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JmsNxn · 02/02/2021, 08:37 AM

Hey, everyone!

Hope everyone is doing well, and doing math, and having fun. I'm having loads of fun working on tetration again. So I added twenty or so pages to my last paper. Now, the major difference in this iteration of the paper is actually something really small. But once you see this small error, it collapses my entire approach in the first iteration.

In my original iteration, the ten or so pages, I assumed that,

$ \psi_{\pi}(t) \le |\psi_y(t)|\\ $

Everywhere! Which turns out to be wrong. So I spent a whole lot of time, after talking to Sheldon, on clarifying everything I know, drawing out every string--and consequently showing more or less the same result. But now it's,

$ \forall \delta>0\exists T \forall t\ge T\,\,\psi_{\pi}(t) \le |\psi_y(t)|\,\,\text{if}\,\,\delta < y < 2\pi - \delta\\ $

It's a curse and a blessing, because it took 20 more pages; I had to rewrite a good amount of the proofs; taught me where I should better explain; reminded me to pull on every thread; and so on... I still see a couple holes in the dam that my finger has to plug for the moment. But I'm so close; I can smell it.

So I present a 29 page paper roughly constructing a tetration function $e \uparrow \uparrow s$ that is holomorphic on $\mathbb{C}/(\mathbb{R} + 2\pi ik)$ and at least continuously differentiable on $\mathbb{R} + 2\pi i k$ excluding singularities.

JmsNxn · 02/04/2021, 04:40 AM

Hey just wanted to remind everyone--there are still a couple bugs and errors. I noticed some more...

sheldonison · (This post was last modified: 02/06/2021, 09:32 PM by sheldonison.)

Thanks James for a superbly written paper, especially the first half of your paper where you carefully prove the convergence of $\phi(s)$ , and that it is entire. Here are a couple of graphs of $\phi$ at the $\pi\\i$ line, and in the complex plane. $\phi(s+1)=\exp(\phi(s)+s)$

The first graph goes from -2, to +7 at the real axis, and -0.5 to 3.5 in the imaginary.

The pi i contour is also interesting, where it is real valued. It starts out negative near zero as the asymptotic behavior of $\phi(s)=\approx\exp(s-1)$ , but it doesn't grow quite as fast, and you get these weird bump around 6+pi*i. The graph goes from -2 to +12. Eventually, the phi(s) becomes large enough negative that exp(phi(s)) becomes really small compared to exp(s), so the phi(s+1) gets close to zero, but slightly negative. Then phi(s+2) is nearly exp(s+1).

Finally, here is phi(s) at pi*i in the complex plane from +4 .. +13 from pi*I-3 to pi*i+1, showing some of the interesting behavior.

JmsNxn · 02/07/2021, 11:11 PM

Hey, so I wanted to try and approach from the path of proving NON-convergence. When you can't prove it does, try and prove it can't. We're going to simplify to the case;

$ -\phi(t+\pi i) = \psi(t)\\ $

Now, I had boiled it down into one thing or another--which is $\frac{\psi(t)}{t} \to 1$ or we get precisely my worst fear; the graphs Sheldon is showing. I was hoping it would level out and not grow too fast, so that it couldn't shrink. But lo and behold, it looks like it grows too fast, causing it to shrink; causing the violent oscillation we see in this graph. I tried running my own pari-gp just to double check (maybe just maybe there was an artifact in Sheldon's computations (not likely though)), and I do notice this weird hump we see in this graph right before the oscillation. Which is truly the death of the method. It has to stay below $t$ but approach $t$ for this thing to work. The moment it breaks this pattern, we get chaos. (And we can notice this in Sheldon's graph, where the tiny hump happens where it just dips below $-t$ causing the insane breakdown.) Now I can't prove anything exactly, but I thought I'd try and frame this post from the worst case scenario.

Define the sequence of numbers $\delta_n = \psi(t+2n)$ for $0 < a < t 2$ that $\delta_N < 1$ (which is assuming the worst possible case is going to happen.)

Then,

$ \delta_{N+1} = e^{t+2N+1 - e^{t+2N - \delta_N}}\\ 0 \le e^{2N+1 - e^{2N+1}} \le |\delta_{N+1}| \le e^{2N+2 - e^{2N-1}} < 1\\ $

which implies that, $\delta_n \to 0$ and it looks double exponential. Now define the sequence of numbers, $\mu_n = \psi(t+2n+1)$ .

$ \mu_n = e^{t+2n - \delta_n}\\ 0 \le e^{2n - 1} \le \mu_n \le e^{2n+1}\\ $

Now it's no hard fact to sandwich,

$ 0 \le e^{2n-1}e^{2n+1 - e^{2n+1}} \le \mu_n \delta_n \le e^{2n+1}e^{2n+2 - e^{2n-1}}\\ 0 \le \mu_n \delta_n \le e^{4n+3-e^{2n-1}} \to 0\\ $

Now, the reason this is so important, and why this is my biggest fear, is that we can now confidently say that, for even $m=2k$

$ \prod_{j=1}^m \psi(t+j) =\prod_{j=1}^k \mu_j \delta_j \to 0\\ $

This kaputs the entire construction of $\tau(s)$ along the line $\Im(s) = \pi$ . Since the crucial philosophy of the proof is to use Banach's Fixed Point Theorem,

$ |\tau_{m+1}(t+\pi i) - \tau_m(t+ \pi i)| \le \frac{|t+m + \pi i|}{|\prod_{j=1}^m \psi(t+j)|} \to \infty\\ $

So unless Sheldon's code is making a rounding error causing us to just dip below $-t$ , and the nested exponentials magnify that error causing oscillation... It looks like this construction will fail on the line $\Im(s)=\pi$ . Which is very unfortunate. I, for one, have faith in Sheldon's code. So I am officially in the not holomorphic camp--it's just $C^{\infty}$ on $(-2,\infty)$ .

Regards, James

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