# Exercise 4.4.2

Use a reccursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n/2) + n^2$. Use the substitution method to verify your answer.

Each level of the tree is $n^2/4^i$, there are $\lg{n}$ levels and 1 leaf. Thus:

$$T(n) = \sum_{i=0}^{\lg{n}-1}\Big(\frac{1}{4}\Big)^i n^2 + 1 < n^2 \sum_{i=0}^{\infty}\Big(\frac{1}{4}\Big)^i + 1 = n^2 \frac{1}{1-1/4} + 1 = \Theta(n^2)$$

We guess $T(n) \le cn^2$ \begin{align} T(n) & \le c(n/2)^2 + n^2 \\ & \le cn^2/4 + n^2 \\ & \le (c/4 + 1)n^2 \qquad (c > 4/3) \\ & \le cn^2 \end{align}