Use substitute algorithm to prove T(n) =4T(n/2) + n^2 compact lower bound

Question

how to use substitute algorithm to prove T(n) =4T(n/2) + n^2 compact lower bound. this is a algorithm of my class home work,but i dont know how to solve it.

Answer 1

Let's assume that $n$ is a power of 2 (otherwise the recurrence isn't really defined all the way), with a base case at $n=1$. Then $$ \begin{align*} T(n) &= n^2 + 4T(n/2) \\ &= n^2 + 4(n/2)^2 + 16T(n/4) \\ &= n^2 + 4(n/2)^2 + 16(n/4)^2 + 64T(n/8) \\ &= \cdots \end{align*} $$ Note that $n^2 = 4(n/2)^2 = 16(n/4)^2$. If $n = 2^m$ then after $m$ steps we will reach $4^mT(n/2^m) = 4^mT(1)$, with $m$ copies of $n^2$ tagging along. Therefore $$ T(n) = mn^2 + 4^mT(1) = n^2\log_2 n + n^2 T(1). $$