Asked but never answered at MSE.

Let $\exp_n(z)$ denote the nth degree Taylor polynomial of $e^z$ :

$\exp_n(z) = 1 + z + z^2/2! + ... + z^n/n! \;$ .

The zeros of $\exp_n(z)$ were studied by Szego in the 1920's and later by others. One of the consequences of Szego's results is that the roots (after division by n) can come arbitrarily close to the imaginary axis.

Question: Is it possible for exp$_n$(z) to have a root that lies precisely on the imaginary axis?


2 Answers 2


Indeed the polynomial $\exp_n(z)$ has no purely imaginary zeros. Write $$ \exp_n(ix) = C_n(x) + i S_n(x) $$ in the obvious notation, with $C_n$ for truncations of cosine, and $S_n$ for truncations of sine. If there is a zero $ix$ for $\exp_n(z)$, then this $x$ must be a common root of $C_n$ and $S_n$, which are both polynomials in ${\Bbb Q}[x]$. Therefore $C_n$ and $S_n$ must have some non-trivial gcd in ${\Bbb Q}[x]$. Now we use the beautiful fact (going back to Schur) that $C_n(x)$ is irreducible, which completes our proof. I learnt of this result of Schur from these lovely notes of Keith Conrad; simply apply Theorem 1 there, or see Corollary 1 which mentions this explicitly.

  • $\begingroup$ I guess one must observe, when $n\ge3$ is odd, that $C_n(x)$ doesn't divide $S_n(x)$. $\endgroup$ Jan 29, 2018 at 8:15

By the fundamental theorem of calculus we know a polynomial of degree $n$ has as many roots in the complex plane. That doesn't mean have any more information of the roots than that, (e.g. is the Galois group $S_n$?). There's also the problem that $e^z$ has no roots in $\mathbb{C}$ and how to account for that.

Using Stirling formula, Laplace transform, Rouché s theorem and things like that, one can so the roots of such polynomials behave like: $$ |z \, e^{1-z} | = 1 $$ I don't know where I first learned of these results maybe in American Mathematical Monthly:

However, I later found many other sources. These polynomials appear whenever you have a holomorphic, or entire function or a Laplace transform or otherwise non-zero function and you need to take some kind of approximation.

This result is still counterintuitive, since setting $z = ix$ and observing $e^{i \pi x} \neq 0$ suggests there should never be zeros.

Varga and Carpenter show:

$$ \theta = |\arg z| \geq \cos^{-1} \left( 1 - \frac{2}{n}\right) $$

this should be enough to rule out a zero being on the imaginary axis.

Using the Cauchy formula, it's possible to write these truncated exponential series as an integral:

$$ (ez)^{-n} p_{n-1}(nz) = \frac{1}{2\pi i} \int_{\{|s|=1\}} \frac{s^{-n} \, e^{n(s-1)}}{s-z} \, ds$$

The exponential polynomials are in fact moments of a distribution.

  • 4
    $\begingroup$ How does that answer the question? $\endgroup$
    – Wojowu
    Jan 27, 2018 at 15:57
  • 6
    $\begingroup$ I may be blind, but I can't see where your answer nor either of the references answers the question of whether or not the polynomials can have imaginary roots $\endgroup$
    – Wojowu
    Jan 27, 2018 at 16:37
  • 3
    $\begingroup$ You might be right. Perhaps you could point out where in the references the question is answered? $\endgroup$
    – Wojowu
    Jan 27, 2018 at 16:44
  • 5
    $\begingroup$ I will try not to sneeze. But OP asked about zeros on the imaginary axis, not about zeros on the real axis. $\endgroup$ Jan 29, 2018 at 0:52
  • 4
    $\begingroup$ While the references you provided have interesting results, these are not sufficient to address the problem. Asymptotics are not enough, because there will be zeros very close to the imaginary axis -- indeed the zeros will be near $\pm in/e$ and on the scale of about $\sqrt n$ near it. For example, the paper by Bleher and Mallison provides such an asymptotic. But such results are insufficient to rule out zeros on a particular line, and on the imaginary axis I would expect these Taylor polynomials to get very small, and yet not hit zero. $\endgroup$
    – Lucia
    Jan 29, 2018 at 4:44

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