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if $a=O(N^2) $ then according to the big oh definition I didn't see why we can't say $a= O(N^4)$ or $= O(N^8)$

CBenni
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goon
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  • One small caveat: you can say this as long as you're referring to the behavior as $N\to\infty$ - which is generally assumed since you're talking about $N$, but not explicit in the notation. Note that saying that $a\in O(x^2)$ as $x\to 0$ does not imply that $a\in O(x^4)$ then! – Steven Stadnicki Feb 04 '13 at 17:03
  • I think the reason many people get confused with "big O" notation is that it's common to write $a(n) = O(n^2)$ when strictly it is more accurate to say that $a(n) \in O(n^2)$ (since $O(n^2)$ is a set, and $a(n)$ is an element of that set). Then since $O(n^2) \subset O(n^4)$ (as sets), it is trivially true that if $a(n)\in O(n^2)$ then also $a(n) \in O(n^4)$. – Chris Taylor Feb 04 '13 at 17:46
  • @ChrisTaylor: That's one way of formalizing the meaning of the big-O notation, but not the only one. See here where I propose a more intuitive (but mostly equivalent) formalization of equalities involving big-O. – hmakholm left over Monica Feb 04 '13 at 20:00
  • how " a∈O(x2) as x→0 " is right? is there a use to the big oh when x → 0 ? – goon Feb 04 '13 at 21:49
  • does what Steven Stadnicki saying has the same logic as when a∈O(x^4) as x→∞ does not imply that a∈O(x^2) then! – goon Feb 04 '13 at 21:52
  • @goon There is a use to big-O notation when talking about approaching 0; it's for measuring how small quantities are, for instance for approximations. One can say that $\sin(x) = x+O(x^3)$ as $x\to 0$, for instance, and that expression makes formal sense. – Steven Stadnicki Feb 04 '13 at 22:49

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Yes, you can say either of those. They don't tell you as much about $a(N)$ as $a(N)=\mathcal O(N^2)$ does, but what little they do say is true.

hmakholm left over Monica
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