How to parameterise a function?

I was trying to use a multi-variable function in commands that take functions as inputs but kept getting errors and was told to parameterise the function but not sure how. The function is f(x,y,z) = x^2 + 3y^2 + 4z^2 - 2xy + 5x - 3y + 2z. I tried doing:

f = @(x,y,z) = x.^2 + 3.*y.^2 + 4.*z.^2 - 2.*x.*y + 5.*x - 3.*y + 2.*z;
ezplot(f)
fminsearch(f,[0,0,0])

and was told to parameterise my functions so that x(1)=x, x(2)=y... Not sure how to continue though. Please help

Answer 1

[Stephen23]

The standard answer is to replace each x, y and z with the corresponding element of the input vector:

g = @(V) = V(1).^2 + 3.*V(2).^2 + 4.*V(3).^2 - 2.*V(1).*V(2) + 5.*V(1) - 3.*V(2) + 2.*V(3);

A more readable alternative is to use two functions, which allows you to keep the original function unchanged. This could be useful if it is important to keep exactly the same equation as some reference, or to use exactly those variables names, or just to maintain readability:

f = @(x,y,z) = x.^2 + 3.*y.^2 + 4.*z.^2 - 2.*x.*y + 5.*x - 3.*y + 2.*z;
g = @(V) f(V(1),V(2),V(3));

and then call the function g in all of your following work. It has the disadvantage that it would likely be a smidgen slower, so do not use this if your need to perform millions of iterations.

[Jennifer Wail]

Thank you that did get me further but now having trouble with fminsearch

 f = @(x) x(1)^2 + 3*x(2)^2 + 4*x(3)^2 - 2*x(1)*x(2) + 5*x(1) - 3*x(2) + 2*x(3)
[x fval] = fminsearch(f,[0,0,0])

Maybe I'll post a second question on that but thank you again you helped me get further on other problems with that

Answer 2

[James Tursa]

Do pretty much exactly what the advice says:

f = @(x) x(1)^2 + 3*x(2)^2 + 4*x(3)^2 - 2*x(1)*x(2) + 5*x(1) - 3*x(2) + 2*x(3);

[Jennifer Wail]

I tried that but now fminsearch gives me index out of bounds error