Filter löschen
Filter löschen

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

8 Ansichten (letzte 30 Tage)
Rounak Saha Niloy
Rounak Saha Niloy am 1 Jan. 2024
Kommentiert: Walter Roberson am 4 Jan. 2024
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 Kommentare
Torsten
Torsten am 2 Jan. 2024
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 am 4 Jan. 2024
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

Melden Sie sich an, um zu kommentieren.

Antworten (2)

Walter Roberson
Walter Roberson am 1 Jan. 2024
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 Kommentar
Dyuman Joshi
Dyuman Joshi am 4 Jan. 2024
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.

Melden Sie sich an, um zu kommentieren.


Torsten
Torsten am 1 Jan. 2024
Bearbeitet: Torsten am 2 Jan. 2024
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])

Produkte


Version

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by