Michaelmas notes

Author: Zentrik

.gitignore

@@ -14,4 +14,7 @@ _bookdown_files
 
 */_main.tex
 
-tikz*.pdf
+tikz*.pdf
+
+.Renviron
+.Rprofile

Differential Equations/Introduction.tex

@@ -45,7 +45,7 @@ \subsection{Limits}\label{limits}}
 
 \begin{itemize}
 \item
-  Informally, if \(\lim_{x \to x_0} f(x) = A\), then \(f(Px)\) can be made arbitrarily close to \(A\) by making \(x\) sufficiently close to \(x_0\)
+  Informally, if \(\lim_{x \to x_0} f(x) = A\), then \(f(x)\) can be made arbitrarily close to \(A\) by making \(x\) sufficiently close to \(x_0\)
 
   \begin{itemize}
   \tightlist
@@ -60,7 +60,7 @@ \subsection{Limits}\label{limits}}
   \item
     for any \(\epsilon > 0\), there exists \(\delta >0\) such that \(|f(x) - A| < \epsilon\) for all \(0 < |x - x_0| < \delta\).
   \item
-    Right hand limit, for example, defined similarly but with \(0 < |x - x_0| < \delta\) replaced with \(0 < x - x_0 < \delta\). A similar procedure can be done for left hand Limits
+    Right hand limit, for example, defined similarly but with \(0 < |x - x_0| < \delta\) replaced with \(0 < x - x_0 < \delta\). A similar procedure can be done for left hand limits
   \end{itemize}
 
   \begin{figure}[h!]
@@ -106,19 +106,15 @@ \subsubsection{Properties}\label{properties}}
 \subsubsection{Proof of uniqueness of limits}\label{proof-of-uniqueness-of-limits}}
 
 Suppose that \(\lim_{x \to x_0} f(x) = A\) \textcolor{red}{and} \(\lim_{x \to x_0} f(x) = B\).
-
-In terms of our epsilon-delta definition, this means that for any \(\epsilon > 0\)h there exists \(\delta_A > 0\) and \(\delta_B > 0\) such that
-
+In terms of our epsilon-delta definition, this means that for any \(\epsilon > 0\) there exists \(\delta_A > 0\) and \(\delta_B > 0\) such that
 \begin{align*}
   &\text{for }0 < |x -x_0| < \delta_A,\ |f(x) - A| < \epsilon / 2 \text{, where } \epsilon / 2 \text{ is an arbitrary positive quantity.} \\
   \color{red}{and} \ &\text{for } 0 < |x -x_0| < \delta_A,\ |f(x) - B| < \epsilon / 2
 \end{align*}
-
 Now let \(\delta = min(\delta_A, \delta_B)\) and consider \(0 < |x -x_0| < \delta\) - follows that
 \begin{align*}
   |A - B| &= |[A - f(x)] - [B - f(x)]| \\
   &\leq |A - f(x)| + |B - f(x)| \\
   &\leq \epsilon
 \end{align*}
-
 Since this holds \textcolor{red}{for all} \(\epsilon > 0\), we must have \(A = B\).

Differential Equations/de.pdf

@@ -3,9 +3,9 @@
 \def\npart {IA}
 \def\nterm {Michaelmas}
 \def\nyear {2021}
-\def\nlecturer {A.\ Chailinor}
+\def\nlecturer {Prof. A. Chailinor}
 \def\ncourse {Differential Equations}
-\def\nauthor{Author}
+
 
 \input{../preamble-dynamic.tex}
 

Differential Equations/de.tex

@@ -312,7 +312,7 @@ \subsubsection{Small n}\label{small-n}}
 \end{lemma}
 
 \begin{proof}
-If $x \in {1, \dots, n} \setminus \{ a_1, \dots, a_k \} \cup \{ b_1, \dots, b_k \}$, $(\sigma \circ \tau) (x) = \sigma \left( \tau(x) \right) = (\tau \circ \sigma)(x)$.
+If $x \in \{1, \dots, n\} \setminus \{ a_1, \dots, a_k \} \cup \{ b_1, \dots, b_k \}$, $(\sigma \circ \tau) (x) = \sigma \left( \tau(x) \right) = (\tau \circ \sigma)(x)$.
 
 Suppose $1 \leq i \leq k - 1$
 \begin{align*}
@@ -378,7 +378,7 @@ \subsubsection{Small n}\label{small-n}}
     \sigma^j(a_1) &= \sigma^i(a_1),\ j > i \geq 1 \\
     \implies \sigma^{j - i}(a_1) &= a_1 \quad \text{\Lightning \ of minimality of j.}
 \end{align*}
-So $\{ a_1, \sigma(a_1), \sigma^2(a_1), \dots, \sigma^{j-1}(a_1) \}$ is a cycle in $\sigma$.\\
+So $\left( a_1, \sigma(a_1), \sigma^2(a_1), \dots, \sigma^{j-1}(a_1) \right)$ is a cycle in $\sigma$.\\
 If $\exists \; b \in X \setminus \{ a_1, \sigma(a_1), \sigma^2(a_1), \dots, \sigma^{j-1}(a_1) \}$.
 Consider $b, \sigma(b), \dots$\\
 Note $\left(b, \sigma(b), \sigma^2(b), \dots, \sigma^{j-1}(b) \right)$ is disjoint from $\left( a_1, \sigma(a_1), \sigma^2(a_1), \dots, \sigma^{j-1}(a_1) \right)$ because $\sigma$ is a bijection.\footnote{If $\sigma^i(b) = \sigma^j(a_1)$ then $b = \sigma^{j - i}(a_1)$ which contradicts $b \in X \setminus \{ a_1, \sigma(a_1), \sigma^2(a_1), \dots, \sigma^{j-1}(a_1) \}$.}\\
@@ -521,7 +521,7 @@ \subsubsection{Small n}\label{small-n}}
 The map
 \begin{align*}
     \operatorname{sgn} : (S_n, \circ) &\to \left( \{ \pm 1 \}, \times \right) \\
-    \sigma *\mapsto \operatorname{sgn}(\sigma)
+    \sigma &\mapsto \operatorname{sgn}(\sigma)
 \end{align*}
 is a well-defined non-trivial (doesn't just map to identity) homomorphism.
 \end{theorem}

Groups/02-dihedral.tex

@@ -9,8 +9,8 @@ \section{Cosets and Lagrange}\label{cosets-and-lagrange}}
 
 \begin{example} ~\vspace*{-1.5\baselineskip}
 \begin{align*}
-    S_3 &= \{ e, \begin{pmatrix}1 & 2 & 3\end{pmatrix}, \begin{pmatrix}1 & 3 & 2\end{pmatrix}, \begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}1 & 3\end{pmatrix}, \begin{pmatrix}2 & 3\end{pmatrix} \} \\
-    H &= \{ \text{id}, \begin{pmatrix}1 & 2 & 3\end{pmatrix}, \begin{pmatrix}1 & 3 & 2\end{pmatrix} \} = A_3 \\
+    S_3 &= \left\{ e, \begin{pmatrix}1 & 2 & 3\end{pmatrix}, \begin{pmatrix}1 & 3 & 2\end{pmatrix}, \begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}1 & 3\end{pmatrix}, \begin{pmatrix}2 & 3\end{pmatrix} \right\} \\
+    H &= \left\{ \text{id}, \begin{pmatrix}1 & 2 & 3\end{pmatrix}, \begin{pmatrix}1 & 3 & 2\end{pmatrix} \right\} = A_3 \\
     \begin{pmatrix}1 & 2\end{pmatrix}H &= \left\{ \begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}1 & 2\end{pmatrix}\begin{pmatrix}1 & 2 & 3\end{pmatrix}, \begin{pmatrix}1 & 2\end{pmatrix}\begin{pmatrix}1 & 3 & 2\end{pmatrix} \right\} \\
     &= \left\{ \begin{pmatrix}1 & 2\end{pmatrix}, \begin{pmatrix}2 & 3\end{pmatrix}, \begin{pmatrix}1 & 3\end{pmatrix} \right\} \\
     \begin{pmatrix}1 & 2 & 3\end{pmatrix}H &= H \ (\text{since H is a subgroup})
@@ -224,7 +224,7 @@ \section{Cosets and Lagrange}\label{cosets-and-lagrange}}
     \implies \exists \; u, v \in \mathbb{Z} \text{ s.t. } a u + v n &= 1 \text{ (Bezout's Theorem)} \\
     \implies au &\equiv 1 \pmod n
 \end{align*}
-Then $\overline{u} \in R_n^*$ is $a^{-1}$
+Then $\overline{u} \in R_n^*$ is $a^{-1}$ as $a^{-1}$ has an inverse $a$ which implies $(a, n) = 1$.
 
 \begin{proof}
 Note $|R_n^*| = \phi(n)$.

Groups/03-cosets.tex

@@ -3,7 +3,7 @@ \section{Direct Products and Small Groups}
 \subsection{Direct Products}
 
 \begin{definition}[(External) direct product of groups]
-  Let $H$ and $K$ be groups. We can construct the (external) \emph{direct product}, $H \times K$, with a set $\{ (h, k),\ h \in H, k \in K \}$ and an operation:
+  Let $H$ and $K$ be groups. We can construct the (external) \emph{direct product}, $H \times K$, with a set $\{ (h, k): h \in H, k \in K \}$ and an operation:
   \begin{align*}
     (h_1, k_1) * (h_2, k_2) &= (h_1 *_H h_2, k_1 *_K k_2) \\
     &= (h_1 h_2, k_1 k_2)
@@ -75,19 +75,19 @@ \subsection{Direct Products}
     \text{Also, } (h, k)^n &= (h^n, k^n) \\
     &= (e_H, e_K) \\
     \implies o(h) \mkern-5mu &\;\mid n \text{ and } o(k) \mid n \text{ by \Cref{lem:five}}. \\
-    \implies \mkern-5mu &\;\mid n.
+    \implies m \mkern-5mu &\;\mid n.
   \end{align*} 
-  Thus we know when $C_m \times C_n \cong C_{mn}$ (Sheet 2, qn 10).
 \end{proof} 
 
+Thus we know when $C_m \times C_n \cong C_{mn}$ (Sheet 2, qn 10).
 Recognising when a group can be written as a direct product of subgroups is trickier.
 
 \begin{proposition}[Direct Product Theorem]
   Let $G$ be a group with subgroups $H$ and $K$, if:
   \begin{enumerate}
     \item each element of $G$ can be written as $hk$, for some $h \in H$, $k \in K$, \label{5.1-1}
     \item $H \cap K = \{ e \}$ \label{5.1-2}
-    \item $hk = kh \; \forall \; h \in H,\ k \in K$ \label{5.1-3}
+    \item $hk = kh \quad \forall \; h \in H,\ k \in K$ \label{5.1-3}
   \end{enumerate} 
   Then $G \cong H \times K$ and we call $G$ the (internal) direct product of $H$ and $K$.
 \end{proposition} 
@@ -190,9 +190,7 @@ \subsection{Small Groups}
     &= b^{-k} a \\
     \text{So } (b^k a) (b^k a) &= b^k b^{-k} a a \\
     &= e.
-  \end{align*} 
-  We can check 
-  \begin{align*}
+    \intertext{We can check}
     \theta : D_{2n} &\to G \\
     r &\mapsto b \\
     t &\mapsto a 
@@ -222,7 +220,7 @@ \subsubsection{Classifying groups of small order}
     \implies G &\cong C_4
     \intertext{Suppose not, then let}
     1 \neq a \in G &\implies o(a) = 2 \\
-    &\implies G \text{ is abelian (qn 7, sheet 1)} \\
+    &\implies G \text{ is abelian (Sheet 1, Q7)} \\
     &\implies C_2 \cong \langle a \rangle \trianglelefteq G.
   \end{align*}
   Let $b \in G \setminus \langle a \rangle$, then $1 \neq b \in G$ so $C_2 \cong \langle b \rangle \trianglelefteq G$. \\
@@ -245,8 +243,8 @@ \subsubsection{Classifying groups of small order}
 \end{lemma} 
 
 \begin{proof}
-  Let $1 \neq g \in G \implies o(g) = 2, 3 \text{ or } 6$ by \Cref{cor:two}.
-  If all non-identity elements have order $2 \implies |G|$ is a $2$-power \Lightning (qn 7, Sheet 1). \\
+  Let $1 \neq g \in G \implies o(g) = 2, 3 \text{ or } 6$ by \nameref{cor:two}.
+  If all non-identity elements have order $2 \implies |G|$ is a $2$-power \Lightning \ (Sheet 1, Q7). \\
   So $\exists \; b \in G, o(b) = 3$ (Note if $o(g) = 6$ then $o(g^2) = 3$). \\
   $C_3 \cong \langle b \rangle \trianglelefteq G$, RHS by \Cref{lem:twelve} (since of index 2).
 

Groups/05-direct.tex

@@ -735,7 +735,7 @@ \subsection{Conjugacy Action}
             If $z \in Z(G)$
             \begin{align*}
                 \implies \operatorname{ccl}_G(z) &= \{ \underbrace{gzg^{-1}}_{gg^{-1} z = z} : g \in G\} \\
-                &= \{z\} \\
+                &= \{z\}
             \end{align*} 
             If $|\operatorname{ccl}_G(z)| = 1$, note $z = e z e^{-1} \in \operatorname{ccl}_G(z)$.
             So $gzg^{-1} = z \ \forall \; g \in G$.

Groups/06-actions.tex

@@ -129,7 +129,7 @@ \subsection{Actions of \texorpdfstring{$\operatorname{GL}_n(\mathbb{C})$}{GLₙ(
         (A, v) &\mapsto Av.
     \end{align*} 
     Note $Iv = v$, $(AB)v = A(B(v))$.
-    This action is faithful: $Av = v \forall \; v \in \mathbb{C}^n \implies A = I_n$ (consider $v = e_i$).
+    This action is faithful: $Av = v \quad \forall \; v \in \mathbb{C}^n \implies A = I_n$ (consider $v = e_i$).
     The action has two orbits
     \begin{align*}
         \operatorname{Orb}_{\operatorname{GL}_n(\mathbb{C})}(\underline{0}) &= \{\underline{0}\} \\
@@ -372,7 +372,7 @@ \subsubsection{In 3 dimensions}
     \begin{align*}
         \det (A - I) &= \det  (A - A A^T) \\
         &= \det A \det (I - A^T) \\
-        &= 1 \cdot \det (I - A)^t \\
+        &= 1 \cdot \det (I - A)^T \\
         &= \det (I - A) \\
         &= (-1)^3 \det (A - I) \\
         &= - \det (A - I) \\