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 Constructing the "analytical" formula for tetration. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 01/17/2011, 01:05 PM (This post was last modified: 01/19/2011, 03:15 AM by mike3.) Hi. I was wondering about the problem of finding "analytical" formulas for tetration, that is, formulas that are more like "formulas" than "procedures" to compute tetration. This includes, e.g. infinite sums and other things like that. A "closed form" formula, that is, one in terms of a finite number of "conventional" functions, is most likely not possible. Now, it seems the Riemann mappings so far are hugely resistant to attempts to describe them, so I was curious on concentrating on the simpler problem of describing the regular iteration, which is what the Riemann mappings would deform. The regular tetrational, or "unipolar superfunction" of $\exp$, can be given by $UT(z) = L + \sum_{n=1}^{\infty} a_n L^{nz}$ with recursively-generated coefficients $a_1 = 1$ $a_n = \frac{B_n(1, 2! a_2, 3! a_3, \cdots, (n-1)! a_{n-1}, 0)}{n! (L^{n-1} - 1)}$. with $B_n$ the "complete" Bell polynomials. The above series is a Fourier series, but can be rearranged into a Taylor series if we so please: \begin{align} UT(z) &= L + \sum_{n=1}^{\infty} \left(a_n \sum_{k=0}^{\infty} \frac{(Ln)^k}{k!} z^k\right)\\ &= L + \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{a_n (Ln)^k}{k!} z^k\\ &= L + \sum_{k=0}^{\infty} \left(\sum_{n=1}^{\infty} a_n (Ln)^k\right) \frac{z^k}{k!} \end{align} The problem, then, is the description of the coefficients $a_n$. The first few are: $a_1 = 1$ $a_2 = \frac{1}{2L - 2}$ $a_3 = \frac{L + 2}{6L^3 - 6L^2 - 6L + 6}$ $a_4 = \frac{L^3 + 5L^2 + 6L + 6}{24L^6 - 24L^5 - 24L^4 + 24L^2 + 24L - 24}$ $a_5 = \frac{L^6 + 9L^5 + 24L^4 + 40L^3 + 46L^2 + 36L + 24}{120L^10 - 120L^9 - 120L^8 + 240L^5 - 120L^2 - 120L + 120}$ ... By pattern recognition, it appears that $a_n = \frac{\sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n,j} L^j}{\prod_{j=2}^{n} j(L^{j-1} - 1)}$. The denominator can be given in true closed form via the "q-Pochhammer symbol" as $a_n = \frac{\sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n,j} L^j}{n! (-1)^{n-1} (L; L)_{n-1}}$. (The appearance of q-analogs is curious. I'm wondering about the connections between q-analogs, tetration, iteration, fractals, chaos, and continuum sums...) The problem, however, is the sequence of "magic" numbers $\mathrm{mag}_{n, j}$ for the numerators. For j starting at 0, we get n = 1: 1 n = 2: 1 n = 3: 2, 1 n = 4: 6, 6, 5, 1 n = 5: 24, 36, 46, 40, 24, 9, 1 n = 6: 120, 240, 390, 480, 514, 416, 301, 160, 64, 14, 1 n = 7: 720, 1800, 3480, 5250, 7028, 8056, 8252, 7426, 5979, 4208, 2542, 1295, 504, 139, 20, 1 n = 8: 5040, 15120, 33600, 58800, 91014, 124250, 155994, 177220, 186810, 181076, 163149, 134665, 102745, 71070, 44605, 24550, 11712, 4543, 1344, 265, 27, 1 n = 9: 40320, 141120, 352800, 695520, 1204056, 1855728, 2640832, 3473156, 4277156, 4942428, 5395818, 5561296, 5433412, 5021790, 4391304, 3625896, 2820686, 2056845, 1398299, 879339, 504762, 260613, 117748, 45178, 13845, 3156, 461, 35, 1 n = 10: 362880, 1451520, 4021920, 8769600, 16664760, 28264320, 44216040, 64324680, 88189476, 114342744, 141184014, 166279080, 187614312, 202901634, 210825718, 210403826, 201934358, 186191430, 164980407, 140216446, 114231817, 88934355, 66047166, 46576620, 31071602, 19460271, 11365652, 6112650, 2987358, 1298181, 488878, 153094, 37692, 6705, 749, 44, 1 ... The first column appears to be factorials, i.e. $\mathrm{mag}_{n, 0} = (n-1)!$, the last is of course $\mathrm{mag}_{n, \frac{(n-1)(n-2)}{2}} = 1$, and the second-to-last looks to be $\mathrm{mag}_{n, \frac{(n-1)(n-2)}{2} - 1} = \frac{(n-2)(n+1)}{2}$ but beyond that I'm not sure. Is there some way to actually prove or disprove these results from the recurrence formula, and even better, to get a "real" (perhaps a sum/product) formula for the coefficients $\mathrm{mag}_{n,j}$? What I'm really after here is some way to analyze that recurrence formula. How would one go about approaching such a problem? Also, $\sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n, j} = \frac{n!(n-1)!}{2^{n-1}} = \mathrm{Polygorial}(n, 3)$ for all its worth (I just plugged 1, 3, 18, 180, 2700 into the oeis.org dictionary.). For all that's worth. I don't know how you could decompose that in a "useful" manner. But again, no proofs, just guesses. « Next Oldest | Next Newest »

 Messages In This Thread Constructing the "analytical" formula for tetration. - by mike3 - 01/17/2011, 01:05 PM RE: Constructing the "analytical" formula for tetration. - by Gottfried - 01/17/2011, 10:10 PM RE: Constructing the "analytical" formula for tetration. - by mike3 - 01/22/2011, 04:00 AM RE: Constructing the "analytical" formula for tetration. - by Gottfried - 01/24/2011, 08:56 AM RE: Constructing the "analytical" formula for tetration. - by mike3 - 01/28/2011, 03:12 AM RE: Constructing the "analytical" formula for tetration. - by Gottfried - 01/28/2011, 02:49 PM RE: Constructing the "analytical" formula for tetration. - by mike3 - 01/28/2011, 09:12 PM RE: Constructing the "analytical" formula for tetration. - by Gottfried - 01/28/2011, 10:42 PM RE: Constructing the "analytical" formula for tetration. - by mike3 - 01/29/2011, 12:10 AM RE: Constructing the "analytical" formula for tetration. - by mike3 - 02/10/2011, 04:20 AM RE: Constructing the "analytical" formula for tetration. - by sheldonison - 02/10/2011, 05:59 AM RE: Constructing the "analytical" formula for tetration. - by mike3 - 02/10/2011, 07:35 AM RE: Constructing the "analytical" formula for tetration. - by tommy1729 - 01/23/2011, 10:59 PM RE: Constructing the "analytical" formula for tetration. - by mike3 - 01/24/2011, 04:34 AM

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