07/01/2022, 09:52 PM

ok this might be a repost , a silly question , an ignored idea or a mistake but anyways

We know fractional iterations do not agree on fixpoints usually ( or always )

Lets say we have fixpoint at 0

and around 0 the function behaves like a polynomial a(z).

Now lets say we have another fixpoint at 1.

Now clearly - using fixpoint formula alone and assuming they must merge from one fix to another ;

1) the derivatives of the fixpoints must match.

2) without a fixpoint in the way of the 2 others that ' misbehaves ' or ' blocks ' the path.

Now even better than 1)

1b) around every fixpoint the function behaves like the polynomial a(z).

---

Now if we want to expand the function easily it works best if the polynomial a(z) is of high degree.

But if the region around 0 is copied to the region around 1 , the same should happen to the region around 1 ; mapping to the region around 2.

This implies that 2 should be fix too.

In fact this implies that our function should be somewhat periodic and have fixpoints at all integers and no other real fixpoints.

Taking this one step further :

the function should be entire.

and

the iterations should be somewhat periodic themselves.

---

This brings us to

f(z) = z + sin(2 pi z)/5

clearly

f(f(z)) = z + sin(2 pi z)/5 + sin(2 pi ( z + sin(2 pi z)/5 ) )/5

what is also of the form z + periodic.

( funny thing z + theta(z) occurs here often )

So how do the iterates of f(z) = z + sin(2 pi z)/5 behave ??

Is the half-iterate of the same type ( z + periodic ) ?

Do the fixpoints sometimes match for noninteger iterates ?

So I wonder about the general idea

f^[t](z) for real t > 0.

and ofcourse f(z) is just a random case.

crazy idea i guess.

regards

tommy1729

We know fractional iterations do not agree on fixpoints usually ( or always )

Lets say we have fixpoint at 0

and around 0 the function behaves like a polynomial a(z).

Now lets say we have another fixpoint at 1.

Now clearly - using fixpoint formula alone and assuming they must merge from one fix to another ;

1) the derivatives of the fixpoints must match.

2) without a fixpoint in the way of the 2 others that ' misbehaves ' or ' blocks ' the path.

Now even better than 1)

1b) around every fixpoint the function behaves like the polynomial a(z).

---

Now if we want to expand the function easily it works best if the polynomial a(z) is of high degree.

But if the region around 0 is copied to the region around 1 , the same should happen to the region around 1 ; mapping to the region around 2.

This implies that 2 should be fix too.

In fact this implies that our function should be somewhat periodic and have fixpoints at all integers and no other real fixpoints.

Taking this one step further :

the function should be entire.

and

the iterations should be somewhat periodic themselves.

---

This brings us to

f(z) = z + sin(2 pi z)/5

clearly

f(f(z)) = z + sin(2 pi z)/5 + sin(2 pi ( z + sin(2 pi z)/5 ) )/5

what is also of the form z + periodic.

( funny thing z + theta(z) occurs here often )

So how do the iterates of f(z) = z + sin(2 pi z)/5 behave ??

Is the half-iterate of the same type ( z + periodic ) ?

Do the fixpoints sometimes match for noninteger iterates ?

So I wonder about the general idea

f^[t](z) for real t > 0.

and ofcourse f(z) is just a random case.

crazy idea i guess.

regards

tommy1729