05/21/2022, 10:37 PM
(05/20/2022, 12:14 PM)tommy1729 Wrote: So we are not stuck but super stuck.
(ok that is a bit of a joke )
Regards
tomm1729
Tommy, you are entirely avoiding the point of this construction.
I don't know how many times I have to tell you that \(\Re(s) < 0\) is beyond our purview. Similarly with \(\Re(s) > 2\). This is in many ways completely unrelated to hyperoperators, we are simply talking about between addition, multiplication and exponentiation. And there, we are not even trying to really solve anything to do with hyperoperators. It just so happens that \(0\) is addition, \(1\) is multiplication and \(2\) is exponentiation. They satisfy a similar equation, yes, but there's no initial value like \(x <s> 1 = x\), which you keep on trying to force into these equations.
In fact, I expect the entire issue you're talking about with successorship at \(-1\), is handled by the fact IT WON'T BE ANALYTIC HERE. It probably won't be analytic along the entire line \(s \in (-\infty,-1]\), probably a larger domain. What I can say, is that for \(x <s> y\), it will be analytic for \(0 \le s \le 2\) and for \(x,y > e\). This is found solely from the implicit function theorem. There's zero question. The trouble is, how do we construct this solution aptly and quickly.
Without a shred of doubt, there's an analytic function \(\phi(x,y,s)\) such that:
\[
x \langle s\rangle y = \exp_{y^{1/y}}^{\circ s}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \phi(x,y,s)\right)\\
\]
and that:
\[
x \langle s\rangle \left( x \langle s+1\rangle y\right) = x \langle s+1\rangle (y+1)
\]
You can actually make a pretty good guess of this function by letting
\[
\phi(x,y,s) \approx -A(s)/\log(y)\,\,\text{for large}\,\,y\to\infty\,\,1\le s \le 2\\
\]
Where \(A(s)\) can be approximated using computations.
Then a newtonian root finder takes care of the strip \(1 \le s \le 2\). Then all you have to do is make sure the pasting is correct, which isn't very hard to do. All we need is \(\phi(x,y,s)\) is analytic at \(1\), and since we still have one degree of freedom at this point, this is pretty simple to do. The main trouble I am having, is how to program this efficiently, construct it effectively mathematically. At this point it's just an implicit solution.
This solution certainly exists, but programming it is very damn hard because we need to take very large \(y\), and at that point the code starts to deteriorate, because we're taking exponentials of things like \(y = 1E20\). And unfortunately at this point \(-1/\log(y)\) isn't small enough to induce an efficient convergence. Doesn't change the fact that this function will exist. The implicit function theorem guarantees it.
I don't care what happens when \(\Re(s) < 0\), we can get there when we get there by iterating the subfunction operator, but I don't care about that. I bet it'll be absolutely disastrous and will give little to no enlightenment. I'm also expecting this to crash and burn exactly at \(s = -1\), and that's okay by me. I don't care about that. I care about looking at the solution between \(0 \le s \le 2\). And even there, complex \(s\) is going to pose a difficult problem, because the domain in \(y\) is very unclear for complex \(s\). But when \(s\) is real, and between here we get the following benefits:
\[
\begin{array}
\phi\\
\phi(x,y,s)\,\,&\text{is small for}\,\,x,y>e,\,\,0 \le s \le 2\\
\phi(x,y,s)\,\, &\text{is real valued}\\
x \langle s \rangle y > y\,\,&\text{so there's no domain issues}\\
\end{array}
\]
It's actually possible to find these solutions pretty accurately for \(y > 1E20\), it looks increasingly like:
\[
\begin{array}
\phi\\
\phi(x,y,s) \approx A(s)y^{-c}\,\,&\text{for}\,\,0\le s \le 1\,\,\text{and some}\,\,c = c(s) > 0\\
\phi(x,y,s) \approx -B(s)/\log(y)\,\,&\text{for}\,\, 1\le s \le 2\\
\end{array}
\]
Once the solution exists for \(y > 1E20\) or somewhere very large, you can pull back using the relationship \(\varphi\) has to bennet.
Again, I can't stress this enough. There is an implicit local solution. The monodromy theorem makes sure it's analytic. FULL STOP. I don't care about \(x<s>1 = x\). Yes, there are thousands of issues with that definition when going backwards. I'm trying to do something very different here.