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 Some "Theorem" on the generalized superfunction MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 05/06/2021, 10:39 PM (05/06/2021, 09:26 PM)Leo.W Wrote: The Law 5 is not so a derivation from the "Cancel Law", it contains more generalization and symmetry. It suggests the f's symmetry should work for every nonzeroth iteration of f, it's more like $f*h=f\to f^t*h=f^t$. [...] The laws of both have something in common, like the anticommunity in multiplication and composition. I suppose these laws are not completely equivalent to those general laws, though there's much similarity, maybe someday we can really combine the multivalued iteration(which can be so sophiscated) with the general laws. Regards, Leo Sure, I guess because the laws you describe are just informal observations. I claim that, if you try to formalize them, you'll come necessarily to the general laws I'm talking about. Now I don't have enough time to touch all the points atm. I'll limit myself to the remark on Law 5. I'm not sure to understand what you mean by symmetry in that context. I won't be too technical, but I'd like to note that some of those sets could be empty so the proof of the cancel law as you state it is tricky. I'm sorry, but the way to go for that law is categorical. Define. the sets $[f,g]$ as the set of all the bijective functions s.t. $\chi f=g\chi$ Cancel Law: $[g,h][f,g]=[f,h]$ This is not correct in general as stated. It needs more fine tuning. Let's ignore the fine details. General Law 5: $[f,g][f,f]=[f,g]$ and $[g,g][f,g]=[f,g]$ Read that as: if $\alpha f=f\alpha$ and $\beta g=g\beta$ then $\beta [f,g]$ and $[f,g]\alpha$ are subsets of $[f,g]$. To prove the other side of the inclusion there are more technical details to check. So even in this case the way to go is the categorical setting. If we do it then the proof follows trivially from cancel Law. Now the key part. Notice that the set $[f,f]$ is a group. It is called the centralizer of $f$ and contains all the functions that commute with f. Among them there is the identity $f^0$ and all the integer iterations $f^{n}$. But the set can be richer than that: it must contain all the fractional iterations, if they exist, and could be even richer and contain functions that are not iterations (think about commutative groups)! The key point to derive Law 5 from the previous is to notice that iterations of $f$ are in $[f,f]$ and iterations of $g$ are in $[g,g]$. We conclude that Corollary: for every n,m $g^m [f,g]$ and $[f,g]f^n$ are subsets of $[f,g]$. Note: n need not to be non-zero in this case. But in the Invariant Law it is crucial. Notice in the end that also the cancel law we are subject to even subtler technical details if we want a formal proof (probably we could only aim at an inclusion.). MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 05/07/2021, 06:37 AM (This post was last modified: 05/08/2021, 09:06 AM by Leo.W. Edit Reason: removed unnecessary repetition ) (05/06/2021, 10:39 PM)MphLee Wrote Wrote: [quote pid='9440' dateline='1620332808'] Leo.W Wrote [...] The laws of both have something in common, like the anticommunity in multiplication and composition. [...] Sure, I guess because the laws you describe are just informal observations. I claim that, if you try to formalize them, you'll come necessarily to the general laws I'm talking about. [...] Note: n need not to be non-zero in this case. But in the Invariant Law it is crucial. Notice in the end that also the cancel law we are subject to even subtler technical details if we want a formal proof (probably we could only aim at an inclusion.). [/quote] Hi, MphLee. It's not very clear if you use bijective to a multivalued function is available? Well, in fact, you'd better consider $g^m [f,g]$ and $[f,g]f^n$ and $[f,g]$ are really equal to each other, since there lies the inverse multivalued function, and m and n could just be generalized into every nonzero complex constant...So this is what I mentioned, it's some kind of a...modern contradiction, you can't apply any laws that are not available to multivalued function to these . Though these laws are informal, I think it's still long before we can gradually accept the concept ''multivalued'', and only then can we get a formal definition of these laws...If it's convinient, you can consider the ''generalized Bottcher equation'' where f(z)=z+sqrt(z) and g(z)=z^2, when you try to generate the series coefficients in zeta, and if you never consider the other branch cut of f, f(z)=z-sqrt(z), the solution is only available in a subset of C, approximately {z|Re(z)>Im(z)^2}, you never get close to the cases beyond that. The law 5 contains 2 parts, since I didn't make a classification. The former part is $g^m [f,g]$ and $[f,g]f^n$ and $[f,g]$ are equal to each other. The latter part is symmetry law's generalization, for every nonzero t, $f*h=f\to f^t*h=f^t$. The symmetry law again, is something with self-contradiction, if you never consider multivalued function, for example, the symmetry cos(-z)=cos(z) never reaches t=-1, since arccos(z)+arccos(-z)=π, for most z, the symmetry of even functions are not available for their inverse if you never consider multivalued function. Of course, this law is not proved in the field of singlevalued functions. I recommend to consider the law of symmetry as an approach to generate a specific branch cut from one to another. If a solution set whose Z(z) is not a constant function has no elements, it's called pseudo-empty set(only my informal definition, maybe other terminology) And it's pretty sure that there lies some pseudo-empty solution set, the most common example, consider f(z)=z and g(z)=a*z+b, obviously that Z(z)=a*Z(z)+b has only a constant solution. The empty solution set lies between most insymmetric functions and symmetric functions. It's unclear to me about how to figure out which of these are empty sets or pseudo-empty sets. And this problem can be something more relevant to corollary theorem. Well the first thing to notice, I think, is one of these are the identity function, so the solution of zeta is Z(z)=f^∞(z), a constant function.(sometimes blows up to infinity)Also, if f or g is constant, a pseudo-empty set. Another case is also easy to check, let f and g be different nonzeroth iteration of the same function, and this solution set may contain only constant solution(maybe infinity) or maybe no solutions at all, otherwise it must contain some function having a pseudo-periodic iteration: $f^{s+t}~=f^s$. Regards, Leo MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 05/07/2021, 09:27 AM Ok, I'm really confused now. I think there is a problem and maybe I know where it lies. We have to reach a consensus on the definitions. Otherwise we will not be able to turn your claims into fully-formal proofs anytime soon. I just make some points that makes the absence of perfect consensus evident. -Yes, a function can not be multivalued and bijective at the same time. -I say bijective because it is the only case where I can see you can prove that those sets are actually equal (when non-empty). In fact when we leave the bijectivity, the inverses are not function but multivalued objects. In that case, I claim, those sets are much richer and inclusion or equality of the sets can fail badly. I see that the way you state the first part of Law 5 is too strong to hold, in any interpretation. It can not be satisfied because composing on the left by a generalized iteration could shrink the set. Doing that on the right could shrink the set in a different manner. So I exclude that you can come up with an equality of the three sets in general. -Mutivalued functions ARE perfectly defined algebraically. Maybe are hard to compute or to manage in your computations but are not really a contradiction. -The generalization of odd function reduces to satisfying hf=fh. If f conjugates the function h(x)=-x then, by definition, f is odd. -Let's call f generalized h-even function if fh=f. This means that f is in the set $[h,{\rm id}]$. Now I can see the for non-zero natual numer of iterations  of f we get an h-even function (easy to prove): $\forall n\in{\mathbb N}_0.f\in [h,{\rm id}]\Rightarrow f^n\in [h,{\rm id}]$ Being h-even is equivalent to not being injective so I agree that if f is h-even function the inverse is not a single valued function. You claim that if we consider the multivalued case we can somehow recover the previous implication when n is a non-zero integer, e.g -3, -2, -1, 1, 2 ... and so on. Ok I can believe that. What I can't see is how you can extend this to generalized iterations of f(when they exists). In the end, I invite you to make formal the definitions. Given a set $X$ and two functions $f,g:X\to X$ I define the set $[f,g]:=\{\chi:X\to X \,\|\, \forall x\in X. \chi(f(x))=g(\chi(x)) \}$ Let me try to formalize what you are talking about. I hope JmsNx can also help me here. You seem to define your set as follows: Given a set $X$, assume $X={\mathbb C}$, given subsets $F_0\subseteq{\mathbb C}$, $F_1\subseteq{\mathbb C}$, $G_0\subseteq{\mathbb C}$, $G_1\subseteq{\mathbb C}$ such that $F_0\cap F_1\neq {\emptyset}$ and $G_0\cap G_1\neq \emptyset$ are  non-empty. Given two possible mutivalud functions  $f: F_0\to F_1$ and  $g:G_0\to G_1$  (to be really precise we should write  $f\subset F_0\times F_1$ and  $g\subset G_0\times G_1$ ) Define the set $\{f;g\}:=\{\zeta: F_0\cap F_1\to {\mathbb C} \,\|\, \exists D\subseteq F_0\cap F_1 \forall z\in D. \zeta(f(z))=g(\zeta(z)) \}$ If this is the case, then the theory of those sets is, I suspect, a billion times more ill behaved and weird than theory of my sets is. I limit myself only with functions well defined on given domains. You consider freely "bundles" of multivalued functions with non-specified domains. It is indeed a more flexible setting for sure. The only way to be more flexible than this is considering non-functional relations. I'm afraid that turning this mess into category theory is beyond my powers atm. It is in fact the realm of differential geometry and sheaf theory. I'm in deep water. ps: I'm not a moderator but I suggest you to remove from your answers the whole citation of my message when it is too lengthy or when it is clear who are you writing to or to what are you referring. We can make the thread more readable and short (less pages). MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 05/07/2021, 11:49 AM (This post was last modified: 05/08/2021, 09:07 AM by Leo.W.) (05/07/2021, 09:27 AM)MphLee Wrote: Ok, I'm really confused now. I think there is a problem and maybe I know where it lies. We have to reach a consensus on the definitions. [...] We can make the thread more readable and short (less pages).Pardon me, I didn't realized that when I simply pushed the reply button    I'll remove it really soon. Your reply is really constructive and meaningful. I never read about algebraic formal definition about multivalued functions, so that in my work I always handle it as ''a set in a set'', which can be a little confusing. Could you please give me some references? I'll be very grateful. The law 5 is indeed strongly-stated, but it has an analogy to the generalization of iteration. I think, in spite of this generalization can cause some monsterous behavior, we can treat it as the methods to move between infinitely many branch cuts or solutions, and sometimes it does help, for instance, consider the case I mentioned before, f(z)=z+sqrt(z) and g(z)=z^2, we can generalize many different branch cuts of f, each one should satisfy most properties which f satisfies.  I only have a informal statement about these multivalued iterations, it's ''choose an appropriate branch cut/value'' to satisfy the equation, and this can be observed through many constructions of zeta. The case f(z)=z^2+v is the one I pay most attention to, in which I use the multivalued property to make the whole plane suitable to the equation in some sense. Also, the multivalued iterations of sine function and exponential function is also tested, you can see supersine and superlog has infinitely many branches, and it's still unproved that you can't choose a branch to fit in the equation in a certain domain. The iteration world is still so wide to explore! Your set definition is gorgeous and already applicable in the various field of single-valued functional iterations, it's very helpful! I'm concerned about more accuate statements of the definition of multivalued functional iterations and can't wait any longer I also have other laws which is unproved nowadays, someday I'll post those and your help would be really appreciated! Regards, Leo JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 05/07/2021, 11:05 PM (This post was last modified: 05/08/2021, 06:12 AM by JmsNxn.) Wow, a very long thread has formed in my absence, lol. I figure I should add a long thread in return, so buckle up, lol. I'd like to add a couple details. Mphlee did a much better job at explaining the concerns I have with this notation. The more I'm rereading through this thread, I've come to the conclusion, in order for Leo's idea to make sense; there's no two ways around it; we need to introduce Riemann Surfaces. As Leo seems dead set on multi-valued functions; it's imperative we introduce a more rigorous framework. Leo makes mention of choosing the "worthy" iteration (lol, which I understand entirely; but he hasn't really given a way to find said worthy iteration; besides sticking to Schroder or Abel about a fixed point). When making the leap to Riemann Surfaces; it's pretty much imperative to use differetial equations. So, I thought we'd take a little detour through differential equations and superfunctions. Let's assume we can develop an iteration, $ F(z,\xi)$ Such that $F(z,F(z',\xi)) = F(z+z',\xi)$. A little unused theorem on this forum, is classifying this superfunction as a flow. $ \frac{d}{dz} F(z,\xi) = \lim_{h\to 0} \frac{F(z+h,\xi) - F(z,\xi)}{h} = \lim_{h\to 0} \frac{F(h,\xi) - \xi}{h} \bullet F(z,\xi) \bullet \xi\\$ I like to call this object, $ \log (F ; \xi) = \lim_{h\to 0} \frac{F(h,\xi) - \xi}{h}$ Which can be referred to as the logarithm of the super function; or traditionally it's known as a generator of a flow map. Now, for every super function there is one logarithm, and for every logarithm there is one superfunction. There is no obvious connection between the logarithm and the initial function $F(1,\xi) = f(\xi)$; but you can derive a formula for it. Call the function, $ h(\xi) = \log (F ; \xi) \\$ And we're reducing the problem to the differential equation, $ y' = h(y)\\$ with the initial condition $y(0) = \xi$. For instance, a tetration function $\text{tet}$ and its super logarithm $\text{slog}$ are completely determined by the map, $ h(\xi) = \lim_{z\to 0} \frac{\text{tet}(z+\text{slog}(\xi)) - \xi}{z}\\$ Where, the solution of the differential equation in a neighborhood of $0$; $ y'(z) = h(y)\\ y(0) = \xi\\ \text{implies}\\ y(z) = \text{tet}(z + \text{slog}(\xi))\\$ And vice versa; so this is an equivalence. Now, the manner I like to view this function is borrowed from my paper The Compositional Integral: The Narrow And The Complex Looking-Glass. If we call $u(s,z)$ the solution to the differential equation, $ u'(s,z) = f(s,u)\,\,\text{and}\,\,u(a) = z\,\,\text{then write}\\ u(s,z) = \int_a^s f(s,z)\,ds\bullet z\\$ To better explain this notation, I refer to the paper; which can be found on arXiv here, https://arxiv.org/abs/2003.05280 But the gist of the notation comes from the Omega notation, which is; Choose an arc $\gamma$ connecting the points $a,s$, and choose a partition of $[0,1]$. We'll call this $\{s_j\}_{j=0}^n$ in which $1 = s_0 \ge s_1 \ge s_2 ... \ge s_{n-1} \ge s_n = 0$. Call the term $\Delta \gamma_j = \gamma(s_{j}) - \gamma(s_{j+1})$. And write, $ \int_\gamma f(s,z)\,ds\bullet z = \lim_{\Delta \to 0} \Omega_{j=0}^{n-1} z + f(\gamma_j,z)\Delta \gamma_j \bullet z\\$ Where, if we call the term $p_{jn}(z) = z + f(\gamma_j,z)\Delta \gamma_j$, this just means, $ \Omega_{j=0}^{n-1} z + f(\gamma_j,z)\Delta \gamma_j \bullet z = p_{0n}(p_{1n}(...p_{(n-1)n}(z)))\\$ Where we are just taking the limit $n\to\infty$. To make a long story short; this thing converges well enough (at least for the purposes of this post). And we get a domain $\mathcal{D} \subset \mathbb{C}$ for which $u(s,z) : \mathcal{U} \times \mathcal{D} \to \mathbb{C}$; and $\mathcal{U}$ is a neighborhood of $a$ and $\overline{\mathcal{D}} = \mathbb{C}$. And this function satisfies the differential equation, $ u'(s,z) = f(s,u)\\ u(a,z) = z\\$ So what does this have to do with super-functions? Much of the paper was written to design a formalism; and that formalism isn't quite complete yet; but you can do some crazy stuff with this notation; and its specific to differential equations/superfunctions. Now; if you'll remember your basics of Riemann Surfaces; a lot of the time we can create a Riemann surface from a differential equation. For that, think of the log map $y' = \frac{1}{s}$; where as we go around and around we spiral around a Riemann surface. The paper I linked to, does not delve into the Riemann surface aspect at all; but it gives a type of language that is very beneficial which would be too hard to explain here. But I'll give a run through. And I swear, it'll get to the point where we define something like a conjugate class $[f,g]$; which I furiously tried to formalize in that paper. Define the function, $ y(s,\xi) = \int_0^s h(\xi)\,ds\bullet \xi$ Then, $ y(0,\xi) = \xi\\ y(s,y(s',\xi)) = y(s+s',\xi)\\$ And this is the equation of a semi-group. And further more; all semi-groups have to look like this for some $h$ (at least analytic ones). But there's more hidden to this notation. The Compositional Integral Theorem; Suppose that $\mathcal{S} \subseteq \mathbb{C}$ is simply connected. Suppose that $\phi(s,z) : \mathcal{S} \times \mathbb{C} \to \mathbb{C}$ is a holomorphic function. For all Jordan curves $\gamma \subset \mathcal{S}$, the following identity is true, $ \int_\gamma \phi(s,z)\,ds\bullet z = z\\$ \\ Which this means; as with Cauchy's integral theorem; a closed integration about a holomorphic function on a simply connected domain is an identity. Now, suppose we have a pole at $s=0$ of $f(s,z)$; and lets take the domain, $ \mathcal{S}_{\delta,\delta'} = \{s \in \mathbb{C}\,|\, \delta' \le |s| \le \delta,\, s \not \in [-\delta,-\delta']\}\\$ Which looks like an annulus about zero with a cut along the negative real axis. This domain is simply connected, call the outer circle $\gamma_\delta$ oriented positively; the inner circle $\gamma_{\delta'}^{-}$ oriented negatively; and the arcs $\rho$ and $\rho^-$ the line $[-\delta,-\delta']$; where they're oriented positively and negatively respectively. Then, $ \int_{\partial S} f(s,z)\,ds\bullet z = z\\$ By The compositional integral theorem. But the path $\partial S = \gamma_{\delta} \bullet \rho \bullet \gamma_{\delta'}^- \bullet \rho^-$. And using the bullet notation, $ z = \int_{\gamma_\delta} f(s,z) ds \bullet \int_{\rho} f(s,z) ds \bullet \int_{\gamma_{\delta'}^-} f(s,z)\,ds\bullet \int_{\rho^{-}} f(s,z)\,ds\bullet z\\$ And you'll begin to see where I'm going with this by the inversion rule. The functional inverse of $\int_\gamma f(s,z)\,ds\bullet z$ is just the integration $\int_\tau f(s,z)\,ds\bullet z$; where $\tau$ is just the opposite orientation of $\gamma$. So what we wind up with, is, $ \int_{\gamma_\delta} f(s,z) ds\bullet z = \int_{\rho^-} \bullet \int_{\gamma_{\delta'}} \bullet \int_{\rho}\\$ So the circle $|s| \le \delta$ and the circle $|s| \le \delta'$; are conjugate similar.  Or rather. $ F(z) = \int_{\gamma_\delta} f(s,z) ds\bullet z\\ G(z) = \int_{\gamma_{\delta'}} f(s,z)\,ds\bullet z\\ \text{then there exists a function}\,P(z)\\ F(z) = P(G(P^{-1}(z)))\\$ Which is how miraculous, $P \in [F,G]$ as Mphlee described. Is $P$ unique? Absolutely not. Because we haven't quite described what $F$ is perfectly. So to that, I introduce The Residual Class.  The Residual Class of a Meromorphic function Let $\mathcal{S}\subseteq \mathbb{C}$ be a simply connected domain. Suppose $f(s,z) : \mathcal{S}/\{\zeta\} \times \mathbb{C} \to \mathbb{C}$ is holomorphic with a singularity at $s = \zeta$. The residual class $\text{Rsd}$ of the function $f$ about the singularity $\zeta$ is given as, $ \text{Rsd}(f, \zeta) \ni \int_\gamma f(s,z) \,ds \bullet z$ For all $\gamma \subset \mathcal{S}$, a Jordan curve containing $\zeta$. And the fact is, $ \forall F,G \in \text{Rsd}(f,\zeta)\,\,\exists P\,\,\text{s.t.}\,\,F(z) = P(G(P^{-1}(z)))\\$ And $P$ can be expressed as an arc $\tau \subset \mathcal{S}$, $ P(z) = \int_\tau f(s,z)\,ds\bullet z\\$ The paper then goes on to use this frantically at developing a formal language; which has its place in Riemann surfaces/differential equations. But note, Leo; I've never once used an abel function or a superfunction to construct these conjugates. And personally, I feel they arise much more naturally than with Abel functions or Schroder functions; which to me seems artificial. Especially if you are going to broach the fields of mechanics and physics; a differential equation will always come out supreme to the physics pedagogy than an iterative process at infinity. It's just the symptom of physics, lol. Anyway; if you want to see how I construct this in much more depth, please read the paper. There isn't much on super-functions; but I definitely graze it often enough. It's written a little loosely; but the rigor is all there. And it's written in a fairly elementary way if you've studied lots of complex analysis. I spent about a year on it; all last year; before and during quarantine, lol. At least corona was good for something. Whew, that was a long post. I'm going to go back to work now. I hope this makes sense, Regards, James EDIT:  I will add, I haven't proven this to its limit yet; as it's fairly tricky; but I've proven things around it. But if we have a function $h$ from before, where this generates a semi-group/iteration/super-function; and we have a function $\phi(s,z)$ which satisfies $\phi(\zeta,z) = h(z)$; then, $ 2\pi i h(z) = \lim_{\delta \to 0} \frac{\int_{\gamma_\delta} \frac{\phi(s,z)}{s-\zeta}\,ds\bullet z - z}{\delta}\\$ But this is tricky to use this how we want... manipulating how you're describing. I do it in a limited manner in that paper. And this gives us a way of relating superfunctions to a differential relationship. But it doesn't do what you're asking your theory to do. That sounds super hard. Especially if you keep with this multi-valued language; you're going to have to talk about riemann surfaces somewhere in your discussion. Regards. Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 05/08/2021, 07:38 AM (This post was last modified: 05/08/2021, 07:50 AM by Leo.W. Edit Reason: typo ) (05/07/2021, 11:05 PM)JmsNxn Wrote: Wow, a very long thread has formed in my absence, lol. I figure I should add a long thread in return, so buckle up, lol. I'd like to add a couple details. Mphlee did a much better job at explaining the concerns I have with this notation. The more I'm rereading through this thread, I've come to the conclusion, in order for Leo's idea to make sense; there's no two ways around it; we need to introduce Riemann Surfaces. As Leo seems dead set on multi-valued functions; it's imperative we introduce a more rigorous framework. Leo makes mention of choosing the "worthy" iteration (lol, which I understand entirely; but he hasn't really given a way to find said worthy iteration; besides sticking to Schroder or Abel about a fixed point). [...] But this is tricky to use this how we want... manipulating how you're describing. I do it in a limited manner in that paper. And this gives us a way of relating superfunctions to a differential relationship. But it doesn't do what you're asking your theory to do. That sounds super hard. Especially if you keep with this multi-valued language; you're going to have to talk about riemann surfaces somewhere in your discussion. Regards. Thank you, James! The theorem you state is related to iteration velocity(or Koenig's function) in wikipedia, this idea is the same to use the differential operator to manipulate the functional iteration, with respect to index t. It's for sure more general. I've read it 3 years ago. It's a wonderful theory and I collected a relative article：https://projecteuclid.org/journals/michi...02009.full Also you may find similar comments in the wikipedia article Iterated functions, it states the iteration velocity is the solution to Julia's equation... And amongst all these theories, the notations are not unified yet, it's kinda messy But honestly I don't see there's a relation between these stuffs and my statements... My statements are more related to the existence of multivalued functional iterations, their conjugacies and transformations between them, and without any proofs on the existence yet. Also my laws include ways to generate and calculate them, so they're more practical in computation. About Riemann Surface, it is more a way to descirbe multivalued functions, and branch cuts is another way, maybe you can consider riemann surface as a set of branch cuts, where each are joint to another. You can say the logarithm function has a riemann surface, and it's okay if you say that logarithm function has infinitely many branch cuts. A branch cut can make us more easily to cope with multivalued functions by considering only a branch of values, which is same as singlevalued. Why I'm so obsessed around the branch cuts is that, it's everywhere! You can see that, if we use the second pairs of fixed points~{2.062+7.589i,2.062-7.589i} instead of the original one~{0.318+1.337i,0.318-1.337i}, the original branch cut, or the original defined logarithm function contains only the first pair. When we use the double-dagger track, and calculate the line segment integral, this original logarithm fails to converge, and we'll fail to arrive at the merged tetration of the second pairs. But we should notice the two branches:{log(z)+2*pi*i,log(z)-2*pi*i} contains the second pairs, then we can use these two branch cuts to calculate the integral, and only then can we manage to merge the second tetration. Choosing a branch cut is difficult to describe, so I only claimed the existence. The choice will be discussed in another thread, I guess. But you can see how it works. The first example is included in #2 Part 2, Section III 1.Miscellaneous Results. It shows if you choose a proper branch, the schroder function can be entire. The same thing happens to Kneser's iterated exponentiation. Most still believe the schroder function is only defined in the upper half plane, but you should notice, while we were generating the schroder function at L~0.318+1.337i, we used the limit $\sigma(z)=\lim_{n\to\infty}s^n(\sigma_0(\log^n(z)))$ where log function has infinitely many branch cuts. We use a slight modification: $\sigma(z)=\lim_{n\to\infty}s^n(\sigma_0(f^n(z)))$, in which f is defined as: if $\Im(z)\ge0$, $f(z)=\log(z)$; if $\Im(z)<0$, $f(z)=\log(z)+2\pi*i$. And then the schroder function is defined only except the points {0,1,e,e^e,...}, so the Kneser's iteration is extended now to the whole complex plane. Another example is when you attempt to generate supersine from the pair ~{7+2i,7-2i} the same thing happens, since arcsin(z) doesn't contain these 2 points, but its another branch cut does. Another example is when you try to generate the merged superfunction of f(z)=z^2+i, the original inverse function g(z)=sqrt(z-i) fails to make the schroder function convergent, so we should modify the branch cut, and finally it turned out as a successful method. And without multivalued function, most iterations are unavailable, whenever you have a pair of f(a)=b and f(b)=a, it's proved that there's no singlevalued second iterative root of f, so do forth ,eighth and so on. Also it's proved similarly, if one have f^n(a)=a but f(a),f^2(a),f^3(a),...,f^(n-1)(a)=/=a, then f have no singlevalued nth iterative root, and hence f^s(f^t)=f^(s+t) is impossible. The multivalued functions play a crucial role in iteration theory, and its multivalued property may mend these holes. And you can check that L~0.883998+6.92228i who satisfies exp(exp(L))=L but exp(L)=/=L, thus tetration has to be multivalued. Another example is also worth thinking, you can see if f(z)=-z+z^2, and g(z)=f(f(z))=z-2*z^3+z^4. You probably say f(z) is a second iterative root of g(z), but if we generate the superfunction of g(z) in the traditional method, it'll result in that g^(1/2)(z)=/=f(z). So it forces you to conclude that there lies many branch cuts of g^(1/2)(z). I considered a way to more rigorously define my declaration in section I, and the description is below. And lastly the three pics show the 3 different branches of half-iteration of tan(z), each satisfies f(f(z))=tan(z) in only a subset of C, wish you enjoy it. Regards, Leo Attached Files Image(s)                 JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 05/08/2021, 07:58 AM Again, Leo. I agree with everything you're saying. What you are saying is fantastically intelligent. But trying to just use "multi-valued" functions will get you nowhere. You have to describe it as a Riemann surface. This is something rarely talked about on this forum or anywhere about tetration. But Tetration defines a Riemann surface, (I forget how to classify it off hand). And the multivalued choice of log; which makes tetration multivalued; is as much a riemann surface, as log is. Again, I'm not saying you are wrong. You are fantastically right. And what you are saying is highly fascinating. But just slow down. I guarantee you, you need Riemann surfaces to explain what you're talking about. Imagine I throw three more variables at you, $ \zeta(f(z,x,y)|h(z,x,y))$ Can you describe how you're multivalued branches behave with these two extra variables? And that they're holomorphic. And that they're well behaved. If you can do that without Riemann surfaces that's incredible. But I guarantee, if you wanna get what you wanna get out of this, read up on Riemann surfaces. This is really fascinating though, Leo. I really appreciate your dialogue. Regards. Leo.W Junior Fellow Posts: 30 Threads: 3 Joined: Apr 2021 05/08/2021, 08:40 AM (This post was last modified: 05/08/2021, 09:21 AM by Leo.W.) (05/08/2021, 07:58 AM)JmsNxn Wrote: Again, Leo. [...] This is really fascinating though, Leo. I really appreciate your dialogue. Regards. It's really nice to communicate! I'm unfamiliar with the word's tone while speaking in English... Maybe I'm speaking in a very serious way, or it makes you uncomfortable or feel like I want to modificate something... not at all~ I apologize for my mistakenly-placed tones or wrong words or sentences... It's not what I meant lol maybe this is cult gap?(sorry again, I'm still learning in slangs and routine dialogues and which words or phrases are to avoid) And I'm rushing because I'm a fan since 2 years ago!!! at that time I really wanna join you all but my poor English confined me  So the words accumulated and I'm over excited and I poured them ...maybe in an inappropriate way, plz forgive my manner I'm more like uh, riemann surfaces are exactly a set of branch cuts, so when I say choosing different branches, it's the same as choosing a value in the riemann surface, or we just focus on a piece, a sheet of the whole riemann surface, therefore it is easier to compute. (About branch cuts) The wikipedia article states "Multi-valued functions are rigorously studied using Riemann surfaces, and the formal definition of branch points employs this concept." So the concepts are replaceable with each other. The riemann surface is more adaptive to differential forms of EDE, and so a quite powerful tool, but not easy to manipulate the composition between them. I'll try to write more about it in my later work. We can treat multivariable iterations as one-variable iterations, you can literally choose a variable as a core, and the other variables are treated as indexes, e.g. $f(x,y,z)=x*y^z$, iteration with respect to the variable y:$(f_{x,z})^2(y)=f(x,f(x,y,z),z)$ And the eigendecompositive equation is now only defined for one-variable?(I added one-variable at the beginning) I have my great gratitude to your patience in communicating with me!  p.s. I'm busy with my final exams recently, during this I'll re-read what you said p.s. Not trying to be modifying or aggresive, but my browser fails to load most of your tex formulae. only my recommendation: Code:$$something$$ can be correctly loaded instead of $$something$$ Regards, Leo JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 05/08/2021, 11:45 PM (This post was last modified: 05/08/2021, 11:53 PM by JmsNxn.) Hey, Leo Yes, absolutely they are interchangeable. The idea of using a multivalued function or a Riemann surface. My point is simply, it's streamlined if you use Riemann surfaces. By which, I mean, certainly it's equivalent; but it'll be a bit slow moving when you want to prove grand things about this theory. I only mean to say it as it's a better choice of language. Using the language of Riemann surfaces will open many doors; especially if you're talking about multi-valued functions. Where it can be kind of stiff; where as with Riemann surfaces, the language is more fluid. That's all I mean. Oh, and don't worry about your language   Everything's fine; I apologize if my tone came off a bit blunt. I didn't mean to sound like that. And you speak very well, so you have nothing to apologize about. I understand that English is my first language, and therefore, I instantly have an advantage. Everything you are saying is very well thought out mathematically; I'd hate for you to feel that I'm discouraging you. I'm very impressed by everything here. I'm simply making the point that this seems more at home with Riemann surfaces, is all; especially if you want to describe all the multivalued branches. It'll just appear more naturally in this language (at least, I feel). But by all means, keep at it. This is fantastic stuff! Regards, James. PS: Good luck on your exam! MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 05/10/2021, 07:34 PM (05/07/2021, 11:05 PM)JmsNxn Wrote: A little unused theorem on this forum, is classifying this superfunction as a flow. [...] Which can be referred to as the logarithm of the super function; or traditionally it's known as a generator of a flow map. Now, for every super function there is one logarithm, and for every logarithm there is one superfunction. There is no obvious connection between the logarithm and the initial function $F(1,\xi) = f(\xi)$; but you can derive a formula for it. Regards.This gave me some Eureka moments... Thank you. Here my I replied Jabotinsky IL and Nixon's program: a first categorical foundation But sadly I can't follow until the end... with Jordan curves and integrals. @Leo I'm glad that you find my reply constructive. We are here to get better and I hope to learn and help others. About the mutivalued vs Riemann. I was also thinking of Riemann surfaces as the right way to go... but I'm to ignorant and I feel the same doubt that maybe Leo is feeling. How can we talk about composition of Riemann surfaces (objects of geometric nature)? Is it possible to talk about iteration of things you can't compose and do no form an (associative)algebraic structure of some kind? @JmsNxn if the answer is yes then my mind could blow... that would be a serious paradigm shift for me. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ « Next Oldest | Next Newest »

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