04/02/2009, 10:39 PM
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