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 Riemann surface equation RB ( f(z) ) = f( p(z) ) tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 05/29/2014, 07:22 PM 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 « Next Oldest | Next Newest »

 Messages In This Thread Riemann surface equation RB ( f(z) ) = f( p(z) ) - by tommy1729 - 05/28/2014, 12:23 PM RE: Riemann surface equation RB ( f(z) ) = f( p(z) ) - by tommy1729 - 05/29/2014, 07:22 PM

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