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Constructing the "analytical" formula for tetration.
#3
@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.



is given by



where the coefficients satisfy the recurrence




where is the nth coefficient of the mth exponentiation (NOT iteration!) of (for iteration, it would be , and is too ambiguous here to be used), i.e. . For , i.e. the decremented exponential, we get, using the binomial theorem (here, is a Stirling number of the 2nd kind),

.

So, the recurrence is

,

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 ? Then the Lagrange inversion formula can be applied to obtain the inverse Schroder function , which then yields the coefficients of a Fourier series for the superfunction of the decremented exponential, which can be topo-conjugated to the desired 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 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 once it is developed.
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RE: Constructing the "analytical" formula for tetration. - by mike3 - 01/22/2011, 04:00 AM

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