The Prime Bet

 Posted on May 31, 2017  |  9 minutes  |  1723 words  |  Markus Shepherd

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? [Read More]
Apéry  Basel problem  Bet  constant  Coprime  Divisor  Euler  gcd  pi  Prime factor  Probability  Pub  Python  Random  Riemann  Wiles  Zeta function  Zhang 

If One at a Time is too Difficult, Try All at Once!

 Posted on December 27, 2014  |  5 minutes  |  926 words  |  Markus Shepherd

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. [Read More]
Analysis  Asymptotics  Combinatorics  Coursera  Derbyshire  Dirichlet  du Sautoy  Number theory  Riemann 

Applying the Explicit Formula

 Posted on November 23, 2014  |  4 minutes  |  649 words  |  Markus Shepherd

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. [Read More]
Approximation  Prime counting function  Prime number theorem  Primes  Riemann  Zeros  Zeta function  Zeta zeros 

How NOT to Earn a Million Dollars

 Posted on November 2, 2014  |  4 minutes  |  771 words  |  Markus Shepherd

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. [Read More]
Bombieri  CMI  Gauss  Littlewood  Logarithm  Mertens  Millennium  Million  Numberphile  Riemann  YouTube  Zagier 

Integral Madness

 Posted on April 13, 2014  |  8 minutes  |  1578 words  |  Markus Shepherd

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 \(\zeta(s)\): [Read More]
Calculus  Euler  Gauss  Legendre  Riemann  Zeros 

In the Beginning, There Was... Euler's Formula!

 Posted on September 29, 2013  |  1 minutes  |  206 words  |  Markus Shepherd

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. [Read More]
Euler  Primes  Product  Riemann  Series