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 Categorical iteration theory andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/19/2007, 07:09 AM (This post was last modified: 10/20/2007, 06:05 PM by andydude.) I've been reading about products lately and I realized that continuous iteration is an example of a categorical product. But since I'm still learning about category theory, I have some questions and observations I'd like to ask and confirm. First, the form in which I'm talking about continuous iteration being a product is in the following sense. Similar to a notation found in dynamical systems, iterates of a function $f : (X \rightarrow X)$ (or equivalently: $f \in X^X$) can be denoted $\Phi^t : X^X \rightarrow X^X$, because the second iterate (for example) of a function from X to X is still a function from X to X. Using this line of thinking, an orbit of the same function can be denoted $\Phi_x : X^X \rightarrow X^T$, because the time space is different from the phase space, and the orbit of f from 0 (for example) would be a function from T to X. The general iterate of the function (usually written $\Phi(t, x) = f^{{\circ}t}(x)$) can then be redefined as an operator of sorts: $\Phi : X^X \rightarrow X^{T \times X}$. Secondly, am I thinking of iteration as a functor between endomorphisms? Or is that another way of looking at it? Or is it simply a morphism between objects in the category of sets? Third, if you put all of the arrows together, along with $\pi_1 : X^{T \times X} \rightarrow X^T$ and $\pi_2 : X^{T \times X} \rightarrow X^X$, then you get almost exactly the definition of a categorical product, except with an X exponentiating every set in the commutative diagram. What I wonder is if this is obvious, or if the non-X version being a categorical product does not impliy the X-exponentiated version is a categorical product? Fourthly, if continuous iteration really is a categorical product, that means that given an orbit map $\Phi_x$ where $x \in X$ and an iterate map $\Phi^t$ where $t \in T$ , the general iterate $\Phi$ is uniquely determined. This makes sense intuitively, which makes me think it is so. Also, given the Abel functional equation, it is possible to reconstruct the general iterate $\exp_b^t(x)$ given only tetration and the super-log, so this makes me think that all functions with a well-defined orbit will also have a unique general iterate $\Phi$. Lastly, it is also known that for analytic iteration to exist uniquely, that the function being iterated must have a hyperbolic fixed point. Does this make iteration fail to be a categorical product? Or does it mean that analytic iteration is only a categorical product over the set of functions with hyperbolic fixed points? This part is really confusing me. Any help or clarification would be most appreciated. Andrew Robbins jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 10/19/2007, 08:21 PM (This post was last modified: 10/19/2007, 08:30 PM by jaydfox.) Quote:Lastly, it is also known that for analytic iteration to exist uniquely, that the function being iterated must have a hyperbolic fixed point Out of curiosity, why did you exclude the possibility of a parabolic fixed point? And what of functions with multiple (non-conjugate) fixed points? Would uniqueness depend on the existence of a single hyperbolic fixed point (or a single pair of conjugates)? And PS, thanks for the links and ideas. Most of my formal and self-training is fairly narrowly concentrated on good old-fashioned number crunching and vector analysis. I have big holes in my understanding of more "basic" topics in topology, group theory, set theory, rings and fields, etc., so it's good to know that between the few of us in this forum, we'll be able to bounce ideas off each other and get guidance. For now, I'm down in the trenches studying your slog, and it's a fascinating thing to study, but I know I need to broaden out a little bit. I'm especially ignorant of the matrix methods (by that, I mean the carleman or bell or whatever matrices, can't remember the names at the moment). ~ Jay Daniel Fox andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/20/2007, 06:15 PM jaydfox Wrote:Quote:Lastly, it is also known that for analytic iteration to exist uniquely, that the function being iterated must have a hyperbolic fixed point Out of curiosity, why did you exclude the possibility of a parabolic fixed point? Well, I suppose I'm drawing experience from $e^x - 1$, and how although this is $C^{\infty}$ (infinitely differentiable) it is not $C^{\omega}$ (analytic). Since $e^x - 1$ is infinitely differentiable, and the general iterate is also infinitely differentiable, because $\eta = e^{1/e}$ is a parabolic fixed point, this allows us to say that the general iterate "perserves the structure" of $e^x - 1$ with respect to infinite differentiability (it doesn't preserve analycity). This is really what I'm interested in, is for what functions can the general iterate be found with similar properties. I suppose I didn't want to exclude parabolic fixed points from the conversation, I just wanted to make sure that I was finding the correct conditions for having an analytic iterate. Andrew Robbins jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 10/20/2007, 06:29 PM I've seen this mentioned a couple times, but why is it that the parabolic case is not analytic, while the hyperbolic is? Is there a thread on this topic that I missed? ~ Jay Daniel Fox andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/20/2007, 07:27 PM (This post was last modified: 10/20/2007, 07:29 PM by andydude.) For a function to be real analytic it must satisfy two properties. (1) it must be infinitely differentiable (iterated real derivatives exist) for all points in the domain, or in other words, there exists a Taylor series expansion, and (2) the Taylor series expansion must converge, or in other words, the radius of convergence must be nonzero. Sometimes there is also a requirement that the function the Taylor series converges to is also the function being differentiated, but this is hard to find a good counterexample for. The general iterate of the function $e^x-1$ has a Taylor series expansion, thus it is infinitely differentiable, but since the radius of convergence is 0 (see this thread), it is not real analytic. The reason why the radius of convergence is 0 is because the root test has no bound. For there to be a nonzero radius of convergence, the root test must be bounded. Actually, to be more specific, its unbounded for non-integer t, for integer t the root test is bounded. For a function to be complex analytic (or holomorphic) it must satisfy essentially the same properties, but over complex derivatives, not real derivatives. In order for a complex derivative to exist a function must satisfy the Cauchy-Riemann equations which are a little more strict than a real derivative. Since the primary difference between real analytic and complex analytic is the kind of derivatives used, there is a tendency to drop the real/complex part and just refer to analycity. In both cases the requirements are (1) infinite differentiability and (2) Taylor series has nonzero radius of convergence for all points in the domain. A special terminology is reserved for cases in which this is almost true except for a finite number of points. A meromorphic function is a function that is holomorphic for all except a finite number of points in the domain. It may be that the iteration of $e^x-1$ (written $DE^t(x)$) is analytic or holomorphic everywhere except x=0. I don't know, but we do know that the general iterate is not analytic at x=0. Also, to be more specific, $DE^t(x)$ is analytic at x=0 for integer t, but is not analytic at x=0 for non-integer t, thus it is not analytic for all (t, x) in the domain $T \times X$. If this is too confusing, then it may be because the coefficients of x in the Taylor series are finite polynomials in t which we don't need to worry about convergence, since they're always "convergent" in some sense. So the Taylor series in x is the only one we need to worry about. In some sense this kind of thinking can also be applied to hyperbolic iteration as well, since the coefficients of x in the Taylor series are functions of t that are representable in closed form. Andrew Robbins « Next Oldest | Next Newest »

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