MA 524 Homework 6 Solutions

Transcription

1 MA 524 Homework 6 Solutos. Sce S(, s the umber of ways to partto [] to k oempty blocks, ad c(, s the umber of ways to partto to k oempty blocks ad also the arrage each block to a cycle, we must have S(, c(,. For equalty to hold, ay block that occurs ay partto of [] to k blocks must be uquely arrageable to a cycle. Sce a block of sze m ca be made to a cycle (m )! ways, ths ca oly be doe f all blocks have sze ether or 2. Ths holds f k (all blocks have sze ) or k (there s oe block of sze 2 ad 2 of sze ). O the other had, f k <, the there exsts a partto to blocks of szes k +,,...,, ad k + > 2, so we caot have equalty ths case. Oe ca also prove ths by ducto usg the recurrece relatos S(, S(, k ) + ks(,, c(, c(, k ) + ( )c(,. Suppose we kow by ducto that S(, c(, wth strct equalty whe 0 < k < 2. The comparg the above recurreces term by term shows that f k, the S(, c(, wth equalty whe k ad strct equalty otherwse. (Whe k, we also have equalty sce S(, c(,.) 2. (a) Sce both sdes are lear maps appled to f, t suffces to prove the clam for f lyg some bass, say f t m for some m. The the left sde evaluates to ( t dt) d ( t m t dt) d ( mt m t dt) d 2 m 2 t m m t m, whle the rght sde evaluates to S(, t k dk dt k tm S(, m(m )(m 2) (m k + )t m S(, (m) k t m. But the result ow follows sce S(, gve the etres of the matrx whch chages bases from {m } 0 ad {(m) } 0.

2 Oe ca also proceed by ducto o. Applyg t d to both sdes of the expresso for dt gves ( t dt) d f S(, t d dt S(, (t k ) (t k dk dt f k k+ dk+ dk f + ktk dtk+ dt f k S(, k ) t k dk dt f + S(, kt k dk k dt f. k The result ow follows usg the recurrece S(, S(, k ) + ks(,. Fally, we gve a combatoral proof. Wrte D d. The D(tf) t Df + f, dt or as operators, Dt td +. I other words, suppose a product of t s ad D s has a t after a D. The ths s equal to the sum of two other terms, oe whch we commute the t to the left past the D, ad oe whch the t ad D ahlate each other. Now suppose that we start wth (td) tdtd td as o the left had sde ad we try to apply the relato Dt td + utl all the t s are to the left of all the D s. Each t wll ether ahlate oe of the D s to ts left, or t wll make t past all the D s. The coeffcet of t k D k s thus the umber of ways whch all but k of the t s ca be matched up wth oe of the D s to ts left. But fact ths s just S(,! To see ths, f the th occurrece of D ahlates the jth (j > ) occurrece of t, place ad j the same block. The result wll be a partto of [] to k blocks (where the smallest umber each block correspods to a t that does ot get ahlated). Coversely, gve a partto of [] to k blocks, let the jth occurrece of t ahlate the th occurrece of D, where s the ext smallest umber the same block as j, f oe exsts. (b) Note that ( t ) t m t m 2 2 t m. t t 2 Hece f Z(S m ) s the cycle dex polyomal for S m, the ( ) g(σ) t Z(S m ) t σ S m t t 2 ). 2

3 Multplyg both sdes by xm ad summg over all m gves m! ( ) ( ) g(σ) xm m! x t exp t t m 0 σ S m ( ) S(, t k k x exp t t k ( ) S(, x k x exp S(, x k x x m S(, m k m S(, x m. m 0 t t 2 t t 2 Takg the coeffcet of x m, t follows that the average value of g(σ) for σ S m s m S(,, whch s the umber of ways to place dstgushable balls to m dstgushable boxes. There s a more tutve proof of ths result usg Bursde s Lemma, whch states that f a group G acts o a set X, the the umber of G-orbts X s equal to the average umber of elemets of X fxed by each elemet of G. To see why ths s true, ote that the umber of G-orbts X s, where O(x) s the O(x) orbt cotag x. But the elemets of O(x) are bjecto wth left cosets of Stab(x) G, where Stab(x) s the subgroup of elemets G that stablze (fx) x. Hece O(x) Stab(x). G But the sum o the rght sde couts pars (g, x) G X such that g(x) x, so t s also equal to the total umber of elemets fxed by g summed over all g G, whch proves the lemma. For ths problem, take G S m ad let X [m], that s, -tuples (x,..., x ) wth each x [m], where σ S m acts by σ(x,..., x ) (σ(x ),..., σ(x )). The the umber of elemets of X that are fxed by ay σ S m s just g(σ) m sce each x must be fxed by σ. Thus the average of g(σ) m s just the umber of orbts X by Bursde s Lemma. But a orbt s the same as a fucto from [] to [m] up to permutatos of m, that s, a placemet of dstgushable objects to m dstgushable boxes. 3

4 3. We clam that for 0 ad 0 k, p( (r ) r k. Ths s easy to prove by ducto o : f 0, ths s gve. Otherwse, p( p(k + ) p( (r ) r k+ (r ) r k (r ) r k. Thus, for 0, p(0) (r ), whle for >, p(0) 0 sce p s a polyomal of degree. It follows that ( ) x ( ) x p(x) p(0) (r ). 0 (Note that f x 0,,...,, the ths sum evaluates to ( + (r )) x r x by the bomal theorem, but ot for x + sce t s mssg a term.) Thus p( + ) 0 ( ) + (r ) 0 r + (r ) (a) We clam a 0 D, the th deragemet umber. To see ths, we eed oly verfy that! D k(. The left sde couts all permutatos of [], ad the terms o the rght sde cout permutatos wth exactly k fxed pots (choose the fxed pots ( ( ways, the choose a deragemet of the other k elemets). Oe could also see ths by drect computato. Recall that E, where E s the shft operator. The whch s the formula for D. a 0 (E ) a 0 ( ) ( ) k E k a 0 k ( ) k! k!(! a k ( ) k!, k! (b) We clam a 0 B, the th Bell umber. To see ths, we eed oly verfy that B + B k(. The left sde couts set parttos of [ + ]. Suppose + les a block wth k other elemets. The there are ( ( ways to choose those elemets, ad B k ways to partto the other k elemets to blocks, whch gves the terms o the rght had sde. Alteratvely, oe ca solve both these parts by relatg the expoetal geeratg fuctos f(x) a x to g(x)! x a 0. Ideed, a! ( k a 0 mples 4

5 f(x) e x g(x). I part (a), a! has geeratg fucto, so the aswer x must have geeratg fucto e x, whch s the geeratg fucto for deragemets. x Smlarly, part (b), a B + has geeratg fucto B (x) e x e ex, where B(x) e ex s the geeratg fucto for the Bell umbers. Thus the aswer must have geeratg fucto e x B (x) e ex B(x). 5