Iteration with two analytic fixed points

[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