比较两个函数的增长率。 （棘手）

Question

I need to compare the growth rate of the following functions:

f(n)=2^n and g(n)=n^log(n) (when n approaches positive infinity).

Is this even possible?

Answer 1

Let n = 2^k . We have:

2^n = 2^(2^k)
n^log(n) = (2^k)^log(2^k) = (2^k)^(k log 2)
         = 2^(k^2 log 2)

Now compare 2^k to k^2 log 2 . This is a basic comparison: 2^k is bigger for all large enough k .

Answer 2

Taking log (base 2) for both the functions, we get log(f(n)) = n where log(g(n)) = (log(n))^2 .

Now, (log(n))^2 = o(n) and log being a monotonically increasing function, we have

g(n) = o(f(n)) , ie, f(n) grows much faster for large values of n .

Here is another way to prove it more rigorously:

Let L = lim{n->inf} g(n) / f(n) = lim{n->inf} n^(log(n))/2^n .

Hence log (L) = lim{n->inf} log^2(n) - n

  ` = lim{n->inf} n*(log^2(n)/n) - 1)`

  ` = lim{n->inf} (n) * lim{n->inf} (log^2(n)/n) - 1)`

  ` = lim{n->inf} (n) * (0-1)`

  ` = lim{n->inf} (-n) = -inf`

=> L = 2^(-inf) = 0

According to the alternative definition of o(n) (small o , see here: https://en.wikipedia.org/wiki/Big_O_notation ),

L = lim{n->inf} g(n) / f(n) = 0

=> g(n) = o(f(n)) .

Here are the figures comparing f(n) and g(n) growth in original and in log scale: