Tetration and p-adic

[Ember Edison]

But the essential result you showed is that:

\[
\lim_{N\to\infty}\,\, ^N a = \overline{a_1a_2\cdots a_N}A\\
\]

Where we repeat to the left; and that means the value is a rational number in p-adic circles; the same way: \(0.\overline{9} = 1\) is a rational number. I'm not the best versed in this shit though; I just know enough to get by if someone starts talking about it . Definitely continue your research though! And you're only doing favours for yourself if you phrase it in p-adic terms

You'd definitely get some eyes on your search for tenure if you proved hands down \(^\infty a \in \mathbb{Q}_p\) for all \(a \in \mathbb{N}\) and \(p\) prime--or even \(a \in \mathbb{Q}\). That's a solid, sexy, result that tenure boards love

This is indeed a result that is very close to the final answer!
The difference between the different "versions" of p-adic is not significant in terms of series representation \( b = \sum\limits_{r \in S} a_r p^r  \).

for example: 
\( \mathbb{Z}_p \to 0\leq r<p \)
\( \mathbb{Q}_p \to S \) is a bounded-below subset of the \( \mathbb{N} \)

...

\( \bar{\mathbb{Q}}_p, \mathbb{C}_p  \to S \) is a more general well-ordered subset of \( \mathbb{Q} \)
\( \Omega_p \to S \) is any well-ordered subset of \( \mathbb{Q} \) with no other restrictions! 

If we can compute the required infinite limit for any \(a_r\), the problem of evaluating \( ^\infty b \) (or my symbol \( sexp(b,\infty) \) ) is solved.

[tommy1729]

Ok suppose we fully understand tetration for p-adic.

What exactly is the consequence of that ?

Does that have implications ?


regards

tommy1729

[JmsNxn]

The difference between the different "versions" of p-adic is not significant in terms of series representation \( b = \sum\limits_{r \in S} a_r p^r  \).

I was pretty sure of this; but I wanted to play it safe. Makes perfect sense though. If a series representation (in the digit series) is 2-adic rational, then it is p-adic rational for all p. I just didn't want to overstep my bounds, lol.

[Ember Edison]

EDIT:!!!!!

Okay, so I can't prove this. But I believe Marco has proven that:

\[
\lim_{N\to \infty}\,\,^N a = A \in \mathbb{Q}_p\\
\]

A (somewhat provocative) simple question: Evaluate A of

\[
a=0\in\mathbb{Q}_2, \lim_{N\to \infty}\,\,^N a = A \in \mathbb{Q}_2\\
\]

What is A? 0 or 1? and \( \mathbb{Q}_3,\mathbb{Q}_5,\mathbb{Q}_7 \)?

[marcokrt]

EDIT:!!!!!

Okay, so I can't prove this. But I believe Marco has proven that:

\[
\lim_{N\to \infty}\,\,^N a = A \in \mathbb{Q}_p\\
\]

A (somewhat provocative) simple question: Evaluate A of

\[
a=0\in\mathbb{Q}_2, \lim_{N\to \infty}\,\,^N a = A \in \mathbb{Q}_2\\
\]

What is A? 0 or 1? and \( \mathbb{Q}_3,\mathbb{Q}_5,\mathbb{Q}_7  \)?

By Definition 1.3 from the mentioned reference "Number of stable digits of any integer tetration" (https://arxiv.org/pdf/2210.07956.pdf), we have that \( V(0):=0 \), since \(^N 0 = 0 \) iff \( N \) is odd and \(^N 0 = 1 \) otherwise. Now, I introduced this definition in the paper, since we want that the "constant congruence speed" of tetration is a function from the nonnegative integers which are not a multiple of \( 10 \) and the whole set \( \mathbb{N}_0 \), but this is not the unique possible solution of the underlined issue. Here is the most interesting point: we may assume that \( V(0) = \infty \) by looking at the solution \( \alpha_{00} \) of the fundamental equation \(y^5=y \) in the commutative ring \( \mathbb{Z}_{10} \) (see page 11, ibid.), since this is the only solution linked to the constant congruence speed of the tetration bases that are congruent to \( 0 \) modulo \( 10 \), and thus its p-adic valuation would always be \( +\infty \). Thus, we would have that \( V(0) = +\infty \) also in different numerical systems, if we assume that the first line of Equation 16 is given by \( v_p(0) \), where \( v_p \) indicates the p-adic valuation of the argument for any given prime number \( p \).
So, the best answer to your question above cannot ignore our preliminary assumption based on the fact that \( V(0) \in \mathbb{Z} \) definition, while it would be not even be an element \( \mathbb{R} \ by construction... otherwise the "constant congruence speed analysis" cannot be taken as a way to solve related problems in the p-adics.

[Ember Edison]

we have that \( V(0):=0 \), since \(^N 0 = 0 \) iff \( N \) is odd and \(^N 0 = 1 \) otherwise. 

In my personal opinion, this conclusion is sufficient. 

Of course further discussion about the definition of V(0) is also interesting!

So, the best answer to your question above cannot ignore our preliminary assumption based on the fact that \( V(0) \in \mathbb{Z} \) definition, while it would be not even be an element \( \mathbb{R} \) by construction... otherwise the "constant congruence speed analysis" cannot be taken as a way to solve related problems in the p-adics.

And the interesting point is that \( V(0)\not\in\mathbb{Q}_p \not\Leftrightarrow V(0)\not\in\mathbb{R} \not\Leftrightarrow V(0)\not\in\mathbb{C}\not\Leftrightarrow V(0)\not\in\Omega_p \), \( (\mathbb{R},\mathbb{C} ) \) unequal we have seen in the standard Tetration, and the rest of the unequality is a simple known fact.

[JmsNxn]

I've definitely lost the plot so far. I'm pretty shit at p-adic stuff. But \(V(0) = -\infty\) seems like a non issue. This is absolutely standard. The same way every valuation satisfies this--this is how we make p-adic norms. Even adding in the fact that \(^N 0 = (N \pmod{2})\).  The reason this happens is because you have assumed that \(0^1 = 0\) and \(0^0 = 1\). And we have a strange oscillation. This is still a singular behaviour though, which is better written with \(V(0) = - \infty\)...

I tend to lean towards Ember's analysis here. As there's room for error about \(V(0)\)... and \(V(0) = -\infty\) seems perfectly natural. Especially because it appears to be acting as a valuation. The same way the degree of a constant polynomial \(\deg( C) = 0\) and the degree of the zero polynomial is \(\deg(0) = -\infty\)... It may seem wrong at first glance, but it describes the algebra perfectly...

I'm out of my depth here though, so correct me if I'm being an idiot! But I see it as being much more natural to just set \(V(0) = -\infty\)....

Regards, James

[marcokrt]

I've definitely lost the plot so far. I'm pretty shit at p-adic stuff. But \(V(0) = -\infty\) seems like a non issue. This is absolutely standard. The same way every valuation satisfies this--this is how we make p-adic norms. Even adding in the fact that \(^N 0 = (N \pmod{2})\).  The reason this happens is because you have assumed that \(0^1 = 0\) and \(0^0 = 1\). And we have a strange oscillation. This is still a singular behaviour though, which is better written with \(V(0) = - \infty\)...

I tend to lean towards Ember's analysis here. As there's room for error about \(V(0)\)... and \(V(0) = -\infty\) seems perfectly natural. Especially because it appears to be acting as a valuation. The same way the degree of a constant polynomial \(\deg( C) = 0\) and the degree of the zero polynomial is \(\deg(0) = -\infty\)... It may seem wrong at first glance, but it describes the algebra perfectly...

I'm out of my depth here though, so correct me if I'm being an idiot! But I see it as being much more natural to just set \(V(0) = -\infty\)....

Regards, James

This is not a real issue, indeed. Let me go back to the origin of the problem for a while and you will clearly see the big picture.
My original goal (and the goal of my trilogy of papers on NNTDM) was to find, in radix-10, the number of new stable digits (i.e., "frozen digits") of any integer tetration \( ^N a \), such that \( a \not\equiv 0 \pmod{10} \), for a unitary increment of \( N \).
So, for this purpose, I simply disregarded the solutions \( \alpha_{00} = 0 = \dots 00000 \) and \( \alpha_{01} = 1 = \dots 000001 \), since we usually cannot see any arbitrarily large number of zeros before \( N \), but we generally agree that \( \dots 000000N = N \), otherwise we would have overcomplicated a solved problem for the two mentioned trivial cases, while the congruence speed is not clearly constant for any \( N \equiv 0 \pmod{10} : N \neq 0 \). It is just a matter of how we decide to define V(N) at the beginning, and this follows from what we are going to study... but after we have finally reached the end of the third paper, we can agree that the next goal may be to focus ourselves on the p-adics as an ending point, rather than as a tool to solve the original problem of the number of stable digits in the integers \( \pmod {10^N} \), so we have to just remove Def. 1.3 from https://arxiv.org/pdf/2210.07956.pdf and set all the \( 15 \) solutions of \( y^5=y \) as a gold standard... it follows that \( V(0) \) is the p-adic order of \( 0 \), which is equal to \( \infty \) (while \( V(1)=0 \) still holds as it has previously been defined).
Now, working backwards from this point, we can also agree that the number of digits that \( ^N 1 \) freezes going from \( N \) to \( N+1 \) is as big as we wish, since \( 1=01=001=0001=\dots \) and also \( 1^1=01=001=0001=\dots \) as \( 1^{(1^1)}=01=001=0001=\dots \), and so forth.

[JmsNxn]

Fuck, I love this, Marco! I'm definitely out of my depth. The only p-adic shit I've done is Fourier p-adic analysis. I'm dumb as fuck with p-adic shit. But that makes perfect sense. I only understood Tate's thesis (Fourier p-adic shit), by thinking of it as a non-archimedean number field. But I have trouble actually making heads or tails of this shit, lmao. I can read it, and understand it. But I'd never be able to prove anything about it. I can't do any exercises....

That makes a lot of sense though, Marco; thanks.