Fibonacci as iteration of fractional linear function

[bo198214]

So what is \(\alpha^{-1}\{f\}(z)\) what is this braces notation?

You still owe an explanation for this ...

nope, the contradicts happen because things like the way u may wish a real-to-real (-1)^x for real x
For most cases even the multiplier is negative, we can still get a superfunction because it's complex-to-complex\
And meanwhile the superfunction only guarantees \(F(z+1)=T(F(z))\) for some T, not \(F(z+t)=T^t(F(z))\) for all real or even complex t. So these examples indeed are superfunctions but wont always allow you to have \(f^s\circ f^t=f^{s+t}\)

Hm, I was always thinking of a superfunction as \(F(t+F^{-1}(z_0))=f^{\circ t}(z_0)\) for some \(z_0\) where \(f^{\circ t}\) is an iteration semigroup.
So if F satisfies \(F(s+1)=f(F(s))\) then the inverse (if invertible) satisfies \(F^{-1}(f(x))=F^{-1}(x)+1\) (Abel function).
But then one can reconstruct \(f^{\circ t}(z)= F(t+F^{-1}(z))\) and it is an iteration semigroup or has the Abelian property as you would say:
\[ f^{\circ s}(f^{\circ t}(z)) = F(s+F^{-1}(F(t+F^{-1}(z)))) = F(s+t+F^{-1}(z)) = f^{\circ s+t}(z) \]
And you say now, because it is not invertible it doesn't work that way anymore?!

I interpret your post as to find a superfunction that is real-to-real and also preserve the property \(f^s\circ f^t=f^{s+t}\),

1. Considering the stuff above I am not sure what it would mean ... You can not derive an iteration \(f^{\circ t}\) from the superfunction so how can you then ask whether it has the Abelian property?
2. *I* didn't have a driving motivation here. I just wanted to present the Fibonacci sequence in connection with iteration of LFTs - that was all. Then people came up - yeah, why not having a real iteration - we don't like complex valued, and so I justed wanted to see whether I can help.
Actually for discussing this I would rather refer to thread Constructing a real valued Fibonacci iteration