Can you square a function of three variables?

Question

If I have a function of three variables and I want to create a new function in which it equals the other function squared, could I literally just square the other function or does this violate any rules? Would this also mean its gradient vector is just squared at a certain point?

Answer 1

HINT

Say you have $f(x,y,z)$ and you would like to define $$g(x,y,z) = f(x,y,z)^2.$$ Then, $$ \vec{\nabla} g = \begin{pmatrix} \partial g/\partial x \\ \partial g/\partial y \\ \partial g/\partial z \end{pmatrix} $$ For example, chain rule implies $$ \frac{\partial g(x,y,z)}{\partial x} = \frac{\partial f(x,y,z)^2}{\partial x} = 2 f(x,y,z) \frac{\partial f(x,y,z)}{\partial x}. $$

Answer 2

could I literally just square the other function

Yes. The square of $f(x,y,z)$ is $[f(x,y,z)]^2$. No violations. For example, if $f(x,y,z) = xyz$, then the square of this function is $g(x,y,z) = (xyz)^2 = x^2y^2z^2$.

Would this also mean its gradient vector is just squared at a certain point?

There may, coincidentally, be specific points $(a,b,c)$ for which $\nabla g(a,b,c)$ is the "square" of $\nabla f(a,b,c)$, i.e., each component of $\nabla g(a,b,c)$ is the square of the corresponding component of $\nabla f(a,b,c)$. But it's not true everywhere.

For the $f$ and $g$ given above, it works trivially for $a=b=c=0$ and it doesn't work for $a=b=c=1$.

Answer 3

A function is not characterized only by the number of independent variables in the domain, but also by the number of dependent variables in the codomain and this make some difference in defining the ''square'' of a function.

If we have a function $f:\mathbb{R}^3 \to \mathbb{R}$ than its value is areal number and we can easily define the square as $f^2(x)=[f(x)]^2$ , i.e. the function whose value is the square of the value of the function $f(x)$.

If we have a function $f:\mathbb{R}^3 \to \mathbb{R}^n$ the value of the function has $n$ components: $f(x,y,z)=(f_1(x,y,z),f_2(xyz),f_3(xyz), \cdots, f_n(x,y,z))$ so we can define the square function as the function that has as components the square of the components: $f^2(x,y,z)=(f_1^2(x,y,z),f_2^2(x,y,z),f_3^2(x,y,z), \cdots, f_n^2(x,y,z))$.

In any case the derivative ( or the derivatives) can be found using the chain rule as in: $$ \frac {\partial f^2(x,y,z)}{\partial x}=2f(x,y,z)\frac {\partial f(x,y,z)}{\partial x} $$