Hi, had literally zero time in the last three months to write or study, but I continue to read all the new exciting posts.
Said that, I think I have a hole in my memory but I remember something of your 2010/2011 work on semi operators, i.e. non-integer hos. One of those, the cheta one, was originally a piecewise fix of Bennet's hyperoperations (aka distributive Hyperoperations) but extended to the real.
Using the superfunction of exp via beta method to build a continuous spectrum of homomorphic abelian groups extending Bennet's sequence seems really interesting.
But here you are doing something different.Since my mind is a bit foggy lately I can't follow it with ease. How exactly can you obtain the Goodstein equation while holding true for \(\omega\) an exp-fix-point the following equation \[x\,<s>\,\omega=\exp^s(\log^s(x)+\omega)?\quad\quad (*)\]
I know you can't write the details right now but even abstractly I don't see the intended direction of the implication.
Let \(\odot_s\) be the standard Bennet hyperoperations for a fixed base \(b\) and extended to the real ranks \(s\) using a tetration function. If we have a sequence of operators \(<s>\) satisfying the property \( (*) \) then why don't we have
\[
\begin{align}
x\,<s>\,\omega&=\exp^s(\log^s(x)+\omega)\\
&=\exp^s(\log^s(x)+\log^s(\omega))\\
&=x \odot_s \omega\\
\end{align}\]?
If the above is true how can \(<s>\) also satisfy the Goodstein equation, i.e. \(<s+1>S=<s><s+1>\)? Isn't \(<3>\) different from tetration?
Note: we have however a chain of holomorphic Goodstein equations. Observe that
\[
\begin{align}
x\odot_{s+1} ({}^{s}b\odot_s y)&=(x\odot_{s+1}{}^{s}b) \odot_s(x \odot_{s+1} y)\\
&=\exp^{s+1}(\log^{s+1}(x)+\log^{s+1}({}^{s}b)) \odot_s(x \odot_{s+1} y)\\
&=x \odot_s(x \odot_{s+1} y)\\
\end{align}\]
Since \({\bf 1}_{0}=0\); \({\bf 1}_{1}=1\); \({\bf 1}_{2}=b\); and \({\bf 1}_{s}=\exp_b^s(0)={}^{s-1}b\); we can define the homomorphic image of the successor as \({\rm succ}_{s+1}(y):={\bf 1}_{s+1}\odot_s y\) and write the previous as \[x\odot_{s+1} {\rm succ}_{s+1}(y)=x \odot_s(x \odot_{s+1} y)\]
Said that, I think I have a hole in my memory but I remember something of your 2010/2011 work on semi operators, i.e. non-integer hos. One of those, the cheta one, was originally a piecewise fix of Bennet's hyperoperations (aka distributive Hyperoperations) but extended to the real.
Using the superfunction of exp via beta method to build a continuous spectrum of homomorphic abelian groups extending Bennet's sequence seems really interesting.
But here you are doing something different.Since my mind is a bit foggy lately I can't follow it with ease. How exactly can you obtain the Goodstein equation while holding true for \(\omega\) an exp-fix-point the following equation \[x\,<s>\,\omega=\exp^s(\log^s(x)+\omega)?\quad\quad (*)\]
I know you can't write the details right now but even abstractly I don't see the intended direction of the implication.
Let \(\odot_s\) be the standard Bennet hyperoperations for a fixed base \(b\) and extended to the real ranks \(s\) using a tetration function. If we have a sequence of operators \(<s>\) satisfying the property \( (*) \) then why don't we have
\[
\begin{align}
x\,<s>\,\omega&=\exp^s(\log^s(x)+\omega)\\
&=\exp^s(\log^s(x)+\log^s(\omega))\\
&=x \odot_s \omega\\
\end{align}\]?
If the above is true how can \(<s>\) also satisfy the Goodstein equation, i.e. \(<s+1>S=<s><s+1>\)? Isn't \(<3>\) different from tetration?
Note: we have however a chain of holomorphic Goodstein equations. Observe that
\[
\begin{align}
x\odot_{s+1} ({}^{s}b\odot_s y)&=(x\odot_{s+1}{}^{s}b) \odot_s(x \odot_{s+1} y)\\
&=\exp^{s+1}(\log^{s+1}(x)+\log^{s+1}({}^{s}b)) \odot_s(x \odot_{s+1} y)\\
&=x \odot_s(x \odot_{s+1} y)\\
\end{align}\]
Since \({\bf 1}_{0}=0\); \({\bf 1}_{1}=1\); \({\bf 1}_{2}=b\); and \({\bf 1}_{s}=\exp_b^s(0)={}^{s-1}b\); we can define the homomorphic image of the successor as \({\rm succ}_{s+1}(y):={\bf 1}_{s+1}\odot_s y\) and write the previous as \[x\odot_{s+1} {\rm succ}_{s+1}(y)=x \odot_s(x \odot_{s+1} y)\]
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)