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Let $P(x)$ be a monic polynomial of degree $d$ with complex coefficients. Let $r_1(P),r_2(P),\dots, r_d(P)$ denote the set of roots, ordered so that $|r_1(P)| \leq |r_2(P)|\leq\dots\leq |r_d(P)|$. Define the map $T$ by:

$$(TP)(x)=x^d+r_1(P)x^{d-1}+r_2(P)x^{d-2}+\dots+r_d(P),$$

i.e. $TP$ is the monic polynomial whose coefficients are the roots of $P$.

Let us call a monic polynomial periodic if $T^KP=P$ for some $K>0$.

The question is: for any $d>0$, does there exist a periodic polynomial of degree $d$, other than the trivial solution $x^d$?

Remark on the definition of T

As pointed out in the comments, the definition of $TP$ is ambiguous if there are two roots $r_i(P)$ and $r_j(P)$ such that $|r_i(P)|=|r_j(P)|$ and $r_i(P)\neq r_j(P)$. If the roots of $P$ have this property, then you may break the ties however you please. For example, if $P(x)=x^3-x$, then it is up to you whether to set $r_2(P)=1$ and $r_3(P)=-1$ or $r_2(P)=-1$ and $r_3(P)=1$. However, either ordering still must have $r_1(P)=0$, since there is no ambiguity there.

Note that the set of polynomials that have this ambiguity has measure zero, so I suspect such considerations will not influence the solution of the problem anyway.

Empirical Evidence

If $d=1$ then the answer is clearly yes (any $P(x)=x-a$ will do the job, with $a\ne 0$). If $d=2$ then $P(x)=x^2+x-2$ is a fixed point of $T$, so in particular is periodic with period 1.

I examined other low degrees by numerical simulation. Note that this requires relaxing the definition of a cycle, since testing for exact equality of floating point numbers is impossible. Thus, for these simulations, the condition $T^KP=P$ was replaced with $\|T^KP-P\|_\infty<\varepsilon$, with $\varepsilon=10^{-10}$. In particular, these simulations can only find polynomials $P$ that are periodic up to some fixed error tolerance.

The simulation was done by first initializing the coefficients of $P$ using values drawn from a standard normal distribution, and then iteratively applying $T$ 1000 times and checking whether the obtained sequence was eventually periodic (up to error $<\varepsilon$). Note that this method might not find all cycles.

The periods found thusly for low degrees are:

$$ \begin{array}{rc} d=3 & \text{possible periods}= 1 ; 11 \\ 4 & 21 \\ 5 & 4 ; 56 \\ 6 & 34 ; 44 \\ 7 & 10 ; 15 ; 26 ; 234 \\ 8 & 3 ; 38 ; 83 ; 292 \\ 9 & 256 ; 311 ; 466 \\ 10 & 275 ; 336 \end{array} $$

Furthermore, for degrees $\leq 8$, all of the simulated sequences eventually became periodic, however this was not true for $d=9$ or $10$ (of course, this does not imply that these sequences never become periodic, just that they did not before the simulation ended).

Crossposted to Mathoverflow:https://mathoverflow.net/questions/364359/a-polynomial-formed-from-the-roots-of-another-polynomial-ad-infinitum

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  • $\begingroup$ Sorry if I’m misreading something, but isn’t $x^d$ trivially periodic? $\endgroup$
    – J.G
    Jun 19, 2020 at 3:44
  • $\begingroup$ @JG, you are correct, I have edited the question $\endgroup$ Jun 19, 2020 at 3:50
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    $\begingroup$ Your $T$ is not well defined. When there are several different roots of $P$ with the same absolute value there is no definite rule of ordering these within the coefficient sequence of $TP$. $\endgroup$ Jun 19, 2020 at 17:57
  • $\begingroup$ @MikeHawk u now have excluded the trivial solution $x^d$. fix ur question. also, nice username $\endgroup$ Jun 20, 2020 at 0:56
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    $\begingroup$ The backward way from $TP$ to $P$ by Vieta formulae looks much more simple to calculate and has no ambiguous cases. On the other hand, to assure that the found orbit $T^{-n}P$, $n\in \Bbb N\cup\{0\}$ provides an answer, we should check that for each polynomial of the orbit the sequence of absolute values of its coefficients is non-decreasing. $\endgroup$ Jun 20, 2020 at 3:41

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