Chapter 8, Problem 1E

Let 2n(equally spaced) points on a circle be chosen. Show that the number ofways to join these points in pairs, so that the resulting nline segments do notintersect, equals the nth Catalan number C_n.

To determine

To show: That the number of ways to join the 2n (equally spaced) points in pairs.

Explanation of Solution

The number of ways to join the 2n (equally spaced) points in pairs, so that the resulting n line segments do not intersect, equals the nth Catalan number $C_n$. Let 2n (equally spaced) points on a circle be chosen.

The $C_0=1$. Now assume that $n\ge 1$. Label the points $1,2,\mathrm{...},2n$ clockwise around the circle.

Let $M=M_n$ denote the set of matchings of the 2n points counted by $C_n$.

For a matching in M let t denote the point matched to point 1. Note that t is even. For $1\le s\le n$ let M(s) denote the set of matchings in M such that point 1 is matched with point 2s.

The sets $\left\{M\left(s\right)\right\}$ partition M, so $\left|M\right|={\displaystyle {\sum }_{s=1}^n\left|M\left(s\right)\right|}$.

For $1\le s\le n$ to compute $\left|M\left(s\right)\right|$. To construct a matching in M(s), there are $C_{s-1}$ ways to match points $2,3,\mathrm{...},2s-1$ and there are $C_{n-s}$ ways to match points $2s+1,2s+2,\mathrm{...},2n$.

Therefore $\left|M\left(s\right)\right|=C_{s-1}{C}_{n-s}$.

By these comments $C_n={\displaystyle {\sum }_{s=1}^n{C}_{s-1}{C}_{n-s}}$.

$C_n=\frac{1}{n+1}\left(\begin{array}{c}2n\\ n\end{array}\right)$

Where, $n=0,1,2,\mathrm{...}$.

Consider the generating function

$g\left(x\right)={\displaystyle \sum _{n=0}^{\infty }{C}_n{x}^n}$

Using the Catalan and obtain

$xg{\left(x\right)}^2=g\left(x\right)-1$

Using the quadratic formula

$g\left(x\right)=\frac{1\pm {\left(1-4x\right)}^{\frac{1}{2}}}{2x}$

In other words

$xg\left(x\right)=\frac{1\pm {\left(1-4x\right)}^{\frac{1}{2}}}{2x}$

Using Newton’s binomial theorem this becomes

$xg\left(x\right)=\frac{1\pm {\displaystyle {\sum }_{n=0}^{\infty }\left(\begin{array}{l}\frac{1}{2}\\ n\end{array}\right){\left(-4\right)}^n{x}^n}}{2}$

For this equation at $x=0$ the left-hand side is zero, so the right-hand side is zero. Therefore,

$\begin{array}{c}xg\left(x\right)=\frac{1-{\displaystyle {\sum }_{n=0}^{\infty }\left(\begin{array}{l}\frac{1}{2}\\ n\end{array}\right){\left(-4\right)}^n{x}^n}}{2}\\ =-\frac{{\displaystyle {\sum }_{n=1}^{\infty }\left(\begin{array}{c}\frac{1}{2}\\ n\end{array}\right){\left(-4\right)}^n{x}^n}}{2}\\ =-\frac{{\displaystyle {\sum }_{n=1}^{\infty }\left(\begin{array}{l}\frac{1}{2}\\ n\end{array}\right){\left(-4\right)}^n{x}^{n-1}}}{2}\\ =-\frac{{\displaystyle {\sum }_{n=0}^{\infty }\left(\begin{array}{c}\frac{1}{2}\\ n+1\end{array}\right){\left(-4\right)}^{n+1}{x}^n}}{2}\end{array}$

Consequently, for $n\ge 1$.

$\begin{array}{c}{C}_n=-\left(\begin{array}{c}\frac{1}{2}\\ n+1\end{array}\right)\frac{{\left(-4\right)}^{n+1}}{2}\\ =\frac{1}{n+1}\left(\begin{array}{l}2n\\ n\end{array}\right)\end{array}$

Note that, $C_n=\left(\begin{array}{l}2n\\ n\end{array}\right)-\left(\begin{array}{l}2n\\ n-1\end{array}\right)$, where, $n=1,2,3,\mathrm{...}$.

Hence, proved.