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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?

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  • $\begingroup$ @martini Thanks, how did you fix thi-- I see, you escaped most characters, did you use a tool for that or do it manually? $\endgroup$ – Tobias Kienzler Oct 22 '13 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 '13 at 15:20
  • $\begingroup$ @martini m-/ I never used that feature until now, always manually linked via []()... Thanks! $\endgroup$ – Tobias Kienzler Oct 22 '13 at 15:21
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    $\begingroup$ You may also be interested in these three questions: one, two, three. $\endgroup$ – Antonio Vargas Nov 5 '13 at 21:59
  • $\begingroup$ @AntonioVargas Thanks, interesting indeed! $\endgroup$ – Tobias Kienzler Nov 6 '13 at 7:32
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The zeros of the scaled functions $f_N(Nz)$ do converge to an airfoil-like curve. See an animation here.

See also these:

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  • $\begingroup$ Awesome links, thanks! $\endgroup$ – Tobias Kienzler Oct 22 '13 at 16:55
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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

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  • $\begingroup$ 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 $2^n-1$ $\endgroup$ – Tobias Kienzler Oct 30 '13 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. $\endgroup$ – Gottfried Helms Oct 30 '13 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. $\endgroup$ – Gottfried Helms Oct 30 '13 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 $\endgroup$ – Tobias Kienzler Oct 30 '13 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) $\endgroup$ – Gottfried Helms Oct 30 '13 at 9:37

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