1 Generating functions, continued
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1 Generatng functons, contnued. Generatng functons and parttons We can make use of generatng functons to answer some questons a bt more restrctve than we ve done so far: Queston : Fnd a generatng functon for the number of ways to dstrbute n balls among 3 boxes f the frst box can contan any number of balls, the second box contans an odd number of balls, and the thrd contans a multple of 5 less than 5. Answer : The procedure for fllng the frst box s known to be represented by the g.f. +x+x 2 + x 3 + x. The second, however, would be x+x3 +x 5 +x 7 + x+x 2 +x 4 +x 6 + ) x, x 2 and the thrd would be + x 5 + x 0 x5 x x. In total, we would have 5, whch s not x 5 x) x 2 ) x 5 ) tremendously amenable to smplfcaton, but we could actually do t wth partal fractons! ths one was done wth computer assstance) 3 2 x) x) 4 + x) x + 5x2 + 4x 3 + 3x 4 + 2x 5 + 2x 6 + x 7 + x 8 The three terms at the begnnng of ths expanson expand to 3 n+2 ) 2 2 x n 33 4 xn 4 x)n So for x > 8, the coeffcent of x n s 3 2 n + ) 33 4 )n 4. Note that we could ve phrased ths another way: Queston 2: Fnd a generatng functon for the number of nonnegatve nteger solutons to x + 2x 2 + ) + 5x 3 n, wth 5x 3 < 30. whch means we can now produce generatng functons f not smple answers to the number of ways to dvde any n nto lnear combnatons. Here s one such queston whch has a certan amount of everyday relevance: x + 5x 2 + 0x x x x 6 n A nonnegatve nteger soluton to ths s equvalent to a way of makng change for n cents n standard Amercan change ncludng the half-dollar and dollar con). The generatng functon for ths can be worked out to be x) x 5 ) x 0 ) x 25 ) x 50 ) x 00 ) Tryng to actually calculate ths, even wth partal fractons, s not recommended. However, usng partal sums and computer algebra systems, we can fnd values for ndvdual entres: Queston 3: How many ways are there to make change for a dollar? Answer 3: From the above, we can get an approxmaton suffcent for workng out the frst hundred terms of the sequence: +x+x 2 + +x 00 )+x 5 +x 0 + +x 00 )+x 0 +x x 00 )+x 25 +x 50 +x 75 +x 00 )+x 50 +x 00 )+x 00 ) whch gves us a 600th-degree polynomal: + x + x 2 + x 3 + x 4 + 2x x x 600. We can thus see that there are 293 ways to make change for a dollar. Page of 7 September 8, 2009
2 Ths all brngs us towards nteger parttons. Above the queston was really how to dvde n nto s, 5s, 0s, 25s, 50s, and 00s. If we look at nteger parttons, we can see that we re really dvdng a number up nto several value classes, e.g. we recall that p5) 7 wth the parttons 5, 4 +, 3+2, 3++, 2++++, 2+2+, and If we allow x to represent the number of s used n a partton, we can see that these parttons are n a one-to-one correspondence wth solutons to x + 2x 2 + 3x 3 + 4x 4 + 5x 5 5, whch would be the x 5 coeffcent of the generatng functon x) x 2 ) x 3 ) x 4 ) x 5 ). Ths gets more troublesome f we wanted a generatng functon for arbtrary partton numbers, but we can at least present a symbolc form, f not an actual formula: p n x) x 2 ) x 3 ) x 4 ) x ) And specfc formulas good enough for fndng specfc coeffcents can be produced by usng partal products and partal sums. Some other neat partton problems: Queston 4: How many ways are there to express n as a sum of dstnct ntegers? Answer 4: Ths s, pleasantly enough, an easer queston than the unrestrcted partto Instead of countng the number of each value that appears wth the expresson +x +x 2 +x 3 +, x we nstead have each number ether appearng once or not at all, so the expresson to represent ths selecton s + x. Thus our total generatng functon s + x) + x 2 ) + x 3 ) + x 4 ) + + x ) and the answer to the queston above s the coeffcent of x n whch does not have a closed form, but s amenable to farly easy symbolc calculaton). Queston 5: How many ways are there to express n as a sum of ntegers no greater than k? Answer 5: Here, nstead of an nfnte number of sub-processes, we only need k of them, so we have for once a generatng functon whch s a fnte product: x) x 2 ) x 3 ) x 4 ) x k ) Now we modfy ths slghtly n a way that wll be useful for us n actually returnng to our permutaton statstcs. Queston 6: How many ways are there to express n as sum of ntegers no greater than k, one of whch s k tself? Answer 6: Ths s as above, except nstead of the kth process beng + x k + x 2k +, x k we have x k + x 2k + xk, so the total generatng functon s x k x k x) x 2 ) x 3 ) x 4 ) x k ) Page 2 of 7 September 8, 2009
3 Ths wll actually brng us to the purpose of creatng a generatng functon for the one statstc whch we handwaved when buldng the twelve-fold way. We could note by observaton that the number of ways to partton n nto nonzero parts of whch the largest s k s equal to the number of ways to partton n nto k nonzero parts, as n ths example for n 8 and k 3: To exhbt ths relatonshp, we have recourse to a vsual technque for presentng parttons: Defnton. The Ferrers dagram for the partton a + a 2 + a a k for a a 2 a 3 a k > 0 conssts of k left-justfed rows of equally-spaced dots wth a dots n the th row, for each. For nstance, here we have a Ferrers dagram for : An alternatve to the Ferrers dagram s a fgure drawn wth boxes nstead of dots. Such a fgure s called a Young dagram, and can be conventonally drawn n one of two styles: wth the largest row on top n what s known as the Englsh-style, shown below on the left) or on bottom whch s French-style, shown below on the rght). Any of these vsualzatons wll work equally well. There are several useful geometrcally-determned propertes of Ferrers dagrams, lsted wthout proof below the proofs are largely straghtforward results of the defnton of the Ferrers dagram). Assumng a Ferrers dagram s assocated wth the partton a + a a k for a a 2 a 3 a k > 0, we have two partcularly useful geometrc propertes. The Ferrers dagram has a heght of k.e. a total of k rows). The dagram has a wdth of a.e. a total of a columns). A Ferrers dagram unquely determnes, and s unquely determned by, a partton. Thus, we may conclude that p k n) s the number of Ferrers dagrams wth n dots of heght k; n Queston 5 we determned the number of Ferrers dagrams of wdth k, and above we hnted that these mght be the same quantty. In fact we can fnd an explct bjecton: Page 3 of 7 September 8, 2009
4 Defnton 2. The transpose or conjugate of a Ferrers dagram s a dagram produced by exchangng the rows and columns of the dagram. Alternatvely, the conjugate of a partton a +a 2 +a 3 + +a k wth a a 2 a k > 0 s a partton b + b 2 + b b a, where b {j : a j }. The second defnton above s a drect defnton of a partton conjugate wthout appeal to geometrc ntuton. Below s an exhbt of conjugaton appled to a Ferrers dagram, or alternatvely to the assocated partton: Note that two successve conjugatons return the orgnal Ferrers dagram; note also that dstnct dagrams have dstnct conjugates. Snce conjugaton swaps heght and wdth, conjugaton llustrates that parttons nto k parts, and parttons wth one part of sze k and other parts of sze k are n fact equnumerous, and snce we had a generatng functon for the latter, we can use t for the former, gvng us a generatng functon for p k n): p k n)x n x k x) x 2 ) x 3 ) x 4 ) x k ) In addton, f we were to conjugate a partton wth a maxmum entry of k, we would get a partton wth a maxmum length of k; ths would correspond to a twelvefold way statstc wthout surjecton, whch we know to be p 0 n) + p n) + p 2 n) + + p k n), whch gves us the not entrely obvous algebrac dentty: k x) x 2 ) x 3 ) x 4 ) x k ) + x x) x 2 ) x 3 ) x 4 ) x ).2 Exponental generatng functons So far, we ve looked at algebrac representatons for dstrbuton of ndstnct objects, and we ve used that to shed a great deal of lght on two rows of our twelvefold way. Can we do the same for dstnct objects? Suppose we have two processes for dstrbutng n dstnct objects; we can thnk of them as sequences a 0, a, a 2, a 3,...) and b 0, b, b 2, b 3,...). We mght ask what the sequence s for dstrbutng n objects usng one or the other but not both) of these processes: ths would be a 0 + b 0, a + b, a 2 + b 2,...). There are plenty of algebrac structures where ths would be smple addton. How about dstrbuton usng these processes alternatngly? When the objects were ndstngushable, recall that ths was a 0 b 0, a 0 b + a b 0, a 0 b 2 + a b + a 2 b 0, ). However, here we must not Page 4 of 7 September 8, 2009
5 only decde how many objects to deal wth usng process A and how many usng process B, but n addton whch objects we deal wth usng each process. If we ncorporate ths nformaton, then we have, a + b, a 2 + 2a b + b 2, a 3 + 3a b 2 + 3a 2 b + b 3 ). The way to dspose of n objects wll end up beng a 0 b n + ) n a b n + + n ) a b n + a n b 0. Ths s actually farly sensble. We can splt our n objects up as dealt wth usng process A and n usng process B n n ) ways; then there are a ways to place the selected objects wth process A and b n ways to place the selected objects wth process B. But now we ask: what useful algebrac object has ths property, that multplyng two assocated wth the sequences a 0, a,...) and b 0, b,...) gves a product assocated wth the sequence a 0 b 0, a 0 b + a b 0, a 0 b 2 + 2a b + a 2 b 0,...)? The answer s a surprsng tweak on the generatng functons we have seen to date: Defnton 3. If a n enumerates a number of ways to place, select, or otherwse arrange n dstnct objects, then fx) anxn s called the exponental generatng functon for a n. Now, note that multplcaton of two generatng functons n fact produces the desred result: ) ) a n b n xn xn a0 b n 0! + a b n! n )! + + a ) n b 0 x n 0! 0! a 0b n +!n )! a b n + + ) x n 0! a nb 0 ) ) ) n n n x n a 0 b n + a b n + + )a n b 0 0 n So ths allows us to construct generatng functons from several ndependent dstrbuton procedures for dstnct tems. As n the case of the ordnary generatng functon, let s start by tryng ths out wth a statstc we already know: Queston 7: Fnd an exponental generatng functon to enumerate the ways to place n dstnct balls n r boxes, so that no box contans more than one ball. Answer 7: Let us start by notng that the enumeraton statstc for such a placement s known to be r n), and we wll judge our procedure a success f t yelds ths expresson as coeffcents. ) ) The generatng functon for puttng balls n a sngle box wll be +. We may ether place no balls n t n only one way), or place the only labeled ball we have on hand n t n only one way). The concatenaton of r such procedures can be performed by multplyng ths quantty by tself r tmes, so our exponental generatng functon here would be +x) r. If we use the bnomal theorem on ths, we would get that the generatng functon s r r n) x n ; however, n an exponental generatng functon we look at coeffcents not of x n but of xn ; t s thus expedent to rewrte ths form as r ) r x n ) n. We can also fnd ways to symbolcally express and answer questons whch would have been nsanely dffcult before: x 0 0! x! Page 5 of 7 September 8, 2009
6 Queston 8: If we have 3 boxes, each of whch can contan no more than 3 balls, how many ways are there to dstrbute n labeled balls among them? Answer 8: Each box can be flled wth ether zero balls n one way, ball n one way, 2 balls n one way, or 3 balls n one way: thus ths procedure has generatng functon ) ) ) ) x 0 x x 2 x x + x2 0!! 2! 3! 6 And concatenaton of the three procedures gves ) 3 + x + x2 + 3x x x x x x x x x9 + 3x + 9 x x x4 4! + 20x5 5! + 50x6 6! + 050x7 7! + 680x8 8! + 680x9 9! The coeffcents on xn for n from 0 to 9 gves the number of ways to dstrbute n dstnct balls for n > 9, of course, such a dstrbuton s mpossble, so there are zero ways to do so). Of course, the real fun comes when our functons are seres nstead of polynomals! As s often the case, we ll start wth a queston to whch we already know the answer. Queston 9: Fnd an exponental generatng functon to enumerate the ways to place n dstnct balls n r boxes, f each box can contan any number of balls. Answer 9: As before, we know ths enumeraton statstc, and t s r n, so we have a reasonable expectaton of gettng that expresson appearng n the coeffcents of the generatng functon. The procedure for fllng one box s, as we saw n the prevous examples, the sum of monomals ndcatng that we can put labeled balls n the box n exactly one way, that s, x!. So, addng up all such monomals, we get that the exponental generatng functon assocated wth puttng balls nto a sngle box s: + x + x2 6 + x4 4! + x5 5! + whch s a seres we can descrbe succnctly wth a functon ths s the well-known Taylor seres for e x, so the exponental generatng functon for the procedure of puttng n dstnct balls n a sngle box freely s e x! Snce we have r boxes to fll, the generatng functon for fllng all x of them s the product of r copes of ths generatng functon, that s, e x ) r e rx. Re-expressng ths as a seres, we see that the generatng functon for ths entre procedure s e rx rx) n r n xn whch meets our expectatons. We ve flled n two cells of row of the twelvefold way. Let s try to do the thrd whch we orgnally needed ncluson-excluson to handle! Queston 0: Fnd an exponental generatng functon to enumerate the ways to place n dstnct balls n r boxes, f each box must contan at least one ball. Answer 0: The procedure for fllng one box s the sum of the same terms as were used n the prevous example, but we now forbd the possblty of leavng the box empty, so our exponental generatng functon assocated wth puttng balls nto a sngle box wll be x + x2 6 + x4 4! + x5 5! + ex ) Page 6 of 7 September 8, 2009
7 so the exponental generatng functon for the process of processng r boxes n ths way s e x ) r. However, f we want to get ths exponental generatng functon nto a state where ndvdual coeffcents can be calculated, we ll need to expand ths a bt further. e x ) r r ) r e x ) ) r r ) r ) r e x r ) r ) r n xn r r ) ) r n xn 0 r ) r ) ) r n x n Page 7 of 7 September 8, 2009
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