deletions | additions       diff --git a/We_take_textbf_Step_1__.tex b/We_take_textbf_Step_1__.tex  index 52795fb..1307753 100644  --- a/We_take_textbf_Step_1__.tex  +++ b/We_take_textbf_Step_1__.tex 
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It's trivial to see that after `unravelling' our recursive formula $i$ times we retrieve the following:  \begin{align*}  R\left(n\right):&= O\left(n\right) + 2R\left(\frac{n}{2}\right)\\  &= iO\left(n\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 equivalently $i=\log_2 n$. We substitute this value into our equation above.