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 slog_b(sexp_b(z)) How does it look like ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/23/2014, 09:15 PM Forgive me if this is an old question or it could be easily derived from previous posts. Or if my memory fails me. But despite many posts on slog_b(z) and sexp_b(z) , I am somewhat puzzled by slog_b(sexp_b(z)). For clarity b is the base and in particular I am (mainly) intrested in bases larger than exp(1/2). Also to avoid confusion slog_b(sexp_b(z)) =/= z. This is the tetration analogue of log_b(exp_b(z)) =/= z. As often in the topic of tetration I feel " lured into deception ". By that I mean that many ideas pop up naturally but none of them seem very solid , intresting or true in retrospect. Also many variations on this question occur naturally , usually in the form of special cases and restrictions. Such as Re(z) > 0. Looking at changing values of b , rather than say z. Or considering different solutions to sexp and how that afffects things. A few notes : 1) analytic continuation is problematic since for REAL z , the equation does in fact hold ! ( and continuation of z = z of course ). This also makes me distrust Taylor series and related calculus. 2) my 2sinh fails here or is meaningless since it is not analytic ! Well at least I think so , because it is then imho not well defined for nonreal imput (z). 3) the chaotic nature of interations z , exp(z) , exp^[2](z) , ... is one of the reason I doubt my own intuition and I feel " lured into deception ". 4) I could probably link or rewrite almost all threads , questions and answers as a note here , since I find that most threads relate to this in one way or another ! 5) One of the " deceptions " was a " nonentire analytic function with all singularies at oo " what makes no sense of course. I however have no systematic way of doing things that automatically avoids such " nonsense ". Maybe Im not the only one who wonders about this. Some of my friends believe this requires new special functions to fully understand. I considered using higher dimensional numbers but that seems like running before you can walk. So I guess the complex plane is the place to start. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/23/2014, 09:15 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/23/2014, 10:46 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/23/2014, 11:36 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/24/2014, 09:50 AM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/24/2014, 10:30 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/25/2014, 04:54 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/25/2014, 10:48 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/26/2014, 08:53 AM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/27/2014, 04:52 AM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/27/2014, 03:27 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/28/2014, 03:12 AM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/24/2014, 10:33 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/24/2014, 10:43 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/27/2014, 03:39 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/28/2014, 09:01 PM RE: slog_b(sexp_b(z)) How does it look like ? - by sheldonison - 04/28/2014, 11:33 PM RE: slog_b(sexp_b(z)) How does it look like ? - by tommy1729 - 04/29/2014, 11:52 AM

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