How to evaluate sym using certain known multiple values?

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I have a 2*3 sym (named "A") as follows:
[ 1, 0, 0]
[-1/(2*(x1/(x2^2 - 10*cos(4*x3*pi) - 10*cos(4*x2*pi) + x3^2 + 21))^(1/2)), (x1*(2*x2 + 40*pi*sin(4*pi*x2)))/(2*(x1/(x2^2 - 10*cos(4*x3*pi) - 10*cos(4*x2*pi) + x3^2 + 21))^(1/2)*(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21)) - (2*x2 + 40*pi*sin(4*pi*x2))*((x1/(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21))^(1/2) - 1), (x1*(2*x3 + 40*pi*sin(4*pi*x3)))/(2*(x1/(x2^2 - 10*cos(4*x3*pi) - 10*cos(4*x2*pi) + x3^2 + 21))^(1/2)*(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21)) - (2*x3 + 40*pi*sin(4*pi*x3))*((x1/(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21))^(1/2) - 1)]
I want to evaluate this 2*3 sym at -
X=[0.394876, 0.963263, 0.173956]
where
x1=X(1);
x2=X(2);
x3=X(3);
How can I do this?
Note, I tried matlabFunction command. This causes certain problems in some certain scenarios. I do not want to use matlabFunction command.

Accepted Answer

Rounak Saha Niloy
Rounak Saha Niloy on 31 Dec 2023
double(subs(A, [x1, x2, x3], X));
Could sort it out. The above code evaluates the sym at X.

More Answers (1)

John D'Errico
John D'Errico on 31 Dec 2023
Easy peasy, though using matlabFunction here is not my recommendation.
syms('x',[1,3])
So x is a vector, with elements [x1,x2,x3].
A = [[ 1, 0, 0]
[-1/(2*(x1/(x2^2 - 10*cos(4*x3*pi) - 10*cos(4*x2*pi) + x3^2 + 21))^(1/2)), (x1*(2*x2 + 40*pi*sin(4*pi*x2)))/(2*(x1/(x2^2 - 10*cos(4*x3*pi) - 10*cos(4*x2*pi) + x3^2 + 21))^(1/2)*(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21)) - (2*x2 + 40*pi*sin(4*pi*x2))*((x1/(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21))^(1/2) - 1), (x1*(2*x3 + 40*pi*sin(4*pi*x3)))/(2*(x1/(x2^2 - 10*cos(4*x3*pi) - 10*cos(4*x2*pi) + x3^2 + 21))^(1/2)*(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21)) - (2*x3 + 40*pi*sin(4*pi*x3))*((x1/(x2^2 - 10*cos(4*pi*x3) - 10*cos(4*pi*x2) + x3^2 + 21))^(1/2) - 1)]];
Now you have X, with the values of x1,x2,x3.
X=[0.394876, 0.963263, 0.173956];
You can simply do this:
double(subs(A,x,X))
ans = 2×3
1.0000 0 0 -3.4478 -50.1285 95.5057
That stuffs the elements of X into x1,x2,x3 respectively. Then the call to double turns the result into numbers.

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