GoKawiilWhat does the Riemann zeta function have to do with the distribution of the primes?
One of the oldest theorems in mathematics is that there are infinitely many primes. Euclid (the geometer!) proved this around 300 BCE, and for centuries afterwards the story ended there. There are infinitely many primes -- so what else is there to ask about them?
However, some infinite sets are more `common' than others. For example, even numbers are pretty common, but powers of 2 are pretty rare: between \(1\) and \(1000,\) there are 500 even numbers, but only ten powers of 2. Thus, an interesting follow up question to Euclid's proof is to ask: can we quantify how common prime numbers are?
One way of expressing the idea of that there are more even numbers than powers of 2 is to note that \[\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} + \cdots = \infty,\] but \[\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^5} + \cdots = 1.\]
In other words, the powers of 2 are so far apart that when we replace \(2^n\) with \(\frac{1}{2^n}\) (turning a large number into a small number), we get a finite sum. But the even numbers are so close together that, even though you're adding terms which are getting smaller and smaller, you have so many terms that the final sum is infinity.
Euler was interested in the corresponding question for primes: is \[\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \cdots,\] the sum of the reciprocals of all the primes, infinite or finite?
Enter: the Riemann zeta function
To compute the infinite sum \[\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7}+ \cdots = \sum_{p\text{ prime}} \frac{1}{p},\] Euler started with what is now called the Riemann zeta function (Riemann enters our story much later... but so many things in math are named after Euler, that we named this object after Riemann despite Euler being the first to discover it).
The Riemann zeta function is the infinite sum \[ \zeta(s)=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac{1}{5^s} + \cdots = \sum_{n=1}^{\infty} \frac{1}{n^s}. \] Euler observed that this infinite sum actually factors as a sum involving primes.
For example, consider the product \[ \bigg(1+\frac1{2^s}+\frac1{4^s}+\cdots\bigg)\bigg(1+\frac1{3^s}+\frac1{9^s}+\cdots\bigg)\bigg(1+\frac1{5^s}+\frac1{25^s}+\cdots\bigg).\] When we distribute out the terms in this sum, we will get terms like \[1 \cdot 1 \cdot 1=1,\ \ \frac{1}{2^s} \cdot 1 \cdot 1=\frac1{2^s},\ \ \frac{1}{2^s} \cdot \frac{1}{3^s} \cdot 1 = \frac{1}{6^s},\ \ \frac{1}{4^s} \cdot \frac{1}{9^s} \cdot \frac{1}{5^s} = \frac{1}{180^s},\] as well as many more. Indeed, when we expand this sum out, we're going to get fractions whose denominators are any number which can be built out of 2, 3, and 5. For example, if we wanted to find the term \(1/75^s\) in the infinite product, we would compute the prime factorization \(75=3\cdot25\) and realize that \[ \frac1{75^s}=1\cdot\frac1{3^s}\cdot\frac1{25^s} \] appears in the expansion.
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