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 Mizugadro, pentation, Book tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 01/16/2015, 12:31 AM Ok I understand now. But I fear I have to tell you your construction is not new. It's just a classical fixpoint method used on an nonstandard function. Your claim that it does not depend on M is thereby true and easy to prove. To give a big hint : suppose an analytic function f has a fixpoint at 0 with f ' (0) = Y > 1. THEN for any real k , as n goes to oo : pent(z) = sexp^[z](x_0) = f^[n] ( Y^(z-k) slog^[n+k](x_0) ). Or said differently for any distinct pair reals k_1,k_2 : f^[n] ( Y^(z-k_1) slog^[n+k_1](x_0) ) = f^[n] ( Y^(z-k_2) slog^[n+k_2](x_0) ). At least if both sides converge. Once you can see that , you will understand. SECOND HINT : plug in koenigs function. *** For those who can still follow , the real question is what if sexp(x-t) = x is a parabolic fixpoint ? Then how do we get GOOD convergeance for sexp(x-t)^[z]. Maybe thats not so hard either , but it seems a logical followup question. *** --- Again for those who still follow , My answer is illuminating but the convergeance speedup is not yet understood. --- I wonder about Eremenko's Conjecture regarding pentation. I rediscovered Eremenko's Conjecture as a kid , guess that explains it. regards tommy1729 « Next Oldest | Next Newest »