Just want to be clear; we're choosing,
\(
\text{Erf}(s) = \frac{2}{\sqrt{\pi}} \int_0^s e^{-x^2}\,dx\\
t(s) = (\text{Erf}(s)+1)/2\\
\)
And then we want to construct,
\(
f(s) = \Omega_{j=1}^\infty e^{t(s-j)z}\,\bullet z\\
\)
Which absolutely is entire in \( s \). I thought I'd translate what you are saying a bit more clearly--just so I understand you well enough.
First of all if \( 1 \in \mathcal{N} \) where \( \mathcal{N} \) is a domain and \( \mathcal{S} \) is a domain,
\(
\sum_{j=1}^\infty ||e^{t(s-j)z} - 1||_{s \in \mathcal{S}, z \in \mathcal{N}} < \infty\\
\)
This is absolutely an entire function. So you're right tommy.
It satisfies the asymptotic equation,
\(
\log f(s+1) = t(s)f(s)\\
\)
But since \( t(s) \to 1 \) as \( \Re(s) \to \infty \); it will produce the asymptotic,
\(
\log f(s+1) \sim f(s) \,\,\text{as}\,\,\Re(s) \to \infty\\
\)
So much of my theorem work could be ported over to this problem; I think it'd be tricky. But it's probably doable.
Tommy, I'd like to propose we call this approach the "multiplicative approach" and mine "the additive approach". Based solely on what the error terms look like in the equation,
\(
\log f(s+1) = A(s)f(s)\,\,\text{is the multiplicative equation}\\
\log f(s+1) = f(s) + B(s)\,\,\text{is the additive equation}\\
\)
I do believe your method will work. And I do believe if you take functions which are similar to the error function they'll produce the same tetration. That's something really surprising I've found from my work. Is that the logistic function is almost arbitrary, any sort of similar mapping will produce the same final tetration function.
I guess my biggest question.... Imagine this produces a THIRD tetration!
I can't really see how to hammer out the kinks on this method. You have to somehow invent a good banach fixed point argument. But on the real-line it converges absolutely; I'd bet a million it's analytic AT LEAST on \( \mathbb{R} \).
Hmmm, Tommy. I'd need you to elaborate to get it all.
\(
\text{Erf}(s) = \frac{2}{\sqrt{\pi}} \int_0^s e^{-x^2}\,dx\\
t(s) = (\text{Erf}(s)+1)/2\\
\)
And then we want to construct,
\(
f(s) = \Omega_{j=1}^\infty e^{t(s-j)z}\,\bullet z\\
\)
Which absolutely is entire in \( s \). I thought I'd translate what you are saying a bit more clearly--just so I understand you well enough.
First of all if \( 1 \in \mathcal{N} \) where \( \mathcal{N} \) is a domain and \( \mathcal{S} \) is a domain,
\(
\sum_{j=1}^\infty ||e^{t(s-j)z} - 1||_{s \in \mathcal{S}, z \in \mathcal{N}} < \infty\\
\)
This is absolutely an entire function. So you're right tommy.
It satisfies the asymptotic equation,
\(
\log f(s+1) = t(s)f(s)\\
\)
But since \( t(s) \to 1 \) as \( \Re(s) \to \infty \); it will produce the asymptotic,
\(
\log f(s+1) \sim f(s) \,\,\text{as}\,\,\Re(s) \to \infty\\
\)
So much of my theorem work could be ported over to this problem; I think it'd be tricky. But it's probably doable.
Tommy, I'd like to propose we call this approach the "multiplicative approach" and mine "the additive approach". Based solely on what the error terms look like in the equation,
\(
\log f(s+1) = A(s)f(s)\,\,\text{is the multiplicative equation}\\
\log f(s+1) = f(s) + B(s)\,\,\text{is the additive equation}\\
\)
I do believe your method will work. And I do believe if you take functions which are similar to the error function they'll produce the same tetration. That's something really surprising I've found from my work. Is that the logistic function is almost arbitrary, any sort of similar mapping will produce the same final tetration function.
I guess my biggest question.... Imagine this produces a THIRD tetration!
I can't really see how to hammer out the kinks on this method. You have to somehow invent a good banach fixed point argument. But on the real-line it converges absolutely; I'd bet a million it's analytic AT LEAST on \( \mathbb{R} \).
Hmmm, Tommy. I'd need you to elaborate to get it all.