# Tetration Forum

Full Version: tetration limit ??
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2 3 4 5
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
Hi!
(e^(1/e))^^oo = e
1.6353 "pentate" oo ~= 3.0886 (30 dimensions of matrix)
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?
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
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.
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?
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

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
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.

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:

No problem, as long as I understand you
bo198214 Wrote:Which code did you use?

"tetration and slog" original by Andrew Robbins is as smoother as "new regular slog".
Pages: 1 2 3 4 5