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+--- Thread: Fibonacci as iteration of fractional linear function (/showthread.php?tid=1607)
RE: Fibonacci as iteration of fractional linear function - Leo.W - 08/12/2022
(08/12/2022, 04:40 PM)bo198214 Wrote:Well bo, if you take abelian property, it'll fail then since this function can be conjugated by diagonal: \(\phi=\frac{\sqrt{5}-1}{2}\)(08/12/2022, 12:05 PM)bo198214 Wrote: However I wonder why it seems to be regular at the right fixed point ...
It's because it is not an iteration at all. It satisfies \(f^{\circ t+1} = \frac{1}{1+f^{\circ t}} \) (which corresponds to the Fibonacci identity) but it does not satisfy \(f^{\circ s+t}=f^{\circ s}\circ f^{\circ t}\), while the original Fibonacci extension is a true (and regular) iteration.
So the question still stands to find a real valued super function \(\frac{1}{1+z}\).
\(f(z)=\frac{1}{1+z},g(z)=\frac{\phi z-\phi-1}{z+1},g^{-1}(f(g(z)))=-(\phi+2)z\)
When you're asking a real valued super function of \(f(z)=\frac{1}{1+z}\)
You're actually transforming a super function problem to another, the relation between the superfunctions of \(f(z)\) and \(s(z)=-(\phi+2)z\) is
\(\alpha^{-1}\{f\}(z)=g(\alpha^{-1}\{s\}(z))\)
And since \(g:\mathbb{R}\to\mathbb{R}\), to find a function \(\alpha^{-1}\{f\}:\mathbb{R}\to\mathbb{R}\), you've equitably to find a real-valued superfunction of \(s(z)\approx-2.618z\)
That is to solve \(R:\mathbb{R}\to\mathbb{R},R(z+1)=-2.618R(z)\)
And as you take abelian property, if you assume \(s^t(z)=R(t+R^{-1}(z))\) you have to solve
\(R:\mathbb{R}\to\mathbb{R},R(z+1)=-2.618R(z)\) for a monotonic R
If it's continuous then it has to be a constant... so no solution if allowing the superfunction to be invertible
If not assumed \(s^t(z)=R(t+R^{-1}(z))\), you still have to work out an abelian iteration mapping R to R for s(z) lol
RE: Fibonacci as iteration of fractional linear function - bo198214 - 08/12/2022
(08/12/2022, 05:21 PM)Leo.W Wrote: Well bo, if you take abelian property, it'll fail then since this function can be conjugated by diagonal: \(\phi=\frac{\sqrt{5}-1}{2}\)Seems really we have to find a common language here:
\(f(z)=\frac{1}{1+z},g(z)=\frac{\phi z-\phi-1}{z+1},g^{-1}(f(g(z)))=-(\phi+2)z\)
When you're asking a real valued super function of \(f(z)=\frac{1}{1+z}\)
You're actually transforming a super function problem to another, the relation between the superfunctions of \(f(z)\) and \(s(z)=-(\phi+2)z\) is
\(\alpha^{-1}\{f\}(z)=g(\alpha^{-1}\{s\}(z))\)
And since \(g:\mathbb{R}\to\mathbb{R}\), to find a function \(\alpha^{-1}\{f\}:\mathbb{R}\to\mathbb{R}\), you've equitably to find a real-valued superfunction of \(s(z)\approx-2.618z\)
That is to solve \(R:\mathbb{R}\to\mathbb{R},R(z+1)=-2.618R(z)\)
And as you take abelian property, if you assume \(s^t(z)=R(t+R^{-1}(z))\) you have to solve
\(R:\mathbb{R}\to\mathbb{R},R(z+1)=-2.618R(z)\) for a monotonic R
So what is \(\alpha^{-1}\{f\}(z)\) what is this braces notation?
What is the Abelian property?
(08/12/2022, 05:21 PM)Leo.W Wrote: If it's continuous then it has to be a constant... so no solution if allowing the superfunction to be invertible
Yeah that was the thing with JmsNxn's superfunction at eta minor: It was not invertible, rather close to periodic.
RE: Fibonacci as iteration of fractional linear function - Leo.W - 08/12/2022
(08/12/2022, 05:38 PM)bo198214 Wrote: So what is \(\alpha^{-1}\{f\}(z)\) what is this braces notation?James did many great works, yeah
What is the Abelian property?
Forgive me. Abelian means commutive, as s and t do in \(f^s\circ f^t=f^{s+t}\)
(08/12/2022, 05:21 PM)Leo.W Wrote: If it's continuous then it has to be a constant... so no solution if allowing the superfunction to be invertible
Yeah that was the thing with JmsNxn's superfunction at eta minor: It was not invertible, rather close to periodic.
but about this problem Idk if anyone can build up such superfunctions?
The eta minor case, a little similar to the kneser's tetration, they won't map \(\mathbb{R}\) to \(\mathbb{R}\) but a sub-domain of real axis. And the fixed point's multiplier were positive, not negative as ours.
It's also, not compatible and comparable to our \(f(z)=\frac{1}{1+z}\) (should take infinity into account though)
you can firstly try to build a \(f^{\frac{1}{2}}\) as an attempt, this would fail by previous analysis. It's equivalent to build a square root of -2.618.
(lmao this expression so funny)
RE: Fibonacci as iteration of fractional linear function - bo198214 - 08/12/2022
(08/12/2022, 05:51 PM)Leo.W Wrote: And the fixed point's multiplier were positive, not negative as ours.
It is negative, that's what I mean with similar to our case.
\(f(x)=b^x\), \(b=\eta_-=e^{-e}\), parabolic fixed point \(z_0=\frac{1}{e}\), \(f'(x)=\log(b)b^x\), \(f'(z_0)=\log(b)z_0=-e\frac{1}{e}=-1\).
RE: Fibonacci as iteration of fractional linear function - Leo.W - 08/12/2022
(08/12/2022, 06:29 PM)bo198214 Wrote:(08/12/2022, 05:51 PM)Leo.W Wrote: And the fixed point's multiplier were positive, not negative as ours.
It is negative, that's what I mean with similar to our case.
\(f(x)=b^x\), \(b=\eta_-=e^{-e}\), parabolic fixed point \(z_0=\frac{1}{e}\), \(f'(x)=\log(b)b^x\), \(f'(z_0)=\log(b)z_0=-e\frac{1}{e}=-1\).
Oh I see u meant eta minor as \(e^{-e}\) I thought it was \(e^{\frac{1}{e}}_-\)
btw negative ones can be generated easily, here's another https://math.eretrandre.org/tetrationforum/showthread.php?tid=1351
I built about tetration base 0.5, at fixed point 0.707.
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.
RE: Fibonacci as iteration of fractional linear function - bo198214 - 08/12/2022
(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
(08/12/2022, 06:36 PM)Leo.W Wrote: btw negative ones can be generated easily, here's another https://math.eretrandre.org/tetrationforum/showthread.php?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.
RE: Fibonacci as iteration of fractional linear function - Leo.W - 08/12/2022
(08/12/2022, 06:47 PM)bo198214 Wrote:nope, the contradicts happen because things like the way u may wish a real-to-real (-1)^x for real x(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
- Yeah thx bo
(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.
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 \(f^s\circ f^t=f^{s+t}\)
I interpret your post as to find a superfunction that is real-to-real and also preserve the property \(f^s\circ f^t=f^{s+t}\), and in that sense it's impossible though, you can take it as extending (-1)^x to a real-to-real function while preserving \((-1)^s(-1)^t=(-1)^{s+t}\)
if my interpretation donot commersurate with your original intention just ignore me lol
It's 2am now I better go sleep
RE: Fibonacci as iteration of fractional linear function - tommy1729 - 08/12/2022
(08/12/2022, 07:00 PM)Leo.W Wrote:(08/12/2022, 06:47 PM)bo198214 Wrote:nope, the contradicts happen because things like the way u may wish a real-to-real (-1)^x for real x(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
- Yeah thx bo
(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.
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 \(f^s\circ f^t=f^{s+t}\)
I interpret your post as to find a superfunction that is real-to-real and also preserve the property \(f^s\circ f^t=f^{s+t}\), and in that sense it's impossible though, you can take it as extending (-1)^x to a real-to-real function while preserving \((-1)^s(-1)^t=(-1)^{s+t}\)
if my interpretation donot commersurate with your original intention just ignore me lol
It's 2am now I better go sleep
agreed.
very much agreed.
and it seems you understand the concept of (local) semi-group homom pretty good !
I bet and wonder about your ideas about what i more or less explained about semi-group homom and the 2sinh method supposedly having it.
regards
tommy1729
RE: Fibonacci as iteration of fractional linear function - bo198214 - 08/12/2022
(08/12/2022, 05:38 PM)bo198214 Wrote: So what is \(\alpha^{-1}\{f\}(z)\) what is this braces notation?
You still owe an explanation for this ...
(08/12/2022, 07:00 PM)Leo.W Wrote: nope, the contradicts happen because things like the way u may wish a real-to-real (-1)^x for real xHm, 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.
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 \(f^s\circ f^t=f^{s+t}\)
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?!
(08/12/2022, 07:00 PM)Leo.W Wrote: 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
RE: Fibonacci as iteration of fractional linear function - tommy1729 - 08/12/2022
(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