We are given two polynomials in the plane $f(x,y)$ and $g(x,y)$. Both define an algebraic curve in the plane.

Let $\hat{h}(x)$ be the Taylor expansion of the function $h$ at a point $(x_0,y_0)$ defined by $f(x,h(x))=0$ (computed by Implicit Function Theorem).

Is it true that, there exists a neighbourhood of $(x_0,y_0)$ such that $$ sign(g(x,y)_{\mid f(x,y)=0})=sign(g(x,y)_{\mid y=\hat{h}(x)}) $$


We have: $ h(x)=h(x_0)+(x-x_0)h'(x_0)+o(x-x_0)$ and $\hat h(x)=h(x_0)+(x-x_0)h'(x_0)$

then $$ g(x,y)_{\mid f(x,y)=0}=g(x,h(x))=g(x,y_0) + \frac{\partial g}{\partial y}(x_0,y_0)((x-x_0)h'(x_0)+o(x-x_0))$$

On the other hand, $$ g(x,y)_{\mid y=\hat h(x)}=g(x,\hat h(x))=g(x,y_0) + \frac{\partial g}{\partial y}(x_0,y_0)((x-x_0)h'(x_0)$$ so my statement would be obvious ?


Assuming $g(x_0,y_0)\neq 0$, this is trivial: by continuity of $g$, there is a neighborhood $U$ of $(x_0,y_0)$ in the plane on which $g$ has constant sign. In particular, inside that neighborhood it has the same sign whether you restrict it to the vanishing set of $f$ or to the graph of $\hat{h}$.

If $g(x_0,y_0)=0$, there is no reason for something like this to be true. For instance, if $g=f$, then $g$ is always $0$ on the vanishing set of $f$, but will not be $0$ on the graph of $\hat{h}$ unless $\hat{h}$ happens to actually be exactly equal to $h$.


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