Final commit?

Author: Zentrik

Methods/04-diffusion.tex

@@ -115,7 +115,7 @@ \subsection{Heat conduction in a finite bar}
 		0 & \text{if } -L \leq x < 0
 	\end{cases}
 \end{align}
-with boundary conditions 
+with boundary conditions
 \begin{align} \label{eq:4.9}
     \theta(L, t) = 1, \quad \theta(-L, t) = 0.
 \end{align}
@@ -149,9 +149,9 @@ \subsubsection{Transforming boundary conditions}
 	\hat \theta(x,0) = H(x) - \frac{x+L}{2L}
 \end{align}
 
-\begin{figure}[h] 
-    \centering 
-    \includegraphics[height=5cm]{04-transformbcs} 
+\begin{figure}[h]
+    \centering
+    \includegraphics[height=5cm]{04-transformbcs}
 \end{figure}
 
 \subsubsection{Seperation of variables}

Methods/Methods.pdf

@@ -320,7 +320,7 @@ \subsection{Infinite product spaces}
 
 Let $(X_n)_{n \in \mathbb N}$ be a sequence of iid r.v.s defined on $(\Omega, \mathcal F, \mathbb P)$ with common law $\mu_{X_n} = m$ for all $n$; this exists by an earlier theorem.
 We define a map $X \colon \Omega \to E$ by $X(\omega) = (X_1(\omega), X_2(\omega), \dots)$.
-This is $\mathcal F$--$\sigma(\mathcal C)$ measurable, since for all $A \in \mathcal C$, we have
+This is $\sigma(\mathcal C)$ measurable, since for all $A \in \mathcal C$, we have
 \begin{align*}
         X^{-1}(A) = \qty{\omega : X_1(\omega) \in A_1, \dots, X_N(\omega) \in A_N} = \bigcap_{n=1}^N X_n^{-1}(A_n) \in \mathcal F
     \end{align*}

ProbAndMeasure/07_ergodic_theory.tex

@@ -104,7 +104,7 @@ \subsection{Distinguishing non-orthogonal states}
 For a state $\ket{\beta}$ orthogonal to both $\ket{\alpha_0}$ and $\ket{\alpha_1}$, we have $\Delta \ket{\beta} = 0$.
 Therefore, $\Delta$ acts nontrivially only in the vector space spanned by $\ket{\alpha_0}$ and $\ket{\alpha_1}$, and hence has at most two nonzero eigenvalues, and its eigenvectors lie in $\vecspan\qty{\ket{\alpha_0}, \ket{\alpha_1}}$.
 
-Now, $\Tr\Delta = 0$ so the eigenvalues are $\delta$ and $-\delta$ for some $\delta \in \mathbb R$.
+Now, $\Tr\Delta = 0$\footnote{$\Tr \Delta = \bra{\alpha_0}\ket{\alpha_0} - \bra{\alpha_1}\ket{\alpha_1}$} so the eigenvalues are $\delta$ and $-\delta$ for some $\delta \in \mathbb R$.
 Let $\ket{p}$ be the eigenvector for $\delta$, and $\ket{m}$ be the eigenvector for $-\delta$, so $\ip{p}{m} = 0$.
 We can write $\Delta$ in its spectral decomposition, giving $\Delta = \delta \op{p}{p} - \delta \op{m}{m}$.
 

ProbAndMeasure/probmeasure.pdf

@@ -589,11 +589,12 @@ \subsection{Shor's algorithm}
 \end{align*}
 where $\omega = 2^{\frac{2\pi i}{M}}$ where $M = 2^m$.
 $S$ is a geometric series with $\alpha = \omega^{cr}$.
-If $\frac{M}{r} \not\in \mathbb Z$, $\alpha^A \neq 1$.
+If $\alpha = 1$ then $c \mid \frac{M}{r}$.
+But if $\frac{M}{r} \not\in \mathbb Z$, this cannot be true.
 We claim that a measurement on $QFT_{2^m} \ket{\mathrm{per}}$ yields an integer $c$ which is close to a multiple of $\frac{M}{r}$ with high probability.
 
 Consider $k\frac{2^m}{r}$ for $k = 0, \dots, r-1$.
-Each of these multiples is within $\frac{1}{2}$ of a unique integer; indeed, $2^m = Br + b$ so $r < 2^m$, giving that $k\frac{2^m}{r}$ cannot be a half integer.
+Each of these multiples is within $\frac{1}{2}$ of a unique integer; indeed, $r < N$ and $2^m > N^2$, giving that $k\frac{2^m}{r}$ cannot be a half integer.
 Consider the values of $c$ s.t. $\abs{c - k \frac{2^m}{r}} < \frac{1}{2}$ for $k = 0, \dots, r-1$.
 % Note that $\omega^{cr} = 1$ if $e^{\frac{2\pi i cr}{M}} = 1$.
 \begin{theorem}