10/06/2022, 12:57 AM
Consider a general entire function f(z) with a repelling fixpoint A and an attracting fixpoint B such that :
1) there is at least one path from A to B by iterations of f(z).
2) the other fixpoints are relatively far away from A and B and the path(s) , or they do not exist.
3) B is actually 0.
4) 0 < norm(f ' (B)) < norm(1/c) < 1
**
Then we can solve the schröder equation
sch( f(z) ) = (1/c) sch(z)
by
sch(z) = sum_k c^k * f^[k](z)
where sum_k means summing over all integer k.
Notice this is not , or not necc , the koenings function ;
I mean not the koenings function developped at point A or point B.
Afterall we know the koenigs does not agree on fixpoints and here we have 2.
On the other hand we only have a part of the neighbourhood of the fixpoints and not the fixpoints themselves agreeing.
But a koenings function should work all around the ( complete ) neighbourhood of its expansion fixpoint.
SO agreeing on 2 fixpoints * and their neighbourhoods * is problematic.
proof is trivial ;
sch(z) /c = sum_k c^(k-1) * f^[k](z) = sum_k c^k * f^[k+1](z) = sch(f(z))
Now the main question :
Can we extend by summability methods and analytic continuation , in a logical and consistant way , the function sch(z) beyond the path(s) from A to B ?
***
Note that the associated superfunction has the same period as c^z , although there are branches etc.
***
Now I noticed that if we have good methods for summability and analytic continuation for double exp functions than
if f(z) is a polynomial ,
We can probably solve this problem for those functions.
***
the idea is that we end up with a superfunction T(z) such that T( - oo + z ) = A AND T( + oo + z) = B = 0.
and a potential 3rd fixpoint lies towards T(+ oo i ) and/or T ( - oo i )
.
Since the koenigs satisfies the semi-group homom near its fixpoint expansion points , this function cannot do the same near its fixpoints ,
so in THAT region it does not satisfy the addition semi-group homom ... but maybe elsewhere it does ...
... unless it is a rule that summability and ana continuation forbid the local creation of the addition semi-group homom ??
Another question is if this ( locally AT LEAST ?? ) agrees with the carleman matrix method expanded somewhere ??
In particular an expansion point on such a path from A to B ? OR maybe rather NOT on that path ?
I considered connections to the gaussian method and the fibonacci numbers , and somewhat green's theorem and convolution ideas but not with a clear result or understanding ...
( i concluded that the infinitesimal ideas i recently had reduce to nothing usefull , consistant or new ideas , only to arrive at old ideas. So that is why im silent about them now )
I could say or speculate more but this will do for now.
What are your ideas ?
Regards
tommy1729
1) there is at least one path from A to B by iterations of f(z).
2) the other fixpoints are relatively far away from A and B and the path(s) , or they do not exist.
3) B is actually 0.
4) 0 < norm(f ' (B)) < norm(1/c) < 1
**
Then we can solve the schröder equation
sch( f(z) ) = (1/c) sch(z)
by
sch(z) = sum_k c^k * f^[k](z)
where sum_k means summing over all integer k.
Notice this is not , or not necc , the koenings function ;
I mean not the koenings function developped at point A or point B.
Afterall we know the koenigs does not agree on fixpoints and here we have 2.
On the other hand we only have a part of the neighbourhood of the fixpoints and not the fixpoints themselves agreeing.
But a koenings function should work all around the ( complete ) neighbourhood of its expansion fixpoint.
SO agreeing on 2 fixpoints * and their neighbourhoods * is problematic.
proof is trivial ;
sch(z) /c = sum_k c^(k-1) * f^[k](z) = sum_k c^k * f^[k+1](z) = sch(f(z))
Now the main question :
Can we extend by summability methods and analytic continuation , in a logical and consistant way , the function sch(z) beyond the path(s) from A to B ?
***
Note that the associated superfunction has the same period as c^z , although there are branches etc.
***
Now I noticed that if we have good methods for summability and analytic continuation for double exp functions than
if f(z) is a polynomial ,
We can probably solve this problem for those functions.
***
the idea is that we end up with a superfunction T(z) such that T( - oo + z ) = A AND T( + oo + z) = B = 0.
and a potential 3rd fixpoint lies towards T(+ oo i ) and/or T ( - oo i )
.
Since the koenigs satisfies the semi-group homom near its fixpoint expansion points , this function cannot do the same near its fixpoints ,
so in THAT region it does not satisfy the addition semi-group homom ... but maybe elsewhere it does ...
... unless it is a rule that summability and ana continuation forbid the local creation of the addition semi-group homom ??
Another question is if this ( locally AT LEAST ?? ) agrees with the carleman matrix method expanded somewhere ??
In particular an expansion point on such a path from A to B ? OR maybe rather NOT on that path ?
I considered connections to the gaussian method and the fibonacci numbers , and somewhat green's theorem and convolution ideas but not with a clear result or understanding ...
( i concluded that the infinitesimal ideas i recently had reduce to nothing usefull , consistant or new ideas , only to arrive at old ideas. So that is why im silent about them now )
I could say or speculate more but this will do for now.
What are your ideas ?
Regards
tommy1729