Typos

Advanced/Random Walks/parts/1 circuits.tex

@@ -23,7 +23,7 @@ In the definitions below, let $X$ be the set of nodes in a circuit.
 
 		Note that this is different than current and resistance, which aren't functions
 		of two arbitrary nodes --- rather, they are functions of \textit{edges}
-		(i.e, two adjecent nodes).
+		(i.e, two adjacent nodes).
 	}
 
 
@@ -136,7 +136,7 @@ It exists only to create a potential difference between the two nodes.
 \end{center}
 
 \problem{}<onecurrents>
-From the circuit diagram above, we immediatly know that $V(A) = 1$ and $V(B) = 0$. \par
+From the circuit diagram above, we immediately know that $V(A) = 1$ and $V(B) = 0$. \par
 What equations related to the currents out of $x$ and $y$ does Kirchoff's law give us? \par
 \hint{Current into $x$ = current out of $x$}
 

Advanced/Random Walks/parts/3 effective.tex

@@ -168,7 +168,7 @@ Using Ohm's law and Kirchoff's law, calculate the effective resistance $R_{\text
 \pagebreak
 
 We can now use effective resistance to simplify complicated circuits. Whenever we see the above constructions
-(resistors in parellel or in series) in a graph, we can replace them with a single resistor of appropriate value.
+(resistors in parallel or in series) in a graph, we can replace them with a single resistor of appropriate value.
 
 
 \problem{}
@@ -243,13 +243,13 @@ If we place $A$ and $B$ at opposing vertices, what is the effective resistance o
 	and $B$ at \texttt{111...1}. We can divide our cube into $n+1$ layers based on how many ones are in each
 	node's binary string, with the $k^\text{th}$ layer having $k$ ones. By symmetry, all the nodes in each layer
 	have the same voltage. This means we can think of the layers as connected in series, with the resistors
-	inside each layer connected in parellel.
+	inside each layer connected in parallel.
 
 	\vspace{2mm}
 
 	There are $\binom{n}{k}$ nodes in the $k^\text{th}$ layer. Each node in this layer has $k$ ones, so
 	there are $n - k$ ways to flip a zero to get to the $(k + 1)^\text{th}$ layer. In total, there are 
-	$\binom{n}{k}(n - k)$ parellel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
+	$\binom{n}{k}(n - k)$ parallel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
 	layer, creating an effective resistance of 
 	$$
 		\frac{1}{\binom{n}{k}(n - k)}