I'd like to add one more point of discussion to this thread, to exemplify that Mphlee is not very far off with his ideas, he's just taking a wrong angle, and a wrong base pair.
Let us write the following expansion:
\[
x [s]_\varphi y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\
\]
We are intrinsically assuming that \(\exp^{\circ s}_b\) is the repelling iteration of \(b^z\)--so that if \(b = \sqrt{2}\), we are choosing the repelling fixed point \(4\). There is always a repelling fixed point for \(b\) in the Shell-thron region, and there is always a singular one nearest to the real line. We are then, using said fixed point and performing a Schroder iteration here.
Now, some important things to note.
This function is holomorphic for \(\Re(x) > e\) and \(\Re(y) > 1\). While the domain in \(s\) is a tad finicky, it is analytic in \(s\), and the best I can say at the moment is:
\[
s \in \mathcal{S} = \{s \in \mathbb{C}\,|\,\Re(s) > -1/2,\, |\Im(s)|< 1\}\\
\]
Whereby this function is holomorphic in \(s\) on \(\mathcal{S}\). The value \(\varphi\) is more difficult, but it is holomorphic in \(\varphi\) on these domains, if we ask something like \(|\varphi| < \delta\), where \(\delta > 0\) and this value isn't too large, but not as tiny as you expect (I find it's usually around \(\delta = 0.5\)).
Now let's do everything Mphlee is describing, but let's do it in \(\varphi\) as opposed to in the actual base equation. For this, we're going to set \(x=3\), because \(x\) doesn't affect much, and has nothing to do with the Goodstein equation. We are in effect, ignoring \(x\), so just set it to \(3\). For this we write the following:
\[
3 [1+s]_\varphi y = \sum_{n,m=0}^\infty g_{nm}(\varphi)s^n(y-e)^m\\
\]
On the domains described:
\[
|g_{nm}(\varphi)| < M\,\,\text{for some constant}\\
\]
And now we can perform something similar to Gottfried's approach (I have this written out much more detailed, just not in latex, pounds and pounds of notes). Let us take this a step further and say there is a:
\[
\varphi(s,y) = \sum_{n,m=0}^\infty a_{nm} s^n(y-e)^m\\
\]
The values \(g_{nm}(\varphi)\) are fully determined, one only needs to study the repelling iteration of exponentials, and Bennet's form. The system of equations \(a_{nm}\) are the unknowns. I have code to do this already, I do not have code for the next part.
Now, per Mphlee's suggestion, we are trying to solve:
\[
3 \langle 1+s \rangle y = 3 [1+s]_{\varphi(s,y)} y\\
\]
Using this knowledge of infinite matrices, that Gottfried uses, that a lot of people use. So the functional equation we ask is:
\[
3 \langle 1+s \rangle \left(3 \langle 2+s \rangle y\right) = 3 \langle 2 +s \rangle (y+1)\\
\]
This reduces into a solution system for \(a_{nm}\). I don't know how to solve it, but I know it can be solved, because I can construct this solution using implicit equations. Additionally, since we are just testing the holomorphy of \(\varphi\), we always have convergent taylor series. Even setting this up, I've been modest about how well these taylor series converge. Bennet's form of semi-operators have GREAT (AND I REPEAT, GREAT) taylor data, I've been modest choosing the domains I chose.
You should definitely notice now, if you haven't yet, that \(\langle 1+s\rangle\) is at least holomorphic for \(|s| < 1\), which in turn gives definition to \(\langle 2+s\rangle\) for \(s \in [-1,0]\) (and its analycity here).
Now, actually bruteforcing this code, I have no fucking clue. I can understand Gottfried's approach, and speak on it; but in no way would I be able to recreate it in Pari. Same way I can understand Sheldon's approach, but I can't code it in myself. I do believe this will probably be the best way to write code for semi-operators, I will not continue this path.
The solution that I have is pretty fucking ugly, as in, it is a mess of symbols. But no less than what is seen here. And that, really Mphlee you're on the right path. This is just a fucking beast of a problem
Another final point I'd like to add is that if we castrate everything to \(O(s^3(y-e)^3)\), then I can get a solution. So instead:
\begin{align}
3 [1+s]_\varphi y &= \sum_{n,m=0}^2 g_{nm}(\varphi)s^n(y-e)^m + O(s^3(y-e)^3)\\
g_{nm}(\varphi) &= a + b\varphi + c\varphi^2 + O(\varphi^3)\\
\varphi(s,y) &= \sum_{n,m=0}^2 a_{nm} s^n(y-e)^m + O(s^3(y-e)^3)\\
\end{align}
Then, this produces a solvable \(3\times 3\) matrix solution thingy, and everything works fine. I've done this and calculated values, everything is good; though it only gives a third order approximation to the solution near \(y = e\) and \(s =1\) (multiplication). And therefore, it does not check the functional equation as much as we'd like (we get like 1-2 digit accuracy (so the first 2 decimals are correct)). But the principle is the same. I simply do not know how to solve this beast, as I'm terrible at matrices, let alone, programming in matrix solutions.
Nonetheless, the math I have says, if we let the order of error to infinity, we're in the clear
Let us write the following expansion:
\[
x [s]_\varphi y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\
\]
We are intrinsically assuming that \(\exp^{\circ s}_b\) is the repelling iteration of \(b^z\)--so that if \(b = \sqrt{2}\), we are choosing the repelling fixed point \(4\). There is always a repelling fixed point for \(b\) in the Shell-thron region, and there is always a singular one nearest to the real line. We are then, using said fixed point and performing a Schroder iteration here.
Now, some important things to note.
This function is holomorphic for \(\Re(x) > e\) and \(\Re(y) > 1\). While the domain in \(s\) is a tad finicky, it is analytic in \(s\), and the best I can say at the moment is:
\[
s \in \mathcal{S} = \{s \in \mathbb{C}\,|\,\Re(s) > -1/2,\, |\Im(s)|< 1\}\\
\]
Whereby this function is holomorphic in \(s\) on \(\mathcal{S}\). The value \(\varphi\) is more difficult, but it is holomorphic in \(\varphi\) on these domains, if we ask something like \(|\varphi| < \delta\), where \(\delta > 0\) and this value isn't too large, but not as tiny as you expect (I find it's usually around \(\delta = 0.5\)).
Now let's do everything Mphlee is describing, but let's do it in \(\varphi\) as opposed to in the actual base equation. For this, we're going to set \(x=3\), because \(x\) doesn't affect much, and has nothing to do with the Goodstein equation. We are in effect, ignoring \(x\), so just set it to \(3\). For this we write the following:
\[
3 [1+s]_\varphi y = \sum_{n,m=0}^\infty g_{nm}(\varphi)s^n(y-e)^m\\
\]
On the domains described:
\[
|g_{nm}(\varphi)| < M\,\,\text{for some constant}\\
\]
And now we can perform something similar to Gottfried's approach (I have this written out much more detailed, just not in latex, pounds and pounds of notes). Let us take this a step further and say there is a:
\[
\varphi(s,y) = \sum_{n,m=0}^\infty a_{nm} s^n(y-e)^m\\
\]
The values \(g_{nm}(\varphi)\) are fully determined, one only needs to study the repelling iteration of exponentials, and Bennet's form. The system of equations \(a_{nm}\) are the unknowns. I have code to do this already, I do not have code for the next part.
Now, per Mphlee's suggestion, we are trying to solve:
\[
3 \langle 1+s \rangle y = 3 [1+s]_{\varphi(s,y)} y\\
\]
Using this knowledge of infinite matrices, that Gottfried uses, that a lot of people use. So the functional equation we ask is:
\[
3 \langle 1+s \rangle \left(3 \langle 2+s \rangle y\right) = 3 \langle 2 +s \rangle (y+1)\\
\]
This reduces into a solution system for \(a_{nm}\). I don't know how to solve it, but I know it can be solved, because I can construct this solution using implicit equations. Additionally, since we are just testing the holomorphy of \(\varphi\), we always have convergent taylor series. Even setting this up, I've been modest about how well these taylor series converge. Bennet's form of semi-operators have GREAT (AND I REPEAT, GREAT) taylor data, I've been modest choosing the domains I chose.
You should definitely notice now, if you haven't yet, that \(\langle 1+s\rangle\) is at least holomorphic for \(|s| < 1\), which in turn gives definition to \(\langle 2+s\rangle\) for \(s \in [-1,0]\) (and its analycity here).
Now, actually bruteforcing this code, I have no fucking clue. I can understand Gottfried's approach, and speak on it; but in no way would I be able to recreate it in Pari. Same way I can understand Sheldon's approach, but I can't code it in myself. I do believe this will probably be the best way to write code for semi-operators, I will not continue this path.
The solution that I have is pretty fucking ugly, as in, it is a mess of symbols. But no less than what is seen here. And that, really Mphlee you're on the right path. This is just a fucking beast of a problem
Another final point I'd like to add is that if we castrate everything to \(O(s^3(y-e)^3)\), then I can get a solution. So instead:
\begin{align}
3 [1+s]_\varphi y &= \sum_{n,m=0}^2 g_{nm}(\varphi)s^n(y-e)^m + O(s^3(y-e)^3)\\
g_{nm}(\varphi) &= a + b\varphi + c\varphi^2 + O(\varphi^3)\\
\varphi(s,y) &= \sum_{n,m=0}^2 a_{nm} s^n(y-e)^m + O(s^3(y-e)^3)\\
\end{align}
Then, this produces a solvable \(3\times 3\) matrix solution thingy, and everything works fine. I've done this and calculated values, everything is good; though it only gives a third order approximation to the solution near \(y = e\) and \(s =1\) (multiplication). And therefore, it does not check the functional equation as much as we'd like (we get like 1-2 digit accuracy (so the first 2 decimals are correct)). But the principle is the same. I simply do not know how to solve this beast, as I'm terrible at matrices, let alone, programming in matrix solutions.
Nonetheless, the math I have says, if we let the order of error to infinity, we're in the clear
