# Natural Deductions

Expository preaching, non-rigorous proofs

## Thursday, 22 April 2010

### Euler Zeta Function

Definition: $\zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \cdots$ (1).

Theorem: $\zeta(s) = \frac{1}{1 - \frac{1}{2^s}} \times \frac{1}{1 - \frac{1}{3^s}} \times \frac{1}{1 - \frac{1}{5^s}} \times \frac{1}{1 - \frac{1}{7^s}} \times \cdots$ (2).

Demonstration: Expand each $\frac{1}{1 - p^s} = 1 + p^s + p^{2s} + p^{3s} + \cdots$ in (2) and multiply out the infinite product. We are left with the sum of reciprocals of every product of powers of primes (to the s), The Fundamental Theorem of Arithmetic shows that this gives each natural number in one way and concludes the proof. $\square$

Equation (1) shows us that $\zeta(1)$ is the harmonic series which diverges, but (2) could not diverge unless there were an Infinitude of Primes. We pointed the zeta laser cannon at the number 1 and fired it, this instantly killed Euclids theorem about the number of primes.

Dirichlet's theorem states on the arithmetic progression (for $(a,k) = 1$) $a, a + k, a + 2k, a + 3k, \cdots$ there are infinitely many primes. Let us recalibrate our weapon so it can operate on these arithmetic progressions rather than $1, 2, 3, 4, \cdots$.

Definition: $\zeta(s,\chi) = \frac{\chi(1)}{1} + \frac{\chi(2)}{2^s} + \frac{\chi(3)}{3^s} + \frac{\chi(4)}{4^s} + \cdots$ (3).

Lemma: $\zeta(s,\chi) = \frac{1}{1 - \frac{\chi(2)}{2^s}} \times \frac{1}{1 - \frac{\chi(3)}{3^s}} \times \frac{1}{1 - \frac{\chi(5)}{5^s}} \times \frac{1}{1 - \frac{\chi(7)}{7^s}} \times \cdots$ (4).

Theorem: Dirichlet's theorem on arithmetic progressions.

Demonstration: Pick $\chi$ so that it corresponds to the arithmetic progression, only finitely many terms are removed from (4) so divergence again lets us conclude the proof. $\square$

How Euler discovered the zeta function - Keith Devlin
Introduction to Analytic Number Theory, Lecture Notes