$T(n) = 2T(n/2) +n\log n$ - Algorithms

Question

According to this http://en.wikipedia.org/wiki/Introduction_to_Algorithms
$T(n)=2T(n/2)+n\log n$ is not case 3 of Master Theorem, can someone explain me why?

And which case of master theorem it is?

Answer 1

It is unimportant to know if something is a case of Master Theorem or not. What is important, though, is solving the problem at hand. In your case, set $n=2^k$. We then get $$T(2^k) = 2T(2^{k-1}) + k2^k \log(2)$$ Defining $T(2^k)$ as $g(k)$, we get \begin{align} g(k) & = 2g(k-1) + k2^{k-1} \log(2) = 2(2g(k-2) +(k-1)2^{k-1}\log(2))+ k2^{k-1} \log(2)\\ & = 4g(k-2) + (k-1)2^k \log(2) + k 2^k \log(2)\\ & = 4(2g(k-3) + (k-2)2^{k-2}\log(2)) + (k-1)2^k \log(2) + k 2^k \log(2)\\ & = 8g(k-3) + (k-2)2^{k}\log(2) + (k-1)2^k \log(2) + k 2^k \log(2)\\ & = 2^k g(0) + \sum_{l=0}^kl \cdot 2^k \log(2)\\ & = 2^k g(0) + k(k-1)2^{k-1} \log(2)\\ & = ng(0) + \dfrac{n \log(n) (\log(n)-1)}{2} \log(2) \end{align}