Adds solution to b

tex/1.6.tex

@@ -16,7 +16,7 @@
 \usepackage[plain]{fancyref}
 \usepackage{luacode}
 \usepackage{titling}
-\usepackage{enumerate}
+\usepackage{enumitem}
 
 % custom deepak packages
 \usepackage{luatrivially}
@@ -43,8 +43,37 @@
 		\rho(x) = \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{x^2}{2}}
 	\end{equation}
 	Then add with its transpose to make symmetric.
+	Let's look at $N = 2$, so $M = \begin{pmatrix}
+		a & b \\
+		b & c
+	\end{pmatrix}$
+	\begin{enumerate}[label=(\alph*), start=2]
+		\item Show that the eigenvalue difference for $M$ is $\lambda = \sqrt{\left(c - a\right)^2 + 4 b^2} = 2 \sqrt{d^2 + b^2}$, where $d = \flatfrac{\left(c-a\right)}{2}$
+	\end{enumerate}
 	\section{Solution} \label{sec:solution}
-	\subsection{Likelihoods} \label{subsec:sola}
+	\subsection{(b) Eigenvalue differences} \label{subsec:solb}
+	\triv the eigenvalues $\nu_{\pm}$ are easy to find.
+	Set up the characteristic equation.
+	\begin{align}
+		0 &= \left(a - \nu\right)\left(c - \nu\right) - b^2 \\
+		&= (ac - b^2) - \left(a + c\right) \nu + \nu^2.
+	\end{align}
+	\thrf
+	\begin{align}
+		\nu_\pm &= \frac{\left(a + c \right) \pm \sqrt{\left(a + c\right)^2 - 4 \left(ac -b^2\right)}}{2} \\
+		\nu_\pm &= \frac{\left(a + c \right) \pm \sqrt{\left(c - a\right)^2 + 4 b^2}}{2} 
+	\end{align}
+	So the difference is $\sqrt{\left(c - a\right)^2 + 4 b^2}$.
+	
+	We have thus $\flatfrac{\lambda^2}{4} = d^2 + b^2$, so the region of between fixed $\lambda$ and $\lambda + \Delta$ is an annulus, around the circle with radius $\flatfrac{\lambda}{2}$.
+	
+	\triv \begin{align}
+		\rho(\lambda) &= \int_{(d, b)} \rho(d, b) \dd{d} \dd{b} \\
+		&\propto \int_{(d, b)} \rho(d, b) \lambda \dd{\lambda} \dd{\phi} \\
+		&\propto 2 \pi \int_{(d, b)} \rho(d, b) \lambda \dd{\lambda}
+	\end{align}
+	The $\propto$ is because some factors of $2$.
+	Important point is that as long as the probability $\rho_M$ is well-enough behaved, then as $\lambda \rightarrow 0$, $\rho(\lambda) \rightarrow 0$.
 	\newpage
 	\listoftodos