Constructing a real valued Fibonacci iteration--its relation to $$1/1+z$$ JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/11/2022, 02:39 AM (This post was last modified: 08/11/2022, 07:00 AM by JmsNxn.) Continuing from the thread Bo posted about mapping the fibonacci iteration to the iteration of an LFT: https://math.eretrandre.org/tetrationfor...p?tid=1607 Which, ultimately, is an isomorphism. Represents a solution to: $$\phi(t) = \phi(t-1) + \phi(t-2)\\$$ And its relations to iterates of $$f(z) = \frac{1}{1+z}\\$$ We are going to try and relate these objects in a more general sense. This means, we are going to consider: $$f^{\circ t}(z) = \frac{\phi(t) +\phi(t-1)z}{\phi(t+1) + \phi(t)z}\\$$ And we are going to ask, when we can produce an iteration $$z \in \mathbb{R}$$ and $$t \in \mathbb{R}$$ that is real valued (disregarding if there are poles). We are first of all, going to assume that $$\phi(t)$$ is entire in $$t$$. We don't care about not entire solutions in $$t$$. And additionally, we will always assume that: $$\phi(0) = 0\,\,\,\phi(1) = 1\,\,\,\phi(t) = \phi(t-1) + \phi(t-2)\\$$ ------------------------------------------------------------------------- To begin, we can write every entire solution to this equation. (This was asked by myself on MathOverflow a long time ago, about the uniqueness of entire linear recurrence relations. I can find the linked article and the reasoning--but until asked I'm too lazy to find it.) Ultimately this isn't that hard to prove for Fibonacci, but it is an advanced result for general linear recurrences. Firstly we take $$\Phi$$ and $$\Psi$$ as Bo described, which are the golden ratio, and its conjugate pair for its minimal polynomial. Then we write: $$\mu^+_k = \log(\Phi) + 2\pi i k\,\,\,\mu^-_j = \log\Psi + 2 \pi i j\,\,\,\,\text{for}\,\,j,k \in \mathbb{Z}\\$$ Then there exists the fundamental solutions of the fibonacci equation, for entire functions, which is written as: $$\phi_{kj}(t) = \frac{e^{\mu^+_k t} + e^{\mu^-_j t}}{\Phi - \Psi}\\$$ This is clearly not real valued, which you can always check. But, each version is a more chaotic version of Bo's iteration of $$\frac{1}{1+z}$$, where: $$f^{\circ t}(z) = \frac{\phi_{kj}(t) + \phi_{kj}(t-1)z}{\phi_{kj}(t+1) - \phi_{kj}(t)z}\\$$ So, this creates a selection of $$\mathbb{Z}^2$$ iterates, which are never real valued. Now we superimpose these solutions on each other. So let's let $$a_{jk} \in \mathbb{C}$$ and additionally, is summable across both indices. Then: $$\phi(t) = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} \phi_{kj}(t)\\$$ While $$\phi(0) = 0$$ and $$\phi(1) = \sum_{j=-\infty}^\infty \sum_{k=-\infty}^\infty a_{jk} = 1$$. This describes every possible entire solution to the fibonacci equation. And this translates to not every iteration of $$f = \frac{1}{1+z}$$, but for sake of brevity, the nicest ones; where the singularities, are in some sense, local. $$f^{\circ t}(z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) - \phi(t)z}\\$$ DISCLAIMER: What I had previously written as $$\theta(z)\phi(z)$$, I mean by the above solution. What I had written before is a valid method, but it proves to be extraneous and non necessary; and additionally would produce Bo's iteration by fact: $$\frac{\theta(t)\phi(t) + \theta(t-1)\phi(t-1)z}{\theta(t+1)\phi(t+1) - \theta(t)\phi(t)z} = \frac{\theta(t)\phi(t) + \theta(t)\phi(t-1)z}{\theta(t)\phi(t+1) - \theta(t)\phi(t)z} = f^{\circ t}(z)\\$$ This means for every fibonacci iteration, there is an iteration of $$\frac{1}{1+z}$$, but for every iteration of $$\frac{1}{1+z}$$ there are many fibonacci iterations. So this isn't a 1-1 correspondence that Bo's described, it's many to one. -------------------------------------------------------------------------------- And largely why I think Bo's solution is unique by construction. Every entire function $$\phi(t)$$ which satisfies the fibonacci equation; must satisfy a strict representation. Let's call Bo's solution, which is the standard from Binet's formula. That: $$\phi_{00}(t) = \frac{\Phi^t - \Psi^t}{\Phi - \Psi}\\$$ Then our solution $$\phi(t)$$ looks like: $$\phi(t) = \sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^\infty a_{jk} \frac{e^{2 \pi i k t} \Phi^t + e^{2 \pi i j t}\Psi^t}{\Phi - \Psi}\\$$ Which equates to: $$\phi(t) = \frac{\theta^+(t) \Phi^t + \theta^-(t) \Psi^t}{\Phi - \Psi}\\$$ And now we ask, where this thing is real valued. In order for that to happen, we can reduce to: $$\overline{\phi(t)} = \phi(\overline{t})\\$$ Which devolves into solving n'th order polynomial solutions for $$t \approx 0$$; and solving the Matrix equation for $$n\times n$$.  By which we are just solving a second order equation in $$\theta^{\pm}$$. Anyone more proefficient in matrix coefficients is welcome to jump in here. This is a strong statement in more scenarios, but for us it relates a standard linear recurrence relationship, to the complex iteration of a function. ------------------------------------------------------------- We should always be able to solve for $$\theta^+$$ and $$\theta^-$$, and by consequence solve for $$a_{jk}$$. Uniqueness is another problem. But this would produce the real valued version of Bo's fibonacci iteration of $$1/1+z$$. By which, we now have: \begin{align} f^{\circ t}(z) &: \mathbb{C} \times \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\\ f^{\circ t}(z) &: \mathbb{R} \times \mathbb{R} \to \mathbb{R}\cup\{\infty\}\\ \end{align} I'm hoping someone else can run the numbers. But I'd call this the "crescent iteration of $$f= 1/1+z$$". This is related, in more ways than one, to the crescent iteration as we think of it. When you take n'th order solutions to Kneser's Taylor series at $$s=0$$, we end up doing something very similar to finding an n'th order expansion of $$\phi(t)$$ at $$t=0$$. These are inextricably linked. JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/11/2022, 03:05 AM I'd also like to add why the beta method doesn't solve this problem instantly. The beta method allows us to make: $$f^{\circ t}(z) : \mathbb{R} \times \mathbb{R} \to \mathbb{R}\cup\{\infty\}\\$$ But, it produces a fibonacci solution $$\phi(t)$$ which has poles all over the place. The theory of "producing the crescent iteration using the beta method" would be to some how use the beta solutions to create an entire $$\phi(t)$$ as above. Which is unique, considering what we are asking of $$\theta^+$$ and $$\theta^-$$. A lot to unpack here, I'm just trying to be as straight forward as possible. bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/11/2022, 03:58 PM Hm, I think uniqueness is exactly the problem here. If we have one real valued solution then we have a lot of real valued solutions via a real valued theta. Maybe one can calculate some, but without having the best, its kinda futile. Btw. this reminds me of an old question that I asked in sci.math.research about the uniqueness of the Fibonacci number extension. Because everyone knows that $$\frac{\Phi^t-\Psi^t}{\Phi-\Psi}$$ *is* an extension, but nobody knows about the *uniqueness* of this extension. https://groups.google.com/g/sci.math.res...WX5ZriAm8I The answer of Waldek Hebisch actually inspired me to write the article about the uniqueness of holomorphic Abel functions. (Which then later turned out to be known by those Perturbed-Fatou-Coordinate guys.) And here in the thread we found also another uniqueness criterion, that it is the regular iteration of the corresponding linear fractional function. So we have a lot of uniqueness criterions for a non-real valued function, damn! bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 08/12/2022, 05:11 PM (08/11/2022, 02:39 AM)JmsNxn Wrote: $$\phi(t) = \frac{\theta^+(t) \Phi^t + \theta^-(t) \Psi^t}{\Phi - \Psi}\\$$ And now we ask, where this thing is real valued. No, we looking for a real-valued superfunction of f (as you did with eta minor). Because the derivatives $$\frac{-1}{(1+x)^2}$$ at the fixed points $$-\Phi,-\Psi$$ are negative, (also remember from here that the fixed point derivatives are the quotient of the eigenvalues, i.e. $$\frac{\Phi}{\Psi}$$ and reciprocal, and $$\Psi$$ is negative.) but at least asymptotically (if not regular) you need to have $$\left(f^{\circ t}\right)'(z_0) = f'(z_0)^t$$, which means in a small vicinity it can not be real valued for non-integer t. That's why I was so surprised about your real valued superfunction at eta minor! tommy1729 Ultimate Fellow Posts: 1,703 Threads: 374 Joined: Feb 2009 08/12/2022, 09:33 PM 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 tommy1729 Ultimate Fellow Posts: 1,703 Threads: 374 Joined: Feb 2009 08/12/2022, 10:44 PM (08/12/2022, 09:33 PM)tommy1729 Wrote: 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 In particular I wonder about real valued fibonacci numbers  1) that satisfy the addition formula Fib(t+s) = Fib(t) * Fib(s+1) + Fib(t-1) * Fib(s). 2) Fib(t) defined as the t ' th fractional derivative of x / (1 - x - x^2 ) ( and ofcourse which fractional derivative ! ) regards tommy1729 JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/12/2022, 11:12 PM (08/12/2022, 05:11 PM)bo198214 Wrote: (08/11/2022, 02:39 AM)JmsNxn Wrote: $$\phi(t) = \frac{\theta^+(t) \Phi^t + \theta^-(t) \Psi^t}{\Phi - \Psi}\\$$ And now we ask, where this thing is real valued. No, we looking for a real-valued superfunction of f (as you did with eta minor). Because the derivatives $$\frac{-1}{(1+x)^2}$$ at the fixed points $$-\Phi,-\Psi$$ are negative, (also remember from here that the fixed point derivatives are the quotient of the eigenvalues, i.e. $$\frac{\Phi}{\Psi}$$ and reciprocal, and $$\Psi$$ is negative.) but at least asymptotically (if not regular) you need to have $$\left(f^{\circ t}\right)'(z_0) = f'(z_0)^t$$, which means in a small vicinity it can not be real valued for non-integer t. That's why I was so surprised about your real valued superfunction at eta minor! Hmmm, I'm confused then. If we have a fibonacci function, say $$\phi(t)$$; which satisfies $$\phi(0) = 0$$ and $$\phi(1) = 1$$; and if it is entire; isn't it necessary that: $$f^{\circ t}(z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) + \phi(t)z}\\$$ Is a fractional iteration? And additionally, the solution I'm suggesting will behave like: $$f'(z_0)^t \theta(t)\\$$ I'm pretty sure that's still okay... but only as a super function; not as an iterate. AHHHHH I see now! This only produces a super function, this doesn't produce a fractional iteration!! So if you went about my route, you'd have: $$F(t,z) = \frac{\phi(t) + \phi(t-1)z}{\phi(t+1) + \phi(t)z}\\$$ Where: $$f(F(t,z)) = F(t+1,z)$$. But you wouldn't have: $$F(s,F(t,z)) = F(t+s,z)\\$$ Yes, okay now I see where my confusion is. Absolutely we can use the regular iteration argument as that forcing our solution to be unique (to both fibonacci, and the iterate of $$1/1+z$$). But when we weaken, leaving the semi-group property, and only consider super functions; then we can make a real valued function super function. It will probably look a lot like $$\eta^-$$ to be honest. I think the only way for me to convince you is to run some code though; so let's get at it! JmsNxn Ultimate Fellow Posts: 985 Threads: 117 Joined: Dec 2010 08/13/2022, 12:05 AM (This post was last modified: 08/13/2022, 12:05 AM by JmsNxn.) It appears you have already beat me to the chase with $$\phi(t) = \frac{\Phi^t - \cos(\pi t)|\Psi|^t}{\Phi-\Psi}\\$$ Which has the exact same shape as I wrote above. The real question is uniqueness then! I think we'll get uncountably infinite real super functions. Jesus! « Next Oldest | Next Newest »

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