TITLE: It is still true that $\dim X \le \dim X^\star$ for infinite-dimensional Banach spaces if $\dim X < 2^{\aleph_0}$?
QUESTION [3 upvotes]: Let $X$ be an infinite-dimensional Banach space and $E^\star$ its topological dual. In this answer, @tree detective showed that if $\dim X \ge 2^{\aleph_0}$, then $\dim X \le \dim X^\star$.
Does the result hold if we remove the assumption $\dim X \ge 2^{\aleph_0}$?

REPLY [6 votes]: A Banach space $X$ of dimension less than $2^{\aleph_0}$ must be finite-dimensional, in which case case $\dim X = \dim X^*$.

H.E. Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298.


text:
\begin{document}
\title[Macphail's Theorem revisited]{Macphail's Theorem revisited}
\author[D. Pellegrino]{Daniel Pellegrino}
\address{Departamento de Matem\'{a}tica \\
Universidade Federal da Para\'{\i}ba \\
58.051-900 - Jo\~{a}o Pessoa, Brazil.}
\email{daniel.pellegrino@academico.ufpb.br and dmpellegrino@gmail.com}
\author[J.. Silva]{Janiely Silva}
\email{janiely.silva@estudantes.ufpb.br}
\thanks{D. Pellegrino is supported by CNPq Grant 307327/2017-5 and Grant 2019/0014
Paraiba State Research Foundation (FAPESQ) and J. Silva is supported by CAPES}
\subjclass[2010]{ 15A60, 40A05}
\keywords{Macphail's Theorem; Dvoretzky--Rogers Theorem; matrices; series; Banach spaces}

\begin{abstract}
In 1947, M. S. Macphail constructed a series in $\ell_{1}$ that converges
unconditionally but does not converge absolutely. According to the literature,
this result helped Dvoretzky and Rogers to finally answer a long standing
problem of Banach Space Theory, by showing that in all infinite-dimensional
Banach spaces, there exists an unconditionally summable sequence that fails to
be absolutely summable. More precisely, the Dvoretzky--Rogers Theorem asserts
that in every infinite-dimensional Banach space $E$ there exists an
unconditionally convergent series ${\textstyle\sum}x^{(j)}$ such that
${\textstyle\sum}\Vert x^{(j)}\Vert^{^{2-\varepsilon}}=\infty$ for all
$\varepsilon>0.$ Their proof is non-constructive and Macphail

's result for $E=\ell_{1}$ provides a constructive proof just for
$\varepsilon\geq1.$ In this note we revisit Machphail's paper and present two
alternative constructions that work for all $\varepsilon>0.$

\end{abstract}
\maketitle

\section{Introduction}

A series in a Banach space is said to be unconditionally convergent if all
rearrangements of that series converge (to the same vector). An old result due
to Dirichlet (1829) asserts that a series $

{\textstyle\sum}

x^{(j)}$ of real or complex scalars is unconditionally convergent precisely
when it is absolutely convergent, i.e., when $

{\textstyle\sum}

\Vert x^{(j)}\Vert$ converges. The same characterization holds for
finite-dimensional Banach spaces, see, for example, \cite[Theorem 1.3.5]{kk}.
In the infinite-dimensional framework it seems that unconditionally convergent
series were first investigated by Orlicz about 100 years after Dirichlet. The
subject attracted the attention of Banach, Mazur and others, who proposed the
following problem (see \cite[Problem 122]{mau} and also \cite{pie} for further
details): in an arbitrary infinite-dimensional Banach space, does there exist
an unconditionally convergent series that does not converge absolutely?

In most of the classical Banach spaces, examples of unconditionally convergent
series that do not converge absolutely were easily obtained. For instance, the
sequence $(j^{-1}e_{j})_{j=1}^{\infty}$ is unconditionally summable in
$\ell_{2}$ but it fails to be absolutely summable. The same example works for
all $\ell_{p}$ spaces for any $p>1,$ but it is useless for $\ell_{1}$. In
1947, Macphail (\cite{mac}) proved that the answer was also positive in
$\ell_{1}$. Macphail's approach has shown that $\ell_{1}$ is essentially the
critical case for all infinite-dimensional Banach spaces; this becomes rather
clear, for instance, in the approach described in \cite[\S 2 and Lemma 1]{rut}
and, according to \cite[pages 2 and 20]{diestel}, inspired Dvoretzky and
Rogers in 1950 (\cite{dr}) to transplant in a highly nontrivial fashion the
construction of Macphail to any infinite-dimensional Banach space and obtain a
definitive solution to the problem.

For a Banach space $E$ and $S$ a finite sequence of vectors $x_{1}
,\ldots,x_{n}$ in $E$, Macphail defined
\[
G(S)=\frac{\sup_{\sigma}\left\Vert \sum_{i\in\sigma}x_{i}\right\Vert
}{\sum_{i=1}^{n}\Vert x_{i}\Vert},
\]
where the supremum runs over all subset $\sigma\subseteq\{1,\ldots,n\}$, and
\[
\mu(E)=\inf G(S),
\]
where the infimum is considered over all finite sequences $S\subset E$. The
parameter $\mu(E)$ is now known as Macphail's constant of $E$. In his paper,
it is first observed that if $\mu(E)=0$, then unconditional convergence does
not imply absolute convergence in $E.$ The main result of \cite{mac} shows
that $\mu(\ell_{1})=0$.

Some alternative proofs of Macphail's result and of the Dvoretzky--Rogers
Theorem appeared later, but they were not constructive (see, for instance,
\cite[page 145]{Kvaratskhelia} and \cite{ara, israel} and the references
therein). We also refer the interested reader to \cite{bayart} for a
non-constructive approach in $\ell_{1}$ using the Kahane--Salem--Zygmund
inequality (see also \cite{mas} for details on the Kahane--Salem--Zygmund
inequality) and \cite{gordon, rut1} for the asymptotic behavior of $\mu
(\ell_{p}^{n})$.

The striking result of Dvoretzky--Rogers, showing that $\mu(E)=0$ for all
infinite-dimensional Banach spaces, attracted the attention of Grothendieck
who referred to the Dvoretzky--Rogers Theorem as the unique decisive result in
the finner metric theory of general Banach spaces known at that time (see
\cite[page 20]{diestel}). Additionally, the result of Dvoretzky and Rogers
answers much more than what is asked in the original problem of Banach's
school. In more precise terms, if $E$ is an infinite-dimensional Banach space,
the Dvoretzky--Rogers Theorem assures the existence of an unconditionally
convergent series $

{\textstyle\sum}

x^{(j)}$ in $X$ such that
\begin{equation}

{\textstyle\sum\limits_{j=1}^{\infty}}

\left\Vert x^{(j)}\right\Vert ^{2-\varepsilon}=\infty\label{qqqq}
\end{equation}
for all $\varepsilon>0.$ However, the proof of the Dvoretzky--Rogers Theorem
do not offer an explicit construction of such sequence. In this note we
revisit Macphail's paper and provide two alternative constructions of such a
sequence in $\ell_{p}$, for all $p\in\lbrack1,2]$ (the case $p>2$ is obvious
and for this reason we do not consider it); unlike Macphail's approach which
is valid for $\varepsilon\geq1,$ the construction presented here completely
satisfies the above statement, i.e., it provides an unconditionally convergent
series $

{\textstyle\sum}

x^{(j)}$ in $\ell_{p}$ satisfying (\ref{qqqq}) for all $\varepsilon>0$.
Similar constructions may be known for experts in the field but we were not
able to find in the literature.

\section{The fisrt approach}

The first alternative construction is valid just for complex scalars and rests
in the properties of some special matrices that date back to the works of
Toeplitz \cite{toe}:

\begin{theorem}
\label{casecomp} Let $p\in\lbrack1,2]$ and $\alpha>1$ a positive integer. For
all positive integers $j$, let
\[
A_{j}:=\left[  1/2+\sqrt{1/4+\log_{\alpha}\left(  \left(  j+1\right)
/2\right)  },1/2+\sqrt{1/4+\log_{\alpha}j}\right]  \cap\mathbb{N}.
\]
The sequence $(x^{(j)})_{j=1}^{\infty}$ defined by
\[
x^{(j)}={\alpha}^{{\frac{(1-k_{j})(k_{j}(2+p)+2p)}{2p}}}\cdot\sum
_{s=1}^{\alpha^{k_{j}(k_{j}-1)}}\left(  \exp\left(  {\left[  j\alpha
^{k_{j}\left(  1-k_{j}\right)  }+\alpha^{k_{j}\left(  1-k_{j}\right)
}-1\right]  2\pi si}\right)  \cdot e_{\alpha^{k_{j}(k_{j}-1)}+s-1}\right)
\]
if $A_{j}\neq\emptyset$, where $k_{j}$ is the unique element of $A_{j}$, and
$x^{(j)}=0$ otherwise, is unconditionally summable in $\ell_{p}$ (over the
complex scalar field) and
\[

{\textstyle\sum\limits_{j=1}^{\infty}}

\left\Vert x^{(j)}\right\Vert ^{2-\varepsilon}=\infty
\]
for all $\varepsilon>0.$
\end{theorem}

The statement may seem somewhat complicated at first glance, but the
construction is fairly simple. We shall first observe that the interval
\[
\left[  1/2+\sqrt{1/4+\log_{\alpha}\left(  \left(  j+1\right)  /2\right)
},\text{ }1/2+\sqrt{1/4+\log_{\alpha}j}\right]
\]
has size smaller than $1$ and thus $A_{j}$ is, in fact, either empty or formed
by a single integer. A straightforward computation shows that when
$j\in\left\{  \alpha^{k\left(  k-1\right)  },\alpha^{k\left(  k-1\right)
}+1,\ldots,2\alpha^{k\left(  k-1\right)  }-1\right\}  $ for a certain positive
integer $k$, we have $A_{j}=\{k\}$; otherwise $A_{j}$ is empty.

\begin{proof}
For each positive integer $n$, consider the $n\times n$ matrix $(a_{rs}
^{(n)})$ with $a_{rs}^{(n)}=\exp(2\pi irs/n);$ it is obvious that
$|a_{rs}^{(n)}|=1$ and one can also check that
\begin{equation}
\sum_{s=1}^{n}a_{rs}^{(n)}\overline{a_{ts}^{(n)}}=n\delta_{rt},\label{i9x}
\end{equation}
where $\delta_{rt}$ denotes the Kronecker delta. In order to prove (\ref{i9x})
note that
\[
\sum_{s=1}^{n}a_{rs}^{(n)}\overline{a_{ts}^{(n)}}=\sum_{s=1}^{n}\exp\left(
2\pi i\frac{rs}{n}\right)  \cdot\exp\left(  -2\pi i\frac{ts}{n}\right)
=\sum_{s=1}^{n}\exp\left(  2\pi i\frac{(r-t)s}{n}\right)  .
\]
If $r=t$, it is obvious that $\sum_{s=1}^{n}a_{rs}^{(n)}\overline{a_{ts}
^{(n)}}=n$. On the other hand, if $r\neq t$, we have
\[
\sum_{s=1}^{n}\exp\left(  2\pi i\frac{(r-t)s}{n}\right)  =\frac{\exp\left(
2\pi i\frac{(r-t)}{n}\right)  -\exp\left(  2\pi i\frac{(r-t)(n+1)}{n}\right)
}{1-\exp\left(  2\pi i\frac{(r-t)}{n}\right)  }
\]
and, recalling that $\exp(\pi i)=-1,$ a straightforward computation gives us
\[
\sum_{s=1}^{n}a_{rs}^{(n)}\overline{a_{ts}^{(n)}}=0.
\]
Let $p^{\ast}$ be the conjugate of $p$, i.e.,
\[
\frac{1}{p}+\frac{1}{p^{\ast}}=1
\]
and let $\ell_{p}^{n}$ denote $\mathbb{C}^{n}$ with the $\ell_{p}$-norm. For
unit vectors $y^{(1)}\in\ell_{p^{\ast}}^{n}$ and $y^{(2)}\in\ell_{\infty}^{n}
$, we have
\begin{equation}
\left\vert \sum_{i_{1},i_{2}=1}^{n}a_{i_{1}i_{2}}^{(n)}y_{i_{1}}^{(1)}
y_{i_{2}}^{(2)}\right\vert \leq n^{\frac{1}{2}+\frac{1}{p}}.\label{0987}
\end{equation}
The proof of (\ref{0987}) follows by Schur's test (also known as Young's
inequality). For the sake of completeness we present a proof here following
the lines of \cite[pages 30 and 31]{djd}. By the H\"{o}lder inequality, we
have
\begin{align*}
\left\vert \sum_{i_{1},i_{2}=1}^{n}a_{i_{1}i_{2}}^{(n)}y_{i_{1}}^{(1)}
y_{i_{2}}^{(2)}\right\vert  &  \leq\sum_{i_{2}=1}^{n}\left\vert \sum_{i_{1}
=1}^{n}a_{i_{1}i_{2}}^{(n)}y_{i_{1}}^{(1)}\right\vert |y_{i_{2}}^{(2)}|\\
&  \leq\left(  \sum_{i_{2}=1}^{n}|y_{i_{2}}^{(2)}|^{2}\right)  ^{\frac{1}{2}
}\left(  \sum_{i_{2}=1}^{n}\left\vert \sum_{i_{1}=1}^{n}a_{i_{1}i_{2}}
^{(n)}y_{i_{1}}^{(1)}\right\vert ^{2}\right)  ^{\frac{1}{2}}\\
&  \leq n^{\frac{1}{2}}\left(  \sum_{i_{2}=1}^{n}\left\vert \sum_{i_{1}=1}
^{n}a_{i_{1}i_{2}}^{(n)}y_{i_{1}}^{(1)}\right\vert ^{2}\right)  ^{\frac{1}{2}
}.
\end{align*}
Since
\[
\left(  \sum_{i_{2}=1}^{n}\left\vert \sum_{i_{1}=1}^{n}a_{i_{1}i_{2}}
^{(n)}y_{i_{1}}^{(1)}\right\vert ^{2}\right)  ^{\frac{1}{2}}=\left(
\sum_{i_{2}=1}^{n}\sum_{\substack{i_{1}=1\\j_{1}=1}}^{n}y_{i_{1}}
^{(1)}\overline{y_{j_{1}}^{(1)}}a_{i_{1}i_{2}}^{(n)}\overline{a_{j_{1}i_{2}
}^{(n)}}\right)  ^{\frac{1}{2}}=\left(  \sum_{\substack{i_{1}=1\\j_{1}=1}
}^{n}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}}\sum_{i_{2}=1}^{n}a_{i_{1}i_{2}
}^{(n)}\overline{a_{j_{1}i_{2}}^{(n)}}\right)  ^{\frac{1}{2}},
\]
by (\ref{i9x}) and by the H\"{o}lder inequality, we have
\begin{align*}
\left\vert \sum_{i_{1},i_{2}=1}^{n}a_{i_{1}i_{2}}^{(n)}y_{i_{1}}^{(1)}
y_{i_{2}}^{(2)}\right\vert  &  \leq n^{\frac{1}{2}}\left(  \sum
_{\substack{i_{1}=1\\j_{1}=1}}^{n}y_{i_{1}}^{(1)}\overline{y_{j_{1}}^{(1)}
}n\delta_{i_{1}j_{1}}\right)  ^{\frac{1}{2}}\\
&  =n\left(  \sum_{i_{1}=1}^{n}|y_{i_{1}}^{(1)}|^{2}\right)  ^{\frac{1}{2}}\\
&  \leq n\left(  \sum_{i_{1}=1}^{n}1\right)  ^{\frac{1}{2}-\frac{1}{p^{\ast}}
}\left(  \sum_{i_{1}=1}^{n}|y_{i_{1}}^{(1)}|^{p^{\ast}}\right)  ^{\frac
{1}{p^{\ast}}}\\
&  =n^{\frac{1}{2}+\frac{1}{p}}.
\end{align*}
Thus, by (\ref{0987}), we obtain
\begin{equation} \label{rr}
\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}\sum\limits_{r=1}
^{n}\left\vert \sum\limits_{s=1}^{n}\varphi_{s}a_{rs}^{(n)}\right\vert
=\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}\sup_{\left\Vert
	\psi\right\Vert _{\infty}\leq1}\left\vert \sum\limits_{r,s=1}^{n}\psi
_{r}\varphi_{s}a_{rs}^{(n)}\right\vert \leq n^{\frac{1}{2}+\frac{1}{p}},
\end{equation}
where $\varphi=(\varphi_{1},\ldots,\varphi_{n})$ and $\psi=(\psi_{1}
,\ldots,\psi_{n})$.

Let $j_{k}:=\alpha^{k(k-1)}$ for all positive integers $k$. Define the
sequence $(x^{(j)})_{j=1}^{\infty}$ as follows: if $j\in\{j_{k},\ldots
,2j_{k}-1\}$, for a certain positive integer $k$, consider
\[
x^{(j)}=j_{k}^{-\left(  \frac{1}{2}+\frac{1}{p}+\frac{1}{k}\right)  }\left(
\sum_{s=1}^{j_{k}}a_{rs}^{(j_{k})}e_{j_{k}+s-1}\right)  ,
\]
where $r=j-j_{k}+1$; otherwise take $x^{(j)}=0$. For all $\varphi\in(\ell
_{p})^{\ast}$ it is plain that
\begin{equation}\label{lim}
\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}

{\textstyle\sum\limits_{j=1}^{\infty}}

\left\vert \varphi\left(  x^{(j)}\right)  \right\vert \leq\sum_{k=1}^{\infty
}\left(  \sup_{\left\Vert \varphi\right\Vert _{p^{{\ast}}}\leq1}\sum_{j=j_{k}
}^{2j_{k}-1}\left\vert \varphi\left(  x^{(j)}\right)  \right\vert \right)  .
\end{equation}

By the definition of $x^{(j)}:=(x_{m}^{(j)})_{m=1}^{\infty}$, we obtain
\begin{align}
\sum_{j=j_{k}}^{2j_{k}-1}\left\vert \varphi\left(  x^{(j)}\right)
\right\vert  &  =\sum_{j=j_{k}}^{2j_{k}-1}\left\vert \sum_{m=1}^{\infty
}\varphi_{m}x_{m}^{(j)}\right\vert \label{88}\\
&  =\sum_{j=j_{k}}^{2j_{k}-1}\left\vert j_{k}^{-\left(  \frac{1}{2}+\frac
{1}{p}+\frac{1}{k}\right)  }\sum_{m=j_{k}}^{2j_{k}-1}\varphi_{m}
a_{[j-j_{k}+1][m-j_{k}+1]}^{(j_{k})}\right\vert \nonumber
\end{align}
for all positive integers $k$. Note that we can assign new indices to the sums
$\sum_{j=j_{k}}^{2j_{k}-1}$ and $\sum_{m=j_{k}}^{2j_{k}-1}$ as follows:
\begin{align*}
\sum_{j=j_{k}}^{2j_{k}-1}\left\vert j_{k}^{-\left(  \frac{1}{2}+\frac{1}
{p}+\frac{1}{k}\right)  }\sum_{m=j_{k}}^{2j_{k}-1}\varphi_{m}a_{[j-j_{k}
+1][m-j_{k}+1]}^{(j_{k})}\right\vert  &  =\sum_{r=1}^{j_{k}}\left\vert
j_{k}^{-\left(  \frac{1}{2}+\frac{1}{p}+\frac{1}{k}\right)  }\sum_{s=1}
^{j_{k}}\varphi_{(j_{k}+s-1)}a_{rs}^{(j_{k})}\right\vert \\
&  =j_{k}^{-\left(  \frac{1}{2}+\frac{1}{p}+\frac{1}{k}\right)  }\sum
_{r=1}^{j_{k}}\left\vert \sum_{s=1}^{j_{k}}\varphi_{(j_{k}+s-1)}a_{rs}
^{(j_{k})}\right\vert .
\end{align*}
Hence, for $n\geq2$, by (\ref{88}) and (\ref{rr}), we have
\begin{align}
\sum_{k=n}^{\infty}\left(  \sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}
\leq1}\sum_{j=j_{k}}^{2j_{k}-1}\left\vert \varphi\left(  x^{(j)}\right)
\right\vert \right)   &  =\sum_{k=n}^{\infty}\left(  \sup_{\left\Vert
\varphi\right\Vert _{p^{\ast}}\leq1}\sum_{j=j_{k}}^{2j_{k}-1}\left\vert
j_{k}^{-\left(  \frac{1}{2}+\frac{1}{p}+\frac{1}{k}\right)  }\sum_{m=j_{k}
}^{2j_{k}-1}\varphi_{m}a_{[j-j_{k}+1][m-j_{k}+1]}^{(j_{k})}\right\vert
\right)  \label{aac}\\
&  =\sum_{k=n}^{\infty}\left(  \sup_{\left\Vert \varphi\right\Vert _{p^{\ast}
}\leq1}j_{k}^{-\left(  \frac{1}{2}+\frac{1}{p}+\frac{1}{k}\right)  }\sum
_{r=1}^{j_{k}}\left\vert \sum_{s=1}^{j_{k}}\varphi_{(j_{k}+s-1)}a_{rs}
^{(j_{k})}\right\vert \right)  \nonumber\\
&  =\sum_{k=n}^{\infty}\left(  j_{k}^{-\left(  \frac{1}{2}+\frac{1}{p}
+\frac{1}{k}\right)  }\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq
1}\sum_{r=1}^{j_{k}}\left\vert \sum_{s=1}^{j_{k}}\varphi_{(j_{k}+s-1)}
a_{rs}^{(j_{k})}\right\vert \right)  \nonumber\\
&  \leq\sum_{k=n}^{\infty}\left(  j_{k}^{-\left(  \frac{1}{2}+\frac{1}
{p}+\frac{1}{k}\right)  }j_{k}^{\frac{1}{2}+\frac{1}{p}}\right)  \nonumber\\
&  =\sum_{k=n}^{\infty}\alpha^{1-k}<\infty.\nonumber
\end{align}
A well-known necessary and sufficient condition for a series $\sum
_{n=1}^{\infty}y_{n}$ to be unconditionally summable (see \cite[Theorem
1.5]{diestel}) is that, given $\delta>0$, there is $n_{\delta}\in\mathbb{N}$
such that
\[
\left\Vert \sum_{n\in M}y_{n}\right\Vert <\delta
\]
whenever $M$ is a finite subset of $\mathbb{N}$ with $\min M>n_{\delta}$.

By (\ref{lim}) and (\ref{aac}) we have
\[
\lim_{n\rightarrow\infty}\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}

{\textstyle\sum\limits_{j=n}^{\infty}}

\left\vert \varphi\left(  x^{(j)}\right)  \right\vert =0.
\]
Hence, given $\delta>0$, there is $n_{\delta}\in\mathbb{N}$ such that $n\geq
n_{\delta}$ implies
\[
\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}

{\textstyle\sum\limits_{j=n}^{\infty}}

\left\vert \varphi\left(  x^{(j)}\right)  \right\vert <\delta.
\]
Then, for all finite sets $M\subset\mathbb{N}$ such that $M\subset\{n_{\delta
},n_{\delta}+1,\ldots\}$, it follows that
\[
\left\Vert \sum_{j\in M}x^{(j)}\right\Vert \leq\sup_{\left\Vert \varphi
\right\Vert _{p^{\ast}}\leq1}\sum_{j\in M}|\varphi(x^{(j)})|\leq
\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}\sum_{j=n}^{\infty
}|\varphi(x^{(j)})|<\delta.
\]
Therefore, $(x^{(j)})_{j=1}^{\infty}$ is unconditionally summable.

On the other hand, when $j\in\{j_{k},\ldots,2j_{k}-1\}$, for a certain
positive integer $k$, we have
\[
\left\Vert x^{(j)}\right\Vert =\left(  j_{k}\cdot\left(  j_{k}^{-\left(
\frac{1}{2}+\frac{1}{p}+\frac{1}{k}\right)  }\right)  ^{p}\right)  ^{\frac
{1}{p}}=j_{k}^{-\left(  \frac{1}{2}+\frac{1}{k}\right)  };
\]
otherwise $\Vert x^{(j)}\Vert=0$. Then, we have
\[

{\textstyle\sum}

\left\Vert x^{(j)}\right\Vert ^{r}=\sum_{k=1}^{\infty}j_{k}\cdot\left(
j_{k}^{-\left(  \frac{1}{2}+\frac{1}{k}\right)  }\right)  ^{r}=\sum
_{k=1}^{\infty}j_{k}^{1-\frac{r}{2}-\frac{r}{k}}=\sum_{k=1}^{\infty}
\alpha^{k(k-1)\left(  1-\frac{r}{2}-\frac{r}{k}\right)  },
\]
that diverges for all $r<2$, because $\lim_{k\rightarrow\infty}\alpha
^{k(k-1)\left(  1-\frac{r}{2}-\frac{r}{k}\right)  }\neq0$ whenever $r<2$.

Note that, considering, for all positive integers $j$,
\[
A_{j}:=\left[  1/2+\sqrt{1/4+\log_{\alpha}\left(  \left(  j+1\right)
/2\right)  },1/2+\sqrt{1/4+\log_{\alpha}j}\right]  \cap\mathbb{N},
\]
each $x^{(j)}$ is precisely
\[
x^{(j)}={\alpha}^{{\frac{(1-k_{j})(k_{j}(2+p)+2p)}{2p}}}\cdot\sum
_{s=1}^{\alpha^{k_{j}(k_{j}-1)}}\left(  \exp\left(  {\left[  j\alpha
^{k_{j}\left(  1-k_{j}\right)  }+\alpha^{k_{j}\left(  1-k_{j}\right)
}-1\right]  2\pi si}\right)  \cdot e_{\alpha^{k_{j}(k_{j}-1)}+s-1}\right)
\]
when $A_{j}\neq\emptyset$, where $k_{j}$ is the unique element of $A_{j}$, and
$x^{(j)}=0$ otherwise. This concludes the proof.
\end{proof}

\section{The second approach}

In this section, following the lines of the proof of Theorem \ref{casecomp},
we borrow an argument credited to Pelczynski to adapt the previous
construction to the real scalar field.

The core of the proof rests on properties of the Walsh system (see \cite[page
8]{wal}), which is the set formed by the following functions:

\begin{align*}
f_{0}(x)  &  =1,~~0\leq x\leq1,\\
f_{1}(x)  &  =
\begin{cases}
1, & 0\leq x<\frac{1}{2},\\
-1, & \frac{1}{2}<x\leq1,
\end{cases}
\\
f_{2}^{(1)}(x)  &  =
\begin{cases}
1, & 0\leq x<\frac{1}{4},~~\frac{3}{4}<x\leq1\\
-1, & \frac{1}{4}<x<\frac{3}{4},
\end{cases}
\\
f_{2}^{(2)}(x)  &  =
\begin{cases}
1, & 0\leq x<\frac{1}{4},~~\frac{1}{2}<x<\frac{3}{4},\\
-1, & \frac{1}{4}<x<\frac{1}{2},~~\frac{3}{4}<x\leq1,
\end{cases}
\\
&  \vdots\\
f_{n+1}^{(2k-1)}(x)  &  =
\begin{cases}
f_{n}^{(k)}(2x), & 0\leq x<\frac{1}{2},\\
(-1)^{k+1}f_{n}^{(k)}(2x-1), & \frac{1}{2}<x\leq1,
\end{cases}
\\
f_{n+1}^{(2k)}(x)  &  =
\begin{cases}
f_{n}^{(k)}(2x), & 0\leq x<\frac{1}{2}\\
(-1)^{k}f_{n}^{(k)}(2x-1), & \frac{1}{2}<x\leq1,
\end{cases}
\end{align*}
where $n\in\mathbb{N}$ and $k=1,2,\ldots,2^{n-1}$.

\begin{theorem}
Let $m$ be a positive integer and $g_{1},\ldots,g_{2^{m}}$ the first
$n:=2^{m}$ functions of the Walsh system. For $i,j=1,\ldots,2^{m}$, let
$a_{i,j}^{(n)}$ the value that each $g_{j}$ takes in the interval
$((i-1)/2^{m},i/2^{m})$. Let $p\in\lbrack1,2]$. For all positive integers $j$,
let
\[
A_{j}:=\left[  1/2+\sqrt{1/4+\log_{2}\left(  \left(  j+1\right)  /2\right)
},1/2+\sqrt{1/4+\log_{2}j}\right]  \cap\mathbb{N}.
\]
The sequence $(x^{(j)})_{j=1}^{\infty}$ defined by
\[
x^{(j)}={2}^{{\frac{( 1-k_{j})( k_{j}(2+p)+2p) }{2p}}}\cdot\sum_{s=1}
^{2^{k_{j}( k_{j}-1) }}\left(  a_{rs} ^{(2^{k_{j}(k_{j}-1)})} \cdot
e_{2^{k_{j}( k_{j}-1) }+s-1}\right)
\]
if $A_{j}\neq\emptyset$, where $r=j-2^{k_{j}(k_{j}-1)}+1$ and $k_{j}$ is the
unique element of $A_{j}$, and $x^{(j)}=0$ otherwise, is unconditionally
summable in $\ell_{p}$ and
\[

{\textstyle\sum\limits_{j=1}^{\infty}}

\left\Vert x^{(j)}\right\Vert ^{2-\varepsilon}=\infty
\]
for all $\varepsilon>0.$
\end{theorem}

\begin{proof}
Recall that $g_{1}=f_{0}$, $g_{2}=f_{1}$, $g_{3}=f_{2}^{(1)}$, $g_{4}
=f_{2}^{(2)}$, and so on. For unit vector $\varphi\in\ell_{p^{\ast}}^{n}$, we
have
\[
\sum\limits_{i=1}^{n}\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}
^{(n)}\right\vert \leq n^{\frac{1}{2}+\frac{1}{p}}
\]
where $\varphi=(\varphi_{1},\ldots,\varphi_{n})$. In fact, for each
$i=1,2,\ldots,n$, by the definition of $g_{j}$ in the interval $((i-1)/n,i/n)$
we obtain
\[
\int_{(i-1)/n}^{i/n}\left\vert \sum_{j=1}^{n}\varphi_{j}g_{j}(t)\right\vert
dt=\frac{1}{n}\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}
^{(n)}\right\vert ,
\]
i.e.,
\begin{equation}
\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}^{(n)}\right\vert
=n\int_{(i-1)/n}^{i/n}\left\vert \sum_{j=1}^{n}\varphi_{j}g_{j}(t)\right\vert
dt, \label{i1}
\end{equation}
Hence, by (\ref{i1})
\begin{equation}
\sum\limits_{i=1}^{n}\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}
^{(n)}\right\vert =n\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}\left\vert \sum
_{j=1}^{n}\varphi_{j}g_{j}(t)\right\vert dt=n\int_{0}^{1}\left\vert \sum
_{j=1}^{n}\varphi_{j}g_{j}(t)\right\vert dt. \label{i2}
\end{equation}
By (\ref{i2}) and by the monotonicity of the $L_{p}$-norms, we have
\begin{equation}
\sum\limits_{i=1}^{n}\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}
^{(n)}\right\vert =n\int_{0}^{1}\left\vert \sum_{j=1}^{n}\varphi_{j}
g_{j}(t)\right\vert dt\leq n\left(  \int_{0}^{1}\left\vert \sum_{j=1}
^{n}\varphi_{j}g_{j}(t)\right\vert ^{2}dt\right)  ^{\frac{1}{2}}. \label{i3}
\end{equation}
Since $\int_{0}^{1}g_{j}(t)g_{k}(t)dt=\delta_{jk}$, for $j,k=1,\ldots,n$, we
have
\begin{equation}
\int_{0}^{1}\left\vert \sum_{j=1}^{n}\varphi_{j}g_{j}(t)\right\vert
^{2}dt=\sum_{j=1}^{n}|\varphi_{j}|^{2}. \label{i4}
\end{equation}
Then, by (\ref{i3}) and (\ref{i4}), we have
\begin{equation}
\sum\limits_{i=1}^{n}\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}
^{(n)}\right\vert \leq n\left(  \sum_{j=1}^{n}|\varphi_{j}|^{2}\right)
^{\frac{1}{2}}. \label{i5}
\end{equation}
On the other hand, by the H\"{o}lder inequality, using that $\left\Vert
\varphi\right\Vert _{p^{\ast}}\leq1$, we obtain
\begin{equation}
\left(  \sum_{j=1}^{n}|\varphi_{j}|^{2}\right)  ^{\frac{1}{2}}\leq\left(
\sum_{j=1}^{n}1\right)  ^{\frac{1}{2}-\frac{1}{p^{\ast}}}\cdot\left(
\sum_{j=1}^{n}|\varphi_{j}|^{p^{\ast}}\right)  ^{\frac{1}{p^{\ast}}}\leq
{n}^{\left(  \frac{1}{2}-\frac{1}{p^{\ast}}\right)  }. \label{i6}
\end{equation}
Therefore, by (\ref{i5}) and (\ref{i6}), we have
\[
\sup_{\left\Vert \varphi\right\Vert _{p^{\ast}}\leq1}\sum\limits_{i=1}
^{n}\left\vert \sum\limits_{j=1}^{n}\varphi_{j}a_{i,j}^{(n)}\right\vert \leq
n\left(  \sum_{j=1}^{n}|\varphi_{j}|^{2}\right)  ^{\frac{1}{2}}\leq n\cdot
{n}^{\left(  \frac{1}{2}-\frac{1}{p^{\ast}}\right)  }=n^{\frac{1}{2}+\frac
{1}{p}}.
\]
The rest of the proof follows the lines of the proof of Theorem \ref{casecomp}
. It suffices to consider $\alpha=2$ in the definition of $j_{k}$ and change
the elements of the complex matrix by the elements of the Walsh system.
\end{proof}

config: arxiv


TITLE: How is the electron $g$-factor determined?
QUESTION [3 upvotes]: I found a description of the experiments at CERN where a Penning trap is used to to determine the g-factor but I could not find the value of the radius of the circle of precession, nor could I (surely my bad!) find a clue in the wikipedia article, they say:

The orbital motion of ions in the radial plane is composed of two
  modes at frequencies which are called the magnetron and the modified cyclotron  frequencies. These motions are similar to the
  deferent and epicycle, respectively, of the Ptolemaic model of the
  solar system.The sum of these two frequencies is the cyclotron
  frequency, which depends only on the ratio of electric charge to mass
  and on the strength of the magnetic field. This frequency can be
  measured very accurately and can be used to measure the masses of
  charged particles.

We also know from another article that if B is one Tesla, the frequency of precession (gyromagnetic ratio) is 1.76*10 ^11 radians per second, but we do not know how long is one radian. In the same article the orbit is depicted as circular, whereas in tne Penning trap article the path is somewhat different.
Or, can you please explain if the value of $r$ (radius and radian) is irrelevant and, if it is not, what is the radius of the precession circle when B= 1 tesla? Can you explain what the value of 1.76 = 1.75882*1.00115965 tells us?

REPLY [3 votes]: As it happens, I used to work on such electron-$g$-factor experiments in Penning-traps, so let me try to shed some light on the issue.
Typially, the symbol $\omega$ in a formula tells you that the corresponding quantity is given in "radians-per-second". This is simply how fast something rotates. You could also use Hertz for such a quantity (typical symbol: $f$), but theoreticians like radians-per-second because otherwise they have to put $\cdot 2 \pi$ everywhere. There is a simple conversion between radians-per-second and Hz:
$$ \omega = 2 \pi f $$
or, for a concrete example,
$$2 \pi \, \frac{\text{rad}}{s} = 1 \, \text{Hz} $$
The symbol $\text{rad}$ in this formula stands for "angle measured in radian", not for radius.
If you want to be super picky, you should describe $\omega$ as an "angular velocity" (how much angle per second), and only $f$ as a "frequency", but most people are lazy and write "the frequency $\omega$", which is technically wrong but understood by all.
The precesion of the electron's magnetic moment in a Penning trap happens at an angular velocity of
$$ \omega_\text{L} = \frac{g}{2}\frac{q}{m}B $$
or, in other words, at a frequency of
$$ f = \frac{g}{4\pi}\frac{q}{m}B  \quad .$$
If the electron were a tiny ball (which it is not, but let's imagine it anyways), this would be how fast the ball is spinning around its own axis.
Unfortunately, there are technical (and physical) limits on how "still" or "motionless" we can keep an electron in a trap. The electron will always move a bit. It turns out, it's best to live with this motion and construct traps in a way that makes the motion as predictable as possible. In the electron-as-a-ball picture, this would be the ball moving in some predictable fashion through your trap (all the while it is of course spinning around its own axis as much as it wants to).
A Penning trap is a special trap where the motion of the electron is a superposition of three simultaneous oscillations: The reduced cyclotron oscillation, the magnetron oscillation, and the axial oscillation. Typically, you try to keep the amplitude of these oscillations small. This helps with measurement accuracy. I don't have values for typical trapped electrons handy right now, but for trapped ions, each oscillation has an amplitude of a few 10 µm.
There is an upside to this electron motion: You can excite the motion to detect the electron's presence and to measure the strength of the magnetic field that the elctron experiences.
This does not yet fully answer your question, and I apologize for that. The full measurement scheme on how to determine the $g$-factor can get arbitrarily complex, depending on how much detail you are interested in. But I hope this answer helps with some of the underlying concepts.


text:
\begin{document}

\title[]
{Congruences for\\
the Almkvist-Zudilin numbers}

\author{Tewodros Amdeberhan}
\address{Department of Mathematics,
Tulane University, New Orleans, LA 70118}
\email{tamdeber@tulane.edu}

\subjclass[2010]{Primary ??}

\date{\today}

\keywords{??}

\begin{abstract}
Given a prime number $p$, the study of divisibility properties of a sequence $c(n)$ has two contending approaches: $p$-adic valuations and superconcongruences. The former searches for the highest power of $p$ dividing $c(n)$, for each $n$; while the latter (essentially) focuses on the maximal powers $r$ and $t$ such that $c(p^rn)$ is congruent to $c(p^{r-1}n)$ modulo $p^t$. This is called supercongruence. In this note, we prove modest supercongruences for certain sequences that have come to be known as the Almkvist-Zudilin numbers and two other naturally related ones.
\end{abstract}

\maketitle

\newcommand{\ba}{\begin{eqnarray}}
\newcommand{\ea}{\end{eqnarray}}
\newcommand{\ift}{\int_{0}^{\infty}}
\newcommand{\nn}{\nonumber}
\newcommand{\no}{\noindent}
\newcommand{\lf}{\left\lfloor}
\newcommand{\rf}{\right\rfloor}
\newcommand{\realpart}{\mathop{\rm Re}\nolimits}
\newcommand{\imagpart}{\mathop{\rm Im}\nolimits}

\newcommand{\op}[1]{\ensuremath{\operatorname{#1}}}
\newcommand{\pFq}[5]{\ensuremath{{}_{#1}F_{#2} \left( \genfrac{}{}{0pt}{}{#3}
{#4} \bigg| {#5} \right)}}

\newtheorem{Definition}{\bf Definition}[section]
\newtheorem{Thm}[Definition]{\bf Theorem}
\newtheorem{Example}[Definition]{\bf Example}
\newtheorem{Lem}[Definition]{\bf Lemma}
\newtheorem{Cor}[Definition]{\bf Corollary}
\newtheorem{Prop}[Definition]{\bf Proposition}
\numberwithin{equation}{section}

\section{Introduction}

\noindent
Let us fix some notational conventions. Denote the set of positive integers by $\mathbb{P}$. For $m\in\mathbb{P}$, let $\equiv_m$ represent congruence modulo $m$. Throughout, assume $p\geq5$ is a prime.

\noindent
The {\em Ap\'ery numbers} $A(n)=\sum_{k=0}^n\binom{n}k^2\binom{n+k}k^2$ were valuable to R. Ap\'ery in his celebrated proof ~\cite{A}
of the irrationality of $\zeta(3)$. Since then these numbers have been a subject of much research. For example, they stand among a host of other sequences with the property
$$A(p^rn)\equiv_{p^{3r}} A(p^{r-1}n)$$
now known as {\em supercongruence} $-$ a term dubbed by F. Beukers ~\cite{B}.

\noindent
At the heart of many of these congruences sits the classical example $\binom{pb}{pc}\equiv_{p^3}\binom{b}c$ which is a stronger variant of the famous Lucas's congruence $\binom{pb}{pc}\equiv_p\binom{b}c$. For a compendium of references on the subject of Ap\'ery-type sequences, see ~\cite{S}.

\noindent
In this paper, true to tradition, we shall investigate similar type of divisibility properties (i.e. supercongruences) of the following three sequences. For  $i\in\{0,1,2\}$ and $n\in\mathbb{P}$, define
\begin{align*} a_i(n):&=\sum_{k=0}^{\lfloor(n-i)/3\rfloor}(-1)^{n-k}\binom{3k+i}k\binom{2k+i}k\binom{n}{3k+i}\binom{n+k}k3^{n-3k-i} \\

\end{align*}
In recent literature, $a_0(n)$ are referred to as the Almkvist-Zudilin numbers.
Our motivation for the present work here emanates from the following claim found in ~\cite{O} (see also ~\cite{CCS}).

\begin{conjecture} For a prime $p$ and $n\in\mathbb{P}$, the Almkvist-Zudilin numbers satisfy
$$a_0(pn)\equiv_{p^3}a_0(n).$$
\end{conjecture}

\smallskip
\noindent
Our main results can be summarized as:

{\em if $p$ is a prime and $n\in\mathbb{P}$, then $a_0(pn)\equiv_{p^3}a_0(n)$ and $a_1(pn)\equiv_{p^2}a_2(pn)\equiv_{p^2}0$.}

\section{Preliminary results}

\begin{lemma}  Let $f(k)=\binom{3k}{k,k,k}\frac{3^{-3k}}{k+1}$ and $g(k)=\binom{3k+1}{k,k,k+1}\frac{3^{-3k}}{k+2}$. We have

\smallskip
(a) $\sum_{j\geq0}\binom{b}j\binom{c}j=\binom{b+c}c$,

\smallskip
(b) $\sum_{k=0}^mf(k)=9\binom{m+2}2f(m+1)$,

\smallskip
(c) $\sum_{k=0}^mg(k)=\frac{9(m+1)(m+3)}8g(m+1)$.
\end{lemma}
\begin{proof} (a) is Vandermode-Chu's identity; (b) is due to $\frac13\binom{m+2}2f(m+1)-\frac13\binom{m+1}2f(m)=f(m)$;
(c) is verified by $\frac{9(m+1)(m+3)}8g(m+1)-\frac{9m(m+2)}8g(m)=g(m)$.  \end{proof}

\begin{corollary} For each prime $p$, we have the following congruences:
$$\sum_{k=0}^{\lfloor(p-1)/3\rfloor}\binom{3k}{k,k,k}\frac{3^{-3k}}{k+1}\,\equiv_p\,\,0 \qquad \text{and}
\qquad \sum_{k=0}^{\lfloor(p-2)/3\rfloor}\binom{3k+1}{k,k,k+1}\frac{3^{-3k}}{k+2}\,\equiv_p\,\,0.$$
\end{corollary}
\begin{proof} By Lemma 2.1(b) we gather $\sum_{k=0}^m\binom{3k}{k,k,k}\frac{3^{-3k}}{k+1}=
\binom{m+2}2\binom{3m+4}{m+1}\binom{2m+3}{m+1}\frac{3^{-3m-1}}{3m+4}$. Put $m=\lfloor\frac{p-1}3\rfloor$.
If $p=3\ell+1\geq11$ then $\binom{3m+4}{m+1}=\binom{p+3}{\ell+1}\equiv_p\binom{3}{\ell+1}\equiv_p0$; if $p=3\ell+2\geq11$ then $\binom{3m+4}{m+1}=\binom{p+2}{\ell+1}\equiv_p\binom{2}{\ell+1}\equiv_p0$. For $p=5, 7$, we have $\binom{2m+3}{m+1}\equiv_p0$. In all cases, the desired congruence holds.

\noindent
For the second congruence, Lemma 2.1(c) gives $\sum_{k=0}^m\binom{3k+1}{k,k,k+1}\frac{3^{-3k}}{k+2}=\frac{m+1}8
\binom{3m+4}{m+1}\binom{2m+3}{m+1}3^{-3m-1}$. Put $m=\lfloor\frac{p-2}3\rfloor$.
If $p=3\ell+1\geq11$ then $\binom{3m+4}{m+1}=\binom{p+3}{\ell+1}\equiv_p\binom{3}{\ell+1}\equiv_p0$; if $p=3\ell+2\geq11$ then $\binom{3m+4}{m+1}=\binom{p+2}{\ell+1}\equiv_p\binom{2}{\ell+1}\equiv_p0$. For $p=5, 7$, we have $\binom{2m+3}{m+1}\equiv_p0$. In all cases, the required congruence remains valid.
\end{proof}

\noindent
{\em Fermat quotients} are numbers of the form $b_p(x)=\frac{x^{p-1}-1}p$ and they played a useful role in the study
of cyclotomic fields and Fermat's Last Theorem, see ~\cite{R}. We need one of their divisibility properties as stated in the
next result, see Lehmer (\cite{L}, page 358).
\begin{lemma} For each prime $p$, it holds that
$$\frac{b_p(3)}2-\frac{p\,b_p(3)^2}4\equiv_{p^2}\sum_{r=1}^{\lfloor p/3\rfloor}\frac1{p-3r}.$$
\end{lemma}

\begin{corollary} Let $H(n)=\sum_{j=1}^n\frac1j$ be the harmonic numbers, $p$ a prime and $n\in\mathbb{P}$, Then,
$$3\left(\frac{3^{(p-1)n}-1}p\right)\equiv_p-(2n)\,H(\lfloor p/3\rfloor).$$
\end{corollary}
\begin{proof} Lemma 2.3 implies $3b_p(3)\equiv_p-2H(\lfloor p/3\rfloor)$. On the other hand, Fermat's little theorem implies $b_p(x^n)\equiv_pn\,b_p(x)$. Taking $x=3$, the proof follows.
\end{proof}

\begin{lemma} For $n\in\mathbb{P}$, we have the identity
$$\sum_{k=1}^n(-1)^k\binom{n}k\binom{n+k}k\frac1k=-2H(n).$$
\end{lemma}
\begin{proof} Given any $y\in\mathbb{P}$, E. Mortenson (\cite{M}, page 990) made application of the Wilf-Zeilberger method to prove the identity
$$\sum_{k=0}^n(-1)^k\binom{n}k\binom{n+k}k\frac1{k+y}=\frac{(-1)^n}y\prod_{j=1}^n\frac{y-j}{y+j}.$$
Actually, the same method offers that the identity is valid for an indeterminate $y$. Now, subtract $\frac1y$ from both sides and take the limit as $y\rightarrow0$. The right-hand side takes the form
$$\frac1{n!}\lim_{y\rightarrow0}\left[\frac{\prod_{j=1}^n(j-y)-\prod_{j=1}^n(j+y)}y\right]=-2\sum_{k=1}^n\frac1k.$$
The conclusion is clear.
\end{proof}

\begin{lemma} Suppose $p$ is a prime and $z$ is a variable. Then,
$$\sum_{k=1}^{\lfloor p/3\rfloor}(-1)^k\binom{\lfloor p/3\rfloor}k\binom{\lfloor p/3\rfloor+k}k\frac{z^k}k
\equiv_p\sum_{k=1}^{\lfloor p/3\rfloor}\binom{3k}{k,k,k}\frac{3^{-3k}z^k}k.$$
\end{lemma}
\begin{proof} The congruence actually holds term-by-term and that is how we proceed. Also, observe that $\binom{n}k\binom{n+k}k=\binom{2k}k\binom{n+k}{2k}$. If $p\equiv_31$, then $\lfloor\frac{p}3\rfloor=\frac{p-1}3$
and hence
\begin{align*} \binom{\frac{p-1}3+k}{2k}
&=\frac{\frac{p-1}3(\frac{p-1}3+k)}{(2k)!}\prod_{j=1}^{k-1}\left(\frac{p-1}3\pm j\right) \\
&\equiv_p\frac{(-1)^k(3k-1)}{3^{2k}(2k)!}\prod_{j=1}^{k-1}(3j\pm1)
=\frac{(-1)^k(3k)!}{3^{3k}(2k)!k!}. \end{align*}
Therefore, we gather that
$$(-1)^k\binom{\frac{p-1}3}k\binom{\frac{p-1}3+k}k=
(-1)^k\binom{2k}k\binom{\frac{p-1}3+k}{2k}\equiv_p\frac{(3k)!}{3^{3k}!k!^3}=\binom{3k}{k,k,k}3^{-3k}.$$
The proof follows by summing over $k$. The case $p\equiv_3-1$ runs analogously. \end{proof}

\begin{corollary} For a prime $p$, we have the congruence
$$pn\sum_{k=1}^{\lfloor p/3\rfloor}\binom{3k}{k,k,k}\frac{3^{-3k}}k\equiv_{p^2}3\left(3^{(p-1)n}-1\right).$$
\end{corollary}
\begin{proof} Follows from combining Corollary 2.4, Lemma 2.5 and Lemma 2.6 with $z=1$.
\end{proof}

\section{Main results on the sequences $a_1(n)$ and $a_2(n)$}

\begin{theorem} For a prime $p$ and $n\in\mathbb{P}$, we have $a_1(pn)\equiv_{p^2} 0$.
\end{theorem}
\begin{proof} Let $k=pm+r$ for $0\leq r\leq p-1$. Note: $3k+1\leq pn$ and $3pm+3r+1\leq pn$. Write
\begin{align*} a_1(pn)=\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=0}^{p-1}
(-1)^{pn-pm-r}&\binom{3pm+3r+1}{pm+r}\binom{2pm+2r+1}{pm+r} \times \\
&\binom{pn}{3pm+3r+1}\binom{pn+pm+r}{pm+r}3^{pn-3pm-3r-1}.\end{align*}
If $t:=3r+1\geq p+1$, it is easy to show that the following terms vanish modulo $p^2$:
$$\binom{3pm+t}{pm+r}\binom{2pm+2r+1}{pm+r}\binom{pn}{3pm+t}=
\binom{3pm+t}{pm+r,pm+r,pm+r+1}\binom{pn}{3pm+t}.$$
Therefore, we may restrict to the remaining sum with $3r+1\leq p$:
\begin{align*} a_1(pn)=\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=0}^{\lfloor(p-1)/3\rfloor}
(-1)^{pn-pm-r}&\binom{3pm+3r+1}{pm+r}\binom{2pm+2r+1}{pm+r} \times \\
&\binom{pn}{3pm+3r+1}\binom{pn+pm+r}{pm+r}3^{pn-3pm-3r-1}.\end{align*}
We need Lucas's congruence $\binom{pb+c}{pd+e}\equiv_p\binom{d}d\binom{c}e$ to arrive at
\begin{align*} a_1(pn)\equiv_p\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=0}^{\lfloor(p-1)/3\rfloor}
(-1)^{pn-pm-r}&\binom{3m}m\binom{3r+1}r\binom{2m}m\binom{2r+1}r\times \\
&\binom{pn}{3pm+3r+1}\binom{n+m}m3^{pn-3pm-3r-1}.\end{align*}
For $0<i<p$, we apply Gessel's congruence $\binom{p}i\equiv_{p^2}(-1)^{i-1}\frac{p}i$ (if $p=3r+1$, know that $k$ is even; in this case, still the corresponding term properly absorbs into the sum below) so that
\begin{align*} \binom{pn}{3pm+3r+1}&=\frac{pn}{3pm+3r+1}\binom{pn-1}{3pm+3r} \\
&=\frac{pn}{3pm+3r+1}\binom{p(n-1)+p-1}{3pm+3r}
\equiv_{p^2}(-1)^r\frac{pn}{3r+1}\binom{n-1}{3m}, \end{align*}
which leads to
\begin{align*} a_1(pn)\equiv_{p^2}pn\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=0}^{\lfloor(p-1)/3\rfloor}
(-1)^{n-m-r}&\binom{3m}m\binom{3r+1}r\binom{2m}m\binom{2r+1}r\times \\
&\frac{(-1)^r}{3r+1}\binom{n-1}{3m}\binom{n+m}m3^{pn-3pm-3r-1}.\end{align*}
Next, we use Fermat's Little Theorem and {\em decouple} the double sum to obtain
\begin{align*} a_1(pn)\equiv_{p^2}n&\sum_{m=0}^{\lfloor n/3\rfloor}(-1)^{n-m}3^{n-3m-1}\binom{3m}m\binom{2m}m
\binom{n-1}{3m}\binom{n+m}m \times \\
&p\sum_{r=0}^{\lfloor(p-1)/3\rfloor}\binom{3r+1}r\binom{2r+1}r\frac{3^{-3r}}{3r+1}.\end{align*}
If we invoke Corollary 2.2, the sum over $r$ is already divisible by $p^2$ which is enough to conclude $a_1(pn)\equiv_{p^2}0$.
\end{proof}

\begin{theorem} For a prime $p$ and $n\in\mathbb{P}$, we have $a_2(pn)\equiv_{p^2} 0$.
\end{theorem}
\begin{proof} The idea is similar to Theorem 0.3, hence it is omitted to avoid needless duplicity.
\end{proof}

\newpage

\section{Main result on the sequence $a_0(n)$}

\noindent
In this section, we prove a weaker version of Conjecture 1.1 which still required a more delicate touch than what has been demonstrated in the previous section for the other two sequences. We also believe that our techniques pave the way in settling the conjecture, fully.

\begin{theorem} For a prime $p$ and $n\in\mathbb{P}$, we have $a_0(pn)\equiv_{p^2} a_0(n)$.
\end{theorem}
\begin{proof}  Let $k=pm+r$ for $0\leq r< p$. Note: $3k=3pm+3r\leq pn$. Using the new parameters,
\begin{align*} a_0(pn)=\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=0}^{p-1}
(-1)^{pn-pm-r}&\binom{3pm+3r}{pm+r}\binom{2pm+2r}{pm+r} \times \\
&\binom{pn}{3pm+3r}\binom{pn+pm+r}{pm+r}3^{pn-3pm-3r}.\end{align*}
If $3r\geq p+1$, a straight-forward argument shows the following terms vanish modulo $p^2$:
$$\binom{3pm+3r}{pm+r}\binom{2pm+2r}{pm+r}\binom{pn}{3pm+3r}=
\binom{3pm+3r}{pm+r,pm+r,pm+r}\binom{pn}{3pm+3r}.$$
Therefore, we may restrict to the remaining sum with $3r\leq p$: thus $a_0(pn)$ becomes
$$\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=0}^{\lfloor p/3\rfloor}
(-1)^{pn-pm-r}\binom{3pm+3r}{pm+r}\binom{2pm+2r}{pm+r}
\binom{pn}{3pm+3r}\binom{pn+pm+r}{pm+r}3^{pn-3pm-3r}.$$

\noindent
Let's isolate the case $r=0$ and subtract $a_0(n)$ from it, then implement the milder form $\binom{pb}{pd}\equiv_p\binom{b}d$ and bring Corollary 2.7 to bear. The outcome is:
\begin{align*} &\sum_{m=0}^{\lfloor n/3\rfloor}
(-1)^{n-m}\binom{3m}m\binom{2m}m
\binom{n}{3m}\binom{n+m}m3^{pn-3pm}-a_0(n) \\
&=\sum_{m=0}^{\lfloor n/3\rfloor}
(-1)^{n-m}\binom{3m}m\binom{2m}m
\binom{n}{3m}\binom{n+m}m3^{n-3m}\left(3^{(p-1)(n-3m)}-1\right) \\
&\equiv_{p^2}p\sum_{m=0}^{\lfloor n/3\rfloor}(-1)^{n-m}\binom{3m}{m,m,m}\binom{n}{3m}
\binom{n+m}m\frac{(n-3m)}{3^{-n+3m+1}}\sum_{k=1}^{\lfloor p/3\rfloor}\binom{3k}{k,k,k}\frac{3^{-3k}}k.
\end{align*}
The remaining sum in $a_0(pn)$, without the term $r=0$ and modulo $p^2$, takes the form
\begin{align} \sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=1}^{\lfloor p/3\rfloor}
(-1)^{n-m-r}&\binom{3m}m\binom{3r}r\binom{2m}m\binom{2r}r
\binom{pn}{3pm+3r}\binom{n+m}m3^{n-3m-3r}. \end{align}
To this sum, we apply Gessel's congruence $\binom{p-1}i\equiv_{p^2}(-1)^i$,  with $0<i<p$, so that
\begin{align*} \binom{pn}{3pm+3r}&=\frac{pn}{3pm+3r}\binom{pn-1}{3pm+3r-1} \\
&=\frac{pn}{3pm+3r}\binom{p(n-1)+p-1}{3pm+3r-1}\equiv_{p^2}(-1)^{r-1}\frac{pn}{3r}\binom{n-1}{3m},
\end{align*}
which converts (4.1), modulo $p^2$, into
\begin{align*} &pn\sum_{m=0}^{\lfloor n/3\rfloor}\sum_{r=1}^{\lfloor p/3\rfloor}
(-1)^{n-m-1}\binom{3m}m\binom{3r}r\binom{2m}m\binom{2r}r
\frac1{3r}\binom{n-1}{3m}\binom{n+m}m3^{n-3m-3r} \\
&\equiv_{p^2} -p\sum_{m=0}^{\lfloor n/3\rfloor}(-1)^{n-m}\binom{3m}{m,m,m}\binom{n}{3m}\binom{n+m}m\frac{(n-3m)}{3^{-n+3m+1}}
\sum_{r=1}^{\lfloor p/3\rfloor}\binom{3r}r\binom{2r}r\frac{3^{-3r}}r.
\end{align*}
However, the two sums (one for $r=0$ above, the other for $r>0$) add up to zero. That means $a_0(pn)-a_0(n)\equiv_{p^2}0$, as desired.
\end{proof}

\bigskip

\smallskip
\noindent
\textbf{Acknowledgements.} The author is grateful to Armin Straub for bringing the conjecture on the Almkvist-Zudilin numbers to his attention.

\bigskip


text:
\begin{document}

\title{\huge{Bounds on Eigenvalues of a Spatial Correlation Matrix}}

\author{Junil Choi and David J. Love\\
\thanks{Junil Choi and David J. Love are with the School of Electrical and Computer Engineering, Purdue University, West Lafayette, IN (e-mail: choi215@purdue.edu, djlove@purdue.edu).}
}

\maketitle

\begin{abstract}
It is critical to understand the properties of spatial correlation matrices in massive multiple-input multiple-output (MIMO) systems.  We derive new bounds on the extreme eigenvalues of a spatial correlation matrix that is characterized by the exponential model in this paper.  The new upper bound on the maximum eigenvalue is tighter than the previous known bound.  Moreover, numerical studies show that our new lower bound on the maximum eigenvalue is close to the true maximum eigenvalue in most cases.  We also derive an upper bound on the minimum eigenvalue that is also tight.  These bounds can be exploited to analyze many wireless communication scenarios including uniform planar arrays, which are expected to be widely used for massive MIMO systems.
\end{abstract}

\begin{keywords}
Spatial correlation matrix,  maximum/minimum eigenvalue, exponential model, massive MIMO.
\end{keywords}
\section{Introduction}\label{sec1}
Spatial correlation in the channel can give both gains and losses depending on the scenario in multiple-input multiple-output (MIMO) systems \cite{exp_model_bruno}.  Spatial correlation is harmful for single-user MIMO systems using multiplexing because the correlation reduces the rank of the communication channel, resulting in a reduced number of parallel paths for spatial multiplexing \cite{exp_model_capa_analysis2}.  On the other hand, spatial correlation is beneficial for multi-user MIMO systems because the strong directivity of channels between a transmitter and users can help to reduce inter-user interference even with simple precoding strategies at the transmitter \cite{exp_model_bruno}.

The most common model for the spatial correlation matrix is the exponential model \cite{exp_model_bruno,exp_model_experiment}.  The exponential model is very simple because the correlation matrix is controlled by one parameter.  Although simple, it has been shown experimentally that the exponential model characterizes uniform linear array (ULA) antenna scenarios well \cite{exp_model_experiment}.  Thus, many works, e.g., capacity analyses in \cite{exp_model_capa_analysis1,exp_capa}, codebook designs for channel state information (CSI) quantization in \cite{tm_correlated5,spatial_codebook1,spatial_codebook2}, and training signal designs for channel estimation in \cite{training2,training5,cl_training_jstsp}, are based on the exponential model for the spatial correlation matrix.

The exponential model is useful for analyzing uniform planar array (UPA) scenarios.  Note that UPA deployments are growing in popularity due to the emergence of massive MIMO systems \cite{massive_mimo3,fdmimo}.  It was shown in \cite{dawei} that the spatial correlation matrix of a UPA can be approximated by the Kronecker product of the spatial correlation matrices corresponding to the vertical and horizontal domain.  In \cite{upa_codebook,dawei}, this approximation was exploited to design codebooks for CSI quantization in a UPA scenario.

Because of the reasons above, we focus on spatial correlation matrices following the exponential model in this paper.  The maximum and minimum eigenvalues of the spatial correlation matrix are important factors because they determine performance in spatially correlated channels \cite{vasan1,vasan2}.  In this paper, we derive new upper and lower bounds on the maximum eigenvalue and an upper bound on the minimum eigenvalue of the correlation matrix.  Although the exact eigenvalues of the exponential model are derived in \cite{bound_old}, the expressions need numerical solutions of the trigonometric function.  Moreover, the bounds derived from the exact expressions are not functions of the number of transmit antennas, which makes it hard to analyze massive MIMO systems with practical numbers of antennas.  Thus, it is desired to have simple and tight upper and lower bounds on extreme eigenvalues expressed with the number of antennas for analyzing the exponential model.

The new upper bound on the maximum eigenvalue, which is based on a novel matrix expansion approach, is tighter than the one from \cite{bound_old}.  Moreover, simulation results show that our lower bound on the maximum eigenvalue is very tight with the true value in general.  The new upper bound on the minimum eigenvalue is tight as well.  All these new bounds are functions of the number of transmit antennas.  The new lower bound on the maximum eigenvalue and upper bound on the minimum eigenvalue are intuitive and simple to be derived; however, we could not find such derivations even after extensive literature search.  Most of the literature adopted the previous bounds in \cite{bound_old} for performance analysis \cite{exp_model_capa_analysis1,cl_training_jstsp} or simply performed numerical studies with the exponential model \cite{exp_capa,tm_correlated5,spatial_codebook1,spatial_codebook2,training2,training5}.

\section{System Model}\label{sec2}
We consider multiple-input single-output (MISO) channels that are spatially correlated at the transmitter side.

Assuming $N_t$ transmit antennas at the transmitter, the input-output relation at baseband is given as
\begin{equation*}
  y=\bh^H \bx + n,
\end{equation*}
where $y$ is the received signal, $\bh\in\mathbb{C}^{N_t}$ is the channel vector, $\bx\in\mathbb{C}^{N_t}$ is the transmitted signal, and $n \sim \cC\cN(0,\sigma^2)$ is additive complex Gaussian noise.  Because we consider spatially correlated channels, $\bh$ is modeled as
\begin{equation*}
  \bh = \bR^{\frac{1}{2}}\bh_w
\end{equation*}
where $\bR=E[\bh\bh^H]$ is the spatial correlation matrix and $\bh_w \sim \cC\cN(\mathbf{0},\bI_{N_t})$ is an i.i.d. complex Gaussian vector.

Depending on antenna structure, $\bR$ can be modeled in many ways.  There is particular interest of antenna deployments using UPAs in massive MIMO systems.  Note that a UPA can support a large number of antennas compactly, which makes massive MIMO practical.  For a UPA, \cite{dawei} showed by eigenvalue and capacity distributions that the spatial correlation matrix of a UPA can be approximated as
\begin{equation}\label{kron_approx}
  \bR\approx \bR_h \otimes \bR_v
\end{equation}
where $\otimes$ is the Kronecker product and $\bR_v$ and $\bR_h$ are the spatial correlation matrices of the horizontal and vertical domains, respectively.

The simplified model in \eqref{kron_approx} makes it possible to adopt any kind of ULA spatial correlation model for $\bR_h$ and $\bR_v$ because a UPA is simply a ULA in each vertical and horizontal domain (assuming co-polarized antennas).  Therefore, it is important to understand the spatial correlation matrix properties of a ULA to understand those of a UPA.

Note that the maximum and minimum eigenvalues of $\bR$ are particularly important in analyzing spatially correlated channels \cite{vasan1,vasan2}.  From \eqref{kron_approx}, we have
\begin{align*}
\bR &\approx \bR_h \otimes \bR_v\\
& = \left(\bU_h \mathbf{\Lambda}_h \bU_h^H\right) \otimes \left(\bU_v \mathbf{\Lambda}_v \bU_v^H\right)\\
& = \left(\bU_h \otimes \bU_v\right) \left(\mathbf{\Lambda}_h \otimes \mathbf{\Lambda}_v\right) \left(\bU_h \otimes \bU_v\right)^H
\end{align*}
where $\bR_h = \bU_h \mathbf{\Lambda}_h \bU_h^H$ and $\bR_v = \bU_v \mathbf{\Lambda}_v \bU_v^H$ denote the eigen-decompositions.  Let $\lambda_k(\bA)$ be the $k$-th largest eigenvalue of the matrix $\bA$.  Then, we have
\begin{align}
\lambda_1(\bR) \approx \lambda_1(\bR_h)\lambda_1(\bR_v),\quad \lambda_{N_t}(\bR) \approx \lambda_{N_t}(\bR_h)\lambda_{N_t}(\bR_v).\label{planar_max_approx}
\end{align}
For these reasons, we focus on the spatial correlation matrix of a ULA in this paper.

We let $\bR$ denote a ULA spatial correlation matrix.  There are many ways to model $\bR$ depending on scenario.  The most common and easy way to model $\bR$ is to rely on the exponential model which is given as\footnote{The field tests from \cite{exp_model_experiment} show that the exponential model characterizes the spatial correlation of ULA very well even with its simplicity.}
\begin{align}\label{exp_model}
\bR_{[i,j]}=\begin{cases}
r^{|i-j|} & \quad \text{if } i\geq j\\
(r^*)^{|i-j|} & \quad \text{if } i< j
\end{cases}
\end{align}
where $^*$ denotes a complex conjugate, $r=ae^{j\theta}$ is the correlation coefficient of $\bR$ with $0\leq a < 1$.  Note that the eigenvalues of $\bR$ only depend on the value of $a$, and $\theta$ only controls the eigenvectors of $\bR$.  Because we are interested in the maximum and minimum eigenvalues of $\bR$, we assume $r=a$ throughout the paper.

\section{Bounds on Eigenvalues}\label{sec3}
We first briefly recall previous results on analyzing the eigenvalues of $\bR$.  We then derive new bounds on the maximum and minimum eigenvalues of $\bR$.

\subsection{Previous Results}\label{prev_bounds}
In \cite{bound_old}, it has been shown that all eigenvalues of $\bR$ can be exactly derived as
\begin{equation}\label{R_eig_exact}
\lambda_i(\bR) = \frac{1-a^2}{1+a^2+2a\cos \phi_i}
\end{equation}
where $\phi_i\neq n\pi$ for any arbitrary integer $n$ are the solutions of the trigonometric equation
\begin{equation*}
\tan N_t \phi_i = \frac{-\sin \phi_i}{\left(\frac{1+a^2}{1-a^2}\right)\cos \phi_i +\frac{2a}{1-a^2}}.
\end{equation*}
From \eqref{R_eig_exact}, it is obvious that\footnote{The bounds in \eqref{max_upper1} and \eqref{min_lower} are reciprocal.  This comes from the fact that $\lambda_1(\bR)$ and $\lambda_{N_t}(\bR)$ have an approximate inverse relation regarding $a$, i.e., $\lambda_1(\bR)$ ($\lambda_{N_t}(\bR)$) increases (decreases) as $a$ grows larger.}
\begin{align}
\lambda_{1}(\bR)&\leq \frac{1+a}{1-a},\label{max_upper1}\\
\lambda_{N_t}(\bR)&\geq \frac{1-a}{1+a}.\label{min_lower}
\end{align}

\noindent \textbf{Remark:} Note that the bounds in \eqref{max_upper1} and \eqref{min_lower} are not functions of the number of antennas $N_t$.  Thus, it is not clear how the extreme eigenvalues would behave as $N_t$ grows large, which is an important aspect in predicting performance of massive MIMO systems \cite{sam}.
\subsection{New Bounds on Eigenvalues}
In the following, we derive an improved upper bound and new lower bound on $\lambda_1(\bR)$ that are both functions of $N_t$.
\begin{lemma}\label{max_bound}
With the exponential model of $\bR$ as in \eqref{exp_model}, the maximum eigenvalue of $\bR$ is bounded as
\begin{equation*}
  \frac{1+a}{1-a}-\frac{2a(1-a^{N_t})}{N_t(1-a)^2}\leq \lambda_1(\bR) \leq \frac{(1+a)(1-a^{N_t-1})}{1-a}
\end{equation*}
when $N_t>1$.
\end{lemma}
\begin{IEEEproof}
We first prove the upper bound.  We extend $\bR$ to an $2(N_t-1)\times 2(N_t-1)$ circulant matrix as
\begin{align*}
\bR_X=
\begin{bmatrix}
1 & a & \cdots & a^{N_t-1} & a^{N_t-2} & a^{N_t-3} & \cdots & a\\
a & 1 & \cdots & a^{N_t-2} & a^{N_t-1} & a^{N_t-2} & \cdots & a^2\\
\vdots & \vdots & \cdots & & \ddots & & & \vdots\\
a & a^2 & \cdots & a^{N_t-2} & a^{N_t-3} & a^{N_t-4} & \cdots & 1\end{bmatrix}.
\end{align*}
Note that $\bR$ is contained in the first $N_t$ rows and $N_t$ columns of $\bR_X$.  Therefore,
\begin{equation*}
  \lambda_{1}(\bR) \leq \lambda_{1}(\bR_X).
\end{equation*}
Because $\bR_X$ is a circulant matrix, the eigenvectors of $\bR_X$ are the columns of the $2(N_t-1)$ point DFT matrix.  If we let $\mathbf{1}_{N}$ be the $N\times 1$ vector with all one entries, it is easy to conclude that the dominant eigenvector of $\bR_X$ is
\begin{equation*}
  \bu_1(\bR_X) = \frac{1}{\sqrt{2(N_t-1)}}\mathbf{1}_{2(N_t-1)}
\end{equation*}
that cophases all of the entries of $\bR_X$ because of the assumption that $a$ is a real positive number.  Then, we have
\begin{align*}
\lambda_{1}(\bR_X) &= \bu_1(\bR_X)^H \bR_X \bu_1(\bR_X)\\
&=\frac{1}{2(N_t-1)}\sum_{k=1}^{2(N_t-1)}\left(\sum_{\ell_1=0}^{N_t-1}a^{\ell_1}+\sum_{\ell_2=1}^{N_t-2}a^{\ell_2}\right)\\

&=\frac{(1+a)(1-a^{N_t-1})}{1-a}.
\end{align*}
Thus,
\begin{equation*}
  \lambda_{1}(\bR) \leq \lambda_{1}(\bR_X)=\frac{(1+a)(1-a^{N_t-1})}{1-a}.
\end{equation*}

To prove the lower bound, we use the definition of the maximum eigenvalue, which follows the general inequality
\begin{equation*}
\lambda_1(\bR) \geq \bff^H \bR \bff
\end{equation*}
with an arbitrary vector $\bff$ satisfying $\|\bff\|_2^2=1$.  Because the elements of $\bR$ are all positive real numbers, the all one vector $\mathbf{1}_{N_t}$ with appropriate normalization would give a good lower bound on $\lambda_1(\bR)$.  Thus, we have
\begin{align*}
\lambda_1(\bR) &\geq \frac{1}{N_t} \mathbf{1}_{N_t}^T \bR \mathbf{1}_{N_t} \\
&=\frac{1}{N_t}\left(2\sum_{\ell=0}^{N_t-1} \sum_{k=0}^{\ell}a^k-N_t\right) \\

&= \frac{1+a}{1-a}-\frac{2a(1-a^{N_t})}{N_t(1-a)^2}
\end{align*}
which finishes the proof.
\end{IEEEproof}
It is obvious that the upper bound in Lemma \ref{max_bound} improves on \eqref{max_upper1} because
\begin{equation*}
  \frac{(1+a)(1-a^{N_t-1})}{1-a}\leq \frac{1+a}{1-a}
\end{equation*}
for arbitrary $0\leq a < 1$.  Moreover, the upper and lower bounds on $\lambda_1(\bR)$ both converge to \eqref{max_upper1} as $N_t\rightarrow \infty$.  This shows that all three bounds become tight when $N_t$ is large.

It is also interesting to analyze the tightness of the upper and lower bounds on $\lambda_1(\bR)$ regarding $a$.  Let $\lambda_1^{\mathrm{diff}}(\bR)$ be the difference of the two bounds in Lemma \ref{max_bound}, which is given as
\begin{equation*}
\lambda_1^{\mathrm{diff}}(\bR)=\frac{(1+a)(1-a^{N_t-1})}{1-a}-\left(\frac{1+a}{1-a}-\frac{2a(1-a^{N_t})}{N_t(1-a)^2}\right).
\end{equation*}
With some algebra, we can show that $\lambda_1^{\mathrm{diff}}(\bR)$ is monotonically increasing with $a$. Moreover, $\lambda_1^{\mathrm{diff}}(\bR)\rightarrow 0$ as $a\rightarrow 0$ and $\lambda_1^{\mathrm{diff}}(\bR)\rightarrow N_t-2$ as $a\rightarrow 1$.  Thus, the two bounds are tight when $a$ is small while the gap becomes large as $a$ increases.\footnote{Note that $\lambda_1^{\mathrm{diff}}(\bR)\rightarrow N_t-2$ is only an asymptotic gap between the two bounds when $a\rightarrow 1$.  The two bounds converge to \eqref{max_upper1} as $N_t\rightarrow \infty$ whenever $a < 1$.}  This is verified numerically in Section \ref{sec_simul}.

Now, we derive an upper bound on the minimum eigenvalue of $\bR$.  The numerical studies in Section \ref{sec_simul} show that the lower bound in \eqref{min_lower} and the new upper bound on $\lambda_{N_t}(\bR)$ are both tight in general.
\begin{lemma}\label{min_upper_bound}
With the exponential model of $\bR$ as in \eqref{exp_model}, the minimum eigenvalue of $\bR$ is upper bounded as
\begin{equation*}
\lambda_{N_t}(\bR)\leq \frac{1-a}{1+a}+\frac{2a(1-(-a)^{N_t})}{N_t(1+a)^2}
\end{equation*}

\end{lemma}
\begin{IEEEproof}
We only prove when $N_t$ is even.  Similar derivation can be shown when $N_t$ is odd.  Using the definition of the minimum eigenvalue, we have the general inequality
\begin{equation*}
\lambda_{N_t}(\bR) \leq \bff^H \bR \bff
\end{equation*}
for an arbitrary vector $\bff$ with $\|\bff\|_2^2=1$.  Let $\tilde{\mathbf{1}}_{N}$ be the $N \times 1$ vector defined as
\begin{equation*}
\tilde{\mathbf{1}}_{N}=\begin{bmatrix}1,-1,1,-1\cdots,(-1)^{N-2},(-1)^{N-1}\end{bmatrix}^T.
\end{equation*}
We can upper bound $\lambda_{N_t}(\bR)$ with $\tilde{\mathbf{1}}_{N_t}$ as
\begin{align*}
  &\lambda_{N_t}(\bR) \leq \frac{1}{N_t} \tilde{\mathbf{1}}_{N_t}^T \bR \tilde{\mathbf{1}}_{N_t} \\
  & = 1 - \frac{2}{N_t}\left(\sum_{k=1}^{\frac{N_t}{2}}\left(N_t-(2k-1)\right)a^{2k-1}\sum_{k=1}^{\frac{N_t}{2}-1}\left(N_t-2k\right)a^{2k}\right).
\end{align*}
We can get the desired result after solving the above series and employing basic algebra.
\end{IEEEproof}

Using these bounds, we can also derive upper and lower bounds on the condition number of $\bR$ and on the (approximated) maximum and minimum eigenvalues of UPA spatial correlation matrix by \eqref{planar_max_approx}.

\section{Numerical Studies}\label{sec_simul}
\begin{figure}[t]
  \centering

  \includegraphics[width=0.9\columnwidth]{maxeigen_r2.eps}\\
  \caption{The plots of $\lambda_1(\bR)$ and its upper and lower bounds according to $N_t$.  The previous upper bound is from \eqref{max_upper1} and the new upper and lower bounds are based on Lemma \ref{max_bound}.}\label{maxeigen_fig}
\end{figure}

First, we plot the maximum eigenvalue of $\bR$, $\lambda_1(\bR)$, and the upper and lower bounds from \eqref{max_upper1} and Lemma \ref{max_bound} according to the number of antennas $N_t$ in Fig. \ref{maxeigen_fig}.  The new upper bound derived in Lemma \ref{max_bound} is tight when $a$ is low to moderate, while the gap between the new upper bound and the true $\lambda_1(\bR)$ becomes large when $a=0.9$.  However, the new upper bound keeps following the curve of $\lambda_1(\bR)$ while the previous upper bound in \eqref{max_upper1} is constant regardless of $N_t$.  It is interesting to point out that the new lower bound is tight for all values of $N_t$ and $a$.  Therefore, the new lower bound can be used as an excellent approximation of $\lambda_1(\bR)$.  As mentioned earlier, the upper and lower bounds in Lemma \ref{max_bound} are tight when $a$ is low to moderate, and the gap becomes large as $a$ approaches one.

In Fig. \ref{mineigen_fig}, we plot the minimum eigenvalue $\lambda_{N_t}(\bR)$, the new upper bound from Lemma \ref{min_upper_bound}, and the previous lower bound in \eqref{min_lower} with $N_t$.  Regarding the minimum eigenvalue, the two bounds are both tight regardless of the values of $a$ and $N_t$.
\begin{figure}[t]
  \centering

  \includegraphics[width=0.9\columnwidth]{mineigen_r1.eps}\\
  \caption{The plots of $\lambda_{N_t}(\bR)$, the previous lower bound from \eqref{min_lower} and the new upper bound from Lemma \ref{min_upper_bound} according to $N_t$.}\label{mineigen_fig}
\end{figure}
\begin{figure}[t]
  \centering

  \includegraphics[width=0.9\columnwidth]{maxeigen_ura.eps}\\
  \caption{The plots of $\lambda_1(\bR)$ for the UPA scenario and its upper and lower bounds with different numbers of $N_h$ and $N_v$.  The correlation coefficients of $\bR_v$ and $\bR_h$ are set to $0.6$ and $0.9$, respectively.}\label{maxeigen_UPA_fig}
\end{figure}

Finally, we plot the maximum eigenvalue of the spatial correlation matrix of UPA given in \eqref{kron_approx} and its upper and lower bounds with different combinations of the numbers of vertical and horizontal domain antennas in Fig. \ref{maxeigen_UPA_fig}.  All bounds are based on the approximations \eqref{planar_max_approx} and derived as in the case of Fig. \ref{maxeigen_fig}.  We set the correlation coefficient of $\bR_h$ as 0.6 and that of $\bR_v$ as 0.9.  It is clear that the new lower bound is very tight for all antenna combinations.  The new upper bound also gives much better tightness compared to the previous upper bound.

\section{Conclusion}
In this paper, we derived new bounds on the maximum and minimum eigenvalues of a spatial correlation matrix that is characterized by the exponential model.  The upper bound on the maximum eigenvalue derived in this paper gives improved tightness than the previous upper bound.  Moreover, using numerical studies, the new lower bound on the maximum eigenvalue is shown to be very tight regardless of the number of antennas and the intensity of spatial correlation.  We also derived a new upper bound on the minimum eigenvalue of the spatial correlation matrix.  It was shown by simulations that the new upper bound and the previous lower bound on the minimum eigenvalue are both tight in general.  The theoretical results derived in this paper can be applied to performance analyses in many wireless communication scenarios including uniform planar array, which are growing in popularity due to the emergence of massive MIMO systems.

\bibliographystyle{IEEEtran}
\bibliography{refs}

\end{document}


TITLE: Two question on ternary Cantor set & Jordan content
QUESTION [0 upvotes]: Is it true that all subsets of the Cantor set have Jordan content zero?

What is the definition of countably generated Boolean algebra?

Does the Boolean algebra of subsets $[0,1]$ which possess Jordan content is countably generated?

Sorry I updated the question.

REPLY [3 votes]: The middle-thirds Cantor set has Jordan measure zero -- it is the intersection of a sequence of finite unions of intervals whose total length converge to zero.
Therefore every subset of it also has Jordan measure zero, since you can find a simple superset of it with arbitrary small content.

If the Boolean algebra of Jordan measurable sets were countably generated, it would itself be countable because then by definition every set in the algebra is generated from some finite sequence of joins, meets and complements from the generator elements. And these sequences of operations can be coded as integers using routine constructions.
However, every interval $[0,\alpha]$ is Jordan measurable, and there are uncountably many such intervals (namely one for each $\alpha\in [0,1]$). Therefore the algebra cannot be countably generated.


TITLE: How can one properly understand the fact that $e^x$ can be differentiated an infinite amount of times?
QUESTION [0 upvotes]: Simply put if I follow the rule derived by the simple proof denoting $e^x$ to be the derivative of $e^x$ then it follows that it should have an infinite number of derivatives. Is this a conceptual mistake ( on my part) or one deeper within the proof ?

REPLY [0 votes]: There is no mistake, with your remark, a trivial induction gives you the existence and value of the n-th derivative for every n, and this imply that it's $C^\infty$ – Tryss [answered in comments]


TITLE: Nontrivial finite group with trivial cohomology in prescribed degree
QUESTION [10 upvotes]: For any non-trivial finite group $G$ there exists some $j > 0$ such that $H^{aj}(G) \neq 0$ for all $a = 1,2,3,\dots$, see e.g. this question: Non-vanishing of group cohomology in sufficiently high degree.
Furthermore, it is not known whether there exists a positive $N$ such that $H^i(G) \neq 0$ for $0 < i \leq N$ implies $G = 1$, see this question Nontrivial finite group with trivial group homologies?.
My question is: Given a positive integer $i$, does there always exist a non-trivial finite group $G$ with $H^i(G) = 0$? (All cohomology groups are meant to be with $\mathbb Z$ coefficients.)

REPLY [4 votes]: The qualifier "finite" is crucial, because you can always construct a general group with prescribed (co)homologies.
For $i$ odd, take any cyclic group $\mathbb{Z}_n$. In fact, its cohomology is trivial in all odd degrees and is isomorphic to the cyclic group in all even degrees. Or take $S_3$. It's a fun exercise to prove a more general result: $H^\text{odd}(G)=0$ if $G$ has periodic cohomology ($\mathbb{Z}_n$ has period 2 and $S_3$ has period 4). In particular, odd cohomology vanishes for any $p$-group $G$ which has a unique $\mathbb{Z}_p$ subgroup (it has period $2|G:\mathbb{Z}_p|$). But the exact opposite occurs in even degrees:
For $i$ even, $G$ cannot be a $p$-group. In fact, its cohomology is nontrivial in all even degrees (can explain why later). I currently cannot say more except for the following: The $p$-primary component of $H^{2k}(G)$ is isomorphic to the set of $G$-invariant elements of $H^{2k}(P)$, where $P$ is a Sylow $p$-subgroup. This makes our attempt more difficult. I will think more.
I briefly thought more: The binary icosahedral group $SL_2(\mathbb{F}_5)$ has trivial cohomology in degrees 2 mod 4. That leaves the case open for $i\in 4\mathbb{Z}$.


TITLE: Function with same derivative
QUESTION [0 upvotes]: If the function $y= e^x$ is such that its derivative is itself $dy/dx = e^x$,
does that mean the second derivative $d^2y/dx^2= e^x$?

REPLY [0 votes]: Not only is the second derivative, but all of them are! We can actually prove it as follows:
We use as a starting point that the derivative of $f(x) = e^x$,
$${df \over dx} = f$$
that is,
$${d \over dx}(e^x) = e^x$$
Now let $k$ be the smallest natural number such that
$${d^k\over dx^k}(e^x) \neq e^x$$
Now let us consider the derivative
$${d^{k-1}\over dx^{k-1}}(e^x)$$
Because $k$ is the smallest natural number for which the derivative of $e^x$ is not $e^x$, the $(k-1)$th derivative of $e^x$ is still $e^x$:
$${d^{k-1}\over dx^{k-1}}(e^x) = e^x;\\
{d^k\over dx^k}(e^x) = {d \over dx}\left({d^{k-1}\over dx^{k-1}}(e^x)\right) = {d\over dx}(e^x) = e^x$$
and therefore the $k$th derivative of $e^x$ cannot be different from $e^x$, if we know that ${d\over dx}(e^x) = e^x$.


TITLE: Is the ring and semi-ring definition of algebra and set linked?
QUESTION [21 upvotes]: "Ring" and "semiring" are concepts defined both in algebra and set theory.
In Algebra
A ring in algebra is a set R equipped with two binary operations + and · called addition and multiplication, that Addition (+) is abelian, Multiplication (⋅) is associative, Multiplication distributes over addition, and Multiplicative identity (1) exists.
A semiring in abstract algebra,  is an algebraic structure similar to a ring, but without the requirement that each element must have an additive inverse.
In Set theory
A ring of sets in measure theory is a family of sets closed under unions and set-theoretic differences.  That is, it obeys the two properties
$$A \setminus B \in \mathcal{R} $$
$$A \cup B \in \mathcal{R}$$
This implies that it is also closed under intersections,
$$A \cap B \in \mathcal{R}$$
A semiring of sets is a non-empty collection S of sets such that

$\emptyset \in S$
If $E \in S$ and $F \in S$ then $E \cap F \in S$.
If $E \in S$ and $F \in S$ then there exists a finite number of
mutually disjoint sets $C_i \in S$ for $i=1,\ldots,n$ such that $E
   \setminus F = \bigcup_{i=1}^n C_i$.

I wonder, are such definitions somehow linked or equivalent? or they are just coincidence that both fields of mathematics used the same terms?

REPLY [4 votes]: This rather long answer is meant to expand on @bonsoon's good answer by presenting the deeper reason why set-theoretic symmetric difference and intersection are precisely the operations we need to prove that measure-theoretic rings of subsets are also rings in the algebraic sense.
(From now on, I use the denomination "algebraic ring" to refer to the ring structure as defined in abstract algebra, and "measure-theoretic ring" to refer to the measure-theoretic definition of "ring of subsets of a set".)

Theorem. Let $A \neq \varnothing$ be a set. A family $\mathcal R$ of subsets of $A$ may form an algebraic ring of subsets of $A$ if and only if it is a measure-theoretic ring of subsets of $A$, i.e. it is such that

$\forall X,Y \in \mathcal R \qquad X \bigtriangleup Y \in \mathcal R$
$\forall X,Y \in \mathcal R \qquad X \cap Y \in \mathcal R$

Remark. I still say "may form" because, in principle, we do not know the set-theoretic operations for which the family $\mathcal R$ may be considered an algebraic ring. The appropriate operations will be clear from the following construction.

Preamble. Consider the set $$\mathbb Z_2 = \{0,1\} $$
If $\mathbb Z_2$ is equipped with the binary operations
\begin{equation}
\begin{split}
+ : \mathbb Z_2 \times \mathbb Z_2 \to \mathbb Z_2 \qquad &(x,y) \mapsto x+y = (x\ \hat +\ y) \mod 2 \\
\cdot : \mathbb Z_2 \times \mathbb Z_2 \to \mathbb Z_2 \qquad &(x,y) \mapsto x\cdot y = (x\ \hat \cdot\ y) \mod 2
\end{split}
\end{equation}
where $\hat +$ and $\hat \cdot$ denote the usual addition and multiplication in the integers, then $(\mathbb Z_2,+,\cdot) $ consitutes a (commutative) algebraic ring.
Consider then the set of all maps from the non-empty set $A$ to $\mathbb Z_2$,
$$
\mathbb Z_2^A = \{f : A \to \mathbb Z_2\}
$$
The generic map $f \in \mathbb Z_2^A$ will send some elements of $A$ to $1 \in \mathbb Z_2$, and the rest to $0 \in \mathbb Z_2$. Call $X\subseteq A$ the set of all values that are mapped to $1$ and denote the map with $\mathbf 1_X$. Consider a similar map $\mathbf 1_Y$ defined analogously from a second subset $Y \subseteq A$. (These maps are called characteristic functions of sets $X$ and $Y$.) We want to induce the above-defined binary operations over $\mathbb Z_2$ upon the set $\mathbb Z_2^A$ in the following way: for all $t \in A$, set
\begin{equation}
\begin{split}
\oplus : \mathbb Z_2^A \times \mathbb Z_2^A \to \mathbb Z_2^A \qquad &(\mathbf 1_X \oplus \mathbf 1_Y)(t) = \mathbf 1_X(t) +\mathbf 1_Y(t) \quad \forall t \in A \\
\odot : \mathbb Z_2^A \times \mathbb Z_2^A \to \mathbb Z_2^A \qquad &(\mathbf 1_X \odot \mathbf 1_Y)(t) = \mathbf 1_X(t)\cdot\mathbf 1_Y(t) \quad \forall t \in A
\end{split}
\end{equation}
This way, $(\mathbb Z_2^A,\oplus,\odot)$ too becomes a (commutative) algebraic ring, where the identity element for $\oplus$ is $\mathbf 1_\varnothing$ (the identically zero map on $A$), and the identity element for $\odot$ is $\mathbf 1_A$.
Relations with set-theoretic operations. Consider two arbitrary subsets $X,Y$ of $A$. We see trivially that
$$
(\mathbf 1_X \oplus \mathbf 1_Y)(t) = \begin{cases}
0 + 0 = 0 & \forall t \in A \setminus (X \cup Y)\\
1 + 0 = 1 & \forall t \in X \setminus Y \\
0 + 1 = 1 & \forall t \in Y \setminus X \\
1 + 1 = 0 & \forall t \in X \cup Y \\
\end{cases}
$$
and that
$$
(\mathbf 1_X \oplus \mathbf 1_Y)(t) = \begin{cases}
0 \cdot 0 = 0 & \forall t \in A \setminus (X \cup Y)\\
1 \cdot 0 = 0 & \forall t \in X \setminus Y \\
0 \cdot 1 = 0 & \forall t \in Y \setminus X \\
1 \cdot 1 = 1 & \forall t \in X \cup Y \\
\end{cases}
$$
Therefore, simply by applying the rules for addition and multiplication in $\mathbb Z_2$, one can check that
$$
(\mathbf 1_X \oplus \mathbf 1_Y)(t) = \mathbf 1_X(t) +\mathbf 1_Y(t) = \mathbf 1_{X \bigtriangleup Y}(t), \qquad \forall t \in A
$$
and that
$$
(\mathbf 1_X \odot \mathbf 1_Y)(t) = \mathbf 1_X(t)\cdot\mathbf 1_Y(t) = \mathbf 1_{X\cap Y}(t), \qquad \forall t \in A
$$
The natural isomorphism. We can now identify $\mathbb Z_2^A$ with $\mathcal P(A)$ through the natural isomorphism $\mathfrak I : \mathbf 1_X \mapsto X$, concluding that $(\mathcal P(A),\bigtriangleup,\cap)$ must be an algebraic ring of subsets of $A$, where the additive operation coincides with set-theoretic symmetric difference and has identity $\varnothing$, and the multiplicative operation coincides with set-theoretic intersection and has identity $A$.
Subrings. The existence of the ring isomorphism $\mathfrak I$ allows us to say that each subring of $(\mathbb Z_2^A, \oplus, \odot)$ is isomorphic to a subring of $(\mathcal P(A), \bigtriangleup, \cap)$, which is then an algebraic ring of subsets of $A$ with respect to $\bigtriangleup$ and $\cap$. Thus we recognize these two operations to be the ones we need to complete our proof.
Characterization of algebraic rings of subsets. We may describe the nature of algebraic rings of subsets of $A$ by studying that of subrings of $\mathbb Z_2^A$. Let $R \subseteq \mathbb Z_2^A$ be non-empty and $(R,\oplus,\odot)$ be a subring of $(\mathbb Z_2^A,\oplus,\odot)$. By the subring criterion (proof here), this is equivalent to saying that $R$ is closed under $\odot$, it contains $\mathbf 1_\varnothing$ and the $\oplus$-inverses of each of its elements. The last requirement means that for all $\mathbf 1_X \in R$, there exists $\mathbf 1_Y \in R$ such that $\mathbf 1_X \oplus \mathbf 1_Y = \mathbf 1_\varnothing$. If $\mathbf 1_X = \mathbf 1_\varnothing$ this surely holds; if $\mathbf 1_X \neq \mathbf 1_\varnothing$, we have
$$
(\mathbf 1_X \oplus \mathbf 1_Y)(t) = \mathbf 1_X(t) + \mathbf 1_Y(t) = \mathbf 1_\varnothing(t) = 0 \quad \forall t \in A
$$
This tells us $\mathbf 1_Y(t)$ should be the additive inverse of $\mathbf 1_X(t)$ in $\mathbb Z_2$ for all $t$ in $A$. By the fact that the additive inverse of all (both) $x \in \mathbb Z_2$ is $x$ itself, it follows that $\mathbf 1_Y(t) = \mathbf 1_X(t)$ for all $t$ in $A$, and so $\mathbf 1_Y = \mathbf 1_X$. This means that any subset of $\mathbb Z_2^A$ may constitute a subring of $(\mathbb Z_2^A, \oplus,\odot)$ as long as it is closed under $\odot$ and it contains the element $\mathbf 1_\varnothing$! When we apply our ring isomorphism $\mathfrak I$ to $R$, we find a family $\mathcal R$ of subsets of $A$ that is closed under $\cap$ and contains $\varnothing$: these conditions are sufficient for it to be an algebraic ring of subsets of $A$.

We may now proceed to the actual proof.
Proof. Let us prove the leftward implication. Let $\mathcal R \subseteq \mathcal P(A)$ be a measure-theoretic ring of subsets of $A$. $\mathcal R$ forms an algebraic ring for the operations $\bigtriangleup$ and $\cap$ if and only if it satisfies the two conditions stated above, viz. it is closed under $\cap$, and it contains $\varnothing$. This is in fact the case: by hypothesis, we see that $\mathcal R$ is closed under intersection; moreover, for any $X \in \mathcal R$, closure under $\bigtriangleup$ provides $X \bigtriangleup X \in \mathcal R$, that is, $\varnothing \in \mathcal R$. Hence we have our thesis.
The rightward implication is trivial: if $(\mathcal R,\bigtriangleup,\cap)$ is an algebraic ring of subsets of $A$, then $\mathcal R$ is closed under both set-theoretic symmetric difference and intersection; therefore it is a measure-theoretic ring of subsets of $A$. $\square$
Remark. Notice that if $\mathcal R$ is a measure-theoretic ring of subsets of a non-empty set $A$, and $A \in \mathcal R$, then $\mathcal R$ is called an algebra of sets. If we look at $\mathcal R$ as an algebraic ring, precisely as a subring of $\mathcal P(A)$, this corresponds to strengthening the condition so that the multiplicative identity of $\mathcal R$ corresponds to that of $\mathcal P(A)$. Some authors actually do require that the multiplicative identity of a subring of a ring be the same as that of the ring in the very definition of subring. If one adopts such a definition of subring (since we have not required here that $A\in \mathcal R$ in general) then it is measure-theoretic algebras, not rings, that have an algebraic ring structure.
Example. Define an $n$-rectangle in $\mathbb R^n$ to be the Cartesian product of $n$ intervals, and call it unbounded if at least one of those intervals is unbounded, bounded otherwise. Define an $n$-multirectangle to be the union of a finite number of $n$-rectangles, and call it unbounded if at least one of the $n$-rectangles is unbounded, bounded otherwise (so all $n$-rectangles are also $n$-multirectangles). The set of all bounded $n$-multirectangles in $\mathbb R^n$ constitutes an algebraic ring of subsets of $\mathbb R^n$, but does not contain $\mathbb R^n$ itself. Instead, if we allow the $n$-multirectangles to be unbounded, we retrieve $\mathbb R^n$ as $\mathbb R^n = (-\infty,+\infty)^n$, and our ring becomes an algebra.


TITLE: Proving residue formula with one pole around $0$
QUESTION [1 upvotes]: $f(z)$ is a function which has a simple pole at $z=0$, has finite amount of poles in the upper halfplane (not on the real axis) and for which holds $\lim \limits _{|z| \to \infty, \Im(z) \ge 0}|zf(z)|=0 $. We want to show by choosing a path, that : $$\int_{-\infty}^{\infty} f(z)dz = 2\pi i( \sum \operatorname{Res} z_{i}) + (\operatorname{Res} 0)\pi i .$$

Planned was to cut out spectacles, that means cut out two arcs with lower arc from $(-\epsilon, \epsilon)$ and upper arc of $(-R,R)$. Then we can pick $\epsilon$ so that $0$ doesn't lie in it but all the other polish points do. So from all those points we get : $2\pi i (\sum \operatorname{Res} z_{i})$.
Then for the $0$ I don't see how to continue.
 Does anybody see a way ? Please do tell.

REPLY [3 votes]: Let $c$ be the residue at $0$. Your main contour going around the origin in an $\epsilon$-arc introduces an error, which you can correct for by adding another integral which goes backwards around the $\epsilon$-arc and then straight through the pole along the real axis.
To evaluate this auxiliary integral, write $f(z) = c\frac{1}{z} + g(z)$. Now $g$ has a removable singularlty at $0$ and no singularites inside the auxiliary contour, so its integral is $0$. You can now handle the auxiliary integral of $\frac{c}{z}$ by symmetry -- $[-\epsilon,0)$ and $(0,\epsilon]$ cancel out, and the arc is just half of an arc all the way around the pole.


TITLE: Why some alpha particle tracks shows curvature *without* applied magnetic field?
QUESTION [6 upvotes]: For the past few months I have been observing alpha tracks with my homemade expansion cloud chamber assembly, trying to replicate early achievements in particle physics using modern, low-cost materials. I've seen some interesting stuffs but I simply couldn't interpret the following phenomena.

These photos have similar scale, about 5 cm in height. Now I know these Y-shaped tracks are due to alphas "colliding" with nucleus of air atoms, but what intrigues me is the tiny but noticeable curvature (pointed by red arrows) of the recoil nucleus.
It is known that due to the mass of alpha particles, it takes incredibly high magnetic field strength to "bend" its path. Recoil nuclei (in this case most likely nitrogen or oxygen) are even more massive, so surely it shouldn't show curvature.
Nobel laureate P.M.S. Blackett published this finding in a 1923 paper (https://royalsocietypublishing.org/doi/abs/10.1098/rspa.1923.0041) *note plate caption number 3. He was working on it then.
I couldn't find a follow-up on that. Was it experimental error, instrumental artefact, or something fundamental?

REPLY [2 votes]: The straight lines are visible because of continuous small angle scatters of the alpha particle with the electrons of the medium .
The original forked scatter is on a nucleus, the apparent curve comes from the scattering of  the low energy  proton on another nucleus.  If the consecutive scatter at the end of a low energy proton is in a different direction , the track struggles, but the thickness of the imaging hides it.
In the link you give there are bends in the other direction also  .


TITLE: Is there a name for this matrix operation?
QUESTION [2 upvotes]: Transforming a matrix by copying each element up to a certain given length ($k$) and then starting on the next row with the second element, and row after that with the third, etc. So each row is shifted by one more. For example:
$$\begin{bmatrix}1\\2\\3\\4\\5\end{bmatrix}
\rightarrow
\begin{bmatrix}1 & 2 & 3\\2 & 3 & 4\\3 & 4 & 5\end{bmatrix}
$$
With a parameter $k=3$ or
$$\begin{bmatrix}1\\2\\3\\4\\5\end{bmatrix}
\rightarrow
\begin{bmatrix}1 & 2 & 3 & 4\\2 & 3 & 4 & 5\end{bmatrix}
$$
With a parameter $k=4$.

REPLY [1 votes]: The operation looks a little too particular to me to have a (well known) name.
The resulting matrix is like a Toeplitz matrix (except that it's constant along the anti-diagonals), could be regarded as some sort of "toeplitzation" (ugh)...
For example, the second example in Matlab/Octave:
>> fliplr(toeplitz([4,5],[4,3,2,1]))
ans =

   1   2   3   4
   2   3   4   5


TITLE: Stationary probability distribution Markov chain obtained by iterations.
QUESTION [0 upvotes]: I have a question regarding Markov chains obtained by random  indep. iterations.
We should consider a Markov chain $\{X_{n}\}^{\infty}_{n=0}$ obtained by random indep. iterations with the functions $f_{1}(x)=x^{2}+17$, $f_{2}(x)=x^{2}-17$ and $f_{3}(x)=0$ where the functions are chosen with equal prob. in each iteration step.
If we suppose $X_{0}=0$. How would I show that $\{X_{n}\}^{\infty}_{n=0}$ has a unique stationary prob. distribution?
The confusing part for me is the third function. If that function would be left out I would get $X_{n+1}=X_{n}^{2}+\xi_{n+1}$, where $\xi_{n+1}=\pm17$ with prob. $1/2$. But how do I see this process with the third function in mind?

REPLY [1 votes]: The states of the Markov chain are $0$ and every iterated image of $0$ by $f_1$ and $f_2$. With the exception of $f_1(f_1(0))=f_1(f_2(0))=17^2+17$ and of $f_2(f_1(0))=f_2(f_2(0))=17^2-17$, it seems that all these images are different.
Thus, the state space of the Markov chain has roughly the structure of a binary tree rooted at $0$, with the exception that the vertices of the second generation are identified two by two. For every $n\geqslant2$, the $2^{n-1}$ vertices of the $n$th generation represent the $n$th iterates of $f_1$ and $f_2$ applied to $0$.
To describe this "almost tree" more precisely, let us label its vertices as $0$, $1'$, $2'$, $1$, $2$, $11$, $21$, $12$, $22$, and so on, thus $0$ points to $1'$ and $2'$, $1'$ and $2'$ point to $1$ and $2$, and, for every nonempty $\{1,2\}$-word $x$, the vertex $x$ points to $1x$ and $2x$.
By symmetry, the stationary distribution of the Markov chain puts the same weight, say $w_n$, on every vertex of the $n$th generation. Recall that the sizes of the successive generation, starting at and including the root, are $1$, $2$, $2$, $4$, $8$, $16$, ..., thus, the usual one-step stationarity yields $$w_0=\frac13\left(w_0+2w_1+\sum_{n=2}^\infty2^{n-1}w_n\right)=\frac13$$ Likewise, $$w_1=\frac13w_0=\frac19\qquad w_2=\frac13(2w_1)=\frac2{27}$$ and, for every $n\geqslant3$, $$w_n=\frac13w_{n-1}=\frac2{3^{n+1}}$$ As a sanity check, note that the total mass of the stationary distribution is indeed $$w_0+2w_1+\sum_{n=2}^\infty2^{n-1}w_n=\frac13+2\frac19+\sum_{n=2}^\infty2^{n-1}\frac2{3^{n+1}}=1$$