Proving that $\pi(2x) < 2 \pi(x) $

Question

In our analytic number theory class we were given the following problem as homework: prove rigorously that for large $x$ the number of primes in $(1,x]$ exceeds that in $(x,2x]$.

In class we proved the prime number theorem, and then proceeded to prove several results such as $\pi(x) = Li(x) +O(x^\theta \ln x)$ and the explicit formula for $\psi_1(x)$.

This is clearly quite intuitive but I'm lost as to what I can use to prove the result. Any help is greatly appreciated.

@mathstudent_113 that's not really a bound on theta then since $\pi(x)\leq x$. Do you mean $1/2\leq \theta <1$? — Arkady, Apr 11 '16 at 23:58
It can be shown that $\pi(ax) \le a\pi(x)$: http://math.stackexchange.com/a/1664810/48606 — Larry Freeman, Apr 12 '16 at 00:12
@LarryFreeman Nice, but I don't really feel the answer is as good as it could be. — Simply Beautiful Art, Apr 12 '16 at 00:16
@Simple Art, intuitively, I agree. If you have an improved result, I would be very interested in seeing it. — Larry Freeman, Apr 12 '16 at 00:17
@Jake The inequality for a given $\theta$ implies to the non-existence of zeros of the Riemann zeta function with real part between $\theta$ and $1$. In particular, no result of such strength is known for $\theta < 1$ (search for keyword "zero-free region" for the strongest known results), and therefore there is no known proof of $\pi(x) = \text{Li}(x) + O(x^\theta \ln x)$ for any $\theta$ that is non-trivial. — Erick Wong, Apr 12 '16 at 16:58
I answered this question here: https://math.stackexchange.com/questions/1426666/a-particular-cases-of-second-hardy-littlewood-conjecture/1427143#1427143 — Peter Humphries, Apr 13 '16 at 17:41

Larry Freeman · Answer 1 · 2016-04-12T16:15:09.197

6

It can be shown that for $x \ge 11$, $\pi(2x) < 2\pi(x)$

For $x=1$, it is not true since $2\pi(1) = 0$ but $\pi(2)=1$
For $x \in \{2,4,10\}$, $2\pi(x)=\pi(2x)$

Pierre Dusart showed that for $x \ge 60184$

$$\frac{x}{\ln{x}-1} < \pi(x) < \frac{x}{\ln {x}-1.1}$$

For the proof, see Theorem 6.9 here.

For $a \ge 2$, $\ln ax - 1.1 = \ln a + \ln x - 1.1 \ge \ln 2 - 1.1 + \ln x > \ln x - 0.41 > \ln x - 1$

So, it follows that for $x \ge 60184, a \ge 2$:

$$\pi(ax) < \frac{ax}{\log{ax}-1.1} < \frac{ax}{\log x - 1} < a\pi(x)$$

By brute force, it can be shown that in all cases where $x < 60184$, $\pi(2x) < 2\pi(x)$.

Here is java code that I used to verify it (adding an image only):

edited Apr 12 '16 at 16:15

answered Apr 12 '16 at 01:46

Larry Freeman

10,433

This is very nice, but then I would have to know how Dusart got his result. Anyway we only need to show for large $x$ so there's no need to verify it for $x<60184$ I think. – Apr 12 '16 at 02:15
2

Here is a link to the Dusart article: http://arxiv.org/PS_cache/arxiv/pdf/1002/1002.0442v1.pdf – Larry Freeman Apr 12 '16 at 16:09

Simply Beautiful Art · Answer 2 · 2016-04-12T00:09:15.603

While this isn't my main area of study, I do think we are stuck with proving $Li(2x)<2Li(x)$ (ignoring $O(x^{\theta}\ln(x))$) for large $x$.

The definition of the offset logarithmic integral is

$$Li(x)=\int_2^x\frac{dt}{\ln(t)}$$

We want to take this and prove that $Li(2x)<2Li(x)$.

Since $\frac{d}{dx}Li(x)=\frac1{\ln(x)}>0$, we see that the offset logarithmic integral has a positive derivative. Since $\frac{d}{dx}\frac1{\ln(x)}=\frac{-1}{x(\ln(x))^2}<0$, we know this function is concave.

This implies that for large enough $x$, $Li(2x)<2Li(x)$ since $Li(x)$ is concave. I'm not quite sure how to do a formal proof or whatever, but I imagine we can go along those lines.

If $Li(2x)<2Li(x)$, I would think

$$\frac{d}{dx}Li(2x)<\frac{d}{dx}2Li(x)$$

$$\frac2{\ln(2x)}<\frac2{\ln(x)}$$

And the last line is much easier to prove.

Just as a question, is there a more formal way to do this? I really just took some thoughts about how the graphs should look and tried to put it into math, but I'm not even sure if that's really the right way to go... — Simply Beautiful Art, Apr 12 '16 at 00:11

Proving that $\pi(2x) < 2 \pi(x) $

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