Caught up bar V&M.

Author: Zentrik

Differential Equations/Higher-Order Linear Differential Equations - Topic 4.pdf

@@ -18,7 +18,7 @@ \section{M\"obius Groups}
 Suppose $\exists$ at least 3 values of $z$ in $\mathbb{C}$ s.t. 
 \begin{align*}
     \frac{az + b}{cz + d} &= \frac{\alpha z + \beta}{\gamma z + \delta} \\
-    ad - bc &\neq 0 \alpha \delta - \beta \gamma \neq 0.
+    ad - bc &\neq 0,\ \alpha \delta - \beta \gamma \neq 0.
 \end{align*}
 Then $\exists \; \lambda \neq 0, \lambda \in \mathbb{C}$ s.t. 
 \begin{align*}
@@ -108,4 +108,260 @@ \section{M\"obius Groups}
         \implies \ker \Phi &= Z
     \end{align*} 
     Finally apply \nameref{thm:six}.
+\end{proof} 
+
+\begin{corollary} \label{cor:7}
+    \begin{align*}
+        \frac{\operatorname{SL}_2(\mathbb{C})}{\{ \pm I \}} &= \mathcal{M}.
+    \end{align*} 
+\end{corollary} 
+
+\begin{proof}
+    Restrict $\phi$ to $\operatorname{SL}_2(\mathbb{C})$
+    \begin{align*}
+        \phi : \operatorname{SL}_2(\mathbb{C}) &\to \mathcal{M} \\
+        \begin{pmatrix}a & b \\c & d\end{pmatrix} &\mapsto \frac{az + b}{cz + d}
+    \intertext{We require $\phi$ to be surjective:}
+        f(z) &= \frac{az + b}{cz + d} \\
+        &= \frac{ \frac{a}{\sqrt{ad - bc}} z + \frac{b}{\sqrt{ad - bc}} }{ \frac{c}{\sqrt{ad - bc}} z + \frac{d}{\sqrt{ad - bc}} }.
+    \end{align*} 
+    And $\ker \phi = \{ \pm I \}$.
+\end{proof} 
+
+\begin{proposition} \label{prp:13}
+    Every M\"obius map can be written as a composition of maps of the following forms:
+    \begin{enumerate}
+        \item $z \mapsto az$, $a \neq 0$; represents a dilation or a rotation \label{itm:1-1}
+        \item $z \mapsto z + b$; a translation \label{itm:1-2}
+        \item $z \mapsto \frac{1}{z}$; inversion. \label{itm:1-3}
+    \end{enumerate} 
+\end{proposition} 
+
+\begin{proof}
+    Let $f(z) = \frac{az + b}{cz + d}$. \\
+    \begin{align*}
+        \text{If } c&= 0;\ z \mapsto \underbrace{\left(\frac{a}{d} \right) z}_{f_1,\ \Cref{itm:1-1}} \mapsto \underbrace{\left( \frac{a}{d} \right) z + \frac{b}{d}}_{f_2,\ \Cref{itm:1-2}} \\
+        f &= f_2 \circ f_1 \\
+        \text{If } c &\neq 0 \\
+        f(z) &= \frac{az + b}{cz + d} = \frac{\left( \frac{a}{c} \right) z + \left( \frac{b}{c} \right)}{z + \left( \frac{d}{c} \right)} \\
+        &= \left( \frac{a}{c} \right) + \frac{\frac{-ad + bc}{c^2}}{z + \frac{d}{c}} \\
+        &= A + \frac{B}{z + \frac{d}{c}},\ B \neq 0 \\
+        z &\underset{f_1,\ \Cref{itm:1-2}}{\mapsto} z + \frac{d}{c} \\
+        &\underset{f_2,\ \Cref{itm:1-3}}{\mapsto} \frac{1}{z + \frac{d}{c}} \\
+        &\underset{f_3,\ \Cref{itm:1-1}}{\mapsto} \frac{B}{z + \frac{d}{c}} \\
+        &\underset{f_4,\ \Cref{itm:1-2}}{\mapsto} A + \frac{B}{z + \frac{d}{c}} \\
+        f &= f_4 \circ f_3 \circ f_2 \circ f_1
+    \end{align*} 
+\end{proof} 
+
+\begin{definition}[Triply transitive action] \label{def:22}
+    A group $G$ acts \emph{triply transitively} on a set $X$ if given $x_1, x_2, x_3 \in X$ all distinct and $y_1, y_2, y_3 \in X$ all distinct there exists $g \in G$ such that $g(x_i) = y_i,\ i = 1, 2, 3$. \\
+    A group $G$ acts \emph{sharply triply transitively} if such a $g$ is unique.
+\end{definition} 
+
+\begin{theorem} \label{thm:16}
+    The action of $\mathcal{M}$ on $\mathbb{C}_\infty$ is sharply triply transitive.
+\end{theorem} 
+
+\begin{proof}
+    Label first triple $\{z_0, z_1, z_\infty\}$ and second triple $\{w_0, w_1, w_\infty\}$.
+    We construct $g \in \mathcal{M}$ s.t.
+    \begin{align*}
+        g : z_0 &\mapsto 0 \\
+        z_1 &\mapsto 1 \\
+        z_\infty &\mapsto \infty.
+    \intertext{First suppose $z_0, z_1, z_\infty \neq \infty$}
+        g(z) &= \frac{(z - z_0) (z_1 - z_\infty)}{(z - z_\infty) (z_1 - z_0)} \\
+        \text{Check: } ``ad - bc" &= (z_0 - z_\infty)(z_1 - z_\infty)(z_1 - z_0) \neq 0.
+    \intertext{If $z_\infty = \infty$}
+        g(z) &= \frac{(z - z_0)}{(z_1 - z_0)}
+    \intertext{If $z_1 = \infty$}
+        g(z) &= \frac{(z - z_0)}{(z - z_\infty)}
+    \intertext{If $z_0 = \infty$}
+        g(z) &= \frac{(z_1 - z_\infty)}{(z - z_\infty)}
+    \intertext{Similarly find $h$ s.t.}
+    g : w_0 &\mapsto 0 \\
+        w_1 &\mapsto 1 \\
+        w_\infty &\mapsto \infty. \\
+    \text{Then } f = h^{-1} g : z_i &\to w_i
+    \end{align*} 
+    Now to prove uniqueness.
+    \begin{align*}
+        \text{Suppose } f' : z_i &\mapsto w_i \\
+        \text{Then } f^{-1} \circ f' : z_i &\mapsto z_i
+        \intertext{Let $g$ be as above}
+        g f^{-1} f' g^{-1} : 0 &\mapsto 0 \implies b = 0 \\
+        : 1 &\mapsto 1 \implies a = d \\
+        : \infty &\mapsto \infty \implies c = 0 \\
+        \implies g f^{-1} f' g^{-1} &= \text{id} \\
+        \implies f^{-1} f' &= \text{id} \\
+        \implies f &= f'.
+    \end{align*} 
+    So, the image of just three points determines the map
+\end{proof} 
+
+\subsection{Conjugacy classes in $\mathcal{M}$}
+Recall $\Phi : \operatorname{GL}_2(\mathbb{C}) \twoheadrightarrow \mathcal{M}$ from proof of \Cref{thm:15} ($\twoheadrightarrow$ means its a surjective homomorphism).
+Suppose $A, B$ are conjugate in $\operatorname{GL}_2(\mathbb{C})$, i.e. $\exists \; P \in \operatorname{GL}_2(\mathbb{C})$ s.t. $PAP^{-1} = B$ then
+\begin{align*}
+    \Phi(P) \Phi(A) \Phi(P)^{-1} &= \phi(PAP^{-1}) \\
+    &= \Phi(B) \in \mathcal{M}
+\end{align*} 
+i.e. $\Phi(A)$ and $\Phi(B)$ are conjugate in $\mathcal{M}$.
+
+Use knowledge of conjugacy classes in $\operatorname{GL}_2(\mathbb{C})$.
+
+\begin{theorem} \label{thm:17}
+    Any non-identity M\"obius map is conjugate to $f(z) = \nu z$ for some $\nu \neq 0, 1$ or to $f(z) = z + 1$.
+\end{theorem} 
+
+\begin{proof}
+    \begin{enumerate}
+        \item
+        \begin{align*}
+            \begin{pmatrix}\lambda & 0 \\0 & \mu\end{pmatrix} \text{ where } \lambda &\neq \mu,\ \lambda \neq 0 \neq \mu. \\
+            \Phi \left( \begin{pmatrix}\lambda & 0 \\0 & \mu\end{pmatrix} \right) &= f,\ f(z) = \frac{\lambda}{\mu} z = \nu z,\ \nu \neq 0, 1. \\
+        \end{align*}
+        \item
+        \begin{align*}
+            \begin{pmatrix}\lambda & 0 \\0 & \mu\end{pmatrix} \text{ where } \lambda &\neq 0 \\
+            \Phi \left( \begin{pmatrix}\lambda & 0 \\0 & \lambda \end{pmatrix} \right) &= \text{id}. \\
+        \end{align*}
+        \item \begin{align*}
+            \begin{pmatrix}\lambda & 1 \\0 & \lambda\end{pmatrix} \text{ where } \lambda &\neq 0 \\
+            \Phi \left( \begin{pmatrix}\lambda & 1 \\0 & \lambda \end{pmatrix} \right) &= f,\ f(z) = \frac{\lambda z + 1}{\lambda} = z + \frac{1}{\lambda} \\
+            \text{i.e. } f &= \Phi \left( \begin{pmatrix}1 & \frac{1}{\lambda} \\0 & 1\end{pmatrix} \right). \\
+            \text{And } \begin{pmatrix}1 & \frac{1}{\lambda} \\0 & 1\end{pmatrix} &\text{ is conjugate to } \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix} \\ 
+            \text{ via } \begin{pmatrix}\lambda & 0 \\ 0 & 1 \end{pmatrix}  \begin{pmatrix}1 & \frac{1}{\lambda} \\0 & 1\end{pmatrix} \begin{pmatrix}\frac{1}{\lambda} & 0 \\ 0 & 1 \end{pmatrix} &= \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}.
+        \end{align*}
+        So $f$ is conjugate to $g$ where $g(z) = z + 1$.
+    \end{enumerate} 
+\end{proof} 
+
+\begin{corollary} \label{cor:8}
+    A non-identity M\"obius map has either 
+    \begin{enumerate}
+        \item 2 fixed points or
+        \item 1 fixed point.
+    \end{enumerate} 
+\end{corollary} 
+
+\begin{proof}
+    Suppose $g f g^{-1} = h$.
+    Then $\alpha$ is a fixed point of $f$ (i.e. $f(\alpha) = \alpha$) $\iff$ $g(\alpha)$ is a fixed point of $h$ (i.e. $h(g(\alpha)) = g(\alpha)$).\footnote{$f(\alpha) = \alpha \iff gf(\alpha) = g(\alpha) \iff h(g(\alpha)) = gfg^{-1}(g(\alpha)) = g(\alpha)$ or $g f = h g \implies h(g(\alpha)) = g(\alpha)$ if $f(\alpha) = \alpha$ and conversely $g(f(\alpha)) = h(g(\alpha)) \implies g(f(\alpha)) = g(\alpha)$.} \\
+    So number of fixed points of $f = $ number of fixed points of $h$ as $g$ is a bijection.
+    By \Cref{thm:17} either, $f$ conjugate to $z \mapsto \nu z$ which has two fixed points: $0, \infty$; or $f$ conjugate to $z \mapsto z + 1$ which has one fixed point: $\infty$.
+\end{proof} 
+
+\subsection{Circles in $\mathbb{C}_\infty$}
+
+A Euclidean circle is the set of points in $\mathbb{C}$ given by some equation $|z - z_0| = r,\ r > 0$.
+A Euclidean line is the set of points in $\mathbb{C}$ given by some equation $|z - a| = |z - b|,\ a \neq b$.
+
+\begin{definition}[Circle in $\mathbb{C}_\infty$]
+    A \emph{circle in $\mathbb{C}_\infty$} is either a Euclidean circle of a set $L \cup \{\infty\}$ where $L$ is a Euclidean line.
+    Its general equation is of the form 
+    \begin{align}
+        Az\bar z + \bar Bz + B\bar z + C = 0, \label{eq:circle}
+    \end{align} 
+    where $A, C \in \mathbb{R}$ and $|B|^2 > AC$.
+\end{definition}
+
+Where $z = \infty$ is a solution iff $A = 0$. \\
+$A = 0$: line \\
+$C = 0$: goes through the origin.
+
+There is a unique circle passing through any 3 distinct points in $\mathbb{C}_\infty$.
+
+\begin{theorem} \label{thm:18}
+    Let $f \in \mathcal{M}$ and $C$ a circle in $\mathbb{C}_\infty$, then $f(C)$ is a circle in $C_\infty$.
+\end{theorem} 
+
+\begin{proof}
+    By \Cref{prp:13}, just need to consider 
+    $f(z) = az,\ z + b \text{ or } \frac{1}{z}.$
+    Let $S_{A, B, C}$ be the circle defined by \Cref{eq:circle}.
+    \begin{align*}
+        f(z) &= az : S_{A, B, C} \mapsto S_{A / a \bar{a}, B / \bar{a}, C} \\
+        f(z) &= z + b: S_{A, B, C} \mapsto S_{A, B - Ab, C + A b \bar{b} - \bar{B}b - B \bar{b}} \\
+        f(z) &= \frac{1}{z} = w: S_{A, B, C} \mapsto A + B w + \bar{B} \bar{w} + C w \bar{w} = 0 = S_{C, \bar{B}, A}.
+    \end{align*} 
+\end{proof} 
+
+\begin{example}
+    Consider the image of $\mathbb{R} \cup \{\infty\}$ (a circle in $\mathbb{C}_\infty$) under 
+    \begin{align*}
+        f(z) &= \frac{z - i}{z + i}.
+    \end{align*} 
+    It is thus a circle in $\mathbb{C}_\infty$ containing $f(0) = -1, f(\infty) = 1, f(1) = -i$ so $f(\mathbb{R} \cup \{\infty\}) = $ unit circle. 
+    Furthermore, complimentary components are mapped to complementary components.
+    \begin{figure}[h] 
+        \centering 
+        \includegraphics[height=5cm]{figures/08-complimentary} 
+    \end{figure}
+\end{example} 
+
+\subsection{Cross-Ratios}
+
+\begin{definition} \label{def:23}
+    The cross-ratio of distinct points $z_1, z_2, z_3, z_4 \in \mathbb{C}$.
+    \begin{align*}
+        [z_1, z_2, z_3, z_4] &= \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_2)(z_3 - z_4)} \\
+        [\infty, z_2, z_3, z_4] &= \frac{(z_2 - z_4)}{(z_3 - z_4)} \\
+        [z_1, \infty, z_3, z_4] &= -\frac{(z_1 - z_3)}{(z_3 - z_4)} \\
+        [z_1, z_2, \infty, z_4] &= -\frac{(z_2 - z_4)}{(z_1 - z_2)} \\
+        [z_1, z_2, z_3, \infty] &= \frac{(z_1 - z_3)}{(z_1 - z_2)} \\
+    \end{align*} 
+\end{definition} 
+
+Note $[0, 1, w, \infty] = \frac{-w}{-1} = w$
+
+\emph{Warning}: different authors use different permutations of $1, 2, 3, 4$ in the definition.
+
+\begin{theorem} \label{thm:19}
+    Given $z_1, z_2, z_3, z_4 \in \mathbb{C}_\infty$ distinct $w_1, w_2, w_3, w_4 \in \mathbb{C}_\infty$ distinct then $\exists \; f \in \mathcal{M}$ s.t. $f(z_i) = w_i \iff [z_1, z_2, z_3, z_4] = [w_1, w_2, w_3, w_4]$.
+    In particular, M\"obius maps preserve cross-ratios $[z_1, z_2, z_3, z_4] = [f(z_1), f(z_2), f(z_3), f(z_4)]$
+\end{theorem} 
+
+\begin{proof} \mbox{}
+    ($\implies$): Suppose $f(z_j) = w_j$ and $z_i, w_i \neq \infty \; \forall \; i$ and $f(z) = \frac{az + b}{cz + d}$, then $cz + j \neq 0 \; \forall \; j$.
+    \begin{align*}
+        \text{So, } w_j - w_k &= f(z_j) - f(z_k) \\
+        &= \frac{(ad - bc) (z_j - z_k)}{(c z_j + d) (c z_k + d)} \\
+        \implies [z_1, z_2, z_3, z_4] &= [w_1, w_2, w_3, w_4] \\
+        &=  [f(z_1), f(z_2), f(z_3), f(z_4)]
+    \end{align*} 
+    Need to check other cases; $z_1 = \infty, w_1 = f(\infty) = \frac{a}{c}$ etc.
+
+    ($\longleftarrow$): Suppose $[z_1, z_2, z_3, z_4] = [w_1, w_2, w_3, w_4]$.
+    Let $g \in \mathcal{M}$ s.t. $g(z_1) = 0, g(z_2) = 1, g(z_4) = \infty$ as triple transitive.
+    Let $h \in \mathcal{M}$ s.t. $h(w_1) = 0, h(w_2) = 1, h(w_4) = \infty$.
+    \begin{align*}
+        g(z_3) &= [0, 1, g(z_3), \infty] \\
+        &= [g(z_1), g(z_2), g(z_3), g(z_4)] \\
+        &= [z_1, z_2, z_3, z_4] \text{ by above} \\
+        &= [w_1, w_2, w_3, w_4] \\
+        &= [h(w_1), h(w_2), h(w_3), h(w_4)] \\
+        &= [0, 1, h(w_3), \infty] \\
+        &= h(w_3).
+    \end{align*} 
+    So $h^{-1}g$ is required map. 
+\end{proof} 
+
+So $[z_1, z_2, z_3, z_4] = f(z_3)$ where $f$ is the unique M\"obius map that sends $z_1 \mapsto 0,\ z_2 \mapsto 1,\ z_4 \mapsto \infty$.
+
+\begin{corollary} \label{cor:9}
+    $z_1, z_2, z_3, z_4$ lie in some circle in $\mathbb{C}_\infty \iff [z_1, z_2, z_3, z_4] \in \mathbb{R}$.
+\end{corollary} 
+
+\begin{proof}
+    Let $C$ be a circle through $z_1, z_2, z_4$. \\
+    Let $g : C \to \mathbb{R} \cup \{\infty\}$ s.t. $g(z_1) = 0,\ g(z_2) = 1,\ g(z_4) = \infty$.
+    \begin{align*}
+        g(z_3) &= [0, 1, g(z_3), \infty] \\
+        &= [g(z_1), g(z_2), g(z_3), g(z_4)] \\
+        &= [z_1, z_2, z_3, z_4] \text{ by \Cref{thm:19}} \\
+        \text{So, } [z_1, z_2, z_3, z_4] \in \mathbb{R} &\iff g(z_3) \in \mathbb{R} \\
+        &\iff z_3 \in C.
+    \end{align*} 
 \end{proof} 

Differential Equations/Mutivariate functions, applications - Topic 5.pdf

@@ -344,7 +344,7 @@ No, e.g. $X'' = X' \cup \mathcal{P}(X') \cup \mathcal{P}(\mathcal{P}(X')) \cup \
 Does every set $Y$ inject into $X, X', X'', \dots$? \
 No, e.g. $Y = X \cup X' \cup X'' \cup \dots$
 
-# Panorama
+## Panorama
 
 - II Logic and Set Theory - We will explore the cardinalities of infinite sets in more details and will learn about the axiomatic system upon which modern mathematics is built.
 

Groups/08-mobius.tex

@@ -0,0 +1,96 @@
+# More about primes (non-examinable)
+
+Bertrand postulated in 1845 that for every $n \in \mathbb{N}$ there is always a prime between $n$ and $2n$ ($n \leq p < 2n$). \
+The primes $2, 5, 11, 23, 47, 89, 179, 359, 719, 1439, 2879$ show it to be true for $n \leq 2^{11}$.
+Bertrand checked it for $n < 3\,000\,000$.
+Chebychev (1850) gave a proof.
+Erd\H{o}s (1932) gave an elementary proof based on the properties of $\binom{2n}{n}$.
+
+::: {.lemma #sone}
+\begin{align*}
+    \binom{2n}{n} \geq \frac{2^{2n}}{2n + 1}.
+\end{align*} 
+:::
+::: {.proof}
+Since $\binom{n}{k + 1} / \binom{n}{k} = \frac{n - k}{k + 1}$, it is evident that $\binom{n}{k}$ increases for $k < \frac{n}{2}$, and decreases for $k > \frac{n}{2}$.
+In particular, $\binom{2n}{n} \geq \frac{2^{2n}}{2n + 1}$, the maximum element is at least as big as the average ($2^{2n}$ is the sum and we have $2n + 1$ elements). 
+:::
+
+::: {.lemma #stwo}
+If $p \leq n$ is a prime dividing $\binom{2n}{n}$, then $p \leq \frac{2n}{3}$.
+:::
+::: {.proof}
+Suppose $\frac{2n}{3} < p \leq n$, then $p \leq n < \frac{3}{2}p < 2p \leq 2n < 3p$, so the numerator and denominator of 
+\begin{align*}
+    \frac{2n (2n - 1) \dots (n + 1)}{n (n - 1) \dots 3 \cdot 2 \cdot 1}
+\end{align*} 
+are divisible by exactly one copy of $p$. ⨳ as it can then not divide it.
+:::
+
+::: {.lemma #sthree}
+If $p$ is a prime and $p^k \mid \binom{2n}{n}$, then $p^k \leq 2n$.
+:::
+::: {.proof}
+The greatest power of $p$ dividing $n! = n (n-1) \dots 3 \cdot 2 \cdot 1$.
+$\left \lfloor \frac{n}{p} \right \rfloor$ is the no. of multiples of $p$ upto $n$, $\left \lfloor \frac{n}{p^2} \right \rfloor$ is the no. of multiples of $p^2$ upto $n$.
+So the greatest power is $\left \lfloor \frac{n}{p} \right \rfloor +\left \lfloor \frac{n}{p^2} \right \rfloor +\left \lfloor \frac{n}{p^3} \right \rfloor \dots = \sum_{i \geq 1}\left \lfloor \frac{n}{p^i} \right \rfloor$.
+:::
+Hence, if $k$ is a power of $p$ dividing $\binom{2n}{n} = \frac{2n !}{(n!)^2}$, then
+\begin{align*}
+    k &\leq \sum_{i \geq 1}\left \lfloor \frac{2n}{p^i} \right \rfloor - 2 \sum_{i \geq 1}\left \lfloor \frac{n}{p^i} \right \rfloor\\
+    &= \sum^l_{i = 1} \left(\left \lfloor \frac{2n}{p^i} \right \rfloor - 2\left \lfloor \frac{n}{p^i} \right \rfloor \right) \text{ where $l$ is the greatest power of $p$ s.t. $p^l \leq 2n$} \\
+    &\leq \sum_{i = 1}^k 1 \text{ since } \left \lfloor 2x \right \rfloor - 2 \left \lfloor x \right \rfloor \leq 1 \\
+    &= l \\
+    \text{so $k \leq l$ and thus $p^k \leq p^l \leq 2n$}.
+\end{align*} 
+
+::: {.lemma #sfour}
+For all $m \in \mathbb{N}, \underset{p \leq m \\ \text{$p$ prime}}{\prod} p \leq 4^m$.
+:::
+::: {.proof}
+By induction on $m$.
+True for $m = 2$.
+If $m > 2$ is even, then
+\begin{align*}
+    \prod_{p \leq m} p = \prod_{p \leq m - 1} p \leq 4^{m - 1} < 4^m.
+\end{align*} 
+If $m = 2k + 1$ is odd, then all primes $k + 2 \leq p \leq 2k + 1$ divide $\binom{2k + 1}{k} = \frac{(2k + 1)!}{k! (k + 1)!} = \frac{(2k + 1) 2k \dots (k + 2)}{k (k - 1) \dots 3 \cdot 2 \cdot 1}$ (as they divide the numerator but not denominator).
+Thus $\prod_{k + 2 \leq p \leq 2k + 1} p \leq \binom{2k + 1}{k} = \binom{2k + 1}{k + 1} \leq \frac{2^{2k + 1}}{2} = 4^k$.
+By the inductive hypothesis,
+\begin{align*}
+    \prod_{p \leq m} p = \prod_{p < k + 1} p \cdot \prod_{k + 2 \leq p \leq 2k + 1} p \leq 4^{k + 1} \cdot 4^k = 4^{2k + 1}
+\end{align*} 
+:::
+
+::: {.theorem #sfive name="Bertrand's Postulate"}
+For all $n \in \mathbb{N}$ s.t. $n \geq 2$, $\exists$ prime $p$ with $n \leq p < 2n$.
+:::
+::: {.proof}
+Clearly the prime factors of $\binom{2n}{n}$ are all $< 2n$.
+Suppose the theorem fails.
+Then all prime factors of $\binom{2n}{n}$ are in fact $< n$. \
+By Lemma \@ref(lem:stwo), they are all $< \frac{2}{3} n$. \
+Consider the prime factorisation of $\binom{2n}{n}$.
+By Lemma \@ref(lem:sthree), each prime contributes $\leq 2n$ to the factorisation.
+Moreover, if $p > \sqrt{2n}$, then $p$ contributes at most $p$ to the factorisation (since $p^2 > 2$).
+Now by Lemma \@ref(lem:sone) and \@ref(lem:sfour)
+\begin{align*}
+    \frac{2^{2n}}{2n + 1} \leq \binom{2n}{n} &\leq \Pi_{p \leq \sqrt{2n}} p \Pi_{\sqrt{2n} < p < \frac{2}{3} n} p \\
+    &\leq (2n)^{\sqrt{2n}} \cdot \Pi_{p < \frac{2}{3} n} p \\
+    \text{by Lemma \@ref(lem:sfour), } \Pi_{p < \frac{2}{3} n} p &\leq 4^{\frac{2}{3} n} \\
+    &\leq (2n)^{\sqrt{2n}} \cdot  4^{\frac{2}{3}n} \\
+    \frac{4^n}{2n + 1} &\leq (2n)^{\sqrt{2n}} \cdot 4^{\frac{2}{3} n},
+\end{align*} which fails when $n$ is large.
+
+How large?
+\begin{align*}
+    4^{\frac{n}{3}} &\leq (2n + 1)(2n)^{\sqrt{2n}} \\
+    \text{and } (2n + 1) &\leq (2n)^2 \leq (2n)^{\sqrt{2n}/3} \text{ for } n \geq 18 \\
+    \text{so } 4^\frac{n}{3} &\leq (2n)^{\frac{4}{3} \sqrt{2n}} \\
+    \text{or } 4^n &\leq (2n)^{4 \sqrt{2n}} \\
+    \text{with } r &= \sqrt{2n}, \text{ this is} \\
+    4^{\frac{r^2}{2}} &\leq r^{8r} \\
+    \text{or } 4^r &\leq r^{16}
+\end{align*} which fails when $r \geq 2^6 = 64$.
+So proof holds when $n \geq 2^{11}$, and also true for smaller values of $n$ as discussed at the beginning of the lecture.
+:::