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    \makebox[\textwidth][l]{\textbf{MATH 6404: Applied [Combinatorics and] Graph Theory} \hfill CU Denver} \\
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    \makebox[\textwidth][l]{Spring 2026 \hfill Instructor: Carlos Mart\'inez}
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\newcommand{\lecturenumber}{8}
\newcommand{\lecturetitle}{Formal Power Series and Generating Functions}
\newcommand{\scribename}{Hanbyul (Han) Lee}
\newcommand{\lecturedate}{Feb 16, 2026}

\begin{document}

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\section{Generating Polynomials}

\begin{definition}[Generating polynomial]
A finite sequence $a_0,a_1,a_2,\dots,a_n$ has the \emph{generating polynomial}
\[
f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n=\sum_{i=0}^{n} a_i x^i.
\]
\end{definition}

\begin{example}[Binomial coefficients]
Let $a_k=\binom{n}{k}$ for $k=0,1,\dots,n$ (note: Pascal's triangle). The corresponding generating polynomial is
\[
f(x)=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n
=\sum_{k=0}^{n}\binom{n}{k}x^k.
\]
\end{example}

\begin{theorem}[Binomial theorem]
For $n\in \mathbb{N}_0$,
\[
\sum_{k=0}^{n}\binom{n}{k}x^k=(1+x)^n.
\]
In particular, this says that $f(x)=\sum_{k=0}^{n}\binom{n}{k}x^k$ has a ``nice'' factorization.
\end{theorem}

\begin{proof}
We prove the identity by induction on $n$.

\noindent\textbf{Base case.} For $n=0$,
\[
\binom{0}{0}x^0 = 1\cdot 1 = (1+x)^0 = 1.
\]

\noindent\textbf{Inductive step.} We use the convention that $\binom{n}{k}=0$ when $k<0$ or $k>n$, so that
\[
\sum_{k=0}^{n}\binom{n}{k}x^k=\sum_{k=-\infty}^{\infty}\binom{n}{k}x^k.
\]
Using the identity $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$, we compute
\begin{align*}
\sum_{k}\binom{n}{k}x^k
&= \sum_{k}\left(\binom{n-1}{k-1}+\binom{n-1}{k}\right)x^k \\
&= \sum_{k}\binom{n-1}{k-1}x^k + \sum_{k}\binom{n-1}{k}x^k \\
&= x\sum_{k}\binom{n-1}{k-1}x^{k-1} + \sum_{k}\binom{n-1}{k}x^k \\
&= x\sum_{k}\binom{n-1}{k}x^{k} + \sum_{k}\binom{n-1}{k}x^k \qquad\text{(re-indexing)}\\
&= x(1+x)^{n-1} + (1+x)^{n-1} \\
&= (1+x)^{n-1}(1+x) \\
&= (1+x)^n,
\end{align*}
where we used the inductive hypothesis to replace $\sum_{k}\binom{n-1}{k}x^{k}$ by $(1+x)^{n-1}$.
\end{proof}

\begin{proof}[Combinatorial proof]
Expanding
\[
(1+x)^n=(1+x)(1+x)\cdots(1+x) \qquad (n\ \text{times}),
\]
the coefficient of $x^k$ counts the number of ways to pick $k$ of the factors from which we choose the $x$ term (and from the remaining factors we choose $1$). Hence there are $\binom{n}{k}$ ways, so the coefficient of $x^k$ is $\binom{n}{k}$.
\end{proof}

\begin{remark}[Some quick consequences]
Evaluating at $x=1$ gives
\[
\sum_{k=0}^{n}\binom{n}{k} = (1+1)^n = 2^n.
\]
Evaluating at $x=-1$ gives
\[
\sum_{k=0}^{n}\binom{n}{k}(-1)^k = 0 \qquad \text{for } n\ge 1.
\]
In particular,
\[
\sum_{\substack{0\le k\le n\\ k\ \mathrm{even}}}\binom{n}{k}
=
\sum_{\substack{0\le k\le n\\ k\ \mathrm{odd}}}\binom{n}{k}.
\]
\end{remark}

\section{Generating Functions}

\begin{definition}[Generating function]
An infinite countable sequence $a_0,a_1,a_2,a_3,\dots$ has the \emph{generating function}
\[
f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots=\sum_{k=0}^{\infty} a_k x^k.
\]
We view $f(x)$ as a \emph{formal power series}: the powers of $x$ are placeholders, and (unlike the generating polynomial case) we do not substitute numerical values for $x$.
\end{definition}

\begin{example}[The all-ones sequence]
Consider $a_0=1,1,1,1,\dots$ (all ones). Its generating function is
\[
f(x)=\sum_{k\ge 0} x^k = 1+x+x^2+x^3+\cdots.
\]
This has a ``nice'' representation as
\[
f(x)=\frac{1}{1-x}.
\]
Indeed, $f(x)$ is the inverse of $(1-x)$ since
\[
f(x)(1-x)=(1+x+x^2+x^3+\cdots)(1-x)=1.
\]
\end{example}

\begin{definition}[Coefficient extraction]
If $A(x)$ is a generating function, we denote
\[
a_n=\left[x^n\right]A(x),
\]
to mean: \emph{extract the coefficient of $x^n$ in $A(x)$.}
\end{definition}

\begin{example}
\[
\left[x^n\right]\left(\sum_{k\ge 0} x^k\right)=1
\qquad\text{for all } n\ge 0.
\]
\end{example}

\begin{remark}
Sometimes, a precise closed-form formula for
\[
a_n=\left[x^n\right]A(x)
\]
can be easily extracted from the form of $A(x)$.
\end{remark}

\begin{remark}[Taylor series and coefficient extraction]
In general, we can extract $\left[x^n\right]A(x)$ using a Taylor series expansion of $A(x)$ around $x=0$:
\[
a_n=\frac{A^{(n)}(0)}{n!}.
\]
This gives the numerical value of $a_n$ (which can be obtained computationally).
\end{remark}

\section{The Algebra of Formal Power Series}

Before going any further, we make the formal setting explicit.

\begin{definition}[Formal power series over $\mathbb{C}$]
\[
\mathbb{C}[[x]]=\left\{\sum_{n\ge 0} a_n x^n : a_n\in \mathbb{C}\ \text{for all } n\ge 0\right\}.
\]
We refer to $\mathbb{C}[[x]]$ as the ``algebra of formal power series.'' (Note: $A(x)\in \mathbb{C}[[x]]$.)
\end{definition}

\begin{remark}[Ring operations]
Let
\[
A(x)=\sum_{n\ge 0} a_n x^n,
\qquad
B(x)=\sum_{n\ge 0} b_n x^n.
\]
\begin{itemize}
\item \textbf{Addition:}
\[
A(x)+B(x)=\sum_{n\ge 0}(a_n+b_n)x^n.
\]
\item \textbf{Scalar multiplication:} for $c\in \mathbb{C}$,
\[
cA(x)=c\left(\sum_{n\ge 0} a_n x^n\right)=\sum_{n\ge 0}(c\,a_n)x^n.
\]
\item \textbf{Multiplication (convolution):}
\[
A(x)\cdot B(x)=\left(\sum_{n\ge 0} a_n x^n\right)\left(\sum_{n\ge 0} b_n x^n\right)
= \sum_{n\ge 0} c_n x^n =: C(x),
\]
where
\[
c_n=\sum_{k=0}^{n} a_k\,b_{n-k}.
\]
\end{itemize}
\end{remark}

\begin{remark}[Inverses and a word of caution]
Recall that an inverse means $S(x)S^{-1}(x)=1$ (by definition of the convolution product).

Some statements that are true \emph{analytically} about a power series (in the sense seen in calculus) may not be true as statements in $\mathbb{C}[[x]]$. For example, $\frac{1}{x}$ makes sense analytically as the inverse of $x$ (since $x\cdot \frac{1}{x}=1$), but it does not make sense in $\mathbb{C}[[x]]$ since the formal power series
\[
x = 0 + x + 0x^2 + 0x^3 + \cdots
\]
has no inverse in $\mathbb{C}[[x]]$.

Indeed, if $x f(x)=1$, then $x f(x)$ has $0$ as its constant term, but the right-hand side of $x f(x)=1$ is $1$.
\end{remark}

\begin{fact}
One can show that $f(x)\in \mathbb{C}[[x]]$ has an inverse if and only if
\[
\left[x^0\right]f(x)\neq 0.
\]
\end{fact}

\end{document}