How can I fit data to a piecewise function, where the breakpoint of the function is also a parameter to be optimised?
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Rahul
el 29 de Sept. de 2025 a las 20:45
Comentada: Rahul
hace alrededor de 4 horas
I have data with x and y values. This data should conform to a function: an assymmetric parabola. Here, the parameters that define the shape of the parabola should be different on either side of the maximum point of the parabola i.e. the breakpoint is where the maximum value of y occurs.
I was hoping to use 'fit' and to define an anonymous function for my data. But I'm not able to work out how to define an anonymous, piecewise function, especially where the breakpoint is one of the parameters to be determined by the fitting procedure, as it is not immediately clear from the data itself where the maximum value of y should occur.
Any help would be appreciated.
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Matt J
el 30 de Sept. de 2025 a las 4:09
Editada: Matt J
el 30 de Sept. de 2025 a las 11:20
Once you've chosen the coefficients of the first parabola [a1,b1,c1], the breakpoint is determined from,
d=-b1/(2*a1)
Only the leading coefficient of the second parabola is a free parameter:
F=@(x) asymParabola(-2,1,0,-0.6,x);
fplot(F,[-10,10]);axis padded %example plot
ft = fittype(@(a1,b1,c1,a2, x) asymParabola(a1,b1,c1,a2, x) )
function y=asymParabola(a1,b1,c1,a2, x)
d=-b1/(2*a1);
b2=-d*2*a2;
c2=polyval([a1,b1,c1],d)-polyval([a2,b2,0],d);
left=(x<=d);
y=x;
y(left)=polyval([a1,b1,c1],x(left));
y(~left)=polyval([a2,b2,c2],x(~left));
end
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Torsten
hace 3 minutos
Editada: Torsten
hace 2 minutos
If the given x,y values have no noise/errors, then the maximum y-value and the maximum of the parabola are one and the same.
Why ? Both parabola can intersect below their respective maxima, and nonetheless the point of intersection can be the maximum y-value of the piecewise function.
But it seems you interpreted the question correctly.
Más respuestas (3)
Walter Roberson
el 29 de Sept. de 2025 a las 21:20
(a1*x.^2 + b1*x + c1) .* (x <= d) + (a2*x.^2 + b2*x + c2) .* (x > d)
Note that for this to work, the coefficients must be constrained to be finite
2 comentarios
Paul
el 30 de Sept. de 2025 a las 0:37
Sounds like both sides of the function should have the same value at x = d, at least that's how interpret the question. If so, then I think the function would look something like
(a1*(x-d).^2 + b1*(x-d) + c) .* (x <= d) + (a2*(x-d).^2 + b2*(x-d) + c) .* (x > d)
Catalytic
el 30 de Sept. de 2025 a las 14:13
Editada: Catalytic
el 30 de Sept. de 2025 a las 14:18
You can also parametrize the model function directly in terms of the break point coordinates (xbreak, ybreak) and two curvature parameters -
F= @(a1,a2,xbreak,ybreak, x) modelFun(a1,a2,xbreak,ybreak, x);
xbreak=3; ybreak=5;
fplot( @(x) F(-2,-0.6,xbreak,ybreak,x), [1,5]);
xline(xbreak,'--')
fType = fittype(F);
function y=modelFun(a1,a2,xbreak,ybreak, x)
X=x-xbreak;
LHS=(X<=0);
RHS=~LHS;
y=X.^2;
y(LHS)=a1.*y(LHS) + ybreak;
y(RHS)=a2.*y(RHS) + ybreak;
end
Matt J
hace alrededor de 11 horas
Editada: Matt J
hace alrededor de 11 horas
Why ? Both parabola can intersect below their respective maxima, and nonetheless the point of intersection can be the maximum y-value of the piecewise function.
F=@(x) asymParabola(-2,1,0,-6,5,-20 ,x);
fplot(F,[-10,-1]);axis padded %example plot
function y=asymParabola(a1,b1,c1, a2, s, rightSlope, x)
%Requirements: a1<0, a2<0, s>=0, m<=0
d=-b1/(2*a1)-s;
c2=polyval([a1,b1,c1],d);
left=(x<=d);
right=~left;
xright=x(right);
y=x;
y(left)=polyval([a1,b1,c1],x(left));
y(right)=a2*(xright-d).^2 + rightSlope*(xright-d) +c2;
end
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