Quaestiones Mathematicæ 30 (2007), 159–165
SECANT AND COSECANT SUMS AND BERNOULLI-NÖRLUND
POLYNOMIALS
PETER J. GRABNER† AND HELMUT PRODINGER*
Abstract. We give explicit formulæ for sums of even powers of secant and cosecant
values in terms of Bernoulli numbers and central factorial numbers.
1. Introduction
We derive explicit formulæ for the secant sum
⌊ N−1
⌋
2
S2m (N) :=
X
k=1
and the cosecant sum
⌋
⌊ N−1
2
C2m (N) :=
X
k=1
1
cos2m kπ
N
1
sin2m kπ
N
.
This research is inspired by the paper [2], where such formulæ were given for m ≤ 6.
Our approach, which uses contour integrals and residues, produces such formulæ quite
effortlessly for any m. The main contribution of the present paper is the identification of
the occurring coefficients as “classical” combinatorial quantities such as central factorial
numbers and Bernoulli numbers.
2. Contour integrals and residues
We consider the secant sum first and start with the contour integral
I
1
1
πN cot(πNz) dz,
2πi RT cos2m πz
(1)
1
1
± iT , 1 − 2N
± iT . By periodicity of the
where RT is the rectangle with corners − 2N
integrand, the integrals along the vertical lines cancel. Furthermore, the integrals along
the horizontal lines tend to 0 when T → ∞, since cot remains bounded and cos tends to
infinity exponentially.
†
This author is supported by the Austrian Science Foundation FWF, project S9605, part of the Austrian
National Research Network “Analytic Combinatorics and Probabilistic Number Theory”.
*
This author is supported by the grant NRF 2053748 of the South African National
Research Foundation.
1
2
P. J. GRABNER AND H. PRODINGER
Thus we have
I
1
1
0=
πN cot(πNz) dz
2m
2πi RT cos πz
⌊ N−1
⌋
2
=2
X
1
cos2m kπ
N
k=1
+ 1 + Res1
z= 2
1
cos2m πz
πN cot(πNz)
by the residue theorem. From this we derive
⌋
⌊ N−1
2
S2m (N) =
X
k=1
1
cos2m kπ
N
1 1
1
πN cot(πNz).
= − − Res1
2m
2 2 z= 2 cos πz
(2)
In [4] the Bernoulli-Nörlund polynomials are introduced by the relation
∞
X tn
ω1 · · · ωk tk ext
=
B (k) (x; ω1 , . . . , ωk ).
(eω1 t − 1) · · · (eωk t − 1) n=0 n! n
(3)
We specialise ω1 = · · · = ωk = 2i, x = ki, and t = πz to obtain
∞
πz k X
(πz)n (k)
Bn (ki; 2i, . . . , 2i).
=
sin πz
n!
n=0
(k)
(k)
(k)
Writing Pn = in Bn (ki; 2i, . . . , 2i) and observing that P2n+1 = 0 we have
∞
X (πz)2n−k
1
(k)
(−1)n P2n .
=
k
(2n)!
sin πz
n=0
We have
Res1
z= 2
1
cos2m
πz
πN cot(πNz) = Res
z=0
1
sin
2m
πz
πN cot(πNz +
(4)
N
π).
2
Notice that
N
cot(πNz + π) =
2
(
cot(πNz)
if N is even,
− tan(πNz) if N is odd.
Thus it is natural to distinguish two cases according to the parity of N.
From [3] we have
π cot πz =
∞
X
π 2n z 2n−1
n=0
π tan πz =
(2n)!
∞
X
π 2n z 2n−1
n=1
(2n)!
(−1)n 4n B2n ,
(−1)n−1 4n (4n − 1)B2n .
(5)
3
Then for even N we have
Res1
z= 2
1
cos2m
πz
πN cot(πNz)
−1
= [z ]
∞
X
(πz)2ℓ−2m
ℓ=0
(2ℓ)!
(−1)
ℓ
(2m)
P2ℓ πN
n=0
z= 2
1
cos2m
πz
= −[z
(2n)!
(−1)n 4n B2n
m (−1)m X 2m
(2m)
P2(m−n) B2n (2N)2n
=
(2m)! n=0 2n
and for odd N
Res1
∞
X
(Nπz)2n−1
πN cot(πNz)
−1
]
∞
X
(πz)2ℓ−2m
ℓ=0
(2ℓ)!
(−1)
ℓ
Summing up, we have for even N
(2m)
P2ℓ πN
∞
X
(πNz)2n−1
(−1)n−1 4n (4n − 1)B2n
(2n)!
n=1
m (−1)m X 2m
(2m)
P2(m−n) B2n (4n − 1)(2N)2n .
=
(2m)! n=1 2n
m 1 (−1)m−1 X 2m
(2m)
S2m (N) = − +
P2(m−n) B2n (2N)2n
2
2(2m)! n=0 2n
(6)
m 1 (−1)m−1 X 2m
(2m)
P2(m−n) B2n (4n − 1)(2N)2n .
S2m (N) = − +
2
2(2m)! n=0 2n
(7)
and for odd N
Equation (2) gives us for even N:
m=1:
m=2:
m=3:
m=4:
m=5:
m=6:
m=7:
m=8:
1 2
N − 32
6
1
28
N 4 + 91 N 2 − 45
90
1
1
4
N 6 + 90
N 4 + 45
N 2 − 568
945
945
8
6
4
4
7
8
8336
1
N + 2835 N + 675 N + 105
N 2 − 14175
9450
1
1
13
82
64
N 10 + 5670
N 8 + 8505
N 6 + 8505
N 4 + 945
N 2 − 54176
93555
93555
12
10
8
6
691
2
31
278
1916
128
N + 93555 N + 141750 N + 178605 N + 212625
N 4 + 2079
N 2 − 365470016
638512875
638512875
14
12
10
8
6
4
2
691
2
311
592
944
N
+
N
+
N
+
N
+
N
+
N
18243225
273648375
66825
1275750
382725
111375
512
155194496
+ 9009
N 2 − 273648375
3617
16
113324
1072
2473
N 16 + 54729675
N 14 + 28733079375
N 12 + 29469825
N 10 + 9568125
N8
325641566250
134432
8533792
1024
274946646272
+ 88409475
N 6 + 1064188125
N 4 + 19305
N 2 − 488462349375
4
P. J. GRABNER AND H. PRODINGER
Equation (2) gives us for odd N:
m=1:
m=2:
m=3:
m=4:
m=5:
m=6:
m=7:
m=8:
1 2
N − 21
2
1 4
N + 31 N 2 − 21
6
1
4
N 6 + 61 N 4 + 15
N 2 − 21
15
17
4
7
8
N 8 + 45
N 6 + 45
N 4 + 35
N 2 − 12
630
31
17
13
82
64
N 10 + 378
N 8 + 135
N 6 + 567
N 4 + 315
N 2 − 21
2835
691
62
527
278
1916
128 2
N 12 + 2835
N 10 + 9450
N 8 + 2835
N 6 + 14175
N 4 + 693
N − 21
155925
10922
691
62
5287
592
944
512
N 14 + 66825
N 12 + 2025
N 10 + 85050
N 8 + 6075
N 6 + 7425
N 4 + 3003
N2
6081075
929569
87376
113324
33232
42041
N 16 + 18243225
N 14 + 7016625
N 12 + 893025
N 10 + 637875
N8
1277025750
134432
8533792
1024 2
+ 1403325
N 6 + 70945875
N 4 + 6435
N − 12
For the cosecant sum, we start with the contour integral
I
1
1
πN cot(πNz)dz,
2m
2πi RT sin πz
−
1
2
(8)
which is again zero and, by summing residues, leads to the equation
⌋
⌊ N−1
2
0=
X
k=1
1
sin2m
kπ
N
1 + (−1)N
1
1
+ Res 2m πN cot(πNz) +
.
2 z=0 sin πz
4
We observe that
⌊
N −1
⌋
2
X
k=1
1
sin2m
kπ
N
+
1 + (−1)N
4
equals the residue that we already calculated for S2m (N) and N even. Thus we have
⌊
C2m (N) =
N −1
⌋
2
X
k=1
1
sin2m
kπ
N
m 1 + (−1)N
(−1)m X 2m
(2m)
2n
P2(m−n) B2n (2N) −
=
.
(2m)! n=0 2n
4
(9)
(2m)
3. Computing P2n
In this section we want to have a closer look at the Laurent series expansion of sin−2m πz.
Our approach is somewhat similar to the one used in [1].
We start with the expansion (5). Differentiating yields
∞
This gives
X (πz)2n−2
1
=
(2n − 1)(−1)n−1 4n B2n .
(2n)!
sin2 πz
n=0
(2)
P2n = −(2n − 1)4n B2n .
(10)
5
Differentiating sin−2m πz twice yields
d2
1
1
1
2
2 2
=
2m(2m
+
1)π
−
4m
π
.
2m
2m+2
2m
dz 2 sin πz
sin
πz
sin πz
(11)
We now write
m
1
sin2m πz
= H2m (z) + R2m (z) =
(2m)
m−ℓ
X (2ℓ − 1)!b
1
2ℓ 4
+ R2m (z),
(2m − 1)!
(πz)2ℓ
(12)
ℓ=1
where H2m is the principal part around z = 0 and R2m denotes the regular part. Since
differentiation preserves principal and regular parts, (11) gives
′′
H2m
(z) = π 2 2m(2m + 1)H2m+2 (z) − 4m2 π 2 H2m (z),
(2m)
which gives the recursion (setting b0
(2m+2)
(2m)
(2m)
+ b2ℓ−2 for 1 ≤ ℓ ≤ m + 1.
(2m)
This recursion shows that the numbers b2ℓ
m
X
(2)
= b2m+2 = 0 and b2 = 1)
(2m)
= m2 b2ℓ
b2ℓ
(13)
(2m)
b2ℓ x2ℓ
are given by
=
ℓ=0
(14)
m−1
Y
(x2 + k 2 ).
(15)
k=0
Thus they are closely related to the central factorial numbers t(n, k) studied in [5, p. 213]:
x
m−1
Y
2
2
(x − k ) =
2m
X
t(2m, 2k + 1)x2k+1
k=0
k=1
(2m)
and a similar expression for odd first argument. This gives b2ℓ = (−1)ℓ+m t(2m, 2ℓ). We
notice that the polynomials in (15) appear mutatis mutandis in [2] as differential operators.
These operators are used to model the recursion (13).
(2m)
In Table 1 we computed the values b2k for small values of m.
(m)
bk
k=2
4
m=2
1
4
1
1
6
4
5
6
8
10
12
14
16
1
8
36
49
14
1
10
576
820
273
30
1
12
14400
21076
7645
1023
55
1
14
518400
773136
296296
44473
3003
91
1
16
25401600
38402064
15291640
2475473
191620
7462
140
1
18
1625702400
2483133696
1017067024
173721912
14739153
669188
16422
204
(m)
Table 1. Table of bk
18
1
for small values of m (compare with [5, Table 6.1, p. 217])
6
P. J. GRABNER AND H. PRODINGER
We now consider the Mittag-Leffler expansion
1
sin2m πz
=
X
H2m (z + n) = H2m (z) +
∞
X
n=1
n∈Z
H2m (z + n) + H2m (z − n) .
(16)
Expanding the last sum into a power series and using (12) yields
1
sin2m πz
∞
= H2m (z) +
m
X (πz)2k
X
4m
1
(2m)
(−1)ℓ−1
4k (−1)k
b2ℓ B2ℓ+2k ,
(2m − 1)! k=0 (2k)!
2ℓ
+
2k
ℓ=1
2k−1 2k
where we have used ζ(2k) = (−1)k−1 2 (2k)!π B2k . This gives

P
2m 2k 4k m−1 (−1)ℓ−1 1 b(2m) B2k−2ℓ for k ≥ m,
ℓ=0
2k−2ℓ 2m−2ℓ
2m
(2m)
P2k =
(−1)k 4k b(2m) / 2m−1
for 0 ≤ k ≤ m − 1.
2m−2k
2k
(17)
Inserting this into (6) and (7) yields for even N
m
m
4m−1 X (−1)ℓ−1 (2m)
4m−1 X (−1)ℓ−1 (2m)
1
2ℓ
S2m (N) =
b2ℓ B2ℓ N −
b2ℓ B2ℓ −
(18)
(2m − 1)! ℓ=1
ℓ
(2m − 1)! ℓ=1
ℓ
2
and for odd N
m
1
4m−1 X (−1)ℓ−1 (2m)
b2ℓ B2ℓ (4ℓ − 1)N 2ℓ −
S2m (N) =
(2m − 1)!
ℓ
2
(19)
ℓ=1
Similarly, we obtain
m
4m−1 X (−1)ℓ−1 (2m)
b2ℓ B2ℓ N 2ℓ
C2m (N) =
(2m − 1)! ℓ=1
ℓ
m
4m−1 X (−1)ℓ−1 (2m)
1 + (−1)N
−
b2ℓ B2ℓ −
. (20)
(2m − 1)! ℓ=1
ℓ
4
References
[1] K Dilcher, Sums of products of Bernoulli numbers, J. Number Theory 60 (1996), 23–41.
[2] N. Gauthier and P. S. Bruckman, Sums of even integral powers of the cosecant and the secant, Fibonacci
Quart. 44 (2007), 264–273.
[3] R.L. Graham, D.E. Knuth, and O. Patashnik, Concrete mathematics: A foundation for computer
science, second ed., Addison Wesley, Reading, MA, 1994.
[4] N. E. Nörlund, Differenzenrechnung, Grundlehren der mathematischen Wissenschaften, vol. XIII,
Springer Verlag, Berlin, 1924.
[5] J. Riordan, Combinatorial identities, John Wiley & Sons Inc., New York, 1968.
7
(P. Grabner) Institut für Mathematik A, Technische Universität Graz, Steyrergasse 30,
8010 Graz, Austria
E-mail address: [email protected]
(H. Prodinger) Department of Mathematics, University of Stellenbosch, 7602 Stellenbosch, South Africa
E-mail address: [email protected]