6
$\begingroup$

Let $f$ be a monic polynomial with bounded integer coefficients and such that all zeros are (in absolute value) greater than $1$. How close can the zeros of $f$ reach $1$ (in absolute value)?

More precisely, let $$ f = a_0 + a_1 X + \cdots + a_{n-1}X^{n-1} + X^n $$ with $a_i \in \mathbb{Z}$ and $\lvert a_i \rvert < M$. Suppose $f(z)=0$ implies $\lvert z \rvert > 1$ (so all zeros have absolute value greater than $1$).

What is an (explicit) positive function $B(n,M)$ with $$\lvert z \rvert - 1 \geq B(n,M)$$ for any zero $z$ of $f$?

If it is easier, then $M=2^n$ can be assumed. Further, I am only interested in the behavior for large $n$, i.e., I want something like $$ \frac{1}{\lvert z \rvert - 1} = O(B(n,2^n)) $$ for $n\to\infty$ and for each zero $z$ of any monic polynomial $f$ with degree at most $n$ and integer coefficients bounded by $2^n$.

$\endgroup$

2 Answers 2

3
$\begingroup$

$\def\conj#1{\overline{#1}}\DeclareMathOperator\Res{Res}$If $z$ is a zero of $f$, then $|z|^2-1=z\conj z-1$ is a zero of the resolvent $g(w)=\Res_z(\conj f(z),z^nf((w+1)/z))$. You can extract a bound on the (integer) coefficients of $g(w)$ from the definition, and then e.g. Cauchy's bound will give you a lower bound on $|z|^2-1$, which in turn implies a lower bound on $|z|-1$. I don’t feel like working it out myself now, but you should get something of the form $B(n,M)\ge M^{-O(n)}$.

$\endgroup$
4
$\begingroup$

Just a brief remark that if $M=2$ and the constant term is $\pm2$, then these are called Garsia numbers. It is known that $z=1$ is a limit point for this set (and some computational results as well). Perhaps, you'll find the following recent paper useful as far as the techniques are concerned.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.