How to write a symbolic function array?

function f=g5(x)
syms q b x
q=[6 7 1;7 11 2;1 2 22]
b=[-1;2;3]
x=transpose(sym('x',[1 3]))
f=1/2*transpose(x)*q*x+transpose(b)*x-2
end

This is my writing function

clc;
clear;
%%gradient descent
syms w
w=[2;2;6]
for i=1:30
    w=w-5*gradient(g5(w))
    %%if norm(gradient(g5(w)))>0.01
        %%break
    %%end
    display(w)
end    

This is the variable I setted trying to find the solution of the function. But the outcome is

w =
 
   152 - 1050*x2 - 150*x3 - 900*x1
- 1050*x1 - 1650*x2 - 300*x3 - 298
 - 150*x1 - 300*x2 - 3300*x3 - 444

I want to insert [2;2;6] into [x1;x2;x3] to get the value of the function. But it seems symbolic variables defined cannot be taken as parameters. Is there any way to change it into independent variables?

[Walter Roberson]

syms q b x

That is equivalent to

q = sym('q');
b = sym('b');
x = sym('x');

and that is going to ovewrite the x that was passed in to the function

function f=g5(x)

so anything you pass in to the function is going to be ignored.

[Zhenwei Yu]

@Walter Roberson

So anyway I can do this? I tried using syms q b only without x. not working though. Or I simply left out function f=g5(x)?

[Walter Roberson]

function f=g5(x)
q = [6 7 1;7 11 2;1 2 22];
b = [-1;2;3];
f = 1/2*x'*q*x + b'*x - 2
end

The result will already be a scalar, so you would not take gradient() or norm() of it.

[Zhenwei Yu]

function f=g5(x)
q = [6 7 1;7 11 2;1 2 22];
b = [-1;2;3];
f = 1/2*x'*q*x + b'*x - 2
end

Not enough input arguments.

Error in g5 (line 4)
f = 1/2*x'*q*x + b'*x - 2

I don't know why but this get an error

@Walter Robertson

[Walter Roberson]

w=[2;2;6];
g5(w)

ans = 512

function f=g5(x)
q = [6 7 1;7 11 2;1 2 22];
b = [-1;2;3];
f = 1/2*x'*q*x + b'*x - 2;
end

Looks okay to me.

[Zhenwei Yu]

@Walter Roberson

Thanks for your help but it seems you may misunderstand my purpose. I got a function 1/2*x'*q*x + b'*x - 2, and I need to calculate its gradient. Something like (3x1+7x2+x3,7/2x1+11/2x2...). And use values in each point of gradient function to make it close to optimal point.

Your function is right and I just tested. But it seems I cannot preserve the unknown variables for further calculation