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Jul 13, 2019 at 19:36 answer added Paata Ivanishvili timeline score: 4
Jul 13, 2019 at 8:35 comment added Dmitry Krachun @fedja Oh, sure, you're right, I got confused. So the answer is $(c+o(1))n^2$ for some $c>0$ arising from the continuous problem, right?
Jul 13, 2019 at 1:25 comment added fedja @DmitryKrachun Why? We are on the plane, not on the line, so the critical power is now $2$, not $1$. The question essentially is what is the capacity of the unit square and what is the equilibrium measure with respect to the kernel $\frac 1{|x-y|}$.
Jul 12, 2019 at 19:55 comment added Sandeep Silwal I asked a similar question here: math.stackexchange.com/questions/3050869/…. For the $1$ dimensional case, you indeed have a $n^2\log(n)$ lower bound and it can be found in the book 'The Cauchy-Schwarz master class', exercise 8.9. However, the technique there does not generalize to even two dimensions.
Jul 12, 2019 at 17:30 comment added Dmitry Krachun @MattF. If, say, the limiting density $\mu$ is continuous with respect to the Lebesgue measure, then the integral of $1/|x-y|$ against $\mu\times \mu$ is infinite. Also, the divergence is logarithmic, so my guess is that the answer is $\Omega(n^2\log{n})$.
Jul 12, 2019 at 17:27 comment added Dmitry Krachun It seems strange: Deterministically, $\operatorname{dist}(p_i, p_j)\leq \sqrt{2}$ which implies roughly $D\geq n^2/3$.
Jul 12, 2019 at 9:43 comment added user44143 Will you share the construction? I would expect there to be some limiting continuous distribution where $E[1/dist]=c$, and then to have $D_n$ of order $cn^2$.
Jul 12, 2019 at 9:27 history edited user44143
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Jul 12, 2019 at 9:11 history edited user44143 CC BY-SA 4.0
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Jul 12, 2019 at 8:50 review First posts
Jul 12, 2019 at 9:13
Jul 12, 2019 at 8:45 history asked Zuo Ye CC BY-SA 4.0