For each integer \(n \ge 1\), let \(P_n(z) = 1 + z + \frac{1}{2!}z^2 + \frac{1}{3!}z^3 + \cdots + \frac{1}{n!}z^n\). Show that all sufficiently large \(n\), the polynomial \(P_n\) has no zeros in \(\abs{z} < 10\), while the polynomial \(P_n(z) - 1\) has exactly 3 zeros there.
Solution
For \[ P_n(z)=\sum_{k=0}^{n}\frac{z^k}{k!}, \] write the exponential tail as \[ R_n(z) = e^z-P_n(z) = \sum_{k=n+1}^{\infty}\frac{z^k}{k!}. \]
1. One has\[ \sup_{\abs{z}\leq10}\abs{R_n(z)} \longrightarrow0 \]as \(n\to\infty\).
For \(\abs{z}\leq10\), \[ \abs{R_n(z)} \leq \sum_{k=n+1}^{\infty}\frac{10^k}{k!}. \] The numerical series \[ \sum_{k=0}^{\infty}\frac{10^k}{k!} = e^{10} \] converges, so its tails tend to zero. The bound is independent of \(z\).
2. On the circle \(\abs{z}=10\),\[ \abs{e^z}\geq e^{-10}. \]
If \(\abs{z}=10\), then \[ \operatorname{Re}z\geq-10. \] Therefore \[ \abs{e^z} = e^{\operatorname{Re}z} \geq e^{-10}. \]
3. For all sufficiently large \(n\), the polynomial \(P_n\) has no zeros in\(\abs{z}<10\).
By step 1, choose \(N_1\) such that for every \(n\geq N_1\), \[ \sup_{\abs{z}\leq10}\abs{R_n(z)} < e^{-10}. \] Then on \(\abs{z}=10\), steps 1 and 2 give \[ \abs{P_n(z)-e^z} = \abs{R_n(z)} < \abs{e^z}. \] Rouché’s theorem implies that \(P_n\) and \(e^z\) have the same number of zeros in \(\abs{z}<10\). The exponential has no zeros, so \(P_n\) has none.
4. The zeros of \(e^z-1\) in \(\abs{z}<10\) are exactly\[ 0, \qquad 2\pi i, \qquad -2\pi i, \]and each is simple.
The equation \[ e^z=1 \] has precisely the solutions \[ z=2\pi i k, \qquad k\in\ZZ. \] The condition \[ \abs{2\pi k}<10 \] holds exactly for \(k=0,\pm1\), since \[ 2\pi<10<4\pi. \] Moreover, \[ \frac{d}{dz}(e^z-1)=e^z, \] which equals \(1\) at every zero, so all three are simple.
5. There is a number \(m>0\) such that\[ \abs{e^z-1}\geq m \]for every \(z\) with \(\abs{z}=10\).
By step 4, none of the zeros of \(e^z-1\) lies on the circle \(\abs{z}=10\). Thus the continuous positive function \[ z\longmapsto\abs{e^z-1} \] has a strictly positive minimum \(m\) on that compact circle.
6. For all sufficiently large \(n\), the polynomial \(P_n-1\) has exactlythree zeros in \(\abs{z}<10\), counting multiplicity.
5. On \(\abs{z}=10\),\[ \begin{aligned} \abs{ (P_n(z)-1)-(e^z-1) } &= \abs{P_n(z)-e^z}\\ &= \abs{R_n(z)}\\ &< \abs{e^z-1}. \end{aligned} \]Rouché’s theorem implies that \(P_n-1\) and \(e^z-1\) have the same number ofzeros in the disk. Step 4 shows that this number is three.
7. For every\[ n\geq\max\{N_1,N_2\}, \]\(P_n\) has no zeros in \(\abs{z}<10\), while \(P_n-1\) has exactly three zerosthere.
This combines steps 3 and 6.
8. Q.E.D.
Step 7 is the required eventual zero-count statement.