deletions | additions       diff --git a/begin_claim_The_algorithm_runs__.tex b/begin_claim_The_algorithm_runs__.tex  index 1dfc0fa..a0524fb 100644  --- a/begin_claim_The_algorithm_runs__.tex  +++ b/begin_claim_The_algorithm_runs__.tex 
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\begin{align*}  R(n) &= iO(n) + \sum_{j=0}^{i-1} 2^j O(\log \frac{n}{2^i}) + 2^iR(\frac{n}{2^i})\\  \text{When our point set contains one point, it's clear that $R(1)$ only requires a constant amount of work.}\\  \text{This occurs when $\frac{n}{2^i}=1\Rightarrow i=\log_2 n$}\\  &= O(n\log n) + \sum_{j=0}^{\log n-1} 2^j O(\log \frac{n}{2^i}) + 2^{\log n}R(1)\\  &= O(n\log n) + n^{\log 2}O(1) + \sum_{j=0}^{\log n-1} 2^j (\log n - \log{2^j})\\   &= O(n\log n) + O(n) + \left(\log n\sum_{j=0}^{\log n-1} 2^j - \sum_{j=0}^{\log n-1} 2^j\log{2^j} \right)\\