second order data fitting using least squares

15 views (last 30 days)
Sangmin
Sangmin on 17 Jan 2019
Edited: Torsten on 18 Jan 2019
hi
I am trying to fitting the data
I want to fitting the data 1 by least square fitting it to a quadratic function around the position of maximum data2(*)
f(x) = a(x-x0)^2 + b(x-x0) + c
where C is an additive constant C = f(x0) = 1.
I used several method (ex data fitting tool...) but failed
If you konw how to solve, pleast let me know
untitled.jpg
  2 Comments
Torsten
Torsten on 17 Jan 2019
x0 is a fitting parameter or set to a fixed value ?
Sangmin
Sangmin on 18 Jan 2019
Edited: Sangmin on 18 Jan 2019
x0 is one of the data with the largest value.
x = [0.81 0.85 0.91 1.00 1.17 1.33 1.36 1.37 1.39 1.40 1.42]
y = [0.58 0.69 0.81 0.93 1 0.91 0.84 0.80 0.74 0.67 0.59]
x0 is maximum point (1.17 1)
I want to fit the quadratic curve through x0

Sign in to comment.

Answers (2)

Torsten
Torsten on 18 Jan 2019
Edited: Torsten on 18 Jan 2019
x = [0.81 0.85 0.91 1.00 1.17 1.33 1.36 1.37 1.39 1.40 1.42];
y = [0.58 0.69 0.81 0.93 1 0.91 0.84 0.80 0.74 0.67 0.59];
x0 = x(5);
y0 = y(5);
xtrans = x - x0;
ytrans = y - y0;
xtrans = xtrans.';
ytrans = ytrans.';
%mat = [sum(xtrans.^4) sum(xtrans.^3);sum(xtrans.^3) sum(xtrans.^2)];
%rhs = [sum(xtrans.^2.*ytrans); sum(xtrans.*ytrans)];
mat = [xtrans.^2 xtrans];
rhs = ytrans;
sol = mat\rhs;
a = sol(1);
b = sol(2);
fun = @(x)a*(x-x0).^2+b*(x-x0)+y0;
yfit = fun(x);
plot(x,y,x,yfit)

Akira Agata
Akira Agata on 18 Jan 2019
Another possible solution:
x = [0.81 0.85 0.91 1.00 1.17 1.33 1.36 1.37 1.39 1.40 1.42]';
y = [0.58 0.69 0.81 0.93 1 0.91 0.84 0.80 0.74 0.67 0.59]';
x0 = x(5);
y0 = y(5);
modelfun = @(a,x) a(1)*(x - x0).^2 + a(2)*(x - x0) + y0;
beta0 = [-1 1]; % Initial guess
mdl = fitnlm(x,y,modelfun,beta0);
xq = linspace(min(x),max(x))';
figure
scatter(x,y)
hold on
plot(xq,predict(mdl,xq))
fitting.png

Categories

Find more on Polynomials 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!