How to evaluate a symbolic expression having `max` and `diff`?

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Rounak Saha Niloy
Rounak Saha Niloy 2024 年 1 月 1 日
コメント済み: Walter Roberson 2024 年 1 月 4 日
I have calculated the jacobian of two functions where variables are x1, x2, x3.
The jacobian is as follows-
JacobianF =
[ diff(max([0, (7*sin(4*pi*x1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x1) + 96*pi*cos(6*pi*x1)*(x3 + sin(6*pi*x1)) + 160*3^(1/2)*pi^2*cos(6*pi*x1)*sin((3^(1/2)*pi*(20*x3 + 20*sin(6*pi*x1)))/3) + 1, 0, 16*x3 + 16*sin(6*pi*x1) + diff(max([0, (7*sin(4*pi*x1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x3) + (80*3^(1/2)*pi*sin((3^(1/2)*pi*(20*x3 + 20*sin(6*pi*x1)))/3))/3]
[diff(max([0, (7*sin(4*pi*x1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x1) - 96*pi*cos((2*pi)/3 + 6*pi*x1)*(x2 - sin((2*pi)/3 + 6*pi*x1)) - 240*2^(1/2)*pi^2*cos((2*pi)/3 + 6*pi*x1)*sin((2^(1/2)*pi*(20*x2 - 20*sin((2*pi)/3 + 6*pi*x1)))/2) - 1, 16*x2 - 16*sin((2*pi)/3 + 6*pi*x1) + diff(max([0, (7*sin(4*pi*x1))/10], [], 2, 'omitnan', ~in(x1, 'real')), x2) + 40*2^(1/2)*pi*sin((2^(1/2)*pi*(20*x2 - 20*sin((2*pi)/3 + 6*pi*x1)))/2), 0]
Now, I need to evaluate this JacobianF at
X = [0.2703 0.6193 0.9370];
where X(1) is x1 and so on.
To evaluate this JacobianF, I have used the following code-
Var_List = sym('x', [1, 3]);
df=double(subs(JacobianF, Var_List, X));
However, I get the following error. What is the cause of this error? How to resolve it and calculate the JacobianF at the specified position?
Error using symengine
Unable to convert expression containing remaining symbolic function calls into double array. Argument must be
expression that evaluates to number.
Error in sym/double (line 872)
Xstr = mupadmex('symobj::double', S.s, 0);
  12 件のコメント
Torsten
Torsten 2024 年 1 月 2 日
Use
min(x,0) = 0.5*(x-abs(x))
as I used
max(x,0) = 0.5*(x+abs(x))
below.
Walter Roberson
Walter Roberson 2024 年 1 月 4 日
Looks like it works for me when y is symbolic.
syms y
b_flat(y, 1, 2, 3)
ans = 
b_flat(y, -10, 5, 17)
ans = 
function Output = b_flat(y,A,B,C)
Output = A+piecewise(0<=floor(y-B),0,floor(y-B))*A.*(B-y)/B-piecewise(0<=floor(C-y),0,floor(C-y))*(1-A).*(y-C)/(1-C);
Output = round(Output*1e4)/1e4;
end

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回答 (2 件)

Walter Roberson
Walter Roberson 2024 年 1 月 1 日
The derivative of max() is not generally defined.
You would probably have more success if you defined in terms of piecewise() instead of in terms of max()
  1 件のコメント
Dyuman Joshi
Dyuman Joshi 2024 年 1 月 4 日
I guess the Sym engine does not have the ability to recognise that the definition of max() can be broken into a piecewise definition, than a derivative can be calculated.
I wonder if that is possible to implement or not.

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Torsten
Torsten 2024 年 1 月 1 日
編集済み: Torsten 2024 年 1 月 2 日
Use
max(x,0) = 0.5*(abs(x)+x)
for real x.
syms x1
f1 = max([0, (7*sin(4*pi*x1))/10], [], 2, 'omitnan', ~in(x1, 'real'))
f1 = 
f2 = 0.5*(abs(7*sin(4*pi*x1)/10)+7*sin(4*pi*x1)/10)
f2 = 
figure(1)
hold on
fplot(f1,[-0.5 0.25])
fplot(f2,[-0.5 0.25])
hold off
df1 = diff(f1,x1)
df1 = 
df2 = diff(f2,x1)
df2 = 
figure(2)
%fplot(df1,[-0.5 0.25])
fplot(df2,[-0.5 0.25])

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