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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.

  1. Differentiate the above equation w.r.t $z$ using the chain rule, to find that $g_v\neq g_w$ at the point in question.
  2. Use the chain rule $D(g\circ T)=Dg\circ T$ for $T$ the linear map defined by $$x\mapsto x-y,y\mapsto y-z,z\mapsto z-x.$$ This means $$D(g\circ T)=Dg \begin{pmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}. $$ Then, the right entry of this product is $\frac{\partial g\circ T}{\partial z}$, which ends up being $g_u-g_w$ and gives the condition $g_u\neq g_w$ at the point of interest.

I think the first method is correct, but I don't understand what's wrong about the second one..

1 Answers 1

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I was helped in one of the chatrooms. My mistake is in calculating the matrix of $T$, which should in fact be $$ \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{pmatrix}.$$