bound for zeros of a polynomial with bounded integer coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:45:56Z http://mathoverflow.net/feeds/question/85324 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85324/bound-for-zeros-of-a-polynomial-with-bounded-integer-coefficients bound for zeros of a polynomial with bounded integer coefficients Daniel Krenn 2012-01-10T10:24:31Z 2012-01-10T12:00:22Z <blockquote> <p>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)?</p> </blockquote> <p>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 $\lvert a_i \rvert &lt; M$. Suppose $f(z)=0$ implies $\lvert z \rvert > 1$ (so all zeros have absolute value greater than $1$). </p> <blockquote> <p>What is an (explicit) positive function $B(n,M)$ with $$\lvert z \rvert - 1 \geq B(n,M)$$ for any zero $z$ of $f$?</p> </blockquote> <p>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 $2^n$.</p> http://mathoverflow.net/questions/85324/bound-for-zeros-of-a-polynomial-with-bounded-integer-coefficients/85326#85326 Answer by Nikita Sidorov for bound for zeros of a polynomial with bounded integer coefficients Nikita Sidorov 2012-01-10T11:14:50Z 2012-01-10T12:00:22Z <p>Just a brief remark that if $M=2$ and the constant term is $\pm2$, then these are called <em>Garsia numbers</em>. It is known that $z=1$ is a limit point for this set (and some computational results as well). Perhaps, you'll find the <a href="http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P36_Garsia.pdf" rel="nofollow">following recent paper</a> useful as far as the techniques are concerned. </p> http://mathoverflow.net/questions/85324/bound-for-zeros-of-a-polynomial-with-bounded-integer-coefficients/85327#85327 Answer by Emil Jeřábek for bound for zeros of a polynomial with bounded integer coefficients Emil Jeřábek 2012-01-10T11:48:59Z 2012-01-10T11:48:59Z <p>$\def\conj#1{\overline{#1}}\DeclareMathOperator\Res{Res}$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 $B(n,M)\ge M^{-O(n)}$.</p>