text:
\begin{document}

\maketitle

\begin{abstract}
It is shown that the area of Voronoi cells for a generalized
Archimedean spiral lattice converges under some scale normalization,
if the angle parameter is badly approximable.
\end{abstract}

\medskip

Keywords:
Voronoi cell,
Delone sets,
spiral,
phyllotaxis.

\section{Introduction}

Phyllotaxis is a term coined by botanists that describes the
organisation of florets in flowers (like daisy) or positions of leaves
along stems of trees or plants.  For a physicist, phyllotaxis can be
seen as the result of a maximum packing of disks on a plane surface.
If disks have some softness and the packing is confined within a
circular border, a phyllotactic solution can give a best packing
fraction \cite{sadoc_2012_a}.  Numerical analysis on the areas of
Voronoi cells of phyllotactic patterns have suggested a large
homogeneity of these areas \cite{sadoc_2013}.  The aim of this paper
is to give a mathematical proof for the convergence of the area of
Voronoi cells for Fermat and Archimedean spiral lattices.

Let $\alpha>0$, $\theta\in \real$.
 The set
\[
 \spiral = \spiral(\alpha,\theta) := \set{z_{j} := j^\alpha
  \rme^{j\theta\rmi} : j\in\zahl_{\ge0}}
\]
is called a (generalized) Archimedean spiral lattice with an exponent
$\alpha$.  It is called a Fermat spiral lattice if $\alpha= \half$.
In botany, the angle parameter $\theta$ is called `divergence'
\cite{coxeter}.  The Voronoi cell of $z_{j}$ in $\spiral$ is given by
\[
V(z_{j}, \spiral)
= \set{\zeta \in \cx:  |\zeta - z_{j}| \le |\zeta - z'|,
 \forall z' \in \spiral}.
\]
The Voronoi tessellation for the Fermat spiral lattice $\alpha=\half$
has been studied as a geometric model of spiral phyllotaxis
\cite{dixon, hotton}.

Sadoc et al. \cite{sadoc_2013} studied the Fermat spiral lattice with
the golden section divergence angle $\theta= 2\pi\tau$, $\tau = \frac{1+\sqrt{5}}{2}$,
and observed the convergence of the area $|V(z_{j},\spiral)|$ as
$j\to\infty$, as well as the existence of the grain boundary, its
quasiperiodicity and self-similarity under the substitution rule.
Yamagishi et al.~\cite{yamagishi_2018_a} proved the existence and the
quasiperiodicity of the grain boundary for the case of the Archimedean
spiral lattice $\alpha=1$, by considering the topological suspension.
The case for general $\alpha$ is addressed in
\cite{gar}.
See Figure \ref{Figure_1_1} for the Voronoi tessellations on $S = S(\alpha, 2\pi\tau)$, with $\alpha = 1, \frac{1}{2}, \frac{1}{4}$.

This paper shows the following theorem.  An irrational number is
called badly approximable if its continued fraction expansion has
bounded partial quotients.

\begin{theorem}
  \label{18}
  Suppose that $\theta/2\pi$ is badly approximable.  Then
  \[
  \lim_{j\to\infty} j^{1-2\alpha}|V(z_{j},\spiral)| = 2\pi\alpha.
  \]
\end{theorem}

\begin{corollary}[Area convergence]
\label{18_1}
  Suppose that $\theta/2\pi$ is badly approximable.
  Then we have
  $\lim_{j\to\infty} |V(z_{j},\spiral)| = \infty$ if $\alpha > \half$,
  $\lim_{j\to\infty} |V(z_{j},\spiral)| = 0$ if $\alpha < \half$,
  and
  \[
  \lim_{j\to\infty} |V(z_{j},\spiral)| = \pi
  \]
  if $\spiral$ is a Fermat spiral lattice $\alpha= \half$.
\end{corollary}

Figure \ref{Figure_1_2} shows the sequences of areas $|V(z_j,S)|$
for $S=S(\alpha,2\pi\tau)$, $\tau=(1+\sqrt{5})/2$, $j=1,2,\dots,10000$,
where $\alpha =1, \half, \frac{1}{4}$.
Figure \ref{Figure_1_3} observes the convergence $j^{1-2\alpha}|V(z_{j},\spiral)| \to 2\pi\alpha$ for
$\alpha = 1, \half, \frac{1}{4}$.
In particular, in the Fermat spiral lattice $\alpha= \half$, the area $|V(z_{j},\spiral)|$ tends to $\pi$.

The outline of this paper is as follows.  Section 2 is a short
introduction to continued fractions and Farey intervals
\cite{hardywright}.  Section 3 recalls the geometry of Voronoi
tessellations on linear lattices $\Lambda = \Lambda(z) = \zahl +
z\zahl$, \cite{hellwig,helical}.  Here we adopt the botanical term
`opposed parastichy pair' and introduce Richards' formula \cite{jean}.
The `rise in phyllotaxis' in botany, or the parastichy transition
\cite{rothen_koch_1989_a}, considers a geodesic flow on the modular
surface $SL(2,\zahl) \backslash \halfplane$, in the vertical direction
by fixing the `divergence' $\re(z)$ of the parameter $z \in
\halfplane$ and taking the limit as $\im(z) \to 0$.  It is a `scenery
flow' on the geometric structures on the torus \cite{arnoux-fisher}.

Section 4 is a technical part.  A basis $\lambda_1, \lambda_2$ of a
linear lattice $\Lambda = \lambda_1 \zahl + \lambda_2 \zahl$ is called
{\em reduced} if $\lambda_2 / \lambda_1 \in \fdom$, where
$\fdom = \set{\zeta \in \halfplane :
  |\zeta|\ge1, -\half \le \re(\zeta) \le \half}$.
It is shown that if $\Lambda$ is a linear lattice with a reduced basis
$\lambda_1, \lambda_2$, and $\phi : \Lambda \to \cx$ is a perturbation
with $|\phi-\id|$ small, then the Voronoi cell $V(\phi(0),
\phi(\Lambda))$ is adjacent to the cell $V(\phi(\lambda),
\phi(\Lambda))$, in the perturbed nonlinear lattice, only if either
$\lambda = \pm \lambda_1, \pm \lambda_2, \pm \lambda_1 \pm \lambda_2$.
This implies that the area $|V(\phi(0),\phi(\Lambda))|$ is close to
the area $|V(0,\Lambda)|$.  Here we assume that $\im(\lambda_2 /
\lambda_1)$ is bounded from above.

The adjacency of Voronoi cells depends on the bifurcation of the
vertices of Voronoi tessellations.  A vertex of Voronoi cell is a
circumcenter of a triangle of sites in $\Lambda$.  If we perturb a
corner of a triangle, then the circumcenter stays on the perpendicular
bisector of the edge for the other two corners of the triangle.

Section 5 is the main part of this paper.  Instead of treating the
discrete family of Voronoi cells, we consider a suspension flow in the
space of Archimedean spiral lattices.  Locally, it can be regarded as
a perturbation of a geodesic flow in the space of linear lattices.  By
Taylor's theorem, the area of the Voronoi cell in the nonlinear
lattice is close to the area of a Voronoi cell in the linear lattice,
which shows Theorem~\ref{18}.

See \cite{pennybacker,barabe} for the history of the study of
phyllotaxis.  The circle packing model of spiral phyllotaxis is
addressed in \cite{rothen_koch_1989_b,yamagishi_2017_a}.  It has been
shown only recently that a Fermat spiral lattice is a Delone set
(relatively dense and uniformly discrete) if and only if its
divergence angle is badly approximable in \cite{akiyama}.  See also
\cite{marklof}.

\begin{figure}
\centering
(a)
\includegraphics[scale=.6]{Fig_1_a.eps}
(b)
\includegraphics[scale=.6]{Fig_1_b.eps}

(c)
\includegraphics[scale=.6]{Fig_1_c.eps}
\caption{Voronoi tessellations for spiral lattices $S = S(\alpha,\theta)$.
(a) The Archimedean spiral lattice with the exponent $\alpha=1$, divergence angle $\theta=2\pi\tau$, where $\tau = (1+\sqrt{5})/2$.  Plot range: $[-1296,1296] \times [-1296,1296]$.
(b) The Fermat spiral lattice $\alpha=1/2$, $\theta=2\pi\tau$.  Plot range: $[-36,36]\times[-36,36]$.
(c) $\alpha=1/4$, $\theta=2\pi\tau$.  Plot range: $[-6,6]\times[-6,6]$.
}
\label{Figure_1_1}
\end{figure}

\begin{figure}
\centering
(a)
\includegraphics[scale=.5]{Fig_2_a.eps}

(b)
\includegraphics[scale=.5]{Fig_2_b.eps}

(c)
\includegraphics[scale=.5]{Fig_2_c.eps}

\caption{
Areas of Voronoi cells $|V(z_j, S)|$, $j=1,2,\dots,10000$.
(a) The Archimedean spiral lattice $S = S(1,2\pi\tau)$, $\tau = (1+\sqrt{5})/2$.
(b) The Fermat spiral lattice $S(1/2, 2\pi\tau)$.
(c) $S(1/4, 2\pi\tau)$.
}
\label{Figure_1_2}
\end{figure}

\begin{figure}
\centering
(a)
\includegraphics[scale=.5]{Fig_3_a.eps}

(b)
\includegraphics[scale=.5]{Fig_3_b.eps}

(c)
\includegraphics[scale=.5]{Fig_3_c.eps}

\caption{
Areas of Voronoi cells, normalized, $j^{1-2\alpha}|V(z_j, S)|$, $1 \le j \le 10000$.
(a) The Archimedean spiral lattice $S(1, 2\pi\tau)$.
(b) The Fermat spiral lattice $S(1/2, 2\pi\tau)$.
(c) $S(1/4, 2\pi\tau)$.
}
\label{Figure_1_3}
\end{figure}

\section{Farey intervals}

Here we recall some basic properties of the Farey series and continued
fractions.  A fraction $\frac{p}{q} \in \rat$ always assumes that
$p\in\zahl$ and $q \in \zahl_{>0}$.  Let $k\in\zahl_{>0}$.  The set of
fractions with denominator $\le k$, $\calF_k = \set{\frac{p}{q} \in
  \rat : p,q \in \zahl, \ 1 \le q \le k}$, is called the \textit{Farey
  series} of order $k$.  We have $\calF_1 = \zahl \subset \calF_2
\subset \calF_3 \subset \dots$, and $\bigcup_{k} \calF_k = \rat$.  An
open interval $(\frac{a}{m}, \frac{b}{n})$ with irreducible fractions
$\frac{a}{m}, \frac{b}{n}$ is called a \textit{Farey interval} of
order $k \in \zahl_{>0}$ if it is a connected component of $\real
\setminus \calF_k$.  A pair of fractions $\frac{a}{m} < \frac{b}{n}$
is called a \textit{Farey pair} if $mb-na=1$.

\begin{lemma}
\label{36}
A pair of irreducible fractions $\frac{a}{m} < \frac{b}{n}$ is a Farey
pair if and only if the interval $(\frac{a}{m}, \frac{b}{n})$ is a
Farey interval of some order $k \in \zahl_{>0}$.  If it is the case,
we have $\max\set{m,n} \le k < m+n$ and
\begin{equation}
\label{1}
\calF_{m+n} \cap
\left(\frac{a}{m}, \frac{b}{n} \right) = \Set{\frac{a+b}{m+n}}.
\end{equation}
\end{lemma}
\begin{proof}
Let $\frac{a}{m} < \frac{b}{n}$ be a Farey pair, and suppose that
$\frac{a}{m} < \frac{p}{q} < \frac{b}{n}$.  Since $q\frac{a}{m}- p =
\frac{qa-pm}{m}<0$ and $q\frac{b}{n}- p = \frac{qb-pn}{n}>0$, we have
$q\frac{a}{m}- p \le -\frac{1}{m}$ and $q\frac{b}{n}- p \ge
\frac{1}{n}$.  Thus
\[
  q =
  \frac{(q\frac{b}{n} - p) - (q\frac{a}{m}- p)}{\frac{b}{n} - \frac{a}{m}}
  \ge \frac{\frac{1}{n} + \frac{1}{m}}{\frac{b}{n} - \frac{a}{m}}
  = m+n.
\]
Moreover we have
\[ \frac{a+b-1}{m+n}
  \le \frac{a}{m}
  < \frac{a+b}{m+n}
  < \frac{b}{n}
  \le \frac{a+b+1}{m+n},
\]
which implies (\ref{1}).

For the converse, let $\frac{a}{m} \in \calF_k$.  Let $b,n \in \zahl$
be a (unique) solution of the equation
\[ mb - na = 1
\]
under the condition $k - m < n \le k$.  Then we have $\max\set{m,n}
\le k < m+n$.  The argument above shows that $(\frac{a}{m},
\frac{b}{n})$ is a Farey interval of order $k$.  See also
\cite[Theorem 28]{hardywright}.
\end{proof}

Let
\begin{equation}
\label{25}
 x
  = a_0+\frac{1}{a_1+\frac{1}{a_2+\cdots}} = [a_0,a_1,a_2,\cdots],
  \ a_0 \in \Z, \ a_i \in \Z_{>0}, \ i\geq 1
\end{equation}
be a continued fraction expansion of $x \in \real$.  We often assume
that $x$ is irrational.  Define the sequences $\set{p_i}_{i\geq -1}$
and $\set{q_i}_{i\geq -1}$ by $p_{-1} = 1$, $q_{-1} = 0$, $p_{0}=
a_0$, $q_{0}=1$, $p_{1}=a_0 a_1 + 1$, $q_{1}=a_1$, and $p_{i+1} =
a_{i+1} p_{i} + p_{i-1}$, $q_{i+1} = a_{i+1} q_{i} + q_{i-1}$,
$i\ge1$.  Let $p_{i,k} = k p_{i} + p_{i-1}$, $q_{i,k} = k q_{i} +
q_{i-1}$ for $i \ge0$, $0 \le k \le a_{i+1}$.  Note that $p_{i,0} =
p_{i-1}$, $q_{i,0} = q_{i-1}$, $p_{i,a_{i+1}} = p_{i+1}$,
$q_{i,a_{i+1}} = q_{i+1}$.  The fraction ${p_i}/{q_i} =
[a_0,a_1,\cdots,a_i]$, $i\ge0$, is called a {\it (principal)
  convergent} of $x$, and ${p_{i,k}}/{q_{i,k}} = [a_0, a_1, \cdots,
  a_i, k]$, $i\ge0$, $0<k<a_{i+1}$, is called an {\it intermediate
  convergent} of $x$.  An induction shows that if $i$ is odd and $0
\le k \le a_{i+1}$, then $(\frac{p_{i,k}}{q_{i,k}}, \frac{p_i}{q_i})$
is a Farey interval and $\frac{p_{i,k}}{q_{i,k}} < x <
\frac{p_i}{q_i}$.  If $i$ is even, $0 \le k \le a_{i+1}$ and
$(i,k)\neq(0,0)$, then $(\frac{p_i}{q_i}, \frac{p_{i,k}}{q_{i,k}})$ is
a Farey interval and $\frac{p_i}{q_i} < x < \frac{p_{i,k}}{q_{i,k}}$.

\begin{lemma}
\label{38}
Let $(\frac{a}{m}, \frac{b}{n})$ be a Farey interval containing $x$.
Then we have either $(\frac{a}{m}, \frac{b}{n}) =(\frac{p_i}{q_i},
\frac{p_{i,k}}{q_{i,k}})$ for some $i$ even and $0 \le k < a_{i+1}$,
or $(\frac{a}{m}, \frac{b}{n}) =(\frac{p_{i,k}}{q_{i,k}},
\frac{p_i}{q_i})$ for some $i$ odd and $0 \le k < a_{i+1}$.
\end{lemma}
\begin{proof}
The interval $(\frac{a}{m}, \frac{b}{n})$ is a connected component of
$\real\setminus\calF_{\max\set{m,n}}$.  Since $\set{q_i}_i$ is an
increasing sequence, there exists $i$ such that $q_i \le \max\set{m,n}
< q_{i+1}$.  If $q_i \le \max\set{m,n} < q_{i,1}$, then the connected
component of $\real\setminus\calF_{\max\set{m,n}}$ containing $x$ is
written as $(\frac{p_{i-1}}{q_{i-1}}, \frac{p_i}{q_i})$ or
$(\frac{p_{i}}{q_{i}}, \frac{p_{i-1}}{q_{i-1}})$.  Otherwise, there
exists $1\le k< a_{i+1}$ such that
$q_{i,k} \le \max\set{m,n} < q_{i,k+1}$,
so the connected component of
$\real\setminus\calF_{\max\set{m,n}}$ containing $x$ is written as
$(\frac{p_{i,k}}{q_{i,k}}, \frac{p_i}{q_i})$ or $(\frac{p_{i}}{q_{i}},
\frac{p_{i,k}}{q_{i,k}})$.
\end{proof}

A fraction $\frac{a}{m}$ is called an approximation to $x\in\real$
on the left, if $\frac{a}{m} < x$ and $0 < mx-a \le qx-p$ whenever
$\frac{p}{q} \le x$ and $0<q\le m$.  A fraction $\frac{b}{n}$
is called an approximation to $x$ on the right, if $\frac{b}{n} > x$
and $0 < b - nx \le p - qx$ whenever $\frac{p}{q} \ge x$ and
$0<q\le n$.  It is clear that a fraction is irreducible
if it is an approximation to some $x$ (on the left or right).

\begin{lemma}
  \label{39}
  Let $x \in \real \setminus \zahl$.  A fraction is an
  approximation to $x$ on the left or right, if and only if it is
  an endpoint of a Farey interval containing $x$.
\end{lemma}

\begin{proof}
  First, we suppose that $\frac{a}{m} < x$ is an approximation to
  $x$ on the left, and show that it is an endpoint of a Farey interval
  containing $x$.  For any fraction $\frac{p}{q} < x$ with $0<p\le
  m$, we have $0 < mx-a \le qx-p$.  This implies that $x -
  \frac{a}{m} \le x - \frac{p}{q}$, and $\frac{p}{q} \le
  \frac{a}{m}$.  So the connected component of $\real \setminus
  \calF_{m}$ containing $x$ has an endpoint $\frac{a}{m}$.  The proof
  for the case $\frac{b}{n} > x$ is similar.

  Conversely, assume that $(\frac{a}{m}, \frac{b}{n})$ is a Farey
  interval containing $x$, and we will show that $\frac{a}{m}$ is an
  approximation to $x$ on the left.  Let $\frac{p}{q}$ be a fraction
  such that $\frac{p}{q} < \frac{a}{m}$, and suppose that $0 < q\le
  m$.  If $q=m$, then we have $p<a$ and $mx-a < qx-p$.  If $q<m$, then
  we have
  \[ \frac{a-p}{m-q} - \frac{b}{n}
   = \frac{qb-np-1}{(m-q)n} \ge 0,
  \]
  which implies that $\frac{a-p}{m-q} \ge \frac{b}{n} > x$ and $mx-a <
  qx-p$.  So $\frac{a}{m}$ is an approximation to $x$ on the left.  We
  can similarly show that $\frac{b}{n}$ is an approximation to $x$ on
  the right.
\end{proof}

\section{Voronoi tessellation for a linear lattice}

Here we introduce some basic properties of Voronoi tessellations for
linear lattices, including Richards' formula, by using the botanical
term `parastichy.'  Let $\Lambda = \Lambda(z) := z\zahl + \zahl$,
where $z = x + \rmi y \in \cx$, $x>0$.  The Voronoi cell of the site
$\lambda \in \Lambda$ is given by
\[
V(\lambda) = V(\lambda, \Lambda)
= \set{\zeta \in \cx:  |\zeta - \lambda| \le |\zeta-\lambda'|,
 \forall \lambda' \in \Lambda}.
\]
The family of Voronoi cells $\calV = \set{V(\lambda)}_{\lambda \in
  \Lambda(z)}$ is a tiling of the plane, that is, $\cx =
\bigcup_{\lambda \in \Lambda} V(\lambda)$, and $\interior(V(\lambda))
\cap \interior(V(\lambda')) = \emptyset$ for $\lambda \neq \lambda'$.

The line segment $\ell(\lambda, \lambda')$ with the endpoints
$\lambda, \lambda' \in \Lambda$, is called a Delaunay edge if it
satisfies the Empty Circumdisk Property, that is, there exists a disk
$K$ such that $K \cap \Lambda = \set{\lambda, \lambda'}$.  For
distinct points $w_1, w_2, w_3 \in \cx$, let $\angle(w_1, w_2, w_3) :=
\Arg(\frac{w_1-w_2}{w_3-w_2})$, where $-\pi < \Arg \le \pi$ denotes
the principal argument.

\begin{lemma}
\label{33}
  Let $\lambda, \lambda' \in \Lambda$ be distinct sites.  The
  following conditions are mutually equivalent.
  \begin{enumerate}
  \item The Voronoi cells $V(\lambda)$, $V(\lambda')$ are
    (edge-)adjacent.
  \item The line segment $\ell(\lambda, \lambda')$ is a Delaunay edge.
  \item $\ell(\lambda, \lambda') \cap \Lambda = \set{\lambda,
    \lambda'}$, and we have
    \[ \angle(\lambda, w, \lambda')
    + \angle(\lambda', w', \lambda) < \pi
    \]
    for any $w, w' \in \Lambda$ whenever $0< \angle(\lambda, w,
    \lambda') <\pi$ and $0< \angle(\lambda', w', \lambda) < \pi$.
  \end{enumerate}
\end{lemma}
\begin{proof}
  See \cite[Lemma 1]{yamagishi_2015_a}.
\end{proof}

In this paper a line segment $\ell(\lambda, \lambda')$ is not called a
Delaunay edge when $V(\lambda)$ is only corner-adjacent to
$V(\lambda')$.  The Delaunay edges do not intersect each other.  A
connected component of the complement of the union of the Delaunay
edges is called a Delaunay polygon.  A Delaunay polygon for the linear
lattice $\Lambda(z)$ is either a rectangle or an acute triangle.

An irreducible fraction $\frac{a}{m}$, $a\in\zahl$, $m\in\zahl_{>0}$,
is called a parastichy index of $\Lambda(z)$ if $V(mz-a)$ is
(edge-)adjacent to $V(0)$.  A pair of parastichy indices $\frac{a}{m},
\frac{b}{n}$ of $\Lambda(z)$ is called an opposed parastichy pair if
$(mx-a)(nx-b)<0$.

\begin{lemma}
  Let $\frac{a}{m}$ be a parastichy index of $\Lambda(z)$.  Then
  $\frac{a}{m}$ is an approximation to $x = \re(z)$ on the left or
  right, or equal to $x$.
\end{lemma}
\begin{proof}
  First suppose that $mx-a>0$.  If $\frac{a}{m}$ is not an
  approximation to $x$ on the left, then there exist $p,q\in \zahl$
  such that $0< q \le m$, and $0< qx-p < mx-a$.  Then we have
  $\frac{\pi}{2}<|\angle(0, qz-p, mz-a)|,
  |\angle(mz-a,(m-q)z-(a-p),0)|$.  By Lemma~\ref{33}, $V(mz-a)$ is not
  adjacent to $V(0)$.  The case $mx-a<0$ is similar.
\end{proof}

\begin{lemma}
  \label{12}
  Suppose that the cell $V(0)$ is adjacent to $V(mz-a)$ and $V(nz-b)$.
  Then $|mb-na| = 0, 1$.
\end{lemma}
\begin{proof}
  Let $\lambda_1 = mz-a$, $\lambda_2 = nz-b$.  We assume $\lambda_1
  \neq \pm \lambda_2$, and show that $mb-na = \pm1$.  The line
  segments $\ell(\lambda, \lambda+ \lambda_1)$, $\ell(\lambda,
  \lambda+ \lambda_2)$ are Delaunay edges for all $\lambda \in
  \Lambda(z)$.  Since Delaunay edges do not intersect, the
  parallelogram $\square(\lambda, \lambda+\lambda_1,
  \lambda+\lambda_1+\lambda_2, \lambda+\lambda_2)$ has no points of
  $\Lambda$ in its interior, or on the edges, except the corners
  $\lambda, \lambda+\lambda_1, \lambda+\lambda_2,
  \lambda+\lambda_1+\lambda_2$, for any $\lambda \in \Lambda$.  So we
  have $\Lambda(z) = z\zahl + \zahl = \lambda_1 \zahl + \lambda_2
  \zahl$.  This implies that the matrix $\begin{pmatrix} m & n \\ a &
    b \end{pmatrix}$ has an inverse matrix with integer components, so
  $|mb-na| = 1$.
\end{proof}

\begin{lemma}
  Let $\frac{a}{m} < \frac{b}{n}$ be an opposed parastichy pair of
  $\Lambda(z)$.  Then we have either $a=p_i$, $m=q_i$, $b=p_{i,k}$,
  $n=q_{i,k}$ for some $i$ even or $a=p_{i,k}$, $m=q_{i,k}$, $b=p_i$,
  $n=q_i$ for some $i$ odd, and $0<k\le a_{i+1}$.
\end{lemma}
\begin{proof}
  The proof follows from Lemma~\ref{38}, since $(\frac{a}{m},
  \frac{b}{n}) \ni x$ is a Farey interval by Lemma~\ref{12}.
\end{proof}

Note that $q_i z - p_i = \fracpart{q_i x} + q_i y \rmi$, $q_{i,k} -
p_{i,k} = \fracpart{q_{i,k} x} + q_{i,k} y \rmi$ for $i\ge0$, $0 \le k
\le a_{i+1}$.  Since $q_{i,k+1} > q_{i,k}$ and
$|\fracpart{q_{i,k+1}x}| < |\fracpart{q_{i,k}x}|$, we have
\begin{equation}
\label{30}
 |\angle(q_{i}z-p_i ,0, q_{i,k+1}z-p_{i,k+1})|
   < |\angle( q_{i}z-p_i, 0, q_{i,k}z-p_{i,k})|
\end{equation}
for $i\ge0$, $0\le k < a_{i+1}$.

\begin{lemma}
$|q_{i,k}z-p_{i,k}| > |q_iz - p_i|$ for $i\ge0$, $0<k<a_{i+1}$.
\end{lemma}
\begin{proof}
Since
\begin{equation}
\label{71}
\frac{q_{i,k}}{q_i} = k + \frac{1}{a_j + \frac{1}{a_{j-1}+\dots}} > 1,
\quad
-\frac{\fracpart{q_{i,k}x}}{\fracpart{q_ix}} = a_{j+1}-k+\frac{1}{a_{j+2}+\dots} >1,
\end{equation}
we have
\[
|q_iz-p_i|^2
= \fracpart{q_ix}^2 + q_i^2y^2
< \fracpart{q_{i,k}x}^2 + q_{i,k}^2y^2
= |q_{i,k}z-p_{i,k}|.
\]
\end{proof}

\begin{lemma}
Let $a,b,c,d,t\in\real_{>0}$.  Then $\im\left(
\frac{-c+td\rmi}{a+tb\rmi} \right) \le
\half(\frac{c}{a}+\frac{d}{b})$.
\end{lemma}
\begin{proof}
$\im\left( \frac{-c+td\rmi}{a+tb\rmi} \right)
= \frac{t(ad+bc)}{a^2+t^2b^2} \le \frac{ad+bc}{2ab}
= \half(\frac{c}{a}+\frac{d}{b})$.
\end{proof}

\begin{lemma}
\label{68}
$\left| \im\left( \frac{q_{i,k}z-p_{i,k}}{q_iz-p_i} \right) \right|
\le 1+\half a_{i+1}$ for $i\ge0$, $0 \le k \le a_{i+1}$,
$(i,k)\neq(0,0)$.
\end{lemma}
\begin{proof}
\begin{align*}
\left| \im\left( \frac{q_{i,k}z-p_{i,k}}{q_iz-p_i} \right) \right|
&= \left| \im\left( \frac{\fracpart{q_{i,k}x}+q_{i,k}y\rmi}{\fracpart{q_ix}+q_iy\rmi} \right) \right| \\

& \le \half \left( \frac{q_{i,k}}{q_i} - \frac{\fracpart{q_{i,k}x}}{\fracpart{q_ix}}
\right) \\
&= \half \left( k + \frac{1}{a_i+\dots}
  + a_{i+1}-k+\frac{1}{a_{i+2}+\dots} \right) \\
&\le 1+\half a_{i+1}.
\end{align*}
\end{proof}

\begin{lemma}
  \label{31}
  Let $i\ge0$, $0\le k<a_{i+1}$, $(i,k)\neq(0,0)$.
  If $\frac{p_{i,k+1}}{q_{i,k+1}}$ is a
  parastichy index for $\Lambda(z)$,
  then $|\angle(q_{i}z-p_i, 0,
  q_{i,k}z - p_{i,k})| > \frac{\pi}{2}$.
\end{lemma}
\begin{proof}
  First suppose that $i$ is even.  Assume that
  $\angle(q_{i,k}z-p_{i,k}, 0, q_{i}z - p_{i}) \le \frac{\pi}{2}$.  In
  the parallelogram $\square(0, q_i z - p_i, q_{i,k+1}z-p_{i,k+1},
  q_{i,k}z-p_{i,k})$, we have
\[ \angle(q_{i,k+1}z-p_{i,k+1}, q_{i,k}z-p_{i,k}, 0) + \angle(0,q_{i}z-p_i,
  q_{i,k+1}z-p_{i,k+1}) > \pi.
\]
   So $p_{i,k+1}/q_{i,k+1}$ is not a parastichy index.
  The case for odd $i$ is shown by a similar argument.
\end{proof}

\begin{lemma}
  \label{32}
  Let $i\ge0$, $0< k \le a_{i+1}$, $(i,k)\neq(0,1)$.  If
  $\frac{p_{i,k-1}}{q_{i,k-1}}$ is a parastichy index for
  $\Lambda(z)$, then $|\angle(q_{i,k}z-p_{i,k}, 0, q_{i}z-p_i)| <
  \frac{\pi}{2}$.
\end{lemma}
\begin{proof}
  First suppose that $i$ is even.  Assume that $\angle(q_{i}z-p_{i},
  0, q_{i,k}z-p_{i,k}) \ge \frac{\pi}{2}$.  In the parallelogram $P =
  \square(0, q_{i}z-p_i, (q_{i}z-p_i) + (q_{i,k}z-p_{i,k}),
  q_{i,k}z-p_{i,k})$, we have
  \[
  \angle(q_{i,k}z-p_{i,k}, 0, q_{i}z-p_i) +
  \angle(q_{i}z-p_i, (q_{i}z-p_i) + (q_{i,k}z-p_{i,k}), q_{i,k}z-p_{i,k}) >
  \pi.
  \]
  In the translated parallelogram
  \[ P - (q_{i}z-p_i) =
  \square(-(q_{i}z-p_i), 0 , q_{i,k}z-p_{i,k},  q_{i,k-1}z-p_{i,k-1}),
  \]
  we have
  \[
  \angle(q_{i,k-1}z-p_{i,k-1},
  -(q_{i}z-p_i), 0) + \angle(0, q_{i,k}z-p_{i,k}, q_{i,k-1}z-p_{i,k-1}) > \pi.
  \]
  So $p_{i,k-1}/q_{i,k-1}$ is not a parastichy index.  The case for
  odd $i$ is similar.
\end{proof}

Let
\begin{equation}
  \label{26}
  \eta_{i,k}:=
  \sqrt{-\frac{(q_ix-p_i)(q_{i,k}x-p_{i,k})}{q_iq_{i,k}}}
\end{equation}
where $i\ge0$, $0 \le k \le a_{i+1}$, $(i,k)\neq(0,0)$.  It is called
Richards' formula \cite[p.29]{jean}.  We have
$\eta_{i,a_{i+1}}=\eta_{i+1,0}$ for $i\ge0$, and $\eta_{i,k} >
\eta_{i,k+1}$ for $i\ge0$, $0\le k < a_{i+1}$, $(i,k)\neq(0,0)$.

\begin{lemma}
\label{58}
  Let $z = x + \rmi y$.  Let $i\ge0$, $0\le k < a_{i+1}$,
  $(i,k)\neq(0,0)$.  If $y = \eta_{i,k}$, then the Voronoi cell $V(0)$
  is a rectangle with two parastichy indices $\frac{p_i}{q_i},
  \frac{p_{i,k}}{q_{i,k}}$.  If $\eta_{i,k+1} < y < \eta_{i,k}$, then
  $V(0)$ is a hexagon with three parastichy indices $\frac{p_i}{q_i},
  \frac{p_{i,k}}{q_{i,k}}, \frac{p_{i,k+1}}{q_{i,k+1}}$.
\end{lemma}
\begin{proof}
  First suppose that $y = \eta_{i,k}$.  Then we have
  \[
  (q_{i,k}x-p_{i,k})(q_i x- p_i) + q_{i,k}y \cdot q_i y = 0,
  \]
  and so $|\angle(q_{i,k}z-p_{i,k}, 0, q_{i}z-p_i)| = \frac{\pi}{2}$.  By
  (\ref{30}) and Lemmas~\ref{31}, \ref{32}, a parastichy index is
  either $\frac{p_i}{q_i}$ or $\frac{p_{i,k}}{q_{i,k}}$.

  Next suppose that $\eta_{i,k} > y > \eta_{i,k+1}$.  Then we have
  $|\angle(q_{i,k+1}z-p_{i,k+1},0, q_{i}z-p_i)| < \frac{\pi}{2} <
  |\angle(q_{i,k}z-p_{i,k}, 0, q_{i}z-p_i)|$.  Lemmas~\ref{31},
  \ref{32} imply that a parastichy index is either $\frac{p_i}{q_i}$,
  $\frac{p_{i,k}}{q_{i,k}}$, or $\frac{p_{i,k+1}}{q_{i,k+1}}$.  The
  Voronoi cell $V(0)$ is not a rectangle, so it is a hexagon and has
  three parastichy indices.
\end{proof}

\begin{lemma}
  \label{20}
  The area of the Voronoi cell $V(0)$ is equal to $y=\im(z)$.
\end{lemma}
\begin{proof}
  By Lemma~\ref{12}, the cell $V(0)$ is adjacent to some $V(mz-a)$,
  $V(nz-b)$, with $|mb-na|=1$.  The cells are parallel translates to
  each other, $V(\lambda) = \lambda + V(0)$, so the area of the cells
  $|V(\lambda)| = |V(0)|$ are equal to the area of the parallelogram
  $H = \square(0, mz-a, (m+n)z-(a+b), nz-b)$,
  \[
  |H| = \left| \det \begin{pmatrix} mx -a & nx -b \\ my & ny
  \end{pmatrix} \right|
  = |(mb - na)y| = y.
  \]
\end{proof}

\begin{figure}
\centering
(a)
\includegraphics[scale=0.6]{Fig_4_a.eps}

(b)
\includegraphics[scale=0.6]{Fig_4_b.eps}

\caption{(a) The set $\fdom \cup (-\fdom)$ of parameters $\lambda_2 / \lambda_1$
for the pairs of generators $\lambda_1, \lambda_2$ of linear lattices.
(b) The half region $\fdom_0$.}
\end{figure}

We fix $x \in \real$ and suppose that $\eta_{i,k} \ge y > \eta_{i,k+1}$.
There exist $\lambda_1, \lambda_2
\in \set{ q_iz-p_i, q_{i,k}z-p_{i,k}, q_{i,k+1}z-p_{i,k+1}}$ such that
\[
  \lambda_{2}/\lambda_{1} \in \fdom \cup (-\fdom),
\]
where
\[
\fdom = \set{w=u+v\rmi \in \cx :
  -\half \le u \le \half, v \ge \frac{\sqrt{3}}{2}, |w|\ge1}.
\]
Denote by
\[
\delta_y := |\lambda_1|,
\]
the length of the shortest vector in $\Lambda(z)$, $z=x+y\rmi$.  The
area of the parallelogram $H = \square(0, \lambda_1, \lambda_1 +
\lambda_2, \lambda_2)$ is equal to $v_y \delta_y^2$, where $v_y :=
|\im(\lambda_2/\lambda_1)|$.  By Lemma~\ref{20} we have $v_y
\delta_y^2 = y$, so
\begin{equation}
\label{72}
\delta_y^2 = \frac{y}{v_y} \le \frac{2}{\sqrt{3}}y.
\end{equation}

\section{Voronoi cells on a perturbation of a linear lattice}

A spiral lattice is locally a perturbation of a linear lattice.  Here
we show that the area of the Voronoi cell in a linear lattice is close
to the area of the Voronoi cell in the perturbed lattice.  A vertex of
a Voronoi tessellation is given as a circumcenter of a triangle.  So
we first consider the perturbation of the circumdisk of a triangle.

\begin{lemma}
\label{50}
Let $f : \cx \to \real$ be a $C^1$ function, and denote by $f(x,y) =
f(x+y\rmi)$ by abuse of notation.  Let $\grad f = (\frac{\partial
  f}{\partial x}, \frac{\partial f}{\partial y})$.  Then we have
$|\grad f|^2 = 4 \frac{\partial f}{\partial z} \frac{\partial
  f}{\partial \bar{z}}$.
\end{lemma}

\begin{proof}
By definition we have $\frac{\partial f}{\partial z} = \half \left(
\frac{\partial f}{\partial x} - \rmi \frac{\partial f}{\partial y}
\right)$, $\frac{\partial f}{\partial \barz} = \half \left(
\frac{\partial f}{\partial x} + \rmi \frac{\partial f}{\partial y}
\right)$.  So we obtain
\[
|\grad f|^2
= \left( \frac{\partial f}{\partial x} \right)^2 +
\left( \frac{\partial f}{\partial y} \right)^2
=  \left( \frac{\partial f}{\partial z}
 + \frac{\partial f}{\partial \barz}\right)^2
 + \left( \frac{\partial f}{\partial z}
 - \frac{\partial f}{\partial \barz}\right)^2
= 4 \frac{\partial f}{\partial z} \frac{\partial f}{\partial \bar{z}}.
\]
\end{proof}

\begin{lemma}
\label{51}
Let $z, z' \in \cx$.  Let $f$ be a real $C^1$ function defined in a
neighborhood of the line segment $\ell(z, z')$.  Suppose that $|\grad
f| \le M_0$ for some $M_0>0$.  Then we have
\[ |f(z') - f(z)|
\le M_0 |z' - z|.
\]
\end{lemma}

\begin{proof}
Let $z(t) = (1-t)z + t z'$, $0 \le t \le 1$.  Then we obtain
\[
 |f(z') - f(z)|
 \le \int_0^1 |\grad f| \; |z'-z|\;dt
 \le M_0 |z'-z|.
\]
\end{proof}

Let $w_0, w_1, w_2 \in \cx$ be distinct.  Assume that
$\im(\frac{w_2-w_0}{w_1-w_0})>0$.  The circumcenter $w_c$ of the
triangle $\triangle(w_0, w_1, w_2)$ is written as
\[
 w_c

  = \frac{w_1+w_2}{2} + \frac{\rmi(w_2-w_1)}{2} f(w_0)
\]
where we denote by
\[ f(w_0)
 := \cot\angle(w_2,w_0,w_1)
 = \frac{\re((\overline{w_1-w_0})(w_2-w_0))}{\im((\overline{w_1-w_0})(w_2-w_0))}
\]
as a function of $w_0$.
A direct calculation shows that
\[ |\grad f| =
\frac{|(w_2-w_1)(w_1-w_0)(w_2-w_0)|}{(\im((\overline{w_1-w_0})(w_2-w_0)))^2},
\]
which is a symmetric function of $w_0, w_1, w_2$.

Let
\[ \fdom_0 = \set{\ww = \uu + \vv\rmi :
  0 \le \uu \le \half, \vv \ge \frac{\sqrt{3}}{2}, |\ww|\ge1}.
\]
Let $U(z, \epsilon) = \set{\zeta \in\cx : |\zeta - z| < \epsilon}$ be
a neighborhood of $z \in \cx$.

\begin{lemma}
\label{52}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.
Let $w_0 \in U(0,\epsilon)$,
$w_1 \in U(1,\epsilon)$,
$w_2 \in U(\ww, \epsilon)$,
and suppose that $\epsilon \le \frac{1}{50}$.
Then we have
\begin{align*}
  |w_1-w_0| &\le 1+2\epsilon , \\
  |w_2-w_0| &\le \frac{2\vv}{\sqrt{3}}(1+2\epsilon) , \\
  |w_2-w_1| &\le \frac{\vv}{\sqrt{3}}(\sqrt{6}+4\epsilon)  ,
\end{align*}
and
\[
 \frac{|(w_2-w_1)(w_1-w_0)(w_2-w_0)|}{(\im((\overline{w_1-w_0})(w_2-w_0)))^2}
  \le \frac{7}{3} .
\]
\end{lemma}
\begin{proof}
Let $w_0 = \epsilon_1 + \rmi \epsilon_2$,
$w_1 = 1 + \epsilon_3 + \rmi \epsilon_4$,
$w_2 = \uu + \epsilon_5 + \rmi (\vv + \epsilon_6)$.
Then we have
\begin{align*}
 |w_1 - w_0|
 &\le |w_1 - 1| + |1- 0| + |0-w_0|
 \le 1+2\epsilon, \\
 \frac{|w_2 - w_0|}{\vv}
  &\le \frac{|w_2-w|}{\vv} + \frac{|w-0|}{\vv} + \frac{|0-w_0|}{\vv}
 \le \frac{2\epsilon}{\sqrt{3}} + \frac{2}{\sqrt{3}}
 + \frac{2\epsilon}{\sqrt{3}}
  = \frac{2}{\sqrt{3}}(1+2\epsilon), \\
 \frac{|w_2 - w_1|}{\vv}
  &\le \frac{|w_2-w|}{\vv} + \frac{|w-1|}{\vv} + \frac{|1-w_1|}{\vv}
  \le   \frac{2\epsilon}{\sqrt{3}} + \sqrt{2} + \frac{2\epsilon}{\sqrt{3}}
  = \frac{\sqrt{6}+4\epsilon}{\sqrt{3}},
\end{align*}
\begin{align*}
  \frac{\im((\overline{w_1-w_0})(w_2-w_0))}{\vv}
  &= \frac{(1+\epsilon_3-\epsilon_1)(\vv+\epsilon_6-\epsilon_2)
    -(\uu+\epsilon_5-\epsilon_1)(\epsilon_4-\epsilon_2)}{\vv} \\
 &\ge (1-2\epsilon) \left(1-\frac{4\epsilon}{\sqrt{3}}\right)
   - \frac{4\epsilon}{\sqrt{3}} \left( \half+2\epsilon \right) \\
 &= 1-(2+2\sqrt{3})\epsilon,
\end{align*}
and
\begin{align*}
\frac{|(w_2-w_1)(w_1-w_0)(w_2-w_0)|}{(\im((\overline{w_1-w_0})(w_2-w_0)))^2}
&\le \frac{2(1+2\epsilon)^2(\sqrt{6}+4\epsilon)}{3(1-(2+2\sqrt{3})\epsilon)^2}
\le \frac{7}{3}
\end{align*}
for $0<\epsilon < \frac{1}{50}$.
\end{proof}

\begin{lemma}
\label{53}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.  Let $w_0 \in U(0,\epsilon)$, $w_1
\in U(1,\epsilon)$, $w_2 \in U(\ww, \epsilon)$.  Suppose that
$\epsilon \le \frac{1}{45+10\vv}$.  Let $w_c$ be the circumcenter of
$\triangle(w_0,w_1,w_2)$.  Then we have $\frac{1}{4} \le \re(w_c) \le
\frac{3}{4}$, $\frac{\vv}{4} \le \im(w_c) \le \frac{3\vv}{4}$.
\end{lemma}
\begin{proof}
The circumcenter $w_{c,0}$ of $\triangle(0,1,\ww)$ is written as
\[ w_{c,0} = \half + \frac{\uu^2-\uu+\vv^2}{2\vv}\rmi,
\]
where we have
\[ \frac{\vv}{3} \le \frac{\vv}{2(1+\uu)} \le \frac{\uu^2-\uu+\vv^2}{2\vv} \le \frac{\vv}{2}
\]
for $\ww \in D$.
Let $w_{c,1}$ be the circumcenter of $\triangle(0,w_1,w_2)$,
and $w_{c,2}$ the circumcenter of $\triangle(0,1,w_2)$.
By Lemmas \ref{51} and \ref{52}, we have
\begin{align*}
|w_{c,2} - w_{c,0}| &\le \frac{|1-0|}{2} \cdot \frac{7}{3} |w_2 - \ww|
 \le \frac{7\epsilon}{6} , \\
|w_{c,1} - w_{c,2}| &\le \frac{|w_2-0|}{2} \cdot \frac{7}{3} |w_1-1|
 \le  \frac{2\vv}{\sqrt{3}} (1+\epsilon)\cdot \frac{7\epsilon}{6} , \\
|w_c - w_{c,1}| &\le \frac{|w_2-w_1|}{2} \cdot \frac{7}{3} |w_0 - 0|
 \le  \frac{\vv}{\sqrt{3}} (\sqrt{6}+4\epsilon) \cdot \frac{7\epsilon}{6}  ,
\end{align*}
which imply that
\[
|w_c - w_{c,0}|
\le \frac{7\epsilon}{6}
   \left(1+\frac{2\vv(1+\epsilon)}{\sqrt{3}}
       + \frac{v(\sqrt{6}+4\epsilon)}{\sqrt{3}} \right)
\le \min\Set{ \frac{\vv}{12}, \frac{1}{4} }
\]
for $\epsilon < \frac{1}{45+10v}$.
This completes the proof.
\end{proof}

A mapping $\varphi : \cx \to \cx$ is called an $\epsilon$-perturbation
if it is a homeomorphisms and $|\varphi(\zeta) -\zeta| < \epsilon$ for
any $\zeta \in \cx$.

\begin{lemma}
\label{54}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.  Let $\Lambda = \zahl + \ww \zahl$ be
a linear lattice.  Let $\varphi : \cx \to \cx$ be an
$\epsilon$-perturbation, where we suppose that $\epsilon \le
\frac{1}{45+10\vv}$.  Then, for any $\lambda \in \Lambda \setminus
\set{0,1,\ww, 1+\ww}$, $\varphi(\lambda)$ is out of the circumdisk $K$
of $\triangle(\varphi(0), \varphi(1), \varphi(\ww))$.
\end{lemma}

\begin{proof}
Denote by $\varphi(0) = \epsilon_1 + \epsilon_2 \rmi$, $\varphi(1) = 1
+ \epsilon_3 + \epsilon_4 \rmi$, $\varphi(\ww) = \uu + \vv\rmi +
\epsilon_5 + \epsilon_6 \rmi$.  Let $w_c = \sigma + \rho\rmi$ be the
circumcenter of $\triangle(\varphi(0), \varphi(1), \varphi(\ww))$.  By
Lemma~\ref{53}, we have $\frac{1}{4} \le \sigma \le \frac{3}{4}$,
$\frac{\vv}{4} \le \rho \le \frac{3\vv}{4}$.  Let $r$ be the radius
of $K$, where we have
\[ r^2 = (\epsilon_1 - \sigma)^2 + (\epsilon_2 - \rho)^2
 = (1+\epsilon_3 - \sigma)^2 + (\epsilon_4 - \rho)^2
 = (\uu + \epsilon_5 - \sigma)^2 + (\vv + \epsilon_6 - \rho)^2.
\]

Let $(j,k) \in \zahl^2 \setminus \set{(0,0), (1,0), (0,1), (1,1)}$.
We are going to show that $|\varphi(j+k\ww) - w_c| > r$.  Denote by
$\varphi(j+k\ww) = j + k\ww + \epsilon_7 + \epsilon_8\rmi$, where $\epsilon_7^2 +
\epsilon_8^2 < \epsilon^2$.

Case (i).  If $j\ge2$ and $k=0$, then
\begin{align*}
|\varphi(j) - w_c|^2 - r^2
&= (j+\epsilon_7 - \sigma)^2 + (\epsilon_8 - \rho)^2
- (1 + \epsilon_3 - \sigma)^2 - (\epsilon_4 - \rho)^2 \\
&= (1+j -2\sigma+\epsilon_7+\epsilon_3)(j-1+\epsilon_7-\epsilon_3)
  + (-2\rho + \epsilon_8 + \epsilon_4)(\epsilon_8-\epsilon_4) \\
 &\ge (\frac{3}{2} - 2\epsilon)(1-2\epsilon)
  - (\frac{3\vv}{2}+2\epsilon) 2\epsilon
> 0.
\end{align*}

Case (ii).  If $j\le-1$ and $k=0$, then
\begin{align*}
|\varphi(j) - w_c|^2 - r^2
&= (j+\epsilon_7 - \sigma)^2 + (\epsilon_8 - \rho)^2
- (\epsilon_1 - \sigma)^2 - (\epsilon_2 - \rho)^2 \\
&= (j -2\sigma+\epsilon_7+\epsilon_1)(j+\epsilon_7-\epsilon_1)
  + (-2\rho + \epsilon_8 + \epsilon_2)(\epsilon_8-\epsilon_2) \\
  &\ge (\frac{3}{2} - 2\epsilon)(1-2\epsilon)
  - (\frac{3\vv}{2}+2\epsilon) 2\epsilon
> 0.
\end{align*}

Case (iii).  If $j\in\zahl\setminus\set{0,1}$ and $k=1$, then
\begin{align*}
&|\varphi(j+\ww) - w_c|^2 - r^2 \\
&= (j+\uu + \epsilon_7 - \sigma)^2 + (\vv+\epsilon_8 - \rho)^2
  - (\uu + \epsilon_5 - \sigma)^2 - (\vv+\epsilon_6 - \rho)^2 \\
&= (j +2\uu-2\sigma+\epsilon_7+\epsilon_5)(j+\epsilon_7-\epsilon_5)
  + (2\vv-2\rho + \epsilon_8 + \epsilon_6)(\epsilon_8-\epsilon_6) \\
&\ge (\frac{1}{2} - 2\epsilon)(2-2\epsilon)
  - (\frac{3\vv}{2}+2\epsilon) 2\epsilon
> 0.
\end{align*}

Case (iv). If $k\ge2$, then
\begin{align*}
& |\varphi(j+k\ww) - w_c|^2 - r^2 \\
&= (j+k\uu + \epsilon_7 - \sigma)^2 + (k\vv+\epsilon_8 - \rho)^2
    - (\uu + \epsilon_5 - \sigma)^2 - (\vv+\epsilon_6 - \rho)^2 \\
&= (j +(k+1)\uu-2\sigma+\epsilon_7+\epsilon_5)(j+(k-1)\uu+\epsilon_7-\epsilon_5) \\
& \qquad  + ((k+1)\vv-2\rho + \epsilon_8 + \epsilon_6)((k-1)\vv+\epsilon_8-\epsilon_6) \\
&\ge - (\uu-\sigma+\epsilon_5)^2 + (\frac{3\vv}{2} - 2\epsilon)(\vv-2\epsilon) \\
&\ge - (\frac{3}{4}+\epsilon)^2 + (\frac{3\vv}{2} - 2\epsilon)(\vv-2\epsilon)
> 0,
\end{align*}
where we used the inequality $(t-\mu)(t-\nu) \ge - \frac{(\mu-\nu)^2}{4}$.

Case (v).  If $k\le-1$, then
\begin{align*}
& |\varphi(j+k\ww) - w_c|^2 - r^2 \\
&= (j+k\uu+\epsilon_7 - \sigma)^2 + (k\vv+\epsilon_8 - \rho)^2
   - (\epsilon_1 - \sigma)^2 - (\epsilon_2 - \rho)^2 \\
&= (j+k\uu -2\sigma+\epsilon_7+\epsilon_1)(j+k\uu+\epsilon_7-\epsilon_1)
  + (k\vv-2\rho + \epsilon_8 + \epsilon_2)(k\vv+\epsilon_8-\epsilon_2) \\
&\ge - (\sigma - \epsilon)^2 + (\frac{3\vv}{2}-2\epsilon)(\vv- 2\epsilon) \\
&\ge - (\frac{3}{4} - \epsilon)^2 + (\frac{3\vv}{2}-2\epsilon)(\vv- 2\epsilon) > 0.
\end{align*}
\end{proof}

\begin{lemma}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.

Let $\Lambda = \zahl + \ww \zahl$ be a linear lattice.  Let $\varphi :
\cx \to \cx$ be an $\epsilon$-perturbation, where we suppose that
$\epsilon \le \frac{1}{45+10\vv}$.  If the Voronoi cell $V(\varphi(0),
\varphi(\Lambda))$ is adjacent to $V(\varphi(j+k\ww),
\varphi(\Lambda))$, then $|j|\le 1$ and $|k| \le 1$.
\end{lemma}

\begin{proof}
For distinct $w_0, w_1, w_2 \in \cx$,
denote the closed circumdisk of $\triangle(w_0, w_1, w_2)$
by $\KK(w_0,w_1,w_2)$.
Let $(j,k) \in \zahl^2$, and suppose that $|j|\ge2$ or $|k|\ge2$.
If $0 \le \uu \le \half$, the union of the closed disks
\begin{align}
K &:=\KK(\varphi(0),\varphi(1), \varphi(\ww))
 \cup \KK(\varphi(0), \varphi(\ww), \varphi(-1+\ww))
 \notag \\
 &\cup \KK(\varphi(0), \varphi(-1+\ww), \varphi(-1))
  \cup \KK(\varphi(0), \varphi(-1), \varphi(-\ww))
 \notag \\
 &\cup \KK(\varphi(0), \varphi(-\ww), \varphi(1-\ww))
 \cup \KK(\varphi(0), \varphi(1-\ww), \varphi(1))
\label{55}
\end{align}
is a closed neighborhood of $\varphi(0)$.  Lemma~\ref{54} applies to
the six circumdisks in (\ref{55}), and we obtain $\varphi(j+k\ww)
\not\in K$.  This implies that the Voronoi cell $V(\varphi(0),
\varphi(\Lambda))$ is not adjacent to $V(\varphi(j+k\ww),
\varphi(\Lambda))$.

If $-\half \le \uu \le 0$, the union of the closed disks
\begin{align}
K &=\KK(\varphi(0),\varphi(1), \varphi(1+\ww))
 \cup \KK(\varphi(0), \varphi(1+\ww), \varphi(-1))
 \notag \\
 &\cup \KK(\varphi(0), \varphi(1), \varphi(-1))
 \cup \KK(\varphi(0), \varphi(-1), \varphi(-1-\ww))
 \notag \\
 &\cup \KK(\varphi(0), \varphi(-1-\ww), \varphi(-\ww))
 \cup \KK(\varphi(0), \varphi(-\ww), \varphi(1))
\label{56}
\end{align}
is a closed neighborhood of $\varphi(0)$.  Since $K \not\ni
\varphi(j+k\ww)$, $V(\varphi(0), \varphi(\Lambda))$ is not adjacent to
$V(\varphi(j+k\ww), \varphi(\Lambda))$.
\end{proof}

Let $R_2 > R_1 > \epsilon > 0$.  Let $U \subset \cx$ be a neighborhood
of the origin such that $U(0,R_2) \subset U$.  A mapping $\varphi : U
\to \cx$ is called a local $(\epsilon, R_1, R_2)$-perturbation if it
is a homeomorphism onto its image, and $|\varphi(\zeta) - \zeta| <
\epsilon$ for $|\zeta| \le R_2$, and $|\varphi(\zeta)| \ge R_1$ for
$|\zeta|\ge R_2$.

\begin{lemma}
\label{60}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.

Let $\Lambda = \zahl + \ww \zahl$ be a linear lattice.  Let $\varphi :
\cx \to \cx$ be a local $(\epsilon, 3+3\vv, 4+4\vv)$-perturbation,
where we suppose that $\epsilon \le \frac{1}{45+10\vv}$.  If
$V(\varphi(0), \varphi(\Lambda))$ is adjacent to $V(\varphi(j+k\ww),
\varphi(\Lambda))$, then $|j|\le 1$ and $|k| \le 1$.
\end{lemma}
\begin{proof}
Let $K$ be a closed neighborhood of $\varphi(0)$, defined in
(\ref{55}) or (\ref{56}).  Then $K \subset U(0, 2+2\vv)$.  If $j+k\ww
\in \Lambda$ and $|j+k\ww| \le U(0,4+4\vv)$, then Lemma~\ref{54}
implies that $V(\varphi(0), \varphi(\Lambda))$ is not adjacent to
$V(\varphi(j+k\ww), \varphi(\Lambda))$.  If $j+k\ww \in \Lambda$ and
$|j+k\ww| \ge U(0,4+4\vv)$, then $|\varphi(j+k\ww)| \ge 3+3\vv$, so
$\varphi(j+k\ww) \not\in K$, and $V(\varphi(0), \varphi(\Lambda))$ is
not adjacent to $V(\varphi(j+k\ww), \varphi(\Lambda))$.
\end{proof}

\begin{lemma}
\label{57}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.

Let $\Lambda = \zahl + \ww \zahl$ be a linear lattice.  Let $\varphi :
\cx \to \cx$ be a local $(\epsilon, 3+3\vv, 4+4\vv)$-perturbation, and
suppose that $\epsilon \le \frac{1}{45+10\vv}$.  Then $V(\varphi(0),
\varphi(\Lambda)) \subset U(0, \frac{3}{4}(1+v))$.
\end{lemma}
\begin{proof}
Denote the circumcenter of $\triangle(w_0, w_1, w_2)$ by $P(w_0, w_1,
w_2)$.  The Voronoi cell $V(\varphi(0), \varphi(\Lambda))$ is a subset
of the quadrilateral with the vertices
$P(\varphi(0),\varphi(1), \varphi(\ww))$,
$P(\varphi(0), \varphi(\ww), \varphi(-1))$,
$P(\varphi(0), \varphi(-1), \varphi(-\ww))$, and
$P(\varphi(0), \varphi(-\ww), \varphi(1))$,
which are all contained in the region
$\set{\zeta : -\frac{3}{4} \le \re(\zeta) \le \frac{3}{4},
  -\frac{3v}{4} \le \im(\zeta)\le \frac{3v}{4}}
  \subset U(0, \frac{3}{4}(1+v))$ by Lemma~\ref{53}.
\end{proof}

Let $d>0$.  Let $W(d) = \set{\zeta \in \cx : -d \le \im(\zeta) \le d}$
be a strip of width $2d$.  Let $H(w) = \set{\zeta \in \cx: |\zeta| \le
  |\zeta-w|}$, $w \in \punccx := \cx \setminus \set{0}$, be a half
plane.  The symmetric difference of two subsets $A,B \subset\cx$ is
denoted by $A \triangle B = (A\setminus B) \cup (B \setminus A)$.  The
area of a subset $A \subset\cx$ is denoted by $|A|$.

\begin{lemma}
  \label{8}
  Let $d,\epsilon>0$, and $w =\uu+ \rmi \vv \in\cx$.  Suppose
  that $|w-1| < \epsilon < \frac{1}{2}$.  Then $|W(d) \cap(H(1)
  \;\triangle\;H(w))| < 4\epsilon d(2d+1)$.
\end{lemma}
\begin{proof}
The perpendicular bisector of the line segment $\ell(0,w)$ intersect
  the horizontal line $\set{\zeta\in\cx: \im(\zeta)=d}$ at the point
  $\frac{w}{2} + \frac{\rmi w}{\uu}(d-\frac{\vv}{2})
   = \frac{\uu}{2} - \frac{\vv}{\uu}(d-\frac{\vv}{2}) + \rmi d$.
  Since $|\uu- 1|<\epsilon$ and $|\vv| < \epsilon$, we have
\[ \left| \frac{\uu}{2} - \frac{\vv}{\uu}(d-\frac{\vv}{2}) - \half \right|
  < \frac{\epsilon}{2} + {2\epsilon} \left( d+\frac{1}{4} \right)
  = \epsilon({2d}+1).
\]
  So we have $W(d)\cap(H(1) \;\triangle\;H(w)) \subset L$, where
  $L:=\set{x+\rmi y: |x-\frac{1}{2}| \le \epsilon({2d}+1),
    |y| \le d}$.  The area of the rectangle $L$ is $|L| =
  2\epsilon({2d}+1) \cdot 2d$.
\end{proof}

\begin{proposition}
\label{66}
Let $\ww = \uu + \vv\rmi \in \fdom_0$.

Let $\Lambda = \zahl + \ww \zahl$ be a linear lattice.
Let $\varphi : \cx \to \cx$ be
a local $(\epsilon, 3+3\vv, 4+4\vv)$-perturbation.
Suppose that $\varphi(0)=0$ and $\epsilon \le \frac{1}{45+10\vv}$.
Then
\[ |V(0, \Lambda) \;\triangle\; V(0, \varphi(\Lambda))|
\le 12 \epsilon(1+v)(5+3v).
\]
\end{proposition}
\begin{proof}
Let $N_1 = \set{\pm 1, \pm \ww, \pm (1+w), \pm(1-\ww)}$.
By Lemmas \ref{60} and \ref{57}, we have
\begin{align*}
V(0, \Lambda) &= \bigcap_{\lambda\in N_1} H(\lambda) \subset U(0, R), \\
V(0, \varphi(\Lambda)) &= \bigcap_{\lambda\in N_1} H(\varphi(\lambda)) \subset U(0, R)
\end{align*}
where $R:= \frac{3}{4}(1+v)$.  So
\[ V(0,\Lambda) \;\triangle\; V(0,\varphi(\Lambda))
 \subset \bigcup_{\lambda \in N_1}((H(\lambda) \;\triangle\; H(\varphi(\lambda))) \cap U(0,R).
\]
By Lemma~\ref{8}, we have
\[
 |(H(\lambda) \;\triangle\; H(\varphi(\lambda))) \cap U(0, R)|
 \le 4\frac{\epsilon}{|\lambda|} \frac{R}{|\lambda|} \left(\frac{2R}{|\lambda|}+1 \right) |\lambda|^2
 \le  4\epsilon R (2R +1).
\]
So we obtain
\[
   |V(0,\Lambda) \;\triangle\; V(0,\varphi(\Lambda))|
   \le 32 \epsilon R(2R+1)
    = 12 \epsilon(1+v)(5+3v).
\]
\end{proof}

\section{Area convergence of Voronoi cells}

In this section, an Archimedean spiral lattice is parameterized, normalized, and then linearized to show the area convergence of Voronoi cells.

Let $\alpha>0$, $\theta \in \real$, $\mu \ge 1$.  The set
\[
  S_\mu := \set{z_{\mu+j} :=
  (\mu+j)^\alpha \rme^{(\mu+j)\theta\rmi} : j\in\zahl, j> -\mu}
\cup \set{0}
\]
is called a parameterized Archimedean spiral lattice with an exponent
$\alpha$.  Note that $S_{\mu} = S_{\mu+1}$
as a point set.  So the $j$-th site $z_{\mu+j}$ in $S_\mu$ can be
regarded as the $0$-th site in $S_{\mu+j}$.  The Voronoi cell of
$z_{\mu+j}$ in $S_\mu$ is given as
\[
  V(z_{\mu+j}, S_\mu)
:= \set{\zeta \in \cx:  |\zeta - z_{\mu+j}| \le |\zeta - z'|,
 \forall z' \in S_\mu}.
\]

The point set
\[
 \Sigma_\mu
 := \Set{\tildez_{\mu,j}:= \left(1+ \frac{j}{\mu}\right)^\alpha
  \rme^{j\theta\rmi} : j\in\zahl, j>-\mu} \cup \set{0}
  = \mu^{-\alpha} \rme^{- \mu \theta \rmi} S_\mu
\]
is
called a normalized spiral lattice.  The Voronoi cell of
$\tildez_{\mu,j}$ in $\Sigma_\mu$ is given as
\[
V(\tildez_{\mu,j},\Sigma_\mu)
:= \set{\zeta \in \cx:  |\zeta - \tildez_{\mu,j}| \le |\zeta - \tildez'|,
 \forall \tildez' \in \Sigma_\mu}.
\]
We have $V(z_{\mu+j}, S_\mu) = \mu^\alpha \rme^{\mu \theta \rmi}
V(z_{\mu,j},\Sigma_\mu)$ and $|V(z_{\mu+j}, S_\mu)| = \mu^{2\alpha}
|V(z_{\mu,j}, \Sigma_\mu)|$.  In particular, by taking $j=0$ we have
\begin{equation}
  \label{13}
  |V(z_{\mu}, S_\mu)| = \mu^{2\alpha} |V(1,\Sigma_\mu)|.
\end{equation}

The linear lattice
\[ \Lambda_\mu := \lambda_\mu \zahl + 2\pi \rmi \zahl,
\quad \lambda_\mu := \frac{\alpha}{\mu} + 2\pi \fracpart{\frac{\theta}{2\pi}} \rmi
\]
is called a linearization of $S_\mu$.  Let $\lambda_{\mu,j}:=\frac{j
  \alpha}{\mu} + 2\pi \fracpart{\frac{j\theta}{2\pi}} \rmi \in
\Lambda_\mu$, $j>-\mu$.  The Voronoi cell of $\lambda_{\mu,j}$ in
$\Lambda_\mu$ is given as
\[ V(\lambda_{\mu,j},\Lambda_\mu) =
 \set{\zeta \in \cx:  |\zeta - \lambda_{\mu,j}| \le |\zeta - \lambda'|,
 \forall \lambda' \in \Lambda_\mu}.
\]

Let $W=\set{s+\rmi t : s>-\mu, -\pi<t<\pi}$.  Define a mapping
$\phi : W \to \cx$ by
\[ \phi(s+\rmi t) = \left( 1+\frac{s}{\mu} \right)^\alpha \rme^{\rmi t}.
\]
 It is a homeomorphism of
$W$ onto $\phi(W) = \cx \setminus \set{\zeta \in \real: \zeta \le 0}$.
We have
\[ \phi(W \cap \Lambda_\mu) \cup \set{0} = \Sigma_\mu
\]
if $\frac{\theta}{2\pi}$ is irrational.  By Taylor's theorem, there
exists a constant $C=C_{\alpha}>0$ which depends only on $\alpha$,
such that
\[ |\phi(s+\rmi t)-(1+\frac{\alpha s}{\mu} +\rmi t)| \le C_\alpha (s^2 + t^2)
\]
for $s + t\rmi \in U(0,1)$.

\bigskip

Suppose that $\frac{\theta}{2\pi}$ is irrational.  We consider the
continued fraction expansion of $x=\frac{\theta}{2\pi}$ as in
(\ref{25}).  We assume that
\begin{equation}
\label{67}
     a_i \le M_1,
\end{equation}
for some constant $M_1>0$.

An integer $j>0$ is called a parastichy number of the linear lattice
$\Lambda_\mu$ if $V(0,\Lambda_\mu)$ is edge-adjacent to
$V(\lambda_{\mu,j},\Lambda_\mu)$.  Let
\begin{equation}
  \label{16}
  \mu_{i,k}
= \frac{\alpha}{2\pi} \sqrt{ - \frac{q_i q_{i,k}}
{\fracpart{\frac{q_i \theta}{2\pi}}
  \fracpart{\frac{q_{i,k}\theta}{2\pi}}}}
\end{equation}
for $i\le1$ and $0 \le k \le a_{i+1}$.  Note that
$\mu_{i,k}<\mu_{i,k+1}$ for $0\le k < a_{i+1}$, and $\mu_{i,a_{i+1}} =
\mu_{i+1,0}$.  If $\mu=\mu_{i,k}$, we have
\[
 q_{i} q_{i,k} \left(\frac{\alpha}{2\pi\mu}\right)^2
+ \fracpart{\frac{q_i\theta}{2\pi}}
\fracpart{\frac{q_{i,k}\theta}{2\pi}} = 0,
\]
and so $|\angle(\lambda_{\mu,q_i},0,\lambda_{\mu,q_{i,k}})| =
\frac{\pi}{2}$.

Let $i\ge1$, $0\le k < a_{i+1}$.  Lemma~\ref{58} implies that if $\mu
= \mu_{i,k}$, then the Voronoi cell $V(0, \Lambda_\mu)$ is a rectangle
with two parastichy numbers $q_i, q_{i,k}$.  If $\mu_{i,k} < \mu <
\mu_{i,k+1}$, then $V(0,\Lambda_\mu)$ is a hexagon with three
parastichy numbers $q_i, q_{i,k}, q_{i,k+1}$.

Suppose that $\mu_{i,k} \le \mu < \mu_{i,k+1}$.  Then there exist
parastichy numbers $m,n \in \set{q_i, q_{i,k}, q_{i,k+1}}$
of $\Lambda_\mu$ such that
\begin{equation}
  \label{21}
  \lambda_{\mu,n}/\lambda_{\mu,m} \in \fdom \cup (-\fdom),
\end{equation}
where $\fdom = \set{w=u+v\rmi \in \cx : -\half \le u \le \half, v \ge
 \frac{\sqrt{3}}{2}, |w|\ge1}$.  Let
\[ \delta_\mu := |\lambda_{\mu,m}|
 = \min\set{ |\lambda_{\mu,q_i}|, |\lambda_{\mu,q_{i,k}}|, |\lambda_{\mu,q_{i,k+1}}|}
\]
be the length of the shortest nonzero vector in $\Lambda_\mu$.
Let $v_\mu := |\im(\lambda_{\mu,n} / \lambda_{\mu,m})|$.
Then we have $|V(0,\Lambda_\mu)| = v_\mu \delta_\mu^2$,
$v_\mu \ge \frac{\sqrt{3}}{2}$,
and
\[ \left( \frac{\delta_\mu}{2\pi} \right)^2
\le \frac{2}{\sqrt{3}} \frac{\alpha}{2\pi\mu}
\]
as in (\ref{72}).
Lemma~\ref{68} implies that
$v_\mu \le M_2 := 1 + \half M_1$.

Let
\[
 \epsilon_\mu := C_\alpha (4+4M_2)^2 \sqrt{ \frac{4\pi\alpha}{\sqrt{3} \mu}}.
\]

\begin{lemma}
\label{70}
Suppose (\ref{67}) and
\begin{equation}
\label{63}
 \mu \ge \frac{4\pi\alpha}{\sqrt{3}} (16C_\alpha(1+M_2))^2.
\end{equation}
Then the mapping $-1+\phi : W \to \cx$ is a local
$(\epsilon_\mu\delta_\mu, (3+3v_\mu)\delta_\mu,
(4+4v_\mu)\delta_\mu)$-perturbation.
\end{lemma}
\begin{proof}
For $|\zeta| \le (4 + 4 v_\mu)\delta_\mu$, we have
\[
 |\phi(\zeta)-1-\zeta|
\le C_\alpha |\zeta|^2 \le C_\alpha (4+4 v_\mu)^2\delta_\mu^2
 \le \epsilon_\mu \delta_\mu.
\]
For $|\zeta| = (4+4 v_\mu)\delta_\mu$, we have
\begin{align*}
 |-1+\phi(\zeta)|
 &\ge |\zeta| - |\phi(\zeta)-1 - \zeta|  \\
 &\ge (4+4 v_\mu)\delta_\mu - C_\alpha(4+4 v_\mu)^2\delta_\mu^2  \\
 &\ge (3 + 3v_\mu)\delta_\mu
\end{align*}
by (\ref{63}).
\end{proof}

\begin{lemma}
\label{73}
Suppose (\ref{67}), (\ref{63}),
and
\begin{equation}
\label{69}
\mu \ge  \frac{4\pi\alpha}{\sqrt{3}} C_\alpha^2 (45+10M_2)^2 (4+4M_2)^4.
\end{equation}
Then there exists a contant $C>0$ depending only on $\alpha$, such that
\begin{equation}
\label{24}
  \frac{|V(0,\Lambda_\mu) \;\triangle\; V(0,\phi(W \cap \Lambda_\mu)-1)|}
       {|V(0,\Lambda_\mu)|}
  \le \frac{C}{\sqrt{\mu}}.
\end{equation}
\end{lemma}
\begin{proof}
By (\ref{69}), we have $\epsilon_\mu \le \frac{1}{45+10v_\mu}$.
Proposition~\ref{66} and Lemma~\ref{70} imply that
\[
 \frac{|V(0,\Lambda_\mu) \;\triangle\; V(0,\phi(W \cap \Lambda_\mu)-1)|}      {\delta_\mu^2}
 \le 12 \epsilon_\mu (1+v_\mu)(5+3v_\mu).
\]
Since $|V(0,\Lambda_\mu)| = v_\mu \delta_\mu^2$, we have
\begin{align*}
 & \frac{|V(0,\Lambda_\mu) \;\triangle\; V(0,\phi(W \cap \Lambda_\mu)-1)|}
 {|V(0,\Lambda_\mu)|} \\

 & \le 12 \epsilon_\mu \frac{(1+v_\mu)(5+3v_\mu)}{v_\mu}
  \le 12 \epsilon_\mu (15+3v_\mu)
 \le 12 \epsilon_\mu (15+3M_2) \\
 & = C \mu^{-1/2} ,
\end{align*}
where $C := 12C_\alpha (15+3M_2)(4+4M_2)^2
\sqrt{\frac{4\pi\alpha}{\sqrt{3}} }$.
\end{proof}

Lemma~\ref{20} implies that
\begin{equation}
\label{59}
  \frac{|V(0,\Lambda_\mu)|}{(2\pi)^2}  = \frac{\alpha}{2\pi\mu}.
\end{equation}
Thus Lemma~\ref{73} reads
\begin{equation}
\label{64}
 |V(1,\Sigma_\mu)|
 = |V(0,\phi(W \cap \Lambda_\mu)-1)|
 = \frac{2\pi\alpha}{\mu} (1 + O(\mu^{-1/2})).
\end{equation}

\begin{theorem}
  \label{19}
  Suppose that the partial quotients $a_i$ in the continued
  fraction expansion of $\theta/2\pi$ is bounded.  Then
  \[
   |V(z_{\mu}, S_\mu)| = 2\pi\alpha \mu^{2\alpha-1}(1 + O(\mu^{-1/2})).
  \]
\end{theorem}
\begin{proof}
This follows from (\ref{13}) and (\ref{64}).
\end{proof}

Theorem~\ref{18} follows immediately from Theorem~\ref{19}.

\section*{Acknowledgement}
The authors thank Shigeki Akiyama and Qinghui Liu for helpful comments
and fruitful discussions.  This work was partially supported by JSPS
Kakenhi Grant Numbers JP15K05011, JP18K13452, and the Research
Institute for Mathematical Sciences, a Joint Usage/Research Center
located in Kyoto University.

config: arxiv


TITLE: Understanding the Carnot cycle
QUESTION [0 upvotes]: My textbook explains the Carnot cycle as follows:

Heat is added to the boiler, where the steam inside expands isothermally at high temperature (T(High))--(both valves closed).
The intake valve is opened (exhaust valve closed) and the steam expands adiabatically. The expanding steam does work by pushing the piston out.
As a result of the expansion, the gas cools to the lower temperature (T(Low)).
The exhaust valve opens (intake valve closed). The piston compresses the cool gas and pushes it out to the exhaust (isothermal compression). The condenser removes heat from the steam to the air outside, and steam condenses to lower-temperature liquid.
A pump moves water to the boiler (adiabatic compression).

I understand step 1, 2, and 3; however, step 4 and 5 I don't grasp. I know that The first steps entail expanding a gas while keeping it at the same temperature, opening the intake valve in order to use the expanding gas to move a piston and do work, and I understand that, according to the first law of thermodynamics, the internal energy of the gas decreases as a result of this work. However, what is the point of step 4 and 5? Why does the gas need to be compressed? Why can't it just go back to the boiler as is? In addition to answering, could someone, in plain language, explain what steps 4 and 5 are for? For example, I am confused about the point that step 4 says that the gas is already cool because it lost energy from doing work. If so, then why is there a lower-temperature reservoir at all. Also, what does "compress" mean in the contexts of step 4 and 5? What is the purpose of this "compression?"

REPLY [1 votes]: Here's a hands-wavy explanation, which might help you're intuition.
The alternative to step 4 and 5 would be to vent the steam to the atmosphere. But the steam is still much hotter than the atmosphere, so a lot of energy will be wasted.  So you want to recapture that energy, and the logical place to put it is back into the boiler.
However, taking it out of the piston is much easier if there's no back pressure, which is achieved by condensing the exhaust gas. This ensures that the gas on the exhaust side has very low pressure and saves energy by allowing the piston to push out the steam more easily.
Once you've recondensed the steam, you've got hot water, which should be returned to the boiler. However, the boiler is at a higher pressure, so you need a pump to push it back in.


TITLE: Find order of consistency of IVP with Butcher Tableau
QUESTION [0 upvotes]: We consider this scalar IVP $\dot{y}(t)=f(t, y(t)), y(0)=y_0$ with Butcher Tableau
\begin{array}
{c|cccc}
0 & 0 & 0 & 0\\
\frac{1}{3} & \frac{1}{3} & 0 & 0\\
\frac{2}{3} &\frac{2}{3} & 0 & 0 \\
\hline
& 0 &\frac{1}{2} &\frac{1}{2}
\end{array}
We're asked to find out the order of consistency $p$.

REPLY [0 votes]: We first prove that this system is autonomization invariant. To do so, we need to check whether the sum of the elements of the right side of the three "upper rows" is equal to the element on the left. That is, whether $0 = 0 + 0 + 0$, $\frac{1}{3} = \frac{1}{3} + 0 + 0$, $\frac{2}{3} = \frac{2}{3} + 0 + 0$, which is the case. If a system is autonomization invariant, we do not need to consider the partial derivatives with respect to time while finding the derivatives we need. That is similar to Find $\alpha$ such that $y_{j+1}=y_j+\frac{h}{2 \alpha}k_1 + h(1- \frac{1}{2 \alpha})k_2$ has order of consistency 2 . There, it was assumed that the system is autonomization invariant, something we should have proved first similar to here.

Now, we need to expand
$$\tau_{j+1} = (\dot{y}(t_j) - \Phi(t_j, y(t_j), 0)) $$$$+ \frac{h}{2}(\ddot{y}(t_j)- 2 \dot{\Phi}(t_j, y(t_j), 0)) + \frac{h^2}{6}(\dddot{y}(t_j)- 3 \ddot{\Phi}(t_j, y(t_j), 0)) $$$$+ ... + O(h^p)$$ until the first term in the parenthesis that is non-zero.
Let $$\dot{y}(t) = f(t, y(t))$$ then
$$\ddot{y}(t) = \frac{df}{dy}(y(t))f(y(t))$$ and
$$\dddot{y}(t) = \frac{d^2f}{dy^2}(y(t))f(y(t))^2 + (\frac{df}{dy}(y(t)))^2f(y(t))$$.
Let
$$\Phi(h) = f(y(t) + \frac{1}{3}hf + \frac{2}{3}hf) $$
$$+ h\frac{1}{2}(\frac{df}{dy}(y(t))(\frac{1}{3} + \frac{2}{3})f(y(t) + \frac{1}{3}hf + \frac{2}{3}hf))
$$
$$+ \frac{h^2}{2} \frac{1}{2}(\frac{d^2f}{dy^2}(y(t)) (\frac{1}{3} + \frac{2}{3})^2 f(y(t) + \frac{1}{3}hf + \frac{2}{3}hf)^2) + ...$$ where
$$\Phi(y(t),0) = f(y(t))$$,
$$\dot{\Phi}(y(t),0)= \frac{1}{2}(\frac{df}{dy}(y(t))f(y(t)))$$
and
$$\ddot{\Phi}(y(t),0)= \frac{h^2}{2} \frac{1}{2}(\frac{d^2f}{dy^2}(y(t)) f(y(t))^2)$$

From that, we can see that $\Phi(h)= \dot{y}(t)$ and $\dot{\Phi}(y(t),0)= \ddot{y}(t)$ but $\ddot{\Phi}(y(t),0) \neq \dddot{y}(t)$. We can thus conclude that the order of consistency $p=2$, and not higher.


TITLE: Every Holomorphic function on a Hartog's figure can be extended holomorphically to the while of $P^n$
QUESTION [1 upvotes]: Let $f$ be a holomorphic function on the Euclidean Hartog's figure that is $$H=\{(z,w)\in P^2 : 1 \gt|z| \gt q_1 \text{or} |w|  \lt q_2 \}$$ where $0 \lt q_i \lt 1$. I need to show that it has a holomorphic extension to whole of $P^2$ where $P^2$ is the unit polydisc.
For a fixed $z$, $$f(z,w)=\sum_{n=-\infty}^{\infty}a_n(z)w^{n}$$, where $$a_n(z)=\frac{1}{2\pi i}\int_{|\alpha|=r} \frac{f(z,\alpha)}{\alpha^{n+1}}d\alpha$$ where $0 \lt r \lt q_2$. Since $f$ depends continuously on $z$ (How do I prove it??) and $f(.,\alpha)$ is holomorphic in $|\alpha| \lt q_2$ we get $$\frac{\partial a_n(z)}{\partial \bar{z}}=\frac{1}{2\pi i}\int_{|\alpha|=r} \frac{\partial f(z,\alpha)}{\partial \bar{z}}.\frac{1}{\alpha^{n+1}}d\alpha=0$$ This shows that $a_n(z)$ is a holomorphic function of $z$ for $|z| \lt 1$.
Now for $|z| \gt q_1$, $a_n(z)=0$ for $n \le -1$ since $f$ is holomorphic in this region. Now since $a_n(z)$ is holomorphic , $a_n(z)=0$ for $n \lt -1$ in the whole $|z| \lt 1$.
 And we are done.
Is this alright??
Thanks for the help!!

REPLY [1 votes]: You don't need the Laurent series expansion of $f$. Just extend your function via the Cauchy integral formula in the variable $z$ and show that the extension is complex-differentiable along each coordinate direction.


TITLE: $X_{n} \rightarrow X$ in probability if and only if $\lim_{n\to\infty} \mathbb{E}(\frac{|X_{n}-X|}{1+|X_{n}-X|})=0$
QUESTION [1 upvotes]: I'm struggling with the next problem:
"Decide reasonably whether the following statement is true or false: Let X be a random variable and $\{X_{n}\}_{n \in \mathbb{N}}$ a family of random variables.
$X_{n} \rightarrow X$ in probability if and only if $\lim_{n\to\infty} \mathbb{E}(\frac{|X_{n}-X|}{1+|X_{n}-X|})=0$ "
My intuition is that the statement is true, and my current approach is:
$\lim_{n\to\infty} \mathbb{E}(\frac{|X_{n}-X|}{1+|X_{n}-X|})=\mathbb{E}(\lim_{n\to\infty}\frac{|X_{n}-X|}{1+|X_{n}-X|})$ because $X_{n}$ and $X$ are measurable by definition of random variable. Now, $\lim_{n\to\infty}|X_{n}-X| = 0$ by convergence in probability, then $\mathbb{E}(\lim_{n\to\infty}\frac{|X_{n}-X|}{1+|X_{n}-X|}) = 0$. The reciprocal is equivalent, following the same argument.
I'm not sure that this proof is complete or it has any flaw. If my intuition failed, the statement could also by false. So I would appreciate any help.
Thanks in advance!

REPLY [2 votes]: First of all you are not mentioning why you may exchange the limit and the expectation. (you may, but why?)
Then you are mixing convergence in probability and almost sure convergence. You cannot follow $$E[\lim_{n\to 0}Y_n] = 0$$ if the limit is only in probability and not almost sure.
Good news: The statement is true. To see this let denote $$Y_n := \frac{|X_{n}-X|}{1+|X_{n}-X|}$$ consider that $Y_n = |Y_n| \le 1$ and show $$ X_n \to X \text{ in probability } \iff Y_n \to 0 \text{ in probability }$$
Then for the "if" part use that for any $\tilde{\varepsilon}$ you have a suitable $\varepsilon$ s.t. $$Y_n = Y_n1_{\{Y_n \ge \varepsilon\}} + Y_n1_{\{Y_n < \varepsilon\}} \le 1_{\{Y_n \ge \varepsilon\}} + \tilde{\varepsilon}1_{\{Y_n < \varepsilon\}}$$
For the "only if" part you have $$E[Y_n] \to 0$$ hence it already follows $$Y_n \to 0 \text{ in probability }$$ and we are done.


TITLE: Finding path integral of Logarithmic function
QUESTION [1 upvotes]: I am revising complex analysis and am stuck on a question.
I have defined a branch of $Log$ on $\Bbb{C}\setminus(-\infty,0]$, that is $$Log(z)=log(|z|)+iargz, argz \in[-\pi,\pi)$$
But how do I find this integral: $$\int_{[1,i]}Log(z) dz$$
I'm guessing I want to use the Fundamental Theorem of Calculus, as I was just asked to state that, but I don't know how, what is the antiderivative of Log?

REPLY [0 votes]: You can do integration by parts (similar as in real analysis), either for determining an antiderivative:
$$
 \int 1 \cdot \operatorname{Log}(z) \, dz = z  \cdot \operatorname{Log}(z) \, dz - \int z \frac 1z \, dz = z  \cdot \operatorname{Log}(z) - z + C
$$
or for computing an integral along the curve:
$$
 \int_\gamma 1 \cdot \operatorname{Log}(z) \, dz = \bigl[ z \cdot \operatorname{Log}(z)\bigr ]_{z=\gamma(0)}^{z=\gamma(1)} - \int_\gamma z \cdot \frac 1z \, dz = \bigl[ z \cdot \operatorname{Log}(z) - z\bigr ]_{z=\gamma(0)}^{z=\gamma(1)}
$$


TITLE: Show that S (a subset of V) is contained in span(S)
QUESTION [0 upvotes]: Let $\text{span}(S) = \lbrace v \in V \mid v\ \text{is a linear combination of vectors in}\ S\rbrace$.
I need to show that $S$ is contained within $\text{span}(S)$. I know if $S$ is nonempty, $0$ is an element of $\text{span}(S)$. Is there any way to build upon that fact in order to prove that $S$ is contained within $\text{span}(S)$? Should we show that $\text{span}(S)$ is a subset of $V$ in order to say that $S$ is contained in $\text{span}(S)$?

Definition 2.1 (Span). Let $V$ be a vector space over a field $F$. Let $S \subseteq V$ be any subset of vectors. The span of $S$, denoted $\text{span}(S)$ is the set of all linear combinations of vectors in $S$. That is
  $$
\text{span}(S) := \{v  \in V \mid v\ \text{is a linear combination of vectors in}\ S\}\\
= \{v \in V \mid \exists c_1, \ldots, c_m \in F,\  v_1, \ldots v_m \in S,\ \text{s.t.}\ v = c_1 v1 + \ldots + c_m v_m \}
$$

Can something similar to this answer the above question?
Let $W$ be a subspace containing $S$. We need to show that $W$ contains
$\text{span}(S)$. This is obvious. Since $W$ is a subspace, it is closed under
addition and scalar multiplication. Therefore, if $W$ contains $S$, it must
contain every linear combinations of elements of $S$. But the set of all
possible linear combinations of elements of $S$ is precisely $\text{span}(S)$.

REPLY [1 votes]: Any element $ s \in S $ is trivially a linear combination of elements from $ S $, since, obviously $ s = 1*s $.
You can imagine span(S) as the set obtained by taking elements of S and "putting them together" in every possible way. Any vector from S can be obtained if you just take it and no other vectors. So it's in span(S) and $ S \subset span(S) $.


TITLE: Using graph theory to find the maximum compatible clique
QUESTION [0 upvotes]: Imagine a person wants to celebrate his birthday.
But some of the guests don't like each other.
Person A says she won't come if Person B or Person C is there.
Person B and Person D say that they won't come if Person E is there.
Person E wont come if Person F is there.
Person F always argues with Person D.
Actually the Birthday child don't like his Friends, but he wants to invite them anyway.
Question: How often does he have to invite at least to the birthday, so there is no dispute.
My work:
As you can read in the title I really want to solve this with graph theory. I draw a directed graph. I have 6 nodes. I draw an Edge whenever a person doesn't like another person. But how can graph theory help me here?
Maybe matching in term of graph theory can help us?

REPLY [4 votes]: Here is the complete graph showing what happens if everyone is invited:

Now delete the incompatible links

Now look for the maximum complete subgraphs:  They are $\{ B, C, D \}$ and $\{ B, C, F \}$.
So the largest party has three people:  $B$ and $C$ with either $D$ or $F$.

In Mathematica:
g = CompleteGraph[6,
  VertexLabels ->
   Rule @@@ Transpose[{Range[6], CharacterRange["A", "F"]}]];

h = EdgeDelete[g,
 {1 <-> 2, 1 <-> 3, 2 <-> 5, 4 <-> 5, 5 <-> 6, 4 <-> 6}];

Select[
    Subgraph[h, #,
    VertexLabels ->
     Rule @@@ Transpose[{Range[6], CharacterRange["A", "F"]}]] & /@
  Subsets[Range[6], {3}], CompleteGraphQ]


TITLE: Does degree of jump determine degrees of relatively r.e. sets?
QUESTION [3 upvotes]: I’m mostly interested in this is the case where $A, \hat{A}$ are r.e. but the general case seems worth asking too.
Suppose I have sets $A >_T \hat{A}$ with $A' \equiv_T \hat{A}'$.  Does this imply that if $C$ is r.e. in $A$ then there is a $\hat{C}$ r.e. in $\hat{A}$ such that $A \oplus C \equiv_T \hat{A} \oplus \hat{C}$?
The following is motivation and can be skipped if it's confusing.
I’m basically wondering if I'm building 2-REA sets whether the jump of the base degree ($A$ or $\hat{A}$ here) completely determine the degrees of the 2-REA sets I can build over that degree (assuming I'm only considering bases comparable to each other) or is it possible to sorta thing out the range of 2-REA degrees that can be built by moving to a smaller base degree.

REPLY [2 votes]: The answer is no.  Every properly n-REA set for n < 3 (I believe Peter Cholak and I have shown this fails at 3 but could always fall apart in write-up) can be extended to a properly n+1 REA set by adding a relative r.e. set.  Now apply this result to a low r.e. set.  You can find that result in a paper by Peter Cholak and Peter Hinman but the result for n=1 was one of the S computability theorists (Shore, Soare, etc) and I don't remember who off the top of my head.


TITLE: Stationary and limiting distributions
QUESTION [1 upvotes]: Consider a CT Markov Process $X=(X_t)_{t\geq0}$ with state space $E\in\mathbb{R}^N$. Are there any general conditions under which a stationary distribution $\pi$ for $X$ is also a limiting distribution and viceversa?
Any useful reference will be much appreciated.

REPLY [2 votes]: For Markov chains, a very useful condition is Harris recurrence,
see https://en.wikipedia.org/wiki/Harris_chain.
This has been generalized to continuous time, see
https://www.jstor.org/stable/3690386?seq=1#metadata_info_tab_contents


TITLE: Saddle points - Show that surface $z = y \sin x$ has infinitely many saddle points.
QUESTION [3 upvotes]: Show that surface $z = y \sin x$ has infinitely many saddle points.
Can someone show me the step-by-step solution for that statement? Detailed explanations will be appreciated. Thank you very much!

REPLY [2 votes]: You first find the critical points of $z=y \sin x$ by finding the points $(x,y)$ for which both partial derivatives are zero. Denoting the partials as $z_x$ and $z_y$, you have $$z_x=y \cos x,$$ $$z_y=\sin x.$$
If these are both zero then $\sin x = 0$ implying $x=n \pi$ for some integer $n$. But then $\cos x=1,-1$ so from $z_x=0$ you see that $y=0$ at any critical point.
So your critical points are all points of the form $(x,y)=(n \pi, 0).$
The next thing one does for checking for saddle points is to compute the following value at each critical point $(x,y)$:
$$J(x,y)=z_{x,x} \cdot z_{y,y} - z_{x,y} z_{y,x},$$
where the double subscripts denote second partials (partials of partials). If this calculation comes out negative, you know you have a saddle point.
I'll leave the rest to you --- you should find that all the above critical points are saddle points.
Note: $J(x,y)$ above is actually the determinant of the "jacobian" at $(x,y)$, where the jacobian is really a matrix. In general one sees whether the jacobian is positive definite, negative definite, or indefinite, or none of these. That is, if the value $J(x,y)$ happens to be $0$ at a critical point, then nothing can be concluded about whether one has a local max, local min, or saddle point.


TITLE: Find X and Y, iid, such that W=X-Y has uniform distribution on (-1,1). It is possible?
QUESTION [1 upvotes]: It is possible to find two random variables X and Y, with X,Y independent and identically distributed, such that w = X-Y  has uniform distribution in (-1,1)? Why?

REPLY [4 votes]: No.  The characteristic function of $X-Y$ is $|\phi|^2$ where $\phi$ is the characteristic function of $X$, but the characteristic function of $U[-1,1]$ is  $\sin(at)/at$ (for some real value of $a$ unimportant here), which takes on negative values.


TITLE: Global regularity for Neumann problem
QUESTION [8 upvotes]: Let $\Omega\subset \mathbb{R}^d $ be a bounded open subset  ($d\in \mathbb{N}$) and denote  $\partial\Omega$  its boundary which we assume to be Lipschitz.  The classical inhomogeneous Neumann problem for Laplace operator associate to data   $f:\Omega\to\mathbb{R}$ and $g: \partial\Omega \to \mathbb{R}$ (measurable functions) consists on finding a function  $u:\Omega\to \mathbb{R}$ satisfying
\begin{equation}\label{eqlocal-Neumann}\tag{$N_1$}
-\Delta u = f  \quad\text{in}~~~ \Omega \quad\quad\text{ and } \quad\quad \frac{\partial u}{\partial \nu}=  g ~~~ \text{on}~~~ \partial \Omega.
\end{equation}

In the standard setting one usually choose $ f$ in $L^2(\Omega)$ or  in the dual space of $H^1(\Omega)$ and $g$ can be choose in the trace spaces of $H^1(\Omega)$ denote by $H^{\frac{1}{2}}(\partial\Omega)$ or in its dual $H^{-\frac{1}{2}}(\partial\Omega)$.
Assume  $f\in L^{2}(\Omega)$ and $g \in H^{1/2}(\partial\Omega)$. We have the following Green-Gauss formula
$$\label{eqgreen-Gauss}
\int_{\Omega} (-\Delta) u v \, \mathrm{d}x  =  \int_{\Omega} \nabla u \cdot \nabla v \, \mathrm{d}x- \int_{\partial \Omega} \gamma_{1} u \gamma_{0}v \, \mathrm{d}\sigma(x), \quad u\in H^{2}(\Omega) ~\hbox{and}~v\in H^{1}(\Omega).
$$
Henceforth, on $\partial \Omega$ we merely write $\gamma_0 v= v$ and  $ \displaystyle\gamma_1 v=\displaystyle  \frac{\partial v}{\partial \nu} $.
Clearly from this Green-Gauss formula, if $u\in H^{2}(\Omega)$ and solves \eqref{eqlocal-Neumann}  then $u $ satisfies the variational problem
$$\label{eqlocalvar-Neumann}\tag{$V_1$}
   \int_{\Omega} \nabla u \cdot \nabla v \, \mathrm{d}x=  \int_{\partial \Omega} f v \, \mathrm{d}x + \int_{\partial \Omega}gv \, \mathrm{d}\sigma(x), \qquad \hbox{for all } ~~v\in H^{1}(\Omega).
$$
In particular if we put $v=1$ the above formulation becomes
$$\label{eqlocalcompatible-Neumann}\tag{$C_1$}
 \int_{\Omega}f\mathrm{d}x+  \int_{\partial \Omega}g\mathrm{d}\sigma(x)=0
$$
which is the compatibility condition.
Vice versa we have the following global regularity result.
Theorem
Assume $\Omega\subset \mathbb{R}^{d}$ is bounded open with $C^2$-boundary. If a  function $ u \in H^{1}(\Omega)$ is solution to \eqref{eqlocal-Neumann} with $f\in L^{2}(\Omega)$ and $g \in H^{1/2}(\partial\Omega)$ then it belongs to $ H^{2}(\Omega).$

Question: In which book or recommendable reference can I find the proof of the above theorem?
   I know that the proof of this Theorem for the corresponding Dirichlet problem has been done in the book by Brezis or by Evans. Patently, both references avoid the inhomogeneous Neumann problem.

Remark
Moreover, observe that  if $u$ solves  \eqref{eqlocal-Neumann} or \eqref{eqlocalvar-Neumann}  so does $\tilde{u} = u+c$ for every $c\in\mathbb{R} $ (that is invariant under additive constant).

REPLY [5 votes]: Proposition 7.7 of chapter 5 in the first volume of "Partial Differential Equations" by M. E. Taylor (Springer 1996) proves the regularity result for $C^\infty$ boundaries. (Notice that any two solutions differ at most by an additive constant.) The argument of the proof should go through for boundaries with $C^{2,1}$ regularity, because then the Neumann trace $\gamma_1:H^2(\Omega)\to H^{1/2}(\partial \Omega)$ is known to be surjective, implying that it suffices to consider only the case $g=0$. I am not sure whether the assumption $\partial\Omega\in C^2$ is sufficient.
Among others the following sources contain detailed results about elliptic boundary problems which in particular apply to the Neumann boundary problem for the Laplace operator: J. Wloka, Partial Differential Equations, CUP, 1987; Chazarain and Piriou, Introduction to the theory of linear partial differential equations, North-Holland 1982, Chapter 5. The latter source uses the method of the Calderon projector to reduce an elliptic boundary value problem to a pseudo-differential equation on the boundary.


TITLE: Wave Operators: Isometry
QUESTION [1 upvotes]: This thread is only Q&A.
Given Hilbert spaces $\mathcal{H}_0$ and $\mathcal{H}$.
Consider Hamiltonians:
$$H_\#:\mathcal{D}H_\#\to\mathcal{H}_\#:\quad H_\#=H_\#^*$$
Denote their evolutions:
$$U_\#(t):=e^{itH_\#}\in\mathcal{B}(\mathcal{H}_\#)$$
Regard an operator:
$$J\in\mathcal{B}(\mathcal{H}_0,\mathcal{H})$$
Assume the limit:
$$\Omega\varphi:=\lim_{t\to\infty}U(t)JU_0(t)\varphi\quad(\varphi\in\mathcal{H}_0)$$
Suppose one has:
$$\{J^*J-1\}U_0(t)\varphi\stackrel{t\to\pm\infty}{\to}0\quad(\varphi\in\mathcal{H})$$

Then one obtains:
  $$\|\Omega\varphi\|=\|\varphi\|\quad(\varphi\in\mathcal{H})$$

How can I check this?

REPLY [0 votes]: By unitarity one has:
$$\|\Omega\varphi\|=\lim_{t\to\infty}\|JU_0(t)\varphi\|$$
It can be written as:
$$\|JU_0(t)\varphi\|^2=\langle\{J^*J-1\}U_0(t)\varphi,U_0(t)\varphi\rangle+\|\varphi\|^2$$
But it is bounded:
$$\|U_0(t)\varphi\|_{t\in\mathbb{R}}\equiv\|\varphi\|_{t\in\mathbb{R}}<\infty$$
So one obtains:
$$\langle\{J^*J-1\}U_0(t)\varphi,U_0(t)\varphi\rangle\stackrel{t\to\infty}{\to}0$$
Concluding the assertion.


TITLE: Why do we have to switch off mobile phones during flight in airplanes?
QUESTION [0 upvotes]: During take offs the flight attendant makes sure that mobile phones are switched off (or maybe turned in airplane mode), why is that necessary?
(I suspect that there is something to do with physics that's why it was posted here.)

REPLY [2 votes]: As others have pointed out, one reason is that the cell towers couldn't really deal with the signals.
The other, and this is more theoretical in nature, is that the working frequencies of the transmitters in the phone, and the CPU and other chips, might give off signals that interfere with the navigational receivers. This isn't entirely crazy, we had a deskside-PDP-11 machine that you could tell was running from the next room because you could hear it on the radio.
That said, the frequencies in question are very widely separated for exactly that reason, and it seems there are no proven examples of this ever happened in spite of millions of hours of phones being left on by mistake.
Update: if you're curious why even having a receiver is an issue, read over the wiki's article on superheterodyne and the ASV radar.


TITLE: Irreducible closed subsets of a scheme corresponds to points (Jim's solution)
QUESTION [4 upvotes]: I have a question concerning Jim's answer to the 1-1 correspondence ofirreducible closed subsets of a scheme and points.
Specifically, I don't know how to verify the following statement:

Now $U \cap Z$ has a unique generic point when $U$ is affine.  Show that if $V$ is also affine and $V \cap Z$ is nonempty then the generic point of $U \cap Z$ lies in $V$ and hence equals the unique generic point of $V \cap Z$.

especially why "the generic point of $U \cap Z$ lies in $V$" implies that the generic point of $U\cap Z$ "hence equals the unique generic point of $V \cap Z$"

REPLY [4 votes]: Recall the definition of a generic point: a point $\eta$ of a scheme $X$ is generic if $\overline{\eta}=X$.
The scenario here is that $Z$ is an irreducible closed subscheme of a scheme $X$, and $U,V$ are affine open subschemes of $Z$.
Recall that from topology, a subset $A\subset B$ is dense if $\overline{A}=B$ or equivalently $A$ intersects every nonempty open subset of $B$.
Let $\eta_U$ be the unique generic point of $U\cap Z$. The closure of $\eta_U$ in $U$ is $U\cap Z$. The the closure of $\eta_U$ in $X$ must contain $U\cap Z$, which means that it must be all of $Z$. This is true for the following reason: suppose it wasn't all of $Z$ - then $Z\setminus (U\cap Z)$ and $\overline{\eta_U}$ would be a decomposition of $Z$ into two proper nontrivial closed subsets, which would give that $Z$ is reducible.
Since $\overline{\eta_U}=Z$, it is a dense subset of $Z$, and thus must be contained in every nonempty open subset of $Z$. In particular, $\eta_U\in V\cap Z$. The closure of $\eta_U$ in $V\cap Z$ is $Z$, so it is also a generic point for $V\cap Z$, and thus must equal the generic point $\eta_V$. This is true because we've shown (or Jim has suggested) that for any affine scheme, there is a unique generic point.


TITLE: Intuitive understanding of the Stieltjes transform
QUESTION [14 upvotes]: I have been using random matrix theory in signal processing and have some trouble understanding what the Stieltjes transform does.
The gist of my work is that I have an $N\times N$ true covariance matrix of the population with $k$ eigenvalues $\lambda_i>1,\ i=\lbrace 1,\ldots,k\rbrace$, corresponding to the signals and the remaining $N-k$ eigenvalues are all $1$, corresponding to the background noise. Assuming Gaussian statistics, I'm deriving the eigenvalue density of the sample covariance matrix of a set of observations from this population, which then is used to detect the signals.
From J. W. Silverstein, “Strong convergence of the empirical dis-
tribution of eigenvalues of large dimensional random ma-
trices”, J. Multivar. Anal. 54, 175–192 (1995), I understand that the Stieltes transform of the density of the sample covariance matrix can be written as
$$m=\sum_{i=1}^{N\to\infty} \frac{1}{\lambda_i (1-c-c z m)-z}$$
where $c$ is the ratio of number of dimensions $N$ to number of observations $M$ as $N,M\to\infty$. Taking it as a polynomial in $m$ and looking for solutions in the imaginary plane, I can retrieve my density using the inverse transform.
So far, all this is well and good and I have my numerical routines that solve this for any model I input and give me the correct density. However, my question is more fundamental, as to the role that Stieltjes transforms plays here. How does one go from the definition in wikipedia
$$S_\rho(z)=\int_I \frac{\rho(x)}{z-x}\ dx$$
where $I$ is the interval of support of $\rho(x)$ to the equation in Silverstein's paper. I understand that Stieltes transforms deal with measures of density and if I had in fact taken the density of the population covariance matrix, i.e., $\rho(\lambda)=\sum_{i=1}^N \delta(\lambda-\lambda_i)$, its Stieltjes transform would've looked like
$$S=\frac{1}{N}\sum_{i=1}^N\frac{1}{z-\lambda_i}$$
So I can see the similarity with the first equation above (barring a sign change that's trivial to take care of), but don't understand how the $(1-c-czm)$ term comes into the picture and how that helps map the density of the eigenvalues of the population covariance matrix to those of the sample.
I've tried reading Terry Tao's blog on the topic, and while that was enlightening, it didn't fix this conceptual gap. It seemed like the original 1965 paper by Marchenko and Pastur dealt with these from a more first principles approach than Silverstein. However, that paper is too dense to tackle, and I'm not sure how using partial differential equations like they do help solve it. Perhaps I'm missing something here because I'm not a mathematician, but I'd appreciate someone helping me fill this gap.

REPLY [20 votes]: Firstly, the equation you attribute to Silverstein (and is sometimes known as the "self-consistent equation" for the Stieltjes transform) is not exact, but only asymptotically valid in the limit $n \to \infty$.  The definition given in Wikipedia is the exact formula.  (Your final formula, by the way, is missing a normalisation factor of either $1/n$ or $1/p$, depending on conventions.)
The self-consistent equation arises by viewing the Stieltjes transform as a trace
$$ S = \frac{1}{n} \hbox{tr}( A - zI )^{-1} = \frac{1}{n} \sum_{j=1}^n ((A-zI)^{-1})_{jj}.$$
It is possible to solve for the $jj^{th}$ component $((A-zI)^{-1})_{jj}$ of the resolvent $(A-zI)$ using the method of Schur complements, in terms of an expression involving the inverse of an $n-1 \times n-1$ minor of $A$, which in turn can be approximated in terms of the eigenvalues of that minor.  The eigenvalues of the minor can in turn be estimated by the eigenvalues of the original matrix by means of the Cauchy interlacing law, and the resulting expression involving these eigenvalues turns out to be a function of the original Stieltjes transform, leading to the (asymptotic) formula mentioned in Silverstein's paper (and also in my blog post you mentioned, in the case of Wigner matrices).
The Stieltjes transform can be viewed as a complexification of the spectral measure.  Indeed, if one looks at the "jump" in the Stieltjes transform as one passes from the upper half plane to the lower half plane, this jump is (up to some factors of $\pi$) essentially the spectral measure.  The real part of the Stieltjes transform is thus essentially the harmonic extension of spectral measure (or, if you like, a smoothed out version of the eigenvalue counting function $N_I$), and the imaginary part is its harmonic conjugate.  As such, it neatly packages the spectral information in a way that can be easily manipulated by the methods of complex analysis.
Another way to view the Stieltjes transform is as a sort of moment generating function for the spectral density.  This is basically the viewpoint taken in free probability (see my later lecture notes on this topic).


TITLE: Morita equivalence and KK-theory
QUESTION [5 upvotes]: Let $A,B,C$ be $C^\ast$-algebras. Suppose $B$ and $C$ to be strongly morita equivalent.
Then $KK(A,B)\cong KK(A,C)$.
Could someone provide a reference or proof of this fact?
I guess the imprimitivity bimodule defines an element in $KK(B,C)$ and Kasparov product with this element is the desired isomorphism, but I don't know how to fill the details. Also I am unsure whether I should add $\sigma$-unitality/separability as hypothesis.

REPLY [2 votes]: In Blackadar's K-Theory for Operator Algebras (first edition), paragraph 13.7.1 is an exercises that guides you to prove that if $A$ and $B$ are strongly Morita equivalent $\sigma$-unital C*-algebras, then $A$ and $B$ are stably isomorphic. Hence use the fact that KK-Theory is stable in both arguments to get to the sought isomorphism.


TITLE: Understanding why its $xy$ and not $2xy$ in the solution of a nonexact differential equation.
QUESTION [0 upvotes]: Here is the problem and its solution:

My question is:
Why it is $xy$ and not $2xy$ in the general solution of the non-exact differential equation? as there is one term $xy$ resulting from the integral of $\bar{M}$ and another term $xy$ resulting from the integral of $\bar{N}.$ Could anyone explain this to me, please?

REPLY [2 votes]: You can aslos solve it this way:
$$(xy+x-1)dx+x^2dy=0$$
$$xydx+x^2dy+(x-1)dx=0$$
$$x(ydx+xdy)+(x-1)dx=0$$
$$x(dxy)+(x-1)dx=0$$
$$dxy+\dfrac {(x-1)}{x}dx=0$$
After integration it gives
$$xy+x-\ln |x|=C$$


TITLE: Project in computer science and mathematics.
QUESTION [0 upvotes]: My background

I'm a third year student. I study mathematics combined with computer science (with focus on modeling, simulations and visualization).  In order to get my degree, I have to make a half-year project where I combine mathematics with computer science.

What I want to do

I am in love with algebra and group theory. But I don't know how it would fit with the interdisciplinary project, where I have to combine it with computer science. Someone suggested cryptology, but I don't know alot about it (any thoughts? and useful references?).

My problem/question

Is it possible to combine computer science with algebra? I also need project ideas (I would also like references).

REPLY [0 votes]: You may consider this old hat, but you could write a solver/simulator for a Rubik's Cube or some other puzzle that can be described with groups.
As for cryptography, there have been proposals for cryptosystems based on braid groups (an overview). I believe they use the conjugacy search problem or something similar as the trap door (i.e. given $a$ and $b$ that are conjugate, find $x$ such that $a=xbx^{-1}$). You might be able to visualize the way the computer manipulates the braids (puts them into normal form and such).


TITLE: Concavity of function implies convex upper contour
QUESTION [1 upvotes]: Today I saw a theorem in class that stated the following:
$f$ is concave $ \Rightarrow\{z \in \mathbb R^n : f(z) \ge c\}$ is convex.
The proof is relatively straight forward and I understand.  However, I have a hard time visualizing this idea to be true.  I try and visualize the idea with a function such as $-x^2$ and I can't seem to see how the upper contour set is convex.  If someone could walk me through a visual example with a graph that would be very helpful.

REPLY [0 votes]: There is a fair chance you may be talking about the set $\{(z, c) \in \mathbb{R}^{n + 1}|f(z) \ge -c\}$ instead. Now, for a proof of the claim, simply use the definition of convexity via epigraphs.
Indeed,
$f$ is a concave function $\iff$ $-f$ is a convex function $\iff$ $epi(-f) := \{(z, c) \in \mathbb{R}^{n + 1}|c \ge -f(z)\} = \{(z, c) \in \mathbb{R}^{n + 1}|f(z) \ge -c\}$ is a convex set.


TITLE: Joint density of the smallest and largest random variables among finite independent random variables with common density
QUESTION [4 upvotes]: I am trying to show the following result.
Let $X_1, \ldots,X_n$ be independent random variables with the common density $f$ and distribution function $F$. If $X$ is the smallest and $Y$ the largest among them, the joint density of the pair $(X, Y)$ for $y>x$ is given by $$n(n-1)f(x)f(y)[F(y)-F(x)]^{n-2}$$
Some thoughts towards a partial solution
Attempt 1:
Given they all share the same density, the Joint density can be calculated as $$f_{X,Y}( x,y) = f_{Y\mid X}( y\mid x) f( x) = f_{X\mid Y}( x\mid y) f( y)$$
So we can choose $x$ in $n$ ways and fixing $x$, we can pick the maximum random variable as $C_{1}^{n-1} = (n-1)$ so this explains $n(n-1)f(x)f(y)$ part but i am unsure why we have the difference of the distribution functions of the $y$ and $x$ times $(n-2)$. i know we have $n-2$ variables to still account for and they are being integrated out. Hence we should have $(n-2)$ terms but why the difference ?
Attempt 2:
The sample space corresponding to $X_1, \ldots,X_n$ is the $n$-dimensional hypercube $\Gamma $ defined by $x_k=f$ and the probabilities equal the $n$-dimensional volume. The natural sample space with the $X_k$ as coordinate variables is the subset $\Omega$ of $\Gamma$ containing all points such that $x_1\leq \cdots \leq x_n$. The hypercube contains $n!$ congruent replicas of the set $\Omega$ and in each the ordered $n$-tuple $(X_1,\ldots,X_n)$ coincides with a fixed permutation of $X_1,\ldots, X_n$.
I am not sure i am getting anywhere with these thoughts. Any help would be much appreciated.

REPLY [4 votes]: If you first find the joint c.d.f. of the pair, then differentiate to get the density, then you're just finding a probability and then differentiating.
$$
\begin{align}
& F_{X,Y}(x,y)=\Pr(X\le x\  \&\  Y\le y) \\[10pt]
& = \Pr(Y\le y) - \Pr(Y\le y\  \&\  X>x) \\[10pt]
& = \Pr(\text{all observations}\le y) - \Pr(y\ge \text{all observations}>x) \\[10pt]
& = (\Pr(X_1\le y))^n - (\Pr(x<X_1\le y))^n \\[10pt]
& = F(y)^n - (F(y)-F(x))^n,\qquad\text{all provided that }x\le y.
\end{align}
$$
Then apply $\dfrac{\partial^2}{\partial x\,\partial y}$.  (The first term vanishes when $\dfrac{\partial}{\partial x}$ is applied.)