# Problem 12.4

## Number of different binary trees

Let $b_n$ denote the number of different binary trees with $n$ nodes. In this problem you will find a formula for $b_n$, as well as an asymptotic estimate.

1. Show that $b = 1$ and that, for $n \ge 1$, $$b_n = \sum_{k=0}^{n-1} b_k b_{n-1-k}$$

2. Referring to Problem 4-4 for the definition of a generating function, let $B(x)$ be the generating function $$B(x) = \sum_{n=0}^{\infty} b_n x^n$$ Show that $B(x) = xB(x)^2 + 1$, and hence one way to express $B(x)$ in closed form is $$B(x) = \frac{1}{2x}(1 - \sqrt{1 - 4x})$$

The Taylor expansion of $f(x)$ around the point $x = a$ is given by

$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^k$$

where $f^{(k)}(x)$ is the $k$th derivative of $f$ evaluated at $x$.

1. Show that $$b_n = \frac{1}{n + 1} \binom{2n}{n}$$ (the $n$th Catalan number) by using the Taylor expansion of $\sqrt{1 - 4x}$ around $x = 0$. (If you wish, instead of using the Taylor expansion, you may use the generalization of the binomial expansion (C.4) to nonintegral exponents $n$, where for any real number $n$ and for any integer $k$, we interpret $\binom{n}{k}$ to be $n(n-1)\ldots(n-k+1)/k!$ if $k \ge 0$, and 0 otherwise).

2. Show that $$b_n = \frac{4^n}{\sqrt{\pi} n^{3/2}} (1 + O(1/n))$$

### a. Calculating $b_0$ and $b_n$

There is exactly one tree with 0 nodes (the empty tree), therefore $b_0 = 1$.

When we construct a tree with $n$ nodes, we have $n$ choices for the root, and the remaining $n - 1$ nodes will be either in the left subtree or the right subtree. This is exactly the given formula:

$$b_n = \sum_{k=0}^{n-1} b_k b_{n-1-k}$$

Where $k$ is the number of elements that are smaller than the chosen root and $n - 1 - k$ is the number of elements larger than the chosen root.

### b. Generating function

This is trippy.

\begin{aligned} xB(x) + 1 &= 1 + x \Big( b_0 x^0 + b_1 x^1 + b_2 x^2 + \ldots \Big)^2 \\ &= 1 + x \Big( b_0^2 + x^0 + (b_0 b_1 + b_1 b_0) x^1 + (b_0 b_2 + b_1 b_1 + b_2 b_0) x^2 + \ldots \Big) \\ &= 1 + x \Big( \sum_{k=0}^{0} b_k b_{0-k} x^0 + \sum_{k=0}^{1} b_k b_{1-k} x^1 + \sum_{k=0}^{2} b_k b_{2-k} x^2 + \ldots \Big) \\ &= 1 + x \Big( \sum_{j=0}^{\infty} \sum_{k=0}^{j} b_k b_{j-k} x^j \Big) \\ &= 1 + x \Big( \sum_{j=0}^{\infty} \sum_{k=0}^{j+1-1} b_k b_{j+1-1-k} x^j \Big) \\ &= 1 + \sum_{j=0}^{\infty} \sum_{k=0}^{j+1-1} b_k b_{j+1-1-k} x^{j+1} \\ &= 1 + \sum_{j=0}^{\infty} b_{j+1} x^{j+1} && \big( \text{because of (a)} \big)\\ &= 1 + \sum_{k=1}^{\infty} b_{k} x^{k} && \big( \text{substituting } k = j + 1 \big)\\ &= b_0 x^0 + \sum_{k=1}^{\infty} b_{k} x^{k} \\ &= \sum_{k=0}^{\infty} b_{k} x^{k} \\ &= B(x) \end{aligned}

Then, to verify the possible solution, we just substitute:

\begin{aligned} xB(x)^2 + 1 &= x \Big( \frac{1}{2x}(1 - \sqrt{1 - 4x}) \Big)^2 + 1 \\ &= \frac{1}{4x}(1 - 2 \sqrt{1 - 4x} + 1 - 4x) + 1 \\ &= \frac{1}{4x}(2 - 2 \sqrt{1 - 4x}) - \frac{4x}{4x} + 1 \\ &= \frac{1}{2x}(1 - \sqrt{1 - 4x}) \\ &= B(x) \end{aligned}

### c. Taylor series expansion

Ugh!

Alright, let's calculate some derivatives for $f(x) = \sqrt{1 - 4x} = (1 - 4x)^{1/2}$.

\begin{aligned} f^{(1)}(x) &= \Big[ (1 - 4x)^{1/2} \Big]' = \frac{1}{2} (1 - 4x)^{-1/2} (1 - 4x)' = -2 (1 - 4x)^{-1/2} = \frac{-2}{(1 - 4x)^{1/2}} \\ f^{(2)}(x) &= \Big[ -2 (1 - 4x)^{-1/2} \Big]' = (- \frac{1}{2})(-2)(1 - 4x)^{-3/2}(1 - 4x)' = -4(1 - 4x)^{-3/2} = \frac{-3}{(1 - 4x)^{3/2}} \\ f^{(3)}(x) &= \Big[ -4 (1 - 4x)^{-3/2} \Big]' = \frac{3 \cdot 4}{2}(1 - 4x)^{-5/2}(1 - 4x)' = -24(1 - 4x)^{-5/2} = \frac{-24}{(1 - 4x)^{5/2}} \\ f^{(4)}(x) &= \Big[ -24 (1 - 4x)^{-5/2} \Big]' = \frac{24 \cdot 5}{2}(1 - 4x)^{-7/2}(1 - 4x)' = -240(1 - 4x)^{-7/2} = \frac{-240}{(1 - 4x)^{7/2}} \\ f^{(5)}(x) &= \Big[ -24 (1 - 4x)^{-7/2} \Big]' = \frac{240 \cdot 7}{2}(1 - 4x)^{-9/2}(1 - 4x)' = -3360(1 - 4x)^{-9/2} = \frac{-3360}{(1 - 4x)^{9/2}} \\ \end{aligned}

Let's observe that for $x = 0$ the denominator is going to be $1$, so we're interested only in the numerator. Let's also notice that the numerator $n_k$ for the $k$-th derivative is:

$$n_k = - 2^k \prod_{i=0}^{k-2} (2k + 1) = - 2^k \frac{(2(k - 1))!}{2^{k-1}(k-1)!} = - \frac{2(2(k - 1))!}{(k-1)!}$$

Thus, the Taylor expansion is:

$$f(x) = \sum_{k=0}^{\infty} - \frac{n_k}{k!}x^k = \sum_{k=0}^{\infty} - \frac{2(2(k-1))!}{k!(k-1)!}x^k$$

Or

$$f(x) = 1 - 2x - 2x^2 - 4x^3 - 10x^4 - 28x^5 - \ldots$$

Substituting that into $B(x)$, we get:

\begin{aligned} B(x) &= \frac{1}{2x}(1 - f(x)) \\ &= \frac{1}{2x} (1 - 1 + 2x + 2x^2 + 4x^3 + 10x^4 + 28x^5 + \ldots) \\ &= 1 + x + 2x^2 + 5x^3 + 14x^4 + \ldots \\ &= \sum_{k=0}^{\infty} \frac{(2k)!}{(k + 1)!k!} x^k \\ &= \sum_{k=0}^{\infty} \frac{1}{k+1} \frac{(2k)!}{k!k!} x^k \\ &= \sum_{k=0}^{\infty} \frac{1}{k+1} \binom{2k}{k} x^k \end{aligned}

Which illustrates that:

$$b_k = \frac{1}{n+1} \binom{2k}{k}$$

### d. Upper bound

We use Stirling's approximation,

$$n! = \sqrt{2 \pi n}\Big(\frac{n}{e}\Big)^n \Bigg(1 + \Theta \Big( \frac{1}{n} \Big) \Bigg) = \sqrt{2 \pi n} n^n e^{-n} (1 + \Theta(1/n))$$

and go on to produce some very ugly math:

\begin{aligned} b_n &= \frac{1}{n+1}\frac{(2n)!}{n!n!} \\ &= \frac{1}{n+1}\frac{ \sqrt{4 \pi n}(2n)^{2n} e^{-2n} (1 + \Theta(1/n)) }{2 \pi n n^{2n} e^{-2n} (1 + \Theta(1/n)) } \\ &= \frac{1}{n+1}\frac{ \sqrt{4 \pi n} 4^{n} n^{2n} (1 + \Theta(1/n)) }{2 \pi n n^{2n} (1 + \Theta(1/n)) } \\ &= \frac{1}{n+1}\frac{ 2 \sqrt{\pi n} 4^{n} (1 + \Theta(1/n)) }{2 \pi n (1 + \Theta(1/n)) } \\ &= \frac{1}{n+1}\frac{ 4^{n} (1 + \Theta(1/n)) }{\sqrt{\pi n} (1 + \Theta(1/n)) } \\ &= \frac{ 4^n }{\sqrt{\pi} n^{3/2} } (1 + \Theta(1/n)) \end{aligned}