How to evaluate sym using certain known multiple values?

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Rounak Saha Niloy
Rounak Saha Niloy 2023-12-31
回答： John D'Errico 2023-12-31
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.

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采纳的回答

Rounak Saha Niloy
Rounak Saha Niloy 2023-12-31
double(subs(A, [x1, x2, x3], X));
Could sort it out. The above code evaluates the sym at X.

更多回答（1 个）

John D'Errico
John D'Errico 2023-12-31
Ran in:
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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