Iteration with two analytic fixed points

[JmsNxn]

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.

This was largely my question. Can we find a Euclidean mapping (entire function) such that \(f(p_0) = p_0\) and \(f(p_1) = p_1\), such that \(f'(p_0) = 1/f'(p_1)\). And that additionally the orbits about a domain \(H\) of \(f\) tend to \(p_0\), and the orbits about a domain \(H\) of \(f^{-1}\) tend to \(p_1\).

I think this is possible, the trouble is, the fractional iterate will not be local about both fixed points. We can't fix a domain \(H\) which contains both fixed points. Even if \(H\) isn't simply connected, there should be a way to find a subdomain that is, that contains both fixed points--contradiction. This again would be my \(B(z)\) argument.

You're invariably getting \(p_1\) (\(p_0\)) to be on the boundary of the basin about \(p_0\) (\(p_1\)). The only way you have holomorphy at both is... if it's the Riemann sphere. Especially because we are making the coordinate change from \( f(z) \mapsto \lambda z\). And \(\lambda z\) technically has \(\infty\) (or \(0\)) at the boundary of the basin. But no holomorphic function on its basin is all of \(\mathbb{C}\), UNLESS it is \(\lambda z\).

This is deeply fascinating. We are definitely tapping the deep core of iteration right now, lmao.

EDIT: I've been devouring the Karlin Mcgregor paper, and I don't think we even need the extraneous paper you asked me to find. Karlin and Mcgregor prove a much more general result hidden as a result about real analysis. It specifically uses the term regular, and this should imply a domain about \([0,1]\) in the complex plane, but it includes \([0,1]\). I'll try to make a better write up of what I mean. But this is a much much better explanation of what I was trying to explain. MUCH MUCH better, lol.

[bo198214]

This was largely my question. Can we find a Euclidean mapping (entire function) such that \(f(p_0) = p_0\) and \(f(p_1) = p_1\), such that \(f'(p_0) = 1/f'(p_1)\). And that additionally the orbits about a domain \(H\) of \(f\) tend to \(p_0\), and the orbits about a domain \(H\) of \(f^{-1}\) tend to \(p_1\).

I think this is possible,

Wait, wait you think it is possible to find an entire function, such that \(f'(p_0) = 1/f'(p_1)\).
I thought exactly the opposite, that such an entire function can not exist (i.e. that exactly this reciprocal fixed point derivations is the criterion whether the regular iterates would coincide ore not).
Then you have to give an example of such an entire function!

[bo198214]

Ok, I even can give a polynomial
\[ p (x)  = x ^ 3 + \frac{\sqrt{5}-3}{2} x^2 - \frac{\sqrt{5}-3}{2} x \]
   
has fixed points 0 and 1 and 
\[ p'(x) = 3x^2 + (\sqrt{5}-3)x - \frac{\sqrt{5}-3}{2} \]

The derivatives at the fixed points are:
\( p'(0) = -\frac{\sqrt{5}-3}{2}, p'(1) = 3 + \frac{\sqrt{5}-3}{2}=\frac{\sqrt{5}+3}{2}\) hence 
\[ p'(0)p'(1) = -\frac{(\sqrt{5}-3)(\sqrt{5}+3)}{4} = 1\]

[JmsNxn]

Ok, I even can give a polynomial
\[ p (x)  = x ^ 3 + \frac{\sqrt{5}-3}{2} x^2 - \frac{\sqrt{5}-3}{2} x \]

has fixed points 0 and 1 and 
\[ p'(x) = 3x^2 + (\sqrt{5}-3)x - \frac{\sqrt{5}-3}{2} \]

The derivatives at the fixed points are:
\( p'(0) = -\frac{\sqrt{5}-3}{2}, p'(1) = 3 + \frac{\sqrt{5}-3}{2}=\frac{\sqrt{5}+3}{2}\) hence 
\[ p'(0)p'(1) = -\frac{(\sqrt{5}-3)(\sqrt{5}+3)}{4} = 1\]

Again bo, not to be the nitpicker that I am. As Milnor would describe it, a polynomial is a map from \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\). It is not a "euclidean mapping" as Milnor describes, because it's not a transcendental entire function.

But additionally, Karlin and Mcgregor shows that:

\[
p^{\circ t}(x)\\
\]

Is not analytic for \(x\) about either fixed point and \(t \in \mathbb{R}\). I imagine you're going to get something a little different here with a branching problem at one of the fixed points.

EDIT: Also, I wasn't able to pull the paper you asked me to pull. It seems I can access the library, but I have limited viewing to mostly only textbooks. For certain journals there are restrictions on who can access it (I assume due to licensing rights), and I presume because I am not actively enrolled at the moment I'm disqualified

[bo198214]

Again bo, not to be the nitpicker that I am. As Milnor would describe it, a polynomial is a map from \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\). It is not a "euclidean mapping" as Milnor describes, because it's not a transcendental entire function.

But James, really, you making up a lot of artificial conditions. Needs to be euclidian, needs to have a vicinity independent of t, etc, etc.
This all is not needed!
Look at the Karlin&McGregor paper, the class of functions they describe contains all *meromorphic functions*.
They use standard regular iteration - the vicinity *depends* on t (and explicitly state that on the 4th line of this paper).
And from *there* they conclude that the only functions inside their class of functions that can have the same regular iterations at both fixed points can only be the linear fractional functions.

And I just wanted to reiterate their argument here applied to the polynomial I gave, why it can not have the same regular iteration at both fixed points (despite reciprocal fixed point derivatives):

So \(z_0=0\) is the attracting fixed point of \(p\) with derivative \(c=\frac{3-\sqrt{5}}{2}\), \( z_\infty=1\) is the repelling fixed point with derivative \(\frac{1}{c}\). Assume we have a representation as \(p(x)=f^{-1}(cf(x))\) with \(f(z_0)=0\) and \(f(z_\infty)\) being a pole. As in my previous post we look at the representation \(p(x)=g^{-1}\left(\frac{1}{c}g(x)\right)\) with \(g=\frac{1}{f}\) where \(g\) has a pole at \(z_0\) and is analytic at \(z_\infty\).
Then \(g^{-1}\) is so to say the multiplicative superfunction it maps \((\infty,0)\) to \((z_0,z_\infty)\) and satisfies:
\[ g^{-1}\left(\frac{1}{c}z\right)=p(g^{-1}(z)) \]
So because \(\frac{1}{c}>1\) we can holomorphically extend \(g^{-1}\) from a small disk around 0 to the whole complex plane via \(g^{-1}\left(\left(\frac{1}{c}\right)^n z\right)=p(g^{-1}(z))\). It's an entire function.
But also we know that \(\lim_{z\to\infty} g^{-1}(z)\) must be \(z_0\) (because g has a pole at \(z_0\))
Then it is a bounded entire function, which is a constant.
And so this construction can not exist.

EDIT: Also, I wasn't able to pull the paper you asked me to pull. It seems I can access the library, but I have limited viewing to mostly only textbooks. For certain journals there are restrictions on who can access it (I assume due to licensing rights), and I presume because I am not actively enrolled at the moment I'm disqualified

maybe I somehow have to reactivate my alumni library access and go to a physical university library (unfortunately online access is only for students and staff of the university) - but this may take a while.

[tommy1729]

Ok, I even can give a polynomial
\[ p (x)  = x ^ 3 + \frac{\sqrt{5}-3}{2} x^2 - \frac{\sqrt{5}-3}{2} x \]

has fixed points 0 and 1 and 
\[ p'(x) = 3x^2 + (\sqrt{5}-3)x - \frac{\sqrt{5}-3}{2} \]

The derivatives at the fixed points are:
\( p'(0) = -\frac{\sqrt{5}-3}{2}, p'(1) = 3 + \frac{\sqrt{5}-3}{2}=\frac{\sqrt{5}+3}{2}\) hence 
\[ p'(0)p'(1) = -\frac{(\sqrt{5}-3)(\sqrt{5}+3)}{4} = 1\]

you got the fixpoints derivatives connection right. 

but how do you know both regulars agree ?

[tommy1729]

Ok, I even can give a polynomial
\[ p (x)  = x ^ 3 + \frac{\sqrt{5}-3}{2} x^2 - \frac{\sqrt{5}-3}{2} x \]

has fixed points 0 and 1 and 
\[ p'(x) = 3x^2 + (\sqrt{5}-3)x - \frac{\sqrt{5}-3}{2} \]

The derivatives at the fixed points are:
\( p'(0) = -\frac{\sqrt{5}-3}{2}, p'(1) = 3 + \frac{\sqrt{5}-3}{2}=\frac{\sqrt{5}+3}{2}\) hence 
\[ p'(0)p'(1) = -\frac{(\sqrt{5}-3)(\sqrt{5}+3)}{4} = 1\]

you got the fixpoints derivatives connection right. 

but how do you know both regulars agree ?

Ok im running ahead of ideas and posts i intended to make myself about this , the semi-group homom and 2sinh method but 

the main idea is the following sufficient condition ( not neccessary condition ! ) :

for fixpoints neighbourhoods to be well described by their regular iterations  ( so well they are equal ) we want f(x) around the fixpoint to be 

1) univalent there until we reach the other fixpoint.

so when fixpoints are 0 and 1 we want a radius around 0 and around 1 connecting them.

2) not having any other fixpoints in the way.

so we want a radius 1 around point 0 (connecting to 1) and around point 1 (connecting to 0).

For them to be univalent in those circles we need f '(z) = 0 to not have zero's there AND not have another fixpoint there.

so lets see

f ' (z) = 0 

this has zero's 0.12 +1/3 i i and 0.12 - 1/3 i approximately ( not using closed form ).
so the radius is small.

the other fixpoint for f(z) = z is

- 0.61803... ( yes " that " number )

***

However according to " flow theory " * what i called it before and will again * we can be more optimistic ;

their is an analytic path / connected region from fixpoint 0 to fixpoint 1.

the path is univalent , has no other fixpoints , the period matches and so it could and should be the same regular iteration.


the logic is that for
 
small h close to fix 0 and small w close to fix 1 and w and h close to eachother ;
 we could say ( by limit )

f^[V](0+h) = 1 - h + O(h^2)

f^[-V](1-w) = 0 + w + O(w^2)

and since the region is univalent and contains no other fix , this makes alot of sense.

therefore we probably have agreement within or close to the open ellips 

[ - 0.611803 , 1.611803 ] for the major axis 

and

[0.5 + 1/3 i, 0.5 - 1/3 i] for the top and bottom of the ellips.


There is probably a way to show this formally for this specific polynomial.

Also as always we take branch cuts appropriately or even convenient.



regards

tommy1729

[tommy1729]

Ok, I even can give a polynomial
\[ p (x)  = x ^ 3 + \frac{\sqrt{5}-3}{2} x^2 - \frac{\sqrt{5}-3}{2} x \]

has fixed points 0 and 1 and 
\[ p'(x) = 3x^2 + (\sqrt{5}-3)x - \frac{\sqrt{5}-3}{2} \]

The derivatives at the fixed points are:
\( p'(0) = -\frac{\sqrt{5}-3}{2}, p'(1) = 3 + \frac{\sqrt{5}-3}{2}=\frac{\sqrt{5}+3}{2}\) hence 
\[ p'(0)p'(1) = -\frac{(\sqrt{5}-3)(\sqrt{5}+3)}{4} = 1\]

you got the fixpoints derivatives connection right. 

but how do you know both regulars agree ?

Ok im running ahead of ideas and posts i intended to make myself about this , the semi-group homom and 2sinh method but 

the main idea is the following sufficient condition ( not neccessary condition ! ) :

for fixpoints neighbourhoods to be well described by their regular iterations  ( so well they are equal ) we want f(x) around the fixpoint to be 

1) univalent there until we reach the other fixpoint.

so when fixpoints are 0 and 1 we want a radius around 0 and around 1 connecting them.

2) not having any other fixpoints in the way.

so we want a radius 1 around point 0 (connecting to 1) and around point 1 (connecting to 0).

For them to be univalent in those circles we need f '(z) = 0 to not have zero's there AND not have another fixpoint there.

so lets see

f ' (z) = 0 

this has zero's 0.12 +1/3 i i and 0.12 - 1/3 i approximately ( not using closed form ).
so the radius is small.

the other fixpoint for f(z) = z is

- 0.61803... ( yes " that " number )

***

However according to " flow theory " * what i called it before and will again * we can be more optimistic ;

their is an analytic path / connected region from fixpoint 0 to fixpoint 1.

the path is univalent , has no other fixpoints , the period matches and so it could and should be the same regular iteration.


the logic is that for
 
small h close to fix 0 and small w close to fix 1 and w and h close to eachother ;
 we could say ( by limit )

f^[V](0+h) = 1 - h + O(h^2)

f^[-V](1-w) = 0 + w + O(w^2)

and since the region is univalent and contains no other fix , this makes alot of sense.

therefore we probably have agreement within or close to the open ellips 

[ - 0.611803 , 1.611803 ] for the major axis 

and

[0.5 + 1/3 i, 0.5 - 1/3 i] for the top and bottom of the ellips.


There is probably a way to show this formally for this specific polynomial.

Also as always we take branch cuts appropriately or even convenient.



regards

tommy1729

for those interested , the analogue idea for 2sinh is essentially similar and simple :

2sinh^[p](0+h) = 1 + h + O(h^2).

for some real p.

where 1 is the exp of 0 !!!

so it "commutes" with exp.

notice 2sinh has "univalent radius" ( as described above )  equal to pi/2. 
pi/2 > 1, so that is good.

also no other fixpoints in the neigbourhood.

So the property of semi-group homom carries over ( from regular at 0 for 2sinh ) to taking exp ... and by induction taking exp any amount of times.
And hence carries over to the whole 2sinh method , hence the tetration has the semi-group homom.

Notice exp is never 0 so the fixpoint at 0 for 2sinh is NOT an issue.

( 2sinh^[A](x) = 0 or exp^[B](x) = 0 is thus not an issue )

regards

tommy1729

[bo198214]

you got the fixpoints derivatives connection right. 

but how do you know both regulars agree ?

They don't, I made a short sketch why it can not in this particular case (or generally for entire functions) from the paper of Karlin&McGregor
just some post ago

[tommy1729]

slightly related and pasted :

ok bo about this fibo thread here and the other ones :

the fibo has 2 eigenvalues say A and B.

then essentially solving

f(x+2) = f(x+1) + f(x)

has the linear property :

every solution is a linear combination of the other solutions.

so 

3 A^x - 2 B^x

solves the equation.

also the real and imag part split up.

re f(x+2) = re ...

So the most general solution is

alpha * A(x) + beta * B(x)

where A and B satisfy :

A(x+1) = A * A(x)

and

B(x+1) = B * B(x).

And those in term can be solved with a 1 periodic theta function each.

So our general solution ON THE COMPLEX PLANE is

alpha * A^(x + theta1(x)) + beta * B^(x + theta(2x))

wich reduces to the form 

A^(x + theta1(x)) + B^(x + theta(2x))

if we igore alpha and beta being potentially 0.

NOTICE 

REALPART ( A^(x + theta1(x)) + B^(x + theta(2x)) )

is also a real solution on the real line.

but not on the complex plane.
( not even if we make analytic continuation to undo the " real part " operator and have imaginary parts )

JUST LIKE a sine is not the equivalent of (-1)^t ; although it works on the real line and has an analytic continuation to the complex 

IT FAILS on the complex plane.

So this is the critisism to taking real parts or using sine and cosine.


***

So now any solution 

A^(x + theta1(x)) + B^(x + theta(2x))

that agrees on the fibonacci sequence can be considered a solution to the generalized fibonacci function , that is analytic and satisfies the equations on the complex plane.

***

And we have uniqueness critertions for the usual exponential ( a TPID which i proved and was ignored but never mind that now )

So that property carries over I think.

***

But the fibonacci satisfy many many equations ...

how about also satisfying the others ??

Is there a kind of semi-group homom ??

I investigated some other equations it ( fibo ) satisfied but they are not all compatible with a fixed generalization  ( A^(x + theta1(x)) + B^(x + theta(2x)) )  or some just for fibo(odd) or fibo(even).

In fact imo the fibo is not the most interesting sequence of its kind.

What about sequences that are strictly increasing ? Much closer to the idea of iterations.

...

My idea is investigating expontial sums such as a given f(x) for instance :

f(x) = a^x + 2 b^x + c^x 

This does not have the issues of polynomials and have already a closed form.

They will often also satisfy fibonacci like equations.
And working backwards is a nice idea.

Also when f(x) has less than 3 fixpoints on the reals we have the nice question

g(g(x)) = f(x).

which seems like a logical next step in tetration.

Notice that the recent nice cubic polynomial you gave was real valued and had two fixpoints where they both had analytic iterations at their two fixpoint.
 
However the cubic had a third real fixpoint which makes it not working near that fixpoint.

That is a typical property of cubics.


But for exponential sums we can make a function exp shaped with 0,1 or 2 fixpoints and fractional iterates analytic at them. 


This would then give us a kind of semi-fibonacci.

And if the superfunction agrees on the regular iterates of both real fixpoints its a really nice thing !!


regards

tommy1729