So I'm having trouble coming up with a general proof to give us \( C^\infty \) pentation (or any hyper-operators), but certainly getting a continuous one is not that hard. We can start by taking,
\(
\frac{d}{dt} \text{slog}(t) = \frac{1}{(\frac{d}{dt} e \uparrow \uparrow t) \bullet \text{slog}(t)} \le \frac{1}{t}\\
\)
So that,
\(
|\text{slog}(a) - \text{slog}(b)| \le \frac{1}{\min(a,b)}|a-b|\\
\)
Then the sequence of convergents,
\(
\tau_{m+1}(t) = \text{slog}(\Phi(t+1) + \tau_m(t+1)) - \Phi(t)\\
\)
Satisfy,
\(
|\tau_{m+1}(t) - \tau_m(t)| \le |\text{slog}(\Phi(t+1) + \tau_m(t+1)) - \text{slog}(\Phi(t+1) + \tau_{m-1}(t+1))|\\
\le\frac{1}{\Phi(t+1)}|\tau_{m}(t+1) - \tau_{m-1}(t+1)|\\
\le \frac{|\tau_1(t+m) - \tau_0(t+m)|}{\prod_{j=1}^m \Phi(t+j)}\\
\le \frac{|t+m|}{\prod_{j=1}^m \Phi(t+j)}\\
\)
Because \( \tau_0 = 0 \) and,
\(
\tau_1(t) = \text{slog}(\Phi(t+1)) - \Phi(t)\\
= \text{slog}(e^t e\uparrow \uparrow \Phi(t)) - \Phi(t)\\
\le \text{slog}(e\uparrow \uparrow \Phi(t) + t) - \Phi(t)\\
\le t\\
\)
This certainly converges uniformly. So we have a continuous function,
\(
e \uparrow \uparrow \uparrow t = \Phi(t+\omega) + \tau(t+\omega)\\
\)
We can continue this for arbitrary order hyper-operators. The trouble comes from proving that this solution is \( C^\infty \). I haven't had the AHA moment yet to prove this. The way I proved tetration is \( C^\infty \) is not very helpful here because we used properties of the logarithm.
Anyway,
\(
e \uparrow^n t : \mathbb{R}^+\to \mathbb{R}^+\\
e \uparrow^n t \,\,\text{is continuous}\\
\)
God, I suck at real analysis, getting \( C^\infty \) might take a while...
\(
\frac{d}{dt} \text{slog}(t) = \frac{1}{(\frac{d}{dt} e \uparrow \uparrow t) \bullet \text{slog}(t)} \le \frac{1}{t}\\
\)
So that,
\(
|\text{slog}(a) - \text{slog}(b)| \le \frac{1}{\min(a,b)}|a-b|\\
\)
Then the sequence of convergents,
\(
\tau_{m+1}(t) = \text{slog}(\Phi(t+1) + \tau_m(t+1)) - \Phi(t)\\
\)
Satisfy,
\(
|\tau_{m+1}(t) - \tau_m(t)| \le |\text{slog}(\Phi(t+1) + \tau_m(t+1)) - \text{slog}(\Phi(t+1) + \tau_{m-1}(t+1))|\\
\le\frac{1}{\Phi(t+1)}|\tau_{m}(t+1) - \tau_{m-1}(t+1)|\\
\le \frac{|\tau_1(t+m) - \tau_0(t+m)|}{\prod_{j=1}^m \Phi(t+j)}\\
\le \frac{|t+m|}{\prod_{j=1}^m \Phi(t+j)}\\
\)
Because \( \tau_0 = 0 \) and,
\(
\tau_1(t) = \text{slog}(\Phi(t+1)) - \Phi(t)\\
= \text{slog}(e^t e\uparrow \uparrow \Phi(t)) - \Phi(t)\\
\le \text{slog}(e\uparrow \uparrow \Phi(t) + t) - \Phi(t)\\
\le t\\
\)
This certainly converges uniformly. So we have a continuous function,
\(
e \uparrow \uparrow \uparrow t = \Phi(t+\omega) + \tau(t+\omega)\\
\)
We can continue this for arbitrary order hyper-operators. The trouble comes from proving that this solution is \( C^\infty \). I haven't had the AHA moment yet to prove this. The way I proved tetration is \( C^\infty \) is not very helpful here because we used properties of the logarithm.
Anyway,
\(
e \uparrow^n t : \mathbb{R}^+\to \mathbb{R}^+\\
e \uparrow^n t \,\,\text{is continuous}\\
\)
God, I suck at real analysis, getting \( C^\infty \) might take a while...
