Problem in Curve fitting
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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?
采纳的回答
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)
times = linspace(x(1),x(end));
plot(x,y,'ko',times,fun(v,times),'b-')
1 个评论
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)),
cond(vander((x-mean(x))/std(x)))
更多回答（2 个）
Walter Roberson
2021-5-20
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 个评论
maryam amiri
2021-5-20
编辑：maryam amiri
2021-5-20
Walter Roberson
2021-5-20
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
2021-5-20
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-')
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
2021-5-20
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.
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.
Girijashankar Sahoo
2021-5-20
0 个投票
check the again, I get your result with same code
3 个评论
maryam amiri
2021-5-20
编辑：maryam amiri
2021-5-20
Girijashankar Sahoo
2021-5-20
v =
-0.0000 0.0000 -0.0019 0.0778 -19.4167 -9.7083
maryam amiri
2021-5-20
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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