• 1 Vote(s) - 5 Average
• 1
• 2
• 3
• 4
• 5
 Pseudoalgebra sheldonison Long Time Fellow Posts: 664 Threads: 23 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 »

 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)