big o - Would this function be O(n^2log_2(n))? -

so given function 65536 n² + 128 n log₂n

and way o(n² log₂n) if

c = 65664, n0 = 2 n ≥ 2 since

c1 = 65536 n1 = 2 when 65536 ≤ c1*n² and
c2 = 128 n2 = 1 when 128 ≤ c2*n

but number i've chosen constant seems bit high, there way check this?

o(65536 n² + 128 n log₂n) same o(n² + n log₂n) since can ignore multiplicative constants. o(n² + n log₂n) equal o(n²) since n² grows faster n log₂n.

also, way, base of logarithms doesn't matter in big-o analysis. logarithms grow @ same rate. after all, log₂n = log n / log 2, , multiplicative constants can factored out. can log n instead of log₂n.

caveat: technically, true statement 65536 n² + 128 n log₂n ∈ o(n² log₂n) because big-o gives an upper bound, not strict one. o(n²) lower upper bound, if makes sense.

that said, should not have come o(n² log₂n). merely result of accidentally turning addition multiplication. rule of thumb, if have multiple things added inside big-o formula, have figure out 1 of them grows fastest , discard others.