For each \(N \in \mathbb{N}\), let \(P_N(z) = \sum_{n=0}^{N} \frac{z^n}{n!}\).
Show that the set \(Z = \{z \in \mathbb{C} \mid P_N(z) = 0 \text{ for some } N \in \mathbb{N}\}\) is discrete (has no accumulation points in \(\mathbb{C}\)).
Find a constant \(c > 0\) such that \(P_N(z)\) has no zeros in \(\{z \in \mathbb{C} \mid |z| < cN\}\). (You may use the inequality \(n! > e^{-n} n^n\).)
Solution
Let \[ R_N(z)=e^z-P_N(z)=\sum_{n=N+1}^\infty\frac{z^n}{n!}. \] Choose \[ c=\frac1{4e}. \] Suppose \(|z|\le cN\). Since \(|z|/(N+j)\le c\) for every \(j\ge1\), \[ |R_N(z)| \le \frac{|z|^{N+1}}{(N+1)!}\frac1{1-c}. \] Using \((N+1)!>e^{-(N+1)}(N+1)^{N+1}\) and \(N/(N+1)<1\), \[ |R_N(z)| <\frac{(ce)^{N+1}}{1-c}. \] On the other hand, \[ |e^z|=e^{\Re z}\ge e^{-|z|}\ge e^{-cN}. \] Therefore \[ \frac{|R_N(z)|}{|e^z|} <\frac{ce}{1-c}\bigl(ce^{1+c}\bigr)^N. \] For \(c=1/(4e)\), \[ ce=\frac14, \qquad ce^{1+c}=\frac14e^{1/(4e)}<\frac13, \] so for every \(N\ge1\), \[ \frac{|R_N(z)|}{|e^z|} <\frac{1/4}{1-1/(4e)}\cdot\frac13<1. \] Hence \(|R_N(z)|<|e^z|\). If \(P_N(z)=0\), then \(e^z=R_N(z)\), a contradiction. Thus \(P_N\) has no zero in \(|z|<cN\).
1. To prove discreteness, let \(K\subset\mathbb C\) be compact and choose \(R\) with \(K\subset\overline{D_R}\). If \(N>R/c\), then every zero of \(P_N\) has modulus at least \(cN>R\), so \(P_N\) has no zero in \(K\). Only finitely many \(N\) can therefore contribute zeros to \(K\), and each such polynomial has finitely many zeros. Hence \(Z\cap K\) is finite for every compact \(K\), so \(Z\) has no accumulation point in \(\mathbb C\).