MaplePrimes - answers and comments on Question, Vector valued functions in maple

Use with(VectorCalculus);

Willem Ottevanger — Fri, 15 Mar 2013 10:04:38 Z

Hi, Is this what you mean?

> restart; with(VectorCalculus);
> F := proc (x, y, z) options operator, arrow; 6*x^2+3*x*y^2-(3/2)*z^3 end proc;
> q := F(px, py, pz);
> H := Hessian(F(x, y, z), [x, y, z]);
> subs(x = px, y = py, z = pz, H);

Hessian

acer — Fri, 15 Mar 2013 10:05:17 Z

Below I use i, i+2, and i^2 for your "point" P(1,2,3), so that it depends on the value of `i`.

Another possibility for referring to such a "point" might be to use a Vector, V, in which case its entries might be referenced instead as V[1], V[2], V[3].

Using expressions,

restart:
F:=6*x^2 + 3*x*y^2 - 1.5*z^3;
H:=VectorCalculus:-Hessian(F,[x,y,z]);
for i from 1 to 3 do
   eval(F,[x=i,y=i+2,z=i^2]);
   eval(H,[x=i,y=i+2,z=i^2]);
end do;

Using operators (procedures),

restart:
F:=(x,y,z)->6*x^2 + 3*x*y^2 - 1.5*z^3:
H:=unapply(VectorCalculus:-Hessian(F(x,y,z),[x,y,z]),[x,y,z]):
for i from 1 to 3 do
   F(i,i+2,i^2);
   H(i,i+2,i^2);
end do;

acer

Variant

Kitonum — Fri, 15 Mar 2013 12:17:17 Z

If you want a vector to be argument in your function, you can use the unapply command:

F:=unapply(6*x[1]^2 + 3*x[2]^2 - 1.5*x[3]^3, x::Vector):

Examples:

F(<1,2,3>);

H:=unapply(Student[VectorCalculus][Hessian](F(<X[1], X[2], X[3]>), [X[1], X[2], X[3]]), X::Vector):

H(<x,y,z>);

H(<1,2,3>);

It is as simple if one does not use a VectorCalculus package

jaytreiman — Sat, 16 Mar 2013 22:20:23 Z

If you want to use a vector with components it is easy to use the ideas in the attached worksheet. Once you get rid of the component labels life is easy.

restart;

Put in the function.

>	F(x,y,z):= 6x^2 + 3xy^2 - 1.5z^3;

(1)

Change the variables to components of a vector w and turn F into a function of a vector or list w.

>	F := unapply(subs({x=w[1],y=w[2],z=w[3]},F(x,y,z)),w);

(2)

Use the standard matrix formula for the Hessian to make the Hessian of F a function of x.

>	HF := unapply(Matrix(3,3,(i,j)->diff(F(x),x[i],x[j])),x);

(3)

One can then simply evaluate F and HF on a list of points or vectors.

(You can also do this with a loop.)

>	Points1 := [<1,1,1>,<2,3,4>,<3,4,5>]; map(F,Points1); map(HF,Points1);

(4)

The gradient is also easy once you have the F.

>	gradientF := unapply(Vector(3,i->diff(F(x),x[i])),x);

(5)

>	map(gradientF,Points1);

(6)

Helpful!

marwaha — Sat, 16 Mar 2013 06:23:01 Z

Hi,

Thanks for the reply. This is helpful but I think I need to understand the 'unapply' command a bit more before I can use it. I did use it in the way you described aboue (although I did not get the 'Example'). I used the unapply command to get a vector valued function but because of the unapply command it can no longer be used for symbolic operations as it becomes merely an operator (I think). For instance if I were to differentiate F within a loop with respect to any other variables, it always returns a zero value.

I'm working with a system of nonic equations for an optimization problem which requires multiple operations. So far I tried using the unapply feature to obtain an operator for evaluating the function at various points and defined another function for other operations.

Is there any way around this?