08/07/2022, 11:44 PM
ok maybe i need a break too and im gonna write nonsense now but here goes anyway :
let x and y be real fixpoints with x =/= y.
assume they agree on the fixpoints and are analytic there and also on an iteration path connecting them.
So we have that one of them is repelling and the other is attracting.
Say the are both nonparabolic fixpoints.
Now f(g(x)) = g(f(x)) = x
Lets then assume we use regular iteration.
And lets assume they can then be matched by the same equation
so
we get two equations that describe the same or should :
g(g(... a^t f(f(...f(z1))))
=
f(f(... b^t g(g(...g(z2))))
where z1 is close to x and z2 is close to y.
and a and b are constants ; the derivatives at the fixpoints.
Now f and g are switched because one is attracting and the other is repelling.
But it should not matter what z1 and z2 are as long as they are on the iteration path from x to y.
so i write simply z.
g(g(... a^t f(f(...f(z1))))
=
f(f(... b^t g(g(...g(z2))))
Now i ignore branches for a while.
not just for simplicity but ...it would be strange if picking branches consistantly would change the equality of those two !?
SO i wont.
NOW it appears a and b should be related !
a^t has a period and so does b^t.
And the periods should somewhat match since they are the same function right ??
so we get a is an integer multiple of b or vice versa, right ?
Or a is an integer multiple of 1/b , right ?
i mean then we get like
period a^t consistant with period b^t or b^(-t).
But remember one was attracting and the other repelling so
a = 1/b ??
Im a bit confused but it seems a and b should be related.
...
So i assume this is an issue/strong condition or something i do not fully understand if we want an analytic at both fixpoints and the semi-group homom.
However we can do a 1-periodic theta mapping if we drop the regular and semi-group homom.
But I do not think that would help.
.... because ...
if we start with z0
f(z0)
f(f(z0))
...
and we end up at fixpoint y then
we should also end up at fixpoint y for noninteger positive real iterations going to infinity.
Like if f^[n](z0) converges to y then
f^[n+pi](z0) should also converge to y.
and by approximating y we should " feel " the periodic nature.
1-periodic theta mapping agrees on integer iterations afterall.
So therefore I am confused and skeptical about the following
let the real fixpoints be 0 < x < y.
let f(z) be a nonzero analytic function analytic within radius r =< (1+x)^2 + (1+y)^2
then the non-integer iterates of f(z) are not analytic within that radius ??
Maybe the way out is that radius is smaller than the period ??
regards
tommy1729
let x and y be real fixpoints with x =/= y.
assume they agree on the fixpoints and are analytic there and also on an iteration path connecting them.
So we have that one of them is repelling and the other is attracting.
Say the are both nonparabolic fixpoints.
Now f(g(x)) = g(f(x)) = x
Lets then assume we use regular iteration.
And lets assume they can then be matched by the same equation
so
we get two equations that describe the same or should :
g(g(... a^t f(f(...f(z1))))
=
f(f(... b^t g(g(...g(z2))))
where z1 is close to x and z2 is close to y.
and a and b are constants ; the derivatives at the fixpoints.
Now f and g are switched because one is attracting and the other is repelling.
But it should not matter what z1 and z2 are as long as they are on the iteration path from x to y.
so i write simply z.
g(g(... a^t f(f(...f(z1))))
=
f(f(... b^t g(g(...g(z2))))
Now i ignore branches for a while.
not just for simplicity but ...it would be strange if picking branches consistantly would change the equality of those two !?
SO i wont.
NOW it appears a and b should be related !
a^t has a period and so does b^t.
And the periods should somewhat match since they are the same function right ??
so we get a is an integer multiple of b or vice versa, right ?
Or a is an integer multiple of 1/b , right ?
i mean then we get like
period a^t consistant with period b^t or b^(-t).
But remember one was attracting and the other repelling so
a = 1/b ??
Im a bit confused but it seems a and b should be related.
...
So i assume this is an issue/strong condition or something i do not fully understand if we want an analytic at both fixpoints and the semi-group homom.
However we can do a 1-periodic theta mapping if we drop the regular and semi-group homom.
But I do not think that would help.
.... because ...
if we start with z0
f(z0)
f(f(z0))
...
and we end up at fixpoint y then
we should also end up at fixpoint y for noninteger positive real iterations going to infinity.
Like if f^[n](z0) converges to y then
f^[n+pi](z0) should also converge to y.
and by approximating y we should " feel " the periodic nature.
1-periodic theta mapping agrees on integer iterations afterall.
So therefore I am confused and skeptical about the following
let the real fixpoints be 0 < x < y.
let f(z) be a nonzero analytic function analytic within radius r =< (1+x)^2 + (1+y)^2
then the non-integer iterates of f(z) are not analytic within that radius ??
Maybe the way out is that radius is smaller than the period ??
regards
tommy1729
