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Problem in Curve fitting

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maryam amiri
maryam amiri on 20 May 2021
Edited: Matt J on 20 May 2021
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?

Accepted Answer

Matt J
Matt J on 20 May 2021
Edited: Matt J on 20 May 2021
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-')
  1 Comment
Matt J
Matt J on 20 May 2021
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

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More Answers (2)

Walter Roberson
Walter Roberson on 20 May 2021
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
  6 Comments
Matt J
Matt J on 20 May 2021
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.

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Girijashankar Sahoo
Girijashankar Sahoo on 20 May 2021
check the again, I get your result with same code
  3 Comments
maryam amiri
maryam amiri on 20 May 2021
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.

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