During a talk I was at today, the speaker mentioned that if you truncate the Taylor series for $e^x - 1$, you'll get lots of roots with nonzero real part, even though the full Taylor series only has pure imaginary roots.

If you plot the roots of truncations of $e^x - 1$ (or check out the ready-made plots in this Mathematica notebook, now available as a PDF) you can see lots of cool features. I'd like to know where they come from! I know there's a vast literature on polynomials, but I'm a total beginner, and I don't know where to start.

Here are a few specific questions:

  1. The roots of a high-degree truncation seem to fall into two categories: roots that lie very close to the imaginary axis, and roots that lie on a C-shaped curve. (Another interpretation is that all of the roots lie on a curve, which has a very sharp kink near the imaginary axis.) Can you write down an equation for the curve?

  2. If you put the roots of a lot of consecutive truncations together on the same plot, you'll see definite "stripes" to the right of the imaginary axis. Once a stripe appears, each higher-degree truncation sticks another root onto the end, making the stripe grow outward. Can you write down equations for the stripes?

  3. If $k$ is odd, the truncation of degree $k$ has no nonzero real roots. If k is even, the truncation of degree $k$ has one nonzero real root. The location of this root depends almost linearly on $k$. Why is the dependence so close to linear? Does it get more linear as $k$ increases, or less?

  4. Can roots be given identities that persist across time? That is, as $k$ increases, can you point to a sequence of roots and say, "those are all the same individual, which was born at $k$ = so-forth, is following such-and-such trajectory, and will grow up to become the root ($2\pi i\cdot$ whatever) of $e^x - 1$"?


I finally got around to googling this a bit, and I immediately came up with http://www.mai.liu.se/~halun/complex/taylor/ which describes the same phenomenon for the exponential function itself. Briefly, if Pn is the Taylor polynomial of ex, then the zeroes of Pn(nx) pile up on the curve |ze1-z|=1, |z|≤1. They credit this discovery to Szegő (1924). See also the paper On the Zeros of the Taylor Polynomials Associated with the Exponential Function by Brian Conrey and Amit Ghosh (The American Mathematical Monthly, 95, No. 6 (1988), pp. 528-533), http://www.jstor.org/stable/2322757 if you have JSTOR access.


I will sketch the answers to all 4 questions. In some cases they are different from the answers for $e^x$.

The integral formula for the remainder in Taylor's theorem is a natural starting point for explaining all these phenomena. Although the Taylor polynomials $T_k(x)$ of $e^x-1$ are initially defined only for nonnegative integers $k$, the integral formula lets one extend the definition to non-integral values of $k$, so that one can watch the zeros of $T_k(x)$ deform continuously as $k$ gradually increases from one integer to the next!

Taylor's theorem says $$f(x) = T_k(x) + \int_0^x \frac{(x-t)^k}{k!} f^{(k+1)}(t)\, dt,$$ which for $f(x)=e^x-1$ leads to $$T_k(x) = e^x - 1 - \int_0^x \frac{(x-t)^k}{k!} e^t \, dt$$ for all $k \in \mathbf{Z}_{\ge 1}$. If we interpret $k!$ as $\Gamma(k+1)$, use the straight line path from $0$ to $x$, and use the negative real axis in the $x$-plane as branch cut, then the right hand side defines an analytic function of two complex variables on the region $$\{ (k,x) \in \mathbf{C}^2 : \text{Re}(k)>-1, x \notin \mathbf{R}_{\le 0} \}.$$ This leads to implicit equations for the "stripes" in Question 2 and the "trajectory" in Question 4.

Let's now consider Question 3, about the real zeros of $T_k(x)$ for $k \in \mathbf{Z}_{\ge 1}$. Since $T_k(x)$ has nonnegative coefficients, all nonzero zeros are negative. Set $x=-y$ and $t=-u$ in $T_k(x)=0$ to obtain $$1-e^{-y} = (-1)^k \int_0^y \frac{(y-u)^k}{k!} e^{-u} \, du.$$ If $k$ is odd, the two sides have opposite sign for $y>0$, so $T_k(x)$ has no nonzero real zeros. If $k$ is even, then $T_k(x) \to +\infty$ as $x \to -\infty$, but $T(x)=x+\cdots<0$ for small negative $x$, so $T_k(x)$ has a negative zero, and by Rolle's theorem there cannot be more than one. Where is it, approximately? Rewrite the equation for even $k$ as $$1-e^{-y} = \frac{y^k}{k!} \int_0^y e^{k \log(1-u/y)} e^{-u} \, du$$ Using Stirling's formula $k! \approx (k/e)^k \sqrt{2\pi k}$, we see that $y \approx k/e$, and then the integral is approximately $\int_0^\infty e^{-eu-u} \, du = 1/(e+1)$, so even more precisely, $$y \approx (k!/(e+1))^{1/k} \approx \frac{k}{e} + \frac{1}{2e} \log k + c + o(1)$$ where $c := \frac{1}{e} \log \left( \frac{\sqrt{2\pi}}{e+1} \right)$. I have skipped the error analysis, but it is not hard. With more work, one could obtain an asymptotic expansion. The negative real zero of $T_k(x)$ is the negative of this $y$. This answers Question 3.

Finally, let's sketch an answer to Question 1, about the locations of the complex zeros. The size of $1-e^{-y}$ is about $1$ if $\text{Re}(y)$ is somewhat positive, and then the analysis is the same as above for the negative real zero, except that now we consider also the other $k^{\text{th}}$ roots with $\text{Re}(y)<0$. So these zeros are very close to a semicircle, as predicted, and the deviations are explained by the fact that the approximation $1/(e+1)$ to the integral above gets replaced by some function of $y/k$. Again, this can be made quantitative, and proved with the aid of Rouché's theorem if desired.

To understand the complex zeros with large positive real part, go back to the first integral equation for $T_k(x)$, substitute $t=x-v$, and divide by $e^x$ to get the equation $$1 - e^{-x} = \int_0^x \frac{u^k}{k!} e^{-u} \, du.$$ To within a factor of $e^{o(k)}$, the absolute value of the integral is $\left| \frac{x^k}{k!} e^{-x} \right|$. So if we take absolute values, use Stirling's formula, and take $k^{\text{th}}$ roots, then up to a factor of $e^{o(1)}=1+o(1)$ we get $$ 1 \approx \frac{|x|}{k/e} \left| e^{x/k} \right| $$ so $z:=x/k$ satisfies $|z e^{1-z}| \approx 1$. Again, this can be made precise.

Above we implicitly used that the order of magnitude of $e^x-1$ is that of $e^x$ if $\operatorname{Re}(x)$ is large and positive, and of order $1$ if $\operatorname{Re}(x)$ is large and negative. In the region in between, the latter can fail when $x$ is close to an integer multiple of $2\pi i$. This leads to zeros close to the segment joining $ik/e$ and $-ik/e$, in which the integral of the previous paragraph can be small.

Final answer to Question 1: If we take the zeros of $T_k(x)$ and normalize by dividing by $k$, they converge to the union of the semicircle $|z|=1/e$ with $\operatorname{Re}(z)\le 0$, the segment joining $i/e$ and $-i/e$, and the part of the curve $|ze^{1-z}|=1$ with $|z| \le 1$ and $\operatorname{Re}(z) \ge 0$.

Note how this differs from Szegö's answer for the Taylor polynomials of $e^x$ mentioned in Harald Hanche-Olsen's post.


I'll make one basic observation: the truncation of $e^x$ to degree $n$ has one real root if $n$ is odd, and none if $n$ is even. Proof: Let $f_n$ be the truncation of $e^x$ to degree $n$. Notice that $f'_n = f_{n-1}$.

Our proof is by induction on $n$; the base case is obvious. Suppose that $f_{2k}$ has no real roots. Then $f_{2k}$ is everywhere positive, so $f_{2k+1}$ is increasing and has at most one real root; being of odd degree, it does have one. Now, suppose that $f_{2k+1}$ has one real root, say at $r$. Then $f_{2k+2}(r) = r^{2k+2}/(2k+2)! > 0 $. But $f_{2k+2}$ has its minimum at $r$, so $f_{2k+2}$ is everywhere positive. This concludes the induction.

How is this relevant to you? Using Rolle's theorem, you can conclude that the truncations of $e^x-1$ have precisely $1$ and either $0$ or $2$ roots respectively.

  • $\begingroup$ Nice! If I'm not mistaken, this proof, combined with Szegő's result that the roots of f_n(nz) accumulate around the curve {|ze^(1-z)| = 1, |z| <= 1} as n goes to infinity, answers question 3 for e^z. $\endgroup$ – Vectornaut Nov 8 '09 at 1:32
  • $\begingroup$ I fixed some typos. $\endgroup$ – David E Speyer Feb 26 '10 at 12:15

Cool Facts/ Generalizations of the above

(This has to do with question 1 which was already answered)

ASYMPTOTICS: Szego curves were studied by Varga and Carpenter for the trigonometric functions in 2000 and 2001. They gave a family of curves $A_n$ which approach a curve like the one described in the post above as $n\to\infty$ the roots. www.math.kent.edu/~varga/pub/paper_230.pdf .

There is actually an earlier paper which does something similar for the roots of the exponential function too.

OTHER SZEGO CURVES: There are MANY other example of Szego curves. For example Bleher and Mallison have explicit descriptions for Szego Curves of Exponential Sums. Which are functions of the form

$$ f(z) = e^{a_1 z} + e^{a_2 z} +\cdots + e^{ a_m z}$$

There is an explicit description of the associated szego curve in terms of a polygon formed by the $a_i$ in the complex plane arxiv.org/abs/math-ph/0605066

There are tons of results on zeros of orthogonal polynomials as well. For example Boyer and Goh have one on Euler polynomials http://arxiv.org/abs/0711.1400v1

I actually did a project on the Szego Curve for the exponential function as an undergraduate and still have some lingering questions. (btw I just posted some pictures and movies of roots of truncated Taylor series and a mathematica notebook for people who are interested in playing around with these things: http://www.math.ucla.edu/~dupuy/psmovies/psmovies.html )

I have a follow-up question: How do you prove that Szego Curves exist for general entire functions? Can one show that they are piecewise subsets of level sets of harmonic functions? I don't even know how to show that a normalization exists in general. It seems that all the math for Szego curves of particular functions use the form of a function to prove their results.

NORMALIZATION PROBLEM: I've thought, but have not been able to prove, that for a general entire function the spurious roots grows like $\lambda_n=|a_n|^{-1/n}$. I will make this more precise. Let $f = a_0 + a_1z+ \cdots $ be an entire function and for simplicity suppose that $f(z)$ has no zeros. Suppose that $z_n$ is any sequence of numbers such that $s_n[f](z_n)=0$. Show that $0<\vert z_n \vert \lambda_n \vert \leq B$ where $B$ is some constant for $N$ sufficiently large.

There are two reasons to suspect that all roots grow like $|a_n|^{-1/n}$. The first is due to the fact that |a_0/a_n| is the product of the roots of the polynomial $s_n[f]$ which makes $|a_n|^{-1/n}$ equal to the multiplicative average asymptotically. The second is that this is the radius of convergence, and the roots of Taylor Series for functions which aren't entire accumulate on the radius of convergence. This statement would be a generalization of this fact.

  • 1
    $\begingroup$ Follow-up questions do best posted as separate questions. $\endgroup$ – j.c. May 20 '10 at 4:07

Not really an answer, just some preliminary thoughts:

Since the truncation fk converges uniformly to f=ex-1 on any compact set, and since the roots of f are simple, we know that for every root of f, if we take large enough k we have a root of fk in its vicinity. This also means that roots for different k can indeed be identified.

Using Stirling's formula one can see that (f-fk)(-ak) is very small when a<1/e. This means that you cannot have a root in this interval. When taking a>1/e the terms start to fluctuate very wildly, so it is not surprising that you will find a root there. Am I right that the constant seems close to 1/e?

  • $\begingroup$ I agree that the constant is near 1/e. $\endgroup$ – Michael Lugo Nov 6 '09 at 13:04

This is obvious, but the roots of the related polynomial $(1+\frac{x}{n})^{n} -1$ are of course much more manageable, all being of the form $n(\omega -1)$ for $\omega$ (not necessarily primitive) $n$-th root of unity. Clearly, $0$ is the only root with zero real part, and when $n$ is even, $-2n$ is the only real root apart from $0,$ while $0$ is the only real root when $n$ is odd.

  • $\begingroup$ This turned out to be a nightmare... $\endgroup$ – user39115 Jun 8 '14 at 11:02

Let x and y belong to R- F(x)=1+x+...+x^n/n! F(y)=1+y+...+y^n/n! F(x)-F(y)<0 If n is even F(x)-F(y)>0 If n is odd In both the cases x>y So F is either monotonically increasing or decreasing so it has almost one root

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    $\begingroup$ How does this answer the question? Please explain more if it does. $\endgroup$ – Joonas Ilmavirta Dec 11 '15 at 6:44

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