如何确定界定n!n³的常数表达式？含T(n)=(n!n+n³)(n²+7logn)背景

To find a constant expression that bounds (n!n^3) (i.e., an upper bound in the context of Big O notation), we can break this down using basic properties of factorials and asymptotic growth:

Step 1: Clarify the Goal #

We need to identify a function (f(n)) and constants (C > 0), (n_0 \geq 1) such that for all (n \geq n_0):
[n!n^3 \leq C \cdot f(n)]
This satisfies the formal definition of Big O, where (f(n)) is our bounding expression and (C) is the constant multiplier.

Step 2: Simple Upper Bound Using (n^n) #

The factorial function (n! = 1 \cdot 2 \cdot \dots \cdot n) is easily upper-bounded by (n^n) (every term in the product is at most (n)):
[n! \leq n^n \quad \text{for all } n \geq 1]

Multiply both sides by (n^3) to get our target expression:
[n!n^3 \leq n^n \cdot n^3 = n^{n+3}]

Here, we can use (C = 1) and (n_0 = 1) since this inequality holds for all (n \geq 1). This gives us the straightforward bound (n!n^3 = O(n^{n+3})).

Step 3: Tighter Bound with Stirling's Approximation #

For a more precise upper bound, we can leverage Stirling's formula, which approximates factorials asymptotically:
[n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n]
This means there exists a constant (K > 0) such that for sufficiently large (n):
[n! \leq K \cdot \sqrt{2\pi n} \left(\frac{n}{e}\right)^n]

Taking (K = e^{1/12} \approx 1.087) (to account for the small error term in Stirling's expansion), we get a safe upper bound:
[n! \leq 3 \cdot \sqrt{n} \left(\frac{n}{e}\right)^n]
(The constant 3 works because (\sqrt{2\pi} \cdot e^{1/12} \approx 2.72), so rounding up gives a buffer.)

Multiply by (n^3) to bound our target expression:
[n!n^3 \leq 3 \cdot \sqrt{n} \left(\frac{n}{e}\right)^n \cdot n^3 = 3 \cdot n^{3.5} \left(\frac{n}{e}\right)^n]

For simplicity, we can use (n^{3.5} \leq n^4) for all (n \geq 1), leading to:
[n!n^3 \leq 3 \cdot n^4 \left(\frac{n}{e}\right)^n]

This gives us the tighter bound (n!n^3 = O\left(n^{3.5} \left(\frac{n}{e}\right)^n\right)) (or (O\left(n^4 \left(\frac{n}{e}\right)^n\right)) for easier writing).

Key Takeaways #

• The simplest bound is (n^{n+3}) with a constant multiplier of 1, valid for all (n \geq 1).
• Stirling's approximation gives a more precise bound that reflects the true asymptotic growth rate of (n!) better than (n^n).
• Both bounds satisfy the Big O definition: the ratio of (n!n^3) to the bound remains bounded by a constant as (n) approaches infinity.

内容的提问来源于stack exchange，提问作者Simon Leung