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

[JmsNxn]

so, if we split the sum into positive parts and negative parts

f(x) = f+(x) + f-(x)

and we use my summation idea for both f+ and f- and then add them

does that give the same thing ??


regards

tommy1729

Could you elaborate more, Tommy?

Breaking:

\[
\zeta_G(s) = \sum_{m=1}^\infty |g(2m-1)|(2m-1)^{-s} - 2^{-s}\sum_{m=1}^\infty |g(2m)|m^{-s}\\
\]

Doesn't add too much to the discussion unless you can describe \(g(2m)\) and \(g(2m-1)\) in a descriptive manner; distinguish their behaviour. When we talk about \(A(x) = \sum_{m\le \lfloor x \rfloor} g(m)\); then we split this into positives and negatives:

\[
\begin{align}
A(x) &= \sum_{1\le m \le \lfloor \frac{x}{2} \rfloor} |g(2m-1)| - |g(2m)|\\
&= \sum_{1\le m \le \lfloor \frac{x}{2} \rfloor} |g(2m-1)| - \sum_{1\le m \le \lfloor \frac{x}{2} \rfloor} |g(2m)|\\
&= A^{\text{odd}}(x) - A^{\text{even}}(x)\\
\end{align}
\]

I don't see anything obvious that could be solved by doing this split; if you can see something: TELL ME! I'd love to see something!

---

To remind everyone of the following graph:

\[
\begin{align}
h(n,k) &= \sum_{j=0}^\infty \frac{\log(n)^j}{j!}j^k\\
g^{(2)}(m) &= (-1)^{m+1} \sum_{n\le m} h\left(n,\frac{\log(m)}{\log(n)}\right) \frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!} \chi_m(n)\\
\chi_m(n) &= 1\,\,\text{if and only if there exists}\,\,k\in \mathbb{N},\,\,n^k = m,\,\,\text{otherwise}, 0\\
\end{align}
\]

The function \(A^{(2)}(x)\) is written as:

\[
A^{(2)}(x) = \sum_{m \le \lfloor x \rfloor} g^{(2)}(m)\\
\]

Here is a graph from \(1 \le x \le 1000\). This took 2 days to compile; please enjoy 

   

This looks exactly like \(O(x \log(x)^2)\)...