1 Counting and Stirling Numbers

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1 1 Coutig ad Stirlig Numbers Natural Numbers: We let N {0, 1, 2,...} deote the set of atural umbers. []: For N we let [] {1, 2,..., }. Sym: For a set X we let Sym(X) deote the set of bijectios from X to X. Permutatios: We defie S Sym([]) ad we call elemets of S permutatios. If π S we may view π as the sequece (π(1), π(2),..., π()) Fallig Factorial: For, N the fallig factorial is () ( 1)( 2)... ( + 1). ( ) ( : For, N we let ) deote the umber of elemet subsets of []. Observatio 1.1 ( )! ( )!! Proof: Costruct a elemet subset K [] by taig a permutatio π S ad the lettig K be the first elemets i the correspodig sequece. The total umber of permutatios is! ad each set of size is obtaied from exactly!( )! permutatios sice the first ad last elemets may be freely permuted amog themselves. (( )) (( : We let Observatio 1.2 (( )) ( + 1 )) deote the umber of multisets with groud set [] ad size. 1 ) Proof: Cosider a sequece of legth + 1 terms each of which is either or ad so that there are exactly copies of ad 1 copies of. For istace. We may associate each such sequece with a multiset with groud set [] ad size by treatig the umber of copies of i betwee the i th ad (i + 1) st copies of as the umber of copies of i i the multiset. (so the example is associated with {1, 2 3, 4}). This is a correspodece, so the total umber of multisets of the give type is equal to the umber of sequeces, but this is just ( ) sice we may choose ay 1 of the + 1 terms to be a. Partitios of Sets: If X is a set, a partitio of X is a set P with the property that A B wheever A, B P are distict, A P A X ad X. If A P we call A a bloc of P. S(, ): For, N we let S(, ) deote the umber of partitios of [] ito blocs.

2 2 Observatio 1.3 S(, ) S( 1, ) + S( 1, 1) Proof: Every partitio of [] ito blocs is either obtaied from a partitio of [ 1] ito blocs by isertig ito oe of the blocs (this ca be doe i ways) or from a partitio of [ 1] ito 1 blocs by addig the ew bloc {}. This correspodece yields the desired equality. Partitios of Numbers: If N a partitio of is a sequece λ (λ 1, λ 2,..., λ ) such that λ 1 λ 2... λ ad i1 λ i. We say that λ is a partitio of ito parts. The Youg Diagram of a partitio of is a collectio of left-aliged boxes so that the umber i the i th row is λ i. p (): For, N we let p () deote the umber of partitios of ito parts. Observatio 1.4 p () p 1 ( 1) + p ( ) Proof: The umber of partitios λ (λ 1,..., λ ) of ito parts is equal to the umber of such partitios with λ 1 plus the umber with λ > 1. The first set is i correspodece with the umber of partitios of 1 (just remove the last elemet), while the secod is i correspodece with the umber of partitios of ito parts (decrease each λ i by 1). Idistiguishable Domai & Codomai: We say that two fuctios f, g : N X are equivalet with N idistiguishable if there exists π Sym(N) so that f π g ad equivalet with X idistiguishable if there exists σ Sym(X) so that σ f g (ad similarly for N ad X idistiguishable). Theorem 1.5 The followig table lists the umber of equivalece classes of fuctios from N to X where N ad X x with the idicated properties: Elts. of N Elts. of X Ay Fuctio Ijectios Surjectios dist. dist. x (x) x!s(, x) (( idist. dist. x )) ( x (( x )) ) x dist. idist. S(, 1) + S(, 2) 1 if x S(, x)... + S(, x) 0 if > x idist. idist. p 1 () + p 2 () 1 if x p x ()... + p x () 0 if > x

3 Proof: The idetities for fuctios with N ad X idistiguishable are fairly straightforward, with the last oe followig from the observatio that every surjectio f : N X yields a partitio of N as {f 1 (1), f 1 (2),... f 1 (x)} ad each such partitio gives rise to exactly x! such fuctios. Whe N is idistiguishable ad X is distiguishable a arbitrary fuctio f : N X correspods to a multiset of size with groud set X where the elemet x X appears exactly f 1 (x) times. If we add the costrait that f is ijective, the we are simply coutig sets istead of multisets. Fially, our surjectios f : N X correspod to multisets where every elemet of X appears at least oce. However, by removig oe copy of each elemet, the umber of such multisets is precisely equivalet to the umber of arbitrary multisets with groud set X ad size x. Whe N is distiguishable ad X is idistiguishable, we have a correspodece betwee partitios of N ito exactly x blocs ad surjectios from N to X. The umber of arbitrary fuctios from N to X is the sum of the umber with rage of size 1, size 2, up to size x (ad so the result follows as before), ad fially ay two ijectios are equivalet so the aswer for this box is 1 if such a ijectio exists ad 0 otherwise. Fially, if both N ad X are idistiguishable, the our surjectios correspod precisely to partitios of the umber ito x parts. The umber of arbitrary fuctios from N to X is the sum of the umber with rage of size 1, size 2, up to size x (ad so the result follows as before), ad the ay two ijectios are equivalet so this box is as give. Propositio 1.6 S(, ) 1! i0 ( 1) i( ) i i Proof: Let N, K be sets with N ad K. For a subset H K let f(h) deote the umber of fuctios from N to K \ H. Now usig the above chart, iclusio-exclusio ad the substitutio i j we fid!s(, ) #{f : N K : f is a surjectio} H K( 1) H f(h) 3 ( ) ( 1) j ( j) j j0 ( ) ( 1) i i i i0

4 Propositio 1.7 x S(, )(x) 4 Proof: The left had side of the above equatio is the total umber of fuctios from N to X where N ad X x. By coutig these fuctios accordig to the size of their rage we get x #{f : N X} #{f : N X : f(n) } ( ) x!s(, ) (x) S(, ) Cycles: If f Sym(X) a cycle of f is a sequece (x 1, x 2,..., x ) with the property that f(x i ) x i+1 for 1 i 1 ad f(x ) x 1. We cosider two cycles equivalet if oe is a cyclic shift of the other. A cycle represetatio of f is a list of cycles of f icludig exactly oe from each equivalece class. c(, ): For, N we let c(, ) deote the umber of permutatios of [] with exactly cycles. Note that c(0, 0) 1 but c(s, 0) c(0, t) wheever s, t > 0. Observatio 1.8 c(, ) ( 1)c( 1, ) + c( 1, 1) Proof: Every permutatio of [] with cycles is either obtaied from a permutatio of [ 1] with cycles by isertig ito ay of the 1 positios immediately followig some umber (which ca be doe i 1 ways) or from a permutatio of [ 1] with 1 cycles by addig a ew cycle (). This correspodece gives the above equatio. s(, ): We defie the Stirlig umber of the first id by s(, ) ( 1) c(, ) Propositio 1.9 (i) (ii) c(, )x x(x + 1)(x + 2)... (x + 1) (x) s(, )x

5 Proof: For (i) we shall cosider the left had side ad the right had side as polyomials i x. Let F (x) deote the right had side ad defie the coefficiets b(, ) by the rule F (x) b(, )x where b(0, 0) 1 ad b(s, 0) b(0, t) 0 wheever s, t > 0. Now we have b(, )x F (x) (x + 1)F 1 (x) 1 1 b( 1, )x +1 + ( 1) b( 1, )x 1 b( 1, 1)x + ( 1) b( 1, )x 1 So we fid that b(, ) b( 1, 1) + ( 1)b( 1, ) for, 1. It follows that the terms b(, ) satisfy the same recurrece as c(, ) ad are equal wheever either iput is zero, so we fid that b(, ) c(, ). This completes the proof of (i). For (ii) we have s(, )x ( 1) c(, )x ( 1) c(, )( x) ( 1) ( x)( x + 1)( x + 2)... ( x + 1) x(x 1)(x 2)... (x + 1) (x) F[x]: For a field F we let F[x] deote the rig of polyomials with idetermiate x ad coefficiets i F. Bases of C[x]: We defie B 1 to be the basis of C[x] give by B 1 {1, x, x 2, x, 3,...} ad B 2 to be the basis of C[x] give by {1, (x), (x) 2, (x) 3,...}. 5

6 6 Propositio 1.10 Regard s {s(, )}, N ad S {S(, )}, N as ifiite matrices. The we have: (i) S is the basis trasformatio matrix from B 2 ad B 1. (ii) s is the basis trasformatio matrix from B 1 to B 2. (iii) (iv) S ad s are iverse matrices. m S(, )s(, m) δ m Proof: Parts (i) ad (ii) follow immediately from Propositios 1.6 ad 1.9. Part (iii) is a immediate cosequece of (i) ad (ii), ad (iv) is a restatemet of (iii).