f(n) 的形式 vs 判定形势
但，此题型过于简单，一般不出现在考题中。
Extended:
Let's set n = 2^m, so m = log(n)
T(n) = 2*T(n^(1/2)) + 1 =>
T(2^m) = 2*T(2^(m/2)) + 1 =>
S(M) = 2*S(M/2)) + 1
通过变量替换，变为了熟悉的、主定理能解决的 形式 => S(m) ~ O(m)
故，S(m) = T(log(n)) ~ O(log(n))
关键点：match 主定理（3）的条件。
a*f(n/b) <= c*f(n) ->
3*f(n/4) <= c*f(n) ->
3*(n/4)*log(n/4) <= c*n*log(n) ->
3/4 * n*log(n/4) <= c*n*log(n)
可见，有c<1时，是满足的。
答案就是O(nlogn)
When f(x) = x*sqrt(x+1)
这里的常数1是个tricky.
去掉它，x^(3/2)
变为x，sqrt(2)*[x^(3/2)]
可见，f(x)~Θ(x^(3/2))
.
so we have a geometric series and thus，
for our upper bound.
In a similar way, we could count just the contribution of the leftmost branches and conclude that .
n)
Extended:
Recurrence Relation T(n)=T(n/8)+T(n/4)+lg(n)
太难：Solution.
(a) T(n) = T(n-1) + cn link
(b) T(n) = T(n-1) + log(n) link
(c) T(n) = T(n-1) + n^2 link
(d) T(n) = T(n-1) + 1/n link
(a) 递归展开，探寻规律：
At the end we use T(2)=T(1)+2c=1+2c. We conclude that
.
可见是O(n^2)
(b) 递归展开，探寻规律：
T(n) = T(n - 1) + log n
= T(n - 2) + log (n - 1) + log n
= T(n - 3) + log (n - 2) + log (n - 1) + log n
= ...
= T(0) + log 1 + log 2 + ... + log (n - 1) + log n
= T(0) + log n!
According to Stirling's formula, 可见是O(n*log(n))
(c) 递归展开，探寻规律：
2
Or simply,
O(n^3)
(d) 这里是个调和级数
This is the nth harmonic number, Hn = 1 + 1/2 + 1/3 + ... + 1/n.
"所有调和级数都是发散于无穷的。但是其拉马努金和存在，且为欧拉常数。"
In fact, Hn = ln(n) + γ + O(1/n)
Hint：比较难，需换两次元。