05/22/2022, 01:29 AM
(05/22/2022, 12:17 AM)tommy1729 Wrote: My apologies James.
But there are reasons why I want more.
I want to find " the " hyperoperator.
Secondly , I was
1) trying to find a way to solve your equations and I felt it was too general bringing me to
2) I wanted to have a uniqueness criterion and doubted your equation has a unique solution.
Which bring me to my questions to you :
How do you think about uniqueness ?
I think there is no uniqueness with your setting.
This might be problematic in the sense that the bundle of functions that satisfy this and do not have closed forms can be best described by the equations and nothing else ??
That is maybe way to pessimistic but I compare with differential equations in many variables with a large number of solutions , where no single solution has a closed form , it is hard to describe all solutions in terms of a given one and we describe the set of all solutions just with the differential equation itself.
***
Another thing what bothers or confuses me is this
x + y and x * y are commutative.
x^y is not.
Should operators between x + y and x*y be commutative or not ?
I guess you say not.
But going from commutative to noncomm and back to commutative and then noncomm again bothers me.
Maybe it is just me.
And maybe im too focused on superfunctions.
But that is all I know ( superfunctions ) when it comes to hyperoperators.
For me asking 2 <s> 2 = 4 seems like a superfunction interpretation too.
And without x <s> 1 = x I feel i have no starting point.
Im not sure what you want is a half-superfunction idea, I think not.
***
What happens when we simply interpolate ?
x <s> y = a * ( x <A> y) + b * ( x <B> y ) + c * ( x <C> y)
where a , b , c , A , B , C are functions of s ? preferably simple functions ?
Could that work ??
***
Finally im not even sure you still want x <3> y = x^^y now , which brings up the question ; how is this related to tetration or ackermann ? And if not , is it not suppose to ?
***
Im not trying to sound hostile to your ideas sorry.
regards
tommy1729
Hey, Tommy. It's not a problem at all, this isn't a solution of analytic inbetween hyper-operators. As long as you understand that, I'm okay. I'll try to answer your questions point by point.
Uniqueness is only existent if you think of this with respect to Bennet's formula. This is not "unique" in a general sense. There are probably many ways to interpolate "hyper-operators" in a meaningful way. The manner this is unique, is that, there is ONE function \(\phi(x,y,s)\) for \(x,y > e\) and \(0 \le s \le 2\) such that:
\[
x \,\langle s \rangle y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \phi(x,y,s)\right)\\\\
\]
Satisfies Goodstein's equation. Remember though, that this is only a small correction which turns the quasi Bennet operators:
\[
x [s] y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y\right)\\
\]
Into satisfying goodstein's equation. So in that sense, yes it is unique. It's the only function we can put in here to satisfy Goodstein's equation. Additionally, we ask that \(\phi(x,y,s) \to 0\) as \(y\to \infty\), this does force uniqueness. This is covered by the implicit function theorem, and the restrictions we are making on the surface in question.
Okay, so the commutative stuff is a no-brainer. I mean. We can't have an analytic function \(a <s> b = b <s> a\) for \(0 \le \Re(s) <2\) and then all of a sudden it becomes noncommutative at \(s= 2\). Think of it this way, the values \(s=0\) and \(s=1\) are these magical values where it happens to be commutative. The operators are in general not commutative, but there are a couple of instances where it then is commutative. I mean, maybe there are beautiful values \(0 \le \Re s_0 \le 2 \) such that this expression is commutative again, but for the moment, we can assume that it is probably non-commutative. Until you find a value that is commutative, I'm going to assume non-commutative. You're trying to add an a priori assumption to the solution.
Okay, so a linear approximation would definitely never work. I'm not sure why you're bringing that up. At least, it doesn't work and has nothing to do with this discussion.
Similarly \( 2<s> 2 = 4\), was largely a test value of how to define these things. I'm not going to concern myself with that anymore. That's on me though, this thread is a mess. This thread is far too fucking disorganized. I apologize, because I've probably confused you. A lot of what I wrote in this thread was discovered on the fly. I have a much better understanding now.
Yes, so \(x\langle 3 \rangle y = x\uparrow\uparrow y\). Yes, I am expecting this to happen. But, the equation/computation/construction is secondary to this. When you let \(s > 2\), you get that \(\phi\) blows up wayyyyyyyyy too fast. Now, this doesn't mean it's incorrect. It just means it's a hopeless affair, and this whole effort becomes useless. I don't care about inbetween exponentiation and tetration, there's nothing to be gained from this avenue.
BUT!!!!
If \(f(y) = x \langle s+1\rangle y\), then yes \(f^{\circ y}(u_0) = x \langle s+2 \rangle y\), for some value \(u_0\) and \(<s+2>\) "between exponentiation and tetration". The thing is though, that this is kind of a meaningless statement, other than \(s \mapsto s+1\) is the same as taking the superfunction. But, we don't have initial values as before. And even further! The equations will begin to break down around here, where \(s=2.5\) we can expect \(\phi\) to be UNBELIEVABLY HUGE, and there's nothing we can do. At least from a computational aspect.
The reason I'm focusing on \(0 \le s \le 2\) is because the value \(\phi\) has very regular growth/is very small/it's feasible to program in.
All in all, think of what I'm trying to do as finding an efficient way of calculating and computing when \(0 \le s \le 2\). I know I haven't explained myself perfectly, but I promise you this is working. I wish I could post all of my code and my experimentations. It's going to be my summer project to explain all this. But Again, we are keeping \(0 \le s \le 2\) because this is where quasi-Bennet operators look a lot like "inbetween addition, multiplication, exponentiation". Technically this would work about tetration, but this avenue of pursuit would be fruitless. It's just that we're really lucky in this strip that \(\phi\) stays very small that we are okay.
As the first big result I have, we can write:
\[
x[s]\left(x[s+1] y\right) = x \langle s+1\rangle_{\phi} y+1\\
\]
And then I can show that \(\phi \sim A(s)/\log(y)\). This only works for \(0 \le s \le 1\). If you start talking about this equation for \(s>1\) it's fucking useless. This identity does not hold. Thats because, \(x[3]y\) DOES NOT LOOK LIKE TETRATION. Trying to find an error here is a fruitless affair. At least from a computational perspective.
So, in conclusion
For \(0 \le s \le 2\) we get that:
\[
x [s] y \approx x \langle s \rangle y\\
\]
And that \(\phi\) is SUPER well behaved here. But here and only here. As soon as we leave this domain a bunch of errors and chaos happens. Which I mean, is much what you're writing by talking about the iterative successorship problem, where \(-1,-2,-3...\) are all successorship, which is obviously a problem. My solution though, doesn't touch these values, and if you try to--expect a branch-cut/non analycity. Honestly, I don't know what's going to happen, but I know it'll be singular of some kind once we start talking about \(s < 0\).
I apologize for how off the rails and unorganized this thread is. But I was transcribing these results in real time. I know what's happening now. And I know how it works now (largely because I've run a lot of computational trials). I don't want to post how all this works until I have a good working model, which I don't yet. So I'm mostly just writing code and trying to find an efficient way of finding \(\phi\), and explaining how this happens.
I apologize if this is all over the place. But I'm confident (not even confident, absolutely sure) there's an implicit solution to this problem. The real problem is constructing the solution. And any attempt I have at constructing this solution is sidelined by poor code, and the inability to sample values for \(y > 1E20\) and at the same time only having decay like \(-1/\log(y)\).
Regards, James