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12 pages
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monomer-dimer, cluster expansion
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%\documentclass[prb,aps,floats,amssymb,preprint,showkeys,showpacs,superscriptaddress]{revtex4}
\documentclass[reqno]{amsart}
\usepackage{graphicx}
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\usepackage{amsmath}
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\usepackage{rotating}
\usepackage{url}
\usepackage{float} 
	\maxdeadcycles=1000
\usepackage{amsthm}
	\newtheorem*{theorem}{Theorem}


\newcommand{\tsum}{\mbox{$\sum$}}

\DeclareMathOperator{\subs}{subs}
\DeclareMathOperator{\floor}{floor}

\begin{document}

%% PG 0 %%
\title[Monomer-Dimer Cluster Expansion]{
	Proof of Convergence for the Lattice Monomer-Dimer Cluster Expansion I, 
	a Simplified Model}

\author[P. Federbush]{P. Federbush\\
Department of Mathematics\\
	University of Michigan \\
	Ann Arbor, MI 48109-1043, USA}


%% PG 1 %%
\begin{abstract}
\renewcommand{\theequation}{\Alph{equation}}
In a set of papers we address the problem of making completely rigorous the development 
of our expression for $\lambda_d(p)$ of the monomer-dimer problem on a $d$-dimensional 
hypercubic lattice
\begin{equation}\label{abstract1}
	\lambda_d(p)=\frac{1}{2}\Big(p\ln(2d)-p\ln(p)-2(1-p)\ln(1-p)-p\Big)
	+\sum_{k=2}a_k(d)p^k
\end{equation}
where $a_k(d)$ is a sum of powers $(1/d)^r$ for
\begin{equation}\label{abstract2}
	k-1\geq r\geq k/2
\end{equation}
In fact as we will point out one has already rigorously established 
the convergence of the sum in \ref{abstract1} for small $p$. It is
the $d$ dependence of $a_k(d)$ that has yet to be rigorously shown. 
The study of these papers establishes the convergence of our cluster 
expansions, resulting in the said $d$ dependence. In this paper we study a cute
little problem, claiming the general case can be dealt with as a 
dressing up of this skeleton problem.
\end{abstract}


\maketitle









%% PG 2 %%
\renewcommand{\theequation}{\arabic{equation}}
\addtocounter{equation}{-2}
\section{The Model Problem}\label{s1}
We start by presenting the model problem. It is interesting in its own right,
and is after all the study of this paper.  $r>0$ is given and real $J_i$ are given,
$i\geq 2$ satisfying
\begin{equation}\label{1}
	|J_i|\leq r^i
\end{equation}
We define
\begin{equation}\label{2}
	Z=Z(N,p)=\sum_{\alpha_i}\prod_i\left((J_ip^iN)^{\alpha_i}\cdot\frac{1}{
	\alpha_i!}\right)
\end{equation}
The $\alpha_i,i\geq 2$ are non-negative integers, and 
the sum over the $a_i$ in \eqref{2} is over all values of the $\alpha_i$,
restricted by 
\begin{equation}\label{3}
	\sum_{i=2}^\infty i\alpha_i\leq \frac{pN}{2}
\end{equation}
We will prove

\begin{theorem}
	There is a $p_0>0$ such that for $0\leq p\leq p_0$
	\begin{equation}\label{4}
		\lim_{N\to\infty}\frac{\ln Z}{N}=\sum_{2}^\infty p^iJ_i
	\end{equation}
\end{theorem}

%% PG 3 %%
The easiest way to prove this theorem (probably) must
be to estimate the corresponding sum to that in \eqref{2} with the
$\alpha_i$ satisfying the complementary inequality
\begin{equation}\label{5}
	\sum_{i=2}^\infty i\alpha_i>\frac{pN}2
\end{equation}
If this part is ``small enough'' then $Z$ becomes
approximately equal
\begin{equation}\label{6}
	Z\cong\prod_i\left(\sum_{\alpha_i=0}^\infty(J_ip^iN)^{\alpha_i}\cdot
	\frac{1}{\alpha_i!}\right)
\end{equation}
and the theorem is easily proved

We follow a much more devious route to the proof, in this paper. 
A route whose steps can all be paralleled in the actual problem.


\section{Introduction}\label{s2}
The cluster expansion approach to $\lambda_d$ of the dimer problem 
on a hypercubic lattice was presented in \cite{1}. A formal argument was 
given for the expansion
\begin{equation}\label{7}
	\lambda_d\sim\frac{1}{2}\ln(2d)-\frac{1}{2}+\sum_ic_i\frac{1}{d^i}
\end{equation}
At present the status of this putative asymptotic 
%% PG 4 %%
expansion is still
not clear. In \cite{4}, working with Friedland, a simple extension of the
expansion formalism of \cite{1} was made to treat the monomer-dimer problem. This 
yielded the expresion
\begin{equation}\label{8}
	\lambda_d(p)=\frac{1}{2}\Big(p\ln(2d)-p\ln p-2(1-p)\ln(1-p)-p\Big)+
	\sum_{k=2}a_k(d)p^k
\end{equation}
where $a_k(d)$ is a sum of powers of $(1/d),(1/d)^r,$ with 
\begin{equation}\label{9}
	k-1\geq r\geq k/2.
\end{equation} 

The situation is a little complicated. The cluster expansion formalism yields
an expression for $a_k(d)$ as a function of the Mayer Series
coefficients of the dimer gas. In \cite{4}, repeated in eq~(12)--(20) 
of \cite{3}, another route was obtained to derive the 
$a_k(d)$ from the Mayer Series coefficients. Although the two routes certainly 
give the same answer, this has not been rigorously established. The
inscrutable algebraic identity that must be proved is detailed in \cite{5}. 
It is child's play to see that the second route leads rigorously
to an expression \eqref{8} for $\lambda_d(p)$ where the sum converges.
%% PG 5 %%
But it is the expression in \eqref{8} derived by the first route (still not
proved rigorously) for which the $a_k(d)$ have the expression in 
powers of $1/d$ described before \eqref{9}. Proof of the convergence of the 
cluster expansion, which we enter into in this paper, 
will show that the two expressions for all $a_k(d)$ are equal, that
the sum in \eqref{8} converges for small $p$, and that the 
$a_k(d)$ have the indicated dependence on $d$.

The limit that must be evaluated to rigorously establish the 
cluster expansion development is presented in Section~\ref{s3}. The relation of the
limit of the model problem, eq~\eqref{4}, to this limit will be clear. In
succeeding sections the theorem of eq~\eqref{4} is proven, taking care 
continuously to carry out steps as they can easily be applied to the general limit 
of Section~\ref{s3}. The basic strategy is to arrange $\ln Z$ as the sum
of terms, chunks. Within the chunks some sums are replaced by 
contour integrals (in many complex variables). A single stationary point
of the integrand dominates in the limit of these integrals.


%% PG 6 %%
\section{The Object of Study}\label{s3}
We must analyze $Z^*$
\begin{equation}\label{10}
	Z^*=\sum_{\alpha_i}\beta(N,\mbox{$\sum$} i\alpha_i)\prod\bar{J}_i^{\alpha_i}
	\frac{N^{\sum\alpha_i}}{\prod(\alpha_i!)}
\end{equation}
This is eq~(5.24) of \cite{2}. Here the $\alpha_i, i\geq2,$ are non-negative 
integers and are restricted by eq~\eqref{3}
\begin{equation}\label{11}
	\beta(N,jN)=e^{NH(p,j)}
\end{equation}
with 
\begin{equation}\label{12}
	\sum i\alpha_i=jN
\end{equation}
and
\begin{align}\label{13}
	H(p,j) & = j\ln p+(1-2j)\ln(1-2j)+j-\frac{p}{2}\left(1-\frac{2j}p\right)\cdot
		\ln\left(1-\frac{2j}p\right) \\ \label{14}
	& \equiv j\ln p+\tilde{H}(p,j)
\end{align}
Eq~\eqref{11} is eq~(5.16) of \cite{2}, eq~\eqref{13} is eq~(5.17) of 
\cite{2}. Eq~\eqref{14} defines $\tilde{H}$. We also define
$\tilde{\beta}$
\begin{equation}\label{15}
	\beta(N,\mbox{$\sum$}i\alpha_i)\equiv p^{\sum i\alpha_i}\tilde{\beta}(N,
	\mbox{$\sum$}i\alpha_i)\equiv p^{\sum i\alpha_i}e^{N\tilde{H}(p,j)}
\end{equation}
$Z^*$ becomes
\begin{equation}\label{16}
	Z^*=\sum_{\alpha_i}\tilde{\beta}(N,\tsum i\alpha_i)\prod (\bar{J}_ip^iN)^{\alpha_i}
	\frac{1}{\prod(\alpha_i!)}
\end{equation}
%% PG 7 %%
Sums are again restricted by \eqref{3}.

One wants to study
\begin{equation}\label{17}
	\lim_{N\to\infty}\frac{\ln Z^*}{N}
\end{equation}
The similarity between eq~\eqref{16} and eq~\eqref{17} and the
pair of equations, eq~\eqref{2} and eq~\eqref{4}, is obvious. The 
$\bar{J}_i$ have a weak dependence on $N$. (They are asymptotically constant.)

The desired limit of \eqref{17} we do not detail now. This limit might be found in
(5.31) and (5.32) of \cite{2}, or in an entirely different form in the discussion
surrounding (24)--(28) in \cite{3}, where another reference is given.





%% PG 8 %%
\section{From Sums to Contour Integrals}\label{s4}
In the next section $Z$ or $Z^*$ from \eqref{2} or
\eqref{16} will be arranged into a sum of terms called chunks. In
some of these chunks there will be a designated set of indices, 
$\mathcal{S}$, such that a portion of the chunk is of the form
\begin{equation}\label{18}
	\prod_{i\in\mathcal{S}}\left(\sum_{i=0}^{m_i}\frac{(J_ip^iN)^{\alpha_i}}
	{\alpha_i!}\right)\beta
\end{equation}
Here if it is $Z$ we are working with $\beta=1$, if it is $Z^*$, $J_i$ becomes
$\bar{J}_i$; from now on such trivial differences will not be 
commented on. In \eqref{18} the other $\alpha_i$ are not summed, having
been assigned certain values and 
$\beta$ may depend on $\alpha_i$. The $J_i$ for $i$ in $\mathcal{S}$ will
be negative, it will be important to control cancellations between positive
and negative terms in evaluating \eqref{18} accurately enough. This motivated
the use of contour integrals. In fact dealing with $Z$ a simpler treatment 
is possible as will be pointed out later, but we want to use
a method that applies to both $Z$ and $Z^*$.

%% PG 9 %%
We set $-a\equiv J_iNp^i$ and note 
\begin{equation}\label{19}
	\sum_{\alpha=0}^n\frac{(-a)^\alpha}{\alpha!}f(\alpha)=
	\frac{1}{2\pi i}\oint_Cdz\frac{\pi}{\sin\pi z}\frac{a^z}{z!}f(z)
\end{equation}
where the contour $C$ is counterclockwise and contains $\{0,1,\dots,n\}$ 
and no other singularities of the integrand. Employing the identity
\eqref{19} in \eqref{18} for all the $\alpha_i$ with $i$ in $\mathcal{S}$
we have converted all the sums in our chunk to a single multivariable 
contour integral. Analytic properties of $\Gamma(z)$ and $\beta$ will be dealt
with later.

The division of $Z$ and $Z^*$ into chunks is to convert sums from having limits as given by 
\eqref{3} to limits as in \eqref{19}
\begin{equation}\label{20}
	0\leq\alpha_i\leq m_i
\end{equation}
In the space of allowed $\alpha_i$ we are fitting hyper-rectangles. This messy procedure
is now illucidated. It is we feel the central idea of the proof. 



%PG 10 %%
\section{Disection into Chunks}\label{s5}
We write this section in the language of $Z^*$ of 
\eqref{16}, changes for $Z$ of \eqref{2} trivial. We first introduce a 
number of parameters. There is $\tilde{M}$ 
\begin{equation}\label{21}
	\tilde{M}=\frac{pN}4
\end{equation}
and $\mu_i$, for $i\geq 2$,
\begin{equation}\label{22}
	\mu_i=\frac{1}{i^{2+\epsilon_1}}
\end{equation}
Each chunk is assigned a ``level'', a non-negative integer, and is either ``free''
or ``boxed''. It requires patience to develop the construction of these chunks.

We let $\mathcal{P}$ be the subset of indices for which
$\bar{J}_i\geq 0$ if $i\in\mathcal{P}$ and $\mathcal{N}$ be the subset
for which $\bar{J}_i<0$. Each chunk has a unique $\alpha_i$ assigned to the
$i\in\mathcal{P}$, say $\alpha_i=t_i$. Thus in a chunk some of the $\alpha_i$
may be summed over, but not the $\alpha_i$ with $i\in\mathcal{P}$. For
any chunk we define
\begin{equation}\label{23}
	R_0=\sum_{i\in\mathcal{P}}it_i
\end{equation}

\subsection*{Level-zero free chunks}
A level zero chunk is uniquely specified by the set of $t_i$, $i\in\mathcal P$. If
%% PG 11 %%
\begin{equation}\label{24}
	R_0\geq \tilde{M}
\end{equation}
then it is a free level-zero chunk. Its precise definition is
\begin{equation}\label{25}
	\prod_{i\in\mathcal P}(\subs(\alpha_i= t_i))\sum_{
	\stackrel{\alpha_i,i\in\mathcal N}{\sum_{i\in\mathcal N}i\alpha_i
	\leq\frac{pN}{2}-R_0}}\tilde{\beta}(N,\sum i\alpha_i)\left(
	\prod_i(\bar{J}_ip^iN)^{\alpha_i}\frac{1}{\prod(\alpha_i!)}\right)
\end{equation}
We are using a Maple-like notation. In \eqref{25} the
$\alpha_i$ for $i\in\mathcal P$ are set equal to $t_i$, and the
remaining $\alpha_i$ are summed over subject to the restriction from
\eqref{3}. As with all the chunks, this chunk is some subsum of the terms in 
\eqref{16}. Different chunks are disjoint, the union of 
all the chunks giving all terms in \eqref{16}.

\subsection*{Level-zero boxed chunks}
Here
\begin{equation}\label{26}
	R_0<\tilde{M}
\end{equation}
We define $C_0$ by
\begin{equation}\label{27}
	C_0=\sup_x\left\{x\mid\sum_{i\in\mathcal N}i\cdot
	\floor(x\mu_i)\leq\frac{pN}{2}-R_0\right\}
\end{equation}
where $\floor(\alpha)$ is the largest integer $\leq \alpha$. We
%% PG 12 %%
then set
\begin{equation}\label{28}
	m_0(i)=\floor(C_0\mu_i)
\end{equation}
The level-zero boxed chunk defined by the $t_i$, $i\in\mathcal P$ is then
\begin{equation}\label{29}
	\prod_{i\in\mathcal P}(\subs(\alpha_i=t_i))\sum_{\alpha_{a_1}=0}^{m_0(a_i)}
	\cdots\sum_{\alpha_{a_s}=0}^{m_0(a_s)}M
\end{equation}
Here $M$ indicates everything after the sums in \eqref{16}. $a_1,\dots,a_s$ are
the indices labelling elements of $\mathcal N$. By using \eqref{27} 
we have
found the biggest box, of a certain shape, we can insert in the
sum. That is we are picking the upper limits in \eqref{29} as large as possible,
with a certain fixed ratio between them

The $R_0\geq\tilde M$ in the free chunk will lead to 
enough smallness in estimates later that
one will not have to study cancellations between
signed terms via the contour integrals of \eqref{19}.
For the boxed chunks one will need to do so.

\subsection*{Level-one chunks}
The level-zero chunks fail to exhaust all terms
%% PG 13 %%
In \eqref{16} because of terms containing some $\alpha_i$ for some 
$i\in \mathcal N$ exceeding the upper limits in \eqref{29}.
We give a subset $\mathcal B_1$ of $\mathcal N$ and to each $i$
in $\mathcal B_1$ we associate a $t_i$ with
\begin{equation}\label{30}
t_i>m_0(i)
\end{equation}
We call the augmented set $\mathcal B_1$ of indices and associated $t_i$,
$\bar{\mathcal B}_1$. We set
\begin{equation}\label{31}
	R_1=\sum_{i\in\mathcal B_1}it_i
\end{equation}
If $R_0+R_1\geq\tilde{M}$ we have the level-one free chunk given as
\begin{equation}\label{32}
	\prod_{i\in\mathcal P\cup\mathcal B_1}(\subs(\alpha_i=t_i))
	\sum_{\stackrel{a_i,i\in\mathcal N_1}{\sum_{i\in\mathcal N_1}i\alpha_i
	\leq\frac{pN}{2}-R_0-R_1}}M
\end{equation}
We have defined $\mathcal N_1=\mathcal N-\mathcal B_1$. To define the boxed chunk
we define
\begin{equation}\label{33}
	C_1=\sup_x\left\{x\mid \sum_{i\in\mathcal N_1}i\floor(x \mu_i)\leq\frac{pN}{2}-
	R_0-R_1\right\}
\end{equation}
and 
\begin{equation}\label{34}
	m_1(i)=\floor(C_1 \mu_i)
\end{equation}
%% PG 14%%
Then the level-one boxed chunk is given as 
\begin{equation}\label{35}
	\prod_{i\in\mathcal P\cup\mathcal B_1}(\subs(\alpha_i=t_i))
	\sum_{\alpha_{a_1}=0}^{m_1(a_1)}\cdots\sum_{\alpha_{a_s}=0}^{m_1(a_s)} M 
\end{equation}
and here the $a_1,\dots,a_s$ label elements of $\mathcal N_1$.

The set of $t_i$ associated to the $i$ in $\mathcal P\cup\mathcal B_1$
uniquely label the level-one chunks.

\subsection*{General level chunks}
We assume we have defined chunks of level-zero
through level-$n$, and we will derive expressions for the 
level-$(n+1)$ chunks. Thus we have
\begin{equation}\label{36}
	\mathcal N=\mathcal N_0\supset\mathcal N_1\supset
	\mathcal N_2\cdots\supset\mathcal N_n
\end{equation}
\begin{equation}\label{37}
	\mathcal B_i\subset \mathcal N_{i-1},\qquad i=1,2,\dots,n
\end{equation}
\begin{equation}\label{38}
	\mathcal N_i=\mathcal N_{i-1}-\mathcal B_i,\qquad i=1,2,\dots,n
\end{equation}
\begin{equation}\label{39}
	\bar{\mathcal B}_i,\qquad i=1,2,\dots,n
\end{equation}
That is, we have an assignment $\alpha_i=t_i$ for 
$i$ in each $\mathcal B_i$. We set
%% PG 15 %%
\begin{equation}\label{40}
	R_k=\sum_{i\in\mathcal B_k}it_i,\qquad k=1,2,\dots,n
\end{equation}
and have $C_0,C_1,\dots, C_n$ with 
\begin{equation}\label{41}
	C_k=\sup_x\left\{x\mid \sum_{i\in\mathcal N_k}i\floor(x\mu_i)\leq
	\frac{pN}{2}-\sum_0^kR_k\right\},\qquad k=0,1,\dots,n
\end{equation}
We set 
\begin{equation}\label{42}
	m_k(i)=\floor(C_k \mu_i),\qquad k=0,1,\dots,n
\end{equation}
and for $i$ in $\mathcal B_k$, $k=2,\dots,n$ one has
\begin{equation}\label{43}
	m_{k-1}(i)<t_i\leq m_{k-2}(i)
\end{equation}
For $k=1$ the analogous condition is given by \eqref{30};
there is in this case no upper bound here imposed. It is not difficult
to see the $C_i$ from a decreasing sequence.

To go to the next level, select the set $\mathcal B_{n+1}\subset \mathcal N_n$
and for $i\in\mathcal B_{n+1}$ require 
\begin{equation}\label{44}
	m_n(i)< t_i\leq m_{n-1}(i)
\end{equation}
The rest follows immediately.

%% PG 16 %%
Let us specify free and boxed chunks of level-$(n+1)$. The free level $n+1$ chunk
arises if
\begin{equation}\label{45}
	\sum_{i=0}^{n+1} R_i\geq \tilde{M}
\end{equation}
and then is given as
\begin{equation}\label{46}
	\prod_{i\in\mathcal P\cup(\mathcal N-\mathcal N_{n+1})}
	(\subs(\alpha_i=t_i))
	\sum_{\stackrel{\alpha_i,i\in\mathcal N_{n+1}}{\sum_{i\in\mathcal N_{n+1}}i\alpha_i
	\leq \frac{pN}{2}-\sum_{i=0}^{n+1}R_i}}M
\end{equation}
If \eqref{45} is not satisfied the boxed chunk is given as
\begin{equation}\label{47}
	\prod_{i\in\mathcal P\cup(\mathcal N-\mathcal N_{n+1})}(\subs(\alpha_i=t_i))
	\sum_{\alpha_{a_1}=0}^{m_{n+1}(a_1)}\cdots\sum_{\alpha_{a_s}=0}^{m_{n+1}(a_s)}M
\end{equation}
where $a_1,\dots,a_s$ are the indices of $\mathcal N_{n+1}$.

%% PG 17 %%
\section{Smallness from high occupation}\label{s6}
In this section we find an upper bound on 
\begin{equation}\label{48}
	\prod_i\left(\frac{(|\bar{J}_i|p^iN)^{\alpha_i}}{\alpha_i!}\right)
\end{equation}
subject to the restriction
\begin{equation}\label{49}
	\sum_ii\alpha_i\geq\tilde{M}
\end{equation}
We use \eqref{3}
\begin{equation}\label{50}
|\bar{J}_i|\leq r^i
\end{equation}
and define
\begin{equation}\label{51}
	Q=\prod_i\left(\frac{((pr)^iN)^{\alpha_i}}{\alpha_i!}\right)
\end{equation}
One then has
\begin{equation}\label{52}
	\ln Q\leq F= N\sum_i[ix_i\ln(pr)-x_i\ln x_i+x_i]
\end{equation}
using the well known inequality $n\ln(n/e)+1\leq \ln n!$,
where we have set
\begin{equation}\label{53}
	\alpha_i=x_iN
\end{equation}
\begin{equation}\label{54}
	\tilde{M}=\tilde{m} N
\end{equation}
Then we have
\begin{equation}\label{55}
	f=\frac{F}{N}=\sum_i[ix_i\ln(pr)-x_i\ln x_i+x_i]
\end{equation}
%% PG 18 %%
We ignore the restriction that $\alpha_i$ must be an integer. We note that 
our upper bound will apply if we restrict the set of
indices in the products of \eqref{48} or \eqref{51}, or
require the $\alpha_i$ to be integers, each of these would lead to a lower upper
bound. 

Now we want to maximize $f$ subject to 
\begin{equation}\label{56}
h=\sum_iix_i=\tilde{m}
\end{equation}
We apply Lagrange multipliers 
\begin{equation}\label{57}
	\frac{d}{dx_i}(f-\lambda h)=0
\end{equation}
getting
\begin{equation}\label{58}
	x_i=(pre^{-\lambda})^i
\end{equation}
and then, from \eqref{56}
\begin{equation}\label{59}
	\sum_{i=2}^\infty{i(pre^{-\lambda})^i=\tilde{m}}
\end{equation}
one gets for $\tilde{m}=cp$ as $p$ goes to zero
\begin{equation}\label{60}
	F\cong-N\tilde{m}\vert\ln(p/\sqrt{\tilde{m}})\vert
\end{equation}

%% PG 19 %%
\section{The Contour Integrals, distorting the contours}\label{s7}
We have displayed the expressions for boxed  chunks of level-zero,
level-one, and general level-$(n+1)$
in eqs. \eqref{29}, \eqref{35}, \eqref{47}. We isolate the portion
of the expression shown in eq.~\eqref{18}, and relabelling indices look
at our present object of study
\begin{equation}\label{61}
	A=\sum_{\alpha_1=0}^{m_1}\cdots\sum_{\alpha_s=0}^{m_s}\prod_{i=1}^s
	\left(\frac{(J_{d_i}p^{d_i}N)^{\alpha_i}}{\alpha_i!}\right)\beta
\end{equation}
Here all the $J_i$ are negative. We set 
\begin{equation}\label{62}
	a_i=-J_{d_i}p^{d_i}N
\end{equation}
and rewrite $A$ as
\begin{equation}\label{63}
	A=\frac{1}{(2\pi i)^s}\oint_{C_1}dz_1\cdots\oint_{C_s}dz_s\prod_{i=1}^s\left(
	\frac{\pi}{\sin\pi z_i}\frac{a_i^{z_i}}{\Gamma(z_i+1)}\right)\beta
\end{equation}
where we take $C_i$ as hugging the interval 
$[-1/2,m_i+1/2]$, encircling it counterclockwise.

We let $g$ be the integrand of \eqref{63}
%% PG 20 %%
so that
\begin{equation}\label{64}
	A=\frac{1}{(2\pi i)^s}\oint_{C_1}dz_1\cdots\oint_{C_s}dz_s g
\end{equation}
and seek a stationary point of $g$, so we solve together
\begin{equation}\label{65}
	\frac{\partial}{\partial z_i} g=0,\qquad i=1,\dots,s
\end{equation}
We now use 
\begin{equation}\label{66}
	\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}
\end{equation}
to write $g$ as
\begin{equation}\label{67}
	g=\prod_{i=1}^s\left(-a_i^{z_i}\Gamma(-z_i)\right)\beta
\end{equation}

We look at the equations \eqref{65} for a stationary point of $Z$,
where then $\beta=1$, getting
\begin{equation}\label{68}
	\frac{\partial}{\partial z_i}(-a_i^{z_i}\Gamma(-z_i))=0
\end{equation}
or taking exponentials and setting $w_i=-z_i$
%% PG 21 %%
\begin{equation}\label{69}
	\frac{d}{dw_i}[-w_i\ln a+\ln\Gamma(w)]=0
\end{equation}
We now use the standard large $w$ approximation
\begin{equation}\label{70}
	\ln n!\cong n\ln n-n
\end{equation}
to get
\begin{equation}\label{71}
	w_i\cong a_i+1
\end{equation}
or 
\begin{equation}\label{72}
	z_i=z_{i0}\cong-a_i-1
\end{equation}
at the stationary point.

In general let the coordinates of the stationary point be $z_i=z_{i0}$.
We then first shift the contour $C_i$ to $C_i'$ where $C_i'$
hugs the interval $[z_{i0},m_i+1/2]$
counterclockwise. One hits no singularities of 
$g$ to distort the contour so. Next we stretch the contour to infinity, plus and minus,
in the imaginary directions, leaving contours $C_i''$. 
%% PG 22 %%
$C_i''$ consists of two line segments parallel to the imaginary axis
\begin{equation}\label{73}
	\begin{gathered}
	\begin{aligned}
	C_i''&=[-\infty i+z_{i0},\infty i+z_{i0}]\\
	&\phantom{==}\mbox{union}\\
	&\phantom{==}[-\infty i+m_i+1/2,\infty i+m_i+1/2]
	\end{aligned}
	\end{gathered}
\end{equation}
Again it will not be hard to show that no singularities
of the integrand are crossed in the distortion, and the segments at $\infty$
to close the $C''_i$ contours contribute nothing.

In treating $Z^*$ and studying 
\begin{equation}\label{74}
	A=\frac{1}{(2\pi i)^s}\oint_{C''_i}dz_1\cdots\oint_{C''_s}dz_sg
\end{equation}
one will have to show the limit in eq~\eqref{17}
picks out just the value of the integrand at the 
center of the first line segment in \eqref{73}.
In studying $Z$ we will use a simpler route to
%% PG 23 %%
study \eqref{61}. But one easily sees that the contour integral method
would yield similar estimates in this case.

%% PG 24 %%
\section{Completion of the Proof for $Z$}\label{s8}
In this section we carry through the proof of the theorem, see eq.~\eqref{4},
for $Z$, much in the way we plan to complete the proof for $Z^*$, see
eq.~\eqref{17}. An important difference is that we do not need use of contour integration.
We divide $Z$ into a sum of three terms
\begin{equation}\label{75}
	Z=Z(N,p)=T_1+T_2+T_3
\end{equation}
$T_1(N,p)$ is the sum of level-zero boxed chunks, $T_2(N,p)$
is the sum of the rest of the boxed chunks, and $T_3(N,p)$ is 
the sum of the free chunks. We will arrive at the theorem by proving, for small
enough $p$, that
\begin{equation}\label{76}
	\lim_{N\to\infty}\frac{\ln T_1}{N}=\sum_2p^iJ_i
\end{equation}
\begin{equation}\label{77}
	\lim_{N\to\infty}\frac{\ln T_2}{N}<\sum_2p^iJ_i
\end{equation}
\begin{equation}\label{78}
	\lim_{N\to\infty}\frac{\ln T_3}{N}<\sum_2p^iJ_1
\end{equation}

%% PG 25 %%
\subsection*{Study of $T_3$}
We may overestimate $T_3$ as follows
\begin{equation}\label{79}
	T_3\leq\sum_{\alpha_i}Q
\end{equation}
where $Q$ is from \eqref{51} and the $\alpha_i$ are restricted by 
\begin{equation}\label{80}
	\sum i\alpha_i\geq \frac{pN}{4}
\end{equation}
Then we write
\begin{equation}\label{81}
	\begin{gathered}
	\begin{aligned}
	\sum_{\alpha_i}Q & =\sum_{\alpha_i}\prod_i\left(\frac{((pr)^iN)^{\alpha_i}}{
	\alpha_i!}\right)\\
	& \leq \sum_{\alpha_i}\prod_i\left(\frac{((pr)^iN)^{\alpha_i/2}}{(\alpha_i/2)!}\right)
	\prod_i\left(\frac{((pr)^iN)^{\alpha_i/2}}{(\alpha_i/2)!}\right)
	\end{aligned}
	\end{gathered}
\end{equation}
all subject to the restriction \eqref{80}. This yields
\begin{equation}\label{82}
	\sum_{\alpha_i}Q\leq AB
\end{equation}
with
\begin{equation}\label{83}
	A=\sup\prod_i\left(\frac{((pr)^iN)^{\alpha_i/2}}{(\alpha_i/2)!}\right)
\end{equation}
and
\begin{equation}\label{84}
B=\sum_{\alpha_i}\prod_i\left(\frac{((pr)^iN)^{\alpha_i/2}}{(\alpha_i/2)!}\right)
\end{equation}
In \eqref{82} and \eqref{83} restriction \eqref{80} is enforced, but getting
a weaker inequality do not
impose it in \eqref{84}.
%% PG 26 %%
For $A$ from \eqref{60} we get 
\begin{equation}\label{85}
	A\cong e^{-N(p/8)|ln(p)/\sqrt{p/8}|}
\end{equation}
Looking at $B$ we note
\begin{equation}\label{86}
\begin{aligned}
	\sum_i\frac{a^{i/2}}{(i/2)!} & =\sum_i\frac{a^i}{i!}+
	\sum_i\frac{a^{1/2+i}}{(1/2+i)!} \\
	& =\sum_i\frac{a^i}{i!}+
	a^{1/2}\sum_i\frac{a^{i}}{(1/2+i)!} \\
	& \leq(1+ca^{1/2})\sum \frac{a^i}{i!}\leq(1+ca^{1/2})e^a
\end{aligned}
\end{equation}
Equation \eqref{78} follows for $p$ small enough with little work.

%% PG 27 %%
\section{Dealing with $Z_1$ and $Z_2$}\label{s9}
In this section we study the methods used to treat $T_1$ and $T_2$ to derive
\eqref{76} and \eqref{77}. We discuss the proof of \eqref{76} alone,
since \eqref{77} requires no new ideas. We write $T_1$, the sum of the level zero
boxed chunks as
\begin{equation}\label{87}
	T_1=\sum_\beta A^\beta B^\beta
\end{equation}
with 
\begin{equation}\label{88}
	A^\beta=\prod_{i\in\mathcal P}\frac{(J_ip^iN)^{t_i}}{t_i!}
\end{equation}
where $t_i=t_i(\beta)$ satisfy
\begin{equation}\label{89}
	R_0=R_0(\beta)=\sum it_i<\frac{pN}{4}
\end{equation}
$C_0(\beta)$ is given by \eqref{27}, and $m_0(i)$ by \eqref{28}, using
which we define
\begin{equation}\label{90}
	B^\beta=\prod_{i\in\mathcal N}\left(\sum_{\alpha_i=0}^{m_0(i)}
	\frac{(J_ip^iN)^{\alpha_i}}{\alpha_i!}\right)
\end{equation}
We now set
\begin{equation}\label{91}
	B_0=\prod_{i\in\mathcal N}(e^{J_ip^iN})
\end{equation}
We write
%% PG 28 %%
\begin{equation}\label{92}
	A^\beta B^\beta=A^\beta B_0+A^\beta E^\beta
\end{equation}
and 
\begin{equation}\label{93}
	\sum A^\beta B^\beta=B_0\sum A^\beta+\sum A^\beta E^\beta
\end{equation}

We study the second term in the right side of \eqref{93} first
\begin{equation}\label{94}
	\left|\sum A^\beta E^\beta\right|\leq \left(\sum A^\beta\right)\sup_\beta
	|E^\beta|
\end{equation}
If we were studying $Z^*$ instead of $Z$ we would have used contour integral techniques,
but in the present case we use simpler methods
\begin{equation}\label{95}
	\sum A^\beta\leq e^{\sum_{i\in\mathcal P}J_ip^iN}
\end{equation}
To study $|E^\beta|$ we first introduce
\begin{equation}\label{96}
	g(a,n)=\sum_{i=n+1}^\infty\left(\frac{|a|^i}{i!}\right)e^{+|a|}
\end{equation}
and 
\begin{equation}\label{97}
	g_i=g(J_ip_i N,m_0(i))
\end{equation}
%% PG 29 %%
In terms of the quantities we have
\begin{equation}\label{98}
	|E^\beta|\leq e^{\sum_{i\in\mathcal N}J_ip^iN}\cdot\left(\prod_{i\in\mathcal N}(1+g_i)-
	1\right)
\end{equation}
The dependence on $\beta$ comes from the dependence of $g_i$ on the $m_0(i)$
which depend on $\beta$.

If $x_i\geq 0$ and $\sum{x_i}\leq 1$ one has the inequality
\begin{equation}\label{99}
	\prod(1+x_i)-1\leq e\sum x_i
\end{equation}
Once we know $\sum g_i\leq 1$ we now have
\begin{equation}\label{100}
	\left|\sum A^\beta E^\beta\right|\leq e^{\sum J_ip^iN}\cdot e
	\cdot \sup_\beta\left(\sum_{i\in\mathcal N}g_i\right)
\end{equation}

Limit \eqref{76} follows from 
\begin{equation}\label{101}
	\lim_{N\to\infty}\frac{1}{N}\ln\left(B_0\sum A^\beta\right)=\sum J_ip^i
\end{equation}
and
\begin{equation}\label{102}
	\lim_{N\to\infty}\frac{\left|\sum A^\beta E^\beta\right|}{B_0\sum A^\beta}=0
\end{equation}
We choose to see \eqref{101} by approximating $\sum A^\beta$ by its largest
term, a justifiable procedure in this case. Note that all terms
in sum are positive
%% PG 30 %%
This is the method used in \cite{1} to treat
\eqref{27}
there, and again in \cite{2} to treat~(5.24) therein.
In both these cases the method is used ``formally''
where it does not really apply since there are positive and negative
terms. It is the work of the present paper and its sequel, part~2, to
show the results in \cite{1} and \cite{2} are none the less correct.

We turn to understanding limit \eqref{102}.
From \eqref{100} we have
\begin{equation}\label{103}
	\frac{\left|\sum A^\beta E^\beta\right|}{B_0\sum A^\beta}\leq
	\frac{e^{\sum J_i p^i N}}{B_0\sum A^\beta}\cdot e\cdot
	\sup_\beta\left(\sum_{i\in\mathcal N}g_i\right)
\end{equation}
From \eqref{101}, now presumed true this becomes
\begin{equation}\label{104}
	\frac{\left|\sum A^\beta E^\beta\right|}{B_0\sum A^\beta}\leq
	e^{N\varepsilon(N)}\cdot e\cdot
	\sup_\beta\left(\sum_{i\in\mathcal N}g_i\right)
\end{equation}
for some $\varepsilon(N)$ that goes to zero with $N$. We may 
deduce an upper bound on $\sum_{i\in\mathcal N}g_i$ as
$\sum_{i\in\mathcal N}\bar g_i$, with $\bar g_i$, $g_i$ computed with
$R_0=\frac{pN}{4}$.

We may get an upper bound for the $\sum_{i\in\mathcal N}\bar g_i$
by looking at
%% PG 31 %%
\begin{equation}\label{105}
	h(r^2p^2N,\gamma pN)
\end{equation}
with 
\begin{equation}\label{106}
	h(a,n)=\frac{a^n/n!}{e^{-a}}
\end{equation}
\begin{equation}\label{107}
	\ln h(r^2p^2N,\gamma pN)\leq -(\gamma pN)\ln
	\left(\frac{\gamma pN}{r^2p^2N}\right)+(\gamma pN)-r^2p^2N
\end{equation}
implying
\begin{equation}\label{108}
	h\leq e^{-\tilde\gamma N}
\end{equation}
with $\tilde\gamma>0$. The first term in the sum for $\bar g_2$ is of the form
$h$, and the sum of all the remaining terms in  $\sum_{i\in\mathcal N}\bar g_i$
are majorized by this term if p is small enough.
This and \eqref{104} yield \eqref{102} and finally
\eqref{76}.











\begin{thebibliography}{00}

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P.~Federbush,
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Physics Letters A, 374 (2009) 131-133.


\bibitem{2}
P.~Federbush and S.~Friedland,
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\bibitem{3}
P.~Butera, P.~Federbush, and M.~Pernici,
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Phys. Rev. E 87, 062113 (2013).


\bibitem{4}
P.~Federbush,
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P.~Federbush,
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\end{thebibliography}




\end{document}




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