Post-class edits

Advanced/Random Walks/parts/2 equivalence.tex

@@ -22,12 +22,12 @@ $$
 
 \problem{}
 Let $x$ be a node in a graph. \par
-Let $B_x$ be the set of $x$'s neighbors, $w(x, y)$ the weight of the edge between nodes $x$ and $y$, and $W_x$
+Let $N_x$ be the set of $x$'s neighbors, $w(x, y)$ the weight of the edge between nodes $x$ and $y$, and $W_x$
 the sum of the weights of all edges connected to $x$.
 
 We saw earlier that the probability function $P$ satisfies the following sum:
 $$
-	P(x) = \sum_{b \in B_x} \biggl( P(b) \times \frac{w(x, b)}{W_x} \biggr)
+	P(x) = \sum_{b \in N_x} \biggl( P(b) \times \frac{w(x, b)}{W_x} \biggr)
 $$
 
 \note{This was never explicitly stated, but is noted in \ref{weightedgraph}.}
@@ -36,7 +36,7 @@ $$
 
 Use Ohm's and Kirchoff's laws to show that the voltage function $V$ satisfies a similar sum:
 $$
-	V(x) = \sum_{b \in B_x} \biggl( V(b) \times \frac{C(x, b)}{C_x} \biggr)
+	V(x) = \sum_{b \in N_x} \biggl( V(b) \times \frac{C(x, b)}{C_x} \biggr)
 $$
 where $C(x, b)$ is the conductance of edge $(x, b)$ and $C_x$ is the sum of the conductances of all edges connected to $x$.
 
@@ -44,16 +44,16 @@ where $C(x, b)$ is the conductance of edge $(x, b)$ and $C_x$ is the sum of the
 \begin{solution}
 	First, we know that 
 	$$
-		\sum_{b \in B_x} I(x, b) = 0
+		\sum_{b \in N_x} I(x, b) = 0
 	$$
 	for all nodes $x$. Now, substitute $I(x, b) = \frac{V(x) - V(b)}{R(x, y)}$ and pull out $V(x)$ terms to get
 	$$
-		V(x) \sum_{b \in B_x} \frac{1}{R(x, b)} - \sum_{b \in B_x} \frac{V(b)}{R(x, b)} = 0
+		V(x) \sum_{b \in N_x} \frac{1}{R(x, b)} - \sum_{b \in N_x} \frac{V(b)}{R(x, b)} = 0
 	$$
 
 	Rearranging and replacing $R(x, b)^{-1}$ with $C(x, b)$ and $\sum C(x, b)$ with $C_x$ gives us 
 	$$
-		V(x) = \sum_{b \in B_x} V(b) \frac{C(x, b)}{C_x}
+		V(x) = \sum_{b \in N_x} V(b) \frac{C(x, b)}{C_x}
 	$$
 \end{solution}
 
@@ -71,7 +71,7 @@ two problems.
 \problem{}<generaleq>
 Let $q$ be a solution to the following equations, where $x \neq a, b$.
 $$
-	q(x) = \sum_{b \in B_x} \biggl( q(b) \times \frac{w(x, b)}{W_x} \biggr)
+	q(x) = \sum_{b \in N_x} \biggl( q(b) \times \frac{w(x, b)}{W_x} \biggr)
 $$
 Show that the maximum and minimum of $q$ are $q(a)$ and $q(b)$ (not necessarily in this order).
 
@@ -80,7 +80,7 @@ Show that the maximum and minimum of $q$ are $q(a)$ and $q(b)$ (not necessarily
 
 	\vspace{2mm}
 
-	Since $q(x)$ is a weighted average of all $q(b), ~b \in B_x$, there exist $y, z \in B_x$ satisfying
+	Since $q(x)$ is a weighted average of all $q(b), ~b \in N_x$, there exist $y, z \in N_x$ satisfying
 	$q(y) \leq q(x) \leq q(z)$. Therefore, none of these can be an extreme point.
 	
 	\vspace{2mm}
@@ -104,7 +104,7 @@ and that $p(x) - q(x) = 0$ for every $x$. \note{Note that $p(x) - q(x) = 0 ~ \fo
 \begin{solution}
 	The equations in \ref{generaleq} for $p$ and $q$ directly imply that
 	$$
-		[p - q](x) = \sum_{b \in B_x} \biggl( [p - q](b) \times \frac{w(x, b)}{W_x} \biggr)
+		[p - q](x) = \sum_{b \in N_x} \biggl( [p - q](b) \times \frac{w(x, b)}{W_x} \biggr)
 	$$
 	Which are the equations from \ref{generaleq} for $(p - q)$.