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Gauss-Lucas Theorem states: "Let f be a polynomial and $f'$ the derivative of $f$. Then the theorem states that the $n-1$ roots of $f'$ all lie within the convex hull of the $n$ roots $\alpha_1,\ldots,\alpha_n$ of $f$."

My Question is: Is there a theorem which states that there exists a permutation $\sigma \in S_n$ that the inner area of the polygon which edges go through the roots of $f$ $$\alpha_{\sigma(1)}\longrightarrow\alpha_{\sigma(2)}\longrightarrow\ldots\longrightarrow\alpha_{\sigma(n)}\longrightarrow\alpha_{\sigma(1)}$$ contains all roots of $f'$?


EDIT (OB) It is not completely clear from the original question wether the OP allowed for self intersections of the polygonal curve with vertices the roots of $f$. The question that has a bounty on its head asks for a polygonal Jordan curve with vertices the roots of $f$ containing the roots of $f'$ further assuming $f$ has simple roots. Roots of $f'$ are allowed to lie on the edges of the polygonal Jordan curve. We further assume $n\geq 3$ and that the roots of $f$ are not all aligned (i.e. not all contained in a real affine line.)

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    *through* the roots, not throw. Also, what exactly do you mean by the frequency polygon? The one from [statistics](http://en.wikibooks.org/wiki/Statistics/Displaying_Data/Frequency_Polygon) does not look related at all.2012-06-29
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    sorry. i corrected my mistake.2012-06-29
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    Is the permutation required to be such that the edges do not cross? Otherwise it's hard to interpret the inner area. In any case, you have to allow the roots on the boundary of the polygon, as the case $n=2$ shows.2012-06-29
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    This is a very interesting question.2012-07-25
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    I started a bounty on this question. To get the reward, either give a counter-example, or produce a proof! **Assume all roots to be _distinct_** (this is not very restrictive, at least when it comes to producing a counter example, since we can modify the constant term of $f$ slightly, so that all its roots are now distinct, without affecting the position of its roots too much, and obviously not the roots of its derivative), and the polygon should be a **polygonal Jordan curve**.2012-07-25
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    @OlivierBégassat If the roots all lie on the same line, we can't have a polygonal Jordan curve through them. Maybe you wanted to say that the roots are in generic position: no three on the same line. And $n\ge 3$.2012-07-25
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    @LeonidKovalev you are right. I'll edit my edit ^^ thanks for your help.2012-07-25
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    I think it might be better to re-write the question from scratch and add an example or further explanation. what is "inner area" and what do you mean by "which edges"?2012-07-28
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    @Maesumi the OP certainly meant to write "whose edges". I might consider rewriting the question and including an example, the only problem is that I don't know how to draw plots. What the OP means by inner area I can explain in a few words: you have a closed polygonal path (in my question it is a Jordan path, so no self intersections) that links all roots of $f$, and there is a compact "inner area" and a non compact "outer area", the same way when you draw a circle, it defines a compact disk $\lbrace |z|\leq 1\rbrace$ and the non compact outer area $\lbrace |z|>1\rbrace$.2012-07-29
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    It certainly can happen that some roots of $f'$ must lie on edges of any such polygonal Jordan curve. For example, take $f(z) = z^4 - z$ with roots $0$ and $a_i$, where the $a_i$ are roots of unity. Then the roots of $f'$ are $2^{-2/3}a_i$.2012-07-30
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    @HansEngler Indeed! The same happens if you consider $(X^2+1)X(X^2-1)(X^2-4)$ for instance.2012-07-30
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    I found an interesting paper: arxiv.org/pdf/1405.06892015-02-02

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