In 1962, Dyson [2] conjectured the following constant term identity.. In recent years, there has been increasing interest in evaluating the coefficients of i=1xbi i , where Pn i=1bi = 0,
Trang 1A New Approach to the Dyson Coefficients
Sabrina X.M Pang College of Mathematics and Statistics Hebei University of Economics and Business Shijiazhuang 050061, P.R China stpangxingmei@heuet.edu.cn
Lun Lv∗
School of Science Hebei University of Science and Technology Shijiazhuang 050018, P.R China klunlv@gmail.com Submitted: Feb 16, 2010; Accepted: Aug 16, 2010; Published: Aug 24, 2010
Mathematics Subject Classifications: 05A30, 33D70
Abstract
In this paper, we introduce a direct method to evaluate the Dyson coefficients
In 1962, Dyson [2] conjectured the following constant term identity
Theorem 1.1 (Dyson’s Conjecture) For nonnegative integers a1, a2, , an,
CT
x
Dn(x, a) = (a1+ a2+ · · · + an)!
a1! a2! · · · an! , where CTxf (x) denotes the constant term and
16i6=j6n
1 − xi
xj
a i
∗ Corresponding author
Trang 2Dyson’s conjecture was proved independently by Gunson [5] and Wilson [11] In
1970, a brief and elegant proof was published by Good [4] Later Zeilberger [13] gave a combinatorial proof
The q-analog of Theorem 1.1 was conjectured by Andrews [1] in 1975, and was first proved, combinatorially, by Zeilberger and Bressoud [14] Recently, Gessel and Xin [3] gave a different proof by using properties of formal Laurent series
In recent years, there has been increasing interest in evaluating the coefficients of
i=1xbi
i , where Pn
i=1bi = 0, in the Dyson product Based on Good’s proof, Kadell [6] gave three non-constant term coefficients Sills and Zeilberger [10] de-scribed an algorithm that automatically conjectures and proves closed-form expressions Later, Sills [9] extended Good’s idea and obtained the closed-form expressions for M being
x s
x r, x s x t
x 2
r , x t x u
x r x s, respectively By virtue of Zeilberger and Sills’ Maple package GoodDyson,
Lv, Xin and Zhou [7] found two closed-form expressions for M that has a square in the numerator Moreover, by generalizing Gessel-Xin’s method [3] for proving the Zeilberger-Bressoud q-Dyson Theorem, Lv, Xin and Zhou [8] established a family of q-Dyson style constant term identities
In this note, we propose a direct calculation approach to evaluating the coefficients
in the Dyson product, and illustrate this approach through the case of M = x2
r/x2
s The applications of our method to other cases like M = x2r
x s x t, M = x r
x s are analogous, and thus omitted More explicitly, we will show that our approach leads to the following theorem Theorem 1.2 (Theorem 1.2 [7]) Let r and s be distinct integers with 1 6 r, s 6 n Then
CT
x
x2
s
x2
r
Dn(x, a) = ar
(1 + a(r))(2 + a(r))
(ar−1) −
n
X
i=1 i6=r,s
ai(1 + a) (1 + a(r)−ai)
Cn(a), (1.1)
where a := a1+ a2+ · · · + an, a(j):= a − aj and Cn(a) := (a1 +a 2 +···+a n )!
a 1 ! a 2 ! ··· a n !
In this section, we will deduce the coefficient for M = x2r
x 2
s By induction on n, we have the following identity,
n
X
k=2
(m + k − 1)!
(k − 2)! =
(m + n)!
(m + 2)(n − 2)!, m, n ∈ N. (2.1)
Let
∆(x1, x2, , xn) :=Y
i<j
(xi−xj) =
xn−11 xn−12 · · · xn−1n
xn−21 xn−22 · · · xn−2
n
. . .
be the Vandermonde determinant in x1, x2, , xn Then [12] presents the following result
Trang 3Lemma 2.1 (Lemma 1-2.12, [12]) For each i = 1, 2, , n, if f (xi) ∈ C((xi)), then we have
∂x 2∂x 3· · ·∂x nf (x1) = ∆(x1, x2, , xn)−1
f (x1) f (x2) · · · f (xn)
xn−21 xn−22 · · · xn−2n
(2.2)
=
n
X
i=1
f (xi) Q
where ∂af (x) := f(x)−f (a)x−a
The following lemma is vital to our approach
Lemma 2.2 (Main Lemma) For n > 2, we have
V1
x1
+ V2
x2
+ · · · + Vn
xn
x1
x2
+ · · · + 1
xn
where Vm :=
n
Q
i=1 i6=m
1 −x m
x i
−1
for m = 1, 2, , n
Proof Let f (xi) = x12
i for i = 1, 2, , n First we claim that
∂x 2∂x 3· · ·∂x nf (x1) = (−1)
n−1
x1x2· · ·xn
1
x1
x2
+ · · · + 1
xn
We prove (2.5) by induction on n Clearly, (2.5) holds when n = 2 Assume that (2.5) holds with n replaced by n − 1 Then we have
∂x2∂x3· · ·∂xnf (x1) = ∂x2
(−1)n−2
x1x3· · ·xn
1
x1 +
1
x3 + · · · +
1
xn
by induction hypothesis
=
(−1) n−2
x 1 x 3 ···x n
1
x 1 +x13 + · · · +x1
n
−
(−1) n−2
x 2 x 3 ···x n
1
x 2 +x13 + · · · +x1
n
x1−x2
=(−1)
n−2
x3· · ·xn
1
x2 1
− 1
x2 2
+ 1
x1x3 −
1
x2x3
+ · · · + 1
x1xn
x2xn
1
x1−x2
=(−1)
n−2
x3· · ·xn
−x1+ x2
x2
1x2 2
x1x2x3 − · · · −
1
x1x2xn
= (−1)
n−1
x1x2· · ·xn
1
x1 +
1
x2 + · · · +
1
xn
Trang 4
Furthermore, it follows by (2.3) that
n
X
i=1
1/x 2 i
Q
j6=i (x i−x j ) =
(−1) n−1
x 1 x 2 · · · x n
1
x 1 + 1
x 2 + · · · + 1
x n
⇔ x 1 x 2· · ·x n
n
X
i=1
1/x 2 i
Q
j6=i (x j−x i ) =
1
x 1
+ 1
x 2
+ · · · + 1
x n
⇔
n
X
i=1
1
x i
Q
j6=i (1 − x i /x j ) =
1
x 1
+ 1
x 2
+ · · · + 1
x n
⇔ V 1
x 1
+V2
x 2
+ · · · + Vn
x n
= 1
x 1
+ 1
x 2
+ · · · + 1
x n
.
This completes the proof
Now we are ready to prove Theorem 1.2 Without loss of generality, we may assume
r = 1 and s = 2 in Theorem 1.2
A new approach to Theorem 1.2 By (2.4) we have
V1−1
x1
+ V2−1
x2
+ · · · + Vn−1
xn
= 0
Multiplying both sides by x 2
V 4 −1 yields
x2
x4 =
(1 − V1)x2 (V4−1)x1 +
1 − V2
V4−1 +
(1 − V3)x2 (V4−1)x3 +
(1 − V5)x2 (V4−1)x5 + · · · +
(1 − Vn)x2 (V4−1)xn. (2.6)
Note that Dn(x, a) = V−a 1
1 V−a 2
2 · · ·V−a n
n , (2.6) implies that
x 2
x 1 x 4
D n (x, a) = x
2
x 1 x 4
n
Y
j=1
V−aj
j
= x2
x 1
(1 − V 1 )x 2
(V 4 − 1)x 1
+1 − V2
V 4 − 1 +
(1 − V 3 )x 2
(V 4 − 1)x 3
+(1 − V5)x2 (V 4 − 1)x 5
+ · · · +(1 − Vn)x2
(V 4 − 1)x n
n
Y
j=1
V−aj
j (2.7)
Multiplying both sides by V4−1 and taking the constant term in the x’s, (2.7) can be rewritten as follows
F (a 1 ) − F (a 1 − 1) = CT
x
x 2
x 1 (V 2 − 1) + x
2
x 1 x 3 (V 3 − 1) + · · · + x
2
x 1 x n
(V n−1)
n
Y
j=1
V−aj
where F (a1) := CTxxx22
Qn j=1V−aj
j For j = 3, 4, , n, observe that
CT
x
x22
x 1 x j
(V j−1)
n
Y
j=1
V−aj
j = CT
x
x22
x 1 x j
D n x , (a 1 , , a j−1 , a j−1, a j+1 , , a n )
− CT
x
x22
x 1 x j
D n (x, a)
=
a 1 + a j−1
1 + a − a 1 − a j
− a 1
a − a 1
− a j−1
1 + a − a j
a j
aCn(a)
−
a 1 + a j
1 + a − a 1 − a j
1 + a − a 1
1 + a − a j
C n (a) by [9, Theorem 1.4]
= −
a 1 a j
(1 + a − a 1 )(1 + a − a 1−a j )+
a 1 a j
a(a − a 1 )
Trang 5
CT
x
x 2
x 1 (V 2 − 1)
n
Y
j=1
V−aj
j = CT
x
x 2
x 1 D n x , (a 1 , a 2 − 1, a 3 , , a n )
− CT
x
x 2
x 1 D n (x, a)
=
− a 1
a − a 1
· a 2
a +
a 1
1 + a − a 1
C n (a) by [9, Theorem 1.1]
=
a 1
1 + a − a 1
− a 1 a 2
a(a − a 1 )
Combining (2.8), (2.9) and (2.10), we obtain the following recurrence
F (a 1 ) − F (a 1−1)
=
a 1
1 + a − a 1
− a 1 a 2
a(a − a 1 )−
n
X
j=3
(1 + a − a 1 )(1 + a − a 1−a j )+
a 1 a j
a(a − a 1 )
C n (a)
=
a 1
1 + a − a 1
− a 1 a 2
a(a − a 1 )−
a 1 (a − a 1 − a 2 ) a(a − a 1 ) −
n
X
j=3
a 1 a j
(1 + a − a 1 )(1 + a − a 1−a j )
C n (a)
=
a 1 (a 1 − 1)
a(1 + a − a 1 )−
n
X
j=3
a 1 a j
(1 + a − a 1 )(1 + a − a 1−a j )
Further noting that F (0) = 0, which can be easily verified, (2.11) finally gives
F (a 1 ) =
a1
X
k=1
k(k − 1)(a − a 1 + k)!
(1 + a − a 1 )(a − a 1 + k)k!−
a1
X
k=1
n
X
j=3
ka j (a − a 1 + k)!
(1 + a − a 1 )(1 + a − a 1−a j )k!
1
a 2 ! · · · a n !
=
a1
X
k=2
(a − a 1 + k − 1)!
(1 + a − a 1 )(k − 2)!−
a1
X
k=1
n
X
j=3
ka j (a − a 1 + k)!
(1 + a − a 1 )(1 + a − a 1−a j )k!
1
a 2 ! · · · a n !
=
a 1 (a 1 − 1) (1 + a − a 1 )(2 + a − a 1 )·
a!
a 1 ! by (2.1) for the case n = a1 and m = a − a1.
−
n
X
j=3
a1
X
k=1
ka j (a − a 1 + k)!
(1 + a − a 1 )(1 + a − a 1 − a j )k!
1
a 2 ! · · · a n !
=
a 1 (a 1 − 1)
(1 + a − a 1 )(2 + a − a 1 )·
a!
a 1 !
−
n
X
j=3
a 1 a j
(1 + a − a 1 )(2 + a − a 1 )(1 + a − a 1 − a j )·
(1 + a)!
a 1 !
a 2 ! · · · a n ! by (2.1)
(1 + a (1) )(2 + a (1) )
(a 1−1) −
n
X
i=3
a i (1 + a) (1 + a (1)−a i )
C n (a).
This completes the proof
Acknowledgments We would like to thank Guoce Xin for valuable suggestions We are also grateful to the referees for helpful comments This work was supported by the National Natural Science Foundation of China (Projects 10926054 and 10901045), the Natural Science Foundation of Hebei Province (Project A2010000828), and Hebei Uni-versity of Science and Technology (Project QD200956)
Trang 6[1] G.E Andrews, Problems and prospects for basic hypergeometric functions, in Theory and Application of Special Functions, ed R Askey, Academic Press, New York,
1975, pp 191–224
[2] F.J Dyson, Statistical theory of the energy levels of complex systems I, J Math Phys 3 (1962), 140–156
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[4] I.J Good, Short proof of a conjecture by Dyson, J Math Phys 11 (1970), 1884 [5] J Gunson, Proof of a conjecture by Dyson in the statistical theory of energy levels,
J Math Phys 3 (1962), 752–753
[6] K.W.J Kadell, Aomoto’s machine and the Dyson constant term identity, Methods Appl Anal 5 (1998), 335–350
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[9] A.V Sills, Disturbing the Dyson conjecture, in a generally GOOD way, J Combin Theory Ser A 113 (2006), 1368–1380
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[11] K.G Wilson, Proof of a conjecture by Dyson, J Math Phys 3 (1962), 1040–1043 [12] G Xin, The ring of Malcev-Neumann series and the residue theorem, Ph.D Thesis, University of Brandeis, May 2004
[13] D Zeilberger, A combinatorial proof of Dyson’s conjecture, Discrete Math 41 (1982), 317–321
[14] D Zeilberger and D M Bressoud, A proof of Andrews’ q-Dyson conjecture, Discrete Math 54 (1985), 201–224
...This completes the proof
Now we are ready to prove Theorem 1.2 Without loss of generality, we may assume
r = and s = in Theorem 1.2
A new approach to Theorem 1.2 By (2.4)... Gunson, Proof of a conjecture by Dyson in the statistical theory of energy levels,
J Math Phys (1962), 752–753
[6] K.W.J Kadell, Aomoto’s machine and the Dyson constant term identity,... by Dyson, J Math Phys (1962), 1040–1043 [12] G Xin, The ring of Malcev-Neumann series and the residue theorem, Ph.D Thesis, University of Brandeis, May 2004
[13] D Zeilberger, A combinatorial