. de 2013
Comentada: Syeda el 9 de Oct. de 2013
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

Respuesta aceptada

Greg Heath
Greg Heath el 9 de Oct. de 2013
The solution is trivial because you have a linear system of equations for the 3 coefficients
A*c = b;
c = A\b
Hope this helps
Thank you for formally accepting my answer
Greg
  3 comentarios
Matt J
Matt J el 9 de Oct. de 2013
Editada: Matt J el 9 de Oct. de 2013
It's really not ideal to fit polynomials bluntly using backslash. That's why MATLAB offers the more robust POLYFIT and why the File Exchange offers a variety of polynomial fitters.
Syeda
Syeda el 9 de Oct. de 2013
yes!! My results are not accurate by using this method.

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Más respuestas (1)

Matt J
Matt J el 8 de Oct. de 2013
Using a nonlinear solver for a linear fitting problem seems like the wrong way to go. A better option might be
  10 comentarios
Matt J
Matt J el 8 de Oct. de 2013
Dunno. Instead of 'x^2*y' maybe you should try either 'x*x*y' or 'y*x^2'.
Syeda
Syeda el 9 de Oct. de 2013
No, it still gives error.
??? Undefined function or method 'polyfitn' for input arguments of type 'cell'

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