Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here is a bad heuristic argument for the prime number theorem. Let n be a positive integer and assume that PNT holds up to n. Then n itself is prime if and only if for each prime p<n the event p|n fails to hold. Assuming that all these events are independent, the probability that n is prime should be around $\prod_{p&lt;n}(1-1/p)$. Minus the logarithm of this is approximately $\sum_{p&lt;n}1/p$. Since the rth prime is about $r\log r$ (by PNT), this is about $\sum_{r&lt;n/\log n}1/r\log r$, which is about $\log\log n$. Therefore, the probability that $n$ is prime is about $\exp(-\log\log n)$, which is $1/\log n$.

Here are a few criticisms of the above argument. The most glaring problem is that for the approximations to be valid, we need our estimates to be correct up to o(1), and they are not. For instance, it is known that $\sum_{p&lt;n}1/p$ is not $\log\log n+o(1)$, but rather $\log\log n+M$, where M is the Meissel-Mertens constant. We can break this failure down into two subfailures. The first is that minus the logarithm of $\prod_{p&lt;n}(1-1/p)$ differs from $\sum_{p&lt;n}1/p$ by a non-zero constant (plus o(1)). The second is that minus the logarithm of $\prod_{p&lt;n}(1-1/p)$ differs from $\log\log n$ by $\gamma+o(1)$, where $\gamma$ is the Euler-Mascheroni constant.

The second problem is more devastating, since it shows that the independence assumption is seriously flawed. (Everything I have said, by the way, is a well known and often pointed out observation.) My question is whether, despite all these problems, some kind of heuristic argument like this can be salvaged. For instance, it's clear that if p and q are two primes that are fairly large and fairly close, then there will be a repulsion between the events p|n and q|n. Can we say heuristically what the effects of these repulsions should be, and thereby understand where the $\gamma$ comes in?

Just to be clear, I am looking for a simple and non-rigorous argument that does not use the zeta function (except perhaps making use of the product formula in a very elementary way, but I'd rather avoid it altogether) that predicts that if PNT holds up to n then the probability that n is prime should be around $1/\log n$. I'm asking the question because I'm pretty sure the answer will be known, and pretty standard, to many people. It just isn't to me.

share|improve this question
1  
Possibly related, T. Tao discussed this on his blog a while ago. terrytao.wordpress.com/2009/09/24/… –  Gjergji Zaimi Feb 26 '10 at 11:18
    
That may indeed be exactly what I am looking for. I'll have to digest it carefully to see. –  gowers Feb 26 '10 at 12:35
add comment

2 Answers

You might look at Courant and Robbins in the section at the end, "The Prime Number Theorem Obtained by Statistical Methods".

Also, there is an article by Montgomery and Wagon, in which they mention the "Mertens Paradox" and extend the argument of Courant and Robbins.

share|improve this answer
add comment

I'm not an expert in this area, but this may be a start.

Rather than $\prod_{p\lt n}$, you can use $\prod_{p\le \sqrt n}$.

$\log\log \sqrt n + \gamma \lt \log\log \sqrt n +\log 2 = \log\log n $

That gets you a little closer, since now you are off by $\log 2 - \gamma \approx 0.116$.

The heuristic probability that $n$ is prime is not

$$\prod_{p\lt n} (1-Pr(p|n))$$

It is the product of probabilities

$$\prod_{p\lt n} (1-Pr(p|n \text{ given no smaller prime divides } n))$$

For $p$ small, the term you get may be close to $(1-1/p)$, but I that's not the case for $p$ large.

For $\sqrt n \lt p$, the term corresponding to $p$ in the product is just $1$.

For $\sqrt[3]n \lt p \le \sqrt n$, if $p$ is the smallest prime dividing $n$, then $n/p$ must be prime, too. Perhaps that means that by strong induction, we should discount these terms by the probability $n/p$ is prime, about $1/\log \frac np$, so that those terms in the product are $(1-1/(p \log \frac np))$.

It looks like you get some sums/integrals if you try to extend this to more terms, and I don't know whether you can expect to get the desired accuracy at the end.

share|improve this answer
    
Gjergji, I think you dropped a factor of x here. –  Michael Lugo Feb 26 '10 at 14:33
    
Yes! And it's already mentioned in the answer, I don't know why I wrote it.. –  Gjergji Zaimi Feb 26 '10 at 14:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.