Higher order inflection points

Question

Consider a smooth plane curve $X\subset\mathbb{P}^2$ of degree $d$. We will say that $x\in X$ is an inflection point of order $s$ if the tangent line $T_xX$, of $X$ at $x\in X$, intersects $X$ in $x\in X$ with multiplicity at least $s$.

For instance, any point of $X$ in an inflection point of order $s = 2$, and the inflection points of order $s = 3$ are the flexes of $X$.

Does there exist a closed formula for the number $I(s)$ of inflection points of order $s$ of $X$. For instance, it is well known that $I(3) = 3d(d-2)$ since flexes are the intersection points of $C$ with its Hessian which is a curve of degree $3(d-2)$.

Furthemore, we must have $I(s) = 0$ for $s > d$ since otherwise $C$ would be reducible.

Answer 1

(The statements you quote are only true if you are working over a field of characteristic zero or $p > d$. I will continue to make that assumption)

The formula is not $I(3)=3d(d-2)$ but rather $\sum_{s>2} (s-2)I(s) = 3d(d-2)$ (changing notation slightly so $I(s)$ counts the points with contact exactly $s$ instead of at least $s$).

The typical curve (non-empty open set of the parameter space) does indeed satisfy $I(3)=3d(d-2)$ and $I(s)=0, s > 3$. I think you can expect many different possibilities under the above constraints but most likely not all of them. I don't this has been worked out in general. For $d=4$ all possibilities and the dimension of the space of curves that satisfy them has been worked out in the PhD thesis of Vermeulen. I have unfortunately misplaced my copy so I can't quote the exact result.

Edit: Found the thesis (Amsterdam 1983). If $d=4$ then $I(3)+2I(4)=24$ and $I(4)$ can take any value between $0$ and $12$, except for $10$ and $11$.