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 Some "Theorem" on the generalized superfunction Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 08/01/2021, 11:23 AM Complement: Another example of multivalued iteration We consider now the second iterative root of the function $f(z)=z+\frac{1}{z}$ And we can observe that the symmetry law $fg=f\to{f^tg=f}$ applies to all of $f(z),f^{1/2}(z),\frac{1}{f(z)},\frac{1}{f^{1/2}(z)}$ Some of these statements that involving set theory are still hypothetical, remaining to be proved (especially the A(z) part) Attached Files Image(s)         JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/02/2021, 08:08 PM (This post was last modified: 08/02/2021, 08:17 PM by JmsNxn.) Very interesting, Leo! This is probably related to the fundamental group of the Riemann Surface of $g$. Where you have a general Riemann surface of the iterates; and there exists 4 canonical projections to $\mathbb{C}$; and we then create an algebra from them. Something like that--it may not be the fundamental group that you're describing; but it definitely borders it. Can't be certain, but it definitely looks like it. This is again, though; why Riemann surfaces are so important. The algebra unveils in a general setting. Very interesting, though. I like the rough work of it you did. Very cool! Regards, James You definitely got me thinking. I think I can write this a tad more general; but I'd prefer starting from $f(z) = z - 1/z$ because it's got simpler features. Then generalize to how much more complex $z+1/z$ is. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/04/2021, 03:58 AM (This post was last modified: 08/04/2021, 04:02 AM by JmsNxn.) Taking a similar approach as Leo; this is largely Leo's discussion; but we're going to switch to $f(z) = z - 1/z$; the reason for this switch is to reduce the group structure. Let's talk about the abel functions of $f$; which produce two branches: $ \alpha_+(f(z)) = \alpha_+(z) + 1\,\,\text{for}\,\, \Im(z) > 0\\ \alpha_-(f(z)) = \alpha_-(z) + 1 \,\,\text{for}\,\, \Im(z) <0\\$ In which: $ \alpha_{+}(z) : \mathbb{H} \to \mathbb{H}\\ \alpha_{+}(\infty) = \infty\\$ Where: $ \mathbb{H} = \{z \in \mathbb{C}\,|\, \Im(z) > 0\}\\$ And vice versa with $\alpha_{-}$.  We can define two half iterations as: $ g_{12}(z) = \alpha_{\pm}^{-1}(\alpha_{\pm}(z) + 1/2)\\$ The beauty of this example is that the domains of definition do not intersect. We have a natural boundary along $\mathb{R}$ which is the julia set of $g$. If we take $g_1(g_2)$ it is nonsense because one exists in a projection to the upper half plane; and the other, the lower half plane. You are doing something like this, Leo. You're just doing it for a much more complicated group structure. But this is where the fundamental group comes in. The fundamental group of the above example is just two non intersecting simply connected domains. In your example it's definitely trickier. I enjoy your posts, Leo; I hope this makes sense. Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 08/05/2021, 04:36 PM (This post was last modified: 08/05/2021, 06:20 PM by Leo.W. Edit Reason: typo ) (08/04/2021, 03:58 AM)Jms Nxn Wrote: Taking a similar approach as Leo; this is largely Leo's discussion; but we're going to switch to $f(z) = z - 1/z$; the reason for this switch is to reduce the group structure. ... I enjoy your posts, Leo; I hope this makes sense.Indeed. I brought up this example because it shows what originally the symmetry law talks about, is that, the functions $f(z),\frac{1}{f(z)}$ by the original symmetry $f(z)=f(z^{-1})\to{f^{-1}(z)=f^{-1}(z)^{-1}\to{f(z)=f(z)^{-1}}}$  and their iterative family $f^t(z),(z\to\frac{1}{f(z)})^t(z)(t\ne0)$ are all sharing the same Riemann surface in a sense that whenever they're considered as multivalued, or sharing each other's iterative family in some certain domain, in this case, $f(z),\frac{1}{f(z)}$ shared their second iterative root, and if you check, they even share their inverse. Or to say, that these 2 functions, are inseperable when considering Riemann surface or multivalued function. However, please let me have some different opinions from that "If we take $g_1(g_2)$ it is nonsense because one exists in a projection to the upper half plane; and the other, the lower half plane." In more precise cases, we should also consider the cases that $g_1,g_2$ are available to be defined outer of their original domain, as analytic continuation, since then their combination makes sense and may derive another piece of the Riemann surface of f. Except for the hadamard functions and q-generalized functions that have poor condition and very hard to re-define beyond their domain.(Anytime we can Analytically continue f(z)'s domain, then we cancontinue such g1(g2(z)) ) We can check out this simply in the construction of the real-natural-tetration, where the 2 Abel functions are taken from the 0.318+1.337i fixed point pair, using the limit definition, we can define the Abel functions only in halfplane, but slightly change the limit definition by using ln(z)+2pi*k*i with k integer, we can then continue the definition and generate every branch cuts of Abel function. I also tried this strategy out in the theta-mapping of the quadratic function $f(z)=z^2+i$, where the original inverse function $f^{-1}(z)=\sqrt{z-i}$ CAN NOT be used to construct the Abel function, so Can not construct the theta-mapping. But changing the branch cuts of inverse of f, it works. I lately worked out the proof, that any polynomials having no multiple fixed points ($f(z)=z+(z-L)^kg(z)$, g is polynomial, k>1 integer then L is called a multiple fixed point), always have a fixed point L which is constructable($\frac{ln(f'(L))}{\pi*i}\notin\mathbb{Q}\setminus{0}$), or is available to generalize the polynomial function's iterations from that point. I wonder if you have any suggestion on this very important aforementioned question: (i'm appreciative) Do any analytic functions, or differentiable Riemann surfaces, have at least one fixed point L which is constructable? And do you already solved the noncontructable cases? I'm like, stuck in the very initial case $f(z)=-z(1-z)$ at $f(0)=0,f'(0)=-1$, I wonder if anyone had already successfully construct a second iterative root? Though if you assume the second iterative root has a Taylor series at z=0, quickly sooner you'll notice it's not helpful, contradicting the fundamental laws of iterated functions. Regards Leo JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/06/2021, 01:35 AM (This post was last modified: 08/06/2021, 04:09 AM by JmsNxn.) (08/05/2021, 04:36 PM)Leo.W Wrote: And do you already solved the noncontructable cases? I'm like, stuck in the very initial case $f(z)=-z(1-z)$ at $f(0)=0,f'(0)=-1$, I wonder if anyone had already successfully construct a second iterative root? Though if you assume the second iterative root has a Taylor series at z=0, quickly sooner you'll notice it's not helpful, contradicting the fundamental laws of iterated functions. Regards Leo This is certainly computable. Start by taking $f^{\circ 2}(z)$ which has a fixed point $f(0)=0,\,f'(0) = 1$. From here we would use Ecalle's construction of the abel function. It will not be holomorphic in a neighborhood of zero--you are correct; zero will be on the boundary. The function $f^{\circ 2}(z)$ has an "attracting petal" and a "repelling petal". Where a Petal will be a domain in $\mathbb{C}$ in which; $ \lim_{n\to\infty} f^{\circ 2n}(z) = 0\,\,\text{for the attracting petal}\\ \lim_{n\to\infty} f^{\circ 2n}(z) \,\text{diverges for the repelling petal}\\$ The value $0 \in \overline{\mathcal{P}_{\pm}}$ where $\mathcal{P}_{\pm}$ is the attracting/repelling petal respectively. I'm going to follow Milnor here and give you a quick sketch on how to construct the Abel function. I highly suggest reading John Milnor's Dynamics in one complex variable. This is found at about page 104; and continues to page 120 or so. I'm only giving a rough rough sketch here; as the arguments are very intricate. To begin; our function: $ f^{\circ 2}(z) = g(z) = z + a z^2+...\\$ The $2$ means we only have 2 petals. One repelling; and one attracting. We first make the substitution $w = \frac{-1}{2az^2} = \frac{c}{z^2}$; and concern ourselves with the attracting petal; where we now have a fixed point at $w = \infty$. I won't walk through the whole proof; just the exact construction method. We're going to find a mapping $\psi(w)$ such that, $ g\circ \psi(w) = \sqrt{\frac{c}{w}}(1+a\frac{c}{w} + o(1/w))\\$ Then taking, $ F(w) = \phi \circ g \circ \psi = w + 1 + o(1)\\ F(w) = w + 1 + O(\frac{1}{sqrt{|w|}})\\$ By composing with the map $\phi(z) = c/z^2$. We've succesfully "changed our variables" to the best possible situation. A large section of the petal will now be mapped to $\Re(w) > R$ for large enough $R > 0$. Choose a point $\hat{w}$ in this half plane, and, $ \beta(w) = \lim_{k\to\infty} F^{\circ k}(w) - F^{\circ k}(\hat{w})\\$ satisfies the Abel equation, $ \beta(F(w)) = \beta(w) + 1\\ \lim_{w \to \infty} \frac{\beta(w)}{w} = 1\\$ And now; we simply change our variables back to $z$ and we've successfully constructed an abel function $\alpha$ on the attractive petal. A similar procedure can be done with the repelling petal by taking the inverse $g^{-1}$. I suggest reading up on Ecalle's method; it's really quite a beautiful construction. There are many fascinating questions. Luckily; the logistic function is rather simple to handle; more complex functions produce much more difficult questions. Specifically; classifying petals; how many there are; how many are repelling/attracting etc... Anyway, for our case here; we can construct a unique Abel function on $\mathcal{P}_{\pm}$ calling $\alpha_\pm$--which is two abel functions. Then, locally we can define, $ h_{\pm}(z) = \alpha^{-1}_{\pm}(\alpha_{\pm}(z) + 1/4)\\$ Where, $ h_+(h_+(z)) = f(z) \,\,\text{for}\,\,z \in \mathcal{P}_+\\ h_-(h_-(z)) = f(z) \,\,\text{for}\,\,z \in \mathcal{P}_-\\$ I hope this helps. You should note that $\mathcal{P}_+ \cap \mathcal{P}_- = \emptyset$. And that these two sets are separated by the julia set. Regards, James  I made an error with how we find the square root of $f$. It's too late to fix now; but $f$ may have two repelling petals ; which would mean we'd have to take another change of variable. The same result pretty much stands; there just might be a change of variable at the end. It'll just amount to a bunch of minus signs is all really. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/06/2021, 09:22 AM (This post was last modified: 08/06/2021, 10:55 AM by JmsNxn.) Also, Leo If you are interested in hard numerical values; I'd turn to fractional calculus. If $f(\xi) = -\xi(1-\xi)$, then in the attracting petal $\mathcal{P}$ about zero, $f^{\circ n}(\xi) \to 0$. Then on this basin, $ f^{\circ 1/2}(\xi) = \frac{d^{-1/2}}{dw^{-1/2}}|_{w=0} \sum_{n=0}^\infty f^{\circ n+1}(\xi) \frac{w^n}{n!}\\ = \frac{1}{\sqrt{\pi}}\int_0^\infty (\sum_{n=0}^\infty f^{\circ n+1}(\xi) \frac{(-x)^n}{n!})x^{-1/2}\,dx\\$ Which is a consequence of Ramanujan's theorem; which is just a representation theorem. If $f = \sin(\xi)$ the above method is the equivalent of doing the Taylor expansion about zero; to make the fractional iterates about 0 (on the real line). I'll explain this better, if you need me to; but most of this comes from my handwritten papers and its difficult to summarize pages. I haven't culminated this research perfectly yet. Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 08/07/2021, 09:13 AM (This post was last modified: 08/10/2021, 08:12 PM by Leo.W. Edit Reason: d*mn I typed erotic mistakenly...lmao ) (08/06/2021, 01:35 AM)JmsNxn Wrote: ...Thank u, James I got the similar idea the other day, just like, since $\left|f'(L)\right|=1$, there must exist a number k that $f^k(z)=g(z),g(L)=L,g'(L)=1$ which is the exotic case, pretty constructable, and we plug in, plug out, so we can get any iterations of f. However, this method you've typed is, I'm afraid, invalid(sorry if this is impolite) for the non-constuctable cases, because I've tested it out many times, for non-constuctive cases, the fundamental law of iteration: $(f^m)^n(z)=f^{mn}(z)$ doesn't hold anymore. A very example, we can construct the more-paid-attention-to problem, which asks for such a function f that $f(f(z))=sin(z)$, by this method, we can quickly test out there lies such function, with $f'(0)=1,f(z)=z-\frac{z^3}{12}-\frac{z^5}{160}+O(z^7)$, but as conjugation stated, $fg=gf\to{f^tg=gf^t}$, so that f must an odd function. This then gives us a hint that -f(z) should also satisfy the equation, as $-f(-f(z))=f(f(z))=sin(z)$, showing that -f(z) is another second iterative root of sin(z). And meanwhile it also shows that your construction will not claim this "another" root. This is because the construction will conserve the property, that, we can calculate by chain rule at the fixed point L, ORIGINALLY:  $\left.\frac{\mathrm{d}f^t(z)}{\mathrm{d}z}\right|_{z=L}=\left(\left.\frac{\mathrm{d}f(z)}{\mathrm{d}z}\right|_{z=L}\right)^t$ In the nonconstructable cases, this will fail because $e^{\frac{m}{n}2\pi{i}t}\ne{} \left(e^{\frac{m}{n}2\pi{i}}\right)^t$ But your method's valid perfectly for exotic cases, because $1^t=1$ So, if you apply the method into practical calculation, the function $h(z)=\alpha^{-1}(\alpha(z)+\frac{1}{4})$ will not generate $f(z)=-z(1-z)$, instead, it gives $h(h(z))=g(z)=z-z^3+\frac{z^4}{2}-\frac{3z^5}{2}+O(z^6)$, another second iterative root of $f(f(z))=z-2z^3+z^4$, and the reason is $h'(0)=g'(0)=1\ne-1$, which breaks the rule aforementioned. Also, we can inspect another fundamental rule, $f^s(f^t(z))=f^{s+t}(z)$, that iterations of the same function are commutative. This doesn't fail, though, and it shows even the 2 different second iterative roots, one is $f(z)=-z(1-z)$, another is a series that has a very small convergent radius, but can be analutically continued to the whole complex plane, $g(z)=z-z^3+\frac{z^4}{2}-\frac{3z^5}{2}+O(z^6)$ that they are commutative. (* Notice that in all calculation process above we treated iterative roots as holomorphic functions  *) Again, a same approach is to solve the Julia equation. We can solve Julia quation for both $f(z)=-z(1-z),f^2(z)=z-2z^3+z^4$, by law 4, we see that they share the same Julia function, denoted here as $\lambda$, then we have $\alpha\left{f^2\right}(z)=\int{\frac{dz}{\lambda(z)}}=\frac{1}{4z^2}+\frac{1}{4z}+\frac{11ln(z)}{8}-O(z)$ The 2 piece of function$\alpha(z),\alpha(f(z))$, as how abel function was defined, should satisfy $\alpha(f(z))=\alpha(z)+\frac{1}{2}$, but this fails in this case, which can be checked easily, you will find 2 graphs of $\alpha(z)+\frac{1}{2},\alpha(f(z))$ are considerably different from each other. In fact, they satisfy $2(\alpha(f(z))-\alpha(z))=\left\{\begin{matrix}1+\frac{11\pi{i}}{4}&\arg(z)>0\\1&\text{nowhere}\\1-\frac{11\pi{i}}{4}&\arg(z)<0\end{matrix} \right.$ due to that abel function contains a term of log(z), which is multivalued. And thankfully there's still another fixed point f(2)=2,f'(2)=3 which is constructable, we can check that, we generate a second root g(z) from z=2, soonly we'll find that $g(2)=2\text{but}g(0)\ne0$, that this second root g(z) didnt conserve the nonconstuctable fixed point 0. That's as far as I've calculated So these results basically show that, if we want to generalize a non-constructable cases' iteration, we can not do this again: 1)find a number k that f^k=g is exotic or g'(L)=1 2)construct g^t, and then take g^(t/k)=f^t Would you like to introduce another method? Thank u ps.I'm focused on the generation of inverse Abel function of f(z)=-z(1-z), which is very challenging, would u like to take consideration into this case? (about the asymptotic of the inverse abel function) Pics showing that h,g,abel function ( sorry the h and g order has been disrupted ) Regards Leo Attached Files Image(s)     Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 08/07/2021, 10:06 AM (This post was last modified: 08/07/2021, 10:12 AM by Leo.W.) (06/09/2021, 11:17 PM)MphLee Wrote: (06/09/2021, 06:00 PM)Leo.W Wrote: You ... equation. Can you expand on this? ... OMG I'm sorry I forgot to answer this... so many things happened recently...(my hometown was struck by heavy flood last month, losing electricity and water, it was hard time ) Quick check: Given $\zeta(f(z))=g(\zeta(z))$ Take derivative $\zeta'(f(z))f'(z)=g'(\zeta(z))\zeta'(z)$ Notice that if we want the equation ONLY contains zeta'(z) without involving any other functions, we should let g'(zeta(z))=constant, so that g(z) is a linear function. And as Law 3 stated, all g can be mapped to g(z)=z+1, which is definitely the Abel equation. or more precisely, $\zeta(f(z))=k\zeta(z)+s$, let $P(z)=\frac{\log\left(\frac{kz+s-z}{Ck-C+s}\right)}{\log(k)},P({C})=0$ which is the superfunction of g(z), we have $P(\zeta(z))=\alpha(z)$ is the Abel function And the derivative of abel function is the reciprocal of Julia function. So the whole RS definition is a derivation from Abel's equation. Regards Leo JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/07/2021, 11:45 AM (This post was last modified: 08/07/2021, 11:53 AM by JmsNxn.) (08/07/2021, 09:13 AM)Leo.W Wrote: (08/06/2021, 01:35 AM)JmsNxn Wrote: ........ Leo, I agree precisely with what you just said. I was glazing over complex details; and speaking a little loose. I suggest reading Milnor though; your half iterate exists and can be constructed. There always exists a half iterate about any fixed poin $|f'(L)| = 1$ where $f'(L) = e^{2\pi i \frac{p}{q}}$. The trouble is; we need to choose the half iterate appropriately; and it maps petals to petals. And there exists multiplie n roots; and multiples of this based on q. Please look at Ecalle; I can't say it any better. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 08/08/2021, 11:22 PM (This post was last modified: 08/08/2021, 11:37 PM by JmsNxn.) Also; I must've missed the challenge to construct an inverse abel function. I'll be hard pressed if the following method doesn't work. If it doesn't, I'll try another. Call $\mathcal{P}$ the set in which, $ f(\xi) = -\xi(1-\xi)\\ \lim_{n\to\infty} f^{\circ n}(\xi) = 0$ Take a neighborhood of zero $\mathcal{U}$ and intersect with $\mathcal{P}$ to get $\mathbb{E} = \mathcal{U} \cap \mathcal{P}$. Change our variables to $g(\xi) = f^{\circ 2}(\xi)$. Then, for some $T \in \mathbb{R}^+$, $ f^{\circ 2z}(\xi) : \mathbb{C}_{\Re(z) > 0,\Im(z) < T} \times \mathbb{E} \to \mathbb{E}\\ f^{\circ 2z}(\xi) = \frac{d^{z-1}}{dw^{z-1}}|_{w=0} \sum_{n=0}^\infty f^{\circ 2(n+1)}(\xi)\frac{w^n}{n!}\\$ Now, if we take the fourth root we will get $h(\xi) = f^{1/2}(\xi) : \mathbb{E} \to \mathbb{E}$; but, you're correct; you'll get $h'(0) = 1$--when we want $h'(0) = \pm i$.  The trick is easier than you're making it. As this converges, let's create a family $f_\lambda^{\circ z}$ for $0 < \lambda < 1$ where $\lambda \to 1$ in the limit and gives us the above iteration; $\frac{d}{d\xi}|_{\xi =0} f_\lambda^{\circ 2z}(\xi) = \lambda^{2z}$. To begin, $ f_\lambda^{\circ 2z(1+\pi i k/\log|\lambda|)}(\xi)$ Is also an iteration, but, $ \frac{d}{d\xi}|_{\xi = 0} f_\lambda^{\circ 2z(1+\pi ik/\log|\lambda|)} = e^{2z \log(\lambda)+2\pi i k z}$ Now, when we take the fourth root there are four options, one negative and one positive; and two imaginary. And voila; if we limit $\lambda \to 1$ we've avoided your problem and we find that through a limit process we can make a square root function which satisfies, $ h'(0) = \pm i\\$ I hope that answers your question; if not I'll go into more detail. I only know how to compute these things through limits; but they again, relate to Ecalle. They might display odd complex behaviour though; or cease to exist at certain points. I tend to find a good heuristic, is to take a limit $\lambda f$ as $\lambda \to 1$; to discover how to control iterates; and choose different iterates. Remember; there are at most n functions about a fixedpoint which are n roots of a function. I've yet to find a function that didn't have exactly n roots about that fixed point though. « Next Oldest | Next Newest »

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