What is O(log(n!)) and O(n!) and Stirling Approximation

Solution 1:

O(n!) isn't equivalent to O(n^n). It is asymptotically less than O(n^n).

O(log(n!)) is equal to O(n log(n)). Here is one way to prove that:

Note that by using the log rule log(mn) = log(m) + log(n) we can see that:

log(n!) = log(n*(n-1)*...2*1) = log(n) + log(n-1) + ... log(2) + log(1)

Proof that O(log(n!)) ⊆ O(n log(n)):

log(n!) = log(n) + log(n-1) + ... log(2) + log(1)

Which is less than:

log(n) + log(n) + log(n) + log(n) + ... + log(n) = n*log(n)

So O(log(n!)) is a subset of O(n log(n))

Proof that O(n log(n)) ⊆ O(log(n!)):

log(n!) = log(n) + log(n-1) + ... log(2) + log(1)

Which is greater than (the left half of that expression with all (n-x) replaced by n/2:

log(n/2) + log(n/2) + ... + log(n/2) = floor(n/2)*log(floor(n/2)) ∈ O(n log(n))

So O(n log(n)) is a subset of O(log(n!)).

Since O(n log(n)) ⊆ O(log(n!)) ⊆ O(n log(n)), they are equivalent big-Oh classes.

Solution 2:

By Stirling's approximation,

log(n!) = n log(n) - n + O(log(n))

For large n, the right side is dominated by the term n log(n). That implies that O(log(n!)) = O(n log(n)).

More formally, one definition of "Big O" is that f(x) = O(g(x)) if and only if

lim sup|f(x)/g(x)| < ∞ as x → ∞

Using Stirling's approximation, it's easy to show that log(n!) ∈ O(n log(n)) using this definition.

A similar argument applies to n!. By taking the exponential of both sides of Stirling's approximation, we find that, for large n, n! behaves asymptotically like n^(n+1) / exp(n). Since n / exp(n) → 0 as n → ∞, we can conclude that n! ∈ O(n^n) but O(n!) is not equivalent to O(n^n). There are functions in O(n^n) that are not in O(n!) (such as n^n itself).