Fibonacci as iteration of fractional linear function
#42
(08/12/2022, 06:47 PM)bo198214 Wrote:
(08/12/2022, 06:36 PM)Leo.W Wrote: Oh I see u meant eta minor as \(e^{-e}\) I thought it was \(e^{\frac{1}{e}}_-\)

That's eta proper (or maybe eta major) \(\eta=e^{\frac{1}{e}}\) - that's how we call it on this forum Smile

(08/12/2022, 06:36 PM)Leo.W Wrote: btw negative ones can be generated easily, here's another https://math.eretrandre.org/tetrationfor...p?tid=1351
I built about tetration base 0.5, at fixed point 0.707.

Well, I think I need a while to digest all your suggestions ...

(08/12/2022, 06:36 PM)Leo.W Wrote: Albeit these superfunctions would oscillate around the fixed point as a limit at infty, thus uninvertible, and thus would not grant you for \(f^s\circ f^t=f^{s+t}\), they're contradicts.
Well, then its not a superfunction, isn't it?! Just don't know whether James superfunction is a real superfunction ... this oscillating behaviour looked very similar to James function.

Have I been using the term "Superfunction" all wrong!?

I always assumed it simply meant a solution:

\[
f(F(t)) = F(t+1)
\]

In that sense my iteration at eta minor is a superfunction. But it's a superfunction that doesn't induce a fractional iteration because it's not \(1-1\); there's no way to pull out an abel function \(F^{-1}(t)\).

What other criteria do we add to the term "superfunction"? Do we necessitate some from of univalence?

I would assume your definition Bo, that we can make:

\[
f^{\circ t}(z_0) = F(t + F^{-1}(z_0))
\]

A little tighter, and stronger. But solving the super function equation doesn't necessitate this; just look at the \(\eta^-\) case. We can have infinite solutions to \(F^{-1}(z_0)\), especially because of how chaotic it is.




People are really piling on Bo here, not sure why...? Everything he's saying is correct. And I see a lot of disagreement which is largely just disagreements on syntax and semantics, and nothing concrete. Btw, when Leo writes \(\alpha^{-1}{f}(z)\), Leo means an inverse abel function of \(f\). I only know that because I've had the luxury of being on this forum whilst you've been away, bo.

To clarify, what Leo is saying is that the semi-group property will fail to hold for many of these solutions; by which he's calling it a "superfunction"--which goes back to the beginning of this post. That we don't always have the luxury of a meaningful inversion (we have a superfunction, but we don't have an abel function). So we can't do the inversion you are asking.

Honestly, I think 90% of the last few posts have been hung up on this distinction.


I think this would be a good time to settle our common terms. Something I wish Mphlee was here for, because him and I settled terms amongst each other. And we always said:

\[
f(F(t)) = F(t+1)\\
\]

means \(F\) is a super function of \(f\).

If:

\[
f^{\circ t}(z_0) = F(t+F^{-1}(z_0))\\
\]

Then \(F\) is an invertible superfunction. By which we'd be able to derive a semi-group property; if \(F^{-1}(z)\) was holomorphic in some significant domain--and these objects send to each other; so that we can write \(f^{\circ t}(z) : D \times H \to H\).



Maybe I'm speaking just from knowing what everyones idiosyncracies are in what they call things, I can understand everyone here. You guys are all in agreement, you're just using different words from each other.
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Messages In This Thread
RE: Fibonacci as iteration of fractional linear function - by JmsNxn - 08/12/2022, 11:47 PM

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