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Consider a square $\{(x,y): 0\le x,y \le 1\}$ divided into $n^2$ small squares by the lines $x = i/n$ and $y = j/n$. For $1\le i \le n$, let $x_i = i/n$ and

$$d_i = \min_{0\le j\le n} \left| \sqrt{1 - x_i^2 } - j/n\right|.$$ Determine $$\lim_{n\to\infty} \sum_{i=1}^n d_i$$ if it exists.

I think this might be proved using Weyl's Theorem (see, e.g., Korner, Fourier Anal., p11 (Cambridge), because as $n$ gets large the the behavior of the curve as it cuts the line joining two vertices is not unlike that of the fractional part of $n\cdot a$, $a$ irrational, as it travels around (say) a unit circle. But I haven't been able to do it. (This was in AMM in 1994 and as far as I know was not answered there.)

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    Did you try writing $d_i = \|\sqrt{n^2 - i^2}\|/n$ where $||x|| = \min_{k\in{\bf Z}}|x-k|$, and then using the Erdös-Turan inequality (http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Tur%C3%A1n_inequality) along with Fourier series for that function (http://math.stackexchange.com/questions/1105648/sum-involving-the-distance-to-the-nearest-integer-function) ? That reduces to bounding $\sum_{0\leq j\leq n}e^{2\pi i k\sqrt{n^2-j^2}}$ for some set of integers $k$.2015-02-03
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    I think I can prove that this converges to 1/4, and follows from the following more general statement. If $f\colon(0,1)\to\mathbb{R}$ is continuously differentiable and $f^\prime(X)$ has absolutely continuous probability distribution for $X$ uniform on $(0,1)$ then, the sequences $nf(i/n)$ ($i=0,1,\ldots,n-1$) become uniformly distributed mod 1, in the limit $n\to\infty$. Equivalently, $\sum_{i=0}^{n-1}\exp(2\pi i h nf(i/n))\to0$ as $n\to\infty$ For each nonzero integer $h$. This applies to the question here by taking $f(x)=\sqrt{1-x^2}$.2015-02-05
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    The method I have in mind for proving my statement uses ideas from Terry Tao's post on equidistribution of polynomial sequences (https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/#vdc) and, in particular, the van der Corput inequality as presented there.2015-02-05
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    @GeorgeLowther: Look forward to seeing it. Will look at the links when I have a chance.2015-02-05

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