deletions | additions       diff --git a/begin_claim_The_algorithm_runs__.tex b/begin_claim_The_algorithm_runs__.tex  index 66aa717..d74f063 100644  --- a/begin_claim_The_algorithm_runs__.tex  +++ b/begin_claim_The_algorithm_runs__.tex 
...
\begin{claim}  The algorithm runs in $O(n\log n)$ $O\left(n\log n\right)$  \end{claim}  \begin{proof}  We obtain the following after `unravelling' our recursive formula $i$ times:  \begin{align*}  R(n) R\left(n\right)  &= O(n) O\left(n\right)  + O(\log n) O\left(\log n\right)  + 2R\left(\frac{n}{2}\right)\\ &= iO(n) iO\left(n\right)  + \sum_{j=0}^{i-1} 2^j O(\log \frac{n}{2^i}) 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 $R(1)$ $R\left(1\right)$  requires a constant amount of work. This occurs when $\frac{n}{2^i}=1$ or equivalently when $i=\log_2 n$. \begin{align*}  R(n) R\left(n\right)  &= iO(n) iO\left(n\right)  + \sum_{j=0}^{i-1} 2^j O(\log \frac{n}{2^i}) O\left(\log \frac{n}{2^i}\right)  + 2^iR\left(\frac{n}{2^i}\right)\\ &= O(n\log n) O\left(n\log n\right)  + \sum_{j=0}^{\log n-1} 2^j O(\log \frac{n}{2^i}) O\left(\log \frac{n}{2^i}\right)  + 2^{\log n}R(1)\\ n}R\left(1\right)\\  &= O(n\log n) O\left(n\log n\right)  + n^{\log 2}O(1) 2}O\left(1\right)  + \sum_{j=0}^{\log n-1} 2^j (\log \left(\log  n - \log{2^j})\\ \log{2^j}\right)\\  &= O(n\log n) O\left(n\log n\right)  + O(n) O\left(n\right)  + \left(\log n\sum_{j=0}^{\log n-1} 2^j - \sum_{j=0}^{\log n-1} 2^j\log{2^j} \right)\\ &= O(n\log n) O\left(n\log n\right)  + \left(\log n \times 2^{\log n} - C \right) \tag{with $C\geq0$}\\ &= O(n\log n) O\left(n\log n\right)  + \left(n\log n - C \right)\\ &\leq O(n\log n) O\left(n\log n\right)  + O(n\log n)\\ O\left(n\log n\right)\\  &\in O(n\log n) O\left(n\log n\right)  \end{align*}