Timeline for Density of the "multiplicative odd numbers"
Current License: CC BY-SA 3.0
7 events
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Dec 14, 2012 at 0:41 | comment | added | George Lowther | @Greg: So, you have $L(x)=\sum_d M(x/d^2)$ giving $\limsup_{x\to\infty}\lvert L(x)/x\rvert\le\limsup_{x\to\infty}\lvert M(x)/x\rvert\sum_d d^{-2}$ (=0). Similarly, $M(x)=\sum_d \mu(d)L(x/d^2)$, so you can go in either direction. | |
Dec 14, 2012 at 0:18 | comment | added | Greg Martin | For example, problem 11(b) in section 6.2 of Montgomery and Vaughan's book "Multiplicative Number Theory I" is to use the method of proof Peter described to prove that $|L(x)| \le Ax \exp({-}c\sqrt{\log x})$ for some positive constants $A$ and $c$. Interestingly, problem 11(c) suggests that it is easy to derive this bound from the corresponding bound for $M(x) = \sum_{n\le x} \mu(n)$, which might be easier to find in the literature. The link between the two functions is given by $\lambda(n) = \sum_{d\colon d^2\mid n} \mu(n/d^2)$. | |
Dec 14, 2012 at 0:00 | comment | added | Peter Humphries | Also, you may be interested in the contents of this paper: staff.science.uu.nl/~dahme104/DistributionOmega.pdf | |
Dec 13, 2012 at 23:59 | comment | added | Peter Humphries | I don't know of a direct reference of such a proof, but this theorem is folklore and it's pretty easy to see why it's true: one can show (say, via comparing Euler products) that $\sum_{n=1}^{\infty}\lambda(n)n^{-s}=\zeta(2s)/\zeta(s)$ for $\Re(s)>1$, then use the fact that this extends meromorphically to the entire complex plane and has no poles in the region $\Re(s)>1-c/\log(|\Im(s)|+2)$. So basically the usual way of proving the prime number theorem, just with a different Dirichlet series. | |
Dec 13, 2012 at 23:37 | vote | accept | Joel Moreira | ||
Dec 13, 2012 at 23:37 | comment | added | Joel Moreira | Thanks! So this seems to be well known among number theorists. Can you give a reference where I can find a proof that $L(x)=o(x)$? | |
Dec 13, 2012 at 22:59 | history | answered | Peter Humphries | CC BY-SA 3.0 |