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I am learning Differential Geometry and someone told me that the second condition of a definition provided in books follows from the first and is hence superfluous. I cannot dispute it, so convince me if he is right.


Suppose two curves $C_1$ and $C_2$ have a regular point P in common. Take a point A on $C_1$ near P and let $D_A$ be the point on $C_2$ closest to A (i.e orthogonal projection of A on $C_2$) then $C_2$ has contact of order $n$ with $C_1$ at P if

$$ \lim_{A\to P} \dfrac{ \mbox{dist}(A,D_A)}{[\mbox{dist}(A,P)]^k}= \begin{cases} c\neq 0, & \text{if } \; k=n+1 \\ 0, & \text{if } \; k\leq n \end{cases}$$

Argument: $$\lim_{A\to P} \frac{AD_A}{(AP)^{n+1}} =c \neq 0$$ then $$\lim_{A\to P} \frac{AD_A}{(AP)^{n}} = \lim_{A\to P} AP \cdot \frac{AD_A}{(AP)^{n+1}}$$

$$ = \lim_{A\to P} AP \cdot \lim_{A\to P} \frac{AD_A}{(AP)^{n+1}} = 0 \cdot c =0$$

The same process can be followed to show that the limit is zero for all $k\leq n$

Note: The above definition is claimed by the authors to be extracted from this original source

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I am suspicious: the argument does not use the fact that $c \neq 0$. Instead you need that $|c| < \infty$. [Added: I think that the definition is an emphatic way of saying "the smallest $k$ such that $\lim_{A \to P} \frac{AD_A}{AP^{k+1}}$ is nonzero".] – Srivatsan Sep 15 '11 at 18:22
@Srivatsan why not? It uses the fact that the limit for $k=n+1$ is $c\neq 0$ to show that the limit would be zero for every $k\leq n$. Re: additional comment, I agree, specially after seeing this argument. I need a confirmation that nothing fishy is going on as this refutes an argument presented in two books. – kuch nahi Sep 16 '11 at 6:38
@Srivatsan Yes. So I guess this person's argument is right. The statement should be that order is the minimum $n$ for which the stated limit is non zero and finite. – kuch nahi Sep 16 '11 at 6:43
Nonzero, yes. But I am not sure about finite. (Also note that the limit is nonzero for $k=n+1$, not $k=n$. So it's more like the the maximum $n$ for which the limit is zero.) – Srivatsan Sep 16 '11 at 6:46
@Srivatsan if it is not finite then we cannot conclude that $0\cdot c =0$ and the argument won't stand. – kuch nahi Sep 16 '11 at 6:48

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