andydude Wrote:PS. Just a side-note, I understand that (\( \mu \)) and (slog) are different functions, but if you swap them in the above equations, then you get the formula for iterated exponentials, which I found interesting.Actually, the connection is very important. In fact, it's actually the reason that we can solve for \( {}^x \check \eta \). Technically, even if you tetrate \( \eta = e^{1/e} \) an infinite number of times, you can't get above the asymptote. So really, the \( {}^x \check \eta \) function is a limiting case of \( \lim_{\epsilon\to 0^+}{}^{(x+k)} (\eta+\epsilon) \), with k an appropriate factor that I'll define in a later post. I have a rough idea of how to explain it, but I don't want to describe it incorrectly and confuse anyone.
\( {}^{x + slog_b(a)}{b} = \exp_b^{[x]}(a) \)
Anyway, once you start exponentially iterating values smaller than eta, you can never get above their associated asymptotes. However, if you start at positive infinity (which means taking a limit), and use iterated logarithms, you can work your way back down to the higher asymptote. That's why I think your notation actually works a little better, even though we're essentially describing the same thing.
By the way, how much of what I'm explaining in my various posts is already known? Some of it I've seen before, and some of it I've seen in an alternative format that masked the meaning, and some of it seems original, but I can't really be sure. I'm wondering if I'm on to anything publishable. I've never been published, so it would be cool if I were on to something.
~ Jay Daniel Fox
