# Exercise 4.4.1

Use a reccursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 3T(\lfloor n/2 \rfloor) + n$. Use the substitution method to verify your answer.

Let's just ignore the floor, it makes no difference whatsoever.

The tree is of depth $\lg{n}$ and has $\Theta(3^{\lg{2}}) = \Theta(2^{\lg{n}})$ leaves. Thus:

\begin{aligned} T(n) &= \sum_{i=0}^{\lg{n}-1}\Big(\frac{3}{2}\Big)^i n + \Theta(n^{\lg3}) \\ &= n\frac{(3/2)^{\lg{n}} - 1}{(3/2) - 1} + \Theta(n^{\lg3}) \\ &= n\Theta(n^{\lg3 - 1}) + \Theta(n^{\lg3}) \\ &= \Theta(n^{\lg3}) \end{aligned}

Let's use substitution. We guess $T(n) \le cn^{\lg3} + 2n$ (and drop the floor):

\begin{aligned} T(n) & \le 3c(n/2)^{\lg3} + 2n/2 + n \\ & \le cn^{\lg3} + 2n \\ & = \Theta(n^{\lg3}) \end{aligned}