# Limits of Probability Measures With General Coefficients

Andrew Yao  Massachusetts Institute of Technology, Cambridge, Massachusetts
<EMAIL_ADDRESS>
###### Abstract.

With operators on formal series in $x_{i}$, $1\leq i\leq N$, which are
symmetric in $N-1$ of the $x_{i}$, probability measures can be studied through
Bessel generating functions. These operators are used with the Dunkl transform
on the Bessel generating functions of a sequence of probability measures
$\\{\mu_{N}\\}_{N\geq 1}$ to get a Law of Large Numbers as $N$, the number of
variables, goes to infinity while $\beta$ is constant. Also, formal power
series are used to express the results. However, we consider when the limits
of partial derivatives involving any number of distinct $x_{i}$ of the
logarithms of the Bessel generating functions at $0$ can be nonzero. Then, the
free cumulant of $\\{\mu_{N}\\}_{N\geq 1}$ of order $k$ is a linear
combination of the limits of the order $k$ partial derivatives.

##  1\. Introduction

###  1.1. Probability Measures

This paper studies probability measures using Bessel generating functions.
However, an important approach used is to view the Bessel generating functions
as formal power series. We are given a positive real number $\theta$ where
$\beta=2\theta$. Here, $\beta=1$, $2$, and $4$ correspond to the GUE, GOE, and
GSE, respectively. The multivariate Bessel function
$B_{(a_{1},\ldots,a_{N})}(x_{1},\ldots,x_{N};\theta)$ for $a_{1}<\cdots<a_{N}$
is based on the $\theta$-corners process, also known as the Gaussian
$\beta$-corners process, with fixed top row $(a_{1},\ldots,a_{N})$
([crystalrandommat], [matrix]), see Definition 5.1. For positive integers $N$,
let $\mathcal{M}_{N}$ denote the set of Borel probability measures on ordered
$N$-tuples $(a_{1},\ldots,a_{N})$ in $\mathbb{R}^{N}$ such that
$a_{1}\leq\cdots\leq a_{N}$. With $\theta$, for $\mu\in\mathcal{M}_{N}$, the
Bessel generating function $G_{\theta}(x_{1},\ldots,x_{N};\mu)$ is

$G_{\theta}(x_{1},\ldots,x_{N};\mu)=\int_{a_{1}\leq\cdots\leq
a_{N}}B_{(a_{1},\ldots,a_{N})}(x_{1},\ldots,x_{N};\theta)\mu(da_{1},\ldots,da_{N}),$

see Definition 5.2.

For a sequence $\\{\mu_{N}\\}_{N\geq 1}$ of probability measures such that
$\mu_{N}\in\mathcal{M}_{N}$ for $N\geq 1$, for $k\geq 1$, we let the random
variable $p_{k}^{N}$ be

(1) $p_{k}^{N}=\frac{1}{N}\sum_{i=1}^{N}\left(\frac{a_{i}}{N}\right)^{k},$

where $(a_{1}\leq\cdots\leq a_{N})$ are distributed according to $\mu_{N}$.
Note that in $p_{k}^{N}$, the $a_{i}$ are scaled by $\frac{1}{N}$. With this,
the moments $\\{m_{k}\\}_{k\geq 1}$ of $\\{\mu_{N}\\}_{N\geq 1}$ are given by
$m_{k}=\lim_{N\rightarrow\infty}\mathbb{E}_{\mu_{N}}(p_{k}^{N})$ for $k\geq
1$. In this paper, we look at the Law of Large Numbers considering the
$p_{k}^{N}$ and moments $\\{m_{k}\\}_{k\geq 1}$. Also, note that in later
results in the paper we look at sequences $\\{\mu_{N}\\}_{N\geq 1}$ of
probability measures satisfying certain conditions, which we discuss later.

######  Definition 1.1.

A sequence $\\{\mu_{N}\\}_{N\geq 1}$ of probability measures such that
$\mu_{N}\in\mathcal{M}_{N}$ for $N\geq 1$ satisfies a Law of Large Numbers if
there exists a sequence $\\{m_{k}\\}_{k\geq 1}$ of real numbers such that

(2)
$\lim_{N\rightarrow\infty}\mathbb{E}_{\mu_{N}}\left(\prod_{i=1}^{s}p_{k_{i}}^{N}\right)=\prod_{i=1}^{s}m_{k_{i}}$

for any positive integer $s$ and positive integers $k_{i}$, $1\leq i\leq s$.

In the above definition, $\\{m_{k}\\}_{k\geq 1}$ are the moments. Later on in
this paper, we consider $\\{\mu_{N}\\}_{N\geq 1}$ such that
$\mu_{N}\in\mathcal{M}_{N}$ is exponentially decaying. The condition for
$\mu\in\mathcal{M}_{N}$ being exponentially decaying is below.

######  Definition 1.2.

A probability measure $\mu$ in $\mathcal{M}_{N}$ is exponentially decaying
with exponent $R>0$ if

$\int_{a_{1}\leq a_{2}\leq\cdots\leq a_{N}}e^{NR\max_{1\leq i\leq
N}|a_{i}|}\mu(da_{1},\ldots,da_{N})$

is finite.

######  Lemma 1.3 ([matrix]*Lemma 2.9).

If $\mu\in\mathcal{M}_{N}$ is exponentially decaying with exponent $R>0$, then
$G_{\theta}(x_{1},\ldots,x_{N};\mu)$ converges for all $(x_{1},\ldots,x_{N})$
in the domain

$\Omega_{R}=\\{(x_{1},\ldots,x_{N})\in\mathbb{C}^{N}||\text{Re}(x_{i})|<R,1\leq
i\leq N\\},$

and is holomorphic in the domain.

A reason we consider $\mu\in\mathcal{M}_{N}$ which are exponentially decaying
is that $G_{\theta}(x_{1},\ldots,x_{N};\mu)$ is convergent and holomorphic in
a neighborhood of the origin, see Lemma 1.3. Soon, we will be able to state
the main result of this paper, Theorem 1.5.

######  Definition 1.4.

For an ordered $N$-tuple $a=(a_{1},a_{2},\ldots,a_{N})$ of nonnegative
integers, let the equivalent partition of $a$, denoted by $\pi(a)$, be the
partition obtained from removing the $a_{i}$ which are $0$ and ordering the
remaining $a_{i}$ in nonincreasing order from left to right.

For a partition $\nu=(a_{1}\geq a_{2}\geq\cdots\geq a_{m})$, let $P(\nu)$
denote the number of distinct permutations of $(a_{1},a_{2},\ldots,a_{m})$.
Also, for a finite ordered list $S$ of positive integers with maximum element
$M$, suppose that $n_{i}$ of the elements of $S$ are $i$ for $1\leq i\leq M$.
Then, if $n=(n_{1},n_{2},\ldots,n_{M})$, let $\sigma(S)$ be $\pi(n)$.

######  Theorem 1.5.

Suppose $\theta$ is a positive real number. Let $\\{\mu_{N}\\}_{N\geq 1}$ be a
sequence of probability measures such that for all $N\geq 1$, $\mu_{N}$ is in
$\mathcal{M}_{N}$ and is exponentially decaying. Assume that for all
partitions $\nu$ with $|\nu|\geq 1$, there exists a real number $c_{\nu}$ such
that

(3)
$\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N}\cdot\frac{\partial}{\partial
x_{i_{1}}}\cdots\frac{\partial}{\partial
x_{i_{r}}}\ln(G_{\theta}(x_{1},\ldots,x_{N};\mu_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq N}=\frac{\ell(\nu)(|\nu|-1)!c_{\nu}}{P(\nu)}$

for all positive integers $i_{1},\ldots,i_{r}$ such that
$\sigma((i_{1},\ldots,i_{r}))=\nu$. Then, $\\{\mu_{N}\\}_{N\geq 1}$ satisfy a
LLN, and for positive integers $k$,

(4) $m_{k}=\sum_{\pi\in NC(k)}\prod_{B\in\pi}\theta^{|B|-1}\left(\sum_{\nu\in
P,|\nu|=|B|}(-1)^{\ell(\nu)-1}P(\nu)c_{\nu}\right).$

Theorem 1.5 generalizes [matrix]*Claim 9.1 and is similar to, but differs in
important places with, [matrix]*Theorem 3.8, where the reverse direction is
also shown. Particularly, in [matrix]*Theorem 3.8, $\theta N\rightarrow\gamma$
as $N\rightarrow\infty$ and the partial derivatives are not scaled; but, in
this paper, $\theta$ is constant and the partial derivatives are scaled by
$\frac{1}{N}$. However, similarly, the moments and LLN satisfaction are
obtained from the value at $x_{i}=0,1\leq i\leq N$ after Dunkl operators are
used on the Bessel generating functions, see Section 5.

Additionally, this paper considers when the limit of any partial derivative
can be nonzero. On the other hand, in [matrix], the limits of partial
derivatives with two or more distinct indices are $0$. This is a reason why
formal power series are needed to prove Theorem 1.5. Note that in Theorem 1.5,
the free cumulant of order $k$ for $k\geq 1$ would be

$c_{k}=\theta^{k-1}\sum_{\nu\in P,|\nu|=k}(-1)^{\ell(\nu)-1}P(\nu)c_{\nu},$

where free cumulants are discussed in [freeprobability]. We see that $c_{k}$
is a linear combination of the limits of the order $k$ partial derivatives.

###  1.2. Formal Power Series

Suppose we have variables $x_{1},\ldots,x_{N}$ and
$\vec{x}=(x_{1},\ldots,x_{N})$. A formal power series $F(x_{1},\ldots,x_{N})$
has form

(5)
$F(x_{1},\ldots,x_{N})=\sum_{\alpha=(\alpha_{1},\ldots,\alpha_{N})\in\mathbb{Z}_{\geq
0}^{N}}c_{F}^{\alpha}\prod_{i=1}^{N}x_{i}^{\alpha_{i}},$

where the $c_{F}^{\alpha}$ are constants. In the paper, we consider formal
power series $F(x_{1},\ldots,x_{N})$ which are symmetrical, or symmetrical in
$N-1$ variables. Also, for simplicity, we usually call a formal power series a
formal series instead.

In Section 5, we see that we can view
$\ln(G_{\theta}(x_{1},\ldots,x_{N};\mu))$ as a symmetrical polynomial, and
afterwards, we can use formal series. In order to find the value at $x_{i}=0$,
$1\leq i\leq N$, after using Dunkl operators on the Bessel generating
function, we look at the constant terms of formal series. Particularly, in
(5), we have that the constant term of $F(x_{1},\ldots,x_{N})$ is
$[1]F(x_{1},\ldots,x_{N})=c_{F}^{0}$, where $0$ is $\alpha$ with
$\alpha_{i}=0$, $1\leq i\leq N$.

###  1.3. Paper Organization

The organization of this paper is as follows. In Section 2, we introduce the
notation for formal power series and Dunkl operators. Next, in Section 3, we
go over results on the limits of operators on formal series with $N$ variables
which are symmetric in $N-1$ variables as $N$ increases to $\infty$.
Afterwards, in Section 4, we prove results important to proving Theorem 1.5,
and in Section 5, we prove Theorem 1.5. Following this, in Section 6, we apply
Theorem 1.5 to the $\beta$-Hermite ensemble. Next, in Appendix A, we state and
prove several combinatorial formulas which are used throughout the paper.
Finally, in Appendix B, we state additional results for formal series.

Acknowledgements. This is a continuation of research done in MIT UROP+. The
author would like to thank Matthew Nicoletti for being the mentor for the
research and suggesting the problem. Also, the author would like to thank
Alexei Borodin and Vadim Gorin for giving comments on the paper.

##  2\. Formal Power Series and Operators

###  2.1. Formal Power Series

Later on in the paper, we look at formal power series in $x_{1},\ldots,x_{N}$.
For $\vec{x}=(x_{1},\ldots,x_{N})$ and a partition $\nu$, let

$M_{\nu}(\vec{x})=\sum_{\begin{subarray}{c}a=(a_{1},\ldots,a_{N}),\\\
a_{i}\in\mathbb{Z}_{\geq 0},1\leq i\leq
N,\pi(a)=\nu\end{subarray}}\prod_{i=1}^{N}x_{i}^{a_{i}}.$

Above, $N=\infty$ is possible. Also, suppose that the set of all partitions is
$P$.

For a symmetric formal series $F(x_{1},\ldots,x_{N})$, we let

$F(x_{1},\ldots,x_{N})=\sum_{\nu\in P,\ell(\nu)\leq
N}c_{F}^{\nu}M_{\nu}(\vec{x}),$

where the $c_{F}^{\nu}\in\mathbb{C}$ are constants. Later on, in Section 4,
when we view the $c_{F}^{\nu}$ as variables, $c_{F}^{\nu}$ has degree $|\nu|$
for $\nu\in P$.

For positive integers $N\geq i$, let $\mathcal{F}_{i}^{N}$ denote the set of
formal series $F(x_{1},\ldots,x_{N})$ which are symmetrical in $x_{j}$ for
$1\leq j\leq N$, $j\not=i$. Suppose that
$F(x_{1},\ldots,x_{N})\in\mathcal{F}_{i}^{N}$. Then, where
$\vec{x}_{i}=(x_{j})_{\begin{subarray}{c}1\leq j\leq N,j\not=i\end{subarray}}$
is $\vec{x}$ with $x_{i}$ removed, we have

$F(x_{1},\ldots,x_{N})=\sum_{d=0}^{\infty}\left(\sum_{\nu\in
P}c_{F}^{d,\nu}M_{\nu}(\vec{x}_{i})\right)x_{i}^{d},$

where the $c_{F}^{d,\nu}\in\mathbb{C}$ are constants.

We can let $c_{F}^{\nu}=0$ for $\nu$ such that $\ell(\nu)\geq N+1$, since
$\vec{x}$ has $N$ variables, and $c_{F}^{d,\nu}=0$ for $\nu$ such that
$\ell(\nu)\geq N$, since $\vec{x}_{i}$ has $N-1$ variables. Particularly, for
$F(x_{1},\ldots,x_{N})\in\mathcal{F}_{i}^{N}$,

(6) $F(x_{1},\ldots,x_{N})=\sum_{d=0}^{\infty}\left(\sum_{\nu\in
P,\ell(\nu)\leq N-1}c_{F}^{d,\nu}M_{\nu}(\vec{x}_{i})\right)x_{i}^{d}.$

######  Definition 2.1.

For a sequence of constants $s=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in P}$ and a
positive integer $i$, let the formal series of $s$ in $\mathcal{F}_{i}^{N}$
for all $N\geq i$ be

$F_{i}(s)(x_{1},\ldots,x_{N})=\sum_{d=0}^{\infty}\sum_{\nu\in P,\ell(\nu)\leq
N-1}c^{d,\nu}M_{\nu}(\vec{x}_{i})x_{i}^{d}.$

Also, where $\vec{x}_{i}=(x_{j})_{j\geq 1,j\not=i}$, let

$F_{i}(s)=\sum_{d=0}^{\infty}\sum_{\nu\in
P}c^{d,\nu}M_{\nu}(\vec{x}_{i})x_{i}^{d}$

as a formal series in $(x_{j})_{j\geq 1}$.

###  2.2. Dunkl Operator

For $N$ variables $x_{1},\ldots,x_{N}$, let $s_{i,j}$ be the operator which
switches $x_{i}$ and $x_{j}$.

######  Definition 2.2.

The Dunkl operators are, for $1\leq i\leq N$,

$\mathcal{D}_{i}:=\frac{\partial}{\partial
x_{i}}+\theta\sum_{\begin{subarray}{c}1\leq j\leq N,\\\
j\not=i\end{subarray}}\frac{1}{x_{i}-x_{j}}(1-s_{i,j}).$

For positive integers $k$, let

(7) $\mathcal{P}_{k}=\sum_{i=1}^{N}\mathcal{D}_{i}^{k}.$

Also, let $\frac{\partial}{\partial x_{i}}$ be denoted by $\partial_{i}$ for
$1\leq i\leq N$. Moreover, for $i\not=j$, let

$\frac{\theta}{x_{i}-x_{j}}(1-s_{i,j})$

be called the switch from $i$ to $j$. We know that the Dunkl operators are
commutative, which is in Proposition 2.3 below. Due to this, the
$\mathcal{P}_{i}$ are commutative as well.

######  Proposition 2.3 ([dunkloperators]*Theorem 1.9).

For positive integers $i,j$, $1\leq i,j\leq N$,
$\mathcal{D}_{i}\mathcal{D}_{j}=\mathcal{D}_{j}\mathcal{D}_{i}$.

##  3\. Limits of Operators On Formal Power Series

In this section, we look at operators on formal power series in $x_{j}$,
$1\leq j\leq N$ such that there exists $i$, $1\leq i\leq N$, for which the
formal power series is symmetrical for $j\not=i$. The main results are Theorem
3.9 and Theorem 3.13, with Theorem 3.9 being a special case of Theorem 3.13.

###  3.1. Basic Results

For a partition $\nu=(a_{1}\geq a_{2}\geq\cdots\geq a_{m}$, let

$S(\nu)=\\{(p_{1},p_{2})|p_{i}=(b_{i,1},\ldots,b_{i,m}),1\leq i\leq
2,b_{1,j}+b_{2,j}=a_{j},1\leq j\leq m\\}.$

If we have partitions $\nu_{1},\nu_{2},\ldots,\nu_{k}$, we let
$\nu=\nu_{1}+\cdots+\nu_{k}$ be the partition formed when
$\nu_{1},\ldots,\nu_{k}$ are joined to each other.

For $1\leq i,j\leq N$, $i\not=j$, let $C_{i,j}$ be the operator such that for
nonnegative integers $a_{k}$, $1\leq k\leq N$,

(8)
$C_{i,j}\left(\prod_{k=1}^{N}x_{k}^{a_{k}}\right)=\begin{cases}x_{i}^{a_{j}-1}\displaystyle\prod_{1\leq
k\leq N,k\not=i,j}x_{k}^{a_{k}}&\text{if $a_{i}=0$ and $a_{j}\geq 1$},\\\
0&\text{ if $a_{i}\geq 1$ or $a_{i}=a_{j}=0$}.\end{cases}$

With this, for positive integers $N\geq i$ and a formal series
$f(x_{1},\ldots,x_{N})\in\mathcal{F}_{i}^{N}$, let the operator
$\mathcal{Q}_{i}^{N}(f(x_{1},\ldots,x_{N}))$ be

(9)
$\mathcal{Q}_{i}^{N}(f(x_{1},\ldots,x_{N}))=\theta\sum_{\begin{subarray}{c}1\leq
j\leq N,\\\
j\not=i\end{subarray}}\frac{d_{i}-C_{i,j}}{N}+f(x_{1},\ldots,x_{N}).$

In Proposition 3.4, we show that $\mathcal{Q}_{i}^{N}(f(x_{1},\ldots,x_{N}))$
is an operator from $\mathcal{F}_{i}^{N}$ to $\mathcal{F}_{i}^{N}$. Sometimes,
for $f(x_{1},\ldots,x_{N})\in\mathcal{F}_{i}^{N}$, we denote
$\mathcal{Q}_{i}^{N}(f(x_{1},\ldots,x_{N}))$ by $\mathcal{Q}_{i}^{N}(f)$, and
for $s=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in P}$, we denote
$\mathcal{Q}_{i}^{N}(F_{i}(s)(x_{1},\ldots,x_{N}))$ by
$\mathcal{Q}_{i}^{N}(F_{i}(s))$. In this paper, for operators
$\mathcal{T}_{i}$, $1\leq i\leq m$, the product

$\prod_{i=1}^{m}\mathcal{T}_{i}$

denotes the $\mathcal{T}_{i}$ being applied from $i=1$ to $m$.

######  Definition 3.1.

For a positive integer $i$, a sequence of formal series
$\\{f_{N}(x_{1},\ldots,x_{N})\\}_{N\geq i}$ is symmetric outside of $i$ if
$f_{N}(x_{1},\ldots,x_{N})\in\mathcal{F}_{i}^{N}$ for all $N\geq i$.

######  Definition 3.2.

Suppose that $i$ is a positive integer and a sequence of formal series
$f=\\{f_{N}(x_{1},\ldots,x_{N})\\}_{N\geq i}$ is symmetrical outside of $i$.
Then, the limit of $f$ outside of $i$ exists if
$\lim_{N\rightarrow\infty}c_{f_{N}}^{d,\nu}$ exists for all $d,\nu$. If the
limit outside of $f$ outside of $i$ exists, the limiting sequence of $f$
outside of $i$ is $s=\\{\lim_{N\rightarrow\infty}c_{f_{N}}^{d,\nu}\\}_{d\geq
0,\nu\in P}$, and the limit of $f$ as $N\rightarrow\infty$ outside of $i$ is

$\lim_{N\rightarrow\infty}f_{N}(x_{1},\ldots,x_{N})=F_{i}(s)=\sum_{d=0}^{\infty}\left(\sum_{\nu\in
P}\left(\lim_{N\rightarrow\infty}c_{f_{N}}^{d,\nu}\right)\cdot
M_{\nu}(\vec{x}_{i})\right)x_{i}^{d}.$

######  Remark 3.3.

For a sequence $\\{f_{N}(x_{1},\ldots,x_{N})\\}_{N\geq i}$ such that
$f_{N}(x_{1},\ldots,x_{N})$ is in $\mathcal{F}_{i}^{N}$ for all $N\geq i$, the
limiting sequence of $f_{N}(x_{1},\ldots,x_{N})$ outside of $i$ being $s$ is
equivalent to the limit of $f_{N}(x_{1},\ldots,x_{N})$ as $N\rightarrow\infty$
outside of $i$ being $F_{i}(s)$.

######  Proposition 3.4.

For $f(x_{1},\ldots,x_{N})$ and $g(x_{1},\ldots,x_{N})$ in
$\mathcal{F}_{i}^{N}$, $\mathcal{Q}_{i}^{N}(f)g(x_{1},\ldots,x_{N})$ is in
$\mathcal{F}_{i}^{N}$, and is

$\displaystyle\sum_{d=0}^{\infty}\sum_{\nu\in P,\ell(\nu)\leq N-1}\left(\theta
c_{g}^{d+1,\nu}-\theta c_{g}^{0,\nu+(d+1)}+\sum_{\begin{subarray}{c}a+b=d,\\\
(p_{1},p_{2})\in
S(\nu)\end{subarray}}c_{f}^{a,\pi(p_{1})}c_{g}^{b,\pi(p_{2})}\right)M_{\nu}(\vec{x}_{i})x_{i}^{d}$
$\displaystyle+\frac{\theta}{N}\cdot\sum_{d=0}^{\infty}\sum_{\nu\in
P,\ell(\nu)\leq
N-1}(-c_{g}^{d+1,\nu}+(\ell(\nu)+1)c_{g}^{0,\nu+(d+1)})M_{\nu}(\vec{x}_{i})x_{i}^{d}.$

###### Proof.

We see that

$g_{1}(x_{1},\ldots,x_{N})=d_{i}g(x_{1},\ldots,x_{N})=\sum_{d=1}^{\infty}\sum_{\nu\in
P}c_{g}^{d,\nu}M_{\nu}(\vec{x}_{i})x_{i}^{d-1}=\sum_{d=0}^{\infty}\sum_{\nu\in
P}c_{g}^{d+1,\nu}M_{\nu}(\vec{x}_{i})x_{i}^{d}.$

Also, let

$g_{2}(x_{1},\ldots,x_{N})=\sum_{1\leq j\leq
N,j\not=i}C_{i,j}g(x_{1},\ldots,x_{N})=\left(\sum_{1\leq j\leq
N,j\not=i}C_{i,j}\right)\left(\sum_{\nu\in
P}c_{g}^{0,\nu}M_{\nu}(\vec{x}_{i})\right).$

Observe that $g_{2}(x_{1},\ldots,x_{N})$ is a formal power series which is
symmetric in $\vec{x}_{i}$. We find the coefficient of
$M_{\nu}(\vec{x}_{i})x_{i}^{d}$. If $\ell(\nu)\geq N$, the coefficient is $0$
in $g_{2}(x_{1},\ldots,x_{N})$. Then, suppose $\ell(\nu)\leq N-1$. For a
monomial $p$ in $M_{\nu}(\vec{x}_{i})$ with variables $(\vec{x}_{i})_{j}$ with
$1\leq j\leq\ell(\nu)$, there are $N-1-\ell(\nu)$ $j$ such that $x_{j}$ is not
in $p$, and these are the $j$ such that $C_{i,j}$ applied to a monomial will
give $px_{i}^{d}$. For a monomial $q$, if

$C_{i,j}q=px_{i}^{d},$

then $q=px_{j}^{d+1}$ and has coefficient $c_{g}^{0,\nu+(d+1)}$ in $g$.
Therefore, the coefficient of $px_{i}^{d}$, and thus
$M_{\nu}(\vec{x}_{i})x_{i}^{d}$, in $g_{2}(x_{1},\ldots,x_{N})$ is
$(N-\ell(\nu)-1)c_{g}^{0,\nu+(d+1)}$. From this,

$g_{2}(x_{1},\ldots,x_{N})=\sum_{d=0}^{\infty}\sum_{\nu\in P,\ell(\nu)\leq
N-1}(N-\ell(\nu)-1)c_{g}^{0,\nu+(d+1)}M_{\nu}(\vec{x}_{i})x_{i}^{d}.$

Next, let

$\displaystyle g_{3}(x_{1},\ldots,x_{N})$
$\displaystyle=f(x_{1},\ldots,x_{N})g(x_{1},\ldots,x_{N})$
$\displaystyle=\sum_{d_{1},d_{2}=0}^{\infty}\sum_{\nu_{1},\nu_{2}\in
P}c_{f}^{d_{1},\nu_{1}}c_{g}^{d_{2},\nu_{2}}M_{\nu_{1}}(\vec{x}_{i})M_{\nu_{2}}(\vec{x}_{i})x_{i}^{d_{1}+d_{2}}.$

For $\nu\in P$ where $\nu$ can be $0$, we find the coefficient of
$M_{\nu}(\vec{x}_{i})x_{i}^{d}$ in $g_{3}(x_{1},\ldots,x_{N})$. If
$\ell(\nu)\geq N$, this coefficient will be $0$. Suppose $\ell(\nu)\leq N-1$,
and let

$q=\prod_{j=1}^{\ell(\nu)}x_{(\vec{x}_{i})_{j}}^{\nu_{j}}.$

Note that if $q_{1}$ and $q_{2}$ are monic monomials such that $q_{1}q_{2}=q$,
then for $1\leq s\leq 2$, if $b_{s,j}$ is the degree of
$x_{(\vec{x}_{i})_{j}}$ in $q_{s}$ for $1\leq j\leq\ell(\nu)$ and
$p(q_{s})=(b_{s,1},\ldots,b_{s,\ell(\nu)})$, $(p(q_{1}),p(q_{2}))\in S(\nu)$.
We see that if $S$ is the set of $(q_{1},q_{2})$ such that $q_{1}$ and $q_{2}$
are monic monomials with $q_{1}q_{2}=q$,

$p:S\rightarrow S(\nu),(q_{1},q_{2})\mapsto(p(q_{1}),p(q_{2}))$

is injective as well as surjective, and therefore is a bijection. Also, the
coefficient of $q_{1}x_{1}^{a}$ and $q_{2}x_{1}^{b}$ is
$c_{f}^{a,\pi(p(q_{1}))}$ in $f$ and $c_{g}^{b,\pi(p(q_{2}))}$ in $g$,
respectively. Then, the coefficient of $qx_{i}^{d}$ is

$\sum_{\begin{subarray}{c}a+b=d,\\\
q_{1}q_{2}=q\end{subarray}}c_{f}^{a,\pi(p(q_{1}))}c_{g}^{b,\pi(p(q_{2}))}=\sum_{\begin{subarray}{c}a+b=d,\\\
(p_{1},p_{2})\in
S(\nu)\end{subarray}}c_{f}^{a,\pi(p_{1})}c_{g}^{b,\pi(p_{2})},$

which is also the coefficient of $M_{\nu}(\vec{x}_{i})x_{i}^{d}$ in
$g_{3}(x_{1},\ldots,x_{N})$. From this,

$g_{3}(x_{1},\ldots,x_{N})=\sum_{d=0}^{\infty}\sum_{\nu\in P,\ell(\nu)\leq
N-1}\left(\sum_{\begin{subarray}{c}a+b=d,\\\ (p_{1},p_{2})\in
S(\nu)\end{subarray}}c_{f}^{a,\pi(p_{1})}c_{g}^{b,\pi(p_{2})}\right)M_{\nu}(\vec{x}_{i})x_{i}^{d}.$

Since $g_{1}(x_{1},\ldots,x_{N})$, $g_{2}(x_{1},\ldots,x_{N})$, and
$g_{3}(x_{1},\ldots,x_{N})$ are in $\mathcal{F}_{i}^{N}$,
$\mathcal{Q}_{i}(f)g(x_{1},\ldots,x_{N})$ is in $\mathcal{F}_{i}^{N}$, with

$\mathcal{Q}_{i}(f)g(x_{1},\ldots,x_{N})=\frac{(N-1)\theta g_{1}(x_{1},\ldots
x_{N})}{N}-\frac{\theta
g_{2}(x_{1},\ldots,x_{N})}{N}+g_{3}(x_{1},\ldots,x_{N}).$

With the expressions we obtained, this completes the proof. ∎

######  Corollary 3.5.

Let $m\geq 0$ be an integer. Suppose $\\{f_{j,N}(x_{1},\ldots,x_{N})\\}_{N\geq
i}$ for $1\leq j\leq m$ and $\\{g_{N}(x_{1},\ldots,x_{N})\\}_{N\geq i}$ are
symmetrical outside of $i$. Also, assume that the limit of
$f_{j,N}(x_{1},\ldots,x_{N})$ for $1\leq j\leq m$ and
$g_{N}(x_{1},\ldots,x_{N})$ as $N\rightarrow\infty$ exist outside of $i$.
Then,

$\lim_{N\rightarrow\infty}\left(\prod_{j=1}^{m}\mathcal{Q}_{i}^{N}(f_{j,N})\right)g_{N}(x_{1},\ldots,x_{N})$

exists outside of $i$. Also, the limit of
$\mathcal{Q}_{i}^{N}(f_{1,N})g_{N}(x_{1},\ldots,x_{N})$ as
$N\rightarrow\infty$ outside of $i$ is equal to

(10) $\sum_{d=0}^{\infty}\sum_{\nu\in P}\lim_{N\rightarrow\infty}\left(\theta
c_{g_{N}}^{d+1,\nu}-\theta
c_{g_{N}}^{0,\nu+(d+1)}+\sum_{\begin{subarray}{c}a+b=d,\\\ (p_{1},p_{2})\in
S(\nu)\end{subarray}}c_{f_{1,N}}^{a,\pi(p_{1})}c_{g_{N}}^{b,\pi(p_{2})}\right)M_{\nu}(\vec{x}_{i})x_{i}^{d}.$

###### Proof.

The $m=0$ case is clear, and for $m=1$, the limit of
$\mathcal{Q}_{i}^{N}(f_{1,N})g_{N}(x_{1},\ldots,x_{N})$ as
$N\rightarrow\infty$ outside of $i$ existing and being equal to (10) follows
from taking the $N\rightarrow\infty$ limit of Proposition 3.4. Then, the rest
of Corollary 3.5 follows from $m=1$ and induction. ∎

######  Proposition 3.6.

Let $i\geq 1$ and $m\geq 0$ be integers. Suppose
$\\{f_{j,N}(x_{1},\ldots,x_{N})\\}_{N\geq 1}$ for $1\leq j\leq m$ and
$\\{g_{N}(x_{1},\ldots,x_{N})\\}_{N\geq 1}$ are symmetrical outside of $i$.
Assume that for $1\leq j\leq m$, $f_{j,N}(x_{1},\ldots,x_{N})$ has a limit as
$N\rightarrow\infty$ with limiting sequence $f_{j}$ outside of $i$. Also,
assume that $g_{N}(x_{1},\ldots,x_{N})$ has a limit as $N\rightarrow\infty$
with limiting sequence $g$ outside of $i$. Then, outside of $i$,

(11)
$\lim_{N\rightarrow\infty}\left(\prod_{j=1}^{m}\mathcal{Q}_{i}^{N}(f_{j,N})\right)g_{N}(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}\left(\prod_{j=1}^{m}\mathcal{Q}_{i}^{N}(F_{i}(f_{j}))\right)F_{i}(g)(x_{1},\ldots,x_{N}).$

###### Proof.

From Corollary 3.5, the limits in both sides exist. Suppose that for $1\leq
j\leq m$, $f_{j}=\\{c_{f_{j}}^{d,\nu}\\}_{d\geq 0,\nu\in P}$, and
$g=\\{c_{g}^{d,\nu}\\}_{d\geq 0,\nu\in P}$. We use induction on $m$, where the
base case $m=0$ is clear. For $m=1$, we can use (10) to show that
$\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(f_{1,N})g_{N}(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(F(f_{1}))F(g)(x_{1},\ldots,x_{N})$.

Assume the statement holds for $m\geq 1$. We want to show the statement holds
for $m+1$. Let

$r_{N}(x_{1},\ldots,x_{N})=\mathcal{Q}_{i}^{N}(f_{1,N})g_{N}(x_{1},\ldots,x_{N}),$

and $r$ be the limiting sequence of
$\mathcal{Q}_{i}^{N}(F_{i}(f_{1}))F_{i}(g)(x_{1},\ldots,x_{N})$ as
$N\rightarrow\infty$ outside of $i$. From $m=1$,
$\lim_{N\rightarrow\infty}r_{N}(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(F_{i}(f_{1}))F_{i}(g)(x_{1},\ldots,x_{N})=F_{i}(r)$,
or $r$ is the limiting sequence of $r_{N}(x_{1},\ldots,x_{N})$ as
$N\rightarrow\infty$ outside of $i$. Then, by the inductive hypothesis,

$\displaystyle\lim_{N\rightarrow\infty}\left(\prod_{j=1}^{m+1}\mathcal{Q}_{i}^{N}(f_{j,N})\right)g_{N}(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}\left(\prod_{j=2}^{m+1}\mathcal{Q}_{i}^{N}(f_{j,N})\right)r_{N}(x_{1},\ldots,x_{N})$
$\displaystyle=\lim_{N\rightarrow\infty}\left(\prod_{j=2}^{m+1}\mathcal{Q}_{i}^{N}(F_{i}(f_{j}))\right)F_{i}(r)(x_{1},\ldots,x_{N})$
$\displaystyle=\lim_{N\rightarrow\infty}\left(\prod_{j=2}^{m+1}\mathcal{Q}_{i}^{N}(F_{i}(f_{j}))\right)\mathcal{Q}_{i}^{N}(F_{i}(f_{1}))F_{i}(g)(x_{1},\ldots,x_{N})$
$\displaystyle=\lim_{N\rightarrow\infty}\left(\prod_{j=1}^{m+1}\mathcal{Q}_{i}^{N}(F_{i}(f_{j}))\right)F_{i}(g)(x_{1},\ldots,x_{N}).$

This completes the proof. ∎

######  Proposition 3.7.

Suppose that $i,j$ are integers such that $1\leq i,j\leq N$ and $i\not=j$.
Then, for integers $k$ with $1\leq k\leq N$ and $k\not=i,j$, if
$f\in\mathcal{F}_{k}^{N}$,

$s_{i,j}\mathcal{Q}_{k}^{N}(f)=\mathcal{Q}_{k}^{N}(f)s_{i,j}$

as operators from $\mathcal{F}_{k}^{N}$ to $\mathcal{F}_{k}^{N}$. Also, if
$f\in\mathcal{F}_{i}^{N}$,

$s_{i,j}\mathcal{Q}_{i}^{N}(f)=\mathcal{Q}_{j}^{N}(s_{i,j}f)s_{i,j}$

as operators from $\mathcal{F}_{i}^{N}$ to $\mathcal{F}_{j}^{N}$.

###### Proof.

This is clear from expanding the operators. ∎

###  3.2. Constant Term

The most important part of the formal series after the $\mathcal{Q}_{i}^{N}$
operators is the constant term, as will be seen in Section 4. For a sequence
of coefficients $s=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in P}$, this corresponds to
$c^{0,0}$. In this subsection, we look at Theorem 3.9 to compute the constant
term following $\mathcal{Q}_{i}^{N}$ operators where there is only one $i$.
Later on, in Theorem 3.13, we look at the constant term following
$\mathcal{Q}_{i}^{N}$ operators where there are any number of distinct $i$.
These values are computed with free cumulants, which are in Definition 3.8.
However, we must go over noncrossing partitions first.

Suppose that we have a partition $\pi$ of a finite, nonempty set $S$ of real
numbers. We let

(12) $\pi=B_{1}\sqcup B_{2}\sqcup\cdots\sqcup B_{m},$

where $B_{i}$, $1\leq i\leq m$, are the blocks of $\pi$, such that the
smallest element of $B_{i+1}$ is greater than the smallest element of $B_{i}$
for $1\leq i\leq m-1$. Also, the length of $\pi$ is $\ell(\pi)=m$.

A partition $\pi$ is noncrossing if for any distinct blocks $B_{1}$ and
$B_{2}$ of $\pi$, there do not exist $a,b\in B_{1}$ and $c,d\in B_{2}$ such
that $a<c<b<d$. Let the set of noncrossing partitions of a finite, nonempty
set $S$ of real numbers be $NC(S)$, and for $k\geq 1$, let
$NC(k)=NC(\\{1,\ldots,k\\})$. A way to represent a partition $\pi$ is the
circular representation, where the elements are spaced around a circle in
order and the convex hulls of the elements of each block of $\pi$ are added.
If $\pi$ is noncrossing, the convex hulls will be disjoint.

Noncrossing partition123456789Crossing partition123456789 Figure 1. Circular
representations of a noncrossing and crossing partition of
$\\{1,2,\ldots,9\\}.$

######  Definition 3.8.

For a sequence of coefficients $s=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in P}$ and a
positive integer $k$, let the free cumulant of order $k$ for $s$ be

(13) $c_{k}(s)=\theta^{k-1}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0,\\\
|\nu|+d=k-1\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c^{d,\nu}.$

A way the free cumulants appear is in (14). Note that these free cumulants
will be used in Section 4 and Section 5 to obtain the free cumulants mentioned
previously for the moments $m_{k}$ in Theorem 1.5.

######  Theorem 3.9.

Let $i$ and $k$ be positive integers. Suppose
$\\{f_{j,N}(x_{1},\ldots,x_{N})\\}_{N\geq i}$ for $1\leq j\leq k-1$ and
$\\{g_{N}(x_{1},\ldots,x_{N})\\}_{N\geq i}$ are sequences of formal series
which are symmetric outside of $i$. For $1\leq j\leq k$, assume that
$f_{j,N}(x_{1},\ldots,x_{N})$ has a limit as $N\rightarrow\infty$ with
limiting sequence $f$ outside of $i$. Also, suppose
$g_{N}(x_{1},\ldots,x_{N})$ has a limit as $N\rightarrow\infty$ with limiting
sequence $g$ outside of $i$. Then,

(14)
$\lim_{N\rightarrow\infty}\left([1]\prod_{j=1}^{k-1}\mathcal{Q}_{i}^{N}(f_{j,N})g_{N}(x_{1},\ldots,x_{N})\right)=\sum_{\begin{subarray}{c}\pi\in
NC(k),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}c_{|B_{1}|}(g)\prod_{i=2}^{m}c_{|B_{i}|}(f).$

###### Proof.

From Corollary 3.5, the limit of
$\prod_{j=1}^{k-1}\mathcal{Q}_{i}^{N}(f_{j,N})g_{N}(x_{1},\ldots,x_{N})$ as
$N\rightarrow\infty$ outside of $i$ exists, and the left hand side of (14) is

$[1]\lim_{N\rightarrow\infty}\prod_{j=1}^{k-1}\mathcal{Q}_{i}^{N}(f_{j,N})g_{N}(x_{1},\ldots,x_{N}),$

We can prove the Theorem with induction. For the base case $k=1$,

$\lim_{N\rightarrow\infty}[1]g_{N}(x_{1},\ldots,x_{N})=c_{g}^{0,0}=c_{1}(g).$

Next, assume that the statement holds for a positve integer $k$. We want to
show the statement holds for $k+1$. From Proposition 3.6, outside of $i$,

$\lim_{N\rightarrow\infty}\prod_{j=1}^{k}\mathcal{Q}_{i}^{N}(f_{j,N})g_{N}(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(F_{i}(f))^{k}F_{i}(g)(x_{1},\ldots,x_{N}).$

Using Corollary 3.5, let $g^{\prime}$ be the limiting sequence of
$\mathcal{Q}_{i}^{N}(F_{i}(f))F_{i}(g)(x_{1},\ldots,x_{N})$ outside of $i$,
with

$c_{g^{\prime}}^{d,\nu}=\theta c_{g}^{d+1,\nu}-\theta
c_{g}^{0,\nu+(d+1)}+\sum_{\begin{subarray}{c}a+b=d,\\\ (p_{1},p_{2})\in
S(\nu)\end{subarray}}c_{f}^{a,\pi(p_{1})}c_{g}^{b,\pi(p_{2})}$

for $d\geq 0$ and $\nu\in P$. By Proposition 3.6,

$\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(F_{i}(f))^{k}F_{i}(g)(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(F_{i}(f))^{k-1}F_{i}(g^{\prime})(x_{1},\ldots,x_{N}).$

However, by the inductive hypothesis,

$\displaystyle[1]\lim_{N\rightarrow\infty}\mathcal{Q}_{i}^{N}(F_{i}(f))^{k-1}F_{i}(g^{\prime})(x_{1},\ldots,x_{N})=\sum_{\begin{subarray}{c}\pi\in
NC(k),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}c_{|B_{1}|}(g^{\prime})\prod_{i=2}^{m}c_{|B_{i}|}(f)$
$\displaystyle=\sum_{\begin{subarray}{c}\pi\in NC(k),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}\theta^{|B_{1}|-1}\left(\sum_{\begin{subarray}{c}\nu\in
P,d\geq 0\\\
|\nu|+d=|B_{1}|-1\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{g^{\prime}}^{d,\nu}\right)\prod_{i=2}^{m}c_{|B_{i}|}(f).$

Also, from Lemma A.2,

$\displaystyle\sum_{\begin{subarray}{c}\pi\in NC(k+1),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}c_{|B_{1}|}(g)\prod_{i=2}^{m}c_{|B_{i}|}(f)$
$\displaystyle=\sum_{\begin{subarray}{c}\pi\in NC(k),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}\left(c_{|B_{1}|+1}(g)+\sum_{j=1}^{|B_{1}|}c_{j}(f)c_{|B_{1}|+1-j}(g)\right)\prod_{i=2}^{m}c_{|B_{i}|}(f).$

Then, for $k+1$, it suffices to show that

$\theta^{|B_{1}|-1}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0\\\
|\nu|+d=|B_{1}|-1\end{subarray}}(-1)^{\ell(\nu)}c_{g^{\prime}}^{d,\nu}P(\nu)=c_{|B_{1}|+1}(g)+\sum_{j=1}^{|B_{1}|}c_{j}(f)c_{|B_{1}|+1-j}(g).$

The left hand side is

$\displaystyle\theta^{|B_{1}|}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0\\\
|\nu|+d=|B_{1}|-1\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{g}^{d+1,\nu}-\theta^{|B_{1}|}\sum_{\begin{subarray}{c}\nu\in
P,d\geq 0\\\
|\nu|+d=|B_{1}|-1\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{g}^{0,\nu+(d+1)}$
$\displaystyle+\theta^{|B_{1}|-1}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0\\\
|\nu|+d=|B_{1}|-1\end{subarray}}\sum_{\begin{subarray}{c}a+b=d,\\\
(p_{1},p_{2})\in
S(\nu)\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{f}^{a,\pi(p_{1})}c_{g}^{b,\pi(p_{2})}.$

However, the right hand side is

$\displaystyle\theta^{|B_{1}|}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0,\\\
|\nu|+d=|B_{1}|\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{g}^{d,\nu}$
$\displaystyle+\theta^{|B_{1}|-1}\sum_{j=1}^{|B_{1}|}\left(\sum_{\begin{subarray}{c}\nu_{1}\in
P,d_{1}\geq 0,\\\
|\nu_{1}|+d_{1}=j-1\end{subarray}}(-1)^{\ell(\nu_{1})}P(\nu_{1})c_{f}^{d_{1},\nu_{1}}\right)\left(\sum_{\begin{subarray}{c}\nu_{2}\in
P,d_{2}\geq 0,\\\
|\nu_{2}|+d_{2}=|B_{1}|-j\end{subarray}}(-1)^{\ell(\nu_{2})}P(\nu_{2})c_{g}^{d_{2},\nu_{2}}\right).$

Observe that

$\sum_{\begin{subarray}{c}\nu\in P,d\geq 0,\\\
|\nu|+d=|B_{1}|\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{g}^{d,\nu}=\sum_{\begin{subarray}{c}\nu\in
P,d\geq 0,\\\
|\nu|+d=|B_{1}|-1\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{g}^{d+1,\nu}+\sum_{\nu\in
P,|\nu|=|B_{1}|}(-1)^{\ell(\nu)}P(\nu)c_{g}^{0,\nu}.$

Here, we want to show that

$\sum_{\nu\in
P,|\nu|=|B_{1}|}(-1)^{\ell(\nu)}P(\nu)c_{g}^{0,\nu}=\sum_{\begin{subarray}{c}\nu\in
P,d\geq 0,\\\
|\nu|+d=|B_{1}|-1\end{subarray}}(-1)^{\ell(\nu)+1}P(\nu)c_{g}^{0,\nu+(d+1)}.$

We look at the coefficient of $c_{g}^{0,\nu}$ with $|\nu|=|B_{1}|\geq 1$. If
$\nu=(a_{1}\geq\cdots\geq a_{m})$ and $R(\nu)$ is the set $\\{a_{i}|1\leq
i\leq m\\}$, the coefficient on the right hand side is

$(-1)^{\ell(\nu)}\sum_{i\in R(\nu)}P(\nu_{i}),$

where $\nu_{i}$ is $\nu$ with $i$ removed. However, in a permutation of $\nu$,
the first integer must be an element $i$ of $R(\nu)$. Then, there are
$P(\nu_{i})$ permutations of the remaining components of $\nu$, and therefore,
for $\nu$ with $|\nu|\geq 1$,

(15) $P(\nu)=\sum_{i\in R(\nu)}P(\nu_{i}).$

With this, the coefficient on the left and right hand side are equal.

In order to complete the proof, it suffices to show that

(16) $\begin{split}&\sum_{\begin{subarray}{c}\nu\in P,d\geq 0,\\\
|\nu|+d=|B_{1}|-1\end{subarray}}\sum_{\begin{subarray}{c}a+b=d,\\\
(p_{1},p_{2})\in
S(\nu)\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c_{f}^{a,\pi(p_{1})}c_{g}^{b,\pi(p_{2})}\\\
&=\sum_{j=1}^{|B_{1}|}\left(\sum_{\begin{subarray}{c}\nu_{1}\in P,d_{1}\geq
0,\\\
|\nu_{1}|+d_{1}=j-1\end{subarray}}(-1)^{\ell(\nu_{1})}P(\nu_{1})c_{f}^{d_{1},\nu_{1}}\right)\left(\sum_{\begin{subarray}{c}\nu_{2}\in
P,d_{2}\geq 0,\\\
|\nu_{2}|+d_{2}=|B_{1}|-j\end{subarray}}(-1)^{\ell(\nu_{2})}P(\nu_{2})c_{g}^{d_{2},\nu_{2}}\right).\end{split}$

We find the coefficient of $c_{f}^{d_{1},\nu_{1}}c_{g}^{d_{2},\nu_{2}}$ on
both sides, where $d_{1}+d_{2}+|\nu_{1}|+|\nu_{2}|=|B_{1}|-1$. On the right
hand side, the coefficient is
$(-1)^{\ell(\nu_{1})+\ell(\nu_{2})}P(\nu_{1})P(\nu_{2})$. For $\nu$ with
$|\nu|=|B_{1}|-d-1$, let $T(\nu)=\\{(p_{1},p_{2})\in
S(\nu)|\pi(p_{1})=\nu_{1},\pi(p_{2})=\nu_{2}\\}$. Then, on the left hand side,
the coefficient is, where $d=d_{1}+d_{2}$,

$\displaystyle\sum_{\begin{subarray}{c}\nu\in P,\\\
|\nu|=|B_{1}|-d-1\end{subarray}}\sum_{\begin{subarray}{c}(p_{1},p_{2})\in
S(\nu),\\\
\pi(p_{1})=\nu_{1},\pi(p_{2})=\nu_{2}\end{subarray}}(-1)^{\ell(\nu)}P(\nu)$
$\displaystyle=\sum_{\begin{subarray}{c}\nu\in P,\\\
|\nu|=|B_{1}|-d-1\end{subarray}}(-1)^{\ell(\nu)}|T(\nu)|\cdot P(\nu)$

Suppose $|T(\nu)|\not=0$. Then, if $\ell(\nu)=\ell(\nu_{1})+\ell(\nu_{2})-k$,
$0\leq k\leq\min(\ell(\nu_{1}),\ell(\nu_{2}))$.

######  Claim 3.9.1.

For $0\leq k\leq\min(\ell(\nu_{1}),\ell(\nu_{2}))$,

$\sum_{\begin{subarray}{c}\nu\in P,|\nu|=|B_{1}|-d-1,\\\
\ell(\nu)=\ell(\nu_{1})+\ell(\nu_{2})-k\end{subarray}}|T(\nu)|\cdot
P(\nu)=\frac{\binom{\ell(\nu_{1})}{k}\binom{\ell(\nu_{2})}{k}}{\binom{\ell(\nu_{1})+\ell(\nu_{2})}{k}}\cdot\frac{(\ell(\nu_{1})+\ell(\nu_{2}))!}{\ell(\nu_{1})!\ell(\nu_{2})!}P(\nu_{1})P(\nu_{2}).$

###### Proof.

Let $l=\ell(\nu_{1})+\ell(\nu_{2})-k$ and $\vec{x}=(x_{1},\ldots,x_{l})$.
Suppose partition $\nu$ has $|\nu|=|B_{1}|-d-1$ and $\ell(\nu)=l$. We see that
the coefficient of any term of $M_{\nu}(\vec{x})$ in
$M_{\nu_{1}}(\vec{x})M_{\nu_{2}}(\vec{x})$ is $|T(\nu)|$. Therefore, the sum
of the coefficients in $M_{\nu_{1}}(\vec{x})M_{\nu_{2}}(\vec{x})$ of terms of
$M_{\nu}(\vec{x})$ is $|T(\nu)|\cdot P(\nu)$, where there are $P(\nu)$ terms
of $M_{\nu}(\vec{x})$.

Suppose that $S$ is the sum of the terms of
$M_{\nu_{1}}(\vec{x})M_{\nu_{2}}(\vec{x})$ which contain all of the $x_{i}$,
$x_{1},\ldots,x_{l}$. If a term $p$ of $S$ contains all of the $x_{i}$, $p$
without its coefficient must be a term of $M_{\nu}(\vec{x})$ for some $\nu\in
P$ with $|\nu|=|B_{1}|-d-1$ and $\ell(\nu)=l$. But, the sum of the
coefficients in $M_{\nu_{1}}(\vec{x})M_{\nu_{2}}(\vec{x})$ of terms of
$M_{\nu}(\vec{x})$ is $|T(\nu)|\cdot P(\nu)$. Therefore, the sum of the
coefficients of $S$ is

$C=\sum_{\begin{subarray}{c}\nu\in P,|\nu|=|B_{1}|-d-1,\\\
\ell(\nu)=\ell(\nu_{1})+\ell(\nu_{2})-k\end{subarray}}|T(\nu)|\cdot P(\nu).$

However, for terms $p_{i}$ in $M_{\nu_{i}}(\vec{x})$, $1\leq i\leq 2$, if
$p_{1}p_{2}$ contains all of the $x_{i}$, then exactly $k$ of the $x_{i}$ must
be in both $p_{1}$ and $p_{2}$. Note that $S$ is the sum of $p_{1}p_{2}$ for
such $(p_{1},p_{2})$, and $C$ is the number of $(p_{1},p_{2})$; we count $C$.
There are

$\frac{(\ell(\nu_{1})+\ell(\nu_{2})-k)!}{k!(\ell(\nu_{1})-k)!(\ell(\nu_{2})-k)!}=\frac{\binom{\ell(\nu_{1})}{k}\binom{\ell(\nu_{2})}{k}}{\binom{\ell(\nu_{1})+\ell(\nu_{2})}{k}}\cdot\frac{(\ell(\nu_{1})+\ell(\nu_{2}))!}{\ell(\nu_{1})!\ell(\nu_{2})!}$

ways to choose the $x_{i}$ for $p_{1}$ and $p_{2}$. For each choice of the
$x_{i}$, there are $P(\nu_{1})$ and $P(\nu_{2})$ possibilities for $p_{1}$ and
$p_{2}$, respectively. Thus,

$\sum_{\begin{subarray}{c}\nu\in P,|\nu|=|B_{1}|-d-1,\\\
\ell(\nu)=\ell(\nu_{1})+\ell(\nu_{2})-k\end{subarray}}|T(\nu)|\cdot
P(\nu)=C=\frac{\binom{\ell(\nu_{1})}{k}\binom{\ell(\nu_{2})}{k}}{\binom{\ell(\nu_{1})+\ell(\nu_{2})}{k}}\cdot\frac{(\ell(\nu_{1})+\ell(\nu_{2}))!}{\ell(\nu_{1})!\ell(\nu_{2})!}\cdot
P(\nu_{1})P(\nu_{2}).$

∎

Afterwards, the coefficient of $c_{f}^{d_{1},\nu_{1}}c_{g}^{d_{2},\nu_{2}}$ on
the left hand side of (16) is, with 3.9.1 and Lemma A.1,

$\displaystyle\sum_{\begin{subarray}{c}\nu\in P,\\\
|\nu|=|B_{1}|-d-1\end{subarray}}(-1)^{\ell(\nu)}|T(\nu)|\cdot P(\nu)$
$\displaystyle=\sum_{k=0}^{\min(\ell(\nu_{1}),\ell(\nu_{2}))}(-1)^{\ell(\nu_{1})+\ell(\nu_{2})-k}\sum_{\begin{subarray}{c}\nu\in
P,|\nu|=|B_{1}|-d-1,\\\
\ell(\nu)=\ell(\nu_{1})+\ell(\nu_{2})-k\end{subarray}}|T(\nu)|\cdot P(\nu)$
$\displaystyle=(-1)^{\ell(\nu_{1})+\ell(\nu_{2})}P(\nu_{1})P(\nu_{2}).$

We are done. ∎

###  3.3. Distinct Indices

Also, we look at the constant term following combinations of
$\mathcal{Q}_{i}^{j}$ with distinct indices $i$ in Theorem 3.13. This is
important for the proof of Theorem 1.5, as will be seen in Section 4 and
Section 5, and for this, we use the operator in (17). But, it is necessary to
consider the free cumulants of $s=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in P}$ which
are symmetrical in the $x_{i}$ beforehand.

######  Proposition 3.10.

Suppose that $i$ is a positive integer, and $s=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in
P}$ is a sequence such that $F_{i}(s)$ is symmetrical with respect to
$(x_{j})_{j\geq 1}$. Then, $c_{k}(s)=0$ for all integers $k\geq 2$.

###### Proof.

Since $F_{i}(s)$ is symmetrical, we have that for all $\lambda\in P$, there
exists a constant $c^{\lambda}$ such that $c^{d,\nu}=c^{\lambda}$ for all
$d,\nu$ such that $\nu+(d)=\lambda$. The free cumulant of $s$ of order $k$ is

$c_{k}(s)=\theta^{k-1}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0,\\\
|\nu|+d=k-1\end{subarray}}(-1)^{\ell(\nu)}P(\nu)c^{d,\nu}=\theta^{k-1}\sum_{\lambda\in
P,|\lambda|=k-1}c^{\lambda}\sum_{\begin{subarray}{c}\nu\in P,d\geq 0,\\\
\nu+(d)=\lambda\end{subarray}}(-1)^{\ell(\nu)}P(\nu),$

and $c_{k}(s)=0$ for $k\geq 2$ follows from (15). ∎

Suppose we have positive integers $i\leq j$, and $k$. For
$F(x_{1},\ldots,x_{j}),G(x_{1},\ldots,x_{j})\in\mathcal{F}_{i}^{j}$, let

(17)
$\mathcal{R}_{i,j}^{k}(F(x_{1},\ldots,x_{j}))G(x_{1},\ldots,x_{j})=\mathcal{Q}_{i}^{j}\left(F(x_{1},\ldots,x_{j})\right)^{k}G(x_{1},\ldots,x_{j})|_{x_{i}=0}.$

######  Lemma 3.11.

Suppose $\lambda$ is a partition of length $m\geq 1$, $N\geq m$ is a positive
integer, $f_{i}\in\mathcal{F}_{N-i+1}^{N-i+1}$ for $1\leq i\leq m$, and
$g\in\mathcal{F}_{N}^{N}$. Then,

$\left(\prod_{i=1}^{m}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i})\right)(g)$

is a symmetric formal series in $x_{1},\ldots,x_{N-m}$.

###### Proof.

This follows from induction on $m$ from $m=1$ to $N$. ∎

######  Lemma 3.12.

Suppose $\lambda$ is a partition of length $m\geq 1$. Let
$\\{f_{i,N}\\}_{N\geq m}$ for $1\leq i\leq m$ and
$\\{g_{N}(x_{1},\ldots,x_{N})\\}_{N\geq m}$ be sequences of formal series such
that $f_{i,N}\in\mathcal{F}_{N-i+1}^{N-i+1}$ and $g_{N}\in\mathcal{F}_{N}^{N}$
for $N\geq m$. Also, suppose $\\{s_{1,N-i+1}f_{i,N}\\}_{N\geq m}$ for $1\leq
i\leq m$ and $\\{s_{1,N}g_{N}\\}_{N\geq m}$, which are symmetrical outside of
$1$, have limits as $N\rightarrow\infty$ outside of $1$. Then, for $i\geq 1$,
the limit from $N=i+m$ to $\infty$ of the formal series

$\left(\prod_{i=1}^{m}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i,N})\right)(g_{N})$

in $x_{1},\ldots,x_{N-m}$ exists outside of $i$.

###### Proof.

We show this with induction on $m$. For the base case $m=1$,

$\lim_{N\rightarrow\infty}\mathcal{Q}_{1}^{N}(s_{1,N}f_{1,N})^{\lambda_{1}}(s_{1,N}g_{N})$

exists outside of $1$ from Corollary 3.5 because $s_{1,N}f_{1,N}$ and
$s_{1,N}g_{N}$ have limits outside of $1$. For some $i\geq 1$, consider $N\geq
i+1$. From Lemma 3.11,
$\mathcal{R}_{N,N}^{\lambda_{1}}(f_{1,N})(g_{N})\in\mathcal{F}_{i}^{N-1}$.

From Proposition 3.7,
$s_{1,N}\mathcal{Q}_{N}^{N}(f_{1,N})^{\lambda_{1}}(g_{N})=\mathcal{Q}_{1}^{N}(s_{1,N}f_{1,N})^{\lambda_{1}}(s_{1,N}g_{N})$.
With this, the coefficient $c^{d,\nu}$ of
$\mathcal{Q}_{N}^{N}(f_{1,N})^{\lambda_{1}}(g_{N})|_{x_{N}=0}$ as a formal
series in $x_{1},\ldots,x_{N-1}$ outside of $i$ will be the coefficient
$c^{0,\nu+(d)}$ of $\mathcal{Q}_{N}^{N}(f_{1,N})^{\lambda_{1}}(g_{N})$ outside
of $N$, and thus the coefficient $c^{0,\nu+(d)}$ of
$\mathcal{Q}_{1}^{N}(s_{1,N}f_{1,N})^{\lambda_{1}}(s_{1,N}g_{N})$ outside of
$1$. Since the limit of
$\mathcal{Q}_{1}^{N}(s_{1,N}f_{1,N})^{\lambda_{1}}(s_{1,N}g_{N})$ exists, the
limit of this coefficient exists from $N=i+1$ to $\infty$. As this holds for
all $d,\nu$, the base case is complete.

Next, assume that the result holds for $m\geq 1$. We want to show that the
result holds for $m+1$. Let

$h_{N-m}=\left(\prod_{i=1}^{m}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i,N})\right)(g_{N}).$

By the inductive hypothesis and because $h_{N-m}$ is symmetric, the limit of
$s_{1,N-m}h_{N-m}=h_{N-m}$ from $N-m=1$ to $\infty$ exists outside of $1$.
Then,

$\left(\prod_{i=1}^{m+1}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i,N})\right)(g_{N})=\mathcal{R}_{N-m,N-m}^{\lambda_{m+1}}(f_{m+1,N})(h_{N-m}).$

is the base case $m=1$ with $f_{m+1,N}$ as $f_{1,N-m}$, $h_{N-m}$ as
$g_{N-m}$, and $N-m$ as $N$, where the conditions are satisfied. Therefore,
from $m=1$, the limit of the above expression outside of $i$ for $i\geq 1$
exists from $N-m=i+1$ to $\infty$, or from $N=i+m+1$ to $\infty$, completing
the proof. ∎

######  Theorem 3.13.

Suppose $\lambda$ is a partition of length $m\geq 1$. Let
$\\{f_{i,N}(x_{1},\ldots,x_{N-i+1})\\}_{N\geq m}$ for $1\leq i\leq m$ and
$\\{g_{N}(x_{1},\ldots,x_{N})\\}_{N\geq m}$ be sequences of formal series such
that $f_{i,N}\in\mathcal{F}_{N-i+1}^{N-i+1}$ and $g_{N}\in\mathcal{F}_{N}^{N}$
for $N\geq m$. Moreover, for $1\leq i\leq m$, assume that
$\\{s_{1,N-i+1}f_{i,N}\\}_{N\geq m}$, which is symmetrical outside of $1$, has
a limit as $N\rightarrow\infty$ with limiting sequence $f_{i}$ outside of $1$.
Also, assume that $\\{s_{1,N}g_{N}\\}_{N\geq m}$, which is symmetrical outside
of $1$, has a limit as $N\rightarrow\infty$ with limiting sequence $g$ outside
of $1$. Then,

(18)
$\begin{split}&\lim_{N\rightarrow\infty}[1]\left(\prod_{i=1}^{m}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i,N})\right)(g_{N})\\\
&=\left(\sum_{\begin{subarray}{c}\pi\in NC(\lambda_{1}+1),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{\ell(\pi)}\end{subarray}}c_{|B_{1}|}(g)\prod_{i=2}^{\ell(\pi)}c_{|B_{i}|}(f_{1})\right)\prod_{i=2}^{m}\left(\sum_{\begin{subarray}{c}\pi\in
NC(\lambda_{i}),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{\ell(\pi)}\end{subarray}}\prod_{i=1}^{\ell(\pi)}c_{|B_{i}|}(f_{i})\right).\end{split}$

###### Proof.

We prove this with induction on $m$. For the base case $m=1$, use Theorem 3.9.
Suppose the result holds for $m\geq 1$. We want to show it holds for $m+1$.

From Lemma 3.11,

$h_{N-m}(x_{1},\ldots,x_{N-m})=\left(\prod_{i=1}^{m}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i,N})\right)(g_{N})$

is symmetric in $x_{1},\ldots,x_{N-m}$ for all $N\geq m$. We denote
$h_{N}(x_{1},\ldots,x_{N})$ by $h_{N}$. Then,

$s_{1,N-m}\mathcal{Q}_{N-m}^{N-m}\left(f_{m+1,N}\right)^{\lambda_{m+1}}h_{N-m}=\mathcal{Q}_{1}^{N-m}\left(s_{1,N-m}f_{m+1,N}\right)^{\lambda_{m+1}}h_{N-m}.$

from Proposition 3.7. Since the switch does not change the constant,

$\displaystyle[1]\mathcal{R}_{N-m,N-m}^{N,\lambda_{m+1}}(f_{m+1,N})\left(\prod_{i=1}^{m}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}(f_{i,N})\right)(g_{N})=[1]\mathcal{Q}_{N-m}^{N-m}\left(f_{m+1,N}\right)^{\lambda_{m+1}}h_{N-m}$
$\displaystyle=[1]\mathcal{Q}_{1}^{N-m}\left(s_{1,N-m}f_{m+1,N}\right)^{\lambda_{m+1}}h_{N-m}.$

We want to compute the limit of this as $N\rightarrow\infty$. With
$f_{N-m}=s_{1,N-m}f_{m+1,N}$ for $N\geq m+1$, outside of $1$, as
$N\rightarrow\infty$, $f_{N}$ has limit $f_{m+1}$, and from Lemma 3.12,
$h_{N}$ has a limit; let $h$ be the limiting sequence of $h_{N}$ as
$N\rightarrow\infty$. Then, from Theorem 3.9 with $N-m$ for $N$,

$\lim_{N\rightarrow\infty}[1]\mathcal{Q}_{1}^{N-m}(f_{N-m})^{\lambda_{m+1}}h_{N-m}=\sum_{\begin{subarray}{c}\pi\in
NC(\lambda_{m+1}+1),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{\ell(\pi)}\end{subarray}}c_{|B_{1}|}(h)\prod_{i=2}^{\ell(\pi)}c_{|B_{i}|}(f_{m+1}).$

But, as $h_{N}$ is symmetric in $x_{i}$, $1\leq i\leq N$, for $N\geq 1$, $h$
is symmetric. Then, by Proposition 3.10, $c_{k}(h)=0$, $k\geq 2$, and
$c_{1}(h)=c_{h}^{0,0}$. From this,

$\displaystyle\lim_{N\rightarrow\infty}[1]\mathcal{Q}_{1}^{N-m}(f_{N-m})^{\lambda_{m+1}}h_{N-m}=c_{h}^{0,0}\left(\sum_{\begin{subarray}{c}\pi\in
NC(\lambda_{m+1}),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{\ell(\pi)}\end{subarray}}\prod_{i=1}^{\ell(\pi)}c_{|B_{i}|}(f_{m+1})\right)$
$\displaystyle=\left(\sum_{\begin{subarray}{c}\pi\in NC(\lambda_{1}+1),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{\ell(\pi)}\end{subarray}}c_{|B_{1}|}(g)\prod_{i=2}^{\ell(\pi)}c_{|B_{i}|}(f_{1})\right)\prod_{i=2}^{m+1}\left(\sum_{\begin{subarray}{c}\pi\in
NC(\lambda_{i}),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{\ell(\pi)}\end{subarray}}\prod_{i=1}^{\ell(\pi)}c_{|B_{i}|}(f_{i})\right)$

using the inductive hypothesis, as needed. ∎

##  4\. Sequences of Operators

###  4.1. Setup

Suppose indices $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ and a positive
integer $N\geq\max(\mathfrak{r})$ are given. For a symmetric formal series
$F(x_{1},\ldots,x_{N})$, let

(19)
$\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=\left(\prod_{j=1}^{k}\left(\mathcal{D}_{i_{j}}+\frac{\partial}{\partial
x_{i_{j}}}F(x_{1},\ldots,x_{N})\right)\right)(1).$

Above, the operators are for formal series. Also, suppose the set
$I_{N}(\lambda)$ consists of indices $l$ of length $|\lambda|$ such that the
first $\lambda_{1}$ indices are $i_{1}$, the next $\lambda_{2}$ indices are
$i_{2}$, and so forth, until the last $\lambda_{m}$ indices are $i_{m}$, where
$1\leq i_{j}\leq N$ for $1\leq j\leq m$.

The main result of Section 4 is Theorem 4.1 below. This is used to prove
Theorem 1.5, see Section 5.3. Note that Theorem 1.5 is a generalization of
Claim 9.1 of [matrix], and some techniques from Section 5 of that paper are
used in the proof of Theorem 4.1.

######  Theorem 4.1.

Suppose $\\{F_{N}(x_{1},\ldots,x_{N})\\}_{N\geq 1}$ is a sequence of symmetric
formal series. Assume that for all partitions $\nu$ with $|\nu|\geq 1$, there
exists a real number $c_{\nu}$ such that

$\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N}\cdot[1]\frac{\partial}{\partial
x_{i_{1}}}\cdots\frac{\partial}{\partial
x_{i_{r}}}F_{N}(x_{1},\ldots,x_{N})=\frac{\ell(\nu)(|\nu|-1)!c_{\nu}}{P(\nu)}$

for all positive integers $i_{1},\ldots,i_{r}$ such that
$\sigma((i_{1},\ldots,i_{r}))=\nu$. Then, for a partition $\lambda$ with
$|\lambda|\geq 1$,

$\displaystyle\lim_{N\rightarrow\infty}\left(\frac{1}{N^{\ell(\lambda)+|\lambda|}}\sum_{l\in
I_{N}(\lambda)}[1]\mathcal{D}_{l}(F_{N}(x_{1},\ldots,x_{N}))\right)$
$\displaystyle=\prod_{i=1}^{\ell(\lambda)}\left(\sum_{\pi\in
NC(\lambda_{i})}\prod_{B\in\pi}\theta^{|B|-1}\left(\sum_{\nu\in
P,|\nu|=|B|}(-1)^{\ell(\nu)-1}P(\nu)c_{\nu}\right)\right).$

Note that for $\nu\in P$ with $|\nu|\geq 1$, if $i_{1},\ldots,i_{r}$ are
positive integers such that $\sigma((i_{1},\ldots,i_{r}))=\nu$, then

$[1]\frac{\partial}{\partial x_{i_{1}}}\cdots\frac{\partial}{\partial
x_{i_{r}}}F_{N}(x_{1},\ldots,x_{N})=\frac{|\nu|!}{P(\nu)}\cdot
c_{F_{N}}^{\nu}.$

After this, the condition of Theorem 4.1 gives

(20)
$\lim_{N\rightarrow\infty}\frac{|\nu|}{\ell(\nu)}\cdot\frac{c_{F_{N}}^{\nu}}{N}=c_{\nu}.$

###  4.2. Sequences

We consider sequences $s$ of $k$ operators over $x_{1},\ldots,x_{N}$, where
$s_{j}$ is the operator at step $j$, $1\leq j\leq k$. Each $s_{j}$ has an
associated index $i_{j}$, where $1\leq i_{j}\leq N$. We say that
$\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ are the indices of $s$. Here are
the possibilities for $s_{j}$:

  1. (1)

(Derivative) $\frac{\partial}{\partial x_{i_{j}}},$ denoted by
$\partial_{i_{j}}$.

  2. (2)

(Switch) $\frac{\theta}{x_{i_{j}}-x_{i}}(1-s_{i_{j},i})$, where $1\leq i\leq
N$, $i\not=i_{j}$. This is the switch from $i_{j}$ to $i$.

  3. (3)

(Term Multiplication) Multiplication by
$c_{F}^{\nu}\partial_{i_{j}}(x_{1}^{a_{1}}\cdots x_{N}^{a_{N}})$ for $a_{i}$,
$1\leq i\leq N$, such that $\nu\in P$ with $|\nu|\geq 1$ and
$\pi((a_{1},\ldots,a_{N}))=\nu$. We say that $c_{F}^{\nu}$ is the constant of
the term multiplication.

  4. (4)

(Change) $\theta(d_{i_{j}}-C_{{i_{j}},i})$, where $1\leq i\leq N$,
$i\not=i_{j}$. This is the change from $i_{j}$ to $i$.

Observe that the term multiplications are equivalent to multiplying by
$\partial_{i_{j}}F(x_{1},\ldots,x_{N})$. Also, if $s_{j}$ is a term
multiplication by

$p=c_{F}^{\nu}\frac{\partial}{\partial x_{i_{j}}}x_{1}^{a_{1}}\cdots
x_{N}^{a_{N}},$

we say that $x_{i}$ is in $s_{j}$ or $s_{j}$ contains $x_{i}$ if $p\not=0$ and
the degree of $x_{i}$ in $p$ is at least $1$.

Next, for $0\leq j\leq k$, where $r(s)_{0}=1$, let

(21) $r(s)_{j}=s_{j}\circ s_{j-1}\circ\cdots\circ s_{1}(1).$

Also, suppose we have indices $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ and
$N\geq\max(\mathfrak{r})$. For sequences $s$ with indices $\mathfrak{r}$ over
$x_{1},\ldots,x_{N}$ such that $r(s)_{k}$ is nonzero and has degree $0$ in the
$x_{i}$, let $D_{N,\mathfrak{r}}(s)$ be the set of $x_{i}$ such that
$i\notin\mathfrak{r}$ which are in term multiplications in $s$.

From now on, view the coefficients $c_{F}^{\nu}$ of $F(x_{1},\ldots,x_{N})$ as
variables rather than constants. We view $c_{F}^{\nu}$ as having degree
$|\nu|$, and from (20), $c_{F}^{\nu}$ has order $N$. Additionally, the
constant term $[1]F(x_{1},\ldots,x_{N})=c_{F}^{0}$ does not matter since we
are taking the derivative. Therefore, we set $c_{F}^{0}=0$, and when we refer
to a coefficient $c_{F}^{\nu}$ with $\nu\in P$, assume $|\nu|\geq 1$.

######  Lemma 4.2.

Suppose $s$ is a sequence of length $k$. For $0\leq j\leq k$,

$r(s)_{j}=\prod_{i=1}^{m}c_{F}^{\nu_{i}}\cdot P(x_{1},\ldots,x_{N}),$

where $s_{j}$ is term multiplication for $j=j_{i}$, $1\leq i\leq m$, with
$j_{i}<j_{i+1}$ for $1\leq i\leq m-1$, the term multiplication $s_{j_{i}}$ has
constant $c_{F}^{\nu_{i}}$, and $P(x_{1},\ldots,x_{N})$ is some homogeneous
integer polynomial which is $0$ or has degree $\sum_{i=1}^{m}|\nu_{i}|-j.$

###### Proof.

We can prove this with induction on $j$. The base case $j=0$ is clear. Assume
the result holds for $j$, $0\leq j\leq k-1$. We want to show the result holds
for $j+1$. Suppose $P(x_{1},\ldots,x_{N})$ and
$P^{\prime}(x_{1},\ldots,x_{N})$ are the polynomials for $r(s)_{j}$ and
$r(s)_{j+1}$, respectively.

If $r(s)_{j}=0$, then $r(s)_{j+1}=0$, and the statement holds. Assume
$r(s)_{j}\not=0$. If $s_{j+1}$ is a derivative, switch, or change, we see that
in $r(s)_{j+1}$, $P^{\prime}(x_{1},\ldots,x_{N})=0$ or
$P^{\prime}(x_{1},\ldots,x_{N})$ is a homogeneous integer polynomial with
degree $\sum_{i=1}^{m}|\nu_{i}|-j-1$, and the $c_{F}^{\nu}$ remain the same.
However, if $s_{j+1}$ is term multiplication by
$c_{F}^{\nu_{m+1}}\partial_{i_{j}}(x_{1}^{a_{1}}\cdots x_{N}^{a_{N}})$,

$r(s)_{j+1}=\prod_{i=1}^{m+1}c_{F}^{\nu_{i}}\cdot\left(\frac{\partial}{\partial
x_{i_{j}}}x_{1}^{a_{1}}\cdots x_{N}^{a_{N}}\right)P(x_{1},\ldots,x_{N}).$

There are now $m+1$ term multiplications, where $j_{m+1}=j+1$ and
$s_{j_{m+1}}=s_{j+1}$ has constant $c_{F}^{\nu_{m+1}}$. Moreover, because
$\pi((a_{1},\ldots,a_{N}))=\nu_{m+1}$, we have that
$P^{\prime}(x_{1},\ldots,x_{N})=(\partial_{i_{j}}x_{1}^{a_{1}}\cdots
x_{N}^{a_{N}})P(x_{1},\ldots,x_{N})$ is $0$ or is a homogeneous integer
polynomial with degree $\sum_{i=1}^{m+1}|\nu_{i}|-j-1$. The induction is
complete. ∎

######  Corollary 4.3.

For indices $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ and
$N\geq\max(\mathfrak{r})$,
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$ is a polynomial in the
$c_{F}^{\nu}$ for $\nu\in P$ with $|\nu|\geq 1$ which is homogeneous of degree
$k$, where $c_{F}^{\nu}$ has degree $|\nu|$ for $\nu\in P$ with $|\nu|\geq 1$.

###### Proof.

This follows from Lemma 4.2. ∎

From Lemma 4.2, if a sequence $s$ of length $k$ has $r(s)_{k}$ have degree $0$
in the $x_{i}$, we know that $r(s)_{k}=d\cdot\prod_{i=1}^{m}c_{F}^{\nu_{i}}$
for nonzero $d$ with the sum of the $|\nu_{i}|$ equal to $k$, where the
$c_{F}^{\nu_{i}}$ are from Lemma 4.2. Then, for such $s$, where $k$ can be any
positive integer, let

(22) $C(s)=d.$

######  Definition 4.4.

For a positive integer $k$, suppose we have a sequence of indices
$\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$. For $N\geq\max(\mathfrak{r})$, let
$T_{N}(\mathfrak{r})$, $T_{N}^{1}(\mathfrak{r})$, and
$T_{N}^{2}(\mathfrak{r})$ denote set of sequences $s$ over
$x_{1},\ldots,x_{N}$ with indices $\mathfrak{r}$ such that $r(s)_{k}$ is
nonzero and has degree $0$ in the $x_{i}$ such that:

  * •

The operators of $s$ in $T_{N}(\mathfrak{r})$ can be any operator
(derivatives, term multiplications, switches, and changes).

  * •

The operators of $s$ in $T_{N}^{1}(\mathfrak{r})$ are only derivatives, term
multiplications, and switches.

  * •

The operators of $s$ in $T_{N}^{2}(\mathfrak{r})$ are only term
multiplications and changes.

Throughout Section 4, $T_{N}(\mathfrak{r})$, $T_{N}^{1}(\mathfrak{r})$, and
$T_{N}^{2}(\mathfrak{r})$ are referred to. Particularly, we have that

$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=[1]\left(\prod_{j=1}^{k}\left(\mathcal{D}_{i_{j}}+\frac{\partial}{\partial
x_{i_{j}}}F(x_{1},\ldots,x_{N})\right)\right)(1)=\sum_{s\in
T_{N}^{1}(\mathfrak{r})}r(s)_{k}.$

###  4.3. Orders

######  Definition 4.5.

Suppose $Q(c^{\nu};N)$ is a polynomial in the $c^{\nu}$, $\nu\in P$,
$|\nu|\geq 1$, with finite degree and finitely many terms, and coefficients
which are functions of $N$. Then, $Q(c^{\nu};N)$ has an order of $N^{k}$ if
for a term

$\prod_{i=1}^{m}c^{\nu_{i}},$

the coefficient $f(N)$ in $Q(c^{\nu};N)$ has $|f(N)|=O(N^{k-m})$ for positive
integers $N$.

If $Q(c^{\nu};N)$ has an order of $N^{k}$, we say that $Q(c^{\nu};N)$ is
$O(N^{k})$. Later, we use $c_{F}^{\nu}$ for $c^{\nu}$, and generally, for $k$
indices $\mathfrak{r}$, if we have sets $S_{N}$ in $T_{N}(\mathfrak{r})$ for
$N\geq 1$, we let

(23) $Q(c_{F}^{\nu};N)=\sum_{s\in S_{N}}r(s)_{k}.$

We get
$Q(c_{F}^{\nu};N)=\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))|_{x_{i}=0,1\leq
i\leq N}$ with $S_{N}=T_{N}^{1}(\mathfrak{r})$ for $N\geq 1$. Later on,
Definition 4.5 and (23) are used with $\mathcal{D}_{\mathfrak{r}}$ as well as
other operators.

######  Proposition 4.6.

Suppose $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$. There exists a constant
$C>0$ such that for all positive integers $N$, for any sequence $s\in
T_{N}(\mathfrak{r})$, $|C(s)|\leq C$.

###### Proof.

For such a sequence $s$, for $0\leq j\leq k$, suppose
$P_{j}(x_{1},\ldots,x_{N})$ is the polynomial for $r(s)_{j}$ in (21). Suppose
$s$ has $m$ term multiplications. Since each other operator decreases the
degree in the $x_{i}$ by $1$, the degree in the $x_{i}$ of
$P_{j}(x_{1},\ldots,x_{N})$ for $0\leq j\leq k$ is at most $k-m$, which is
less than $k$. Using this, Proposition 4.6 can be seen by, for $0\leq j\leq
k-1$, bounding the factor $s_{j+1}$ changes the sum of the absolute values of
the coefficients from $P_{j}(x_{1},\ldots,x_{N})$ to
$P_{j+1}(x_{1},\ldots,x_{N})$. Note that $P_{k}(x_{1},\ldots,x_{N})=C(s)$. ∎

######  Proposition 4.7.

For a positive integer $k$ and indices $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq
k}$, for $N\geq\max(\mathfrak{r})$,
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$ has order $N^{k}$.

###### Proof.

We know from Corollary 4.3 that the expression will be a polynomial in the
$c_{F}^{\nu}$ which is homogeneous of order $k$. Suppose that for a positive
integer $m$, for $1\leq i\leq m$, $\nu_{i}\in P$ such that $|\nu_{i}|\geq 1$,
and the sum of the $|\nu_{i}|$ is $k$. Note that the number of possible
$\nu_{1},\ldots,\nu_{m}$, $1\leq m\leq k$, is finite. Then, if we show that
for $N\geq\max(\mathfrak{r})$, the coefficient of

$p=\prod_{i=1}^{m}c_{F}^{\nu_{i}}$

is of order $N^{k-m}$, we will be done by Definition 4.5. For
$N\geq\max(\mathfrak{r})$, suppose the coefficient of $p$ is $r$; we want to
show that $|r|=O(N^{k-m}).$ Let $T$ be the set of sequences $s$ such that
$r(s)_{k}=dp$, where $d\not=0$ is a real number. From Proposition 4.6, there
exists $C>0$ such that for all $N$, for all $s\in T$, $|C(s)|\leq C$. Then,

$r=\sum_{s\in T}C(s),|r|=\Bigg{|}\sum_{s\in T}C(s)\Bigg{|}\leq\sum_{s\in
T}|C(s)|\leq C\cdot|T|$

with the triangle inequality, and it suffices to show that $|T|=O(N^{k-m})$.

Suppose $s\in T$. We know that exactly $m$ of the $s_{i}$ must be term
multiplications, and as seen in the proof of Proposition 4.6, the degree in
the $x_{i}$ of $r(s)_{j}$ for $0\leq j\leq k$ is at most $k-m$.

######  Claim 4.7.1.

Suppose $s\in T$. If $x_{i}\in D_{N,\mathfrak{r}}(s)$, suppose the first term
multiplication $x_{i}$ is in is $s_{j^{\prime}}$. Then, $s_{j}$ must be a
switch from $i_{j}$ to $i$ for some $j$, $j^{\prime}<j\leq k$.

###### Proof.

Observe that, in $r_{j^{\prime}}(s)$, $x_{i}$ is in all of the terms, and must
be removed because $r_{k}(s)$ cannot contain $x_{i}$. For the sake of
contradiction, assume the statement does not hold. If $j^{\prime}=k$, then
$r_{k}(s)=r_{j^{\prime}}(s)$ contains $x_{i}$, a contradiction since
$r_{k}(s)=dp$ for a nonzero real number $d$. Therefore, $j^{\prime}<k$. Below,
we view $r_{j}(s)$ for $1\leq j\leq k$ as a polynomial in the $x_{i}$, and a
term in $r_{j}(s)$ is a product of the $x_{i}$ with its coefficient.

Consider $s_{j}$, $j^{\prime}<j\leq k$. We see that if $s_{j}$ is a
derivative, term multiplication, or switch, all nonzero terms of $r_{j}(s)$
will have $x_{i}$. Then, in $r_{k}(s)\not=0$, the terms will contain $x_{i}$,
a contradiction. ∎

Let $S=\\{x_{i_{j}}|1\leq j\leq k\\}$. If $|D_{N,\mathfrak{r}}(s)|=d$, from
4.7.1, we see that we can find one switch from $i_{j}$ to $i$ for each
$x_{i}\in D_{N,\mathfrak{r}}(s)$, giving $d$ switches in total, where $0\leq
d\leq k-m$. Also, the number possibilities for $D_{N,\mathfrak{r}}(s)$, the
number subsets of $\\{x_{1},\ldots,x_{N}\\}\backslash S$ with size $d$, is at
most $N^{d}$. Next, suppose $X\subset\\{x_{1},\ldots,x_{N}\\}\backslash S$,
$|X|=d$. Because the variables outside of $\mathfrak{r}$ are symmetrical, the
number of $s\in T$ such that $D_{N,\mathfrak{r}}(s)=X$ is the same for all
such $X$; let this number be $T_{d}$.

In $s$, there are $m$ term multiplications, known as $\alpha(s)$ operators.
Also, there are $d\leq k-m$ operators which are switches from $i_{j}$ to $i$,
where $x_{i}\in D_{N,\mathfrak{r}}(s)$ and the $i$ are distinct, known as
$\beta(s)$ operators. The other $k-m-d$ operators, known as $\gamma(s)$
operators, can be derivatives or any switch. In this proof, there can be
overcounting of $N_{d}$, with $s\in T$ counted multiple times.

For $d$, $0\leq d\leq k-m$, there are

$\frac{k!}{m!d!(k-m-d)!}$

possible groupings of the $s_{j}$ into $\alpha(s),\beta(s)$, and $\gamma(s)$
operators. For a term multiplication, or a $\alpha(s)$ operator, suppose the
term before the derivative is $q$. We know that all $x_{i}$ in $q$ must be in
$S\cup D_{N,\mathfrak{r}}(s)$. Also, the total degree of $q$ after the
derivative is at most $k-m$. We look at degrees of the $x_{i}$ in $q$. Note
that for $x_{i_{j}}$, where $\partial_{i_{j}}q\not=0$, the degree must be at
least $1$ and at most $k-m+1$, giving $k-m+1$ possibilities. For the other
$x_{i}$, the degree must be at least $0$ and at most $k-m$, also giving
$k-m+1$ possibilities. With this, the number of possibilities for the
$\alpha(s)$ operators is at most

$(k-m+1)^{(k+|D_{N,\mathfrak{r}}(s)|)m}=(k-m+1)^{(k+d)m}.$

On the other hand, for the $\beta(s)$ operators, there being $d$ in total,
there are $d!$ possible orderings. Finally, each of the other $\gamma(s)$
operators, there being $k-m-d$ in total, have $N$ possibilities, giving a
total of $N^{k-m-d}$. Then, for a constant $C_{d}$ not depending on $N$,

$T_{d}\leq\frac{k!}{m!d!(k-m-d)!}\cdot(k-m+1)^{(k+d)m}\cdot d!\cdot
N^{k-m-d}=C_{d}\cdot N^{k-m-d}.$

From this, for $0\leq d\leq k-m$, the number of $s\in T$ such that
$|D_{N,\mathfrak{r}}(s)|=d$ is at most $N^{d}\cdot T_{d}\leq C_{d}\cdot
N^{k-m}$, where there are at most $N^{d}$ choices for $D_{N,\mathfrak{r}}(s)$.
Therefore,

$|T|\leq\left(\sum_{d=0}^{k-m}C_{d}\right)N^{k-m},$

and $|T|=O(N^{k-m})$, as desired. ∎

###  4.4. Remainders

######  Proposition 4.8.

Suppose $s=\\{a_{\nu}(N)\\}_{\nu\in P}$ is a sequence of variables over
positive integers $N$ such that for each $\nu\in P$, there exists a constant
$C>0$ such that $|a_{\nu}(N)|\leq CN$ for all positive integers $N$. Then, if
$Q(c^{\nu};N)$ is order $N^{k}$,

$\lim_{N\rightarrow\infty}\frac{Q(a_{\nu}(N);N)}{N^{k+1}}=0,$

where we let $c^{\nu}=a_{\nu}(N)$ for all $\nu\in P$.

###### Proof.

This is clear from considering each term of $Q(a_{\nu}(N);N)$, with there
being a finite number of terms. ∎

Suppose we have a polynomial $R$ in the $c_{F}^{\nu}$ which is homogeneous of
degree $k$ with order $N^{k}$. For $N\geq 1$, we can let
$c_{F}^{\nu}=c_{F_{N}}^{\nu}$, and from the conditions of Theorem 4.1, we see
that for each $\nu\in P$, if $a_{\nu}(N)=c_{F_{N}}^{\nu}$, there exists a
constant $C>0$ such that $|a_{\nu}(N)|\leq CN$ for all $N\geq 1$. Afterwards,
we use Proposition 4.8 on $R$ with $c_{F}^{\nu}=a_{\nu}(N)$ for $N\geq 1$ to
get that

(24)
$\lim_{N\rightarrow\infty}\frac{R(a_{\nu}(N);N)}{N^{k+1}}=\lim_{N\rightarrow\infty}\frac{R(c_{F_{N}}^{\nu};N)}{N^{k+1}}=0.$

We use (24) in the proof of Theorem 4.1 in Subsection 4.6 to show that as
$N\rightarrow\infty$, such remainders $R$ go to $0$. Particularly, (24) can be
used with various results from Subsections 4.4 and 4.5.

######  Proposition 4.9.

Suppose that $\lambda$ is a partition with $\ell(\lambda)=m$ and
$|\lambda|=k$. Also, suppose $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$, where
the first $\lambda_{1}$ indices are $1$, the next $\lambda_{2}$ indices are
$2$, and so forth, until the last $\lambda_{m}$ indices are $m$. Then, for
$N\geq m$,

$\frac{1}{N^{m}}\sum_{l\in
I_{N}(\lambda)}[1]\mathcal{D}_{l}(F(x_{1},\ldots,x_{N}))=[1]D_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))+R,$

where $R$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and
order $N^{k-1}$.

###### Proof.

Suppose that in $l\in I_{N}(\lambda)$, the first $\lambda_{1}$ indices are
$i_{1}$, the next $\lambda_{2}$ indices are $i_{2}$, and so forth, until the
last $\lambda_{m}$ indices are $i_{m}$, where $1\leq i_{j}\leq N$ for $1\leq
j\leq m$. From Proposition 4.7, for all $l\in I_{N}(\lambda)$,
$\mathcal{D}_{l}(F(x_{1},\ldots,x_{N}))$ has order $N^{k}$. Note that if $l$
has all $i_{j}$ distinct, $[1]\mathcal{D}_{l}(F(x_{1},\ldots,x_{N}))$ is
equal, by symmetry, to $[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$.
Consider the set $S$ of other $l$, where some of the $i_{j}$ are equal. Let

$R^{\prime}=\frac{1}{N^{m}}\sum_{l\in
S}[1]\mathcal{D}_{l}(F(x_{1},\ldots,x_{N})).$

From Corollary 4.3, $R^{\prime}$ is a polynomial in the $c_{F}^{\nu}$ which is
homogeneous of degree $k$.

Note that there are $N^{m}-O(N^{m-1})$ $l$ where all of the $i_{j}$ are
distinct, and dividing by $N^{m}$ will give

$\left(1-O\left(\frac{1}{N}\right)\right)[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N})).$

Also, $|S|=O(N^{m-1})$ and each
$\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$ is order $N^{k}$, so
$R^{\prime}$ is $\frac{1}{N^{m}}\cdot O(N^{m-1})\cdot N^{k}=O(N^{k-1})$. Then,
we get

$R=R^{\prime}-O\left(\frac{1}{N}\right)[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$

is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and order
$N^{k-1}$, as desired. ∎

######  Lemma 4.10.

Suppose that $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ are indices. For
$N\geq\max(\mathfrak{r})$, let $H$ be the set of sequences $s$ in
$T_{N}^{1}(\mathfrak{r})$ which satisfy the following conditions:

  * •

For $1\leq j\leq k$, if $s_{j}$ is the switch from $i_{j}$ to $i$,
$i\notin\mathfrak{r}$.

  * •

There do not exist integers $i$, $j_{1}$, and $j_{2}$, $1\leq i\leq N$, $1\leq
j_{1},j_{2}\leq k$, $j_{1}\not=j_{2}$, such that $s_{j_{1}}$ is the switch
from $i_{j_{1}}$ to $i$ and $s_{j_{2}}$ is the switch from $i_{j_{2}}$ to $i$.

Then,

$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=\sum_{s\in
H}r(s)_{k}+R,$

where $R$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and
order $N^{k-1}$.

###### Proof.

Suppose that $Q$ is the set $s\in T_{N}^{1}(\mathfrak{r})$ such that at least
one of the conditions is not followed. Then,

$R=\sum_{s\in Q}r(s)_{k}.$

Since $Q\subset T_{N}^{1}(\mathfrak{r})$, $R$ is a polynomial in the
$c_{F}^{\nu}$ which is homogeneous of degree $k$. It suffices to show that the
coefficient in $R$ of each term

$p=\prod_{i=1}^{m}c_{F}^{\nu_{i}}$

is order $N^{k-m-1}$ from Definition 4.5. We consider the set $T$ of $s\in Q$
which end with $p$. Following the proof of Proposition 4.7, let the
coefficient of $p$ be $r$. We want to show that $r$ is $O(N^{k-m-1})$, and we
know that $|C(s)|\leq C$ for a $C>0$ independent of $N$. Therefore, it
suffices to show that $|T|=O(N^{k-m-1})$.

In $s\in T$, there are $m$ term multiplications, or $\alpha(s)$ operators.
Also, there are $d$, $d\leq k-m$, $\beta(s)$ operators which are switches from
$i_{j}$ to $i$ where $x_{i}\in D_{N,\mathfrak{r}}(s)$, such that each
$x_{i}\in D_{N,\mathfrak{r}}(s)$ is in exactly one $\beta(s)$ operator. The
other $k-m-d$ operators are $\gamma(s)$ operators, and are derivatives or
switches.

Consider when one of the switches is from $i_{j}$ to $i\in\mathfrak{r}$. Since
$\mathfrak{r}$ and $D_{N,\mathfrak{r}}(s)$ are disjoint, the switch must be a
$\gamma(s)$ operator, and the number of possibilities for the $\gamma(s)$
operators is $O(N^{k-m-d-1})$. With the number of possibilities for
$D_{N,\mathfrak{r}}(s)$ being at most $N^{d}$ and for the other choices being
bounded, the total number of possible $s$ is then $O(N^{k-m-1})$.

For the next case, suppose there are two switches with the same $i$. Then, we
could have a $\beta(s)$ and $\gamma(s)$ switch or two $\gamma(s)$ switches
with the same $i$. For the first case, there are at most $N$ ways to pick $i$.
Afterwards, the number of possibilities for $D_{N,\mathfrak{r}}(s)$ is at most
$N^{d-1}$ and the $\gamma(s)$ switches is $O(N^{k-m-d-1})$. Where there are
$O(1)$ possibilities for the other choices, this gives $O(N^{k-m-1})$
possibilities for the first case. Next, for the second case, there are at most
$N$ ways to choose $i$. Afterwards, the number of possibilities for
$D_{N,\mathfrak{r}}(s)$ is at most $N^{d}$ and the $\gamma(s)$ switches is
$O(N^{k-m-d-2})$. Where there are $O(1)$ possibilities for the other choices,
this gives $O(N^{k-m-1})$ possibilities for the second case. Therefore,
$|T|=O(N^{k-m-1})$, as desired. ∎

###  4.5. Sequences For Different Operators

Next, we consider the operators $\mathcal{Q}_{i}^{N}$ from Section 3.
Particularly,

$N\mathcal{Q}_{i}^{N}\left(\frac{\partial_{i}F}{N}\right)=\theta\sum_{\begin{subarray}{c}1\leq
j\leq N,\\\ j\not=i\end{subarray}}(d_{i}-C_{i,j})+\frac{\partial}{\partial
x_{i}}F(x_{1},\ldots,x_{N})$

is used. In Theorem 4.13, we show that we can replace each $\mathcal{D}_{i}$
with $N\mathcal{Q}_{i}^{N}\left(\frac{\partial_{i_{j}}F}{N}\right)$.
Afterwards, in Theorem 4.14, we use the $\mathcal{R}_{i,j}^{k}$ operators to
evaluate the resulting expression. Similarly to before, for
$\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ and $N\geq\max(\mathfrak{r})$, we
let

(25)
$\mathcal{Q}_{\mathfrak{r}}^{N}(F(x_{1},\ldots,x_{N}))=N^{k}\left(\prod_{j=1}^{k}\mathcal{Q}_{i_{j}}^{N}\left(\frac{\partial_{i_{j}}F}{N}\right)\right)(1)=\sum_{s\in
T_{2}^{k}(\mathfrak{r})}r(s)_{k}.$

######  Proposition 4.11.

Let $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$. For $N\geq\max(\mathfrak{r})$,
$[1]\mathcal{Q}^{N}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$ has order $N^{k}$.

###### Proof.

We can use the same proof as Proposition 4.7, but with no derivatives and each
switch replaced by a change, with the switch from $i$ to $j$ replaced by the
change from $i$ to $j$. ∎

######  Lemma 4.12.

Suppose that $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$. For
$N\geq\max(\mathfrak{r})$, suppose $H$ is the set of sequences $s$ in
$T_{N}^{2}(\mathfrak{r})$ which satisfy the following conditions:

  * •

For $1\leq j\leq k$, if $s_{j}$ is the change from $i_{j}$ to $i$,
$i\notin\mathfrak{r}$.

  * •

There do not exist integers $i$, $j_{1}$, and $j_{2}$, $1\leq i\leq N$, $1\leq
j_{1},j_{2}\leq k$, $j_{1}\not=j_{2}$, such that $s_{j_{1}}$ is the change
from $i_{j_{1}}$ to $i$ and $s_{j_{2}}$ is the change from $i_{j_{2}}$ to $i$.

Then,

$[1]\mathcal{Q}_{\mathfrak{r}}^{N}(F(x_{1},\ldots,x_{N}))=\sum_{s\in
H}r(s)_{k}+R,$

where $R$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and
order $N^{k-1}$.

###### Proof.

We can use the same proof as Lemma 4.10, but with no derivatives and each
switch replaced by a change, with the switch from $i$ to $j$ replaced by the
change from $i$ to $j$. ∎

######  Theorem 4.13.

Suppose that $k$ is a positive integer and $\mathfrak{r}=\\{i_{j}\\}_{1\leq
j\leq k}$. For $N\geq\max(\mathfrak{r})$,

(26)
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=[1]\mathcal{Q}_{\mathfrak{r}}^{N}(F(x_{1},\ldots,x_{N}))+R,$

where $R$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and
order $N^{k-1}$.

###### Proof.

We note that in $T_{N}^{1}(\mathfrak{r})$, the sequences can have derivatives,
but cannot in $T_{N}^{2}(\mathfrak{r})$, where $T_{N}^{1}(\mathfrak{r})$ and
$T_{N}^{2}(\mathfrak{r})$ correspond to the left and right hand side of (26),
respectively. Then, let $T_{N}^{3}(\mathfrak{r})$ be the set of $s\in
T_{N}^{1}(\mathfrak{r})$ which do not contain derivatives.

Suppose that $H$ is the set of $s$ in $T_{N}^{1}(\mathfrak{r})$ satisfying the
Lemma 4.10 conditions. From the Lemma,

$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=\sum_{s\in
H}r(s)_{k}+R_{1},$

where $R_{1}$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$
and order $N^{k-1}$. Consider $H_{1}=H\cap T_{N}^{3}(\mathfrak{r})$. We see
that $H_{1}$ is the set of $s\in T_{N}^{1}(\mathfrak{r})$ satisfying the
conditions of Lemma 4.10 with only switches and term multiplications, and

$\sum_{s\in H}r(s)_{k}=\sum_{s\in H_{1}}r(s)_{k}+R_{2},R_{2}=\sum_{s\in
H\cap(T_{N}^{1}(\mathfrak{r})\backslash T_{N}^{3}(\mathfrak{r}))}r(s)_{k}.$

Also, let $H_{2}$ be the set of $s$ in $T_{N}^{2}(\mathfrak{r})$ satisfying
the Lemma 4.12 conditions. By the Lemma,

$[1]\mathcal{Q}_{\mathfrak{r}}^{N}(F(x_{1},\ldots,x_{N}))=\sum_{s\in
H_{2}}r(s)_{k}+R_{3},$

where $R_{3}$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$
and order $N^{k-1}$.

######  Claim 4.13.1.

$R_{2}$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and
order $N^{k-1}$

###### Proof.

We bound the number of $s\in T_{N}^{1}(\mathfrak{r})\backslash
T_{N}^{3}(\mathfrak{r})$, the $s\in T_{N}^{1}(\mathfrak{r})$ which have a
derivative, such that $r(s)_{k}=dp$, where

$p=\prod_{i=1}^{m}c_{F}^{\nu_{i}}.$

Looking at the proof of Proposition 4.7, the derivatives of $s$ only appear in
$\gamma(s)$ operators. Where $|D_{N,\mathfrak{r}}(s)|=d$, the number of ways
to choose the $\gamma(s)$ operators such that there is at least $1$ derivative
is $O(N^{k-m-d-1})$. Then, we see that the number of $s$ in
$T_{N}^{1}(\mathfrak{r})\backslash T_{N}^{3}(\mathfrak{r})$, and therefore
$H\cap(T_{N}^{1}(\mathfrak{r})\backslash T_{N}^{3}(\mathfrak{r}))$, ending
with $p$ is $O(N^{k-m-1})$, as needed. ∎

For a sequence $s$ in $H_{1}$, let $c(s)$ be the sequence in $H_{2}$ where if
$s_{j}$ is the switch from $i_{j}$ to $i$, then $c(s)_{j}$ is the change from
$i_{j}$ to $i$, and all other operators are the same. We have that $s\in
H_{1}$ have switches and term multiplications, while $s\in H_{2}$ have changes
and term multiplications. Clearly, $c$ is a bijection from $H_{1}$ to $H_{2}$.

######  Claim 4.13.2.

For all $s\in H_{1}$, $r(s)_{k}=r(c(s))_{k}$.

###### Proof.

Consider $s\in T_{N}(\mathfrak{r})$ with term multiplications, switches, and
changes such that:

  * •

For $1\leq j\leq k$, if $s_{j}$ is the switch or change from $i_{j}$ to $i$,
$i\notin\mathfrak{r}$.

  * •

There do not exist integers $i$, $j_{1}$, and $j_{2}$, $1\leq i\leq N$, $1\leq
j_{1},j_{2}\leq k$, $j_{1}\not=j_{2}$, such that $s_{j_{1}}$ is the switch or
change from $i_{j_{1}}$ to $i$ and $s_{j_{2}}$ is the switch or change from
$i_{j_{2}}$ to $i$.

Suppose that $s_{j}$, $1\leq j\leq k$, is a switch from $i_{j}$ to $i$ and
$s_{j^{\prime}}$, $j<j^{\prime}\leq k$, are not changes. It suffices to show
that if we make $s_{j}$ the change from $i_{j}$ to $i$, $r(s)_{k}$ is the
same. Then, for $s\in H_{1}$, if the switches of $s$ are
$s_{j_{1}},s_{j_{2}},\ldots,s_{j_{r}}$, $1\leq j_{1}<\cdots<j_{r}\leq k$, we
could make $s_{j_{u}}$ into a change from $u=1$ to $r$ and get $c(s)$. This
would give $r(c(s))_{k}=r(s)_{k}$.

Suppose $q$ is a term of $r(s)_{j-1}$. We look at $s_{j}(q)$, which is a sum
of terms or $0$. Suppose $q^{\prime}$ is a term of $s_{j}(q)$ and the degree
of $x_{i}$ in $q^{\prime}$ is at least $1$. Afterwards, where
$j<j^{\prime}\leq k$, using that $s\in H_{1}$ is the switch from
$i_{j^{\prime}}$ to $i^{\prime}\not=i$ or a term multiplication. In
$r(s)_{k}$, the degree of $x_{i}$ must be $0$, but to decrease the degree of
$x_{i}$, $s_{j^{\prime}}$, $j^{\prime}>j$, will make all terms $0$. Therefore,
$q^{\prime}$ will not contribute to $r(s)_{k}$. So, the only term of
$s_{j}(q)$ which will contribute to $r(s)_{k}$ will be $q^{\prime}$ with
degree of $x_{i}$ equal to $0$, and since $s_{j}$ is the switch from $i_{j}$
to $i$, in $s_{j}(q)$, there will be at most one such $q^{\prime}$. With this,
we can set

$s_{j}(q)=q^{\prime}$

if such $q^{\prime}$ exists, and $s_{j}(q)=0$ if not.

Assume the degree of $x_{i}$ and $x_{i_{j}}$ are both at least $1$ in $q$.
Then, in $s_{j}(q)$, the degree in $x_{i}$ of all terms will be at least $1$,
so $q$ does not contribute to $r(s)_{k}$. Now, replace $s_{j}$ with the change
from $i_{j}$ to $i$. Note that in $\theta(d_{i_{j}}-C_{i_{j},i})(q)=\theta
d_{i_{j}}(q)$, the degree of $x_{i}$ is at least $1$, and because of this,
using the logic from above, after replacing $s_{j}$ there are also no
contributions to $r(s)_{k}$. Therefore, in this case, $r(s)_{k}$ is unchanged.

Next, suppose the degree of $x_{i}$ is $0$ and $x_{i_{j}}$ is at least $1$ in
$q$. In this case, $q^{\prime}$ exists and will be $\theta d_{i_{j}}(q)$. But,
since the degree of $x_{i_{j}}$ is at least $1$,

$\theta(d_{i_{j}}-C_{i_{j},i})(q)=\theta d_{i_{j}}(q)=q^{\prime}=s_{j}(q).$

Therefore, we can replace $s_{j}$ with $\theta(d_{i_{j}}-C_{i_{j},i})$ and
$r(s)_{k}$ will be the same.

Consider when the degree of $x_{i_{j}}$ in $q$ is $0$. If the degree of
$x_{i}$ is at least $1$ in $q$, then $q^{\prime}$ will exist, and be $-\theta
C_{i_{j},i}(q)$. Otherwise, if the degree of $x_{i}$ is $0$,
$s_{j}(q)=0=-\theta C_{i_{j},i}(q)$. However, since the degree of $x_{i_{j}}$
in $q$ is $0$,

$\theta(d_{i_{j}}-C_{i_{j},i})(q)=-\theta C_{{i_{j}},i}(q)=s_{j}(q).$

Therefore, after replacing $s_{j}$ with $\theta(d_{i_{j}}-C_{i_{j},i})$,
$r(s)_{k}$ is unchanged here as well. ∎

Since $c:H_{1}\rightarrow H_{2}$ is a bijection, with 4.13.2,

$\sum_{s\in H_{1}}r(s)_{k}=\sum_{s\in H_{2}}r(s)_{k},$

and we are done. ∎

######  Theorem 4.14.

Suppose that $\lambda$ is a partition with $\ell(\lambda)=m$ and
$|\lambda|=k$. Then, for $N\geq m$, if, for $1\leq i\leq N$,
$\vec{X}_{i}=(x_{1},\ldots,x_{i},0,\ldots,0)$ is $\vec{x}$ with $x_{j}=0$ for
$i+1\leq j\leq N$,

$\displaystyle[1]N^{k}\prod_{i=1}^{m}\mathcal{Q}_{i}^{N}\left(\frac{\partial_{i}F}{N}\right)^{\lambda_{i}}(1)$
$\displaystyle=[1]\prod_{i=1}^{m}\left((N-i+1)^{\lambda_{i}}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}\left(\frac{\partial_{N-i+1}F(\vec{X}_{N-i+1})}{N-i+1}\right)\right)(1)+R,$

where $R$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$ and
order $N^{k-1}$.

###### Proof.

By symmetry,

$[1]N^{k}\prod_{i=1}^{m}\mathcal{Q}_{i}^{N}\left(\frac{\partial_{i}F}{N}\right)^{\lambda_{i}}(1)=[1]N^{k}\prod_{i=1}^{m}\mathcal{Q}_{N-i+1}^{N}\left(\frac{\partial_{N-i+1}F}{N}\right)^{\lambda_{i}}(1).$

Here, $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$ consists of $k$ indices, the
first $\lambda_{1}$ being $N$, the next $\lambda_{2}$ being $N-1$, and so
forth, until the last $\lambda_{m}$ are $N-m+1$. Let $T$ be the set of
sequences $s$ with indices $\mathfrak{r}$ over $x_{1},\ldots,x_{N}$ consisting
of term multiplications and changes such that:

  * •

For $1\leq j\leq m-1$, after $\lambda_{1}+\cdots+\lambda_{j}$ operators, the
next $\lambda_{j+1}$ operators cannot have changes from $N-j$ to $i$ and term
multiplications which contain $x_{i}$, $N-j+1\leq i\leq N$.

  * •

$r(s)_{k}$ has degree $0$ in the $x_{i}$.

Note that it is possible for $s\in T$ to have $r(s)_{k}=0$.

Suppose $0\leq d\leq m$. For a sequence $s$ of term multiplications and
changes, let $r_{d}(s)_{i}$, $0\leq i\leq k$, be as follows. Set
$r_{d}(s)_{0}=1$, and for $1\leq i\leq k$,
$i\not=\lambda_{1}+\cdots+\lambda_{j}$, $1\leq j\leq d$, let
$r_{d}(s)_{i}=s_{i}(r_{d}(s)_{i-1})$. Also, for $1\leq j\leq d$, let

$r_{d}(s)_{\lambda_{1}+\cdots+\lambda_{j}}=(s_{\lambda_{1}+\cdots+\lambda_{j}}(r_{d}(s)_{\lambda_{1}+\cdots+\lambda_{j}-1}))|_{x_{N-j+1}=0}.$

Note that $r_{0}(s)_{i}=r(s)_{i}$, $0\leq i\leq k$.

######  Claim 4.14.1.

For $0\leq i\leq k$ and any sequence $s$ of term multiplications and changes,
there exists a positive integer $n$ such that for $0\leq d\leq m$,
$r_{d}(s)_{i}$ is homogeneous in $x_{1},\ldots,x_{N}$ with degree $n$ or
$r_{d}(s)_{i}=0$.

###### Proof.

This is clear by induction from $i=0$ to $k$. ∎

######  Claim 4.14.2.

For all $s\in T$, $r_{d}(s)_{k}$ is the same for $0\leq d\leq m$.

###### Proof.

To show this, we prove that $r_{d}(s)_{k}=r_{d-1}(s)_{k}$, $1\leq d\leq m$.
Observe that $r_{d}(s)_{i}=r_{d-1}(s)_{i}$ for $1\leq
i\leq\lambda_{1}+\cdots+\lambda_{d}-1$, and

$r_{d-1}(s)_{k}-r_{d}(s)_{k}=s_{k}\circ\cdots\circ
s_{\lambda_{1}+\cdots+\lambda_{d}+1}(r_{d-1}(s)_{\lambda_{1}+\cdots+\lambda_{d}}-r_{d-1}(s)_{\lambda_{1}+\cdots+\lambda_{d}}|_{x_{N-d+1}=0}),$

where each term of
$r_{d-1}(s)_{\lambda_{1}+\cdots+\lambda_{d}}-r_{d-1}(s)_{\lambda_{1}+\cdots+\lambda_{d}}|_{x_{N-d+1}=0}$
contains $x_{N-d+1}$. But, if $\lambda_{1}+\cdots+\lambda_{d}+1\leq i\leq k$,
$s_{i}$ does not have a $x_{N-d+1}$ since $s\in T$, and $s_{i}$ cannot remove
$x_{N-d+1}$, unless all terms become $0$. Then, all nonzero terms of
$r_{d}(s)_{k}-r_{d-1}(s)_{k}$ contain $x_{N-d+1}$. Since $s\in T$,
$r_{0}(s)_{k}=r(s)_{k}$ has degree $0$ in the $x_{i}$, so $r_{d-1}(s)_{k}$ and
$r_{d}(s)_{k}$, and hence $r_{d}(s)_{k}-r_{d-1}(s)_{k}$, have degree $0$ in
the $x_{i}$ from 4.14.1. But, all nonzero terms of
$r_{d}(s)_{k}-r_{d-1}(s)_{k}$ contain $x_{i}$, and therefore,
$r_{d}(s)_{k}-r_{d-1}(s)_{k}=0$. ∎

######  Claim 4.14.3.

$[1]\prod_{i=1}^{m}\left((N-i+1)^{\lambda_{i}}\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}\left(\frac{\partial_{N-i+1}F(\vec{X}_{N-i+1})}{N-i+1}\right)\right)(1)=\sum_{s\in
T\cap T_{N}^{2}(\mathfrak{r})}r(s)_{k}.$

###### Proof.

The left hand side of the above expression is

$\sum_{s\in T}r_{m}(s)_{k}=\sum_{s\in T}r(s)_{k}=\sum_{s\in T\cap
T_{N}^{2}(\mathfrak{r})}r(s)_{k},$

since, by 4.14.2, $r_{m}(s)_{k}=r(s)_{k}$ for $s\in T$, and $r(s)_{k}\not=0$
if and only if $s\in T\cap T_{N}^{2}(\mathfrak{r})$. ∎

Suppose that $T^{\prime}$ is the set of $s\in T_{N}^{2}(\mathfrak{r})$
satisfying the Lemma 4.12 conditions. From the Lemma,

$[1]N^{k}\prod_{i=1}^{m}\mathcal{Q}_{N-i+1}^{N}\left(\frac{\partial_{N-i+1}F}{N}\right)^{\lambda_{i}}(1)=\sum_{s\in
T^{\prime}}r(s)_{k}+R_{1},$

where $R_{1}$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$
and order $N^{k-1}$.

######  Claim 4.14.4.

The set $T^{\prime}$ is a subset of $T$.

###### Proof.

It suffices to show that if $s\in T^{\prime}$, for
$d>\lambda_{1}+\cdots+\lambda_{j}$, $s_{d}$ does not contain $x_{i}$,
$N-j+1\leq i\leq N$. For the sake of contradiction, assume $s\in T^{\prime}$
and $s_{d}$ contains $x_{i}$, $N-j+1\leq i\leq N$, for some
$d>\lambda_{1}+\cdots+\lambda_{j}$. From the conditions of Lemma 4.12, $x_{i}$
must be in a term multiplication. After, $x_{i}$ will be in all terms of
$r(s)_{d}$. But, $x_{i}$ cannot be removed by an operator unless all the terms
become $0$, a contradiction to $s\in T_{N}^{2}(\mathfrak{r})$. ∎

Since $T^{\prime}\subset T_{N}^{2}(\mathfrak{r})$ as well, $T^{\prime}\subset
T\cap T_{N}^{2}(\mathfrak{r})$. Again, we consider the coefficient of

$p=\prod_{i=1}^{m}c_{F}^{\nu_{i}}.$

Let $U$ and $U^{\prime}$ be the set of $s$ in $T_{N}^{2}(\mathfrak{r})$ and
$T^{\prime}$, respectively, such that $r(s)_{k}=dp$, $d\not=0$. Then, $U\cap
T$ will be the set of $s$ in $T\cap T_{N}^{2}(\mathfrak{r})$ such that
$r(s)_{k}=dp$, $d\not=0$. We show that $|U\backslash U^{\prime}|$ is
$O(N^{k-m-1})$; where $U^{\prime}\subset U\cap T$, $|U\cap T\backslash
U^{\prime}|$ will be $O(N^{k-m-1})$ as well. But, as in Lemma 4.12, this
follows from the proof of Lemma 4.10 without derivates and using changes
instead of switches. After, with Proposition 4.6,

$\sum_{s\in T\cap T_{N}^{2}(\mathfrak{r})}r(s)_{k}=\sum_{s\in
T^{\prime}}r(s)_{k}+R_{2},$

where $R_{2}$ is a homogeneous polynomial in the $c_{F}^{\nu}$ with degree $k$
and order $N^{k-1}$, completing the proof. ∎

###  4.6. Proof of Theorem 4.1

By Proposition 4.9, as $N\rightarrow\infty$,

$\lim_{N\rightarrow\infty}\left(\frac{1}{N^{m+|\lambda|}}\sum_{l\in
I_{N}(\lambda)}[1]\mathcal{D}_{l}(F_{N}(x_{1},\ldots,x_{N}))\right)=\lim_{N\rightarrow\infty}\frac{1}{N^{|\lambda|}}[1]\mathcal{D}_{\mathfrak{r}}(x_{1},\ldots,x_{N}),$

using (24) with $k=|\lambda|-1$. Afterwards, from Theorem 4.13 and Theorem
4.14,

$\lim_{N\rightarrow\infty}\frac{1}{N^{|\lambda|}}[1]\mathcal{D}_{\mathfrak{r}}(x_{1},\ldots,x_{N})=\lim_{N\rightarrow\infty}[1]\prod_{i=1}^{m}\left(\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}\left(\frac{\partial_{N-i+1}F_{N}(\vec{X}_{N-i+1})}{N-i+1}\right)\right)(1),$

where (24) is used again.

To evaluate this, note that

(27)
$s_{1,N-i+1}\frac{\partial_{N-i+1}F_{N}(\vec{X}_{N-i+1})}{N-i+1}=\frac{\partial_{1}F_{N}(\vec{X}_{N-i+1})}{N-i+1},$

and the coefficient of $x_{1}^{d}M_{\nu}(x_{2},\ldots,x_{N-i+1})$ in (27) is
$\frac{(d+1)c_{F_{N}}^{\nu+(d+1)}}{N-i+1}$ for $\nu\in P$ with $\ell(\nu)\leq
N-i$. Here, we can have $\nu=0$. But, from (20),

$\lim_{N\rightarrow\infty}\frac{(d+1)c_{F_{N}}^{\nu+(d+1)}}{N-i+1}=\frac{(d+1)(\ell(\nu)+1)c_{\nu+(d+1)}}{|\nu|+d+1},$

and the limiting sequence of (27) outside of $1$ for $1\leq i\leq m$ is
$f=\\{c^{d,\nu}\\}_{d\geq 0,\nu\in P}$, where
$c^{d,\nu}=\frac{(d+1)(\ell(\nu)+1)c_{\nu+(d+1)}}{|\nu|+d+1}$. From Theorem
3.13,

$\displaystyle\lim_{N\rightarrow\infty}[1]\prod_{i=1}^{m}\left(\mathcal{R}_{N-i+1,N-i+1}^{\lambda_{i}}\left(\frac{\partial_{N-i+1}F_{N}(\vec{X}_{N-i+1})}{N-i+1}\right)\right)(1)=\prod_{i=1}^{m}\left(\sum_{\begin{subarray}{c}\pi\in
NC(\lambda_{i})\end{subarray}}\prod_{B\in\pi}c_{|B|}(f)\right).$

With this, $\\{\mu_{N}\\}_{N\geq 1}$ satisfy a LLN, where

$\lim_{N\rightarrow\infty}\mathbb{E}_{\mu_{N}}\left(\prod_{i=1}^{m}p_{\lambda_{i}}^{N}\right)=\prod_{i=1}^{m}m_{\lambda_{i}},m_{k}=\sum_{\pi\in
NC(k)}\prod_{B\in\pi}c_{|B|}(f)$

for $k\geq 1$. Suppose that for $\nu\in P$, for $d\geq 1$, $N_{d}(\nu)$ of the
components of $\nu$ are $d$. Then, if $\nu^{\prime}+(d)=\nu$,
$\ell(\nu)P(\nu^{\prime})=N_{d}(\nu)P(\nu)$, and for $l\geq 1$,

$\displaystyle c_{l}(f)$
$\displaystyle=\theta^{l-1}\sum_{\begin{subarray}{c}\nu^{\prime}\in P,d\geq
0,\\\
|\nu^{\prime}|+d=l-1\end{subarray}}(-1)^{\ell(\nu^{\prime})}P(\nu^{\prime})\frac{(d+1)(\ell(\nu^{\prime})+1)c_{\nu^{\prime}+(d+1)}}{|\nu^{\prime}|+d+1}$
$\displaystyle=\theta^{l-1}\sum_{\nu\in
P,|\nu|=l}(-1)^{\ell(\nu)-1}\frac{\ell(\nu)c_{\nu}}{|\nu|}\sum_{\begin{subarray}{c}\nu^{\prime}\in
P,d\geq 1,\\\ \nu^{\prime}+(d)=\nu\end{subarray}}dP(\nu^{\prime})$
$\displaystyle=\theta^{l-1}\sum_{\nu\in
P,|\nu|=l}(-1)^{\ell(\nu)-1}\frac{P(\nu)c_{\nu}}{|\nu|}\sum_{\begin{subarray}{c}\nu^{\prime}\in
P,d\geq 1,\\\ \nu^{\prime}+(d)=\nu\end{subarray}}dN_{d}(\nu)$
$\displaystyle=\theta^{l-1}\sum_{\nu\in
P,|\nu|=l}(-1)^{\ell(\nu)-1}P(\nu)c_{\nu},$

which gives (4), as required.

##  5\. Probability Measures

###  5.1. Bessel Generating Functions

A Gelfand-Tsetlin pattern with $N$ rows is a set of real numbers
$\\{\lambda_{i}^{j}\\}_{1\leq i\leq j\leq N}$ such that
$\lambda_{i}^{j+1}\leq\lambda_{i}^{j}\leq\lambda_{i+1}^{j+1}$ for $1\leq i\leq
j\leq N-1$. We let $\mathcal{G}_{N}$ denote the set of Gelfand-Tsetlin
patterns with $N$ rows.

######  Definition 5.1.

The multivariate Bessel function for $a_{1}<\cdots<a_{N}$ is

$\displaystyle B_{(a_{1},\ldots,a_{N})}(x_{1},\ldots,x_{N};\theta)=$
$\displaystyle\int_{\begin{subarray}{c}d=\\{y_{i}^{k}\\}\in\mathcal{G}_{N},\\\
y_{i}^{N}=a_{i},1\leq i\leq
N\end{subarray}}\left[\exp\left(\sum_{k=1}^{N}x_{k}\cdot\left(\sum_{i=1}^{k}y_{i}^{k}-\sum_{j=1}^{k-1}y_{j}^{k-1}\right)\right)\right]\cdot$
$\displaystyle f_{(a_{1},\ldots,a_{N})}(d)\prod_{1\leq i\leq k\leq
N-1}dy_{i}^{k},$

where for $d=\\{y_{i}^{k}\\}\in\mathcal{G}_{N}$ such that
$y_{i}^{N}=a_{i},1\leq i\leq N$,

$\displaystyle f_{(a_{1},\ldots,a_{N})}(d)=$
$\displaystyle\prod_{k=1}^{N}\frac{\Gamma(k\theta)}{\Gamma(\theta)^{k}}\prod_{1\leq
i<j\leq N}(a_{j}-a_{i})^{1-2\theta}\cdot$
$\displaystyle\prod_{k=1}^{N-1}\left[\prod_{1\leq i<j\leq
k}(y_{j}^{k}-y_{i}^{k})^{2-2\theta}\prod_{a=1}^{k}\prod_{b=1}^{k+1}|y_{a}^{k}-y_{b}^{k+1}|^{\theta-1}\right].$

######  Definition 5.2.

The Bessel generating function of $\mu\in\mathcal{M}_{N}$ is defined as

(28) $G_{\theta}(x_{1},\ldots,x_{N};\mu):=\int_{a_{1}\leq\cdots\leq
a_{N}}B_{(a_{1},\ldots,a_{N})}(x_{1},\ldots,x_{N};\theta)\mu(da_{1},\ldots,da_{N}).$

Observe that $f_{(a_{1},\ldots,a_{N})}(d)$ is the probability density of the
$\theta$-corners process or $\beta$-corners process with top row
$a_{1}<\cdots<a_{N}$ in [crystalrandommat], [matrix]. Note that this is also
referred to as the orbital $\beta$ process in [betaprocess]. A scaled
multivariate Bessel function and the Bessel generating function for $\theta=1$
are studied in [Airy]. The following result is well known.

######  Theorem 5.3.

For any symmetric polynomial $P(x_{1},\ldots,x_{N})$ and each $N$-tuple of
reals $(a_{1}<a_{2}<\cdots<a_{N})$,

(29)
$P(\mathcal{D}_{1},\ldots,\mathcal{D}_{N})B_{(a_{1},\ldots,a_{N})}=P(a_{1},\ldots,a_{N})B_{(a_{1},\ldots,a_{N})}.$

From Lemma 1.3, if $\mu\in\mathcal{M}_{N}$ is exponentially decaying, then the
BGF of $\mu$ converges in an open neighborhood of the origin in
$\mathbb{C}^{N}$. Additionally, from (29),

$\mathcal{P}_{k}B_{(a_{1},\ldots,a_{N})}=\left(\sum_{i=1}^{N}a_{i}^{k}\right)B_{(a_{1},\ldots,a_{N})},$

which can be used to prove the following Proposition.

######  Proposition 5.4 ([matrix]*Proposition 2.11).

For a positive integer $s$, let $k_{1},\ldots,k_{s}$ be positive integers.
Suppose $\mu\in\mathcal{M}_{N}$ is exponentially decaying. Then,

(30)
$\left(\prod_{i=1}^{s}\mathcal{P}_{k_{i}}\right)G_{\theta}(x_{1},\ldots,x_{N};\mu)\bigg{|}_{x_{i}=0,1\leq
i\leq
N}=\mathbb{E}_{\mu}\left(\prod_{i=1}^{s}\left(\sum_{j=1}^{N}a_{j}^{k_{i}}\right)\right).$

###  5.2. Setup

In Theorem 1.5, let
$F_{N}(x_{1},\ldots,x_{N})=\ln(G_{N}(x_{1},\ldots,x_{N};\theta))$ for $N\geq
1$. Note that for $\nu\in P$ with $|\nu|\geq 1$, if $i_{1},\ldots,i_{r}$ are
positive integers such that $\sigma((i_{1},\ldots,i_{r}))=\nu$, then

$\frac{\partial}{\partial x_{i_{1}}}\cdots\frac{\partial}{\partial
x_{i_{r}}}\ln(G_{N}^{\theta})\bigg{|}_{x_{i}=0,1\leq i\leq
N}=\frac{|\nu|!}{P(\nu)}\cdot c_{F_{N}}^{\nu}.$

Also, from Proposition 5.4, for exponentially decaying
$\mu\in\mathcal{M}_{N}$, for a partition $\lambda$ with $|\lambda|\geq 1$,

(31)
$\begin{split}\mathbb{E}_{\mu}\left(\prod_{i=1}^{\ell(\lambda)}p_{\lambda_{i}}^{N}\right)&=\mathbb{E}_{\mu}\left(\prod_{i=1}^{\ell(\lambda)}\left(\frac{1}{N}\sum_{j=1}^{N}\left(\frac{a_{j}}{N}\right)^{\lambda_{i}}\right)\right)\\\
&=\frac{1}{N^{\ell(\lambda)+|\lambda|}}\left(\prod_{i=1}^{\ell(\lambda)}\mathcal{P}_{\lambda_{i}}\right)G_{\theta}(x_{1},\ldots,x_{N};\mu)\bigg{|}_{x_{i}=0,1\leq
i\leq N}.\end{split}$

To get that $\\{\mu_{N}\\}_{N\geq 1}$ satisfy a LLN, we use (31) with
$\mu=\mu_{N}$ and take the $N\rightarrow\infty$ limit.

######  Proposition 5.5 ([matrix]*Lemma 5.2).

Suppose that $F$ is a $(k+1)$-times continuously differentiable function on a
neighborhood of $(0,\ldots,0)\in\mathbb{C}^{N}$, with Taylor series expansion

$F(x_{1},\ldots,x_{N})=\sum_{\nu,|\nu|\leq k,\ell(\nu)\leq
N}c_{F}^{\nu}M_{\nu}(\vec{x})+O(\left\lVert x\right\rVert^{k+1}).$

Then, for any $\lambda=(\lambda_{1},\ldots,\lambda_{m})$ with $|\lambda|=k$,

$\left(\prod_{i=1}^{m}\mathcal{P}_{\lambda_{i}}\right)\exp(F(x_{1},\ldots,x_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq
N}=\left(\prod_{i=1}^{m}\mathcal{P}_{\lambda_{i}}\right)\exp(\tilde{F}(x_{1},\ldots,x_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq N},$

where

$\tilde{F}(x_{1},\ldots,x_{N})=\sum_{\nu,|\nu|\leq k,\ell(\nu)\leq
N}c_{F}^{\nu}M_{\nu}(\vec{x}).$

Suppose $\mu\in\mathcal{M}_{N}$ is exponentially decaying. Then, from Lemma
1.3, $G_{\theta}(x_{1},\ldots,x_{N};\mu)$ and therefore
$F(x_{1},\ldots,x_{N})=\ln(G_{\theta}(x_{1},\ldots,x_{N};\mu))$ are symmetric
as and holomorphic in a neighborhood of the origin. Moreover,
$F(0,\ldots,0)=\ln(G_{\theta}(0,\ldots,0;\mu))=0$.

###  5.3. Proof of Theorem 1.5

Suppose we have a symmetric polynomial $\tilde{F}(x_{1},\ldots,x_{N})$ with
$\tilde{F}(0,\ldots,0)=0$. We have that

$\displaystyle\frac{\partial}{\partial
x_{i}}R(x_{1},\ldots,x_{N})\exp(\tilde{F}(x_{1},\ldots,x_{N}))$
$\displaystyle=\left(\frac{\partial}{\partial
x_{i}}R(x_{1},\ldots,x_{N})+R(x_{1},\ldots,x_{N})\frac{\partial}{\partial
x_{i}}\tilde{F}(x_{1},\ldots,x_{N})\right)\exp(\tilde{F}(x_{1},\ldots,x_{N})).$

Moreover, because $\tilde{F}(x_{1},\ldots,x_{N})$ is symmetric,

$s_{i,j}(R(x_{1},\ldots,x_{N})\exp(\tilde{F}(x_{1},\ldots,x_{N})))=(s_{i,j}R(x_{1},\ldots,x_{N}))\exp(\tilde{F}(x_{1},\ldots,x_{N})).$

Also, since $\tilde{F}(0,\ldots,0)=0$,
$R(x_{1},\ldots,x_{N})\exp(\tilde{F}(x_{1},\ldots,x_{N}))$ at $x_{i}=0$,
$1\leq i\leq N$ is $R(0,\ldots,0)$. From above, where
$\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$

$\displaystyle\left(\prod_{j=1}^{k}\mathcal{D}_{i_{j}}\right)\exp(\tilde{F}(x_{1},\ldots,x_{N}))$
$\displaystyle=\left(\prod_{j=1}^{k}\left(\mathcal{D}_{i_{j}}+\frac{\partial}{\partial
x_{i_{j}}}\tilde{F}(x_{1},\ldots,x_{N})\right)\right)(1)\cdot\exp(\tilde{F}(x_{1},\ldots,x_{N}))$
$\displaystyle=\mathcal{D}_{\mathfrak{r}}(\tilde{F}(x_{1},\ldots,x_{N}))\cdot\exp(\tilde{F}(x_{1},\ldots,x_{N})),$

and

$\left(\prod_{j=1}^{k}\mathcal{D}_{i_{j}}\right)\exp(\tilde{F}(x_{1},\ldots,x_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq N}=[1]\mathcal{D}_{\mathfrak{r}}(\tilde{F}(x_{1},\ldots,x_{N})).$

Note that since $\tilde{F}(x_{1},\ldots,x_{N})$ is a polynomial,
$\mathcal{D}_{\mathfrak{r}}(\tilde{F}(x_{1},\ldots,x_{N}))$ is also a
polynomial. Also, for a partition $\lambda$ with $\ell(\lambda)=m$, we have
that

(32)
$\left(\prod_{i=1}^{m}\mathcal{P}_{\lambda_{i}}\right)\exp(\tilde{F}(x_{1},\ldots,x_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq N}=\sum_{l\in
I_{N}(\lambda)}[1]\mathcal{D}_{l}(\tilde{F}(x_{1},\ldots,x_{N})).$

Next, let $\lambda$ be a partition of length $m$ with $|\lambda|\geq 1$.
Suppose $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq|\lambda|}$, where the first
$\lambda_{1}$ indices are $1$, the next $\lambda_{2}$ are $2$, and so forth,
until the last $\lambda_{m}$ are $m$. For Theorem 1.5, we look at the limit of
(31) for $\mu=\mu_{N}$ as $N\rightarrow\infty$. For $N\geq 1$, we let
$F_{N}(x_{1},\ldots,x_{N})=\ln(G_{\theta}(x_{1},\ldots,x_{N};\mu_{N})$. For
$N\geq 1$, because $\mu_{N}$ is exponentially decaying, we see that
$F_{N}(x_{1},\ldots,x_{N})$ is symmetric and holomorphic in a neighborhood of
the origin. Then, we can use Proposition 5.5 on $F_{N}(x_{1},\ldots,x_{N})$.

Suppose that $\tilde{F}_{N}(x_{1},\ldots,x_{N})$ is the polynomial for
$F_{N}(x_{1},\ldots,x_{N})$ from Proposition 5.5 with $k=|\lambda|$. By
Proposition 5.5, (31), and (32),

$\displaystyle\mathbb{E}_{\mu_{N}}\left(\prod_{i=1}^{m}p_{\lambda_{i}}^{N}\right)$
$\displaystyle=\frac{1}{N^{m+|\lambda|}}\left(\prod_{i=1}^{m}\mathcal{P}_{\lambda_{i}}\right)\exp(F_{N}(x_{1},\ldots,x_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq N}$
$\displaystyle=\frac{1}{N^{m+|\lambda|}}\left(\prod_{i=1}^{m}\mathcal{P}_{\lambda_{i}}\right)\exp(\tilde{F}_{N}(x_{1},\ldots,x_{N}))\bigg{|}_{x_{i}=0,1\leq
i\leq N}$ $\displaystyle=\frac{1}{N^{m+|\lambda|}}\sum_{l\in
I_{N}(\lambda)}[1]\mathcal{D}_{l}(\tilde{F}_{N}(x_{1},\ldots,x_{N})).$

Where $\tilde{F}_{N}(x_{1},\ldots,x_{N})$, a symmetric polynomial, is a
symmetric formal series for $N\geq 1$, with the conditions in Theorem 1.5, we
can use Theorem 4.1 on $\\{\tilde{F}_{N}(x_{1},\ldots,x_{N})\\}_{N\geq 1}$.
Then, with the above expression, we are done.

##  6\. Eigenvalue Distribution of Random Matrices

From [AGZ], for all $\beta>0$, consider the measure, known as the
$\beta$-Hermite ensemble, on $\mathcal{M}_{N}$ with probability density

(33) $d_{N,\beta}(x_{1},\ldots,x_{N})=C_{N,\beta}\prod_{1\leq i<j\leq
N}|x_{i}-x_{j}|^{\beta}\prod_{i=1}^{N}e^{-\frac{\beta x_{i}^{2}}{4}},$

where

$C_{N,\beta}=(2\pi)^{-\frac{N}{2}}\left(\frac{\beta}{2}\right)^{\frac{\beta
N(N-1)}{4}+\frac{N}{2}}\Gamma\left(\frac{\beta}{2}\right)^{N}\left(\prod_{i=1}^{N}\Gamma\left(\frac{i\beta}{2}\right)\right)^{-1}.$

In the paper, we scale $d_{N,\beta}(x_{1},\ldots,x_{N})$ by $\sqrt{N}$ to get
the probability density

(34) $d^{s}_{N,\beta}(x_{1},\ldots,x_{N})=\frac{C_{N,\beta}}{N^{\frac{\beta
N(N-1)}{4}+\frac{N}{2}}}\prod_{1\leq i<j\leq
N}|x_{i}-x_{j}|^{\beta}\prod_{i=1}^{N}e^{-\frac{\beta x_{i}^{2}}{4N}}$

on $\mathcal{M}_{N}$. For $\beta=1,2,$ and $4$, $d_{N,\beta}$ is the
probability density of the ordered eigenvalues of the GUE, GOE, and GSE,
respectively. Also, from [AGZ], there exist random tridiagonal matrices with
eigenvalue distribution being the $\beta$-Hermitian ensemble for $\beta>0$.

Suppose $\\{\mu_{N}\\}_{N\geq 1}$ are a sequence of probability measures such
that $\mu_{N}$ is in $\mathcal{M}_{N}$ and has density $d^{s}_{N,\beta}$. In
Proposition 6.3, we see that if $\beta>0$ is fixed, Theorem 1.5 on
$\\{\mu_{N}\\}_{N\geq 1}$ gives the moments as well as LLN satisfaction. This
is an example of how the results of this paper can be used on probability
measures.

######  Proposition 6.1 ([betaprocess]*Corollary 3.7).

Let $a_{1},\ldots,a_{N},y_{1},\ldots,y_{N},c$ be $2N+1$ arbitrary complex
numbers. Then,

$B_{(ca_{1},\ldots,ca_{N})}(y_{1},\ldots,y_{N})=B_{(a_{1},\ldots,a_{N})}(cy_{1},\ldots,cy_{N}).$

######  Proposition 6.2.

Suppose $\mu\in\mathcal{M}_{N}$ is exponentially decaying. For $c>0$, let
$c\mu$ be $\mu$ scaled by $c$. Then,

$G_{\theta}(x_{1},\ldots,x_{N};c\mu)=G_{\theta}(cx_{1},\ldots,cx_{N};\mu).$

###### Proof.

The probability density of $c\mu$ is
$c\mu(a_{1},\ldots,a_{N})=\frac{1}{c^{N}}\mu\left(\frac{a_{1}}{c},\ldots,\frac{a_{N}}{c}\right)$.
With Proposition 6.1, where $a^{\prime}=(a_{1}^{\prime}\leq\cdots\leq
a_{N}^{\prime})$, $a_{i}^{\prime}=\frac{a_{i}}{c}$, $1\leq i\leq N$,

$\displaystyle G_{\theta}(x_{1},\ldots,x_{N};c\mu)$
$\displaystyle=\int_{a=(a_{1}\leq\cdots\leq
a_{N})}B_{a}(x_{1},\ldots,x_{N};\theta)c\mu(a_{1},\ldots,a_{N})da_{1}\cdots
da_{N}$ $\displaystyle=\int_{a^{\prime}=(a_{1}^{\prime}\leq\cdots\leq
a_{N}^{\prime})}B_{ca^{\prime}}(x_{1},\ldots,x_{N};\theta)c^{N}c\mu(ca_{1}^{\prime},\ldots,ca_{N}^{\prime})da_{1}^{\prime}\cdots
da_{N}^{\prime}$ $\displaystyle=\int_{a^{\prime}=(a_{1}^{\prime}\leq\cdots\leq
a_{N}^{\prime})}B_{a^{\prime}}(cx_{1},\ldots,cx_{N};\theta)\mu(a_{1}^{\prime},\ldots,a_{N}^{\prime})da_{1}^{\prime}\cdots
da_{N}^{\prime}$ $\displaystyle=G_{\theta}(cx_{1},\ldots,cx_{N};\mu).$

∎

######  Proposition 6.3.

Suppose $\theta>0$ is fixed. Consider the sequence $\\{\mu_{N}\\}_{N\geq 1}$
of probability measures such that for all positive integers $N$,
$\mu_{N}\in\mathcal{M}_{N}$ and $\mu_{N}$ has probability density
$d_{N,2\theta}^{s}$ from (34). Then, $\\{\mu_{N}\\}_{N\geq 1}$ satisfy a LLN,
with $m_{2k-1}=0$ and $m_{2k}=\frac{1}{k+1}\binom{2k}{k}$ for $k\geq 1$.

###### Proof.

For $N\geq 1$, consider $\frac{\mu_{N}}{\sqrt{N}}$, which has density
$d_{N,2\theta}$ in (33). It is easy to see that $\mu_{N}$ and
$\frac{\mu_{N}}{\sqrt{N}}$ are exponentially decaying. From
[betaprocess]*Proposition 4.2,

(35)
$G_{\theta}\left(x_{1},\ldots,x_{N};\frac{\mu_{N}}{\sqrt{N}}\right)=\exp\left(\frac{1}{2\theta}\sum_{i=1}^{N}x_{i}^{2}\right).$

Then, from Proposition 6.2 and (35),

$G_{\theta}(x_{1},\ldots,x_{N};\mu_{N})=G_{\theta}\left(\sqrt{N}x_{1},\ldots,\sqrt{N}x_{N};\frac{\mu_{N}}{\sqrt{N}}\right)=\exp\left(\frac{N}{2\theta}\sum_{i=1}^{N}x_{i}^{2}\right).$

Afterwards, by Theorem 1.5, $\\{\mu_{N}\\}_{N\geq 1}$ satisfy a LLN with
$c_{(2)}=\frac{1}{\theta}$ and $c_{\nu}=0$ for other $\nu\in P$, $|\nu|\geq
1$. For $k\geq 1$, $m_{2k-1}=0$ and $m_{2k}=|T(k)|$, where $T(k)$ is the set
of $\pi\in NC(2k)$ which have all blocks of size $2$. But,
$|T(k)|=\frac{1}{k+1}\binom{2k}{k}$, the $k$th Catalan number. ∎

From Proposition 6.3, if $(a_{1}\leq\cdots\leq a_{N})$ are distributed with
density $d_{N,\beta}(x_{1},\ldots,x_{N})$, for positive integer $s$ and
positive integers $k_{i}$, $1\leq i\leq s$,

$\lim_{N\rightarrow\infty}\mathbb{E}_{\mu_{N}}\left(\prod_{i=1}^{s}\left(\frac{1}{N}\sum_{i=1}^{N}\left(\frac{a_{i}}{\sqrt{N}}\right)^{k_{i}}\right)\right)=\prod_{i=1}^{s}m_{k_{i}},$

with the moments $m_{k}$ given in Proposition 6.3. Sometimes, as seen above,
for sequences $\\{\mu_{N}\\}_{N\geq 1}$ of exponentially decaying probability
measures, for $N\geq 1$, we can scale $\mu_{N}$ by an appropriate power of $N$
and use Proposition 6.2 to satisfy the conditions of Theorem 1.5.

##  Appendix A Combinatorial Formulas

######  Lemma A.1.

For all nonnegative integers $a$, $b$, and $m$ such that $m\leq a$,

(36)
$\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b}{k-m}}{\binom{a+b}{k}}(-1)^{k}=\frac{(-1)^{m}a!b!}{(a+b)!}.$

###### Proof.

We use induction on $m$, from $m=a$ to $0$. The $m=a$ base case is clear.
Suppose the statement holds for $m+1$, $0\leq m<a$. We show the statement
holds for $m$ with induction on $b$, where the base case $b=0$ is clear. For
the inductive step, assume (36) for $m$ holds for $b\geq 0$. We want to show
(36) for $m$ holds for $b+1$. Note that

$\displaystyle\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b+1}{k-m}}{\binom{a+b+1}{k}}(-1)^{k}=\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b}{k-m}}{\binom{a+b}{k}}\cdot\frac{(b+1)(a+b+1-k)}{(a+b+1)(b+1-k+m)}\cdot(-1)^{k}$
$\displaystyle=\frac{b+1}{a+b+1}\left(\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b}{k-m}}{\binom{a+b}{k}}\left(1+\frac{a-m}{b+1-k+m}\right)(-1)^{k}\right)$
$\displaystyle=\frac{(-1)^{m}a!(b+1)!}{(a+b+1)!}+\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b}{k-m}}{\binom{a+b}{k}}\cdot\frac{(a-m)(-1)^{k}}{b+1-k+m},$
$\displaystyle=\frac{(-1)^{m}a!(b+1)!}{(a+b+1)!}+\frac{a-m}{b+1}\cdot\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b+1}{k-m}}{\binom{a+b}{k}}(-1)^{k}$
$\displaystyle=\frac{(-1)^{m}a!(b+1)!}{(a+b+1)!}+\frac{a-m}{b+1}\cdot\left(\sum_{k=m}^{a}\frac{\binom{a}{k}\binom{b}{k-m}}{\binom{a+b}{k}}(-1)^{k}+\sum_{k=m+1}^{a}\frac{\binom{a}{k}\binom{b}{k-m-1}}{\binom{a+b}{k}}(-1)^{k}\right)$
$\displaystyle=\frac{(-1)^{m}a!(b+1)!}{(a+b+1)!}$

with the inductive hypotheses, completing the proof. ∎

######  Lemma A.2.

For a positive integer $k$,

$\sum_{\begin{subarray}{c}\pi\in NC(k+1),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}b_{|B_{1}|}\cdot\prod_{i=2}^{m}a_{|B_{i}|}=\sum_{\begin{subarray}{c}\pi\in
NC(k),\\\ \pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}\left(b_{|B_{1}|+1}+\sum_{j=1}^{|B_{1}|}a_{j}b_{|B_{1}|+1-j}\right)\cdot\prod_{i=2}^{m}a_{|B_{i}|}.$

###### Proof.

Suppose that for $k\geq 1$, $NC^{\prime}(k)=NC(\\{1,3,4,\ldots,k+1\\})$. For
$\pi\in NC^{\prime}(k)$, define the set $S(\pi)$ of partitions $\lambda$ in
$NC(k+1)$ as follows. The set $S(\pi)$ consists of partitions $\lambda$ such
that $\lambda$ is $\pi$ with $2$ added to $B_{1}$, or $\lambda$ is $\pi$ with
$B_{1}^{\prime}$ and $B_{1}^{\prime\prime}$ instead of $B_{1}$, where
$B_{1}^{\prime}$ is the last $j-1$ elements of $B_{1}$ with $1$, and
$B_{1}^{\prime\prime}$ is the remaining elements of $B_{1}$ with $2$ added,
where $1\leq j\leq|B_{1}|$. Clearly, $S(\pi)\subset NC(k+1)$.

######  Claim A.2.1.

The union of the $S(\pi)$ for $\pi\in NC^{\prime}(k)$ is $NC(k+1)$ and the
$S(\pi)$ are disjoint for $\pi\in NC^{\prime}(k)$.

###### Proof.

For $\lambda\in NC(k+1)$, suppose the decomposition into blocks is

$C_{1}\sqcup C_{2}\sqcup\cdots\sqcup C_{m},$

following the same conditions as the decomposition in (12). Note that $2\in
C_{1}$ or $2\in C_{2}$.

If $2\in C_{1}$, let $\pi=B_{1}\sqcup\cdots\sqcup B_{m}\in NC^{\prime}(k)$
have $B_{i}=C_{i}$ for $2\leq i\leq m$ and $B_{1}=C_{1}\backslash\\{2\\}$. We
see that $\pi$ with $2$ added to $B_{1}$ is $\lambda$, so $\lambda\in S(\pi)$.
Suppose $\lambda\in S(\pi)$, $\pi=B_{1}\sqcup\cdots\sqcup B_{m}\in
NC^{\prime}(k)$. However, $C_{1}$ of $\lambda$ contains a $2$ and since the
only element of $S(\pi)$ with first block containing $2$ is $\pi$ with $2$
added to $B_{1}$, we know that $\pi$ must be $\lambda$ with $2$ removed from
$C_{1}$. So, $\pi$ is unique, and $\lambda$ is not in $S(\pi)$ for more than
one $\pi\in NC^{\prime}(k)$.

Suppose that $2\in C_{2}$. Then, let $\pi=B_{1}\sqcup\cdots\sqcup B_{m-1}\in
NC^{\prime}(k)$ have $B_{1}=\left(C_{1}\cup C_{2}\right)\backslash\\{2\\}$ and
$B_{i-1}=C_{i}$ for $3\leq i\leq m$. Since $\lambda$ is noncrossing, $\pi$ is
also noncrossing. Note that $1\in C_{1}$ and $2\in C_{2}$. Suppose $j_{1}\in
C_{1}$ and $j_{2}\in C_{2}$, where $j_{1},j_{2}>2$. Because $\lambda$ is
noncrossing, we must have that $j_{1}>j_{2}.$ Therefore, the elements of
$C_{1}$ which are not $1$ are greater than all of the elements of $C_{2}$. We
see that, where $1\leq j=|C_{1}|\leq|B_{1}|$, $C_{1}$ is the last $j-1$
elements of $B_{1}$ with $1$ added, and $C_{2}$ is the remaining elements of
$B_{1}$ with $2$ added. Therefore, $\lambda\in S(\pi)$.

Assume $\lambda\in S(\pi)$, where $\pi=B_{1}\sqcup\cdots\sqcup B_{m}\in
NC^{\prime}(k)$. Since $\lambda$ does not have $2$ in the first block,
$\lambda$ is $\pi$ with $B_{1}^{\prime}$ and $B_{1}^{\prime\prime}$ instead of
$B_{1}$, where $B_{1}^{\prime}$ is the last $j-1$ elements of $B_{1}$ with $1$
added, and $B_{1}^{\prime\prime}$ is the remaining elements of $B_{1}$ with
$2$ added, for some $1\leq j\leq|B_{1}|$. Then, the first and second block of
$\lambda$ would be $C_{1}=B_{1}^{\prime}$ and $C_{2}=B_{1}^{\prime\prime}$,
respectively. But, in $\pi$,

$B_{1}=(B_{1}^{\prime}\cup B_{1}^{\prime\prime})/\\{2\\}=(C_{1}\cup
C_{2})/\\{2\\},$

and $B_{i}=C_{i+1}$ for $i\geq 2$. Thus, $\pi$ is unique. ∎

With A.2.1,

$\displaystyle\sum_{\begin{subarray}{c}\pi\in NC(k+1),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}b_{|B_{1}|}\cdot\prod_{i=2}^{m}a_{|B_{i}|}$
$\displaystyle=\sum_{\pi\in NC^{\prime}(k)}\sum_{\begin{subarray}{c}\lambda\in
S(\pi),\\\ \lambda=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}b_{|B_{1}|}\cdot\prod_{i=2}^{m}a_{|B_{i}|}$
$\displaystyle=\sum_{\begin{subarray}{c}\pi\in NC^{\prime}(k),\\\
\pi=B_{1}\sqcup\cdots\sqcup
B_{m}\end{subarray}}\left(b_{|B_{1}|+1}+\sum_{j=1}^{|B_{1}|}a_{j}b_{|B_{1}|+1-j}\right)\cdot\prod_{i=2}^{m}a_{|B_{i}|},$

giving Lemma A.2 since $|\\{1,3,4,\ldots,k+1\\}|=|\\{1,2,3,\ldots,k\\}|=k$. We
are done. ∎

######  Lemma A.3.

For all positive integers $n$,

(37) $\sum_{\pi\in
NC(n)}\prod_{B\in\pi}r_{|B|}=\frac{1}{n+1}\cdot[z^{-1}]\left(\frac{1}{z}+\sum_{j=1}^{\infty}r_{j}z^{j-1}\right)^{n+1}.$

###### Proof.

For nonnegative integers $m_{1},m_{2},\ldots,m_{n}$ such that
$\sum_{k=1}^{n}k\cdot m_{k}=n$, let the number of $\pi\in NC(n)$ with $m_{k}$
blocks of size $k$, $1\leq k\leq n$ be $N(m_{1},m_{2},\ldots,m_{n})$. From
[KREWERAS],

$N(m_{1},m_{2},\ldots,m_{n})=\frac{n!}{\left(n-\sum_{k=1}^{n}m_{k}+1\right)!\displaystyle\prod_{k=1}^{n}m_{k}!}.$

Then,

(38) $\sum_{\pi\in
NC(n)}\prod_{B\in\pi}r_{|B|}=\sum_{\begin{subarray}{c}d,m_{k}\geq 0,1\leq
k\leq n,\\\ d+\sum_{k=1}^{n}m_{k}=n+1,\\\ \sum_{k=1}^{n}k\cdot
m_{k}=n\end{subarray}}\frac{n!}{d!}\cdot\prod_{k=1}^{n}\frac{r_{k}^{m_{k}}}{m_{k}!}.$

We have that the right hand side of (37) is

$\frac{1}{n+1}\cdot\sum_{\begin{subarray}{c}j_{i}\geq 0,1\leq i\leq n+1,\\\
j_{1}+\cdots+j_{n+1}=n\end{subarray}}\prod_{i=1}^{n+1}r_{j_{i}}=\sum_{\begin{subarray}{c}m_{k}\geq
0,0\leq k\leq n,\\\ \sum_{k=0}^{n}m_{k}=n+1,\\\ \sum_{k=1}^{n}k\cdot
m_{k}=n\end{subarray}}\frac{n!}{m_{0}!}\cdot\prod_{k=1}^{n}\frac{r_{k}^{m_{k}}}{m_{k}!},$

which is (38) with $m_{0}$ as $d$. We are done. ∎

##  Appendix B Polynomial Coefficients

######  Lemma B.1.

Suppose the indices are $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$. For a term

$p=\prod_{i=1}^{m}c_{F}^{\nu_{i}},$

there exists a polynomial $f(x)$ such that the coefficient of $p$ in
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$ is $f(N)$ for
$N\geq\max(\mathfrak{r})$.

###### Proof.

Suppose $N\geq\max(\mathfrak{r})$. If the sum of the $|\nu_{i}|$ is not $k$,
from Corollary 4.3, the coefficient is $f(N)=0$. Now, suppose that the sum of
the $|\nu_{i}|$ is $k$. Let $S$ be the set of $x_{i_{j}}$ and $T$ be the set
of $s\in T_{N}^{1}(\mathfrak{r})$ such that $r(s)_{k}=dp$ for a nonzero $d$.
Then, the ending coefficient of $p$ will be the sum of $C(s)$ for $s\in T$.

Also, for $s\in T$, let $X(s)$ be the set $x_{i}$ such that
$i\notin\mathfrak{r}$ and $x_{i}$ appears in switches or term multiplications
of $s$. Since each $x_{i}\in X(s)$ from term multiplications must be
eliminated by at least one switch by 4.7.1, and each other $x_{i}\in X(s)$
appears in at least one switch, because there are at most $k$ switches,
$|X(s)|\leq k$.

For each $A\subset\\{x_{1},\ldots,x_{N}\\}\backslash S$, where the number of
$s\in T$ such that $X(s)=A$ is finite, let

$g(A)=\sum_{s\in T,X(s)=A}C(s).$

Suppose $D$ is an integer, $0\leq D\leq k$. By symmetry, $g(A)$ is the same
for all $A\subset\\{x_{1},\ldots,x_{N}\\}\backslash S$, such that $|A|=D$. Let
this value be $r_{D}$, where $r_{D}$ is independent of $N$. Since
$0\leq|X(s)|\leq k$, we see that the coefficient of $p$ is $f(N)$ for

$f(x)=\sum_{0\leq D\leq k}r_{D}\cdot\binom{x-|S|}{D}.$

where $\binom{N-|S|}{D}$ is the number of $A$ such that $|A|=D$. We are done.
∎

Suppose $k$ is a positive integer. Let $S_{k}$ be the set of all multisets $q$
of partitions with size at least $1$ such that the sum of $|\nu|$ for $\nu\in
q$ is $k$. Additionally, let

$\mathfrak{P}_{k}=\Bigg{\\{}\prod_{\nu\in q}c_{F}^{\nu}\Bigg{|}q\in
S_{k}\Bigg{\\}},$

where if

$p=\prod_{\nu\in q}c_{F}^{\nu},$

the length of $p$ is $\ell(p)=|q|$.

For indices $\mathfrak{r}$ of length $k$, for $p$ in $\mathfrak{P}_{k}$, let
the coefficient of $p$ in
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))$ be
$f_{p,\mathfrak{r}}(N)$ for $N\geq\max(\mathfrak{r})$. Note that for
$N\geq\max(\mathfrak{r})$,

(39)
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=\sum_{p\in\mathfrak{P}_{k}}f_{p,\mathfrak{r}}(N)p$

from Lemma B.1. Also, from Proposition 4.7, $f_{p,\mathfrak{r}}(x)$ has degree
at most $k-\ell(p)$. Let the coefficient of $x^{k-\ell(p)}$ in
$f_{p,\mathfrak{r}}(x)$ be $s_{\mathfrak{r}}(p)$. Note that
$s_{\mathfrak{r}}(p)$, which can be $0$, is the leading order coefficient of
$f_{p,\mathfrak{r}}(x)$.

######  Proposition B.2.

For $k\geq 1$, consider indices $\mathfrak{r}=\\{i_{j}\\}_{1\leq j\leq k}$.
Where $\lambda=\sigma(\mathfrak{r})$ and $m=\ell(\lambda)$,

(40)
$\sum_{p\in\mathfrak{P}_{k}}s_{\mathfrak{r}}(p)p=\prod_{i=1}^{m}\left(\sum_{\pi\in
NC(\lambda_{i})}\prod_{B\in\pi}\theta^{|B|-1}\left(\sum_{\nu\in
P,|\nu|=|B|}(-1)^{\ell(\nu)-1}\frac{|\nu|P(\nu)}{\ell(\nu)}c_{F}^{\nu}\right)\right).$

###### Proof.

Consider the indices $\mathfrak{r}^{\prime}$ of length $k$, where the first
$\lambda_{1}$ indices are $1$, the next $\lambda_{2}$ are $2$, and so forth,
until the last $\lambda_{m}$ indices are $m$. From commutativity of the
$\mathcal{D}_{i}$ and symmetry in the $x_{i}$,
$[1]\mathcal{D}_{\mathfrak{r}}(F(x_{1},\ldots,x_{N}))=[1]\mathcal{D}_{\mathfrak{r}^{\prime}}(F(x_{1},\ldots,x_{N}))$
for all $N\geq\max(\mathfrak{r})$. Therefore,
$f_{p,\mathfrak{r}}(x)=f_{p,\mathfrak{r}^{\prime}}(x)$ and
$s_{\mathfrak{r}}(p)=s_{\mathfrak{r}^{\prime}}(p)$ for all
$p\in\mathfrak{P}_{k}$.

Suppose $c_{F}^{\nu}$ is a real for all $\nu\in P$, $|\nu|\geq 1$ and
$F_{N}(x_{1},\ldots,x_{N})$ has $c_{F_{N}}^{\nu}=Nc_{F}^{\nu}$ for $\nu\in P$
with $1\leq\ell(\nu)\leq N$ and $c_{F_{N}}^{\nu}=0$ for $\nu\in P$ with
$\ell(\nu)\geq N+1$. Then, with $a_{\nu}(N)=c_{F_{N}}^{\nu}$, from Proposition
4.8, Proposition 4.9, and Theorem 4.1 for $\lambda$,

$\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N^{k}}[1]\mathcal{D}_{\mathfrak{r}^{\prime}}(F_{N}(x_{1},\ldots,x_{N}))=\lim_{N\rightarrow\infty}\left(\frac{1}{N^{k+m}}\sum_{l\in
I_{N}(\lambda)}[1]\mathcal{D}_{l}(F_{N}(x_{1},\ldots,x_{N}))\right)$
$\displaystyle=\prod_{i=1}^{m}\left(\sum_{\pi\in
NC(\lambda_{i})}\prod_{B\in\pi}\theta^{|B|-1}\left(\sum_{\nu\in
P,|\nu|=|B|}(-1)^{\ell(\nu)-1}\frac{|\nu|P(\nu)}{\ell(\nu)}c_{F}^{\nu}\right)\right),$

where $c_{\nu}=\frac{|\nu|}{\ell(\nu)}c_{F}^{\nu}$. However, substituting
$c_{F_{N}}^{\nu}=Nc_{F}^{\nu}$ for $c_{F}^{\nu}$ in (39) gives

$\frac{1}{N^{k}}[1]\mathcal{D}_{\mathfrak{r}^{\prime}}(F_{N}(x_{1},\ldots,x_{N}))=\sum_{p\in\mathfrak{P}_{k}}\frac{f_{p,\mathfrak{r}^{\prime}}(N)p}{N^{k-\ell(p)}},$

where the $c_{F}^{\nu}$ and thus $p\in\mathfrak{P}_{k}$ are constants. Since
$f_{p,\mathfrak{r}^{\prime}}(x)$ has degree at most $k-\ell(p)$ from
Proposition 4.7,
$\lim_{N\rightarrow\infty}\frac{f_{p,\mathfrak{r}^{\prime}}(N)}{N^{k-\ell(p)}}=s_{\mathfrak{r}^{\prime}}(p)p=s_{\mathfrak{r}}(p)p$.
Taking the limit $N\rightarrow\infty$ gives

$\lim_{N\rightarrow\infty}\frac{1}{N^{k}}[1]\mathcal{D}_{\mathfrak{r}^{\prime}}(F_{N}(x_{1},\ldots,x_{N}))=\sum_{p\in\mathfrak{P}_{k}}s_{\mathfrak{r}^{\prime}}(p)p=\sum_{p\in\mathfrak{P}_{k}}s_{\mathfrak{r}}(p)p.$

Since the above expression is a finite polynomial in the $c_{F}^{\nu}$,
varying the $c_{F}^{\nu}$ to be any reals gives the result. ∎

## References