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I am trying to solve the following problem from Cohen's Basic Techniques of Combinatorial Theory:

A collection of $n$ lines in the plane are are said to be in general position if no two are parallel and no three are concurrent. Let $a_n$ be the number of regions into which $n$ lines in general position divide the plane. How big is $a_n$?

The recurrence relationship for $a_n$ is immediate, namely

$$a_{n+1}=a_n+n.$$

I have been trying to apply generating functions to solve this recurrence, but I am making a mistake somewhere. My attempt is as follows:

Let $G(X)=\sum_{k\geq 1} a_k X^k$. Multiplying both sides of the recurrence by $X^k$ and summing over all valid $k$ gives

$$\sum_{k\geq 1}a_{k+1}X^k=\sum_{k\geq 1}a_kX^k+\sum_{k\geq 1}kX^k.$$

Using the fact that $\frac{1}{(1-x)^2}=\sum_{k\geq 1} kX^k$, I get

$$\sum_{k\geq 1}a_{k+1}X^k=G(X)+\frac{1}{(1-x)^2}.$$

The left hand side is equal to

$$\frac{G(X)-a_1}{X}=\frac{G(X)-2}{X}$$

since $a_1=2$. Solving for $G(X)$, I get

$$G(X)=\frac{2-4X+3x^2}{(1-x)^3}.$$

Expanding with partial fractions gives

$$G(X)=\frac{3}{1-x}-\frac{2}{(1-x)^2}+\frac{1}{(1-x)^3}.$$

Transforming this into power series gives

\begin{align} G(X)&=3\sum_{k\geq 0}X^k-2\sum_{k\geq 0} (k+1)X^k+\sum_{k\geq 0}(k+1)(k+2)X^k\\ &=\sum_{k\geq 0} (3-2(k+1)+(k+1)(k+2))X^k, \end{align} which is not correct. I have tried starting over, but I seems to be making a pretty serious mistake. Any thoughts?

Solution:

The real recurrence relation is $a_{n+1}=a_n+(n+1)$. Let $G(X)=\sum_{k\geq 1} a_k X^k$. Multiplying both sides of the recurrence by $X^k$ and summing over all valid $k$ gives

$$\sum_{k\geq 1}a_{k+1}X^k=\sum_{k\geq 1}a_kX^k+\sum_{k\geq 1}(k+1)X^k.$$

Solving in terms of $G(X)$ gives

$$G(X)=\frac{x(2-2x+x^2)}{(1-x)^3}.$$

By expanding with partial fractions we see

$$G(X)=\frac{x(2-2x+x^2)}{(1-x)^3}=\frac{x}{1-x}+\frac{x}{(1-x)^3}.$$

Using the power series identities $\frac{1}{2}\sum_{k\geq 1}k(k-1)X^{k-1}=\frac{x}{(1-x)^3}$ and $\sum_{k\geq 1}X^k=\frac{x}{1-x}$. Substituting these power series in gives

\begin{align} G(X)&=\frac{1}{2}\sum_{k\geq 1}k(k-1)X^{k-1}+\sum_{k\geq 1}X^k\\ &=\frac{1}{2}\sum_{k\geq 1}k(k+1)X^{k}+\sum_{k\geq 1}X^k \quad \dagger\\ &=\sum_{n\geq 1} (\binom{k+1}{2}+1)X^k. \end{align}

Therefore, by equating coefficients, we have $$a_n=\binom{n+1}{2}+1.$$

The step $\dagger$ follows from the fact that the first coefficient in the power series is $0$.

Note: This is NOT homework, I am merely working through exercises during summer vacation. I'm sure there are other ways to approach the problem, but I am trying to practice using generating functions.

Holdsworth88
  • 8,818

2 Answers2

8

As an alternative to generating functions (which is usually a very nice way to solve problems), let me mention the following. Assume no three lines are concurrent.

Each time a line is added, it adds a region for each region it passes through, i.e. it divides that region in two. The number of regions through which a line passes is equal to the number of lines crossed plus one. Each time the line crosses another line, it creates a point of intersection, so we could count the number of lines crossed plus one to be the number of points of intersection added plus the number of lines added. Thus, the number of regions is one (the plane) plus the number of points of intersection plus the number of lines.

If we have $n$ lines, there are $\binom{n}{2}$ points of intersection. Thus, the number of regions is $$ \binom{n}{2}+\binom{n}{1}+\binom{n}{0} $$

robjohn
  • 345,667
4

Your recurrence is off by $1$: it should be $a_n=a_{n-1}+n$ for $n>0$, and you could make life a little easier by setting $a_0=1$. Then if you set $a_n=0$ for $n<0$, you can write simply $$a_n=a_{n-1}+n+[n=0]\;,$$ where the last term is an Iverson bracket. Now you get

$$\begin{align*} G(x)&=\sum_na_nx^n\\ &=\sum_na_{n-1}x^n+\sum_nnx^n+1\\ &=xG(x)+\frac{x}{(1-x)^2}+1\;, \end{align*}$$

so

$$\begin{align*} G(x)&=\frac{x}{(1-x)^3}+\frac1{1-x}\\ &=\sum_n\binom{n+2}2x^{n+1}+\sum_nx^n\\ &=\sum_n\binom{n+1}2x^n+\sum_nx^n\;, \end{align*}$$

and $$a_n=\binom{n+1}2+1\;.$$

Note that $\sum_kkx^k=\frac{x}{(1-x)^2}$, not $\frac1{(1-x)^2}$.

Brian M. Scott
  • 616,228