Two Coefficients of the Dyson Product

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Two Coeffcents of the Dyson Product rxv:07.460v mth.co 7 Nov 007 Lun Lv, Guoce Xn, nd Yue Zhou 3,,3 Center for Combntorcs, LPMC TJKLC Nnk Unversty, Tnjn 30007, P.R. Chn lvlun@cfc.nnk.edu.cn gn@nnk.edu.cn 3 nkzhouyue@gml.com November, 007 Abstrct In ths pper, the closed-form epressons for the coeffcents of r nd r s s t n the Dyson product re found by pplyng n etenson of Good s de. As consequences, we fnd severl nterestng Dyson style constnt term denttes. Mthemtcs Subject Clssfcton. Prmry 05A30, secondry 33D70. Key words. Dyson conjecture, Dyson product, constnt term Introducton For nonnegtve ntegers,,..., n, defne D n, :, Dyson product j j n where :,..., n nd :,..., n. Dyson conjectured the followng constnt term dentty n 96. Theorem. Dyson s Conjecture. D n, + + + n!.!! n! where f mens to tke the constnt term n the s of the seres f.

Dyson s conjecture ws frst proved ndependently by Gunson5 nd by Wlson0. Lter n elegnt recursve proof ws publshed by Good4. Andrews conjectured the q-nlog of the Dyson conjecture whch ws frst proved, combntorlly, by Zelberger nd Bressoud n 985. Recently, Gessel nd Xn 3 gve very dfferent proof by usng propertes of forml Lurent seres nd of polynomls. Good s de hs been etended by severl uthors. The current nterest s to evlute the coeffcents of monomls M of degree 0 n the Dyson product, where M : n 0 b wth n 0 b 0. Kdell 6 outlned the use of Good s de for M to be n, n n nd. Along ths lne, Zelberger nd Slls 9 presented cse n study n epermentl yet rgorous mthemtcs by descrbng n lgorthm tht utomtclly conjectures nd proves closed-form. Usng ths lgorthm, Slls 8 guessed nd proved closed-form epressons for M to be s nd tu r s. These results nd r, st r ther q-nloges were recently generlzed for M wth squre free numertor by Lv, Xn nd Zhou 7 by etendng Gessel-Xn s Lurent seres method 3 for provng the q-dyson Theorem. The cses for M hvng squre n the numertor re much more complcted. By etendng Good s de, we obtn closed forms for the smplest cses M r nd s M r s t. In dong so, we guess these two formuls smultneously, wrtten s sum nsted of sngle product. Our mn results re stted s follows. Theorem.. Let r nd s be dstnct ntegers wth r,s n. Then D n, r s r r + r + r r,s + + r where : + + + n, j : j nd C n : + + + n!!! n!. Theorem.3. Let r,s nd t be dstnct ntegers wth r,s,t n. Then + r s t D n, r r + r + r where, r nd C n re defned s Theorem.. r,s,t + + r C n,. C n,. The proofs wll be gven n Secton. In Secton 3, we construct severl nterestng Dyson style constnt term denttes. Proof of Theorem. nd Theorem.3 Good s proof 4 of the Dyson Conjecture uses the recurrence D n, D n,e k, k

where e k : 0,...,0,,0,...,0 s the kth unt coordnte n-vector. It follows tht the followng recurrence holds for ny monoml M of degree 0. M D n, k M D n,e k. Thus f we cn guess formul, then we cn prove t by checkng the ntl condton, the recurrence nd the boundry condtons. Ths s the so clled Good-style proof. In our cse, denote by F L r,s, resp. G L r,s,t, the left-hnd sde of. resp.., nd by F R r,s, resp. G R r,s,t, the rght-hnd sde of. resp... Wthout loss of generlty, we my ssume r,s nd t 3 n Theorems. nd.3,.e., we need to prove tht F L F R, G L G R, where F L : F,, nd we use smlr nottons for F R,G L nd G R.. Intl Condton We cn esly verfy tht F L 0 F R 0 0, G L 0 G R 0 0.. Recurrence We need to show tht F R nd G R stsfy the recurrences In order to do so, we defne F R G R F R e k,. k G R e k.. k H : + + C n, + H : + + C n, + H : + + + C n, 3,4,...,n. Then F R H + n 3 H nd G R H + n 4 H. Therefore to show. nd., t suffces to show the followng: 3

Lemm.. For ech,,...,n, we hve the recurrence H n k H e k. Proof.. For H, H e k + + C ne + + C ne k k k + + + k + C n k + + + + C n + + C n H.. For H, H e k + + + C ne + k k + k + + + + + k + + + + + + C n + + + C n H. + + C ne k C n 3. For H wth 3,...,n, wthout loss of generlty, we my ssume 3. H 3 e k k 3 + + + 3 C 3 ne + + 3 C ne 3 + + + 3 C 3 ne 3 + + 3 C ne k 3 + + + 3 C 3 n+ + 3 C n 3 3 + + + 3 C n+ 3 3 + 3 C n + + 3 + 3 + 3 3 k4 3 + + 3 3 + + + + 3 C n H 3. Ths completes the proof. 4 C n

.3 Boundry Condtons Now we consder the boundry condtons. For ny k wth k n, D n,,..., k,0, k+,..., n D n k, k where k :,..., k, k+,..., n. Thus we hve n k k D n,,..., k,0, k+,..., n P k D n k, k,.3 k where P k s gven by P k : k n k k 0, k ; n + 3 + n 3 + j j, k ; 3 <j n,, otherwse. Tkng the constnt term n the s of.3, we obtn F L,..., k,0, k+,..., n 0, k ; + + n + j j D 3 3 n,, k ; 3 <j n D k n k, k, otherwse. By Theorem. nd 8, Theorem., we hve 3 D n, C n, D n, + C n. So we obtn the followng boundry condtons lso recurrences F L,..., k,0, k+,..., n 0, k ; C + n + n FL,, + j G L,,j,, k ; 3 3 <j n F L k, otherwse. 5

We need to show tht F R,..., k,0, k+,..., n stsfes the sme boundry condtons. More precsely, the condtons by replcng ll F L by F R nd ll G L by G R : F R,..., k,0, k+,..., n 0, k ; C + n + n FR,, + j G R,,j,, k ; 3 3 <j n F R k, otherwse..4 Smlr computton for M 3 yelds the boundry condtons: G L,..., k,0, k+,..., n 0, k ; C + n 3 F L,3, n 4 G L,3,,, k ; C + n 3 F L,, 3 n 4 G L,,, 3, k 3; G L k, otherwse, so we need to prove the boundry condtons for G R,..., k,0, k+,..., n : G R,..., k,0, k+,..., n 0, k ; C + n 3 F R,3, n 4 G R,3,,, k ; C + n 3 F R,, 3 n 4 G R,,, 3, k 3; G R k, otherwse..5 These re summrzed by the followng lemm. Lemm.. If k 0 wth k,,...,n, then F R,..., k,0, k+,..., n stsfes the boundry condtons.4 nd G R,..., k,0, k+,..., n stsfes the boundry condtons.5. Proof. We only prove the frst prt for brevty nd smlrty. Snce the cses k,3,...,n re strghtforwrd, we only prove the cse k. Notethtdurng theproofof thslemm, we hve + 3 + + n 3 + + n becuse 0. 6

3 Snce n 3 j3 j j + j n 3 j3 n j + j j3 we hve F R,, + + 3 3 j + j 3 j3 j3 3 j + j 3 + n j + j j3 j j + + j 3 C n + 3 +,.6 + n j n j + + + j3 j + 3 j3 j 3 C + n n + 3 + + C n by.6 3 + + C n j λ + + j j3 j3 3 3 j + j + j + j 3,.7 where λ : C + + n. Observe tht 3 <j n j k3 k.8 nd 3 <j n j n k3 k,j 3 <j n k3 k3 k + k j n k k3 + k k + k 3 <j n n j k + k n 3 n 3 k3 3 <j n k + k j + 3 3 j3 j + 3 <j n j + j j + j j + by.8 +..9 7

Thus we obtn tht j G R,,j, 3 <j n Observe tht nd + + + + + k3 3 <j n 3 <j n + + + 3 j + j C n 3 <j n j n k3 k,j k3 k,j + + + k k3 n k n + + k + + + C n +λ + + k + + k k + k C n C n + 3 C n + + + + k + + k k3 n k + k3 3 j j + j + + j j3 k3 3 C n by.9 3 k k + 3 k + k..0 + k + k k + 3 k + k3 k3 k 3 + + + + + 3 3 + + + + + 3 3 + + + 3 3 + + +. 3 + + + + + + + + + + + + +.. 8

Therefore by.7,.0,. nd., we hve + C n + F R,, + j G R,,j, 3 3 <j n F R,0, 3,..., n. Tht s to sy F R,0, 3,..., n stsfes boundry condtons.4..4 The Proof Now we cn prove Theorems. nd.3. Wthout loss of generlty, we my ssume r,s nd t 3 n Theorems. nd.3. Proof of Theorems. nd.3. We prove by nducton on n for the two theorems smultneously. Clerly,. nd. hold when n,3. Assume tht. nd. hold wth n replced by n. Thus for k,,...,n. nd. gve F L r,s, k F R r,s, k, G L r,s,t, k G R r,s,t, k. ThtstosyF L,..., k,0, k+,..., n ndf R,..., k,0, k+,..., n resp. G L,..., k,0, k+,..., n nd G R,..., k,0, k+,..., n stsfy the sme boundry condtons. Addtonlly F L nd F R resp. G L nd G R hve the sme ntl condton nd recurrence. It follows tht F L F R resp. G L G R. 3 Severl Dyson Style Constnt Term Identtes By lnerly combnng Theorems. nd.3, we obtn smple formuls. Proposton 3.. Let r,s,t,u, nd v be dstnct ntegers n {,,...,n}. Then s t u v D n, 0, 3. r r + s u s v r s t D n, r+ r + r s,t D n, + r + r s C n, 3. + r C n. 3.3 It s worth mentonng tht 3.3 follows from 3. nd 3., snce s u s v + t u t v s t + s u t v + s v t u. A consequence of Proposton 3. s the followng: 9

Corollry 3.. Let I:{,,..., m } be m-element subset of {,,...,n} nd r n s postve nteger wth r I. Then we hve m j j j Proof. Observe tht r m j j j D n, r+ + r j I + r j C n. + 4 3 + + m m + + m m + The corollry then follows by 3. nd 3.3. m k m k k l l. l l k Dscussons: Aswehveseenntheproof,weneedtoguesstheformulsofF R ndg R smultneously. Ths s unlke the coeffcents for M s t / u nd M s t / u v, whch hve resonble product formuls nd re equl! The net cses should be M wth r s or 3 r n the numertor, both hvng three cses for the denomntor. The dffculty s: guess three coeffcents smultneously; obtn enough dt. The study of the q-nloges of these formuls wll be n completely dfferent route nd wll not be dscussed n ths pper. Acknowledgments. We would lke to cknowledge Zelberger nd Slls Mple pckge GoodDyson, whch s vlble from the web stes http://www.mth.rutgers.edu/ ~zelberg nd http://mth.georgsouthern.edu/~slls. Ths work ws supported by the 973 Project, the PCSIRT project of the Mnstry of Educton, the Mnstry of Scence nd Technology nd the Ntonl Scence Foundton of Chn. References G. E. Andrews, Problems nd prospects for bsc hypergeometrc functons, n Theory nd Applcton of Specl Functons, ed. R. Askey, Acdemc Press, New York, 975, pp. 9 4. F. J. Dyson, Sttstcl theory of the energy levels of comple systems I, J. mth. Phys. 3 96, 40 56. 3 I. M. Gessel nd G. Xn, A short proof of the Zelberger-Bressoud q-dyson theorem, Proc. Amer. Mth. Soc. 34 006, 79 87. 0

4 I. J. Good, Short proof of conjecture by Dyson, J. Mth. Phys. 970, 884. 5 J. Gunson, Proof of conjecture by Dyson n the sttstcl theory of energy levels, J. Mth. Phys. 3 96, 75 753. 6 K. W. J. Kdell, Aomoto s mchne nd the Dyson constnt term dentty, Methods Appl. Anl. 5 998, 335 350. 7 Lun Lv, Guoce Xn nd Yue Zhou, A Fmly of q-dyson Style Constnt Term Identtes, submtted, rxv:0706.009. 8 A. V. Slls, Dsturbng the Dyson conjecture, n generlly GOOD wy, J. Combn. Theory Ser. A 3 006, 368 380. 9 A. V. Slls nd D. Zelberger, Dsturbng the Dyson conjecture n Good wy, Eperment. Mth. 5 006, 87 9. 0 K. G. Wlson, Proof of conjecture by Dyson, J. Mth. Phys. 3 96, 040 043. D. Zelberger nd D. M. Bressoud, A proof of Andrews q-dyson conjecture, Dscrete Mth. 54 985, 0 4.