Qs on extension of continuous iterations from analytic functs to non-analytic Leo.W Fellow   Posts: 86 Threads: 7 Joined: Apr 2021 06/18/2022, 10:38 AM (This post was last modified: 06/18/2022, 10:40 AM by Leo.W.) Idk if it's time to talk about continuous iterations for non-analytic functions that can still be defined on the whole complex plane, but these ideas did disastrously harm to my brain lol. I worked out some cases last year, these are my results. Consider any function g(x):R->R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ring-shaped slice has its own rotation angle, which can be written as: $f(z)=ze^{ig(\|z\|)}$ Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps: $\text{For real t, }f^t(z)=ze^{itg(\|z\|)}$ Remember abs(e^(it))=1 for real t, $\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\|z\|)}=ze^{itg(\|z\|)}e^{isg(\|z\|)}=\bigg(ze^{itg(\|z\|)}\bigg)e^{isg\big(\|ze^{itg(\|z\|)}\|\big)}=f^s(f^t(z))$ it's easy to prove, and also satisfies the geometric intuition that the t-th iteration of a rotation is a rotation whose angular velocity is t times the original rotation's. Another idea, came by trying to figure out whether there's continuous iteration of the conjugate function conj(z) We start by conj(z)=abs(z)*e^(-i*arg(z)), consider a set of functions $f_u(z)=\|z\|e^{iu\arg(z)}$ where u is real, Then $\text{For real t and u^t, }f_u^t(z)=\|z\|e^{iu^t\arg(z)}$ Because $\text{For real s and t, }f_u^{s+t}(z)=\|z\|e^{iu^{s+t}\arg(z)}=\|ze^{iu^tz}\|e^{iu^s\arg(\|z\|e^{iu^tz})}=f_u^s(f_u^t(z))$ if u^t and u^s are both real. But since u=-1, t=1/2, u^t is not real, we cannot find a functional square root of conj(z) through this. Do you have more examples, I basically based all this on the Polar decomposition of complex numbers and don't have so much time to take more investigation into it. Regards, Leo  MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/18/2022, 11:00 AM (This post was last modified: 06/18/2022, 11:28 AM by MphLee.) I guess you can do the same but instead of rotating you applying scaling by a non-negative scalar $k\in\mathbb R$. Given an arbitrary function $k:{}S^1 \to \mathbb R^+$ that assigns to every point on the unitary circle a real number we can define $f_k:\mathbb C\to\mathbb C$ defined as $f_k(z)=k_{{\rm arg}(z)}\cdot z$. Obviously we can extend it as $f_k^t(z)=k^t_{{\rm arg}(z)}\cdot z$. If we define $k^t:{}S^1\to \mathbb R^+$ as the composition $(k^t)_\theta:=(k_\theta)^t$ we have $f^t_k=f_{k^t}$ If you look carefully on the construction what we have done is consider a partition of $\bigcup _{i\in I}X_i=\mathbb C$ were we define on each class $X_i$ a function $f_i:X_i\to X_i$ that we know how to continuously iterate, i.e. how to extend it to a group action of $\mathbb R$ over it. We then just paste all the functions/actions $f_i$ into an endofunction of $\mathbb C$. $f=\coprod_{i\in I}f_i:\bigcup _{i\in I}X_i\to \bigcup _{i\in I}X_i$ Since the pasting of the functions has no information about how the various classes/fibers $X_i$ paste together into the topological/analytical structure of the total space $\mathbb C$ we can end up with non-analytic $f:\mathbb C\to \mathbb C$ that can be continuously iterated. Addendum: note that the construction I have given provides an infinite amount of examples and how to build them. In my and your case we use the polar decomposition of $\mathbb C\setminus\{0\}\simeq S^1\times \mathbb R^+$, i.e. as modulus and argument, to obtain a partition of $\mathbb C$. The complex can be presented as a bundle ${\rm arg}:\mathbb C\to S^1$, i.e. $\bigcup_{\theta\in S^1}\mathbb R^+e^{i\theta}\to S^1$, or as the bundle $|\cdot|:\mathbb C\to\mathbb R^+$, i.e. $\bigcup_{r\in \mathbb R^+}rS^1\to \mathbb R^+$. All of this is very tied to decomposition techniques one sees in the spectral decomposition of linear operators to fiber bundles. The resulting map $f$ is, in fact, a bundle endomorphism. In fact, I suspect that to make sure that your pasting is $\mathcal C^0$ ,$\mathcal C^1$,$\mathcal C^\infty$ or $\mathcal C^\omega$ you have to ask not only for a bundle that decompose the domain, but an additional structure on it (parallel transport? a connection?) MSE MphLee Mother Law $(\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ Leo.W Fellow   Posts: 86 Threads: 7 Joined: Apr 2021 06/18/2022, 12:05 PM (This post was last modified: 06/18/2022, 12:13 PM by Leo.W.) (06/18/2022, 11:00 AM)MphLee Wrote: I guess you can do the same but instead of rotating you applying scaling by a non-negative scalar $k\in\mathbb R$. Given an arbitrary function $k:{}S^1 \to \mathbb R^+$ that assigns to every point on the unitary circle a real number we can define $f_k:\mathbb C\to\mathbb C$ defined as ... Since the pasting of the functions has no information about how the various classes/fibers $X_i$ paste together into the topological/analytical structure of the total space $\mathbb C$ we can end up with non-analytic $f:\mathbb C\to \mathbb C$ that can be continuously iterated. Yes you're perfectly right about this! And thanks for your post I now know how to type formula using \+( and \+) in texts, helped a lot. I'd thought about the partitions, and your partitions worked out very well, em, I don't think the functions have to distinguish each self in $\mathcal{C}^0,\mathcal{C}^1$ or etc.,  maybe globally $\mathcal{C}^0$ is what I meant mostly, (thus it won't allow someone to approximate any iteration by fixed point or other analytic methods) This inspires me more, For $\mathcal{C}^0$, one can cut the $\mathbb{C}$ with parellel lines and the function can be for each slice, move it in the direction of these parellel lines, which is an action of addition, or $f_k^t(z)=z+tk_{a*Im(z)+b*Re(z)}$ however I don't quite understand all the terms (spectral decomposition of linear operators to fiber bundles especially) in topology, but I'll learn them as soon as possible. But partition won't work out for more cases, like abs(z), sgn(z), arg(z), log(abs(arg(z)+sgn(z))) and more wierder functions, I'm still stuck about the conj(z), as its transformation cannot be described by countably infinitely many partitions $X_i$ and each has an $f:X_i\to X_i$. I wonder how one can get a continuous iteration without endomorphism, since it breaks the symmetry $f:X_i\to X_i$, or even a more complex question, like their iterations to complex heights, since then you can't use eigen decomposition or linear operators, like the cases $k\in \mathbb{C},f_k^t(z)=k_{\arg(z)}z$ Anyway thank u Regards, Leo  MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/18/2022, 02:55 PM (This post was last modified: 06/18/2022, 03:12 PM by MphLee.) (06/18/2022, 12:05 PM)Leo.W Wrote: I'd thought about the partitions, and your partitions worked out very well, em, I don't think the functions have to distinguish each self in $\mathcal{C}^0,\mathcal{C}^1$ or etc.,  maybe globally $\mathcal{C}^0$ is what I meant mostly, (thus it won't allow someone to approximate any iteration by fixed point or other analytic methods)I believed the point was to obtain a non $\mathcal{C}^\omega$ (analytic) $f:X\to X$ s.t. we know how to continuously iterate it $f^t$. My method was just to chop $X$ into a bunch of $X_i$ and define on every piece a function $f_i:X_i\to X_i$ that we know how to $\mathbb R$-iterate. What I was saying is that given all the $f_i$ the pasting $f:=\coprod_{i\in I}f_i$ needs not to have any nice properties, even if all the $f_i$ are, say, $\mathcal C^n$ for some $n\in \mathbb N\cup \{\infty,\omega\}$. What I mean is that if we want the gluing $f:=\coprod_{i\in I}f_i$  to carry some continuity/analytical property of the $f_i$ we need the bundle $X\to I$, to have some additional structure, e.g. a way to measure how the fibers $X_i$ sits in $X$ as $i\in I$ varies in a continuous way. Quote:This inspires me more, For $\mathcal{C}^0$, one can cut the $\mathbb{C}$ with parallel lines and the function can be for each slice, move it in the direction of these parellel lines, which is an action of addition, or $f_k^t(z)=z+tk_{a*Im(z)+b*Re(z)}$ I don't think that the global function you are describing is $\mathcal{C}^0$. Let me restate abstractly the underlying concept. Assume we want to present $\mathbb C$ as a bundle of lines, each of the same shape of the real line and all parallel. This reduces to a decomposition of $\mathbb C$ as a real vector space along its two dimensions. So given a base for $\mathbb C$, say $\{v,w\}$, we can express every complex $z$ as $kv+jw$ for $k,j\in\mathbb R$. This means that $\mathbb C\simeq \mathbb R\times\mathbb R$ Now this defines a partition of the complex plane into a bundle of parallel lines as $\mathbb C=\bigcup_{p\in \mathbb Rv}p+\mathbb R w$, where $\{v,w\}$ is a base, where $p+\mathbb Rw=\{z\,|\,\exists k\in \mathbb R,\, z=p+kw\}$ are all the lines parallel to the line $\mathbb R w$ and $p=jv$. Each complex number can be expressed in this base using the change of coordinate functions associated with that base $z=\varphi_0(z)v+\varphi_1(z)w$. The bundle we are using is $\varphi_0:\mathbb C=\bigcup_{p\in \mathbb Rv}p+\mathbb R w\to \mathbb R$. The classes are sets of the form $jv+\mathbb R w$. On each of these we will move by sliding, as you were proposing. So the function on the piece $f_j:jv+\mathbb R w\to jv+\mathbb R w$ is just addittion by some element of $\mathbb R w=\{kw\,|\, k\in\mathbb R\}$ that varies on $j$. $f_j(z)=f_j(jv+kw)=(jv+kw)+ \alpha_j w=z+\alpha_j w$ Thus the pasting looks as $f(z)=f_{\varphi_0(z)}(z)=(\varphi_0(z)v+\varphi_1(z)w)+ \alpha_{\varphi_1(z)} w=\varphi_0(z)v+(\varphi_1(z)+ \alpha_{\varphi_1(z)})w$ And its iteration is $f^t(z)=\varphi_0(z)v+(\varphi_1(z)+ t\alpha_{\varphi_1(z)})w$ (In)Sanity check: is this a group action? Identity: $f^0(z)=\varphi_0(z)v+(\varphi_1(z)+ 0\alpha_{\varphi_1(z)})w=\varphi_0(z)v+\varphi_1(z)w=z$; homomorphism proposition: unwind the definition $f^t(f^s(z))=f^t(f_{\varphi_0(z)}^s(z))=f_{\varphi_0(f_{\varphi_0(z)}^s(z))}^t (f_{\varphi_0(z)}^s(z))$. Now assume wlog that $\varphi_0(z)=j$, then by definition also $\varphi_0( f_{\varphi_0(z)}^s(z) )=\varphi_0 ( jv+(\varphi_1(z)+s\alpha_j )w )=j$ We conclude that $f^t(f^s(z))= f_j^t(f_j^s(z))$ that, by initial assumption satisfies $f_j^{t+s}(z)$. $\square$ Why this is not continuous? Yes, you analysis guys can find the prof easily, but I'd observe the structural reason for it. Since each $\alpha_j w$ may be oriented in both ways, and they are totally arbitrary when $j$ varies, we have no way to assume that "nearness" in the neighboring fibers get preserved after the transformation. I.e. the various given a neighborhood $U\subseteq \mathbb R$ of $j$ the action $f_u$ on the $z'$ near $z=jv+kw$ could scatter badly all the $f_u(z')$ very far from $f_j(z)$. Quote:however I don't quite understand all the terms (spectral decomposition of linear operators to fiber bundles especially) in topology, but I'll learn them as soon as possible. I was sloppy, sorry. I meant that this relates many topics like spectral decomposition of linear operators and fiber bundles. Quote:But partition won't work out for more cases, like abs(z), sgn(z), arg(z), log(abs(arg(z)+sgn(z))) and more wierder functions, That's quite the problem! Given an arbitrary function $f:X\to X$ can we partition $X$ is pieces $X_i$ were we know how to iterate the restrictions $f|_{X_i}$? If we can do this then it's like diagonalizing, and the set of indices of the partition is much like the spectrum of a matrix. I think of this like abstract eigentheory. Some function will be non abstractly diagonalizable, obviously. The key here is, whenever this partition technique can work, is to find the "optimal" components of $X$ that are closed under $X$. This is not the precise statement. I believe that given a $f:X\to X$ one has to start from the partition on $X$ generated by the relation $x\sim_f y$ iff $f^m(x)=f^n(y)$ for some $m,n \in\mathbb N$. But even this, is not sufficient. We must merge into this a study of group and monoid actions in order to find the right orbit... and even then it could fail. That's only my opinion btw. Quote:I'm still stuck about the conj(z), as its transformation cannot be described by countably infinitely many partitions $X_i$ and each has an $f:X_i\to X_i$. I wonder how one can get a continuous iteration without endomorphism, since it breaks the symmetry $f:X_i\to X_i$, or even a more complex question, like their iterations to complex heights, since then you can't use eigen decomposition or linear operators, like the cases $k\in \mathbb{C},f_k^t(z)=k_{\arg(z)}z$ Anyway thank u I think we can adapt the partition technique to conj since conjugation respects the concentric circle-partition you were describing. Idk how, I have not time to investigate this but... take a circle of radius $r$, i.e. consider a function $f_r:rS^1\to rS^1$. We need it to be idempotent and reflect each point vertically, as conjugation does. But I believe you can interpolate it as you like, almost. Maybe playing with $f_r^t(re^{i\theta})=re^{iK_r(t,\theta)}$ as a function of the angle on the circle where $K_r(t, -):{}S^1\to S^1$ are an endofunctions of the circle that satisfy $K_r(1,\theta)=-\theta$ and $K_r(s+t,\theta)=K_r(s,K_r(t,\theta))$. MSE MphLee Mother Law $(\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/18/2022, 06:49 PM (This post was last modified: 06/18/2022, 07:07 PM by MphLee.) Something similar is doable for $|\cdot|:\mathbb C\to\mathbb C$. This is idempotent because it decompose into $|\cdot|:\mathbb C\to\mathbb R^+\to \mathbb C$ where the second arrow is an inclusion. So every eq. classes of the polar decomposition maps to a single point, the fixed point. It means that we cannot extend it to a $\mathbb Z$ iteration, hence $\mathbb R$ iterating it is impossible. Only iteration where the time $f^t$ belongs to a monoid (has noninvertible elements) is possible. The time we can iterate on could be the monoid of non-negative reals $\mathbb R_0$ quotiented by the relation $t\sim s$ iff $t,s\in [1,\infty)$ and if $s=t$ otherwise. This amounts to collapsing together all the reals bigger than one (cauz of idempotency of the complex norm), i.e. we are considering the closed unit interval $[0,1]$ equipped with the operation $s{\bar {+}}t={\rm min}(s+t,1)$. Call that object $I=\mathbb R_0/\sim$. We can $I$-iterate the function $|\cdot|:\mathbb C\to\mathbb C$. On each class of the partition $rS^1=\{re^{i\theta}\}$ we consider the endomap $f_r(z)=f_r(re^{i\theta})=r$. The way i'd go is by selecting a family of continuous paths $\gamma(-,\theta): [0,1]\to (-\pi,\pi]$ s.t $\gamma(0,\theta)=\theta$ and $\gamma(1,\theta)=0$ and such that $\gamma(s{\bar {+}}t,\theta)=\gamma(s,\gamma(t,\theta))$ where we use the operation $s{\bar {+}}t={\rm min}(s+t,1)$. Take a whole family $\gamma_r (t,\theta)$ and define $f_r^t:rS^1\to rS^1$ as $f^t_r(re^{i\theta})=|z|e^{ i\gamma_r(t,\theta) }$ where $t\in I$ is in the quotient monoid we have defined before. Consider then $f$ as $f^t(z)=f^t_{{\rm arg}(z)}(z)$ Questions The question is thus reduced to the existence of at least one of such "homotopies" $\gamma(-,\theta): [0,1]\to S^1$. How to do it? Do they exists? Maybe we can consider $\gamma(t,\theta)=(1-t)\theta$ but then $\gamma(t, \gamma(s,\theta)=(1-t)(1-s)\theta=(1-(t+s)+ts)\theta =\gamma(t+s,\theta)+ts\theta$. Something is clearly off. Also, we lose continuity at -1, because all the points in the upper half plane get shrunk clockwise, and the one in the bottom go counterclockwise. What about using the same rotation everywhere? We place our discontinuity at $1\in S^1$? What if instead we just subtract angles? Consider $\beta(t,\theta)={\rm max}(\theta-t2\pi, 0)$ then le'ts check the iteration property: $\beta(t,\beta(s,\theta))={\rm max}({\rm max}(\theta-s2\pi, 0)-t2\pi, 0)$ by distributivity we get $\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-s2\pi-t2\pi, -t2\pi), 0)$, by some algebra $\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-(s+t)2\pi, -t2\pi), 0)$, now there are two cases: if $\theta-(s+t)2\pi< -t2\pi\leq 0$ then the overall result is zero because $t\in [0,1]$, otherwise the first term wins so $\beta(t,\beta(s,\theta))={\rm max}( {\rm max}(\theta-(s+t)2\pi, 0)=\beta(t+s,\theta)$ To complete the proof that we have an $I$-iteration just note how when $1R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ring-shaped slice has its own rotation angle, which can be written as: $f(z)=ze^{ig(\|z\|)}$ Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps: $\text{For real t, }f^t(z)=ze^{itg(\|z\|)}$ Remember abs(e^(it))=1 for real t, $\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\|z\|)}=ze^{itg(\|z\|)}e^{isg(\|z\|)}=\bigg(ze^{itg(\|z\|)}\bigg)e^{isg\big(\|ze^{itg(\|z\|)}\|\big)}=f^s(f^t(z))$ it's easy to prove, and also satisfies the geometric intuition that the t-th iteration of a rotation is a rotation whose angular velocity is t times the original rotation's. Let h(z) be analytic with fixpoint 1. g(z) = norm(z) and h ' (1) = r * e^i then h^[t](z) = h^[n] [ r^t f^[t] (h^[-n](z)) ] is that an analytic solution ? seems just the same as the regular fixpoint method ( koenings function ). the idea is that the continu map is asymptotic to a linear map near the fixpoint 1 and that should be sufficient to make the whole formula in the limit analytic. maybe a dumb question sorry. And not suggesting to actually apply such formula's. just for theory... if not analytic somehow , complex-continu solutions to tetration ?? hmm been a while since we talked about that ... the disctinction between non-analytic C^oo and complex-continu fascinates me. regards tommy1729 MphLee Long Time Fellow    Posts: 374 Threads: 30 Joined: May 2013 06/21/2022, 07:58 PM Excuse me Leo. My posts are full of typos and inaccuracies, but I claim the ideas are correct. When I get some time I'll go back fixing the formulae. MSE MphLee Mother Law \((\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ tommy1729 Ultimate Fellow     Posts: 1,918 Threads: 414 Joined: Feb 2009 06/25/2022, 10:59 PM (06/18/2022, 10:38 AM)Leo.W Wrote: Idk if it's time to talk about continuous iterations for non-analytic functions that can still be defined on the whole complex plane, but these ideas did disastrously harm to my brain lol. I worked out some cases last year, these are my results. Consider any function g(x):R->R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ring-shaped slice has its own rotation angle, which can be written as: $f(z)=ze^{ig(\|z\|)}$ Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps: $\text{For real t, }f^t(z)=ze^{itg(\|z\|)}$ Remember abs(e^(it))=1 for real t, $\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\|z\|)}=ze^{itg(\|z\|)}e^{isg(\|z\|)}=\bigg(ze^{itg(\|z\|)}\bigg)e^{isg\big(\|ze^{itg(\|z\|)}\|\big)}=f^s(f^t(z))$ it's easy to prove, and also satisfies the geometric intuition that the t-th iteration of a rotation is a rotation whose angular velocity is t times the original rotation's. Another idea, came by trying to figure out whether there's continuous iteration of the conjugate function conj(z) We start by conj(z)=abs(z)*e^(-i*arg(z)), consider a set of functions $f_u(z)=\|z\|e^{iu\arg(z)}$ where u is real, Then $\text{For real t and u^t, }f_u^t(z)=\|z\|e^{iu^t\arg(z)}$ Because $\text{For real s and t, }f_u^{s+t}(z)=\|z\|e^{iu^{s+t}\arg(z)}=\|ze^{iu^tz}\|e^{iu^s\arg(\|z\|e^{iu^tz})}=f_u^s(f_u^t(z))$ if u^t and u^s are both real. But since u=-1, t=1/2, u^t is not real, we cannot find a functional square root of conj(z) through this. Do you have more examples, I basically based all this on the Polar decomposition of complex numbers and don't have so much time to take more investigation into it. In 2 dimensions its all fun and games. But What could be useful is an analogue in higher dimensions. Since conformal maps barely exist in higher dimensions , nor analytic BUT multivariable taylors do , this might be interesting. Rotations are associated with quaternions … or more numbers. So there might be a link. Regards  tommy1729 JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 06/26/2022, 05:37 AM (This post was last modified: 06/26/2022, 05:37 AM by JmsNxn.) I'm very fucking confused in this thread. What is $||z||$? How exactly do you know that the semi-group functional equation is being satisfied? I'm not doubting you, I am giving you the benefit of the doubt. But I'm still confused. Mphlee just made me more confused. Could you perhaps draw out the argument more, Leo? I'd really appreciate it. tommy1729 Ultimate Fellow     Posts: 1,918 Threads: 414 Joined: Feb 2009 06/28/2022, 02:17 PM (06/26/2022, 05:37 AM)JmsNxn Wrote: I'm very fucking confused in this thread. What is $||z||$? How exactly do you know that the semi-group functional equation is being satisfied? I'm not doubting you, I am giving you the benefit of the doubt. But I'm still confused. Mphlee just made me more confused. Could you perhaps draw out the argument more, Leo? I'd really appreciate it. He means the norm of the complex Value z.  So the formula works  Yay  Regards  Tommy1729 « Next Oldest | Next Newest »

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