Possible Duplicate:
Solving or approximating recurrence relations for sequences of numbers
I know that the solution for $T(n) = 2 T(n/2) + O(n)$ is $ T(n) = O(n \log(n))$
But how do you get to that point? I don't understand when it says put t into the equation repeatedly until it drops out...
Any help?
It means "open" the recursion.
For simplicity - denote $O(n)$ as $c\cdot n$:
$$ \begin{align} T(n) &= 2T(n/2) + cn \\ &= 2(2T(n/4) + cn/2) + cn\\ &= 2 (2 (2T(n/8) + cn/4) + cn/2) + cn\\ & \vdots \end{align}$$
It might give you intuition, but it is NOT a proof. To prove it, you will need mathematical induction or the master theorem.
Proving with induction (assuming O(n) component is n for simplicity):
Claim: T(n) <= n*logn + n
Base:
T(1) = 1 (assumption)
Assumption: the claim is correct for all k < n.
Proof:
T(n) = 2T(n/2) + n = (assumption) <= 2* (n/2 * log(n/2) + n/2) + n
= n*log(n/2) + 2n = n*(log(n)-log(2)) + 2n = (assuming base 2 for log)
= n*(log(n) -1 ) + 2n = nlogn -n + 2n = n*logn +n
QED
