Continuously Iterating Modular Arithmetic Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   07/14/2022, 10:58 AM (This post was last modified: 08/14/2022, 11:35 PM by Catullus.) How do you continuously iterate the function $\dpi{110} \mod(x,k)$? Like what is $\dpi{110} \mod^{.5}(x,1)$, or $\dpi{110} \mod^i(x,1)$? Also, another name for $\dpi{110} \mod(x,1)$ is $\dpi{110} \mod(x,1)$, which is short for the fractional part of x. The fractional part of π is π-3, but π-3 is not a fraction. So, what is up with that? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ MphLee Long Time Fellow Posts: 367 Threads: 28 Joined: May 2013 07/14/2022, 01:14 PM (This post was last modified: 07/14/2022, 01:14 PM by MphLee.) Let's do some order. Mod functions are in the realm of ring theory. I assume you can guess what a ring is. It is good place where you have stuff that behaves as if they were numbers. You can add and subtract that stuff and you can multiply it in a way that distributes over addition. Given a ring $$R$$ we can study its ideals $$I\subseteq R$$. Ideals are special subsets of the ring that behaves as if they were linear subspaces of $$R$$ (lines, planes and so on) but somewhat more abstractly than these. Now, an ideal $$I$$ in $$R$$ kinda traces out a figure in $$R$$ and we can use it to "fold" $$R$$ in such a way that the folded ring $$R/I$$ has all the points in the figure $$I\subseteq R$$ collapsed to zero, so they disappears. More technically, an ideal defines an equivalence relation over the ring. This folding is called quotienting out $$R$$ by the ideal $$I$$. Given an ideal $$I\subseteq R$$ there exists a canonical (unique map) going from the original ring to its quotiented by $$I$$ $$\pi_I:R\to R/I$$ Sending $$r\in R$$ to $$[r]_I:=r+I$$. Ideals of a ring contains the information about divisibility of stuff inside $$R$$: multiples and divisors. Archetypal example. Take the ring of integers $$\mathbb Z$$: all of its ideals are of the form $$n\mathbb Z$$: so they are $$0\mathbb Z, 1\mathbb Z, 2\mathbb Z,...,n\mathbb Z,...$$. These ideals define the relation of being equivalent modulo something. Given the ideals $$0\mathbb Z=\{0\},\quad 1\mathbb Z=\mathbb Z=\{...,\,-2,\,-1,\,0,\,1,\,2,\,3,\,...\},\quad 2\mathbb Z=\{...,\,-2,\,0,\,2,\,4\,...\},\quad...,\quad n\mathbb Z=\{...,\,-n,\,0,\,n,\,2n\,...\},\quad ...$$ We obtain quotient rings $$\mathbb Z/0\mathbb Z\simeq \mathbb Z,\quad \mathbb Z/1\mathbb Z\simeq 0,\quad \mathbb Z/2\mathbb Z,\quad...,\quad\mathbb Z/n\mathbb Z,\quad ...$$ And we have canonical ring homomorphisms $$a\mapsto \pi_n(a)=[a]_n:=a+n\mathbb Z$$ $$\pi_n:\mathbb Z\to \mathbb Z/n\mathbb Z$$ Observe. what I've defined here are the real "mod functions": $$\pi_n(a)=a\, {\rm mod}\, n$$. They are ring theoretically in nature. Also notice that $$\pi_1$$ is a bit special since it is a constant function, thus the only of the mod/projection morphisms to not be a morphism of unital rings\/fields: the field $$\mathbb F_1=\mathbb Z/1\mathbb Z$$ with one elements does not exists. About iterating mods. so this makes impossible to iterate the mod funtions since they are not endofucntions. Sure, you can post-compose them with the identifications $$i_n:\mathbb Z/n\mathbb Z \to \mathbb Z$$ but you will obtain idempotent functions, thus trivial dynamics. The proof goes showing that $$i_n$$ inverts $$\pi_n$$ on the right $$\pi_n(i_n([a]_n))=\pi_n(a)=[a]_n$$, we conclude that the composition $$i_n\circ \pi_n:\mathbb Z\to \mathbb Z$$ is idempotent. All of this only to say that $${\rm mod}\, 1$$ is not the fractional part in general. It is an abuse of notation and an attempt to extend a concept belonging to integers to the real numbers. Now someone could say: but $$\mathbb R$$ is a field, hence a ring, so we can mod it out by $$1\mathbb Z=\mathbb Z$$ and obtain reals number mod 1. The answer is NO. This is all an abuse of notation: the integers are not an ideal of the real numbers. Real numbers are a field and fields have not "linear subspaces", using the heuristic I used at the beginning. This happens because fields have only trivial ideals... i.e. they are 1 dimensional, and their dimension 1 cannot be decomposed further... That means that the operation of quotienting out the integers from the real is not ring theoretic but group theoretic... so it is not the modulo operation anymore. Not properly. The additive group $$\mathbb R$$ of real numbers can be quotiented by its subgroup of integers giving as a result the circle group $$\mathbb R/\mathbb Z\simeq S^1$$ and the canonical quotient surjective morphism of groups $$\pi_\mathbb Z:\mathbb R\to \mathbb R/\mathbb Z\simeq S^1$$ Even in this case you can think of it as folding the line so that all the integers get folded over the zero... like winding the real line. An yes, the equivalence relation induced by the quotiend reminds the modulo 1 eq. relation. $$\pi_\mathbb Z(x)=\pi_\mathbb Z(y) \quad \iff \quad y-x \in\mathbb Z$$ i.e. $$x$$ and $$y$$ have the same non-integer part. Iterating the fractional part. Again, it turns out that if you post-compose $$\pi_\mathbb Z:\mathbb R\to S^1$$ to the obvious map $$i:{}S^1\to \mathbb R$$ that identifies the non-integer part in the interval $$[0,1)$$ you have a right inverse, and again we conclude that the non-integer part $$\{x\}=(i\circ \pi_\mathbb Z)(x)$$ is idempotent: not interesting dynamics. Note. the non-integer part of $$\pi$$ is still a real number, but if you look for the fractional part properly, then you have to consider $$\pi:\mathbb Q\to \mathbb Q/\mathbb Z$$. But again, your problem can't arise here since $$\pi\notin \mathbb Q$$. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 07/15/2022, 07:07 AM (This post was last modified: 07/15/2022, 07:51 AM by Catullus.) Oh, alrighty then. Thank you very much for for your excellent reply. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ MphLee Long Time Fellow Posts: 367 Threads: 28 Joined: May 2013 07/15/2022, 01:59 PM (This post was last modified: 07/15/2022, 02:00 PM by MphLee.) It was a pleasure Catullus. This is really an interesting topic. If you want you can try to share your own thoughts on it, it doesn't matter if you have not results or proofs. It is enough that you applied to it with passion, maybe you have questions or you have a personal way to approach the problem and you look for tips and observations. Anyways... if you really want something cool to think about, I suggest you to go deeper instead into the functions $$F_a(n)=[a]_n=a \,{\rm mod}\, n$$ for varying $$n$$. Only understanding those is very challenging... maybe iterating them would be an even greater challenge. Just some hints: every function $$[a]_n\in \mathbb Z/n\mathbb Z$$ belongs to a different ring... so how we obtain $$F_a$$? Just imagine $$F_a$$ to take inputs in $$\mathbb Z$$ and give output in the union of all the "modular rings" $$\bigcup _{n\in\mathbb N}\mathbb Z/n\mathbb Z$$. Why it is interesting? Because the zeros of $$F_a$$ are where $$a$$ has it's prime decomposition, i.e. $$F_a$$ detects $$a$$ prime divisors. If instead you wanto to continue investigating functions $$\pi_n(a)=a\, {\rm mod} \,n$$ then you should begin to study what it means to iterate idempotent functions. A function is idempotent iff $$f(f(x))=f(x)$$... this means that it has all fixed points or pre-periodic point. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   07/19/2022, 01:15 AM (This post was last modified: 07/19/2022, 01:23 AM by Catullus.) How do you continuously iterate the extension of $\dpi{110} mod(k,x)$ to non natural numbers? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow Posts: 1,058 Threads: 121 Joined: Dec 2010 07/19/2022, 01:38 AM It's idempotent Catullus. There is no continuous iteration. if $$f^{\circ n}(x) = f(x)$$ then that's it, there's no fractional iteration.... Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 07/19/2022, 02:09 AM By iterating the extension of $\dpi{110} mod(k,x)$ to non natural numbers, I meant doing stuff like $\dpi{110} mod^{.5}(1,x)$ or $\dpi{110} mod^i(1,x)$. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow Posts: 1,058 Threads: 121 Joined: Dec 2010 07/19/2022, 02:34 AM (07/19/2022, 02:09 AM)Catullus Wrote: By iterating the extension of $\dpi{110} mod(k,x)$ to non natural numbers, I meant doing stuff like $\dpi{110} mod^{.5}(1,x)$ or $\dpi{110} mod^i(1,x)$.Oh, I apologize, you mean iterate in the modulus. That is a very very difficult question. Honestly, I'd expect that people like Andrew Wiles would have trouble answering that. It is far too connected to the nature of prime numbers, and fields, and field classes, and etc etc... I doubt anyone has the answer for you. I'm hopefully not sounding like a broken record, but you'd probably get a fields medal if you could solve that. MphLee Long Time Fellow Posts: 367 Threads: 28 Joined: May 2013 07/19/2022, 07:59 AM @James. This times James I don't fully agree since I was able to do something on it... and it seems beautiful... not very hard but the implication it has points to a greater field of study that in my opinion was already fully explored by Grothendieck and algebraic geometers. Keywords should be rational points of algebraic curves, rational functions over the torus and scheme theory. @Catullus. Anyways Catullus, I'm not gonna share an inch of it. No time atm, have to do something more important and tbh I'd like to ear first some of your proposal to attack the problem, or I'd like you to share some of your preliminary analysis of the problem you are asking solutions to... I'm not sure to understand why you ask this, why do you find interesting? Why should I invest time on it exposing the material if you do not invest time researching/playing with your questions? MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 1,058 Threads: 121 Joined: Dec 2010 07/19/2022, 09:08 AM (This post was last modified: 07/19/2022, 09:33 AM by JmsNxn.) (07/19/2022, 07:59 AM)MphLee Wrote: @James. This times James I don't fully agree since I was able to do something on it... and it seems beautiful... not very hard but the implication it has points to a greater field of study that in my opinion was already fully explored by Grothendieck and algebraic geometers. Keywords should be rational points of algebraic curves, rational functions over the torus and scheme theory. @Mphlee, well maybe you should get a fields' medal. Let's see that deep Grothendiek construction. Besides, I meant, there's no analytic iteration--it must be some grothendiek shit. And to fully describe it; you'd definitely get a fields medal... I mean, this question borders the deepest shit within elliptic function theory, and modular function theory. You describe a function that satisfies these kind of modular iterations.... nah bro. You ain't done it. Restricting to the iterations to $$x \text{mod} y$$ and observing $$x,y \in \mathbb{N}$$. Where we iterate in $$y$$ in some form, is a very deep problem. To describe how it could work is different, than proving it behaves in such a manner. It's part of a program (forget the name) on iteration/natural numbers. It's probably harder than the 3n+1 problem... Like you can't tell me you've solved: $$x\,\text{mod} \left(x \text{mod} p\right) = x \text{mod} y\\$$ Where $$p = p(x,y) \in \mathbb{N}$$ and is effectively a modular square root. It's just not done... Sure it becomes: $$x \text{mod} p = y\\$$ But ffs, even this is a deep problem in number theory. Let alone, if we guess as $$p\to\infty$$ in the initial equation. and we talk about prime pairs and prime powers. Trust me, Mphlee. I have a strong background in analytic number theory. If you could answer these questions, the fields medal people would s*** your d***. Lol. « Next Oldest | Next Newest »

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