deletions | additions       diff --git a/WSPD_Approximation_For_our_approximation__.md b/WSPD_Approximation_For_our_approximation__.md  index 0e33a0e..766314d 100644  --- a/WSPD_Approximation_For_our_approximation__.md  +++ b/WSPD_Approximation_For_our_approximation__.md 
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What is the radius of the circle centered at `i'`? By the triangle inequality if we set the radius to be \(|i,i'| + L_r\) where \(L_r\) was the optimal radius then we cover all of the same points. The same thing applies for `j'`. As such, we have that our diameter is given by the following \(d=\left(|i,i'| + L_r \right) + \left(|j,j'| + R_r \right) + |i',j'|\).  We substitute in our WSPD bounds of \(|i,i'| \leq \frac{2}{s}|i,j|\) (resp. \(|j,j'| \leq \dots\) ) and \(|i',j'| \leq (1 + \frac{4}{s}) |i,j|\). This gives us that \(\text{diam}(i',j') \leq \frac{2}{s}|i,j| + L_r + \frac{2}{s}|i,j| + R_r + (1 + \frac{4}{s}) |i,j| \) which simplifies to \(|i,j| \left( \frac{2}{s} + \frac{2}{s} + 1 + \frac{4}{s} \right) + L_r + R_r \) = \left((\frac{s+8}{s}) |i,j| + L_r + R_r\right)\).