• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 slog_b(sexp_b(z)) How does it look like ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/27/2014, 03:39 PM I came to reconsider an old idea ; Use more than one index for the branches. In its simplest general case something similar to : $f^{-1}(f^{1}(z)) = z + f_n(z) + g_m(z) + h_{nm}(z)$ Where m,n are integers that depend on z. Apart from maybe double periodic functions Im not sure when this is useful ... 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

Users browsing this thread: 1 Guest(s)