2

The PNT states that $$\pi(x) \sim \cfrac{x}{\log x} \qquad (x\rightarrow\infty).$$ Let's define a function $M$ to be $$M(x) := \cfrac{\pi(x)-x/\log x}{\lvert\pi(x)-x/\log x\rvert},$$ which returns either $1$ or $-1$, depending on what function $\pi(x)$ or $x/\log x$ is bigger.

enter image description here

In this picture we can clearly see that $\pi(x)$ seems to be much larger than $x/\log x$, where $x$ is approximately greater than $50$. The graph gets bigger and bigger but I think I have heard of a number, let's call it $\Xi$ that satisfies $M(\Xi) = -1$. And not only that - I think some author stated that for $x\in\mathbb{R}, M(x)$ is infinitely many times $-1$, $1$.

1) What is the number $\xi$ called?

2) How can we proof that $M(x)$ changes sign infinitely many often?

Daniel
  • 432
  • 2
    That limit is not $\frac{x}{\log(x)}$. Do you mean $\lim_{x\to \infty }\pi(x)\log(x)/x=1$? – Mark Viola Jul 26 '20 at 17:52
  • I just multiplied both sides by x/logx – Daniel Jul 26 '20 at 17:57
  • 2
    @Daniel If $f(x)$ is a function of $x$, it doesn't make sense to say that its limit as $x \to +\infty$ is some function of $x$. The limit should be a constant. – Anonymous Jul 26 '20 at 17:58
  • 2
    You cannot remove the expression from the limit. After all, $x$ is a dummy variable of the limit. – Mark Viola Jul 26 '20 at 17:58
  • To answer your question, Wikipedia says that we have $\pi(x) > (1 + 1/\log x)(x/\log x)$ for $x \geq 599$. So the graph of $\pi(x)$ is above that of $x/\log x$ at least for all $x \geq 599$. – Anonymous Jul 26 '20 at 18:00
  • 1
    The way you've rewritten it still doesn't make sense. Both sides are $+\infty$. This is almost trivial, not a theorem that was only proved at the end of the 19th century. You just need to know there are infinitely many prime numbers. – Anonymous Jul 26 '20 at 18:05
  • 1
    $\pi (x)> \frac{x}{\ln x}$ for $x \geq 17$ is well known bound. – Sil Jul 26 '20 at 18:10
  • 4
    @Sil is right, a more interesting thing is to look at the log integral function $li(x)$. Littlewood proved that $\pi(x)-li(x)$ changes sign infinitely often. Check out Skewes numbers. – tassle Jul 26 '20 at 18:28
  • There is a cross-over point with $x\ge10$, namely $\pi(x)=x/\ln x=6$ for $x\approx16.9988873523$. (Hence the requirement $x\ge17$ in the result cited by @Sil.) – Barry Cipra Jul 26 '20 at 18:48
  • @BarryCipra, Yes, and also just before the prime $13$, at $12.7132067888676321\ldots$. (The crossing before prime $11$ occurs already at $8.61316945644139859\ldots$, so below $10$.) – Jeppe Stig Nielsen Jul 26 '20 at 19:38
  • @JeppeStigNielsen, very nice! – Barry Cipra Jul 26 '20 at 19:46

1 Answers1

2

(In this post, I will commit the usual abuse of notation/terminology and use $f(x)$ for both the function $f$ and the value of that function at $x$.)

For $x\ge 17$ the graph of $\pi(x)$ is always above the graph of $\frac{x}{\log x}$. See comments to the question for details.

Last last integer $x$ for which $\pi(x) < \frac{x}{\log x}$, is $x=10$. The last real numbers for which this inequality holds, are the open interval $16.99888735\ldots < x < 17$. If you plot $\pi(x)$ with horizontal segments only (no vertical segments at the jumps), the last intersection between the two graphs is therefore $16.99888735\ldots$.

As mentioned also in the comments to the question, if you consider $\mathrm{Li}(x)=\int_2^x \frac1t \mathrm{d}t$ instead of $\frac{x}{\log x}$, then it is true that the graphs of $\pi(x)$ and $\mathrm{Li}(x)$ cross infinitely often (for history, read about Skewes's number).

Jeppe Stig Nielsen
  • 5,109
  • 21
  • 29