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+--- 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