NTIC Taking the PNT Further

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.

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Figure 25.1.1. Prime counting function along with Gauss and another

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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

$\begin{equation*} Li(x)-\frac{1}{2}Li(\sqrt{x})\text{.} \end{equation*}$

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This was indeed a better approximation (in red in the graphic above). You can try it interactively below.

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@interact
def _(n=(1000,(1000,10^6))):
   P = plot(prime_pi,n-1000,n, color='black', legend_label=r'$\pi(x)$')
   P += plot(Li,n-1000,n, color='green', legend_label='$Li(x)$')
   P += plot(lambda x: Li(x) - .5*Li(sqrt(x)), n-1000,n, color='red', legend_label=r'$Li(x)-\frac{1}{2}Li(\sqrt{x})$')
   show(P)

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This second estimate seems better. One might think one could keep adding and subtracting

$\begin{equation*} \frac{1}{n}Li(x^{1/n}) \end{equation*}$

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to get even closer, with this start to the pattern.

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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:

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@interact
def _(n=(1000,(1000,10^6)),k=(3,[1..10])):
   P = plot(prime_pi,n-1000,n, color='black', legend_label=r'$\pi(x)$')
   P += plot(Li,n-1000,n, color='green', legend_label='$Li(x)$')
   F = lambda x: sum([Li(x^(1/j))*moebius(j)/j for j in [1..k]])
   P += plot(lambda x: Li(x) - .5*Li(sqrt(x)),n-1000,n, color='red', legend_label=r'$Li(x)-\frac{1}{2}Li(\sqrt{x})$')
   P += plot(F,n-1000,n, color='blue', legend_label=r'$\sum_{j=1}^{%s} \frac{\mu(j)}{j} Li(x^{1/j})$'%k)
   show(P)

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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?