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Aymen Hafeez

I’ve written about the Riemann zeta function in other posts and touched briefly on the product formula representation. Here, I wanted to go through a derivation of the product formula. Like most maths related subjects I write about it won’t be particularly mathematically rigorous, but will aim more to convey the intuition behind the product formula representation.

The Riemann zeta function is defined as follows for complex numbers ss with real part greater than 11:

ζ(s)=∑n=1∞1ns=1+12s+13s+14s+15s+⋯ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} = 1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + \frac{1}{5^{s}} + \cdots

The infinite series converges when the real part of ss exceeds 11. The Euler product formula expresses ζ(s)\zeta(s) as a product of terms involving prime numbers:

ζ(s)=∏p prime11−p−s\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}

To derive this we start with the infinite series representation and writing each term as its prime factorisation:

12s=(12s)⋅113s=(13s)⋅114s=(12s)⋅(12s)15s=(15s)⋅116s=(13s)⋅(12s), \begin{aligned} \frac{1}{2^{s}} &= \left( \frac{1}{2^{s}} \right) \cdot 1 \\ \frac{1}{3^{s}} &= \left( \frac{1}{3^{s}} \right) \cdot 1 \\ \frac{1}{4^{s}} &= \left( \frac{1}{2^{s}} \right) \cdot \left( \frac{1}{2^{s}} \right) \\ \frac{1}{5^{s}} &= \left( \frac{1}{5^{s}} \right) \cdot 1 \\ \frac{1}{6^{s}} &= \left( \frac{1}{3^{s}} \right) \cdot \left( \frac{1}{2^{s}} \right), \\ \end{aligned}

and so on. This is just using the fundamental theorem of arithmetic. Rewriting the series with these terms gives us

ζ(s)=(1⋅1)+(12s⋅1)+(13s⋅1)+(12s⋅12s)+(15s⋅1)+(13s⋅12s)+⋯ \zeta(s) = (1 \cdot 1) + \left( \frac{1}{2^{s}} \cdot 1 \right) + \left( \frac{1}{3^{s}} \cdot 1 \right) + \left( \frac{1}{2^{s}} \cdot \frac{1}{2^{s}} \right) + \left( \frac{1}{5^{s}} \cdot 1 \right)+ \left( \frac{1}{3^{s}} \cdot \frac{1}{2^{s}} \right) + \cdots

Next, we can group the terms by the prime numbers involved:

ζ(s)=(1+12s+122s+123s+⋯ )⋅(1+13s+132s+133s+⋯ )⋅(1+15s+152s+153s+⋯ )+⋯ \zeta(s) = \left( 1 + \frac{1}{2^{s}} + \frac{1}{2^{2s}} + \frac{1}{2^{3s}} + \cdots \right) \cdot \left( 1 + \frac{1}{3^{s}} + \frac{1}{3^{2s}} + \frac{1}{3^{3s}} + \cdots \right) \cdot \left( 1 + \frac{1}{5^{s}} + \frac{1}{5^{2s}} + \frac{1}{5^{3s}} + \cdots \right) + \cdots

Here we see that each term being multiplied is a geometric series. Recall that in general a geometric series can be written as

∑n=0∞arn=a1−r \sum_{n=0}^{\infty} a r^{n} = \frac{a}{1 - r}

And so, each factor of the above product can be simplified:

1+12s+122s+123s+⋯=11−12s1+13s+132s+133s+⋯=11−13s1+15s+152s+153s+⋯=11−15s\begin{aligned} 1 + \frac{1}{2^{s}} + \frac{1}{2^{2s}} + \frac{1}{2^{3s}} + \cdots &= \frac{1}{1 - \frac{1}{2^{s}}} \\ 1 + \frac{1}{3^{s}} + \frac{1}{3^{2s}} + \frac{1}{3^{3s}} + \cdots &= \frac{1}{1 - \frac{1}{3^{s}}} \\ 1 + \frac{1}{5^{s}} + \frac{1}{5^{2s}} + \frac{1}{5^{3s}} + \cdots &= \frac{1}{1 - \frac{1}{5^{s}}} \\ \end{aligned}

Substituting the simplified terms back into the product we had above gets us the product formula.

ζ(s)=(11−12s)⋅(11−13s)⋅(11−14s)⋅(11−15s)⋯\begin{aligned} \zeta(s) = \left( \frac{1}{1 - \frac{1}{2^{s}}} \right) \cdot \left( \frac{1}{1 - \frac{1}{3^{s}}} \right) \cdot \left( \frac{1}{1 - \frac{1}{4^{s}}} \right) \cdot \left( \frac{1}{1 - \frac{1}{5^{s}}} \right) \cdots \end{aligned} ζ(s)=∏p prime11−p−s \zeta(s) = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}

We can also understood this more simply starting with the Fundamental Theorem of Arithmetic, which states that any integer n≥1n \geq 1 can be expressed as

n=2ϵ23ϵ35ϵ5⋯ , n = 2^{\epsilon_2}3^{\epsilon_3}5^{\epsilon_5} \cdots,

where ϵ2,ϵ3,ϵ5⋯\epsilon_2,\epsilon_3,\epsilon_5 \cdots are non-negative integers. Raising both sides to the power of −s-s,

n−s=2−ϵ2s3−ϵ3s5−ϵ5s⋯ , n^{-s} = 2^{-\epsilon_2s}3^{-\epsilon_3s}5^{-\epsilon_5s} \cdots,

and taking the sum,

∑n−s=∑2−ϵ2s3−ϵ3s5−ϵ5s⋯ , \sum n^{-s} = \sum 2^{-\epsilon_2s}3^{-\epsilon_3s}5^{-\epsilon_5s} \cdots, ∑n−s=(1+2−s+2−2s+⋯ )(1+3−s+3−2s+⋯ )(1+5−s+5−2s+⋯ ) \sum n^{-s} = (1 + 2^{-s} + 2^{-2s} + \cdots)(1 + 3^{-s} + 3^{-2s} + \cdots)(1 + 5^{-s} + 5^{-2s} + \cdots)

And so, we see that the terms on the right hand side, for each prime pp, equate to

1+p−s+p−2s+⋯=(1−p−s)−1,1 + p^{-s} + p^{-2s} + \cdots = (1 - p^{-s})^{-1},

giving us

∑n−s=∏p prime11−p−s \sum n^{-s} = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}

This is just such a cool result. And again, while this derivation is by no means mathematically rigorous I think it provides a good high-level overview for the intuition behind the Euler product for the Riemann zeta function.