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

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The discussion focuses on calculating the sum of lengths, \( S_n \), for a regular \( n \)-gon inscribed in a unit circle. The formula derived for \( S_n \) involves summing the distances between vertices, leading to \( S_n = n \sum_{r=1}^n \sin(r\pi/n) \). As \( n \) approaches infinity, the limit \( \lim_{{n}\to{\infty}}\frac{S_n}{n^2} \) is approximated as \( \frac{2}{\pi} \) using power series expansions for sine and cosine. The discussion also notes that an additional term in the summation does not affect the result since it evaluates to zero. Overall, the thread provides a mathematical exploration of geometric properties related to regular polygons.
lfdahl
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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 $$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}$$
 
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$$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}=0$$

(?)
 
(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 $$S_n = \text{im}\,\frac{2n(1-e^{-ir\pi/n})}{(1-e^{ir\pi/n})(1-e^{-ir\pi/n})} = \frac{n\sin(\pi/n)}{1 - \cos(\pi/n)}.$$
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(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: $$\frac{S_n}{n^2} \approx \frac{n(\pi/n)}{n^2\bigl(\frac{\pi^2}{2n^2} \bigr)} = \frac2\pi.$$
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Last edited:
greg1313 said:
$$\lim_{{n}\to{\infty}}\frac{S_n}{n^2}=0$$

(?)
Hi, greg1313!(Wave)

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

Concerning the index $r$: Shouldn´t $r$ have the maximum value $n-1$ and not $n$??
 
lfdahl said:
A small question:

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

You´re right!(Blush)
 
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