In an article about Metcalfe's law, I've read that $n(n-1)/2$ is asymptotically proportional to $n^2$.
What does this mean?
PS:
I did find asymptotically optimal but I'm not sure if they mean the same thing.
2 Answers
$\begingroup$Proportional to $n^2$ when $n$ tends to infinity. In other words, the larger $n$, the closer the expression at hand is to $cn^2$, for some constant $c$.
In Landau's notation, $\dfrac {n(n-1)}2=\Theta(n^2)$.
$\endgroup$ 1 $\begingroup$The limit calculation$$ \lim_{n \to \infty}\frac{\frac{n(n-1)}{2}}{n^2} = \frac{1}{2} $$says that for large $n$ the expression is approximately proportional to $n^2$ with proportionality constant $1/2$.
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