Edits

Advanced/Lattices/main.tex

@@ -1,7 +1,7 @@
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[
-	nosolutions,
+	solutions,
 	singlenumbering
 ]{../../resources/ormc_handout}
 

Advanced/Lattices/parts/0 intro.tex

@@ -15,17 +15,22 @@ $$
 for integer coefficients $a_i$.
 
 \problem{}
-Which of the following generate $\mathbb{Z}^3$?
+Which of the following generate $\mathbb{Z}^2$?
 \begin{itemize}
 	\item $\{ (1,2), (2,1) \}$
 	\item $\{ (1,0), (0,2) \}$
 	\item $\{ (1,1), (1,0), (0,1) \}$
 \end{itemize}
 
+\begin{solution}
+	Only the last.
+\end{solution}
+
 \vfill
 
 \problem{}
-Find a set of vectors that generates $\mathbb{Z}^2$.
+Find a set of vectors that generates $\mathbb{Z}^2$. \\
+$\{ (0, 1), (1, 0) \} doesn't count.$
 
 \vfill
 
@@ -40,8 +45,6 @@ Find a set of vectors that generates $\mathbb{Z}^n$.
 \definition{}
 A \textit{fundamental region} of a lattice is the parallelepiped spanned by a generating set. The exact shape of this region depends on the generating set we use.
 
-\vfill
-
 \problem{}
 Draw two fundamental regions of $\mathbb{Z}^2$ using two different generating sets. Verify that their volumes are the same.
 

Advanced/Lattices/parts/1 minkowski.tex

@@ -3,6 +3,11 @@
 \theorem{Blichfeldt's theorem}
 Let $X$ be a finite connected region. If the volume of $X$ is greater than $1$, $X$ must contain two distinct points that differ by an element of $\mathbb{Z}^n$. In other words, there exist distinct $x, y \in X$ so that $x - y \in \mathbb{Z}^n$.
 
+\vfill{4mm}
+
+Intuitively, this means that you can translate $X$ to cover two lattice points at the same time.
+
+
 \problem{}
 Draw a region in $\mathbb{R}^2$ with volume greater than 1 that contains no lattice points. Find two points in that region which differ by an integer vector.
 \hint{Area is two-dimensional volume.}
@@ -33,6 +38,10 @@ The following picture gives the idea for the proof of Blichfeldt's theorem. Expl
 \problem{}
 Let $X$ be a region $\in \mathbb{R}^2$ of volume $k$. How many integral points must $X$ contain after a translation?
 
+\begin{solution}
+	$\lceil k \rceil$
+\end{solution}
+
 \vfill
 
 \definition{}
@@ -75,7 +84,7 @@ Let $K$ be a region in $\mathbb{R}^2$ satisfying \ref{mink}. Scale this region b
 
 \begin{itemize}
 	\item How does the volume of $K'$ compare to $K$?
-	\item Show that the sum of any two points in $K'$ lies in $K$
+	\item Show that the sum of any two points in $K'$ lies in $K$ \hint{Use convexity.}
 	\item Apply Blichfeldt's theorem to $K'$ to prove Minkowski's theorem in $\mathbb{R}^2$.
 \end{itemize}
 

Advanced/Lattices/parts/2 orchard.tex

@@ -12,7 +12,7 @@ Show that if $r < \frac{1}{\sqrt{R^2 + 1}}$, you have at least one directon with
 \hint{Take a look at the ray through the point $(R, 1)$ and calculate the distance from the closest integer points to the ray.}
 
 \begin{solution}
-	Consider the ray from the origin through the point $(R, 1)$. Clearly, the two closest lattice points are $(1, 0)$ and $(R − 1, 1)$. They are equally far from the ray so let's calculate the distance from $(1, 0)$ to our ray. Call this distance $\delta$. Consider the triangle with vertices $(0, 0)$, $(1, 0)$, and $(R, 1)$. Then the area of this triangle is $12$. On the other hand, the area is also given by $\frac{1}{2} \delta \sqrt{R^2 + 1}$. So, $\delta = \frac{1}{\sqrt{R^2+1}}$. Therefore, if $r < \frac{1}{\sqrt{R^2+1}}$, we will have a clear line of sight given by this ray.
+	Consider the ray from the origin through the point $(R, 1)$. Clearly, the two closest lattice points are $(1, 0)$ and $(R - 1, 1)$. They are equally far from the ray so let's calculate the distance from $(1, 0)$ to our ray. Call this distance $\delta$. Consider the triangle with vertices $(0, 0)$, $(1, 0)$, and $(R, 1)$. Then the area of this triangle is $\frac{1}{2}$. On the other hand, the area is also given by $\frac{1}{2} \delta \sqrt{R^2 + 1}$. So, $\delta = \frac{1}{\sqrt{R^2+1}}$. Therefore, if $r < \frac{1}{\sqrt{R^2+1}}$, we will have a clear line of sight given by this ray.
 \end{solution}
 
 
@@ -39,6 +39,12 @@ Show that there is no line of sight through the orchard if $r > \frac{1}{R}$. Yo
 Prove that there exists a rational approximation of $\sqrt{3}$ within $10^{-3}$ with denominator at most $501$. Come up with an upper bound for the smallest denominator of a $\epsilon$-close rational approximation of any irrational number $\alpha > 0$. Your bound can have some dependence on $\alpha$ and should get smaller as $\alpha$ gets larger. \\
 \hint{Use the orchard.}
 
+\begin{solution}
+	Take the line through the origin of slope $\sqrt{3}$. We would like an orchard for which $r = 10^{-3}$ gives no line of sight, since this will guarantee an integer point within a distance of $10^{-3}$. Then by our previous problem, we can take $10^{-3} > \frac{1}{R}$, so take $R > 1000$. Now since this line intersects the boundary of the orchard at $( \frac{R}{2}, \frac{\sqrt{3R}}{2} )$, we have that the $x$-coordinate is at most $\frac{R}{2} = 501$. Then we have that our lattice point $(x, y)$ satisfies $\sqrt{3x} - 10^{-3} < y < \sqrt{3x} + 10^{-3}$, so $\sqrt{3} - 10^{-3} < \frac{x}{y} < \sqrt{3} + \frac{10^{-3}}{x}$. Therefore, $\frac{y}{x}$ is a rational approximation that is $10^{-3}$-close to $\sqrt{3}$ and has denominator at most $501$. Notice that we got closer than we need to.
+
+	Repeating this same process, our upper bound for the denominator of an $\epsilon$-close approximation of $\alpha$ is $\frac{\text{cos}({\text{atan}({\alpha})})}{\epsilon} = \frac{1}{\sqrt{\alpha^2 + 1 \epsilon}}$.
+\end{solution}
+
 \vfill
 \pagebreak