So, I've thought of working on a single case rather than working on the more general problem. I started by looking at:
\[
2 <s> 4 = 2<s+1> 3\\
\]
Which births from:
\[
\begin{align}
2<s> \left(2 <s+1> 2\right) &= 2 <s+1> (2+1)\\
2 <s> 4 &= 2<s+1> 3\\
\end{align}
\]
As a reminder of the notation, we write:
\[
x <s>_\varphi y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\
\]
If we omit the \(\varphi\), it is assumed that it is the unique implicit \(\varphi\) which solves Goodstein's equation. In the case above, we assume that:
\[
2<s>2 = 4\\
\]
Therefore \(\varphi = 0\), because that's the only value which satisfies the equation. From here, we are discussing the surface of values \(\Phi \subset \mathbb{C}^3\), such if \(\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)\), then
\[
2 <s>_{\varphi_1} \left(2 <s+1>_{\varphi_2} 2\right) = 2 <s+1>_{\varphi_3} 3\\
\]
As we are assuming that \(\varphi_2 = 0\), to ensure that \(2 <s> 2 = 4\) everywhere, we are reduced to a path along surface the surface \(\Phi\), \(\boldsymbol{\varphi} = (\varphi_1,0,\varphi_3)\).
So to ensure we have the correct functional equation we get the following:
\[
\varphi_1 = \log^{\circ s}_{\sqrt{2}}\left(2<s+1>_{\varphi_3} 3\right) - 6\\
\]
Now, we are allowed to move \(\varphi_3\) freely, so long as \(\varphi_3 = 0\) when \(s=0,1\). There is only one value of \(\varphi_3\) though, that we actually want. But we can still play around with possible values. Upon doing this, we get:
\[
2 <s>_{\varphi_1} 4 = 2 <s+1>_{\varphi_3} 3\\
\]
For the moment it appears that \(\varphi_3\) is arbitrary, but that's because we haven't introduced a non constant "exponent". Safe to say, though, that there is some function \(\varphi_3(s)\) that will be the actual function we want. This will be on a moving surface \(\Phi(s)\), which changes as \(s\) moves.
To make things more interesting, we look at:
\[
2 <s+1> (2 + w) = 4 + A(s)w + \mathcal{O}(w^2)\\
\]
Therein \(A = 2\) for \(s = 0\). Whereupon
\[
2 <s> \left( 2 <s+1> (2 + w)\right) = 2 <s> (4 + A(s)w + \mathcal{O}(w^2))\\
\]
Then,
\[
2<s>(4 + A(s)w + \mathcal{O}(w^2)) = 2 <s+1> 3 + w\\
\]
Since we're going to try and only care about the first term in \(w\), and make some kind of a differential equation, we can drop the pesky \(\mathcal{O}(w^2)\). And since we've already solved for \(\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)\) when \(w=0\) (\(\varphi_2 = 0\)).
But in addition to that, we have the functional equation:
\[
\varphi_1 = \log^{\circ s}_{(4+Aw)^{1/(4+Aw)}}\left(2<s+1>_{\varphi_3} (3+w)\right) - 6-A(s)w\\
\]
And with this equation:
\[
2 <s>_{\varphi_1} \left(2 <s+1>_{\varphi_2} 2 + w\right) = 2 <s+1>_{\varphi_3} 3+w\\
\]
Now we consider,
\[
\begin{align}
\varphi(2,4+A(s) w,s) &= \varphi_1\\
\varphi(2,2+w,s+1) &= \varphi_2 = \mathcal{O}(w)\\
\varphi(2,3+w,s+1) & = \varphi_3\\
\end{align}
\]
So the correct answer, and the exact \(\varphi_3\) we would need, is the value such that,
\[
\begin{align}
\lim_{s\to 0} \lim_{w\to 0} \frac{\partial}{\partial w}\varphi_3 &= \lim_{s\to 0} \lim_{w\to 1} \frac{\partial}{\partial w} \varphi_2\\
\lim_{s\to 1} \lim_{w\to 0} \frac{\partial}{\partial w}\varphi_3 &= \lim_{s\to 1} \lim_{w\to 1} \frac{\partial}{\partial w} \varphi_2\\
\end{align}
\]
We mean this derivative in the complex sense too. This limit formula is mainly important at the \(0,1\) endpoints, but it should hold everywhere. It's nice to use this limit formula, because we've reduced the question to making sure that we're only worried about the end points (addition and multiplication and exponentiation). Ensuring the derivatives work here is crucial, and with derivatives we get close to almost having our answer.
This gives us an avenue of solving for \(A\), which then gives us an avenue to solve for \(\varphi_2\), then \(\varphi_1,\varphi_3\) are related as above, then we are simply choosing the correct value of \(\varphi_3\), such you can paste them all together with no discontinuities, which is certainly possible, just from looking at it like an initial value problem ODE.
This effectively reduces the problem into solving for \(\varphi_2\) and \(\varphi_3\), where \(\varphi_1\) is a function of both. So it effectively reduces the problem by one degree. Which makes sense considering that we're trying to do calculus on a surface \(\Phi \subset \mathbb{C}^3\), which has complex-dimension 2.
I'm having not much luck testing these things because my code is just too damn slow. Unless Sheldon has a speedup, I'm not sure how to make the code faster.
Some additional notes, which help visualize the problem is that; there always exists a \(\boldsymbol{\varphi}\) if we assume that two of the components are zero.
So if I assume that \(\varphi_2 = 0\) and \(\varphi_3 = 0\), then we always have a solution \(\varphi_1\) such that:
\[
2 <s>_{\varphi_1} \left(2<s>_0 2+w\right) = 2 <s+1>_0 3 + w
\]
This would not be the solution we'd want though, as it wouldn't paste together properly. There's only one function that equals the correct equation pasted together.
But this is very important. Because if you fix \(s\), then there exists a value \(\varphi_1\) such that \(\boldsymbol{\varphi} = (\varphi_1,0,0)\) which solves the equation. Then, if you move \(\varphi_1\) by \(\epsilon\), you move \(\varphi_2\) and/or \(\varphi_3\) by \(\epsilon'\). So we can with \(s\) fixed, have a surface \(\Phi \subset \mathbb{C}^3\) (which means \(\Phi\) is of complex-dimension \(2\)), of possible values. Moving \(s\) moves the shape of the surface. When \(s=0,1,2\), we're reduced to a point mass at \((0,0,0)\). As we move \(s \sim 0 \to 1\) the surface grows in a ball shaped mass, and retracts back to the point mass. This is done in \(\mathbb{C}^3\), so it's a surface in this space...
I'm still working heavily on this, but it might be a while before I can sort it out. The code is pretty fucking awful, and I need to find a way to improve it to do more accessible tests. I hope Sheldon can help, but he's moved away from tetration, so I might be on my own, lol.
Regards, James
\[
2 <s> 4 = 2<s+1> 3\\
\]
Which births from:
\[
\begin{align}
2<s> \left(2 <s+1> 2\right) &= 2 <s+1> (2+1)\\
2 <s> 4 &= 2<s+1> 3\\
\end{align}
\]
As a reminder of the notation, we write:
\[
x <s>_\varphi y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)\\
\]
If we omit the \(\varphi\), it is assumed that it is the unique implicit \(\varphi\) which solves Goodstein's equation. In the case above, we assume that:
\[
2<s>2 = 4\\
\]
Therefore \(\varphi = 0\), because that's the only value which satisfies the equation. From here, we are discussing the surface of values \(\Phi \subset \mathbb{C}^3\), such if \(\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)\), then
\[
2 <s>_{\varphi_1} \left(2 <s+1>_{\varphi_2} 2\right) = 2 <s+1>_{\varphi_3} 3\\
\]
As we are assuming that \(\varphi_2 = 0\), to ensure that \(2 <s> 2 = 4\) everywhere, we are reduced to a path along surface the surface \(\Phi\), \(\boldsymbol{\varphi} = (\varphi_1,0,\varphi_3)\).
So to ensure we have the correct functional equation we get the following:
\[
\varphi_1 = \log^{\circ s}_{\sqrt{2}}\left(2<s+1>_{\varphi_3} 3\right) - 6\\
\]
Now, we are allowed to move \(\varphi_3\) freely, so long as \(\varphi_3 = 0\) when \(s=0,1\). There is only one value of \(\varphi_3\) though, that we actually want. But we can still play around with possible values. Upon doing this, we get:
\[
2 <s>_{\varphi_1} 4 = 2 <s+1>_{\varphi_3} 3\\
\]
For the moment it appears that \(\varphi_3\) is arbitrary, but that's because we haven't introduced a non constant "exponent". Safe to say, though, that there is some function \(\varphi_3(s)\) that will be the actual function we want. This will be on a moving surface \(\Phi(s)\), which changes as \(s\) moves.
To make things more interesting, we look at:
\[
2 <s+1> (2 + w) = 4 + A(s)w + \mathcal{O}(w^2)\\
\]
Therein \(A = 2\) for \(s = 0\). Whereupon
\[
2 <s> \left( 2 <s+1> (2 + w)\right) = 2 <s> (4 + A(s)w + \mathcal{O}(w^2))\\
\]
Then,
\[
2<s>(4 + A(s)w + \mathcal{O}(w^2)) = 2 <s+1> 3 + w\\
\]
Since we're going to try and only care about the first term in \(w\), and make some kind of a differential equation, we can drop the pesky \(\mathcal{O}(w^2)\). And since we've already solved for \(\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)\) when \(w=0\) (\(\varphi_2 = 0\)).
But in addition to that, we have the functional equation:
\[
\varphi_1 = \log^{\circ s}_{(4+Aw)^{1/(4+Aw)}}\left(2<s+1>_{\varphi_3} (3+w)\right) - 6-A(s)w\\
\]
And with this equation:
\[
2 <s>_{\varphi_1} \left(2 <s+1>_{\varphi_2} 2 + w\right) = 2 <s+1>_{\varphi_3} 3+w\\
\]
Now we consider,
\[
\begin{align}
\varphi(2,4+A(s) w,s) &= \varphi_1\\
\varphi(2,2+w,s+1) &= \varphi_2 = \mathcal{O}(w)\\
\varphi(2,3+w,s+1) & = \varphi_3\\
\end{align}
\]
So the correct answer, and the exact \(\varphi_3\) we would need, is the value such that,
\[
\begin{align}
\lim_{s\to 0} \lim_{w\to 0} \frac{\partial}{\partial w}\varphi_3 &= \lim_{s\to 0} \lim_{w\to 1} \frac{\partial}{\partial w} \varphi_2\\
\lim_{s\to 1} \lim_{w\to 0} \frac{\partial}{\partial w}\varphi_3 &= \lim_{s\to 1} \lim_{w\to 1} \frac{\partial}{\partial w} \varphi_2\\
\end{align}
\]
We mean this derivative in the complex sense too. This limit formula is mainly important at the \(0,1\) endpoints, but it should hold everywhere. It's nice to use this limit formula, because we've reduced the question to making sure that we're only worried about the end points (addition and multiplication and exponentiation). Ensuring the derivatives work here is crucial, and with derivatives we get close to almost having our answer.
This gives us an avenue of solving for \(A\), which then gives us an avenue to solve for \(\varphi_2\), then \(\varphi_1,\varphi_3\) are related as above, then we are simply choosing the correct value of \(\varphi_3\), such you can paste them all together with no discontinuities, which is certainly possible, just from looking at it like an initial value problem ODE.
This effectively reduces the problem into solving for \(\varphi_2\) and \(\varphi_3\), where \(\varphi_1\) is a function of both. So it effectively reduces the problem by one degree. Which makes sense considering that we're trying to do calculus on a surface \(\Phi \subset \mathbb{C}^3\), which has complex-dimension 2.
I'm having not much luck testing these things because my code is just too damn slow. Unless Sheldon has a speedup, I'm not sure how to make the code faster.
Some additional notes, which help visualize the problem is that; there always exists a \(\boldsymbol{\varphi}\) if we assume that two of the components are zero.
So if I assume that \(\varphi_2 = 0\) and \(\varphi_3 = 0\), then we always have a solution \(\varphi_1\) such that:
\[
2 <s>_{\varphi_1} \left(2<s>_0 2+w\right) = 2 <s+1>_0 3 + w
\]
This would not be the solution we'd want though, as it wouldn't paste together properly. There's only one function that equals the correct equation pasted together.
But this is very important. Because if you fix \(s\), then there exists a value \(\varphi_1\) such that \(\boldsymbol{\varphi} = (\varphi_1,0,0)\) which solves the equation. Then, if you move \(\varphi_1\) by \(\epsilon\), you move \(\varphi_2\) and/or \(\varphi_3\) by \(\epsilon'\). So we can with \(s\) fixed, have a surface \(\Phi \subset \mathbb{C}^3\) (which means \(\Phi\) is of complex-dimension \(2\)), of possible values. Moving \(s\) moves the shape of the surface. When \(s=0,1,2\), we're reduced to a point mass at \((0,0,0)\). As we move \(s \sim 0 \to 1\) the surface grows in a ball shaped mass, and retracts back to the point mass. This is done in \(\mathbb{C}^3\), so it's a surface in this space...
I'm still working heavily on this, but it might be a while before I can sort it out. The code is pretty fucking awful, and I need to find a way to improve it to do more accessible tests. I hope Sheldon can help, but he's moved away from tetration, so I might be on my own, lol.
Regards, James