commutive polyomials ? A(B) = B(A) ?? tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 08/06/2022, 12:24 PM Im still not even fully understanding polynomial compositions !! Here is one related question : https://math.stackexchange.com/questions...olynomials any ideas ? regards tommy1729 bo198214 Administrator Posts: 1,628 Threads: 105 Joined: Aug 2007 08/06/2022, 05:03 PM (This post was last modified: 08/06/2022, 05:34 PM by bo198214.) 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 tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 08/06/2022, 08:52 PM 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 bo198214 Administrator Posts: 1,628 Threads: 105 Joined: Aug 2007 08/07/2022, 06:25 AM 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 ... bo198214 Administrator Posts: 1,628 Threads: 105 Joined: Aug 2007 08/07/2022, 06:48 AM 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 $x^2$. JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 08/07/2022, 09:03 AM 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. « Next Oldest | Next Newest »

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