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Let see this post as a complement of Roots of polynomials with bounded integer coefficients.

Let $k$ be a positive integer. Let $R_k$ be the set of positive roots of (nonzero) polynomials with integer coefficients in $[-k,k]$:

$$R_k:=\left\{r > 0 \ \middle| \ \sum_{i=0}^na_ir^i=0,n\in\mathbb N,a_i\in\mathbb Z,|a_i|\leq k, a_n \neq 0\right\}.$$

Let $s_k$ be the supremum of $R_k$. The post mentioned above states (without proof) that $s_k=k+1$.
Question: What is the (detailed) proof of this statement?

Below is the proof that $s_k \ge k+1$ (so it remains to prove that $s_k \le k+1$):

Let $n$ be a positive integer. Consider the polynomial $$X^n-k(1+X+X^2+\dots+X^{n-1}) = X^n-k\frac{X^n-1}{X-1} = \frac{X^{n+1}-X^n-kX^n+k}{X-1},$$ then $r$ is a root iff $r^{n+1}+k = (k+1)r^n$, iff $r + \frac{k}{r^n} = k+1$, so that if $r_n$ is its largest positive root then $\lim_{n \to \infty}r_n=k+1$. $\square$

Sebastien Palcoux
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1 Answers1

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Suppose $a_nx^n+\cdots+a_0=0$. Then $$ |x|^n\le\frac{k}{|a_n|}(|x|^{n-1}+\cdots+1)\le k\frac{|x|^n-1}{|x|-1}\le \frac{k|x|^n}{|x|-1}$$

$$\therefore\ |x|^{n+1}-|x|^n\le k|x|^n$$ $$\therefore\ |x|\le k+1$$

Chrystomath
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