How to fit a smooth curve on discrete data.

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Raj Arora
Raj Arora el 19 de Oct. de 2022
Editada: Torsten el 20 de Oct. de 2022
Hello all, I want to produce an equation that can develop a continous smooth curve (does not matter whether it follow any distribution or any plot) which connect the data given below. Using that equation I can interpolate data in between but I want a smoooth curve not a discrete curve. Can anyone please help me with this.
x = [3, 2.5, 2, 1.5, 1, 0.5]
y = [1.8, 1.75, 1.71, 1.55, 0.8, 1.25]
plot(x,y)

Respuesta aceptada

Torsten
Torsten el 19 de Oct. de 2022
Maybe something like this:
x = [3, 2.5, 2, 1.5, 1, 0.5];
y = [1.8, 1.75, 1.71, 1.55, 0.8, 1.25];
plot(x,y)
xx = linspace(x(1),x(end),100);
yy = interp1(x,y,xx,'cubic');
hold on
plot(xx,yy)
hold off
  4 comentarios
Raj Arora
Raj Arora el 20 de Oct. de 2022
Alex can you tell me the procedure how you come up with this equaiton and values of p1 p2 p3 p4?
Torsten
Torsten el 20 de Oct. de 2022
Editada: Torsten el 20 de Oct. de 2022
Alex's idea is to approximate your data by a curve. This curve does not pass exactly through each of your data points, but gives a good approximation over the complete interval with a small number of coefficients.
x = [3, 2.5, 2, 1.5, 1, 0.5];
y = [1.8, 1.75, 1.71, 1.55, 0.8, 1.25];
fun = @(p,x) p(1)./(p(2)+(x-p(3)).^2)+p(4) ;
fun1 = @(p) fun(p,x)-y;
p0 = [-0.1;0.1;0.9;2] ;
sol = lsqnonlin(fun1,p0)
Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
sol = 4×1
-0.1412 0.1199 0.8609 1.8143
hold on
plot(x,y,'o')
xx = x(1):-0.01:x(end);
plot(xx,fun(sol,xx))
hold off

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