Quake edits

src/Advanced/Fast Inverse Root/parts/5 float.tex

@@ -6,10 +6,10 @@ In base 10, we interpret place value as follows:
 \begin{itemize}
 	\item $0.1 = 10^{-1}$
 	\item $0.03 = 3 \ \times 10^{-2}$
-	\item $0.0008 = 8 \times 10^{-4}$
+	\item $0.0208 = 2 \times 10^{-2} + 8 \times 10^{-4}$
 \end{itemize}
 
-\footnotetext{this is a misnomer, but that's ok.}
+\footnotetext{\say{decimal} is a misnomer, but that's ok.}
 
 \vspace{5mm}
 
@@ -24,7 +24,7 @@ We can do the same in base 2:
 
 \problem{}
 Rewrite the following binary decimals in base 10: \par
-\note{You may leave your answer as a fraction}
+\note{You may leave your answer as a fraction.}
 \begin{itemize}
 	\item $\texttt{1011.101}$
 	\item $\texttt{110.1101}$

src/Advanced/Fast Inverse Root/parts/6 approximate.tex

@@ -1,4 +1,4 @@
-\section{Approximation}
+\section{Integers and Floats}
 
 \generic{Observation:}
 For small values of $a$, $\log_2(1 + a)$ is approximately equal to $a$. \par
@@ -10,16 +10,23 @@ We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $
 
 TODO: why? Graphs.
 
-\problem{}
+\problem{}<convert>
 Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$. \par
 
 \vspace{5mm}
 
 Namely, show that
 \begin{equation*}
-	\log_2(x_f) ~=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
+	\log_2(x_f) ~=~ \frac{x_i}{2^{23}} - 127 + \varepsilon
 \end{equation*}
-for some error term $\varepsilon$
+for some correction term term $\varepsilon$
+
+\vspace{5mm}
+
+Before you start, make sure you understand what we're trying to do: \par
+We're finding an expression for $\log_2(x_f)$ in terms of $x_i$ and an correction term $\varepsilon$. \par
+That is, we're finding a closed-form expression that connects the two interpretations \par
+(\texttt{float} and \texttt{int}) of the bit string $x$.
 
 
 \begin{solution}

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

@@ -32,7 +32,11 @@ If we rewrite this using notation we're familiar with, we get the following:
 \vspace{2mm}
 
 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??
+\begin{itemize}
+	\item How does Quake approximate $\frac{1}{\sqrt{x}}$ by simply subtracting and dividing by two?
+	\item What's special about $6240089$?
+\end{itemize}
+
 
 \problem{}
 Using basic log rules, rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2(x)$.
@@ -45,39 +49,84 @@ Using basic log rules, rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2(x)$.
 
 
 \vfill
+\pagebreak
 
+\generic{Setup:}
+We are now ready to show that $\texttt{Q\_sqrt} \approx \frac{1}{\sqrt{n_f}}$. \par
+For convenience, let's call the bit string of the inverse square root $r$. \par
+In other words,
+\begin{equation*}
+	r_f := \frac{1}{\sqrt{n_f}}
+\end{equation*}
+This is the value we want to approximate.
 
-\problem{}
-Find the exact value of $\kappa$ in terms of $\varepsilon$. \par
-\note{Remember, $\varepsilon$ is the correction term in the approximation $\log_2(1 + a) = a + \varepsilon$.}
+\problem{}<finala>
+Find an approximation for $\log_2(r_f)$ in terms of $n_i$ and $\varepsilon$ \par
+\note{Remember, $\varepsilon$ is the correction constant in our approximation of $\log_2(1 + a)$.}
 
 \begin{solution}
-	Say $g_f = \frac{1}{\sqrt{n_f}}$ (i.e, $g_f$ is the value we want to compute). We then have:
-	\begin{align*}
-		\log_2(g_f)
-		&=\log_2(\frac{1}{\sqrt{n_f}}) \\
-		&=\frac{-1}{2}\log_2(n_f) \\
-		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right)
-	\end{align*}
-	But we also know that
-	\begin{align*}
-		\log_2(g_f)
-		&=\frac{g_i}{2^{23}} + \varepsilon - 127
-	\end{align*}
-	So,
-	\begin{align*}
-		\frac{g_i}{2^{23}} + \varepsilon - 127
-		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right) \\
-		\frac{g_i}{2^{23}}
-		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} \right) + \frac{3}{2}(\varepsilon - 127) \\
-		g_i
-		&=\frac{-1}{2}\left(n_i \right) + 2^{23}\frac{3}{2}(\varepsilon - 127) \\
-		&=2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
-	\end{align*}
-
-	and thus $\kappa = 2^{23}\frac{3}{2}(\varepsilon - 127)$.
+	\begin{equation*}
+		\log_2(r_f)
+		~=~ \log_2(\frac{1}{\sqrt{n_f}})
+		~=~ \frac{-1}{2}\log_2(n_f)
+		~\approx~ \frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right)
+	\end{equation*}
 \end{solution}
 
 \vfill
 
+\problem{}<finalb>
+Let's call the \say{magic number} in the code above $\kappa$, so that
+\begin{equation*}
+	\texttt{Q\_sqrt}(n_f) = \kappa - (n_i \div 2)
+\end{equation*}
+Use problems \ref{num:convert} and \ref{num:finala} to show that $\texttt{Q\_sqrt}(n_f) \approx r_i$. \par
+
+\begin{solution}
+	From \ref{convert}, we know that
+	\begin{equation*}
+		\log_2(r_f) \approx \frac{r_i}{2^{23}} + \varepsilon - 127
+	\end{equation*}
+
+	\note{
+		Our approximation of $\log_2(1+a)$ uses a fixed correction constant, \par
+		so the $\varepsilon$ here is equivalent to the $\varepsilon$ in \ref{finala}.
+	}
+
+	Combining this with the result from \ref{finala}, we get:
+	\begin{align*}
+		\frac{r_i}{2^{23}} + \varepsilon - 127
+		&~\approx~\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right) \\
+		\frac{r_i}{2^{23}}
+		&~\approx~\frac{-1}{2}\left( \frac{n_i}{2^{23}} \right) + \frac{3}{2}(\varepsilon - 127) \\
+		r_i
+		&~\approx~\frac{-1}{2}\left(n_i \right) + 2^{23}\frac{3}{2}(\varepsilon - 127)
+		~=~ 2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
+	\end{align*}
+
+	\vspace{2mm}
+
+	This is exactly what we need! If we set $\kappa$ to $\frac{2^{24}}{3}(\varepsilon - 127)$, then
+	\begin{equation*}
+		r_i ~\approx~ \kappa - (n_i \div 2) ~=~ \texttt{Q\_sqrt}(n_f)
+	\end{equation*}
+\end{solution}
+
+\vfill
+
+\problem{}
+What is the exact value of $\kappa$ in terms of $\varepsilon$? \par
+\hint{Look at \ref{finalb}. We already found it!}
+
+\begin{solution}
+	This problem makes sure our students see that
+	$\kappa = \frac{2^{24}}{3}(\varepsilon - 127)$. \par
+	See the soluton to \ref{finalb}.
+\end{solution}
+
+\makeatletter
+\if@solutions\else
+	\vspace{2cm}
+\fi\makeatother
+
 \pagebreak