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# 10^-6n^2 and n^2 both are O(n^2) how?

 0 if we multiply very very small number in n^2 then big O of n^2 and (very very small)*n^2 is n^2,why and how? asked 21 Oct '18, 00:53 1 accept rate: 0% That's literally how big O notation is defined. Any constant does not matter for the asymptotic behavior. Wikipedia defines the term pretty well: https://en.wikipedia.org/wiki/Big_O_notation (21 Oct '18, 02:07) algmyr7★ big O is more of "how much slower code gets with larger input" rather than just how slow it is. (21 Oct '18, 02:13)
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question asked: 21 Oct '18, 00:53

question was seen: 108 times

last updated: 21 Oct '18, 02:13