1

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 Answers2

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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, $\dfrac {n(n-1)}2=\Theta(n^2)$.

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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 $1/2$.

Ethan Bolker
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