(04/02/2009, 09:56 PM)bo198214 Wrote: 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.

we reduce to (1+1/f(n))^((1+1/f(n))^n) = Q

in essense we only need to understand the relation between n and f(n).

further switch f(n) and n to get

(1+1/n)^((1+1/n)^f(n)) = Q

ln(1+1/n) * (1+1/n)^f(n) = ln(Q)

replace ln(Q) by Q

f(n) = ln(Q/ln(1+1/n)) / ln(1+1/n)

done.

but lim n-> oo sexp_(1+f(n))[slog_(1+f(n))[n] + 1/2] = n + C

0 < C

seems harder and not so related at first.

worse , it might have problems stating it like above ... because our n needs to be after the second fixpoint and our superfuntions need to be defined at their second fixpoint ... which " evaporates " at oo as n goes to oo.

and hence our superfunctions become valid and defined > q_n where lim q_n = oo !!

if f(n) does not grow to fast this might be ok , but on the other hand to arrive at C at our RHS f(n) seems to need some fast growing rate.

so f(n) is strongly restricted and C must be unique and existance is just assumed.

i do not know anything efficient to compute f(n) apart from numerical *curve-fitting* upper and lower bounds as described above.

or another example , actually the original OP rewritten :

lim n-> oo sexp_(1+f(n))[slog_(1+f(n))[n] + 1/2] = C

0 < C

now we must take the first fixpoint approaching 1 .. or the second ??

it seems easiest to take the first fixpoint , if we take the second we have the same problem of the " evaporating ' fixpoint as above.

on the other hand , we dont know the radiuses for bases 1+f(n) expanded at their first or second fixpoint.

again , its hard to find f(n) and C despite they are probably strongly restriced - even unique -.

another idea that might make sense is that there exists a function g(n) such that

but lim n-> oo sexp_(1+f(n))[slog_(1+f(n))[n] + 1/2] = g(n)

0 < g(n) < n

and that g(n) gets closer and closer towards the end of the radius of one of the fixpoint expansions as n grows.

and that might be inconsistant with the other equations/ideas above.

so many questions.

regards

tommy1729