feat: removes some trivs and thrfs for wider consumption

tex/1.14.tex

@@ -132,7 +132,7 @@
 		\pdv{\log P}{\sigma} &= - \frac{N}{\sigma} - \frac{S + N \left(\mean{x} - \mu \right)^2}{2} \frac{-2}{\sigma^3} \\
 		\pdv{\log P}{\sigma} &= - \frac{N}{\sigma} + \frac{S + N \left(\mean{x} - \mu \right)^2}{\sigma^3}
 	\end{align}
-	\triv we can now sub $\mu = \mean{x}$, giving us
+	Naturally we can now sub $\mu = \mean{x}$, giving us
 	\begin{align}
 		\pdv{\log P}{\sigma} &= - \frac{N}{\sigma} + \frac{S}{\sigma^3}.
 	\end{align}

tex/1.2.tex

@@ -19,7 +19,7 @@
 \usepackage{enumerate}
 
 % custom deepak packages
-\usepackage{luatrivially}
+% \usepackage{luatrivially}
 \usepackage{subtitling}
 
 \begin{luacode*}
@@ -27,7 +27,7 @@
 \end{luacode*}
 
 \title{Problem 1.2}
-\subtitle{Probability distributions: gory version}
+\subtitle{Probability distributions}
 \author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
 % want empty date
 \predate{}

tex/1.6.tex

@@ -54,13 +54,13 @@
 	\end{enumerate}
 	\section{Solution} \label{sec:solution}
 	\subsection{(b) Eigenvalue differences} \label{subsec:solb}
-	\triv the eigenvalues $\nu_{\pm}$ are easy to find.
+	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
+	Therefore
 	\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} 
@@ -69,7 +69,7 @@
 
 	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}
+	We see that \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}
@@ -82,13 +82,13 @@
 	
 	\subsubsection{diagonal elements}
 	For the diagonal elements, say $a$, we have a double of a Gaussian variable.
-	\thrf for instance $\rho(a) = \rho_g(\flatfrac{a}{2})$.
+	Therefore for instance $\rho(a) = \rho_g(\flatfrac{a}{2})$.
 	\begin{align}
 		\rho(a) &\propto \frac{1}{\sqrt{2 \pi}} e^{-\flatfrac{a^2}{8}},
 	\end{align}
 	which means that $\sigma_a = 2$.
 	
-	\subsubsection{off-diagonal}
+	\subsubsection{off-diagonal} \label{subsec:offdiag}
 	For off diagonal elements, say $b$, we have two independent Gaussians being summed.
 	\begin{align}
 		\rho(b) &= \int_{-\infty}^\infty \dd{x} \rho_g(x) \rho_g(b - x) \\
@@ -103,8 +103,10 @@
 
 	\subsubsection{standard deviation of difference}
 	We want to show that the standard deviation of $d = \flatfrac{\left(c - a\right)}{2}$ is equal to the standard deviation of $b$, $\sqrt{2}$.
-	\triv the numerator gets added in quadrature, $\sigma_d \propto \sqrt{2} \sigma_a$, denominator rescales by $\frac12$, becomes $\sqrt{2}$.
-	
+	As we see, the numerator gets added in quadrature, $\sigma_d \propto \sqrt{2} \sigma_a$, denominator rescales by $\frac12$, becomes $\sqrt{2}$.
+	This exactly parallels the derivation in \cref{subsec:offdiag}.
+	In fact, you can pretty easily set up a homomorphism between sums and differences of variables with distributions of Gaussians of other means and standard deviations and distributions that sum in the ways implied by these expressions.\todo{What's the nice categorical view of this stuff? Morphisms of the category of probability spaces to some measurable space.}
+
 	\subsection{(d) Find the formula for probability distirbution of $\lambda$}
 	
 	The two boys $d$ and $b$ are independent Gaussians, with standard deviations $\sqrt{2}$.