# Roots of polynomials with bounded integer coefficients

For $a\in\mathbb N$, let $R_a\subseteq\mathbb{R}$ be the set of real roots of polynomials whose coefficients are integers with absolute value at most $a$.

$$R_a=\left\{r\in\mathbb{R}\middle|\sum_{i=0}^na_ir^i=0,n\in\mathbb N,a_i\in\mathbb Z,|a_i|\leq a\right\}$$

(Note that the degrees of the allowed polynomials aren't bounded.)

I want to know which numbers can be arbitrarily well approximated by such roots. In particular, is $R_a$ dense on some interval for any $a$?

Discussion:

If $r\in R_a$ and $r\neq 0$ then $1/r\in R_a$, since by reversing the order of the coefficients of the polynomial of which $r$ was a root one obtains a polynomial of which $1/r$ is a root.

In the comments, mathworker21 noted that by considering the polynomial $1 - aX -\dots-aX^n$ one can show that every nonzero $r\in R_a$ satisfies $|r|>\frac{1}{a+1}$. Therefore we also have $|r|<a+1$, so $R_a$ is bounded.

We could also ask about the set $R_a^\mathbb C$ of the roots of such polynomials in $\mathbb C$. John Baez has a page about such sets here, which contains some beautiful pictures.

He's also written a paper, which gives a reference to a paper by Thierry Bousch in which it is proved that the roots of the Littlewood polynomials (those with coefficients $\pm1$ only) are dense on the annulus $2^{-1/4}<|z|<2^{1/4}$.

Since $\{-1,1\}\subseteq\{-1,0,1\}$, this answers the complex analogue of my question: the set $R_1^\mathbb C$ is dense on some ball. Unfortunately the paper is written in French, so I can't tell if the proof generalises to the reals. (The real result isn't an immediate corollary of the complex version, because there might be a sequence in $R_a^\mathbb C$ tending to $x\in\mathbb R$ without there being such a sequence in $R_a^\mathbb C\cap\mathbb R$.)

• Isn't $R_a$ finite and hence not dense on any interval? You may be interested to know that there are more precise bounds. See en.wikipedia.org/wiki/… – Rob Arthan Apr 23 '18 at 10:00
• @RobArthan Even though the coefficients are bounded the degree of the polynomial isn't (I meant to add a note to that effect, but forgot). So there are potentially infinitely many roots. In fact the polynomials $X^n - aX^{n-1} - \dots-aX - a$ have roots that tend to $a+1$ as $n$ tends to infinity, which shows that $R_a$ is infinite. – Oscar Cunningham Apr 23 '18 at 10:08
• I don't think $R_1$ is dense in $[-1,1]$. I don't think you can get nonzero reals really close to $0$. The first term will dominate. For example, if the constant term is $1$, then the value of the polynomial at some $\epsilon > 0$ is at least $1-\epsilon-\epsilon^2-\epsilon^3-\dots$ which is bounded away from $0$. Similarly if the constant term is $-1$. If the constant term is $0$, just go to the first nonzero term and repeat this argument. – mathworker21 Apr 23 '18 at 10:14
• @mathworker21 Good point! Working out the details gives that $R_a\cap\left[-\frac{1}{a+1},\frac{1}{a+1}\right]$ only contains $0$. So $R_a$ certainally isn't dense on any interval that meets $\left(-\frac{1}{a+1},\frac{1}{a+1}\right)$. – Oscar Cunningham Apr 23 '18 at 10:31
• It could be described as incredible that if the result of John Baez could be valid for real roots, the author has not highlighted it with all the importance that such a "corollary" would have. – Piquito Apr 23 '18 at 15:31

of Odlyzko and Poonen implies that there is an interval in which $R_1$ is dense. Namely, there is some $\delta>0$ so that the interval $[-\phi^{-1}-\delta,-\phi^{-1}]$, where $$\phi=\frac{1+\sqrt{5}}{2}$$ is the golden ratio, is contained in the closure of $R_1$. See the bottom of page 318 to page 319 (see also page 330 for specific values of $\delta$ that work, and page 325 for remarks on results specific to $R_1$). In fact, the density is achieved already by numbers which are roots of polynomials with coefficients all $0$ or $1$ (no $-1$'s necessary).