3 ms·
If I understand correctly, f(n) << g(n) is equivalent to f(n) = o(g(n)), not f(n) = O(g(n)).
by gvx 6y ago
If I understand correctly, f(n) << g(n) is equivalent to f(n) = o(g(n)), not f(n) = O(g(n)).
- dfdz 6y agoIt depends on the author/field. In analysis sometimes f(n) <<g(n) means f(n) = o(g(n)), but other times (especially in discrete math) f(n) << g(n) means f(n) = O(g(n)). Due to this ambiguity, I think the notation f(n) ~< g(n) or in latex f(n) \lesssim g(n) is more clear