01/03/2023, 01:26 PM
Thank you for your useful insightful reply James.
I wanted to add that we can use the same trick for 0 or 1 fixpoint as well.
The 0 fixpoint case is a bit complicated and relates to quite old posts i made.
Usually mixing sums and limits and sometimes summability methods.
That case might not even be analytic.
However I wanted to share the beauty of the 1 fixpoint case.
WLOG lets set the fixpoint at 0 to make things easy.
sketch of the idea :
I will use the example f(x) := 5x + x^3 for which we want to find the super and abel function.
( and thereby we get close to taking a half-iterate , but there is a theta function issue )
Finding the superfunction S(x) such that S(x+1) = f(S(x)) follows by using the classical koenigs function around the fixpoint 0.
So we are left to find
F(x) such that F(f(x)) = c F(x)
( from which by taking log_c(x) we find the abel function ... notice however this abel might not be the inverse of the koenigs function !! Hence not necc giving the half-iterate by combining this abel with the koenigs function. aka the theta function issue mentioned above )
notice f ' (0) = 5.
Also notice n iterates of f(x) grow about x^(3^n)
So we get for appropriate c :
F(x) = sum_k c^k asinh( f^[k](x) )
where the sum runs over all integer k.
notice asinh is close to id(x) near 0.
and asin(x^(3^n)) is close to 3^n.
so if for instance c is around 1/4,
our sum behaves a bit like sum_n (4/5)^n + (3/4)^n
where the sum runs over positive integers n.
hence it converges !
and we get F(x) such that F(f(x)) = c F(x).
So we see this idea is part of a large family.
I guess you like that.
regards
tommy1729
I wanted to add that we can use the same trick for 0 or 1 fixpoint as well.
The 0 fixpoint case is a bit complicated and relates to quite old posts i made.
Usually mixing sums and limits and sometimes summability methods.
That case might not even be analytic.
However I wanted to share the beauty of the 1 fixpoint case.
WLOG lets set the fixpoint at 0 to make things easy.
sketch of the idea :
I will use the example f(x) := 5x + x^3 for which we want to find the super and abel function.
( and thereby we get close to taking a half-iterate , but there is a theta function issue )
Finding the superfunction S(x) such that S(x+1) = f(S(x)) follows by using the classical koenigs function around the fixpoint 0.
So we are left to find
F(x) such that F(f(x)) = c F(x)
( from which by taking log_c(x) we find the abel function ... notice however this abel might not be the inverse of the koenigs function !! Hence not necc giving the half-iterate by combining this abel with the koenigs function. aka the theta function issue mentioned above )
notice f ' (0) = 5.
Also notice n iterates of f(x) grow about x^(3^n)
So we get for appropriate c :
F(x) = sum_k c^k asinh( f^[k](x) )
where the sum runs over all integer k.
notice asinh is close to id(x) near 0.
and asin(x^(3^n)) is close to 3^n.
so if for instance c is around 1/4,
our sum behaves a bit like sum_n (4/5)^n + (3/4)^n
where the sum runs over positive integers n.
hence it converges !
and we get F(x) such that F(f(x)) = c F(x).
So we see this idea is part of a large family.
I guess you like that.
regards
tommy1729