Generalisation of discriminant to roots of given multiplicity

Question

Let $p$ be a polynomial over the complex numbers. The discriminant of $p$ is zero if and only if it has a multiple root. I was wondering: For $n\geq 3$, does there exist a function $f_n$ such that $f_n(p)$ is vanishes if and only if $p$ has a root with multiplicity at least $n$?

Answer 1

Question: "Let p be a polynomial over the complex numbers. The discriminant of p is zero if and only if it has a multiple root. I was wondering: For n≥3, does there exist a function fn such that fn(p) is vanishes if and only if p has a root with multiplicity at least n?"

Answer: You may be aware of the "Sylvester matrix" calculating the "discriminant polynomial" $D(f):=Discr(f(x),f'(x))$ of a degree $n$ polynomial

$$f(x)=a_nx^n+\cdots + a_0\in k[x]$$

where $k$ is a field. The discriminant polynomial $D(f)$ is a homogeneous poynomial in the coefficients of $f(x)$. It has the property that $D(f)=0$ iff $f(x)$ has a multiple root. If $D(f)=D(f')=0$ it follows $f(x)$ has a root of multiplicity $\geq 3$ and so on. Hence you may use $D(f)$ repeatedly.

Note: Im uncertain if there is one function $f_n$ with the property you describe for any field $k$, but if $k \subseteq \mathbb{R}$ is a subfield of the field of real numbers you may define

$$f_n:=D(f)^2+D(f')^2+\cdots +D(f^{(n-2)})^2$$

and $f_n$ will have the property you ask for since $f_n=0$ iff $D(f)=\cdots =D(f^{(n-2)})=0$.