# is it correct?

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• June 5th 2006, 08:27 PM
malaygoel
Quote:

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

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

$\frac{d^2}{2^n}=2\left(\frac{1}{2^{(n-1)}}\right)$
$\frac{d^2}{2^n}=\frac{2}{2^{(n-1)}}$
$\frac{2^2}{2^n}=\frac{2}{2^{(n-1)}}$
$\frac{4}{2^n}=\frac{2}{2^{(n-1)}}$
$\frac{2}{2^{(n-1)}}=\frac{2}{2^{(n-1)}}$

but why would you say the diameter is 2?

It is given in the question that O1 has radius of 1 unit
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