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 An intuitive idea log*^[n](F(exp*^[n](z0))) tommy1729 Ultimate Fellow Posts: 1,605 Threads: 363 Joined: Feb 2009 03/03/2021, 12:57 AM Thinking about my 2sinh method, I came up with an intuitive iteration that might be interesting. let z0 = a0 + b0 i  with e < a0 , 0 =< b0 < 1.  Now define exp*(z) as abs( exp(re(z))*cos(im(z)) ) + abs( exp(re(z))*sin(im(z)) ) i. notice exp*(positive real) = exp(positive real).  Now define log* as the appropriate log branches. Let z_n = z_n(z0) = exp*^[n](z0). Now consider for various functions F ; log*^[n](F(exp*^[n](z0))) Looks alot like the 2sinh method and many others. BUT PERHAPS BETTER BEHAVED ?? Also how fast does this sequence z_n grow ? Is log*^[n](F(exp*^[n](z0))) analytic ? equal to the 2sinh method for F an iteration of 2sinh ?? The idea is to avoid the need for analytic continuations and just keep iterating without issues. Is log*^[n](2sinh(exp*^[n](z0))) = exp(z0) , exp*(z0) or conjugate exp(z0) ?? Another old sketchy idea but worth reconsidering imo regards tommy1729 « Next Oldest | Next Newest »

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