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November 8th, 2018, 01:16 PM   #11
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Cool

I have published my paper now:

https://arxiv.org/abs/1810.07877


The manuscript is not tailored for publication, so I've seen some people/reviewers become frustrated for not understanding the argumentation, but mostly due to their own incompetence. Others have found the text to be pretty understandable though. The proof of the limits wasn't presented with rigor, but that was pointed out in the text, that's how Mathematics is done anyway (and if you disagree with me, you probably wouldn't disagree with Alon Amit).

Since I didn't write the manuscript with publication in mind, I made it simple and sort of monotonous, the intent now is just to make the new results known, it's unlikely it's fit for publication as is. It's about the novelty, it's not about the presentation.

The contents of this paper present formulae for the harmonic numbers of any order, for the partial sums of two Fourier series which include them, and as a consequence, a new proof for the Basel problem, as well as a proof for the closed-forms of $\displaystyle \zeta(2k)$, integer $\displaystyle k>=1$. It also has new integral representations for $\displaystyle \zeta(2k+1)$, which appear in the formulas of $\displaystyle H_{2k+1}(n)$.

Assuming the closed-forms of $\displaystyle \zeta(2k)$, we are led to conclude the limits of the integrals that appear on $\displaystyle H_{2k}(n)$, and vice-versa.

I am now thinking whether it's possible to extend the reasoning employed for the harmonic numbers to any harmonic progression, in principle it should be.

Last edited by jrsousa2; November 8th, 2018 at 01:20 PM.
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November 8th, 2018, 01:29 PM   #12
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Quote:
Originally Posted by jrsousa2 View Post
I've seen some people/reviewers become frustrated for not understanding the argumentation, but mostly due to their own incompetence.
Shinichi Mochizuki is that you?
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November 8th, 2018, 01:33 PM   #13
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Quote:
Originally Posted by Maschke View Post
Shinichi Mochizuki is that you?
Definitely possible, but am not gonna say why I think so. Lol.
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November 8th, 2018, 02:00 PM   #14
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Btw, let me place the final results here as a shortcut

One of the three possible formulas for the harmonic numbers is:
$\displaystyle \sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\pi\int_{0} ^{1} (1-u)\cot{\pi u}\left(1-\cos{2\pi n u}\right)\,du$

But more generally, for the odd powers we get:
$\displaystyle H_{2k+1}(n)=\frac{1}{2n^{2k+1}}-\frac{(-1)^{k}(2\pi)^{2k+1}}{2}\int_{0}^{1}\sum_{j=0}^{k} \frac{B_{2k-2j}\left(2-2^{2k-2j}\right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n)\,du$

where $\displaystyle f(u,n)=\cot{\pi u}\left[1-\cos{2\pi n(1-u)}\right]$.
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