Consider the following functions: f(n) = n², if n is even log n, if n is odd…

Key idea: f(n) and g(n) swap between n² and log n depending on whether n is even or odd, so their ratio behaves very differently on the two parity subsequences.

• For even n: f(n) = n² and g(n) = log n, so f(n)/g(n) = n² / log n → ∞ as n → ∞. Hence f(n) is not O(g(n)).

• For odd n: f(n) = log n and g(n) = n², so f(n)/g(n) = log n / n² → 0 as n → ∞. Hence f(n) is not Ω(g(n)).

Conclusion: Because along one infinite subsequence the ratio f(n)/g(n) → ∞ and along another it → 0, there are no constant multiplicative bounds that hold for all sufficiently large n. Therefore none of the standard asymptotic relations (big-O, Theta, little-o, etc.) hold between f and g; the correct statement is that the functions cannot be compared using standard asymptotic notation.