-4
$\begingroup$

Does there exists a good asymptotic formula for $$A(x) := \prod_{p\leq x}(1-\frac 1p).$$

By using a heuristic argument one can guess: $$A(x) \sim \frac{1}{2\,\mathrm{ln}(x)}.$$

Here is the argument:

$$\frac{x^2}{2\, \mathrm{ln}(x) } \sim \pi(x^2)-\pi(x) = \sum_{p_1<p_2<\dots<p_k\leq x}(-1)^k\left\lfloor \frac{x^2}{p_1p_2\cdots p_k}\right\rfloor \\ \stackrel{??}{\simeq}\sum_{p_1<p_2<\dots<p_k\leq x}(-1)^k\left( \frac{x^2}{p_1p_2\cdots p_k}\right) = x^2 \prod_{p\leq x}(1-\frac{1}{p}).$$

Is this conjectural asymptotic formula true? If so, can this heuristic argument be completed to an exact proof? The later question is more important for me!

Thanks!

$\endgroup$

closed as off-topic by user9072, Lucia, Boris Bukh, Stefan Kohl, Alex Degtyarev Mar 8 '15 at 18:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Community, Lucia, Boris Bukh, Stefan Kohl, Alex Degtyarev
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The constant is wrong. See en.wikipedia.org/wiki/Mertens%27_theorems $\endgroup$ – Emanuele Tron Mar 8 '15 at 8:14
  • $\begingroup$ Nice! But I need to know how to handle the fractional parts in the first sum. $\endgroup$ – Mostafa Mar 8 '15 at 8:24
  • $\begingroup$ You asked "Is this formula true? If so, can this be completed to an exact proof?": the premise about the formula is false, so no point worrying of turning a heuristic for a wrong result into a rigorous proof! Anyway, what bothers me the most is that you use the Prime Number Theorem in the very first asymptotic equivalence and still you worry about not using it in the follow-up. $\endgroup$ – Emanuele Tron Mar 8 '15 at 8:30
  • $\begingroup$ @EmanueleTron The main reason for the question is that one can easily relate the number of twin primes less than n to a similar sum and if we have good estimates on the fractional parts, their density can be related to $\frac 12 \prod_{2< p\leq \sqrt{x}} (1-2p^{-1})$ . $\endgroup$ – Mostafa Mar 8 '15 at 8:36
  • 2
    $\begingroup$ I don't know what you mean by "directly". Mertens' theorem readily gives the sum of the error terms. The idea of sieve theory is to introduce weights that cut down the number of terms in the sieving process. This is also how Zhang proved his spectacular theorem that there are infinitely many prime pairs with distance at most 70 million apart. Since then, 70 million was lowered to 246. This is a pretty good approximation to the twin prime conjecture, although still very far away from it. Number theory is hard. $\endgroup$ – GH from MO Mar 8 '15 at 14:48
7
$\begingroup$

Mertens' Theorem says (page 65, An Introduction to Sieve Methods and Their Applications, Cojocaru and Murty):

$$ \displaystyle\prod_{p< x}\left(1-\frac 1p\right)=\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right), $$

where $\gamma$ is the Euler-Mascheroni constant.

$\endgroup$

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