n)$ also an upper bound?

Current License: CC BY-SA 3.0

14 events

when toggle format	what		by	license	comment
Apr 5, 2017 at 9:31	vote	accept	H A Helfgott
Apr 5, 2017 at 1:45	answer	added	Peter Humphries		timeline score: 17
Apr 4, 2017 at 23:51	comment	added	H A Helfgott		Don't forget the condition $n\leq x$. That becomes $n\leq x/2$ when you sum over even $n$ and the divide them by $2$.
Apr 4, 2017 at 23:30	comment	added	Gerhard Paseman		I may have made a mistake. Since I count only square free n, I write it as the sum over odd square free n as term(n) + term(2n), where term(2n) is -term(n)/3 + mu(n)log(2)/3sigma(n). Of course, the above is over all odd numbers; for your sum there are missing terms which I am hoping will be dwarfed by the approximation to T(A). Gerhard "Crosses Fingers Against Large Oscillations" Paseman, 2017.04.04.
Apr 4, 2017 at 21:52	comment	added	H A Helfgott		Peter Humphries: I think the explicit formula would be just $$\rho(y) = \zeta(2) + \sum_s F(s) y^{s-1} + O(1/y)$$, where $s$ ranges over all non-trivial zeroes of the Riemann zeta function, and $$F(s) = \frac{1}{\zeta(s) (s-1)^2} \prod_p \left(1 + \frac{1}{(p+1)(p^{s+1}-1)}\right)$$.
Apr 4, 2017 at 21:29	comment	added	H A Helfgott		I'm not sure I see this. I can see the following: for $S_{\text{odd}}(x) = \sum_{\ŧext{$n$ odd: n\leq x}} (\mu(n)/\sigma(n)) \log(x/n)$, we have $\rho(x) = S_\text{odd}(x) - (1/3) S_\text{odd}(x/2)$. How do you get the identities you mention?
Apr 4, 2017 at 20:16	comment	added	Gerhard Paseman		Let us denote S(A) the subsum where n is restricted to belong to A, and T(A) similarly except log (x/n) is replaced by 1. I get S(all numbers) = 2* S(odd numbers) / (2+1) + (log 2)T(odd numbers)/(2+1). Am I right, and does this help? (One can pull a prime p out to get coefficients p/(p+1) and (log p)/(p+1) similarly.) Gerhard "Hopefully One Doesn't Become Zero" Paseman, 2017.04.04.
Apr 4, 2017 at 20:03	comment	added	H A Helfgott		Gerhard Paseman: yes!
Apr 4, 2017 at 20:02	comment	added	Gerhard Paseman		For those of us slow on the uptake , $\mu(n)$ is the Moebius function and $\sigma(n)$ is sum of divisors? Gerhard "Making Sure Of These Things" Paseman, 2017.04.04.
Apr 4, 2017 at 19:56	comment	added	Peter Humphries		Also it's not surprising that the error term for $\check{m}(x)$ is often much smaller than $O(x^{-1/2})$; it's the same reason that $\|M(x)\|/\sqrt{x}$ is usually smaller than $1$, namely that you have an expansion of the normalised error term of these summatory functions as (essentially) almost periodic functions of the form $\sum_{\gamma} c_{\gamma} x^{i\gamma}$, and "most" of the time, there will be a ton of cancellation in these sums. You can make this precise via the Rubinstein-Sarnak method of determining the limiting logarithmic distributions of these normalised error terms.
Apr 4, 2017 at 19:52	comment	added	Peter Humphries		Can you write up the details on the asymptotic expression for $\rho(x)$?
Apr 4, 2017 at 19:50	comment	added	H A Helfgott		Well, at the coarsest level: $\check{m}(x)$ tends to $1$, and it is not hard to show from this that $\rho(x)$ tends to $\zeta(2)$. If GRH holds, then the error term should be $O(1/sqrt(x))$ in either case. Confusingly, it seems smaller in practice.
Apr 4, 2017 at 19:44	comment	added	Peter Humphries		What is the asymptotic expression?
Apr 4, 2017 at 19:31	history	asked	H A Helfgott	CC BY-SA 3.0