Iteration with two analytic fixed points
#34
(08/09/2022, 12:07 PM)bo198214 Wrote: 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 Sad
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RE: Iteration with two analytic fixed points - by JmsNxn - 08/11/2022, 12:01 AM

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