Ok so basically I discussed the very simple case where the natural boundary is the unit circle and our special function is a sum of iterations.
And it only had one fixpoint and that was very attracting.
And perhaps I should mention that for |x| < 1 , x^v for real v always maps to smaller values.
So even x^(2^v(t)) where v(t) is a real valued function associated with the super by t iterations ( or even not associated ) will remain within the unit circle.
Other functions and their random supers might not behave so nice; staying within the bounds of integer iterates ( when doing real iterates ) is a must but not a given.
( although this might be a local uniqueness idea !? )
The boundary of the filled julia exactly matches the natural boundary so they are certainly related.
Also the filled julia of x^2 has only one component , what justifies the reflection idea.
But this clearly is a special case.
Functions like f1(x) = x + x^2 + x^4 + x^8 + ... or f2(x) = x + x^3 + x^9 + x^27 + ...
Fall into this category.
Notice that f3(x) = f1(x) + f2(x) does not satisfy equations unless perhaps f3(x) = f3(1/x).
SO already that is more complicated.
Many special functions with simple natural boundaries can be created by using sums, compositions, products over double sums , summing over primes 1 mod 3 etc
We have recently seen such functions of type lambert, zeta, theta etc.
sometimes we can just plug in values and sometimes not.
But without further ado, I want to go more into the type of iteration sum functions in the spirit of this thread.
***
let
f(z) = (z^2 + z^3)/2
Then define the opa function
opa(z) = z + f(z) + f(f(z)) + f(f(f(z))) + ...
The taylor series gets quite complicated.
The function f(z) seems more contracting than z^2 because of the average of different arguments.
So we expect the filled julia set to be larger than the filled unit circle.
Indeed it is.
Ofcourse many questions arise !!
the julia has only one component which is good.
but f(z) has multiple fixpoints and hence also multiple "opinions " about fractional iterations.
This makes the iterations more complicated.
The riemann mapping from the natural boundary of opa(z) to the unit circle is also not trivial I think.
And the functional equations and reflection formulas are more complex or debatable.
YET f(z) is one of the simplest cubics.
And thus opa(z) is one of the simplest iteration sums.
This topic is getting wildly complicated.
analytic number theory or analytic combinatorics might relate or not.
This makes sense though :
opa( (z^2 + z^3)/2 ) = opa(z) - z
but it is probably not a uniqueness case as we have seen from the first post.
So alot to think about.
But I will add a picture of a truncated opa(z) for inspiration.
White is associated to where the function opa(z) " diverges after the boundary ".
Maybe you can see the fixpoints of f(z) at -2,0,1 ...
regards
tommy1729
And it only had one fixpoint and that was very attracting.
And perhaps I should mention that for |x| < 1 , x^v for real v always maps to smaller values.
So even x^(2^v(t)) where v(t) is a real valued function associated with the super by t iterations ( or even not associated ) will remain within the unit circle.
Other functions and their random supers might not behave so nice; staying within the bounds of integer iterates ( when doing real iterates ) is a must but not a given.
( although this might be a local uniqueness idea !? )
The boundary of the filled julia exactly matches the natural boundary so they are certainly related.
Also the filled julia of x^2 has only one component , what justifies the reflection idea.
But this clearly is a special case.
Functions like f1(x) = x + x^2 + x^4 + x^8 + ... or f2(x) = x + x^3 + x^9 + x^27 + ...
Fall into this category.
Notice that f3(x) = f1(x) + f2(x) does not satisfy equations unless perhaps f3(x) = f3(1/x).
SO already that is more complicated.
Many special functions with simple natural boundaries can be created by using sums, compositions, products over double sums , summing over primes 1 mod 3 etc
We have recently seen such functions of type lambert, zeta, theta etc.
sometimes we can just plug in values and sometimes not.
But without further ado, I want to go more into the type of iteration sum functions in the spirit of this thread.
***
let
f(z) = (z^2 + z^3)/2
Then define the opa function
opa(z) = z + f(z) + f(f(z)) + f(f(f(z))) + ...
The taylor series gets quite complicated.
The function f(z) seems more contracting than z^2 because of the average of different arguments.
So we expect the filled julia set to be larger than the filled unit circle.
Indeed it is.
Ofcourse many questions arise !!
the julia has only one component which is good.
but f(z) has multiple fixpoints and hence also multiple "opinions " about fractional iterations.
This makes the iterations more complicated.
The riemann mapping from the natural boundary of opa(z) to the unit circle is also not trivial I think.
And the functional equations and reflection formulas are more complex or debatable.
YET f(z) is one of the simplest cubics.
And thus opa(z) is one of the simplest iteration sums.
This topic is getting wildly complicated.
analytic number theory or analytic combinatorics might relate or not.
This makes sense though :
opa( (z^2 + z^3)/2 ) = opa(z) - z
but it is probably not a uniqueness case as we have seen from the first post.
So alot to think about.
But I will add a picture of a truncated opa(z) for inspiration.
White is associated to where the function opa(z) " diverges after the boundary ".
Maybe you can see the fixpoints of f(z) at -2,0,1 ...
regards
tommy1729