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
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
