What is the simplified Big-O notation for
3
⋅
𝑁
3
+
𝑂
(
2
𝑁
2
)
?

Question

What is the simplified Big-O notation for 
3
⋅
𝑁
3
+
𝑂
(
2
𝑁
2
)
 ?

BlackTom AI · Accepted Answer

Because the cubic term grows faster than the quadratic term, the expression simplifies to O(N^3).

1) Relevant knowledge points and concepts
- Big-O notation describes an asymptotic upper bound and ignores constant factors and lower-order terms.
- If f(N) = aN^k + lower-order terms, the dominating term is the one with largest exponent k; f(N) ∈ O(N^k).
- O(2N^2) denotes some function bounded above by a constant multiple of N^2 for large N; it is itself O(N^2).

2) Complete reasoning process (step by step)
Let the expression be F(N) = 3·N^3 + O(2N^2). Interpret O(2N^2) as a function g(N) for which there exist constants C > 0 and N0 such that for all N ≥ N0, g(N) ≤ C·2N^2 (equivalently g(N) ≤ C' N^2 for some C').
For simplicity, take a specific bound g(N) ≤ K·N^2 for large N.

3) Intermediate calculation steps
For N ≥ 1 we have N^2 ≤ N^3. Therefore g(N) ≤ K·N^2 ≤ K·N^3.
So for N sufficiently large:
F(N) = 3N^3 + g(N) ≤ 3N^3 + K·N^3 = (3 + K)·N^3.
This shows F(N) is bounded above by a constant multiple of N^3 for large N, hence F(N) ∈ O(N^3).

4) Why the correct answer is correct
The cubic term 3N^3 dominates the quadratic term asymptotically. Constants (like 3 or 2) are irrelevant in Big-O. Therefore the simplified Big-O is O(N^3).

5) Why the other options are incorrect
- O(N^2): too small; 3N^3 grows faster than any constant·N^2, so F(N) is not bounded by N^2 for large N.
- O(3⋅N^3): equivalent to O(N^3) (constants are ignored), but not the simplest conventional form; the simplest form is O(N^3).
- O(N^3+N^2): correct in a literal sense but not simplified; since N^3 dominates N^2, this reduces to O(N^3).

6) Conclusion
The simplified Big-O notation is O(N^3), which matches the provided answer O(N^3).

What is the simplified Big-O notation for 3 ⋅ 𝑁 3 + 𝑂 ( 2 𝑁 2 ) ?单项选择题

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