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tetration limit ??
#1
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
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#2
Hi! Smile
(e^(1/e))^^oo = e
1.6353 "pentate" oo ~= 3.0886 (30 dimensions of matrix) Smile
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#3
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?
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#4
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
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#5
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:


I dont see what useful function f that could be.
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#6
nuninho1980 Wrote:Hi! Smile
(e^(1/e))^^oo = e
1.6353 "pentate" oo ~= 3.0886 (30 terms) Smile

Hello!
How do you tetrate 1.6353^^1.6353?
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#7
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:


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
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#8
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. Wink
my cpu c2d e6600 is slow. but the future the gpu (CUDA Big Grin) will help to the cpu for calculate very fast. my gpu geforce 9800gt maybe will support! Wink
if you don't know CUDA then you see http://www.nvidia.com/object/cuda_what_is.html. Smile

note: sorry for bad english. Wink
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#9
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. Wink

No problem, as long as I understand you Smile
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#10
bo198214 Wrote:Which code did you use?

"tetration and slog" original by Andrew Robbins is as smoother as "new regular slog". Wink
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