text:
\begin{document}

\title[The best constant in the Khintchine inequality]{The best
constant in the Khintchine inequality for slightly dependent
random variables}

\author[O.~Herscovici]{Orli Herscovici}
\address{O.~Herscovici\\Department of Mathematics,
University of Haifa,
3498838  Haifa, Israel}
\email{orli.herscovici@gmail.com}

\author[S.~Spektor]{Susanna Spektor}
\address{S.~Spektor\\Department of Mathematics and Statistics
Sciences, PSB,
Sheridan College Institute of Technology and Advanced Learning,
4180 Duke of York Blvd., Mississauga, ON L5B 0G5}
\email{susanna.spektor@sheridancollege.ca}

\maketitle

\begin{abstract}

We use combinatoric techniques to evaluate the best constant
in the Khintchine inequality  under condition that the sum of the
Rademacher random variables  is fixed. An asymptotic expansion
of the constant, in case when the sample size of the random
variables is growing, is provided.  We also investigated an
asymptotic behaviour of growing samples where the relation
between numbers of positive and negative Rademacher random
variables is fixed.

\medskip

\noindent 2010 Classification: 46B06, 60C05

\noindent Keywords:
Khintchine inequality, dependent Rademacher random variables,
best constant, asymptotic expansion
\end{abstract}

\setcounter{page}{1}

\section{Introduction}

The classical Khintchine inequality states that  for any
$p\in (0, \infty)$ there exists constants $A_p$ and $B_p$, such that
\[
A_p\left(\sum_{i=1}^Na_i^2\right)^{1/2}\leq
\EE\left(\left|\sum_{i=1}^Na_i\varepsilon_i\right|^p\right)^{1/p}
\leq B_p \left(\sum_{i=1}^Na_i^2\right)^{1/2},
\]
for arbitrary $N \in \N$. Here, for $i=1,\ldots, N$, $a_i \in \R$ and
$\{\varepsilon_i\}$ is a sequence of Rademacher random variables,
i.e. mutually independent random variables with distribution
$\PP(\varepsilon_i=1)=\PP(\varepsilon_i=-1)=\dfrac 12$.

The computation of the best possible constants, $A_p$ and $B_p$,
has attracted a lot of interest.   It took work of many
mathematicians to settle all possible cases for the moment $p$.
We refer to \cite{Haagerup,KK, LO, NO, SSz} for the importance of
the inequality and historical accounts. In the past decade the
Khintchine-type inequalities for different kinds of random variables
were investigated, for example: in case of exponential family
\cite{ENT}, for symmetric discrete uniform random variables
\cite{HT}, Steinhaus variables \cite{K}, rotationally invariant random
vectors \cite{KK},  to name a few.

In many problems of Analysis and Probability it is important to
consider random vectors with dependent coordinates, for example,
so-called
log-concave random vectors, which in general have dependent
coordinates, but  whose behaviour is similar to that of Rademacher
random vector or to the Gaussian random vector (see e.g. \cite{G}
and references there in)
 or k-wise independent random variables in \cite{PS}.

Recently, the Khintchine inequality for slightly dependent
Rademacher random variables, established in \cite{SS, SS2}, have
found its application in statistics and data science \cite{KMS}.
However, the moment comparison in those inequalities have been obtained
with the non-optimal constant, $C_p$, (for more details see
Remark \ref{remark2} below):
\begin{align}\label{classic}
\EE_M \left|\sum_{i=1}^{N}a_i\varepsilon_i\right|^p \leq
 C_p^p \, \|a\|_{2},
\end{align}
where $p\geq 2$ and by $\EE_M$ we  denote an expectation with
condition that
\begin{align}\label{2}
\sum_{i=1}^{N}\varepsilon_i=M, \quad -N\leq M\leq N.
\end{align}

\bigskip

In the present paper, we introduce a combinatorial method which
enables us to compute the best constant $C_p$ in \eqref{classic}.
We will stick to the case when $0 \leq M \leq N$ in \eqref{2}. The
case when  $M<0$ can be treated similarly.

\bigskip

Our main result is the following theorem.

\begin{manualtheorem}{3.1}
Let $\varepsilon_i$, \, $1\leq i\leq N$,  be Rademacher random
variables satisfying condition $\sum_{i=1}^N\varepsilon_i=M$,
where $0\leq M\leq N$.
Let $a =(a_1, \ldots, a_N) \in \R^{N}$. Then for any \, $p \in \N$,
\begin{align*}
\EE_M\left(\left|\sum_{i=1}^N\varepsilon_ia_i\right|^{2p}\right)\leq
C_{2p}^{2p}||a||^{2p}_2,
\end{align*}
where
\begin{align}
C_{2p}^{2p}=\frac{2^NN^p}{\binom{N}{\frac{N+M}{2}}
\binom{2p+N-1}{2p}}\sum_{m=0}^p\binom{p-m+\frac{N-M}{2}-1}{p-m}
\binom{2m+M-1}{2m}.\nn
\end{align}
\end{manualtheorem}

\bigskip

Let us note here that condition \eqref{2} with $M=0$ requires even
number of elements. We called it balanced case.
We show that in this case the coefficient
$C_{2p}^{2p}$ has  a simpler form (see Corollary~\ref{balanced}):
\begin{align*}
C_{2p}^{2p}=\left(\frac{N}{2}\right)^{p+1}
\cdot \frac{\sqrt{\pi}\,\Gamma\left(\frac{N}{2}\right)}{\Gamma(p+
\frac{N}{2}+\frac{1}{2})}\cdot\frac{(2p)!}{2^{p} p!}.
\end{align*}
Moreover, we provide its upper and asymptotic bounds (see Proposition~\ref{boundbalanced}):
\begin{align}
C_{2p}^{2p}\leq \frac{2^{N}N^{p}}{(N+1)^{p}}
\cdot\dfrac{\left(\frac N2!\right)^2}{N!} \cdot\frac{(2p)!}{2^pp!}\sim
e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\cdot\frac{(2p)!}{2^pp!}.\nn
\end{align}

The reader might be interested in the behaviour of the best
constant $C_{2p}^{2p}$ for the growing $N$ when the sum of the
Rademacher random variables,  $M$, is fixed. This result is
presented in Section~4.

We have also investigated an asymptotic behaviour of growing
samples  where the ratio between number of positive and
negative Rademacher random variables is fixed, i.e. the sum, $M$,
of those variables is not fixed, but depending on the sample size,
$N$. We are not incorporating those results in the Introduction, for
the reader's simplicity. They are presented in the Section 5 of this
paper.

\begin{remark}\label{remark2}
 The proof of the
restricted Khintchine inequality in \cite{SS} uses a  weak
dependency (\ref{2}), which doubles the variance by comparing
two sets of data. Thus, the constant became of the order
$C_{2p}^{2p}=\frac{(2p)!}{p!}$.  Our current constant, unlike in
\cite{SS}, has an order of
$C_{2p}^{2p}=C(N,M)\frac{(2p)!}{2^p p!}$.
In the case of the Classical
Khintchine Inequality, the constant
$C_{2p}^{2p}=\frac{(2p)!}{2^p p!}$ is of the same order as our
current result  (see for example \cite{Garlin}).

\end{remark}

\textbf{Applications:} The Khintchine inequality provides moment
bounds which can be often adapted to the tail bounds on the tail
probability of a test statistic. As an immediate application of our
results in statistics is that the constant $C_{2p}^{2p}$ would
improve sub-Gaussian bounds on the tail probability of the test
statistic in \cite{KMS} for small  values of $N$.

Another application of our inequality with the best constant is that it
is used instead of a bootstrap estimator procedure. That minimizes
the computation time for A/B testing platforms, for example
\cite{Adam}.

\bigskip

The paper is organized as following. In the next section we provide
the necessary  combinatorial results. In Section~3, we will establish
the best constant of the Khintchine inequality \eqref{classic}.
Section~4 is devoted to the asymptotic expansion of the best
constant in our Khintchine type inequality with fixed sum value
$M$. In Section~5 we show asymptotic bounds of the best
constant in case of proportionally growing samples with
respectively growing sum value $M$.

\section{Combinatorial approach to Rademacher random
variables}

In this section  we will consider a case when the Rademacher
random variables, in the sample of size $N$, satisfy the following
condition
\begin{align}
\sum_{i=1}^N\varepsilon_i=M,\quad\text{with } 0\leq M \leq N.
\label{posM}
\end{align}

\begin{lemma}\label{Pdif}
Let $\varepsilon_i$, $i\leq N$,  be Rademacher random variables
satisfying condition \eqref{posM} and let $p_1+\ldots + p_N=2p$, \,
$ p_i\in \{0, \ldots, 2p\}$. Then,
\begin{align}
\PP_{Dif}&=\PP_{+}-\PP_{-}=\dfrac{\sum_{m=0}^p
\binom{p-m+\ell-1}{p-m}\binom{2m+M-1}{2m}}
{\binom{2p+N-1}{2p}},\nn
\end{align}
where
\begin{align}
\PP_{+}&=\PP\left(\left\{\prod_{i=1}^N\varepsilon_i^{p_i}=1\right\}
\bigcap \left\{\sum_{i=1}^{N}\varepsilon_i=M>0\right\}\right),\nn\\
\PP_{-}&=\PP\left(\left\{\prod_{i=1}^N\varepsilon_i^{p_i}=-1\right\}
\bigcap \left\{\sum_{i=1}^{N}\varepsilon_i=M>0\right\}\right).\nn
\end{align}
\end{lemma}

\begin{proof}

From the condition \eqref{posM} we can immediately conclude that
the number of  positive ($\varepsilon_i=1$) variables is by $M$
greater  than the number of  negative ($\varepsilon_i=-1$)
variables.
Let us suppose that there are $M+\ell$ positive  and $\ell$ negative
variables. Obviously, $M+2\ell=N$, which is equivalent to
$\ell=\frac{N-M}{2}$. Note, that both $N$ and $M$ are either even
or odd.

A renumeration of variables $\varepsilon_i^{p_i}$ gives that
\begin{align*}
\prod_{i=1}^{N}\varepsilon_i^{p_i}&=\prod_{i=1}^{\ell}(-1)^{p_i}
\prod_{j=\ell+1}^{2\ell+M}1^{p_j}=
\left\{
\begin{array}{rl}
1,& p_1+\ldots+p_\ell \text{ is even},\\
-1, & p_1+\ldots+p_\ell \text{ is odd}.
\end{array}\right.
\end{align*}

Denote now by $\TO$ the number of all solutions of
$p_1+\ldots+p_N=2p$ for which the sum of the first $\ell$ integers
$p_i$ is odd and by $\TE$ -- the number of all solutions for which
the sum of the first $\ell$ integers $p_i$ is even.

We can write now, that
\begin{align}
\PP_{Dif}=\frac{\TE-\TO}{T},\label{pdifMpos}
\end{align}
where $T=\binom{2p+N-1}{2p}$ is the number of weak
compositions of $2p$ into $N$ parts.
To find $\TE-\TO$, we divide the sequences summing to $2p$ into
classes and sum over each class separately.
We know that
$2p=p_1+\ldots+p_{\ell}+\ldots+p_{2\ell}+\ldots+p_{2\ell+M}.$
 Therefore for a given sequence $(p_1,\ldots,p_{2\ell+M})$ there
 exist a class
 $c=(c_1, \ldots, c_{\ell}, p_{2\ell+1}, \ldots,p_{2\ell+M})$, where
 $c_j,p_i \in \{0, \ldots, 2p\}$. Obviously, any sequence
 $(p_1,\ldots,p_{2\ell+M})$ belongs to unique class $c$, and all
 sequences in the same class $c$ satisfy $c_j=p_j+p_{\ell+j}$ for
 $j\in\{1,\ldots,\ell\}$. For each such class $c$ we consider the
 difference $(\TE-\TO)_c$. Then,
 \[
 \TE-\TO=\sum_{\textit{over all $c$}}(\TE-\TO)_c.
 \]
We associate the even numbers $p_i$ with $1$ and the odd
numbers $p_i$ with $-1$. We have now,
\begin{align*}
(\TE-\TO)_c&=\sum_{\substack{p_j+p_{2\ell-j+1}=c_j\\ \text{for }
1\leq j\leq\ell}}(-1)^{p_1+\ldots+p_{\ell}}
=\prod_{j=1}^{\ell}\sum_{p_j+p_{2\ell-j+1}=c_j}(-1)^{p_j}.
\end{align*}

For any $c_j>0$ there are $c_j+1$ options for $p_j$. Therefore,
$\sum_{p_j+p_{2\ell-j+1}=c_j}(-1)^{p_j}=(-1)^0+(-1)^1+\cdots+
(-1)^{c_j}$, which means that
\begin{align*}
&\sum_{p_j+p_{2\ell-j+1}=c_j}(-1)^{p_j}=
\begin{cases}
$1$, \quad \text{if $c_j$ is even,  $ j=1, \ldots, \ell$};\\
$0$,  \quad \text{if l $c_j$ is odd $ j=1, \ldots, \ell$}.
\end{cases}
\end{align*}
So, we have that
\begin{align*}
&(\TE-\TO)_c=
\begin{cases}
$1$, \quad \text{if all $c_j$  are even,  $ j=1, \ldots, \ell$};\\
$0$,  \quad \text{if not  all $c_j$  are even\, $\, j=1, \ldots, \ell$}.
\end{cases}
\end{align*}
Note, the classes in which not all of $c_j$ are even would not
change the number $\TE-\TO$. Therefore, for given class $c$ the
number $(\TE-\TO)_c$ would be equal to the number of sequences
$c=(c_1, \ldots, c_{\ell},p_{2\ell+1},\ldots,p_{2\ell+M})$, where all
$c_j=2z_j$ are even. This is the number of all possible ways of
choosing $z_j\in\{0,\ldots,p\}$, such that
\begin{align}\label{eq1}
2p=2z_1+\ldots +2z_{\ell}+p_{2\ell+1}+\ldots+p_{2\ell+M}.
\end{align}
Suppose that
\begin{align}\label{eq2}
p_{2\ell+1}+\ldots+p_{2\ell+M}=2m.
\end{align}
Therefore, we have to solve the equation
\begin{align}\label{eq3}
2z_1+\ldots +2z_{\ell}=2p-2m,
\end{align}
and the number of its solutions equals $\binom{p-m+\ell-1}{p-m}$.
The same will happen for any other configuration of
$p_{2\ell+1},\ldots,p_{2\ell+M}$ with
$p_{2\ell+1}+\ldots+p_{2\ell+M}=2m$. The number of such
configurations equals the number of weak compositions of $2m$
into $M$ parts given by $\binom{2m+M-1}{2m}$. Thus, we obtain
\begin{align*}
(\TE-\TO)\Big|_{\substack{c=(2z_1, \ldots, 2z_{\ell},p_{2\ell+1},
\ldots,p_{2\ell+M})\\ 2z_1+\ldots +2z_{\ell}=2p-2m \\ p_{2\ell+1}+
\ldots+p_{2\ell+M}=2m}}=\binom{p-m+\ell-1}{p-m}
\binom{2m+M-1}{2m}.
\end{align*}
Finally, summing over $m$, where $0\leq m\leq p$, we get
\begin{align}
\label{TeToposM}
\TE-\TO&=\sum_{m=0}^p\binom{p-m+\ell-1}{p-m}
\binom{2m+M-1}{2m}.
\end{align}

By using the fact that $\ell=\frac{N-M}{2}$ and substituting
\eqref{TeToposM} into \eqref{pdifMpos}, we obtain the Lemma's
statement.
\end{proof}

\begin{remark}
Note, the case when $-N \leq M < 0$ can be calculated similarly.
\end{remark}

\begin{corollary}\label{remark3}
If Rademacher random variables satisfy condition that \,
$\sum_{i=1}^N\varepsilon_i=N$, then
\begin{align}
\PP_{Dif}=1.\nn
\end{align}
\end{corollary}
\begin{proof}
In this case $\ell=0$, which means that instead of equations
\eqref{eq1}-\eqref{eq3}, we have to solve the equation
$p_1+\ldots+p_N=2p$. Thus, we obtain that
$\TE-\TO=\binom{2p+N-1}{2p}$. Dividing by $T$ as in
\eqref{pdifMpos} completes the proof.
\end{proof}
\begin{corollary}\label{corPdif-bal}
If Rademacher random variables satisfy \,
$\sum_{i=1}^N\varepsilon_i=0$, then
\begin{align*}
\PP_{Dif}=\frac{\binom{p+\frac{N}{2}-1}{p}}{\binom{2p+N-1}{2p}}.
\end{align*}
\end{corollary}
\begin{proof}
In this case $\ell=\frac{N}{2}$. Thus, instead of equations
\eqref{eq1}-\eqref{eq3}, we have to solve
$2p=2z_1+\ldots +2z_{\frac{N}{2}}$. We obtain that
$\TE-\TO=\binom{p+\frac{N}{2}-1}{p}$. Division by $T$, as in
\eqref{pdifMpos}, completes the proof.
\end{proof}

\begin{lemma}\label{Product}
Let $\varepsilon_i$, \,$i\leq N$,  be Rademacher random variables
satisfying condition \eqref{posM} and let $p_1+\ldots + p_N=2p$, \,
$p_i\in \{0, \ldots, 2p\}$.
Denote by  $\EE_M$  an expectation with condition \eqref{posM}.
Then,
\begin{align}
\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)=
\frac{{2^N}}{\binom{N}{\frac{N+M}{2}}\binom{2p+N-1}{2p}}
\sum_{m=0}^p\binom{p-m+\frac{N-M}{2}-1}{p-m}
\binom{2m+M-1}{2m}.
\end{align}
\end{lemma}

\begin{proof}

Denote $D=\{i: \varepsilon_i=1\}$ and $D^c=\{i: \varepsilon_i=-1\}$.
Note, the cardinalities $\card(D)=\frac{N+M}{2}$ and
$\card(D^c)=\frac{N-M}{2}$.
We have then
\begin{align*}
\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)&=1\times
\PP_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}=1\right)-1\times
\PP_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}=-1\right)\nn\\
&=\frac{\PP_{Dif}}{\PP\left(\sum_{i=1}^{N}\varepsilon_i=M\right)}.
\end{align*}
The $\PP_{Dif}$ have been calculated in Lemma \ref{Pdif}.
Let us find ${\PP\left(\sum_{i=1}^{N}\varepsilon_i=M\right)}$. The
event
$$
\left\{\sum_{i=1}^{N}\varepsilon_i=M\right\}=\biguplus
\{\varepsilon_i=1, \forall i \in D \quad \& \quad  \varepsilon=-1,
\forall i \in D^c\}.
$$
Thus,
\begin{align}\label{e41}
\PP\left(\sum_{i=1}^{N}\varepsilon_i=M\right)=\frac{1}{2^N}
\binom{N}{\frac{N+M}{2}}.
\end{align}

Combining result of Lemma~\ref{Pdif} and \eqref{e41} completes
the proof.
\end{proof}
\begin{corollary}\label{corE-unbal}
If Rademacher random variables satisfy \,
$\sum_{i=1}^N\varepsilon_i=N$, then
\begin{align}
\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)=2^N.\nn
\end{align}
\end{corollary}
\begin{proof}
It follows from the Corollary~\ref{remark3}.
\end{proof}
\begin{corollary}\label{corE-bal}
If Rademacher random variables satisfy \,
$\sum_{i=1}^N\varepsilon_i=0$, then
\begin{align*}
\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)=
\frac{2^N\binom{p+\frac{N}{2}-1}{p}}{\binom{N}{\frac{N}{2}}
\binom{2p+N-1}{2p}}.
\end{align*}
\end{corollary}
\begin{proof}
It follows from the Corollary~\ref{corPdif-bal}.
\end{proof}

\section{Proof of the Main Theorem}
The main result of this paper is the following.
\begin{theorem}\label{mainTHM}
Let $\varepsilon_i$, \, $i\leq N$,  be Rademacher random variables
satisfying condition \eqref{posM}.
Let $a =(a_1, \ldots, a_N) \in \R^{N}$. Then for any $p \in \N$,
\begin{align*}
\EE_M\left(\left|\sum_{i=1}^N\varepsilon_ia_i\right|^{2p}\right)\leq
C_{2p}^{2p}||a||^{2p}_2,
\end{align*}
where
\begin{align}
C_{2p}^{2p}=\frac{2^NN^p}{\binom{N}{\frac{N+M}{2}}
\binom{2p+N-1}{2p}}\sum_{m=0}^p
\binom{p-m+\frac{N-M}{2}-1}{p-m}\binom{2m+M-1}{2m}.
\label{fullC2p}
\end{align}
\end{theorem}

\begin{proof}
Using multinomial theorem, due to linearity of conditional
expectation, we obtain
\begin{align}\label{multi}
\EE_M\left|\sum_{i=1}^N\varepsilon_ia_i\right|^{2p}
&=\sum_{\substack{p_1+\ldots + p_N=2p\\
                 p_i\in \{0, \ldots, 2p\}}}\dfrac{(2p)!}{p_1!\ldots p_N!}
                 a_1^{p_1}\ldots a_N^{p_N}\EE_M\left(
                 \prod_{i=1}^N\varepsilon_i^{p_i}\right)\nn\\
&=\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)\cdot
(a_1+\ldots+a_N)^{2p}.
\end{align}

The conditional expectation in \eqref{multi} have been computed in
Lemma~\ref{Product}. Therefore, we need to estimate the second
term of the product in \eqref{multi}. For any $a=(a_1,\ldots,a_N)$ it
holds that $a_1+\ldots+a_N\leq|a_1|+\ldots+|a_N|\equiv||a||_1$
and $||a||_2\leq ||a||_1\leq\sqrt{N}||a||_2$. Therefore we obtain
\begin{align*}
(a_1+\ldots+a_N)^{2p}\leq||a||_1^{2p}\leq N^p||a||^{2p}_2,
\end{align*}
and, respectively,
\begin{align*}
\EE_M\left|\sum_{i=1}^N\varepsilon_ia_i\right|^{2p}\leq
\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)\cdot N^p
||a||^{2p}_2.
\end{align*}
It follows that
\begin{align}
C_{2p}^{2p}&=\EE_M\left(\prod_{i=1}^N\varepsilon_i^{p_i}\right)
\cdot N^p.\nn
\end{align}
Substitution the expression for the conditional expectation,
obtained in Lemma~\ref{Product}, completes the proof.
\end{proof}
\begin{corollary}[Balanced Case]\label{balanced}
If \, $\sum_{i=1}^N\varepsilon_i=0$, then
\begin{align}\label{bal}
C_{2p}^{2p}=\left(\frac{N}{2}\right)^{p+1}
\cdot \frac{\sqrt{\pi}\,\Gamma\left(\frac{N}{2}\right)}{\Gamma(p+
\frac{N}{2}+\frac{1}{2})}\cdot\frac{(2p)!}{2^{p} p!}.
\end{align}
\end{corollary}
\begin{proof}
It follows from the Corollary~\ref{corE-bal} that
\begin{align}
C_{2p}^{2p}&=\frac{2^N\binom{p+\frac{N}{2}-1}{p}}{\binom{N}
{\frac{N}{2}}\binom{2p+N-1}{2p}}\cdot N^p\nn\\
&=\frac{2^{N-1}\left(\frac N2\right)!\left(p+\frac N2-1\right)!}
{\left(2p+N-1\right)!}\cdot N^p \cdot\frac{(2p)!}{p!}.\nn
\end{align}
Using the fact that $x!=\Gamma(x+1)=x \Gamma(x)$, we obtain
$(p+\frac{N}{2}-1)!=\Gamma(p+\frac{N}{2})$ and
$(2p+N-1)!=\Gamma(2p+N)$. Applying duplication formula $
\Gamma(2x)=\pi^{-\frac{1}{2}}2^{2x-1}\Gamma(x)\Gamma(x+
\frac{1}{2})$
to $\Gamma(2p+N)$, we obtain \eqref{bal}.

\end{proof}

\begin{corollary}\label{imbalanced}
If \, $\sum_{i=1}^N\varepsilon_i=N$, then
\begin{align}
C_{2p}^{2p}&=2^NN^p.\nn
\end{align}
\end{corollary}
\begin{proof}
It follows immediately from the Corollary~\ref{corE-unbal}.
\end{proof}

\section{Asymptotic expansion of the best constant }
The expression for the best constant $C_{2p}^{2p}$ obtained in
the Theorem~\ref{mainTHM} provides the exact value of
$C_{2p}^{2p}$, but does not show its order or behaviour for
growing $N$.
In this Section we address these issues for cases where the sum
of the Rademacher random variables in a sample is fixed and not
dependent on the sample's size $N$.

First of all, we consider a balanced case, when
$\sum_{i=1}^N\varepsilon_i=0$. In such a case the number of
positive Rademacher random variables in a sample equals to the
number of negative Rademacher random variables.

\subsection{Balanced case, $M=0$.}

\begin{proposition}[Balanced Case: Upper Bound]
\label{boundbalanced}
The upper bound of \eqref{bal} is
\begin{align}
C_{2p}^{2p}\leq \frac{2^{N}N^{p}}{(N+1)^{p}}
\cdot\dfrac{\left(\frac N2!\right)^2}{N!} \cdot\frac{(2p)!}{2^pp!}\sim
e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\cdot\frac{(2p)!}{2^pp!},\nn
\end{align}
as $N\rightarrow\infty$.
\end{proposition}
\begin{proof}

In order to approximate \eqref{bal}, we repeatedly apply formula
$\Gamma(x+1)=x\Gamma(x)$ to obtain:
\begin{align*}
\Gamma\left(\frac N2 +p+\frac 12\right)
&=\prod_{i=0}^{p-1}\left(\frac{N+1}{2}+i\right)\Gamma\left(
\frac{N}{2}+\frac{1}{2}\right).
\end{align*}
It is easy to see that
\begin{align*}
\prod_{i=0}^{p-1}\left(\frac{N+1}{2}+i\right)\geq \left(
\frac{N+1}{2}\right)^p.
\end{align*}

Now, using the fact that
$\Gamma\Big(n+\frac{1}{2}\Big)=\dfrac{(2n)!\sqrt{\pi}}{2^{2n}n!}$,
we get
\begin{align}\nn
C_{2p}^{2p}&\leq \Big(\frac{N}{2}\Big)^{p+1}
\cdot \frac{2^N(\frac{N}{2})!\Gamma\left(\frac{N}{2}\right)}
{\left(\frac{N+1}{2}\right)^pN!}\cdot\frac{(2p)!}{2^pp!}\\\nn
&=\frac{N^{p}}{(N+1)^p}
\cdot\frac{2^N\left(\frac N2!\right)^2}{N!} \cdot\frac{(2p)!}{2^pp!}.
\end{align}

Let us consider now an asymptotic behaviour of the constant
$C_{2p}^{2p}$ as $N\rightarrow\infty$.

Using the fact that
\begin{align*}
\frac{N^p}{(N+1)^p}=\left(1-\frac{1}{N}\right)^{-p}=\left(1-\frac{1}{N}
\right)^{-N\cdot\frac{p}{N}}\sim e^{-\frac{p}{N}}
\end{align*}

 and asymptotic approximation of the central binomial coefficient
 $\binom{2n}{n}\sim\frac{2^{2n}}{\sqrt{\pi n}}$ (see
 \cite{Elezovic2014, Luke1969}), we obtain:
 \begin{align}
C_{2p}^{2p}\sim e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\cdot
\frac{(2p)!}{2^pp!},\nn
\end{align}
which completes the proof.
\end{proof}

It is easy to see from
Corollary~\ref{imbalanced} that  if $M=N$ then
$C_{2p}^{2p}=2^NN^p$ and no estimate needed for this particular
case.

\subsection{Imbalanced Case, $\mathbf{0 < M < N}$.}

In this subsection we consider an  asymptotic behaviour of   the
constant $C_{2p}^{2p}$  when sample of  Rademacher random
variables is growing while its  sum $M$ is fixed.
\begin{theorem}[Asymptotics of growing sample with given $M$]
\label{ratiothm}
Let $\varepsilon_1, \ldots, \varepsilon_N$ be Rademacher random
variables, i.e. such that
$P(\varepsilon_i=1)=P(\varepsilon_i=-1)=1/2$, with condition that
$\sum_{i=1}^N\varepsilon_i=M$, where $M$ is fixed. Then, for any
integer $p\geq 2$
\begin{align}
C_{2p}^{2p}&\sim \sqrt{\frac{\pi (N^2-M^2)}{2N}}\,
\frac{e^{-\frac{Mp}{N}}}{(M-1)!}
\frac{(2p)!}{2^p}
\cdot\sum_{m=0}^p\frac{(2m+M-1)!}{(p-m)!(2m)!}\left(\frac{2}{N-M}
\right)^m,\label{asymp-gen}
\end{align}
when $N\rightarrow\infty$.
\end{theorem}

\begin{proof}
We start from the equation~\eqref{fullC2p} of the
Theorem~\ref{mainTHM}. After simplification on the binomial
coefficients, it can be written as
\begin{align}
C_{2p}^{2p}
&=\frac{{2^NN^p(2p)!\left(\frac{N+M}{2}\right)!\frac{N-M}{2}}}
{N\cdot(2p+N-1)!(M-1)!}\sum_{m=0}^p\frac{(p-m+\frac{N-M}{2}-1)!
(2m+M-1)!}{(p-m)!(2m)!}.\label{eq50}
\end{align}
Note, that for any fixed $p$
\begin{align}
\lim_{N\rightarrow\infty}\frac{(N-1)!N^{2p}}{(2p+N-1)!}=1.
\label{eq51}
\end{align}
By
applying the Stirling's approximation formula for
$N\rightarrow\infty$ with fixed $p$, $M$, and $0\leq m\leq p$,
 we obtain
\begin{align}
N!&\sim\sqrt{2\pi N}\left(\frac{N}{e}\right)^N,\label{eq52}\\
\left(\frac{N+M}{2}\right)!&
\sim\sqrt{\pi (N+M)}\cdot e^\frac{M}{2}\left(\frac{N}{2e}
\right)^{\frac{N}{2}},\label{eq53}\\
(p-m+\frac{N-M}{2}-1)!&\sim\sqrt{\frac{2\pi}{p-m+\frac{N-M}{2}}}
\left(\frac{p-m+\frac{N-M}{2}}{e}\right)^{p-m+\frac{N-M}{2}}\nn\\
&\sim \sqrt{\frac{2\pi}{\frac{N-M}{2}}}\frac{\left(
\frac{N-M}{2}\right)^{p-m+\frac{N-M}{2}}}{e^\frac{N-M}{2}}\nn\\
&\sim \sqrt{\frac{2\pi}{\frac{N-M}{2}}}\frac{N^\frac{N-M}{2}N^pe^{-
\frac{Mp}{N}}}{2^\frac{N-M}{2}e^\frac{N}{2}2^p}\left(\frac{2}{N-M}
\right)^m.\label{eq54}
\end{align}

Combining \eqref{eq51}-\eqref{eq54} with \eqref{eq50}, we get
\begin{align}
C_{2p}^{2p}
&\sim\frac{{2^N(2p)!\frac{N-M}{2}}}{N^p(M-1)!}
\frac{\sqrt{\pi (N+M)}\cdot e^\frac{M}{2}\left(\frac{N}{2e}
\right)^{\frac{N}{2}}}{\sqrt{2\pi N}\left(\frac{N}{e}\right)^N}\nn\\
&\cdot\sum_{m=0}^p\frac{(2m+M-1)!}{(p-m)!(2m)!}
\sqrt{\frac{2\pi}{\frac{N-M}{2}}}\frac{N^\frac{N-M}{2}N^pe^{-
\frac{Mp}{N}}}{2^\frac{N-M}{2}e^\frac{N}{2}2^p}\left(\frac{2}{N-M}
\right)^m.\nn
\end{align}

After simplification of the last expression, we obtain
\begin{align}
C_{2p}^{2p}
&\sim\sqrt{\pi\frac{N^2-M^2}{2N}}
\frac{e^{-\frac{Mp}{N}}}{(M-1)!}\frac{(2p)!}{2^p}\sum_{m=0}^p
\frac{(2m+M-1)!}{(p-m)!(2m)!}\left(\frac{2}{N-M}\right)^m,\nn
\end{align}
which completes the proof.
\end{proof}
\begin{corollary}
Let $\varepsilon_1, \ldots, \varepsilon_N$ be Rademacher random
variables, i.e. such that
$P(\varepsilon_i=1)=P(\varepsilon_i=-1)=1/2$, with condition that
$\sum_{i=1}^N\varepsilon_i=1$. Then, for any integer $p\geq 2$
\begin{align}
C_{2p}^{2p}&\sim e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\,
\frac{(2p)!}{2^pp!}
,\label{asymp-corM1}
\end{align}
when $N\rightarrow\infty$.
\end{corollary}
\begin{proof}
If $M=1$, then, by substituting this value into \eqref{eq50}, we
obtain that, for any fixed $p$ and for $N\rightarrow\infty$,
\begin{align}
C_{2p}^{2p}&\sim e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\,
\frac{(2p)!}{2^p}
\cdot\sum_{m=0}^p\frac{1}{(p-m)!}\left(\frac{2}{N-1}\right)^m,\nn\\
&\leq e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\,
\frac{(2p)!}{2^p}
\cdot\sum_{m=0}^p\frac{p^m}{p!}\left(\frac{2}{N-1}\right)^m,\nn\\
&\sim e^{-\frac{p}{N}}\sqrt{\frac{\pi N}{2}}\,
\frac{(2p)!}{2^pp!},\nn
\end{align}
That completes the proof.
\end{proof}

\section{Asymptotic of proportionally growing sample}

In the previous section we have obtained a constant $C_{2p}^{2p}$
for Khintchine type inequality in the case when sum of samples of
Rademacher random variables of given size $N$ is fixed and is
equal to $M$. We also studied an asymptotic behaviour of this
constant when $N\rightarrow\infty$.

In this section we investigate an asymptotic behaviour of growing
samples where the ratio between number of positive and number
of negative Rademacher random variables is fixed. In this settings
the sum, $M$,  of Rademacher random  variables is not fixed and
depends on the sample size.

\begin{theorem}[Asymptotics of proportionally growing sample]
\label{ratiothm}
Let $\varepsilon_1, \ldots, \varepsilon_N$ be  Rademacher random
variables, i.e.  such that
$P(\varepsilon_i=1)=P(\varepsilon_i=-1)=1/2$, with condition that
$\sum_{i=1}^N\varepsilon_i =M$.
Let a number of negative $\varepsilon_i$ is $n$ and a number of
positive $\varepsilon_i$ is $\alpha n$, for some fixed real
$\alpha>1$.
Then, for any integer $p\geq 2$
\begin{align}
C_{2p}^{2p}&\sim \sqrt{\frac{2\pi n\alpha}{\alpha+1}}\,
\frac{\alpha^{\alpha n}\, 2^{(\alpha+1)n}}{(\alpha+1)^{(\alpha+1)n}}
\frac{(2p)!}{(\alpha+1)^p}
\cdot\sum_{m=0}^p\frac{(\alpha-1)^{2m}n^m}{(p-m)!(2m)!},
\label{asymp1}
\end{align}
when $n\rightarrow\infty$.

\end{theorem}
\begin{proof}
It is easy to see that $N=(\alpha+1)n$ and $M=(\alpha-1)n$.
By substituting these values into equation~\eqref{fullC2p} of the
Theorem~\ref{mainTHM}, we obtain that
\begin{align}
C_{2p}^{2p}&=\frac{2^{(\alpha+1)n}(\alpha+1)^{p-1}n^p(\alpha n)!
(2p)!}{(2p+(\alpha+1)n-1)!((\alpha-1)n-1)!}\sum_{m=0}^p
\frac{(p-m+n-1)!(2m+(\alpha-1)n-1)!}{(p-m)!(2m)!}.\label{eq41}
\end{align}
Note, that for any fixed $p$,
\begin{align}
\lim_{n\rightarrow\infty}\frac{((\alpha+1)n)!((\alpha+1)n)^{2p-1}}
{(2p+(\alpha+1)n-1)!}=1.
\label{eq42}
\end{align}

Now, we evaluate the asymptotic for each  of the factorials
containing $n$ by
applying the Stirling's approximation formula $n!\sim\sqrt{2\pi n}
\left(\frac{n}{e}\right)^n$. It is easy to show that
\begin{align}
\frac{(\alpha n)!}{((\alpha+1)n)!((\alpha-1)n-1)!}&\sim
\sqrt{\frac{n}{2\pi }}\sqrt{\frac{\alpha(\alpha-1)}{\alpha+1}}
\left(\frac{\alpha^\alpha e^\alpha}{(\alpha+1)^{\alpha+1}
(\alpha-1)^{\alpha-1}n^\alpha}\right)^n. \label{eq46}
\end{align}
Let us consider factorials under summation.
\begin{align}
(p-m+n-1)!&\sim\sqrt{\frac{2\pi}{p-m+n}}\left(\frac{p-m+n}{e}
\right)^{p-m+n}\nn\\
&\sim\sqrt{\frac{2\pi}{p-m+n}}\left(\frac{n}{e}\right)^{p-m+n}e^{p-m}
\left(1+\frac{p-m}{n}\right)^{p-m}\nn\\
&=\sqrt{\frac{2\pi}{p-m+n}}\left(\frac{n}{e}\right)^{n}n^{p-m}\left(1+
\frac{p-m}{n}\right)^{p-m}.\label{leftGamma}
\end{align}
Similarly,
\begin{align}
(2m+(\alpha-1)n-1)!\sim\sqrt{\frac{2\pi}{2m+(\alpha-1)n}}
\left(\frac{(\alpha-1)n}{e}\right)^{(\alpha-1)n}((\alpha-1)n)^{2m}
\left(1+\frac{2m}{(\alpha-1)n}\right)^{2m}.\label{rightGamma}
\end{align}

Taking into the account that $p$ is  fixed, with $0\leq m\leq p$, and
that $n\rightarrow\infty$, results \eqref{leftGamma} and
\eqref{rightGamma} can be simplified further as following:
\begin{align}
(p-m+n-1)!&\sim\sqrt{\frac{2\pi}{n}}\left(\frac{n}{e}\right)^{n}n^{p-m},
\label{leftGamma1}\\
(2m+(\alpha-1)n-1)!&\sim\sqrt{\frac{2\pi}{(\alpha-1)n}}
\left(\frac{(\alpha-1)n}{e}\right)^{(\alpha-1)n}((\alpha-1)n)^{2m}.
\label{rightGamma1}
\end{align}
Applying  \eqref{eq42}, \eqref{eq46}, \eqref{leftGamma1} and
\eqref{rightGamma1} to \eqref{eq41}, we obtain
\begin{align}
C^{2p}_{2p}&\sim 2^{(\alpha+1)n}(\alpha+1)^{p-1}n^p (2p)!\nn\\
&\sqrt{\frac{n}{2\pi }}\sqrt{\frac{\alpha(\alpha-1)}{\alpha+1}}
\left(\frac{\alpha^\alpha e^\alpha}{(\alpha+1)^{\alpha+1}
(\alpha-1)^{\alpha-1}n^\alpha}\right)^n\frac{1}{((\alpha+1)n)^{2p-1}}
\nn\\
&\cdot\sum_{m=0}^p\frac{((\alpha-1)n)^{2m}}{(p-m)!(2m)!}\sqrt{\frac{2\pi}{n}}
\left(\frac{n}{e}\right)^{n}n^{p-m}
\cdot\sqrt{\frac{2\pi}{(\alpha-1)n}}\left(\frac{(\alpha-1)n}{e}
\right)^{(\alpha-1)n}\nn.
\end{align}
Simplification of the last expression completes the proof.
\end{proof}

\bigskip

Now, we give an upper bound on asymptotic behaviour of the
$C_{2p}^{2p}$ for proportionally growing sample.
 We can state the following proposition.
\begin{proposition}
Let $\varepsilon_1, \ldots, \varepsilon_N$ be  Rademacher random
variables, i.e.  such that
$P(\varepsilon_i=1)=P(\varepsilon_i=-1)=1/2$, with condition that
$\sum_{i=1}^N\varepsilon_i =M$.
Let a number of negative $\varepsilon_i$ is $n$ and a number of
positive $\varepsilon_i$ is $\alpha n$, for some fixed real
$\alpha>1$. Then for any $p\geq 2$, the coefficient $C_{2p}^{2p}$
has the following upper bound.

\begin{align}
C_{2p}^{2p}\leq \sqrt{\frac{2\pi n\alpha}{\alpha+1}}\,
\frac{\alpha^{\alpha n}\, 2^{(\alpha+1)n}}{(\alpha+1)^{(\alpha+1)n}}
\frac{(\alpha-1)^{2p}n^p}{(p+1)^p}
\frac{(2p)!}{(\alpha+1)^pp!},\nn
\end{align}
where $n\rightarrow\infty$.
\end{proposition}
\begin{proof}
Denote by $\widetilde{C}_{2p}^{2p}$ the right hand side of
\eqref{asymp1}.

It is easy to see that $\frac{1}{(p-m)!(2m)!}=\frac{1}{(p+m)!}
\binom{p+m}{2m}$,
and $\frac{1}{(p+m)!}\leq\frac{1}{p!(p+1)^m}$.

Therefore, we obtain the following upper bound for the constant
$\widetilde{C}_{2p}^{2p}$.
\begin{align}
\widetilde{C}_{2p}^{2p}&\leq\sqrt{\frac{2\pi n\alpha}{\alpha+1}}\,
\frac{\alpha^{\alpha n}\, 2^{(\alpha+1)n}}{(\alpha+1)^{(\alpha+1)n}}
\frac{(2p)!}{(\alpha+1)^p}
\cdot\sum_{m=0}^p\binom{p+m}{2m}\frac{(\alpha-1)^{2m}n^m}{p!
(p+1)^m}\label{eq47}\\
&\sim\sqrt{\frac{2\pi n\alpha}{\alpha+1}}\,
\frac{\alpha^{\alpha n}\, 2^{(\alpha+1)n}}{(\alpha+1)^{(\alpha+1)n}}
\frac{(\alpha-1)^{2p}n^p}{(p+1)^p}
\frac{(2p)!}{(\alpha+1)^pp!},\nn
\end{align}
and the proof is complete.
\end{proof}
\begin{corollary} If $\alpha\rightarrow 1$ and $n\rightarrow\infty$
then
\begin{align}
C_{2p}^{2p}\leq \sqrt{\pi n}\,
\frac{(2p)!}{2^pp!}.\nn
\end{align}
\end{corollary}
\begin{proof}
Applying $\alpha\rightarrow 1$ to the \eqref{eq47}, we obtain
\begin{align}
C_{2p}^{2p}&\leq \lim_{\alpha\rightarrow 1} \sqrt{\frac{2\pi n\alpha}
{\alpha+1}}\,
\frac{\alpha^{\alpha n}\, 2^{(\alpha+1)n}}{(\alpha+1)^{(\alpha+1)n}}
\frac{(2p)!}{(\alpha+1)^p}
\cdot\sum_{m=0}^p\binom{p+m}{2m}\frac{(\alpha-1)^{2m}n^m}{p!
(p+1)^m}\nn\\
&=\sqrt{\pi n}\,
\frac{(2p)!}{2^pp!},\nn
\end{align}
and the proof is complete.
\end{proof}
\begin{remark}
Let us note here that, in case when $\alpha = 1$, the number of
positive and number of negative Rademacher random variables in
a sample are equal and equal to $n=\frac{N}{2}$. Thus, this upper
bound is comparable with the upper bound evaluated in
Proposition~\ref{boundbalanced}.
\end{remark}

We have shown that, in  case when the ratio of positive and
negative Rademacher random variables in a sample is given
explicitly, the
 proportionally growing samples imply that a sum of the  variables
 in the sample is changing as $N$ grows. Now, we will consider
 another aspect of such grow, namely, when the ratio between the
 sum of the Rademacher random variables, $M$, and their sample
 size  ,$N$, is given explicitly.

\begin{proposition}
Let $\varepsilon_1, \ldots, \varepsilon_N$ be  Rademacher random
variables, i.e.  such that
$P(\varepsilon_i=1)=P(\varepsilon_i=-1)=1/2$, with condition that
$\sum_{i=1}^N\varepsilon_i =M$, with
$M=\beta N, \quad 0<\beta<1$.
Then, for any integer $p\geq 2$,
\begin{align}
C_{2p}^{2p}&\sim\sqrt{\frac{\pi N}{2}}(1-\beta^2)^\frac{N+1}{2}
\left(\frac{1+\beta}{1-\beta}\right)^\frac{\beta N}{2}
\frac{(1-\beta)^p(2p)!}{2^p}\sum_{m=0}^p
\frac{2^m\beta^{2m}N^m}{(1-\beta)^m(p-m)!(2m)!}.\nn
\end{align}
when $N\rightarrow\infty$.

\end{proposition}

\begin{proof}
By substituting $M=\beta N$ into equation~\eqref{eq50}, we obtain
that
\begin{align}
C_{2p}^{2p}&=\frac{2^{N}N^p\left(\frac{(1+\beta) N}{2}\right)!(2p)!
\left(\frac{(1-\beta)N}{2}\right)}{N\cdot(2p+N-1)!(\beta N-1)!}
\sum_{m=0}^p\frac{\left(p-m+\frac{(1-\beta)N}{2}-1\right)!
(2m+\beta N-1)!}{(p-m)!(2m)!}.\label{eq61}
\end{align}
Approximation of the factorial $(2p+N-1)!$ can be obtained from
\eqref{eq51}-\eqref{eq52}.
By
applying the Stirling's approximation formula
$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, it is easy to show that
\begin{align}
\frac{\left(\frac{(1+\beta) N}{2}\right)!}{(2p+N-1)!(\beta N-1)!}&\sim
\frac{1}{2\cdot\beta^{\beta N}N^{2p-1}}\sqrt\frac{\beta(1+\beta)N}
{\pi}\left(\frac{(1+\beta)e}{2N}\right)^\frac{(1+\beta)N}{2}.\label{eq66}
\end{align}
Let us consider factorials under summation.
\begin{align}
(p-m+\frac{(1-\beta)N}{2}-1)!&\sim \sqrt{\frac{2\pi}{p-m+\frac{(1-
\beta)N}{2}}}\nn\\
&\cdot\left(\frac{{(1-\beta)N}}{2e}\right)^{\frac{(1-\beta)N}{2}}
\left(\frac{(1-\beta)N}{2}\right)^{p-m}\left(1+\frac{p-m}{\frac{(1-
\beta)N}{2}}\right)^{p-m}.\label{leftGamma6}
\end{align}
Similarly,
\begin{align}
(2m+\beta N-1)!\sim\sqrt{\frac{2\pi}{2m+\beta N}}\left(\frac{\beta N}
{e}\right)^{\beta N}(\beta N)^{2m}\left(1+\frac{2m}{\beta N}
\right)^{2m}.\label{rightGamma6}
\end{align}

Taking into the account that $p$ is  fixed, with $0\leq m\leq p$, and
that $N\rightarrow\infty$, results \eqref{leftGamma6} and
\eqref{rightGamma6} can be simplified further as following:
\begin{align}
(p-m+\frac{(1-\beta)N}{2}&-1)!\nn\\&\sim\sqrt{\frac{4\pi}{N(1-\beta)}}
\left(\frac{(1-\beta)N}{2e}\right)^{\frac{(1-\beta)N}{2}}\left(\frac{(1-
\beta)N}{2}\right)^{p-m},\nn
\\
(2m+\beta N-1)!&\sim\sqrt{\frac{2\pi}{\beta N}}\left(\frac{\beta N}{e}
\right)^{\beta N}(\beta N)^{2m}.\nn
\end{align}
Applying  \eqref{eq51}-\eqref{eq52}, \eqref{eq66},
\eqref{leftGamma6} and \eqref{rightGamma6} to \eqref{eq61}, we
obtain
\begin{align}
C^{2p}_{2p}\sim2^{N-1}&N^p(2p)!(1-\beta)\frac{1}
{2\cdot\beta^{\beta N}N^{2p-1}}\sqrt\frac{\beta(1+\beta)N}{\pi}
\left(\frac{(1+\beta)e}{2N}\right)^\frac{(1+\beta)N}{2}\nn\\
&\cdot\sum_{m=0}^p\frac{(\beta N)^{2m}}{(p-m)!(2m)!}
\sqrt{\frac{4\pi}{N(1-\beta)}}\left(\frac{(1-\beta)N}{2e}\right)^{\frac{(1-
\beta)N}{2}}\nn\\
&\cdot\left(\frac{(1-\beta)N}{2}\right)^{p-m}\sqrt{\frac{2\pi}{\beta N}}
\left(\frac{\beta N}{e}\right)^{\beta N}\nn.
\end{align}

Simplification of the last expression completes the proof.
\end{proof}

\vspace{0.5cm}
\textbf{Acknowledgement}.
We are thankful to A.B.~Kashlak from the University of Alberta for
his valuable suggestions and very useful discussions.

The research of the first author was supported by the Israel
Science Foundation (grant No. 1144/16).

config: arxiv


TITLE: Can I use the power rule to integrate a square root in the denominator?
QUESTION [3 upvotes]: $$\int { \frac { dx }{ \sqrt { 3-4{ x }^{ 2 } }  }  } $$

Can I use the power rule to integrate this integral if first I transform the $ \sqrt{(3-4x^2)}$ to $(3-4x^2)^{-1/2}$$?$ Which techniques can I use?

REPLY [2 votes]: This is a very natural question. The short answer is no.
But there is a better answer to give you, which is that you are capable of checking your antiderivatives. Namely, if you believe that
$$ \int (3 - 4x^2)^{-1/2} dx \stackrel{?}{=} 2(3 - 4x^2)^{1/2},$$
then you can check this by verifying that
$$ \frac{d}{dx} 2 (3 - 4x^2)^{1/2} = (3 - 4x^2)^{-1/2}.$$
(Or if you think the antiderivative is a little bit different, you can check that one instead). Perfoming that check here, you will see that this relationship does not hold.


TITLE: On the automorphisms of the Klein Quartic
QUESTION [14 upvotes]: I am trying to solve a problem from Miranda's book, Algebraic Curves and Riemann Surfaces.  On page 84, problem K gives the Klein curve $X$ as a smooth projective plane curve defined by the equation $xy^3+yz^3+zx^3=0$.  The problem asks us to show that this Riemann surface, of genus g=3, realizes the Hurwitz bound by finding 168 automorphisms of $X$.
I have found an automorphism subgroup of order 3 (cyclically permuting the variables), and one of order 7 (multiplying the coordinates by appropriate 7th roots of unity), but I just can't figure out how to get a subgroup of order 8, or 4, or 2.  Could somebody please give me a hint.
I've spent the last little while reading up on this subject, and most of the discussions involve using a heptagonal tiling etc.. Given where this problem is placed in the book, I can't appeal to that kind of reasoning, so I'm asking for a way to find these automorphisms directly from the defining equation of $X$.
Any help or hint would be appreciated.  Even an involution :)

REPLY [1 votes]: Elkies essay "The Klein Cubic in Number Theory" lists the following symmetry of the Klein cubic:
$$
\frac{1}{\sqrt{-7}}
\begin{bmatrix}
\zeta - \zeta^6 & \zeta^2 - \zeta^5 & \zeta^4 - \zeta^3 \\
\zeta^2 - \zeta^5 & \zeta^4 - \zeta^3 & \zeta - \zeta^6 \\
\zeta^4 - \zeta^3 & \zeta - \zeta^6 & \zeta^2 - \zeta^5  \\
\end{bmatrix}$$
where $\zeta$ is a primitive $7$-th root of unity. Together with the $21$ symmetries you've already found, this generates the full group of $168$.
I'd like to give a conceptual explanation of why this should work, but I don't know if I can; I'll give it some thought.