MATLAB Answers

Fitting Curve With an Inverse Which Fits a Polynomial

90 ビュー (過去 30 日間)
Ephraim Bryski
Ephraim Bryski 2020 年 10 月 5 日
コメント済み: Ameer Hamza 2020 年 10 月 5 日
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!

採用された回答

Ameer Hamza
Ameer Hamza 2020 年 10 月 5 日
編集済み: Ameer Hamza 2020 年 10 月 5 日
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 件のコメント
Ameer Hamza
Ameer Hamza 2020 年 10 月 5 日
I am glad to be of help!
Yes, symbolic mathematics is much slower as compared to numerical equivalent.

サインインしてコメントする。

その他の回答 (0 件)

製品


リリース

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by