Info

Cette question est clôturée. Rouvrir pour modifier ou répondre.

Polyfit does not seem to be operating correctly 2011b(7.13.0.564)

1 vue (au cours des 30 derniers jours)
Jack
Jack le 19 Juil 2012
Clôturé : MATLAB Answer Bot le 20 Août 2021
figure(500)
x = (0: 0.1: 2.5)';
y = erf(x);
[p,S,mu] = polyfit(x,y,6);
BestFit = polyval(p,x);
plot(x,y,'o',x,BestFit,'x');
grid on;
p =
0.0017 -0.0092 0.0016 0.0708 -0.1747 0.1822 0.9230
Correct Solution:
p =
0.0084 -0.0983 0.4217 -0.7435 0.1471 1.1064 0.0004
This is the example from the help page for polyfit.

Réponses (1)

Star Strider
Star Strider le 19 Juil 2012
If you request ‘S’ and ‘mu’ from ‘polyfit’, you need to provide them to ‘polyval’ as well.
BestFit = polyval(p,x,S,mu);
should work. To get everything you asked for from ‘polyfit’ and ‘polyval’, you can also get confidence intervals on the fitted values:
[BestFit, delta] = polyval(p,x,S,mu);
  2 commentaires
Jack
Jack le 19 Juil 2012
Consistency between the input and output arguments does produce consistent results from polyval, but asking for mu from polyfit produces incorrect results for p. Why should the results for p depend on the other requested outputs?
Star Strider
Star Strider le 20 Juil 2012
Modifié(e) : Star Strider le 20 Juil 2012
Requesting ‘S’ and ‘mu’ from ‘polyfit’ asks it to center and scale the x-vector before doing the fit so the fit is more accurate in some situations. Without the additional scaling parameters, ‘polyval’ calculates y-values for an uncentered and unscaled x-vector different from those ‘polyfit’ used to calculate the fit. The polynomial fit parameters are correct in this context, and will of course be different from polynomial fit parameters calculated from an uncentered and unscaled x-vector.

Cette question est clôturée.

Produits

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by