commit

PS3/doc.tex

@@ -360,65 +360,65 @@ S_u = [1, s; s, 1];
 
             \item %3b
 
-            To find the E-\textbf{k} relationship in terms of cosine functions, we must first
+            To find the \(E\)-\(\vec{k}\) relationship in terms of cosine functions, we must first
             find \( h_0 \) in terms of trigonometric functions.
             We will require the following trigonometric identities to do so:
 
             \begin{align}
-                sin(\alpha \pm \beta) = sin(\alpha)cos(\beta) \pm sin(\beta)cos(\alpha)
+                \sin(\alpha \pm \beta) = \sin(\alpha)\cos(\beta) \pm \sin(\beta)\cos(\alpha)
                 \label{eq:sin_sum}
             \end{align}
 
             \begin{align}
-                cos(\alpha \pm \beta) = cos(\alpha)cos(\beta) \mp sin(\alpha)sin(\alpha)
+                \cos(\alpha \pm \beta) = \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\alpha)
                 \label{eq:cos_sum}
             \end{align}
 
             \begin{align}
-                1 = sin^2(\theta) cos^2(\theta)
+                1 = \sin^2(\theta) \cos^2(\theta),
                 \label{eq:sum_squares}
             \end{align}
 
-            As well as the following general sine/cosine properties:
+            as well as the following general sine/cosine properties:
             \begin{align}
-                cos(\theta) &= cos(-\theta)
+                \begin{split}
-                sin(\theta) &= -sin(-\theta)
+                \cos(\theta) &= \cos(-\theta) \\
+                \sin(\theta) &= -\sin(-\theta).
                 \label{eq:neg_sin_cos}
+                \end{split}
             \end{align}
 
             First we can rewrite the exponentials in \( h_0 \) using Euler's identity:
 
             \begin{align*}
-                h_0 = -t_c \left( 1 + cos(-k_x a - k_y b) + i sin(-k_x a - k_y b) + cos(-k_x a + k_y b) + i sin(-k_x a + k_y b) \right)
+                h_0 = -t_c \left( 1 + \cos(-k_x a - k_y b) + i \sin(-k_x a - k_y b) + \cos(-k_x a + k_y b) + i \sin(-k_x a + k_y b) \right)
             \end{align*}
 
             Using identities (\ref{eq:sin_sum}) and (\ref{eq:cos_sum}), we can expand this relationship further.
 
-            
-            % TODO no idea how to format this...
             \begin{align*}
-                h_0 = -t+c ( 1 &+ cos(-k_x a)cos(-k_y b) - sin(-k_x a)sin(-k_y b) + i sin(-k_x a)cos(-k_y b) + i sin(-k_y b)cos(-k_x a) \\
+                h_0 = -t+c \big( 1 &+ \cos(-k_x a)\cos(-k_y b) - \sin(-k_x a)\sin(-k_y b) + i \sin(-k_x a)\cos(-k_y b) + i \sin(-k_y b)\cos(-k_x a) \big. \\
-                               &+ cos(k_y b)cos(k_x a) + sin(k_y b)sin(k_x a) + i sin(k_y b)cos(k_x a) - i sin(k_x a)cos(k_y b) )
+                           \big. &+ \cos(k_y b)\cos(k_x a) + \sin(k_y b)\sin(k_x a) + i \sin(k_y b)\cos(k_x a) - i \sin(k_x a)\cos(k_y b) \big)
             \end{align*}
 
             Here we can use the properties from (\ref{eq:neg_sin_cos}) to reduce this equation. Since \( \left| h_0 \right| ^2 = h_0 h^*_0 \),
             we also obtain a simple expression for \( h^*_0 \).
 
             \begin{align*}
-                h_0 &= -t_c ( 1 + 2cos(k_x a)cos(k_y b) -  2 i sin(k_x a)cos(k_y b)) \\
+                h_0 &= -t_c \left( 1 + 2\cos(k_x a)\cos(k_y b) -  2 i \sin(k_x a)\cos(k_y b)\right) \\
-                h^*_0 &= -t_c ( 1 + 2cos(k_x a)cos(k_y b) + 2 i sin(k_x a)cos(k_y b))
+                h^*_0 &= -t_c \left( 1 + 2\cos(k_x a)\cos(k_y b) + 2 i \sin(k_x a)\cos(k_y b)\right)
             \end{align*}
             
             We can now find \( \left| h_0 \right| ^2 = h_0 h^*_0 \):
 
             \begin{align*}
-                h_0 h^*_0 = 1 + 4cos(k_x a)cos(k_y b) + 4cos^2(k_y b)
+                h_0 h^*_0 = 1 + 4\cos(k_x a)\cos(k_y b) + 4\cos^2(k_y b)
             \end{align*}
 
-            From this, we can finally obtain an expression for \( E(k) \):
+            From this, we can finally obtain an expression for \( E(\vec{k}) \):
 
             \begin{align}
-                E(k) = E_0 \pm t_c \sqrt{1 + 4cos(k_x a)cos(k_y b) 4 cos^2(k_y b)}
+                E(\vec{k}) = E_0 \pm t_c \sqrt{1 + 4\cos(k_x a)\cos(k_y b) 4 \cos^2(k_y b)}
             \end{align}
 
         \end{enumerate}
@@ -491,7 +491,11 @@ S_u = [1, s; s, 1];
 
             \item %4e
 
+            To generate the points, we used a simple MATLAB script, found in Appendix (A).
+
 
         \end{enumerate}
 
+    \app
+
 \end{document}