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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}{10}
\newcommand{\lecturetitle}{Power Series Continued and Graph Coloring}
\newcommand{\scribename}{Hunter Pearson}
\newcommand{\lecturedate}{February 23, 2026}

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\section{Review of Last Lecture}
Last time we saw $$F(x) = \sum_{n \geq 0} f_nx^n$$ where $f_0 = 0$, $f_1 = 1$, and $f_n = f_{n-1} + f_{n-2}$

We obtained $$F(x) = \frac{x}{1-x-x^2}$$

\section{Closed-form Formula for Fibonacci Numbers}

If we use partial fractions, we need to factor the denominator of $F(x)$.

Recall: $$ax^2 + bx+c = 0 \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Thus for $1-x-x^2$, its two roots are $$r_1 = \frac{-1 + \sqrt{5}}{2} \hspace{1 cm} r_2 = \frac{-1 -\sqrt{5}}{2}$$

Note that $r_1 \cdot r_2 = -1$

Then we get \begin{align*}
    1 - x - x^2 &= -(x - r_1)(x-r_2)\\
    &= -\frac{-1}{r_1 \cdot r_2}(x - r_1)(x-r_2)\\
    &= \frac{(x - r_1)}{r_1}\frac{(x-r_2)}{r_2}\\
    &= (\frac{x}{r_1}-1)(\frac{x}{r_2}-1)\\
    &= (1-\frac{x}{r_1})(1-\frac{x}{r_2})\\
\end{align*}

This implies $$F(x) = \frac{x}{(1-\frac{x}{r_1})(1-\frac{x}{r_2})} = 
\frac{A}{1-
\frac{x}{r_1}} + \frac{B}{1-\frac{x}{r_2}}$$

We need to find A and B. First, we can see that this this is equivalent to $$x = A(1-\frac{x}{r_2}) + B(1-\frac{x}{r_1})$$

Now there are two cases:

Case $x = r_1$: \begin{align*}
    r_1 &= A(1-\frac{r_1}{r_2}) + B(1-\frac{r_1}{r_1})\\
    &= A(1-\frac{r_1}{r_2}) + B(0)\\
    &= A(1-\frac{r_1}{r_2})\\
    \implies A &= \frac{r_1}{1-\frac{r_1}{r_2}}\\
    &= \frac{1}{\sqrt{5}}
\end{align*}

Case $x = r_2$: \begin{align*}
    r_2 &= A(1-\frac{r_2}{r_2}) + B(1-\frac{r_2}{r_1})\\
    &= B(1-\frac{r_2}{r_1}\\
    \implies B &= \frac{r_2}{1-\frac{r_2}{r_1}}\\
    &= \frac{-1}{\sqrt{5}}
\end{align*}

Together these imply that $$F(x) = \frac{1}{\sqrt{5}} \left( \sum_{n \geq 0} \frac{1}{r_1^n}x^n \right) - \frac{1}{\sqrt{5}} \left(\sum_{n \geq 0} \frac{1}{r_2^n}x^n \right)$$

One can check that $$\frac{1}{r_1} = \frac{1}{\frac{-1+\sqrt{5}}{2}} = \frac{2}{-1 + \sqrt{5}} \cdot \frac{1 + \sqrt{5}}{1 + \sqrt{5}} = \frac{1+\sqrt{5}}{2}$$
and $$\frac{1}{r_2} = \dots = \frac{1-\sqrt{5}}{2}$$

This gives us $$F(x) = \frac{1}{\sqrt{5}} \left( \sum_{n \geq 0} \left(\frac{1+\sqrt{5}}{2}\right)^nx^n \right) - \frac{1}{\sqrt{5}} \left(\sum_{n \geq 0} \left( \frac{1-\sqrt{5}}{2} \right)^nx^n \right)$$

Recall that $F(x) = f_0 + f_1x + f_2x^2 + f_3x^3 + \dots$, which implies $$f_n = \frac{1}{\sqrt{5}} \left( \frac{1 + \sqrt{5}}{2} \right)^n - \frac{1}{\sqrt{5}} \left( \frac{1-\sqrt{5}}{2} \right)^n$$

This is an integer!

\section{Catalan Generating Function}

Recall $c_0 = 1$ and $c_n = \sum_{j=1}^n c_{j-1}c_{n-j}$ for $n \geq 1$.

\subsection*{Step 1} The above implies that $$c_nx^n = \left( \sum_{j=1}^n c_{j-1}c_{n-j} \right) x^n$$

Which implies $$\sum_{n \geq 1} c_nx^n = \sum_{n \geq 1} \left( \sum_{j=1}^n c_{j-1}c_{n-j} \right) x^n$$

\subsection*{Step 2}

Let $C(x) = \sum\limits_{n \geq 0} c_nx^n$

The left hand side of the equation from step one gives us $$\sum_{n \geq 1} c_nx^n = C(x) - c_1 = C(x)-1$$

The right hand side of the equation from step one gives us \begin{align*}
    \sum_{n \geq 1} \left( \sum_{j=1}^n c_{j-1}c_{n-j} \right) x^n &= x\left(\sum_{n \geq 1} \left( \sum_{j=1}^n c_{j-1}c_{n-j} \right) x^{n-1}\right)\\
    &= x\left(\sum_{n \geq 1} \left( \sum_{j=0}^{n-1} c_{j}c_{n-j-1} \right) x^{n-1}\right)\\
    &= x\left(\sum_{n \geq 0} \left( \sum_{j=0}^{n} c_{j}c_{n-j} \right) x^{n}\right)\\
    &= xC(x)C(x)
\end{align*}

\subsection*{Step 3}

So $C(x)-1 = xC(x)^2$.

Solving for $C(x)$ using the quadratic formula gives us $C(x) = \frac{1 \pm \sqrt{1-4x}}{2x}$

\vspace{.5 cm}

{\color{vermillion}There are two issues: \begin{itemize}
    \item There seems to be two generating functions, and this cannot be.
    \item $\frac{1}{x}$ does not make sense as a formal power series, as $x$ has no inverse.
\end{itemize}}

This raises a question: Which sign should we use, $+$ or $-$?

Idea: Choose the sign such that the whole numerator $1 \pm \sqrt{1 - 4x}$ has no constant term. In other words, we want $1 \pm \sqrt{1 - 4x} = w_0 + w_1x + w_2x^2 + w_3x^3 + \dots$ where $w_0 = 0$.

If we can do this, then we shift the sequence one step to the left, i.e. $\frac{1 \pm \sqrt{1 - 4x}}{x} = w_1 + w_2x + w_3x^2 + w_4x^3 + \dots$.
This will take care of both problems.

There is a generalization of the binomial theorem where for any $n \in \mathbb{Q}$, we have \begin{equation}(1+x)^n = \sum_{k \geq 0} \binom{n}{k}x^k,\end{equation}
where the binomial coefficient is again generalized to work when $n$ is rational.
We will not cover this generalization, but note that $\binom{n}{0} = 1$, even if $n = \frac{1}{2}$, $k = 0$.

Since $\sqrt{1 - 4x} = (1-4x)^{\frac{1}{2}}$, we can use this generalization to find that $(1 - 4x)^{\frac{1}{2}} = 1 + w_1x + w_2x^2 + \dots$. Then $1-(1-4x)^\frac{1}{2}= 0-w_1x - w_2x^2 - w_3x^3 - \dots$, which is what we needed.

This implies that we should choose the minus sign in order to solve our issues.

Thus $C(x) = \frac{1-\sqrt{1-4x}}{2x}$

\subsection*{Step 4}

One can recover $c_n = \frac{1}{n+1}\binom{2n}{n}$ by using $C(x)$ and equation (1)

\section{Graph Coloring and Chromatic Polynomials}

Let $G=(V,E)$ be a graph.

A vertex coloring of $G$ is a mapping $c:V \rightarrow S$, where $S$ is a finite set of colors.

For example, $S = \{\text{red, black, blue}\}$. Sometimes it is easier to label these colors as numbers instead. i.e let $S = \{1,2,3\}$.

A coloring is proper if for all $e = \{u , v\} \in E$, it holds that $c(u) \not = c(v)$.

\subsection*{Some Examples}
\begin{figure}[h]
    \centering
\end{figure}
\begin{figure}[h]
    \centering
    \begin{minipage}{.333\textwidth}
    \centering
        \begin{tikzpicture}[scale = 2]
            \node [draw, shape = circle] (m) at (0,.75) {$u$};
            \node [draw, shape = circle] (x) at (0,0) {$x$};
            \node [draw, shape = circle] (v) at (1,.75) {$v$};
            \node [draw, shape = circle] (w) at (1,0) {$w$};
            \draw[-, thick] (m) to (v);
            \draw[-, thick] (m) to (x);
            \draw[-, thick] (x) to (v);
            \draw[-, thick] (w) to (v);
        \end{tikzpicture}
        \caption{Original Graph}
        \label{fig:Orig}
    \end{minipage}%
    \begin{minipage}{.333\textwidth}
        \centering
        \begin{tikzpicture}[scale = 2]
            \node [draw, shape = circle, fill = skyblue] (m) at (0,.75) {$u$};
            \node [draw, shape = circle, fill = reddishpurple] (x) at (0,0) {$x$};
            \node [draw, shape = circle, fill = yellow] (v) at (1,.75) {$v$};
            \node [draw, shape = circle, fill = reddishpurple] (w) at (1,0) {$w$};
            \draw[-, thick] (m) to (v);
            \draw[-, thick] (m) to (x);
            \draw[-, thick] (x) to (v);
            \draw[-, thick] (w) to (v);
        \end{tikzpicture}
        \caption{A Proper Coloring}
        \label{fig:Proper}
    \end{minipage}%
    \begin{minipage}{.333\textwidth}
      \centering
        \begin{tikzpicture}[scale = 2]
            \node [draw, shape = circle, fill = skyblue] (m) at (0,.75) {$u$};
            \node [draw, shape = circle, fill = reddishpurple] (x) at (0,0) {$x$};
            \node [draw, shape = circle, fill = reddishpurple] (v) at (1,.75) {$v$};
            \node [draw, shape = circle, fill = yellow] (w) at (1,0) {$w$};
            \draw[-, thick] (m) to (v);
            \draw[-, thick] (m) to (x);
            \draw[-, thick] (x) to (v);
            \draw[-, thick] (w) to (v);
        \end{tikzpicture}
        \caption{Not A Proper Coloring}
        \label{fig:non Proper}
    \end{minipage}
\end{figure}

\subsection*{Chromatic Number}

The chromatic number of $G$, denoted $\chi(G)$ is the minimum cardinality of the set of colors $S$ such that a proper coloring exists.

i.e. in $c:V \rightarrow S$, what is the minimum $|S|$ such that a proper coloring is possible.

\subsubsection*{From Our Example}
\begin{enumerate}
    \item $\chi (G)  \leq 3$ as we produced a proper coloring with $|S| = 3$ different colors.
    \item Also, we have the 3 cycle $u - v - x -u$. Any 3 cycle needs at least 3 colors. Thus $\chi(G) \geq 3$
\end{enumerate}

Thus since $3 \leq \chi(G) \leq 3$, we have $\chi(G) = 3$.

Note: In general, for any $k \in \mathbb{N}$, $\chi(K_n) = n$, where $K_n$ is the complete graph on n nodes.

\subsection*{Counting Proper Colorings}

How many proper colorings of $G$ with $t$ colors are there?

We write $P(G,t)$ to denote the number of proper colorings on $V$ using t colors.

$P(G.t) = |\{c:V \rightarrow [t]|\text{ $c$ is proper}\}|$

Note: By definition of $\chi(G)$, $P(G,t) = 0$ for $0 \leq t < \chi(G)$ and $P(G,t) > 0$ for $t \geq \chi(G)$

\subsubsection*{From Our Example}

\begin{itemize}
    \item $u$ has t options
    \item $v$ has $t-1$ options (color must be different than the color of $u$)
    \item $x$ has $t-2$ options (color must be different than the color of $v$ and $u$, which are themselves different)
    \item $w$ has $t-1$ options (color must be different than color of $v$)
\end{itemize}

So $P(G,t) = t(t-1)(t-2)(t-1) = t^4 - 4t^3 + 5t^2 - 2t$

\subsection{Next Time}
\begin{itemize}
    \item We will show $P(G,t)$ is indeed a polynomial no matter what $G$ is.
    \item We will see what graph theoretic information about $G$ is embedded in $P(G,t)$.
\end{itemize}




\end{document}