As Sheldon has thoroughly convinced me of non-holomorphy of my tetration. I thought I'd provide the proof I have that it is \( C^{\infty} \) on the line \( (-2,\infty) \). I sat on this proof and didn't develop it much because I was too fixated on the holomorphy part. But, I thought it'd be nice to have a proof of \( C^{\infty} \).
Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that,
\(
\tau^{(k)}(t) : (-2,\infty) \to \mathbb{R}\,\,\text{for}\,\,k<K\\
\sum_{m=1}^\infty ||\tau^{(k)}_{m+1}(t) - \tau^{(k)}_{m}||_{a \le t \le b} <\infty\\
\)
Where,
\(
\tau^{(k)}_0(t) = 0\\
\tau^{(k)}_m(t) = \frac{d^k}{dt^k} \log(1+\frac{\tau_{m-1}(t+1)}{\phi(t+1)})\\
\)
And \( ||...||_{a\le t \le b} \) is the sup-norm across some interval \( [a,b] \subset (-2,\infty) \). As a forewarning, this is going to be very messy...
Now to begin we can bound,
\(
||\phi^{(j)}(t)||_{a\le t \le b} \le M\,\,\text{for}\,\,j\le k\\
\)
And that next,
\(
\phi^{(k)}(t+1) + \tau_m^{(k)}(t+1) = \frac{d^k}{dt^k} e^{\phi(t) +\tau_{m+1}(t)}\\
= \sum_{j=0}^k \binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)})\\
= \sum_{j=0}^{k-1} \binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)}) + \phi(t+1)e^{-t}(\frac{d^k}{dt^k} e^{\tau_{m+1}(t)})
\)
Now,
\(
\frac{d^k}{dt^k} e^{\tau_{m+1}(t)} = e^{\tau_{m+1}(t)}(\tau_{m+1}^{(k)}(t) + \sum_{j=0}^{k-1} a_j \tau_{m+1}^{(j)}(t))\\
\)
So, we ask you to put on your thinking cap, and excuse me if I write,
\(
\tau_{m}^{(k)}(t+1) = A_m + C_m\tau_{m+1}^{(k)}(t)\\
\)
And by the induction hypothesis,
\(
\sum_{j=1}^\infty ||A_{j+1} - A_j||_{a\le t \le b} < \infty\\
\sum_{j=1}^\infty ||C_{j+1} - C_j||_{a\le t \le b} < \infty\\
\)
Which is because these terms are made up of finite sums and products of \( \tau_m^{(j)} \) and these are said to be summable. Now the proof is a walk in the park.
\(
\tau_{m}^{(k)}(t) = \frac{\tau_{m-1}^{(k)}(t+1) - A_{m-1}}{C_{m-1}}... = \sum_{j=0}^{m-1} (\prod_{k=0}^{m-1-j} C_{m-1-k}^{-1}) A_j\\
\)
Where, we've continued the iteration and set \( \tau_0 = 0 \) and \( \tau_1 = 0 \) for \( k>1 \), and \( \tau_1 = 1 \) for \( k=1 \) (but we're tossing this away because we know it's differentiable). Therefore,
\(
\sum_{m=1}^\infty ||\tau_{m+1}^{(k)}(t) - \tau_m^{(k)}(t)||_{a \le t \le b} < \infty\\
\)
Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code.
EDIT: I'll do it properly as I correct my paper and lower my expectations of the result.
***********************
As to the second part of this post--now that we have \( C^\infty \) out of the way, we ask if we can continue this iteration and get pentation. Now, \( \text{slog} \) will certainly be \( C^{\infty} \) and \( \frac{d}{dt}e \uparrow \uparrow t > 0 \) so it's a well defined bijection of \( \mathbb{R} \to (-2,\infty) \). So, first up to bat is to get another phi function,
\(
\Phi(t) = \Omega_{j=1}^\infty e^{t-j} e \uparrow \uparrow x \bullet x = e^{t-1} e \uparrow \uparrow (e^{t-2}e \uparrow \uparrow (e^{t-3} e \uparrow \uparrow ...))
\)
This will be \( C^\infty \) (it'll be a bit trickier to prove because we aren't using analytic functions, but just bear with me). And it satisfies the equation,
\(
\Phi(t+1) = e^t (e \uparrow \uparrow \Phi(t))\\
\)
By now, I think you might know where i'm going with this.
\(
e \uparrow^3 t = \lim_{n\to\infty} \text{slog} \text{slog} \cdots (n\,\text{times})\cdots\text{slog} \Phi(t+\omega_1 + n)\\
\)
And now I'm going to focus on showing this converges... Wish me luck; after being trampled by this holomorphy I thought I'd stick to where things are nice--no nasty dips to zero and the like...
Now, the idea is to apply Banach's fixed point theorem, but it's a bit more symbol heavy now. We will go by induction on the degree of the derivative. So let's assume that,
\(
\tau^{(k)}(t) : (-2,\infty) \to \mathbb{R}\,\,\text{for}\,\,k<K\\
\sum_{m=1}^\infty ||\tau^{(k)}_{m+1}(t) - \tau^{(k)}_{m}||_{a \le t \le b} <\infty\\
\)
Where,
\(
\tau^{(k)}_0(t) = 0\\
\tau^{(k)}_m(t) = \frac{d^k}{dt^k} \log(1+\frac{\tau_{m-1}(t+1)}{\phi(t+1)})\\
\)
And \( ||...||_{a\le t \le b} \) is the sup-norm across some interval \( [a,b] \subset (-2,\infty) \). As a forewarning, this is going to be very messy...
Now to begin we can bound,
\(
||\phi^{(j)}(t)||_{a\le t \le b} \le M\,\,\text{for}\,\,j\le k\\
\)
And that next,
\(
\phi^{(k)}(t+1) + \tau_m^{(k)}(t+1) = \frac{d^k}{dt^k} e^{\phi(t) +\tau_{m+1}(t)}\\
= \sum_{j=0}^k \binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)})\\
= \sum_{j=0}^{k-1} \binom{k}{j} (\frac{d^{k-j}}{dt^{k-j}} \phi(t+1)e^{-t})(\frac{d^j}{dt^j} e^{\tau_{m+1}(t)}) + \phi(t+1)e^{-t}(\frac{d^k}{dt^k} e^{\tau_{m+1}(t)})
\)
Now,
\(
\frac{d^k}{dt^k} e^{\tau_{m+1}(t)} = e^{\tau_{m+1}(t)}(\tau_{m+1}^{(k)}(t) + \sum_{j=0}^{k-1} a_j \tau_{m+1}^{(j)}(t))\\
\)
So, we ask you to put on your thinking cap, and excuse me if I write,
\(
\tau_{m}^{(k)}(t+1) = A_m + C_m\tau_{m+1}^{(k)}(t)\\
\)
And by the induction hypothesis,
\(
\sum_{j=1}^\infty ||A_{j+1} - A_j||_{a\le t \le b} < \infty\\
\sum_{j=1}^\infty ||C_{j+1} - C_j||_{a\le t \le b} < \infty\\
\)
Which is because these terms are made up of finite sums and products of \( \tau_m^{(j)} \) and these are said to be summable. Now the proof is a walk in the park.
\(
\tau_{m}^{(k)}(t) = \frac{\tau_{m-1}^{(k)}(t+1) - A_{m-1}}{C_{m-1}}... = \sum_{j=0}^{m-1} (\prod_{k=0}^{m-1-j} C_{m-1-k}^{-1}) A_j\\
\)
Where, we've continued the iteration and set \( \tau_0 = 0 \) and \( \tau_1 = 0 \) for \( k>1 \), and \( \tau_1 = 1 \) for \( k=1 \) (but we're tossing this away because we know it's differentiable). Therefore,
\(
\sum_{m=1}^\infty ||\tau_{m+1}^{(k)}(t) - \tau_m^{(k)}(t)||_{a \le t \le b} < \infty\\
\)
Of which, I've played a little fast and loose, but filling in the blanks would just require too much tex code.
EDIT: I'll do it properly as I correct my paper and lower my expectations of the result.
***********************
As to the second part of this post--now that we have \( C^\infty \) out of the way, we ask if we can continue this iteration and get pentation. Now, \( \text{slog} \) will certainly be \( C^{\infty} \) and \( \frac{d}{dt}e \uparrow \uparrow t > 0 \) so it's a well defined bijection of \( \mathbb{R} \to (-2,\infty) \). So, first up to bat is to get another phi function,
\(
\Phi(t) = \Omega_{j=1}^\infty e^{t-j} e \uparrow \uparrow x \bullet x = e^{t-1} e \uparrow \uparrow (e^{t-2}e \uparrow \uparrow (e^{t-3} e \uparrow \uparrow ...))
\)
This will be \( C^\infty \) (it'll be a bit trickier to prove because we aren't using analytic functions, but just bear with me). And it satisfies the equation,
\(
\Phi(t+1) = e^t (e \uparrow \uparrow \Phi(t))\\
\)
By now, I think you might know where i'm going with this.
\(
e \uparrow^3 t = \lim_{n\to\infty} \text{slog} \text{slog} \cdots (n\,\text{times})\cdots\text{slog} \Phi(t+\omega_1 + n)\\
\)
And now I'm going to focus on showing this converges... Wish me luck; after being trampled by this holomorphy I thought I'd stick to where things are nice--no nasty dips to zero and the like...
