# sample variance of regular polygon upon superimposition of vertices

Given, the vertices of a regular polygon, the centroid here would be the sample mean of the vertices and we assume it to be at the origin. The distance from each vertex to centroid is $\frac{s}{2\sin(\frac{180}{n})}$ where $n$ is number of sides and $s$ is the length of side. My definition of variance of vertices (loose notation, as I havent defined any probability law or sampling law) is sum of squares of distance from each vertex to the center divided by the total number of vertices.

Question: What would be the 'above defined' variance if one vertex was made to be superimposed on another adjacent vertex, i.e an edge was dropped? (Edge is dropped from a visual perspective only, and note that the number of vertices is still the same in the denominator for computing the variance.) Similarly, what would be the variance if two edges were dropped by superimposing two vertices on two corresponding adjacent vertices? and so on, what would be the variances after this process is done with three vertices? Can this sample variance be parametrized either as a formula or as an approximation, given as input the number of edges that are dropped (as in via superimposition), the number of sides originally before the dropping process, and the length of each side? Do note again, that the total number of points is not reduced and remains the same, and the edge is only visually seeming like dropped although in reality the edge exists with zero distance.

On dimension, let's stick to 2D coordinates being used for describing each vertex of the regular polygon. Just in case you need some more structure or standardization of notation, you may assume that all the regular polygons that we start with prior to the point-superposition/edge dropping process are inscribed in a unit-circle.

Also, I ask this question because I'd like to do this: Let V be the variance of the initial regular polygon. Then say the edge dropping was done to get a polygon with a given configuration of say k out of the total t points superimposing at P_1. Then I compute the variance of the new set of points, which will be lesser than V. Then I'd like to scale by \alpha the rest of the vertices t-k so that the variance of all the points equals V. Then I'd like to compare the sum of squares of distances of all pairs of points in initial polygon vs the same in the configuration after the k, t-k configuration was reached and see how much they differ by.

• Where do the vertices end-up after "superposition"? The midpoint of the dropped edge? – Blue Jul 23 '15 at 23:55
• Let me illustrate with an equilateral triangle ABC. Vertex A gets superimposed on vertex B. Then in next step, Vertex C gets superimposed on Vertex B as well as A, which are at the same location. By now all vertices are at the same location. Does this answer your question? The 'mid-point' in your question got me confused. – qlinck Jul 24 '15 at 0:00
• OK. (I was thinking of "coalescing" vertices at an intermediary location.) The tricky part in all this would seem to be that order is important: eg, superimposing $A$ onto $B$, and then $A$-and-$B$ onto $C$ puts those three vertices at location $C$, which could be a very different situation than if $A$ and $C$ went to location $B$, or if $B$ and $C$ went to location $A$. Ignoring symmetries, it appears that we'd have to tabulate vertex variances for every possible ordering of "edge-drops". Tedious. Maybe someone at Mathematica.StackExchange could automate this and see if any patterns emerge. – Blue Jul 24 '15 at 0:31
• But since it is a regular polygon, i thought the computed variance was invariant to orientation or order? isn't that so..am thinking why. Note Var(X+a)=Var(X). Moreover there is no scaling in terms of Var(aX)=a^2Var(X) in this case here. – qlinck Jul 24 '15 at 0:41
• The polygon may start off regular, but once you start dropping edges, it (more often than not) becomes irregular. Among other things, this affects where the centroid moves, which affects the variance. Take pentagon $ABCDE$, and superimpose $C$, $D$, $E$ at position $D$. You get (visually) an isosceles triangle containing the circle's center; superimposing $C$, $D$, $E$ at position $C$ (or $E$), you get a very different isosceles triangle not containing the center. The centroids differ, as do the calculated variances, so the order of combining $C$, $D$, $E$ together does indeed matter. – Blue Jul 24 '15 at 1:18

As noted in the comments, we can cut to the chase by simply tabulating where the various superimposed vertices accumulate at the end of the edge-dropping process. For a regular $n$-gon, we can assign integer weights $p_i$ (with $i=0, 1, \dots n-1$) that count the vertices that land on vertex $P_i$. Of course, we must have $\sum_i p_i = n$. In what follows, we take the polygon to be centered at the origin, $O$, with its vertices on the unit circle, so that $|P_i| = 1$.

The centroid of the final figure with multiply-counted vertices is given by $$K = \frac{1}{n}\,\sum_i p_i P_i$$

With this, we can compute the "vertex variance" as follows:

\begin{align} n V_n = \sum_i p_i\;|P_i - K|^2 &= \sum_i p_i\;(P_i - K)\cdot(P_i - K ) \\[4pt] &= \sum_i p_i\;|P_i|^2 - 2 \sum_i p_i P_i \cdot K + \sum_i p_i |K|^2 \\[4pt] &= \left(\sum_i p_i\right) - 2 \left(\sum_i p_i P_i\right) \cdot K + \left(\sum_i p_i\right) |K|^2 \\[4pt] &= n - 2 \left( n K \right)\cdot K + n |K|^2 \\[4pt] &= n \left( 1 - |K|^2 \right) \end{align}

So, \begin{align} n^2 V_n = n^2 - n^2 |K|^2 &= \left( \sum_i p_i \right)^2 - \left|\;\sum_i p_i P_i\;\right|^2 \\[6pt] &= \sum_i p_i^2 + 2\sum_{j<k} p_j p_k - \sum_i p_i^2 |P_i|^2- 2\sum_{j<k} p_j p_k P_j\cdot P_k \\[6pt] &= 2\sum_{j<k} p_j p_k ( 1 - P_j\cdot P_k ) \end{align} Now, since $P_j$ and $P_k$ are unit vectors, we have $$P_j\cdot P_k = \cos \angle P_j O P_k = \cos\left(\frac{2\pi}{n}(j-k)\right)$$ and since $1-\cos 2\theta = 2\sin^2 \theta$, we have finally

$$V_n = \frac{4}{n^2}\;\sum_{j < k} p_j p_k \sin^2\left(\frac{\pi}{n}(j-k)\right) \tag{\star}$$

(I get the sense that there must be a far shorter route to the result, but this is the best I have right now.)

Notice that the coefficients on a given $p_j p_k$ match for those $\overline{P_j P_k}$ in various "families" of the polygon. For instance, the sides of the polygon are segments $\overline{P_j P_k}$ where $|j-k| = 1$ (with index arithmetic done modulo $n$), so all corresponding $p_j p_k$ terms have coefficient $\sin^2(\pi/n)$; the "skip-one" diagonals have endpoints with $|j-k|=2$, so corresponding terms have coefficient $\sin^2(2\pi/n)$; and so forth.

Here are some examples of $(\star)$ for small $n$:

\begin{align} V_2 &= p_0 p_1 \\[6pt] V_3 &= \frac{1}{3} \left( p_0 p_1 + p_1 p_2 + p_2 p_0 \right) \\[6pt] V_4 &= \frac{1}{8} \left( p_0 p_1 + p_1 p_2 + p_2 p_3 + p_3 p_0 \right) +\frac{1}{4} \left( p_0 p_2 + p_1 p_3 \right) \\[6pt] V_5 &= \frac{5-\sqrt{5}}{50} \left( p_0 p_1 + p_1 p_2 + p_2 p_3 + p_3 p_4 + p_4 p_0 \right) \\[4pt] &+ \frac{5+\sqrt{5}}{50} \left( p_0 p_2 + p_1 p_3 + p_2 p_4 + p_3 p_0 + p_4 p_1 \right) \\[6pt] V_6 &= \frac{1}{36} \left( p_0 p_1 + p_1 p_2 + p_2 p_3 + p_3 p_4 + p_4 p_5 + p_5 p_0 \right) \\[4pt] &+ \frac{1}{12} \left( p_0 p_2 + p_1 p_3 + p_2 p_4 + p_3 p_5 + p_4 p_0 + p_5 p_1\right) + \frac{1}{9} \left( p_0 p_3 + p_1 p_4 + p_2 p_5 \right) \end{align}

(Double-check that each value reduces to $1$ when all $p_i = 1$.)

• Can you clarify: Is $O$ the origin in the angle?Also, just to confirm what is the $j-k$ notation denoting in the $sin$ function...are they indices themselves as in the sigma? – qlinck Jul 24 '15 at 14:23
• @qlinck: I've adjusted the answer to note that $O$ is the origin (and the center of the polygon), as well as that the vertices are on the unit circle. My apologies for being a little lazy about those key points. As for $j-k$: yes, those are indices. So, for instance, the summation in equation $(\star)$ with $n=10$ would have terms like $p_0 p_1 \sin^2((0-1)\pi/10)$ and $p_2 p_8 \sin^2 ((2-8)\pi/10)$ and $p_3 p_5 \sin^2 ((3-5)\pi/10)$, etc. – Blue Jul 24 '15 at 18:18
• ok thanks! Good answer.. Will give the bounty! It will allow me to award the bounty in 4 hours. – qlinck Jul 24 '15 at 18:58