f(x) = f(g(x))

[tommy1729]

to make a long story short

for f(z) an entire function

it seems that to find at least 1 function g(z) =/= z such that

f(z) = f(g(z))

we can *do* the following :

consider a branch of f^[-1](z) , call it branch A.

the branch below it is branch B.

the domain of branch A is nonempty and connected and the domain of B is also nonempty and connected.

we can map dom A or dom B uniquely and bijectively to a unit circle.

RMT1(dom(A)) = unit circle RMT2(dom(B)) = unit circle.

hence by the Riemann Mapping Theorem (RMT), we can map the dom A to the dom B and visa versa by :

RMT2^[-1] [RMT1(dom(A))] = dom B

RMT1^[-1] [RMT2(dom(B))] = dom A

hence RMT2^[-1] [RMT1(z)] or RMT1^[-1] [RMT2(z)] is an invariant.

these 2 solutions however can equal id(z) when dom(A) = dom(B).

further , these invariants have branches too.

we need to find and express the domains A and B.

and we need to find the riemann mappings.

and even when we have done that , we still dont have a property of the invariant ... we dont even know if its cylic under iteration , if all the other invariants are interations of it , if it is entire , if it has singularities or branches , ...

but i thought id mention that once again the Riemann Mapping Theorem is involved and how it works !!

note :

for instance if one would try to find the period of a function by using this , it wont be that simple :

if the invariant turns out to be x + C you have your period.

but to find x + C you (probably*) first need to know that dom B is just dom A +/- C !!

this is circular !!

( * to construct the Riemann Mappings , at least by this method )

well maybe there is a way around that , but the general case f(z) = f(g(z)) probably doesnt have a simple way around such problems ...

nevertheless , we know g(z) relates to the superfunction , so we(?) are not giving up on this

if i am missing something trivial about this subject , plz inform me.

regards

tommy1729