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