en>SporkBot: Repair duplicate template args

2014-12-23T05:03:59Z

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	~~{{Unreferenced\|date=December 2009}}~~		There is nothing to write about myself really.<br>Yes! Im a part of this site.<br>I really hope Im useful at all<br><br>My blog :: Fifa 15 Coin Generator [[http://Xn----9Sbiy2Ae2C.Xn--P1ai/index.php/component/k2/item/75-%D0%B3%D0%BB%D0%B0%D0%B2%D0%BD%D1%8B%D0%B9-%D0%B2%D1%80%D0%B0%D1%87 Http://Xn----9Sbiy2Ae2C.Xn--P1Ai/]]
	~~{{Redirect\|Distinguished element\|sets with pre-defined distinguished elements\|Pointed set}}~~

	In enumerative [[combinatorics\|combinatorial]] [[mathematics]], [[identity (mathematics)\|identities]] are sometimes established by arguments that rely on singling out one '''"distinguished element"''' of a set.

	~~==Definition==~~

	~~Let~~ <~~math~~>~~\mathcal{A}</math> be~~ a ~~family~~ of ~~subsets of the set <math>A</math> and let <math>x \in A</math> be a distinguished element of set <math>A</math>~~. ~~Then suppose there is a predicate~~ <~~math~~>P(X,x)</math> that relates a subset <math>X\subseteq A</math> to <math>x</math>. Denote <math>\mathcal{A}(x)</math> to be the set of subsets <math>X</math> from <math>\mathcal{A}</math> for which <math>P(X,x)</math> is true and <math>\mathcal{A}-x</math> to be the set of subsets <math>X</math> from <math>\mathcal{A}</math> for which <math>P(X,x)</math> is false, Then <math>\mathcal{A}(x)<~~/math~~> ~~and~~ <~~math~~>\mathcal{A}-x</math> are disjoint sets, so by the method of summation, the cardinalities are additive<ref name=Petkovsek2002>{{cite journal\|last=Petkovšek\|first=Marko\|coauthors=Tomaž Pisanski\|title=Combinatorial Interpretation of Unsigned Stirling and Lah Numbers\|journal=University of Ljubljana preprint series\|date=November 2002\|volume=40\|issue=837\|pages=1-6\|url=http:~~//www.imfm.si/preprinti/PDF/00837.pdf\|accessdate=12 July 2013}}</ref>~~

	:~~<math>\|\mathcal{A}\| = \|\mathcal{A}(x)\| + \|\mathcal{A}-x\|</math>~~

	~~Thus the distinguished element allows for a decomposition according to a predicate that is a simple form of a~~ [[~~divide and conquer algorithm]]. In combinatorics, this allows for the construction of [[recurrence relation]]s. Examples are in the next section.~~

	~~==Examples==~~
	* The [[binomial coefficient]] <math>{n \choose k}</~~math> is the number of size-''k'' subsets of a size-''n'' set. A basic identity, one of whose consequences is that these are precisely the numbers appearing in [[Pascal's triangle]], states that:~~

	~~::<math>{n \choose k-1}+{n \choose k}={n+1 \choose k}.<~~/~~math>~~

	~~:'''Proof:''' In a size~~-~~(''n'' + 1) set, choose one distinguished element. The set of all size~~-~~''k'' subsets contains: (1) all size~~-~~''k'' subsets that ''do'' contain the distinguished element, and (2) all size~~-~~''k'' subsets that ''do not'' contain the distinguished element~~. ~~If a size-''k'' subset of a size~~-~~(''n'' + 1) set ''does'' contain the distinguished element, then its other ''k'' − 1 elements are chosen from among the other ''n'' elements of our size~~-~~(''n'' + 1) set. The number of ways to choose those is therefore <math>{n \choose k-1}<~~/~~math>~~. If a size-''k'' subset ''does not'' contain the distinguished element, then all of its ''k'' members are chosen from among the other ''n'' "non-distinguished" elements. The number of ways to choose those is therefore <math>{n \choose k}</~~math>.~~

	*The number of subsets of any size-''n'' set is 2<sup>''n''</~~sup>.~~

	~~:'''Proof:''' We use [[mathematical induction]]. The basis for induction is the truth of this proposition in case ''n'' = 0. The [[empty set]] has 0 members and 1 subset, and 2<sup>0<~~/sup> = 1. The induction hypothesis is the proposition in case ''n''; we use it to prove case ''n'' + 1. In a size-(''n'' + 1) set, choose a distinguished element. Each subset either contains the distinguished element or does not. If a subset contains the distinguished element, then its remaining elements are chosen from among the other ''n'' elements. By the induction hypothesis, the number of ways to do that is 2<sup>''n''</~~sup>. If a subset does not contain the distinguished element, then it is a subset of the set of all non~~-~~distinguished elements. By the induction hypothesis, the number of such subsets is 2<sup>''n''</sup>. Finally, the whole list of subsets of our size~~-~~(''n'' + 1) set contains 2<sup>''n''</sup> + 2<sup>''n''<~~/~~sup> = 2<sup>''n''+1<~~/~~sup> elements.~~

	* Let ''B''<sub>''n''</sub> be the ''n''th [[Bell number]], i.e., the number of [[partition of a set\|partitions of a set]] of ''n'' members. Let ''C''<sub>''n''</sub> be the total number of "parts" (or "blocks", as combinatorialists often call them) among all partitions of that set. For example, the partitions of the size-~~3 set {''a'', ''b'', ''c''} may be written thus:~~

	~~::<math>\begin{matrix}abc \\ a/bc \\ b/ac \\ c/ab \\ a/b/c \end{matrix}</math>~~

	~~:We see 5 partitions, containing 10 blocks, so ''B''<sub>3</sub> = 5 and ''C''<sub>3</sub> = 10. An identity states:~~

	~~::<math>B_n+C_n=B_{n+1}.\,</math>~~

	~~:'''Proof:''' In a size~~-~~(''n'' + 1) set, choose a distinguished element. In each partition of our size~~-~~(''n'' + 1) set, either the distinguished element is a "singleton", i.e~~., the set containing ''only'' the distinguished element is one of the blocks, or the distinguished element belongs to a larger block. If the distinguished element is a singleton, then deletion of the distinguished element leaves a partition of the set containing the ''n'' non-distinguished elements. There are ''B''<sub>''n''</sub> ways to do that. If the distinguished element belongs to a larger block, then its deletion leaves a block in a partition of the set containing the ''n'' non-~~distinguished elements. There are ''C''<sub>''n''<~~/~~sub> such blocks.~~

	~~==See also==~~
	* [[Combinatorial principles]]
	* [[Combinatorial proof]]

	~~==References==~~
	~~{{reflist}}~~

	~~{{DEFAULTSORT:Method Of Distinguished Element}}~~
	~~[[Category:Combinatorics~~]]

en>Omnipaedista: see Talk:Circle_progression#Nomenclature_and_Chromatic_Version

2013-12-15T14:00:01Z

see Talk:Circle_progression#Nomenclature_and_Chromatic_Version

Show changes

en>Hyacinth: /* See also / Rhythm changes

2012-08-27T03:42:23Z

Ragtime progression - Revision history

en>SporkBot: Repair duplicate template args

en>Omnipaedista: see Talk:Circle_progression#Nomenclature_and_Chromatic_Version

en>Hyacinth: /* See also */ *Rhythm changes

en>Hyacinth: /* See also / Rhythm changes