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2 fixpoint failure
#1
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
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#2
i modified a bit
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