Finding the gradient and the hessian of the same function

If I have the function

fv = @(x) x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));

What would be the best way to find the hessian of said function?

Answer 1

[KSSV]

Read about gradient, hessian.

[Howie]

When I try computing the hessian, I get the following

fv = @(x) x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));


hessian(fv,[x(1),x(2)])

Unrecognized function or variable 'x'.

jacobian(gradient(fv))

[KSSV]

syms x y 
fv =  x^2 + x*y + (3/2)*y^2 - 2*log(x) - log(y);
hessian(fv,[x,y])

[KSSV]

If you want x(1) and x(2) like:

syms x [1 2] 
fv =  x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));
hessian(fv,x)

[Howie]

And how would we find the gradient?

[KSSV]

In the same way...read the doc.

[Howie]

syms x [1 2] 
fv =  x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));
%hessian(fv,x)

[x(1),x(2)] = gradient(fv)

Error using sym/gradient
Too many output arguments.

I get an error when trying to use gradient function

[Torsten]

Same way means:

syms x [1 2] 
fv =  x(1)^2 + x(1)*x(2) + (3/2)*x(2)^2 - 2*log(x(1)) - log(x(2));
g = gradient(fv,x)

g =