Riemann on Understanding the Riemann Hypothesis
https://riemannhypothesis.info/tags/riemann/
Recent content in Riemann on Understanding the Riemann HypothesisHugo -- gohugo.iomarkus@recommend.games (Markus Shepherd)markus@recommend.games (Markus Shepherd)Wed, 31 May 2017 05:41:25 +0000The Prime Bet
https://riemannhypothesis.info/2017/05/the-prime-bet/
Wed, 31 May 2017 05:41:25 +0000markus@recommend.games (Markus Shepherd)https://riemannhypothesis.info/2017/05/the-prime-bet/Let’s say you sit in a pub, minding your own business, when all of a sudden a stranger walks up to you and offers you a bet:
We’ll choose two positive integers at random. If they have any divisor in common (other than \(1\)) I’ll pay you a dollar, else you’ll pay me a dollar. Are you in?
Apart from the question what kind of establishments you frequent, you should be wondering: is this a good bet for you?If One at a Time is too Difficult, Try All at Once!
https://riemannhypothesis.info/2014/12/if-one-at-a-time-is-too-difficult-try-all-at-once/
Sat, 27 Dec 2014 12:00:16 +0000markus@recommend.games (Markus Shepherd)https://riemannhypothesis.info/2014/12/if-one-at-a-time-is-too-difficult-try-all-at-once/In the past months, I spent as much time as I had on taking online courses at Coursera. One particularly interesting course, both from a mathematical and computational point of view, is Analytic Combinatorics which applies combinatorics (i.e., the art of counting) to the analysis of algorithms by finding formulae, exact or asymptotic, for their running time.
It is notoriously difficult to find exact formulae for general combinatorial constructs. Typically, we want to know how many objects, e.Applying the Explicit Formula
https://riemannhypothesis.info/2014/11/applying-the-explicit-formula/
Sun, 23 Nov 2014 10:47:17 +0000markus@recommend.games (Markus Shepherd)https://riemannhypothesis.info/2014/11/applying-the-explicit-formula/It’s quite some time since we arrived at Riemann’s main result, the explicit formula
\[ J(x)=\mathrm{Li}(x)-\sum_{\Im\varrho>0}\left(\mathrm{Li}(x^\varrho)+\mathrm{Li}(x^{1-\varrho})\right)+\int_x^\infty\frac{\mathrm{d}t}{t(t^2-1)\log t}-\log2, \]
where \(J(x)\) is the prime power counting function introduced even earlier. It’s high time we applied this!
First, let’s take a look at \(J(x)\) when calculating it exactly:
You see how this jumps by one unit at prime values (\(2\), \(3\), \(5\), \(7\), \(11\), \(13\), \(17\), \(19\)), by half a unit at squares of primes (\(4\), \(9\)), by a third at cubes (\(8\)), and by a quarter at fourth powers (\(16\)), but is constant otherwise.How NOT to Earn a Million Dollars
https://riemannhypothesis.info/2014/11/how-not-to-earn-a-million-dollars/
Sun, 02 Nov 2014 12:06:30 +0000markus@recommend.games (Markus Shepherd)https://riemannhypothesis.info/2014/11/how-not-to-earn-a-million-dollars/I recently spent some time on the formidable website Numberphile which explains mathematical ideas, some important, some recreational, in short and accessible videos. Definitely worth checking out. One of the videos that is most relevant to us explains the Riemann Hypothesis:
As mentioned before, it’s not easy to explain the details and the beauty of the Riemann Hypothesis in few words, but I think the video definitely succeeds in compressing the essentials into 17 minutes.Integral Madness
https://riemannhypothesis.info/2014/04/integral-madness/
Sun, 13 Apr 2014 15:25:54 +0000markus@recommend.games (Markus Shepherd)https://riemannhypothesis.info/2014/04/integral-madness/We’ve seen the calculus version
\[ J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log\zeta(s)x^s\frac{\mathrm{d}s}{s}, \]
of the Euler product, and we know how to express \(\xi(s)\) as a product over its roots
\[ \xi(s)=\xi(0)\prod_\varrho\left(1-\frac{s}{\varrho}\right), \]
where
\[ \xi(s) = \frac{1}{2} \pi^{-s/2} s(s-1) \Pi(s/2-1) \zeta(s) \newline = \pi^{-s/2} (s-1) \Pi(s/2) \zeta(s). \]
High time we put everything together – the reward will be the long expected explicit formula for counting primes!First, let’s bring the two formulae for \(\xi(s)\) together and rearrange them such that we obtain a formula for \(\zeta(s)\):In the Beginning, There Was... Euler's Formula!
https://riemannhypothesis.info/2013/09/in-the-beginning-there-was-eulers-formula/
Sun, 29 Sep 2013 17:39:48 +0000markus@recommend.games (Markus Shepherd)https://riemannhypothesis.info/2013/09/in-the-beginning-there-was-eulers-formula/I will start this blog the way Bernhard Riemann started his paper: with Euler’s product formula that John Derbyshire called the golden key:
\[ \zeta(s)=\sum_{n\ge1}n^{-s}=\prod_{p}(1-p^{-s})^{-1} \]
This holds for any complex number \(s\) with \(\Re s > 1\). If you look up a proof in any modern textbook, you will find a number technical rearrangements that end up in an examination of the absolute convergence on both sides. But actually, the formula is nothing but a fancy way of writing out the Sieve of Eratosthenes.