Exercise 2.3.3

Use mathematical induction to show that when $n$ is an exact power of 2, the solution of the recurrence

$$ T(n) = \begin{cases} 2 & \text{if } n = 2, \\ 2T(n/2) + n & \text{if } n = 2^k, \text{for } k > 1. \end{cases} $$

is $T(n) = n\lg{n}$

Boy, this is going to be a lot of $\LaTeX$.

First, let's establish a function on which we're going to perform induction:

$$ F(k) = T(2^k) $$

We want to prove that:

$$ F(k) = 2^k \lg{2^k} $$

Base. It is simple enough:

$$ F(1) = T(2) = 2 = 2\lg2 = 2^1\lg{2^1} $$

Step. We assume that:

$$ F(k) = 2^k \lg{2^k} $$

We prove it for $k + 1$:

$$ \begin{align} F(k + 1) & = T(2^{k+1})= 2T(\frac{2^{k+1}}{2}) + 2^{k+1} = \\ & = 2T(2^k) + 2^{k+1} = 2 \cdot 2^k \lg{2^k} + 2^{k+1} = \\ & = 2^{k+1}(\lg{2^k} + 1) = 2^{k+1}(\lg{2^k} + \lg2) = \\ & = 2^{k+1}\lg{2^{k+1}} \end{align} $$