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irrational iterate of 2z(1-z) - BenStandeven - 08/13/2010
From examining the superfunction for 4z(1-z) about z=0, (one of the few elementary examples for a non-Mobius function), we get: So [one of] the iterate of is . Putting these functions into the "Mandlebrot" form by conjugating with , we get that the iterate of is . I was hoping to include some pictures of the Julia sets of these two functions, but I don't have ready access to a program that can draw general cubic sets (such as the old Autodesk Chaos program). So I'll try again in the morning, using Fractint. RE: irrational iterate of 2z(1-z) - BenStandeven - 08/14/2010
(08/13/2010, 06:47 AM)BenStandeven Wrote: I was hoping to include some pictures of the Julia sets of these two functions, but I don't have ready access to a program that can draw general cubic sets (such as the old Autodesk Chaos program). So I'll try again in the morning, using Fractint. Here is the Julia set of : [attachment=731] And here is the set for : [attachment=732] They don't seem to be displaying on my machine, though. RE: irrational iterate of 2z(1-z) - tommy1729 - 08/09/2014
First Tommy-Ben Conjecture : let a,b,c be positive integers. Let X,Y be positive irrational numbers that are linearly independant but they are NOT algebraicly independant. Nomatter what X,Y are , there is NO non-MÃ¶bius closed form function f(x) such that f^[a + b X + c Y](x) is also a closed form for every a,b,c. --- Ben's OP was an example where f^[a + b X](x) had a closed form for every a,b. ( X was lb(3) ) Im very very convinced of this conjecture. --- Second Tommy-Ben conjecture : Let a_i be positive integers and X_i be linear independant positive irrational numbers. Let n be an integer > 0. If f^[a + a_1 X + a_2 X_2 + ... + a_n X_n](x) is a closed form for every a_i then the superfunction of f is a composition of at least 2 functions with an addition rule and the X_i are all of the form log(A_i)/log© where the A_i are integers and C is a constant. regards tommy1729 |