## More symmetry and Another Product

We’ve seen that $$\zeta(s)$$ satisfies the functional equation $\zeta(1-s)=2^{1-s}\pi^{-s}\cos(\pi s/2)\Pi(s-1)\zeta(s).$ (Well, it still needs to be proved, but let’s just assume it’s correct for now.) The goal of this post is an even more symmetrical form that will yield the function $$\xi(s)$$ which we can develop into an incredibly useful product expression. On our wish list for $$\xi(s)$$ we find three items: It’s an entire function, i.e., a function that’s holomorphic everywhere in $$\mathbb{C}$$ without any poles. [Read More]

## 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: [Read More]

## 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. [Read More]

## In the Beginning, There Was... Euler's 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. [Read More]