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+\documentclass{article}
+
+% set up telugu
+\usepackage{fontspec}
+\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu]
+\usepackage{polyglossia}
+\setdefaultlanguage{english}
+\setotherlanguage{telugu}
+
+%other packages
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{physics}
+\usepackage{siunitx}
+\usepackage{todonotes}
+\usepackage[plain]{fancyref}
+\usepackage{luacode}
+\usepackage{titling}
+\usepackage{enumerate}
+
+% custom deepak packages
+\usepackage{luatrivially}
+\usepackage{subtitling}
+
+\begin{luacode*}
+	math.randomseed(31415926)
+\end{luacode*}
+
+\newcommand{\kb}{k_{\mathrm{B}}}
+
+\title{Problem 3.8}
+\subtitle{Microcanonical energy fluctuations}
+\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
+% want empty date
+\predate{}
+\date{}
+\postdate{}
+
+% !TeX spellcheck = en_GB
+\begin{document}
+	\maketitle
+	
+	This problem talks about about weakly coupling subsystems and looking at fluctuations.
+	An earlier result is that
+	\begin{equation}
+		\sigma_{E_1}^2 = - \flatfrac{\kb}{\left(\pdv[2]{S_1}{E_1} + \pdv[2]{S_2}{E_2}\right)}
+	\end{equation}
+	
+	\begin{enumerate}[(a)]
+		\item Show that
+		\begin{equation}
+			\frac{1}{\kb} \pdv[2]{S}{E} = - \frac{1}{\kb T} \frac{1}{N c_v T},
+		\end{equation}
+		where $c_v$ is the inverse of the total specific heat at constant volume.
+		The specific heat $c_v$ is the energy needed per particle to change the temperature by one unit:
+		\begin{equation}
+			N c_v = \left.\left(\pdv{E}{T}\right)\right|_{V, N}
+		\end{equation}
+	\end{enumerate}
+
+	\section{Solution} \label{sec:solution}
+	\subsection*{(a)}
+	We start with the general case, keeping volume and particle number constant.
+	To start, we use $\frac{1}{T} = \left. \pdv{S}{E} \right|_{V, N}$, so 
+	\begin{align}
+		\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= \frac{1}{\kb} \left. \pdv{}{E} \frac{1}{T} \right|_{V, N} \\
+		&= \frac{1}{\kb} \left. \pdv{}{E} \frac{1}{T} \right|_{V, N} \\
+		&= - \frac{1}{\kb} \left. \frac{1}{T^2} \pdv{T}{E} \right|_{V, N},
+	\end{align}
+	and we use the Physicist's Theorem to invert the derivative, giving us $\flatfrac{1}{N c_v} = \left.\left(\pdv{T}{E}\right)\right|_{V, N}$, so
+	\begin{align}
+		\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= - \frac{1}{\kb} \frac{1}{T^2} \frac{1}{N c_v} \\
+		\frac{1}{\kb} \left.\pdv[2]{S}{E} \right|_{V, N} &= - \frac{1}{\kb T} \frac{1}{N c_v T},
+	\end{align}
+	as desired.
+
+	\newpage
+	\listoftodos
+
+\end{document}