Gu_5
Consider two functions where $f(n) = \log n$ and $g(n) = \sqrt{n}$.
Based on the definition of Big Oh ($O$), which statement accurately describes the relationship between $f(n)$ and $g(n)$?
Choose the correct option that reflects the asymptotic upper bound of $f(n)$ with respect to $g(n)$:

Question

BlackTom AI · Accepted Answer

We start by restating the problem to ensure clarity: we are given f(n) = log n and g(n) = sqrt(n), and we need to determine which statement correctly describes the asymptotic upper bound of f(n) with respect to g(n) in Big O terms.

Option 1: f(n) = O(g(n)) because as n approaches infinity, the growth rate of f(n) is slower than the growth rate of g(n), indicating that f(n) can be bounded above by g(n) with a constant multiplier.
- This aligns with the standard growth rates: logarithmic functions grow slower than any root function. Since log n grows much more slowly than sqrt(n), there exist constants C > 0 and n0 such that for all n ≥ n0, log n ≤ C * sqrt(n). This makes f(n) = O(g(n)). The reasoning reflects the intuition that a log function is dominated by a square-root function in the long run.

Option 2: g(n) = O(f(n)) because as n approaches infinity, the growth rate of g(n) is slower than the growth rate of f(n), indicating that g(n) can be bounded above by f(n) with a constant multiplier.
- This is inaccurate because sqrt(n) grows faster than log n, not slower. In formal terms, there is no constants C and n0 such that sqrt(n) ≤ C * log n for all n ≥ n0, because the ratio sqrt(n)/log n tends to infinity. Therefore, g(n) is not O(f(n)).

Option 3: Both f(n) = O(g(n)) and g(n) = O(f(n)) because they can each be bounded above by the other with appropriate constants for sufficiently large n.
- This would imply that f and g have the same asymptotic growth rate, which is false here. As discussed, log n grows strictly slower than sqrt(n), so f(n) is O(g(n)) but g(n) is not O(f(n)). The statement of mutual Big-O is incorrect for these particular functions.

In summary, the reasoning for each option hinges on comparing growth rates: logarithmic growth versus square-root growth. The key takeaway is that log n grows slower than sqrt(n), making f(n) = O(g(n)) true, while the reverse and mutual-embedding claims do not hold.

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