Algebraic vs. analytic definition of the multiplicity of a polynomial's root

Question

Let $f(x) = a(x - c_1)^{d_1}(x - c_2)^{d_2} \dots (x - c_n)^{d_n}$ be a polynomial over the complex numbers ($n, d_i \in \{1, 2, \dots\}$, $a \in \mathbb{C}\setminus \{0\}$), where the roots $c_1, c_2, \dots, c_n$ are pairwise distinct. It is known that for every $i \in \{1, 2, \dots, n\}$ $c_i$ is a root of $f^{(m)}$ for $m \in \{0, 1, \dots, d_i - 1\}$. Is it possible that $c_i$ is a root of $f^{(d_i)}$?

Answer 1

If I understand the question correctly:

You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$:

$$f(x)=f(c)+(x-c)f'(c)+\dots+\frac{(x-c)^nf^{(n)}(c)}{n!}$$

Thus, if $c$ is a root of $f^{(k)}$ for $k \in \{0, \dots d\}$, then

$$f(x)=\frac{(x-c)^{d+1}f^{d+1}(c)}{(d+1)!}+\dots+\frac{(x-c)^nf^{(n)}(c)}{n!}$$

$$f(x)=(x-c)^{d+1}\left(\frac{f^{d+1}(c)}{(d+1)!}+\dots+\frac{(x-c)^{n-d-1}f^{(n)}(c)}{n!}\right)$$

And there is a factor $(x-c)^{d+1}$ in $f$, hence the multiplicity of the root $c$ is at least $d+1$.