# Exercise 15.2.3

Use the substitution method to show that the solution to the recurrence (15.6) is $\Omega(2^n)$.

The recurrence is:

$$P(n) = \begin{cases} 1 & \text{if } n = 1, \\ \sum_{k=1}^{n-1} P(k) P(n - k) & \text{if } n \ge 2 \end{cases}$$

Our guess is that $P(m) \ge c \cdot 2^m$.

Trying to quantify $P(n)$, we get:

\begin{aligned} P(n) &= \sum_{k=1}^{n-1} P(k) P(n - k) \\ &\ge \sum_{k=1}^{n-1} c \cdot 2^k \cdot c \cdot 2^{n-k} \\ &= c^2 \sum_{k=1}^{n-1} 2^k 2^{n-k} \\ &= c^2 \sum_{k=1}^{n-1} 2^n \\ &= c^2 (n - 1) 2^n \\ &\ge c \cdot 2^{n} \end{aligned}

(when $c \ge 1$ and $n \ge 2$)