Fitting Curve With an Inverse Which Fits a Polynomial

69 visualizzazioni (ultimi 30 giorni)
Hi. I have 8 data points with x and y values. I would like to input new y values and interpolate x values.
I am able to input new x values and interpolate y values. I can fit the points with a sixth order polynomial for y vs. x which is valid in the range. However, I cannot fit a polynomial for x vs. y
One approach is to solve the polynomial for each y value; however, I have thousands of y values I want to interpolate for, so it would be extremely computationally intensive.
Does anyone know a faster approach? Thanks!

Risposta accettata

Ameer Hamza
Ameer Hamza il 5 Ott 2020
Modificato: Ameer Hamza il 5 Ott 2020
The inverse of a polynomial is not a polynomial, so you cannot simply interpolate the inverse function. Following shows two approaches
1) fzero()
x = linspace(0, 2, 8);
y = 5*x.^6 + 3*x.^5; % y varies from 0 to 416.
pf = polyfit(x, y, 6);
y_pred = @(x) polyval(pf, x);
% find x, when y = 100;
y_val = 100;
x_val = fzero(@(x) y_pred(x)-y_val, rand);
2) Polynomial root finding. This method gives all possible solutions
x = linspace(0, 2, 8);
y = 5*x.^6 + 3*x.^5; % y varies from 0 to 416.
pf = polyfit(x, y, 6);
% find x, when y = 100;
y_val = 100;
pf(end) = pf(end)-y_val;
x_vals = roots(pf);
x_vals = x_vals(imag(x_vals)==0); % if you only want real roots.
  2 Commenti
Ephraim Bryski
Ephraim Bryski il 5 Ott 2020
I used the fzero() approach; it works perfectly and is much faster than solve(). Thank you!
Ameer Hamza
Ameer Hamza il 5 Ott 2020
I am glad to be of help!
Yes, symbolic mathematics is much slower as compared to numerical equivalent.

Accedi per commentare.

Più risposte (0)

Categorie

Scopri di più su Polynomials in Help Center e File Exchange

Prodotti


Release

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by