• 1 Vote(s) - 5 Average
• 1
• 2
• 3
• 4
• 5
 Pseudoalgebra tommy1729 Ultimate Fellow Posts: 1,438 Threads: 349 Joined: Feb 2009 10/19/2016, 08:47 AM So originally i tried to work from " the inside " like $Exp_b^{[1/2]} ( g(b,x) )$ but from " the outside " like $h ( Exp_b^{[1/2}] (x) )$ we got already the following result. ( i Will omit x sometimes , since it Goes to oo ) For $0 < t < 1 , 0 < e/b < 1$ $Exp_b^[t] = Exp^[t] ^ z$ Now z > 1 must be true. $z = ln Exp_b^[t] / ln Exp^[t]$ Simplify $z = ln(b) Exp_b^{[t-1]} /Exp^{[t-1]}$ Since z > 1 and $Exp_b^{[t-1]} / Exp^{[t-1]} < 1$ we get $1 < z < ln(b)$ and $1/ln(b) < Exp_b^{[t-1]} /Exp^{[t-1]} < 1$. -- Notice for integer n > 0 we get by the above and induction $Exp_b^{[t-1-n]} / Exp^{[t-1-n]}$ ~~ $1/ln(b)$ I assume it holds for n = 0 , that would imply that powers dominate bases for subexponential tetration. In other words Conjecture for p > 1 : $( Exp^{[t]} )^p > Exp_b^{[t]}$ -- However we need much better understanding and approximations. We are not close to answering semiexp_q * semiexp_s ~ semiexp_d ^ R For a given pair (q,s) and a desired best fit (d,R). I considered the base change but without succes. The approximation slog - slog_b ~~ constant is insufficient. See also http://math.stackexchange.com/questions/...ase-a-e1-e Although that might be hard to read. Regards Tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Pseudoalgebra - by tommy1729 - 10/05/2016, 12:21 PM RE: Pseudoalgebra - by tommy1729 - 10/08/2016, 12:22 PM RE: Pseudoalgebra - by tommy1729 - 10/13/2016, 02:32 AM RE: Pseudoalgebra - by tommy1729 - 10/19/2016, 08:47 AM RE: Pseudoalgebra - by sheldonison - 10/23/2016, 09:17 PM

Users browsing this thread: 1 Guest(s)