# Prove that given root of a polynomial is zero by approximation

Let $$\alpha$$ be a root of a polynomial $$a_nx^n + \ldots + a_1x$$ with integral coefficients.

I would like to determine $$\varepsilon > 0$$ depending on $$a_1, \ldots, a_n$$ so that $$|\alpha| < \varepsilon$$ implies $$\alpha = 0$$.

Is it possible to give a "formula" for such an $$\varepsilon$$ without refering to the complete list of roots?

• Root separation: en.wikipedia.org/wiki/… – Matt F. Feb 11 at 16:24
• @MattF. Thank you, that is what I am searching for. Do you know any good source (besides the one mentioned on Wikipedia)? – J. Fabian Meier Feb 11 at 20:44
• The accepted solution does not even use the fact that the coefficients are integral. – jarauh Feb 12 at 7:55

Of course we must assume some $$a_j \ne 0$$. Say $$a_j$$ is the one with least index. Then you want $$\varepsilon$$ such that $$p(x) = a_n x^{n-j} + \ldots + a_j \ne 0$$ for $$|x| < \varepsilon$$. You may use inequalities such as $$|p(x)| \ge |a_j| - \sum_{k=j+1}^{n} |a_k| |x|^{k-j} \ge |a_j| - m \sum_{k=j+1}^n |a_k|$$
where $$m = \max(|x|^{n-j}, |x|)$$. Thus $$p(x) \ne 0$$ if $$m < |a_j|/\sum_{k=j+1}^n |a_k|$$.
• In your first sentence, the 1 in $a_1$ is a typo? – Gerry Myerson Feb 11 at 20:38
Abbott, John, Bounds on factors in $$\Bbb Z[x]$$, J. Symb. Comput. 50, 532-563 (2013). ZBL1295.12010.: