commutive polyomials ? A(B) = B(A) ??
#1
Im still not even fully understanding polynomial compositions !!


Here is one related question :


https://math.stackexchange.com/questions...olynomials


any ideas ?


regards

tommy1729
Reply
#2
I thought calculating the Schröder iteration for formal powerseries is doing that (for the case c_0=0 and |c_1|!=1).
So you just take that algorithm and see whether it stops after a while if you want to know whether the polynomial has a polynomial that it commutes with.
Though I am not sure what's the actual question.

Here is the link to the recursive calculation of the powerseries coefficients:
https://math.eretrandre.org/tetrationfor...3#pid10733
Reply
#3
Hold on Bo.

Polynomials have many fixpoints.

And there is also the 1-periodic theta mapping.

So just because the regular koenigs function based on a selected fixpoint does not result in another polynomial that commutes with the given one does not disprove its existance.

Secondly suppose we have a given polynomial A(x) of degree j.

Then proving that no other polynomial commutes with it requires checking it for all degrees ...

That is a supertask, since there are infinitely many integers.

That adresses the issue with finding a polynomial B(x) such that A(B(x)) = B(A(x) WHEN we are given an A(x).

The question in the link is more subtle :

A,B,C are distinct polynomials where none are given.

We want to FIND these 3 polynomials.

And they must satisfy 

A(B(x)) = B(A(x))

A(C(x)) = C(A(x))

B(C(x)) = C(B(x))

and the polynomials A,B,C must have rational coefficients and have distinct degrees.

Now read the question in the link again.

I think that should clarify all.

In the comment mick conjectured x^7  + x + 1 does not commute with polynomials of even degree.
( which seems close to what you referred to )

regards

tommy1729
Reply
#4
I think I don't understand the question, say I take \(A=x^3\), \(B=x^9\) and \(C = x^{27}\)
then \(A\circ B = B\circ A = x^{27}\), \(A\circ C = C\circ A = x^{81}\) and \(B\circ C = C\circ B = x^{243}\).
You can take any initial polynomial A and just set \(B=A^{\circ m}\) and \(C=A^{\circ n}\) for some odd m,n ...
Reply
#5
Ah, no I see, the question was really formulated in bad way.
He *additionally* wants to have that it does not commute with any even-degree polynomial D, that's why he excludes my previous answer which would commute with say \(x^2\).
Reply
#6
No way you have polynomial solutions to this. I understand implicit solutions, no way you get a polynomial solution. Or "for most polynomial sets" you don't have polynomial solutions. If you do have solutions then first of all, bo is right (map a common fixed point between all three functions (A,B,C map \(\widehat{\mathbb{C}}\) to itself polynomially), this is just done from a change of variables). Take the Schroder coordinate and compare the \(A,B,C\) solutions based on their multipliers. You're done.

God knows when this is polynomial. But bo is right to look at the Schroder coordinate.
Reply




Users browsing this thread: 1 Guest(s)