diff --git a/Advanced/Generating Functions/parts/00 introduction.tex b/Advanced/Generating Functions/parts/00 introduction.tex
index 96fca76..adf9871 100755
--- a/Advanced/Generating Functions/parts/00 introduction.tex	
+++ b/Advanced/Generating Functions/parts/00 introduction.tex	
@@ -4,7 +4,7 @@
 Say we have a sequence $a_0, a_1, a_2, ...$. \par
 The \textit{generating function} of this sequence is defined as follows:
 \begin{equation*}
-	A(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + a_3x+3 + ...
+	A(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + a_3x^3 + ...
 \end{equation*}
 
 Under some circumstances, this sum does not converge, and thus $A(x)$ is undefined. \par
@@ -44,6 +44,7 @@ Assuming $|x| < 1$, show that
 \begin{equation*}
 	\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...
 \end{equation*}
+\hint{use some clever algebra. What is $x \times (1 + x + x^2 + ...)$? }
 
 \begin{solution}
 	Let $S = 1 + x + x^2 + ...$ \par
diff --git a/Advanced/Generating Functions/parts/01 fibonacci.tex b/Advanced/Generating Functions/parts/01 fibonacci.tex
index af82307..3c99478 100755
--- a/Advanced/Generating Functions/parts/01 fibonacci.tex	
+++ b/Advanced/Generating Functions/parts/01 fibonacci.tex	
@@ -50,7 +50,7 @@ we used to define the Fibonacci numbers.
 \pagebreak
 
 
-\problem{}
+\problem{}<fibo>
 Using the problems on the previous page, find $F(x)$ in terms of $x$.
 
 \begin{solution}
@@ -77,7 +77,7 @@ A \textit{rational function} $f$ is a function that can be written as a quotient
 That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials.
 
 \problem{}
-Solve this equation for $F(x)$, expressing it as a rational function.
+Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational function.
 
 \begin{solution}
 	\begin{align*}
@@ -98,20 +98,20 @@ Solve this equation for $F(x)$, expressing it as a rational function.
 \pagebreak
 
 
-\problem{}<pfd>
-Now that we have a rational function for $F(x)$, \par
-find a closed-form expression for its coefficients.
-
-\vspace{2mm}
-
-Do this using \textit{partial fraction decomposition:} \par
-We can break up a rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
+\definition{}
+\textit{Partial fraction decomposition} is an algebreic technique that works as follows: \par
+If $p(x)$ is a polynomial and $a$ and $b$ are constants,
+we can rewrite the rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
 \begin{equation*}
-	F(x) = \frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b}
+	\frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b}
 \end{equation*}
 where $c$ and $d$ are constants.
 
 
+\problem{}<pfd>
+Now that we have a rational function for $F(x)$, \par
+find a closed-form expression for its coefficients using partial fraction decomposition.
+
 \begin{solution}
 	\begin{align*}
 		F(x)