Section25.4Connecting to the Primes

The rest of this final chapter is devoted to seeing why this might be related to the distribution of prime numbers, such as Von Koch's result showing the RH is equivalent to a bound on the error \(\left|\pi(x)-Li(x)\right|\).

We'll pursue this in three steps.

Our first step is to see the connection between \(\pi(x)\) and \(\mu(n)\).
Then we'll see the connection between these and \(\zeta\).
Finally, we'll see how the zeros of \(\zeta\) come into play.

Subsection25.4.1Connecting to Moebius

Let's begin by defining a new function.

The function above is one Riemann called \(f\), but which we (following Edwards and Derbyshire) will call \(J(x)\). It is very similar to \(\pi(x)\) in its definition, so it's not surprising that it looks similar.

Definition25.4.1

We define \[J(x)=\pi(x)+\frac{1}{2}\pi(\sqrt{x})+\frac{1}{3}\pi(\sqrt[3]{x})+\frac{1}{4}\pi(\sqrt[4]{x})+\cdots=\sum_{k=1}^\infty \frac{1}{n}\pi\left(x^{1/n}\right)\]

This looks like it's infinite, but it's not actually infinite. For instance, we can see on the graph that \[J(20)=\pi(20)+\frac{1}{2}\pi(\sqrt{20})+\frac{1}{3}\pi(\sqrt[3]{20})+\frac{1}{4}\pi(\sqrt[4]{20})=8+\frac{2}{2}+\frac{1}{3}+\frac{1}{4}=9\frac{7}{12}\] because \(\sqrt[5]{20}\approx 1.8\) and \(\pi(\sqrt[5]{20})\approx\pi(1.8)=0\), so the sum ends there.

Okay, so we have this new function. Yet another arithmetic function. So what?

Ah, but what have we been doing to all our arithmetic functions to see what they can do, to get formulas for them? We've been Moebius inverting them, naturally! In this case, Moebius inversion could be really great, since it would give us something about the thing being added; namely, \(\pi(x)\).

The only thing standing in our way is that \[J(x)=\sum_{k=1}^\infty \frac{1}{n}\pi\left(x^{1/n}\right)\] is not a sum over divisors. But it turns out that, just like when we took the limits of the sum over divisors \(\sum_{d\mid n}\frac{1}{d}\), we got \(\sum_{n=1}^\infty \frac{1}{n}\), we can do the same thing with Moebius inversion.

Fact25.4.2

If \(\sum_{n=1}^\infty f(x/n)\) and \(\sum_{n=1}^\infty g(x/n)\) both converge absolutely, then \[g(x)=\sum_{n=1}^\infty f(x/n)\Longleftrightarrow f(x)=\sum_{n=1}^\infty \mu(n)g(x/n)\; .\]

For us, we've just defined \(g=J\) with \(f(x/n)=\frac{1}{n}\pi\left(x^{1/n}\right)\), and so we get the very important result that \[\pi(x)=\sum_{n=1}^\infty \mu(n)\frac{J(x^{1/n})}{n}=J(x)-\frac{1}{2}J(\sqrt{x})-\frac{1}{3}J(\sqrt[3]{x})-\frac{1}{5}J(\sqrt[5]{x})+\frac{1}{6}J(\sqrt[6]{x})+\cdots\]

Remark25.4.3

If that last use of Moebius inversion looked a little sketchy, it does to me too, but I cannot find a single source where it's complained about that \(f(x/n)=\frac{1}{n}\pi\left(x^{1/n}\right)\) is really a function of \(x\) and \(n\), not just \(x/n\). In any case, the result is correct, via a somewhat different explanation of this version of inversion in a footnote in Edwards.