'What is O(log(n!)) and O(n!) and Stirling Approximation
What is O(log(n!)) and O(n!)? I believe it is O(n log(n)) and O(n^n)? Why?
I think it has to do with Stirling Approximation, but I don't get the explanation very well.
Could someone correct me if I'm wrong (about O(log(n!) = O(n log(n)))? And if possible the math in simpler terms? I don't think I will need to prove that in reality I just want an idea of how this works.
Solution 1:[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:[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).
Sources
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Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Austin |
| Solution 2 |