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So I cooked up this expression for $1/\pi$, but I'm not immediately able to prove it, although computation shows it to be true. Perhaps somebody can think of a neat proof!

$$\mathrm{Let}~~S_n=\{\omega \in \mathbb{C}: w^n=1\}.\quad \mathrm{Let}~~ a_n=\frac{1}{n}~\sup_{T~ \subseteq S_n}\left|\sum_{\omega \in T}\omega \right|.$$

$$\mathrm{Then}~~ a_n \to \frac{1}{\pi} ~\mathrm{as}~~ n \to \infty.$$

(It seems that the value of $a_n$ can usually be attained by picking all of those $n^{th}$ roots of unity in the upper half plane, for example. The limit would follow from a proof of this observation.)

Thanks, and enjoy!

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@amWhy: A semantic quibble: \mathrm is intended for setting math upright (as in $\det{A}$), while \text is the command to write text in a math environment. Also, you can get the spacing right by inserting it into the text environment \text{Let } S_n instead of tildes. –  t.b. Jul 8 '11 at 2:35

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up vote 4 down vote accepted

A sketch. Consider a given $T$ and the corresponding sum $S$ as a vector in the complex plane. Let $L$ be its perpendicular through the origin. If there are any elements of $T$ on the wrong side of $L$, omitting them increases the length of $S$, so WLOG every element of $T$ lies in the upper half plane. It's not hard to see that if $T$ isn't maximal, adding another element of $S_n$ in the upper half plane increases the length of $S$, so the conclusion follows.

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I was writing up the same idea. The symmetry of the roots means you just need the maximum over $[0,\frac{\pi}{n}]$ and the difference over this range will get very small. So pick any direction you like (you have chosen the positive imaginary axis) and take the points that help. –  Ross Millikan Jul 8 '11 at 2:44
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That's most of the proof really. Selecting $ T $ in the upper half plane will lead to $ a_n \to (1/2) \int_0^{1} e^{\pi i x} dx = 1/\pi $ as it's just a Riemann sum approximation (the factor of $ 1/2 $ comes from the fact that $ n $ is twice-ish the number of roots in the upper half plane.) –  anon Jul 8 '11 at 2:54
    
Cool! If anyone feels like writing up a complete solution, I'll accept it as the answer. If not, I'll accept Quiaochu's sketch. –  Bruno Joyal Jul 14 '11 at 18:05

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