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this is a question from a Temple prelim exam, and i'm trapped in it! We have $p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}$ and we have to prove that $\forall r>0 \quad \exists N\in\mathbb{N}$ s.t. $p_n(z)$ has no zeros in $B_r(0)$ for $n>N$.

I tried to use Rouche's theorem observing that $1+\frac{z^n}{n!}$ has the nth roots of $n!$ as complex zeros, which are outside a fixed ball for n sufficiently large, but i'm having trouble proving that $|1+\frac{z^n}{n!}|>|z+\cdots+\frac{z^{n-1}}{(n-1)!}|$ in $|z|=r$ (if this is true). If anyone could help..

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For $r=1$, we do not have $|1+\frac 1{n!}| > |1 + \dots + \dots \frac 1{(n-1)!}$, so this is not the way, because $r$ has to be arbitrary (perhaps this is true for sufficiently small $r$, though). – Patrick Da Silva Apr 13 '12 at 22:28
Someone here in SE provided a great animation that showed the zeroes converged to a curve much like half a leminiscate. – Pedro Tamaroff Apr 14 '12 at 2:26
up vote 5 down vote accepted

Here's another approach using Rouche's Theorem:

Let $f(z) = \frac{p_n(z)}{e^Z}$, and $g(z) = 1$. Note that $z \mapsto e^z$ has no zeros. Since convergence is uniform on compact sets (in this case $\bar B_r(0)$), we can choose $n$ large enough so that $|f(z) -1| <1$, $\forall z \in \bar B_r(0)$ (clearly $f$ has no zeros or poles on the circle of radius $r$). Then by Rouche's Theorem, $Z_f-P_f = Z_g-P_g$ (zeros and poles of $f,g$). Since $P_f=Z_g=P_g=0$, we have $Z_f = 0$, from which the result follows.

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Yes, this would be a nice approach using Rouche's theorem. You essentially use the same idea as mine but finish your proof by using Rouche's theorem instead of doing it by hand (because if you think about it, $|p_n(z)/e^z - 1| < 1$ is essentially the same as $|p_n(z) - e^z| < |e^z|$), which I think is neater. (I have not done much complex analysis so perhaps this is a more standard approach, but I don't know.) Good job =) +1 – Patrick Da Silva Apr 14 '12 at 0:17

It was proved by K. S. K. Iyengar in 1938 [1] that every zero of the polynomial

$$s_n(z) = \sum_{k=0}^{n} \frac{z^k}{k!}$$

lies in the annulus

$$\frac{n}{e^2} < |z| < n.$$

A much stronger result was proved in 1966 by Buckholtz [2], namely that the rescaled polynomials $s_n(n z)$ have no zeros in the region

$$ \left\{ z \in \mathbb{C} \,\colon \left|z e^{1-z}\right| \leq 1 \,\,\text{ and }\,\, |z| \leq 1 \right\}. $$

As the proof is short and sweet I will include it here.

Proof: Let $z \in \mathbb{C}$ with $|z| \leq 1$ and $\left|z e^{1-z}\right| \leq 1$. Then

$$\begin{align*} \left|1 - e^{-nz} s_n(nz)\right| &= \left|e^{-nz} \sum_{k=n+1}^{\infty} \frac{n^k z^k}{k!} \right| \\ &= \left|\left(z e^{1-z}\right)^n e^{-n} \sum_{k=n+1}^{\infty} \frac{n^k z^{k-n}}{k!} \right| \\ &\leq e^{-n} \sum_{k=n+1}^{\infty} \frac{n^k}{k!} \\ &= 1 - e^{-n} s_n(n) \\ &< 1. \end{align*}$$

Having $s_n(nz) = 0$ here would contradict this inequality.


This bound is sharp in the sense that the limit points of the zeros of $s_n(n z)$ are precisely the points in the set

$$ \left\{ z \in \mathbb{C} \,\colon \left|z e^{1-z}\right| = 1 \,\,\text{ and }\,\, |z| \leq 1 \right\}. $$

This is known as the Szegő curve.


[1] K. S. K. Iyengar, A note on the zeros of $\sum_{r=0}^{n} \frac{x^r}{r!} = 0$, The Mathematics Student 6 (1938), pp. 77-78.

(As the above paper apparently doesn't exist electronically I have transcribed it here.)

[2] J. D. Buckholtz, A characterization of the exponential series, The American Mathematical Monthly 73 (1966), no. 4, part II, pp. 121–123.

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Wow! Impressive. +1 – Patrick Da Silva Apr 14 '12 at 0:20
Nice........... – copper.hat Apr 14 '12 at 0:50
The name for the inner bound is the Szego curve. – anon Apr 14 '12 at 2:29
Thanks @anon, I should have mentioned that. – Antonio Vargas Apr 14 '12 at 2:32
...I linked to a few more related papers here. – J. M. Apr 14 '12 at 2:45

Using complex analysis, you can prove that the convergence of the series $$ \sum_{k=0}^{\infty} \frac{z^k}{k!} $$ is uniform over any disk centered at $0$. Therefore, given $r$, for every $\varepsilon > 0$, there exists $N$ such that for all $n > N$ and all $|z| < r$, we have $|p_n(z) - e^z| < \varepsilon$. Therefore it suffices to put $e^z$ "away from zero in $B_r(0)$" to win this.

Now given $z = x + iy$, you know that $e^z = e^x(\cos y + i \sin y)$ by Euler's formula, hence $|e^z| > \varepsilon$ if and only if $e^x > \varepsilon$, which means $x > \log(\varepsilon)$. So given a good choice $\varepsilon$ this could work. Take an arbitrary $r > 0$, and let $\varepsilon(r) = e^{-r}$. Therefore, for $x > -r$ and $n > N$, we have $|e^z| > \varepsilon$ and $|p_n(z) - e^z| < \varepsilon$, which means $$ |p_n(z)| \ge |e^z| - |e^z - p_n(z)| = |e^x| - |e^z - p_n(z)|> \varepsilon - \varepsilon = 0. $$

Here's how I started : an attempt of a sketch of a proof (I don't know if the details work out very well and short, but I gave it a try)

$$ \sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z = \lim_{n \to \infty} \left( 1 + \frac zn \right)^n $$ Therefore it seems reasonable to expect (because the expansion of the polynomial on the right is quite similar to the series on the left) that to put lower bounds on the size of the LHS, we could use estimates on the RHS. This would give $$ \left| \sum_{k=0}^n \frac{z^k}{k!} \right| \cong \left| \left( 1 + \frac zn \right)^n \right| = \left| 1 + \frac zn \right|^n $$ which you can easily bound below if you have choice on $n$.

For instance, $$ \left(1 + \frac zn \right)^n = \sum_{k=0}^n \binom nk z^k = \sum_{k=0}^n \frac{n!}{k!(n-k)!}z^k = \sum_{k=0}^n \frac{n!}{(n-k)!} \frac{z^k}{k!} $$

After a while, I believed using the uniform convergence of the series was probably a better way to go and I left this approach behind, because the large coefficients in the latter sum $(n!/(n-k)!)$ made me lose faith in it ; I didn't believe I could prove the $\cong$ part easily without using the convergence in the series, so I just let it go.

Hope that helps,

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I left the "wrong idea" part at the beginning to leave you with how I came up with the proof. Not using the fact, at first glance, that the series converges to $e^z$ seemed completely foolish to me. (Perhaps there is another idea, but I believe this is the first that should come to mind.) – Patrick Da Silva Apr 13 '12 at 22:25
+1: I suggest you move the correct part to the top, and the wrong part to the bottom. – Aryabhata Apr 13 '12 at 22:37
@Aryabhata : I was sick of editing this answer over and over to make it correct, and now you ask me to edit it again? XD But it is a good idea. Thanks – Patrick Da Silva Apr 13 '12 at 22:45

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