# upper bound on product of distances from points on a circle

Let C be a circle of radius 1 in the complex plane with n points on the boundary. Provide an upper bound on the product of the distances of a given point on the circle to the other n points. The goal is to provide an answer based on how densely the point are positioned on the circle.

The goal is to characterize this based on how densely the points are placed. I am not sure of the correct notion of density to use for the unit circle but on the real line the correct notion would be something like this first described by Arne Beurling: For a discrete set on the real line Λ denote by n(r) the smallest number of points in any interval [x,x+r], r>0. The lower uniform density is defined by $l.u.d.(\Lambda)=\lim_{r\rightarrow\infty}\frac{n(r)}{r}$.

So the goal is to give an answer of the form

maximum product of distances ≤ a function of the correct notion of l.u.d. on the circle.

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## 1 Answer

Taking logs, it seems you are interested in a bound on $$\sum_1^n\log|a-e^{2\pi i\theta_k}|$$ where $a$ is your given point on the unit circle. It is natural to estimate this by comparing the sum (or, better, the average, $(1/n)\sum_1^n\log|a-e^{2\pi i\theta_k}|$) to the integral, $$\int_0^1\log|a-e^{2\pi i\theta}|d\theta$$ Koksma's Theorem bounds the difference between the average and the integral in terms of the discrepancy of the $n$ points, a concept similar to that lower uniform density you mention. Details can be found, for instance, in Kuipers and Niederreiter, Uniform Distribution of Sequences.

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thanks this was a very useful answer is the modulus of continuity of f(θ)=log|a−e2πiθ| easy to calculate? Based on page 145 of kuipers and Niederreiter this is equal to M(h)=sup|θ2−θ1|≤h|f(θ2)−f(θ1)| for 0≤h≤1. In this case this is equal to M(h)=sup|θ2−θ1|≤hlog|a−e2πiθ2||a−e2πiθ1| –  mohi Jun 14 '13 at 2:18
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