Let $g:\mathbb R^3\to \mathbb R$ be smooth and $T$ be a linear isomorphism of $\mathbb R^3$. I'm trying to determine conditions on $g$ such that the fiber $(g\circ T)^{-1}(0)$ is a graph of a smooth function about a point $p$.
The obvious way to do this is to differentiate $g\circ T$ w.r.t $z$ using the chain rule, which gives one answer.
I think there's another possibility, of looking at the right entry of $Dg\cdot A$, where $A$ is the matrix representing $T$ in the standard basis. Thing is, this method gives me a different answer. Can anyone help me out?
The following is a theorem of $ZF$:
The axiom of choice holds if and only if for every infinite set $A$, there exists a bijection of $A$ with $A\times A$. (i.e. $|A|=|A|^2$)
Let us overview the theorem of Zermelo, namely if the axiom of choice holds then $\kappa=\kappa^2$ for every infinite...
Let $g:\mathbb R^3\to \mathbb R$, $g=g(u,v,w)$ be a function and $(x_0,y_0,z_0)$ a point satisfying $$g(x-y,y-z,z-x)=0.$$ I need to give conditions on $g$ to enable expressing $z$ as a function of $x,y$.
Using different methods I get different results.
Differentiate the above equation w.r.t $z...
