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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... Thanks! $\endgroup$ – Tobias Kienzler Oct 22 '13 at 15:21