On extension to "other" iteration roots bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 08/18/2022, 06:32 AM (This post was last modified: 08/18/2022, 06:32 AM by bo198214.) Not to disappoint you either, Leo.W, but *you* missed one thing! Your explanations might be well and right, however did not address what I was pointing to. For me Abelian property was just $$f^s\circ f^t=f^t\circ f^s$$, because everywhere else Abelian refers to commutativity. And I just wanted to point out to you that there was more to a continuous iteration - which in turn you pointed out to me! LoL. Only now when I look back at at one of your answers, I see you meant already $$f^s\circ f^s = f^{s+t}$$ by "Abelian property". However I never encountered that naming in some literature. Even when I googled it now, I didn't see clear definitions (On Wikipedia e.g. there is a chapter "Abelian Property" but it is not explicitly defined and they rather use the name "translation functional equation" for non-integers. I always used the name iteration (semi)group - which though is also quite cumbersome to use, because it refers to more than just the equation. So from which literature did you obtain this definition? JmsNxn Ultimate Fellow Posts: 1,179 Threads: 123 Joined: Dec 2010 08/18/2022, 06:45 AM I'd love to just jump in here, and respond on Mphlee's behalf. Who designed how $$f \circ g = g \circ f$$ means they belong to the same categorical group, but does not admit that we have the semi-group property. It simply means the preimages, and that the images, agree in a fairly good manner. But it does not mean the semi-group property is followed; at best it means the super function property is defined. I know this might not mean much, but if you read Mphlee's work this is very much his disagreement with some of the things Leo has said. I think you are expressing Mphlee's concern analytically, bo. Again, though, this is just off of Mphlee and his description of the class of functions which are commutative--and their relationship to fixed points. Leo.W Fellow Posts: 86 Threads: 7 Joined: Apr 2021 08/28/2022, 01:32 PM (08/18/2022, 06:32 AM)bo198214 Wrote: Not to disappoint you either, Leo.W, but *you* missed one thing! Your explanations might be well and right, however did not address what I was pointing to. For me Abelian property was just $$f^s\circ f^t=f^t\circ f^s$$, because everywhere else Abelian refers to commutativity. And I just wanted to point out to you that there was more to a continuous iteration - which in turn you pointed out to me! LoL. Only now when I look back at at one of your answers, I see you meant already $$f^s\circ f^s = f^{s+t}$$ by "Abelian property". However I never encountered that naming in some literature. Even when I googled it now, I didn't see clear definitions (On Wikipedia e.g. there is a chapter "Abelian Property" but it is not explicitly defined and they rather use the name "translation functional equation" for non-integers. I always used the name iteration (semi)group - which though is also quite cumbersome to use, because it refers to more than just the equation. So from which literature did you obtain this definition? I'm sorry if I abused this term *I just thought Abelian can be used as an adjective word* to replace commutative. The point is that the iterations are defined in the way $$f^s\circ f^s = f^{s+t}$$ when you say that f^s is an iteration, I don't see why would iterations not guarantee $$f^s\circ f^s = f^{s+t}$$...? And if you're not referring to iterations, then I personally think $$f_s$$ is a lil better than $$f^s$$ as notation. I'm sorry I pointed out to you, I think there's some misunderstanding. The solutions g(z) to the functional equation $$f(g(z))=g(f(z))$$ given f(z) is denoted $$[f,f]$$ as MphLee, but not all solutions are iterations, or continuous iterations, your $$f^s$$ can be interpreted as elements in $$[f,f]$$, I barely think that it is iteration only after $$f^sf^t=f^{s+t}$$ because that will make sense, otherwise we can give such solutions in any forms but doesn't provide us with any useful info, like, what about replacing multiplication with another bunch of solutions of $$f(z+1)=f(z)+1$$ But, yeah, you're right, there's more to continuous iteration  I had no intention to disrupt your research or something else, I just love communication with you fantastic guys I consider that my ability to arrange what things I intended to say is just sh*t, and without termilogies, lol, maybe my English is so pretentiously spoken, plz forgive me for that, I'm still learning and struggling with GRE and TOEFL and all these stuffs Quote:But it does not mean the semi-group property is followed; at best it means the super function property is defined. I know this might not mean much, but if you read Mphlee's work this is very much his disagreement with some of the things Leo has said. I'd love to know what the disagreement is against, James And I wonder where can I browse his works? I want to learn about how much has he did and study more Regards, Leo bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 08/28/2022, 03:38 PM (08/28/2022, 01:32 PM)Leo.W Wrote: The point is that the iterations are defined in the way $$f^s\circ f^s = f^{s+t}$$ when you say that f^s is an iteration, I don't see why would iterations not guarantee $$f^s\circ f^s = f^{s+t}$$...? And if you're not referring to iterations, then I personally think $$f_s$$ is a lil better than $$f^s$$ as notation. Then why do you speak about "Abelian property" and "Abelian-ity" at all, why don't you call it just "iteration"? For me, if someone says "iteration of a function", it means the iteration with natural numbers, perhaps extended to negative integers as iterated inverse function. And the thing is in this case you can really say "the iteration". Btw. the notation $$f^t$$ is similarly badly chosen because you can not distinguish it from power - particularly if you work in a calculus that does not have the "(x)" appended - i.e. if I write $$f\circ g$$ and not $$f(g(x))$$, e.g. formal powerseries calculus. That's why I typically write $$f^{\circ s}$$ and I saw that also in literature. But with non-integer iterations you can not talk about "the" iteration, that's why a notation that does not include the type of iteration is also misleading. I tried to work around that with a notation like $$f^{\mathfrak{R}t}$$ for regular iteration, but sure this is a rather clumsy approach and also not used in literature. In older papers you quite often encounter the simple $$f_s$$ or even $$f(s,x)$$ with $$f(s+t,x)=f(s,f(t,x))$$. JmsNxn Ultimate Fellow Posts: 1,179 Threads: 123 Joined: Dec 2010 08/30/2022, 03:29 AM (This post was last modified: 08/30/2022, 03:29 AM by JmsNxn.) (08/28/2022, 01:32 PM)Leo.W Wrote: I'd love to know what the disagreement is against, James And I wonder where can I browse his works? I want to learn about how much has he did and study more Oh not quite a disagreement. I think you have just misunderstood me. But when you create a class of functions such that $$f(g) = g(f)$$, you do not essentially describe fractional iterates. Because alternative classes of these functions exist. So it is not fundamental, in the sense that you write $$f(g) = g(f)$$--it doesn't entirely determine the space. It restricts values on the space, but doesn't solve it. This is a standard issue theorem in category theory (I know it from abstract algebra, personally). I'm just trying my best to reexplain the initial partial objections Mphlee had. He explained it much better, but you have to read the 100 or so pages Mphlee has posted on this forum. I'm not trying to be a stick in the mud, Leo   I love your work! But we need to always understand the limitations. « Next Oldest | Next Newest »

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