Originally Posted by **Quick**

it might be that I'm tired, buy I can't seem to think where you got the 2^(-1/2) from. However, you can see our equations differ,

mine= $\displaystyle \frac{d^2}{2^n} $ yours= $\displaystyle 2\left(\frac{1}{2^{(n-1)}}\right) $

However, there is a certain value of d that would make our equations the same.

$\displaystyle \frac{d^2}{2^n}=2\left(\frac{1}{2^{(n-1)}}\right) $

$\displaystyle \frac{d^2}{2^n}=\frac{2}{2^{(n-1)}} $

$\displaystyle \frac{2^2}{2^n}=\frac{2}{2^{(n-1)}} $

$\displaystyle \frac{4}{2^n}=\frac{2}{2^{(n-1)}} $

$\displaystyle \frac{2}{2^{(n-1)}}=\frac{2}{2^{(n-1)}} $

but why would you say the diameter is 2?