Week 5-6: The Binomial Coefficients
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1 Wee 5-6: The Biomial Coefficiets March 6, Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers are 1, 5, 12, 22,, defied as the umbers of poits of dilated petagos The a a for 1 with a 0 1 The a , 1 The -go umbers are 1,, 3 3, 6 8, The umbers ( 3 ( 1( 2 6 ( 3 are tetrahedral umbers, ie, ( 3 is the umber of lattice poits of the tetrahedro 3 ( defied by 3 ( {(x, y, z : x 0, y 0, z 0, x + y + z 3} Theorem 12 The umber of odecreasig coordiate paths from (0, 0 to (m, with m, 0 equals ( m + m 2 Biomial Theorem Theorem 21 (Biomial Expasio For iteger 1 ad variables x ad y, ( (x + y x y, (1 + x 0 0 ( x 1
2 3 Biomial Idetities Defiitio 31 For ay real umber α ad iteger, defie the biomial coefficiets ( α 0 if < 0 1 if 0 α(α 1 (α + 1/! if > 0 Propositio 32 (1 For real umber α ad iteger, ( ( α α 1 + ( α 1 1 (2 For real umber α ad iteger, ( ( α α 1 α 1 (3 For oegative itegers m,, ad such that m +, ( m + ( ( m i i Propositio 33 For itegers, 0, ( i0 m0 ( m Proof Applyig the Pascal formula agai ad agai, we have ( ( ( ( ( ( ( ( ( Note that ( Multiomial Theorem ( Theorem 41 (Multiomial Expasio For ay positive iteger, ( (x 1 + x x 1, 2,, , 2,, 0 2 x 1 1 x 2 2 x,
3 where the coefficiets ( 1, 2,, are called multiomial coefficiets Proof! 1! 2!! (x 1 + x x (x 1 + x x (x 1 + x x }{{} u 1 u 2 u (u i x 1, x 2, x, 1 i { } umber of permutatios of the multiset { 1 x 1, 2 x 2,, x } , 2,, , 2,, 0 ( 1, 2,, x 1 1 x 2 2 x 5 Newto Biomial Theorem Theorem 51 (Newto s Biomial Expasio Let α be a real umber If 0 x < y, the ( α (x + y α x y α, where ( α 0 α(α 1 (α + 1! Proof Apply the Taylor expasio formula for the fuctio (x+y α of two variables Corollary 52 If z < 1, the ( α (1 + z α 0 1 (1 z ( α α The idetity 0 ( α z, ( z ( α + 1 z 0 ( α + 1 ( 1 is called the reciprocity law of biomial coefficiets 3
4 Proof Apply the Taylor expasio formula I particular, sice 1 1 z i0 zi, we have 1 (1 z ( i z z i 1 ( i 1 + +i i 0 1 ( + 1 z This shows agai that the umber of oegative iteger solutios of the equatio x 1 + x x z i equals the biomial coefficiet ( Uimodality of Biomial Coefficiets Defiitio 61 A sequece s 0, s 1, s 2,, s is said to be uimodal if there is a iteger (0 such that s 0 s 1 s s +1 s Theorem 62 Let be a positive iteger The sequece of biomial coefficiets ( ( ( (,,,, is a uimodal sequece More precisely, if is eve, ( ( ( ( ( < < < > > > ; 0 1 /2 1 ad if is odd, ( ( < 0 1 ( < < Proof Note that the quotiet ( /( 1 ( 1/2 The uimodality follows immediately ( ( + 1/2 ( > > > 1 { 1 if ( + 1/2 1 if ( + 1/2 (
5 A sequece s 0, s 1,, s of positive umbers is said to be log-cocave if s 2 i s i 1 s i+1, i 1,, 1 The coditio implies that the sequece log s 1, log s 2,, log s are cocave, ie, log s i (log s i 1 + log s i+1 /2 Propositio 63 If a sequece (s i is log-cocace, the it is uimodal Proof Assume the sequece is ozero The coditio s 2 i s i 1s i+1 is equivalet to s i 1 s i s i s i+1 If there exists a i 0 such that s i0 s i0 +1, ie, s i0 s i0 +1 1, the s i 1 s i s 0 s 1 s i0 s i0 +1 If there exists a i 0 such that s i0 1 s i0, ie, s i 0 1 s i0 ie, s i0 1 s i0 s 1 s 1 for all i i 0, ie, 1, the the s i 1 s i 1 for all i i 0, Now for the odecreasig umbers s i s i+1, there exists a idex i 0 such that s i0 1 1 s i 0 s i0 s i0 +1 It follows that s 0 s 1 s i0 s i0 +1 s The sequece s i ( i of biomial coefficiets is log-cocave I fact, s 2 i s i 1 s i+1 ( i + 1(i + 1 i( i > 1, i 1,, 1 Give a graph G with vertices A colorig of G with t colors is said to be proper if o two adjacet vertices receive the same color The umber of proper colorigs turs out to be a polyomial fuctio of t, called the chromatic polyomial of G, deoted χ(g, t, ad it ca be writte as the form χ(g, t ( 1 a t 0 Cojecture 64 (Log-Cocavity Cojecture The coefficiets of the above chromatic polyomial satisfies the log-cocave equality: a 2 a 1 a +1 Whe the iequalities are strict iequalities, it is called the Strict Log-Cocavity Cojecture 5
6 A cluster of a set S is a collectio A of subsets of S such that o oe is cotaied i aother A chai is a collectio C of subsets of S such that for ay two subsets, oe subset is always cotaied i aother subset For example, for S {a, b, c, d}, the collectio { } A {a, b}, {b, c, d}, {a, c}, {a, d} is a cluster; while the collectio { } C, {b, d}, {a, b, d}, {a, b, c, d} is a chai I more geeral cotext, a cluster is a atichai of a partially ordered set ( Theorem 65 (Sperer Every cluster of a -set S cotais at most /2 subsets of S Proof Let S {1, 2,, } We actually prove the followig stroger result by iductio o : The power set P (S ca be partitioed ito disjoit chais C 1, C 2,, C m with ( m /2 If so, the for each cluster A of S, Cosequetly, For 1, For 2, A A C i 1 for all 1 i m A m m C i A C i m i1 ( /2 ( 1 0 1, ( /2 ( 2 1 2, i1 {1} {1} {1, 2}, ( /2 For 3, ( /2 ( 3 1 3, {2} {1} {1, 2} {1, 2, 3}, {2} {2, 3}, {3} {1, 3} 6
7 ( For 4, /2 ( The 6 chais ca be obtaied i two ways: (i Attach a ew subset at the ed to each chai of the chai partitio for 3 (this ew subset is obtaied by appedig 4 to the last subset of the chai; (ii delete the last subsets i all chais of the partitio for 3 ad apped 4 to all the remaiig subsets {1} {1, 2} {1, 2, 3} {1, 2, 3, 4}, {2} {2, 3} {2, 3, 4}, {3} {1, 3} {1, 3, 4}, {4} {1, 4} {1, 2, 4}, {2, 4}, {3, 4} Note that the chai partitio satisfies the properties: (i Each chai is saturated i the sese that o subset ca be added i betwee ay two cosecutive subsets; (ii i each chai the size of the begiig subset plus the size of the edig subset equals A chai partitio satisfyig the two properties is called a symmetric chai partitio The above chai partitios for 1, 2, 3, 4 are symmetric chai partitios Give a symmetric chai partitio for the case 1; we costruct a symmetric chai partitio for the case : For each chai A 1 A 2 A i the chai partitio for the case 1, if 2, do A 1 A 2 A A {}, ad A 1 {} A 2 {} A 1 {}; if 1, do A 1 A 1 {} It is clear that the chais costructed form a symmetric chai partitio I fact, the chais costructed are obviously saturated Sice A 1 + A 1, the A 1 + A {} A 1 + A + 1, ad whe 2, A 1 {} + A 1 {} A 1 + A A 1 + A + 1 Now for each chai B 1 B 2 B l of the symmetric chai partitio for the case, sice B 1 B l, we must have B 1 /2 B l (otherwise, if B l < /2 the B 1 + B 2 <, or if B 1 > /2 the B 1 + B l > By defiitio of /2 ad /2, we have B 1 /2 /2 B l This meas that B 1 B 2 B l cotais exactly oe /2 -subset ( ad exactly oe /2 -subset Note that the umber of /2 -subsets of S is /2 ad the umber of /2 -subsets of S is It follows that the umber of chais i the costructed ( /2 7
8 symmetric chai partitio is ( /2 ( ( /2 /2 Thus every cluster of the power set P (S has size less tha or equal to cluster P /2 (S is of size ( /2 The The proof of the Specer theorem actually gives the costructio of clusters of maximal size Whe eve, there is oly oe such cluster, P (S : the collectio of all 2 2-subsets of S; ad whe odd, there are exactly two such clusters, P 1 2 P (S : the collectio of all 2 -subsets of S, ad +1 (S : the collectio of all 2 -subsets of S Example 61 (a Let S {1} The 1 ad ( ( There are two clusters: ad {1} (b Let S {1, 2} The 2 ad ( There is oly oe cluster of maximal size: { {1}, {2} } (c Let S {1, 2, 3} The 3 ad ( ( There are two clusters of maximal size: { } { } {1}, {2}, {3} ad {1, 2}, {1, 3}, {2, 3} (d Let S {1, 2, 3, 4} The 4 ad ( There is oly oe cluster of maximal size: { } {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4} (e Let S {1, 2, 3, 4, 5} The 5 ad ( ( There are two clusters of maximal size: { } {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}, { {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5 7 Dilworth Theorem Let (X, be a fiite partially ordered set A subset A of X is called a atichai if ay two elemets of A are icomparable I cotrast, a chai is a subset C of X whose ay two elemets are comparable Thus a chai is a liearly ordered subset of X It is 8
9 clear that ay subset of a chai is also a chai, ad ay subset of a atichai is also a atichai The importat coectio betwee chais ad atichais is: A C 1 for ay atichai A ad chai C Example 71 Let X {1, 2,, 10} The divisibility maes X ito a partially ordered set The subsets {2, 3, 5, 7}, {2, 5, 7, 9}, {3, 4, 5, 7}, {3, 4, 7, 10}, {3, 5, 7, 8}, {3, 7, 8, 10}, {4, 5, 6, 7, 9}, {4, 6, 7, 9, 10}, {5, 6, 7, 8, 9}, {6, 7, 8, 9, 10} are atichais, they are actually maximal atichais; while the subsets {1, 2, 4, 8}, {1, 3, 6}, {1, 3, 9}, {1, 5, 10}, {1, 7} are chais ad they are actually maximal chais Let (X, be a fiite poset We are iterested i partitioig X ito disjoit uio of atichais ad partitioig X ito disjoit uio of chais Let A be a atichai partitio of X ad let C be a chai of X Sice o two elemets of C ca be cotaied i ay atichai i A, the A C Similarly, for ay chai partitio C ad a atichai A of X, there are o two elemets of A belogig to a chai of C, we the have C A Theorem 71 Let (X, be a fiite poset, ad let r be the largest size of a chai The X ca be partitioed ito r but o fewer atichais I other words, mi { A : A is a atichai partitio } max { C : C is a chai } Proof It is eough to show that X ca be partitioed ito r atichais Let X 1 X ad let A 1 be the set of all miimal elemets of X 1 Let X 2 X 1 A 1 ad let A 2 be the set of all miimal elemets of X 2 Let X 3 X 2 A 2 ad let A 3 be the set of all miimal elemets of X 3 Cotiuig this procedure we obtai a decompositio of X ito atichais A 1, A 2,, A p By the previous argumet we always have p r O the other had, for ay a p A p, there is a elemet a p 1 A p 1 such that a p 1 < a p Similarly, there is a elemet a p 2 A p 2 such that a p 2 < a p 1 Cotiuig this process we obtai a chai a 1 < a 2 < < a p Sice r is the largest size of a chai, we the have r p Thus p r The followig dual versio of the theorem is ow as the Dilworth Theorem 9
10 Theorem 72 (Dilworth Let (X, be a fiite poset Let s be the largest size of a atichai The X ca be partitioed ito s, but ot less tha s, chais I other words, mi { C : C is a chai partitio } max { A : A is a atichai } Proof It suffices to show that X ca be partitioed ito s chais We proceed by iductio o X Let X For 1, it is trivially true Assume that 2 Let A mi be the set of all miimal elemets of X, ad A max the set of all maximal elemets of X Both A mi ad A max are maximal atichais We divide the situatio ito two cases Case 1 A mi ad A ma are the oly maximal atichais of X Tae a elemet x A mi ad a elemet y A ma such that x y (possibly x y Let X X {x, y} If X, the X {x, y} ad x < y, thus s 1 ad x < y is the required chai partitio Assume X, the X has oly the maximal atichais A mi {x} ad A max {y} The largest size of atichais of X is s 1 Sice X 1, by iductio the set X ca be partitioed ito s 1 chais C 1,, C s 1 Set C s {x y} The collectio {C 1,, C s } is a chai partitio of X Case 2 The set X has a maximal atichai A {a 1, a 2,, a s } of size s such that A A mi ad A A ma Let A A + {x X : x a i for some a i A}, {x X : x a i for some a i A} The sets A + ad A satisfy the followig properties: 1 A + X (Sice A mi A, ie, there is a miimal elemet ot i A; this miimal elemet caot be i A +, otherwise, it is larger tha oe elemet of A by defiitio 2 A X (Sice A max A, ie, there is a maximal elemet ot i A; this maximal elemet caot be i A, otherwise, it is smaller tha oe elemet of A by defiitio 3 A A + A (It is always true that A A + A For each x A + A, there exist a i, a j A such that a i x a j by defiitio, the a i a j, which implies a i a j so that i j, thus x a i a j A 4 A A + X (Suppose there is a elemet x A A +, the x is either ahead or behid ay member of A, thus A {x} is a atichai of larger size tha A Sice A ad A + are smaller posets havig the maximal atichai A of size s, the by iductio, by iductio A ca be partitioed ito s chais C1, C 2,, C s with the maximal elemets a 1, a 2,, a s respectively, ad A + ca be partitioed ito s chais C 1 +, C+ 2,, C+ s with the miimal elemets a 1, a 2,, a s respectively Thus we obtai a partitio of X ito s chais C 1 C+ 1, C 2 C+ 2,, C s C + s 10
11 Example 72 Let X {1, 2,, 20} be the poset with the partial order of divisibility The the subset {1, 2, 4, 8, 16} is a chai of maximal size The set X ca be partitioed ito five atichais {1}, {2, 3, 5, 7, 11, 13, 17, 19}, {4, 6, 9, 10, 14, 15}, {8, 12, 18, 20}, {16} However, the size of the atichai {2, 3, 5, 7, 11, 13, 17, 19} of size 8 is ot maximal I fact, {4, 6, 7, 9, 10, 11, 13, 15, 17, 19} is a atichai of size 10 The set X ca be partitioed ito te chais {1, 2, 4, 8, 16}, {3, 6, 12}, {5, 10, 20}, {7, 14}, {9, 18}, {11}, {13}, {15} {17}, {19} This meas that {4, 6, 7, 9, 10, 11, 13, 15, 17, 19} is a atichai of maximal size Example 73 Let X {(i, j Z 2 : 0 i, j 3, } be a poset whose partial order is defied by (i, j (, l if ad oly if i ad j l The size of the logest chai is 7 For istace, (0, 0 < (1, 0 < (1, 1 < (1, 2 < (2, 2 < (2, 3 < (3, 3 is a chai of legth 7 The the followig collectio of subsets {(0, 0}, {(1, 0, (0, 1}, {(2, 0, (1, 1, (0, 2}, {(3, 0, (2, 1, (1, 2, (0, 3}, {(3, 1, (2, 2, (1, 3}, {(3, 2, (2, 3}, {(3, 3} is a atichai partitio of X The maximal size of atichai is 4 ad the poset X ca be partitioed ito 4 disjoit chais: (0, 0 < (0, 1 < (0, 2 < (0, 3 < (1, 3 < (2, 3 < (3, 3, (1, 0 < (1, 1 < (1, 2 < (2, 2 < (3, 2, (2, 0 < (2, 1 < (3, 1, (3, 0 {(0, 0, (0, 1, (0, 2, (0, 3, (1, 3, (2, 3, (3, 3}, {(1, 0, (1, 1, (1, 2, (2, 2, (3, 2}, {(2, 0, (2, 1, (3, 1}, {(3, 0} 11
12 The Hasse diagram of the poset X is (3, 3 (3, 2 (2, 3 (3, 1 (2, 2 (1, 3 (3, 0 (2, 1 (1, 2 (0, 3 (2, 0 (1, 1 (0, 2 (1, 0 (0, 1 (0, 0 Fidig a atichai of maximal size for a poset is a difficult problem So far there is o caoical way to do this job 12
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