08/12/2022, 09:34 PM

(08/12/2022, 08:32 PM)tommy1729 Wrote:(08/12/2022, 07:11 AM)bo198214 Wrote:(08/12/2022, 01:16 AM)tommy1729 Wrote: 1) This identity is like hundreds of years old and mentioned a thousand times and even on wiki.Well, I mean you reading this thread, seeing the people asking for a real extension, obviously knowing the formula and didn't say a thing (before)?

(08/12/2022, 01:16 AM)tommy1729 Wrote: 2) the fibonacci sequence clearly is not an iteration since we have 0,1,1,2,... the occurence of 1 twice makes it not an iteration.So what, we were looking in the context of LFT here where it is an iteration. Anyways you can also consider it an iteration in 2d vectors.

I also wanted to follow up how the corresponding LFTs look.

(08/12/2022, 01:16 AM)tommy1729 Wrote: 3) that identity does not satisfy the recursion f(x+1) = f(x) + f(x-1). It is just a lame cos used for a dubious unmotivated interpolation.The equation \(f(x+2)=f(x+1)+f(x)\) boils down to \(\Phi^2=\Phi+1\) and

\begin{align}

\cos(\pi t + 2\pi)|\Psi|^2 &= \cos(\pi t + \pi)|\Psi| + \cos(\pi t)\\

\cos(\pi t)\Psi^2 &= - \cos(\pi t) |\Psi| + \cos(\pi t)\\

\Psi^2 &= \Psi + 1

\end{align}

And \(\Phi\) and \(\Psi\) are exactly the solutions of \(x^2=x+1\). It's even written in "the wiki" as you call it that it satisfies the Fibonacci identity. So your reasoning rather seems lame, dubious and unmotivated ...

ok , i take part 3 back although i still have some issues with it.

regards

tommy1729

ok bo about this fibo thread here and the other ones :

the fibo has 2 eigenvalues say A and B.

then essentially solving

f(x+2) = f(x+1) + f(x)

has the linear property :

every solution is a linear combination of the other solutions.

so

3 A^x - 2 B^x

solves the equation.

also the real and imag part split up.

re f(x+2) = re ...

So the most general solution is

alpha * A(x) + beta * B(x)

where A and B satisfy :

A(x+1) = A * A(x)

and

B(x+1) = B * B(x).

And those in term can be solved with a 1 periodic theta function each.

So our general solution ON THE COMPLEX PLANE is

alpha * A^(x + theta1(x)) + beta * B^(x + theta(2x))

wich reduces to the form

A^(x + theta1(x)) + B^(x + theta(2x))

if we igore alpha and beta being potentially 0.

NOTICE

REALPART ( A^(x + theta1(x)) + B^(x + theta(2x)) )

is also a real solution on the real line.

but not on the complex plane.

( not even if we make analytic continuation to undo the " real part " operator and have imaginary parts )

JUST LIKE a sine is not the equivalent of (-1)^t ; although it works on the real line and has an analytic continuation to the complex

IT FAILS on the complex plane.

So this is the critisism to taking real parts or using sine and cosine.

***

So now any solution

A^(x + theta1(x)) + B^(x + theta(2x))

that agrees on the fibonacci sequence can be considered a solution to the generalized fibonacci function , that is analytic and satisfies the equations on the complex plane.

***

And we have uniqueness critertions for the usual exponential ( a TPID which i proved and was ignored but never mind that now )

So that property carries over I think.

***

But the fibonacci satisfy many many equations ...

how about also satisfying the others ??

Is there a kind of semi-group homom ??

I investigated some other equations it ( fibo ) satisfied but they are not all compatible with a fixed generalization ( A^(x + theta1(x)) + B^(x + theta(2x)) ) or some just for fibo(odd) or fibo(even).

In fact imo the fibo is not the most interesting sequence of its kind.

What about sequences that are strictly increasing ? Much closer to the idea of iterations.

...

My idea is investigating expontial sums such as a given f(x) for instance :

f(x) = a^x + 2 b^x + c^x

This does not have the issues of polynomials and have already a closed form.

They will often also satisfy fibonacci like equations.

And working backwards is a nice idea.

Also when f(x) has less than 3 fixpoints on the reals we have the nice question

g(g(x)) = f(x).

which seems like a logical next step in tetration.

Notice that the recent nice cubic polynomial you gave was real valued and had two fixpoints where they both had analytic iterations at their two fixpoint.

However the cubic had a third real fixpoint which makes it not working near that fixpoint.

That is a typical property of cubics.

But for exponential sums we can make a function exp shaped with 0,1 or 2 fixpoints and fractional iterates analytic at them.

This would then give us a kind of semi-fibonacci.

And if the superfunction agrees on the regular iterates of both real fixpoints its a really nice thing !!

regards

tommy1729