about polyval and polyfit

y=[22.9000
   23.2000
   23.0000
   24.4000
   26.7000
   25.0000
   25.5000
   26.3000
   26.1000
   26.2000
   27.0000
   26.3000
   28.1000
   26.4000
   27.8000]';
x1=1:15;
x2=11:25;
x3=22:36;
x4=11111:11125;
y1=polyval(polyfit(x1,y,14),x1);
y2=polyval(polyfit(x2,y,14),x2);
y3=polyval(polyfit(x3,y,14),x3);
y4=polyval(polyfit(x4,y,14),x4);

then,why y1,y2,y3,y4 are quite different?

I think,theoretically they are exactly y.

Thanks for any help.

Answer 1

[Jan]

A solution is described in the appearing warning message: center and scale the data. As the message claims, this is explained in "help polyfit".

y = [22.9, 23.2, 23, 24.4, 26.7, 25, 25.5, 26.3, 26.1, ...
     26.2, 27, 26.3, 28.1, 26.4, 27.8]
x1 = 1:15;
p1 = polyfit(x1, y, 14);   % ==> Warning appears.

% Let POLYFIT scale and center the data:
[p1, s1, u1] = polyfit(x1, y, 14);  % ==> no warning

yy = polyval(p1, x1, s1, u1);
max(abs(y-yy))   % >> 9.2122e-012

The resulting difference is 2500*eps. This is in the expected range, because polynomials of order 14 are extremly sensitive to changes in the points at the edges.

[xu]

Thanks for your reply.I know what you mean.But my question is, if I fit y (there are 15 points in it)with a polynomial p(x) of degree 14----then there are 15 undetermined coefficients,and each pair(xi,yi) can generate one equation,so finally we have 15 equations and 15 unknowns.I think the coefficients of p(x) can be determined through solving the 15 equations.Consequently,if I evaluate p(x) at x,the result should be the same with y.Am I right? Thanks.

[John D'Errico]

No, you are not right! You are forgetting that the arithmetic will be done in floating point, using doubles. Raise 15 to the 14th power. Now, raise 11125 to the 14th power. Both numbers are larger than the largest number that can be represented as a double anyway. But even so, numbers this large will completely destroy the ability to do linear algebra (as polyfit must do) on the arrays it will build. Polynomial models of that order are famously, foolishly poor things to build anyway. Doing so without centering and scaling the data is just criminal abuse of data.

Answer 2

[Wayne King]

Hi XU,

Keep in mind that polyfit() is fitting the best polynomial in the least-squares sense. You are trying to fit a polynomial with the same y-values but for vastly different values of x, so why do you expect the fit to be exactly the same?

Further, these are badly-conditioned fits. Reduce the order of your polynomial to 3 (for example) and see what happens for y1 to y3. (y4 is still badly conditioned).

y=[22.9000 23.2000 23.0000 24.4000 26.7000 25.0000 25.5000 26.3000 26.1000 26.2000 27.0000 26.3000 28.1000 26.4000 27.8000]; 
x1=1:15; 
x2=11:25; 
x3=22:36; 
x4=11111:11125; 
y1=polyval(polyfit(x1,y,3),x1); 
y2=polyval(polyfit(x2,y,3),x2);
y3=polyval(polyfit(x3,y,3),x3);

[xu]

Thanks for your reply.I know what you mean.But my question is, if I fit y (there are 15 points in it)with a polynomial p(x) of degree 14----then there are 15 undetermined coefficients,and each pair(xi,yi) can generate one equation,so finally we have 15 equations and 15 unknowns.I think the coefficients of p(x) can be determined through solving the 15 equations.Consequently,if I evaluate p(x) at x,the result should be the same with y.Am I right? Thanks.

[Walter Roberson]

No, you are not right.

x^14 will have a precision at least 14 bits less than x itself will. You get big errors when you create a 14 degree polynomial for a non-trivial range. You would get higher precision if you created the horner version of the polynomial.

What might be *algebraically* true really doesn't matter, because real floating point arithmetic does not obey the rules of algebra. In floating point arithmetic, there are _many_ values x such that (1+x)-1 evaluates to 0. For example, (1 + 1E-100)-1 is *not* going to be 1E-100.

Answer 3

[xu]

polyfit by this way,i.e.[p1, s1, u1] = polyfit(x1, y, 14),is more satisfactory when x is not so regular.But I don't know why.