Fibonacci as iteration of fractional linear function

[bo198214]

the fibonacci addition formula was explictly recently mentioned by me in one of the recent related threads.

Where? And from what principle did you arrive at it?

if we add ( to the addition formula ) the condition of boundedness on a region we have uniqueness of the fibo.

We have that already for the normal Fibonacci identity AFAIK

this relates to the analogue of the uniqueness of the gamma function , and was mentioned and proved ( but still not accepted or ignored ) by the TPID I PROVED  about exp functions and their uniqueness.

Yeah, just reading it, but this assumption:

the general solution is b^(z + theta(z)) where theta (z) is an entire 1 periodic function.

needs to be proven first.
We know that if we have two solutions S and T that \(S^{-1}(T(z+1))=S^{-1}(T(z))+1\) and I already omit considerations of invertibility and domain of definition. Dunno why you know about a general form of solution (but please if you want to discuss this, open a new thread!)
And please the TPID thread is supposed to stay clean (as stated in the first post)!
Answers or proofs go into separate threads.

this relates to james ideas of fractional integrals , integral transforms and ramanujan. or kouznetsov.

Yes, everything is connected

as for the real part with sine or cosine , you see that it does not work out nice ... but my other argument was that it fails on the complex plane.

and that is probably true for all those "sine replacements of (-c)^t".

You started a topic about those sine replacements working for iterating - x.
Nice pictures and it works for the reals ... but probably fails for the complex.

and the reason is they are just taking the real part of the actual complex solution.

taking real part is not analytic in a sense ( even if the sine or cosine is )

As I see it - it is not so much taking the real part of something, but you replace something that satisfies \(h(z+1)=c\; h(z)\) with something of the same property. Or maybe you can even consider it as the property \(h(z+1)=1+h(z)\).

because they ARE two exponentials with base the eigenvalue.

Actually you can write it as a linear fraction with only *one* exponential, with the quotient of the two Eigenvalues as base.

more interesting is how tribonacci might be related to analogues of the linear fraction iterations ??

Well, you first have to come up with this analogon.
I don't see any.
I would rather vote for a quadronacci, has more chance to find a representation, because there are also no triternions, only quaternions