A Generalization of the Chu-Vandermonde Convolution and some Harmonic Number Identities

0.24 MB·

by  M. J. Kronenburg

A Generalization of the Chu-Vandermonde
Convolution and some Harmonic Number Identities
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0
M.J. Kronenburg
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r
a
M Abstract
AgeneralizationoftheChu-Vandermondeconvolutionispresentedandproved
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withtheintegralrepresentationmethod. Thisidentitycanbetransformedinto
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anotheridentity,whichhasasspecialcasestwoknownidentities. Anotheriden-

tity that is closely related to this identity is presented and proved. Some cor-
]
O respondingharmonicnumberidentitiesarederived,whichhaveasspecialcases
C someknownharmonicnumberidentities. Foronecombinatorialsumarecursion
formula is derived and used to compute a few examples.
.
h
t Keywords: binomial coefficient, combinatorial identities, harmonic number.
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MSC 2010: 05A10, 05A19
m
[
1 A Generalization of the Chu-Vandermonde
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v Convolution
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The following theorem is a generalization of the Chu-Vandermonde convolution.
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Theorem 1.1.
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1. (cid:88)n (cid:18)a(cid:19)(cid:18) b (cid:19)(cid:18)k(cid:19)(cid:18)n−k(cid:19) (cid:18)a+b−c−d(cid:19)(cid:18)a(cid:19)(cid:18)b(cid:19)
0 = (1.1)
k n−k c d n−c−d c d
7 k=0
1
: Proof. Applying the trinomial revision identity [10, 11, 13] to the summand twice,
v
the identity simplifies to:
i
X
n (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
r (cid:88) a−c b−d a+b−c−d
a = (1.2)
k−c n−d−k n−c−d
k=0
Usingtheintegralrepresentationmethodforcombinatorialsums[4,5], thisbecomes:
1
(cid:88)∞ (1+x)a−c (1+y)b−d
Res Res
x xk−c+1 y yn−d−k+1
k=0
(1+x)a−c(1+y)b−d (cid:88)∞ (cid:16)y(cid:17)k
=Res Res
x y x−c+1yn−d+1 x
k=0
(1+x)a−c(1+y)b−dxc−1
=Res Res
x y yn−d+1(1−y/x) (1.3)
(1+x)a−c(1+y)b−dxc
=Res Res
y x yn−d+1(x−y)
(1+y)a+b−c−d
=Res
y yn−c−d+1
(cid:18) (cid:19)
a+b−c−d
=
n−c−d
Forthisidentity,thelowerlimitofthesummationmaybereplacedbymax(n−b,c)
and the upper limit by min(a,n−d). The special case c=d=0 reduces to the Chu-
Vandermonde convolution:
n (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) a b a+b
= (1.4)
k n−k n
k=0
and the special case d = 0 was already known in literature [8]. The special case
a=b=n reduces to:
(cid:88)n (cid:18)n(cid:19)2(cid:18)k(cid:19)(cid:18)n−k(cid:19) (cid:18)2n−c−d(cid:19)(cid:18)n(cid:19)(cid:18)n(cid:19)
= (1.5)
k c d n c d
k=0
The following identity replaces a binomial coefficient by its symmetry equivalent [10,
11, 13]:
(cid:18) (cid:19) (cid:18) (cid:19)
n −n+k−1
=(−1)k (1.6)
k k
Replacing a by a−p and b by b−q, and using (1.6), and then taking a = b = −1,
the identity transforms to:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+k q+n−k k n−k n+p+q+1 p+c q+d
= (1.7)
p q c d n−c−d c d
k=0
The special case c=d=0 reduces to [6, 8, 16]:
n (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) p+k q+n−k n+p+q+1
= (1.8)
p q n
k=0
2
and the special case p=q =0 reduces to [8, 10, 11, 15]:
n (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) k n−k n+1
= (1.9)
c d c+d+1
k=0
Another theorem that is closely related to theorem 1.1 is the following.
Theorem 1.2.
min(a,b−m)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k m+k a+b−c−d a b
= (1.10)
k m+k c d m+a−d c d
k=0
Proof. Applying the trinomial revision identity [10, 11, 13] to the summand twice,
the identity simplifies to:
min(a,b−m)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) a−c b−d a+b−c−d
= (1.11)
k−c k+m−d m+a−d
k=0
Usingtheintegralrepresentationmethodforcombinatorialsums[4,5], thisbecomes:
(cid:88)∞ (1+x)a−c (1+y)b−d
Res Res
x xk−c+1 yyk+m−d+1
k=0
(1+x)a−c(1+y)b−d (cid:88)∞ (cid:18) 1 (cid:19)k
=Res Res
x y x−c+1ym−d+1 xy
k=0
(1+x)a−c(1+y)b−dxc−1
=Res Res
x y ym−d+1(1−1/(xy))
(1+x)a−c(1+y)b−dxc (1.12)
=Res Res
y x ym−d+1(x−1/y)
(1+1/y)a−c(1+y)b−d
=Res
y ym+c−d+1
(1+y)a+b−c−d
=Res
y ym+a−d+1
(cid:18) (cid:19)
a+b−c−d
=
m+a−d
For this identity, the lower limit of the summation may be replaced by
max(c,d−m). The special case m=0 and c=d=0 reduces to:
min(a,b)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) a b a+b
= (1.13)
k k a
k=0
3
which is equivalent to the special case n = b of (1.4). The special case m = 0 and
a=b=n reduces to:
(cid:88)n (cid:18)n(cid:19)2(cid:18)k(cid:19)(cid:18)k(cid:19) (cid:18)2n−c−d(cid:19)(cid:18)n(cid:19)(cid:18)n(cid:19)
= (1.14)
k c d n−d c d
k=0
The special cases c = d = 0, c = 1, d = 0 and c = d = 1 of this formula are well
known [8].
2 Harmonic Number Identities
Thedefinitionofthegeneralizedharmonicnumberswithnonnegativeintegern,com-
plex order m and complex offset c, is [12, 14]:
n
(cid:88) 1
H(m) = (2.1)
c,n (c+k)m
k=1
from which follows that H(m) =0, and for notation H(m) =H(m). From this defini-
c,0 n 0,n
tion follows for nonnegative integer c:
H(m) =H(m) −H(m) (2.2)
c,n c+n c
The classical harmonic numbers are:
H =H(1) (2.3)
n 0,n
Using d/dxΓ(x) = Γ(x)ψ(x) where ψ(x) is the digamma function, and using
ψ(x+n+1)−ψ(x+1) = H(1) [1], these harmonic numbers are linked to bino-
x,n
mial coefficients:
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
d x+y x+y x+y
= H(1) = (H −H ) (2.4)
dx n n x+y−n,n n x+y x+y−n
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
d n n n
= H(1) = (H −H ) (2.5)
dx x+y x+y x+y,n−2(x+y) x+y n−(x+y) x+y
For the generalized harmonic numbers (2.1):
d
H(m) =−mH(m+1) (2.6)
dx x+y,n x+y,n
When differentiating finite summation terms, care must be taken that the differen-
tiated symbol is not present in the summation limits. Because the argument of a
classical harmonic number cannot be negative, these harmonic numbers impose con-
straints on the parameters. When there are additional constraints on the parameters
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they are mentioned.
Differentiating (1.1) to a, the following identity for a≥n results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k n−k
H
k n−k c d a−k
k=0 (2.7)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
n−c−d c d a+b−n a+b−c−d a−c
Differentiating to b, the following identity for b≥n results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k n−k
H
k n−k c d b−n+k
k=0 (2.8)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
n−c−d c d a+b−n a+b−c−d b−d
Replacing k by n−k and interchanging a with b and c with d, these two identities
are equivalent. Differentiating to a and b, the following identity for a≥n and b≥n
results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k n−k
H H
k n−k c d a−k b−n+k
k=0
(cid:18)a+b−c−d(cid:19)(cid:18)a(cid:19)(cid:18)b(cid:19) (2.9)
= [H(2) −H(2)
n−c−d c d a+b−n a+b−c−d
+(H −H +H )(H −H +H )]
a+b−n a+b−c−d a−c a+b−n a+b−c−d b−d
The special case a=b=n and c=d=0 reduces to the known identities [2, 3, 8, 14,
16]:
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)
H = (2H −H ) (2.10)
k k n n 2n
k=0
(cid:88)n (cid:18)n(cid:19)2H H =(cid:18)2n(cid:19)[H(2)−H(2)+(2H −H )2] (2.11)
k k n−k n n 2n n 2n
k=0
Differentiating (1.1) to c, the following identity for b≤n results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b n−k a+b−d b
H = (H −H +H ) (2.12)
k n−k d k n−d d n−d a+b−d a
k=0
Differentiating to d, the following identity for a≤n results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k a+b−c a
H = (H −H +H ) (2.13)
k n−k c n−k n−c c n−c a+b−c b
k=0
Replacing k by n−k and interchanging a with b and c with d, these two identities
are equivalent. Differentiating to c and d, the following identity for a≤n and b≤n
5
results:
n (cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b
H H
k n−k k n−k
k=0 (2.14)
(cid:18) (cid:19)
a+b
= [H(2)−H(2) +(H −H +H )(H −H +H )]
n n a+b n a+b a n a+b b
In (1.1) replacing k by c+k and n by n+c, and differentiating to c, the following
identity for a−c≥n results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b c+k n−k
H
c+k n−k c d a−c−k
k=0 (2.15)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
n−d c d a+b−n−c a+b−c−d a−c
Replacing n with n+d and differentiating to d, the following identity for b−d ≥ n
results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k n+d−k
H
k n+d−k c d b−d−n+k
k=0 (2.16)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
n−c c d a+b−n−d a+b−c−d b−d
Replacing k by n−k and interchanging a with b and c with d, these two identities
are equivalent. Differentiating to c and d, the following identity for a−c ≥ n and
b−d≥n results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b c+k n+d−k
H H
c+k n+d−k c d a−c−k b−d−n+k
k=0
(cid:18)a+b−c−d(cid:19)(cid:18)a(cid:19)(cid:18)b(cid:19) (2.17)
= [H(2) −H(2)
n c d a+b−n−c−d a+b−c−d
+(H −H +H )(H −H +H )]
a+b−n−c−d a+b−c−d a−c a+b−n−c−d a+b−c−d b−d
Differentiating (1.7) to p, the following identity results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+k q+n−k k n−k
H
p q c d p+k
k=0 (2.18)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
n+p+q+1 p+c q+d
= (H −H +H )
n−c−d c d n+p+q+1 p+q+c+d+1 p+c
The special case p=q =0 is found in [15], and the special case c=d=0 is found in
[16]. Differentiating to q, the following identity results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+k q+n−k k n−k
H
p q c d q+n−k
k=0 (2.19)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
n+p+q+1 p+c q+d
= (H −H +H )
n−c−d c d n+p+q+1 p+q+c+d+1 q+d
6
Replacing k by n−k and interchanging p with q and c with d, these two identities
are equivalent. Differentiating to p and q the following identity results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+k q+n−k k n−k
H H
p q c d p+k q+n−k
k=0
(cid:18)n+p+q+1(cid:19)(cid:18)p+c(cid:19)(cid:18)q+d(cid:19) (2.20)
= [H(2) −H(2)
n−c−d c d p+q+c+d+1 n+p+q+1
+(H −H +H )(H −H +H )]
n+p+q+1 p+q+c+d+1 p+c n+p+q+1 p+q+c+d+1 q+d
The special case p = q = 0 is found in [15], and the special case c = d = 0 is found
in [16]. In (1.7) replacing k by c+k and n by n+c, and differentiating to c, the
following identity results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+c+k q+n−k c+k n−k
H
p q c d p+c+k
k=0
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
n+p+q+c+1 p+c q+d
= (H −H +H )
n−d c d n+p+q+c+1 p+q+c+d+1 p+c
(2.21)
Replacing n with n+d and differentiating to d, the following identity results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+k q+n+d−k k n+d−k
H
p q c d q+n+d−k
k=0
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
n+p+q+d+1 p+c q+d
= (H −H +H )
n−c c d n+p+q+d+1 p+q+c+d+1 q+d
(2.22)
Replacing k by n−k and interchanging p with q and c with d, these two identities
are equivalent. Differentiating to c and d the following identity results:
n (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) p+c+k q+n+d−k c+k n+d−k
H H
p q c d p+c+k q+n+d−k
k=0
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
n+p+q+c+d+1 p+c q+d
= [H(2) −H(2)
n c d p+q+c+d+1 n+p+q+c+d+1
+(H −H +H )(H −H +H )]
n+p+q+c+d+1 p+q+c+d+1 p+c n+p+q+c+d+1 p+q+c+d+1 q+d
(2.23)
Differentiating (1.10) to a, the following identity for a≥b−m results:
b−m(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k m+k
H
k m+k c d a−k
k=0 (2.24)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
m+a−d c d m+a−d a+b−c−d a−c
7
Differentiating to b, the following identity for b≥a+m results:
a (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k m+k
H
k m+k c d b−m−k
k=0 (2.25)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
m+a−d c d b−m−c a+b−c−d b−d
Differentiating to c, the following identity for d≥m results:
min(a,b−m)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b m+k
H
k m+k d k
k=0 (2.26)
(cid:18) (cid:19)(cid:18) (cid:19)
a+b−d b
= (H −H +H )
m+a−d d a a+b−d b−m
Differentiating to d, the following identity with m=d results:
min(a,b−d)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k d+k
H
k d+k c d k
k=0 (2.27)
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
a+b−c−d a b
= (H −H +H )
a c d a a+b−c−d b−d
Differentiating to a and c, the following identity for a≥b−m and d≥m results:
b−m(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b m+k
H H
k m+k d k a−k
k=0
(cid:18)a+b−d(cid:19)(cid:18)b(cid:19) (2.28)
= [H(2)−H(2)
m+a−d d a a+b−d
+(H −H +H )(H −H +H )]
a a+b−d m+a−d a a+b−d b−m
Differentiating to a and d, the following identity for a≥b−d and m=d results:
b−d(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k d+k
H H
k d+k c d k a−k
k=0
(cid:18)a+b−c−d(cid:19)(cid:18)a(cid:19)(cid:18)b(cid:19) (2.29)
= [H(2)−H(2)
a c d a a+b−c−d
+(H −H +H )(H −H +H )]
a a+b−c−d a−c a a+b−c−d b−d
Differentiating to b and c, the following identity for b≥a+m and d≥m results:
a (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b m+k
H H
k m+k d k b−m−k
k=0
(cid:18)a+b−d(cid:19)(cid:18)b(cid:19) (2.30)
= [H(2) −H(2)
m+a−d d b−m a+b−d
+(H −H +H )(H −H +H )]
b−m a+b−d b−d b−m a+b−d a
8
Differentiating to b and d, the following identity for b≥a+d and m=d results:
a (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
(cid:88) a b k d+k
H H
k d+k c d k b−d−k
k=0
(cid:18)a+b−c−d(cid:19)(cid:18)a(cid:19)(cid:18)b(cid:19) (2.31)
= [H(2) −H(2)
a c d b−d a+b−c−d
+(H −H +H )(H −H +H )]
b−d a+b−c−d b−d−c b−d a+b−c−d a
3 A Recursion Formula for a Combinatorial Sum
Using a recursion formula, a rational function P (n) is found such that:
m
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)
km = P (n) (3.1)
k n m
k=0
From (1.5) or (1.14) with d=0 we have:
(cid:88)n (cid:18)n(cid:19)2(cid:18)k(cid:19) (cid:18)2n−m(cid:19)(cid:18)n(cid:19)
= (3.2)
k m n m
k=0
This formula is rewritten as:
(cid:88)n (cid:18)n(cid:19)2m(cid:89)−1 (cid:18)2n(cid:19)m(cid:89)−1(n−j)2
(k−j)= (3.3)
k n 2n−j
k=0 j=0 j=0
Now the following is used:
m−1 m (cid:20) (cid:21)
(cid:89) (cid:88) m
(k−j)= (−1)m−j kj (3.4)
j
j=0 j=0
where(cid:2)a(cid:3)istheStirlingnumberofthefirstkind[10]. ThenitisclearthatP (n)has
b m
the following recursion formula:
m(cid:89)−1(n−k)2 m(cid:88)−1 (cid:20)m(cid:21)
P (n)= − (−1)m−k P (n) (3.5)
m 2n−k k k
k=0 k=0
The following are a few examples:
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)
= (3.6)
k n
k=0
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)n
k = (3.7)
k n 2
k=0
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(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19) n3
k2 = (3.8)
k n 2(2n−1)
k=0
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)n3(n+1)
k3 = (3.9)
k n 4(2n−1)
k=0
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)n3(n3+n2−3n−1)
k4 = (3.10)
k n 4(2n−1)(2n−3)
k=0
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)n4(n+1)(n2+2n−5)
k5 = (3.11)
k n 8(2n−1)(2n−3)
k=0
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)n3(n6+3n5−13n4−15n3+30n2+8n−2)
k6 = (3.12)
k n 8(2n−1)(2n−3)(2n−5)
k=0
(cid:88)n (cid:18)n(cid:19)2 (cid:18)2n(cid:19)n4(n+1)(n5+5n4−15n3−35n2+70n−14)
k7 = (3.13)
k n 16(2n−1)(2n−3)(2n−5)
k=0
TheMathematica(cid:13)R [17]programusedtocomputetheexpressionsisgivenbelow:
P[0]=1;
P[m_]:=P[m]=Factor[Simplify[Product[(n-k)^2/(2n-k),{k,0,m-1}]
-Sum[StirlingS1[m,k]P[k],{k,0,m-1}]]]
References
[1] G.E.Andrews,R.Askey,R.Roy,SpecialFunctions,CambridgeUniversityPress,
1999.
[2] X. Chen, W. Chu, The Gauss F (1)-summation theorem and harmonic number
2 1
identities, Integral Transforms Spec. Funct. 20 (2009) 925-935.
[3] W. Chu, L. De Donno, Hypergeometric series and harmonic number identities,
Adv. in Appl. Math. 34 (2005) 123-137.
[4] R.V.Churchill, J.W.Brown, Complex Variables and Applications, McGraw-Hill,
1984.
[5] G.P. Egorychev, Integral Representation and the Computation of Combinatorial
Sums, Translations of Mathematical Monographs, 59, Amer. Math. Soc., 1984.
[6] H.W.Gould,SomeGeneralizationsofVandermonde’sConvolution,Amer. Math.
Monthly 63 (1956) 84-91.
[7] H.W. Gould, Final Analysis of Vandermonde’s Convolution, Amer. Math.
Monthly 64 (1957) 409-451.
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