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What is the maximum total number of zeroes a univariate polynomial $f\in\mathbb{C}[z]$ of degree $d$, together with all of its derivatives, can have at $k$ given points of $\mathbb{C}$?

I am interested in this question for constant $k$, say $k=3$.

One trivial bound is $d \choose 2$, an other is $kd$.

Does there exist an example of a polynomial with more than $k+d$ zeroes in total of it and its derivatives?

In other words, if $p$ is a polynomial and $x_1,\ldots x_k$ are his zeroes then I am asking how many $i,j$ there exist such that $p^{(i)}(x_j)=0$

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    $\begingroup$ Dear @Klim Efremenko: my hunch is that, although the question is still a bit unclear, the classic Gauss–Lucas theorem (en.wikipedia.org/wiki/Gauss%E2%80%93Lucas_theorem) can be used to completely answer your question. What I don't understand about your question is your last sentence: how can a polynomial possibly have more than $k$ zeros if only $k$ points have been prescribed? Do you mean counting with multiplicity? $\endgroup$ – Peter Heinig Mar 7 '18 at 11:41
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    $\begingroup$ I am not counting multiplicities of zero here. $\endgroup$ – Klim Efremenko Mar 7 '18 at 11:49
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    $\begingroup$ I think you mean $\binom{d+1}2$ rather than $\binom d2$. $\endgroup$ – Emil Jeřábek 3.0 Mar 7 '18 at 13:17
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The answer is yes. Here is an example for $k=3$. For $m>0$, take $f(z)=z^m(z^2-1)^m$. Then $d=3m$. Of course, $-1$, $0$ and $1$ are roots of $f^{(i)}$ for $0\le i\le m-1$, so we have $3m=d$ roots. However, $f^{(i)}$ is an odd function for $i=m+1,m+3,\ldots, 3m-1$, producing $m$ more roots in $0$. So in this example, the count of roots is at least $\frac{4}{3}d$, which is bigger than $d+3=d+k$ once $m>3$.

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  • $\begingroup$ Thanks!!! Do you know if someone studied this question? Like what is the best upper bound one can get for the number of zeroes? In fact, for k=2 I can show upper bound of d+1 $\endgroup$ – Klim Efremenko Mar 7 '18 at 21:36
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    $\begingroup$ No, I haven't ever seen this or some related question ever being discussed. I guess Gauß-Lucas will indicate that the extreme cases may arise when all roots are on a line, so without loss of generality $f(z)$ has only real roots. $\endgroup$ – Peter Mueller Mar 7 '18 at 22:19

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