Symbolic method (combinatorics): Difference between revisions

Revision as of 05:50, 15 September 2004

Revision as of 07:29, 15 September 2004

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===Sequence===

The ''sequence construction'', denoted by <math>\mathcal{A} = \mathfrak{G}\{\mathcal{B}\}</math> is defined as

:<math>\mathfrak{G}\{\mathcal{B}\} = \mathcal{E} + \mathcal{B} + (\mathcal{B} \times \mathcal{B}) + (\mathcal{B} \times \mathcal{B} \times \mathcal{B}) + \ldots</math>

In other words, a sequence is the neutral element, or an element of <math>\mathcal{B}</math>, or an ordered pair, ordered triple, etc. This leads to the relation

:<math>A(z) = 1 + B(z) + B(z)^{2} + B(z)^{3} + \ldots = \frac{1}{1 - B(z)}</math>

===Set===

The ''set'' (or ''powerset'') ''construction'', denoted by <math>\mathcal{A} = \mathfrak{P}\{\mathcal{B}\}</math> is defined as

:<math>\mathfrak{P}\{\mathcal{B}\} = \prod_{\beta \in \mathcal{B}}(\mathcal{E} + \{\beta\})</math>

which leads to the relation

:<math>\begin{matrix}A(z) & = & \prod_{\beta \in \mathcal{B}}(1 + z^{|\beta|}) \\

& = & \prod_{n=1}^{\infty}(1 + z^{n})^{B_{n}} \\

& = & \exp \left ( \ln \prod_{n=1}^{\infty}(1 + z^{n})^{B_{n}} \right ) \\

& = & \exp \left ( \sum_{n = 1}^{\infty} B_{n} \ln(1 + z^{n}) \right ) \\

& = & \exp \left ( \sum_{n = 1}^{\infty} B_{n} \cdot \sum_{k = 1}^{\infty} \frac{(-1)^{k}z^{nk}}{k} \right ) \\

& = & \exp \left ( \sum_{k = 1}^{\infty} \frac{(-1)^{k}}{k} \cdot \sum_{n = 1}^{\infty}B_{n}z^{nk} \right ) \\

& = & \exp \left ( \sum_{k = 1}^{\infty} \frac{(-1)^{k} B(z^{k})}{k} \right )

\end{matrix}</math>

[[Category:combinatorics]]