What is the Sum of Lengths for a Regular n-gon Inscribed in a Unit Circle?

[lfdahl]

Let $S_n$ be the sum of lengths of all the sides and all the diagonals of a regular $n$-gon inscribed in a unit circle.

(a). Find $S_n$.

(b). Find $$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}$$

 

[Greg]

Spoiler

$$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}=0$$

(?)

 

[Opalg]

(a). Find $S_n$.

[sp]With the vertices at the points $e^{2\pi ij/n}$ on the unit circle, the distance between $e^{2\pi ij/n}$ and $e^{2\pi ik/n}$ is $$ \begin{aligned} \bigl| e^{2\pi ij/n} - e^{2\pi ik/n} \bigr| &= \sqrt{(e^{2\pi ij/n} - e^{2\pi ik/n}) (e^{-2\pi ij/n} - e^{-2\pi ik/n})} \\ &= \sqrt{2 - 2\cos(2\pi(j-k)/n)} \\&= 2\sin(|j-k|\pi/n), \end{aligned}$$ (the mod signs coming from the fact that we want the positive square root). If we sum that for $j$ and $k$ going from $1$ to $n$ then we will count each diagonal twice (once starting from the $e^{2\pi ij/n}$ end and once from the $e^{2\pi ik/n}$ end).

Therefore $$S_n = \sum_{j=1}^n\sum_{k=1}^n \sin(|j-k|\pi/n).$$ For each fixed $j$, the sum over $k$ will include each of the numbers $\sin\frac {r\pi}n \ (1\leqslant r \leqslant n)$ exactly once. Therefore $$S_n = n\sum_{r=1}^n \sin(r\pi/n) = n\, \text{im} \sum_{r=1}^n e^{ir\pi/n} = n\,\text{im} \frac2{1-e^{ir\pi/n}},$$(geometric series). Take the imaginary part to get $$S_n = \text{im}\,\frac{2n(1-e^{-ir\pi/n})}{(1-e^{ir\pi/n})(1-e^{-ir\pi/n})} = \frac{n\sin(\pi/n)}{1 - \cos(\pi/n)}.$$
[/sp]

(b). Find $$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}$$

[sp]Do this the lazy way, replacing $\sin(\pi/n)$ and $\cos(\pi/n)$ by the start of their power series: $$\frac{S_n}{n^2} \approx \frac{n(\pi/n)}{n^2\bigl(\frac{\pi^2}{2n^2} \bigr)} = \frac2\pi.$$
[/sp]

 

[lfdahl]

Spoiler

$$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}=0$$

(?)

Spoiler

Hi, greg1313!(Wave)

In fact $$\lim_{{n}\to{\infty}}\frac{S_n}{n^2} > 0$$
Please cf. Opalg´s solution.

 

[lfdahl]

(a). Find $S_n$.

[sp]With the vertices at the points $e^{2\pi ij/n}$ on the unit circle, the distance between $e^{2\pi ij/n}$ and $e^{2\pi ik/n}$ is $$ \begin{aligned} \bigl| e^{2\pi ij/n} - e^{2\pi ik/n} \bigr| &= \sqrt{(e^{2\pi ij/n} - e^{2\pi ik/n}) (e^{-2\pi ij/n} - e^{-2\pi ik/n})} \\ &= \sqrt{2 - 2\cos(2\pi(j-k)/n)} \\&= 2\sin(|j-k|\pi/n), \end{aligned}$$ (the mod signs coming from the fact that we want the positive square root). If we sum that for $j$ and $k$ going from $1$ to $n$ then we will count each diagonal twice (once starting from the $e^{2\pi ij/n}$ end and once from the $e^{2\pi ik/n}$ end).

Therefore $$S_n = \sum_{j=1}^n\sum_{k=1}^n \sin(|j-k|\pi/n).$$ For each fixed $j$, the sum over $k$ will include each of the numbers $\sin\frac {r\pi}n \ (1\leqslant r \leqslant n)$ exactly once. Therefore $$S_n = n\sum_{r=1}^n \sin(r\pi/n) = n\, \text{im} \sum_{r=1}^n e^{ir\pi/n} = n\,\text{im} \frac2{1-e^{ir\pi/n}},$$(geometric series). Take the imaginary part to get $$S_n = \text{im}\,\frac{2n(1-e^{-ir\pi/n})}{(1-e^{ir\pi/n})(1-e^{-ir\pi/n})} = \frac{n\sin(\pi/n)}{1 - \cos(\pi/n)}.$$
[/sp]

(b). Find $$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}$$

[sp]Do this the lazy way, replacing $\sin(\pi/n)$ and $\cos(\pi/n)$ by the start of their power series: $$\frac{S_n}{n^2} \approx \frac{n(\pi/n)}{n^2\bigl(\frac{\pi^2}{2n^2} \bigr)} = \frac2\pi.$$
[/sp]

Thankyou, Opalg, for an elegant solution!(Cool)

A small question:

Spoiler

Concerning the index $r$: Shouldn´t $r$ have the maximum value $n-1$ and not $n$??

 

[Opalg]

A small question:

Spoiler

Concerning the index $r$: Shouldn´t $r$ have the maximum value $n-1$ and not $n$??

[sp]It doesn't really matter, because the extra term when $r=n$ is $\sin(n\pi/n) = \sin\pi = 0$. (Bigsmile)
[/sp]

 

[lfdahl]

[sp]It doesn't really matter, because the extra term when $r=n$ is $\sin(n\pi/n) = \sin\pi = 0$. (Bigsmile)
[/sp]

Spoiler

You´re right!(Blush)