08/11/2022, 07:01 AM
(08/11/2022, 12:01 AM)JmsNxn Wrote: 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.
(08/11/2022, 12:01 AM)JmsNxn Wrote: 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.

