deletions | additions       diff --git a/begin_claim_The_algorithm_runs__.tex b/begin_claim_The_algorithm_runs__.tex  index d74f063..4678240 100644  --- a/begin_claim_The_algorithm_runs__.tex  +++ b/begin_claim_The_algorithm_runs__.tex 
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&= iO\left(n\right) + \sum_{j=0}^{i-1} 2^j O\left(\log \frac{n}{2^i}\right) + 2^iR\left(\frac{n}{2^i}\right)  \end{align*}  When our point set contains one point, it's clear that $R\left(1\right)$ requires a constant amount of work. This occurs when $\frac{n}{2^i}=1$ or equivalentlywhen  $i=\log_2 n$. \begin{align*}  R\left(n\right) &= iO\left(n\right) + \sum_{j=0}^{i-1} 2^j O\left(\log \frac{n}{2^i}\right) + 2^iR\left(\frac{n}{2^i}\right)\\