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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.

Posts: 27

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Joined: Apr 2009

(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

:

And here is the set for

:

They don't seem to be displaying on my machine, though.

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Joined: Feb 2009

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