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(10/18/2022, 05:40 AM)Shanghai46 Wrote: Also, there's no real nice property we would use to easely extend fractional iterations and ranks of hyperoperations. Like associativity or commutativity.
@Shanghai
I agree Shanghai. The absence of such a condition relating fractional ranks seems to be the biggest obstacle. I tried to begin a little exploration of the problem and its possible solutions here [2022] but I'm still far from a decent solution. Probably the real condition needed is nonalgebraic but topological/differential, as bo198214 is suggesting here when he says
Quote:Maybe there is no obstacle it is more about how to define this smoothness I was mentioning above.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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(10/21/2022, 07:46 PM)bo198214 Wrote:
 I think of addition has rank 1, multiplication has rank 2, exponentiation has rank 3, and I would continue in a similar fashion to higher ranks. For noninteger ranks I would assume some smoothness moving operation k into operation k+1.
[...]
 Maybe there is no obstacle it is more about how to define this smoothness I was mentioning above.
How'd you go about it? As simply as you are stating it? I.e. something like \(f(s,y):=b[s]y\) s.t. \(f\in \mathcal C^\infty (\mathbb R^2)\)? Or you mean that a more sophisticate condition involving smoothness or the parameter \(s\) should be developed?
Btw, maybe I'd better start a new thread but, extending your suggestion, maybe we could ask for local representability as a powerseries.... since I see you have a good fluency in the math of formal powerseries I'd like to ask you about formal Ackermannfunctions here.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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11/15/2022, 03:52 AM
(This post was last modified: 12/19/2022, 04:34 AM by Catullus.)
(10/15/2022, 12:53 PM)MphLee Wrote: 1) What do you think when you ear the term rank in the context of hyperoperations?
2) What do you think is the deep meaning and importance of the rank parameter?
3) What is, in your opinion, the main obstacle to solving the problem of noninteger ranks?
4) How do you see the mathematician of the future surpass this obstacle, if they ever manage to do it at all. In which field of mathematics do you see the key to the solution residing in? 1) One plus the amount of times the super function operation is applied to addition.
2) The deep meaning and importance of the rank parameter is the hyper recursion being used that iterates the super function operation itself, and how chains of repeatedly applying one operator form the next. In fact, the recursion is so strong that \(3[n]3\) \([>]\)\(\text{PRA}\). This hyper recursion can be used to generate super huge numbers!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! The rank parameter is also important because diagonalizing over the rank parameter could be used to define fundamental sequences for ordinals such as \(\def\ {\omega}\varphi(\ ,0,0)\), which is equal to \(\ [\ ]\ \). Via ordinal hierarchies such as Hardy Hierarchy, defining fundamental sequences for ordinals can be used to generate super huge numbers!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Please see this webpage for how ordinal tetration could be defined. The method used on that webpage for working with ordinal tetration could be generalized to higher ordinal hyperoperations.
3) The hyper recursion that is used to form the hyperoperation sequence.
4) It was pointed out that a uniqueness criterion similar to this one for tetration could be used to do continuous iteration on lots of functions. With that method, it could be possible to do continuous iteration on the super function function.
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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(10/18/2022, 07:31 AM)Gottfried Wrote: 1) For me as an old teacher in statistics the term "rank" has a special connotation: "ranks" are only positive and "rank_a  rank_b" has no specific meaning other than "rank_b" might be larger, equal or smaller than "rank_a". Like the table of football teams  there is better/equal/worse and ranks from 1 to 18 (say 18 teams in a league) but no negative rank. So a generalization of hyperoperations on the base of imagination of something like "ranks" makes it somehow odd in my personal neuronal wiring... . A better concept is perhaps that of "index": there is no coconnotation of positiveness, only that of integral numbers. But here I've been "educated" already by literature of L. Euler and later mathematicians, who introduced "fractional indexes" for instance fractional bounds and/or indexes for infinite sums, showing that such constructions might have a meaningful evaluation.
One concept that formed/modified my idea of "hyperoperations" has been that of Markus Müller in the 90ies, who introduced as a teenager the idea of fractional indices for hyperoperations like "1do nothing", "2increase / 1/2decrease", "3add,1/3subtract" , "4multiply, 1/4divide" and so on even respecting the multiple inverses when higher indexes are considered. This is already completely different from any notion of "rank"  but somehow attractive on its own.
2) The "deep meaning"  hmmm. My take so far, and I've at best a glimpse of a deeper meaning, is that the hyperoperations are growing by the idea of iteration. Like multiplication is basically iteration of addition, and so on. So the more basic machine is the idea of "iteration" . In contrast, I've not not yet arrived at such an wider idea like one of a (continuous?) "fieldofhyperoperations". The basic engine of the paradigma of "iteration" might be inherently too weak to model such a contiuous space of hyperoperations  and even if we are able to proceed to fractional iteration in some way. Maybe. Maybe... :) (I remember some answer of Qiaochu Yuan in MSE a couple of years ago [1] [2] [3], where he criticized this notion of "hyperoperationsbyiteration" in a very lucide way; the third link is the one which initially rose my interest in his answers on this subject) But my practical approaches are  >>sigh<< (perhaps)  so far based on hyperoperations with index formed by the idea of iteration.
3) This follows from 2). Perhaps that basic paradigma of "iteration" as the parameter in the hyperoperationshierarchy is not perfect/precise enough. We might be able to shape it some way up towards the operation of tetration as "iterated exponentiation" indeed as a useful tool for computation of some processes in reality, but this might be holy ( :) I mean: plagued by holes). For a short reminder: iteration of a clockarithmetic doesn't make it an object which could ever include inversion ...
4) Could not say more than vague inspirations, not yet ready in a form that could be operationable in any way. Let's see ....
@Gottfried: Excuse me for the long delay. I hadn't enough time to share my comments to your great answers. Really interesting indeed!
1) You are right in some sense. Index seems more appropriate but also too vague. Rank is really cemented in my mind as something related to hyperoperations... mainly because as binary operations can be arranged into a hierarchy based on "computation hardness/complexity" a position of an operation inside such a hierarchy should be termed it's rank inside the hierarchy... but also the various levels of the hierarchy should be called ranks? This choice is supported by well established logicsettheoretic usage.
I like a lot the linguistics and terminological problems in mathematics and the issues you bring to the discussion are really interesting.
Also the term hyperoperations is problematic because there is a completely unrelated branch of mathematics, at least 30 years old and now much more mainstream than our beloved hyperoperations, that deals with multivalued operations and claims the term.
Also rank is used in many other places: in linear algebra it is the vector space dimension of the image of a linear map... but then also the dimension can be generalized to be noninteger.
Maybe a totally new term should have been coined for the "grade" of an hyperoperation. But how can we fix this?
2/3) What do you have in mind when you say " The basic engine of the paradigma of "iteration" might be inherently too weak [...] Perhaps that basic paradigma of "iteration" as the parameter in the hyperoperationshierarchy is not perfect/precise enough."? I kinda feel the same way... it is a vague and mathematical unsubstantiated feeling but I'd love to know if you can expand on this.
Also Qiaochu Yuan makes some good points that any foundational work on hyperoperations needs to be pair with Devlin's critique of "multiplication as repeated addition"...
@Catullus: Thanks for the contribution.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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(12/18/2022, 09:08 AM)MphLee Wrote: @Catullus: Thanks for the contribution. You're welcome.
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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I forgot to share my craziest idea for what ranks are. We live in a Universe of ages or ranks. First few ages pretty boring, but once the age of tetration kicks in, things begin to get interesting. The Big Bang wasn't the beginning of the Universe, just the beginning of the current age.
Daniel
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12/20/2022, 02:23 AM
(This post was last modified: 12/20/2022, 02:44 AM by JmsNxn.)
Personally, I've always used the term "degree". I do not like index, because it implies a banalityand eschews the fact that this indexing is done off of complexity of the iteration. Where as degree, implies we go up a level similarly to polynomials. This also dates back to older work I had done, where we write:
$$
P(z) = \sum_{j=0}^n a_j \left(\alpha \uparrow^j z\right)\\
$$
Where now \(P\) is of "degree n". Which I wrote as \(\deg(P) = n\). If you apply the super function operator \(\uparrow\), then \(\deg (\uparrow P) = \deg(P) + 1\). Where then, by triviality: \(\deg(\alpha \uparrow^n z) = n\). But I think "degree", might be a tad too general. We don't really say "the degree of \(x^n\) is \(n\)"despite it being correct, it rolls off the tongue wrong. Instead, we call \(n\) the power of \(x\).
In that regard, I really like "rank". Where the "rank" of the hyperoperator plays the role of the power of \(x\).But when combining different hyperoperators together; we take the "max of the ranks" to get the "degree".
Also, I thought I'd add. I could never fully define this \(\deg\) function, but I could justify its existence very well; and It tides a bit next to "rank" but is an independent idea.
Any function \(f_0\) is a finite combination of \(\alpha \cdot z\), \(\alpha + z\), under composition and addition is \(\deg f = 0\). From this, any function \(f_n = \uparrow f_{n1}\), where \(\deg(f_n) = n\) is found by taking some \(deg(f_{n1}) = n1\) and writing:
$$
f_n(z) = \frac{d^z}{dw^z}\Big{}_{w=0} \sum_{j=0}^\infty f_{n1}^{\circ j}(z)\frac{w^j}{j!}\\
$$
I could never totally show this \(\deg\) was well defined. But it is well defined for the bounded analytic hyper operators. And it spits out the right answer. Things might get more tricky for higher levels. And it only works as a hierarchy on superpositions and compositions of hyperoperators in the bounded case.
So all in all, I support the term "rank"; though I use it to represent the rank of a hyperoperator; the rank of how deep of a recursion you need; the rank of how many Mellin transforms we have to do. Where then, degree, is a general term to deal with more functions, but still preserving the main tenets of "rank".
The "rank" of a hyperoperator is \(n\) is the same as the power of a monomial \(x^n\) is \(n\). The "degree" of a function \(f\), is the same as the degree of a polynomial \(\sum_{j=0}^n a_j x^j\); where it may have no obvious "rank" (it is not a monomial).
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thank you James, for your precious insight. One day I'll give back something to the community.
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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(02/20/2023, 02:41 PM)MphLee Wrote: thank you James, for your precious insight. One day I'll give back something to the community.
You definitely have. Do not sell yourself short!
