필터 지우기
필터 지우기

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

조회 수: 1 (최근 30일)
Rounak Saha Niloy
Rounak Saha Niloy 2024년 1월 1일
댓글: Walter Roberson 2024년 1월 4일
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);
  댓글 수: 12
Torsten
Torsten 2024년 1월 2일
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 2024년 1월 4일
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

댓글을 달려면 로그인하십시오.

답변 (2개)

Walter Roberson
Walter Roberson 2024년 1월 1일
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()
  댓글 수: 1
Dyuman Joshi
Dyuman Joshi 2024년 1월 4일
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.

댓글을 달려면 로그인하십시오.


Torsten
Torsten 2024년 1월 1일
편집: Torsten 2024년 1월 2일
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])

카테고리

Help CenterFile Exchange에서 Number Theory에 대해 자세히 알아보기

제품


릴리스

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by