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 Explicit formula for the tetration to base $$e^{1/e}$$? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 02/12/2015, 10:27 AM Hi. I was wondering about this. I saw this: http://arxiv.org/abs/1105.4735 made by two forum members here. It gives asymptotic formulae for the tetration and "upper super-exponential" to base $e^{1/e]$, also called "$\eta$". In particular, the formula is given for the tetration at base $\eta$ with $F(z) = e \left(1 - \frac{2}{z} \left(1 + \sum_{m=1}^{\infty} \frac{P_m(-\log(z))}{(3z)^m}\right)\right)$ so that $^z \eta$ is just a constant shift of this function. I'm not sure if this is a convergent formula or just an asymptotic, as the paper says asymptotic analysis determines the coefficients. The first few polynomials $P_n$ are given by $P_1(t) = t$ $P_2(t) = t^2 + t + 1/2$ $P_3(t) = t^3 + \frac{5}{2}t^2 + \frac{5}{2}t + \frac{7}{10}$ $P_4(t) = t^4 + \frac{13}{3}t^3 + \frac{45}{6}t^2 + \frac{53}{10}t + \frac{67}{60}$ $P_5(t) = t^5 + \frac{77}{12}t^4 + \frac{101}{6}t^3 + \frac{83}{4}t^2 + \frac{653}{60}t + \frac{2701}{1680}$ Now, is there some kind of explicit (at least a finite summation/product etc. is what I mean) formula for these polynomials? Thanks to Richard Stanley on mathoverflow.net here: http://mathoverflow.net/questions/57627/...lso-how-ca an explicit formula was found for the polynomial coefficients for the regular iteration at the base $e$, in particular if $F(z+1) = \exp(F(z))$ and $F(z) = \sum_{n=0}^{\infty} a_n L^{nz}$ then $a_0 = L$ $a_1 = 1$ and for $n > 1$, $a_n = \frac{P_n(L)}{\prod_{j=2}^{n} j (L^{j-1} - 1)}$ where $P_n(L) = \sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n,j} L^j$ are polynomials, and the coefficients are given explicitly by $\mathrm{mag}_{n, {{n-1} \choose {2}} - j} = \sum_{S} \beta_n(S)$ where $S$ ranges over all subsets of $\{1, 2, \cdots, n-2\}$ such that $\sum_{s \in S} s = j$. And $\beta_n(S) = \sum_{T \subseteq S} (-1)^{|S - T|} \alpha_n(T)$ where $\alpha_n(T)$ is given for the given subset $T$ by ordering its elements in the order $t_1 < t_2 < \cdots < t_k$, as $\alpha_n(T) = \prod_{l=0}^{k-1} S(n - t_l, n - t_{l+1})$. where the $S$-numbers are just the familiar Stirling second-kind numbers and $t_0 = 0$. So is there an explicit formula for the coefficients, or, perhaps more elegantly, the polynomials (e.g. as sums of multiplied simpler polynomials, for example) in the $\eta$ tetration? Also, are these formulas of any interest? I don't know if anyone saw them on the MathOverflow site. I'm the original poster of the question linked there, btw. Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 02/13/2015, 02:26 PM (This post was last modified: 02/13/2015, 02:31 PM by Gottfried.) (02/12/2015, 10:27 AM)mike3 Wrote: Hi. I was wondering about this. I saw this: http://arxiv.org/abs/1105.4735 made by two forum members here. It gives asymptotic formulae for the tetration and "upper super-exponential" to base $e^{1/e]$, also called "$\eta$". In particular, the formula is given for the tetration at base $\eta$ with $F(z) = e \left(1 - \frac{2}{z} \left(1 + \sum_{m=1}^{\infty} \frac{P_m(-\log(z))}{(3z)^m}\right)\right)$ so that $^z \eta$ is just a constant shift of this function. I'm not sure if this is a convergent formula or just an asymptotic, as the paper says asymptotic analysis determines the coefficients. The first few polynomials $P_n$ are given by $P_1(t) = t$ $P_2(t) = t^2 + t + 1/2$ $P_3(t) = t^3 + \frac{5}{2}t^2 + \frac{5}{2}t + \frac{7}{10}$ $P_4(t) = t^4 + \frac{13}{3}t^3 + \frac{45}{6}t^2 + \frac{53}{10}t + \frac{67}{60}$ $P_5(t) = t^5 + \frac{77}{12}t^4 + \frac{101}{6}t^3 + \frac{83}{4}t^2 + \frac{653}{60}t + \frac{2701}{1680}$ Now, is there some kind of explicit (at least a finite summation/product etc. is what I mean) formula for these polynomials?MeToo ... ;-) I second this inquiry; I too was unable to detect the procedere here. Quote:(...) an explicit formula was found for the polynomial coefficients for the regular iteration at the base $e$, in particular if $F(z+1) = \exp(F(z))$ (...) So is there an explicit formula for the coefficients, or, perhaps more elegantly, the polynomials (e.g. as sums of multiplied simpler polynomials, for example) in the $\eta$ tetration? Also, are these formulas of any interest? I don't know if anyone saw them on the MathOverflow site. I'm the original poster of the question linked there, btw. As I commented in your MO-question: without having a more elegant, non-recursive formulation for the coefficients, they are interesting as they occur in the schroeder-function for the function $g_b(x) = \exp_b(x)-1$ (and I find it surprising, that they are present in these two families of functions: $\exp_b(x)$ and $\exp_b(x) -1$ ) . I've a bit additional context for that coefficients which, in my opinion, allows to speculate about some more elegant formula to compute them, but I didn't find such formula yet (and didn't enter this problem deeply because I felt you were not interested in that broader context when I posted my answer) Here is the link to a small treatize where I observed that magic-numbers in the formal powerseries of the Schroeder functions. http://go.helms-net.de/math/tetdocs/APT.htm (I hope I do not spoil your patience/your interest. But if so, then please pardon me, and I can delete this posting G.) For instance, I had the sets [1,5,6,6] and [1,6,5,6] (one of that two sets is in your partial sets of magic numbers) in the matrix A4 and subsequently the according subsets equal to yours in A2,A3,A5 and so on. The "context" here is, that they are at the edges of the Ak-matrices and also the set of the left edge occurs in the schroeder-function and the set of the right edge in the inverse schroeder-function (see at the next part of the article) But if we look at the rows of the Ak-matrices instead of at the columns, then we find simply the corresponding sets of the Stirling numbers 2nd kind (factorially scaled) and that of the first kind, one at the top, one at the bottom. To arrive from one to the other there might be some specific transferoperation through the rows; and in case one could identify such a transferoperation (and this is "elegant") then we have a computation method for your magic numbers because they are at the left and the right borders of each row. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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