Riemann surface equation RB ( f(z) ) = f( p(z) ) - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: Riemann surface equation RB ( f(z) ) = f( p(z) ) ( /showthread.php?tid=875) |

Riemann surface equation RB ( f(z) ) = f( p(z) ) - tommy1729 - 05/28/2014
Let f(z) be an analytic function with a riemann surface that has branches. Let RB be shorthand for going from one specific Riemann Branch to another specific one. Then the equation that holds locally or globally : RB ( f(z) ) = f( p(z) ) with p(z) a degree 1 or 2 polynomial. fascinates me. For instance f(2z+1) is another branch of f(z). That is fascinating. But how to solve such a thing ? in particular RB ( f(z) ) = f(z + C) This relates to tetration and dynamics. Not sure if its in the books. Btw do not confuse with a simple invariant : p (f(z)) is another branch of f(z). which is different !! regards tommy1729 RE: Riemann surface equation RB ( f(z) ) = f( p(z) ) - tommy1729 - 05/29/2014
It is tempting to do the following by definition : solve f(z) = g(z) solve p(z) = q(z) solve RB ( f(z) ) = solve f( p(z) ) => RB solve f(z) = solve f( p(z) ) => RB g(z) = q( g(z) ) and then work with that q invariant. Not sure if that is both formal and usefull ... Also there are branch issues perhaps ... g(z) = q ( g(z) ) we might continue like ... so g(z) is the superfunction of T(z) where q(T(z)) is another branch of T(z). In other words T(p(z)) = T(z). And then we get the weird fact that we have a similar equation. SO if Q(z) is a solution then it appears superf(Q^[-1](z)) is also a solution ! When done twice : Q(z) -> superf(Abel(Q^[-1](z))) or when done C times : superf(Abel^[C](Q^[-1](z))). This © reminds me of the " half superfunction " We have talked about before !! Hmm. regards tommy1729 |