Problem in Curve fitting

Hi,

I have a sets of data [x,y] that I want to fit with a function F(v,x) where v contains six free parameters.

x=[70,75,80,83,90,100]; y=[1,1,0.97,0.95,0.9,0];

I found the best fitted curve by cftool for this data set (polynomial degree 5):

but the result is different when I use lsqcurvefit.

v0=[0,0,0,0,0,0];
fun = @(v,x)v(1)*x.^5 + v(2)*x.^4 + v(3)*x.^3 + v(4)*x.^2 + v(5)*x + v(6);
x=[70,75,80,83,90,100];y=[1,1,0.97,0.95,0.9,0];
v=lsqcurvefit(fun,v0,x,y);
times = linspace(x(1),x(end));
plot(x,y,'ko',times,fun(v,times),'b-')

this is the result:

It seems lsqurvefit did not fitted the curve to the points.

any idea that why it does not work for me?

Answer 1

[Matt J]

Although polyfit is the better tool here, both polyfit and lsqcurvefit will be challenged by the scaling of your xdata, which is making the problem highly ill-conditioned. Rescaling helps considerably, as shown below,

v0=[0,0,0,0,0,0];
fun = @(v,x)v(1)*x.^5 + v(2)*x.^4 + v(3)*x.^3 + v(4)*x.^2 + v(5)*x + v(6);
x=[70,75,80,83,90,100];y=[1,1,0.97,0.95,0.9,0];
x=(x-mean(x))/std(x);

[v,fval,~,exitflag]=lsqcurvefit(fun,v0,x,y)

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

v = 1×6

-0.0381 -0.0852 -0.0060 0.0310 -0.0643 0.9500

fval = 5.8273e-19

exitflag = 1

times = linspace(x(1),x(end));
plot(x,y,'ko',times,fun(v,times),'b-')

[Matt J]

Another way to see the need for scaling is to look its effect on the condition number of the Vandermonde matrix,

x=[70,75,80,83,90,100];

cond(vander(x)),

ans = 2.2962e+15

cond(vander((x-mean(x))/std(x)))

ans = 230.1084

Answer 2

[Walter Roberson]

fun = @(v,x)v(1)*x.^5 + v(2)*x.^4 + v(3)*x.^3 + v(4)*x.^2 + v(5)*2 + v(6);
                                                            ^^^^^^

Should be

v(5)*x 

[maryam amiri]

that was a mistake. I fixed it. still the same result.

[Walter Roberson]

I see what you mean.

In R2021a (at least), lsqcurvefit() is quite sensitive to initial conditions.

I suggest using polyfit(), as that produces the correct answer (to within round-off)

[maryam amiri]

Thanks. polyfit() worked.

x=[70,75,80,83,90,100];y=[1,1,0.97,0.95,0.9,0];
v=polyfit(x,y,5);
times = linspace(x(1),x(end));
plot(x,y,'ko',times,fun(v,times),'b-')

[Matt J]

In R2021a (at least), lsqcurvefit() is quite sensitive to initial conditions.

Possibly, but note that the sensitivity can't have anything to do with local minima. It is a convex problem and, in theory, lsqcurvefit should converge to the solution from any initial point.

[Walter Roberson]

However, it stops when it thinks the residue is good enough, or if it gets too very small step sizes.

It is a convex problem

Your v = -0.0381 -0.0852 -0.0060 0.0310 -0.0643 0.9500 has two sign changes, so the function itself is not convex.

[Matt J]

However, it stops when it thinks the residue is good enough, or if it gets too very small step sizes.

Yes, the ill-conditioning of the problem does cause one of these lsqcurvefit stopping criteria to be triggered prematurely, and where it stops will indeed depend on the initial point.

Your v = -0.0381 -0.0852 -0.0060 0.0310 -0.0643 0.9500 has two sign changes, so the function itself is not convex.

Yes, the polynomial being fitted is surely not convex as a function of x as we can also see from the plots. However, the least squares objective is convex as a function of v, which is why, in theory, lsqcurvefit should be globally convergent for this problem.

Answer 3

[Girijashankar Sahoo]

check the again, I get your result with same code

[maryam amiri]

There was a mistake in code. I fixed it. But the result still the same.

could you send the result for v?

[Girijashankar Sahoo]

v =

-0.0000 0.0000 -0.0019 0.0778 -19.4167 -9.7083

[maryam amiri]

v=[ -0.0000 0.0000 -0.0019 0.0778 -19.4167 -9.7083];

x=[75:100];

y = v(1)*x.^5 + v(2)*x.^4 + v(3)*x.^3 + v(4)*x.^2 + v(5)*x + v(6);

z=plot(x,y,'b');

still has problem.