如果我们考虑C_1和C_2，使得C_1
(n^C_2)*log(n) grows faster than (n^C_1)
这是因为(n^C_1)增长速度比(n^C_2)慢(显然)。
also, for values of n larger than 2 (for log in base 2), log(n) grows faster than
1.
in fact, log(n) is asymptotically greater than any constant C,
because log(n) -> inf as n -> inf
if both (n^C_2) is asymptotically than (n^C_1) AND log(n) is asymptotically greater
than 1, then we can certainly say that
(n^2)log(n) has greater complexity than (n^1.5)
我们认为log(n)是一个“缓慢增长”的函数，但它仍然比1增长得更快，这是关键所在。
coder101在评论中提出了一个有趣的问题，实际上是：
is n^e = Ω((n^c)*log_d(n))?
where e = c + ϵ for arbitrarily small ϵ
让我们做一些代数题。
n^e = (n^c)*(n^ϵ)
so the question boils down to
is n^ϵ = Ω(log_d(n))
or is it the other way around, namely:
is log_d(n) = Ω(n^ϵ)
为了做到这一点，让我们找到一个满足 n^ϵ > log_d(n) 的 ϵ 值。
n^ϵ > log_d(n)
ϵ*ln(n) > ln(log_d(n))
ϵ > ln(log_d(n)) / ln(n)
因为我们确切地知道，这是事实。
ln(n) * c > ln(ln(n)) (1)
as n -> infinity
We can say that, for an arbitrarily small ϵ, there exists an n large enough to
satisfy ϵ > ln(log_d(n)) / ln(n)
because, by (1), ln(log_d(n)) / ln(n) ---> 0 as n -> infinity.
有了这个知识，我们可以说：
is n^ϵ = Ω(log_d(n))
for arbitrarily small ϵ
which means that
n^(c + ϵ) = Ω((n^c)*log_d(n))
for arbitrarily small ϵ.
一个通俗易懂的说法是：
n^1.1 > n * ln(n)
for some n
also
n ^ 1.001 > n * ln(n)
for some much, much bigger n
and even
n ^ 1.0000000000000001 > n * ln(n)
for some very very big n.
