Let $N$ be a big integer number and consider the equation :

$$ x^{N} + a_{N-1} x^{N-1} + ...+a_{1} x + o(\frac{1}{N})=0,$$ where $o(h)$ is by definition a term such that $\lim_{h \to 0} o(h)/h =0$. Assume that all coefficients $a_j$ have a big norm : $|| a_j||\approx N^{j-1}$ except $a_{1}$ which is asymptotically a nonzero constant.

Let $x_s$ be a complex root of above polynomial with the smallest norm. I want to show something similar to $|| x_s || \approx o(\frac{1}{N}) $ or any upper bound like $|| x_s || \leq \frac{1}{N} $ or even smaller than that. It's obvious that if we don't have the term $o(\frac{1}{N})$ then smallest norm root of the polynomial $$ x^{N} + a_{N-1} x^{N-1} + ...+a_{1} x=0$$ is $x_s=0$. So intuitively makes sense to say that for the first polynomial $|| x_s ||$ is also small but I can't show how small it is .

I do not expect someone gives me the exact solution of this problem because I understand we need more details, but please let me know how you tackle with these kind of questions.

  • 3
    $\begingroup$ very strange definition of $O(h)$ $\endgroup$ – Fedor Petrov Dec 8 '18 at 7:32
  • $\begingroup$ i've changed the $O$ into the usual $o$. $\endgroup$ – Liviu Nicolaescu Dec 8 '18 at 10:49

Denote $x=y/N$, your equation in terms of $y$ rewrites as $N^{-1}\sum_{k=0}^{N} (a_k N^{1-k})y^{k}=0$, you need a small root of such a polynomial in $y$. Denote $a_kN^{1-k}=b_k$, then $b_1$ is asymptotically constant, $b_0$ tends to 0 (I understand you so, please use $o$ instead of $O$ if it is the case), other $b_i$ are bounded (and also $b_N$ is very small but we do not use this.) You may use Rouchet theorem for the circle $|y|=c$ for small $c$ and the functions $f(z)=b_1z$, $g(z)=\sum_{j\ne 1} b_j z^j$. The function $f$ has exactly one root inside the circle $|y|=c$ and $|f|>|g|$ on the circle (for any fixed $c>0$ this is so if $b_0$ is small enough). Thus you get a root of your polynomial $f+g$ smaller than $c$.


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