polynomial doesnt fit with original data

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Habib
Habib on 6 May 2013
Hi, I give Matlab 2009a vectors X and Y. I plot(X,Y) and I want to fit a polynomial on it. on the plot, I can see a good fit with 6th order polynomial. Then I save the coefficients of polynomial (say vector A) and again plot(polyval(A,X). But they dont match with the original curve. They are not even close!
these are X and Y:
1000 0.003719
1100 0.004066
1200 0.004507
1300 0.005044
1400 0.005683
1500 0.006432
1600 0.0073
1700 0.0083
1800 0.009449
1900 0.01077
2000 0.01227
2100 0.014
2200 0.01597
2300 0.01824
2400 0.02087
2500 0.02391
2600 0.02749
2700 0.03175
2800 0.03693
2900 0.04343
3000 0.05208
3100 0.06516
3129.9 0.07103
and I got A from basic fitting tools as below:
Coefficients:
p1 = 0.00097657
p2 = 0.001743
p3 = -0.00084778
p4 = -0.00046857
p5 = 0.0064581
p6 = 0.01318
p7 = 0.013886
Norm of residuals =
0.0011309
I must mention that when I was fitting I got this warning: "polynomial is badly conditioned" but the fit looks good.
what am I doing wrong. I could see this problem in Excel too.
  • I need to have a function (Y=f(X)) for my data. If I shouldn't use polynomial fitting, do you have other suggestions?

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Accepted Answer

Image Analyst
Image Analyst on 6 May 2013
I don't know what you did, but when I did it, it looks great:
xy = [1000 0.003719
1100 0.004066
1200 0.004507
1300 0.005044
1400 0.005683
1500 0.006432
1600 0.0073
1700 0.0083
1800 0.009449
1900 0.01077
2000 0.01227
2100 0.014
2200 0.01597
2300 0.01824
2400 0.02087
2500 0.02391
2600 0.02749
2700 0.03175
2800 0.03693
2900 0.04343
3000 0.05208
3100 0.06516
3129.9 0.07103]
% Extract x and y.
x = xy(:, 1);
y = xy(:, 2);
% Get the fit coefficients.
[coeffs, S, mu] = polyfit(x, y, 6);
% Get the fitted values.
fittedY = polyval(coeffs, x, S, mu);
% Plot everything.
plot(x, y, 'r*', 'LineWidth', 2);
hold on;
plot(x, fittedY, 'bd-', 'LineWidth', 2, 'MarkerSize', 15);
grid on;
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
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Habib
Habib on 7 May 2013
For polynomials analytically d/dx= n*x^(n-1). but when some correcting factors (mu and S) are attached to the polynomial how can we take them into account for taking derivative? Is there a tutorial for that? I dont know. If there is a method please let me know.
Habib
Habib on 7 May 2013
I just taught of another solution for taking derivative. I can take numerical derivative. suppose that i want to take the deviative at x0:
[coeffs, S, mu] = polyfit(x, y, 6);
dx=1 % a small value compared to x values
A= polyval(coeffs, x0+dx, S, mu);
B= polyval(coeffs, x0-dx, S, mu);
der=(A-B)/(2*dx) % derivative with enough precision at x0
This way I can take the derivative of my data set at any x0, say 2231.5. do you agree?. If I had used the original data set, the derivative of all points between 2200 and 2300 would be the same and there would be a sudden break at provided x points in the original data set.
I will examine it tomorrow.

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More Answers (1)

John D'Errico
John D'Errico on 6 May 2013
Sigh. This happens over and over again. Fit a linear model, then a quadratic. It does better. So a cubic model must be better yet. Why not fit a 100th order model? It should do really well.
The problem is with high order polynomial models is they DON'T do well. They have miserable problems with conditioning, especially when your data is like this. Don't believe me? Don't understand why?
What is 3000^6?
3000^6
ans =
7.29e+20
Yeah. Its huge. The problem is a double precision number in MATLAB (or almost ANY environment) only has about 16 digits of precision. So when you do arithmetic on numbers with such a wide dynamic range, you end up with ill-conditioned systems of equations. (Why do you think you got that warning message?)
You CAN improve things a bit by transforming the problem, centering and scaling as suggested. See the difference between these two models with your data:
polyfit(x,y,6)
Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and
scaling as described in HELP POLYFIT.
> In polyfit at 76
ans =
1.0486e-20 -1.1933e-16 5.5538e-13 -1.3466e-09 1.7944e-06 -0.001241 0.35052
polyfit((x - 2000)/1000,y,6)
ans =
0.010486 0.0065015 -0.0087511 0.00106 0.01436 0.016792 0.012122
The latter problem is far better conditioned. If you want evidence...
cond(bsxfun(@power,x,0:6))
ans =
8.3329e+23
cond(bsxfun(@power,(x-2000)/1000,0:6))
ans =
91.758
Even with that, the model is simply not that great. It is not even monotone o=ver the data.
P = polyfit((x - 2000)/1000,y,6);
roots(polyder(P))*1000 + 2000
ans =
2681.7 + 533.47i
2681.7 - 533.47i
1047.5 + 0i
1536.3 + 398.64i
1536.3 - 398.64i
So the polynomial has a minimum at 1047.5, inside the data range.
Far better is to use a tool designed to deal with curve fitting of data like this, and to do so intelligently. I'd suggest my SLM tools.
http://www.mathworks.com/matlabcentral/fileexchange/24443-slm-shape-language-modeling
  1 Comment
Habib
Habib on 7 May 2013
Thanks John. your guide was very helpful. I have to go deep through concerns of curve fitting. Otherwise I will some day end up with big problem in computing. I downloaded your tool.

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