Post-class edits
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@ -6,8 +6,8 @@ We would like to compute the probability of our particle stopping at node $A$. \
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\vspace{2mm}
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In other words, we want a function $P(n): N \to [0, 1]$ that returns the probability that our particle stops at $A$,
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where $N$ is the set of nodes in $G$.
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In other words, we want a function $P: \text{Nodes} \to [0, 1]$ that maps each node of the graph
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to the probability that our particle stops at $A$.
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\begin{center}
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\begin{tikzpicture}
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@ -92,11 +92,17 @@ Find $P(x)$ in terms of $P(v_1), P(v_2), ..., P(v_n)$.
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\problem{}
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How can we use \ref{oneunweighted} to find $P(n)$ for any $n$?
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In general, how do we find $P(n)$ for any node $n$?
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\begin{solution}
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If we write an equation for each node other than $A$ and $B$, we have a system of $|N| - 2$
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linear equations in $|N| - 2$ variables.
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\vspace{2mm}
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We still need to show that this system is nonsingular, but
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that's outside the scope of this handout. This could
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be offered as a bonus problem.
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\end{solution}
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\vfill
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@ -1,10 +1,12 @@
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\section{Circuits}
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An \textit{electrical circuit} is a graph with a few extra properties,
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called \textit{current}, \textit{voltage}, and \textit{resistance}.
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called \textit{current}, \textit{voltage}, and \textit{resistance}. \par
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In the definitions below, let $X$ be the set of nodes in a circuit.
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\begin{itemize}[itemsep=3mm]
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\item \textbf{Voltage} is a function $V(n): N \to \mathbb{R}$ that assigns a number to each node of our graph. \par
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\item \textbf{Voltage} is a function $V: X \to \mathbb{R}$ that assigns a number to each node of our graph. \par
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In any circuit, we pick a \say{ground} node, and define the voltage\footnotemark{} there as 0. \par
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We also select a \say{source} node, and define its voltage as 1. \par
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@ -15,15 +17,18 @@ called \textit{current}, \textit{voltage}, and \textit{resistance}.
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\footnotetext{
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In the real world, voltage is always measured \textit{between two points} on a circuit.
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Voltage is defined as the \textit{difference} in electrical charge between two points.
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Here, all voltages are measured with respect to our \say{ground} node.
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Hence, voltage is a function of two nodes.
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This detail isn't directly relevant to the problems in this handout, so you mustn't worry about it today. \par
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Just remember that the electrical definitions here are a significant oversimplification of reality.
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\vspace{2mm}
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Note that this is different than current and resistance, which aren't functions
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of two arbitrary nodes --- rather, they are functions of \textit{edges}
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(i.e, two adjecent nodes).
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}
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\item \textbf{Current} is a function $I(e^\rightarrow): N \times N \to \mathbb{R}$ that assigns a number to each
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\textit{oriented edge} $e^\rightarrow$ in our graph. An \say{oriented edge} is just an ordered pair of nodes $(n_1, n_2)$. \par
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\item \textbf{Current} is a function $I: X^2 \to \mathbb{R}$ that assigns a number to each
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\textit{oriented edge} in our graph. An \say{oriented edge} is just an ordered pair of nodes $(n_1, n_2)$. \par
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\vspace{1mm}
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@ -31,7 +36,7 @@ called \textit{current}, \textit{voltage}, and \textit{resistance}.
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Naturally, $I(a, b) = -I(b, a)$.
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\item \textbf{Resistance} is a function $R(e): N \times N \to \mathbb{R}^+_0$ that represents a certain edge's
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\item \textbf{Resistance} is a function $R: X^2 \to \mathbb{R}^+_0$ that represents a certain edge's
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resistance to the flow of current through it. \par
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Resistance is a property of each \textit{link} between nodes, so order doesn't matter: $R(a, b) = R(b, a)$.
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\end{itemize}
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@ -50,10 +55,14 @@ the case! Any circuit obeys \textit{Ohm's law}, stated below:
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$$
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V(a, b) = I(a,b) \times R(a,b)
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$$
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\note{
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$V(a, b)$ is the voltage between nodes $a$ and $b$. If this doesn't make sense, read the footnote below. \\
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In this handout, it will be convenient to write $V(a, b)$ as $V(a) - V(b)$.
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}
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This handout uses two notations for voltage: two-variable $V(a, b)$ and one-variable $V(a)$. \par
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The first represents the voltage between points $a$ and $b$, better reflecting reality (see the footnote below).
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The second measures the voltage between $a$ and ground, and is more convenient to use in equations.
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\textbf{Try to use the single-variable notation in your equations.}
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Convince yourself that $V(a, b) = V(a) - V(b)$.
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\vfill
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\definition{Kirchoff's law}
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@ -64,25 +73,46 @@ Formally, we can state this as follows:
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\vspace{2mm}
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Let $x$ be a node in our circuit and $B_x$ the set of its neighbors. We than have
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Let $x$ be a node in our circuit and $N_x$ the set of its neighbors. We than have
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$$
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\sum_{b \in B_x} I(x, b) = 0
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\sum_{b \in N_x} I(x, b) = 0
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$$
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which must hold at every node \textbf{except the source and ground vertices.} \par
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\hint{Keep this exception in mind, it is used in a few problems later on.}
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\vfill
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\pagebreak
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\begin{instructornote}
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Be aware that some students may not be comfortable with these concepts from physics,
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nor with the circuit notation on the next page.
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\vspace{2mm}
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It may be a good idea to give the class a quick lecture on this topic,
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explaining the basics of electonic circuits and circuit diagrams.
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\vspace{2mm}
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Things to cover:
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\begin{itemize}
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\item All the definitions on the previous page, in detail.
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\item What's an Ohm, an Amp, a Volt?
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\item Measuring voltage. Why is $V(a, b) = V(a) - V(b)$?
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\item What does the $\Omega$ in the picture below mean?
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\item Circuit symbols in the diagram below.
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\end{itemize}
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\vspace{2mm}
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You could also draw connections to the graph flow handout,
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if the class covered it before.
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\end{instructornote}
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Consider the circuit below. This the graph from \ref{firstgraph}, turned into a circuit by:
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Consider the circuit below. \textbf{This the graph from \ref{firstgraph}}, turned into a circuit by:
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\begin{itemize}
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\item Replacing all edges with $1\Omega$ resistors
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\item Attaching a 1 volt battery between $A$ and $B$
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@ -107,7 +137,8 @@ It exists only to create a potential difference between the two nodes.
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\problem{}<onecurrents>
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From the circuit diagram above, we immediatly know that $V(A) = 1$ and $V(B) = 0$. \par
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What equations related to the currents out of $x$ and $y$ does Kirchoff's law give us?
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What equations related to the currents out of $x$ and $y$ does Kirchoff's law give us? \par
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\hint{Current into $x$ = current out of $x$}
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\vfill
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@ -22,12 +22,12 @@ $$
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\problem{}
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Let $x$ be a node in a graph. \par
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Let $B_x$ be the set of $x$'s neighbors, $w(x, y)$ the weight of the edge between nodes $x$ and $y$, and $W_x$
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Let $N_x$ be the set of $x$'s neighbors, $w(x, y)$ the weight of the edge between nodes $x$ and $y$, and $W_x$
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the sum of the weights of all edges connected to $x$.
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We saw earlier that the probability function $P$ satisfies the following sum:
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$$
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P(x) = \sum_{b \in B_x} \biggl( P(b) \times \frac{w(x, b)}{W_x} \biggr)
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P(x) = \sum_{b \in N_x} \biggl( P(b) \times \frac{w(x, b)}{W_x} \biggr)
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$$
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\note{This was never explicitly stated, but is noted in \ref{weightedgraph}.}
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@ -36,7 +36,7 @@ $$
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Use Ohm's and Kirchoff's laws to show that the voltage function $V$ satisfies a similar sum:
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$$
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V(x) = \sum_{b \in B_x} \biggl( V(b) \times \frac{C(x, b)}{C_x} \biggr)
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V(x) = \sum_{b \in N_x} \biggl( V(b) \times \frac{C(x, b)}{C_x} \biggr)
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$$
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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 $x$.
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@ -44,16 +44,16 @@ where $C(x, b)$ is the conductance of edge $(x, b)$ and $C_x$ is the sum of the
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\begin{solution}
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First, we know that
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$$
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\sum_{b \in B_x} I(x, b) = 0
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\sum_{b \in N_x} I(x, b) = 0
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$$
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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
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$$
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V(x) \sum_{b \in B_x} \frac{1}{R(x, b)} - \sum_{b \in B_x} \frac{V(b)}{R(x, b)} = 0
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V(x) \sum_{b \in N_x} \frac{1}{R(x, b)} - \sum_{b \in N_x} \frac{V(b)}{R(x, b)} = 0
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$$
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Rearranging and replacing $R(x, b)^{-1}$ with $C(x, b)$ and $\sum C(x, b)$ with $C_x$ gives us
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$$
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V(x) = \sum_{b \in B_x} V(b) \frac{C(x, b)}{C_x}
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V(x) = \sum_{b \in N_x} V(b) \frac{C(x, b)}{C_x}
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$$
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\end{solution}
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@ -71,7 +71,7 @@ two problems.
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\problem{}<generaleq>
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Let $q$ be a solution to the following equations, where $x \neq a, b$.
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$$
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q(x) = \sum_{b \in B_x} \biggl( q(b) \times \frac{w(x, b)}{W_x} \biggr)
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q(x) = \sum_{b \in N_x} \biggl( q(b) \times \frac{w(x, b)}{W_x} \biggr)
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$$
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Show that the maximum and minimum of $q$ are $q(a)$ and $q(b)$ (not necessarily in this order).
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@ -80,7 +80,7 @@ Show that the maximum and minimum of $q$ are $q(a)$ and $q(b)$ (not necessarily
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\vspace{2mm}
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Since $q(x)$ is a weighted average of all $q(b), ~b \in B_x$, there exist $y, z \in B_x$ satisfying
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Since $q(x)$ is a weighted average of all $q(b), ~b \in N_x$, there exist $y, z \in N_x$ satisfying
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$q(y) \leq q(x) \leq q(z)$. Therefore, none of these can be an extreme point.
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\vspace{2mm}
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@ -104,7 +104,7 @@ and that $p(x) - q(x) = 0$ for every $x$. \note{Note that $p(x) - q(x) = 0 ~ \fo
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\begin{solution}
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The equations in \ref{generaleq} for $p$ and $q$ directly imply that
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$$
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[p - q](x) = \sum_{b \in B_x} \biggl( [p - q](b) \times \frac{w(x, b)}{W_x} \biggr)
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[p - q](x) = \sum_{b \in N_x} \biggl( [p - q](b) \times \frac{w(x, b)}{W_x} \biggr)
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$$
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Which are the equations from \ref{generaleq} for $(p - q)$.
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@ -38,20 +38,20 @@ out of $A$ is equal to the current flowing into $B$.
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\problem{}
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Using Kirchoff's law, show that the following equality holds. \par
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Remember that we assumed Kirchoff's law holds only at nodes other than $A$ and $B$. \par
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\note[Note]{As before, $B_x$ is the set of neighbors of $x$. Naturally, $B_B$ is the set of neighbors of $B$.}
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\note[Note]{As before, $N_x$ is the set of neighbors of $x$.}
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$$
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\sum_{b \in B_A} I(S, b) = \sum_{b \in B_B} I(b, B)
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\sum_{b \in N_A} I(A, b) = \sum_{b \in N_B} I(b, B)
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$$
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\begin{solution}
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Add Kirchoff's law for all vertices $x \neq A$ to get
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$$
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\sum_{\forall x} \biggl( ~ \sum_{b \in B_x } I(x, b) \biggr) = 0
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\sum_{\forall x} \biggl( ~ \sum_{b \in N_x } I(x, b) \biggr) = 0
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$$
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This sum counts both $I(x, y)$ and $I(x, y)$ for all edges $x, y$, except $I(x, y)$ when $x$ is
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$A$ or $B$. Since $I(a, b) + I(b, a) = 0$, these cancel out, leaving us with
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$$
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\sum_{b \in B_A} I(A, b) + \sum_{b \in B_B} I(B, b) = 0
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\sum_{b \in N_A} I(A, b) + \sum_{b \in N_B} I(B, b) = 0
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$$
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\vspace{2mm}
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@ -61,7 +61,7 @@ $$
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\vfill
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If we call this current $I_A = \sum_{b \in B_A} I(A, b)$, we can pretend that the box contains only one resistor,
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If we call this current $I_A = \sum_{b \in N_A} I(A, b)$, we can pretend that the box contains only one resistor,
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carrying $I_A$ units of current. Using this information and Ohm's law, we can calculate the
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\textit{effective resistance} of the box.
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