how to find the two constants that suits the best non linear fitting

Dear All,

I have a series of three different parameters (y, x & t, please see the excel file attached)), that have to fit the following model :

y= a*sqrt(x)*exp(b*t).

My question is the following, how do I find the best values for the a and b, by the best I mean that corresponds to the best fitting for the points. Is there a function in Matlab that does this automatically like for example the polyfit that gives you a, b and c (for a second degree polynomial, a*x^2 +b*x+c), if not then please can someone help me ?

Thanking you in advance,

Le Fou

Answer 1

[Star Strider]

Fitting two variables with either lsqcurvefit or nlinfit with two independent variables requires that you combine the independent variables into one variable and then address them separately in your objective function.

This provides a decent fit:

D = matfile('matlab_1.mat');
x = D.x;
t = D.t;
y = D.y;

xt = [x t];

fxt = @(b,xt) b(1).*sqrt(xt(:,1)).*exp(b(2).*xt(:,2));

x0 = [0.01; 0.01];

B = lsqcurvefit(fxt, x0, xt, y);

ye = fxt(B,xt);

figure(1)
plot3(x, t, y, '.b')
hold on
plot3(x, t, ye, '.r')
hold off
grid on
xlabel('X')
ylabel('T')
zlabel('Y')
legend('Data', 'Regression Fit', 'Location', 'NE')

It is necessary to combine x and t into one matrix, xt here. The function you are fitting (I called it ‘fxt’) then uses x=xt(:,1) and t=xt(:,2). After that, everything is relatively routine. I got a reasonably good fit with the ‘x0’ values I chose, producing a=0.53 and b=0.0013 with respect to the variables in your original equation.

[LE FOU]

Hi Star Strider,

Is there a possible solution with the following matlab function fitnlm ?

I also have a question, why you gave x0 its value x0 = [0.01; 0.01] ?

Thanking you in advance,

[Star Strider]

My pleasure!

You could use fitnlm to solve it. The code for that is:

D = matfile('matlab_1.mat');
x = D.x;
t = D.t;
y = D.y;
xt = [x t];
fxt = @(b,xt) b(1).*sqrt(xt(:,1)).*exp(b(2).*xt(:,2));
x0 = [0.01; 0.01];
mdl1 = fitnlm(xt, y, fxt, x0)

producing:

mdl1 = 

Nonlinear regression model:
    y ~ b1*sqrt(x1)*exp(b2*x2)

Estimated Coefficients:
          Estimate        SE        tStat       pValue  
          ________    __________    ______    __________

      b1     0.52962      0.024469    21.645    3.5645e-95
      b2    0.001324    2.8403e-05    46.615             0

Number of observations: 2440, Error degrees of freedom: 2438
Root Mean Squared Error: 6.44
R-Squared: 0.697,  Adjusted R-Squared 0.697
F-statistic vs. zero model: 4.47e+03, p-value = 0

The choice of the initial parameter estimates is sometimes a matter of trial and error if, as in this situation, I have no idea what your data are or what process you are modeling. In this problem, those initial estimates provided the best final fit of the few I tried. They depend on the data and the function you are fitting.

[LE FOU]

ok thanks a lot it is solved

[Star Strider]

My pleasure!

Answer 2

[Stephen23]

Try lsqnonline or lsqcurvefit .

You should have a look at the examples too.

[LE FOU]

I am not a professional user, can you please give me more help, all I need is to find the best a & b

[LE FOU]

I need to use the following model because it is the one that fits best my data, but what I don't know is how to find the best values of a and b, that will allow me to get the best fit, please see the attachment, I already have the values of y, x and t:

y= a*sqrt(x)*exp(b*t)

Thanks

[Stephen23]

Please read the examples that are in the link I gave. MATLAB documentation gives complete, working examples, which you can copy and try yourself.

If the link did not work, have a look at this example:

http://www.mathworks.com/help/optim/ug/nonlinear-curve-fitting-with-lsqcurvefit.html

Answer 3

[Matt J]

Your model is loglinear

               log(y)=  A + .5*log(x) + b*t

where A=log(a). It might be enough to solve these linear equations, if your errors aren't too large,

     [logX,T]=ndgrid(log(x)/2,t);

      rhs=log(y(:))  - logX(:);
      lhs=reshape( [ones(size(T)),  T] ,  [],2);

      p=lhs\rhs;
      A=p(1);
      a=exp(A);
      b=p(2);

The above assumes that your y-data is indexed y(x,t).

If nothing else, it should provide a systematic initial guess for the solvers the others have recommended. You could also use fminspleas ( Download ) which can take advantage of the fact that y is linear in a.

    fun=@(b,xd) reshape(  sqrt(x(:))*exp(b*t(:).')  ,[],1);

    [b,a]=fminspleas({fun},p(2),[],y(:));

[LE FOU]

Hi Matt J

my data is not indexed y(x,t), so the solution above does not work.

what about the second one, when I am trying to apply it it is telling me that p is not defined, please check the error below

Undefined function 'p' for input arguments of type 'double'.

thanks,

[Matt J]

my data is not indexed y(x,t),

So, it is indexed y(t,x)? If so, it's just a 1-line modification,

   [T,logX]=ndgrid(t,log(x)/2);

what about the second one, when I am trying to apply it it is telling me that p is not defined

The "second one" is not separate from the first. p was computed using the loglinear model in the first code segment I gave you. If you already have a good intial guess for b, though, you can give that instead. Additionally, if y is indexed y(t,x), then you must modify fun,

 fun=@(b,xd) reshape(  exp(b*t(:))*sqrt(x(:).')  ,[],1);