Non-trivial extension of max(n,1)-1 to the reals and its iteration. - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Non-trivial extension of max(n,1)-1 to the reals and its iteration. (/showthread.php?tid=868) Non-trivial extension of max(n,1)-1 to the reals and its iteration. - MphLee - 05/16/2014 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/789520/extensions-of-an-operatornamemaxn-1-1-to-the-reals-with-and-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? RE: Non-trivial extension of max(n,1)-1 to the reals and its iteration. - tommy1729 - 05/16/2014 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 RE: Non-trivial extension of max(n,1)-1 to the reals and its iteration. - MphLee - 05/16/2014 (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. RE: Non-trivial extension of max(n,1)-1 to the reals and its iteration. - MphLee - 05/17/2014 Update: I voted for the closure of the question on MSE because it is not clear at all.