I need to prove that, for a curve $\alpha$ at canonical local form, if a plane $P$ is s.t.:
1) $P$ constains the tangent in $0$;
2) For all interval $(-s,s)$ we have points at both sides of $P$.
Then, $P$ is the osculating plane.
Well, the book gives us a proof that follow these steps:
1) $P$ is at form $y=0$ or $z=cy$ (OK!);
2) $y=0$ is the retificant plane, so there's an interval s.t. the points of curve are all at the same side, so $P:z=cy$ (OK!);
3) At canonical form, if $|s|$ is suficiently small, $y(s)>0$ and $z(s)$ has the same signal of $s$ (OK!);
4) Then, $c$ must be positive and negative, so $c=0$.
I do not understand the step 4.
Many thanks!
We know that $y(0)=z(0)=y'(0)=z'(0)=0$.
From 3) we know that $y=\alpha s^2+O(s^3)$ and $z=\beta s^3 + O(s^4)$. Otherwise, if $z$ had a quadratic term, there would be an interval where $z(s)>0$.
So $|z/y| = |\frac\beta\alpha s|+O(s^2)$, thus for every $|c|>0$ there is interval lying at one side. So we need to conclude that $c=0$: