- Thread starter
- Admin
- #1

http://mathhelpboards.com/math-notes-49/what-riemann-hypothesis-11812.html

- Thread starter MarkFL
- Start date

- Thread starter
- Admin
- #1

http://mathhelpboards.com/math-notes-49/what-riemann-hypothesis-11812.html

- Admin
- #2

- Jan 26, 2012

- 4,029

The summary Balarka is writing seems to be at a good level for those working at the undergrad level but isn't overwhelming to read either. If I don't follow some part I am at least aware of the topic that needs more study. I've skimmed the whole thread but will be reading in detail very soon. Well done, Balarka .

- Mar 22, 2013

- 573

However, the only _assumed_ result I have used is that Perron's formula. If anyone goes through the post rigorously and finds curiosity about Perron, I can prove it for them. It requires a bit contour manipulation and some hard-core estimation. In it's most general and powerful form, it is

$$\sum_{n \leq x} \frac{a_n}{n^s} = \frac1{2\pi i} \int_{c - iT}^{c + iT} \sum_{n=1}^\infty \frac{a_n}{n^{s+\omega}} \frac{x^\omega}{\omega} \mathrm{d}\omega + O \left ( \frac{x^c}{T(\sigma + c - 1)^\alpha} \right) + O \left ( \frac{f(2x) x^{1-\sigma} \log x}{T} \right ) + O \left ( \frac{f(N) x^{1-\sigma}}{T|x - N|} \right ) $$

Where $a_n = O(f(n))$, $\psi(n)$ not decreasing, $\sum_{n \leq x} |a_n|/n^\sigma = O((\sigma - 1)^{-\alpha})$ in the neighborhood of $\sigma = 1$, some $c > 0$ such that $\sigma + c > 1$ and a noninteger $x$ with $N = \text{int}(x)$.

Letting $T \to \infty$ and $s = 1$, however, one arrives at the weaker formula shown in wiki, weaker in the sense that the absolute convergence may not be verified anymore, neither of the sum nor of the integral (the zeta zero sum in the explicit formula for $\psi(x)$ a couple posts back is, for example, not absolutely converging and it'd be a capital mistake to compute the sum without ordering the roots, see Riemann's trap )

I've also explained some elementary version of RH, like the moebius-RH : $M(x) = O(x^{1/2+\epsilon})$ and it's probabilistic interpretation, the $\psi$-RH : $|\psi - x| \leq k \cdot \sqrt{x}^{1/2} \log^2 x$ for some $k > 0$ which is in fact the tightest bound, following Littlewood's result. I feel a relation to prime gaps (can be probabilistically explained!) would have been interesting but unfortunately I haven't studied it at all. There are a few plots and a table (and _will be_ computational results) too, so some intuition is present. In the whole, I believe the intuitive and the unintuitive part cancels each other out

That reminds me, I think I have to add a Venn diagram to the analytic continuation post : a lot of hubbub about analytic continuation is going round. Some are getting the wrong idea, and some never looking back at it. While analytic continuation is a delicate business in complex analysis (indeed, they are the starting point of Riemannsurfaces) I have really used nothing than basics of calculus in that post.

Finally, I'd like to have a few people reading it thoroughly and asking me anytime about anything that isn't explained well in the thread.

- Mar 22, 2013

- 573

The solutions are expected to be posted a few weeks later, it maybe further delayed on request from members (due to reasons like trying a problem in there, etc). Solution/ideas about any of the problems are most certainly welcome.

Pointing out any kind of typo (there are a lot of them!) are welcome, especially $\LaTeX$ ones.

Any kind of questions regarding the content will definitely be answered by the author (that's me).