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