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+--- Thread: tetration limit ?? (/showthread.php?tid=262)
tetration limit ?? - tommy1729 - 04/01/2009
maybe this is the dumbest tetration question ever , but here goes.
i was thinking about a " tetration limit ".
as you all know , im particular intrested in half iterates of exponential functions.
we all know
lim n -> oo ( 1 + f(n) ) ^ n = e
if f(n) = 1/n
now this is the typical exponential limit.
but what is the typical half iterate exponential limit ?
lim n -> oo F [( 1 + f(n) ) ; n ] = Q = ???
where F stands for the half iterate exponential of base ( 1 + f(n) )
and f(n) and Q are the unknown function and unknown value.
regards
tommy1729
RE: tetration limit ?? - nuninho1980 - 04/01/2009
Hi!
(e^(1/e))^^oo = e
1.6353 "pentate" oo ~= 3.0886 (30 dimensions of matrix)
RE: tetration limit ?? - bo198214 - 04/02/2009
tommy1729 Wrote:lim n -> oo F [( 1 + f(n) ) ; n ] = Q = ???
where F stands for the half iterate exponential of base ( 1 + f(n) )
and what is "; n"? Power, iteration, or, or, or?
RE: tetration limit ?? - tommy1729 - 04/02/2009
bo198214 Wrote:tommy1729 Wrote:lim n -> oo F [( 1 + f(n) ) ; n ] = Q = ???
where F stands for the half iterate exponential of base ( 1 + f(n) )
and what is "; n"? Power, iteration, or, or, or?
n is an integer
F [( 1 + f(n) ) ; F [( 1 + f(n) ) ; n ] ] = (1 + f(n) ) ^ n
F has two arguments seperated by " ; " and has the form
F[ base ; z ] and is the half iterate of base ^ z.
clear ?
regards
tommy1729
RE: tetration limit ?? - bo198214 - 04/02/2009
tommy1729 Wrote:n is an integer
F [( 1 + f(n) ) ; F [( 1 + f(n) ) ; n ] ] = (1 + f(n) ) ^ n
F has two arguments seperated by " ; " and has the form
F[ base ; z ] and is the half iterate of base ^ z.
Perhaps then you should start with the simpler case of the double iterate. And look what a suitable function f you would find that:
\( \lim_{n\to\infty} (1+f(n))^{(1+f(n))^{n}} = Q \)
I dont see what useful function f that could be.
RE: tetration limit ?? - bo198214 - 04/02/2009
nuninho1980 Wrote:Hi!
(e^(1/e))^^oo = e
1.6353 "pentate" oo ~= 3.0886 (30 terms)
Hello!
How do you tetrate 1.6353^^1.6353?
RE: tetration limit ?? - tommy1729 - 04/02/2009
bo198214 Wrote:tommy1729 Wrote:n is an integer
F [( 1 + f(n) ) ; F [( 1 + f(n) ) ; n ] ] = (1 + f(n) ) ^ n
F has two arguments seperated by " ; " and has the form
F[ base ; z ] and is the half iterate of base ^ z.
Perhaps then you should start with the simpler case of the double iterate. And look what a suitable function f you would find that:
\( \lim_{n\to\infty} (1+f(n))^{(1+f(n))^{n}} = Q \)
I dont see what useful function f that could be.
first of all , i have to comment that i find you are changing subject a little bit.
what you call a simpler case might be a harder case or an unrelated case.
no offense.
( i think your limit like question is to " sensitive ". i wont explain that )
but i will try to give it a go anyways ...
(1 + f(n)) ^ (1 + f(n)) ^ n = Q
i try to use the identity lim (1 + 1 / g(n)) ^ g(n) = e
( for many g(n) )
so : 1 + 1/ ((1 + f(n)) ^n) = 1 + f(n)
so f(n) = (1 + f(n)) ^ - n
which leads to
f(z) = ( 1 + f(z) ) ^ - z
and Q = e
... maybe ...
however note that ( 1 + 1/n + 1/n^3 ) ^ n = e too , so some flexibility can be added to the equation for f(z) ...
at first sight id estimate f(z) around a / log(z) + b/ log(z)^c for some reals a, b and c.
another method might be taking the log of both sides getting an expression for log(Q) and then using l'hospital.
maybe that is more succesfull.
maybe both give a working result but different !?!
this might be bo's objection somewhat hidden.
but i dont think such an issue occurs in my OP.
( informally : in general limits with double exponential speed tends to converge to a finite numbers less then limits of slower functions )
regards
tommy1729
RE: tetration limit ?? - nuninho1980 - 04/03/2009
bo198214 Wrote:Hello!
How do you tetrate 1.6353^^1.6353?
my pc evaluated it, running "maple 12" w/ code v1.1 about "tetration and slog".
x^^y = z
if the "z" isn't to close the point fixed (may to infinity) then you reduce any case decimal "x" and/or "y"
but it's bit difficult to close the point fixed (3.08~3.10).
the tetration is slower than slog. but I will try more smooth for calcuate slog.
my cpu c2d e6600 is slow. but the future the gpu (CUDA
) will help to the cpu for calculate very fast. my gpu geforce 9800gt maybe will support! if you don't know CUDA then you see http://www.nvidia.com/object/cuda_what_is.html.
note: sorry for bad english.
RE: tetration limit ?? - bo198214 - 04/03/2009
nuninho1980 Wrote:bo198214 Wrote:Hello!
How do you tetrate 1.6353^^1.6353?
my pc evaluated it, running "maple 12" w/ code v1.1 about "tetration and slog".
Which code did you use?
Quote:
note: sorry for bad english.
No problem, as long as I understand you
RE: tetration limit ?? - nuninho1980 - 04/03/2009
bo198214 Wrote:Which code did you use?
"tetration and slog" original by Andrew Robbins is as smoother as "new regular slog".