Playing around with the incomplete/finite exponential series

$$f_N(x) := \sum_{k=0}^N \frac{z^k}{k!} \stackrel{N\to\infty}\longrightarrow e^z$$

for some values on alpha (e.g. solve sum_(k=0)^19 z^k/(k!) = 0 for z), I made a few observations:

  • The sum of the roots of $f_N$ are $-N$
  • The product of the roots of $f_N$ are $(-1)^N\cdot N!$
  • Their imaginary part seems to lie between $\pm10$
  • The zeros seem to form an interesting shape:

Patterns for $N=17, 18, 19$

Now the sum and product part are clear, since

$$\begin{align} f_N(x) &= \frac1{N!}\left(z^N + N z^{N-1} + N(N-1)z^{N-2} + ... + N!\right) \\ &= \frac1{N!}(z-z_{N0})(z-z_{N1})\cdots(z-z_{NN}) \\ &= \frac1{N!}\left(z^N - \left(\sum_{k=0}^Nz_{Nk}\right) z^{N-1} + ... + (-1)^N\prod_{k=0}^N z_{NK}\right) \end{align}$$

and since $e^z=0 \Leftarrow \Re z\to-\infty$ it is clear that the roots tend towards real parts with negative infinity, but I'm still intrigued by the questions

what ($N$-dependent) curve do the zeros of $f_N(z)$ lie on, does that curve maintain its shape for varying $N$ and merely translate or also deform; and what other properties of the zeros (e.g. absolute value) can be derived?

  • $\begingroup$ @martini Thanks, how did you fix thi-- I see, you escaped most characters, did you use a tool for that or do it manually? $\endgroup$ Oct 22, 2013 at 15:19
  • $\begingroup$ I used the "add a link" feature of the edit/answer window here on SE. Just click on the "chain" and paste the link there ... $\endgroup$
    – martini
    Oct 22, 2013 at 15:20
  • $\begingroup$ @martini m-/ I never used that feature until now, always manually linked via []()... Thanks! $\endgroup$ Oct 22, 2013 at 15:21
  • 1
    $\begingroup$ You may also be interested in these three questions: one, two, three. $\endgroup$ Nov 5, 2013 at 21:59
  • $\begingroup$ @AntonioVargas Thanks, interesting indeed! $\endgroup$ Nov 6, 2013 at 7:32

2 Answers 2


The zeros of the scaled functions $f_N(Nz)$ do converge to an airfoil-like curve. See an animation here.

See also these:

  • $\begingroup$ Awesome links, thanks! $\endgroup$ Oct 22, 2013 at 16:55
  • $\begingroup$ Perhaps an interesting relation: there is a proposition of a function as 2-fold iteration $f_c(z)=c^z \text{ with } c,z \in \mathbb C$ and a method to find periodic-points $f^{o2}_c(z)=z$ with a function called "HyperW()" or $HW()$ in an article as a generalization of the Lambert-W. The method to actually approximate periodic points means to solve the truncated taylor-series of $f^{o2}_c(z)$ . This is a strong relation to the subject here. I've done some investigation. See question math.stackexchange.com/q/3706909 and one answer math.stackexchange.com/a/3713978 . $\endgroup$ Jun 11, 2020 at 10:03

One more picture; here I rescale the radial distances from the origin to their logarithm; the roots of the polynomials $f_{16},f_{32},f_{64},f_{128}$ are shown, the magenta line is that for $f_{16}$. I find it interesting, that the radial positions fit nicely together, see the straight lines from the origin (the roots do not exactly match with the lines but remarkably good)

enter image description here

  • $\begingroup$ So in other words, you plot on a log-log scale, right? Wow, that's an interesting observation! Also interesting: The amount of roots between two such lines appears to be of the form $2^n-1$ $\endgroup$ Oct 30, 2013 at 8:49
  • $\begingroup$ Well, that's not quite the correct transformation. In fact, if I do separately rescale the x-axis and the y-axis, then the curves approximate squares (you might try this using an asinh-scaling instead, which becomes quite similar to a log-log-scaling for values >1). No, here I compute the polar coordinates of the roots, rescale the distance to the origin $\sqrt{\Re(\rho)^2+\Im(\rho)^2}= |\rho| $ by logarithmizing and recalc the rectangular coordinates from that. $\endgroup$ Oct 30, 2013 at 9:10
  • $\begingroup$ @tobias:the number of roots between two such lines plus that on one line is just 2^<index-of-line> because the number of complex roots is just doubling with each step. $\endgroup$ Oct 30, 2013 at 9:14
  • $\begingroup$ Ah true, it's too early in the morning... I wonder if the roots really don't lie exactly on these line or whether that's numerical accuracy... Also, the real roots seem to be very regularly spaced $\endgroup$ Oct 30, 2013 at 9:22
  • $\begingroup$ @Tobias: Yepp. Do you have Pari/GP? It was no problem to get the 128 roots by the simple call <pre> polroots(Pol(exp(x) + O( x^2^7)))) </pre> in just a second.(One should increase the default powerseries precision and the computing precision sufficiently high) $\endgroup$ Oct 30, 2013 at 9:37

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