Section 25.1 Taking the PNT Further
Recall Gauss’ approximating function for \(\pi(x)\text{,}\) the logarithmic integral function (Definition 21.2.2). Let’s remind ourselves just how well it performs.
As we can see, it wasn’t too bad of an estimate. But, as mathematicians, we hope we could get a little closer. At the end of Subsection 21.3.1 we tried (among several other things) the fairly weird amended function
\begin{equation*}
Li(x)-\frac{1}{2}Li(\sqrt{x})\text{.}
\end{equation*}
This was indeed a better approximation (in red in the graphic above). You can try it interactively below.
This second estimate seems better. 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 \(\mu(2)\text{,}\) not from alternating powers of \(-1\text{.}\) You may try it interactively below:
From anything one can see in the preceding interact, this set of approximations doesn’t seem to add any of accuracy beyond \(k=3\text{.}\) In fact, at \(x=1000000\text{,}\) 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 \(\sum_{j=1}^\infty\frac{\mu(j)}{j}Li(x^{1/j})\text{.}\) More importantly, both of these are clearly not integers, so this type of analysis alone will not yield a computable, exact formula for \(\pi(x)\text{.}\) So here are some questions we might raise.
- Where does the Moebius \(\mu\) in that approximation come from anyway?
- Since this wasn’t enough, what else is involved in the error\begin{equation*} \left|\pi(x)-Li(x)\right|\text{?} \end{equation*}
- Are there connections with things other than just \(\pi(x)\text{?}\)
- What does this have to do with winning a million dollars?
