Polynomial $p_n(z) = (1 + \frac{z}{n})^n - 1$ has a property that all its zeros lie on the circle of radius $n$. It is easy to see because $$\frac{z}{n} = e^{\frac{i2\pi k}{n}} - 1$$

So we can "fit" zeros of polynomials of degree $n$ on the curve of degree $2$ (circle). Is it possible to do similar things i.e. find the curve of smaller than $n$ order where all zeros are placed, with the polynomial $$q_n(z) = \sum^n_{k = o}\frac{z^k}{k!} - 1$$

As both $p_n(z)$ and $q_n(z)$ $\to$ $e^z - 1$ as $n \to \infty$


1 Answer 1


There is no known explicit formula for the curve(s) on which the zeros of $q_n(z)$ lie. They can be approximated, however.

Since most of the zeros travel outward to infinity at a rate of $\approx n$, it's common to scale them by $1/n$. This means we are now looking at the zeros of the polynomial $Q_n(z) \stackrel{\text{def}}{=} q_n(nz)$.

The zeros of $Q_n(z)$ accumulate on a set of curves in the complex plane. These curves are:

  • the circle $|z| = 1/e$ in the left half-plane ($\operatorname{Re} z < 0$),

  • the segment of the imaginary axis $[-i/e,i/e]$, and

  • the curve $|ze^{1-z}| = 1$ in the right half-plane ($\operatorname{Re} z > 0$).

See, for instance, this question on MathOverflow.

It is also possible to find better approximations to the curves on which the zeros lie. See, for example, the method in the paper by Carpenter, Varga, and Waldvogel Asymptotics for the zeros of the partial sums of $e^z$, I. In particular, the sections where they calculate the curves $D_n$. Though this curve is specifically for the zeros of the polynomials $Q_n(z)+1$, it should still yield a good approximation for the zeros of $Q_n(z)$ in the right half-plane.

The paper is available from Varga's website here: PDF link.

In the left half-plane the zeros approach the semicircle $|z| = 1/e$, and, I showed in this paper that those zeros which aren't too close to the imaginary axis satisfy

$$ |z| = \frac{1}{e} + \frac{\log n}{en} + O\left(\frac{1}{n}\right) $$

as $n \to \infty$. To see this, take $a=0$, $b=1$, and $\varphi(z) = 1$ in the integral in $F(z)$ at the top of Section 3. This makes $\xi = \mu = \nu = 0$, and the claim then comes from Theorem 3.1.a.

  • $\begingroup$ Thanks for the references. $\endgroup$
    – hOff
    Jun 3, 2015 at 12:21
  • $\begingroup$ Sure thing. Feel free to ask if you have any questions. I skipped lots of details, especially in the part about the circle $|z| = 1/e$. $\endgroup$ Jun 3, 2015 at 12:37
  • $\begingroup$ Hmm, there is a typo in my paper. This is a special case which is just outside of what I considered, so I have made a small correction to the last part this answer to reflect what I believe is true but I haven't worked through all the details. $\endgroup$ Jun 7, 2015 at 19:32

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