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 Constructing the "analytical" formula for tetration. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 01/22/2011, 04:00 AM (This post was last modified: 01/22/2011, 10:11 AM by mike3.) @Gottfried: That's interesting, your matrix seems to be related to the iteration of the decremented exponential. I'm thinking that perhaps that would be a better line of attack here. The regular Schroder function for the fixed point at 0 of a function with such a fixed point, i.e. $f(z) = \sum_{n=1}^{\infty} f_n z^n$ is given by $\chi(z) = \sum_{n=1}^{\infty} \chi_n z^n$ where the coefficients $\chi_n$ satisfy the recurrence $\chi_1 = 1$ $\chi_n = \frac{1}{a_1 - a_1^n} \sum_{m=1}^{n-1} \chi_m {f^{\cdot m}}_n$ where ${f^{\cdot m}}_n$ is the nth coefficient of the mth exponentiation (NOT iteration!) of $f$ (for iteration, it would be ${f^{\circ m}}_n$, and ${f^m}_n$ is too ambiguous here to be used), i.e. ${f^{\cdot m}}_n = \frac{1}{n!} \left[\frac{d^n}{dx^n} f(x)^m\right]_{x = 0}$. For $f(z) = e^{uz} - 1$, i.e. the decremented exponential, we get, using the binomial theorem (here, $\left\{{n \atop m}\right\}$ is a Stirling number of the 2nd kind), ${f^{\cdot m}}_n = u^n \frac{m!}{n!} \left\{{n \atop m}\right\}$. So, the recurrence is $\chi_n = \frac{u^{n-1}}{1 - u^{n-1}} \sum_{m=1}^{n-1} \frac{m!}{n!} \left\{{n \atop m}\right\} \chi_m$, which looks to be much more "linear" than the weird Bell-polynomials recurrence. Could it be possible to come up with an explicit formula for the $\chi_n$? Then the Lagrange inversion formula can be applied to obtain the inverse Schroder function $\chi^{-1}$, which then yields the coefficients of a Fourier series for the superfunction of the decremented exponential, which can be topo-conjugated to the desired $UT$ function (although it may be offset from the original, but at least we would have a formula with explicit, non-recursive coefficients). ADD: I tried some expansions of those $\chi_n$ coefficients. I noticed that the numerators seemed to contain coefficients that look like the second column of your matrices (the -1, -5, -6, -5 thing). Could there be some simpler way to transform that second column into the last column? If so, then we might be able to extract a simpler formula from the explicit formula for $\chi_n$ once it is developed. « 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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