Big O basically drops constants

O(2n+1) = O(2n) = O(n)

o(2n+1) > o(2n) > o(n)

Both define upper boundary, but big O can read like "the function is generally proportional to ___" and little o can read like "the function is never more than _____"

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bobmarleybiceps an hour ago | parent [-]

that's not the convention I know... (from CLRS algorithms book).

Big O is sort of like <=. little o is sort of like <.

https://dl.acm.org/doi/10.1145/1008328.1008329

	▲	npinsker an hour ago \| parent [-]
		yes, GP is slightly off, but the idea of <= and < is accurate. O(n) is: "there exists a K such that, for sufficiently large n, Kn is always larger". o(n) is: "for every K, for sufficiently large n, Kn is always larger". (So o(2n+1) and o(2n) and o(n) are the same thing.)