From 7c1ed606a095be2d7fb2a39c399110162e2082e9 Mon Sep 17 00:00:00 2001 From: fargolo Date: Fri, 13 Oct 2023 10:35:35 -0300 Subject: [PATCH] Fix latex double bars in markdown --- docs/src/index.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/docs/src/index.md b/docs/src/index.md index 422f387..aa65ba3 100644 --- a/docs/src/index.md +++ b/docs/src/index.md @@ -24,9 +24,9 @@ Given any constructable combinatorial structure, one can use a set of operators # Introduction -For newcomers, this an analytic approach to combinatorial problems. Modelling this type of problem often relies on intuitive arguments. Analytic combinatorics describe such situations with a grammar of operators: Sum, Cartesian product, Sequence, Multiset, Powerset and Cycle. Such operators yield an algebraic expression (e.g. ``P(z)``), called the generating function, which is directly related to the problem via complex analysis. We are generally interested in the coefficients of its series expansion. That is, let the series expansion of ``P(z)`` be ``T(P(z)) = \\sum_{n=1}^{\\infty} a_n x^n``. Then, the values of ``a^n`` correspond to the counts of objects of size ``n`` in this combinatorial class. +For newcomers, this an analytic approach to combinatorial problems. Modelling this type of problem often relies on intuitive arguments. Analytic combinatorics describe such situations with a grammar of operators: Sum, Cartesian product, Sequence, Multiset, Powerset and Cycle. Such operators yield an algebraic expression (e.g. ``P(z)``), called the generating function, which is directly related to the problem via complex analysis. We are generally interested in the coefficients of its series expansion. That is, let the series expansion of ``P(z)`` be ``T(P(z)) = \sum_{n=1}^{\infty} a_n x^n``. Then, the values of ``a^n`` correspond to the counts of objects of size ``n`` in this combinatorial class. -For instance, the number of binary words (e.g. abababbabab...) of size n is given by ``W_n = 2^n``. Using the sequence operator (``SEQ(A) \\implies A(z) = \\frac{1}/{1-z}``) , we find the generating function: ``W = SEQ(Z+Z) \\implies W(z) = \\frac{1}{1 - 2z}``. ``T(W(z)) = 1 + 2z + 4z^2 + 8z^3 + ...``. +For instance, the number of binary words (e.g. abababbabab...) of size n is given by ``W_n = 2^n``. Using the sequence operator (``SEQ(A) \implies A(z) = \frac{1}/{1-z}``) , we find the generating function: ``W = SEQ(Z+Z) \implies W(z) = \frac{1}{1 - 2z}``. ``T(W(z)) = 1 + 2z + 4z^2 + 8z^3 + ...``. This approach can be used to solve complex problems in a systematic way.