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+--- Thread: 2 fixpoint failure (/showthread.php?tid=534)
2 fixpoint failure - tommy1729 - 11/09/2010
so why isnt there an entire superfunction S(x) of a nontrivial entire function F(x) where F(x) has two fixpoints , and the superfunctions agree on both fixpoints ?
i think i have an idea :
to find a superfunction we usually solve
f ( F(x) ) = a ( f(x) )
where 'a' is a linear function.
that works well for 1 fixpoint because 'a' is lineair and has thus only 1 fixpoint itself.
so
f ( F(x) ) = b ( f(x) )
with 'b' being a function with 2 (distinct) fixpoints is logical for 2 (distinct) fixpoints expansions.
BUT
b^[r] might cycle different from F(x)^[r]
thus b^[r] is entire (in r) IFF F(x) is.
hence no entire solution for this one ...
b must be a moebius function.
and hence f(x) must have poles , thus be at best meromorphic over the complex plane.
but we might consider the equation f ( F(x) ) = 'moebius' ( f(x) ) to be unsatifying :
we dont know how to solve it !!
we can however say that it must have a solution IFF we have the property that there is agreement on both fixpoint expansions.
then again , thats very nonconstructive and maybe not so usefull. ( without a method to solve the equation )
( modified for clarity , i realized it was not so clear )
a penny for your thoughts.
regards
tommy1729
RE: 2 fixpoint failure - tommy1729 - 11/13/2010
i modified a bit