NTIC Connecting to the Primes

Section 25.4 Connecting to the Primes

The last few sections of this final chapter are devoted to seeing why the Riemann Hypothesis might be related to the distribution of prime numbers. In this, we will loosely follow the very interesting exposition of Prime Obsession by John Derbyshire, [E.4.1].

For motivation, think of Von Koch's result Theorem 25.2.4 connecting the RH to a bound on the error between \(\pi(x)\) and the log integral. Our goal is more detailed, however.

We'll pursue this connection in three steps.

Our first step is to see the connection between \(\pi(x)\) and \(\mu(n)\) (25.4.1).
Then we'll see the connection between these and \(\zeta\) (25.5).
Finally, we'll see how the zeros of \(\zeta\) come into play (25.6).

Subsection 25.4.1 Connecting to Moebius

Let's begin by defining a new function. Here is its graph.

Figure 25.4.1. The J function on two ranges

If you are reading this online, evaluate the following cell to define it, as well as to plot a bit of it in any range you prefer.

Riemann called this function \(f\text{.}\) Following [E.4.4] and [E.4.1], we will call it \(J(x)\text{.}\) It is very similar to \(\pi(x)\) in its definition, so it's not surprising that it looks similar.

Definition 25.4.2.

We define

\begin{equation*} 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_{n=1}^\infty \frac{1}{n}\pi\left(x^{1/n}\right) \end{equation*}

This looks like it's an infinite sum, but for any given \(x\text{,}\) it is finite. For instance, let's calculate \(J(20)\text{:}\)

\begin{equation*} 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} \end{equation*}

because \(\sqrt[5]{20}\approx 1.8\) and \(\pi(\sqrt[5]{20})\approx\pi(1.8)=0\text{,}\) so the sum ends there, and we can see that on the graph.

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! (Recall Section 23.2.) In this case, Moebius inversion could be really great, since it would give us information about the thing being added, which is the all-important \(\pi(x)\text{.}\)

The only thing standing in our way is that

\begin{equation*} J(x)=\sum_{n=1}^\infty \frac{1}{n}\pi\left(x^{1/n}\right) \end{equation*}

is not a sum over divisors. But it turns out that, just like when we took the limits of the sum over divisors \(\sum_{d\mid n}\frac{1}{d}\text{,}\) we got \(\sum_{n=1}^\infty \frac{1}{n}\text{,}\) we can do the same thing with Moebius inversion.

Fact 25.4.3.

If \(\sum_{n=1}^\infty f(x/n)\) and \(\sum_{n=1}^\infty g(x/n)\) both converge absolutely, then

\begin{equation*} g(x)=\sum_{n=1}^\infty f(x/n)\Longleftrightarrow f(x)=\sum_{n=1}^\infty \mu(n)g(x/n)\text{.} \end{equation*}

We can use this by setting \(g=J\) with \(f(x/n)=\frac{1}{n}\pi\left(x^{1/n}\right)\text{.}\) Applying this, we achieve a very important result writing \(\pi(x)\) in terms of \(J\text{:}\)

\begin{equation*} \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\text{.} \end{equation*}

Remark 25.4.4.

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 \(n\text{,}\) not just \(x/n\text{.}\) In any case, the result is correct, via a somewhat different explanation of this version of inversion in a footnote in Edwards' discussion of this matter in [E.4.4].