Update 4.4.md

Corrections made to exercise 4.4-4 solution.

docs/Chap04/4.4.md

@@ -109,29 +109,35 @@
 
 - The subproblem size for a node at depth $i$ is $n - i$.
 
-    Thus, the tree has $n$ levels and $2^{n - 1}$ leaves.
+    Thus, the tree has $n + 1$ levels ($i = 0, 1, 2, \dots, n$) and $2^n$ leaves.
 
     The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, n - 1$, is $2^i$.
 
+    The $n$-th level has $2^n$ leaves each with cost $\Theta(1)$, so the total cost of the $n$-th level is $\Theta(2^n)$.
+
+    Adding the costs of all the levels of the recursion tree we get the following:  
+
     $$
     \begin{aligned}
-    T(n) & = \sum_{i = 0}^{n - 1} 2^i \\\\
-         & = \frac{2^n - 1}{2 - 1} \\\\
-         & = 2^n - 1 \\\\
+    T(n) & = \sum_{i = 0}^{n - 1} 2^i + \Theta(2^n)\\\\
+         & = \frac{2^n - 1}{2 - 1} + \Theta(2^n)\\\\
+         & = 2^n - 1 + \Theta(2^n)\\\\
          & = \Theta(2^n).
     \end{aligned}
     $$
 
-- We guess $T(n) \le c2^n + n$,
+- We guess $T(n) \le c2^n - d$,
 
     $$
     \begin{aligned}
-    T(n) & \le 2 \cdot c2^{n - 1} + (n - 1) + 1 \\\\
-         & =   c2^n + n \\\\
-         & =   O(2^n).
+    T(n) & \le 2(c2^{n - 1} - d) + 1 \\\\
+         & = c2^n - 2d + 1 \\\\
+         & \le c2^n - d
     \end{aligned}
     $$
 
+    Where the last step holds for $d \ge 1$. Thus $T(n) = O(n^2)$.
+
 ## 4.4-5
 
 > Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n - 1) + T(n / 2) + n$. Use the substitution method to verify your answer.