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Minor cleanup

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Mark 2023-10-04 09:29:46 -07:00
parent 10177fa934
commit 0744a88c00
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1 changed files with 5 additions and 4 deletions
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9
Advanced/Random Walks/parts/3 effective.tex
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@ -250,9 +250,10 @@ If we place $A$ and $B$ at opposing vertices, what is the effective resistance o
There are $\binom{n}{k}$ nodes in the $k^\text{th}$ layer. Each node in this layer has $k$ ones, so 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 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 $(k + 1)^\text{th}$ $\binom{n}{k}(n - k)$ parellel connections from the $k^\text{th}$ layer to the $(k + 1)^\text{th}$
layer, creating an effective resistance of $(\binom{n}{k}(n - k))^{-1}$. layer, creating an effective resistance of
$$
\vspace{2mm} \frac{1}{\binom{n}{k}(n - k)}
$$
The total effective resistance is therefore The total effective resistance is therefore
$$ $$
@ -263,7 +264,7 @@ If we place $A$ and $B$ at opposing vertices, what is the effective resistance o
To calculate the limit as $n \rightarrow \infty$, note that To calculate the limit as $n \rightarrow \infty$, note that
$$ $$
\binom{n}{k}(n - k) = \frac{n!}{(n - k - 1)! \times k!} = n \binom{n-1}{k}. \binom{n}{k}(n - k) = \frac{n!}{(n - k - 1)! \times k!} = n \binom{n-1}{k}
$$ $$
So, the sum is So, the sum is
$$ $$