Non-trivial extension of max(n,1)-1 to the reals and its iteration. MphLee Long Time Fellow Posts: 367 Threads: 28 Joined: May 2013 05/16/2014, 01:37 PM (This post was last modified: 05/16/2014, 02:40 PM by MphLee.) I was very interested in the problem of extending $max(n,1)-1$ to the reals (that is actual the "incomplete" predecessor function $S^{\circ (-1)}$ over the naturals) I've asked the same question on MSE but it was a bit ignored...I hope because it is trivial! http://math.stackexchange.com/questions/...its-fracti The question is about extensions of $A(n):=max(n,1)-1$ to the reals with some conditions Quote:A-$A(x)=max(x,1)-1$ only if $x \in \mathbb{N}$ B-$A(x)$ is not discontinuous I just noticed that I've made a lot of errors in my MathSE question, I'll fix it in this post (and later on mathSE) From successor and inverse successor we can define the subtraction in this way $x-0:=x$ and $x-(y+1):=S^{\circ(-1)}(x-y)$ with $A$ , that is a modified predecessor function we could define its iteration using an "esotic subtraction" that is "incomplete" for naturals and is "complete" for reals (like we are cutting all the negative integers) Quote:$x -^* 0:=x$ $x-^* (n+1)=A(x-^* n)$ In this way we have Quote:$x-^*1=A(x)$ and $x-^*n=A^{\circ n}(x)$ How we can go in order to extend $x-^*y$ to real $y$? For example what can we know about $x-0,5$ ? for example $(x-^*0.5)-^*1=(x-^*1)-^*0.5=x-^*1.5$ if we put $x=0$ $(0-^*0.5)-^*1=(0-^*1)-^*0.5=0-^*1.5$ then if $0_^*0.5=\alpha$ $\alpha-^*1=0-^*0.5=0-^*1.5$ if $\alpha$ is not a natural number $\alpha-^*1=\alpha=0-^*1.5$ !?? what is going on here? If $\alpha$ is natural $max(\alpha,1)-1=\alpha=0-^*1.5$ in this case we should have that $\alpha=0$. What do you think about this? 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: 1,742 Threads: 382 Joined: Feb 2009 05/16/2014, 09:21 PM Your function is equal to (x + abs(x))/2. abs(x) can be written as sgn(x) x. sgn(x) is well approximated by tanh(100x). This gives that your function is very close to (x+tanh(100x) x)/2 The problem with your function is that it has all positive integers as fixpoints. Too many fixpoints to have half-iterates valid everywhere. Im not sure if you want an interpolation or an approximation like I just gave. Also the reason you get little response is probably because your mainly asking " what makes this question more intresting " ? If you know what I mean. Asking what properties to look for or asking what questions to ask is similar. You have to decide what you want to do , want to see solved or what properties you desire. Otherwise it sounds weird. Kinda like asking for " a special integer ". Math is like driving a car without a map. You dont know where you will end up. But if you want to end up somewhere you have too start , stop and drive. I hope my metaphor is understood. I assume you are still young. You dont have to tell me about your age but I suspect it. Hope you dont mind me saying. regards tommy1729 MphLee Long Time Fellow Posts: 367 Threads: 28 Joined: May 2013 05/16/2014, 10:20 PM (This post was last modified: 05/17/2014, 11:48 AM by MphLee.) (05/16/2014, 09:21 PM)tommy1729 Wrote: Your function is equal to (x + abs(x))/2. abs(x) can be written as sgn(x) x. sgn(x) is well approximated by tanh(100x). This gives that your function is very close to (x+tanh(100x) x)/2 I've plotted this and is the same as $max(0,x)$ or the same as $lim_{h \rightarrow 0^+}h(ln(e^{0/h}+e^{x/h}))=max(0,x)$ ... I know this but I was loking for a function that coincides with $max(1,x)-1=max(0,x-1)$ only for the naturals... If I did not understand something of your formula of approximation tell me. Quote:The problem with your function is that it has all positive integers as fixpoints. Too many fixpoints to have half-iterates valid everywhere. Ok..I don't get this (I'm not good with analysis and the iteration theory)... but help to to understand pls. A fixpoint is a x such that $A(x)=x$ in the case of $A(0)=0$... is it the only fixpoint of A? Maybe you talk about the fact that $A^{n}(0)=0$ and $A^{n}(m)=max(0,m-n)$ for every $n$. This is really a big problem for the real iteration problem? Quote:Im not sure if you want an interpolation or an approximation like I just gave. Also the reason you get little response is probably because your mainly asking " what makes this question more intresting " ? If you know what I mean. Asking what properties to look for or asking what questions to ask is similar. You have to decide what you want to do , want to see solved or what properties you desire. Otherwise it sounds weird. Kinda like asking for " a special integer ". Math is like driving a car without a map. You dont know where you will end up. But if you want to end up somewhere you have too start , stop and drive. I hope my metaphor is understood. Yea I know, the question is a bit unclear and is because I don't exatly know what to ask..or how to ask it. I'm a bit confused about this problem but it is very interesting for me and I think it can be important for oher things I'm doing. Quote:I assume you are still young. You dont have to tell me about your age but I suspect it. Hope you dont mind me saying. regards tommy1729 Don't worry I'm enough young (but not very very young) Maybe you feel more my lack of knowledge on some really basic topics...I don't study math at school probably thats why. 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: 367 Threads: 28 Joined: May 2013 05/17/2014, 07:10 PM Update: I voted for the closure of the question on MSE because it is not clear at all. 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   06/09/2022, 11:06 PM (This post was last modified: 11/15/2022, 04:49 AM by Catullus.) (05/16/2014, 01:37 PM)MphLee Wrote: Quote:A-$A(x)=max(x,1)-1$        only if $x \in \mathbb{N}$ B-$A(x)$ is not discontinuous$$\forall n\in\Bbb N\max(n,1)-1=n-1$$ Defining $$A(x)$$ as $$x-\sin(x*\pi)-1$$ might work. Can someone please tell me a special integer? 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 06/10/2022, 09:07 AM This will work, but is not really what I was looking for in my question. Back in the days my mind was really foggy on this problem because I wasn't able to phrase the real problem I had in mind. Here indeed your solution  $$(x-1)-\sin(x\pi)$$ is correct. But probably I was looking for somethign like a smooth approximation of $$T(x)\sim \max(x,1)-1$$  and then the family $$T_\theta=T(x)+\theta(x)$$ for $$\theta(x+1)=\theta(x)$$ and $$\theta(n)=0$$ for each $$n\in\mathbb Z$$. So Tommy's solution fits better since $$\max(x,1)-1=\lim_{k\to\infty} \frac{(x-1)+(x-1){\rm tanh}(k(x-1))}{2}$$ Here Tommy's approximation for $$k=6$$ Here then $$T_\theta(x)=\frac{(x-1)+(x-1){\rm tanh}(k(x-1))}{2} + \lambda \sin(x\pi)$$ for $$\lambda=0.2$$ After all this years I have to think more about this and see if those approximations can be useful for my endgame... One day I'll post something about it. 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   06/10/2022, 10:17 AM (06/10/2022, 09:07 AM)MphLee Wrote: After all this years I have to think more about this and see if those approximations can be useful for my endgame...Your endgame? Endgame of what? 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 06/10/2022, 11:04 AM Excuse me, I mean that I'm not sure anymore that those approximations are useful for the purpose of my research. I'm, and I was, researching about a far reaching generalization of hyperoperations. I don't know your background so I'll make it self-contained. I'm studying special functions of the kind $${\bf g}:J\to X$$ and maps of kind $$\rho:X\to \mathbb N$$ for $$J$$ a dynamical system. That means that fixing a $$\rho$$ of that kind we can send each $$\bf g$$ to a map $$\rho{\bf g}:J\to \mathbb N$$. Here in this post, 8 years ago, I was initiating the study, with scarce success, of functions $$f:J\to \mathbb N$$ that are of the form $$f=\rho{\bf g}$$. 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 06/11/2022, 06:22 AM (This post was last modified: 11/15/2022, 04:51 AM by Catullus.) (06/10/2022, 09:07 AM)MphLee Wrote: But probably I was looking for somethign like a smooth approximation of $$T(x)\sim \max(x,1)-1$$  and then the family $$T_\theta=T(x)+\theta(x)$$ for $$\theta(x+1)=\theta(x)$$ and $$\theta(n)=0$$ for each $$n\in\mathbb Z$$.Log(sqrt(2),sqrt(2)^x+sqrt(2))-1 is a smooth approximation of max(x,1)-1. 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 06/15/2022, 10:59 PM Catullus, this option was already pointed back then. It is the common representation of the tropical operation. max, as as the "Litvinov-Maslov dequantization" of the bennet preaddition. 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