edits

src/Advanced/Fast Inverse Root/parts/7 quake.tex

@@ -14,23 +14,25 @@ float Q_rsqrt( float number ) {
 }
 \end{lstlisting}
 
-This code defines
-a function \texttt{Q\_rsqrt} that consumes a float named
+This code defines a function \texttt{Q\_rsqrt} that consumes a float named
 \texttt{number} and approximates its inverse square root (in other words, \texttt{Q\_rsqrt} computes $1/\sqrt{\texttt{number}}$).
 
 \vspace{2mm}
 
-If we rewrite this using the notation we're familiar with, we get the following:
+If we rewrite this using notation we're familiar with, we get the following:
 \begin{equation*}
-	\texttt{Q\_sqrt}(n_i) = 6240089 - (n_i \div 2) \approx \frac{1}{\sqrt{n_f}}
+	\texttt{Q\_sqrt}(n_f) = 6240089 - (n_i \div 2) \approx \frac{1}{\sqrt{n_f}}
 \end{equation*}
-$6240089$ is the decimal value of hex \texttt{0x5f3759df}. \par
-It is a magic number hard-coded into the function.
+
+\note{
+	\texttt{0x5f3759df} is $6240089$ in hexadecimal. \par
+	It is a magic number hard-coded into \texttt{Q\_sqrt}.
+}
 
 \vspace{2mm}
 
-Our goal in this section is to understand why this works. \par
-How are we able to approximate $\frac{1}{\sqrt{x}}$ by only subtracting and dividing by two??
+Our goal in this section is to understand why this works: \par
+How does Quake approximate $\frac{1}{\sqrt{x}}$ by simply subtracting and dividing by two??
 
 \problem{}
 Using basic log rules, rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2(x)$.