Fitting implicit function with data

Dear all

I want to fit an implicit function with data I have. The function is in terms of X and Y(2 variables) and has 8 constants (x(1)..x(8)) to be found.

 f=x(1).*(abs(((x(2).*(X-x(3)))+x(4).*(Y-x(5))))).^(x(6))+x(7).*(abs((X-x(3)-(Y-x(5))))).^(x(6))-x(8).^(x(6));

and my data are:

 X = [0 0.41 1.2 2.01 3.5 5 7.1 10.1 12.06 14.2 17.1 17.4 17.9 17.1 14.2 10.1 7.1 5 3.5 2.01 0 -0.8 -3.5 -5.4 -9.5 -12.1 -15.5 -18.5 -22.3 -25.5 -26.7 -27.3 -26.7 -22.5 -18.8 -15.5 -12.1 -9.5 -5.4 -1.7 0];
 Y = [17.9 17.7 17.5 18.3 17.87 17.48 16.74 16.7 15.8 15.3 9.02 8.6 0 -4.6 -13.8 -24.1 -26.9 -26.7 -26.5 -26.2 -27.3 -27.9 -26.9 -25.2 -25.1 -22.8 -20.2 -20.1 -10.6 -4.6 -0.9 0 4 8 10.5 11.3 13.8 16.2 17 17.6 17.9];

Can anyone help me to find the constants?

[Mohsen]

I did as follows but needs to be more accurate, How can I pass it through specific points?

function obj= fit_simp(x,X,Y)

X = [0 0.41 1.2 2.01 3.5 5 7.1 10.1 12.06 14.2 17.4 17.9 17.1 14.2 10.1 7.1 5 3.5 2.01 0 -0.8 -3.5 -5.4 -9.5 -12.1 -15.5 -18.5 -22.3 -25.5 -26.7 -27.3 -26.7 -22.5 -18.8 -15.5 -12.1 -9.5 -5.4 -1.7 0];

Y = [17.9 17.7 17.5 18.3 17.87 17.48 16.74 16 15 13.5 8.6 0 -4.6 -13.8 -24.1 -26.9 -26.7 -26.5 -26.2 -27.3 -27.5 -26.7 -26 -25.1 -22.8 -20.2 -17 -10.6 -4.6 -0.9 0 4 8 10.5 11.3 13.8 16.2 17 17.6 17.9];

obj = @(x) x(1).*(abs(((x(2).*(X-x(3)))+x(4).*(Y-x(5))))).^(x(6))+x(7).*(abs((X-x(3)-(Y-x(5))))).^(x(6))-x(8).^(x(6));

x0=[0.001    0.1880  -1.44    0.108   -4    9.984    2.9064e-10    2.1175];

options = optimset('Display','iter','TolFun',1e-8)

[x,resnorm,residual,exitflag,output]=lsqnonlin(obj,x0,[-1],[13],options)

obj_new= @(X,Y) x(1).*(abs(((x(2).*(X-x(3)))+x(4).*(Y-x(5))))).^(x(6))+x(7).*(abs((X-x(3)-(Y-x(5))))).^(x(6))-x(8).^(x(6))

scatter(X,Y)

hold on

ezplot(obj_new,[-30,30,-30,30])

Answer 1

[John D'Errico]

Sorry, but it seems a foolish goal to fit that curve with your function. There are far easier ways to accomplish a curve fit.

Some basic rules of curve fitting:

- If you are using a complex model that was suggested to you by a friend/colleague/boss/etc because it was able to fit their data, or they know of someone who used that model, then you are probably using the wrong model. Consider a spline instead.

- If you have no physical reasons for having chosen the model you are using, EXCEPT that it MIGHT fit IF you throw in enough terms, then you are probably using the wrong model. Consider a spline instead.

- If you are using a complicated model where the parameters have no meaning to you in context, so the resulting nonlinear fit does not converge (possibly) because you cannot guess good starting values, then you are using the wrong model. Consider a spline instead.

Were it my choice, I would just convert to polar coordinates, then fit r(theta) using a tool like a least squares spline model, designed to be a periodic function.

theta = atan2(Y,X);
[thetas,tags] = sort(theta);
r = sqrt(X.^2 + Y.^2);
rs = r(tags);

For example, using my SLM toolbox,I might try this:

slm = slmengine(thetas,rs,'plot','on','endconditions','periodic','knots',15);

thetahat = linspace(-pi,pi,200);
rhat = slmeval(thetahat,slm);
xhat = rhat.*cos(thetahat);
yhat = rhat.*sin(thetahat);
plot(X,Y,'-o',xhat,yhat,'r:')
grid on

[Mohsen]

Thanks, but I need to fit the data with the function I mentioned. It has a physical meaning.

[John D'Errico]

So then fit the model. Wanting it to be more accurate is a waste of time though. The model will be as accurate as it is. If it is not sufficiently accurate, then the model is not a good approximation to the system that produced that data, or you need to get better data. In some cases, you may need to get better starting values.

Forcing the model to pass through specific points (which will take a fair amount of effort here) will just make the model fit worse overall, not better.

[Tien Tran]

Hi John

How to get constants and the function of spline?