I thought I'd write a quick proof sketch that Tommy's method will probably work. There are obviously kinks which need to be worked out; and gaps, but I think I'll use this as further evidence. As far as I'm concerned I'm just translating Tommy, I take no real credit. I'm just trying to write it in a more approachable manner, similar to the \( \beta \)-tetration.
We begin with the function,
\(
A(s) = \frac{1}{\sqrt{\pi}}\int_{-\infty}^s e^{-x^2}\,dx = \frac{1}{\sqrt{\pi}}\int_{-\infty}^0 e^{-(s+x)^2}\,dx\\
\)
Thereore,
\(
\sum_{j=1}^\infty |A(s-j)| < \infty\\
\)
And additionally,
\(
\sum_{j=1}^\infty |e^{A(s-j)} - 1| < \infty\\
\)
So that,
\(
\sum_{j=1}^\infty ||e^{A(s-j)z} - 1 || < \infty\\
\)
For a supremum norm across arbitrary compact sets \( s \in \mathcal{S} \) and \( z \in \mathcal{K} \). By my work in infinite compositions, this means that,
\(
\text{Tom}(s) = \Omega_{j=1}^\infty e^{A(s-j)z}\,\bullet z\\
\)
converges to a holomorphic function in \( s \in \mathbb{C} \) and \( z \in \mathbb{C} \), where the function is constant in \( z \) so we can drop the variable. This function is very important, because it's an ENTIRE asymptotic solution to tetration. I've only ever managed to make an almost everywhere holomorphic asymptotic solution (a similar kind of function, but littered with singularities). This means that,
\(
\log \text{Tom}(s+1) \sim \text{Tom}(s)\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\, |\arg(s)| < \pi/4\\
\)
\(
\text{Tom}(s+1) = e^{\displaystyle A(s)\text{Tom}(s)}\\
\)
Now, there definitely is a secret lemma, something I haven't found, which makes all of this work. But for the moment. Remember that,
\(
\log\beta_\lambda(s+1) \sim \beta_\lambda(s)\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\, |\arg(\lambda s)| < \pi/2\\
\)
For my proof of the beta method, this is 90% of the work/requirements, in which we can describe a tetration,
\(
F_\lambda(s) = \lim_{n\to\infty} \log^{\circ n}\beta_\lambda(s+n)\\
\)
So assuming this is the crucial ingredient; this description is enough for,
\(
\text{tet}_{\text{Tom}}(z+x_0) = \lim_{n\to\infty} \log^{\circ n}\text{Tom}(z+n)\\
\)
Will, and definitely should converge.
My God Tommy! We're really opening pandora's box with these tetrations. I don't want to prove that this thing converges because I believe it's yours to show. But I can see it clearly! Fantastic work, Tommy! Absolutely Fantastic!
Deep Sincere Regards, James
PS
Absolutely fucking fantastic, Tommy!
Hmmmmmmmmmm, I think this might be the beta method in disguise the more I think about it,
\(
y(s) = \log \text{Tom}(s+1) = A(s)\text{Tom}(s)\\
\)
Then,
\(
\log y(s+1) = y(s) + \log A(s+1)\\
\)
So, I believe that the multiplicative/additive case should be isomorphic. So the question is more, how much does,
\(
\log A(s+1)\,\,\text{looks like}\,\,-\log(1+e^{-s/\sqrt{1+s}})\\
\)
If the error decays well enough; you are just reproducing the \( \beta \) method, Tommy.
Hmmmmmm. This is still beyond beautiful.
Tommy, the more I think about it; this is the \( \beta \) method. And I think I can prove it...
We begin with the function,
\(
A(s) = \frac{1}{\sqrt{\pi}}\int_{-\infty}^s e^{-x^2}\,dx = \frac{1}{\sqrt{\pi}}\int_{-\infty}^0 e^{-(s+x)^2}\,dx\\
\)
Thereore,
\(
\sum_{j=1}^\infty |A(s-j)| < \infty\\
\)
And additionally,
\(
\sum_{j=1}^\infty |e^{A(s-j)} - 1| < \infty\\
\)
So that,
\(
\sum_{j=1}^\infty ||e^{A(s-j)z} - 1 || < \infty\\
\)
For a supremum norm across arbitrary compact sets \( s \in \mathcal{S} \) and \( z \in \mathcal{K} \). By my work in infinite compositions, this means that,
\(
\text{Tom}(s) = \Omega_{j=1}^\infty e^{A(s-j)z}\,\bullet z\\
\)
converges to a holomorphic function in \( s \in \mathbb{C} \) and \( z \in \mathbb{C} \), where the function is constant in \( z \) so we can drop the variable. This function is very important, because it's an ENTIRE asymptotic solution to tetration. I've only ever managed to make an almost everywhere holomorphic asymptotic solution (a similar kind of function, but littered with singularities). This means that,
\(
\log \text{Tom}(s+1) \sim \text{Tom}(s)\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\, |\arg(s)| < \pi/4\\
\)
\(
\text{Tom}(s+1) = e^{\displaystyle A(s)\text{Tom}(s)}\\
\)
Now, there definitely is a secret lemma, something I haven't found, which makes all of this work. But for the moment. Remember that,
\(
\log\beta_\lambda(s+1) \sim \beta_\lambda(s)\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\, |\arg(\lambda s)| < \pi/2\\
\)
For my proof of the beta method, this is 90% of the work/requirements, in which we can describe a tetration,
\(
F_\lambda(s) = \lim_{n\to\infty} \log^{\circ n}\beta_\lambda(s+n)\\
\)
So assuming this is the crucial ingredient; this description is enough for,
\(
\text{tet}_{\text{Tom}}(z+x_0) = \lim_{n\to\infty} \log^{\circ n}\text{Tom}(z+n)\\
\)
Will, and definitely should converge.
My God Tommy! We're really opening pandora's box with these tetrations. I don't want to prove that this thing converges because I believe it's yours to show. But I can see it clearly! Fantastic work, Tommy! Absolutely Fantastic!
Deep Sincere Regards, James
PS
Absolutely fucking fantastic, Tommy!
Hmmmmmmmmmm, I think this might be the beta method in disguise the more I think about it,
\(
y(s) = \log \text{Tom}(s+1) = A(s)\text{Tom}(s)\\
\)
Then,
\(
\log y(s+1) = y(s) + \log A(s+1)\\
\)
So, I believe that the multiplicative/additive case should be isomorphic. So the question is more, how much does,
\(
\log A(s+1)\,\,\text{looks like}\,\,-\log(1+e^{-s/\sqrt{1+s}})\\
\)
If the error decays well enough; you are just reproducing the \( \beta \) method, Tommy.
Hmmmmmm. This is still beyond beautiful.
Tommy, the more I think about it; this is the \( \beta \) method. And I think I can prove it...