06/15/2015, 01:00 AM
(This post was last modified: 06/15/2015, 03:07 AM by sheldonison.)
(06/11/2015, 10:27 AM)sheldonison Wrote: ....\( \lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[\pi\sqrt{\frac{2\eta\cdot n}{e}} -2 \right] \approx 388.7874 \)
\( \lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[\pi\sqrt{\frac{2\eta\cdot n}{e}} + \text{slog}_e(\frac{n}{e}-1) + C + \mathcal{O}(\theta) \right] -n = 0\;\;\;\;C \approx -2 - \text{slog}_e(388.7874/e-1) \)
I was about to post a closely related question about Pi in the Mandelbrot set; it takes about \( \pi \sqrt{n} \) iterations to escape near the parabolic cusp at c=0.25. Then I found this paper about the occurrence of Pi in the Mandelbrot set; although I haven't finished reading their paper, but I am sure the same mechanisms can be used to justify the result, that it takes \( \pi \sqrt{\frac{2\eta \cdot n}{e}} \) iterations to "escape" for sexp_{eta+1/n}, and for iterating \( z \mapsto \exp(z)-1+\frac{1}{n} \), it takes \( \pi\sqrt{2n} \) iterations.
http://www.doc.ic.ac.uk/~jb/teaching/jmc...elbrot.pdf
- Sheldon

