This is mainly aimed at Caleb because it might be new to him , but ofcourse to all people 
Although it is not a natural boundary at all I can not help think about this
pxp(z) = (1 + a_0 z)(1 + a_1 z^2)(1 + a_2 z^3)...
where pxp(z) clearly has the unit circle as " boundary " since it has a dense set of zero's there.
But pxp(z) is actually just exp(z).
This is a product expansion for exp(z) valid for |z| < 1.
I will put the pxp paper by Gottfried in an attachement.
Now what is fun is the analogue ; we have a unit boundary ( kinda ) , an infinite expression and even a functional equation :
exp(- x) = 1/exp(x)
hence
pxp(- z) = 1/pxp(z).
Then again it does not map the inside to the outside of the boundary and vice versa.
Just a comment ...
And a reminder that we have not even considered products yet.
***
I want to say (again)
f(x) = x + x^2 + x^4 + x^8 + ...
now x^(2^n) is the super of the function g(x) := x^2.
and x^(2^(c i)) = x^(-1) is thus a function that commutes with it ( iterates of g(x) )
( for appropriate c )
Notice how the imaginary direction of the super is linked to the reflection.
more general :
IF a function is an infinite sum of iterates ,
and it has a natural boundary that is equal to the edge of its filled julia ,
and the filled julia set has only one component.
and the filled julia set has only one fixed point where the dynamics is analytically conjugate to an irrational rotation ( siegel disk) or an attracting c z or c z^n (for constants c and n ).[url=https://en.wikipedia.org/wiki/Irrational_rotation][/url]
then the imaginary direction of the super is linked to the reflection if a reflection formula exists.
Then again , here I have so many restrictions that makes a nice reflection formula unlikely.
Remember In previous posts I wrote the reflection formulas that have nice properties and make most sense.
But say moebius transforms can not map to any edge of some julia.
However maybe a riemann mapping can map things to a circle and save the day.
Notice there are two reflection functions :
L_1( f(z) ) = f( L_2 (z) )
after a riemann mapping we might get
L_1 ( f(z) ) = f*( 1/z )
where f* relates to f by the riemann mapping.
or something like that.
Just some thoughts.
regards
tommy1729
Although it is not a natural boundary at all I can not help think about this
pxp(z) = (1 + a_0 z)(1 + a_1 z^2)(1 + a_2 z^3)...
where pxp(z) clearly has the unit circle as " boundary " since it has a dense set of zero's there.
But pxp(z) is actually just exp(z).
This is a product expansion for exp(z) valid for |z| < 1.
I will put the pxp paper by Gottfried in an attachement.
Now what is fun is the analogue ; we have a unit boundary ( kinda ) , an infinite expression and even a functional equation :
exp(- x) = 1/exp(x)
hence
pxp(- z) = 1/pxp(z).
Then again it does not map the inside to the outside of the boundary and vice versa.
Just a comment ...
And a reminder that we have not even considered products yet.
***
I want to say (again)
f(x) = x + x^2 + x^4 + x^8 + ...
now x^(2^n) is the super of the function g(x) := x^2.
and x^(2^(c i)) = x^(-1) is thus a function that commutes with it ( iterates of g(x) )
( for appropriate c )
Notice how the imaginary direction of the super is linked to the reflection.
more general :
IF a function is an infinite sum of iterates ,
and it has a natural boundary that is equal to the edge of its filled julia ,
and the filled julia set has only one component.
and the filled julia set has only one fixed point where the dynamics is analytically conjugate to an irrational rotation ( siegel disk) or an attracting c z or c z^n (for constants c and n ).[url=https://en.wikipedia.org/wiki/Irrational_rotation][/url]
then the imaginary direction of the super is linked to the reflection if a reflection formula exists.
Then again , here I have so many restrictions that makes a nice reflection formula unlikely.
Remember In previous posts I wrote the reflection formulas that have nice properties and make most sense.
But say moebius transforms can not map to any edge of some julia.
However maybe a riemann mapping can map things to a circle and save the day.
Notice there are two reflection functions :
L_1( f(z) ) = f( L_2 (z) )
after a riemann mapping we might get
L_1 ( f(z) ) = f*( 1/z )
where f* relates to f by the riemann mapping.
or something like that.
Just some thoughts.
regards
tommy1729
