To determine whether $\log(n!) \in \Theta(n \log n)$, let's analyze the asymptotic behavior of $\log(n!)$ and see if it can be bounded both above and below by $n \log n$ up to constant factors.

Step 1: Approximate $\log(n!)$ Using Stirling's Approximation

Stirling's approximation gives us: $n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$ Taking the logarithm of both sides, we get: $\log(n!) \approx \log(\sqrt{2 \pi n}) + \log \left( \frac{n}{e} \right)^n$ Breaking this down further: $\log(n!) \approx \frac{1}{2} \log(2 \pi n) + n \log \left( \frac{n}{e} \right)$ This simplifies to: $\log(n!) \approx \frac{1}{2} \log(2 \pi n) + n \log n - n$ For large $n$, the $n \log n$ term dominates, and we can approximate: $\log(n!) \approx n \log n$

Step 2: Establish Big-Theta Bounds

For $\log(n!)$ to be in $\Theta(n \log n)$, it needs to be asymptotically bounded both above and below by $n \log n$.

1. Upper Bound: From Stirling's approximation, we know that $\log(n!) \approx n \log n$, so there exists a constant $c_1$ such that: $\log(n!) \leq c_1 n \log n$ for sufficiently large $n$.

2. Lower Bound: Similarly, $\log(n!) \approx n \log n$ implies there exists a constant $c_2$ such that: $\log(n!) \geq c_2 n \log n$ for sufficiently large $n$.

Since both bounds hold, we can conclude: $\log(n!) \in \Theta(n \log n)$

Final Answer

Yes, $\log(n!) \in \Theta(n \log n)$.

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Would you like further details or have any questions?

Here are some related questions:

1. How does Stirling’s approximation work, and why is it useful?
2. How can we apply asymptotic notation for other factorial functions?
3. What other functions can be bounded with $\Theta(n \log n)$?
4. How do logarithmic functions compare with other polynomial functions in asymptotic analysis?
5. Can we use Big-O instead of Big-Theta for similar approximations?

Tip: Stirling’s approximation is very helpful in estimating the growth rates of factorials and is a valuable tool in combinatorics and algorithm analysis.