A simple question about solving the polynomial
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
This might be a very simple question to the programming gurus. any kind of help will be largely appreciated!!
x=[100 330 380 500 550];
y=[100 450 500 600 700];
p=polyfit(x,y,1); % first degree polynomial curve fitting
yFit=polyval(p,10); % return the value of y (named yFit) evaluated at 10 for example
My question is how can I returen the value of x evaluated at a value of y?
0 commentaires
Réponses (2)
Geoff
le 22 Mar 2012
Well, there's two things here. Did you even want a polynomial in terms of x? If not, why not initially solve in terms of y?
Otherwise, if you want to express that single answer in terms of either y or x, remember that your vector 'p' contains the solved coefficients for a line:
y = Ax + B
So you tell us how you would find the value of x.
3 commentaires
Geoff
le 23 Mar 2012
Well, I reminded you of the formula for a line. By calling polyfit() you found the values for A and B. That just leaves you to rearrange the equation for x.
John D'Errico
le 23 Mar 2012
You can't mean algebra? We have really big computers for that kind of stuff!
Walter Roberson
le 23 Mar 2012
X_At_Particular_Y = interp1(y, x, The_Particular_Y, 'linear');
and conversely
Y_At_Particular_X = interp1(x, y, The_Particular_X, 'linear');
Warning: this formulation will only work if both x and y are strictly increasing or strictly decreasing.
If one of your variables is not strictly monotonic, then there are multiple locations for projecting that variable on to the other axis.
0 commentaires
Voir également
Catégories
En savoir plus sur Polynomials dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!