Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently working through Analytic Combinatorics (free online) and i'm stuck at a very interesting example. After introducing some admissible constructions (combinatorial sum, cartesian product, sequence, powerset, multiset) and talking about how to define structures recursively, there's an example I can't really wrap my head around.

Here's the part i don't understand (page 35 in the pdf):

page36 page37

The book uses $\{\epsilon\}$ to denote the neutral class consisting of a single element of size zero. Could someone explain how this fits into the recursive definition? Without the term $\{\epsilon\}$ it would have made sense to me.

Thanks a lot!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

If you had $\mathcal{T}=\mathcal{T}\times\nabla\times\mathcal{T}$, every triangulation would have to contain a triangle. Using the iterative interpretation of the recursive definition discussed on page 32, you’d have $\mathcal{T}^{[0]}=\varnothing$, $\mathcal{T}^{[1]}=\{\nabla\}$, $\mathcal{T}^{[2]}=\{\nabla+\nabla+\nabla\}$, and so on, with no way to get the triangulation of a square. Including the $\{\epsilon\}$ term allows you to get things like $\epsilon+\nabla+\nabla$ in $\mathcal{T}^{[2]}$, which gives you the triangulation of a square. This is what they’re getting at in the parenthetical remark in this sentence:

Then, a triangulation decomposes into its root triangle and two subtriangulations (that may well be “empty”) appearing on the left and right sides of the root triangle; ...

The $\{\epsilon\}$ term allows for those empty subtriangulations.

share|improve this answer
    
thanks you very much, that cleared it up! –  ifubic Sep 20 '12 at 21:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.