Is O((log n)!) polynomial?

Question

I wanna find O((log n)!) . I think O((log n)!) and O(n!) are equal! Because I think when n is infinite (log n)! and n! are equal. Is it true? If yes how can I show that? If not is the O((log n)!) polynomial?

Answer 1

I think your follow up question is whether (logn)! is polynomially bounded. It is obviously not a polynomial itself. Stirling's Approximation gives us

n!≤en^[n+1/2]*e^(−n)

So,

(log n)!≤e(log n)^[1/2+log n]*e^(−log n)

Now (log n)^log n=(e^loglogn)^logn=e^[(logn)⋅(loglogn)]

So, the order of growth is approximately e^[(logn)(loglogn)−logn] =n^[(loglogn)−1]

This is unfortunately not bounded by any polynomial, since loglogn will eventually exceed any positive integer.

For example, compare (log n)! with n^2 .

At n=e^10,(log n)!=3480 , while (e^10)2≈4.85×108

At n=e^100,(log n)!≈10157 , while e^200≈1086

Answer 2

Since the proper math has already been done, let me add a more intuitive explanation of why O(log(n)!) > O(n^c) for any given c . We will assume the logarithm is base 2, and for simplicity choose c as 10. (The argument would work just as well with different numbers of general values).

So, why will log(n)! grow ever larger than n^10 ? Let's take a look at both functions' values at the powers of 2, more specificaly, how much greater they grew compared to the last power of 2. ( n = 2^p from now on)

log(2^p)! = p * log(2^(p-1))! , (2^p)^10 = 2^10 * (2^(p-1))^10 . This may seem complicated, but it tells us that the log(n)! function will multiply it's value at each p th power of 2 by p , but n^10 will multiply it's value only by 1024 , so the log(n)! will grow ever larger eventually.

Also, log(n)! grows slower than any exponential, similar argument can be made by observing how much both function multiply their value when n grows by 1.

Answer 3

lets get back to some basic mathematics:
we know that if log a > log b, then a>b :(log base is greater than 1)
click here to get more on this

Now we know that log(N!)=NLogN ( see here for proof )

and holding same argument,we get, log((log N)!)=logN logLogN
Since,
log(N!) is of polynomial degree ,and log((log N)!) is of logarithmic order,
clearly, O(N!) >O((logN)!)
hope this helps.

Answer 4

By using Stirling's approximation stating that $ n! ~ \\sqrt{2 \\pi n} \\frac{n}{e}^n (1 + O(frac{1}{n})) $, and using the same formula for $ \\log{n} $, one can check that for $ n! $ the leading factor is $ n ^ n $ while for $ \\log{n}! $ the same factor becomes $ \\log{n} ^ \\log{n} $. By using the fact that $ \\ln{n} \\leq n - 1 $, I believe you can easily show that $ O(\\log{n}!) $ is less than $ O(n!) $.