https://riemannhypothesis.blog/tags/gauss/ Recent content in Gauss on Understanding the Riemann Hypothesis Hugo -- gohugo.io en markus@recommend.games (Markus Shepherd) markus@recommend.games (Markus Shepherd) Fri, 05 Aug 2016 15:38:11 +0000 https://riemannhypothesis.blog/2016/08/primes-from-a-different-world/ Fri, 05 Aug 2016 15:38:11 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/2016/08/primes-from-a-different-world/ Every textbook on number theory will begin with a treatise on prime numbers; every treatise on prime numbers will begin by emphasising their importance as building blocks or atoms of our number system: every integer can be expressed as a product of prime numbers in one way and one way only. Six is two times three and there is no other way to decompose it.1 Euclid proved this over two thousand years ago and it is so fundamental (hence the name fundamental theorem of arithmetic) to our thinking about numbers that we take it for granted. https://riemannhypothesis.blog/2014/11/how-not-to-earn-a-million-dollars/ Sun, 02 Nov 2014 12:06:30 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/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. https://riemannhypothesis.blog/2014/04/integral-madness/ Sun, 13 Apr 2014 15:25:54 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/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 \(\zeta(s)\): https://riemannhypothesis.blog/2013/11/are-primes-independent/ Sun, 24 Nov 2013 14:32:23 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/2013/11/are-primes-independent/ The question may sound silly, but I hope it will become apparent that it’s very reasonable to ask. What we will examine here is the probabilistic interpretation of the prime distribution. So, essentially we ask: “What’s the probability that a randomly chosen number is prime?” Those familiar with some basic probability theory know the notion of independency in this context, so the question I’m basically interested in here is if the probability to find a prime is independent of the preceding or following numbers.