Start proofreading result section
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Janita Willumsen
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@ -85,5 +85,20 @@
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% Appendix
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\subfile{sections/appendices}
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\end{document}
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% Method
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% Problem 1: OK
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% Problem 2: OK
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% Problem 3: OK - missing theory of phase transition
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% Problem 7:
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% Problem 9:
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% Results
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% Problem 4:
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% Problem 5:
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% Problem 6:
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% Problem 8:
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% Problem 9:
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%
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%
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@ -127,5 +127,20 @@ and susceptibility
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\frac{\chi}{N} &= \frac{4}{N k_{B} T} \bigg( \frac{3e^{8 \beta J} + e^{-8 \beta J} + 3}{(\cosh(8 \beta J) + 3)^{2}} \bigg) \ .
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\end{align*}
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\section{Change in total system energy}\label{sec:delta_energy}
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When we consider the change in energy after flipping a single spin, we evaluate
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$\Delta E = E_{\text{after}} - E_{\text{before}}$. We find the $3^{2}$ values as
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\begin{align*}
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\Delta E = -8J - (-8J) = 0 \\
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\Delta E = -8J - 0 = -8J \\
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\Delta E = -8J - 8J = -16J \\
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\Delta E = 0 - (-8J) = 8J \\
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\Delta E = 0 - 0 = 0 \\
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\Delta E = 0 - 8J = -8J \\
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\Delta E = 8J - (-8J) = 16J \\
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\Delta E = 8J - 0 = 8J \\
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\Delta E = 8J - 8J = 0,
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\end{align*}
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where the five distinct values are $\Delta E = \{-16J, -8J, 0, 8J, 16J\}$.
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\end{document}
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@ -177,6 +177,9 @@ Boltzmann constant we derive the remaining units, which can be found in Table
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\label{tab:units}
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\end{table}
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\subsection{Phase transition and critical temperature}\label{subsec:phase_critical}
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% P9 critical temperature
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\subsection{The Markov chain Monte Carlo method}\label{subsec:mcmc_method}
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Markov chains consist of a sequence of samples, where the probability of the next
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sample depend on the probability of the current sample. Whereas the Monte Carlo
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@ -202,8 +205,9 @@ At each step of flipping a spin, the change in energy is evaluated as
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\begin{align*}
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\Delta E &= E_{\text{after}} - E_{\text{before}} \ .
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\end{align*}
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We have to evaluate
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Since the total system energy only takes three different values, the change in
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energy can take $3^{2}$ values. However, there are only five distinct values
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$\Delta E = \{-16J, -8J, 0, 8J, 16J\}$, which we find in Appendix \ref{sec:delta_energy}.
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\begin{figure}[H]
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\begin{algorithm}[H]
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\caption{Metropolis-Hastings Algorithm}
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@ -226,13 +230,22 @@ We have to evaluate
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\EndProcedure
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\end{algorithmic}
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\end{algorithm}
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\caption{Algo}
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\end{figure}
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\subsection{Phase transition and critical temperature}\label{subsec:phase_critical}
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We can avoid computing the Boltzmann factor
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\begin{equation}
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p(\mathbf{s} | T) = e^{-\beta \Delta E}
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\label{eq:boltzmann_factor}
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\end{equation}
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at every spin flip, by using a look up table (LUT) with the possible values. We use
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the change in energy $\Delta E$ as a key for the resulting value of the exponential
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function in a hash map.
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\subsection{Implementation}\label{subsec:implementation}
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% P2
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% P3 boundary condition and if-tests
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To avoid the overhead of if-tests, and take advantage of the parallelization, we
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define an index for every edge case. That is, for a spin at a given boundary index
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we use a pre-set index (...), to avoid if-tests and reduce overhead and runtime.
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% P7 parallelization
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\subsection{Tools}\label{subsec:tools}
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The Ising model and MCMC methods are implemented in C++, and parallelized using
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@ -244,17 +257,3 @@ addition, while optimizing our implementation we used the profiler
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\end{document}
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% Method
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% Problem 1: OK
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% Problem 2:
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% Problem 3:
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% Results
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% Problem 4:
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% Problem 5:
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% Problem 6:
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% Problem 7:
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% Problem 8:
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% Problem 9:
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%
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%
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@ -2,5 +2,108 @@
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\begin{document}
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\section{Results}\label{sec:results}
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\subsection{Burn-in time}\label{subsec:burnin_time}
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We start with a lattice where $L = 20$, to study the burn-in time, that is the
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number of Monte Carlo cycles necessary for the system to reach an equilibrium.
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We consider two different temperatures $T_{1} = 1.0 J/k_{B}$ and $T_{2} = 2.4 J/k_{B}$,
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where $T_{2}$ is close to the critical temperature. We can use the correlation
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time $\tau \approx L^{d + z}$ to determine time, where $d$ is the dimensionality
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of the system and $z = 2.1665 \pm 0.0012$ \footnote{This value was determined by
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Nightingale and Blöte for the Metropolis algorithm.}
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% Need to include a section of Onsager's analytical result.
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We show the numerical estimates for temperature $T_{1}$ of $\langle \epsilon \rangle$
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in Figure \ref{fig:burn_in_energy_1_0} and $\langle |m| \rangle$ in Figure
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\ref{fig:burn_in_magnetization_1_0}. For temperature $T_{2}$, the numercal estimate
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of $\langle \epsilon \rangle$ is shown in Figure \ref{fig:burn_in_energy_2_4} and
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$\langle |m| \rangle$ in Figure \ref{fig:burn_in_magnetization_2_4}. The lattice
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is initialized in both an ordered and an unordered state. We observe that for
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$T_{1}$ there is no change in either expectation value with increasing number of
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Monte Carlo cycles, when we start with an ordered state. As for the unordered
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lattice, we observe a change for the first 5000 MC cycles, where it stabilizes.
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The approximated expected energy is $-2$ and expected magnetization is $1.0$,
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which is to be expected for temperature 0f $1.0$. T is below the critical and the
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pdf using $T = 1.0$ result in
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\begin{align*}
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p(s|T=1.0) &= \frac{1}{e^{-\beta \sum E(s)}} e^{-\beta E(s)} \\
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&= \frac{1}{e^{-(1/k_{B}) \sum E(s)}} e^{-(1/k_{B}) E(s)} \ .
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\end{align*}
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% Burn-in figures
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_1_0.pdf}
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\caption{$\langle |m| \rangle$ as a function of time, for $T = 1.0 J / k_{B}$}
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\label{fig:burn_in_magnetization_1_0}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_energy_1_0.pdf}
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\caption{$\langle \epsilon \rangle$ as a function of time, for $T = 1.0 J / k_{B}$}
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\label{fig:burn_in_energy_1_0}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_magnetization_2_4.pdf}
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\caption{$\langle |m| \rangle$ as a function of time, for $T = 2.4 J / k_{B}$}
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\label{fig:burn_in_magnetization_2_4}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/burn_in_time_energy_2_4.pdf}
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\caption{$\langle \epsilon \rangle$ as a function of time, for $T = 2.4 J / k_{B}$}
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\label{fig:burn_in_energy_2_4}
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\end{figure}
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% Histogram figures
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/pd_estimate_1_0.pdf}
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\caption{Histogram $T = 1.0 J / k_{B}$}
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\label{fig:histogram_1_0}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/pd_estimate_2_4.pdf}
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\caption{Histogram $T = 2.4 J / k_{B}$}
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\label{fig:histogram_2_4}
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\end{figure}
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% Phase transition figures
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\begin{figure}
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/1M/energy.pdf}
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\caption{$\langle \epsilon \rangle$ for $T \in [2.1, 2.4]$, 1000000 MC cycles.}
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\label{fig:phase_energy}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/1M/magnetization.pdf}
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\caption{$\langle |m| \rangle$ for $T \in [2.1, 2.4]$, 1000000 MC cycles.}
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\label{fig:phase_magnetization}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/1M/heat_capacity.pdf}
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\caption{$C_{V}$ for $T \in [2.1, 2.4]$, 1000000 MC cycles.}
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\label{fig:phase_heat}
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\end{figure}
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/1M/susceptibility.pdf}
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\caption{$\chi$ for $T \in [2.1, 2.4]$, 1000000 MC cycles.}
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\label{fig:phase_susceptibility}
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\end{figure}
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% Critical temp regression figure
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\begin{figure}[H]
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\centering
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\includegraphics[width=\linewidth]{../images/phase_transition/fox/wide/1M/linreg.pdf}
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\caption{Linear regression, where $\beta_{0}$ is the intercept $T_{c}(L = \infty)$ and $\beta_{1}$ is the slope.}
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\label{fig:linreg}
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\end{figure}
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\end{document}
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