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\begin{document}
\section*{Math 222 Fall 2022, Lecture 30: Generating functions}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/22fco}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/22fco}}}

\setcounter{section}{5}

\section{A brief introduction to generating functions}

We have previously seen how the concept of polynomials can be used to prove
combinatorial identities. Namely, we have used the polynomial identity trick
(Lecture 14, Corollary 2.2.5) to generalize identities from $x\in\mathbb{N}$
to $x\in\mathbb{R}$ (under certain conditions). The underlying idea is that if
two polynomials $P$ and $Q$ (in a single indeterminate, with real
coefficients) satisfy $P\left(  x\right)  =Q\left(  x\right)  $ for all
$x\in\mathbb{N}$, then they are identical as polynomials (and thus $P\left(
x\right)  =Q\left(  x\right)  $ also holds for all $x\in\mathbb{R}$).

Today I will show another way to use polynomials -- this time, without
computing any of their values. That is, we will just regard polynomials as
\textquotedblleft formal expressions\textquotedblright\ of the form
$a_{0}+a_{1}X+a_{2}X^{2}+\cdots+a_{n}X^{n}$, which are added and subtracted
coefficientwise and multiplied by the standard rule (distributivity and
$\left(  aX^{n}\right)  \cdot\left(  bX^{m}\right)  =abX^{n+m}$). We will not
substitute any actual numbers for the indeterminate $X$, but merely manipulate
polynomials algebraically and then extract certain coefficients. This doesn't
sound like a particularly deep or useful activity, but it turns out to give
short proofs to various binomial identities that are otherwise far from trivial.

This activity is actually a simple instance of the technique of
\textquotedblleft\textbf{generating functions}\textquotedblright. In general,
a generating function doesn't have to be a polynomial, but can be any
\textbf{formal power series} (i.e., roughly speaking, an \textquotedblleft
infinite polynomial\textquotedblright, or \textquotedblleft polynomial with
infinitely many nonzero coefficients\textquotedblright, such as $1X+2X^{2}%
+3X^{3}+\cdots=\sum_{k\in\mathbb{N}}kX^{k}$). A proper introduction into this
technique would work at this general level (such introductions can be found,
e.g., in \cite[Chapter 3]{21s} or \cite[Chapter 7 (in the 1st edition)]%
{Loehr-BC}\footnote{I am deliberately citing the 1st edition of this text,
since the 2nd edition does a worse job properly defining formal power series.}
or \cite{Sambal22}; see also \cite{Niven69} for a short overview). Here,
however, we will content ourselves with polynomials.

\subsection{Notations and the binomial formula revisited}

\begin{convention}
In this whole chapter, $X$ shall denote an indeterminate, and we will be
working mostly with polynomials in a single indeterminate $X$ with rational coefficients.
\end{convention}

\begin{definition}
Let $P$ be a polynomial in the indeterminate $X$. Let $n\in\mathbb{N}$. Then,
the $X^{n}$-coefficient of $P$ (that is, the coefficient before $X^{n}$ in
$P$) will be denoted by $\left[  X^{n}\right]  P$.
\end{definition}

(Yes, this is one more meaning for square brackets we are introducing.
Fortunately, it does not clash with the notation $\left[  n\right]  $ for the
set $\left\{  1,2,\ldots,n\right\}  $, nor with the notation $\left[
\mathcal{A}\right]  $ for the truth value of $\mathcal{A}$.)

For example,%
\begin{align*}
\left[  X^{3}\right]  \left(  \left(  1+X\right)  ^{5}\right)   &  =\left[
X^{3}\right]  \left(  1+5X+10X^{2}+10X^{3}+5X^{4}+1\right)
=10\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  X^{8}\right]  \left(  \left(  1+X\right)  ^{5}\right)   &  =0.
\end{align*}


Let us recall how two polynomials are multiplied:

\begin{proposition}
\label{prop.gf.PQ=}Let%
\begin{align*}
P  &  =a_{0}X^{0}+a_{1}X^{1}+a_{2}X^{2}+\cdots=\sum_{m\in\mathbb{N}}a_{m}%
X^{m}\ \ \ \ \ \ \ \ \ \ \text{and}\\
Q  &  =b_{0}X^{0}+b_{1}X^{1}+b_{2}X^{2}+\cdots=\sum_{m\in\mathbb{N}}b_{m}X^{m}%
\end{align*}
be two polynomials.\footnotemark\ Then,%
\begin{equation}
PQ=\sum_{m\in\mathbb{N}}\left(  \sum_{i=0}^{m}a_{i}b_{m-i}\right)  X^{m}.
\label{eq.prop.gf.PQ=.PQ=}%
\end{equation}
In other words, for each $m\in\mathbb{N}$, the coefficient of $X^{m}$ in the
polynomial $PQ$ is given by%
\begin{equation}
\left[  X^{m}\right]  \left(  PQ\right)  =\sum_{i=0}^{m}a_{i}b_{m-i}%
=\sum_{i=0}^{m}\left(  \left[  X^{i}\right]  P\right)  \cdot\left(  \left[
X^{m-i}\right]  Q\right)  . \label{eq.prop.gf.PQ=.XnPQ=}%
\end{equation}

\end{proposition}

\footnotetext{The sums are infinite, but are supposed to contain only finitely
many nonzero addends. For example, when we write $P=\sum_{m\in\mathbb{N}}%
a_{m}X^{m}$, we assume that all sufficiently large $m\in\mathbb{N}$ satisfy
$a_{m}=0$. (If we drop this requirement, then we end up with \textbf{formal
power series} rather than polynomials; for now let's not go there.)}Depending
on how you define polynomials, Proposition \ref{prop.gf.PQ=} is either the
definition of the product $PQ$ or an easy consequence of that definition. It
is precisely what is obtained when you multiply the equalities $P=\sum
_{m\in\mathbb{N}}a_{m}X^{m}=\sum_{i\in\mathbb{N}}a_{i}X^{i}$ and $Q=\sum
_{m\in\mathbb{N}}b_{m}X^{m}=\sum_{j\in\mathbb{N}}b_{j}X^{j}$ and expand (using
distributivity):%
\begin{align*}
PQ  &  =\left(  \sum_{i\in\mathbb{N}}a_{i}X^{i}\right)  \left(  \sum
_{j\in\mathbb{N}}b_{j}X^{j}\right)  =\underbrace{\sum_{i\in\mathbb{N}}%
\ \ \sum_{j\in\mathbb{N}}}_{=\sum_{\left(  i,j\right)  \in\mathbb{N}%
\times\mathbb{N}}}\underbrace{a_{i}X^{i}b_{j}X^{j}}_{=a_{i}b_{j}X^{i+j}}\\
&  =\sum_{\left(  i,j\right)  \in\mathbb{N}\times\mathbb{N}}a_{i}b_{j}%
X^{i+j}=\sum_{m\in\mathbb{N}}\ \ \sum_{\substack{\left(  i,j\right)
\in\mathbb{N}\times\mathbb{N};\\i+j=m}}a_{i}b_{j}\underbrace{X^{i+j}%
}_{\substack{=X^{m}\\\text{(since }i+j=m\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split up the sum}\\
\text{according to the value of }i+j
\end{array}
\right) \\
&  =\sum_{m\in\mathbb{N}}\ \ \sum_{\substack{\left(  i,j\right)  \in
\mathbb{N}\times\mathbb{N};\\i+j=m}}a_{i}b_{j}X^{m}=\sum_{m\in\mathbb{N}%
}\underbrace{\left(  \sum_{\substack{\left(  i,j\right)  \in\mathbb{N}%
\times\mathbb{N};\\i+j=m}}a_{i}b_{j}\right)  }_{\substack{=\sum_{i=0}^{m}%
a_{i}b_{m-i}\\\text{(since the pairs }\left(  i,j\right)  \in\mathbb{N}%
\times\mathbb{N}\text{ satisfying }i+j=m\\\text{are precisely the pairs
}\left(  0,m\right)  \text{, }\left(  1,m-1\right)  \text{, }\ldots\text{,
}\left(  m,0\right)  \text{,}\\\text{that is, the pairs }\left(  i,m-i\right)
\text{ for }i\in\left\{  0,1,\ldots,m\right\}  \text{)}}}X^{m}\\
&  =\sum_{m\in\mathbb{N}}\left(  \sum_{i=0}^{m}a_{i}b_{m-i}\right)  X^{m}.
\end{align*}
This proves (\ref{eq.prop.gf.PQ=.PQ=}). The equality
(\ref{eq.prop.gf.PQ=.XnPQ=}) easily follows from this.

Addition of polynomials is simpler: It is coefficientwise. That is, for any
two polynomials $P$ and $Q$ and any $n\in\mathbb{N}$, we have%
\begin{equation}
\left[  X^{n}\right]  \left(  P+Q\right)  =\left(  \left[  X^{n}\right]
P\right)  +\left(  \left[  X^{n}\right]  Q\right)  . \label{eq.gf.XnP+Q}%
\end{equation}


\bigskip

Recall the binomial formula (Lecture 7, Theorem 1.3.19), which states that%
\begin{equation}
\left(  x+y\right)  ^{n}=\sum_{k=0}^{n}\dbinom{n}{k}x^{k}y^{n-k}
\label{eq.gf.binform}%
\end{equation}
for any $n\in\mathbb{N}$ and any real numbers $x$ and $y$. Since we proved
this formula by formal algebraic manipulations, we see instantaneously that it
holds not only when $x$ and $y$ are real numbers, but also more generally when
$x$ and $y$ are two polynomials (and the same proof that we gave applies
verbatim in this case). This has a few useful consequences:

\begin{lemma}
\label{lem.gf.binform1}Let $n\in\mathbb{N}$. Then:

\textbf{(a)} We have%
\[
\left(  1+X\right)  ^{n}=\sum_{m\in\mathbb{N}}\dbinom{n}{m}X^{m}.
\]


\textbf{(b)} We have%
\[
\left(  1-X\right)  ^{n}=\sum_{m\in\mathbb{N}}\left(  -1\right)  ^{m}%
\dbinom{n}{m}X^{m}.
\]


\textbf{(c)} We have
\[
\left(  1-X^{2}\right)  ^{n}=\sum_{m\in\mathbb{N}}\left(  -1\right)
^{m}\dbinom{n}{m}X^{2m}.
\]

\end{lemma}

\begin{proof}
\textbf{(a)} We have seen that the binomial formula (\ref{eq.gf.binform})
holds for any polynomials $x$ and $y$. Hence, we can apply this formula to
$x=X$ and $y=1$. We thus obtain\footnote{Recall that the symbol
\textquotedblleft$\overset{0}{=}$\textquotedblright\ means \textquotedblleft
equal because the two sums differ only in addends which are
zero\textquotedblright. In other words, we write \textquotedblleft%
$a\overset{0}{=}b$\textquotedblright\ if $a$ and $b$ are two sums that differ
only in addends that are zero. As we know, any two such sums $a$ and $b$ are
necessarily equal.}%
\begin{align*}
\left(  X+1\right)  ^{n}  &  =\sum_{k=0}^{n}\dbinom{n}{k}X^{k}%
\underbrace{1^{n-k}}_{=1}=\sum_{k=0}^{n}\dbinom{n}{k}X^{k}\\
&  =\sum_{m=0}^{n}\dbinom{n}{m}X^{m}\overset{0}{=}\sum_{m\in\mathbb{N}}%
\dbinom{n}{m}X^{m}%
\end{align*}
(since every integer $m>n$ satisfies $\dbinom{n}{m}=0$ and therefore
$\dbinom{n}{m}X^{m}=0X^{m}=0$). Since $X+1=1+X$, this proves Lemma
\ref{lem.gf.binform1} \textbf{(a)}. \medskip

\textbf{(b)} This is similar to part \textbf{(a)}. Namely, apply the binomial
formula (\ref{eq.gf.binform}) to $x=-X$ and $y=1$. We thus obtain%
\begin{align*}
\left(  -X+1\right)  ^{n}  &  =\sum_{k=0}^{n}\dbinom{n}{k}\underbrace{\left(
-X\right)  ^{k}}_{=\left(  -1\right)  ^{k}X^{k}}\underbrace{1^{n-k}}_{=1}%
=\sum_{k=0}^{n}\dbinom{n}{k}\left(  -1\right)  ^{k}X^{k}=\sum_{k=0}^{n}\left(
-1\right)  ^{k}\dbinom{n}{k}X^{k}\\
&  =\sum_{m=0}^{n}\left(  -1\right)  ^{m}\dbinom{n}{m}X^{m}\overset{0}{=}%
\sum_{m\in\mathbb{N}}\left(  -1\right)  ^{m}\dbinom{n}{m}X^{m}.
\end{align*}
Since $-X+1=1-X$, this proves Lemma \ref{lem.gf.binform1} \textbf{(b)}%
.\medskip

\textbf{(c)} This is similar to part \textbf{(b)}. Namely, apply the binomial
formula (\ref{eq.gf.binform}) to $x=-X^{2}$ and $y=1$. We thus obtain%
\begin{align*}
\left(  -X^{2}+1\right)  ^{n}  &  =\sum_{k=0}^{n}\dbinom{n}{k}%
\underbrace{\left(  -X^{2}\right)  ^{k}}_{=\left(  -1\right)  ^{k}X^{2k}%
}\underbrace{1^{n-k}}_{=1}=\sum_{k=0}^{n}\dbinom{n}{k}\left(  -1\right)
^{k}X^{2k}=\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}X^{2k}\\
&  =\sum_{m=0}^{n}\left(  -1\right)  ^{m}\dbinom{n}{m}X^{2m}\overset{0}{=}%
\sum_{m\in\mathbb{N}}\left(  -1\right)  ^{m}\dbinom{n}{m}X^{2m}.
\end{align*}
Since $-X^{2}+1=1-X^{2}$, this proves Lemma \ref{lem.gf.binform1} \textbf{(c)}.
\end{proof}

\subsection{A new proof of the Chu--Vandermonde identity}

Now let us see some applications.

Recall the Chu--Vandermonde identity (Lecture 15, Theorem 2.3.1), which states
that
\begin{equation}
\dbinom{a+b}{n}=\sum_{k=0}^{n}\dbinom{a}{k}\dbinom{b}{n-k}
\label{eq.gf.vandermonde}%
\end{equation}
for all $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$. (We have renamed $x$ and $y$
as $a$ and $b$ here.)

Here is a new proof of this identity:

\begin{proof}
[New proof of the Chu--Vandermonde identity.]As in Lecture 15, we WLOG assume
that $a,b\in\mathbb{N}$ (since we can derive the general case from this case
using the \textquotedblleft polynomial identity trick\textquotedblright).

Now, Lemma \ref{lem.gf.binform1} \textbf{(a)} yields
\[
\left(  1+X\right)  ^{a}=\sum_{m\in\mathbb{N}}\dbinom{a}{m}X^{m}%
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \left(  1+X\right)
^{b}=\sum_{m\in\mathbb{N}}\dbinom{b}{m}X^{m}.
\]
Hence, the formula (\ref{eq.prop.gf.PQ=.PQ=}) for the product of two
polynomials (applied to $P=\left(  1+X\right)  ^{a}$, $a_{m}=\dbinom{a}{m}$,
$Q=\left(  1+X\right)  ^{b}$ and $b_{m}=\dbinom{b}{m}$) yields%
\[
\left(  1+X\right)  ^{a}\cdot\left(  1+X\right)  ^{b}=\sum_{m\in\mathbb{N}%
}\left(  \sum_{i=0}^{m}\dbinom{a}{i}\dbinom{b}{m-i}\right)  X^{m}.
\]
Hence, the coefficient of $X^{n}$ in the polynomial $\left(  1+X\right)
^{a}\cdot\left(  1+X\right)  ^{b}$ is%
\[
\sum_{i=0}^{n}\dbinom{a}{i}\dbinom{b}{n-i}=\sum_{k=0}^{n}\dbinom{a}{k}%
\dbinom{b}{n-k}.
\]
On the other hand, the polynomial $\left(  1+X\right)  ^{a}\cdot\left(
1+X\right)  ^{b}$ can be rewritten as follows:%
\[
\left(  1+X\right)  ^{a}\cdot\left(  1+X\right)  ^{b}=\left(  1+X\right)
^{a+b}=\sum_{m\in\mathbb{N}}\dbinom{a+b}{m}X^{m}%
\]
(by Lemma \ref{lem.gf.binform1} \textbf{(a)}). Hence, the coefficient of
$X^{n}$ in this polynomial is $\dbinom{a+b}{n}$.

Now, we have computed this coefficient in two different ways, obtaining
$\sum_{k=0}^{n}\dbinom{a}{k}\dbinom{b}{n-k}$ from our first way and
$\dbinom{a+b}{n}$ from our second. But of course, the answers have to be
equal. Thus, we obtain%
\[
\dbinom{a+b}{n}=\sum_{k=0}^{n}\dbinom{a}{k}\dbinom{b}{n-k}.
\]
This proves the Chu--Vandermonde identity.
\end{proof}

Note the strategy that we followed in our above argument: We have proved an
equality between two numbers (in our case, (\ref{eq.gf.vandermonde})) by

\begin{itemize}
\item identifying these two numbers as corresponding coefficients of two
polynomials (in our case, $\left(  1+X\right)  ^{a}\cdot\left(  1+X\right)
^{b}$ and $\left(  1+X\right)  ^{a+b}$)

\item and then showing that the two polynomials are equal.
\end{itemize}

The argument looks almost magical, seeing how it transformed the
trivial-looking equality $\left(  1+X\right)  ^{a}\cdot\left(  1+X\right)
^{b}=\left(  1+X\right)  ^{a+b}$ into the nontrivial Chu--Vandermonde identity
(\ref{eq.gf.vandermonde}). This illustrates the power of polynomials (and,
more generally, generating functions). A polynomial is not just a convenient
\textquotedblleft package\textquotedblright\ for its coefficients, but also
provides ways to manipulate them \textquotedblleft as a
collective\textquotedblright\ that are not easily available when one works
with single coefficients by themselves. A simple manipulation of polynomials
can thus \textquotedblleft encode\textquotedblright\ rather complicated
arguments on the level of single coefficients. \medskip

\begin{fineprint}
To remain fully honest (and dispel the magic somewhat), I should mention that
the trivial-looking equality $\left(  1+X\right)  ^{a}\cdot\left(  1+X\right)
^{b}=\left(  1+X\right)  ^{a+b}$ is not \textbf{entirely} trivial: It is a
consequence of the \textquotedblleft law of exponents\textquotedblright%
\ $P^{a}P^{b}=P^{a+b}$, which in turn is proved (by induction on $a$) using
associativity of multiplication. Thus, our proof tacitly relied on the fact
that the multiplication of polynomials is associative. This fact is easy to
prove (see \cite[proof of Theorem 3.2.6]{21s} for a proof), but not completely
obvious (its proof requires a bit of sum manipulation). So the above proof is
not entirely effortless, at least if you count in the effort needed to prove
this fact.

The magic of the above proof is thus not one of immaculate conception. Rather,
the concept of a polynomial has allowed a simple result (such as associativity
of multiplication for polynomials) to cast a long shadow. This kind of
\textquotedblleft amplification\textquotedblright\ of results is a hallmark of
abstract algebra.
\end{fineprint}

\subsection{The alternating Chu--Vandermonde identity}

So we have learnt that simple (even obvious) identities between polynomials
can turn into nontrivial identities between their coefficients if you focus on
a single coefficient. This can often be reverse-engineered: When you want to
prove a nontrivial identity between two numbers, you might wonder whether
these two numbers could be the $X^{n}$-coefficients of two equal polynomials
whose equality is much more obvious.

This is the technique of \textbf{generating functions}. In its general form,
instead of using polynomials, it uses formal power series (essentially
\textquotedblleft polynomials\textquotedblright\ that can have infinitely many
nonzero coefficients). For the examples we shall explore in today's lecture,
polynomials will suffice.

\bigskip

Let us try our hands at another application of the same technique. This time,
we shall prove a binomial identity that we haven't already proved in a
different way:

\begin{theorem}
[\textquotedblleft alternating Chu--Vandermonde identity\textquotedblright%
]\label{thm.gf.altvdm}Let $n\in\mathbb{N}$ and $x\in\mathbb{R}$. Then,%
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{x}{k}\dbinom{x}{n-k}=%
\begin{cases}
\left(  -1\right)  ^{n/2}\dbinom{x}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}%
\end{cases}
\]

\end{theorem}

\begin{proof}
Since both sides are polynomial functions in $x$, it suffices to prove this
equality for $x\in\mathbb{N}$ (by the \textquotedblleft polynomial identity
trick\textquotedblright\ from Lecture 14). So let us WLOG assume that
$x\in\mathbb{N}$, and let us rename $x$ as $a$. Thus, we must prove that%
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{a}{k}\dbinom{a}{n-k}%
\overset{?}{=}%
\begin{cases}
\left(  -1\right)  ^{n/2}\dbinom{a}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}%
\end{cases}
\]


We try to prove this in the same way as we just proved the Chu--Vandermonde
identity: We interpret both sides as the $X^{n}$-coefficients of polynomials,
and we show that the polynomials are equal.

What is the polynomial whose $X^{n}$-coefficient is
\[%
\begin{cases}
\left(  -1\right)  ^{n/2}\dbinom{a}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd}%
\end{cases}
\]
for each $n\in\mathbb{N}$ ? I claim that this polynomial is $\left(
1-X^{2}\right)  ^{a}$. Indeed, Lemma \ref{lem.gf.binform1} \textbf{(c)} yields%
\begin{align*}
\left(  1-X^{2}\right)  ^{a}  &  =\sum_{m\in\mathbb{N}}\left(  -1\right)
^{m}\dbinom{a}{m}X^{2m}\\
&  =\dbinom{a}{0}X^{0}-\dbinom{a}{1}X^{2}+\dbinom{a}{2}X^{4}-\dbinom{a}%
{3}X^{6}\pm\cdots\\
&  =\sum_{\substack{m\in\mathbb{N}\\\text{is even}}}\left(  -1\right)
^{m/2}\dbinom{a}{m/2}X^{m},
\end{align*}
so that the $X^{n}$-coefficient of this polynomial $\left(  1-X^{2}\right)
^{a}$ is%
\begin{equation}
\left[  X^{n}\right]  \left(  \left(  1-X^{2}\right)  ^{a}\right)  =%
\begin{cases}
\left(  -1\right)  ^{n/2}\dbinom{a}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}%
\end{cases}
\label{pf.thm.gf.altvdm.1}%
\end{equation}


On the other hand, what is the polynomial whose $X^{n}$-coefficient equals the
sum%
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{a}{k}\dbinom{a}{n-k}\text{ ?}%
\]
The shape of this sum is a dead giveaway; other than the letter used for the
summation index (which is $k$, not $i$ here), this is precisely the sum on the
right hand side of (\ref{eq.prop.gf.PQ=.XnPQ=}) obtained when $P=\left(
1-X\right)  ^{a}$ and $Q=\left(  1+X\right)  ^{a}$. Indeed, if we define the
two polynomials%
\[
P:=\left(  1-X\right)  ^{a}=\sum_{m\in\mathbb{N}}\left(  -1\right)
^{m}\dbinom{a}{m}X^{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.gf.binform1} \textbf{(b)}}\right)
\]
and
\[
Q:=\left(  1+X\right)  ^{a}=\sum_{m\in\mathbb{N}}\dbinom{a}{m}X^{m}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.gf.binform1} \textbf{(a)}%
}\right)  ,
\]
then (\ref{eq.prop.gf.PQ=.XnPQ=}) (applied to $m=n$) yields%
\begin{align}
\left[  X^{n}\right]  \left(  PQ\right)   &  =\sum_{i=0}^{n}\left(  -1\right)
^{i}\dbinom{a}{i}\cdot\dbinom{a}{n-i}=\sum_{i=0}^{n}\left(  -1\right)
^{i}\dbinom{a}{i}\dbinom{a}{n-i}\nonumber\\
&  =\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{a}{k}\dbinom{a}{n-k}.
\label{pf.thm.gf.altvdm.2}%
\end{align}
However, these polynomials $P$ and $Q$ satisfy%
\[
PQ=\left(  1-X\right)  ^{a}\left(  1+X\right)  ^{a}=\left(
\underbrace{\left(  1-X\right)  \left(  1+X\right)  }_{=1-X^{2}}\right)
^{a}=\left(  1-X^{2}\right)  ^{a}.
\]
Thus, we can rewrite (\ref{pf.thm.gf.altvdm.2}) as
\[
\left[  X^{n}\right]  \left(  \left(  1-X^{2}\right)  ^{a}\right)  =\sum
_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{a}{k}\dbinom{a}{n-k}.
\]
Comparing this with (\ref{pf.thm.gf.altvdm.1}), we obtain
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{a}{k}\dbinom{a}{n-k}=%
\begin{cases}
\left(  -1\right)  ^{n/2}\dbinom{a}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd,}%
\end{cases}
\]
which is precisely what we needed to prove. Thus, we have established Theorem
\ref{thm.gf.altvdm}.
\end{proof}

\begin{fineprint}
A more conventional proof of Theorem \ref{thm.gf.altvdm} can be found in
\cite[\S 7.30.2]{detnotes}.

\bigskip

The following neat identity is a simple consequence of Theorem
\ref{thm.gf.altvdm}:

\begin{corollary}
\label{cor.gf.altvdm.nk2}Let $n\in\mathbb{N}$. Then,%
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}^{2}=%
\begin{cases}
\left(  -1\right)  ^{n/2}\dbinom{n}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}%
\end{cases}
\]

\end{corollary}

\begin{proof}
Apply Theorem \ref{thm.gf.altvdm} to $x=n$, and simplify the left hand side by
noticing that%
\[
\dbinom{n}{k}\underbrace{\dbinom{n}{n-k}}_{\substack{=\dbinom{n}{k}\\\text{(by
Lecture 6,}\\\text{Theorem 1.3.9)}}}=\dbinom{n}{k}\dbinom{n}{k}=\dbinom{n}%
{k}^{2}.
\]

\end{proof}
\end{fineprint}

\bigskip

\subsection{Dixon's identity}

Our next goal is to show a deeper binomial identity:

\begin{theorem}
[Dixon's identity]\label{thm.gf.dixon}Let $a,b,c\in\mathbb{N}$. Then,%
\begin{equation}
\sum_{k\in\mathbb{Z}}\left(  -1\right)  ^{k}\dbinom{b+c}{c+k}\dbinom{c+a}%
{a+k}\dbinom{a+b}{b+k}=\dfrac{\left(  a+b+c\right)  !}{a!b!c!}.
\label{eq.thm.gf.dixon.eq}%
\end{equation}

\end{theorem}

Two comments are worth making here:

\begin{itemize}
\item It is easy to show that the sum on the LHS of (\ref{eq.thm.gf.dixon.eq})
is well-defined (i.e., only finitely many addends are nonzero). In fact, for
example, the $\dbinom{b+c}{c+k}$ factor is $0$ whenever $k\notin\left\{
-c,-c+1,\ldots,b\right\}  $.

\item The RHS of (\ref{eq.thm.gf.dixon.eq}) is the multinomial coefficient
$\dbinom{a+b+c}{a,b,c}$. (See Lecture 22, \S 2.10 for the definition of
multinomial coefficients.)
\end{itemize}

How can we prove Theorem \ref{thm.gf.dixon}? There are various proofs (see,
e.g., \cite{Ward91}), but none of them is easy. Let me show a mysterious proof
using polynomials.\footnote{This proof is a particular case of I. J. Good's
beautiful proof of the \textbf{Dyson conjecture} (which is a generalization of
Theorem \ref{thm.gf.dixon}, although not in a very obvious way). See Good's
short note \cite{Good70} or Andrews's more detailed writeup \cite[\S 3.3]%
{MacMah78} for the latter proof.}

First, for the sake of brevity, let us denote the LHS of
(\ref{eq.thm.gf.dixon.eq}) by $F\left(  a,b,c\right)  $. That is, we set%
\begin{equation}
F\left(  a,b,c\right)  :=\sum_{k\in\mathbb{Z}}\left(  -1\right)  ^{k}%
\dbinom{b+c}{c+k}\dbinom{c+a}{a+k}\dbinom{a+b}{b+k}
\label{eq.thm.gf.dixon.F=def}%
\end{equation}
for all $a,b,c\in\mathbb{N}$. Thus, Dixon's identity (\ref{eq.thm.gf.dixon.eq}%
) can be rewritten as%
\begin{equation}
F\left(  a,b,c\right)  =\dfrac{\left(  a+b+c\right)  !}{a!b!c!}.
\label{eq.thm.gf.dixon.eqF}%
\end{equation}


Our strategy for proving this identity will be to show that both of its sides
satisfy the same recurrence and have the same base values. The base values are
an easy matter:

\begin{lemma}
\label{lem.gf.dixon.0}Dixon's identity (\ref{eq.thm.gf.dixon.eqF}) is true if
$a=0$ or $b=0$ or $c=0$.
\end{lemma}

\begin{proof}
WLOG assume that $a=0$ (since the cases $b=0$ and $c=0$ are analogous). Then,%
\begin{align*}
F\left(  a,b,c\right)   &  =\sum_{k\in\mathbb{Z}}\left(  -1\right)
^{k}\dbinom{b+c}{c+k}\dbinom{c+a}{a+k}\dbinom{a+b}{b+k}\\
&  =\sum_{k\in\mathbb{Z}}\left(  -1\right)  ^{k}\dbinom{b+c}{c+k}\dbinom{c}%
{k}\dbinom{b}{b+k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a=0\right) \\
&  =\sum_{\substack{k\in\mathbb{Z};\\k<0}}\left(  -1\right)  ^{k}\dbinom
{b+c}{c+k}\underbrace{\dbinom{c}{k}}_{\substack{=0\\\text{(since }k<0\text{)}%
}}\dbinom{b}{b+k}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  -1\right)  ^{0}}_{=1}%
\underbrace{\dbinom{b+c}{c+0}}_{=\dbinom{b+c}{c}}\underbrace{\dbinom{c}{0}%
}_{=1}\underbrace{\dbinom{b}{b+0}}_{=\dbinom{b}{b}=1}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{k\in\mathbb{Z};\\k>0}}\left(
-1\right)  ^{k}\dbinom{b+c}{c+k}\dbinom{c}{k}\underbrace{\dbinom{b}{b+k}%
}_{\substack{=0\\\text{(since }k>0\\\text{and thus }b+k>b\text{)}}}\\
&  =\dbinom{b+c}{c}=\dfrac{\left(  b+c\right)  !}{b!c!}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{an easy consequence of}\\
\text{the factorial formula for BCs}%
\end{array}
\right) \\
&  =\dfrac{\left(  a+b+c\right)  !}{a!b!c!}\ \ \ \ \ \ \ \ \ \ \left(
\text{an easy consequence of }a=0\right)  .
\end{align*}
In other words, (\ref{eq.thm.gf.dixon.eqF}) is true. This proves Lemma
\ref{lem.gf.dixon.0}.
\end{proof}

Next, let us find a recurrence relation that both sides of
(\ref{eq.thm.gf.dixon.eqF}) should satisfy. For the RHS, this is rather easy:

\begin{lemma}
\label{lem.gf.dixon.rec}Let $a,b,c$ be positive integers. Then,%
\[
\dfrac{\left(  a+b+c\right)  !}{a!b!c!}=\dfrac{\left(  \left(  a-1\right)
+b+c\right)  !}{\left(  a-1\right)  !b!c!}+\dfrac{\left(  a+\left(
b-1\right)  +c\right)  !}{a!\left(  b-1\right)  !c!}+\dfrac{\left(
a+b+\left(  c-1\right)  \right)  !}{a!b!\left(  c-1\right)  !}.
\]

\end{lemma}

\begin{proof}
Upon bringing all fractions to a common denominator (using the identity
$n!=\left(  n-1\right)  !\cdot n$, which holds for every $n>0$), this
simplifies to $a+b+c=a+b+c$, which is indeed true.

(Alternatively, Lemma \ref{lem.gf.dixon.rec} is a particular case of
Proposition 2.10.4 \textbf{(d)} in Lecture 22.)
\end{proof}

Now to the hard part: We must show that the LHS $F\left(  a,b,c\right)  $ of
(\ref{eq.thm.gf.dixon.eqF}) satisfies the same recurrence as the RHS. In other
words, we must prove that
\[
F\left(  a,b,c\right)  =F\left(  a-1,b,c\right)  +F\left(  a,b-1,c\right)
+F\left(  a,b,c-1\right)
\]
for every three positive integers $a,b,c$.

Now is the time to pull the polynomial rabbit out of our hat. This time, we
will use polynomials in \textbf{three} indeterminates $X,Y,Z$. We shall use
the notation $\left[  X^{i}Y^{j}Z^{k}\right]  P$ for the $X^{i}Y^{j}Z^{k}%
$-coefficient of a polynomial $P$. For example, $\left[  X^{1}Y^{0}%
Z^{1}\right]  \left(  \left(  2-X\right)  \left(  2-Y\right)  \left(
2-Z\right)  \right)  =2$. For another example, for any $a,b,c\in\mathbb{N}$,
we have%
\[
\left[  X^{a}Y^{b}Z^{c}\right]  \left(  \left(  X+Y+Z\right)  ^{a+b+c}\right)
=\dbinom{a+b+c}{a,b,c}%
\]
by the multinomial formula (Lecture 23, Theorem 2.10.5).

For any $a,b,c\in\mathbb{N}$, we define a polynomial%
\begin{equation}
P_{a,b,c}:=\left(  Y-Z\right)  ^{b+c}\cdot\left(  Z-X\right)  ^{c+a}%
\cdot\left(  X-Y\right)  ^{a+b}. \label{eq.gf.dixon.Pabc=}%
\end{equation}
This is a polynomial in $X,Y,Z$ (so we might call it $P_{a,b,c}\left(
X,Y,Z\right)  $, but we would soon tire of carrying the $X,Y,Z$ arguments
around). The main reason why this polynomial is useful is that $F\left(
a,b,c\right)  $ turns out to be one of its coefficients (up to sign):

\begin{lemma}
\label{lem.gf.dixon.coeff}For any $a,b,c\in\mathbb{N}$, we have%
\[
F\left(  a,b,c\right)  =\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}%
Y^{2b}Z^{2c}\right]  \left(  P_{a,b,c}\right)  .
\]

\end{lemma}

\begin{proof}
Let us first observe the following general fact: If $n,m\in\mathbb{N}$ and
$k\in\mathbb{Z}$ are arbitrary, then%
\begin{equation}
\dbinom{n+m}{n-k}=\dbinom{n+m}{m+k}. \label{pf.lem.gf.dixon.coeff.symmetry}%
\end{equation}
(Indeed, this follows easily from the symmetry of BCs (Lecture 6, Theorem
1.3.9)\footnote{In more detail: Let $n,m\in\mathbb{N}$ and $k\in\mathbb{Z}$.
Then, the symmetry of BCs (Lecture 6, Theorem 1.3.9) yields%
\[
\dbinom{n+m}{n-k}=\dbinom{n+m}{\left(  n+m\right)  -\left(  n-k\right)
}=\dbinom{n+m}{m+k}.
\]
This proves (\ref{pf.lem.gf.dixon.coeff.symmetry}).}.)

As we already said, the binomial formula (\ref{eq.gf.binform}) holds not just
for numbers, but also for polynomials. This includes multivariate polynomials
(for the same reasons). Thus, in particular, we can apply (\ref{eq.gf.binform}%
) to $n=b+c$, $x=Y$ and $y=-Z$. We thus obtain%
\begin{align}
\left(  Y-Z\right)  ^{b+c}=  &  \sum_{k=0}^{b+c}\dbinom{b+c}{k}Y^{k}\left(
-Z\right)  ^{b+c-k}=\sum_{i=0}^{b+c}\dbinom{b+c}{i}Y^{i}\left(  -Z\right)
^{b+c-i}\nonumber\\
\overset{0}{=}  &  \sum_{i\in\mathbb{Z}}\dbinom{b+c}{i}Y^{i}\left(  -Z\right)
^{b+c-i} \label{pf.lem.gf.dixon.coeff.1}%
\end{align}
(since every $i\in\mathbb{Z}\setminus\left\{  0,1,\ldots,b+c\right\}  $
satisfies $\dbinom{b+c}{i}=0$).\ \ \ \ \footnote{You may wonder about the
negative powers of $Y$ that are contained in this sum (in the addends with
$i<0$). However, these powers can be ignored, since they are getting
multiplied with $0$ anyway (since $\dbinom{b+c}{i}=0$). The same holds for the
negative powers of $-Z$ (which are contained in the addends with $i>b+c$).}

Similarly,%
\begin{equation}
\left(  Z-X\right)  ^{c+a}=\sum_{j\in\mathbb{Z}}\dbinom{c+a}{j}Z^{j}\left(
-X\right)  ^{c+a-j} \label{pf.lem.gf.dixon.coeff.2}%
\end{equation}
and%
\begin{equation}
\left(  X-Y\right)  ^{a+b}=\sum_{k\in\mathbb{Z}}\dbinom{a+b}{k}X^{k}\left(
-Y\right)  ^{a+b-k}. \label{pf.lem.gf.dixon.coeff.3}%
\end{equation}


Now, the definition of $P_{a,b,c}$ yields%
\begin{align}
P_{a,b,c}  &  =\left(  Y-Z\right)  ^{b+c}\cdot\left(  Z-X\right)  ^{c+a}%
\cdot\left(  X-Y\right)  ^{a+b}\nonumber\\
&  =\left(  \sum_{i\in\mathbb{Z}}\dbinom{b+c}{i}Y^{i}\left(  -Z\right)
^{b+c-i}\right)  \cdot\left(  \sum_{j\in\mathbb{Z}}\dbinom{c+a}{j}Z^{j}\left(
-X\right)  ^{c+a-j}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  \sum_{k\in\mathbb{Z}}\dbinom{a+b}{k}%
X^{k}\left(  -Y\right)  ^{a+b-k}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.gf.dixon.coeff.1}), (\ref{pf.lem.gf.dixon.coeff.2}) and
(\ref{pf.lem.gf.dixon.coeff.3})}\right) \nonumber\\
&  =\underbrace{\sum_{i\in\mathbb{Z}}\ \ \sum_{j\in\mathbb{Z}}\ \ \sum
_{k\in\mathbb{Z}}}_{=\sum_{\left(  i,j,k\right)  \in\mathbb{Z}^{3}}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \underbrace{\dbinom{b+c}{i}Y^{i}\left(  -Z\right)
^{b+c-i}\dbinom{c+a}{j}Z^{j}\left(  -X\right)  ^{c+a-j}\dbinom{a+b}{k}%
X^{k}\left(  -Y\right)  ^{a+b-k}}_{=\left(  -1\right)  ^{\left(  b+c-i\right)
+\left(  c+a-j\right)  +\left(  a+b-k\right)  }\dbinom{b+c}{i}\dbinom{c+a}%
{j}\dbinom{a+b}{k}X^{c+a-j+k}Y^{a+b-k+i}Z^{b+c-i+j}}\nonumber\\
&  =\sum_{\left(  i,j,k\right)  \in\mathbb{Z}^{3}}\underbrace{\left(
-1\right)  ^{\left(  b+c-i\right)  +\left(  c+a-j\right)  +\left(
a+b-k\right)  }}_{\substack{=\left(  -1\right)  ^{i+j+k}\\\text{(since
}\left(  b+c-i\right)  +\left(  c+a-j\right)  +\left(  a+b-k\right)
\\=2a+2b+2c-\left(  i+j+k\right)  \equiv i+j+k\operatorname{mod}2\text{)}%
}}\dbinom{b+c}{i}\dbinom{c+a}{j}\dbinom{a+b}{k}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ X^{c+a-j+k}Y^{a+b-k+i}Z^{b+c-i+j}\nonumber\\
&  =\sum_{\left(  i,j,k\right)  \in\mathbb{Z}^{3}}\left(  -1\right)
^{i+j+k}\dbinom{b+c}{i}\dbinom{c+a}{j}\dbinom{a+b}{k}X^{c+a-j+k}%
Y^{a+b-k+i}Z^{b+c-i+j}. \label{pf.lem.gf.dixon.coeff.4}%
\end{align}
What is the coefficient of $X^{2a}Y^{2b}Z^{2c}$ in this polynomial? This
coefficient is clearly%
\[
\sum_{\substack{\left(  i,j,k\right)  \in\mathbb{Z}^{3}%
;\\c+a-j+k=2a;\\a+b-k+i=2b;\\b+c-i+j=2c}}\left(  -1\right)  ^{i+j+k}%
\dbinom{b+c}{i}\dbinom{c+a}{j}\dbinom{a+b}{k},
\]
since the addend for any given triple $\left(  i,j,k\right)  \in\mathbb{Z}%
^{3}$ contributes to this coefficient if and only if the exponents $c+a-j+k$,
$a+b-k+i$ and $b+c-i+j$ equal $2a$, $2b$ and $2c$, respectively. In other
words, we have%
\begin{align}
&  \left[  X^{2a}Y^{2b}Z^{2c}\right]  \left(  P_{a,b,c}\right) \nonumber\\
&  =\sum_{\substack{\left(  i,j,k\right)  \in\mathbb{Z}^{3}%
;\\c+a-j+k=2a;\\a+b-k+i=2b;\\b+c-i+j=2c}}\left(  -1\right)  ^{i+j+k}%
\dbinom{b+c}{i}\dbinom{c+a}{j}\dbinom{a+b}{k}. \label{pf.lem.gf.dixon.coeff.5}%
\end{align}


Let us now simplify the sum on the RHS of this equality. The three equations
$c+a-j+k=2a$, $a+b-k+i=2b$ and $b+c-i+j=2c$ are dependent: More precisely, the
third of them follows from the first two (since the sum of all three equations
is $2a+2b+2c=2a+2b+2c$). Thus, we can remove the third equation, transforming
the sum into%
\[
\sum_{\substack{\left(  i,j,k\right)  \in\mathbb{Z}^{3}%
;\\c+a-j+k=2a;\\a+b-k+i=2b}}\left(  -1\right)  ^{i+j+k}\dbinom{b+c}{i}%
\dbinom{c+a}{j}\dbinom{a+b}{k}.
\]
The remaining two equations $c+a-j+k=2a$ and $a+b-k+i=2b$ uniquely determine
$i$ and $j$ in terms of $k$: namely,%
\begin{align*}
i  &  =2b-\left(  a+b-k\right)  =b-a+k\ \ \ \ \ \ \ \ \ \ \text{and}\\
j  &  =\left(  c+a+k\right)  -2a=c-a+k.
\end{align*}
Thus, instead of summing over all triples $\left(  i,j,k\right)  \in
\mathbb{Z}^{3}$ satisfying these two equations, we can just as well sum over
all $k\in\mathbb{Z}$ and substitute $b-a+k$ for $i$ and $c-a+k$ for $j$ in the
sum. Hence,
\begin{align*}
&  \sum_{\substack{\left(  i,j,k\right)  \in\mathbb{Z}^{3}%
;\\c+a-j+k=2a;\\a+b-k+i=2b}}\left(  -1\right)  ^{i+j+k}\dbinom{b+c}{i}%
\dbinom{c+a}{j}\dbinom{a+b}{k}\\
&  =\sum_{k\in\mathbb{Z}}\left(  -1\right)  ^{\left(  b-a+k\right)  +\left(
c-a+k\right)  +k}\dbinom{b+c}{b-a+k}\dbinom{c+a}{c-a+k}\dbinom{a+b}{k}\\
&  =\sum_{k\in\mathbb{Z}}\underbrace{\left(  -1\right)  ^{\left(  b-a+\left(
a-k\right)  \right)  +\left(  c-a+\left(  a-k\right)  \right)  +\left(
a-k\right)  }}_{\substack{=\left(  -1\right)  ^{a+b+c-3k}=\left(  -1\right)
^{a+b+c+k}\\\text{(since }-3k\equiv k\operatorname{mod}2\text{)}%
}}\underbrace{\dbinom{b+c}{b-a+\left(  a-k\right)  }}_{=\dbinom{b+c}{b-k}%
}\underbrace{\dbinom{c+a}{c-a+\left(  a-k\right)  }}_{=\dbinom{c+a}{c-k}%
}\dbinom{a+b}{a-k}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have
substituted }a-k\text{ for }k\text{ in the sum}\right) \\
&  =\sum_{k\in\mathbb{Z}}\underbrace{\left(  -1\right)  ^{a+b+c+k}}_{=\left(
-1\right)  ^{a+b+c}\left(  -1\right)  ^{k}}\ \ \underbrace{\dbinom{b+c}{b-k}%
}_{\substack{=\dbinom{b+c}{c+k}\\\text{(by
(\ref{pf.lem.gf.dixon.coeff.symmetry}))}}}\ \ \underbrace{\dbinom{c+a}{c-k}%
}_{\substack{=\dbinom{c+a}{a+k}\\\text{(by
(\ref{pf.lem.gf.dixon.coeff.symmetry}))}}}\ \ \underbrace{\dbinom{a+b}{a-k}%
}_{\substack{=\dbinom{a+b}{b+k}\\\text{(by
(\ref{pf.lem.gf.dixon.coeff.symmetry}))}}}\\
&  =\left(  -1\right)  ^{a+b+c}\underbrace{\sum_{k\in\mathbb{Z}}\left(
-1\right)  ^{k}\dbinom{b+c}{c+k}\dbinom{c+a}{a+k}\dbinom{a+b}{b+k}%
}_{\substack{=F\left(  a,b,c\right)  \\\text{(by (\ref{eq.thm.gf.dixon.F=def}%
))}}}\\
&  =\left(  -1\right)  ^{a+b+c}F\left(  a,b,c\right)  .
\end{align*}
\newline Now, (\ref{pf.lem.gf.dixon.coeff.5}) becomes
\begin{align*}
&  \left[  X^{2a}Y^{2b}Z^{2c}\right]  \left(  P_{a,b,c}\right) \\
&  =\sum_{\substack{\left(  i,j,k\right)  \in\mathbb{Z}^{3}%
;\\c+a-j+k=2a;\\a+b-k+i=2b;\\b+c-i+j=2c}}\left(  -1\right)  ^{i+j+k}%
\dbinom{b+c}{i}\dbinom{c+a}{j}\dbinom{a+b}{k}\\
&  =\sum_{\substack{\left(  i,j,k\right)  \in\mathbb{Z}^{3}%
;\\c+a-j+k=2a;\\a+b-k+i=2b}}\left(  -1\right)  ^{i+j+k}\dbinom{b+c}{i}%
\dbinom{c+a}{j}\dbinom{a+b}{k}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since, as we said, the third equation }b+c-i+j=2c\\
\text{under the summation sign follows from the other two}%
\end{array}
\right) \\
&  =\left(  -1\right)  ^{a+b+c}F\left(  a,b,c\right)  .
\end{align*}
Solving this for $F\left(  a,b,c\right)  $, we obtain%
\[
F\left(  a,b,c\right)  =\dfrac{\left[  X^{2a}Y^{2b}Z^{2c}\right]  \left(
P_{a,b,c}\right)  }{\left(  -1\right)  ^{a+b+c}}=\left(  -1\right)
^{a+b+c}\cdot\left[  X^{2a}Y^{2b}Z^{2c}\right]  \left(  P_{a,b,c}\right)  .
\]
This proves Lemma \ref{lem.gf.dixon.coeff}.
\end{proof}

The next lemma is a straightforward computation:

\begin{lemma}
\label{lem.gf.dixon.-1}Let $Q$ be a polynomial in $X,Y,Z$ that is divisible by
$\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)  $ (that is, $Q$
can be written as $\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)
R$ for some polynomial $R$). Then,%
\[
\dfrac{X^{2}Q}{\left(  X-Y\right)  \left(  Z-X\right)  }+\dfrac{Y^{2}%
Q}{\left(  Y-Z\right)  \left(  X-Y\right)  }+\dfrac{Z^{2}Q}{\left(
Z-X\right)  \left(  Y-Z\right)  }=-Q.
\]

\end{lemma}

\begin{fineprint}
It would be easier to restate this lemma in the form
\[
\dfrac{X^{2}}{\left(  X-Y\right)  \left(  Z-X\right)  }+\dfrac{Y^{2}}{\left(
Y-Z\right)  \left(  X-Y\right)  }+\dfrac{Z^{2}}{\left(  Z-X\right)  \left(
Y-Z\right)  }=-1.
\]
However, this requires working with \textbf{rational functions} (i.e., formal
ratios of polynomials) instead of polynomials. To keep our arguments maximally
elementary, we prefer to stay among the polynomials, and thus we only divide
polynomials by other polynomials if the former are divisible by the latter.
\end{fineprint}

\begin{proof}
[Proof of Lemma \ref{lem.gf.dixon.-1}.]Bringing the fractions to a common
denominator, we find%
\begin{align*}
&  \dfrac{X^{2}Q}{\left(  X-Y\right)  \left(  Z-X\right)  }+\dfrac{Y^{2}%
Q}{\left(  Y-Z\right)  \left(  X-Y\right)  }+\dfrac{Z^{2}Q}{\left(
Z-X\right)  \left(  Y-Z\right)  }\\
&  =\dfrac{X^{2}\left(  Y-Z\right)  Q+Y^{2}\left(  Z-X\right)  Q+Z^{2}\left(
X-Y\right)  Q}{\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)  }\\
&  =\dfrac{\left(  X^{2}\left(  Y-Z\right)  +Y^{2}\left(  Z-X\right)
+Z^{2}\left(  X-Y\right)  \right)  Q}{\left(  Y-Z\right)  \left(  Z-X\right)
\left(  X-Y\right)  }\\
&  =\dfrac{-\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)
Q}{\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since a straightforward expansion of both sides}\\
\text{verifies that }X^{2}\left(  Y-Z\right)  +Y^{2}\left(  Z-X\right)
+Z^{2}\left(  X-Y\right) \\
=-\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)
\end{array}
\right) \\
&  =-Q.
\end{align*}

\end{proof}

The next lemma is about how the coefficients of a polynomial get shifted when
that polynomial is multiplied by a power of $X$:

\begin{lemma}
\label{lem.gf.dixon.shift}Let $P$ be a polynomial in $X,Y,Z$. Let
$u,i,j,k\in\mathbb{N}$. Then,%
\[
\left[  X^{i}Y^{j}Z^{k}\right]  P=\left[  X^{i+u}Y^{j}Z^{k}\right]  \left(
X^{u}P\right)  .
\]

\end{lemma}

\begin{proof}
Let us first prove the analogous fact about polynomials in one variable: If
$P$ is a polynomial in a single variable $X$, and if $u,i\in\mathbb{N}$, then%
\begin{equation}
\left[  X^{i}\right]  P=\left[  X^{i+u}\right]  \left(  X^{u}P\right)  .
\label{pf.lem.gf.dixon.shift.1}%
\end{equation}


Indeed, write $P$ in the form $P=\sum_{n\in\mathbb{N}}a_{n}X^{n}$. Then,
$\left[  X^{i}\right]  P=a_{i}$. However, from $P=\sum_{n\in\mathbb{N}}%
a_{n}X^{n}$, we obtain%
\[
X^{u}P=X^{u}\sum_{n\in\mathbb{N}}a_{n}X^{n}=\sum_{n\in\mathbb{N}}%
a_{n}\underbrace{X^{u}X^{n}}_{=X^{u+n}=X^{n+u}}=\sum_{n\in\mathbb{N}}%
a_{n}X^{n+u},
\]
which is clearly a polynomial whose $X^{i+u}$-coefficient is $a_{i}$. In other
words, $\left[  X^{i+u}\right]  \left(  X^{u}P\right)  =a_{i}$. Comparing this
with $\left[  X^{i}\right]  P=a_{i}$, we obtain $\left[  X^{i}\right]
P=\left[  X^{i+u}\right]  \left(  X^{u}P\right)  $. Thus,
(\ref{pf.lem.gf.dixon.shift.1}) is proved.

The case of three variables $X,Y,Z$ is analogous; the only difference is that
there are now two further variables $Y$ and $Z$ that passively tag along.
\end{proof}

Now to the crucial step:

\begin{lemma}
\label{lem.gf.dixon.recF}Let $a,b,c$ be positive integers. Then,%
\[
F\left(  a,b,c\right)  =F\left(  a-1,b,c\right)  +F\left(  a,b-1,c\right)
+F\left(  a,b,c-1\right)  .
\]

\end{lemma}

\begin{proof}
Since $a,b,c$ are positive integers, we have $a,b,c\geq1$. Thus, the exponents
$b+c$, $c+a$ and $a+b$ in (\ref{eq.gf.dixon.Pabc=}) are $\geq2$ each. Hence,
they are $\geq1$ each. Therefore, the polynomial $P_{a,b,c}$ is divisible by
$\left(  Y-Z\right)  \left(  Z-X\right)  \left(  X-Y\right)  $.

The definition of $P_{a-1,b,c}$ yields%
\begin{align}
P_{a-1,b,c}  &  =\left(  Y-Z\right)  ^{b+c}\cdot\underbrace{\left(
Z-X\right)  ^{c+a-1}}_{=\dfrac{\left(  Z-X\right)  ^{c+a}}{Z-X}}%
\cdot\underbrace{\left(  X-Y\right)  ^{a-1+b}}_{=\dfrac{\left(  X-Y\right)
^{a+b}}{X-Y}}\nonumber\\
&  =\left(  Y-Z\right)  ^{b+c}\cdot\dfrac{\left(  Z-X\right)  ^{c+a}}%
{Z-X}\cdot\dfrac{\left(  X-Y\right)  ^{a+b}}{X-Y}\nonumber\\
&  =\dfrac{\left(  Y-Z\right)  ^{b+c}\cdot\left(  Z-X\right)  ^{c+a}%
\cdot\left(  X-Y\right)  ^{a+b}}{\left(  X-Y\right)  \left(  Z-X\right)
}\nonumber\\
&  =\dfrac{P_{a,b,c}}{\left(  X-Y\right)  \left(  Z-X\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.gf.dixon.Pabc=})}\right)  .
\label{pf.lem.gf.dixon.recF.1}%
\end{align}


Lemma \ref{lem.gf.dixon.coeff} yields%
\[
F\left(  a,b,c\right)  =\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}%
Y^{2b}Z^{2c}\right]  \left(  P_{a,b,c}\right)  .
\]
However, we can also apply Lemma \ref{lem.gf.dixon.coeff} to $a-1$ instead of
$a$ (since $a\geq1$, so that $a-1\in\mathbb{N}$). Thus, we obtain
\begin{align*}
F\left(  a-1,b,c\right)   &  =\underbrace{\left(  -1\right)  ^{a-1+b+c}%
}_{=-\left(  -1\right)  ^{a+b+c}}\cdot\underbrace{\left[  X^{2\left(
a-1\right)  }Y^{2b}Z^{2c}\right]  \left(  P_{a-1,b,c}\right)  }%
_{\substack{=\left[  X^{2\left(  a-1\right)  +2}Y^{2b}Z^{2c}\right]  \left(
X^{2}P_{a-1,b,c}\right)  \\\text{(by Lemma \ref{lem.gf.dixon.shift}, applied
to }P=P_{a-1,b,c}\\\text{and }i=2\left(  a-1\right)  \text{ and }j=2b\text{
and }k=2c\text{ and }u=2\text{)}}}\\
&  =-\left(  -1\right)  ^{a+b+c}\cdot\left[  \underbrace{X^{2\left(
a-1\right)  +2}}_{=X^{2a}}Y^{2b}Z^{2c}\right]  \left(  X^{2}%
\underbrace{P_{a-1,b,c}}_{\substack{=\dfrac{P_{a,b,c}}{\left(  X-Y\right)
\left(  Z-X\right)  }\\\text{(by (\ref{pf.lem.gf.dixon.recF.1}))}}}\right)  \\
&  =-\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}Y^{2b}Z^{2c}\right]
\left(  X^{2}\cdot\dfrac{P_{a,b,c}}{\left(  X-Y\right)  \left(  Z-X\right)
}\right)  \\
&  =-\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}Y^{2b}Z^{2c}\right]
\dfrac{X^{2}P_{a,b,c}}{\left(  X-Y\right)  \left(  Z-X\right)  }.
\end{align*}
Similarly,%
\begin{align*}
F\left(  a,b-1,c\right)   &  =-\left(  -1\right)  ^{a+b+c}\cdot\left[
X^{2a}Y^{2b}Z^{2c}\right]  \dfrac{Y^{2}P_{a,b,c}}{\left(  Y-Z\right)  \left(
X-Y\right)  }\ \ \ \ \ \ \ \ \ \ \text{and}\\
F\left(  a,b,c-1\right)   &  =-\left(  -1\right)  ^{a+b+c}\cdot\left[
X^{2a}Y^{2b}Z^{2c}\right]  \dfrac{Z^{2}P_{a,b,c}}{\left(  Z-X\right)  \left(
Y-Z\right)  }.
\end{align*}
Summing these three equations together, we find%
\begin{align*}
&  F\left(  a-1,b,c\right)  +F\left(  a,b-1,c\right)  +F\left(
a,b,c-1\right)  \\
&  =-\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}Y^{2b}Z^{2c}\right]
\dfrac{X^{2}P_{a,b,c}}{\left(  X-Y\right)  \left(  Z-X\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}%
Y^{2b}Z^{2c}\right]  \dfrac{Y^{2}P_{a,b,c}}{\left(  Y-Z\right)  \left(
X-Y\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}%
Y^{2b}Z^{2c}\right]  \dfrac{Z^{2}P_{a,b,c}}{\left(  Z-X\right)  \left(
Y-Z\right)  }\\
&  =-\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}Y^{2b}Z^{2c}\right]
\underbrace{\left(  \dfrac{X^{2}P_{a,b,c}}{\left(  X-Y\right)  \left(
Z-X\right)  }+\dfrac{Y^{2}P_{a,b,c}}{\left(  Y-Z\right)  \left(  X-Y\right)
}+\dfrac{Z^{2}P_{a,b,c}}{\left(  Z-X\right)  \left(  Y-Z\right)  }\right)
}_{\substack{=-P_{a,b,c}\\\text{(by Lemma \ref{lem.gf.dixon.-1}, applied to
}Q=P_{a,b,c}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  \left[  X^{i}Y^{j}Z^{k}\right]  U\right)  +\left(
\left[  X^{i}Y^{j}Z^{k}\right]  V\right)  =\left[  X^{i}Y^{j}Z^{k}\right]
\left(  U+V\right)  \\
\text{for any }i,j,k\in\mathbb{N}\text{ and any two polynomials }U,V
\end{array}
\right)  \\
&  =-\left(  -1\right)  ^{a+b+c}\cdot\underbrace{\left[  X^{2a}Y^{2b}%
Z^{2c}\right]  \left(  -P_{a,b,c}\right)  }_{=-\left[  X^{2a}Y^{2b}%
Z^{2c}\right]  P_{a,b,c}}\\
&  =\left(  -1\right)  ^{a+b+c}\cdot\left[  X^{2a}Y^{2b}Z^{2c}\right]  \left(
P_{a,b,c}\right)  =F\left(  a,b,c\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Lemma \ref{lem.gf.dixon.coeff}}\right)  .
\end{align*}
This proves Lemma \ref{lem.gf.dixon.recF}.
\end{proof}

We are now ready to prove Theorem \ref{thm.gf.dixon}:

\begin{proof}
[Proof of Theorem \ref{thm.gf.dixon}.]We must prove (\ref{eq.thm.gf.dixon.eq}%
). As we recall, the equality (\ref{eq.thm.gf.dixon.eqF}) is just a
restatement of (\ref{eq.thm.gf.dixon.eq}), so it suffices to prove
(\ref{eq.thm.gf.dixon.eqF}) instead.

The proof is completely straightforward at this point: The LHS $F\left(
a,b,c\right)  $ of (\ref{eq.thm.gf.dixon.eqF}) satisfies a recurrence (Lemma
\ref{lem.gf.dixon.recF}), but the RHS $\dfrac{\left(  a+b+c\right)  !}%
{a!b!c!}$ satisfies the same recurrence (by Lemma \ref{lem.gf.dixon.rec}).
Furthermore, the starting values (i.e., the values when one of $a,b,c$ is $0$)
are also the same (by Lemma \ref{lem.gf.dixon.0}). It thus follows that the
LHS and the RHS are always equal, so that (\ref{eq.thm.gf.dixon.eqF}) holds.

\begin{fineprint}
For the sake of completeness, here is the proof in more detail:

We shall prove (\ref{eq.thm.gf.dixon.eqF}) by induction on $a+b+c$.

\textit{Base case:} If $a+b+c=0$, then $a=b=c=0$, and thus
(\ref{eq.thm.gf.dixon.eqF}) holds by Lemma \ref{lem.gf.dixon.0}.

\textit{Induction step:} Let $m$ be a positive integer. Assume (as the
induction hypothesis) that (\ref{eq.thm.gf.dixon.eqF}) holds whenever
$a+b+c=m-1$. We must now prove (\ref{eq.thm.gf.dixon.eqF}) in the case when
$a+b+c=m$.

So let $a,b,c\in\mathbb{N}$ satisfy $a+b+c=m$. We must prove
(\ref{eq.thm.gf.dixon.eqF}). If $a=0$ or $b=0$ or $c=0$, then
(\ref{eq.thm.gf.dixon.eqF}) follows directly from Lemma \ref{lem.gf.dixon.0}.
Thus, we WLOG assume that none of $a,b,c$ is $0$. Hence, $a,b,c$ are positive
integers. Thus, Lemma \ref{lem.gf.dixon.recF} yields%
\begin{equation}
F\left(  a,b,c\right)  =F\left(  a-1,b,c\right)  +F\left(  a,b-1,c\right)
+F\left(  a,b,c-1\right)  . \label{pf.thm.gf.dixon.rec}%
\end{equation}
However, from $a+b+c=m$, we obtain $\left(  a-1\right)
+b+c=\underbrace{a+b+c}_{=m}-1=m-1$. Thus, by our induction hypothesis,
(\ref{eq.thm.gf.dixon.eqF}) holds for $a-1$ instead of $a$. In other words, we
have
\[
F\left(  a-1,b,c\right)  =\dfrac{\left(  \left(  a-1\right)  +b+c\right)
!}{\left(  a-1\right)  !b!c!}.
\]
Similarly,%
\begin{align*}
F\left(  a,b-1,c\right)   &  =\dfrac{\left(  a+\left(  b-1\right)  +c\right)
!}{a!\left(  b-1\right)  !c!}\ \ \ \ \ \ \ \ \ \ \text{and}\\
F\left(  a,b,c-1\right)   &  =\dfrac{\left(  a+b+\left(  c-1\right)  \right)
!}{a!b!\left(  c-1\right)  !}.
\end{align*}
In light of these three equalities, we can rewrite (\ref{pf.thm.gf.dixon.rec})
as%
\begin{align*}
F\left(  a,b,c\right)   &  =\dfrac{\left(  \left(  a-1\right)  +b+c\right)
!}{\left(  a-1\right)  !b!c!}+\dfrac{\left(  a+\left(  b-1\right)  +c\right)
!}{a!\left(  b-1\right)  !c!}+\dfrac{\left(  a+b+\left(  c-1\right)  \right)
!}{a!b!\left(  c-1\right)  !}\\
&  =\dfrac{\left(  a+b+c\right)  !}{a!b!c!}\ \ \ \ \ \ \ \ \ \ \left(
\text{by Lemma \ref{lem.gf.dixon.rec}}\right)  .
\end{align*}
In other words, (\ref{eq.thm.gf.dixon.eqF}) holds for our $a,b,c$. This
completes the induction step. Thus, by induction, we have proved
(\ref{eq.thm.gf.dixon.eqF}), and with it Theorem \ref{thm.gf.dixon}.
\end{fineprint}
\end{proof}

Oof. I don't remember where I learned the above proof, nor do I have a good
explanation of \textquotedblleft where it comes from\textquotedblright\ (or,
more pragmatically, how to find such proofs). However, it is an impressive
exhibition of the power of polynomials. \medskip

\begin{fineprint}
Let us mention the following consequence of Theorem \ref{thm.gf.dixon}
(similar to Corollary \ref{cor.gf.altvdm.nk2}, but significantly harder):

\begin{corollary}
\label{cor.gf.altvdm.nk3}Let $n\in\mathbb{N}$. Then,%
\[
\sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}^{3}=%
\begin{cases}
\left(  -1\right)  ^{n/2}\dfrac{\left(  3n/2\right)  !}{\left(  n/2\right)
!^{3}}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}%
\end{cases}
\]

\end{corollary}

\begin{proof}
[Proof sketch.]When $n$ is odd, the sum $\sum_{k=0}^{n}\left(  -1\right)
^{k}\dbinom{n}{k}^{3}$ is easily seen to be $0$ since its addends cancel out
in pairs (namely, $\left(  -1\right)  ^{k}\dbinom{n}{k}^{3}$ cancels $\left(
-1\right)  ^{n-k}\dbinom{n}{n-k}^{3}$). So we WLOG assume that $n$ is even.
Then,%
\begin{align*}
&  \sum_{k=0}^{n}\left(  -1\right)  ^{k}\dbinom{n}{k}^{3}\\
\overset{0}{=}  &  \sum_{k\in\mathbb{Z}}\left(  -1\right)  ^{k}\dbinom{n}%
{k}^{3}\\
=  &  \sum_{k\in\mathbb{Z}}\underbrace{\left(  -1\right)  ^{n/2+k}}_{=\left(
-1\right)  ^{n/2}\left(  -1\right)  ^{k}}\dbinom{n}{n/2+k}^{3}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }n/2+k\\
\text{for }k\text{ in the sum}%
\end{array}
\right) \\
=  &  \left(  -1\right)  ^{n/2}\sum_{k\in\mathbb{Z}}\left(  -1\right)
^{k}\underbrace{\dbinom{n}{n/2+k}^{3}}_{\substack{=\dbinom{n/2+n/2}{n/2+k}%
^{3}\\=\dbinom{n/2+n/2}{n/2+k}\dbinom{n/2+n/2}{n/2+k}\dbinom{n/2+n/2}{n/2+k}%
}}\\
=  &  \left(  -1\right)  ^{n/2}\underbrace{\sum_{k\in\mathbb{Z}}\left(
-1\right)  ^{k}\dbinom{n/2+n/2}{n/2+k}\dbinom{n/2+n/2}{n/2+k}\dbinom
{n/2+n/2}{n/2+k}}_{\substack{=\dfrac{\left(  n/2+n/2+n/2\right)  !}{\left(
n/2\right)  !\left(  n/2\right)  !\left(  n/2\right)  !}\\\text{(by Theorem
\ref{thm.gf.dixon}, applied to }a=n/2\text{ and }b=n/2\text{ and
}c=n/2\text{)}}}\\
=  &  \left(  -1\right)  ^{n/2}\dfrac{\left(  n/2+n/2+n/2\right)  !}{\left(
n/2\right)  !\left(  n/2\right)  !\left(  n/2\right)  !}=\left(  -1\right)
^{n/2}\dfrac{\left(  3n/2\right)  !}{\left(  n/2\right)  !^{3}}.
\end{align*}
Thus, Corollary \ref{cor.gf.altvdm.nk3} is proved.
\end{proof}

With Corollary \ref{cor.gf.altvdm.nk2} and Corollary \ref{cor.gf.altvdm.nk3},
we have paid off all our proof debts from Lecture 15, Remark 2.3.5. As we
said, neither $\sum_{k=0}^{n}\dbinom{n}{k}^{3}$ nor $\sum_{k=0}^{n}\left(
-1\right)  ^{k}\dbinom{n}{k}^{4}$ seem to have explicit expressions, so the
series of binomial identities we have proved has come to a natural end.

And so has this lecture. More applications of generating functions can be
found in \cite[Chapter 3]{21s}, \cite[Chapter 7 (in the 1st edition)]%
{Loehr-BC}, \cite{Sambal22}, \cite{Wilf04}, \cite{Stanley-EC1} and many other
texts on combinatorics (as well as probability theory, where they are perhaps
even more central).
\end{fineprint}

\begin{thebibliography}{99999999}                                                                                         %


\bibitem[21s]{21s}Darij Grinberg, \textit{Algebraic Combinatorics (Drexel
Spring 2021 Math 701 lecture notes)}, 4 May 2021.\newline\url{https://www.cip.ifi.lmu.de/~grinberg/t/21s/lecs.pdf}

\bibitem[Good70]{Good70}\href{https://doi.org/10.1063/1.1665339}{I. J. Good,
\textit{Short Proof of a Conjecture by Dyson}, Journal of Mathematical Physics
\textbf{11}, no. 6, 1970, p. 1884.}

\bibitem[Grinbe15]{detnotes}Darij Grinberg, \textit{Notes on the combinatorial
fundamentals of algebra}, 15 September 2022.\newline%
\url{http://www.cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf} \newline The
numbering of theorems and formulas in this link might shift when the project
gets updated; for a \textquotedblleft frozen\textquotedblright\ version whose
numbering is guaranteed to match that in the citations above, see
\url{https://github.com/darijgr/detnotes/releases/tag/2022-09-15c} .

\bibitem[Loehr11]{Loehr-BC}%
\href{http://www.math.vt.edu/people/nloehr/bijbook.html}{Nicholas A. Loehr,
\textit{Bijective Combinatorics}, Chapman \& Hall/CRC 2011.}

\bibitem[MacMah78]{MacMah78}%
\href{https://libgen.is/book/index.php?md5=5F1C2A86B1A6F1A434CD43DEF9F0BF93}{Percy
Alexander MacMahon, \textit{Collected Papers: Volume I (Combinatorics)},
edited by George E. Andrews, MIT Press 1978.}

\bibitem[Niven69]{Niven69}Ivan Niven, \textit{Formal Power Series}, The
American Mathematical Monthly \textbf{76}, No. 8 (Oct., 1969), pp.
871--889.\newline\url{https://www.maa.org/programs/maa-awards/writing-awards/formal-power-series}

\bibitem[Sambal22]{Sambal22}\href{https://arxiv.org/abs/2205.00879v2}{Benjamin
Sambale, \textit{An invitation to formal power series}, arXiv:2205.00879v2}.

\bibitem[Stanle11]{Stanley-EC1}Richard P. Stanley, \textit{Enumerative
Combinatorics, volume 1}, Second edition, version of 15 July 2011. Available
at \url{http://math.mit.edu/~rstan/ec/} .\newline See
\url{http://math.mit.edu/~rstan/ec/} for errata.

\bibitem[Ward91]{Ward91}%
\href{https://www.maths.tcd.ie/pub/ims/bull27/bull27_46-54.pdf}{James Ward,
\textit{100 Years of Dixon's Identity}, IMS Bulletin \textbf{27} (1991), pp.
46--54.}

\bibitem[Wilf04]{Wilf04}Herbert S. Wilf, \textit{generatingfunctionology}, 2nd
edition 2004.\newline\url{https://www.math.upenn.edu/~wilf/DownldGF.html}
\end{thebibliography}


\end{document}