Fitting Curve With an Inverse Which Fits a Polynomial

조회 수: 45 (최근 30일)
Ephraim Bryski
Ephraim Bryski 2020년 10월 5일
댓글: Ameer Hamza 2020년 10월 5일
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!

채택된 답변

Ameer Hamza
Ameer Hamza 2020년 10월 5일
편집: Ameer Hamza 2020년 10월 5일
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.
  댓글 수: 2
Ephraim Bryski
Ephraim Bryski 2020년 10월 5일
I used the fzero() approach; it works perfectly and is much faster than solve(). Thank you!
Ameer Hamza
Ameer Hamza 2020년 10월 5일
I am glad to be of help!
Yes, symbolic mathematics is much slower as compared to numerical equivalent.

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Polynomials에 대해 자세히 알아보기

제품


릴리스

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by