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 Fibonacci as iteration of fractional linear function tommy1729 Ultimate Fellow Posts: 1,699 Threads: 373 Joined: Feb 2009 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 JmsNxn Ultimate Fellow Posts: 977 Threads: 114 Joined: Dec 2010 08/12/2022, 11:47 PM (This post was last modified: 08/12/2022, 11:57 PM by JmsNxn.) (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 (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. bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/12/2022, 11:57 PM (This post was last modified: 08/12/2022, 11:57 PM by bo198214.) Quote: (08/12/2022, 05:38 PM)bo198214 Wrote: (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!? No, James, everything is alright. I was just absent minded when saying that. bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/15/2022, 05:47 PM (This post was last modified: 08/15/2022, 05:49 PM by bo198214.) I realize that my investigation of the relationship between Fibonacci Extensions and corresponding LFTs are barely thorough. Just for recapitulation We took the LFT $$f(z)=\frac{1}{1+z}$$ realized that $$f^{\circ n}(z)=\frac{\phi_{n-1}z+\phi_n}{\phi_n z+\phi_{t+1}}$$, where $$\phi_n$$ is the $$n$$-th Fibonacci number. So there could be a relation between non-integer iteration of $$f$$ and non-integer extension of $$\phi$$. However I just took the iteration of the LFT and plugged in the Fibonacci extension and claimed that it was the regular iteration of the LFTs. But I din't prove anything and just wondered that the alternative (real valued) Fibonacci extension didn't work (not an iteration group). Being an iteration group (though I say "group" I always also include semi-group in the meaning, it is just too long to write it out each time) means the following: $$f^{\circ s}\circ f^{\circ t} = f^{\circ s+t}, \quad f^{\circ 1} = f$$ So lets have a look at corresponding LFTs \begin{align} f^{\circ s}(f^{\circ t}(z)) &= \frac{\phi_s+\phi_{s-1}\frac{\phi_t + \phi_{t-1}z}{\phi_{t+1} + \phi_t z}}{\phi_{s+1}+\phi_s\frac{\phi_t + \phi_{t-1}z}{\phi_{t+1} + \phi_t z}}\\ &=\frac{\phi_s(\phi_{t+1} + \phi_t z)+\phi_{s-1}(\phi_t + \phi_{t-1}z)}{\phi_{s+1}(\phi_{t+1} + \phi_t z)+\phi_s(\phi_t + \phi_{t-1}z)}\\ &=\frac{\phi_s\phi_{t+1} + \phi_{s-1}\phi_t + (\phi_s\phi_t+\phi_{s-1}\phi_{t-1})z} {\phi_{s+1}\phi_{t+1} + \phi_s\phi_t + (\phi_{s+1}\phi_t+\phi_s\phi_{t-1})z} \end{align} So being an iteration group in terms of Fibonacci extension means the following peculiar identity: $$\phi_s\phi_{t+1} + \phi_{s-1}\phi_t = \phi_{s+t}$$ which reduces to the Fibonacci identity for t=1! So lets call this the extended Fibonacci identity. So lets check this at least for the standard Fibonacci extension $$\phi_t = \frac{\Phi^t - \Psi^t}{\Phi-\Psi}$$ \begin{align} \phi_s\phi_{t+1} + \phi_{s-1}\phi_t &= \frac{\Phi^s-\Psi^s}{\Phi-\Psi}\frac{\Phi^{t+1}-\Psi^{t+1}}{\Phi-\Psi} +\frac{\Phi^{s-1}-\Psi^{s-1}}{\Phi-\Psi}\frac{\Phi^t-\Psi^t}{\Phi-\Psi}\\ (\phi_s\phi_{t+1} + \phi_{s-1}\phi_t)(\Phi-\Psi)^2 &= \Phi^s\Phi^{t+1}-\Phi^s\Psi^{t+1}-\Psi^s\Phi^{t+1} + \Psi^s\Psi^{t+1} + \Phi^{s-1}\Phi^t - \Phi^{s-1}\Psi^t - \Psi^{s-1}\Phi^t + \Psi^{s-1}\Psi^s\\ &=\Phi^{s+t+1}-\Phi^s\Psi^t\Psi-\Psi^s\Phi^t\Phi + \Psi^{s+t+1}+ \Phi^{s+t-1} - \frac{1}{\Phi}\Phi^s\Psi^t - \frac{1}{\Psi}\Psi^s\Phi^t + \Psi^{s+t-1}\\ \frac{1}{\Phi}&=-\Psi\\ &=\Phi^{s+t+1} + \Psi^{s+t+1}+ \Phi^{s+t-1}  + \Psi^{s+t-1}\\ &=\Phi\Phi^{s+t} + \Psi\Psi^{s+t} - \Psi\Phi^{s+t} - \Phi\Psi^{s+t}\\ &=(\Phi-\Psi)\Phi^{s+t} - (\Phi-\Psi)\Psi^{s+t}\\ &=(\Phi-\Psi)^2\phi_{s+t} \end{align} So it is indeed true! The standard extension corresponds to an iteration group, and it is the regular one, because it is analytic at the fixed points. But - and now, Tommy, your aversion against the real-valued Fibonacci extensions becomes justified - this identity is not true for the real-valued Fibonacci extension $$\phi'_t=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}$$. Though it even satisfies what Leo.W. calls the Abelian property!: $$f^{\circ s}\circ f^{\circ t} = f^{\circ t}\circ f^{\circ s}$$ or in terms of Fibonacci $$\phi'_s\phi'_{t+1} + \phi'_{s-1}\phi'_t = \phi'_t\phi'_{s+1} + \phi'_{t-1}\phi'_s$$ But for example s=0.5 and t=0.5 do not sum up to 1 for the real-valued extension. So who accepts the challenge and finds a real valued solution to the extended Fibonacci identity?! I think one can not find a single valued solution because this would mean to have a single valued real iteration close to a fixed point of negative multiplier. See also this thread. I will continue here with finding a real (and multi) valued iteration group. But time is already too limited, so one later time. Sorry for not answering your replies, has also to be postponed. bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/15/2022, 07:11 PM And this is the most crooked thing I ever saw with iteration pictures     tommy1729 Ultimate Fellow Posts: 1,699 Threads: 373 Joined: Feb 2009 08/15/2022, 11:47 PM (08/15/2022, 05:47 PM)bo198214 Wrote: I realize that my investigation of the relationship between Fibonacci Extensions and corresponding LFTs are barely thorough. Just for recapitulation We took the LFT $$f(z)=\frac{1}{1+z}$$ realized that $$f^{\circ n}(z)=\frac{\phi_{n-1}z+\phi_n}{\phi_n z+\phi_{t+1}}$$, where $$\phi_n$$ is the $$n$$-th Fibonacci number. So there could be a relation between non-integer iteration of $$f$$ and non-integer extension of $$\phi$$. However I just took the iteration of the LFT and plugged in the Fibonacci extension and claimed that it was the regular iteration of the LFTs. But I din't prove anything and just wondered that the alternative (real valued) Fibonacci extension didn't work (not an iteration group). Being an iteration group (though I say "group" I always also include semi-group in the meaning, it is just too long to write it out each time) means the following: $$f^{\circ s}\circ f^{\circ t} = f^{\circ s+t}, \quad f^{\circ 1} = f$$ So lets have a look at corresponding LFTs \begin{align} f^{\circ s}(f^{\circ t}(z)) &= \frac{\phi_s+\phi_{s-1}\frac{\phi_t + \phi_{t-1}z}{\phi_{t+1} + \phi_t z}}{\phi_{s+1}+\phi_s\frac{\phi_t + \phi_{t-1}z}{\phi_{t+1} + \phi_t z}}\\ &=\frac{\phi_s(\phi_{t+1} + \phi_t z)+\phi_{s-1}(\phi_t + \phi_{t-1}z)}{\phi_{s+1}(\phi_{t+1} + \phi_t z)+\phi_s(\phi_t + \phi_{t-1}z)}\\ &=\frac{\phi_s\phi_{t+1} + \phi_{s-1}\phi_t + (\phi_s\phi_t+\phi_{s-1}\phi_{t-1})z} {\phi_{s+1}\phi_{t+1} + \phi_s\phi_t + (\phi_{s+1}\phi_t+\phi_s\phi_{t-1})z} \end{align} So being an iteration group in terms of Fibonacci extension means the following peculiar identity: $$\phi_s\phi_{t+1} + \phi_{s-1}\phi_t = \phi_{s+t}$$ which reduces to the Fibonacci identity for t=1! So lets call this the extended Fibonacci identity. So lets check this at least for the standard Fibonacci extension $$\phi_t = \frac{\Phi^t - \Psi^t}{\Phi-\Psi}$$ \begin{align} \phi_s\phi_{t+1} + \phi_{s-1}\phi_t &= \frac{\Phi^s-\Psi^s}{\Phi-\Psi}\frac{\Phi^{t+1}-\Psi^{t+1}}{\Phi-\Psi} +\frac{\Phi^{s-1}-\Psi^{s-1}}{\Phi-\Psi}\frac{\Phi^t-\Psi^t}{\Phi-\Psi}\\ (\phi_s\phi_{t+1} + \phi_{s-1}\phi_t)(\Phi-\Psi)^2 &= \Phi^s\Phi^{t+1}-\Phi^s\Psi^{t+1}-\Psi^s\Phi^{t+1} + \Psi^s\Psi^{t+1} + \Phi^{s-1}\Phi^t - \Phi^{s-1}\Psi^t - \Psi^{s-1}\Phi^t + \Psi^{s-1}\Psi^s\\ &=\Phi^{s+t+1}-\Phi^s\Psi^t\Psi-\Psi^s\Phi^t\Phi + \Psi^{s+t+1}+ \Phi^{s+t-1} - \frac{1}{\Phi}\Phi^s\Psi^t - \frac{1}{\Psi}\Psi^s\Phi^t + \Psi^{s+t-1}\\ \frac{1}{\Phi}&=-\Psi\\ &=\Phi^{s+t+1} + \Psi^{s+t+1}+ \Phi^{s+t-1}  + \Psi^{s+t-1}\\ &=\Phi\Phi^{s+t} + \Psi\Psi^{s+t} - \Psi\Phi^{s+t} - \Phi\Psi^{s+t}\\ &=(\Phi-\Psi)\Phi^{s+t} - (\Phi-\Psi)\Psi^{s+t}\\ &=(\Phi-\Psi)^2\phi_{s+t} \end{align} So it is indeed true! The standard extension corresponds to an iteration group, and it is the regular one, because it is analytic at the fixed points. But - and now, Tommy, your aversion against the real-valued Fibonacci extensions becomes justified - this identity is not true for the real-valued Fibonacci extension $$\phi'_t=\frac{\Phi^t+\cos(\pi t)(-\Psi)^t}{\Phi-\Psi}$$. Though it even satisfies what Leo.W. calls the Abelian property!: $$f^{\circ s}\circ f^{\circ t} = f^{\circ t}\circ f^{\circ s}$$ or in terms of Fibonacci $$\phi'_s\phi'_{t+1} + \phi'_{s-1}\phi'_t = \phi'_t\phi'_{s+1} + \phi'_{t-1}\phi'_s$$ But for example s=0.5 and t=0.5 do not sum up to 1 for the real-valued extension. So who accepts the challenge and finds a real valued solution to the extended Fibonacci identity?! I think one can not find a single valued solution because this would mean to have a single valued real iteration close to a fixed point of negative multiplier. See also this thread. I will continue here with finding a real (and multi) valued iteration group. But time is already too limited, so one later time. Sorry for not answering your replies, has also to be postponed. yes ! that is what i was aiming at in part. this combines many ideas from the regular posters here. the fibonacci addition formula was explictly recently mentioned by me in one of the recent related threads. it is way more general than the fibonacci standard recursion. in fact it is almost analogue to solving f(2x) = f(x)^2 or f(x+1) = e f(x) where both have f(x) = exp(x) as a solution. notice the fibo addition formula has less freedom in its solutions. especially no " if f is a solution so is constant times f ". if we add ( to the addition formula ) the condition of boundedness on a region we have uniqueness of the fibo. 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. this relates to james ideas of fractional integrals , integral transforms and ramanujan. or kouznetsov. basically all no surprise : fibo is just in essense iterating 2 exponentials ; the eigenvalues. 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 ) So im still very skeptical and the idea is that the fibo will just have the same uniqueness conditions as the exponentials do ... because they ARE two exponentials with base the eigenvalue. --- more interesting is how tribonacci might be related to analogues of the linear fraction iterations ?? if fibo is essentially iterations of LF then what is tribonacci essentially an iteration of ???????? --- fractional derivates remains a road of possible interest though. --- regards tommy1729 bo198214 Administrator Posts: 1,594 Threads: 101 Joined: Aug 2007 08/17/2022, 12:53 PM (08/15/2022, 11:47 PM)tommy1729 Wrote: 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? Quote: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 Quote: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: (06/16/2022, 11:05 PM)tommy1729 Wrote: 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. (08/15/2022, 11:47 PM)tommy1729 Wrote: this relates to james ideas of fractional integrals , integral transforms and ramanujan. or kouznetsov.Yes, everything is connected Quote: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 $$h(z+1)=1+h(z)$$. (08/15/2022, 11:47 PM)tommy1729 Wrote: 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. Quote: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 tommy1729 Ultimate Fellow Posts: 1,699 Threads: 373 Joined: Feb 2009 08/17/2022, 11:09 PM tommy1729 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? ** I mentioned it for instance here : https://math.eretrandre.org/tetrationfor...p?tid=1616 post 6. Quote: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 **  yes but the original fibo extented to the reals satisfies that. it can be shown in many ways. but the matrix eigenvalues concept does. Quote: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: (06/16/2022, 11:05 PM)tommy1729 Wrote: 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. ** what are you talking about ? We know all superfunctions agreeing on 1 point are related by a 1-periodic theta function. ** (08/15/2022, 11:47 PM)tommy1729 Wrote: this relates to james ideas of fractional integrals , integral transforms and ramanujan. or kouznetsov.Yes, everything is connected Quote: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 $$h(z+1)=1+h(z)$$. ** not sure what to say ... ** (08/15/2022, 11:47 PM)tommy1729 Wrote: 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. Quote: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 **  yeah it was a vague idea. but i think it might be interesting. regards tommy1729 ** Gottfried Ultimate Fellow Posts: 872 Threads: 128 Joined: Aug 2007 09/14/2022, 08:05 AM (This post was last modified: 09/14/2022, 09:50 AM by Gottfried.) (08/11/2022, 06:57 PM)bo198214 Wrote: Just to see what we are talking about with this fib2 variant: $$\text{fib}_2(t) = \frac{\Phi^t - \left|\Psi\right|^t}{\Phi-\Psi}$$ The black dots are the real Fibonacci numbers. Maybe it is also an option to somehow theta-distort the fib2 to fit fib. But then again, if we have one solution then there are too many more. Just to add the reference to an interesting discussion of the Binet/complex and the real/cos()-concepts    in MSE: https://math.stackexchange.com/questions...or-complex       Sheldon answers on my question about this, and relates the Schroeder and the Kneser-attempt to the two methods. He gave two answers, perhaps the second answer is the more powerful one... An answer to a later, related question gives also a plot for the real-to-real function at https://math.stackexchange.com/a/798221 Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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