https://riemannhypothesis.blog/tags/euler/ Recent content in Euler on Understanding the Riemann Hypothesis Hugo -- gohugo.io en markus@recommend.games (Markus Shepherd) markus@recommend.games (Markus Shepherd) Wed, 31 May 2017 05:41:25 +0000 https://riemannhypothesis.blog/2017/05/the-prime-bet/ Wed, 31 May 2017 05:41:25 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/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? 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/2014/04/from-zeta-to-j-and-back-and-yet-again-back/ Sun, 06 Apr 2014 14:13:42 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/2014/04/from-zeta-to-j-and-back-and-yet-again-back/ We know a lot about the \(\zeta\) and \(\xi\)-functions, we’ve learnt all about the different prime counting functions, most notably \(J(x)\), so it’s high time we found a connection between the two. Probably not too surprisingly, the crucial link is our good friend, the Euler product \[ \zeta(s)=\prod_{p}(1-p^{-s})^{-1}. \] What we want to develop now is a version of this product that will suit us to find a formula that magically can count primes. https://riemannhypothesis.blog/2013/10/does-the-euler-product-converge/ Tue, 08 Oct 2013 18:36:28 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/2013/10/does-the-euler-product-converge/ Usually, I don’t care too much about convergence as a general overview of the argument is what I aim at here, and otherwise I’ll just trust that things “behave well”. But some words concerning convergence won’t harm. It’s a well known fact that the harmonic series (which we shortly touched in the previous post) \(\sum1/n\) diverges. I think the best (though not easiest) proof of this to compare it with the corresponding integral: https://riemannhypothesis.blog/2013/10/euler-product-revisited/ Sun, 06 Oct 2013 22:24:34 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/2013/10/euler-product-revisited/ From the previous post we know that the harmless looking series \(\sum n^{-s}\) can be extended to the product \(\prod (1-p^{-s})^{-1}\). At first sight, this does not seem terribly helpful, and it actually makes the rather easy series more complicated. So what’s the big deal? It’s what has actually been suppressed in the above notation: The sequences we run through. The series runs over all natural numbers (or positive integers, if you prefer), the product runs through all prime numbers. https://riemannhypothesis.blog/2013/09/in-the-beginning-there-was-eulers-formula/ Sun, 29 Sep 2013 17:39:48 +0000markus@recommend.games (Markus Shepherd) https://riemannhypothesis.blog/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.