# A Polynomial Formed from the Roots of Another Polynomial ad infinitum

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).

• Sorry if I’m misreading something, but isn’t $x^d$ trivially periodic?
• 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 $TP$. Jun 19, 2020 at 17:57
• @MikeHawk u now have excluded the trivial solution $x^d$. fix ur question. also, nice username Jun 20, 2020 at 0:56
• 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. Jun 20, 2020 at 3:41