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+--- Thread: Regular "pentation"? (/showthread.php?tid=446)
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RE: Regular "pentation"? - mike3 - 05/07/2010
(05/07/2010, 03:17 AM)andydude Wrote: I think I finally understand what you're talking about here. There is an interval that plays the same role in pentation as \( (e^{-e}, e^{1/e}) \) plays in tetration. I talked about the upper bound of this interval here, but whether or not there is a lower bound to this interval is unknown.
Yes, without an extension to real/complex heights for \( b \le e^{-e} \) for tetration, determining the lower bound (if any) does not seem possible.
pentation? (Based on "polynomial" tetration) - Gottfried - 03/29/2011
[update:] after I've written the following I got aware, that this should be better threaded under something like "polynomial tetration/pentation" since the basic powerseries for the tetration as I used it here was derived from the "polynomial tetration" using the diagonalization of the truncated real (square) Bellmatrix and its diagonalization - *not* of the triangular Bellmatrix of the regular tetration
While usually I handle the tetration having a series which has the height-parameter in the exponents of its single terms, I played a bit around with a powerseries-representation, and especially a powerseries-epansion for b^^h for b=4 and beginning at x=0 (instead of x=1 as usual). So that powerseries gives b^^0 = 0, b^^1=1,b^^2=4 and so on. Obviously that powerseries has no constant term. It begins like
PHP Code:
b^^h= 0.938409102074*h - 0.167927302891*h^2 + 0.287554406742*h^3 - 0.151567019585*h^4 +...
Having a powerseries without constant term cries for iteration, so this allows to implement pentation to base 4 as an iteration of the current powerseries. Moreover, I have then a two-parametric formula (as usually in x and h in exp_b°h(x) if base b is fixed).
If I set x=1 and p="pentation-height"=1 I get the tetration beginning at 0 with height 1 (which gives 1) and this "iterated 1 time". If I iterate this 2 times, I get the following powerseries
PHP Code:
((b^^h)^^^2 = exp_4°h(0))^^^2 = 0.880611642856*h - 0.305463247599*h^2 + 0.560396635817*h^3 - 0.482701855960*h^4 +...
Hmm. Again, if I tetrate 0 to the height 1/2 I get
PHP Code:
b^^0.5 = 0.457214343478
If I "do it two times" (???), p=2 I arrive at
PHP Code:
(b^^0.5)^^^2 = 0.417108029167
PHP Code:
(b^^0.5)^^^0.5 = 0.478298158357
Does someone have a pentation-implementation which could provide comparable/concurring values? Well, you may see: I'm not even feeling that I understood what that computation (and the change of that parameters) tells me conceptually at all...Gottfried
RE: pentation? (Based on "polynomial" tetration) - BenStandeven - 04/04/2011
(03/29/2011, 09:50 AM)Gottfried Wrote: [update:] after I've written the following I got aware, that this should be better threaded under something like "polynomial tetration/pentation" since the basic powerseries for the tetration as I used it here was derived from the "polynomial tetration" using the diagonalization of the truncated real (square) Bellmatrix and its diagonalization - *not* of the triangular Bellmatrix of the regular tetration
While usually I handle the tetration having a series which has the height-parameter in the exponents of its single terms, I played a bit around with a powerseries-representation, and especially a powerseries-epansion for b^^h for b=4 and beginning at x=0 (instead of x=1 as usual). So that powerseries gives b^^0 = 0, b^^1=1,b^^2=4 and so on. Obviously that powerseries has no constant term. It begins likeand for instance for b^^0.5 = 0.457214343478 .PHP Code:b^^h= 0.938409102074*h - 0.167927302891*h^2 + 0.287554406742*h^3 - 0.151567019585*h^4 +...
That isn't going to give the standard version of pentation, then. You're iterating b^^(x-1) instead of b^^x.