Showing $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$

Question

The expression I have here is part of the upper bound given to the minimum distance of any 2 points out of $n$ points embedded in the unit sphere $\mathbb{S}^2$ in $\mathbb{R}^3$. If you have $n$ points on this unit sphere then we must have 2 points with a distance $d$ such that $$d \le \sqrt{4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)}$$

This result was proven by Fejes Tóth. What I wish to show is that $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$.

We can see in the limit as $n \to \infty$, $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \to 0^+$. Also using basic numerical simulation for many and large $n$ I can visually verify that this statement seems to be true. The point now would be to show that for any $n$, then there exists some constant $k$ such that $$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{k}{n}$$

This would be good enough. I'll be working on it here just thought the community may have some ideas :-)

Answer 1

One approach is to use L'hopital's rule. For your purposes, it suffices to show that the following limit exists. $$\lim_{n \rightarrow \infty} {4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \over {1 \over n}}$$ This is a ${0 \over 0}$ limit, for which L'hopital applies. But it would be complicated to check in this form. Instead, you can let $x = {n \over n - 2}$, making ${1 \over n} = {1 \over 2} - {1 \over 2x}$, and the limit becomes

$$\lim_{x \rightarrow 1} {4-\csc^2\left( \frac{\pi x}{6} \right) \over {1 \over 2} - {1 \over 2x} }$$ This is again a ${0 \over 0}$ limit, but it's easier to use L'hopital, and you get that the limit equals ${f'(1) \over g'(1)}$, where $f(x) = 4-\csc^2\left( \frac{\pi x}{6} \right)$ and $g(x) = {1 \over 2} - {1 \over 2x}$. Since $g'(1) = {1 \over 2} \neq 0$, you don't even have to calculate $f'(1)$ since any value will give that the limit exists.

Answer 2

We wish to show that $$f(x)=\left(4-\frac kx\right)\sin^2\frac{\pi x}{6(x-2)}\le1$$ for all positive real $x$. This is evidently the case when $4-k/x\le0$ so we consider $x>k/4$ with $$f'(x)=\frac k{x^2}\sin^2\frac{\pi x}{6(x-2)}-\left(4-\frac kx\right)\frac{24\pi}{(6x-12)^2}\sin\frac{\pi x}{6(x-2)}\cos\frac{\pi x}{6(x-2)}.$$ We claim that $f'(x)>0$ for all $x>k/4$. This is equivalent to showing that $$\tan\frac{\pi x}{6(x-2)}-\frac{2\pi x(4x-k)}{3k(x-2)^2}>0.$$ The second term is at most $8\pi/3k$ so the inequality holds whenever $$\frac x{x-2}>\frac6\pi\arctan\frac{8\pi}{3k}\implies x>2+\frac2{\frac6\pi\arctan\frac{8\pi}{3k}-1}.$$ This is true whenever $k>8\pi/\sqrt3$ so $$4-\csc^2\left( \frac{n \pi}{6(n-2)} \right) \le \frac{\frac{8\pi}{\sqrt3}+\epsilon}{n}$$ for any $\epsilon>0$. It follows that $d<3.81n^{-1/2}$.

Answer 3

By Taylor, we have $$\csc \left(\frac{\pi n}{6 (n-2)}\right)=2-\frac{2 \pi }{\sqrt{3} n}+\frac{\frac{7 \pi ^2}{9}-\frac{4 \pi }{\sqrt{3}}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=4-\frac{8 \pi }{\sqrt{3} n}+\frac{8 \pi \left(5 \pi -6 \sqrt{3}\right)}{9 n^2}+O\left(\frac{1}{n^3}\right)$$

$$4-\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=\frac{8 \pi }{\sqrt{3} n}-\frac{8 \left(\pi \left(5 \pi -6 \sqrt{3}\right)\right)}{9 n^2}+O\left(\frac{1}{n^3}\right)$$ $$4-\csc^2 \left(\frac{\pi n}{6 (n-2)}\right)=\frac{8 \pi }{\sqrt{3} n}+O\left(\frac{1}{n^2}\right)$$

Edit

Another way to do it using @Dzoooks' hint. $$\frac{\pi n}{6 (n-2)}=\frac{\pi (n-2)+2\pi}{6 (n-2)}=\frac \pi 6+\frac{\pi }{3 (n-2)}=\frac \pi 6+a$$ $$\csc \left(a+\frac{\pi }{6}\right)=2-2 \sqrt{3} a+7 a^2-\frac{23 a^3}{\sqrt{3}}+O\left(a^4\right)$$ $$\csc^2\left(a+\frac{\pi }{6}\right)=4-8 \sqrt{3} a+40 a^2+\left(-\frac{92}{\sqrt{3}}-28 \sqrt{3}\right) a^3+O\left(a^4\right)$$ $$4-\csc^2\left(a+\frac{\pi }{6}\right)=8 \sqrt{3} a-40 a^2+\left(\frac{92}{\sqrt{3}}+28 \sqrt{3}\right) a^3+O\left(a^4\right)$$ Now, let $a=\frac{\pi }{3 (n-2)}$ and, continuing with Taylor series, the same result.