Find the asymptotic tight bound in
$$ T(n) = 4T\left(\frac{n}{2}\right) + n^{2}\log n. $$
where $ \log n= \log _{2}n $ and $T(1) = 1$.
I should solve this using all three common methods: iteration, master theorem and substitution.
It is highly likely that this kind of recurrence equation will be in my test in two days. Thank you very much in advance!
On
We can compute exact formulas for your recurrence and not just at powers of two. Suppose we first solve the recurrence $$T(n) = 4 T(\lfloor n/2\rfloor) + n^2 (1+\lfloor \log_2 n\rfloor)$$ with $T(0)=0.$
Then let $$n= \sum_{k=0}^{\lfloor \log_2 n\rfloor} d_k 2^k$$ be the binary representation of $n$ and unroll the recursion to get the following exact formula for $T(n):$ $$T(n) = \sum_{j=0}^{\lfloor \log_2 n\rfloor} 4^j (1+\lfloor \log_2 n\rfloor -j) \left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} d_k 2^{k-j} \right)^2 \\= \sum_{j=0}^{\lfloor \log_2 n\rfloor} (1+\lfloor \log_2 n\rfloor -j) \left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} d_k 2^k \right)^2.$$
Now to get an upper bound on this consider a string of one digits, which gives $$T(n)\le \sum_{j=0}^{\lfloor \log_2 n\rfloor} (1+\lfloor \log_2 n\rfloor -j) \left(\sum_{k=j}^{\lfloor \log_2 n\rfloor} 2^k \right)^2 = \sum_{j=0}^{\lfloor \log_2 n\rfloor} (1+\lfloor \log_2 n\rfloor -j) (2^{1+\lfloor \log_2 n\rfloor}-2^j)^2 \\ = \sum_{j=1}^{1+\lfloor \log_2 n\rfloor} j (2^{1+\lfloor \log_2 n\rfloor}-2^{1+\lfloor \log_2 n\rfloor-j})^2 =2^{2(1+\lfloor \log_2 n\rfloor)} \sum_{j=1}^{1+\lfloor \log_2 n\rfloor} j \times (1-2^{-j})^2.$$ The dominant term in the sum is $j$ which contributes $\sum_{j=1}^{1+\lfloor \log_2 n\rfloor} j$ to give $$\frac{1}{2} 2^{2(1+\lfloor \log_2 n\rfloor)} (1+\lfloor \log_2 n\rfloor)(2+\lfloor \log_2 n\rfloor).$$ For a lower bound consider a one followed by zeros to obtain $$T(n)\ge \sum_{j=0}^{\lfloor \log_2 n\rfloor} (1+\lfloor \log_2 n\rfloor -j) 2^{2\lfloor \log_2 n\rfloor} = 2^{2\lfloor \log_2 n\rfloor}\sum_{j=1}^{1+\lfloor \log_2 n\rfloor} j.$$ Simplifying the sum term we get $$\frac{1}{2} 2^{2\lfloor \log_2 n\rfloor} (1+\lfloor \log_2 n\rfloor)(2+\lfloor \log_2 n\rfloor)$$ which is exact this time compared to the upper bound which was asymptotic.
The conclusion is that by taking the two bounds together we obtain $$T(n)\in\Theta\left(2^{2\lfloor \log_2 n\rfloor} (1+\lfloor \log_2 n\rfloor)(2+\lfloor \log_2 n\rfloor)\right) = \Theta\left(n^2 \times \log^2 n\right).$$
A variety of Master Theorem computations can be found at this MSE link. In fact there is another very similar solution to this problem at this MSE link II.
You're likely to get a comment about asking multiple questions as one big one, but to get you started, I'll do an iterative expansion.
Let $n=2^k$, since strictly speaking those are the only values for which your recurrence makes sense. Then your recurrence takes the form $$ T(2^k) = 2^2T(2^{k-1})+2^{2k}k $$ Now let's iterate $$\begin{align} T(2^k) &= 2^2[T(2^{k-1})]+2^{2k}k \\ &= 2^2[2^2T(2^{k-2})+2^{2(k-1)}(k-1)]+2^{2k}k\\ &\qquad= 2^4T(2^{k-2})+2^{2k}(k-1)+2^{2k}k\\ &= 2^4[2^2T(2^{k-3})+2^{2(k-2)}(k-2)]+2^{2k}(k-1)+2^{2k}k\\ &\qquad= 2^6T(2^{k-3})+2^{2k}(k-2)+2^{2k}(k-1)+2^{2k}k \end{align}$$ and the general pattern appears to be $$\begin{align} T(2^k) &= 2^{2j}T(2^{k-j})+2^{2k}(k-j+1)+2^{2k}(k-k+2)+\cdots+2^{2k}k\\ &= 2^{2j}T(2^{k-j})+2^{2k}((k-j+1)+(k-j+2)+\cdots k) \end{align}$$ assuming we've correctly guessed the pattern, we let $j=k$ and obtain $$ T(2^k)= 2^{2k}T(2^0)+2^{2k}(1+2+3+\cdots k)=(2^k)^2+(2^k)^2\frac{k(k+1)}{2} $$ Finally, recall that we set $n=2^k$ so $k=\lg n$ so we have the solution $$ T(n) =n^2+n^2\frac{\lg n(\lg n+1)}{2}=\Theta(n^2\lg^2n) $$ Now all you have to do is complete the remaining two parts. (grins)