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How would I rank the following functions by order of their growth; i.e satisfying g1 = Ω(g2), g2 = Ω(g3), etc.
I am having trouble starting this, could some please get me started? Or possibly help me understand this a little more?
Here are the functions I need to rank:
(sqrt(2))^(logn), n^2, n!, (logn)!, (3/2)^n, n^3, log^(2) n, log(n!), 2^(2n), n^(1/logn), lnlnn, n*2^n, n^(loglogn), lnn, 1, 2^(logn), (logn)^(logn), e^n, 4^(logn), (n+1)!, sqrt(logn), 2^sqrt(2logn), n, 2^n, nlogn, 2^(2^(n+1)), 2^(100^(100))
Some are trivial: n, n^2, n^3, but after that I am not sure how they rank. Could someone please help me? I'm not looking for an answer, but some understanding?
2^(2^(n+1)) seems to be the biggest of the functions above.
Some of these you can get by playing with exponents. For example, (sqrt(2))^(logn) = 2^[(1/2)*lgn] = sqrt(n)
also (logn)^(logn) = 2^(lglgn*lgn) = n^lglgn, which is bigger that n to any constant power, but smaller than all the plain exponentials (like 2^n)
and 4^lgn = 2^(2*lgn) = n^2
and n^(1/lgn) = 2^[lgn*(1/lgn)] = 2, so it is a constant, just like the very large constant 2^(100^(100))
(logn)! is a tricky one, using Sterling's formula n! ~ sqrt(2pi*n)(n/e)^n, you can see that this is a little less than lgn^lgn
log(n!) is much smaller, it is nlgn
keep in mind that for exponentials, constants matter, that is:
(3/2)^n << 2^n << n2^n << e^n << 2^(2n) = 4^n << n! << (n+1)!
Hmmm, I think the only other tricky one left is 2^sqrt(2lgn). I do not know a way to simplify this, but you can compare it to other functions, by taking the lg of each. For example, you can see that lgn << 2^sqrt(2lgn) << sqrt(n), because the lg of them are:
lglgn < sqrt(2lgn) < (1/2)lgn
I think that should allow you to rank the rest. Good luck!