03/16/2009, 01:36 PM
(This post was last modified: 03/16/2009, 02:13 PM by sheldonison.)
Gottfried Wrote:....When rereading my article about "pascalmatrix tetrated" (*1) and playing/checking some computations, it came to my attention, that I already have a formula for sexp with fixed (integer) height parameter, but variable with the base. So feeding log(2) as parameter to the series may give 2^^3, and feeding log(3) gives then 3^^3 . In all my recent consideration the base-parameter was deeply calculated in the coefficients and the height-parameter was isolated and thus subject to change; here the height-parameter is hidden in the coefficients and the base-parameter is explicit and can be subject to change.
We need simply the first column of the tetrated pascalmatrix, scale it by reciprocal factorials and use it as coefficients for the powerseries:
\( {b\^\^}^h = \sum_{k=0}^{\infty} \frac{{P\^\^}^h_{k,0}}{k!}\log(b)^k \)
....
Gottfried,
Does your equation work for values of a,p > e^(1/e)?
Assuming it does, my next question would be does your equation need a \( b\^\^(h+\theta(h)) \) term?
Below is my base conversion equation, (which is more or less the same as Jay's earlier results), where n is an integer, so sexp(n) is always well defined and \( \theta(x) \) is a small 1-cyclic sinusoid transfer function, equal to zero for integers. \( \theta(x) \) is important if the desired sexp function is required to have all odd derivatives have positive values for all x>-2, otherwise the 5th order derivatives showed negative values. In my post, I tried to characterize \( \theta(x) \), where one base=e and the other base is a little larger than e^(1/e).
\( \text{sexp}_a(x + \theta(x)) =
\text{ } \lim_{n \to \infty}
\text{log}_a^{\circ n}(\text{sexp}_b (x + \text{slog}_b(\text{sexp}_a(n))) \)
