Polynominal Fitting Constrained in 1 Coefficient
2 views (last 30 days)
Show older comments
x = [0;7.8438;15.6493;23.4767;31.2822;39.0986;46.9123];
y = [1;0.8990;0.7102;0.5747;0.4717;0.3766;0.2956];
p = polyfit(x,y,5); % coefficient-set p in polynominal function f(x) = p(6)*x^5 + p(5)*x^4 + ... + p(1)
y_hat = polyval(p,x);
Here y_hat(1) is 1.0001, i.e. f(0) = 1.0001
I would like to fit a 5deg polynominal to y,x subject to the constraint that f(0) = 1 exactly.
I suspect its a case for fmincon, any other ideas? (I don't want to use splines)
0 Comments
Accepted Answer
Walter Roberson
on 19 Dec 2012
f(0) = 1 exactly implies that p(1) = 1 exactly. So subtract that 1 from each y value.
y1 = y - 1;
Now, hand-waving time: subtract the p(1) from the polynomial, and factor the x. You get x * (p(6) * x^4 + p(5) * x^3 + p(4) * x^2 + p(3) * x + p(2)). So (wave, wave) divide your y1 by x,
y1divx = y1(2:end) ./ x(2:end); %remove the (0,1) pair
leaving (wave, wave) p(6) * x^4 + p(5) * x^3 + p(4) * x^2 + p(3) * x + p(2) to be fit by a 4th order polynomial:
p1divx = polyfit(x(2:end), y1div, 4);
y1divx_hat = polyval(p1divx, x);
y1_hat = y1divx_hat .* x + 1; %undo division by x and subtraction of 1
More Answers (0)
See Also
Categories
Find more on Get Started with Curve Fitting Toolbox in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!