Riemann’s Zeta and Random Walks

The Riemann Zeta function is a function of a complex variable $s$ that analytically continues the sum of the Dirichlet series

$\zeta(s) = \sum\limits_{n=1}^{\infty} \frac{1}{n^s}$

In 1737, the revered Swiss mathematician, Leonhard Euler, discovered the fundamental connection between the Zeta function and prime numbers. The proof is as follows, given

$\zeta (s)=\sum\limits _{n=1}^{\infty }{\frac {1}{n^{s}}} = 1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{5^{s}}}+\ldots$ $(1.1)$

we will multiply both sides of the equation by ${\frac {1}{2^{s}}}$

${\frac {1}{2^{s}}}\zeta (s) = {\frac {1}{2^{s}}}+{\frac {1}{4^{s}}}+{\frac {1}{6^{s}}}+{\frac {1}{8^{s}}}+{\frac {1}{10^{s}}}+\ldots$ $(1.2)$

then subtract $(1.2)$ from $(1.1)$ to remove all factors of 2

$\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)=1+{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+{\frac {1}{9^{s}}}+{\frac {1}{11^{s}}}+{\frac {1}{13^{s}}}+\ldots$

Repeating this for factors of 3

$\frac {1}{3^{s}}\left(1-{\frac {1}{2^{s}}}\right)\zeta (s) = {\frac {1}{3^{s}}}+{\frac {1}{9^{s}}}+{\frac {1}{15^{s}}}+{\frac {1}{21^{s}}}+{\frac {1}{27^{s}}}+{\frac {1}{33^{s}}}+\ldots$

$\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{2^{s}}}\right)\zeta (s)=1+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+{\frac {1}{11^{s}}}+{\frac {1}{13^{s}}}+{\frac {1}{17^{s}}}+\ldots$

and so on. If we continue this process to infinity for $\frac{1}{p^s}$ , where $p$ is a prime, the expression reduces to

$\ldots \left(1-{\frac {1}{11^{s}}}\right)\left(1-{\frac {1}{7^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{2^{s}}}\right)\zeta (s) = 1$

which can be rearranged to produce

$\zeta(s) = \frac{1}{\left(1-{\frac {1}{2^{s}}}\right)\left(1-{\frac {1}{3^{s}}}\right)\left(1-{\frac {1}{5^{s}}}\right)\left(1-{\frac {1}{7^{s}}}\right)\left(1-{\frac {1}{11^{s}}}\right)+\ldots } = \prod\limits_{\forall p\:\in\:\mathrm{P}}^{\:} \frac{1}{1-p^{-s}}$ $(1.3)$

So why are we doing all of this? Well, it turns out the reciprocal of the Zeta function has some remarkable properties of its own.

Consider the inverse based on Euler’s prime product formula

$\frac{1}{\zeta(s)} = \prod\limits_{\forall p\:\in\:\mathrm{P}}^{\:} \left(1-p^{-s}\right) = \left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\ldots$

expanding this expression we have

$\frac{1}{\zeta(s)} = 1 + \sum_{n \text{ prime}} \left({\frac{-1} {n^s} }\right) + \sum_{n \mathop = p_1 p_2} \left({ \frac{-1}{p_1^s} \frac{-1} {p_2^s} }\right) + \sum_{n \mathop = p_1 p_2 p_3} \left({ \frac {-1} {p_1^s} \frac {-1} {p_2^s} \frac{-1}{p_3^s} }\right) + \ldots$

which can be rearranged to demonstrate

$\frac{1}{\zeta(s)} = 1 + \frac{-1}{2^s} + \frac{-1}{3^s} + \frac{0}{4^s} + \frac{-1}{5^s} + \frac{1}{6^s} + \frac{-1}{7^s} + \frac{0}{8^s} + \frac{0}{9^s} + \frac{1}{10^s} + \frac{-1}{11^s} + \ldots$ $(1.4)$

On close examination one can see that the numerator in $(1.4)$ actually corresponds to the values of the Möbius function $\mu(x)$ . Indeed, the reciprocal of the Zeta function can be formally defined by