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The Gauss-Lucas theorem relates the location of zeros of a polynomial to the location of zeros of its derivative:

Suppose $f(z)\in \mathbb{C}[z]$ is a non-constant polynomial with roots $\alpha_1,\ldots,\alpha_n$, and let $$K = K(\alpha_1,\ldots,\alpha_n) \subset \mathbb{C}$$ denote the convex hull of these points. Then all roots of $f'(z)$ lie inside $K$.

If we instead consider a polynomial with real coefficients, Rolle's theorem gives a stronger condition on the location of zeros of the derivative:

Suppose $f(x)\in\mathbb{R}[x]$ is a non-constant polynomial with real roots $\alpha_1\leq \cdots \leq \alpha_n$, counted with multiplicity. Then for any $i < j$, the closed interval $$ I = [\alpha_i, \alpha_j] \subset \mathbb{R} $$ contains some root of $f'(x)$.

The second statement in "local" in the sense that knowing only two roots of $f$ gives us some information about where the roots of $f'$ lie. In the first statement, knowing a subset of the roots will not in general determine the convex hull.

Question: For $f(z)\in\mathbb{C}[z]$, is there any information we get about the location of roots of $f'(z)$ if we know only the locations of (say) 3 non-collinear* roots of $f(z)$?

Some guesses which are false: Given three roots $\alpha_1, \alpha_2, \alpha_3$ of $f(z)$,

  1. $f'(z)$ must contain a root inside the triangle spanned by $\alpha_1, \alpha_2, \alpha_3$.

  2. $f'(z)$ must contain a root inside the circle passing through $\alpha_1, \alpha_2, \alpha_3$.

*as suggested by Gerry Myerson, the case when $\alpha_i$ are collinear may require separate analysis. But even this case is unresolved up to my understanding.

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  • $\begingroup$ I will think more about your question. There is a paper with some partial results which I will try to find. Meanwhile, I will point you to a slightly related question which you may find interesting which I asked on this site some time ago: mathoverflow.net/questions/189245/… $\endgroup$ – Trevor J Richards Mar 22 '17 at 21:40
  • $\begingroup$ [edited to make note of the degenerate case when roots are collinear] $\endgroup$ – Harry Richman Jul 28 '17 at 4:39
  • $\begingroup$ Could you give an indication of Gerry Myerson's explanation that information about colinear roots gives no information about the zeros of the derivative? $\endgroup$ – Trevor J Richards Jul 28 '17 at 16:20
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    $\begingroup$ @CarlSchildkraut Yes, if $p(z)$ has zeros at the third roots of unity, and the other zeros of $p$ are sufficiently far away, then $p$ will have two critical points arbitrarily close to $0$. On the other hand, if the other roots of $p$ are close to (or in) the unit disk, then $p$ will have at least two, or all, of its critical points in the unit disk. The question is what happens in the "intermediate case", when at least some of the other zeros of $p$ are (relatively) close to, but not in, the unit disk. $\endgroup$ – Trevor J Richards Jul 29 '17 at 17:43
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    $\begingroup$ I would assume the following paper on approximate Gauss-Lucas theorems (see already page 1) is relevant: arxiv.org/pdf/1706.05410.pdf $\endgroup$ – Andreas Rüdinger Jul 29 '17 at 19:58
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Following is information from Marden's Geometry of the Zeros, Sections 25 and 26. Note: Some further information can be found on my paper "Approximate Gauss--Lucas Theorems" (also linked in the comments).

Theorem: (Kakeya) For $1\leq p\leq n$, there is a function $\varphi(n,p)$ such that if a degree $n$ polynomial has $p$ zeros lying in a disk with radius $R$, then that polynomial has at least $p-1$ critical points lying in the concentric disk with radius $R\cdot\varphi(n,p)$.

Thus for your example, let $R$ denote one half the diameter of the set of three zeros, and let $C$ denote the disk containing the three zeros, with radius $R$. Then the concentric disk with radius $R\cdot\varphi(n,p)$ contains at least two critical points of the polynomial. Note that in general, for the known bounds, the degree $n$ of the polynomial must be known.

Various estimates on this function $\varphi$ have been given:

Kakeya: $\varphi(n,2)=\csc(\pi/n)$.

Biernacki: $\varphi(n,n-1)\leq(1+1/n)^{1/2}$.

Marden: $\varphi(n,p)\leq\csc(\pi/(n-p+1))$ and $\varphi(n,p)\leq\displaystyle\prod_{k=1}^{n-p}[(n+k)/(n-k)]$.

Richards: $\varphi(n,p)\leq1+\dfrac{8(n-p)^2}{p-8(n-p)^2}$ (subject to the assumption that $p>8(n-p)^2$). (This result is in the linked paper, definitely not in Marden's awesome book!)

Complete citations may be found in my linked paper. Sorry I did not think of this earlier as the answer, I was working on trying to find the smallest radius which would catch at least one critical point, not all $n-p$. And in any case I have no excuse for my (now deleted) earlier answer!

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  • $\begingroup$ Worth noting given OP's final remark: From the standpoint of finding a disk which will contain $2$ (or in general, $p-1$) critical points close to the known zeros, colinearity does not seem to play any role. $\endgroup$ – Trevor J Richards Aug 3 '17 at 2:10

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