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Let $v$ be the vector $\sum_{i \leq n} e_i$ and $w$ be $\sum_{i \leq n} i e_i$. Using dot products, the angle between $v$ and $w$ can easily be shown* to approach $\frac \pi 6$ as $n \to \infty$.

What is the geometric understanding of this? Can this be proved using synthetic geometry - or at least explained intuitively?

*Proof:

$$\begin{align*} \theta_n &= \arccos \frac {|v \cdot w|} {|v||w|} \\ &= \arccos \frac {\sum_{i \leq n} i} {\sqrt{\sum_{i \leq n} 1^2 } \sqrt{\sum_{i \leq n} i^2}} \\ &= \arccos \frac {n(n+1)}{2} \frac{\sqrt 6}{\sqrt{n}\sqrt{n(n+1)(2n+1)}} \\ &= \arccos \frac {\sqrt 6} 2 \sqrt{\frac {n+1}{2n+1}} \\ &\to \frac \pi 6 \end{align*}$$

SRobertJames
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  • I doubt you can find an easy geometric explanation of this. – TheBestMagician2 May 01 '23 at 00:06
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    @TheBestMagician2 It's hard to imagine that $\pi/6$ just happened. One odd thing that may spark some ideas: $\pi/6$ reminds me of $\pi^2/6$, and $|w| = 1^2 + 2^2 + 3^2....$ reminds me of $\zeta(2)$, and there indeed are geometric proofs of $\zeta(2) = \pi^2/6$ – SRobertJames May 01 '23 at 01:10
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    Hmm. If, instead, you let $w=(x_1,x_2,\ldots,x_n)$ be a random vector with the coordinates $x_i$ drawn uniformly and indepedently from the interval $[0,1]$, then the expected value of the inner product is $E(\sum_i x_i)=n/2$ and the expected value of the squared norm is $E(\sum_ix_i^2)=n/3$. This would lead to the same "expected" value for the inner product ($\sqrt3/2$). Scare quotes needed, because the expected value of a ratio is not the ratio of expected values :-) Anyway, thinking of the vector $\dfrac1nw$ as having uniformly distributed random coefficients in $[0,1]$ may give something. – Jyrki Lahtonen May 01 '23 at 03:33
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    (cont'd) Standard results imply that when $n\to\infty$ the squared norm divided by $n$, $\frac1n\sum_ix_i^2$, is within $\epsilon$ of $1/2$ with a probability $\to1$. For any $\epsilon>0$. That might get rid of the scare quotes in the limit? – Jyrki Lahtonen May 04 '23 at 05:30
  • @JyrkiLahtonen This is great. Would you consider expanding your comments into an answer? – SRobertJames May 05 '23 at 02:19
  • Thanks. But I'm undecided. After all, essentially I'm replacing your use of sums with integrals. And you are looking at exactly Riemann sums related to the very same integrals. The real explanation could still come from the geometry of that high dimensional cube :-) – Jyrki Lahtonen May 05 '23 at 02:37
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    @SRobertJames Slight error in the proof (but not in the final result): before taking the limit the factor of the first square root is in fact $\sqrt{\frac{3}{2}}$, it doesn't change the final result because in fact when taking the limit the other square root tends to $\sqrt{\frac{1}{2}}$ – Amit May 18 '23 at 23:00
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    @Amit Thanks for catching that. I corrected it - can you take a look? – SRobertJames May 19 '23 at 00:42
  • It's correct now – Amit May 19 '23 at 07:18

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