Recall Gauss' approximating function for \(\pi(x)\), the logarithmic integral function (Definition 21.2.2). Let's remind ourselves how it did.

It wasn't too bad of an estimate. But, as mathematicians, we hope we could get a little closer. At that time, among several other things, we tried this fairly weird amended function: \begin{equation*}Li(x)-\frac{1}{2}Li(\sqrt{x})\; .\end{equation*} And this was indeed a better approximation.

So one might think one could keep adding and subtracting \begin{equation*}\frac{1}{n}Li(x^{1/n})\end{equation*} to get even closer, with this start to the pattern.

As it turns out, that is not quite the right pattern. In fact, the minus sign comes from \(\mu(2)\), not from \((-1)^{2+1}\), as usually is the case in series which begin by alternating!

This set of approximations doesn't really add a lot of accuracy beyond \(k=3\). In fact, at \(x=1000000\), taking the approximation with the sum \(\sum_{j=1}^3\frac{\mu(j)}{j}Li(x^{1/j})\) is essentially the same as going all the way to infinity in \(\sum_{j=1}^\infty\frac{\mu(j)}{j}Li(x^{1/j})\), and both of these are clearly not integers. This alone will not yield a computable, exact formula for \(\pi(x)\). So here are some questions we might raise.

• So where does the Moebius \(\mu\) in that approximation come from anyway?

• What else is there to the error \begin{equation*}\left|\pi(x)-Li(x)\right|\end{equation*} since this one wasn't enough?

• Are there connections with things other than just \(\pi(x)\)?

• What does this have to do with winning a million dollars?