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 

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 

From Zeta to J and Back (And Yet Again Back)

 Posted on April 6, 2014  |  5 minutes  |  924 words  |  Markus Shepherd

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. [Read More]
Euler  Functions  Fundamental Theorem  Golden Key  Logarithm  Newton 

Does the Euler Product Converge?

 Posted on October 8, 2013  |  4 minutes  |  697 words  |  Markus Shepherd

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: [Read More]
Euler  Product  Series 

Euler Product Revisited

 Posted on October 6, 2013  |  4 minutes  |  737 words  |  Markus Shepherd

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. [Read More]
Euler  Product  Series 

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