Width deviation of convex polygons

The width of an image along various directions is basic information in the Image Processing Technique. We are interested in the deviation of the widths of compact convex sets in the plane along a random direction. Although it is an important and useful quantity both practically and theoretically, as long as we know, there is no serious study on it.

Let $\Omega=\mathbb{R}/2\pi\mathbb{Z}=(-\pi,\pi]$ and $\mathbb{P}$ be the normalized Lebesgue measure on $\Omega$. Consider the probability space $(\Omega,\mathbb{P})$; we write $\Omega$ for short. Let a compact convex set $T$ in the complex plane $\mathbb{C}$ be given. We identify $T$ with its boundary. We discuss the length of the orthogonally projected shadow of $T$ by the light from a random direction $\omega$, say $X_{T}(\omega)$. It is also interpreted as the width of $T$ along the orthogonal direction of $\omega$, that is, $\omega\pm\frac{\pi}{2}$.

We are interested in the uniformity (or its contrary) of the deviation of $X_{T}(\omega)$ with respect to $\omega\in\Omega$. For this purpose, we consider the deviation rate of the random variable $X_{T}$, say $\delta(X_{T})$, defined by

$$\delta(X_{T})=\frac{\sqrt{\mathbb{E}(X^{2}_{T})-\mathbb{E}(X_{T})^{2}}}{\mathbb{E}(X_{T})},$$

(1)

where $\mathbb{E}(~{})$ is the expectation of the random variables. Clearly, $\delta(X_{T})$ is invariant among the similarity images of $T$. It is also clear that $\delta(X_{T})=0$ if and only if $T$ is a closed curve of constant width (c.f. [5]). Hence, $\delta(X_{T})$ measures the degree how far $T$ is from convex bodies of constant width. For the ellipse $T=\{(x,y);~{}(x^{2}/a^{2})+(y^{2}/b^{2})=1\}$ with the perimeter $L$, it is not difficult to see that

$$\delta(X_{T})=\frac{\sqrt{2\pi^{2}(a^{2}+b^{2})-L^{2}}}{L}.$$

In this paper, we study the deviation rate of the convex $n$-gons $T$ with $n\geq 2$. We decompose $X_{T}$ as the sum of random variables coming from its edges, say $X_{T}=\sum_{j=1}^{n}X_{\alpha_{j}}$. Throughout this paper, the branch of the argument of a complex number is chosen to be $(-\pi,\pi]$. For a complex number $\alpha\in\mathbb{C}$ with $\arg(\alpha)=\theta$, define a random variable $X_{\alpha}$ on $\Omega$ by

$$X_{\alpha}(\omega)=|\alpha|\sin(\theta-\omega)_{+},$$

where $x_{+}=\max\{x,0\}$. In other words, $X_{\alpha}(\omega)$ is the length of the orthogonally projected shadow of the right side of the vector $\vec{O\alpha}$ by the light from the $\omega$ direction (the left side of the vector is permeable and makes no shadow).

Let $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}~{}(n\geq 2)$ be a sequence of distinct complex numbers. They are arranged in the counter clockwise order forming a convex $n$-gon if and only if

$$(*)~{}\arg(\alpha_{2}-\alpha_{1})\leq\arg(\alpha_{3}-\alpha_{2})\leq\cdots\\ \leq\arg(\alpha_{n+1}-\alpha_{n})\leq\arg(\alpha_{2}-\alpha_{1})+2\pi~{}~{}({\rm mod}~{}2\pi),$$

where we always consider the suffix $j$ related to the $n$-gon $T$ in modulo $n$, so that $\alpha_{n+1}=\alpha_{1},~{}\alpha_{0}=\alpha_{n}$, etc. In this case, the convex $n$-gon with vertices $\alpha_{1},\cdots,\alpha_{n}$ is denoted by $T=T(\alpha_{1},\cdots,\alpha_{n})$.

The above $T$ is non-degenerate (i.e. all the vertices are extremal points) if and only if “$<$” holds everywhere in the above inequalities. If this is not the case, then we identify $T=T(\alpha_{1},\cdots,\alpha_{n})$ with $T(\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{m})$, where $\{\alpha^{\prime}_{1},\cdots,\alpha^{\prime}_{m}\}$ are the set of extremal points among $\{\alpha_{1},\cdots,\alpha_{n}\}$ arranged in the counter clockwise order.

Define a random variable $X_{T}$ by

$$X_{T}(\omega)=\sum_{j=1}^{n}X_{\alpha_{j+1}-\alpha_{j}}(\omega),$$

which agrees with the former explanation as to the length of the shadow of $T$ (see Figure 1).

The set of convex $n$-gons $T=T(\alpha_{1},\cdots,\alpha_{n})$ can be identified with the sequence of $n$ distinct complex numbers $(\alpha_{1},\cdots,\alpha_{n})$ satisfying $(*)$. We denote the space consisting of these $(\alpha_{1},\cdots,\alpha_{n})$ by $\Theta_{n}$. We consider the usual structures coming from $\mathbb{C}^{n}$ on $\Theta_{n}$. Then, $\cup_{k=2}^{n-1}\Theta_{k}$ is considered as the boundary of $\Theta_{n}$ in the above sense.

Denote the factor space of $\Theta_{n}$ divided by the similarity equivalence by $\tilde{\Theta}_{n}$. Then it is a compact space. Most of our notions like the regular $n$-gon are notions in $\tilde{\Theta}_{n}$ rather than $\Theta_{n}$. The deviation rate $\delta(X_{T})$ can be considered as a continuous and piecewise smooth functional on $\tilde{\Theta}_{n}$. Hence, it has the minimum and the maximum in $\Theta_{n}$.

It turned out that the solution of the minimization problem of $\delta(X_{T})$ is related to the convex bodies of constant width. A Reuleaux body¹¹1In literature, it is often referred to Reuleaux ”polygon”, though its boundary is not linear. In this paper, we use the word ”body” to distinguish it from a genuine polygon. is a convex body of constant width whose boundary consists of a finite number of circular arcs with the center in it and the radius equal to the width.

The reader can consult [5, 6], but for the self-containedness, we review the Reuleaux body here briefly.

Let $D$ be a Reuleaux body with width $r$. Then, $C=\partial D$ consists of a finite number of circular arcs, say $C_{1},\cdots,C_{p}$, of radius $r$. The endpoints of the circular arcs are called essential if the 2 neighboring circular arcs have different centers, and hence cannot be joined into a single circular arc.

We assume that the circular arcs $C_{1},\cdots,C_{p}$ are arranged in the counter clockwise order and all the endpoints are essential. Let $A_{1},A_{2}$ be the endpoints of $C_{1}$; $A_{2},A_{3}$ be the endpoints of $C_{2}$, $\cdots$; $A_{p},A_{p+1}$ be the endpoints of $C_{p}$ (the suffices of $C$ or $A$ are considered modulo $p$ so that $A_{p+1}=A_{1}$). Note that a center of any one of the circular arcs is an essential endpoint of some other circular arcs on $C$, since otherwise, we have 2 points in $C$ with a distance larger than $r$. Conversely any endpoint is a center of some circular arc since there must be the same number of circular arcs and centers.

Hence, the center of $C_{1}$ is one of $A_{3},\cdots,A_{p}$. Let it be $A_{k}$. Since $D$ is strictly convex, the tangent vector of $C$ at $c\in C$ to the counter clockwise direction rotates counter clockwise as $c$ moves. The center of the circular arc containing $c\in C$ is the other intersection of the normal line to the tangent vector at $c$ with $C$. This intersection moves at the endpoints of the circular arcs to the counter clockwise direction since the normal lines from the same point rotate to the counter clockwise direction. Hence, the center of this circular arc is the next end point to that before. This means that the center of $C_{1}$ is $A_{k}$, the center of $C_{2}$ is $A_{k+1}$, $\cdots$.

On the other hand, the center of the circular arc $C_{k}$ must be $A_{2}$ since otherwise, either it is $A_{1}$ or there is $l\neq 1,2$ such that $\overline{A_{k}A_{l}}=r$. The latter is impossible since if so, by the convexity and the assumption on $C$, the circular arc $C_{1}$ can be extended to $A_{l}$ contradicting the assumption that both $A_{1}$ and $A_{2}$ are essential end points. The former is impossible since if so, then we have a contradiction that $\overline{A_{2}A_{k+1}}>r$. Thus, we have

$$A_{2}=\mbox{the center of }C_{k}=A_{2k-1},$$

and hence, $2\equiv 2k-1~{}(\mbox{mod}~{}p)$. This implies that $p$ is odd.

Consider the diagonals connecting 2 points in $C$, one of the center and an interior point of the circular arc centered by it. Any of 2 diagonals with different centers always intersect just at one point. Let $\Lambda_{j}$ be the set of angles from $A_{j}$ to the points in the circular arc centered by $A_{j}$. We always consider the angles in modulo $2\pi$. Then, by the above argument, $\Lambda_{j}$ and $\Lambda_{l}$ with $j\neq l$ are essentially disjoint. The same thing holds for $\Lambda_{j}$ and $\Lambda_{l}+\pi$. Let $\Lambda=\cup_{j=1}^{p}\Lambda_{j}$. Then, the union is essentially disjoint. Moreover, $\Lambda$ and $\Lambda+\pi$ is also essentially disjoint. It also holds that $\Lambda\cup(\Lambda+\pi)=(-\pi,\pi]$. This is because the Lebesgue measure of $\Lambda$ is $\pi$, since the integration of the exterior angle of $C$ is $2\pi$, which counts the angles in $\Lambda_{j}~{}(j=1,\cdots,p)$ twice, once at $C_{j}$, once at its center, where the exterior angle jumps just the amount of angles in $\Lambda_{j}$.

Thus, it holds that the circular arcs $C_{1},\cdots,C_{p}$ can be embedded by parallel translations by $\vec{v}_{1},\cdots,\vec{v}_{p}\in\mathbb{C}$ into a circle $\mathbb{S}$ of radius $r$ so that

$$\tilde{\Lambda}:=\cup_{j=1}^{p}(C_{j}+\vec{v}_{j})\subset\mathbb{S}$$

satisfies that

	$\displaystyle\#((C_{j}+\vec{v}_{j})\cap(C_{k}+\vec{v}_{k}))<\infty~{}\mbox{for any}~{}j\neq k,$
	$\displaystyle\#(\tilde{\Lambda}\cap R_{\pi}\tilde{\Lambda})<\infty~{}\mbox{and}~{}\tilde{\Lambda}\cup R_{\pi}\tilde{\Lambda}=\mathbb{S},$		(2)

where $R_{\pi}$ is the rotation on $\mathbb{S}$ by angle $\pi$, and “$\#$” implies the number of elements in a set.

We write a Reuleaux $p$-body to indicate the number $p$ of circular arcs with different centers. In particular, a Reuleaux $p$-body is called regular if all the circular arcs have the same length. A regular Reuleaux 3-body is well known as “Reuleaux triangle”.

A Reinhardt $n$-gon ([3], [9]) is an equilateral convex $n$-gon that can be inscribed in a Reuleaux body, containing all the essential endpoints of the circular arcs. It is referred to as a Reinhalt polygon if $n$ is not necessarily specified. If the lengths of the circular arcs of a Reuleaux $p$-body have ratio $n_{1}:n_{2}:\dots:n_{p}$ with integer $n_{i}$’s, we can divide the circular arcs into $n_{i}\ (i=1,\dots,p)$ parts of equal length and take the convex-hull of all the division points to get a Reinhardt $n$-gon, where $n=n_{1}+\cdots+n_{p}$, which is specially called a Reinhardt $(n_{1},\cdots,n_{p})$-gon.

In particular, let $p\geq 3$ be an odd integer and take a regular Reuleaux $p$-body $D$. We divide all the circular arcs of $D$ into $q$ parts of equal length. Then, the convex hull of all the division points forms a Reinhardt $pq$-gon, which is specially a Reinhardt $(\underbrace{q,\cdots,q}_{\mbox{$p$ times}})$-gon, see Figure 2. Denote $q^{p}=(\underbrace{q,\cdots,q}_{\mbox{$p$ times}})$ for short. Note that Reinhardt $1^{p}$-gon is the regular $p$-gon.

A Reinhardt polygon gives the solution of three optimization problems on convex $n$-gons when $n$ is not a power of $2$:

Problem 1.

Maximize the perimeter for a fixed diameter [9],

Problem 2.

Maximize the width for a fixed diameter [2],

Problem 3.

Maximize the width for a fixed perimeter [1].

Given a cyclic integer vector $(n_{1},n_{2},\dots,n_{p})$, there exists a unique way to construct a Reinhardt $(n_{1},n_{2},\dots,n_{p})$-gon when this is possible. We reproduce a necessary and sufficient condition of Reinhardt [9] that a given integer cyclic vector $(n_{1},n_{2},\dots,n_{p})$ forms a Reinhardt $(n_{1},n_{2},\dots,n_{p})$-gon as a byproduct of our result (Theorem 6 and 7). In particular, when $n$ is not a power of $2$, we have a Reinhardt $(n/p)^{p}$-gon for any odd divisor $p>1$. Many interesting properties on Reinhardt polygons are discussed in [6, 7, 3, 4]. When $n$ is odd, the regular $n$-gon is one of the Reinhardt polygons but it is not unique when $n$ is a composite. Reinhardt $n$-gon is unique if and only if $n=p$ or $2p$ with a prime $p$. A Reinhardt polygon may have no symmetry at all (see Figures in [3]).

In this paper, we prove the following results.
(1) The deviation rate for the compact convex sets $T$ attains the maximum if and only if $T$ is the 2-gon (Theorem 2).
(2) For the integer $n\geq 2$ which is not a power of $2$, the minimum of $\delta(X_{T})$ for $T\in\Theta_{n}$ is

$$\nu_{n}:=\sqrt{\frac{\pi}{4n\tan(\frac{\pi}{2n})}+\frac{\pi^{2}}{8n^{2}\sin^{2}(\frac{\pi}{2n})}-1}.$$

(3)

Theorem 6 gives a complete characterization of the shapes that attain the minimum value $\nu_{n}$. As a consequence, we show that the minimum shape is nothing but a Reinhardt $n$-gon (Theorem 7).

The minimum values for odd $n\geq 3$ is strictly decreasing in $n$.
(3) For even $m\geq 4$, regular $m$-gon is far from the minimum shape. Let $n(<m)$ be the odd number such that either $n=\frac{m}{2}$ or $n=\frac{m}{2}+1$. If $n=\frac{m}{2}$, then $\delta(X_{T_{m}})=\delta(X_{T_{n}})$ holds, and if $n=\frac{m}{2}+1$, then $\delta(X_{T_{m}})>\delta(X_{T_{n}})$ holds, where $T_{m},~{}T_{n}$ are the regular $m$-gon and $n$-gon, respectively (Theorem 4).

(4) Let $T$ be the regular triangle. We call a truncation of it a quadrangle obtained by cutting off a vertex by a line near it. It is called a parallel truncation if the line is parallel to the opposite side, otherwise a non-parallel truncation (Figure 3). We prove that a parallel truncation of the regular triangle increases the deviation rate, while a non-parallel truncation decreases it (Theorem 8).

Take any convex polygon $T=T(\alpha_{1},\cdots,\alpha_{n})$. We prove that $\delta(X_{T})\leq\sqrt{(\pi^{2}/8)-1}$, and that the equality holds only when $T$ is 2-gon. Let

$$\beta_{j}=\alpha_{j+1}-\alpha_{j}=r_{j}e^{\mathbf{i}\theta_{j}}~{}(j=1,\cdots,n)$$

and let $\theta_{jk}\in(-\pi,\pi]$ satisfies that

$$\theta_{jk}\equiv\theta_{j}-\theta_{k}~{}({\rm mod}~{}2\pi)~{}(j,k=1,\cdots,n).$$

It is sufficient to prove that $\frac{\mathbb{E}(X_{T}^{2})}{\mathbb{E}(X_{T})^{2}}\leq\frac{\pi^{2}}{8}$. Hence, it is sufficient to prove that $I:=\frac{\pi^{3}}{2}\mathbb{E}(X_{T})^{2}-4\pi\mathbb{E}(X_{T}^{2})\geq 0$. Then by Theorem 1, we have

$\displaystyle I=\frac{\pi^{3}}{2}\mathbb{E}(X_{T})^{2}-4\pi\mathbb{E}(X_{T}^{2})=\frac{\pi^{3}}{2}\left(\sum_{j=1}^{n}\mathbb{E}(X_{j})\right)^{2}-4\pi\sum_{j,k=1}^{n}\mathbb{E}(X_{j}X_{k})$

	$\displaystyle=\frac{\pi^{3}}{2}\left(\sum_{j=1}^{n}\frac{r_{j}}{\pi}\right)^{2}-\sum_{j,k=1}^{n}r_{j}r_{k}V(\theta_{j}-\theta_{k})=\sum_{j,k=1}^{n}r_{j}r_{k}\left(\frac{\pi}{2}-V(\theta_{j}-\theta_{k})\right)$
	$\displaystyle=\sum_{j,k=1}^{n}r_{j}r_{k}\left(\frac{\pi}{2}-(\pi-|\theta_{jk}|)\cos\theta_{jk}-|\sin\theta_{jk}|\right).~{}~{}~{}~{}(\#)$

Since $r_{j}r_{k}\cos\theta_{jk}$ is the inner product between $\vec{O\beta_{j}}$ and $\vec{O\beta_{k}}$, we have

$$\sum_{j,k=1}^{n}r_{j}r_{k}\cos\theta_{jk}=\left(\sum_{j=1}^{k}\vec{O\beta_{j}},~{}\sum_{j=1}^{k}\vec{O\beta_{j}}\right)=(\vec{0},\vec{0})=0.$$

This implies that in the above equality $(\#)$, the constant in the coefficient of $\cos\theta_{jk}$ can be changed anyway keeping the equality. Hence, we have

$$I=\sum_{j,k=1}^{n}r_{j}r_{k}\left(\frac{\pi}{2}-\left(\frac{\pi}{2}-|\theta_{jk}|\right)\cos\theta_{jk}-|\sin\theta_{jk}|\right).$$

Thus, to prove $I\geq 0$, it is sufficient to prove that

$$\left(\frac{\pi}{2}-|x|\right)\cos x+|\sin x|\leq\frac{\pi}{2}$$

for any $x\in(-\pi,\pi]$, which can be verified easily. Moreover, the equality holds if and only if $x=0$ or $\pi$, which implies that $T$ is 2-gon.

For a general compact convex set $S$ with $C=\partial S$, replacing the above $I$ by

$$I=\int\int_{C\times C}\left(\frac{\pi}{2}-\left(\frac{\pi}{2}-\theta(u,v)\right)\cos\theta(u,v)-|\sin\theta(u,v)|\right)dudv,$$

where $\theta(u,v)$ is the angle between the tangent lines of $C$ at $u$ and $v$, we have the same statement, which completes the proof. $\Box$

Let

$$\mathcal{H}=\{\alpha\in\mathbb{C};~{}\Im(\alpha)>0,~{}\mbox{or,}~{}\Im(\alpha)=0~{}\mbox{and}~{}\Re(\alpha)\geq 0\}$$

be the upper half plane endowed with the quotient topology of $\mathbb{C}$ by identifying $z$ and $-z$. For $\alpha\in\mathbb{C}$, we define

$$\iota(\alpha)=\left\{\begin{array}[]{cc}\alpha&(\alpha\in\mathcal{H})\\ -\alpha&(\alpha\notin\mathcal{H})\end{array}\right..$$

For a finite set of nonzero complex numbers $S=\{\alpha_{1},\cdots,\alpha_{n}\}$, define $\iota(S)\subset\mathcal{H}$ by

$$\iota(S)=\mbox{Abbreviation}\{\iota(\alpha_{1}),\cdots,\iota(\alpha_{n})\},$$

where $\mbox{Abbreviation}\{\alpha_{1}^{\prime},\cdots,\alpha_{n}^{\prime}\}$ is the set of complex numbers obtained by replacing any of $\alpha_{j}^{\prime},\alpha_{k}^{\prime}$ with $\arg(\alpha_{j}^{\prime})=\arg(\alpha_{k}^{\prime})$ by $\alpha_{j}^{\prime}+\alpha_{k}^{\prime}$.

A nonzero element in $\mathcal{H}$ is sometimes called a pre-edge. A sequence of pre-edges $\beta_{1},\cdots,\beta_{m}\in\mathcal{H}$, say $\mathcal{B}=\mathcal{B}(\beta_{1},\cdots,\beta_{m})$ is called a pre-edge bundle (of size $m$) if

$$0\leq\arg(\beta_{1})<\arg(\beta_{2})<\cdots<\arg(\beta_{m})<\pi.$$

Denote by $\Xi_{m}$ the set of pre-edge bundles of size $m$.

For a convex $n$-gon $T=T(\alpha_{1},\cdots,\alpha_{n})$, we define its asymmetrization $\iota(T)$ as the pre-edge bundle $\mathcal{B}=\mathcal{B}(\beta_{1},\cdots,\beta_{m})$ such that

$$\iota(\{\alpha_{2}-\alpha_{1},~{}\alpha_{3}-\alpha_{2},~{}\cdots,~{}\alpha_{n+1}-\alpha_{n}\})=\{\beta_{1},\beta_{2},\cdots,\beta_{m}\}.$$

In this case, $T$ is called a realization of $\mathcal{B}$. Let $T=T(\alpha_{1},\cdots,\alpha_{n})$ be a convex $n$-gon and $\mathcal{B}=\mathcal{B}(\beta_{1},\cdots,\beta_{m})$ be its asymmetrization. Then,

	$\displaystyle U=$
	$\displaystyle T\left(\gamma,\gamma+\frac{1}{2}\beta_{1},\cdots,\gamma+\frac{1}{2}(\beta_{1}+\cdots+\beta_{m}),\gamma+\frac{1}{2}(\beta_{2}+\cdots+\beta_{m}),\cdots,\gamma+\frac{1}{2}\beta_{m}\right)$

is another realization of $\mathcal{B}$ than $T$, where $\gamma=-\frac{1}{4}(\beta_{1}+\cdots+\beta_{m})$. We call $U$ the symmetrization of $T$ (or $\mathcal{B}$).

In this case, $U$ is symmetric, that is, $U=T(\gamma_{1},\cdots,\gamma_{2k})$ with even size $2k$ and

$$\gamma_{k+1}=-\gamma_{1},~{}\gamma_{k+2}=-\gamma_{2},~{}\cdots,~{}\gamma_{2k}=-\gamma_{k}$$

holds. For a symmetric $U=T(\gamma_{1},\cdots,\gamma_{2k})$, it holds that

$$X_{U}(\omega)=2|\gamma_{j}|\sin(\arg(\gamma_{j})-\omega)\\ \mbox{for any}~{}\omega~{}\mbox{with}~{}\arg(\gamma_{j-1}-\gamma_{j})\leq\omega\leq\arg(\gamma_{j}-\gamma_{j+1}).$$

(4)

This representation of $X_{U}$ is called the diagonal representation.

Let $\beta$ be a pre-edge with $\arg(\beta)=\theta$. Define a random variable $\tilde{X}_{\beta}$ as

$$\tilde{X}_{\beta}(\omega)=\frac{|\beta|}{2}|\sin(\theta-\omega)|,$$

and for a pre-edge bundle $\mathcal{B}=\mathcal{B}(\beta_{1},\cdots,\beta_{m})$, let $\tilde{X}_{\mathcal{B}}=\sum_{j=1}^{m}\tilde{X}_{\beta_{j}}$.

A regular $n$-gon is denoted by $T_{n}~{}(n=2,3,\cdots)$. A pre-edge bundle $R_{m}=\mathcal{B}(\beta_{1},\cdots,\beta_{m})$ is said to be regular if

$$|\beta_{1}|=\cdots=|\beta_{m}|~{}\mbox{and}~{}\arg(\beta_{j+1})-\arg(\beta_{j})\equiv\pi/m~{}({\rm mod}~{}\pi)~{}(j=1,\cdots,m).$$

By Theorem 3, to prove $(i)$, it is sufficient to prove that

$$\delta(X_{T_{2m}})=\nu_{m}~{}~{}(m=1,2,\cdots)$$

for $T_{2m}=T(e^{\mathbf{i}\pi/2m},e^{\mathbf{i}3\pi/2m},\cdots,e^{\mathbf{i}\pi(1+2(2m-1))/2m})$. Using the diagonal representation (4), we have

	$\displaystyle\mathbb{E}(X_{T_{2m}})=\frac{2m}{2\pi}\int_{-\pi/2}^{-\pi/2+\pi/m}2\sin(\frac{\pi}{2m}-\omega)d\omega$
	$\displaystyle=\frac{2m}{\pi}\int_{\pi/(2m)}^{-\pi/(2m)}\sin(\frac{\pi}{2}-\omega)d\omega=\frac{2m}{\pi}\int_{-\pi/(2m)}^{\pi/(2m)}\cos\omega d\omega=\frac{4m}{\pi}\sin\frac{\pi}{2m}$

	$\displaystyle\mathbb{E}(X^{2}_{T_{2m}})=\frac{2m}{2\pi}\int_{-\pi/2}^{-\pi/2+\pi/m}4\sin^{2}(\frac{\pi}{2m}-\omega)d\omega$
	$\displaystyle=\frac{4m}{\pi}\int_{\pi/(2m)}^{-\pi/(2m)}\sin^{2}(\frac{\pi}{2}-\omega)d\omega=\frac{4m}{\pi}\int_{-\pi/(2m)}^{\pi/(2m)}\cos^{2}\omega d\omega$
	$\displaystyle=\frac{2m}{\pi}\sin\frac{\pi}{m}+2$

Hence,

	$\displaystyle\delta(X_{T_{2m}})=\sqrt{\frac{\mathbb{E}(X^{2}_{T_{2m}})}{\mathbb{E}(X_{T_{2m}})^{2}}-1}=\sqrt{\frac{\pi\sin\frac{\pi}{m}}{8m\sin^{2}\frac{\pi}{2m}}+\frac{\pi^{2}}{8m^{2}\sin^{2}\frac{\pi}{2m}}-1}$
	$\displaystyle=\sqrt{\frac{\pi}{4m\tan(\frac{\pi}{2m})}+\frac{\pi^{2}}{8m^{2}\sin^{2}(\frac{\pi}{2m})}-1}.$