(05/30/2014, 04:10 PM)JmsNxn Wrote: Hi, everyone. This is a continuation of my last thread http://math.eretrandre.org/tetrationforu...hp?tid=847
I don't have the time to explain too much now, but I realize a mistake I made and it causes for an inaccurate result. Take \( 0<\sigma <1 \)
\( \frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \G(s) \frac{w^{-s}}{(^{-s} e)}\,ds = \vartheta(-w) \)
Quite remarkably:
\( \vartheta(-w) = \sum_{n=0}^\infty \frac{(-w)^n}{(^n e)n!} + k(w) \)
where I've recently calculated that:
\( k(w) = \lim_{n\to\infty} \frac{w^n}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i \infty} \G(s-n) \frac{w^{-s}}{(^{n-s} e)}\,ds \)
Where I have been unable to calculate if this converges to a limit. If it does, (and I think it does), we are good.
I'm not sure if I can show recursion with this new transform but I'll try my best to work on this.
We recall the important property, of recovering tetration: \( [\frac{d^z}{dw^z} \vartheta(w)]_{w=0} = \frac{1}{^z e} \), for \( \Re(z) > -1 \)
Two issues I see here:
1. The integral for \( k(w) \) involves a complex tetration. Yet, we don't have complex tetration in the first place if we are trying to construct it -- so how does this work?
2. Given the chaotic behavior of tetration I've mentioned, I don't think this integral converges. The following is a graph of the reciprocal Kneser tetration \( \frac{1}{^z e} \) on the complex plane:
The brightness gives the magnitude (brighter = bigger), and the hue the phase. The lines show effective (i.e. when shifted by the \( n - s \) in the tower) integration paths for a \( \sigma \) value near \( \frac{1}{2} \) and increasing \( n \). Grey areas are areas where the function could not be computed due to arithmetic overflow. It will have both very large and very small values in those areas. (When I say "very large" I mean HUGE -- the integer parts of the magnitudes of most of these numbers are so big that all the matter in the Universe could not give anywhere even close to enough stuff to write down their digits.)
I didn't include the factor of \( \Gamma \) or the \( w^{-z} \) factor in the plot, because these things don't decay strongly enough to suppress the monster growth of reciprocal tetration in the complex plane (also, to include the \( \Gamma \) factor in this specific formula would require the creation of a separate graph for each \( n \), and I don't want to clog this post up with graphs!). The areas of huge values will remain regardless, so this should still be a useful illustration of the problem.
I'd be extremely surprised if an integral through all that mess was well-behaved enough to have some kind of limit.
This seems to be a common problem plaguing all of these tetration integral formulas. I think any such formula, with nothing to cancel out the rapid growth, that takes a limit of integrals in the right half-plane with integration paths extending further and further right and crossing that region, or which relies on such a limit, is doomed to failure.
I did, however, have an idea for a possible solution: what about trying to use your fractional calculus not for the tetration function, but for the super-logarithm? The principal branch of the super-logarithm \( \mathrm{slog}(z) \) is defined with
\( \mathrm{slog}(1) = 0 \)
\( \mathrm{slog}(e^z) = \mathrm{slog}(z) + 1 \)
\( \mathrm{slog}(z) \) is holomorphic for all \( \Re(z) > \Re(L) \), where \( L = -W_0(-1) \) is a principal fixed point of the logarithm.
Henryk Trappmann showed that a function satisfying these conditions (actually, the range of holomorphism need only be the "sickle" between the two principal fixed points of the logarithm, but the above is simpler to state) is unique. Kneser gave the construction which shows it exists.
This function is very well-behaved in the region given, compared to the tetration. It looks to be asymptotically exponentially bounded -- heck to be asymptotically bounded by a linear function there. That is,
\( |\mathrm{slog(z)}| < K|z| \) for some \( K > 0 \) and \( \Re(z) > R > \Re(L) \) for any \( R \).
The only rub is that the non-extended version (i.e. only what you can derive from the first two criteria above) is not defined at every positive integer, rather it is defined at every power tower \( 1 \), \( e \), \( e^e \), \( e^{e^e} \). ... and takes on integer values at each of those. I'm not sure how that impacts the use of your methods.
With the super-logarithm in hand, you then have \( \mathrm{slog}^{-1}(z) =\ ^z e \) (functional inverse)
