1

At the beginning I will explain some concepts and at the end I will ask the specific question.

Let's consider some operator $ \overset {s}{\underset{k=x}{ \lower 3pt { \LARGE\Xi}}} $ on function $f (k)$ which in case of divergence take its analytic value;

$\displaystyle \overset {s}{\underset{k=x}{ \lower 6pt { \Huge\Xi}}} f(k)=\overbrace { \sum_{k_{s}=x} ^{\infty} ... \sum_{k_{2}=x}^{\infty} \sum_{k_1=x}^{\infty}}^{s}f (k_s*...*k_2 * k_1)$

If $\mu (k)$ is the Möbius function we get identity $ \overset {-1}{\underset{k=x}{ \lower 3pt { \LARGE\Xi}}} f (k)=\sum\limits_{k=x}^{\infty}f (k) \mu (k+1-x) $

Notice that power function $k^{-z}$ is 'simple' function for $\overset {s}{\underset{k=x}{ \lower 3pt { \LARGE\Xi}}} $ operator because of distribute powers through the parentheses. In that case we get $\displaystyle \overset {s}{\underset{k=1}{ \lower 3pt { \LARGE\Xi}}} k^{-z}={\zeta (z)}^s$

And now question (I know that beginning was actually unnecessary, but I think it's pretty interesting and worth of mentioning).

Is there some way to write $ \overset {s}{\underset{k=1}{ \lower 3pt { \LARGE\Xi}}} f (k) $ in simple form?

And in particular what is $\displaystyle \overset {-1}{\underset{k=1}{ \lower 6pt { \Huge\Xi}}} q^k $ equal to. This could be actually very useful.

Jean Marie
  • 81,803
Wreior
  • 396
  • Just a comment about typesetting. Perhaps you should ask in TeX-stackexchange how to produce that big operator $\Xi$ properly. Note that it should be larger in display style than in inline mode. – amrsa Jan 14 '23 at 14:29
  • Yep I know. I was making the size of it manualny. Here is code for it XD \overset {s}{\underset{k=x}{ \lower 3pt { \LARGE\Xi}}} – Wreior Jan 14 '23 at 14:37
  • Specifically: How to create my own math operator with limits? – L. F. Jan 14 '23 at 16:38
  • If it converges absolutely then $\sum_{m\in \Bbb{Z}{\ge 1}^k} f(\prod{j=1}^k m_j) = \sum_{n\ge 1} f(n) d_k(n)$ where $\zeta(s)^k = \sum_{n\ge 1} d_k(n)n^{-s}$, $d_k(n)=\sum_{m\in \Bbb{Z}{\ge 1}^k,\prod{j=1}^k m_j=n} 1$. What do you expect more? – reuns Jan 14 '23 at 16:54
  • Some nicer form. If you have $ \overset {2}{\underset{k=1}{ \lower 3pt { \LARGE\Xi}}}q^k$ you don't write it with this $d_k(n )$ stuff but you can tell it's lambert series and by using this equation https://math.stackexchange.com/questions/4572175/infinite-series-as-integral-representation you can find its form convergent for all x. I'm looking for analogos of that generalisation in that case https://math.stackexchange.com/questions/4586393/integral-representation-for-series-of-any-order where you can actualy write Lambert function too – Wreior Jan 14 '23 at 17:15

0 Answers0