Combinatorial Networks Week 1, March 11-12

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1 1 Nots on March 11 Combinatorial Ntwors W 1, March Th Pigonhol Principl Th Pigonhol Principl If n objcts ar placd in hols, whr n >, thr xists a box with mor than on objcts 11 Thorm Givn a simpl graph on n vrtics, thr ar two vrtics of th sam dgr Proof: Lt G b a simpl graph on n vrtics Thr no harm in assuming that G has no vrtx of dgr 0, thn th dgr of any vrtx must b in {1,, n 1} Thus by P-P(Th Pigonhol Principl), thr must b two vrtics of th sam dgr Exampls 1 Substs with divisions Lt [n] {1,,, n} Considr all th substs S [n], such that no distinct i, j S satisfying i j What is th maximal numbr of S? It s obvious that T {n + 1,, n + n} with n lmnts satisfis th dsird condition W claim that this maximal numbr is n indd Assum that thr is a subst S of [n] satisfying th condition but S n + 1 For any odd a [n], lt C a {a, 0} [n] Sinc [n] a [n],odd C a and thr ar n such sts totally, by P-P, thr xist distinct i, j S C a for som odd a [n] Hnc ithr i j or j i, contradicts with th dfinition of S 13 Thorm For any x R and intgr n > 0, thr is a rational numbr p/q, such that x p/q < 1/nq Proof: Exrcis 14 Thorm(Erdös-Szrs) For any squnc of lngth mn + 1 distinct ral numbrs a 0, a 1, a mn, thr is an incrasing subsqunc of lngth m + 1 or a dcrasing subsqunc of lngth n + 1 Proof: For any 0 i mn, lt t i dnot th maximum lngth of an incrasing subsqunc starting with a i Assum all t i {1,,, m} For 1 j m, dfin S j {i : t i j} By P-P, thr is j 0 such that S j0 n + 1 Lt i 0 < i 1 < < i n S j0 W claim that a i0 a i1 a in, so it s a dcrasing subsqunc of lngth n + 1 If it dosn t hold, say, a i0 > a i1, by using th incrasing subsqunc of lngth j 0 starting with a i1, w gt an incrasing subsqunc of lngth j starting with a i0 Contradiction! 1

2 Exrcis Find a squnc of lngth mn such that thr is no incrasing subsqunc of lngth m + 1 nor dcrasing subsqunc of lngth n Doubl Counting 15 Lmma For any simpl graph G, v V d(v) E(G) Proof: For any vrtx v and any dg, dfin i(v, ) 1 if v, ls i(v, ) 0 Not that i(v, ) i(v, ) v V E E v V It implis d(v) E(G) E v V 16 Thorm Lt t(n) b th numbr of divisors of n and t(n) 1 t(j), n Thn t(n) H(n), as n, whr H(n) Proof: For any intgr i, j, if i j, dfin d(i, j) 1, ls dfin d(i, j) 0 Thn w hav j t(j) d(i, j) Sinc j1 d(i, j) j1 1 i j1 d(i, j), w hav as n It implis t(j) j1 t(n) 1 n n n i n 1 i + O(n), t(j) H(n) + O(1) H(n) j1

3 13 Binomial Thorm (1 + x) n x i i Dfinition: For intgr, and polynomial f(x), lt [x ]f b th cofficint in th trm in f(x) 17 Fact For any polynomials f 1 (x),, f (x), lt thn for any intgr n, [x n ]f f(x) f i (x), i 1++i n j1 [x ij ]f j 18 Fact ( ) n n ( ) n i Proof: ( ) n [x n ](1 + x) n n i+jn i+jn i ([x i ](1 + x) n )([x j ](1 + x) n ) i ) ) j ) 19 Fact For all positiv intgr n, ) n ) n n! n Proof: Sinc ln x dx ln 1 +1 ln x dx, 3

4 w gt That is, which implis Not that Thus, ln n 1 ln x dx ln n! n+1 ln x dx ) n+1 n ln n! ln ln +1 n + 1 ) n n! + 1 ) n+1 ( ) ( ), ) n+1 ) n+1 4 n) ) n ) n n! n Rmar: Stirling s Formula n! ) n, πn as n 110 Thorm For 1 n, Proof: For any x (0, 1], not that x ) x i x (1 + x) n and 1 + x x From thm w can gt )x i (1 + x)n x nx x 4

5 At last, lt x /n, and substitut it into th formula abov, thn w gt as dsird ), 111 Corollary ) ( n ) ) Nots on March 1 1 Binomial Thorm For any intgr n > 0, and any ral x, (1 + x) n 0 x Nwton s Binomial Thorm For any ral r, and any ral x ( 1, 1), (1 + x) n 0 x Hr, ( ) r r(r 1)(r + 1)! 3 Corollary Lt x ( 1, 1), and r n, whr intgr n > 0, (1 + x) n 0 ( ) n x (1 x) n 1 Inclusion-Exclusion + 1 ( 1) x x 4 Thorm(Inclusion-Exclusion) For substs A 1,, A n X, X \ 0 A i I [n]( 1) I i I A i 5

6 Proof: For any subst A X, dfin its charactristic function f A (x), whr f A (x) 1 if x A, ls f A (x) 0, thn f A (x) A x X Considr n F (x) (1 f Ai (x)) f Ai (x) I [n]( 1) I i I Not that i I f A i (x) is th charactristic function of i I A i, and that F (x) is th charactristic function of X \ n A i, sinc F (x) 1 if and only if x / A i for all i 1,,, n, and ls F (x) 0 So by what hav obsrvd bfor, 5 Corollary X \ A i x X x X F (x) I [n]( 1) I i I I [n]( 1) I x X i I I [n]( 1) I i I f Ai (x) f Ai (x) A i A i X X \ I [n]( 1) I +1 A i i I Dfinition: A drangmnt π : [n] [n] is a bijction(prmutation) such that π(i) i for all i [n] 6 Thorm Lt D n b th st of all drangmnt from [n] to [n], thn D n n! ( 1) i i! Proof: Lt X b th st of all th bijctions from [n] to [n], and for ach i [n], lt A i b th st {π X : π(i) i} 6

7 Sinc D n X \ n A i, and for ach I [n], i I A i (n I )!, thn by Inclusion-Exclusion, w gt D n I [n]( 1) I i I A i ( 1) (n )! n! 0 ( 1) i i! 7 Corollary D n n!, as n Exrcis Lt ϕ(n) b th numbr of intgrs m [n] rlativly prim to n If n p a1 1 pat t, whr a 1,, a t ar positiv intgrs and p 1,, p t ar diffrnt prims, thn t ) ϕ(n) n (1 1pi 8 Thorm Suppos that m, n ar positiv intgrs with m n, thn th numbr of surjctions from [m] to [n] is ( 1) (n ) m 0 Proof: Lt X {f : [m] [n]}, and A i {f : [m] [n] \ {i}} for ach i [n] Thn X \ A i {all surjctions from [m] to [n]} By Inclusion-Exclusion, X \ A i I [n]( 1) I i I Gnrating Function A i ( 1) (n ) m Dfinition: Givn {a n } n 0, f(x) n 0 a nx n is calld th gnrating function of {a n } n 0 Addtion: f(x) + g(x) n 0 (a n + b n )x n ; Multiplying: f(x)g(x) n 0 c nx n, whr c n i+jn a ib j 0 7

8 Dfinition: A triangulation of n-gon, is that to join th vrtics to divid this n-gon into triangls with intrscting only at vrtics Lt b n 1 is th numbr of triangulations of n-gon whr n 3 and b 1 1, b 0 0 Ths numbrs b 0, b 1, b,, ar calld Catalan numbrs 9 Thorm For all n 1, b n 1 n ( ) n n 1 Proof: By th dfinition of triangulation and b n 1, whr n 3, b n 1 b i b n i+1 i3 Sinc b 1 1 and b 0 0, it implis for, b b i b i Lt f(x) b th gnrating function of {b } 0, that is, f(x) b x x + b x x b i b i x x + f(x)f(x) Thus f (x) f(x) + x 0, and sinc b 0 f(0) 0, which implis f(x) 1 1 4x By Nwton s Binomial Thorm, f(x) 1 1 ( 1 ) ( 4) x ( )!!( 1)! x Hnc b 0 1 ( )!!( 1)! 1 ( ) 1 Exrcis if p is odd, if p is vn, Lt p b a positiv intgr, prov that A i I [n],1 I p( 1) I +1 A i ; i I A i I [n],1 I p( 1) I +1 A i i I 8