What are the properties of the roots of the incomplete/finite exponential series?

Question

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?

Answer 1

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

See also these:

• Zeros of truncated Taylor series by Jonas (see the references at the end) (2013)

• Zeroes of the partial sums of the exponential function by Zemke (2009)

• On the zeroes of the nth partial sum of the exponential series by Zemyan (2005)

• The zeros of the partial sums of the exponential series by Walker (2003)

Answer 2

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)