Discussion on "tetra-eta-series" (2007) in MO

[Caleb]

In MO the user Caleb Briggs brings to discussion an old attempt of mine to the series \( T(x,2) = 1^{1^x} - 2^{2^x} + 3^{3^x} - \cdots + \cdots \) .
I've tried this on 2007, and only later learned some techniques with which I might have assessed this with more success. But my basic observations and also the computations that I'd been able to do (over a small range of the exponent \(x\) ) come out to be correct.      

For the friends of visual data - there are some nice pictures to see there.

Here is the link : 

https://mathoverflow.net/questions/43666...eta-series   

Here the link to my fiddlings:

http://go.helms-net.de/math/tetdocs/Tetra_Etaseries.pdf  

Have fun...

Gottfried

@James :

I meant splitting the positive and negative parts of  \( T(x,2) = 1^{1^x} - 2^{2^x} + 3^{3^x} - \cdots + \cdots \)  and then using

 https://math.eretrandre.org/tetrationfor...p?tid=1688

my summability method on it.



regards

tommy1729

OHHHHH!

That makes much more sense Tommy  

So you're saying, let's continuum sum:

\[
E(z,s) = \sum_{j=1}^z (2j)^{(2j)^{-s}}\\
\]

And:

\[
P(z,s) = \sum_{j=1}^z (2j-1)^{(2j-1)^{-s}}\\
\]

Where then; Gottfried's function equals:

\[
\zeta_G(s) = \lim_{z\to\infty} \left(P(z,s) - E(z,s)\right)\\
\]

I mean, I don't see any reason this shouldn't work. Any continuum sum method works. But I'm not sure if this allows us to take \(-1 < \Re(s) \le 0\). Especially considering there is a pole at \(s=0\)--which should muddy the waters incredibly. I think the key though, may be somewhere in here. But we want to first find that value \(C \in \mathbb{R}\), such that:

\[
\zeta_G(s) - \frac{C}{s} = H(s)\\
\]

Where \(H(s)\) is holomorphic at \(s=0\). Because I don't think your summability method works in neighborhoods of poles (it's equivalent to the fractional calculus indefinite sum). And when there's a pole in the indefinite sum, it means an extra residue is added in the contour integral representation. Which muddies the waters so to speak.

But for \(\Re(s) > 0\)--your expression should work perfectly fine. And works for any indefinite sum. The function:

\[
f(z,s) = z^{z^{-s}}\\
\]

Is in the Ramanujan space \(f(z,s) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})\) for \(\rho, \tau \in \mathbb{R}^+\), and \(|\tau| < \pi/2\). But only for \(\Re(s) > 0\). I think this will suffer the same problem as my approach for \(-1 < \Re(s) < 0\). And I'm not sure yet how to handle this.

But:

\[
\sum_{j=1}^z f(j,s)\\
\]

Is perfectly indefinitely summable.

yes.

My method was designed for entire function, and those nasty residue issues are one of the reasons.

Things can get complicated with combining summability methods, analytic continuation , poles , singularities and continuum sums.

One of the nightmares is they disagree on where to take branch cuts. 
Then what ?

Im sure there are ways around these issues, but care should be taken.

I just proposed the method , not sure it even works well here or get nice results.
In particular the same result.

Poles were debates in old times when it came to continuum sums , analytic continuation and summability methods so your knowledge intuition or memory is correct , conscious or unconscious.
Cauchy principle value and +/- sign debates ring a bell.

I would not be surprised if Caleb found a solution.


regards

tommy1729

I'm currently working on a solution when looking at residues-- it looks like the question gets very difficult, but also very fascinating. In a few days I will probably finish writing a very long post that explores the whole poles problem.