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 Pentation's definitional ambiguity Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 10/17/2009, 09:54 PM (This post was last modified: 10/18/2009, 02:51 AM by Base-Acid Tetration.) An important moral question is facing us, my fellow friends of hyper-operations. As we all know, tetration, having log^n singularites for all n, has an infinite number of branches. When we define pentation as a tetration-iterational (superfunction of tetration), we face ambiguity of definition: WHICH branch of tetration SHOULD we mean? Restricting tet(z) to its principal branch (defined for the complex plane excluding (-infinity, -2]) seems the most "correct" definition, but at the same time the most unnatural. The same problem goes for hexation since the superlogarithm has two conjugate branch points, penta-logarithm will have a penta-logarithmic singularity at fixed point of tetration. (It is not known whether pentation has any complex fixed points.) This is a highly subjective question that won't be settled by any amount of attempted rigor, so I propose a split up classification of pentations that we will study. I think we should define "proper pentation" as the function that has pen(z+1)= Tet(pen(z)), where Tet = mainly prinicipal branch of tetration. Or can't we study properties that will apply for all pentations? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/22/2009, 07:54 AM (10/17/2009, 09:54 PM)Base-Acid Tetration Wrote: As we all know, tetration ... has an infinite number of branches. Yes, but what kind of infinity? I'm still not convinced that the branches of tetration are even countable. Granted, we can assign the main branch as 0, then count out which branchpoint we want, then index each branch around that branchpoint. This would give us 2 indices for each branch, but then there is the issue of branches above and below everywhere else. Are these indexable in such a manner? or do we need to add more indices to account for these? (10/17/2009, 09:54 PM)Base-Acid Tetration Wrote: This is a highly subjective question that won't be settled by any amount of attempted rigor, so I propose a split up classification of pentations that we will study. I think we should define "proper pentation" as the function that has pen(z+1)= Tet(pen(z)), where Tet = mainly prinicipal branch of tetration. Or can't we study properties that will apply for all pentations? Hehe, even tetration is a highly subjective function. It is very recent that tetration approaches seem to be hinting at a unified analytic continuation of all approaches. And we still aren't sure if this is really the case. In the beginning it was a lot of guesses and hunches, which led to many different tetrations, and I personally had a lot of fun studying the differences between all tetrations (and I still do). So I think it would also be a lot of fun studying all pentations too. Andrew Robbins Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 10/22/2009, 11:40 AM what are the andydude Wrote:branches above and below everywhere else? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/22/2009, 08:27 PM (10/22/2009, 11:40 AM)Base-Acid Tetration Wrote: what are the andydude Wrote:branches above and below everywhere else? For example, suppose we know that ${}^{b}a = c$, then we also know that ${}^{b+1}a = a^c = a^{c + 2 \pi i k/\ln(a)}$, so we can work backwards and say that for every k, there exists a branch such that ${}^{b}a = c + \frac{2 \pi i k}{\ln(a)}$ is this the same as the indexing scheme above? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 10/22/2009, 09:45 PM (This post was last modified: 10/22/2009, 09:47 PM by Base-Acid Tetration.) andydude Wrote:for every k, there exists a branch such that ${}^{b}a = c + \frac{2 \pi i k}{\ln(a)}$ is this the same as the indexing scheme above? Then I would think the number of branches is still countable, though i doubt it because a[4]b 's cardinality is that of the continuum. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/24/2009, 09:27 AM (This post was last modified: 10/24/2009, 09:47 AM by bo198214.) (10/22/2009, 09:45 PM)Base-Acid Tetration Wrote: Then I would think the number of branches is still countable, though i doubt it because a[4]b 's cardinality is that of the continuum. Potentially there uncountably many branches. Branches are usually determined by the non-homotopic curves that lead to a point. E.g. take the logarithm. There is one branching point at 0. Paths are homotopic iff they wind the same number around 0. log has a branch for each winding number which is countably many. Now consider a function with two branch points. For each branch point we have a winding number and so we have $\mathbb{Z}\times\mathbb{Z}$ branches which is still countable. For $n$ branch points we have $\mathbb{Z}^n$ branches. But for infinitely many branch points, which is the case for tetrations, they have a branch point at every integer $\le 2$, we have $\mathbb{Z}^\omega$ many branches. And we know that already $2^\omega$ is uncountable. On the other hand this is only under the assumption that branches dont coincide. E.g. consider $\sqrt{z}$. It has a branch point at 0. But only two branches. But even if we assume that each branch point gives only two branches we still have $2^\omega$, i.e. uncountably many branches. If however more branches coincide it can lessen the number of different branches. !!!EDIT!!!: After reading mike3's posting I see what is the error in the above reasoning: A path with finite length can not wind around infinitely many branchpoints of tetration. It always needs to choose a finite number out the infinitely many branchpoints to wind around. And thatswhy the number of branches must be countable. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 10/24/2009, 01:02 PM (This post was last modified: 10/24/2009, 01:26 PM by Base-Acid Tetration.) (10/24/2009, 09:27 AM)bo198214 Wrote: A path with finite length can not wind around infinitely many branchpoints of tetration. Why does the path have to be of finite length? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/24/2009, 01:20 PM (10/24/2009, 01:02 PM)Base-Acid Tetration Wrote: Why does the path have to be of finite length? Doesnt have a curve from A to B always finite length? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 10/24/2009, 01:23 PM (This post was last modified: 10/24/2009, 01:26 PM by Base-Acid Tetration.) (10/24/2009, 01:20 PM)bo198214 Wrote: (10/24/2009, 01:02 PM)Base-Acid Tetration Wrote: Why does the path have to be of finite length? Doesnt have a curve from A to B always finite length? Why does it have to be a curve from A to B? CAn't it go off to infinity, winding around the inifnite number of points along the way, in the infinite limit? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/24/2009, 02:01 PM (This post was last modified: 10/24/2009, 02:02 PM by bo198214.) (10/24/2009, 01:23 PM)Base-Acid Tetration Wrote: Why does it have to be a curve from A to B? CAn't it go off to infinity, winding around the inifnite number of points along the way, in the infinite limit? Why do the branches depend on the path? This is because you do an analytic continuation along the path. You start with some point $z_0$ where your function is defined, e.g. $z_0=0$ for tetration. And then you choose a path (not crossing any singularity) from $z_0$ to $z$ where you want to determine the function's value at $z$. You continue the function along this path (i.e. cover it with powerseries expansions with overlapping disks of convergence) until you reach $z$ and then you have your value at $z$. One can show that if you choose to homotopic paths, i.e. two paths that you can continuously deform into each other without crossing singularities, then the resulting continuation value is the same. E.g. if you continue the logarithm from $z_0=1$ to $z=-1$ you can do this with a path through the northern halfplane and you arrive at $log(-1)=i \pi$ or you can do it with a path through the southern halfplane and you arrive at $-i\pi$. Any two paths through the northern halfplane would give you the same value $i\pi$. You can not deform the northern path into the southern path as you would have to cross 0. Thatswhy both values may be different. So the non-homotopic paths from $z_0$ to $z$ determine the possibly different values at $z$ and hence it branches. Thatswhy you need paths between two points of the complex plane, and not some limit of paths or paths going to infinity or so. « Next Oldest | Next Newest »

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