Iteration with two analytic fixed points

[bo198214]

Actually I already found the flaw.
I was looking at it on the complex plane and found out that \(\arctan(c\tan(z))\) has branch points at \(\pm i\text{artanh}(1/c)+\pi k\) for \(c>1\) and \(\pm i\text{artanh}( c ) + \frac{\pi}{2} + \pi k\) for \(0< c< 1\). (Takes to long for just quickly explaining it, you have to look at the conformal map of tan, and then you see that i/c is mapped (by cz) to the branch point of arctan at i. So there the trouble begins.)

This means these branch points come close to the real line (assume c>1) at \(x_{2k}=\frac{\pi}{2} 2k\) by \(\text{artanh}(c^{-t})\) for \(t\to \infty\) and at \(x_{2k+1}=\frac{\pi}{2}(2k+1)\) by \(\text{artanh}( c^{t} )\) for \(t\to -\infty\).
So for two consecutive fixed points \(x_{2k}\) and \(x_{2k+1}\) one of them can not be contained in an analytic vicinity independent of t (and for two nonconsecutive fixed point the belt it tightened between them).
This time though not because a pole moves in, but a branch point moves in when \(t \to \pm\infty\).

Mhm, so there seems something true about the condition of a domain independent of t.
Though I wonder whether for any non-entire function there can be two fixed points in a domain D such that \(f^{\circ n}\) is holomorphic on D for all \(n\in\mathbb{N}\) (nothing to do with non-integer iterations).
Maybe you have a proof for this - as a first step. (I mean if there is such domain D then the function is already entire)
Looks like branch points and poles are "sucked in" by the fixed point and hence come arbitrarily near.

[JmsNxn]

Though I wonder whether for any non-entire function there can be two fixed points in a domain D such that \(f^{\circ n}\) is holomorphic on D for all \(n\in\mathbb{N}\) (nothing to do with non-integer iterations).
Maybe you have a proof for this - as a first step. (I mean if there is such domain then D then the function is already entire)

Okay, so I'm good for 3 cases. These are the same three cases Milnor calls:

\(D\) is Hyperbolic; which means it's a simply connected domain biholomorphic to the unit disk.

\(D\) is Euclidean; which means it's a simply connected domain isomorphic to \(\mathbb{C}\)

\(D\) is spherical; which means it's a simply connected domain isomorphic to \(\widehat{\mathbb{C}}\)

Unfortunately I don't know much advanced shit outside these options.

But I do know, That usually it will behaved like the spherical case. But it will be on more complex Riemann surfaces than the Riemann sphere. So you can do dynamics on very exotic Riemann spheres, and technically it relates to iterations on \(\arcsin\) or whatever. But at that point, you better pull out your Riemann surface calculus tools. Because shit gets crazy.

That's the limit of my knowledge. I know a lot about the first three cases though

[bo198214]

For me it appears roughly like this.
If we have this domain D with the two fixed points contained, then either one fixed point is attracting and attracts some singularity that would then be inside D.
Or points are pushed through between the fixed points (like between the two repelling fixed points in e^x) and then a singularity would also take this path (in this case one might need the *continuous* iteration so that the trajectory cuts the domain D.)
Maybe you were pointing in that direction with your B(z) sets.

[JmsNxn]

Yes, I think we're really close to finding the exact language. I'm sure we can find it. But, I hope you understand when I mean "we can't iterate about two fixed points". Though I did say it a little off hand and incorrectly. There is a good amount of truth to this statement. But it's specific to how we phrase that statement. And we have to phrase it perfectly. I think that's the key to explaining multiple fixed points though; we need a strong statement that restricts what multiple fixed point iterations look like.

I'm really excited. Sorry Bo, I've been up all night. And I've been really wired all night. Can't talk anymore. It's 0700 for me and I haven't slept yet. I've been thinking about this shit too much

[bo198214]

Yeah, I also need a pause. Just summarising:

We can have a real analytic function with two fixed points (say left is attracting and right is repelling) such that the analytic continuation of the regular iteration at the first fixed point is the regular iteration at the second fixed point.
In the complex case we can even have vicinities independent of t around each fixed point such that there is a connecting continuation for each t.

However it seems in no case one can have a continuation path  independent of t, or: One can not have a domain D containing both fixed points, such that all the iterations of an iteration semi group are analytic on D.

[tommy1729]

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

[bo198214]

As to the fixed points, yes, they alwasys need to be reciprocals of each other (like in the linear fractional case),
because then the period \(2\pi i/\log( c)\) is either positive or negative when you set c=a, or c=b.
Good observation!

[bo198214]

One can verify that also by direct calculation.
Assume we have these fixed points \(z_0\) and \(z_\infty\) where the Kœnigs/Schröder function f is 0 and \(\infty\) (has a pole) respectively.
Let \(\ell_c(z)=cz\) and \(\nu(z)=1/z\), then we take the derivative:
\begin{align}
(f^{-1}\circ \ell_c \circ f)' &= \frac{1}{f'\circ f^{-1}\circ \ell_c\circ f}\cdot (\ell_c'\circ f)\cdot f'\\
(f^{-1}\circ \ell_c \circ f)'(z_0) &= \frac{1}{f'(z_0)} c f'(z_0) = c
\end{align}

However for \(z_\infty\) we can not use this derivation because  \(f'(z_0)=\infty\)? We need to find a safer way. But we found already the trick to take the inverse of \(f\) which is analytic at \(z_\infty\). So let \(f=\nu\circ g\) then 
\[f^{-1}\circ \ell_c \circ f = g^{-1} \circ \nu \circ \ell_c \circ \nu\circ g = g^{-1}\circ \ell_{1/c} \circ g\]
and as above:
\[(f^{-1}\circ \ell_c \circ f )'(z_\infty) = (g^{-1}\circ \ell_{1/c} \circ g)'(z_\infty) = \frac{1}{c} \]

This is amazing in the sense that as long as the Kœnigs/Schröder function has an analytic 0 and a pole then the corresponding t-iterations have \(c^t\) and \(c^{-t}\) as fixed point derivations. So one could assume that this is the normal situation. However we consider it rather the exception, why?
Because the regular Kœnigs/Schröder function at one fixed point is often not analytic/a pole at the other fixed point.
Though we could have \(\lim_{z\to z_\infty} f^{-1}(c^t f(z)) = z_\infty \) for all t, the limit \(\lim_{z\to z_\infty} (f^{-1}\circ \ell_{c^t} \circ f)'(z) \) would be some other number for discrete t for those functions.
Quite mysterious I would say.

On the other hand if the fixed point derivations are not reciprocal, like with \(\exp_{\sqrt{2}}'(2) = \log(2)\) and \(\exp_{\sqrt{2}}'(4) = 2\log(2) \) then we can surely say that the regular iteration at one fixed point can not be the regular iteration at the other for all t (or just a continuous interval) because as Tommy pointed out: the superfunctions have different periods then.

[bo198214]

OMG, I just found out Sheldon asked that same question already on MSE Jul 2013!
https://math.stackexchange.com/questions...-iteration
And he got an answer!
Maybe we should read 
MR0729400 S. Dubuc, Étude théorique et numérique de la fonction de Karlin-McGregor, J. Analyse Math. 42 (1982/83), 15–37.

However I don't have access to university anymore, maybe JmsNxn can procuree this article.

However this article deals with an older article
Samuel Karlin and James McGregor, Embedding iterates of analytic functions with two fixed points into continuous groups, Trans. Amer. Math. Soc. 132 (196, 137–145. MR 224790, DOI 10.1090/S0002-9947-1968-0224790-2

which is freely accessible here (I also append it to this post).

However when skimming through this article. He only considers a very particular group of holomorphic function that do not allow branches for example.
From this functions he concludes that the only ones where the iterations coincide at both fixed points are the linear fractional functions.

I think we already a step ahead in our discussion here

[JmsNxn]

....
MR0729400 S. Dubuc, Étude théorique et numérique de la fonction de Karlin-McGregor, J. Analyse Math. 42 (1982/83), 15–37.

However I don't have access to university anymore, maybe JmsNxn can procuree this article.

I'll load up my U of T library account. U of T's great because they've made it so that alumni can always have access to the library. You just need an active u of t email, which I have. U of T's one of the rare universities that doesn't deactivate accounts. I'll get to it! Worse comes to worse I'll have to request it or something.