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Fix $\epsilon >0$. Suppose $p_{n}(z) = z^{n}+a_{n-1}z^{n-1}+ \cdots + a_{0} \in \mathbb{Z}[x]$ is irreducible and has all positive real roots. Show that independently of $n$ except for finitely many explicitly computable exceptions that $|a_{n-1}| \geq (2-\epsilon)n$.

The sum of the roots is $-\frac{a_{n-1}}{a_n}$ which is $-a_{n-1}$. So $-a_{n-1} >0$ which means that $a_{n-1} \leq 0$. So $|a_{n-1}| = -a_{n-1}$. The problem can then be rephrased as follows: Show that the sum of the roots of the above polynomial is $ \geq (2-\epsilon)n$. Could we use a proof by contradiction? I think irreducibility is an important property here.

Source: Computational excursions in analysis and number theory by P Borwein

  • 0
    This is not true as stated. For example, z-1 satisfies the premises but not the conclusion.2011-01-03
  • 0
    The correct statement is: for all $\epsilon>0$ there is $N>0$ such that for all $n \geq N$ and polynomials as stated, $|a_{n-1}| \geq (2-\epsilon)n$.2011-01-03
  • 3
    What is the source of the problem?2011-01-03

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