Coefficients of Tetrational Function mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 03/28/2010, 03:25 AM Hi. I'm curious: Is there any ideas about how a closed form or even a series formula, recurrence, etc. could be found for the Taylor coefficients of the tetrational function at 0 developed from, e.g. the Abel iteration, or the Cauchy integral (which seem to give the same thing though it's not proven), to base e? I.e. $\mathrm{tet}(z) = \sum_{n=0}^{\infty} a_n z^n$ and what is $a_n$? We have $a_0 = 1$ and then $a_1 = 1.0917...$, etc. but is there any way that coefficient $a_1$ for example could be expressed in terms of elementary or "well-known" (used by the general maths. community, that is -- e.g. gamma, polylog, zeta, erf, etc.) non-elementary special functions and known mathematical constants (like e, pi, eulergamma, etc.)? Even if it can only be done with an infinite expansion like an infinite sum or something involving those? Or is this function simply so darned exotic that it will utterly refuse and defy any and all attempts to try to relate it to known mathematical functions? Ztolk Junior Fellow Posts: 21 Threads: 4 Joined: Mar 2010 04/02/2010, 02:49 PM Just to be clear, you're looking for the Taylor expansion of the infinite-order tetration? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 04/02/2010, 03:02 PM (This post was last modified: 04/02/2010, 03:04 PM by Base-Acid Tetration.) (04/02/2010, 02:49 PM)Ztolk Wrote: Just to be clear, you're looking for the Taylor expansion of the infinite-order tetration? no, not $f(z) = {}^\infty z$, but he means $f(z) = {}^z e$. andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 04/28/2010, 09:11 PM (This post was last modified: 04/29/2010, 05:04 AM by andydude.) [update] BTW, I'm not talking about tetration, I'm talking about the superlogarithm here. The equivalent coefficient of tetration at a similar point would be $\frac{1}{a_1}$, so keep that in mind while reading this. [/update] (03/28/2010, 03:25 AM)mike3 Wrote: is there any way that coefficient $a_1$ for example could be expressed in terms of elementary or "well-known" ... special functions?No one knows. I would certainly like to find out. Proving that would amount to showing that $a_1$ is both irrational and transcendental. (03/28/2010, 03:25 AM)mike3 Wrote: Even if it can only be done with an infinite expansion like an infinite sum or something involving those? Or is this function simply so darned exotic that it will utterly refuse and defy any and all attempts to try to relate it to known mathematical functions? Yes, certainly. I can tell you exactly what $a_1$ is in complicated form (according to natural/intuitive tetration): $a_1 = \lim_{n\to\infty}\frac{d_{n1}}{d_{n0}}$ where $d_{nk} = \det(\mathbb{A}_x[e^x]_{nk})$ and $\mathbb{A}_x[e^x]_{nk}$ is the kth Cramer's rule matrix associated with the n*n truncated Abel matrix of the function $(x \mapsto e^x)$. Now think about this. Determinants (det) are just a bunch of additions and multiplcations. So this would involve infinite sums and infinite products in the limit as n goes to infinity. But each and every step is a well known function, so this is kind of what you are describing. However, this form is not a form that lends itself to standard analysis, and in that sense it is "so darned exotic" as you put it. One of the many things I have been trying to do is put this into a form that does lend itself to standard analysis, but so far I have been less than successful. One thing that I have been successful at is turning the numerator and denominator of $a_1$ into a single infinite summation (whose coefficients I have found expressions for (n < 6)), but since they don't work for (n=7) I gave up. If you are interested in these coefficients, I uploaded them awhile ago here (they are under Robbins_draft.zip in the file Robbins_detcoeff.pdf). I also found a much better way to compute the 4,21,160 sequence, which I submitted to OEIS (and yes I'm credited). So my hope for this research was to find coefficients such that (for all n): $\frac{d_{n1}}{d_{n0}} = \frac{\sum_{k=0}^{binom(n,2)} C_{k1n}\ln(b)^k}{\sum_{k=0}^{binom(n,2)} C_{k0n}\ln(b)^k}$ and I continue to hope until I find it is impossible. I still think that this approach is the most likely to succeed in finding a closed form. Andrew Robbins « Next Oldest | Next Newest »

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