2

I know that a generating function for the Zeta function at the even integers already exists, but how about the Zeta function at the odd integers?

I've done some research, and found some alternative formulas for the harmonic numbers that allowed me to create a generating function for $\zeta(2k+1)$, but I'd like to know if it'd be new.

That would be nice if the answer is no, after all, it gets me really frustrated when I find out that all my discoveries in Math are actually just rediscoveries. And that's been the case in like 99% of the times I found something.

Just found out the answer is yes, but my formula is different anyway, less bad.

Theoretical Economist
  • 6,472

3 Answers3

5

Yes, it exists and it is now new. $$\sum_{n\geq 1}\zeta(2n+1)z^{2n+1} = \sum_{n\geq 1}\sum_{m\geq 1}\left(\frac{z}{m}\right)^{2n+1}=\sum_{m\geq 1}\frac{\frac{z^3}{m^3}}{1-\frac{z^2}{m^2}}=\sum_{m\geq 1}\frac{z^3}{m(m-z)(m+z)}$$ equals $$ -\gamma z+\psi(1+z)+\psi(1-z)=-\frac{z}{2}\left(H_z+H_{-z}\right)$$ (where $\gamma$ is the Euler-Mascheroni constant and $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$) due to $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}. $$ Are you interested in the exponential generating function?

Jack D'Aurizio
  • 353,855
  • @HansMusgrave: I am not getting what the issue is. Is there a generating function for $\zeta(2n+1)$? Yes, the shown one. Is it well-known? Yes. Are there other OGFs? No, the OGF is unique. – Jack D'Aurizio Aug 19 '18 at 15:50
2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\verts{z} < 1}$:

\begin{align} \sum_{n = 1}^{\infty}\zeta\pars{2n + 1}z^{n} & = \sum_{n = 1}^{\infty} \bracks{\zeta\pars{2n + 1} - 1}\pars{\pm z^{1/2}}^{2n} + \sum_{n = 1}^{\infty}z^{n} \\[5mm] & = \sum_{n = 1}^{\infty} \bracks{\zeta\pars{2n + 1} - 1}\pars{\pm z^{1/2}}^{2n} + {z \over 1 - z} \end{align}

The first sum can be evaluated with the A & S $\ds{\mathbf{\color{black}{6.3.15}}}$ identity. Namely,

\begin{align} &\sum_{n = 1}^{\infty}\zeta\pars{2n + 1}z^{n} \\[2mm] = &\ \bracks{\!\!{1 \over 2\pars{\pm\root{z}}} - {1 \over 2}\,\pi\cot\pars{\!\pi\bracks{\pm\root{z}}\!}\! - {1 \over 1 - z} + 1 - \gamma - \Psi\pars{\! 1 \pm \!\root{z}\!}\!\!} \\[2mm] & \phantom{\bracks{A}}+ {z \over 1 - z} \\[5mm] = &\ \pm\,{1 \over 2\root{z}} \mp {1 \over 2}\,\pi\cot\pars{\pi\root{z}} - \gamma - \Psi\pars{1 \pm \root{z}} \end{align}

where $\ds{\gamma}$ is the Euler-Mascheroni Constant and $\ds{\Psi}$ is the Digamma Function. By adding the expressions for both signs $\ds{~\pm~}$and dividing by two:

$$ \bbx{\sum_{n = 1}^{\infty}\zeta\pars{2n + 1}z^{n} = -\gamma - {\Psi\pars{1 + \root{z}} + \Psi\pars{1 - \root{z}} \over 2}} $$

Felix Marin
  • 89,464
0

Based on this answer, it wouldn't be much work to construct your generating function from what has been known at least as far back as 2013 (and probably much further).

My experience with research in classical mathematics is that not only are the interesting results known, they've been known for so long that the results aren't digitized. This can introduce somewhat of a challenge in proving novelty for a new idea.

I'm usually extremely open to the DIY approach, but starting out in mathematical research is one area where having a professional opinion would be valuable. Many (definitely not all -- be respectful) professors are willing to sit down and talk about research for hours, especially if it has anything whatsoever to do with their interests. Such conversations can be a good way to help gauge whether an idea is novel or not and hopefully to spark interest in other ideas as well.

That said, novelty isn't everything. If your research doesn't tie in to the current body of literature well, it will likely need to be more exciting than just another formula (not to disparage the result. I don't know the field well enough) to be publishable. Reading through current papers and using those as starting points is likely to be a fruitful area of exploration. Assuming recent papers represent the cutting edge, anything beyond that which you discover is almost certainly going to be both new and relevant to what other researchers (i.e., your target audience) are interested in.

Hans Musgrave
  • 1,979