Is log(n!) = Θ(n·log(n))?

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Determining if log(n!) Equals Θ(n·log(n))

Understanding the relationship between logarithmic functions and factorials is fundamental in computer science, especially when analyzing algorithm complexities. In this context, determining whether log(n!) is in the order of Θ(n·log(n)) helps in comprehending the efficiency of algorithms that involve factorial computations.

Understanding log(n!)

The expression log(n!) refers to the logarithm of the factorial of n. The factorial of a number n (denoted as n!) is the product of all positive integers up to n. Mathematically, it is expressed as:

[ n! = 1 \times 2 \times 3 \times \dots \times n ]

Taking the logarithm of both sides gives:

[ \log(n!) = \log(1) + \log(2) + \log(3) + \dots + \log(n) ]

Understanding Θ(n·log(n))

The notation Θ(n·log(n)) represents the asymptotic tight bound of a function. It means that the function grows at the same rate as n multiplied by the logarithm of n for large values of n. In simpler terms, Θ(n·log(n)) describes a function whose growth rate is proportional to n·log(n).

Is log(n!) = Θ(n·log(n))?

Yes, log(n!) is in Θ(n·log(n)). This can be demonstrated using Stirling's Approximation, which provides an approximate value for factorials and is especially useful for large n.

Applying Stirling's Approximation

Stirling's Approximation states that:

[ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n ]

Taking the logarithm of both sides:

[ \log(n!) \approx \log\left(\sqrt{2\pi n} \left( \frac{n}{e} \right)^n\right) ] [ \log(n!) \approx \log(\sqrt{2\pi n}) + \log\left( \left( \frac{n}{e} \right)^n \right) ] [ \log(n!) \approx \frac{1}{2}\log(2\pi n) + n\log\left( \frac{n}{e} \right) ] [ \log(n!) \approx \frac{1}{2}\log(2\pi) + \frac{1}{2}\log(n) + n\log(n) - n\log(e) ] [ \log(n!) \approx n\log(n) - n + \frac{1}{2}\log(n) + \text{constants} ]

As n becomes large, the dominant term in the expression is ( n\log(n) ). The other terms grow slower compared to ( n\log(n) ) and thus become insignificant in the asymptotic analysis.

Therefore:

[ \log(n!) = \Theta(n\log(n)) ]

Conclusion

Through Stirling's Approximation, it is clear that log(n!) grows proportionally to n·log(n) as n becomes large. This relationship is essential when analyzing the time complexity of algorithms that involve factorial computations or related logarithmic operations.