Range for values of cyclotomic polynomials, where $x$ is replaced by the golden ratio $0.61...$ ? And is it dense?

155 Views Asked by At

This is a recreational math question.

I just played with the cyclotomic polynomials; and replacing $x$ by $1$,$-1$,$I$ gives some interesting patterns; setting $x=2$ seems to give some (uninteresting to me) pattern of unbounded values (of course related to the Mersenne-numbers $2^n-1$).

More interesting seems it to me when I replace $x=g $ where $g \approx 0.618...$ is the golden ratio.
Up to the $5000$'th cyclotomic polynomial it seems that the range of occuring values is bounded to the interval $g^2 \ldots g^{-2}$.
[edit] It seems the occurence of values $g^2$ and $g^{-2}$ are artifacts due to properties of $g$; if I use another value $0 \lt x \lt 1$ it is $1-x$ and ${1 \over 1-x}$ which in the case of $x=g$ becomes accidentally identical with the squares.[/edit]
Q1: Is it true, that the range is bounded? And is it easy to show it, why?

Looking at the first ,say 12, digits only it seems, that there are values which occur more often, or let's more precisely say, small intervals instead of values; it looks like some clustery structure like accumulation points, for instance $0.5,1,g$ itself and some other special values.

Q2: Does the range $g^2 ... g^{-2}$ become dense, when we increase the order of the polynomials up to infinity?


[Update] To make the properties (accumulation-points, bounds) better visible I show the values to $12$ digits for the first $120$ cyclotomic polynomials evaluated at $x=0.1$
$$ \small \small \begin{array} {r|lllllll} 1..5& -0.900000000000 & 1.10000000000 & 1.11000000000 & 1.01000000000 & 1.11110000000 \\ 6..10& 0.910000000000 & 1.11111100000 & 1.00010000000 & 1.00100100000 & 0.909100000000 \\ 11..15& 1.11111111110 & 0.990100000000 & 1.11111111111 & 0.909091000000 & 0.900909910000 \\ 16..20& 1.00000001000 & 1.11111111111 & 0.999001000000 & 1.11111111111 & 0.990099010000 \\ ..& 0.900900990991 & 0.909090909100 & 1.11111111111 & 0.999900010000 & 1.00001000010 \\ ..& 0.909090909091 & 1.00000000100 & 0.990099009901 & 1.11111111111 & 1.09889011000 \\ ..& 1.11111111111 & 1.00000000000 & 0.900900900910 & 0.909090909091 & 0.900009090091 \\ ..& 0.999999000001 & 1.11111111111 & 0.909090909091 & 0.900900900901 & 0.999900009999 \\ & 1.11111111111 & 1.09890098901 & 1.11111111111 & 0.990099009901 & 0.999000000999 \\ & 0.909090909091 & 1.11111111111 & 0.999999990000 & 1.00000010000 & 0.999990000100 \\ & 0.900900900901 & 0.990099009901 & 1.11111111111 & 0.999999999000 & 0.900009000099 \\ & 0.999900009999 & 0.900900900901 & 0.909090909091 & 1.11111111111 & 1.00999898990 \\ & 1.11111111111 & 0.909090909091 & 0.999000000999 & 1.00000000000 & 0.900009000090 \\ & 1.09890109889 & 1.11111111111 & 0.990099009901 & 0.900900900901 & 1.09998889011 \\ & 1.11111111111 & 0.999999999999 & 1.11111111111 & 0.909090909091 & 0.999990000000 \\ & 0.990099009901 & 0.900000090009 & 1.09890109890 & 1.11111111111 & 0.999999990000 \\ & 1.00000000000 & 0.909090909091 & 1.11111111111 & 1.00999899000 & 0.900009000090 \\ & 0.909090909091 & 0.900900900901 & 0.999900009999 & 1.11111111111 & 1.00099999900 \\ .. & 0.900000090000 & 0.990099009901 & 0.900900900901 & 0.909090909091 & 0.900009000090 \\ .. & 1.00000000000 & 1.11111111111 & 0.999999900000 & 0.999000000999 & 0.999999999900 \\ 101..105& 1.11111111111 & 1.09890109890 & 1.11111111111 & 0.999900009999 & 1.10998878900 \\ 106..110& 0.909090909091 & 1.11111111111 & 1.00000000000 & 1.11111111111 & 1.09998900010 \\ 111..115& 0.900900900901 & 0.999999990000 & 1.11111111111 & 1.09890109890 & 0.900009000090 \\ 116..120& 0.990099009901 & 0.999000000999 & 0.909090909091 & 0.900000090000 & 1.00010000000 \end{array} $$ (Of course, we get rational values, nicely separated, for this; however since $g$ is irrational the range shall be nearly randomly scattered, thus the question of dense-ity might not be trivial) [/update]

1

There are 1 best solutions below

2
On

From numerical experiments, it looks to me like when $0 < t < 1$, $$\limsup_{n \to \infty} \Phi_n(t) = \dfrac{1}{1-t}$$

Here's a link to an animation showing $(1-t) \Phi_n(t)$ for $2 \le n \le 1000$ and $0.01 < t < 0.99$. There seems to be interesting structure here, to say the least.