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Suppose that $n$ lines are drawn on a plane in such a way that no lines are parallel and no three of them intersect at a point. Let $r_n$ be the number of regions in the plane is divided into after drawing the $n$ lines.

a) Find a recursive formula for the sequence {$r_n$}, $n\geq0$. Prove (without using induction) that your formula is correct.

b) Find a close formula for the generating function of the sequence {$r_n$}.

So for part a), I've found the recursion formula by drawing out the plane and the first few lines.

Since I got $n=0, r_0=1$ ; $n=1, r_1=2$ ; $n=2, r_2=4$ ; $n=3, r_3=7$ ; $n=4, r_4=11$

I'm saying the recursion formula is $r_n=r_{n-1}+n$. I'm not sure how I can prove this without induction. I think I need some way to show that adding an $n$th line will add $n$ regions but I have no idea how to go about this.

For b) I set $R(x)=r_0+r_1x+r_2x^2+...+r_nx^n$. Then,

$$R(x)=r_0+r_1x+...+r_nx^n$$

$$=r_0+r_1x+\sum_{k\geq0}(r_{k-1}+k)x^k$$

$$=r_0+r_1x+\sum_{k\geq0}r_{k-1}x^k+\sum_{k\geq0}kx^k$$ $$=r_0+r_1+(r_1x^2+r_2x^3+...)+(2x^2+3x^3...)$$ $$=r_0+r_1+xF(x)+\sum_{k\geq0}kx^k$$

Solving for $R(x)$ gives me

$$R(x)=\frac{1+2x+\sum_{k\geq0}x^k}{1-x}$$

I feel like I'm missing something at this step, since I'm not sure what to do with the $$\sum_{k\geq0}kx^k.$$ I thought about using the identity $$\sum_{k\geq0}x^k=\frac{1}{1-x}$$ but I don't know if I'm allowed to do that since we $k$ as a coefficient, so it's not like a constant I can just pull out.

Any thoughts on how I can keep going in the right direction for either a) or b) would be really appreciated!

Karen M.
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2 Answers2

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Good news is you're in the right track!

For a) when you add the $n$th line, imagine the line is coming from infinity to intersect the first line. On its way, it is cutting the region in two pieces, and we previously had one, so is creating one new region indeed. In fact, you can see each intersection corresponds to adding a new region. That line when it intersects the $n-1$th line, also goes for ever and is creating one more region which corresponds to intersecting at the infinity!

For b) Let $f(x) = \sum_{n=0}^{\infty}r_nx^n$ be the generating function of our recursion $r_n = r_{n-1} + n.$ Simply plugging yields $$f(x) = \sum_{n=0}^{\infty}r_nx^n = r_0 + \sum_{n=1}^{\infty}r_nx^n = r_0 + \sum_{n=1}^{\infty}(r_{n-1} + n)x^n$$

which is

$$r_0 + x \underbrace{\left(\sum_{n=1}^{\infty}r_{n-1}x^{n-1} \right)}_{=\sum_{n=0}^\infty r_nx^n} + \sum_{n=1}^\infty nx^n=r_0 + xf(x) + \dfrac{x}{(1-x)^2}$$

so $f(x) = 1 + xf(x) + \dfrac{x}{(1-x)^2},$ thus $f(x) = \dfrac{x^2-x+1}{(1-x)^3}.$

Note: As you said, $\sum_{n=0}^\infty x^n = \dfrac{1}{1-x}.$ By taking derivative from both sides and multiplying both sides to $x,$ we will get $\sum_{n=1}^\infty nx^n = \dfrac{x}{(1-x)^2}.$

Ehsan M. Kermani
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  • Thanks so much! I could figure out intuitively that adding an $n$th line gives n more regions but I couldn't visualize it past a certain point. Corresponding intersections to regions makes way more sense. And I hadn't thought of taking the derivative to get the sum of $nx^n$, that helps a ton :) – Karen M. Feb 13 '15 at 09:01
  • @KarenMatsuo, Glad to hear that :) – Ehsan M. Kermani Feb 13 '15 at 09:08
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if L intersect whit the L',divides two neighbor region to four region ,so it adds tow region.we count each region twice except two region at infinity.so it adds $(2(n-1)+2)/2=n$region

ali
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  • I hadn't thought about corresponding the intersections of the new line with the new regions. Thank you! – Karen M. Feb 13 '15 at 09:11