o(g(n)) = { the set of all f such that for any positive constant c there exists a positive constant n0 satisfying 0 <= f(n) <= cg(n) for all n >= n0 }.So, for example, n2 is o(n3) but not o(n2).
Since g always grows strictly faster than f, we have
f(n) = o(g(n)) implies limn->oo (f(n)/g(n)) = 0.If g is asyptotically non-negative, the implication will also run the other way, i.e., if the limit is 0, then f(n) = o(g(n)) is true.
w(g(n)) = { the set of all f such that for any positive constant c there exists a positive constant n0 satisfying 0 <= f(n) <= cg(n) for all n >= n0 }.So, for example, n3 is w(n2) but not w(n3).
As before, since f always grows strictly faster than g, we have
f(n) = o(g(n)) implies limn->oo (g(n)/f(n)) = 0.
(nk). This is true even if the lower order terms are multiplied by something
in o(n), such as log n (Is this true if we replace o with O?).
)n)
for any positive
. Your book has
a nice proof of this using limits.
Some logarithms are use more frequently:
(ln n).