41 ms·
The o(1) here represents a decreasing function in m, such as 1/m. (Note: This is little o, not big o.) Therefore, for any k > 1: m^(1+o(1)) grows more slowly t
by theemathas 4y ago
The o(1) here represents a decreasing function in m, such as 1/m. (Note: This is little o, not big o.)
Therefore, for any k > 1: m^(1+o(1)) grows more slowly than m^k but faster than m