text:
\begin{document}
\begin{abstract}
		We determine the weak limit of the distribution of the random variables ``height'' and ``range'' on the set
		of $p$-watermelons without wall restriction as the number of steps tends to infinity.
		Additionally, we provide asymptotics for the moments of the random variable ``height''.
\end{abstract}
\maketitle

\section{Introduction}

The model of \emph{vicious walkers} was originally introduced by Fisher~\cite{MR751710} as a model
for wetting and melting processes. In general, the vicious walkers model is concerned with $p$ random
walkers on a $d$-dimensional lattice. In the lock step model, at each time step all of the walkers
move one step in one of the allowed directions, such that at no time any two random walkers share the
same lattice point.

A configuration that attracted much interest amongst mathematical physicists and combinatorialists is
the \emph{watermelon configuration}, which is the model underlying this paper (see Figure~\ref{fig:wm:nowall}
for an example). This configuration can be studied with or without the presence of an impenetrable wall.
By tracing the paths of the vicious walkers through the lattice we can identify the (probabilistic)
vicious walkers model with certain sets of non-intersecting lattice paths. It is exactly this
equivalent point of view that we adopt in this paper.
We proceed with a precise definition. A \emph{$p$-watermelon of length $2n$} is a set of $p$ lattice
paths in $\Z^2$ satisfying the following conditions:
\begin{itemize}
		\item the $i$-th path starts at position $(0,2i)$ and ends at $(2n,2i)$,
		\item the paths consist of steps from the set $\left\{ (1,1),(1,-1) \right\}$ only and
		\item the paths are \emph{non-intersecting}, that is, at no time any two path share the same lattice point.
\end{itemize}
An example of a $4$-watermelon of length $16$ is shown in Figure~\ref{fig:wm:nowall} (for the moment, the dashed lines and the
labels should be ignored).
\begin{figure}[width=8cm]
		\begin{center}
				\includegraphics{watermelon_nowall.eps}

		\end{center}
		\caption{Example of a $4$-watermelon of length $16$ without wall, height $11$, depth $4$ and range $15$}
		\label{fig:wm:nowall}
\end{figure}

Since its introduction, the vicious walkers model has been studied in numerous papers.
While early results mostly analyse the vicious walkers model in the continuum limit, there are nowadays
many results for certain configurations  directly based on the lattice path description given above.
With the increasing number of results it became clear that vicious walkers are very important objects
in mathematical areas far beyond its original scope.
For example, Guttmann, Owczarek and Viennot~\cite{MR1651492} related the star and watermelon
configurations to the theory of Young tableaux and integer partitions.
Later, Krattenthaler, Guttmann and Viennot~\cite{MR1801472} proved new, exact as well as asymptotic,
results for the number of certain configurations of vicious walkers.

The vicious walkers model is also very closely related to random matrix theory, as can be seen
from articles by, e.g.,  Baik~\cite{MR1773413}, Johansson~\cite{MR1900323} and Nagao and Forrester~\cite{MR1877962}.
Later, Katori and Tanemura~\cite{MR2029612} and Gillet~\cite{gillet} studied the diffusion scaling
limit of certain configurations of vicious walkers, namely stars and watermelons, respectively.

In 2003, Bonichon and Mosbah~\cite{MR2022577} presented an algorithm for uniform random generation of
watermelons, which relies on the counting results by Krattenthaler, Guttmann and Viennot~\cite{MR1801472}.
Amongst other things, Bonichon and Mosbah studied the parameter height on the set of watermelons (with and without wall).

In this paper we rigorously analyse the following two parameters on the set of $p$-watermelons:
\begin{itemize}
		\item The \emph{height} of a watermelon is the maximum ordinate reached by its top most branch.
		\item The \emph{range} of a watermelon is the difference of the maximum of its top most branch and
				the minimum of its bottom most branch (the \emph{depth} of the watermelon).
\end{itemize}
The $4$-watermelon depicted in Figure~\ref{fig:wm:nowall} has the height $11$ and the range $11+4=15$.

Katori et. al.~\cite{katori2} and Schehr et. al.~\cite{schehr} studied the parameter ``height'' in the continuous limit,
and recovered the leading terms for some of the asymptotics proved in
\cite{watermelons:withwall,watermelons:withoutwall}. Additionally, Schehr et. al. gave some arguments concerning
the behaviour of the parameter ``height'' as the number of walkers tends to infinity.

Now, consider the set $\mathfrak m_{n}^{(p)}$ of $p$-watermelons of length $2n$, endowed with the uniform probability measure.
We can then speak of the random variables ``height'', denoted by $H_{n,p}$, and ``range'', denoted by $R_{n,p}$, on this set.
We determine the weak limits of $H_{n,p}$ and $R_{n,p}$ as the number $n$ of steps tends to infinity (see Theorem~\ref{thm:height:weak_limit}
and Theorem~\ref{thm:range_weaklimit}, respectively).
Additionally, we determine asymptotics for the moments of $H_{n,p}$.
In particular, we prove that the $s$-th moment of the random variable ``height'' behaves
like $\kappa_sn^{s/2}+\tau_sn^{(s-1)/2}+O\left( n^{s/2-1} \right)$ for some explicit numbers $\kappa_s$ and
$\tau_s$, see Theorem~\ref{thm:height_moments}.

Techniques similar to those applied in this paper can also be used to analyse the random variable height
on the set of $p$-watermelons under the presence of an impenetrable wall.
For details we refer to~\cite{watermelons:withwall}.

The paper is organised as follows. The next section contains some well known results that are
needed in the subsequent sections. In Section~\ref{sec:height} we consider the random variable ``height'',
and we determine the weak limit as well as asymptotics for all moments.
In the last section, we determine the weak limit of the random variable ``range''.

\section{Preliminaries}\label{sec:preliminaries}
In this section we collect several results which will be needed in the two subsequent sections. All these
results are either well known in the literature and/or can easily be derived by means of standard techniques. We,
therefore, remain very brief, give only a few comments on the proofs and in each case refer to the corresponding literature for details.

We start with an exact enumeration result for the total number of watermelons confined to a horizontal strip.
\begin{lemma}
		The number $m_{n,h,k}^{(p)}$ of $p$-watermelons without wall, length $2n$, height $<h$ and depth $>-k$ is given by
		\begin{align*}
			m_{n,h,k}^{(p)}

				&=  \det_{0\le i,j<p}\left( \sum_{\ell\in\Z}\left( \binom{2n}{n+\ell(h+k)+i-j}-\binom{2n}{n+\ell(h+k)+h-i-j} \right) \right).
		\end{align*}
		The total number $m_{n}^{(p)}$ of $p$-watermelons is given by
		\begin{align*}
			m_{n}^{(p)} &= \det_{0\le i,j<p}\left( \binom{2n}{n+i-j} \right).
		\end{align*}
\end{lemma}
This lemma follows immediately from the well-known Lindstr\"om--Gessel--Viennot formula
(see \cite[Corollary 3]{gesselviennot} or \cite[Lemma 1]{MR0335313}), together with an iterated
reflection principle.

\begin{remark}
		Since any $p$-watermelon without wall and length $2n$ has depth $>-n-1$, we see that the number of watermelons with height $<h$
		and no restriction on the depth is given by $m_{n,h,n+1}^{(p)}$. For the sake of convenience, this quantity will also be denoted by
		$m_{n,h}^{(p)}$. In this special case, the determinantal expression above simplifies to
		\begin{align*}
				m_{n,h}^{(p)} &= \det_{0\le i,j<p}\left( \binom{2n}{n+i-j}-\binom{2n}{n+h-i-j} \right).
		\end{align*}
\end{remark}
\begin{lemma}\label{lem:total_num_asymptotics}
We have
\begin{equation*}
m_{n}^{(p)}
= \left(\frac{2}{n}\right)^{\binom{p}{2}}\binom{2n}{n}^p\left(\prod_{i=0}^{p-1}i!\right) \left( 1+O(n^{-1}) \right)
\end{equation*}
as $n\to\infty$.
\end{lemma}
\begin{proof}[Proof (Sketch)]
		The result is established from the closed form expression for $m_{n}^{(p)}$, viz
		\[
		m_{n}^{(p)} = \det_{0\le i,j<p}\left( \binom{2n}{n+i-j} \right)
		= \binom{2n}{n}^p\left(\prod_{i=0}^{p-1}i!\frac{(2n+i)!}{(2n)!}\left( \frac{n!}{(n+i)!} \right)^2\right).

		\]
		For details on the evaluation of this (and many more) determinant, we refer to \cite{MR1701596}.
\end{proof}
\begin{lemma}
For $|m-z|\le n^{5/8}$, $z$ bounded, and arbitrary $N>1$ we have the asymptotic expansion
		\begin{multline} \label{eq:approx_binom_quot}
				\frac{\binom{2n}{n+m-z}}{\binom{2n}{n}}
				= e^{-m^2/n}
				\sum_{u=0}^{4N+1}\left(\frac{z}{\sqrt n} \right)^{u}\frac{1}{u!}H_u\left( \frac{m}{\sqrt n} \right)\\
				+e^{-m^2/n}\sum_{u=0}^{4N+1}\left(\frac{z}{\sqrt n} \right)^{u}
					\sum_{l=1}^{3N+1}n^{-l}\sum_{k=0}^{u-1}
					\sum_{r=1}^{2l}F_{r,l}\binom{2r}{u-k}\frac{(-1)^{u-k}}{k!}H_k\left( \frac{m}{\sqrt n}  \right)
					\left( \frac{m}{\sqrt n}  \right)^{2r+k-u} \\
					+O\left( e^{-m^2/n}n^{-1-2N} \right)
		\end{multline}
		as $n\to\infty$.
		Here, the $F_{r,l}$ are some constants the explicit form of which is of no importance in the sequel, and
		$H_k(z)$ denotes the $k$-th Hermite polynomial, that is,
		\begin{align}
				\frac{H_k(z)}{k!} &= \sum_{m\ge 0}\frac{(-1)^{k-m}}{(k-m)!}\frac{(2z)^{2m-k}}{(2m-k)!},\qquad k\ge 0.
				\label{eq:hermitepolynomial}
 		\end{align}
		\label{lem:approx_binom_quot}
\end{lemma}
The lemma above follows from Stirling's approximation for the factorials.
For a detailed proof we refer to \cite[Lemma 6]{watermelons:withwall}.

\section{Height}\label{sec:height}
In this section we derive asymptotics for the distribution as well as for the moments of the random variable $H_{n,p}$.
As mentioned before, the number of $p$-watermelons with length $2n$ and height $<h$ is given by $m_{n,h}^{(p)}=m_{n,h,n+1}^{(p)}$.
Consequently, we have for the distribution of $H_{n,p}$
\begin{equation}\label{eq:height_cdf}
\P\left\{ H_{n,p}+1 \le h \right\} = \frac{m_{n,h}^{(p)}}{m_{n}^{(p)}}.
\end{equation}
\begin{theorem}\label{thm:height:weak_limit}
		For each fixed $t\in(0,\infty)$ we have the asymptotics
		\begin{equation}
				\P\left\{ \frac{H_{n,p}+1}{\sqrt n}\le t \right\}=
				\frac{2^{-\binom{p}{2}}}{\prod_{j=0}^{p-1}j!}
				\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}\left( t \right)e^{-t^2} \right)
				+O\left( n^{-1/2}e^{-t^2} \right)
		\end{equation}
		as $n\to\infty$, where $H_a(x)$ denotes the $a$-th Hermite polynomial.
\end{theorem}
\begin{proof}
	Set $\vec x=(x_0,\dots,x_{p-1})$ and $\vec y=(y_0,\dots,y_{p-1})$, and consider the more general quantity
	\begin{align*}
		m_{n,h}^{(p)}(\vec x,\vec y) &= \det_{0\le i,j<p}\left( \binom{2n}{n+x_i-y_j}-\binom{2n}{n+h-x_i-y_j} \right).
	\end{align*}
	Factoring $\binom{2n}{n}$ out of each row of the determinant above and replacing each
	entry

	with its asymptotic expansions as given in Lemma~\ref{lem:approx_binom_quot}, we find the asymptotics
	\[
	m_{n,h}^{(p)}(\vec x,\vec y) = \binom{2n}{n}^p\left( D_N(\vec x,\vec y)+O\left(e^{-h^2/n}n^{-1-2N}\right) \right),
	\qquad n\to\infty,
	\]
	where
	\[
	D_N(\vec x,\vec y)=
	\det_{0\le i,j<p}\left(
	\sum_{u=0}^{4N+1}\left(
		\left(\frac{y_j-x_i}{\sqrt n}\right)^uT_{u;N}(0,n)
		-\left(\frac{y_j+x_i}{\sqrt n}\right)^uT_{u;N}(h,n)
		\right)\right)
	\]
	and $N>0$ is an arbitrary integer. Here, $T_{u;N}(h,n)$ is given by (see Lemma~\ref{lem:approx_binom_quot})
	\[
	T_{u;N}(h,n)=e^{-h^2/n}\left(
		\frac{H_u(h/\sqrt n)}{u!}+\sum_{l=1}^{3N+1}n^{-l}\sum_{k=0}^{u-1}
					\sum_{r=1}^{2l}F_{r,l}\binom{2r}{u-k}\frac{(-1)^{u-k}}{k!}H_k\left( \frac{h}{\sqrt n}  \right)
					\left( \frac{h}{\sqrt n}  \right)^{2r+k-u}
	\right).
	\]

	The quantity $D_N(\vec x,\vec y)$ is seen to be polynomial in the $x_i$'s and $y_j$'s.
	This polynomial is divisible by the factors $(x_j-x_i)$ and $(y_j-y_i)$ for $0\le i<j<p$, for if $x_j=x_i$ then
	the $j$-th and the $i$-th row are equal and, therefore, the determinant is zero (if $y_j=y_i$ then the $j$-th and $i$-th column are equal).
	Hence,
	\begin{equation*}

	D_N(\vec x,\vec y)=
		n^{-\binom{p}{2}}

		\frac{\prod\limits_{0\le i<j<p}(x_j-x_i)(y_j-y_i)}{\prod\limits_{0\le j<p}j!^{2}}
		\chi(n,h)
		+O(n^{-1/2}e^{-h^2/n})
		,\qquad n\to\infty.
	\end{equation*}
	Here, the error term is determined by noting that every power of $x_j$ and $y_j$ entails a factor
	of $n^{-1/2}$, as can be seen from the definition of $D_N(\vec x,\vec y)$ above.
	The unknown coefficient $\chi(n,h)$
	can now be determined by comparing coefficients on both sides of the equation above.
	Comparing the coefficients of $\prod_{j=0}^{p-1}x_j^jy_j^j$, we obtain
	(after some simplifications) the equation
	\[
	\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}\left( \frac{h}{\sqrt n} \right)e^{-h^2/n} \right)
	=\chi(n,h).
	\]
	If we specialise by setting $x_j=y_j=j$, then we see that
	\[ m_{n,h}^{(p)}=n^{-\binom{p}{2}}\binom{2n}{n}^p
	\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}\left( \frac{h}{\sqrt n} \right)e^{-h^2/n} \right)
	+O\left(n^{-1/2}e^{-h^2/n}  \right).
	\]

	Setting $h=t\sqrt n$ and replacing $m_{2n}^{(p)}$ with its asymptotic equivalent as given by
	Lemma~\ref{lem:total_num_asymptotics}, we obtain the result.
\end{proof}

\begin{figure}[width=8cm]
		\begin{center}
				\includegraphics{cdfheight.eps}

		\end{center}
		\caption{Comparison of the c.d.f. of the random variable ``height'' on the set of $3-$watermelons of length $500$
		without wall (dotted curve) and the limiting distribution as given by Theorem~\ref{thm:height:weak_limit}.}
		\label{fig:cdf:height}
\end{figure}

Let us now turn our attention to the moments of the distribution of $H_{n,p}$. Clearly, we have for $s\in\N$,
\begin{equation}\label{eq:moments_exact}
	\E\left(H_{n,p}^{s}\right)=\sum_{h\ge 1}h^{s}\frac{m_{n,h+1}^{(p)}-m_{n,h}^{(p)}}{m_{n}^{(p)}}
=\sum_{h\ge 1}\left( h^s-(h-1)^s \right)\frac{m_{n}^{(p)}-m_{n,h}^{(p)}}{m_n^{(p)}}.
\end{equation}
The dominant terms of the asymptotics for the moments are going to be expressed by linear combinations of certain infinite exponential sums.
Asymptotics for these sums are to be determined now.

\begin{lemma}
		For $\nu\ge0$ and $\mu>0$ define
		\[ f_{\nu,\mu}(n)=\sum_{h\ge 1}h^\nu e^{-\mu h^2/n}. \]
		This sum admits the asymptotic series expansion
		\[
		f_{\nu,\mu}(n) \approx \frac{1}{2}\Gamma\left(\frac{ \nu+1}{2}\right)\left( \frac{n}{ \mu} \right)^{(\nu+1)/2}
		+\sum_{m\ge 0}\left( \frac{ \mu}{n} \right)^m \frac{(-1)^{\nu+m}B_{2m+\nu+1}}{(2m+\nu+1)!m!},
		\]
		as $n\to\infty$, where $\Gamma$ denotes the gamma function and $B_m$ is the $m$-th Bernoulli number defined via
		$\sum_{j\ge 0}B_jt^j/j!=t/(e^t-1)$.
		\label{lem:f_nu,mu}
\end{lemma}
\begin{proof}[Proof (Sketch)]
		Asymptotics for sums of this form can often be obtain by means of Mellin transform techniques. For a
		detailed overview of Mellin transforms, harmonic sums and asymptotics, we refer to \cite{MR1337752}.

		We proceed with a sketch of the proof. The inverse Mellin transform gives
		\begin{align*}
				f_{\nu,\mu}(n)=\sum_{h\ge 1}h^\nu e^{-\mu h^2/n} =&
				\sum_{h\ge 1}\frac{h^\nu}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\left( \frac{\mu h^2}{n} \right)^{-z}dz \\
				=&\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\left( \frac{\mu}{n} \right)^{-z}\zeta(2z-\nu)dz.
		\end{align*}
		The integrand has simple poles at $z=(\nu+1)/2$ and $z=0,-1,-2,\dots$ corresponding to the poles of the
		zeta and the gamma function, respectively.
		The result is now obtained by pushing the line of integration to the left and taking into account the residues.

		For the sake of convenience, we mention the evaluations
 		\begin{align*}
				\Res_{z=-m} \Gamma(z) &= \frac{(-1)^m}{m!}, \qquad m=0,1,2,\dots, \\
				\Res_{z=1} \zeta(z) &= 1 \\
				\zeta(-m) &= B_{m+1}\frac{(-1)^m}{m+1}, \qquad m=0,1,2,\dots,
		\end{align*}
		where $B_m$ denotes the $m$-th Bernoulli number defined via  $\sum_{j\ge 0}B_j t^j/j!=t/(e^t-1)$.
\end{proof}

The rest of this section is devoted to the proof of Theorem~\ref{thm:height_moments} below, which gives the final
expression for the asymptotics of the moments. In order to present the proof of this theorem in a clear fashion
we split it into a series of lemmas. For a more detailed overview of the proof, we refer directly to the proof of
Theorem~\ref{thm:height_moments}.

As a first step, we prove in Lemma~\ref{lem:height_moments} a preliminary asymptotic expression for the moments of
the height distribution. The presented compact form of the asymptotics makes use of certain linear operators
that are going to be defined now.
\begin{definition}\label{def:Xi10}
	Let $\Xi_1$ and $\Xi_0$ denote the linear operators

	defined by
	\begin{align*}
			\Xi_1\left(h^{\nu}e^{-\mu h^2}\right) &=
				\frac{1}{2}\Gamma\left(\frac{ \nu+1}{2}\right)\left( \frac{1}{ \mu} \right)^{(\nu+1)/2} \\
			\Xi_0\left(h^{\nu}e^{-\mu h^2}\right) &=
			(-1)^\nu\frac{B_{\nu+1}}{(\nu+1)!},
	\end{align*}
	where $B_k$ denotes the $k$-th Bernoulli number.
\end{definition}
By Lemma~\ref{lem:f_nu,mu} we have
\[
	f_{\nu,\mu}(n) = \Xi_1\left( h^{\nu}e^{-\mu h^2} \right)n^{(\nu+1)/2}+\Xi_0\left(  h^{\nu}e^{-\mu h^2} \right)
	+O(n^{-1}),\qquad n\to\infty,
\]
so that $\Xi_1$ and $\Xi_0$ yield the coefficients of the first two terms in the asymptotic expansion of $f_{\nu,\mu}(n)$.

The preliminary expression for the asymptotics of the moments can now be proven in pretty much the same way as in
Theorem~\ref{thm:height:weak_limit}.
\begin{lemma}\label{lem:height_moments}
	For $s\in\N$, $s\ge 1$, the $s$-th moment of the random variable ``height'' satisfies the
	asymptotics
	\begin{equation}\label{eq:height_moments}
			\E\left( H_{n,p}^s\right)=
			s\Xi_1\left(\kappa_{p}h^{s-1}\right)n^{s/2}
			-\Xi_1\left(\binom{s}{2} \kappa_{p}h^{s-2}+\tau_{p}h^{s-1}\right)n^{(s-1)/2}
			+\Xi_0(\kappa_{p})

			+O\left( n^{s/2-1} \right)
	\end{equation}
	as $n\to\infty$, where
	\[
		\kappa_p=1-\frac{2^{-\binom{p}{2}}}{\prod\limits_{0\le j<p}j!}
		\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}\left( h \right)e^{-h^2} \right)
	\]
	and
	\[
	\tau_p=(p-1)\frac{2^{-\binom{p}{2}}}{\prod\limits_{0\le j<p}j!}
				\det_{0\le i,j<p}\left(
			\begin{cases}
				(-1)^iH_{i+j}(0)-H_{i+j}\left(h\right)e^{-h^2} & \textrm{if $i<p-1$} \\
				(-1)^pH_{p+j}(0)-H_{p+j}\left(h\right)e^{-h^2} & \textrm{if $i=p-1$}
			\end{cases}
		\right).

	\]
Here, $H_k(z)$ denotes the $k$-th Hermite polynomial.
\end{lemma}

\begin{proof}
	Recall the exact expression for the $s$-th moment of the random variable ``height'' (see Equation~\eqref{eq:moments_exact}),
	\begin{equation}\label{eq:height_moments_exact_proof}
	\E\left( H_{n,p}^s \right)=\sum_{h=1}^{n+2p-2}\left( h^s-(h-1)^s \right)\frac{m_n^{(p)}-m_{n,h}^{(p)}}{m_n^{(p)}}.
	\end{equation}
	Asymptotics for this quantity can be obtained in pretty much the same way as Theorem~\ref{thm:height:weak_limit}.
	Compared to the problem of determining asymptotics for \eqref{eq:height_cdf}, the main difference
	now is the summation over $h$.

	We consider the more general quantity
	\begin{align*}
		m_n^{(p)}(\vec x,\vec y)-m_{n,h}^{(p)}(\vec x,\vec y) =
		\det_{0\le i,j<p}\left( \binom{2n}{n+x_i-y_j}\right)
		-\det_{0\le i,j<p}\left( \binom{2n}{n+x_i-y_j}-\binom{2n}{n+h-x_i-y_j} \right),
	\end{align*}
	where $\vec x=(x_0,\dots,x_{p-1})$ and $\vec y=(y_0,\dots,y_{p-1})$.
	As a first step, we pull $\binom{2n}{n}$ out of each row of the determinants above.
	Now, we restrict the range of summation in \eqref{eq:height_moments_exact_proof}
	to $1\le h\le n^{1/2+\varepsilon}$ for some $\varepsilon>0$. This truncation is justified by
	Stirling's formula, which shows that
	\[
	\frac{\binom{2n}{n+\alpha}}{\binom{2n}{n}}=O\left( e^{-n^{2\varepsilon}} \right),\qquad n\to\infty,
	\]
	whenever $|\alpha|\ge n^{1/2+\varepsilon}$.
	This implies that the total contribution of all summands in \eqref{eq:height_moments_exact_proof} satisfying
	$h>n^{1/2+\varepsilon}$ is exponentially small as $n\to\infty$ and, therefore, negligible.
	In all the remaining summands we replace all the quotients of binomial coefficients with their
	asymptotic expansions as given in Lemma~\ref{lem:approx_binom_quot}.
	Finally, we re-extend the range of summation to $h\ge 1$, which, again, introduces an exponentially small
	error term.
	This gives the asymptotics
	\[
		\E\left( H_{n,p}^s \right)
		=\sum_{h\ge 1}\left(\left( h^s-(h-1)^s \right)
		\left(\frac{\binom{2n}{n}^p}{m_n^{(p)}}D_N(\vec e,\vec e)
		+O\left(e^{-h^2/n}n^{\binom{p}{2}-1-2N}\right)\right)
		\right), \qquad n\to\infty,
	\]
	where $\vec e=(0,1,\dots,p-1)$.
	Here, the structure of the error term is a consequence of Lemma~\ref{lem:total_num_asymptotics}, and
	the quantity $D_N(\vec x,\vec y)$ is defined by
	\begin{multline}\label{eq:D_N_shape}
	D_N(\vec x,\vec y)=
	\det_{0\le i,j<p}\left(
	\sum_{u=0}^{4N+1}\left(
		\left(\frac{y_j-x_i}{\sqrt n}\right)^uT_{u;N}(0,n)\right)\right) \\
	-\det_{0\le i,j<p}\left(
	\sum_{u=0}^{4N+1}\left(
		\left(\frac{y_j-x_i}{\sqrt n}\right)^uT_{u;N}(0,n)
		-\left(\frac{y_j+x_i}{\sqrt n}\right)^uT_{u;N}(h,n)
		\right)\right),
	\end{multline}
	where $N>0$ is an arbitrary integer and
	\[
	T_{u;N}(h,n)=e^{-h^2/n}\left(
		\frac{H_u(h/\sqrt n)}{u!}+\sum_{l=1}^{3N+1}n^{-l}\sum_{k=0}^{u-1}
		\sum_{r=1}^{2l}F_{r,l}\binom{2r}{u-k}\frac{H_k\left( h/\sqrt n  \right)}{k!}
					\left(- \frac{h}{\sqrt n}  \right)^{2r+k-u}
	\right).
	\]

	As a consequence of Lemma~\ref{lem:f_nu,mu}, we see (after expanding the term $(h-1)^s$) that
	\[
	\sum_{h\ge 1}\left(\left( h^s-(h-1)^s \right)O\left(e^{-h^2/n}n^{\binom{p}{2}-1-2N}\right)\right)
	= O\left( n^{\binom{p}{2}-2N+(s-1)/2} \right),
	\]
	which is negligible for sufficiently large $N$.	Hence, we have the asymptotics
	\[
		\E\left( H_{n,p}^s \right)
		=\frac{\binom{2n}{n}^p}{m_n^{(p)}}\sum_{h\ge 1}\Big(\left( h^s-(h-1)^s \right)
		D_N(\vec e,\vec e)	\Big)
		+O\left(n^{\binom{p}{2}-2N+(s-1)/2}\right)	, \qquad n\to\infty.
	\]
	It remains to determine the part of $D_N(\vec x,\vec y)$ that gives the dominant contribution to
	the asymptotics above.
	First, we note that $D_N(\vec x, \vec y)$ is a polynomial in the $x_i$'s and $y_i$'s.
	Obviously, $D_N(\vec x,\vec y)$ is equal to zero whenever $x_i=x_j$ or $y_i=y_j$ for some $i\neq j$, for if
	$x_i=x_j$ ($y_i=y_j$) then the $i$-th and $j$-th rows (columns) of the determinants involved in the definition
	of $D_N(\vec x,\vec y)$ are equal, and, therefore, the determinants are equal to zero.
	This implies that $D_N(\vec x,\vec y)$ is of the form
	\[
	D_N(\vec x,\vec y)=
		n^{-\binom{p}{2}}
		\frac{\prod\limits_{0\le i<j<p}(x_j-x_i)(y_j-y_i)}{\prod\limits_{0\le j<p}j!^{2}}
		\left( \chi(n,h)+\sum_{j=0}^{p-1}\left( \xi_j(n,h)\frac{x_j}{\sqrt n}+\eta_j(n,h)\frac{y_j}{\sqrt n} \right)
		+O\left( n^{-1}e^{-h^2/n} \right) \right)
	\]
	as $n\to\infty$.
	By comparing coefficients of $\prod_{j=0}^{p-1}x_j^jy_j^j$ on both sides of the equation above,
	we have already seen (see Theorem~\ref{thm:height:weak_limit}) that
	\[
	\chi(n,h)=
	\det_{0\le i,j<p}\left( (-1)^{i}H_{i+j}(0)\right)-
	\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}\left( \frac{h}{\sqrt n} \right)e^{-h^2/n} \right)
	.
	\]
	Analogously we can determine $\xi_k(n,h)$.
	By comparing the coefficients of $x_k\prod_{j=0}^{p-1}x_j^jy_j^j$ on both sides of the
	equation above we obtain the equations
	\[ 0=\xi_{k}(n,h)-\xi_{k+1}(n,h),\qquad k<p-1, \]
	and
	\[
	\xi_{p-1}(n,h) =
		-\frac{1}{p}
		\det_{0\le i,j<p}\left(
			\begin{cases}
				(-1)^iH_{i+j}(0)-H_{i+j}\left( \frac{h}{\sqrt n} \right)e^{-h^2/n} & \textrm{if $i<p-1$} \\
				(-1)^pH_{p+j}(0)-H_{p+j}\left( \frac{h}{\sqrt n} \right)e^{-h^2/n} & \textrm{if $i=p-1$}
			\end{cases}
		\right).
	\]
	Note, that the coefficient of $x_k\prod_{j=0}^{p-1}x_j^jy_j^j$ in the first determinant of \eqref{eq:D_N_shape} is
	equal to zero, which is easily seen to be true for $k<p-1$, and for $k=p-1$ this is seen to be true
	by a series of column and row operations that yield a new matrix consisting of two non-square blocks.
	Similar expressions (with $i$ and $j$ interchanged) can be found for the $\eta_k(n,h)$, $0\le k<p$.

	Noting that $H_{i+j}(0)$ is non-zero if and only if $i+j$ is even we deduce that
	$(-1)^{i}H_{i+j}(0)=(-1)^jH_{i+j}(0)$, which implies
	\[ \xi_{p-1}(n,h)=\eta_{p-1}(n,h), \]
	and also
	\[
	\det_{0\le i,j<p}\left( (-1)^{i}H_{i+j}(0)\right) = \det_{0\le i,j<p}\left( (-1)^{(i+j)/2}H_{i+j}(0)\right)
	=2^{\binom{p}{2}}\prod_{j=0}^{p-1}j!.
	\]
	Here, the last equality has been proven in Lemma~\ref{lem:detEval}.

	If we specialise to $x_j=y_j=j$, $0\le j<p$, then we obtain
	\[
		D_N(\vec e,\vec e)=
		n^{-\binom{p}{2}}
		\left( \chi(n,h)+2\binom{p}{2}\xi_{p-1}(n,h)n^{-1/2} \right)+O\left(n^{-1} e^{-h^2/n} \right),
		\qquad n\to\infty,
	\]
	where $\vec e = (0,1,\dots,p-1)$.

	Choosing $N$ large enough and expanding the term $h^s-(h-1)^s$ in the asymptotics for
	$\E\left( H_{n,p}^s \right)$ above, we obtain with the help of Lemma~\ref{lem:f_nu,mu}
	the asymptotics
	\[
	\E\left( H_{n,p}^s \right)=
	\frac{\binom{2n}{n}^p}{m_n^{(p)}}
	\sum_{h\ge 1}\left( sh^{s-1}-\binom{s}{2}h^{s-2} \right)D_N(\vec e,\vec e)
	+O\left( n^{s/2-1} \right),
	\qquad n\to\infty,
	\]
	and replacing $D_N(\vec e,\vec e)$ with its asymptotic expansion as given above proves the lemma.

\end{proof}

\begin{lemma}\label{lem:detEval}
Let $H_k(x)$ denote the $k$-th Hermite polynomial as defined by Equation~\eqref{eq:hermitepolynomial}.
We have the determinant evaluation
\begin{equation}
\det_{0\le i,j<p}\left( (-1)^{(i+j)/2}H_{i+j}(0) \right)=
2^{\binom{p}{2}}\prod_{j=0}^{p-1}j!.
\end{equation}
\end{lemma}
\begin{proof}
	The determinant under consideration is a Hankel determinant. Therefore, we can hope to evaluate it
	with the help of orthogonal polynomials (for details see~\cite[Section 2.7]{MR1701596}).
	It is well known (see, e.g., \cite[page 105]{MR0106295}) that for $k\in\N$ we have
	\[ H_{2k+1}(0)=0 \qquad\textrm{and}\qquad H_{2k}(0)=(-1)^k\frac{(2k)!}{k!}. \]
	Consequently, we obtain
	\[
	\det_{0\le i,j<p}\left( (-1)^{(i+j)/2}H_{i+j}(0) \right)=
	2^{\binom{p}{2}}\det_{0\le i,j<p}\left( \frac{1+(-1)^{i+j}}{2}\frac{2^{(i+j)/2}}{\sqrt\pi}\Gamma\left(\frac{i+j+1}{2} \right) \right).
	\]

	The $(i,j)$-th entry of the determinant on the right hand side above is seen to be precisely the $(i+j)$-th moment with
	respect to the Gaussian weight $w(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ on $\R$, that is,
	\[
	\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^ke^{-x^2/2}dx =
	\frac{1+(-1)^{k}}{2}\frac{2^{k/2}}{\sqrt\pi}\Gamma\left(\frac{k+1}{2} \right),\qquad k=0,1,2\dots
	\]

	The family of monic orthogonal polynomials associated with the weight $w(x)$ is given by
	\begin{equation}\label{eq:orth_poly_h}
			2^{-n/2}H_n\left( \frac{x}{\sqrt{2}} \right),\qquad n=0,1,2,\dots
	\end{equation}
	where $H_n(x)$ denotes the $n$-th Hermite polynomial as defined by Equation~\eqref{eq:hermitepolynomial}.
	The three term recursion relation for the orthogonal polynomials~(\ref{eq:orth_poly_h}) is seen to be (cf. \cite[p.105]{MR0106295})
	\[ 2^{-(n+1)/2}H_{n+1}\left( \frac{x}{\sqrt 2} \right)
	=x2^{-n/2}H_{n}\left( \frac{x}{\sqrt 2} \right)-n2^{-(n-1)/2}H_{n-1}\left( \frac{x}{\sqrt 2} \right),
	\qquad n=1,2,\dots, \]
	with the initial values $H_0\left(\frac{x}{\sqrt 2} \right)=1$ and $2^{-1/2}H_1\left(\frac{x}{\sqrt 2} \right)=x$.
	Now, an application of \cite[Theorem 11]{MR1701596}) shows that

	\[
	\det_{0\le i,j<p}\left( \frac{1+(-1)^{i+j}}{2}\frac{2^{(i+j)/2}}{\sqrt\pi}\Gamma\left(\frac{i+j+1}{2} \right) \right)
	=\prod_{j=0}^{p-1}j!,
	\]
	which proves the claim.
\end{proof}
\begin{lemma}\label{lem:Xi1_property}
Let $\mu>0$ denote a real number.
The operator $\Xi_1$ from Definition~\ref{def:Xi10} satisfies the relation
\begin{equation}\label{eq:Xi1_relation}
\Xi_1\left( \frac{d}{dh}\left( h^\nu e^{-\mu h^2} \right) \right)=
\begin{cases}
	-1 & \textrm{if $\nu=0$} \\
	0 & \textrm{if $\nu>0$}.
\end{cases}
\end{equation}
\end{lemma}
\begin{proof}
	For $\nu=0$ the claim follows immediately from the definition of the operator $\Xi_1$.
	For $\nu>0$ we calculate
	\[
		\Xi_1\left( h^{\nu+1}e^{-\mu h^2} \right)=\frac{\nu}{2\mu}\Xi_1\left( h^{\nu-1}e^{-\mu h^2} \right),
	\]
	from which the claims follows upon multiplying by $2\mu$ and rearranging the terms.
\end{proof}

The next result is not obvious at all, and, on the contrary, is a quite surprising fact.
\begin{lemma}\label{lem:kappa_tau_relation}
Let $\kappa_p$ and $\tau_p$ denote the determinants defined in Lemma~\ref{lem:height_moments}.
We have the relation
\begin{equation}\label{eq:kappa_tau_relation}
(p-1)\frac{d}{dh}\kappa_p=\tau_p,\qquad p\ge 1.
\end{equation}
\end{lemma}
\begin{proof}
For the sake of convenience we set
\[ C=2^{-\binom{p}{2}}\left( \prod_{j=0}^{p-1}j! \right)^{-1}. \]
The derivative of a $p\times p$ determinant is the sum $p$ determinants, where the $j$-th addend is equal to
the original determinant with the $j$-th row replaced by its derivative. Hence,
\[
\frac{d}{dh}\kappa_p = C\left( \sum_{j=0}^{p-2}M_j \right)+CM_{p-1},
\]
where
\[
M_i=
\det\left(\
\begin{pmatrix}
	\H_{0,0} & \cdots & \H_{0,p-1} \\ \vdots & \ddots & \vdots \\ \H_{i-1,0} & \cdots & \H_{i-1,p-1} \\
	-H_{i+1}(h)e^{-h^2} & \cdots & -H_{i+p}(h)e^{-h^2} \\
	\H_{i+1,0} & \cdots & \H_{i+1,p-1} \\ \vdots & \ddots & \vdots \\ \H_{p-1,0} & \cdots & \H_{p-1,p-1}
\end{pmatrix}
\ \right)
,
\qquad
\H_{i,j}=(-1)^iH_{i+j}(0)-H_{i+j}(h)e^{-h^2}.
\]

We want to mention that $(p-1)CM_{p-1}$ is equal to the expression for $\tau_p$ except for the constant terms
in the last row of the determinant.

For $0\le i<p-1$ the quantity $M_i$ can also be represented by the expression
\[
M_i=
\det\left(\
\begin{pmatrix}
	\H_{0,0} & \cdots & \H_{0,p-1} \\ \vdots & \ddots & \vdots \\ \H_{i-1,0} & \cdots & \H_{i-1,p-1} \\
	\H_{i+1}& \cdots & \H_{i+p} \\
	(-1)^{i+1}H_{i+1}(0) & \cdots & (-1)^{i+1}H_{i+p}(0) \\
	\H_{i+2,0} & \cdots & \H_{i+2,p-1} \\ \vdots & \ddots & \vdots \\ \H_{p-1,0} & \cdots & \H_{p-1,p-1}
\end{pmatrix}
\ \right),
\qquad 0\le i<p-1,
\]
which is more convenient to work with.

The Laplace expansion for determinants with respect to the row $j+1$, $0\le j<p-1$, gives
\[ M_j = \sum_{k=0}^{p-1}(-1)^{j+1}H_{j+1+k}(0)M_{j,k}, \qquad 0\le j<p-1, \]
where $M_{j,k}$ denotes the minor of $M_j$ obtained by removing row $j+1$ and column $k$, i.e.,
\[
M_{j,k}= \det
\left(\
\begin{pmatrix}
\H_{0,0} & \cdots & \H_{0,k-1} & \H_{0,k+1} & \cdots & \H_{0,p-1} \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
\H_{i-1,0} & \cdots & \H_{i-1,k-1} & \H_{i-1,k+1} & \cdots & \H_{i-1,p-1} \\
\H_{i+1,0} & \cdots & \H_{i+1,k-1} & \H_{i+1,k+1} & \cdots & \H_{i+1,p-1} \\
\H_{i+2,0} & \cdots & \H_{i+2,k-1} & \H_{i+2,k+1} & \cdots & \H_{i+2,p-1} \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
\H_{p-1,0} & \cdots & \H_{p-1,k-1} & \H_{p-1,k+1} & \cdots & \H_{p-1,p-1}
\end{pmatrix}
\ \right).
\]

Now, consider the sum
\[
\sum_{j=0}^{p-2}M_j =
\left(\sum_{j=0}^{p-2}\sum_{k=0}^{p-2}(-1)^{j+1}H_{j+1+k}(0)M_{j,k}\right)
+\sum_{j=0}^{p-2}(-1)^{j+1}H_{j+p}(0)M_{j,p-1}.
\]
The first sum on the right hand side in fact is equal to zero as is going to be shown now.
First, note that
\[ M_{j,k} = M_{k,j} \]
since the matrices involved are transposes of each other.
Recalling that $H_k(0)$ is non zero if and only if $k$ is an even number we deduce that
\[
(-1)^{j+1}H_{j+1+k}(0)M_{j,k}=-(-1)^{k+1}H_{k+1+j}(0)M_{k,j},
\]
and both expressions correspond to different addends of the double sum above ($j+1+k$ has to be even).
This shows that the value of the double sum is indeed equal to zero.

For the second sum we have
\begin{align*}
\sum_{j=0}^{p-2}(-1)^{j+1}H_{j+p}(0)M_{j,p-1}
&= -\sum_{j=0}^{p-2}(-1)^{p}H_{j+p}(0)M_{p-1,j} \\
&= \det_{0\le k,l<p}\left(
\begin{cases}
(-1)^kH_{k+l}(0)-H_{k+l}(h)e^{-h^2} & \textrm{if $k < p-1$} \\
(-1)^pH_{p+l} & \textrm{if $k=p-1$}
\end{cases}
\right),
\end{align*}
which proves the lemma.
\end{proof}

We are now able to to state and prove the final expression for the asymptotics of the moments.
\begin{theorem}\label{thm:height_moments}
	The expected value of the random variable $H_{n,p}$ satisfies the asymptotics
	\begin{equation}\label{eq:height_moments_expectedvalue}
		\E\left( H_{n,p}\right) =  \Xi_1\left( \kappa_p \right)\sqrt{n}+p-\frac{3}{2}+O\left( n^{-1/2} \right),
		\qquad n\to\infty,
	\end{equation}
	and for $s\in\N$, $s\ge 2$, we have the asymptotics
	\begin{equation}\label{eq:height_moments_higher}
		\E\left( H_{n,p}^s\right) = s\Xi_1(\kappa_ph^{s-1})n^{s/2}
		+(s-1)\left( p-1-\frac{s}{2} \right)\Xi_1\left( \kappa_ph^{s-2} \right)n^{(s-1)/2}
		+O\left( n^{s/2-1} \right),
		\qquad n\to\infty.
	\end{equation}
	Here, $\kappa_p$ is defined by
	\[
	\kappa_p=1-\frac{2^{-\binom{p}{2}}}{\prod\limits_{0\le j<p}j!}
	\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}\left( h \right)e^{-h^2} \right),
	\]
	where $H_k(z)$ denotes the $k$-th Hermite polynomial.
\end{theorem}
\begin{proof}
	As a first step we need to establish some simple facts concerning the quantity $\kappa_p$.
	To be more precise, we have to show that $\kappa_p$ is an even function with respect to $h$ that has
	no constant term, i.e., is of the form
	\[ \kappa_p = \sum_{k=0}^{K}\sum_{m=1}^{M}\lambda_{k,m}h^{2k}e^{-mh^2} \]
	for some numbers $K$, $M$ and some constants $\lambda_{k,m}$.

	It is obvious from the definition of the Hermite polynomials (see Equation~\eqref{eq:hermitepolynomial})
	that the $k$-th Hermite polynomial is an even (odd) polynomial whenever $k$ is even (odd).
	This also implies the equality $(-1)^iH_{i+j}(0)=(-1)^jH_{i+j}(0)$.
	Now, replacing $h$ by $-h$ in the definition of $\kappa_p$, factoring $(-1)^i$ out of the $i$-th row and
	$(-1)^j$ out of the $j$-th row we see that the expression remains unaltered. Hence, $\kappa_p$ is an even function of $h$.
	The constant term of $\kappa_p$ is seen to be equal to
	\[
	1-\frac{2^{-\binom{p}{2}}}{\prod_{j=0}^{p-1}j!}\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0) \right)
	=1-\frac{2^{-\binom{p}{2}}}{\prod_{j=0}^{p-1}j!}\det_{0\le i,j<p}\left( (-1)^{(i+j)/2}H_{i+j}(0) \right)=0,
	\]
	where the last equality is a consequence of Lemma~\ref{lem:detEval}.
	This proves the claimed form of $\kappa_p$.

	We are now going to prove the asymptotics~(\ref{eq:height_moments_higher}). Therefore, we assume that $s>1$.
	The properties of $\kappa_p$ established above together with Lemma~\ref{lem:Xi1_property} imply the equation
	\[ \Xi_1\left( \frac{d}{dh}\left( \kappa_ph^{s-1} \right) \right)=0, \]
	and the product rule for the derivative together with Lemma~\ref{lem:kappa_tau_relation} show that
	\[ \Xi_1\left( \tau_ph^{s-1} \right)=-(s-1)(p-1)\Xi_1\left( \kappa_ph^{s-2} \right). \]
	The asymptotics~(\ref{eq:height_moments_higher}) is now obtained from the asymptotics~(\ref{eq:height_moments})
	upon noting that the $\Xi_0$-term is negligible for $s\ge2$.

	Finally, we prove the asymptotics~(\ref{eq:height_moments_expectedvalue}) and, therefore, assume $s=1$.
	For the sake of simplicity we set
	\[ C = 2^{-\binom{p}{2}}\left( \prod_{j=0}^{p-1}j! \right)^{-1}. \]
	From Lemma~\ref{lem:kappa_tau_relation} and Lemma~\ref{lem:Xi1_property} we deduce that
	\[
	\Xi_1\left( \tau_p \right)=(p-1)\Xi_1\left( \frac{d}{dh}\kappa_p \right)
	=-(p-1)\Xi_1\left( C\frac{d}{dh}\chi(h) \right),
	\]
	where
	\[ \chi(h)=\det_{0\le i,j<p}\left( (-1)^iH_{i+j}(0)-H_{i+j}(0)e^{-h^2} \right). \]
	This last determinant can be evaluated to a closed form expression with the help of Lemma~\ref{lem:detEval}.
	Factoring $1-(-1)^je^{-h^2}$ out of each column of the determinant we see that
	\begin{align*}
		\chi(h)
		&=\left( \prod_{j=0}^{p-1}\left( 1-(-1)^je^{-h^2} \right) \right)\det_{0\le i,j<p}\left( (-1)^{(i+j)/2}H_{i+j}(0) \right) \\
		&=\frac{1}{C}\left(1-e^{-2h^2}\right)^{\floor{p/2}}\left(1-e^{-h^2}\right)^{\ceil{p/2} -\floor{p/2}}.
	\end{align*}
	Now, an application of Lemma~\ref{lem:Xi1_property} shows that
	\[ \Xi_1\left( \frac{d}{dh}\chi(h) \right)=-1, \]
	which implies
	\[ \Xi_1\left( \tau_p \right)=1-p. \]

	The last step of the proof is the evaluation of the quantity $\Xi_0(\kappa_p)$. Recalling that $\kappa_p$ is
	an even function with respect to $h$ as well as the fact that all odd Bernoulli numbers except for $B_1$ are zero,
	i.e., $B_{2\nu+1}=0$, $\nu\ge 1$, we deduce the equation
	\[
	\Xi_0(\kappa_p)
	=\Xi_0\left( 1-C\chi(h) \right)
	=\Xi_0\left( 1-\left(1-e^{-2h^2}\right)^{\floor{p/2}}\left(1-e^{-h^2}\right)^{\ceil{p/2} -\floor{p/2}} \right).
	\]
	The definition of $\Xi_0$ reveals that $\Xi_0\left( h^\nu e^{-\mu h^2} \right)$ is independent of $\mu$.
	Consequently, we see that
	\[ \Xi_0(\kappa_p) = B_1=-\frac{1}{2}. \]
	This proves the asymptotics~(\ref{eq:height_moments_expectedvalue}) and completes the proof of the theorem.
\end{proof}
\begin{table}
\caption{
This table gives the coefficient of the dominant asymptotic term of $\E H_{n,p}^s$ as $n\to\infty$ for small values of $s$ and $p$
(see  Theorem~\ref{thm:height_moments}).}
\begin{tabular}{c|r|r|r}
\label{tab:moment_height_dominant_coef}
		$s\kappa_{s}^{(p)}$ & \centering{$s=1$} & $s=2$ & $s=3$\\
		\hline
		$p=1$ & $\frac{1}{2}\sqrt\pi=0.88\dots$ & $1$ & $\frac{3}{4}\sqrt\pi=1.32\dots$\\
		$p=2$ & $\frac{2+\sqrt 2}{4}\sqrt\pi=1.51\dots$ & $\frac{5}{2}$ & $\frac{3(12+\sqrt 2)}{16}\sqrt\pi=4.45\dots$\\
		$p=3$ & $\frac{72+45\sqrt 2-16\sqrt 3}{96}\sqrt\pi=1.99\dots$ & $\frac{25}{6}$ & $\frac{1584+315\sqrt 2-32\sqrt 3}{385}\sqrt\pi=9.11\dots$\\
		$p=4$ & $\frac{10368+17091\sqrt 2-3776\sqrt 3}{20736}\sqrt{\pi}=2.39\dots$ & $\frac{1915}{324}$ & $\frac{520992+165969\sqrt 2-29824\sqrt{3}}{82944}\sqrt\pi=15.04\dots$\\
\end{tabular}
\end{table}
Table~\ref{tab:moment_height_dominant_coef} shows the constant of the dominant asymptotic term as $n\to\infty$ for the
$s$-th moment of the height distribution for small values of $s$ and $p$.

\section{Range}\label{sec:range}
We determine the asymptotics for $n\to\infty$ of
\begin{equation}\label{eq:range_cdf_exact}
\P\left\{ R_{n,p}\le r \right\}
	=\frac{1}{m_{n}^{(p)}}\sum_{h=2p-2}^{r}\left( m_{n,h+1,r-h+1}^{(p)}-m_{n,h,r-h+1}^{(p)} \right).

\end{equation}
Note that $m_{n,h+1,r-h+1}^{(p)}-m_{n,h,r-h+1}^{(p)}$ is the number of watermelons with height exactly $h$ and range $\le r$.

\begin{theorem}\label{thm:range_weaklimit}
	For each fixed $t\in(0,\infty)$ we have the asymptotics
	\begin{equation}
			\P\left\{ \frac{R_{n,p}+1}{\sqrt n}\le t \right\}\to
			\frac{2^{-\binom{p}{2}}}{\prod_{i=0}^{p-1}i!}
			\int_0^t\left(\left. \frac{d}{dz}T_p(z,w)\right|_{z=t} \right)dw,
			\qquad n\to\infty,
	\end{equation}
	where
\[ T_p(z,w)=\det_{0\le i,j<p}\left( (-1)^i\left(\sum_{\ell\in\Z}H_{i+j}(\ell z)e^{-(\ell z)^2} \right)
-\left(\sum_{\ell\in\Z}H_{i+j}\left( \ell z+w \right)e^{-(\ell z+w)^2}\right)\right).
\]

Here, $H_a$ denotes the $a$-th Hermite polynomial.
\end{theorem}
\begin{proof}
Since $m_{n,2p-2,k}^{(p)}=0$ for any $k$, Equation~\eqref{eq:range_cdf_exact} can be rewritten as
\[
\P\left\{ R_{n,p}\le r \right\}
	=\frac{m_{n,r+1,1}^{(p)}}{m_{n}^{(p)}}
	+\frac{1}{m_{n}^{(p)}}\sum_{h=2p-1}^{r}\left( m_{n,h,r-h+2}^{(p)}-m_{n,h,r-h+1}^{(p)}  \right).
\]
The first term on the right-hand side is negligible. To see this, we note that $m_{n,r+1,1}$ is equal to the number of $p$-watermelons
with wall and height $\le r$, which is of order $\binom{2n}{n}^pn^{-p^2}$ as $n\to\infty$ (see \cite{watermelons:withwall} for details),
whereas $m_n^{(p)}$ is of order $\binom{2n}{n}^pn^{-\binom{p}{2}}$ (see Lemma~\ref{lem:total_num_asymptotics}).

Asymptotics for the sum on the right-hand side can now be established in a fashion analogous to the proof
of Theorem~\ref{thm:height:weak_limit}.  A more detailed presentation of these techniques can also be found
in \cite[Theorem 2]{watermelons:withwall}.
We find the asymptotics
\[
	\P\left\{ R_{n,p}\le r \right\} \sim
	\frac{\binom{2n}{n}^pn^{-\binom{p}{2}}}{m_{n}^{(p)}}
	\sum_{h=2p-1}^{r}\left(T_p\left( \frac{r+2}{\sqrt n},\frac{h}{\sqrt n} \right)- T_p\left( \frac{r+1}{\sqrt n},\frac{h}{\sqrt n} \right) \right)
\]
as $n\to\infty$, where
\[
T_p(t,w)=\det_{0\le i,j<p}\left(
			(-1)^i\left(\sum_{\ell\in\Z}H_{i+j}(\ell t)e^{-(\ell t)^2} \right)
			-\left(\sum_{\ell\in\Z}H_{i+j}\left( \ell t+w \right)e^{-(\ell t+w)^2}\right)
\right).
\]

Now, Taylor series expansion shows that
\[
T_p\left( \frac{r+2}{\sqrt n},\frac{h}{\sqrt n} \right)- T_p\left( \frac{r+1}{\sqrt n},\frac{h}{\sqrt n} \right)
=\frac{1}{\sqrt n}T_p'\left( \frac{r+1}{\sqrt n},\frac{h}{\sqrt n} \right)+O\left( n^{-1} \right),\qquad n\to\infty,
\]
where $T'$ denotes the derivative of $T$ with respect to its first argument.
Setting $r+1=t\sqrt n$ we see that
\[
\sum_{h=2p-1}^{r}\left(T_p\left( \frac{r+2}{\sqrt n},\frac{h}{\sqrt n} \right)- T_p\left( \frac{r+1}{\sqrt n},\frac{h}{\sqrt n} \right) \right)
\sim \sum_{h=2p-1}^{r}\frac{1}{\sqrt n}T_p'\left( \frac{r+1}{\sqrt n},\frac{h}{\sqrt n} \right)
\to \int_0^{t}T'\left( t,w \right)dw
\]
as $n\to\infty$.

\end{proof}
\begin{remark}
	For the special case $p=1$ we recover a well-known fact originally proven by
	Chung~\cite{MR0467948} and Kennedy~\cite{MR0402955}. Namely, the equality of the distributions
	of the height of Brownian excursions and the range of Brownian bridges. This result also
	follows from a more general relation between excursions an bridges proved by Vervaat~\cite{0392.60058}.

	In fact, for $p=1$ we have
	\[ \left.\frac{d}{dz}T_1(z,w)\right|_{z=t}=-\sum_{\ell\in\Z}2\ell^2te^{-(\ell t)^2}+2\sum_{\ell\in\Z}\ell(\ell t+w)e^{-(\ell t+w)^2}, \]
	which shows that
	\[
	\P\left\{ \frac{R_{n,1}+1}{\sqrt n}\le t \right\}\to \sum_{\ell\in\Z}\left( 1-2(\ell t)^2 \right)e^{-(\ell t)^2},
	\qquad n\to\infty,
	\]
	by Theorem~\ref{thm:range_weaklimit}.
	This shows that the distribution of the range of $1$-watermelons without wall weakly converges to the limiting distribution
	of the height of $1$-watermelons with wall (see \cite{watermelons:withwall}).
\end{remark}

\bibliographystyle{plain}
\bibliography{pmelons}

\end{document}


TITLE: Completing the square
QUESTION [0 upvotes]: $2x^2 - 6x - 5$
My workings are
$2 ( x^2 - 3x + (3/2)^2 - (3/2)^2 - 2.5 ) = 0$
$ 2 ( x - 1.5)^2 - 4.75 = 0$
$ (x-1.5)^2 = 2.375 $
From here I go on to find X which is not the correct answer .. Can anyone help me on where I have gone wrong ? Thanks in advance !

REPLY [0 votes]: You divide by 2 twice. $ 2( (x-3/2)^2-4.75)=0 \rightarrow (x-3/2)^2=4.75$


TITLE: how to find convolution between $f(x)=e^{-2x^2}$ and $g(x)=e^{-2x^2}$?
QUESTION [2 upvotes]: I wanted to find the convolution between the two function. By definition I get,
$$(f*g)(x)=\int_{\mathbb R}e^{-2(x-y)^2}e^{-2x^2}dy=\int_{-\infty}^{\infty}e^{-4x^2+4xy-2y^2}dy $$
But I am not able to calculate the given integral.

REPLY [2 votes]: Another way:
$$
\int_{-\infty}^{\infty}e^{-4x^2+4xy-2y^2}dy
= \int_{-\infty}^{\infty}e^{-2(y-x)^2-2x^2}dy
= e^{-2x^2} \int_{-\infty}^{\infty}e^{-2(y-x)^2}dy
$$
Here,
$$
\int_{-\infty}^{\infty}e^{-2(y-x)^2}dy
= \{ z = y-x \}
= \int_{-\infty}^{\infty}e^{-2z^2}dz
= \{ u = \sqrt{2}z \}
= \int_{-\infty}^{\infty}e^{-u^2} \frac{du}{\sqrt{2}}
= \sqrt{\frac{\pi}{2}}
$$
Thus,
$$
(f*g)(x) = \sqrt{\frac{\pi}{2}} e^{-2x^2}.
$$


TITLE: Galaxy Cluster Data
QUESTION [0 upvotes]: I have a modification to ${\rm MOND}$ that appears to work much better than 'Standard' ${\rm MOND}$ at the galactic scale.  I want to test this model now on larger structures.  Does anyone know where I can find kinematic data on galaxy clusters, preferably some 'gold' standard that people have been examining for years.  I know that the missing mass is a classic problem and that galaxy clusters are a problem for ${\rm MOND}$ as well as $\Lambda{\rm CDM}$.  I need basically the total mass of the cluster, the radial location of each galaxy in the cluster, and the line of sight velocity of the galaxy.

REPLY [0 votes]: As near as I can tell, the 'gold' standard for Galaxy Clusters is the Coma Cluster.  It's where Dark Matter was first detected and still appears to have a lot of papers dedicated to it's kinematics and evolution.  I found one fellow who has taken the time to download the NED database and clean it up himself.  It can be found here:
http://pw2.netcom.com/~ahighe/a1656.htm
There's also the NED database.  If you enter the cluster's designation (e.g. A 1656), the cluster's approximate radius (in arcmin), you can get a pretty good set of data in ASCII, HTML or XML form:
https://ned.ipac.caltech.edu/forms/nearname.html
Hope this helps someone with a similar need.


TITLE: Could anyone explain how my textbook did this simplification?
QUESTION [0 upvotes]: The book is talking about proving Pascal's triangle increases until the middle, until which point it decreases.

Theorem 4.2 refers to the Multiplicative formula
I just don't understand how that left side could simplify to $1$. And how do they rearrange it?

REPLY [2 votes]: The numerator of the left side is $n(n-1)\cdots (n-k+1)$.
The numerator of the right side is $n(n-1) \cdots (n-k+1)(n-k)$.
Dividing both sides by $n(n-1)\cdots (n-k+1)$, the left's numerator becomes $1$, while the right's numerator becomes $(n-k)$.
Similarly, the denominator on the left side is $k(k-1)\cdots 1$.
The denominator on the right side is $(k+1)k(k-1)\cdots 1$.
Multiplying both sides by $k(k-1)\cdots 1$, the left's denominator becomes $1$, while the right's denominator becomes $k+1$.
This results in: $1/1 = (n-k)/(k+1)$.
Does this clear things up?


TITLE: How to find the derivative of $x{^2/3} - (x - 1)^{1/3}$ by the limit calculation?
QUESTION [2 upvotes]: I had to solve an exercise and was going to ask for help here, but as I typed, I spotted an error myself. I felt sorry to throw away all that MathJax stuff I typed for more than an hour, so I decided to post the question anyway, since it still can be useful to someone. If someone has any suggestions, corrections, I would be very grateful.
This is the task itself, which I'd like to quote, at least for search engines to index, if someone else has this problem.

Graph the curves in Exercises 39-48.
a. Where do the graphs appear to have vertical tangent lines?
b. Confirm your findings in part (a) with limit calculations …

The problem itself:
I can't solve exercise 45: $y = x^{2/3} - (x - 1)^{1/3}$.
I would be grateful if someone told me how to solve this exactly, but any help or hint will be greatly appreciated.
I need to find the derivative of this function, but by limit calculations. I know, the derivative should be $\frac{2}{3x^{1/3}} - \frac{1}{3(x- 1)^{2/3}}$, but I can't find it through the limit:
$$
\lim_{h \to 0} \frac{(x + h)^{2/3} - (x - 1 + h)^{1/3} - x^{2/3} + (x- 1)^{1/3}}{h}
$$

Tried approach 1:
I tried to split the fraction in four ones, multiply both the nominator and denominator by a power of nominator necessary to get rid of the cubic root, like:
$$
\frac{x + h}{h(x + h)^{1/3}}
$$
But this didn't help, since dangling 'h's were still remaining in the expressions and I needed to combine the four fractions into a single one again, thus multiplying the nominators by cubic roots again.

Tried approach 2:
This one actually worked, and I'll post it as an answer.

REPLY [2 votes]: Let's split the limit fraction in two ones:
$$
\lim_{h \to 0} \frac{(x + h)^{2/3} - (x - 1 + h)^{1/3} - x^{2/3} + (x- 1)^{1/3}}{h} = \\
\lim_{h \to 0} (\frac{(x + h)^{2/3} - x^{2/3}}{h} - \frac{(x- 1)^{1/3} - (x - 1 + h)^{1/3} }{h})
$$
For simplicity let's deal with the fractions without the limit notation:
$$
\frac{(x + h)^{2/3}- x^{2/3}}{h} \text{let's call it the left part} - \frac{(x- 1)^{1/3} - (x - 1 + h)^{1/3}}{h} \text{let's call it the right part}
$$
I multiplied both the left and right parts to make the nominators become differences of the cubes (to get rid of the cubic roots).
The left part turned to be this:
$$
\require{cancel}
\frac{(x + h)^2 - x^2}{h((x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3})} = \\
\frac{\cancel{x^2} + 2\cancel{h}x + \cancelto{h}{h^2} - \cancel{x^2}}{\cancel{h}((x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3})} = \\
\frac{2x + h}{(x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3}}
$$
the right part:
$$
\require{cancel}
\frac{\cancel{x} - \cancel{1} - \cancel{x} + \cancel{1} - \cancelto{1}{h}}{\cancel{h}((x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + h)^{1/3} + (x - 1 + h)^{2/3})} = \\
\frac{- 1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + h)^{1/3} + (x - 1 + h)^{2/3}}
$$
Since the h tends to 0, the limit is:
$$
\require{cancel}
\lim_{h \to 0}(\frac{2x + \cancelto{0}{h}}{(x + \cancelto{0}{h})^{4/3} + x^{2/3}(x + \cancelto{0}{h})^{2/3} + x^{4/3}} - \frac{1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + \cancelto{0}{h})^{1/3} + (x - 1 + \cancelto{0}{h})^{2/3}}) = \\
\frac{2x}{x^{4/3} + x^{2/3}x^{2/3} + x^{4/3}} - \frac{1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1)^{1/3} + (x - 1)^{2/3}} = \\
\frac{2\cancel{x}}{3\cancelto{x^{1/3}}{x^{4/3}}} - \frac{1}{3(x - 1)^{2/3}} = \\
\frac{2}{3x^{1/3}} - \frac{1}{3(x - 1)^{2/3}}
$$


TITLE: Calculating magnitude of a complex fraction
QUESTION [1 upvotes]: I am having a hard time determining the magnitude of a complex fraction.
$$G(j\omega) = \frac{\omega^{2}_{n}}{j\omega(j\omega + 2\zeta \omega_{n})}$$
I understand that $$|G(j\omega)| = \sqrt{Re\{G(j\omega)\}^{2}+Im\{ G(j\omega)\}^{2}}$$
How would I break $G(j\omega)$ up into real and imaginary parts
Update:
$ \omega_{n}, \zeta$ are pure values
$j$ is an imaginary number
$\omega$ is the independent variable

REPLY [1 votes]: Assuming that $\omega, \omega_n,\zeta \in \mathbb{R}$, and using that $\,|z|^2 = z \cdot \bar z\,$:
$$
\big|G(j\omega)\big|^2 = \frac{\omega^{2}_{n}}{j\omega(j\omega + 2\zeta \omega_{n})} \cdot \frac{\omega^{2}_{n}}{-j\omega(-j\omega + 2\zeta \omega_{n})} = \frac{\omega_n^{\,4}}{\omega^2 \left(4 \zeta^2\omega_n^{\,2}+\omega^2\right)}
$$


TITLE: EW theory vertices
QUESTION [1 upvotes]: I'm trying to undertand the following vertex:
Initial state of up and anti-down quarks with finalk state made of $W^+$ boson. Does it go with left or right projector? I think that from Lagrangian it should go with left projector but the vertex with $e^+$ and $W^-$ in initial state and $\bar{\nu}_e$ in the final one goes with right projector and this is not read from Lagrangian.
Accordingly to my professor, the case R-positron with $W^-$ to give R-antineutrino has a vertex that goes with $\gamma^\mu P_R$, so what I want to know is how extract this vertex if the Lagrangian does not contain that term, just $P_R\gamma^\mu P_L = \gamma^\mu P_L$

REPLY [2 votes]: It would be helpful if you just wrote the Lagrangian terms you are looking at.
Destroying a Left-handed u and R antidown to create a W+ corresponds to the vertices
$$
W_\mu^- \overline {d} P_R\gamma^\mu P_L u + W_\mu^+ \overline {u} P_R\gamma^\mu P_L d ,
$$
while destroying a R-positron and a W- to yield a R-antineutrino to
$$
W_\mu^-\overline {e} P_R\gamma^\mu P_L \nu +  W_\mu^+ \overline{\nu}  P_R\gamma^\mu P_L e ~~~,
$$
where you focus on the first term in each line.
Note half the species, the R leptons/quarks and L anti leptons/antiquarks are simply missing from these couplings.  (In an, impossible, notional world with no masses, these components would be missing everywhere and all spinors would be Weyl, and projectors would be superfluous.)
Further note L is in no way privileged over R: it is simply a convention of us mooring chirality on leptons/quarks instead of antileptons/antiquarks. (Sometimes this convention is subverted in QFT texts or GUT arraying.)


TITLE: Are $x^y$ and $x^z$ "Powers" of $x$?
QUESTION [2 upvotes]: Or would "powers of $x$" in those examples refer to $y$ and $z$, respectively?

REPLY [1 votes]: In the expression "$x^y$", $y$ is the "exponent"/"power" and $x$ is the "base".
But if $x$ is an arbitrary real number, I would not generally call $x^y$ "a power of $x$", though I might call it "$x$ raised to a power".

"Powers of $n$" where $n$ is a particular positive integer (or rarely, perhaps a positive rational) refers to the values $n^m$ where $m$ is an integer. In an integer-only context, the "powers of $n$" might require $m$ to be nonnegative or even positive. But there are also contexts where we allow $m$ to be a negative integer (the famous film(s) Powers of Ten is one example). Curiously, a phrase like "negative powers of $n$" is common, because context makes it clear that it must be the exponent $m$ which is negative.
It's difficult to provide a definitive source for the meaning of this phrase because it is not often carefully defined in textbooks. The English Wikipedia page for "Power of two" has some references, though.


TITLE: Finding volume of a revolution
QUESTION [0 upvotes]: I want to find the volume of the revolution that occurs when the region bounded by $y = x^2$ and $y = 1$ is revolved around the line $y=2$. The problem is that it is not solid and I cannot understand how the disk method could be used. How can one attack the problem?
Thanks in advance

REPLY [0 votes]: The disk method can be used here. $V = \displaystyle \int_{-1}^1 \pi\left((2-x^2)^2-1^2\right)dx$


text:
\begin{document}

\

\baselineskip .7cm

\vs{27mm}

\begin{center}

{\LARGE \bf A Novel View on the Physical Origin of E$_8$}

\vs{3mm}

Matej Pav\v si\v c

Jo\v zef Stefan Institute, Jamova 39,
1000 Ljubljana, Slovenia

e-mail: matej.pavsic@ijs.si

\vs{6mm}

{\bf Abstract}

\end{center}

We consider a straightforward extension of the 4-dimensional spacetime $M_4$
to the space of extended events associated with strings/branes, corresponding
to points, lines, areas, 3-volumes, and 4-volumes in $M_4$. All those objects
can be elegantly represented by the Clifford numbers $X\equiv x^A \gamma_A \equiv
x^{a_1 ...a_r} \gamma_{a_1 ...a_r},~r=0,1,2,3,4$.
This leads to the concept of the so-called Clifford
space ${\cal C}$,
a 16-dimensional manifold whose tangent space at every point is the Clifford algebra
${\cal C \ell }(1,3)$. The latter space besides an algebra is also a vector
space whose elements
can be rotated into each other in two ways: (i) either by the action of
the rotation matrices of SO(8,8) on the components $x^A$ or (ii) by the
left and right action of the Clifford numbers $R=$exp$\,[\alpha^A \gam_A]$ and
$S=$exp$\,[\beta^A \gam_A]$ on $X$. In the latter case, one does not
recover all possible rotations of the group SO(8,8). This
discrepancy between the transformations (i) and (ii) suggests that one should
replace the tangent space ${\cal C \ell }(1,3)$ with a vector space
$V_{8,8}$ whose basis elements are generators of the Clifford
algebra ${\cal C \ell }(8,8)$, which contains the Lie algebra
of the exceptional group E$_8$ as a subspace. E$_8$ thus arises
from the fact that, just as in the spacetime $M_4$ there are  $r$-volumes
generated by the tangent vectors of the spacetime, there are $R$-volumes,
$R=0,1,2,3,...,16$, in the Clifford space ${\cal C}$, generated by the tangent vectors
of ${\cal C}$.

\section{Introduction}

The unification of fundamental interactions has turned out to be an
elusive goal. Amongst the numerous approaches that have been
pursued, Kaluza-Klein and string/brane theories are of
particular importance. While Kaluza-Klein theories\,\ci{Kaluza} introduce
extra dimensions into the game, string/brane theories\ci{strings}
incorporate the extended nature of physical objects. Although
very promising, these theories have encountered serious
problems that have remained unresolved, and, so far, they
have not led to a realistic model. In order to develop such a model, one has
to find a grand unification group, e.g., SU(5), SO(10),
E$_6$ or E$_8$, which fits into a more fundamental theory
that also incorporates the Poincar\'e group of spacetime.
For the time being, a lot of research is devoted to explaining
the properties of these groups regardless of their
possible deeper origin in physics.

Last year, Garrett Lisi \ci{Lisi} proposed that one could unify
all the fields of the Standard model and gravity by means
of the exceptional group E$_8$. Lisi's approach differs
from others \ci{e8} in attempting to explain generations in terms
of triality. It has attracted considerable
attention and criticism. Regardless of
whether the particular model due to Lisi will
turn out to be correct or not, it would be desirable
to find a possible physical basis for the group E$_8$.
In other words, we would like to know whether there
exists a particular physical arrangement that leads
to E$_8$. Is it due to the existence of higher-
dimensional dynamics? We will show that we do not
need extra dimensions of spacetime in order to
arrive at E$_8$. We can remain in the 4-dimensional spacetime
$M_4$ and consider not only events (points) but
also oriented lines, areas, 3-volumes and 4-volumes
in $M_4$. All these objects can be elegantly
represented by the Clifford numbers $X^A \gam_A
\equiv X^{a_1 a_2 ... a_r} \gam_{a_1 a_2 ... a_r}$, where $r=0,1,2,3,4$ and
$a=0,1,2,3$. This leads to the concept of the
so-called {\it Clifford space}\,\ci{Pezzaglia}--\ci{PavsicArena},
a 16-dimensional manifold whose tangent space is
the Clifford algebra\footnote{
In the introduction, we used a simplified notation
for various spaces, namely $M_4,~V_4,~ {\cal C \ell}(4),~
V_{16}$ and ${\cal C \ell}(16)$, even though Minkowski space
$M_4$ has signature (1,3) and should be written
more precisely as $M_{1,3}$, $V_4$ as $V_{1,3}$,
${\cal C \ell}(4)$ as ${\cal C \ell}(1,3)$, $V_{16}$ as
$V_{8,8}$ and ${\cal C \ell}(16)$ as ${\cal C \ell}(8,8)$.}
${\cal C \ell} (4)$.

In these papers it has been assumed that the arena for physics is
Clifford space ${\cal C}$ and that physical
objects are described by Clifford algebra valued
fields on ${\cal C}$. At every point $X \in {\cal C}$,
a field $\Phi$ can be transformed according to\,\ci{PavsicKaluza,PavsicKaluzaLong}
\be
    \Phi \rightarrow \Phi' = {\rm R} \Phi {\rm S}
\lbl{1.1}
\ee
where R and S are in general two different Clifford
numbers\,\ci{PavsicKaluza,PavsicKaluzaLong} to be specified later. For the
time being, let us only say that the transformation
(\ref{1.1}) in general reshuffles the grade $r$
 of the components $\Phi = \phi^A \gam_A$. Its
 action on $\phi^A$ or $\gam_A$ can be just a
 {\it rotation}. Altogether (if $D=2^n$ is the dimension of
 ${\cal C \ell}(n)$)  there are $D(D-1)/2$
 independent rotations, 120 if $n=4$, but they cannot
 all be generated by the 16 independent $\gam_A$ which
 close ${\cal C \ell}(4)$. However, we can perform
 rotations in ${\cal C \ell}(4)$ by means of the
 rotation matrices acting on $\phi^A$ according to
 \be
      \phi'^A = {L^A}_B \phi^B
\lbl{1.2}
\ee
The existence of 120 rotational degrees of freedom
implies the presence of extra degrees of freedom
besides those spanned by the basis $\{\gam_A \}$.

At this point, we will employ the fact that the
Clifford algebra ${\cal C \ell}(4)$ is a vector space.
Eq.\,(\ref{1.1}) performs only 32 independent
rotations in the vector space (16 due to
the left and 16 due to the right transformations).
Therefore, instead of the basis $\{\gam_A \}$, we may
use an alternative basis $\{\Gam_A \}$, $A=1,2,...,16$,
whose elements $\Gam_A$ generate the Clifford
algebra ${\cal C \ell}(16)$. The new basis elements
$\Gam_A$ span a 16-dimensional space $V_{16}$ which,
as a vector space, is isomorphic to ${\cal C \ell}(4)$.
Rotations in $V_{16}$ are generated by bivectors
$\Gam_{AB} \equiv {\cal C \ell}(16)$; they act on
a vector $\phi^A \Gam_A \in V_{16}$ and belong to
the Lie algebra of the group SO(16). The latter
group has not only a vector representation but also
spinor representations. In our set up
spinors come from the Clifford
algebra ${\cal C \ell}(16)$, namely, from its left
(right) minimal ideals. The Lie algebra so(16),
together with the space ${\rm S}_{16}^+$ of real,
positive chiral ${\cal C \ell}(16)$ spinors, forms
the Lie algebra ${\rm e}_8$ of ${\rm E}_8$.

So, we have found that e$_8$ arises from the spacetime
$M_4$ itself, from which one can construct a Clifford algebra.
The latter algebra, as a vector space, is isomorphic
to $V_{16}$, and its basis vectors $\Gam_A$ can generate the
Clifford algebra ${\cal C \ell}(16)$, whose subspace
is e$_8$:
\be
   V_4 \rightarrow {\cal C \ell}(4) \sim V_{16}
   \rightarrow {\cal C \ell}(16) \supset e_8
\lbl{1.3}
\ee

In the following, we will explore this procedure in more
detail.

\section{Generalizing spacetime}

Special and general relativity are valid in the spacetime manifold
$M_4$. Choosing a point ${\cal P} \in M_4$, the tangent
space at that point is a vector space
$V_4 \equiv T_{\cal P} (M_4)$, whose basis vectors can be
taken to be generators $\gam_a,~a=0,1,2,3$, of the Clifford
algebra, satisfying
\be
   \gam_a \cdot \gam_b \equiv \mbox{$\frac{1}{2}$}
   (\gam_a \gam_b + \gam_b \gam_a) = \eta_{ab}
\lbl{2.1}
\ee
where $\eta_{ab}= {\rm diag}\, (1,-1,-1,-1)$ is the Minkowski
metric tensor.

While the symmetrized product (\ref{2.1}) gives the metric,
the antisymmetrized (wedge) product
\be
   \gam_a \wg \gam_b \equiv \mbox{ $\frac{1}{2}$}
   [\gam_a,\gam_b] = \mbox{ $\frac{1}{2}$}
   (\gam_a \gam_b - \gam_b \gam_a)
\lbl{2.2}
\ee
gives a bivector. In general, the antisymmetrized product
of $r$ vectors
\be
   \gam_{a_1 a_2 ... a_r} \equiv \gam_{a_1} \wg \gam_{a_2}
   \wg ... \wg \gam_{a_r} \equiv \frac{1}{r!}\,
   [\gam_{a_1},\gam_{a_2},...,\gam_{a_r}] \equiv \gam_A
\lbl{2.2a}
\ee
gives an $r$-vector. The objects $\gam_{a_1 a_2 ... a_r}
\equiv \gam_A$ form a basis of the Clifford algebra
${\cal C \ell}(1,3)$, whose elements are Clifford numbers
called {\it polyvectors} and which are given by a superposition
\be
    X = X^{a_1 ... a_r}\gam_{a_1 ... a_r} \equiv X^A \gam_A
\lbl{2.3}
\ee
We take $a_1 < a_2 < ... < a_r$ in order to avoid multiple
occurrences of the same term.

The components $X^{a_1 ... a_r} \equiv X^A$ are holographic
projections of oriented $r$-areas onto the coordinate
$r$-planes, spanned by $\gam_{a_1 ... a_r} \equiv \gam_A$.

In the spacetime manifold $M_4$, in which we choose a point
${\cal P}_0$ as the origin, the vectors $x = x^a
\gam_a \in V_4 ({\cal P}_0) = T_{{\cal P}_0}(M_4)$
are in one-to-one correspondence with other points
${\cal P} \in M_4$, at least within a region $B\subset M_4$
around ${\cal P}_0$.

Similarly, the polyvectors $X=X^A \gam_A \in {\cal C \ell}(1,3)$
are in one-to-one correspondence with the points ${\cal E}$ of a
16-dimensional manifold ${\cal C}$, called a {\it Clifford space},
in which we choose a point ${\cal E}_0$ as the origin,
such that $\gam_A \in T_{{\cal E}_0} ({\cal C})$ are tangent
polyvectors at ${\cal E}_0$. In other words, vectors of the
tangent space at ${\cal E}_0 \in {\cal C}$ can be mapped
into the points ${\cal E}\in {\cal C}$ of a region
around ${\cal E}_0$. A tangent space to a point
${\cal E} \in {\cal C}$ is the vector space defined by the Clifford algebra
${\cal C \ell}(1,3)$.

The metric is given by the scalar product
\be
G_{AB} = \gam_A^\ddg * \gam_B \equiv \langle
    \gam_A^\ddg  \gam_B \rangle_0
\lbl{2.4}
\ee
Here $\ddg$ reverses the order of the 1-vectors ---for example,
    $ (\gam_{a_1} \gam_{a_2} .... \gam_{a_r})^\ddg =
     \gam_{a_r} ... \gam_{a_2} \gam_{a_1}$ ---
and is necessary for the consistent raising and lowering
of indices\,\ci{PavsicMaxwell,PavsicKaluzaLong}.
With the definition (\ref{2.4}), the signature of ${\cal C}$ is
$(8,8)$.

Since a Clifford space is such a straightforward
generalization of spacetime, its has been assumed
\ci{Pezzaglia}--\ci{PavsicArena} that physics also
has to be generalized. According to such a novel theory,
the arena for physics is not the spacetime $M_4$
but the Clifford space ${\cal C}$ (also called ${\cal C}$-space).
The fundamental building blocks of physical space are then
not points of spacetime but extended $r$-dimensional
cells ($r$-cells), or extended events described by
$r$-vectors. For definiteness, we may imagine that
extended events are associated with $p$-branes in
the quenched mini\-superspace description \ci{AuriliaQuenched,
AuriliaFuzzy}, but, in principle, extended events can
be related to whichever fundamental extended objects that can
be described by $r$-vectors.

\section{Group SO(8,8) acting in Clifford space}
\subsection{Left and right transformations of
a polyvector}

A polyvector $X\in {\cal C \ell}(1,3)$ can be transformed
into another polyvector $X' \in {\cal C \ell}(1,3)$
by
\be
   X' = {\rm R}\, X \, {\rm S}
\lbl{3.1}
\ee
where R, S $\in {\cal C \ell}(1,3)$. Requiring that the
norm of $X$ be invariant
\be
   |X'|^2 \equiv {X'}^{\,\ddg}*X' \equiv \, \langle {X'}^{\,\ddg}
   X'\rangle_0\, = \, \langle{\rm S}^\ddg X^\ddg {\rm R}^\ddg {\rm R}
   X {\rm S}\rangle_0 = \langle X^\ddg X \rangle_0
\lbl{3.2}
\ee
we have
\be
   {\rm R}^\ddg {\rm R} =1 ~, ~~~~~~~~~~~{\rm S}^\ddg {\rm S}\,= 1
\lbl{3.3}
\ee

Writing
\be
    {\rm R} = {\rm e}^{\frac{1}{2}\alpha^A \gam_A} \, ,\qquad ~~
   {\rm S} = {\rm e}^{\frac{1}{2}\beta^A \gam_A}\, ,
\lbl{3.4}
\ee
we have from eq.\,(\ref{3.3}) that
\be
               \alpha^* \gam_A^\ddg = -\,\gam_A \alpha^A
\lbl{3.5}
\ee
where we assume that reversion also acts on the imaginary
unit $i$, which we consider a bivector $i=e_q e_p$ of the
phase space spanned by $e_q$ and $e_p$ \ci{PavsicBook}.
The parameters $\alpha^A,~\beta^A,~A=1,2,...,16$
are, therefore, either real or imaginary because, depending on
the grade of $\gam_A$, we have:
\be
    \gam_A^\ddg = \gam_A ~~~ {\rm if}~ A \in \{A\}_+\, ,~~~
   \gam_A^\ddg = -\, \gam_A ~~~ {\rm if} ~A \in \{A\}_-
\lbl{3.6}
\ee

As an example, let us consider the case
\be
   {\rm R} = {\rm e}^{\frac{1}{2}\alpha \gam_1 \gam_2}=
   {\rm cos} \, \frac{\alpha}{2}+ \gam_1 \gam_2 \,
   {\rm sin} \, \frac{\alpha}{2} \; , ~~~
  {\rm S} = {\rm e}^{\frac{1}{2}\beta \gam_1 \gam_2}=
   {\rm cos} \, \frac{\beta}{2}+ \gam_1 \gam_2 \,
   {\rm sin} \, \frac{\beta}{2}
\lbl{3.7}
\ee
and examine how various Clifford numbers $X=X^C \gam_C$
transform under (\ref{3.1}).

(i) If $X= X^1 \gam_1 + X^2 \gam_2$, then the transformation
(\ref{3.1}) gives
\be
   X' = X^1 \left ( \gam_1 \, {\rm cos}\, \mbox{$\frac{\alpha-\beta}{2}$}\,+
   \gam_2 \, {\rm sin}\, \mbox{$\frac{\alpha-\beta}{2}$} \right ) +
   X^2 \left ( -\gam_1 \, {\rm sin}\, \mbox{$\frac{\alpha-\beta}{2}$}\,+
   \gam_2 \, {\rm cos}\, \mbox{$\frac{\alpha-\beta}{2}$} \right )
\lbl{3.8}
\ee

(ii) If $X= X^3 \gam_3 + X^{123}\, \gam_{123}$, then
\be
    X' = X^3 \left ( \gam_3 \, {\rm cos}\, \mbox{$\frac{\alpha+\beta}{2}$}\,+
   \gam_{123} \, {\rm sin}\, \mbox{$\frac{\alpha+\beta}{2}$} \right ) +
   X^{123} \left ( -\gam_{2} \, {\rm sin}\, \mbox{$\frac{\alpha+\beta}{2}$}\,+
   \gam_{123} \, {\rm cos}\, \mbox{$\frac{\alpha+\beta}{2}$} \right )
\lbl{3.9}
\ee

(iii) If $X = s\, {\underline 1} + X^{12}\, \gam_{12}$, then
\be
     X' = s \left ( {\underline 1} \, {\rm cos}\, \mbox{$\frac{\alpha+\beta}{2}$}\,+
   \gam_{12} \, {\rm sin}\, \mbox{$\frac{\alpha+\beta}{2}$} \right ) +
   X^2 \left ( -{\underline 1} \, {\rm sin}\, \mbox{$\frac{\alpha+\beta}{2}$}\,+
   \gam_2 \, {\rm cos}\, \mbox{$\frac{\alpha+\beta}{2}$} \right )
\lbl{3.10}
\ee

(iv) If $X= {\tl X}^1 \, \gam_5 \gam_1 + {\tl X}^2 \, \gam_5 \gam_2$,
then
 \be
   X' = {\tl X}^1 \left ( \gam_5 \gam_1 \, {\rm cos}\,
   \mbox{$\frac{\alpha-\beta}{2}$}\,+
   \gam_5 \gam_2 \, {\rm sin}\, \mbox{$\frac{\alpha-\beta}{2}$} \right ) +
   {\tl X}^2 \left ( -\gam_5 \gam_1 \, {\rm sin}\,
    \mbox{$\frac{\alpha-\beta}{2}$}\,+
  \gam_5 \gam_2 \, {\rm cos}\, \mbox{$\frac{\alpha-\beta}{2}$} \right )
\lbl{3.10a}
\ee

The usual rotations of vectors or pseudovectors are reproduced
if the angle $\beta$ for the right transformation is
equal to minus the angle $\alpha$ for the left transformation, i.e.,
if
\be
\beta = -\alpha
\lbl{3.11}
\ee
Then all other transformations which mix the grade vanish.

Condition (\ref{3.11}) is a prerequisite for the consistency of
ordinary rotations and Lorentz transformations. If we release the
condition (\ref{3.11}) and allow for arbitrary independent
$\alpha$ and $\beta$, in particular $\alpha=\beta$, then
we go outside the Lorentz group in $M_4$ and arrive at
rotations in Clifford space.

In the example (\ref{3.7}), the generator $\gam_1 \gam_2$
changes sign under reversion, so the parameters $\alpha$ and
$\beta$ are real. If we consider, for example, the
rotations generated by $\gam_1$, then, since $\gam_1^\ddg =
\gam_1$, the transformation parameter must be imaginary,
which we write explicitly as
  \be
   {\rm R} = {\rm e}^{\frac{1}{2}\,i\, \alpha \gam_1}=
   {\rm ch} \, \frac{\alpha}{2}+ i\,  \gam_1 \,
   {\rm sh} \, \frac{\alpha}{2} \,,  ~~~
  {\rm S} = {\rm e}^{\frac{1}{2}\, i\, \beta \gam_1}=
   {\rm ch} \, \frac{\beta}{2}+ i \,\gam_1 \,
   {\rm sh} \, \frac{\beta}{2}
\lbl{3.11a}
\ee
Taking $X= s {\ul 1}$ or $X = X^1 \gam_1$, we have
\be
    {\rm R}\, s {\ul 1}\, {\rm S} = s \left ({\ul 1} \, {\rm ch}\,
    \mbox{$\frac{\alpha+\beta}{2}$} + i \, \gam_1 \, {\rm sh}\,
    \mbox{$\frac{\alpha+\beta}{2}$}\right )
\lbl{3.11b}
\ee
\be
    {\rm R}X^1 \, \gam_1 {\ul 1}\, {\rm S} = X^1
    \left ( -i \, {\ul 1} \, {\rm ch}\,
    \mbox{$\frac{\alpha+\beta}{2}$} +  \gam_1 \, {\rm sh}\,
    \mbox{$\frac{\alpha+\beta}{2}$}\right )
\lbl{3.11c}
\ee
We see that $i \gam_1$ generates rotations in the plane
$(i\, {\ul 1}, \gam_1)$ or $({\ul 1}, i\gam_1)$.

For infinitesimal transformations we have
\be
  X' = {\rm e}^{\frac{1}{2}\alpha^A \gam_A }X^C \gam_C\,
       {\rm e}^{\frac{1}{2}\beta^B \gam_B } \approx
       1 + \delta X
\lbl{3.12}
\ee
where
\bear
\delta X &=& \delta X^C \gam_C = \mbox{ $\frac{1}{2}$}
   (\alpha^A \gam_A \gam_C + \beta^A \gam_C \gam_A) X^C \nonumber \\
  &=& \mbox{$\frac{\alpha^A + \beta^A}{4}$} \, \{\gam_A,\gam_C \}X^C
   \,+ \, \mbox{$\frac{\alpha^A - \beta^A}{4}$} \, [\gam_A,\gam_C] X^C \nonumber \\
   &=& \mbox{$\frac{\alpha^A + \beta^A}{4}$} \,f_{AC}^D \gam_D X^C
   \,+ \, \mbox{$\frac{\alpha^A - \beta^A}{4}$} \, C_{AC}^D \gam_D X^C
\lbl{3.13}
\ear
where $\{ \gam_A,\gam_C\}= f_{AC}^D \, \gam_D$ and $[\gam_A,\gam_C]=
 C_{AC}^D \,\gam_D$. Introducing
 \be
    {\epsilon^D}_C =\mbox{$\frac{\alpha^A + \beta^A}{4}$} \,f_{AC}^D \gam_D
    \,+\, \mbox{$\frac{\alpha^A - \beta^A}{4}$} \, C_{AC}^D \gam_D
\lbl{3.14}
\ee
we find
\be
   \delta X^D = {\epsilon^D}_C\, X^C
\lbl{3.15}
\ee
which denotes infinitesimal rotations of the ${\cal C}$-space
coordinates $X^C$ if ${\epsilon^D}_C$ is antisymmetric, and other
linear transformations otherwise. Eq.\,(\ref{3.14}) implies that
there are only 32 independent components of ${\epsilon^D}_C$,
of which 16 are due to $\alpha^A$, and 16 to $\beta^A$.

A single generator, e.g., $\gam_{12}$ in examples
(\ref{3.7})--(\ref{3.10a}), generates rotations not in a single
plane but in several planes of ${\cal C}$-space at once.
An additional complication is that, in
order to satisfy the requirement (\ref{3.3}) for the
invariance of the norm, some transformation parameters
must be real, and some must be imaginary. Therefore, some rotations
are within ${\cal C \ell}(1,3)$ while others are between
${\cal C \ell}(1,3)$ and  $i\,{\cal C \ell}(1,3)$, as
illustrated in (\ref{3.11a})--(\ref{3.11c}). Our space
is then enlarged to
$\mathbb{C}\otimes {\cal C \ell}(1,3)$.

So, we have extended real polyvectors to complex ones.
In quantized theories, physical particles are described by
fields that, in general, are complex valued. In our case,
these fields are $\Phi = \phi^A \gam_A$, whose values are
in a complexified Clifford algebra. The quadratic form is
\be
   \Phi^\ddg *\Phi = {\phi^\ddg}^A \phi^B G_{AB} =
   \phi_{(R)}^A \phi_{(R)}^B \, G_{AB} +
    \phi_{(I)}^A \phi_{(I)}^B \, G_{AB}
\lbl{3.16}
\ee
where $G_{AB}$ is defined according to eq.\,(\ref{2.4}).

The real components $\phi_{(R)}^A$ can be transformed amongst
themselves by rotations of the group SO(8,8), and the
same is true for the imaginary components:
\be
   \phi_{(R)}^{'A} = {L^A}_B \phi_{(R)}^B \, , \quad ~~~~
    \phi_{(I)}^{'A} = {L^A}_B \phi_{(I)}^B
\lbl{3.18}
\ee
There are also mixed transformations between
$\phi_{(R)}^{A}$ and $\phi_{(I)}^{A}$. The Lie algebra
of SO(8,8) has 120 independent generators. Altogether,
the group of transformations that preserve the quadratic
form (\ref{3.16}) has 120 dimensions due to rotations
of $\phi_{(R)}^{A}$, 120 dimensions due to rotations
of $\phi_{(I)}^{A}$, and $16 \times 16 =256$ dimensions
belonging to mixed transformations between $\phi_{(R)}^{A}$
and $\phi_{(I)}^{A}$, which sums to 496. This, of course,
is in agreement with the dimension $32 \times 31/2 = 496$
of the SO(16,16) acting on $(\phi_{(R)}^{A},\phi_{(I)}^{A})$.

\subsection{Vector space $V_{8,8}$}

We have seen that, amongst left and right transformations
of a real polyvector $X$, there are norm-preserving
transformations, i.e., rotations that preserve or
change the grade of the elements $\gam_A \in
{\cal C}(1,3)$. Therefore, besides the rotations in the
planes $(\gam_\mu , \gam_\nu )$ and
$(\gam_5 \gam_\mu , \gam_5 \gam_\nu )$, we also have
rotations in the planes $(\gam_\mu , \gam_{\rho \sigma})$
spanned by vectors and bivectors, etc. Such grade
-mixing rotations generated by (\ref{3.1}) do not act in
a single plane but in several planes at once, as
illustrated in examples (\ref{3.7})--(\ref{3.10a}).
Moreover, there are planes in ${\cal C \ell}(1,3)$ in
which rotations cannot be generated by a transformation
(\ref{3.1}). For instance, a rotation in the plane
$({\ul 1},\gam_1)$ does not occur in (\ref{3.1}).
Instead, one has a rotation in the plane
$({\ul 1},i\gam_1)$, which is not in ${\cal C \ell}(1,3)$,
but in $\mathbb{C}\otimes {\cal C \ell}(1,3)$.

A basis $\gam_A \in {\cal C \ell}(1,3)$ cannot generate
all possible rotations of the group SO(8,8)
acting within ${\cal C \ell}(1,3)$. However, we
can construct the SO(8,8) rotations directly, without
using $\gam_A$ as generators:
\be
   X' = X'^A \gam_A = {L^A}_B X^B \gam_A = X^A \gam'_A
\lbl{3.19}
\ee
where
\be
     X'^A =  {L^A}_B X^B \, , ~~~~~~\gam'_A = {L^B}_A \gam_B
\lbl{3.20}
\ee
So, either the components $X^A$ or the basis (poly)vectors $\gam_A$
can be rotated by a matrix ${L^A}_B$ of SO(8,8).

If we consider ${\cal C \ell}(1,3)$ as a vector space
$V_{8,8}$ and forget about the algebra, then,
instead of $\gam_A \in {\cal C \ell}(1,3)$, we can use any
other elements that span $V_{8,8}$.
Let us choose as our basis vectors the generators
$\Gam_A,~A=1,2,...,16$ of ${\cal C \ell}(8,8)$. The generators of
rotations in $V_{8,8}$ are then the 120 independent
bivectors $\Gam_{AB} = \Gam_A \wg \Gam_B \in {\rm so}(8,8) \subset
{\cal C \ell}(8,8)$.

Our space is thus the 16-dimensional vector space $V_{8,8}$,
spanned by the vectors
$\Gam_A \in {\cal C \ell}(8,8)$. An element $X=X^A \Gam_A
\in V_{8,8}$ is rotated according to
\be
  X' = {\rm e}^{\frac{1}{4}\alpha^{AB} \Gam_{AB}}\, X
  {\rm e}^{-\frac{1}{4}\alpha^{AB} \Gam_{AB}}
\lbl{3.21}
\ee
By also taking into account the ${\cal C \ell}(8,8)$
spinors, we arrive at the Lie algebra e$_8$ as a subspace
of ${\cal C \ell}(8,8)$.

\section{Vector space $V_8^{(1)} \oplus V_8^{(2)}$}

Now, we will describe an alternative procedure that also leads
from ${\cal C \ell}(1,3)$ to e$_8$. We can split
${\cal C \ell}(1,3)\sim V_{8,8}$ into two subspaces
$V_{2,6}$ and $V_{6,2}$ spanned by the following basis
elements:
\be
   V_{2,6}\;:~~~\{\gam_{A_1}\} = \{1,\gam_\mu, \gam_{0r}\}\; ,
  ~~ A_1 = 1,2,...,8 \nonumber
\ee
 \be
   V_{6,2}\;:~~~\{\gam_{A_2}\} = \{\gam_5,\gam_5 \gam_\mu,
   \gam_5 \gam_{0r}\}\;,
   ~~A_2 = 9,10,...,16
\lbl{4.1}
\ee
where the three elements $\gam_5 \gam_{0r}$, $r=1,2,3$, correspond to
the three elements $\gam_{rs}$. Every element
$\Phi\in {\cal C \ell}(1,3)$ can thus be split into two parts
in the following way:
\be
   \Phi = \phi^A \gam_A =
   \phi^{A_1} \gam_{A_1} + \phi^{A_2} \gam_{A_2}
\lbl{4.2}
\ee
For convenience, we perform the split ${\cal C \ell}(1,3)
\sim V_{8,8} = V_{2,6} \oplus V_{6,2}$. Alternative splits,
such as  $V_{8,8} = V_{0,8} \oplus V_{8,0}=
V_{1,7} \oplus V_{7,1}$, etc., are also possible in principle.
For simplicity, we will from now on use the symbol $V_8$ for all
these possible spaces and consider the generic
split
\be
    {\cal C \ell}(1,3) \sim V_{8,8} =
    V_8^{(1)} \oplus  V_8^{(2)}
\lbl{4.3}
\ee

The basis elements $\gam_A \equiv \gam_{a1...a_r}$, $r=0,1,2,3,4$,
given by eq.\,(\ref{2.2a}), are $r$-vectors of
${\cal C \ell}(1,3)$. Instead of a basis consisting of
16 $r$-vectors, we can consider a basis that consists of
16 independent spinors $\xi_{\tl A} \equiv \xi_{\alpha i}$,
$\alpha = 1,2,3,4$, $i=1,2,3,4$, that span four independent
left minimal ideals of ${\cal C \ell}(1,3)$ \ci{Teitler,PavsicKaluzaLong}.
Then, $\Phi$ can be expanded according to
\be
  \Phi = \psi^{\tl A} \xi_{\tl A} = \psi^{\alpha 1} \xi_{\alpha 1} +
  \psi^{\alpha 2} \xi_{\alpha 2} +\psi^{\alpha 3} \xi_{\alpha 3}+
  \psi^{\alpha 4} \xi_{\alpha 4}
\lbl{4.4}
\ee
Four linearly independent spinors of one minimal left
ideal form a 4-dimensional representation of ${\cal C \ell}(1,3)$.
In terms of $\xi_{\alpha i}$, for a fixed $i$, say $i=1$,
every element of ${\cal C \ell}(1,3)$ can be represented
as a $4\times 4$ matrix. In terms of all 16 basis elements
$\xi_{\tl A}$, every element $\Phi \in {\cal C \ell}(1,3)$
can be represented as a $16 \times 16$ matrix according to
\ci{PavsicKaluzaLong}
\be
    \langle {\xi^{\tl A}}^\ddg \Phi \xi_{\tl B} \rangle_S =
    {(\Phi)^{\tl A}}_{\tl B} = {\delta^i}_j \otimes
    {(\Phi)^\alpha}_\beta
\lbl{4.5}
\ee
where $\alpha,~\beta$ are the 4-spinor indices of a given
left minimal ideal and the bracket with subscript $S$ denotes
the normalized scalar part\,\ci{PavsicKaluzaLong}.
The above formula explicitly shows
that the 16-dimensional representation of ${\cal C \ell}(1,3)$ is
reducible: the matrices are block-diagonal, with the blocks
being just $4 \times 4$ matrices. In general, instead of
16 spinors $\xi_{\tl A} \equiv \xi_{\alpha i}$, one can use some
other set of 16 linearly independent objects $q_A$,
each of which is a superposition of $\xi_{\tl A}$. A matrix
\be
   {(\Phi)^A}_B = \langle {q^{A}}^\ddg \Phi q_{B} \rangle_S
\lbl{4.5a}
\ee
then may not be block-diagonal, but, of course, it can be
transformed back into block-diagonal form. If $q_A = \gam_A$,
then $\langle {\gam^{A}}^\ddg \Phi \gam_{B} \rangle_S$ is
a representation of a Clifford number $\Phi$ in the
Clifford algebra basis $\gam_A$.

Let us now generalize from
{\it reducible} $16 \times 16$ matrices to
{\it arbitrary} $16 \times 16$ matrices. An arbitrary
$16 \times 16$ matrix can be expanded in terms of 256
linearly independent $16 \times 16$ matrices that represent
a basis of the Clifford algebra ${\cal C \ell}(8)$ (or one of
its non-compact forms), where $256 = 2^8$ is the dimension
of ${\cal C \ell}(8)$ and $16= 2^{8/2}$ is the dimension of
a corresponding space that forms an irreducible representation
of ${\cal C \ell}(8)$.

Let us now return to the split (\ref{4.3}). Instead of the
basis elements $\gam_{A_1},~ \gam_{A_2} \in {\cal C \ell}(1,3)$
that, according to (\ref{4.5}) and (\ref{4.5a}), can be
represented by {\it reducible} (block diagonal) $16 \times 16$
matrices, let us now consider new basis elements
$\Gam_{\bar A}^{(1)}$ and $\Gam_{\bar A}^{(2)}$, ${\bar A}=1,2,...,8$,
that are 1-vectors of ${\cal C \ell}(8)$, represented by
$16 \times 16$ matrices that, in general, are not reducible
to a block diagonal form (\ref{4.5}). Every element of
$V_8^{(1)}$ can now be expanded in terms of $\Gam_{\bar A}^{(1)}$,
and analogously for $V_8^{(2)}$ and $\Gam_{\bar A}^{(2)}$.

Besides the 1-vectors
$\Gam_{\bar A}^{(1)}\in V_8^{(1)} \subset {\cal C \ell}(V_8^{(1)})$
and
$\Gam_{\bar A}^{(2)}\in V_8^{(2)} \subset {\cal C \ell}(V_8^{(2)})$,
which are the Clifford algebra generators of the two copies
of ${\cal C \ell}(8)$, let us consider also 2-vectors
$$\Gam_{{\bar A}{\bar B}}^{(1)} \in {\rm so}(V_8^{(1)})~~ {\rm and}
~~\Gam_{{\bar A}{\bar B}}^{(2)} \in {\rm so}(V_8^{(2)})$$
The latter objects are generators of rotation in
$V_8^{(1)}$ and $V_8^{(2)}$, respectively.

In addition to the rotations in each of the spaces
$V_8^{(1)}$, $V_8^{(2)}$, there are also transformations
that mix those two spaces, and they are given by the generators
$\Gam_{\bar A}^{(1)} \otimes\Gam_{\bar A}^{(2)}$. Altogether,
there are 28 generators of ${\rm so}(V_8^{(1)})$, 28 generators
of ${\rm so}(V_8^{(2)})$, and $8 \times 8 =64$ generators
$\Gam_{\bar A}^{(1)} \otimes\Gam_{\bar A}^{(2)}$, which sums
to a total of 120 generators. This is the dimension of
so(16) that can be decomposed according to \ci{BaezOctonions,CastroE8}
\be
    {\rm so} (16) =  {\rm so}(V_8^{(1)})\, \oplus \, {\rm so}(V_8^{(2)})
    \, \oplus \,  V_8^{(1)} \otimes  V_8^{(2)}
\lbl{4.6}
\ee

In addition to 1-vectors and 2-vectors in ${\cal C \ell}(8)$, we
also have spinors as elements of left (right) minimal ideals
of ${\cal C \ell}(8)$. The dimension of a given minimal left ideal
of ${\cal C \ell}(8)$ is $2^{8/2} = 16$. There are 16 linearly
independent basis spinors $\zeta_{ A},~{ A} =1,2,...,16$,
and they span the spinor space $S_8$ that decomposes into
two invariant subspaces $S_8^+$ and $S_8^-$ forming,
respectively, the left-handed and right-handed spinor
representation of so(8). From the basis spinors of
${\cal C \ell}(V_8^{(1)})$ and ${\cal C \ell}(V_8^{(2)})$,
we can form the following objects that live in
${\cal C \ell}(16)={\cal C \ell}(8)\otimes {\cal C \ell}(8)$:
\be
    \zeta_{\bar A}^{+(1)} \otimes  \zeta_{\bar B}^{+(2)}\; ,
   \qquad\qquad \zeta_{\bar A}^{-(1)} \otimes  \zeta_{\bar B}^{-(2)}
\lbl{4.7}
\ee
\be
    \zeta_{\bar A}^{-(1)} \otimes  \zeta_{\bar B}^{+(2)}\; ,
   \qquad\qquad \zeta_{\bar A}^{+(1)} \otimes  \zeta_{\bar B}^{-(2)}
\lbl{4.8}
\ee
where ${\bar A},~{\bar B} = 1,2,...,8$. There are
$64 + 64+64+64 = 256 = 2^{16/2}$ basis spinors of
${\cal C \ell}(16)$. Let us retain only 126 of them, say
those of eq. (\ref{4.7}) that form a positive chiral representation
(left handed) of so(16). The objects
 $\zeta_{\bar A}^{+(1)} \otimes  \zeta_{\bar B}^{+(2)}$ and
 $\zeta_{\bar A}^{-(1)} \otimes  \zeta_{\bar B}^{-(2)}$
 span the 128-dimensional space $S_{16}^+$ of positive chiral
 spinors of ${\cal C \ell}(16)$.

 As it has been discussed
 in \ci{BaezOctonions,CastroE8}, the space
 \be
       {\rm so}(V_8^{(1)})\, \oplus \, {\rm so}(V_8^{(2)})
    \, \oplus \,  V_8^{(1)} \otimes  V_8^{(2)}
    \, \oplus \,
    S_{8}^{+(1)} \otimes  S_{8}^{+(2)}
    \, \oplus \, S_{8}^{-(1)} \otimes  S_{8}^{-(2)} = {\rm e}_8
\lbl{4.9}
\ee
is the Lie algebra e$_8$ of E$_8$. Its dimension is exactly
right, namely,  $120+64+64=248$, and eq.\,(\ref{4.9}) is,
in fact, a decomposition of
\be
    {\rm e}_8 = {\rm so} (16) \oplus {\rm S}_{16}^+
\lbl{4.10}
\ee
where  $S_{16}^+$ forms the 128-dimensional irreducible
representation of so(16). In eq.\,(\ref{4.9}),
$V_8^{(1)}$ and $V_8^{(2)}$ are the vector representation
spaces for the two copies of so(8), whereas
$ S_{8}^{+(1)},~  S_{8}^{+(2)}$ and
$S_{8}^{-(1)},~S_{8}^{-(2)}$ are the corresponding spinor
representation spaces.

\section{Discussion and conclusion}

The exceptional group E$_8$ is somewhat enigmatic. Its smallest
representation is the 248-dimensional adjoint representation, and the
space in which $E_8$ acts as the isometry group cannot be determined
without first defining $E_8$\,\ci{BaezOctonions}.
Eqs.\,(\ref{4.9}) and (\ref{4.10}) demonstrate the curious fact
that e$_8$ consists of the Lie algebra of rotations so(16) in a
16-dimensional vector space and the corresponding spinor space
$S_{16}^+$ on which the generators of so(16) act. Both, the
generators of so(16) and the spinors of $S_{16}^+$ are
elements of ${\cal C \ell}(16)$, which contains e$_8$ as a
subspace.

${\cal C \ell}(16)$ is indeed a big space, probably too
big to be useful for the unification of the currently known
interactions and particles because it predicts too many
new particles and interactions. Therefore, it seems more reasonable
to stay with (\ref{4.9}), which also describes e$_8$ but with the
corresponding spaces being subspaces of ${\cal C \ell}(V_8^{(1)})$
or ${\cal C \ell}(V_8^{(2)})$. The dimension of the latter is $2^8=256$,
which is modest in comparison with the dimensionality
$2^{16} = 256\times 256$ of ${\cal C \ell}(16)$. Although
${\cal C \ell}(16)= {\cal C \ell}(8)\otimes {\cal C \ell}(8)$,
such a product is not necessary for the construction of
e$_8$ according to eq.\,(\ref{4.9}). Only the direct products
of the vector and spinor subspaces of the two copies of
${\cal C \ell}(8)$, but not of so(8), are necessary.
We have arrived at ${\cal C \ell}(8)$ quite naturally by
starting from the 4-dimensional spacetime $M_4$. From its tangent
vectors, we can generate the 16-dimensional Clifford algebra
${\cal C \ell}(1,3)$, to which we can
ascribe a physical interpretation
as the space of oriented $r$-vectors associated with strings and
branes\,\ci{Castro}--\ci{PavsicArena}, \ci{PavsicMaxwell}--\ci{AuriliaFuzzy}.
We consider ${\cal C \ell}(1,3)$ as a vector space
$V_{8,8}$ in which one can perform 120 independent rotations.
However, there are only 32 independent basis elements $\gam_A$ of
${\cal C \ell}(1,3)$, which, since they form an algebra, can
generate only 32 independent rotations in ${\cal C \ell}(1,3)$,
corresponding to left and right transformations in eq.\,(\ref{3.1}).
The fact that the $\gam_A$ form an algebra is now an undesirable
property because the product of two such gammas gives another
gamma in the same set. Thus we have concentrated on the vector
nature of $V_{8,8}$ and considered instead of $\{\gam_A \}$
another basis $\{\Gam_A \}$, whose objects do not close
an algebra and can thus form an outer product
$\Gam_{AB}=\Gam_A \wg \Gam_B$.
Next, it has turned out to be convenient to split
$V_{8,8}$ into two subspaces $V_8^{(1)}$ and $V_8^{(2)}$,
with bases $\{ \Gam_{\bar A}^{(1)} \}$ and
$\{ \Gam_{\bar A}^{(2)} \}$, ${\bar A} = 1,2,...,8$.
The bivectors $\Gam_{{\bar A}{\bar B}}^{(1)}$ of\,
${\cal C \ell}(V_8^{(1)})$ and $\Gam_{{\bar A}{\bar B}}^{(2)}$ of
${\cal C \ell}(V_8^{(2)})$ are generators of rotations in
the corresponding vector spaces. We have also considered
spinors of the Clifford algebras ${\cal C \ell}(V_8^{(1)})$ and
${\cal C \ell}(V_8^{(2)})$. Thus we have arrived at e$_8$ expressed
in the form (\ref{4.9}) by using only the elements of the two
copies of ${\cal C \ell}(8)$ and the direct product of the
vector and spinor representation spaces. Even if the decomposition
(\ref{4.9})
is a known result, it is fascinating to see how it fits into a straightforward
extension of 4-dimensional spacetime $M_4$ to the 16-dimensional
space of $r$-volumes. The latter space, as shown
in refs.\,\ci{Castro}--\ci{PavsicArena},\ci{PavsicMaxwell}--\ci{AuriliaFuzzy},
can be associated with strings
and branes, fundamental physical objects that, in many respects,
are very promising, although not the only ones considered in the
literature, for unification and quantum gravity.

\vs{6mm}

{\bf Acknowledgement}

This work was supported by the Ministry of High Education, Science
and Technology of Slovenia.

config: arxiv


TITLE: Determine asymptotic distribution and efficiency of an estimator
QUESTION [2 upvotes]: I have the following problem:
Let $X_1,..., X_n$ be a sample of independent, identically distributed random variables, with density $$f_{\theta}(x)=\begin{cases}
e^{\theta-x}, & \text{if } x\geq \theta\\
0, & \text{elsewhere}\\
\end{cases}$$
Let $\hat\theta_n$ be the maximum-likelihood estimator (MLE) of $\theta$. I am asked to find the asymptotic distribution of $\sqrt n (\hat\theta_n-\theta)$ and $\hat\theta_n$ efficiency.
So far, I computed the MLE, which is $\hat\theta_n=min\{x_i\}$, for $i=0,...,n.$. I also know that if I find that $\sqrt n (\hat\theta_n-\theta)$ converges in distribution to $N(0,I(\theta))$, we would have both the asymptotic distribution and efficiency of the estimator.
Central Limit Theorem could be a way to do it, but I'm really struggling on it.
Any hints or solutions would be appreciated.

REPLY [1 votes]: To answer the question you are asked, find the distribution function of $T=\sqrt n(\hat\theta_n-\theta)$.
Now $\hat\theta_n$ has CDF $$P(\hat\theta_n\le t)=\begin{cases} 1-e^{-n(t-\theta)} &,\text{ if }t\ge\theta \\ 0&,\text{ if }t<\theta \end{cases}$$
So CDF of $T$ should be
\begin{align}
F_n(t)&=P\left(\sqrt n(\hat\theta_n-\theta)\le t\right)
\\&=P\left(\hat\theta_n\le \frac{t}{\sqrt n}+\theta\right)
\\&=\begin{cases}1-e^{-\sqrt nt}&,\text{ if }t\ge 0 \\ 0&,\text{ if }t<0\end{cases}
\end{align}
Since $n(\hat\theta_n-\theta)\sim \mathsf{Exp}(1)$, it is clear that $T$ is exponential with mean $1/\sqrt n$.
Taking limit as $n\to\infty$, show that $F_n(t)$ converges to another distribution function. This would give you a degenerate asymptotic distribution of the MLE $\hat\theta_n$.
To arrive at a non-degenerate limiting distribution of the MLE, an appropriate scaling of $\hat\theta_n$ is $T'=n(\hat\theta_n-\theta)$. We know $T'$ is exactly distributed as $\mathsf{Exp}(1)$, which is naturally also the asymptotic distribution.

Maximum likelihood estimators are expected to be (asymptotically) efficient in most cases.
Formally, if $T_0$ is the uniformly minimum variance unbiased estimator (UMVUE) of $\theta$, then (large sample) efficiency of $\hat\theta_n$ is defined by $\mathrm e=\frac{\operatorname{Var}(T_0)}{\operatorname{Var}(\hat\theta_n)}$. Find $T_0$ and hence conclude $\hat\theta_n$ is indeed efficient.


TITLE: Bipartite graphs, degrees and neighbors
QUESTION [1 upvotes]: Let $G = (X \cup Y, E)$ be a bipartite graph with no isolated vertices and let $S \subseteq X$. Suppose $|S| = |N(S)|$ (neighboors of $S$) and that $\forall \{x,y\} \in E$, where $x\in X$ and $y\in Y$, $\deg(x) \ge \deg(y)$.
It is not obvious to me why $\sum_{x\in S}{\deg(x)} \ge \sum_{y\in N(S)}{\deg(y)}$. It seems obvious for my colleagues but they can't explain it to me. Can anyone provide a formal proof or clear intuition ?

REPLY [0 votes]: The stated result is wrong, at least in its current form form. Consider for instance $S=\{x_1,x_2\}$, $X=S\cup\{x_3\}$, $Y=\{y_1,y_2\}$ and edges $(x_1,y_1)$, $(x_1,y_2)$ $(x_3,y_1)$, $(x_3,y_2)$.
Remark that $x_2$ is isolated.
Then all assumptions are verified, but the conclusion is not: sum of degrees is $2$ in $S$ and $4$ in $N(S)$.


text:
\subsection{Full recovery of harmonic oscillations}
\label{sec:sines}

To illustrate the results obtained in Section~\ref{sec:shift-inv-rho-bounds}, let us consider the problem of \textit{full recovery} of (ordinary) harmonic oscillations. Namely, we are asked to estimate on $\Dom_N$ a harmonic oscillation
\[
x_t = \sum_{k=1}^s \alpha_k e^{i \omega_k t}, \quad t \in \Dom_N, \quad N \in \Z_+,
\]
without the knowledge of frequencies $\omega_1, ..., \omega_s$. Since estimation is required on the entire $\Dom_N$, we will measure the statistical performance of an estimator $\wh x$ of $x$ via the mean-square error\footnotemark:
\footnotetext{The results of this section can be generalized, in a straightforward manner, for the risk measured by the width of the confidence interval for~$\ell_2$-loss. We omit this generalization in order to simplify the presentation.}
\[
\Risk(\wh x, x) := (2N+1)^{-1/2} \left[\E \| \wh x - x \|_{N,2}^2\right]^{1/2}.
\]
Note that if the frequencies were known, the ordinary least-squares estimator would satisfy
\[
\sup_{\S_{\omega_1, ..., \omega_s}} \Risk(\wh x, x) \le C\sigma \sqrt{\frac{s}{2N+1}},
\]
where $\S_{\omega_1, ..., \omega_s}$ is the set of harmonic oscillations corresponding to the fixed tuple of frequencies $\omega_1, ..., \omega_s$. On the other hand, when the frequencies are unknown, one has the lower bound~\cite[Theorem 2]{recht2}
\begin{equation}
\label{eq:sines-lower}
\sup_{x \in \S^{(s)}} \Risk(\wh x, x) \ge c\sigma \sqrt{\frac{s \log(N+1)}{2N+1}},
\end{equation}
where~$\S^{(s)}$ is the set of all harmonic oscillations with no more than $s$ frequencies. Moreover, the bound~\eqref{eq:sines-lower} is in fact attained on a subspace with \textit{separated} frequencies in the sence of~\eqref{eq:min-sep}, indicating that the general case is not harder, from the statistical viewpoint, than that of well-separated frequencies . As such, one could hope to match~\eqref{eq:sines-lower} by some adaptive in $\omega_1, ..., \omega_s$ estimator, whether the frequencies are restricted to be well-separated or not. However, to the best of our knowledge, the only estimator known to match~\eqref{eq:sines-lower}, called \textit{Atomic Soft Thresholding} (AST) and studied in~\cite{recht1,recht2}, only does so in the case of well-separated frequencies, see~\cite[Theorem 1]{recht2}. As such, the question whether the lower bound~\eqref{eq:sines-lower} can be matched in the general case, is still open.

A crucial step towards bridging this gap has been taken in~\cite{harchaoui2015adaptive} where it was suggested to use one-sided version of a uniform-fit estimator jointly with Proposition~\ref{th:sines-old}, exploting that~\eqref{eq:sines-old-rho} holds for (generalized) harmonic oscillations without any frequency separation assumptions. In particular, fitting a ``left'' uniform-fit estimator $\wh \vphi \in \C_N^+(\Z)$ on $\Dom_N^+$, and a ``right'' estimator $\wh \vphi \in \C_N^{-}(\Z)$ on $\Dom_N^-$, and using the bound $C\rho^3 \sqrt{\log N}$ on the price of adaptation for such estimators\footnotemark, one obtains for such construction the correct in~$\sigma$ and~$N$ rate
\[
\sup_{x \in S^{(s)}} \Risk(\wh x, x) \le C \sigma \sqrt{\frac{s^{12} \log (N+1)}{2N+1}}.
\]
As we see, the price to pay is a polynomial factor in $s$, and additional assumption $N \ge cs^2 \log s$ needed for Proposition~\ref{th:sines-old}.
\footnotetext{This bound holds for both two-sidfed and one-sided uniform-fit estiamtors, see~\cite{harchaoui2015adaptive}.}

Using the results presented in ths paper, we immeduately improve the dependence on $s$ in the above bound by replacing one-sided uniform-fit estimators with estimators of the form~\eqref{opt:l2con}, used together with Theorem~\ref{th:l2con-pred} and the improved bound~\eqref{eq:sines-rho} instead of~\eqref{eq:sines-old-rho}:
\begin{equation}
\label{eq:sines-unilateral}
\sup_{x \in S^{(s)}} \Risk(\wh x, x) \le C \sigma \sqrt{\frac{s^{4} \log^2 (N+1)}{2N+1}}.
\end{equation}
Note that while this estimators requries the knowledge of~$s$ in advance (AST does not), this requirement can be circumvented by using~\eqref{opt:l2pen-pred++} instead of~\eqref{opt:l2con-pred}, at the expense of an additional logarithmic factor. Moreover, \eqref{eq:sines-unilateral} can be further improved if the signal frequencies are restricted to be well-separated. In this case, using Proposition~\ref{th:sepsines}, the same estimator satisfies
\begin{equation}
\label{eq:sines-unilateral-sep}
\sup_{x \in S^{(s)}_{\sep}} \Risk(\wh x, x) \le C \sigma \sqrt{\frac{s^{2} + s \log(N+1)}{2N+1}}
\end{equation}
where the supremum is taken over the set of harmonic oscillations with no more than $s$ frequencies and pairwise separation at least $4 \pi/(N+1)$.

Finally, let us describe how the results obtained so far can be combined, resulting in the state-of-the-art estimator of harmonic oscillations which improves over the bound~\eqref{eq:sines-unilateral} in the general case while still preserving~\eqref{eq:sines-unilateral-sep}. Namely, consider the following procedure:
\begin{enumerate}
\item
Pick some $M \le N$, and divide the observation domain~$\Dom_N$ into the large central subdomain~$\Dom_M$ and the smaller subdomains~$\Dom^+ := \Dom_{N-M}^M$ and~$\Dom^- := \Dom_{N-M}^{-N}$.
\item
Estimate the signal on~$\Dom_M$ with a two-sided filter~$\wh\vphi \in \C_{N-M}(\Z)$, on~$\Dom^+$ with a one-sided filter~$\wh\vphi^+ \in \C_{M+N}^+(\Z)$, and on~$\Dom^-$ with a one-sided filter~$\wh\vphi^- \in \C_{M+N}^-(\Z)$.
\item
Choose $M$ to minimize the total bound on~$\Risk$ over $\Dom_N$.
\end{enumerate}
Direct calculations lead to the following choice of $M$:
\begin{equation*}

C_1 s \log(N+1) \le \frac{M+1}{N-M+1} \le C_2 s \log(N+1),
\end{equation*}
and the resulting bound is
\begin{equation}
\label{eq:sines-composite}
\sup_{x \in S^{(s)}} \Risk(\wh x, x) \le C \sigma \sqrt{\frac{s^{3} \log (N+1) + s^{2} \log^2 (N+1)}{2N+1}}.
\end{equation}
These calculations are provided in Appendix~\ref{sec:prmisc}.


TITLE: convergence functions calcultions
QUESTION [0 upvotes]: Show that

$\displaystyle\int_1^{\infty} \frac{1}{x^7e^x}$
and
$\displaystyle\int_1^{\infty} \frac{\sin(x)}{x^2}$

are convergence.
I have no idea how to solve the first one but I think I know the second:
$$\begin{align}
\int_1^{\infty} \frac{\sin(x)}{x^2}
&= \int_1^{\infty} \frac{\sin(x)}{x \cdot x}
\\&=\int_1^{\infty} \frac{1}{x}
\\&=\lim\limits_{x \to M} \int_1^{M} \frac{1}{M}
\\&=\left[\frac{1}{x^2}\right]^M
\\&=\frac{1}{M^2}-\frac{1}{1}
\\&=\frac{1}{\infty^2}-1
\\&=0-1= 1
\end{align}$$
Am I right? And how to solve the first one?
Thanks

REPLY [0 votes]: For 1, since $e^{-x}<1$ on $(1,\infty)$,
$$\left|\frac{1}{x^7e^x}\right|<{1\over x^7},$$
and $\int_1^\infty{1\over x^7}$ obviously converges.
For 2, since $|\sin x|\le1$,
$$\left|{\sin x\over x^2}\right|\le{1\over x^2},$$
and $\int_1^\infty {1\over x^2}$ obviously converges.


TITLE: C^\infty versus semiclassical wavefront sets
QUESTION [5 upvotes]: Zworski states that if $u$ is a compactly supported distribution, independent of the semiclassical parameter $h$, then the relationship between the $C^\infty$ and semiclassical wavefront sets of $u$ is given by:
$WF_h(u)=(\mathrm{supp}(u)\times\{0\})\cup WF(u)$.
Would someone please explain this relationship?  Thanks!

REPLY [2 votes]: Heuristically, this result follows because the semiclassical wavefront set measures oscillations at frequency $\frac{\xi}{h}$. To understand this, observe that since $u$ does not depend on $h$, if $u$ is smooth, it has small high frequency oscillations, and hence, for $\xi\neq 0$, and hence points where $u$ is smooth and $\xi\neq 0$ do not appear in $WF_h(u)$.  However, where $u$ is not smooth, $u$ has high frequency oscillations and so these points do appear.
Below you will find a proof of the statement.
To see that $WF_h(u)\cap \mathbb{R}^d\times \{0\}=$ supp $(u)\times\{0\}$ observe that
$$\mathcal{F}_h(\chi u)(0)=\langle u, \chi\rangle \neq 0$$
for $\chi$ supported near $x_0\in $ supp $(u)$ and
$$\mathcal{F}_h(\chi u)(0)=0 $$
for $\chi$ supported near $x_0\notin $ supp $(u)$.
To see the second part of $WF_h(u)$, simply use the characterization of $WF(u)$ that for $\xi\neq 0$, $(x_0,\xi_0)\notin WF(u)$ if and only if for $\chi$ supported sufficiently close to $x_0$, all $N>0$ and $\xi$ in a conic neighborhood of $\xi_0$,
$$|\mathcal{F}(\chi u)(\xi)|\leq C_N\langle \xi \rangle ^{-N}.$$
Thus, $(x_0,\xi_0)\notin WF(u)$ if and only if for all $N>0$ and all $\xi$ in a neighborhood of $\xi_0$,
$$|\mathcal{F}_h(\chi u)(\xi)|\leq C_N\langle \xi/h\rangle^{-N}.$$
But, since $\xi_0\neq 0$, this gives $(x_0,\xi_0)\notin WF(u)$ if and only if for all $N>0$ and $\xi$ in a neighborhood of $\xi_0$,
$$|\mathcal{F}_h(\chi u)(\xi)|\leq C_Nh^N$$
and hence if and only if $(x_0,\xi_0)\notin WF_h(u)$.