is it possible to differentiate max function?

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Frank Oosterveld
Frank Oosterveld el 3 de Dic. de 2018
Editada: Walter Roberson el 17 de Feb. de 2023
Hi,
Is it possible with the basic matlab functions to numerically differentiate a max function? This is what I tried, but did not work: 'Error using symengine, Input arguments must be convertible to floating-point numbers.'
r_par = 100;
R = 3;
P = @(x1,x2) r_par/2*(max(0,x1^2 + x2^2 - R^2))^2
gP = gradient(P, [x1, x2]);
gF = matlabFunction(gP);
x = [1 2];
x1 = x(1);
x2 = x(2);
ans = P(x1,x2)

Respuestas (1)

Walter Roberson
Walter Roberson el 3 de Dic. de 2018
Editada: Walter Roberson el 3 de Dic. de 2018
max() is not defined for symbolic . recode in terms of piecewise
Max = @(aa,bb) piecewise(aa>bb,aa,bb)
Note that matlabFunction can only handle piecewise if you tell it to write to aa file and even then the result will not be vectorized .
  6 comentarios
Alec Jacobson
Alec Jacobson el 17 de Feb. de 2023
Editada: Alec Jacobson el 17 de Feb. de 2023
yeah, I wouldn't mind needing to specify the ComparisonMethod or if it determined that based on assumptions to deal with real vs. complex input (e.g., through an error for complex).
Meanwhile, the frustrating thing I'm finding about piecewise is that it gets confused with float inputs:
a = 1;
b = 2;
piecewise(a<b,a,b)
Maybe it's only defined for symbolic? That makes it difficult to prototype mixing numeric and symbolic code (which I see as a huge advantage of matlab vs. maple etc.)
Walter Roberson
Walter Roberson el 17 de Feb. de 2023
Editada: Walter Roberson el 17 de Feb. de 2023
In the time since I wrote the original answer, MATLAB added a symbolic max() and min() as place-holders that wait until the question is decidable.
syms aa bb
max(aa, bb)
ans = 
So now you can
syms x1 x2
r_par = 100;
R = 3;
P = @(x1,x2) r_par/2*(max(0,x1^2 + x2^2 - R^2))^2
P = function_handle with value:
@(x1,x2)r_par/2*(max(0,x1^2+x2^2-R^2))^2
But you cannot call gradient() because that looks for numeric inputs, so you call diff()
gP = [diff(P, x1), diff(P, x2)]
gP = 
and then you can see that matlabFunction gets confused and thinks that x1 and x2 are the stride counts in numeric diff calls...
matlabFunction(gP)
ans = function_handle with value:
@(x1,x2)[diff(max([0.0,x1.^2+x2.^2-9.0],[],2,'omitnan'),x1).*max([0.0,x1.^2+x2.^2-9.0],[],2,'omitnan').*1.0e+2,diff(max([0.0,x1.^2+x2.^2-9.0],[],2,'omitnan'),x2).*max([0.0,x1.^2+x2.^2-9.0],[],2,'omitnan').*1.0e+2]

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