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 Pseudoalgebra tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 10/05/2016, 12:21 PM (This post was last modified: 10/08/2016, 11:54 AM by tommy1729.) Consider iterations of exponentials of a fixed height. Also called growth. For instance semi-exponentials. In algebra the main thing is Sum and product. Kinda. When we consider asymptotics i call it pseudoalgebra. So for semiexp we get the natural questions such as The best fit ( given by the symbol = ) $Exp_2^{[0.5]}(x) Exp_3^{[0.5]}(x) Exp_5^{[0.5]}(x) = Exp_y^{[0.5]}(x)$ Where y is the value we seek and x > 1. This is - for clarity - an asymptotic equation for bases ( 2,3,5,y). It reminds me of base change and others. How about these ? How to find such identities ? Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 10/08/2016, 12:22 PM (This post was last modified: 10/08/2016, 12:27 PM by tommy1729.) @ means approximation. Lemma $Exp_a^[b] (x) @ Exp^[b] ( @ Ln(a) x)$ From there we get $Exp_q^{[1/2]}(x) Exp_s^{[1/2]}(x) @ Exp_d^{[1/2]}(x)$ Where d is $Exp(d) = @ Ln^{[1/2]}( Exp^{[1/2]}( Ln(q) x) Exp^{[1/2]} ( Ln(s) x) ) / x$ ( notice this can be rewritten with 1 semi-exp and 2 semi-logs too ) But this is not the full story ofcourse. We need proofs. Perhaps consider other ways to handle the issue. And a qualitative understanding of the formula for d such as d ~ (qs)^2 or the alike. I wonder if you would have done it differently ? Also a table would be nice. Still alot of work to do. Regards Tommy1729 The master tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 10/13/2016, 02:32 AM Im afraid the strategy fails. For exp_b^[a] <*> and a < 1 we get <*> @ = exp^[a] ( T x ) Where T Goes to 1 as x grows and for a >= 1 , T Goes to ln(b). Proof sketch S commutes with exp. S(T x) = ln S ( T b^x) / ln(B). = S ln ( T b^x) / ln(B) = S ( ln T + ln B x ) / ln B Maybe. Still thinking ... Regards Tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 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 sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 10/23/2016, 09:17 PM (This post was last modified: 10/23/2016, 09:25 PM by sheldonison.) (10/05/2016, 12:21 PM)tommy1729 Wrote: .... So for semiexp we get the natural questions such as The best fit ( given by the symbol = ) $Exp_2^{[0.5]}(x) Exp_3^{[0.5]}(x) Exp_5^{[0.5]}(x) = Exp_y^{[0.5]}(x)$ Mick's question on mathstack exchange is related to this post. In my answer, I considered $\exp_2^{0.5}(x)\;$ and $\exp_e^{0.5}(x)\;$. See math.stackexchange.com If one uses the analytic solution for the half iterates of base_2 and the half iterate of base_e (ignoring the conjectured nowhere analytic basechange type solutions), the fractional exponentials are not at all well ordered. If a\exp_b^{0.5}(x)\;$ as x grows super exponentially large. There are more details in my post, but if g(x)=0, then $\;\exp_e^{0.5}(x)=\exp_2^{0.5}(x)\;\;$ $g(x) = \text{slog}_e(\text{sexp}_2(x+0.5))-\text{slog}_e(\text{sexp}_2(x))-0.5$ Consider the 2nd peak for base2 occurs near 5.668, where the half iterate base_2 is larger than the half iterate base_e. x2=sloge(sexp2(5.668 +0.5))= 5.03973936018302 xe=sloge(sexp2(5.668 ))+0.5 =5.03945210684265 The question is how much larger is x2 than xe? We can take the logarithm twice of both numbers, and compare sexp_e(x2-2) vs sexp_e(xe-2), and they differ by about +418960.3! This very large difference can be compared to the difference in log(log(2^x)) vs log(log(e^x)) which is always log(log(2)) or -0.3665 - Sheldon « Next Oldest | Next Newest »

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