text:
\begin{document}
  \title{Dependence of multivariate extremes}

\author{C. Viseu \footnote{Instituto Superior de Contabilidade e
Administra\c c\~ ao de Coimbra, Coimbra,
Portugal.\newline
E-mail:cviseu@iscac.pt}\\  Instituto Superior de Contabilidade\\
e Administra\c c\~ ao de Coimbra\\ Portugal \and L. Pereira, A.P.
Martins, H. Ferreira \footnote{Departamento de
Matem\'atica, Universidade da Beira Interior, 6200 Covilh\~a,
Portugal.
E-mail: lpereira@ubi.pt; amartins@ubi.pt;
helena.ferreira@ubi.pt}\\Departamento de Matem\'atica \\
Universidade da Beira Interior\\ Portugal}
\date{}
\maketitle

\noindent {\bf Abstract:} We give necessary and sufficient conditions
for two sub-vectors of a random vector with a multivariate extreme value distribution,
corresponding to the limit distribution of the maximum of a multidimensional stationary sequence with extremal index, to be independent or totally dependent.
Those conditions involve first relations between the multivariate extremal
indexes of the sequences and secondly a coefficient that measure the strength of dependence between both sub-vectors.
The main results are illustrated with an auto-regressive sequence and a 3-dependent sequence.

\section{Introduction}
\label{sec:1}

Multivariate Extreme Value Analysis is frequently applied in the context of modeling environmental data, for which the phenomenon of dependence is often intrinsic. This paper focuses on the characterization of total dependence and of independence of two multivariate extreme value distributions.

Let ${\bf{X}}=\{{\bf{X}}_{n}^{(d)}=(X_{n,1},\ldots
,X_{n,d}\}_{n\geq 1}$ be a
$d-$dimensional stationary sequence with common distribution
function (d.f.) $Q({\bf{x}}^{(d)})=Q(x_{1},\ldots ,x_{d}),$
$\mathbf{x}^{(d)}\in
\mathbb{R}^{d},$ and  ${\bf{M}}_{n}=({M}_{n,1},\ldots ,{M}_{n,d}
)$ the vector of pointwise maxima, where
${M}_{n,i}$ is the maximum of $i$-th component of $\bf X$.
Denote $\widehat{\mathbf{M}}_{n}=(\widehat{M}_{n,1}, \ldots
,\widehat{M}_{n,d})$ the corresponding vector of pointwise maxima of the
associated $d$-dimensional sequence, $\widehat{\mathbf{X}}=\{\widehat{\mathbf{X}}_{n}^{(d)}\}_{n\geq
1}$, of independent and identically distributed
(i.i.d.) random vectors having the same distribution function
$Q.$

In this multivariate setting operations among vectors are defined componentwise, that is
for each $d>1$ and $\mathbf{a}^{(d)}, \mathbf{b}^{(d)}\in\mathbb{R}^d$, $\mathbf{a}^{(d)}\leq\mathbf{b}^{(d)}$, if and only if $a_{j}\leq b_{j}$, for all $j=1,2,\ldots,d$.

If there exist sequences $\{(a_{n,1}>0, \ldots,a_{n,d}>0)\} _{n\geq 1}$ and $\{
(b_{n,1},\ldots ,b_{n,d})\}_{n\geq 1},$ such
that for
$\mathbf{u}(\mathbf{x}^{(d)})=\{\mathbf{u}_{n}(\mathbf{x}^{(d)})=(
a_{n,1}x_{1}+b_{n,1},\ldots,a_{n,d}x_{d}+b_{n,d}) \} _{n\geq
1},$
$$P\left( \widehat{\mathbf{M}}_n\leq \mathbf{u}_{n}(\mathbf{x}
^{(d)})\right) =P\left( \underset{j=1}{\overset{d}{\bigcap
}}\left\{ \widehat{M}_{n,j}\leq a_{n,j}x_{j}+b_{n,j}\right\} \right)
\xrightarrow [n\rain]{}G(\mathbf{x}^{(d)}),\ \mathbf{x}^{(d)}\in
\mathbb{R}^{d},$$ where $G$ is a d.f. with non-degenerate margins,
then $Q$ is said to be in the max-domain of
attraction of $ G$ ($Q\in D(G)$) and $G$ is said to be a Multivariate Extreme
Value (MEV) distribution function.

We will assume, without loss of generality, that the univariate
marginal distributions of $G$ are equal to $F$.

It is well known that the relationship between the d.f. $G(\mathbf{x}^{(d)})$, $\mathbf{x}^{(d)}\in
\mathbb{R}^{d},$ and its marginal distributions $F(x_j),j=1,\ldots
,d,$ can be characterized by its copula function, $D_G,$ which exhibits a
number of interesting properties, namely its stability equation:
\begin{equation}
D_{G}^{t}(y_{1},\ldots ,y_{d})=D_{G}(y_{1}^{t},\ldots ,y_{d}^{t}),\text{ }\forall t>0\text{  and  }\left(
y_{1},\ldots ,y_{d}\right) \in \left[ 0,1\right] ^{d}.  \label{eq1}
\end{equation}

If the stationary sequence ${\bf {X}}$ satisfies some mixing conditions, $D(\mathbf{u}_n(\mathbf{x}^{(d)}))$ of
Hsing (\cite{Hsing}) or
 $\Delta (\mathbf{u}_n(\mathbf{x}^{(d)}))$ of Nandagopalan (\cite{Nandagopalan}), and
\begin{equation*}
P\left( \mathbf{M}_n\leq \mathbf{u}_n(\mathbf{x}^{(d)})\right)
\xrightarrow [n\rain]{}H(\mathbf{x}^{(d)}),\ \mathbf{x}^{(d)}\in
\mathbb{R}^{d},
\end{equation*}
where $H$ is a d.f. with non-degenerate components, then $H$ is also a MEV d.f..
The MEV d.f $H$ and $G$ can be related through
the multivariate extremal index function, $\theta({\boldsymbol{\tau}}^{(d)})=\theta\left(\tau_1,\ldots,\tau_d\right)$ introduced by Nandagopalan (\cite{Nandagopalan}), which is a measure of clustering among the extreme values of a multivariate stationary sequence.

\begin{definition}
A $d-$dimensional stationary sequence $\mathbf{X}$,
is said to have multivariate extremal index $\theta^{\mathbf{X}}
({\boldsymbol{\tau}}^{(d)}) \in \left[
0,1\right] ,$ if for each $
\boldsymbol{\tau}^{(d)}=\left( \tau_1,\ldots ,\tau_d\right) \in \mathbb{R}_+^{d}$ there exists
$\mathbf{u}_{n}^{(
\boldsymbol{\tau}^{(d)})}=(u_{n,1}^{(\tau_1)},\ldots
,u_{n,d}^{(\tau_{d})}),$ $n\geq 1,$ satisfying
\begin{equation*}
nP(X_{1,j}>u_{n,j}^{(\tau_j)})\xrightarrow [n\rain]{}\tau_j, \ j=1,2,\ldots,d, \quad
P\left(\widehat{\mathbf{M}}_{n}\leq \mathbf{u}_{n}^{(
\boldsymbol{\tau}^{(d)})}\right) \xrightarrow [n\rain]{}G ({\boldsymbol{\tau}}^{(d)}) \quad \textrm{and}
\end{equation*}
\begin{equation*}
P\left(\mathbf{M}_{n}\leq \mathbf{u}_{n}^{({\boldsymbol{\tau}}^{(d)})}\right)
\xrightarrow [n\rain]{}G^{\theta^{\mathbf {X}} ({\boldsymbol{\tau}}^{(d)})} ({\boldsymbol{\tau}}^{(d)}).
\end{equation*}
\end{definition}

As in one dimension, the extremal index is a key parameter when relating the properties of extreme values of a stationary sequence to those of independent random vectors from the same $d-$dimensional marginal distribution. However, unlike the one dimensional case, it is not a constant for the whole process, but instead depends on the vector ${\boldsymbol{\tau}}^{(d)}$.

It is now clear that the existence of $\theta^{\mathbf{X}}
(\boldsymbol{\tau}^{(d)})$ allows us to write
\begin{equation*}
H({\bf x}^{(d)})=G^{\theta^{\bf X}({\boldsymbol{\tau}}^{(d)})}({\bf x}^{(d)}) \quad \textrm{with} \quad \tau_j\equiv\tau_j(x_j)=-\log F(x_j), \ j=1,2,\ldots,d.
\end{equation*}

Taking $d=p+q$, it follows, as a consequence of the definition of multivariate extremal index, that the sequences ${\bf{X}}^{(p)}=\{ \textbf{X}_n^{(p)}=(X_{n,1},\ldots ,X_{n,p})\}_{n\geq
1}$ and $\mathbf{X}^{(q)}=\{\mathbf{X}_n^{(q)}=(X_{n,p+1},\ldots
,X_{n,p+q})\}_{n\geq 1}$ have, respectively, extremal indexes
$$\theta^{\mathbf{X}^{(p)}}(\boldsymbol{\tau}^{(p)})=\lim_{{\tau_j\rightarrow 0^+}\atop{j=p+1,\ldots ,p+q}}\theta^{\mathbf{X}}(\boldsymbol{\tau}^{(p+q)})\quad \textrm{and}\quad \theta^{\mathbf{X}^{(q)}}(\boldsymbol{\tau}^{(q)})=\lim_{{\tau_j\rightarrow 0^{+}}\atop{j=1,\ldots ,p}}\theta^{\mathbf{X}}(\boldsymbol{\tau}^{(p+q)}).$$

In the notation of the extremal index we shall omit the sequence, whenever it is clear by
the context and the argument of the function.

Hereinafter, let ${\bf{Y}}=(Y_1,\ldots,Y_{p+q})$ and $\widehat{\bf {Y}}=(\widehat{Y}_1,\ldots,\widehat{Y}_{p+q})$ be, respectively,\linebreak  two random vectors
with distribution functions $G^{\theta^{\mathbf{X}}\left({\boldsymbol{\tau}}^{(p+q)}\right)}$ and $G$, where
$\mathbf{Y}^{(p)}=(Y_1,\ldots,Y_p)$ and $\mathbf{Y}^{(q)}=(Y_{p+1},\ldots,Y_{p+q})$ denote two sub-vectors of ${\bf{Y}}$ and
$\widehat{\mathbf{Y}}^{(p)}=(\widehat{Y}_1,\ldots,\widehat{Y}_p)$ and $ \widehat{\mathbf{Y}}^{(q)}=(\widehat{Y}_{p+1},\ldots,\widehat{Y}_{p+q})$
two sub-vectors of $\widehat{\mathbf {Y}}$.

In section 2 we discuss conditions under which $\mathbf{Y}^{(p)}$ and $\mathbf {Y}^{(q)}$ are independent or totally dependent. These conditions are established first by relations between the extremal indexes  $\theta^{\mathbf{X}}(\boldsymbol{\tau}^{(p+q)})$, $\theta^{\mathbf{X}^{(p)}}(\boldsymbol{\tau}^{(p)})$ and $\theta^{\mathbf{X}^{(q)}}(\boldsymbol{\tau}^{(q)})$ and secondly by a coefficient that measures the strength of dependence between $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$.

The main results are illustrated in section 3 with an auto-regressive sequence and a 3-dependent sequence.

\section{Main results}
\label{sec:2}

If the d.f. $Q$ belongs to the domain of attraction of a MEV
distribution, $G$, and ${\bf{X}}$ has extremal index $\theta({\boldsymbol{\tau}}^{(p+q)}),$ ${\boldsymbol{\tau}}^{(p+q)}=(\tau_1,\dots,\tau_{p+q})\in \mathbb{R}_+^{p+q},$ then we
have
\begin{equation}
G^{\theta (\mathbf{\boldsymbol{\tau}}^{(p)})}(\mathbf{x}^{(p)})G^{\theta (\boldsymbol{\tau}^{(q)})}(\mathbf{x}^{(q)})\leq G^{\theta (\boldsymbol{\tau}^{(p+q)})}(\mathbf{x}^{(p+q)})\leq \min \{ G^{\theta
(\boldsymbol{\tau}^{(p)})}(\mathbf{x}^{(p)}),G^{\theta (\boldsymbol{\tau}^{(q)})}(\mathbf{x}^{(q)})\} ,  \label{eq2}
\end{equation}
for each $\mathbf{x}^{(p+q)}\in \mathbb{R}^{p+q}$ and $\tau_j=-\log F(x_j),$ $j=1,\ldots ,p+q.$

 The inequality on the right holds true for every
multivariate distribution, while the inequality on the left is a property
of MEV distributions. The lower bound corresponds to the case where
$\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are independent and the
upper bound corresponds to the case where $\mathbf{Y}^{(p)}$ and
$\mathbf{Y}^{(q)}$ are totally dependent.

From (\ref{eq2}) we obtain the following bounds for the multivariate extremal index function
$\theta(\boldsymbol{\tau}^{(p+q)}),\boldsymbol{\tau}^{(p+q)} \in
\mathbb{R}_+^{p+q}$.
\begin{equation}
\frac{\max \{ \theta (\boldsymbol{\tau}^{(p)})\gamma (\boldsymbol{\tau}
^{(p)}),\theta (\boldsymbol{\tau}^{(q)})\gamma (
\boldsymbol{\tau}^{(q)})\}}{\gamma (\boldsymbol{\tau}^{(p+q)})}\leq\theta({\boldsymbol{\tau}}^{(p+q)})\leq
\frac{\theta ({\boldsymbol{\tau}}^{(p)})\gamma ({\boldsymbol{\tau}}^{(p)})+\theta ({\boldsymbol{\tau}}^{(q)})\gamma (
{\boldsymbol{\tau}}^{(q)})}{\gamma ({\boldsymbol{\tau}}^{(p+q)})} , \label{eq3}
\end{equation}
where
\begin{equation*}
\gamma ({\boldsymbol{\tau}}^{(p+q)})=-\log G(F^{-1}(e^{-\tau_1}),\ldots ,F^{-1}(e^{-\tau_{p+q}}))=\lim_{n\rightarrow\infty}nP\left({\bf X}_1^{(p+q)} \not\leq u_n^{({\boldsymbol{\tau}}^{(p+q)})}\right),
\end{equation*}
\begin{equation*}
\gamma({\boldsymbol{\tau}}^{(p)})=\!\!\!\!\lim_{{\tau_j\rightarrow
0^+}\atop{j=p+1,\ldots ,p+q}} \gamma ({\boldsymbol{\tau}}^{(p+q)})\ \ \textrm{and}\ \
\gamma({\boldsymbol{\tau}}^{(q)})=
\lim_{{\tau_j\rightarrow 0^+}\atop{j=1,\ldots
,p}}\gamma ({\boldsymbol{\tau}}^{(p+q)}).
\end{equation*}

\noindent The next result follows naturally from these bounds.

\begin{proposition}
Suppose that $Q\in D(G)$ and $\mathbf{X}$ has
extremal index $\theta (\boldsymbol{\tau}^{(p+q)}),$
$\boldsymbol{\tau}^{(p+q)}\in \mathbb{R}_+^{p+q}.$

\noindent {\bf{(i)}} If $\ \widehat{\mathbf{Y}}^{(p)}$ and $\widehat{\mathbf{Y}}^{(q)}$  are independent, then $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are independent if and only if
\begin{equation*}
\theta (\boldsymbol{\tau}^{(p+q)})=\frac{\theta (
\boldsymbol{\tau}^{(p)})\gamma (\boldsymbol{\tau}^{(p)})+\theta (
\boldsymbol{\tau}^{(q)})\gamma (\boldsymbol{\tau}^{(q)})}{
\gamma (\boldsymbol{\tau}^{(p)})+\gamma (\boldsymbol{\tau}^{(q)})},\ \boldsymbol{\tau}^{(p+q)}\in \mathbb{R}_+^{p+q}.
\end{equation*}

\noindent {\bf{(ii)}} If $\ \widehat{\mathbf{Y}}^{(p)}$ and
$\widehat{\mathbf{Y}}^{(q)}$ are totally dependent, then
$\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are totally dependent if
and only if $$\theta(\boldsymbol{\tau}^{(p+q)})=\frac{\max \{
\theta(\boldsymbol{\tau}^{(p)})\gamma(\boldsymbol{\tau}^{(p)}),
\theta(\boldsymbol{\tau}^{(q)})\gamma(\boldsymbol{\tau}^{(q)})\}
}{\max \{\gamma( \boldsymbol{\tau}^{(p)}),\gamma(\boldsymbol{\tau}^{(q)})\} } ,\ \boldsymbol{\tau}^{(p+q)}\in
\mathbb{R}^{p+q}.
$$
\end{proposition}

The necessary and sufficient conditions for $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$
to be independent or totally dependent given in the previous result demand the evaluation of the extremal index function $\theta(\boldsymbol{\tau}^{(p+q)}),$ in each point $\boldsymbol{\tau}^{(p+q)}\in
\mathbb{R}_+^{p+q}.$ Nevertheless this task can be simplified with the characterizations, given in
Ferreira (\cite{Ferreira}), for independence and total dependence of the multivariate marginals of a
MEV distribution. These characterizations are essential to prove the following propositions which guarantee that the independence or total dependence between $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ only depends on the value of the extremal
index in some points.

\begin{proposition}
Suppose that $Q\in D(G)$ and the sequence $\mathbf{X}=\{\mathbf{X}_{n}^{(p+q)}\} _{n\geq 1}$ has extremal index $\theta(\boldsymbol{\tau}^{(p+q)}),$ $\boldsymbol{\tau}^{(p+q)}\in
\mathbb{R}_+^{p+q}.$

The sub-vectors $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are independent if and only if
\begin{equation}
\label{eq4}
\theta (\mathbf{1}^{(p+q)})=\frac{\theta (\mathbf{1}^{(p)})\gamma (\mathbf{1}^{(p)})+
\theta (\mathbf{1}^{(q)})\gamma (\mathbf{1}^{(q)})}{\gamma (\mathbf{1}^{(p+q)})},
\end{equation}
where $\mathbf{1}^{(k)}=(1,\ldots,1),\ k>1$, denotes the
$k$-dimensional unitary vector.
\end{proposition}

\bdem
Suppose that $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are independent. Since (\ref{eq3}) holds for all
\linebreak $\mathbf{\boldsymbol{\tau}}^{(p+q)}\in \mathbb{R}_+^{p+q}$, we have
in particular for $ \boldsymbol{\tau}^{(p+q)}=(\tau,\ldots ,\tau)\in \mathbb{R}_+^{p+q}$, with \linebreak $\tau \equiv \tau(x)=-\log F(x),\ x
\in \mathbb{R},$
$$
\theta(\boldsymbol{\tau}^{(p+q)})=\frac{\theta(\boldsymbol{\tau}^{(p)})\gamma(\boldsymbol{\tau}^{(p)})+\theta(\boldsymbol{\tau}^{(q)})\gamma (\boldsymbol{\tau}^{(q)})}{\gamma(\boldsymbol{\tau}^{(p+q)})}.
$$
\noindent Now attending to the fact that $\theta (c\boldsymbol{\tau}^{(k)})=\theta
(\boldsymbol{\tau}^{(k)})$ for each $\boldsymbol{\tau}^{(k)}\in
\mathbb{R}_+^k,$ $k>1$ and $c>0$, we can write
$$
\theta(\boldsymbol{\tau}^{(p+q)})=\theta (\mathbf{1}^{(p+q)}),\quad \theta(\boldsymbol{\tau}^{(p)})=\theta(\mathbf{1}^{(p)}),\quad \theta
(\boldsymbol{\tau}^{(q)})=\theta(\mathbf{1}^{(q)}),
$$
and  from (\ref{eq1}), for all
$\boldsymbol{\tau}^{(p+q)}=(\tau ,\ldots ,\tau )\in
\mathbb{R}_+^{p+q}$,
\begin{eqnarray}
\gamma({\boldsymbol{\tau}}^{(p+q)}) &=&-\log G(F^{-1}(e^{-\tau }),\ldots ,F^{-1}(e^{-\tau }))=-\log D_{G}(e^{-\tau},\ldots ,e^{-\tau}) \nonumber \\
&=&-\log D_G^{\tau}(e^{-1},\ldots ,e^{-1})=\tau \gamma (\mathbf{1}^{(p+q)}),
\label{eq5}
\end{eqnarray}
$\gamma(\boldsymbol{\tau}^{(p)})=\tau \gamma
(\mathbf{1}^{(p)})\text{ and  }\gamma(\boldsymbol{\tau}^{(q)})=\tau
\gamma (\mathbf{1}^{(q)}).$ Equality (\ref{eq4}) is now
straightforward.

On the other hand  if (\ref{eq4}) is verified, then for
$\mathbf{x}^{(p+q)}=(x,\ldots ,x)$ we have
\begin{eqnarray*}
G_{\mathbf{Y}}(\mathbf{x}^{(p+q)}) &=&G^{\theta (\mathbf{1}^{(p+q)})}(\mathbf{x}^{(p+q)})=D_{G}^{\theta (\mathbf{1}^{(p+q)})}(e^{-\tau},\ldots
,e^{-\tau})\\
&=&D_{G}^{\theta(\mathbf{1}^{(p+q)})\tau}(e^{-1},\ldots
,e^{-1}) =\exp(-\tau \gamma (\mathbf{1}^{(p+q)})\theta (\mathbf{1}^{(p+q)}))\\
&=&\exp (-\tau(\theta(\mathbf{1}^{(p)})\gamma (\mathbf{1}^{(p)})+\theta
(\mathbf{1}^{(q)})\gamma(\mathbf{1}^{(q)})))=G_{\mathbf{Y}^{(p)}}(\mathbf{x}^{(p)})G_{\mathbf{Y}^{(q)}}(\mathbf{x}^{(q)})
\end{eqnarray*}
and from Proposition 2.1 (\cite{Ferreira}) we conclude that $\mathbf{Y}^{(p)}$ and
$\mathbf{Y}^{(q)}$ are independent.\edem

\begin{proposition}
Suppose that $Q\in D(G)$, $\mathbf{X}=\{\mathbf{X}_{n}^{(p+q)}\} _{n\geq 1}$ has extremal index\linebreak $\theta(
\boldsymbol{\tau}^{(p+q)}),$ $\boldsymbol{\tau}^{(p+q)}\in \mathbb{R}_+^{p+q}.$

\noindent {\bf{(i)}} If $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are totally dependent then
there exists  $\boldsymbol{\tau}^{(p+q)}\in \mathbb{R}_+^{p+q}$ with \linebreak
$\tau_j\equiv \tau_j(x_j)=-\log F(x_j), x_j \in \mathbb{R},\  j=1,\ldots,p+q$, such that
$$\gamma(\boldsymbol{\tau}^{(p)})\theta(\boldsymbol{\tau}^{(p)})=\gamma(\boldsymbol{\tau}^{(q)})\theta(
\boldsymbol{\tau}^{(q)})=\theta_1\tau_1\ldots=\theta_{p+q}\tau_{p+q}=d>0$$
and $\theta(\boldsymbol{\tau}^{(p+q)})=\gamma\left(\frac{\boldsymbol{\tau}^{(p+q)}}{d}\right)^{-1}$.

\noindent {\bf{(ii)}} If there exists $\boldsymbol{\tau}^{(p+q)}\in \mathbb{R}_+^{p+q}$ with
$\tau_j\equiv \tau_j(x_j)=-\log F(x_j),\ x_j \in \mathbb{R},\  j=1,...,p+q$, such that
\begin{equation*}
\gamma (\boldsymbol{\tau}^{(p+q)})\theta (\boldsymbol{\tau}^{(p+q)})=\theta_1\tau_1\ldots=\theta_{p+q}\tau_{p+q}=d>0,
\end{equation*}
then $\mathbf{Y}^{(p)}$ and $\mathbf{Y}^{(q)}$ are totally dependent.
\end{proposition}

\bdem
{\bf{(i)}} From Proposition 2.1 (\cite{Ferreira}), if $\mathbf{Y}^{(p)}$ and
$\mathbf{Y}^{(q)}$ are totally dependent, then there exists
$\mathbf{\boldsymbol{\tau}}^{(p+q)}\in \mathbb{R}_+^{p+q}$ such that
$$
\theta(\boldsymbol{\tau}^{(p)})\gamma(\boldsymbol{\tau}^{(p)})=\theta(\boldsymbol{\tau}^{(q)})\gamma(
\boldsymbol{\tau}^{(q)})=\theta (\boldsymbol{\tau}^{(p+q)})\gamma(\boldsymbol{\tau}^{(p+q)})=d=\theta_1\tau_1=\ldots=\theta_{p+q}\tau_{p+q},
$$
with $d \in ]0,1[.$ Hence
\begin{eqnarray*}
\theta(\boldsymbol{\tau}^{(p+q)}) &=&\frac{d}{\gamma(\boldsymbol{\tau}^{(p+q)})}=
\frac{d}{-\log D_G(\exp (-\tau_1),\ldots ,\exp (-\tau_{p+q}))}\\
&=&\frac{1}{-\log D_{G}\left( \exp \left( -\frac{\tau_{1}}{d}\right),\ldots ,\exp \left(-\frac{\tau_{p+q} }{d}\right)\right)}=\frac{1}{\gamma\left(\frac{\boldsymbol{\tau}^{(p+q)}}{d}\right)}.
\end{eqnarray*}
\edem

Another way to look at issues concerning independence or total dependence is to use parameters that measure the strength of dependence between ${\bf{Y}}^{(p)}$ and ${\bf{Y}}^{(q)}$. We therefore define, in the following result, the dependence structure of ${\bf{Y}}^{(p)}$ and ${\bf{Y}}^{(q)}$ through the coefficient $\epsilon^{({\bf{Y}}^{(p)},{\bf{Y}}^{(q)})}$ of Ferreira (\cite{Ferreira}). This coefficient emerged from the extremal coefficient of $\bf Y$, $\epsilon^{\bf{Y}}$, defined in Martins and Ferreira (\cite{Martins}) as
$$G^{\theta(\textbf{1}^{(p+q)})}(\textbf{x}^{(p+q)})=F^{{\epsilon}^{\mathbf{Y}}}(x), \ \ x\in \mathbb{R},$$
and the relationship
$$P\left({\bf{Y}}^{(p)}\leq{\textbf{x}}^{(p)},{\bf{Y}}^{(q)}\leq{\textbf{x}}^{(q)}\right)=\left(G_{\mathbf{Y}}^{(p)}({\textbf{x}}^{(p)})
G_{\mathbf{Y}}^{(q)}({\textbf{x}}^{(q)})\right)^{\frac{\epsilon^{\mathbf{Y}}}{\epsilon^{{\bf{Y}}^{(p)}}+\epsilon^{{\bf{Y}}^{(q)}}}}.$$
It is then defined as
$\epsilon^{\left({\bf{Y}}^{(p)},{\bf{Y}}^{(q)}\right)}=\frac{\epsilon^{\mathbf{Y}}}{\epsilon^{{\bf{Y}}^{(p)}}+\epsilon^{{\bf{Y}}^{(q)}}}$ and has the following interesting pro\-per\-ties.

\begin{proposition}$\ $\\

\noindent {\bf{(i)}} $\epsilon^{({\bf{Y}}^{(p)},{\bf{Y}}^{(q)})}=\frac{\theta({\bf{1}}
^{(p+q)})\gamma ({\bf{1}}^{(p+q)})}{\theta({\bf{1}}^{(p)})\gamma(
{\bf{1}}^{(p)})+\theta({\bf{1}}^{(q)})\gamma({\bf{1}}^{(q)})}$

\noindent {\bf{(ii)}} $\epsilon^{({\bf{Y}}^{(p)},{\bf{Y}}^{(q)})}=1$ if and only if ${\bf{Y}}^{(p)}$
and ${\bf{Y}}^{(q)}$ are independent.

\noindent {\bf{(iii)}} If $\ {\bf{Y}}^{(p)}$ and ${\bf{Y}}^{(q)}$  are totally dependent, then
$\epsilon^{({\bf{Y}}^{(p)},{\bf{Y}}^{(q)})}= \frac{\max\{\epsilon^{{\bf{Y}}^{(p)}},\epsilon^{{\bf{Y}}^{(q)}}\}}{\epsilon^{{\bf{Y}}^{(p)}}+\epsilon^{{\bf{Y}}^{(q)}}}.$

\end{proposition}

\bdem
\noindent {\bf{(i)}} Since
$$
\epsilon^{({\bf{Y}}^{(p)},{\bf{Y}}^{(q)})}=\frac{\theta({\bf{1}}
^{(p+q)})\log G({\bf x}^{(p+q)})}{\theta({\bf{1}}^{(p)}) \log G({\bf x}^{(p)})+\theta({\bf{1}}^{(q)})\log G({\bf x}^{(q)}))}
$$
the result follows from (\ref{eq5}).

\noindent {\bf{(ii)}} It is an immediate consequence of (i) and Proposition 2.

\noindent{\bf{(iii)}} If ${\bf{Y}}^{(p)}$ and ${\bf{Y}}^{(q)}$ are totally dependent then from (\ref{eq2}), we have
\begin{equation*}
\epsilon^{({\bf{Y}}^{(p)},{\bf{Y}}^{(q)})} = \frac{-\log \min \{G^{\theta (\bf{1}^{(p)} )}({\bf{x}}^{(p)}),G^{\theta ({\bf{1}}^{(q)})}({\bf{x}}^{(q)})\}}
{\theta({\bf{1}}^{(p)}) \log G({\bf{x}}^{(p)})+ \theta({\bf{1}}^{(q)}) \log G({\bf{x}}^{(q)})}=\frac{\max\{\epsilon^{{\bf{Y}}^{(p)}},\epsilon^{{\bf{Y}}^{(q)}}\}}{\epsilon^{{\bf{Y}}^{(p)}}+\epsilon^{{\bf{Y}}^{(q)}}}.
\end{equation*}
\edem

\section{Examples}
\label{sec:7}

\begin{example}
{\rm{\noindent Let $\{ Y_n \} _{n\geq 1}$ be a sequence of i.i.d.
random variables with common d.f. $F$ and consider an
auto-regressive sequence of maxima $ \{X_n\}_{n\geq 1}$ defined by
$$X_n=\max\{Y_n,Y_{n+1}\},\ n\geq 1,$$
with marginal distribution function $F^2.$

Let $\{u_n^{(\tau_i)}\} _{n\geq 1},\ i=1,\ldots,p,$ and $\{v_n^{(\tau'_j)}\} _{n\geq 1},j=p+1,\ldots ,p+q,$
be sequences of real numbers such that $\lim_{n\to \infty}n(1-F^{2}(u_{n}^{(\tau _{i})}))=\tau _{i}$ and $\lim_{n\to \infty}nF^2(-v_n^{(\tau'_j)})) =\tau_{j}^{\prime}.$

The sequences $\{X_n\}_{n\geq 1}$ and $\{-X_n\}_{n\geq 1}$ have, respectively, extremal indexes $\theta_1=1/2$ and $\theta_2=1.$

\noindent For sequences $\mathbf{X}_{n}^{(p+q)}=\left\{
\begin{array}{lcl}
X_{n,i}=X_{n}&,& i=1,\ldots p\\
X_{n,i}=-X_{n}&,& i=p+1,\ldots ,p+q
\end{array}
\right.$, $\mathbf{X}_{n}^{(p)}=(X_{n},\ldots ,X_{n})$ and $\mathbf{X}_n^{(q)}=(-X_n,\ldots ,-X_n)$, we have
\begin{eqnarray*}
\lim_{n\to \infty}P(\mathbf{M}_{n}^{(p)}\leq \mathbf{u}_n^{(\boldsymbol{\tau}^{(p)})})&=&\exp
\left(-\frac{1}{2}\underset{1\leq j\leq p}{\max}\tau_{j}\right),\\
\lim_{n\to \infty}P(\widehat{\mathbf{M}}_{n}^{(p)}\leq \mathbf{u}_n^{(\boldsymbol{\tau}^{(p)})})&=&\exp \left( -\underset{1\leq j\leq p}{
\max }\tau _{j}\right) , \\
\lim_{n\to \infty}P\left( \mathbf{M}_{n}^{(q)}\leq (v_{n}^{(\tau _{p+1}^{\prime })},\ldots
,v_{n}^{(\tau _{p+q}^{\prime })})\right) &=&\exp \left( -\underset{
p+1\leq j\leq p+q}{\max }\tau _{j}^{\prime }\right) .
\end{eqnarray*}
Since the order statistics maximum and minimum are asymptotically
independent (\cite{Davis},\cite{Pereira}) we obtain
\begin{equation*}
{P( \mathbf{M}_n\leq(u_n^{(\tau_1)},\ldots ,u_n^{(\tau
_p)},v_n^{(\tau_{p+1}^{\prime })},\ldots ,v_n^{(\tau _{p+q}^{\prime
})}) ) }\xrightarrow [n\rain]{}\exp \left( -\frac{1}{2}\underset{1\leq
j\leq p}{\max }\tau _{j}-\underset{p+1\leq j\leq
p+q}{\max }\tau _{j}^{\prime }\right)
\end{equation*}
and consequently  $\gamma(\boldsymbol{\tau}^{(p+q)})=\gamma(\boldsymbol{\tau}^{(p)})+\gamma(\boldsymbol{\tau}^{(q)})=\underset{1\leq j\leq p}{\max }\tau _{j}+\underset{
p+1\leq j\leq p+q}{\max }\tau _{j}^{\prime }$   and   \linebreak
$\theta(\boldsymbol{\tau}^{(p+q)})\gamma(\boldsymbol{\tau}^{(p+q)}) =\frac{1}{2}\ \underset{1\leq j\leq p}{\max }\tau _{j}+\underset{p+1\leq j\leq p+q}{\max}
\tau_{j}^{\prime }.$ Therefore
\begin{equation*}
\theta (\boldsymbol{\tau}^{(p+q)})=\frac{\theta (\boldsymbol{\tau
}^{(p)})\gamma ( \boldsymbol{\tau}^{(p)})+\theta (\boldsymbol{\tau
}^{(q)})\gamma (\boldsymbol{\tau} ^{(q)})}{\gamma (\boldsymbol{\tau
}^{(p)})+\gamma (\boldsymbol{\tau}^{(q)})}.
\end{equation*} }}
\end{example}

\begin{example}
{\rm{\noindent Let ${\bf{U}}=\{U_n\}_{n\geq 1}$ be a sequence of i.i.d. random variables with common d.f. $H$ in the domain of attraction of the extreme value distribution $F$, and independent of the i.i.d. chain ${\bf{J}}=\{J_n\}_{n\geq 1}$ such that $P(J_1=0)=P(J_1=1)=1/2$.

Let us consider a stationary 1-dependent sequence ${\bf{Z}}=\{Z_n\}_{n\geq 1},$ defined as $Z_n=U_n$ if $J_n=0$ and $Z_n=U_{n+1}$ otherwise, and let ${\bf{v}}=\{v_n\}_{n\geq1}$ be a sequence of normalized levels to $\bf{Z}$, and consequently also to $\bf{U}$.

We can now define a 3-dependent stationary sequence
${\bf{X}}=\{{\bf{X}}_n=(X_{n,1},X_{n,2},X_{n,3})\}$ as
$$(X_{n,1},X_{n,2},X_{n,3})=(Z_n,Z_{n+2},Z_{n+1}), \ \ n\geq 1,$$
with common distribution function
$$T(x_1,x_2,x_3)=\frac{1}{2}\prod_{i=1}^3H(x_i)+\frac{1}{4}H(x_1)H(\min\left\{x_2,x_3\right\})+\frac{1}{4}H(x_2)H(\min\left\{x_1,x_3\right\})$$
belonging to the domain of attraction of
$$G(x_1,x_2,x_3)=\left\{
\begin{array}{lcl}
F(x_1)F(x_2)F^{\frac{1}{2}}(x_3) & , & \ x_1<x_3\wedge x_2<x_3 \\
F(x_1)F^{\frac{3}{4}}(x_2)F^{\frac{3}{4}}(x_3) & , &  \ x_1<x_3\wedge x_3\leq x_2 \\
F^{\frac{3}{4}}(x_1)F(x_2)F^{\frac{3}{4}}(x_3) & , &  \ x_3\leq x_1\wedge x_2< x_3 \\
F^{\frac{3}{4}}(x_1)F^{\frac{3}{4}}(x_2)F(x_3) & , &  \ x_3\leq x_1\wedge x_3 \leq x_2
\end{array}
\right.$$

Now applying Proposition 2.1 of Smith and Weissman (1996) (\cite{smith}) to the sequence
$U^{\bf{X}}=\{\max\{X_{n,1},X_{n,2},X_{n,3}\}=\max\{Z_n,Z_{n+1},Z_{n+2}\}\}_{n\geq1}$ which verifies the condition $D^{(k)}(v_n), \ k=2$, of Chernick et al. (\cite{Chernick}), we easily obtain
$$\theta^{\bf{X}}(1,1,1)=\lim_{n\rightarrow \infty}\frac{P\left(\max\left\{Z_1,Z_2,Z_3\right\}>v_n\geq \max \left\{Z_2,Z_3,Z_4\right\}\right)}{P\left(\max\left\{Z_1,Z_2,Z_3\right\}>v_n\right)}=\frac{3}{10}.$$

 For random vectors $\widehat{{\bf{Y}}}=\left(\widehat{Y}_1,\widehat{Y}_2,\widehat{Y}_3\right)$ and ${\bf{Y}}=\left(Y_1,Y_2,Y_3\right)$  with d.f. $G_{\widehat{{\bf{Y}}}}\equiv G$ and $G_{\bf{Y}}\equiv G^{\theta (\boldsymbol{\tau}^{(3)})}$, ${\bf{Y}}^{(2)}=\left(Y_1,Y_2\right)$ and ${\bf{Y}}^{(1)}=Y_3$ we obtain $\epsilon^{\widehat{{\bf{Y}}}}=\frac{5}{2}$, $\epsilon^{\bf{Y}}=\frac{3}{4}$, $\epsilon^{{\widehat{\bf{Y}}}^{(2)}}=2$, $\epsilon^{{\widehat{{\bf{Y}}}}^{(1)}}=1$, $\epsilon^{{{\bf{Y}}}^{(1)}}=\frac{3}{4}=\epsilon^{{{\bf{Y}}}^{(2)}}$. Consequently  $\epsilon^{\left({\bf{Y}}^{(2)},{\bf{Y}}^{(1)}\right)}=\frac{1}{2},$ $\epsilon^{\left(\widehat{\bf{Y}}^{(2)},\widehat{\bf{Y}}^{(1)}\right)}=\frac{5}{6}$ and from Proposition 4 we conclude that neither ${\bf{Y}}^{(1)}$ and ${\bf{Y}}^{(2)}$ nor $\widehat{\bf{Y}}^{(1)}$ and $\widehat{\bf{Y}}^{(2)}$ are independent.

Nevertheless, there exists $\boldsymbol{\tau}^{(3)}=(1,1,1)$ such that $\gamma(\boldsymbol{\tau}^{(3)})\theta(\boldsymbol{\tau}^{(3)})=\theta_1\tau_1=\theta_2\tau_2=\theta_3\tau_3=\frac{3}{4}$ and attending to Proposition 3 we can say that ${\bf{Y}}^{(1)}$ and ${\bf{Y}}^{(2)}$ are totally dependent.}}
\end{example}


TITLE: Uniform Probability Distribution CDF and Probability
QUESTION [1 upvotes]: Suppose a value $x$ is chosen at random in the interval $[0,10]$. In other words, $x$ is an observed value of a random variable $X \sim \mathrm{UNIF}(0,10)$. The value $x$ divides the interval $[0,10]$ into two sub-intervals.
a.) Find the CDF of the length of the shorter sub-interval.
b.) What is the probability that the ration of lengths of the shorter to the longer sub-interval is less than $\frac{1}{4}$
I have no idea to start with this problem. The book says the answer to a.) is $\mathrm{UNIF}(0,5)$ but I am not sure how they found this as a CDF.

REPLY [2 votes]: When the cut falls in $[0;5]$, the shorter-length will be uniformly distributed over $[0;5]$ and when the cut falls in $[5;10]$, the shorter-length be uniformly distributed over $[10-10;10-5]$... so...
Alternatively:
Let $X$ the cut-point and $S$ be the shorter-length.   Now for any length $s: 0\leq s\leq 5$ then $S=s$ means either $X=s$ or $X=10-s$.   So the probability density of $S$ is:
$$\begin{align}f_{S}(s) & = (\;f_X(s)+f_X(10-s)\;)\quad \big[0\leq s\leq 5\big] \\[2ex] & = \tfrac 1 5\quad \big[0\leq s\leq 5\big]\end{align}$$
Which is the probability density function of $\mathcal U[0;5]$
Hence: $\Pr(0\leq S\leq s) = \frac s 5 \quad \big[0\leq s\leq 5\big]$

Now you wish to find $\Pr(0\leq \frac{S}{10-S}\leq \frac 1 4)$


TITLE: An integral about radial function.
QUESTION [1 upvotes]: If $f$ is a radial function and $f(x)=f_0(|x|)$, I want to show $$\int_{\mathbb{R}^d}f(u)e^{-2\pi ix \cdot u}du=\int_{0}^{\infty}f_0(s)\left(\int_{\mathbb{S}^{d-1}} e^{-2\pi i |x| s(x' , u')}du'\right)s^{d-1}ds.$$
My trying is taking $s=|u|$ and applying the spherical coordinate transformation.  I don't know where is the $|x|s$ in the power of $e$ in the RHS comes from. Besides, where should I use the function $f$ is radial in the calculation?
Is there any special integral formulas about radial functions used here?
Thanks a  lot!

REPLY [1 votes]: The general form (I think that it is in Folland's blue book) of the change of variables is:
$$\int_{\mathbb{R^d}}g(x)dx = \int_0^\infty  \left( \int_{S^{d-1}} g(r\omega) d\omega \right) r^{d-1} dr = \int_{S^{d-1}}  \left( \int_0^\infty g(r\omega) r^{d-1}dr \right) d\omega,$$
where $\omega$ denotes a direction on the sphere and $d\omega$ denotes the sphere surface measure, and $r$ denotes the radius. (Note: the $r^{d-1}$ is a scaling factor from integrating on spheres of radius $r$.) So, $r\omega$ denotes an arbitrary point in $\mathbb{R}^d$. In particular, for $y \in \mathbb{R}^d$ nonzero, $y = |y|\frac{y}{|y|} = r\omega$.
Keep in mind. If $f$ is radial, then $f(r\omega)$ only depends on $r$.


TITLE: How to you find out what a matrix does to an equation.
QUESTION [1 upvotes]: Lets say I have an equation of a plane, $$x-3y+2z=0 $$
and I get matrix to transform it with say a 3x3 matrix with just a-i as place holders for the values in the matrix.
How would I find what the image of the plane is under the matrix transformation, also If there would be a different method for a function of one variable please let me know, I really dont seem to understand how to apply matrix transformations to equations.

REPLY [0 votes]: We can write the plane $x - 3 y + 2 z = 0$ as
$$\pmatrix{1 & -3 & 2} \pmatrix{x\\y\\z} = 0,$$
that is, as the kernel of the covector $\omega \in (\Bbb R^3)^*$ given in the standard basis by $\pmatrix{1 & -3 & 2}$. An (invertible) matrix $A \in GL(3, \Bbb R)$ acts on covectors by $A \cdot \omega := \omega \circ A^{-1}$ (i.e., via the contragredient representation), so the transformed plane is
$$\pmatrix{1 & -3 & 2} A^{-1} \pmatrix{x\\y\\z} = 0.$$
If one writes $A$ in terms of its components $a_{ij}$ and expands, one recovers an equation of the form $a x + b y + c z = 0$. Since we may scale both sides of the equation without changing the plane, and multiplying both sides by $\det A$ replaces the above equation with
$$\pmatrix{1 & -3 & 2} \operatorname{adj}(A) \pmatrix{x\\y\\z} = 0 ,$$
where $\operatorname{adj}(A)$ denotes the adjugate matrix of $A$, which is typically nicer than the inverse.


TITLE: How to compute $\lim\limits_{n\rightarrow \infty} \{(2+\sqrt{3})^{n}\}$, where $\{x\}$ is the fractional part of $x$?
QUESTION [6 upvotes]: I've stumbled upon the following problem:
$$\lim_{n\rightarrow \infty} \{(2+\sqrt{3})^{n}\}$$
where "{}" notates the fractional part.
I've never studied this kind of problem, there exists any reference that I could read about
this type of problem?

REPLY [13 votes]: Hint: Note, for all integers $n \ge 0$, that
$$f(n) = (2 + \sqrt{3})^n + (2 - \sqrt{3})^n \tag{1}\label{eq1A}$$
is an integer, which you can fairly easily show by expanding both terms and collecting the powers of $\sqrt{3}$ using the Binomial theorem. Also, note $0 \lt 2 - \sqrt{3} \lt 1$.


TITLE: How do collisions lead to phase change of the excited atom?
QUESTION [2 upvotes]: In collisional broadening the excited atoms interact with the surrounding atoms and de-excitation of the excited atom takes place. How does this lead to reduced lifetime? Why is this de-excitation non-radiative? How does the phase change?
Thanks in advance.

REPLY [0 votes]: The simplest way to put this is, when an atom hits another, it will cause a little bit of damage to each of the atoms. But with Newtons third law of motion, for every action there is an equal and opposite reaction, the atoms will bounce and transfer their energy to another atom through a collision and soon lots of atoms are damaged and moving (that's what make the vibrations). This would shorten the life of the molecule, which is then shortening the life of whatever all of the molecules together makes up. That could just harm the object a little bit but if the vibration was big enough, then the blow would have set more sets of atoms on a collision course weakening more areas shortening the life a lot more. Also I am looking for some green laser pointers in stores and not online so if you know a store with some or just green laser diode modules, then please let me know at calebeil@hpsvikings.org. Warning, if anybody spams me I will report them, and block them. Thank you!


TITLE: Finding $X$ and $\mathcal F$ such that $E(X)<\infty$ and $E(E(X|\mathcal F)^2)=\infty$
QUESTION [0 upvotes]: I'm  looking for an example of some random variable $X$ and a sigma-algebra $\mathcal F$ such that $E(X)<\infty$ and $E(E(X|\mathcal F)^2)=\infty$, with the requirement that $E(X|\mathcal F) \neq X$ a.s

A natural way would be to look for $X$ and $Y$ such that $E(X)<\infty$, $E(Y)<\infty$, $E(X^2)=E(Y^2)=\infty$ and $E(X|Y)=a Y + b$. I can't come up with such $X$ and $Y$ off the top of my head ...

REPLY [0 votes]: Let Y and Z be i.i.d with mean 0 and infinite variance. Let X=Y+Z. Then E(X|Y)=Y  because EZ|Y)=EZ=0. Of course $EX^2$=$\infty$


text:
\begin{document}

\begin{abstract}
  In this paper we construct a new family of lattice packings for
  superballs in three dimensions (unit balls for the $l^p_3$ norm)
  with $p \in (1, 1.58]$. We conjecture that the family also exists
  for $p \in (1.58, \log_2 3 = 1.5849625\ldots]$. Like in the densest
  lattice packing of regular octahedra, each superball in our family
  of lattice packings has $14$ neighbors.
\end{abstract}

\maketitle

\markboth{M.~Dostert and F.~Vallentin}{New dense superball packings in three dimensions}

\section{Introduction}

Finding dense packings of spheres in $n$-dimensional Euclidean space
is one of the most central problems in discrete geometry. In this
paper, we consider lattice packings of superballs in dimension
three. \textit{Superballs} are unit balls for the $\ell^p_n$ norm:
\[
\mathcal{B}^p_n = \{ x \in \mathbb{R}^n : \|x\|_p \leq 1\} \;
\text{where} \; \|x\|_p = \left(\sum_{i = 1}^n |x_i|^p\right)^{1/p}.
\]

We distinguish between three different kinds of packings which are
increasingly restrictive: congruent packings, translative packings,
and lattice packings. Let $\mathcal{K}$ be a convex body in
$\mathbb{R}^n$. By $\mathcal{K}^\circ$ we denote its topological
interior. We consider the \textit{special orthogonal group}
\[
\mathrm{SO}(n) = \{ A \in \mathbb{R}^{n \times n} : AA^{\sf T} = I_n,\; \det A = 1\},
\]
and $I_n$ denotes the identity matrix.

A \textit{congruent packing} of $\mathcal{K}$ is defined as
\[
\mathcal{P} = \bigcup_{i \in \mathbb{N}} \left(x_i + A_i
  \mathcal{K}\right) \text{ with } (x_i, A_i) \in \mathbb{R}^n \times
\mathrm{SO}(n),
\]
where
\[
(x_i + A_i \mathcal{K}^\circ) \cap (x_j + A_j \mathcal{K}^\circ) =
\emptyset \; \text{ whenever } \; i \neq j.
\]
\textit{Translative packings} are congruent packings without
rotations, i.e.\ all matrices $A_i$ are equal to $I_n$. Moreover, if
the set $\{x_i : i \in \mathbb{N}\}$ forms a lattice (a discrete
subgroup of $\mathbb{R}^n$), then we call this packing a
\textit{lattice packing}.

The \textit{(upper) density} of a congruent packing $\mathcal{P}$ is
defined as
\[
\delta(\mathcal{P}) = \limsup_{r \rightarrow \infty} \sup_{c \in
  \mathbb{R}^n} \frac{\vol(B(c,r) \cap \mathcal{P})}{\vol B(c,r)},
\]
where $B(c,r) = \{x \in \mathbb{R}^n : \|x-c\|_2 \leq r\}$ is the
Euclidean ball of radius $r$ with center $c$.

Jiao, Stillinger, and Torquato \cite{Jiao2009a, Jiao2011a} determined
four families of dense superball packings via computer
simulations. They divided the possible values of $p$ into four
regimes:
\[
p \in [1, \infty) = [1, \log_2 3 = 1.5849\ldots] \cup [\log_2 3, 2]
\cup [2, 2.3018\ldots] \cup [2.3018\ldots, \infty).
\]
For every regime they found a family of lattices which defines dense
lattice packings for $\mathcal{B}^p_3$. The family for the first
regime $[1,\log_2 3]$ they call the $\mathbb{O}_1$ lattices, and
$\mathbb{O}_0$, $\mathbb{C}_0$, $\mathbb{C}_1$ correspondingly for the
second, third, and fourth regime. In their approach, they even allowed
for congruent packings, but it turned out that the densest congruent
packings they obtained were lattice packings. They conjectured that
they found the densest congruent packings of $\mathcal{B}^p_3$ for all
values of $p$.

However, Ni, Gantapara, de Graaf, van Roij, and Dijkstra
\cite{Ni2012a} found, also via computer simulations, denser lattice
packings for superballs in the second regime. Also for $p = 1.4$,
which falls into the first regime, they give one lattice for which
they claim (but see Section~\ref{sec:Findings})) that it is denser
than the corresponding $\mathbb{O}_1$ lattice. They report (see page
8830 in \cite{Ni2012a}) that due to numerical instabilities they could
not investigate values $p < 1.4$.

In this paper, we use a method of Minkowski \cite{Minkowski1904a} to
determine locally densest lattice packings of
$\mathcal{B}^p_3$. Minkowski applied this method to determine the densest
lattice packing of regular octahedra $\mathcal{B}^1_3$. Starting from
Minkowski's lattice we found a family of new lattice packings for the
first regime which is denser than the $\mathbb{O}_1$ lattices, see
Table~\ref{table:FirstRegime}.  Here, each superball in our family of
lattice packings has $14$ neighbors, like in the densest lattice
packing of regular octahedra. We also found new lattice packings in
the second regime which are denser than the $\mathbb{O}_0$ lattices,
see Table~\ref{table:SecondRegime}. For the third and fourth regime
the densest lattices we found are the $\mathbb{C}_0$ lattices,
respectively the $\mathbb{C}_1$ lattices.

{\small
\begin{table}[htb]
\begin{center}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{ccc}
$p$ & \textsl{packing density of $\mathbb{O}_1$ lattices} & \textsl{packing density of new family} \\[0.5ex]
$1$  & $18/19 = 0.94736\ldots$ & $0.94736\ldots$ \\
$1.1$  & $0.90461\ldots$ & $0.90913\ldots$ \\
$1.2$  & $0.87121\ldots$ & $0.87861\ldots$ \\
$1.3$  & $0.84516\ldots$ & $0.85375\ldots$ \\
$1.4$  & $0.82497\ldots$ & $0.83284\ldots$ \\
$1.5$  & $0.80948\ldots$ & $0.81395\ldots$ \\
$\log_2 3$ & $0.79594\ldots$ & $0.79594\ldots$ \\
\end{tabular}
\\[0.3cm]
\end{center}
\caption{Packing density of $\mathbb{O}_1$ lattices (see
  \cite{Jiao2009a, Jiao2011a}) and of our new family (see
  Section~\ref{sec:Findings} and Section~\ref{sec:Family}). When $p = 1$ we obtained the lattice
  which determines the densest lattice packing of regular octahedra
  (see \cite{Minkowski1904a}) and when $p = \log_2 3$ we obtain the
  body centered cubic lattice.}
\label{table:FirstRegime}
\end{table}
}

{\small
\begin{table}[htb]
\begin{center}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{ccc}
$p$ & \textsl{packing density of $\mathbb{O}_0$ lattices} & \textsl{packing density of new lattices} \\[0.5ex]
$\log_2 3$ & $0.79594\ldots$ & $0.79594\ldots$ \\
$1.6$ & $0.79084\ldots$  & $0.79084\ldots$ \\
$1.7$ & $0.76567\ldots$  & $0.76610\ldots$ \\
$1.8$ & $0.75126\ldots$  & $0.75303\ldots$ \\
$1.9$ & $0.74364\ldots$  & $0.74550\ldots$ \\
$2$ & $\pi/\sqrt{18} =  0.74048\ldots$ & $0.74048\ldots$ \\
\end{tabular}
\\[0.3cm]
\end{center}
\caption{Packing density of $\mathbb{O}_0$ lattices (see
  \cite{Jiao2009a, Jiao2011a}) and of the new lattices (see
  \cite{Ni2012a} and see Section~\ref{sec:Findings}). When $p = 2$ we
  obtain the face centered cubic lattice.}
\label{table:SecondRegime}
\end{table}
}

The structure of the paper is as follows: We first explain Minkowski's
method and how we approached it computationally in
Section~\ref{sec:Determination}. Section~\ref{sec:Findings} contains a
report on some of our findings, in particular new lattice packings in
the first and second regime.  Section~\ref{sec:Family} provides a
computer-assisted proof proving that the family of new dense lattice
packings exists for $p \in [1,1.58]$. We end with
Section~\ref{sec:Conjectures} by posing some conjectures and open
problems.

\section{Determination of lattice packings by local optimization}
\label{sec:Determination}

Let $\Lambda$ be a lattice so that
\[
\mathcal{P} = \bigcup_{x \in \Lambda} (x + \mathcal{K})
\]
forms a lattice packing of the convex body $\mathcal{K}$, i.e.\ for
all distinct lattice vectors $x, y \in \Lambda$ we have
\begin{equation}
\label{eq:packingcondition}
(x + \mathcal{K}^\circ) \cap (y + \mathcal{K}^\circ) = \emptyset.
\end{equation}
Then, $\Lambda$ is called a \textit{packing lattice}. From
\eqref{eq:packingcondition} it immediately follows that a packing
lattice is characterized by
\[
\Lambda \cap (\mathcal{K} - \mathcal{K})^\circ = \emptyset, \; \text{ where }
\mathcal{K} - \mathcal{K} = \{x - y : x, y \in \mathcal{K}\}
\]
is the \textit{difference body} of $\mathcal{K}$. In other words,
$\Lambda$ is a packing lattice for $\mathcal{K}$ if and only if the
condition
\[
\|x\|_{\mathcal{K} - \mathcal{K}} \geq 1 \; \text{ for all } x \in \Lambda \setminus \{0\}
\]
holds; here
\[
\|x\|_{\mathcal{L}} = \inf\left\{ \lambda : \frac{1}{\lambda} x \in \mathcal{L}\right\}
\]
is the \textit{Minkowski norm} of $x$ defined by a centrally symmetric
convex body $\mathcal{L}$.

The \textit{general linear group} of degree $n$ over a ring $R$ is defined as
\[
\GL_n(R) = \{ B \in R^{n \times n} : \exists A \in R^{n \times n}: AB = BA = I_n\}.
\]
A matrix $B \in \GL_n(\mathbb{R})$ with linearly independent column
vectors $b_1, \ldots, b_n$ specifies a lattice $\Lambda$ by taking all
integral linear combinations of $b_1, \ldots, b_n$:
\[
\Lambda = \mathbb{Z} b_1 + \cdots + \mathbb{Z} b_n = B\mathbb{Z}^n.
\]
Two matrices $B, B' \in \GL_n(\mathbb{R})$ determine the same lattice
if and only if there is a matrix $U \in \GL_n(\mathbb{Z})$ such that
$BU = B'$. Matrix $B$ also gives a fundamental domain $F$ of $\Lambda$ by
\[
F = \left\{ \sum_{i=1}^n \alpha_i b_i : \alpha_i \in [0,1], i = 1, \ldots, n\right\}
\]
Then, the volume of a fundamental domain of $\Lambda$ is
$\vol(\mathbb{R}^n / \Lambda) = \vol F = |\det(B)|$. The density of a
lattice packing
$\mathcal{P} = \bigcup_{x \in \Lambda} (x + \mathcal{K})$ is
\[
\delta(\mathcal{P}) = \frac{\vol \mathcal{K}}{\vol(\mathbb{R}^n /
  \Lambda)} =\frac{\vol \mathcal{K}}{|\det B|}.
\]
So one can state the problem of finding a densest lattice packing of a
convex body $\mathcal{K}$ as the following minimization problem:
\[
\begin{split}
\text{minimize } & |\det B| \\
\text{so that } & B \in \GL_3(\mathbb{R}) / \GL_3(\mathbb{Z})\\
& \|Bu\|_{\mathcal{K} - \mathcal{K}} \geq 1 \text{ for all } u \in \mathbb{Z}^n\setminus\{0\}.
\end{split}
\]
It follows from Mahler's selection theorem that the minimum is
attained, see \cite[Theorem 30.1]{Gruber2007a}.

Two distinct translates $x + \mathcal{K}$ and $y + \mathcal{K}$ are
called \textit{neighbors} in a lattice packing $\mathcal{P}$ if they
have a nonempty intersection. The number of neighbors coincides for
all translates. How many neighbors can $\mathcal{K}$ have at most?
Minkowski showed that $\mathcal{K}$ has at most $3^n-1$ neighbors and
if $\mathcal{K}$ is strictly convex, then the number of neighbors is
bounded by $2^{n+1}-2$, see \cite[Theorem
30.2]{Gruber2007a}. Swinnerton-Dyer proved that when a lattice $\Lambda$
achieves a density which is locally maximal, then $\mathcal{K}$ has at
least $n(n+1)$ neighbors, see \cite[Theorem 30.3]{Gruber2007a}.

Let $B \in \GL_3(\mathbb{R})$ be a matrix defining a locally densest
lattice packing $\Lambda = B\mathbb{Z}^3$ of $\mathcal{K}$. Minkowski
\cite{Minkowski1904a} (see also \cite[\S 32]{Gruber1987a},
\cite{Betke2000a}, \cite{Haus1999a}) showed that after performing a
suitable $\GL_n(\mathbb{Z})$-transformation to $B$ we can reduce to
the following three cases.

\noindent \textbf{Case (I):} $Bu + \mathcal{K}$ with $u \in \mathcal{U}^1$ are neighbors of $\mathcal{K}$, where
\[
\mathcal{U}^1 = \{\pm (1,0,0), \pm (0,1,0), \pm (0,0,1), \pm (1,-1,0), \pm (0, 1,-1), \pm (1,0,-1)\},
\]
but $\pm (-1,1,1), \pm (1,-1,1), \pm (1,1,-1)$ are not.

\noindent \textbf{Case (II):} $Bu + \mathcal{K}$, with $u \in \mathcal{U}^2$, are neighbors of $\mathcal{K}$, where
\[
\mathcal{U}^2 = \{\pm (1,0,0), \pm (0,1,0), \pm (0,0,1), \pm (1,1,0), \pm (0, 1,1), \pm (1,0,1)\},
\]
but $\pm (1,1,1)$ are not.

\noindent \textbf{Case (III):} $Bu + \mathcal{K}$, with $u \in \mathcal{U}^3$, are neighbors of $\mathcal{K}$, where
\[
\mathcal{U}^3 = \mathcal{U}^2 \cup \{\pm (1,1,1)\}.
\]

\begin{figure}[htb]
  \centering
    \includegraphics[scale=0.12]{log23Packing.png}
    \hspace*{15mm}
    \includegraphics[scale=0.13]{log23Ball.png}
  \put(-57,27){\small \textcolor{red}{$\bf{b_3}$}}
  \put(-115,38){\small \textcolor{red}{$\bf{b_2+b_3}$}}
   \put(-23,78){\small \textcolor{red}{$\bf{b_1}$}}
   \put(-14,33){\small \textcolor{red}{$\bf{b_1 + b_3}$}}
  \put(-68,47){\small \textcolor{red}{$\bf{b_1 + b_2 + b_3}$}}
  \put(-65,97){\small \textcolor{red}{$\bf{b_1 + b_2}$}}
  \put(-83,80){\small \textcolor{red}{$\bf{b_2}$}}
  \caption{On the left: A part of a lattice packing for $B^{\log_2 3}_3$ satisfying Case (III). On the right: Seven of fourteen neighbors of the packing. Contact points are labeled in red. }
  \label{fig:Neighbors}
\end{figure}

From now on, we are interested in lattice packings of superballs in three
dimensions. We perform a rescaling by setting
$\mathcal{K} = \frac{1}{2}\mathcal{B}^p_3$ so that
$\|x\|_{\frac{1}{2}\mathcal{B}^p_3 - \frac{1}{2} \mathcal{B}^p_3} =
\|x\|_p$.
Note that scaling $\mathcal{K}$ only scales the packing lattices but
does not effect the packing density.

For each of the three cases, Case (I), (II), and (III), we can
numerically find critical points of the following finite nonlinear
optimization problem
\begin{equation}
\label{eq:nonlinear}
\begin{split}
\text{minimize } & \det B \\
\text{so that } & B \in \GL_3(\mathbb{R})\\
& \det B > 0\\
& \|Bu\|_p = 1 \text{ for all } u \in \mathcal{U}^i,
\end{split}
\end{equation}
for $i = 1, 2, 3$, in order to identify candidates for locally densest
lattice packings.

After finding a matrix $B$ which is a feasible solution of this
optimization problem, we have to check whether $B$ indeed defines a
packing lattice for $\mathcal{B}^p_3$. For this it suffices to verify
that $\|Bu\|_p \geq 1$ holds for only finitely many vectors
$u \in \mathbb{Z}^n \setminus \{0\}$ as the following lemma shows; see
also Dieter~\cite{Dieter1975a}.

\begin{lemma}
  Suppose $p,q \in [1,\infty)$ satisfy the relation
  $\frac{1}{p} + \frac{1}{q} = 1$ and suppose that a matrix
  $B \in \GL_n(\mathbb{R})$ is given. If
  $u = (u_1, \ldots, u_n) \in \mathbb{Z}^n$ is such that
  $\|Bu\|_p \leq \mu$ for some nonnegative number $\mu$, then
\[
|u_i| \leq \|B^{-1}_i\|_q \; \mu \quad \text{ for all } i = 1, \ldots, n,
\]
where $B^{-1}_i$ is the $i$-th row vector of $B^{-1}$.
\end{lemma}

\begin{proof}
We apply the triangle inequality and H\"older's inequality, and get
\[
|u_i| = |B^{-1}_i Bu| \leq \sum_{j=1}^n |B^{-1}_{i,j} (Bu)_j|
\leq \|B^{-1}_i\|_q \|Bu\|_p \leq \|B^{-1}_i\|_q \; \mu.
\qedhere
\]
\end{proof}

If $1 < p < 2$ and if $B \in \GL_3(\mathbb{R})$ satisfies the equality
conditions of Case~(III)
\[
\|Bu\|_p = 1 \text{ for all } u \in \mathcal{U}^3,
\]
then this extra check is not necessary, as the next lemma shows. This is
a consequence of Hanner's inequality \cite{Hanner1956a}
\begin{equation}
\label{eq:Hanner}
\|x + y\|^p_p + \|x - y\|^p_p \geq (\|x\|_p + \|y\|_p)^p + \left| \|x\|_p - \|y\|_p \right|^p,
\end{equation}
which holds for all $1 < p < 2$ and all $x, y \in \mathbb{R}^n$.

\begin{lemma}
\label{lem:Hanner}
If $1 < p < 2$ and if $B \in \GL_3(\mathbb{R})$ satisfies
$\|Bu\|_p = 1$ for all $u \in \mathcal{U}^3$, then $\|Bu\|_p \geq 1$
for all $u \in \mathbb{Z}^n \setminus \{0\}$.
\end{lemma}

\begin{proof}
  As a start we consider $u = (1,-1,0)$. We write $u$ as the sum of
  two vectors $v, w$ in $\mathcal{U}^3$: $u = (1,0,0) +
  (0,-1,0)$. Then by Hanner's inequality~\eqref{eq:Hanner} we get
\[
\begin{split}
\|Bu\|^p_p & \; = \; \|Bv + Bw\|^p_p \\
& \; \geq \; (\|Bv\|_p + \|Bw\|_p)^p + \left| \|Bv\|_p - \|Bw\|_p \right|^p -  \|Bv - Bw\|^p_p \\
& \; \geq \; 2^p - 1 \; \geq \; 1.
\end{split}
\]
The vectors $u = (1,0,-1), (0,1,-1)$ can be treated similarly.

Now we consider $u = (-1,1,1)$. We write $u$ as $u = v + w$ with
$u = (0,1,1)$, $w = (-1,0,0)$ and apply Hanner's inequality to get
$\|Bu|^p_p \geq 2^p - 1$. Again, the vectors $u = (1,-1,1), (1,1,-1)$
can be treated similarly.

So we have $\|Bu\|_p \geq 1$ for all
$u \in \mathbb{Z}^3 \setminus \{0\}$ with $|u_i| \leq 1$ and we
proceed by induction. For the inductive step assume that
$\|Bu\|_p \geq 1$ for all $u \in \mathbb{Z}^3 \setminus \{0\}$ with
$|u_i| \leq M$. Let $u \in \mathbb{Z}^3$ with $|u_i| \leq M+1$ for all
$i$ and $|u_i| = M+1$ for some
$i$. Define $v, w \in \mathbb{Z}^3$ componentwise by
\[
v_i = \left\lceil \frac{u_i}{2} \right\rceil \; \text{ and } \; w_i =
\left\lfloor \frac{u_i}{2} \right\rfloor .
\]
Then, $|v_i|, |w_i| \leq M$. Hence, by the induction hypothesis,
$Bv\|_p \geq 1$, and $\quad \|Bw\|_p \geq 1$. Since
$v_i - w_i \in \{0,1\}$, we have $\|Bv-Bw\|_p = 1$. Hanner's
inequality~\eqref{eq:Hanner} implies the desired inequality
$\|Bu\|_p^p \geq 2^p - 1 \geq 1$.
\end{proof}

\section{Numerical findings}
\label{sec:Findings}

In order to explore dense lattice packings of superballs in three
dimensions we numerically found critical points of the nonlinear
minimization problem~\eqref{eq:nonlinear}. We randomly chose a matrix
$B \in \GL_3(\mathbb{R})$, where we chose the matrix entries randomly
according to the normal distribution $N(0,1)$ with mean~$0$ and
variance~$1$. Then we applied Newton's method to $B$ to find a
critical point in the neighborhood of $B$. By applying this procedure
to $10,000$ randomly chosen starting points, we obtained a set of
packing lattices. For the implementation we used the function
\verb|root| of the python package \verb|scipy.optimize|, see
\cite{SciPy2018a}. We determined feasible packing lattices for each of
these three cases.

\subsection{First regime}

The highest packing densities over all obtained solutions we found for
$p = 1, 1.1, 1.2, 1.3, 1.4, 1.5$ belong to Case (III). We list them in
Table~\ref{table:CaseIII} and we analyze them in detail in
Section~\ref{sec:Family}. The density we obtained for $p = 1.4$
coincides with the value reported in \cite{Ni2012a}, but the
basis $e_1, e_2, e_3$ given in~\cite[page 8829]{Ni2012a} does not
give a packing lattice since the $\ell^p_3$-norm of
$2e_1-e_2-e_3$ is too small.

{\small
\begin{table}[htb]
\[
L_{1} =
\begin{pmatrix}
-0.333333333333 & 0.166666666667 & 0.5\\
0.5 & -0.333333333333 & 0.166666666667\\
0.166666666667 & 0.5 & -0.333333333333\\
\end{pmatrix}
\]
\[
L_{1.1} =
\begin{pmatrix}
-0.364125450067 & 0.193419513868 & 0.539049770666\\
0.539049770666 & -0.364125450067 & 0.193419513867\\
0.193419513867 & 0.539049770666 & -0.364125450068\\
\end{pmatrix}
\]
\[
L_{1.2} =
\begin{pmatrix}
-0.392613644302 & 0.22381214158 & 0.569113821114\\
0.569113821115 & -0.392613644298 & 0.223812141583\\
0.223812141575 & 0.569113821114 & -0.392613644306\\
\end{pmatrix}
\]
\[
L_{1.3} =
\begin{pmatrix}
-0.419839537546 & 0.260336714788 & 0.589023079183\\
0.589023079194 & -0.419839537534 & 0.260336714788\\
0.260336714754 & 0.589023079183 & -0.419839537578\\
\end{pmatrix}
\]
\[
L_{1.4} =
\begin{pmatrix}
-0.446984776893 & 0.307534456657 & 0.595696355817\\
0.595696355844 & -0.446984776872 & 0.307534456649\\
0.307534456588 & 0.595696355814 & -0.446984776962\\
\end{pmatrix}
\]
\[
L_{1.5} =
\begin{pmatrix}
-0.475292821919 & 0.375983627555 & 0.580059051165\\
0.580059051205 & -0.475292821888 & 0.375983627545\\
0.375983627482 & 0.58005905116 & -0.475292821997\\
\end{pmatrix}
\]
\caption{Matrices $L_p \in \GL_3(\mathbb{R})$ giving the densest
  lattice packing of $\mathcal{B}^p_n$ we found. They all belong to
  Case (III). }
\label{table:CaseIII}
\end{table}
}

\subsection{Second regime}

The highest packing density over all obtained solutions we found for
$p = 1.6, 1.7, 1.8, 1.9, 2.0$ belong to Case (I). We list them in
Table~\ref{table:CaseI}. We were not able to identify a pattern. The
density we obained for $p = 1.7$ coincides with the value reported in
\cite{Ni2012a}, but in \cite{Ni2012a} a corresponding
matrix is not given.  For more computational results we refer to the
thesis~\cite{Dostert2017a} of the first author.

{\small
\begin{table}[htb]
\[
L_{1.6} =
\begin{pmatrix}
-0.000274732684343 & 0.00144215026174 & -0.999980951403\\
0.509783945989 & 0.500572697171 & -0.499408517681\\
0.509509213365 & -0.499408255119 & -0.500850623006\\
\end{pmatrix}
\]
\[
L_{1.7} =
\begin{pmatrix}
0.458033772615 & 0.556735224273 & 0.553047497039\\
-0.530691863753 & 0.577007354869 & 0.459643833129\\
-0.0936769789857 & -0.0202829019484 & 0.988672468644\\
\end{pmatrix}
\]
\[
L_{1.8} =
\begin{pmatrix}
-0.330208442415 & -0.696395141028 & 0.551458413193\\
0.624661955256 & -0.637870063365 & 0.316559795406\\
0.237379400053 & -0.0588146621535 & 0.954027742247\\
\end{pmatrix}
\]
\[
L_{1.9} =
\begin{pmatrix}
-0.325366212309 & -0.0828873750566 & 0.930867632285\\
0.230698700286 & 0.676231149106 & 0.66666839749\\
-0.697856599406 & 0.59207566084 & 0.335768664213\\
\end{pmatrix}
\]
\[
L_2 =
\begin{pmatrix}
0.000000000000 & 0.707106781187 & 0.707106781187\\
0.707106781623 & 0.00000000000 & 0.70710678075\\
0.707106781623 & 0.70710678075 & 0.000000000000 \\
\end{pmatrix}
\]
\caption{Matrices $B_p \in \GL_3(\mathbb{R})$ giving the densest
  lattice packing of $\mathcal{B}^p_n$ we found. They all belong to
  Case (I). $L_2$ determines the face centered cubic lattice which
  defines a densest sphere packing in three dimensions.}
\label{table:CaseI}
\end{table}
}

\section{A new family of lattice packings}
\label{sec:Family}

When looking at the numerical solutions we found in the first regime,
one immediately comes to the conclusion that the new found lattices
belong to a family of lattices which one can easily parametrize.

Consider
\[
L(x,y,z) \in \GL_3(\mathbb{R}) \quad \text{with} \quad L(x,y,z) =
\begin{pmatrix}
-x & y & z\\
z & -x & y\\
y & z & -x
\end{pmatrix},
\]
where $x, y, z$ are chosen so that
\[
\|L(x,y,z)u\|_p = 1 \text{ for all } u \in \mathcal{U}^3.
\]
This implies
\[
\begin{split}
& \|L(x,y,z)(1,0,0)\|^p_p = |-x|^p + |y|^p + |z|^p = 1\\
& \|L(x,y,z)(1,1,0)\|^p_p = |-x+y|^p + |z-x|^p + |y + z|^p = 1\\
& \|L(x,y,z)(1,1,1)\|^p_p = 3|-x+y+z|^p = 1.
\end{split}
\]
We also have the inequalities
\[
z \geq x \geq y \geq 0
\]
Thus, $x,y,z$ has to satisfy the following nonlinear system:
\begin{equation}
\label{eq:nonlinearsystem}
\begin{split}
& z \geq x \geq y \geq 0\\
& x^p + y^p + z^p = 1 \\
& (x-y)^p + (z-x)^p + (y+z)^p = 1\\
& 3(-x +y +z)^p = 1\\
\end{split}
\end{equation}

The family of lattices starts at $p = 1$ with
$x = 1/3, y = 1/6, z = 1/2$, which defines the densest lattice packing
of regular octahedra found by Minkowski. The family ends at
$p = \log_2 3 = 1.5849625\ldots$ with $x = y = z = \frac{1}{2}$ which
defines the body centered cubic lattice.

We want to prove that the family indeed exists. When the nonlinear
system~\eqref{eq:nonlinearsystem} has a solution,
Lemma~\ref{lem:Hanner} ensures that the members of the family are
packing lattices for the corresponding superball
$\frac{1}{2}B^p_3$. We will apply the following theorem of Cohn,
Kumar, and Minton \cite[Theorem 3.1]{Cohn2016a}, which is an effective
implicit function theorem.

\begin{theorem}
\label{thm:ExistenceTheorem}
Let $V$ and $W$ be finite-dimensional normed vector spaces over
$\mathbb{R}$, and suppose that $f: B(x_0, \epsilon) \rightarrow W$ is
a $C^1$ function, where $x_0 \in V$ and $\varepsilon > 0$. Suppose
also that $T:W \rightarrow V$ is a linear operator such that
\begin{align}
\label{eq:ExistenceTheorem}
\|Df(x) \circ T - id_W\| < 1 - \frac{\|T\| \cdot |f(x_0)|}{\varepsilon}
\end{align}
for all $x \in B(x_0, \varepsilon)$. Then there exists
$x_* \in B(x_0, \varepsilon)$ such that $f(x_*) = 0$. Moreover, in
$B(x_0, \varepsilon)$, the zero locus $f^{-1}(0)$ is a $C^1$
submanifold of dimension $\dim V - \dim W$.
\end{theorem}

In the theorem, $\|\cdot\|$ denotes the operator norm, $Df(x)$ is the
Jacobian of $f$ at $x$, and $id_W$ is the identity operator on $W$,
and $B(x_0, \varepsilon) \subseteq V$ is the \textsl{open} ball with
center $x_0$ and radius $\varepsilon$, where the distance is measured
using the norm of $V$.

\begin{theorem}
\label{thm:Existence}
The nonlinear system~\eqref{eq:nonlinearsystem} has a unique solution
$(x_*(p), y_*(p), z_*(p))$ for all $p \in [1,1.58]$. In particular,
the matrix $L(x_*(p),y_*(p),z_*(p))$ defines a packing lattice for
$\frac{1}{2} \mathcal{B}^p_3$.
\end{theorem}

\begin{proof}
Our proof is computer assisted. We use interval arithmetic as implemented in
the free open source mathematics software system SageMath~\cite{SageMath}.

Define the function $f_p : B((x_0, y_0, z_0), \varepsilon) \to \mathbb{R}^3$ by
\[
f_p(x,y,z) =
\begin{pmatrix}
x^p + y^p + z^p - 1\\
(x-y)^p + (z-x)^p + (y+z)^p - 1\\
3(-x + y + z)^p - 1
\end{pmatrix}.
\]
The Jacobian $Df_p(x,y,z)$ of $f_p$ equals
\[
p
\begin{pmatrix}
x^{p-1} & y^{p-1} & z^{p-1}\\
(x-y)^{p-1}-(z-x)^{p-1} & -(x-y)^{p-1}+(y+z)^{p-1} & (z-x)^{p-1}+(y+z)^{p-1}\\
-3(-x+y+z)^{p-1} & 3(-x+y+z)^{p-1} & 3(-x+y+z)^{p-1}
\end{pmatrix}.
\]
As norms we choose the $\ell^\infty_3$ norm for the domain, as well as
for the codomain. Then the $\ell^\infty_3$ operator norm is the maximum
of the $\ell^1_3$ norms of the rows of the considered matrix.

Now we subdivide the interval $[1.1.58]$ in smaller subintervals and
for every subinterval we choose an $\varepsilon > 0$ and a starting
point $(x_0,y_0,z_0)$ which we found by the numerical solution of the
nonlinear system~\eqref{eq:nonlinearsystem}.

Our choice of $\varepsilon > 0$ and using the $\ell^{\infty}_3$ norm
ensures that the ball $B((x_0,y_0,z_0), \varepsilon)$ lies in the
region $z \geq x \geq y \geq 0$.  We set $T = Df_p(x_0,y_0,z_0)^{-1}$
and verify that inequality~\eqref{eq:ExistenceTheorem} is satisfied
for all $p$ in the subinterval and for all
$(x,y,z) \in B((x_0,y_0,z_0), \varepsilon)$. Then, the assumptions of
Theorem~\ref{thm:ExistenceTheorem} are satisfied and we can conclude
that the nonlinear system~\eqref{eq:nonlinearsystem} has a unique
solution $(x_*(p), y_*(p), z_*(p))$ for all $p$ in the
subinterval. The verification of~\eqref{eq:ExistenceTheorem} uses
interval arithmetic. Our Sage function \texttt{verify} is only a few
lines long and can be found in Appendix~\ref{sec:a} of this paper.

For example, we choose the subinterval $[1,1.01]$, the starting point
$(x_0,y_0,z_0) = (\frac{1}{3}, \frac{1}{6}, \frac{1}{2})$, and
$\varepsilon = 0.03$. For $T$ we choose
\[
T =
\begin{pmatrix}
\frac{1}{2} & 0 & -\frac{1}{6}\\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{6}\\
 0 & \frac{1}{2} & 0\\
\end{pmatrix}.
\]
Then, Sage computes
\[
\|Df_p(x,y,z)  T - I_3\| \in [0.00000000000000000,0.035585437892437462]
\]
and
\[
1 - \frac{\|T\| \cdot |f_p(x_0,y_0,z_0)|}{\varepsilon} \in [0.60895579575438163,1.0000000000000000]
\]
for all $(x,y,z) \in B((x_0,y_0,z_0), \varepsilon)$ and for all
$p \in [1,1.01]$. This shows that~\eqref{eq:nonlinearsystem} has a
unique solution for each $p \in [1,1.01]$.

More examples of our choises can be found in Appendix~\ref{sec:b}; all
choices can be found as ancillary file from the \texttt{arXiv.org} e-print archive.
\end{proof}

We conjecture that the family of lattices also exists for
$p \in (1.58, \log_2 3 = 1.5849625007\ldots]$. We could enlarge the interval for which
Theorem~\ref{thm:Existence} holds by increasing the precision and by
using smaller subintervals. For example, running the following Sage
code shows that the lattice exists for
$p \in [1.5849625, 1.5849625+10^{-10}]$.
\begin{verbatim}
verify(1.5849625, 0.499999999842, 0.499999124646, 0.500000875038,
0.0000002, 0.0000000001)
\end{verbatim}
However, when $p$ approaches $\log_2 3$, then the Jacobian approaches
\[
Df_{\log_2 3}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right) =
\log_3 2
\begin{pmatrix}
\frac{2}{3} & \frac{2}{3} & \frac{2}{3}\\
0 & 1 & 1\\
-2 & 2 & 2\\
\end{pmatrix}
\]
which is singular. Furthermore, there is no $\varepsilon > 0$ so that
the ball $B((\frac{1}{2},\frac{1}{2},\frac{1}{2}), \varepsilon)$ is
contained in the region $z \geq x \geq y$. Currently, we do not know
how to modify the approach to be able to handle these two
difficulties.

\section{Conjectures and open problems}
\label{sec:Conjectures}

Based on Section~\ref{sec:Findings} we pose the
following conjectures and open problems:

\begin{enumerate}
\item The family of lattices determined by~\eqref{eq:nonlinearsystem}
  exists for all $p \in (1.58, \log_2 3]$.
\item The lattices we found in the first regime, and the ones found by
  Jiao, Stillinger, and Torquato \cite{Jiao2009a} in the third and
  fourth regime give the densest lattice packings of superballs for
  the corresponding $p$.
\item It would be interesting to develop a better understanding of the
  densest known lattices in the second regime.
\item Is there a value of $p \neq \infty$ for which the upper bound
  for translative packings of superballs determined
  in~\cite{Dostert2017b} matches the corresponding lower bound?
\item For $p > \log_2 3$ there are no lattices which fall into Case
  (II) or into Case (III).
\item What is the largest value of $p \leq 2$ so that the kissing
  number of $\mathcal{B}^p_3$ superballs is strictly larger than $12$?
\end{enumerate}

\section*{Acknowledgements}

We thank Crist\'obal Guzm\'an and Philippe Moustrou for helpful
discussions.

This material is based upon work supported by the
National Science Foundation under Grant No. DMS-1439786 while the
first author was in residence at the Institute for Computational and
Experimental Research in Mathematics in Providence, RI, during the
"Point Configurations in Geometry, Physics and Computer Science"
semester program. The second author was partially supported by the
SFB/TRR 191 ``Symplectic Structures in Geometry, Algebra and
Dynamics'', funded by the DFG. This project has received funding from
the European Union's Horizon 2020 research and innovation programme
under the Marie Sk\l{}odowska-Curie agreement number~764759.

config: arxiv


TITLE: Is there a strategy to switching $\infty$ limits to $-\infty$ limits?
QUESTION [2 upvotes]: I asked recently in this question how to use the definition of
$$e:= \lim_{x\to\infty}\left( 1+\frac1x \right)^x$$
to show that
$$\lim_{x\to -\infty}\left( 1+\frac1x \right)^x = e.$$
A helpful answer said that
$$\left(1+\frac1{\eta}\right)^\eta = \left(\frac1{\left(1+\frac1{-(\eta+1)}\right)^{-(\eta+1)}}\right)^{-\eta/(\eta+1)}$$
and $-(\eta + 1) \xrightarrow[]{\eta \to -\infty} \infty$.
It struck me as a brilliant way to solve the problem, and I wondered how one might come up with it. Is the rearrangement done in this answer an instance of a more general strategy, or is it more or less just fiddling until one gets the right form?

REPLY [1 votes]: In general, if $h(x)=1+O(x^{-2})$ then $\lim_{x\to\infty} h(x)^x = 1$.
Now, let $h(x)=\left(1-\frac{1}{x^2}\right)=\left(1-\frac{1}{x}\right)\left(1+\frac 1x\right)$.
Since $h(x)^x\to 1$ as $x\to\infty$ and $(1+1/x)^x\to e$ as $x\to\infty$ we have that $\left(1-\frac 1x\right)^x\to \frac{1}{e}$ as $x\to\infty$.
But then $\left(1-\frac1x\right)^{-x}\to e$ as $x\to\infty$. Replacing $x$ with $-x$, we see that:
$$\lim_{x\to -\infty} \left(1+\frac{1}{x}\right)^x=e$$
So all you have to prove is the first claim.


TITLE: Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$
QUESTION [4 upvotes]: Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$

REPLY [3 votes]: Note that
$$
1+2\sin\theta\cos\theta=\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=
(\cos\theta+\sin\theta)^2
$$
Therefore
\begin{align}
\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}
&=\frac{(\cos\theta+\sin\theta)^2}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}
\\[6px]
&=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}
\\[6px]
&=\frac{\cos\theta(1+\tan\theta)}{\cos\theta(1-\tan\theta)}
\\[6px]
&=\frac{1+\tan\theta}{1-\tan\theta}
\end{align}


TITLE: Show that the shortest distance between two opposite edges of a tetrahedron is $\frac{2a}{\sqrt{6}}$.
QUESTION [4 upvotes]: The tetrahedron is formed by the planes
    $ y+z=0$,
    $z+x=0$,
    $x+y=0$, and
    $x+y+z=a$
I'm not able to visualise the sides that will constitute the tetrahedron, so not able to figure out which edges will lie opposite to each other. Please help me understand the formation of the tetrahedron.

REPLY [1 votes]: The "easy" way to visualize the tetrahedron is to do a coordinate change:
$$X=z+y\\Y=z+x\\Z=x+y\\x+y+z=(X+Y+Z)/2$$ In this new coordinate system the sides are the $X=0$, $Y=0$, $Z=0$, and the plane that intersects each axis at $2a$. Your original tetrahedron is similar to this one, rotated $45^\circ$ two times


TITLE: closed operator, projection
QUESTION [1 upvotes]: Let $A: D \subset X \to X$ be a closed linear operator. X is a Banach space. Furthermore we have $\gamma: [0,1] \to \mathbb{C}$, $\gamma$ is a $C^1$ curve and $\gamma \subset \rho(A)$, where $\rho(A)$ is the resolvent set of A.
Define $P = \frac{1}{2 \pi i} \int_{\gamma} R(z) dz$.
Now, I'm trying to show that $PX$ is a subset of $D$ and $AP = \frac{1}{2 \pi i} \int_{\gamma} zR(z) dz$.
I think for the second part, one has to use the identity $AR(z) = zR(z) - id$.
Thanks for any help!

REPLY [1 votes]: Once you know that $P$ is a projection, both of the desired properties follow from the closedness of $A$ as well as the fact that the considered integral is a limit (in norm) of Riemann sums. You can find details in Vol. IV of Reed and Simon's Methods of Modern Mathematical Physics (Section XII.2, Theorem XII.5 (c)).


TITLE: Find all integers $m$ and $n$ such that $n\cos\frac{\pi}{m}= \sqrt{8+\sqrt{32+\sqrt{768}}}$
QUESTION [2 upvotes]: Find all integers $m$ and $n$ such that:
  $$n\cos\frac{\pi}{m}= \sqrt{8+\sqrt{32+\sqrt{768}}}$$

I tried to find the relation of $m$ and $n$, but it's difficult for me. I need help! Thanks!

REPLY [3 votes]: $$8+\sqrt{32+\sqrt{768}}=8+\sqrt{8(\sqrt3+1)^2}=8\left(1+\dfrac{\sqrt3+1}{2\sqrt2}\right)=8(1+\cos(60-45)^\circ)=16\cos^2(7.5^\circ)$$