Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 802791, 7 pages
http://dx.doi.org/10.1155/2013/802791
Research Article
Algorithms for Some Euler-Type Identities for
Multiple Zeta Values
Shifeng Ding and Weijun Liu
Department of Mathematics, Central South University, Changsha, Hunan 410083, China
Correspondence should be addressed to Shifeng Ding; isgibt@yahoo.com.cn
Received 21 December 2012; Accepted 11 January 2013
Academic Editor: C. Conca
Copyright © 2013 S. Ding and W. Liu. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
𝑠
𝑠
𝑠
Multiple zeta values are the numbers defined by the convergent series 𝜁(𝑠1 , 𝑠2 , . . . , 𝑠𝑘 ) = ∑𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑘 >0 (1/𝑛11 𝑛22 ⋅ ⋅ ⋅ 𝑛𝑘𝑘 ), where 𝑠1 , 𝑠2 ,
. . . , 𝑠𝑘 are positive integers with 𝑠1 > 1. For 𝑘 ≤ 𝑛, let 𝐸(2𝑛, 𝑘) be the sum of all multiple zeta values with even arguments whose
weight is 2𝑛 and whose depth is 𝑘. The well-known result 𝐸(2𝑛, 2) = 3𝜁(2𝑛)/4 was extended to 𝐸(2𝑛, 3) and 𝐸(2𝑛, 4) by Z. Shen and
T. Cai. Applying the theory of symmetric functions, Hoffman gave an explicit generating function for the numbers 𝐸(2𝑛, 𝑘) and
then gave a direct formula for 𝐸(2𝑛, 𝑘) for arbitrary 𝑘 ≤ 𝑛. In this paper we apply a technique introduced by Granville to present an
algorithm to calculate 𝐸(2𝑛, 𝑘) and prove that the direct formula can also be deduced from Eisenstein’s double product.
1. Introduction
Recently, some identities similar to (3) have also been established. Given two positive integers 𝑛 and 𝑘 (suppose 𝑛 ≥ 𝑘),
define a number 𝑍(𝑛, 𝑘) by
The multiple zeta sums,
𝜁 (𝑠1 , 𝑠2 , . . . , 𝑠𝑚 ) =
1
𝑠𝑚 ,
𝑠1 𝑠2
𝑛
𝑛
𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑘 >0 1 2 ⋅ ⋅ ⋅ 𝑛𝑚
∑
(1)
are also called Euler-Zagier sums, where 𝑠1 , 𝑠2 , . . . , 𝑠𝑚 are
positive integers with 𝑠1 ≥ 2. Clearly, the Riemann zeta
function 𝜁(𝑠),
∞
1
,
𝑠
𝑛
𝑛=1
𝜁 (𝑠) = ∑
𝑠 > 1,
𝑗1 +𝑗2 =𝑛
𝑗1 ,𝑗2 ≥1
∑
𝑗1 +𝑗2 +⋅⋅⋅+𝑗𝑘 =𝑛
𝑗1 ,𝑗2 ,...,𝑗𝑘 ≥1
𝜁 (2𝑗1 ) 𝜁 (2𝑗2 ) ⋅ ⋅ ⋅ 𝜁 (2𝑗𝑘 ) .
𝐸 (2𝑛, 𝑘) =
∑
𝜁 (2𝑗1 , 2𝑗2 , . . . , 2𝑗𝑘 ) .
𝑗1 +⋅⋅⋅+𝑗𝑘 =𝑛
𝑗1 ,𝑗2 ,...,𝑗𝑘 ≥1
2𝑛 + 1
𝜁 (2𝑛) .
2
(3)
(4)
Then, for 𝑘 ∈ {3, 4, . . . , 9}, the value of 𝑍(𝑛, 𝑘) is known [1–5].
Following [6], for 𝑘 ≤ 𝑛, let 𝐸(2𝑛, 𝑘) be the sum of all
multiple zeta values with even arguments whose weight is 2𝑛
and whose depth is 𝑘; that is,
(2)
is the case 𝑚 = 1 in (1). The multiple zeta functions have
attracted considerable interest in recent years.
For Riemann’s zeta function 𝜁(𝑠), Euler proved the following identity:
∑ 𝜁 (2𝑗1 ) 𝜁 (2𝑗2 ) =
𝑍 (𝑛, 𝑘) =
(5)
In [7], Gangl et al. proved the following identities:
3
𝐸 (2𝑛, 2) = 𝜁 (2𝑛) ,
4
𝑛−1
for 𝑛 ≥ 2,
1
∑ 𝜁 (2𝑟 + 1, 2𝑛 − 2𝑟 − 1) = 𝜁 (2𝑛) ,
4
𝑟=1
for 𝑛 ≥ 2.
(6)
(7)
2
Journal of Applied Mathematics
Recently, using harmonic shuffle relations, Shen and Cai
proved the following results in [8]:
5
1
𝐸 (2𝑛, 3) = 𝜁 (2𝑛) − 𝜁 (2) 𝜁 (2𝑛 − 2) ,
8
4
for 𝑛 ≥ 3,
5
35
𝐸 (2𝑛, 4) = 𝜁 (2𝑛) − 𝜁 (2) 𝜁 (2𝑛 − 2) ,
64
16
(8)
𝑛≥𝑘≥1
sin (𝜋√1 − 𝑠√𝑡)
√1 − 𝑠 sin (𝜋√𝑡)
.
𝜁 (2𝑛) 2𝑘 − 1
(
)
𝑘
22(𝑘−1)
−
[(𝑘−1)/2]
∑
𝑗=1
𝑘𝑎𝑘(𝑁) = − ∑ 𝑏𝑘−𝑗 𝑎𝑗(𝑁) ,
𝜁 (2𝑗) 𝜁 (2𝑛 − 2𝑗) 2𝑘 − 2𝑗 − 1
(
),
𝑘
22𝑘−3 (2𝑗 + 1) 𝐵2𝑗
(10)
where 𝐵2𝑗 is the 2𝑗th Bernoulli number.
In this paper we use a technique introduced by Granville
[9] to present an elementary recursion algorithm to calculate
𝐸(2𝑛, 𝑘), we also give some direct formula for 𝐸(𝑛, 𝑘) for
arbitrary 𝑛 ≥ 𝑘. Our algorithm may be of some interest if
we note that it is obtained through an elementary analytic
method and that the statement of the algorithm is fairly
simple.
∞
1
𝑏𝑗 ≜ ∑
𝑟=1
𝑟 ≠ 𝑁
(𝑁)
𝑎𝑘−1
=
𝜋𝑘−1
𝑘!(2𝑁)𝑘−1
2𝑛−2𝑘+2
.
(12)
Theorem 2. Given a positive integer 𝑁, we have
sin (𝜋√𝑁2 − 𝑥)
𝑥
2
𝑁+1
. (13)
)
=
2𝑁
∏ (1 + 2
(−1)
𝑟 − 𝑁2
𝜋𝑥√𝑁2 − 𝑥
𝑟=1
∞
𝑟 ≠ 𝑁
When 𝑘 is not large, we may use the following recursion
algorithm to calculate 𝑎𝑘(𝑁) then use Theorem 1 to get the
formula for 𝐸(2𝑛, 𝑘).
(15)
𝑘𝜋
1
− cos ( )
2𝑗
2
(2𝑁𝜋)
×
[(𝑘−1)/2]
∑
𝑗=0
(𝑁)
=
𝑎𝑘−1
[(𝑘−1)/2]
1
22𝑘−2 𝑘!
⋅
(11)
𝑁=1 𝑁
∀𝑗 = 1, 2, 3, . . . .
[𝑘/2]
{
(−1)𝑗 (𝑘 + 2𝑗)!
𝑘𝜋
{sin ( 2 ) ∑ (2𝑗)! (𝑘 − 2𝑗)!
𝑗=0
{
⋅
𝑟 ≠ 𝑁
(𝑁)
𝑎𝑘−1
,
Theorem 4. For 𝑘 ≥ 1, one has
𝑥
(𝑁) 𝑘−1
) = 𝑎0(𝑁) + 𝑎1(𝑁) 𝑥 + ⋅ ⋅ ⋅ + 𝑎𝑘−1
𝑥 + ⋅⋅⋅.
∏ (1 + 2
𝑟 − 𝑁2
𝑟=1
∞
− 𝑟2 )
In [6], Hoffman established an interesting result [6,
Lemma 1.3] to obtain his formula (10) for 𝐸(2𝑛, 𝑘). This
lemma might be deduced from the theory of Bessel functions.
Using the expressions for the Bessel functions of the first kind
with a half integer index, we may deduce from the generating
function (13) a direct formula for 𝑎𝑘(𝑁) .
∞
𝐸 (2𝑛, 𝑘) = ∑
𝑗
(𝑁2
⋅
Then, for any two positive integers 𝑛 and 𝑘 with 𝑛 ≥ 𝑘, one has
(14)
where 𝑏1 , 𝑏2 , . . . , are the numbers defined by
2. Statements of the Theorems
Theorem 1. Let 𝑁 denote a positive integer. Let 𝑎0(𝑁) , 𝑎1(𝑁) ,
𝑎2(𝑁) , . . . , be a series of numbers defined by
for 𝑘 ≥ 1,
𝑗=0
(9)
Based on this generating function, some formulas for 𝐸(2𝑛, 𝑘)
for arbitrary 𝑛 ≥ 𝑘 are given. For example, Hoffman obtained
that
𝐸 (2𝑛, 𝑘) =
𝑘−1
𝑎0(𝑁) = 1;
for 𝑛 ≥ 4.
In [6], applying the theory of symmetric functions,
Hoffman established the generating function for the numbers
𝐸(2𝑛, 𝑘). He proved that
1 + ∑ 𝐸 (2𝑛, 𝑘) 𝑡𝑛 𝑠𝑘 =
Theorem 3. The coefficients 𝑎0(𝑁) , 𝑎1(𝑁) , . . . , 𝑎𝑘(𝑁) , . . . , can be
calculated recursively by the following formulas:
∑
𝑗=0
1
𝑁2𝑘−2−2𝑗
(−1)𝑗 (𝑘 + 2𝑗 + 1)!
(2𝑗 + 1)! (𝑘 − 2𝑗 − 1)!
(16)
}
1
,
2𝑗+1 }
(2𝑁𝜋)
}
(−1)𝑗 (2𝜋)2𝑗 (2𝑘 − 1 − 2𝑗)!
(𝑘 − 1 − 2𝑗)! (2𝑗 + 1)!
(17)
.
To deduce (17) from (16), we only need to write the
(𝑁)
, respectively, according to whether 𝑘 is odd
expression of 𝑎𝑘−1
or even, and use [(𝑘 − 1)/2] − 𝑗 (if 𝑘 is odd) or [𝑘/2] − 𝑗 (if 𝑘 is
even) to replace 𝑗. In the two cases, we will get the expression
(𝑁)
. By Theorem 1, we have
(17) for 𝑎𝑘−1
𝐸 (2𝑛, 𝑘)
=
[(𝑘−1)/2]
∑
𝑗=0
(−1)𝑗 𝜋2𝑗 (2𝑘 − 1 − 2𝑗)!
𝜁 (2𝑛 − 2𝑗) ,
22𝑘−2−2𝑗 𝑘! (𝑘 − 1 − 2𝑗)! (2𝑗 + 1)!
(18)
which reproduces Hoffman’s formula (10).
Journal of Applied Mathematics
3
(𝑁)
= 0, for all 𝑚 ≥ 𝑁, and
Then, we have 𝑃0(𝑁) = 𝑄0(𝑁) ≜ 1, 𝑄𝑚
3. Proofs of the Theorems
Proof of Theorem 1. The left side of (12) is
∑
𝑗1 +⋅⋅⋅+𝑗𝑘 =𝑛
𝑗1 ,...,𝑗𝑘 ≥1
=
(𝑁)
=
𝑃𝑗−1
𝜁 (2𝑗1 , 2𝑗2 , . . . , 2𝑗𝑘 )
2
𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑗−1 >𝑁 (𝑛1
−
2
𝑁2 ) ⋅ ⋅ ⋅ (𝑛𝑗−1
1
2𝑗
2𝑗
2𝑗
1
2
𝑘
𝑗1 +⋅⋅⋅+𝑗𝑘 =𝑛 𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑘 >0 𝑛1 𝑛2 ⋅ ⋅ ⋅ 𝑛𝑘
(19)
(𝑁)
𝑄𝑘−𝑗
=
1
∑
2
𝑁>𝑛𝑗+1 >𝑛𝑗+2 >⋅⋅⋅>𝑛𝑘 ≥1 (𝑛𝑗+1
− 𝑁2 ) ⋅ ⋅ ⋅ (𝑛𝑘2 − 𝑁2 )
𝑗1 ,...,𝑗𝑘 ≥1
=
∑
− 𝑁2 )
,
∀1 < 𝑗 ≤ 𝑘,
∑
∑
1
∑
∑
,
∀1 ≤ 𝑗 < 𝑘.
1
2𝑗1 2𝑗2
𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑘 >0 𝑗1 +⋅⋅⋅+𝑗𝑘 =𝑛 𝑛1 𝑛2
𝑗1 ,...,𝑗𝑘 ≥1
(25)
2𝑗
⋅ ⋅ ⋅ 𝑛𝑘 𝑘
.
Consider the sum (22). For 𝑗 ∈ {1, 2, . . . , 𝑘}, we treat each
sum in (22) with respect to 𝑛𝑗 as follows:
The second sum in (19) is the coefficient of 𝑥2𝑛 in the formal
power series
1
1
∏ 2
2𝑛−2𝑘+2
𝑛
− 𝑛𝑗2
𝑛
𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑘 >0 𝑗
1≤𝑚≤𝑘 𝑚
∑
𝑚 ≠ 𝑗
𝑗
∞
∑(
𝑗=1
𝑗
∞
𝑥2 ∞ 𝑥2
𝑥2
)
(
)
⋅
⋅
⋅
(
)
∑
∑
2
𝑛12 𝑗=1 𝑛22
𝑗=1 𝑛𝑘
𝑗
=
∞
∑
[
𝑛𝑗 =𝑘−𝑗+1
[
2𝑘
𝑥
= 2
(𝑛1 − 𝑥2 ) (𝑛22 − 𝑥2 ) ⋅ ⋅ ⋅ (𝑛𝑘2 − 𝑥2 )
𝑘
= ∑(
𝑗=1
𝑘
𝑗=1
1
𝑥2𝑘
).
∏ 2
𝑛𝑗2 − 𝑥2 1≤𝑚≤𝑘 𝑛𝑚
− 𝑛𝑗2
1
).
2 − 𝑛2
𝑛𝑗2𝑛−2𝑘+2 1≤𝑚≤𝑘 𝑛𝑚
𝑗
∏
=
(21)
𝑘
∏
(26)
∞
1
(𝑛 ) (𝑛 )
𝑃 𝑗𝑄 𝑗
2𝑛−2𝑘+2 𝑗−1 𝑘−𝑗
𝑛
𝑛𝑗 =𝑘−𝑗+1 𝑗
∞
∑
1
(𝑁) (𝑁)
𝑃𝑗−1
𝑄𝑘−𝑗
2𝑛−2𝑘+2
𝑁
1
𝑃(𝑁) 𝑄(𝑁) .
2𝑛−2𝑘+2 𝑗−1 𝑘−𝑗
𝑁
𝑁=1
1
∏
(22)
𝑓𝑁 (𝑥) = ∏ (1 +
(𝑁)
= 0 for 1 ≤ 𝑁 <
In the last step, 𝑁 begins with 1 since 𝑄𝑘−𝑗
𝑘 − 𝑗.
It follows that the sum (22) becomes that
∞
Now, consider the function
∑
𝑥
).
𝑟2 − 𝑁2
𝑁=1
(23)
We partition 𝑓𝑁(𝑥) into two parts. Let
𝑟>𝑁
=
∑
∞
𝑚 ≠ 𝑗
∏ (1 +
1
)
− 𝑛𝑗2
= ∑
1
.
2 − 𝑛2
2𝑛−2𝑘+2
𝑛
𝑗
𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑘 >0 𝑗=1 𝑛𝑗
1≤𝑚≤𝑘 𝑚
𝑟=1
𝑟 ≠ 𝑁
2
𝑛1 >𝑛2 >⋅⋅⋅>𝑛𝑗−1 >𝑛𝑗 1≤𝑚<𝑗 𝑛𝑚
∑
𝑁=𝑘−𝑗+1
𝑚 ≠ 𝑗
∑
∏
1
)]
2 − 𝑛2
𝑛
𝑗
𝑛𝑗 >𝑛𝑗+1 >⋅⋅⋅>𝑛𝑘 ≥1 𝑗<𝑚≤𝑘 𝑚
]
𝑚 ≠ 𝑗
1
∑
×(
Hence, the sum (19) is
∑
×(
(20)
It follows that the coefficient of 𝑥2𝑛 earlier is
∑(
1
𝑛𝑗2𝑛−2𝑘+2
𝑁2𝑛−2𝑘+2
.
(27)
(𝑁)
(𝑁)
(𝑁) (𝑁)
+ 𝑃1(𝑁) 𝑄𝑘−2
+ ⋅ ⋅ ⋅ + 𝑃𝑘−1
𝑄0 in (27)
Clearly, the sum 𝑃0(𝑁) 𝑄𝑘−1
𝑘−1
is the coefficient of 𝑥 in the Cauchy product of
[𝑃0(𝑁) +𝑃1(𝑁) 𝑥+𝑃2(𝑁) 𝑥2 + ⋅ ⋅ ⋅] ⋅ [𝑄0(𝑁) + 𝑄1(𝑁) 𝑥+𝑄2(𝑁) 𝑥2 + ⋅ ⋅ ⋅] ;
(28)
𝑥
) = 𝑃0(𝑁) + 𝑃1(𝑁) 𝑥 + 𝑃2(𝑁) 𝑥2 + ⋅ ⋅ ⋅ ,
𝑟2 − 𝑁2
𝑥
) = 𝑄0(𝑁) + 𝑄1(𝑁) 𝑥 + 𝑄2(𝑁) 𝑥2 + ⋅ ⋅ ⋅ .
∏ (1 + 2
2
𝑟
−
𝑁
1≤𝑟<𝑁
(𝑁)
(𝑁)
(𝑁) (𝑁)
𝑃0(𝑁) 𝑄𝑘−1
+ 𝑃1(𝑁) 𝑄𝑘−2
+ ⋅ ⋅ ⋅ + 𝑃𝑘−1
𝑄0
(24)
that is, it is the coefficient of 𝑥𝑘−1 in the power series
(𝑁) 𝑘−1
𝑓𝑁 (𝑥) = 𝑎0(𝑁) + 𝑎1(𝑁) 𝑥 + ⋅ ⋅ ⋅ + 𝑎𝑘−1
𝑥 + ⋅⋅⋅.
(29)
4
Journal of Applied Mathematics
Therefore, the sum (27) is
∞
∑
𝑁=1 𝑁
Proof of Theorem 3. Taking logarithms of both sides of (32),
we get that
(𝑁)
𝑎𝑘−1
2𝑛−2𝑘+2
.
(30)
∞
log (𝑓𝑁 (𝑥)) = ∑ log (1 −
𝑟=1
𝑟 ≠ 𝑁
The proof is completed.
Remark 5. If we take 𝑥 to be a complex variable, then the
series
∞
𝑓𝑁󸀠 (𝑥)
−1
= ∑ 2
𝑓𝑁 (𝑥)
𝑁
−
𝑟2 − 𝑥
𝑟=1
(31)
𝑟 ≠ 𝑁
𝑟 ≠ 𝑁
∞ ∞
is absolutely and uniformly convergent for 𝑥 in any compact
set in the complex plane; thus, the function
∞
𝑥
)
𝑟2 − 𝑁2
𝑓𝑁 (𝑥) = ∏ (1 +
𝑟=1
𝑟 ≠ 𝑁
𝑟=1 𝑗=0 (𝑁
𝑟 ≠ 𝑁
sin (𝜋𝑧) = 𝜋𝑧∏ (1 −
𝑟=1
𝑧
),
𝑟2
𝑧 ∈ C.
(33)
(34)
𝑧
≜ lim ∏ (1 −
).
𝑛→∞
𝑟
+
𝜔
𝑟=−𝑛
2
𝑧 − 2𝜔𝑧
)
𝑟2 − 𝜔2
2
−1
)
∞
= − ∑ 𝑏𝑗+1 𝑥𝑗 ,
𝑗=0
where we denote
𝑏𝑗+1 = ∑
𝑟=1
𝑟 ≠ 𝑁
1
(𝑁2
− 𝑟2 )
𝑗+1
,
for 𝑗 = 0, 1, 2, . . . .
(40)
The order of the summation can be changed since the series
2
∑∞𝑟=1 (−1/(𝑁2 −𝑟2 −𝑥)) is dominated by ∑∞
𝑛=1 (𝐿/𝑛 ) for some
positive constant 𝐿. From (39), we get that
∞
(41)
or
∞
∞
∞
𝑘=1
𝑖=0
𝑗=0
(42)
Write out the Cauchy product in the right side of (42), then
compare the coefficient of 𝑥𝑘−1 on both sides. We get that
𝑧 − 2𝜔𝑧
sin [𝜋 (𝜔 − 𝑧)]
𝑧 −1
) .
(1 − ) (1 − 2
sin (𝜋𝜔)
𝜔
𝑁 − 𝜔2
𝑘−1
∞
sin [𝜋 (𝑁 − 𝑧)]
𝑧2 − 2𝑁𝑧
) = 2𝑁2 (−1)𝑁
.
∏ (1 − 2
2
𝑟
−
𝑁
𝜋
− 𝑧) (𝑧2 − 2𝑁𝑧)
(𝑁
𝑟=1
𝑟 ≠ 𝑁
(36)
We write 𝑧2 −2𝑁𝑧 = −𝑥. Or equivalently, let 𝑧 = 𝑁±√𝑁2 − 𝑥.
Then, we get
𝑟=1
𝑟 ≠ 𝑁
𝑗+1
∑ 𝑘𝑎𝑘(𝑁) 𝑥𝑘−1 = − (∑𝑎𝑖(𝑁) 𝑥𝑖 ) ( ∑ 𝑏𝑗+1 𝑥𝑗 ) .
Now, let 𝜔 → 𝑁. We get
∞
− 𝑟2 )
𝑗=0
(35)
∏ (1 +
𝑟=1
𝑟 ≠ 𝑁
𝑗=0
𝑓𝑁󸀠 (𝑥) = −𝑓𝑁 (𝑥) ∑𝑏𝑗+1 𝑥𝑗 ,
Let 𝑁 ≥ 1 be temporarily fixed. By (34), for 𝜔 ∉ Z we have
=
1
(𝑁2
𝑟 ≠ 𝑁
𝑛
𝑟=1
𝑟 ≠ 𝑁
∞
∞
∞
∞
𝑧
sin [𝜋 (𝜔 − 𝑧)]
= ∏ (1 −
)
sin (𝜋𝜔)
𝑟+𝜔
𝑟=−∞
∏ (1 −
− 𝑟2 )
= − ∑ (𝑥𝑗 ∑
Similar to Euler’s formula, Eisenstein studied a product of two
variables and proved that for (𝜔, 𝑧) ∈ (C\Z)×C the following
formula holds (see [10, page 17]):
∞
𝑗+1
2
(39)
(32)
Proof of Theorem 2. First we recall Euler’s classical formula
2
𝑥𝑗
= − ∑∑
is analytic in the complex plane. Hence, it may be expanded
as a Taylor series.
∞
(38)
By Remark 5, the series may be differentiated term-by-term;
hence, we have
∞
𝑥
∑ 2
𝑟
−
𝑁2
𝑟=1
𝑥
).
𝑁2 − 𝑟2
sin (𝜋√𝑁2 − 𝑥)
𝑥
2
𝑁+1
. (37)
)
=
2𝑁
(−1)
𝑟2 − 𝑁2
𝜋𝑥√𝑁2 − 𝑥
𝑘𝑎𝑘(𝑁) = − ∑ 𝑏𝑗+1 𝑎𝑖(𝑁) = − ∑ 𝑏𝑘−𝑗 𝑎𝑗(𝑁) .
𝑗=0
𝑖+𝑗=𝑘−1
𝑖,𝑗≥0
(43)
Proof of Theorem 4. We now study the the generating function
𝑓𝑁 (𝑥) = 2𝑁2 (−1)𝑁+1
sin (𝜋√𝑁2 − 𝑥)
.
(44)
= 1.
(45)
𝜋𝑥√𝑁2 − 𝑥
We may use L’Hospital’s rule to verify that
lim 2𝑁2 (−1)𝑁+1
𝑥→0
sin (𝜋√𝑁2 − 𝑥)
𝜋𝑥√𝑁2 − 𝑥
Journal of Applied Mathematics
5
Now we expand out 𝑓𝑁(𝑥). We have
We may apply Hoffman’s result [6, Lemma 1.3] to get the
direct formula for
(𝑘) 󵄨󵄨
󵄨󵄨
sin√𝑥
) 󵄨󵄨󵄨
.
(
(54)
√𝑥 𝑥 󵄨󵄨󵄨𝑥=(𝑁𝜋)2
𝑚
𝑓𝑁 (𝑥) =
=
𝑚 2𝑚
2
2𝑁2 (−1)𝑁+1 ∞ (−1) 𝜋 (𝑁 − 𝑥)
∑
𝑥
(2𝑚 + 1)!
𝑚=0
2𝑁2 (−1)𝑁+1 ∞ (−1)𝑚 𝜋2𝑚
∑
𝑥
𝑚=0 (2𝑚 + 1)!
Here, we use some simple properties of the Bessel functions
of the first kind to give its direct expression.
𝑚
𝑚
× ∑ ( ) 𝑁2𝑚−2𝑘 (−1)𝑘 𝑥𝑘
𝑘
(46)
Lemma 6. Let 𝑘 ≥ 0 be an integer and let 𝑥 > 0. Then one has
𝜋
𝑑𝑘 sin √𝑥
) = √ (−1)𝑘 2−𝑘 𝑥−(2𝑘+1)/4 𝐽𝑘+1/2 (√𝑥) , (55)
(
𝑘
√𝑥
2
𝑑𝑥
𝑘=0
∞
(−1)𝑘
2𝑘
𝑘=0 𝑁
= 2𝑁2 (−1)𝑁+1 ∑
where 𝐽𝑘+1/2 (𝑥) denotes the Bessel function of the first kind of
index 𝑘 + 1/2.
∞
(−1)𝑚 (𝑁𝜋)2𝑚 𝑚 𝑘−1
( )𝑥 .
𝑘
(2𝑚 + 1)!
𝑚=𝑘
×∑
The Bessel functions with a half-integer index can be
represented by elementary functions. The following lemma
is well known.
By (11) and (13), we have
Lemma 7. Let 𝑘 ≥ 0 be an integer, and let 𝑥 > 0. Then, one has
(𝑁)
𝑎𝑘−1
= 2𝑁2−2𝑘 (−1)𝑁+𝑘+1
∞
(−1)𝑚 (𝑁𝜋)2𝑚 𝑚
( ),
𝑘
(2𝑚 + 1)!
𝑚=𝑘
×∑
∀𝑘 = 1, 2, 3, . . . .
(47)
𝐽𝑘+1/2 (𝑥)
=√
Consider the function
∞
sin (𝑁𝜋√𝑥)
(−1)𝑚 (𝑁𝜋)2𝑚 𝑚
𝑥 .
= ∑
𝑁𝜋√𝑥
𝑚=0 (2𝑚 + 1)!
[𝑘/2]
(−1)𝑗 (𝑘 + 2𝑗)!
2 1 {
𝑘𝜋
)
sin
(𝑥
−
∑
𝜋 √𝑥 {
2 𝑗=0 (2𝑗)! (𝑘 − 2𝑗)!
{
(48)
⋅
Clearly, the sum in (47) can be rewritten as
[(𝑘−1)/2]
× ∑
(𝑘) 󵄨󵄨
∞
1 sin (𝑁𝜋√𝑥)
(−1)𝑚 (𝑁𝜋)2𝑚 𝑚
)
( )= (
𝑘
𝑘!
(2𝑚 + 1)!
𝑁𝜋√𝑥
𝑥
𝑚=𝑘
∑
󵄨󵄨
󵄨󵄨 , (49)
󵄨󵄨
󵄨󵄨𝑥=1
where ()(𝑘)
𝑥 means the 𝑘th derivative of a function with respect
to 𝑥.
We denote 𝜙(𝑥) = sin√𝑥/√𝑥. Then, we have
𝜙 (𝑁2 𝜋2 𝑡) =
sin (𝑁𝜋√𝑡)
𝑁𝜋√𝑡
,
(50)
𝑗=0
𝑑𝑡𝑘
=
󵄨
𝑑𝑘 𝜙 (𝑥) 󵄨󵄨󵄨
󵄨󵄨
(𝑁𝜋)2𝑘
𝑑𝑥𝑘 󵄨󵄨
󵄨𝑥=𝑁2 𝜋2 𝑡
,
(𝑁)
𝑎𝑘−1
=
𝜋𝑘−1
𝑘!(2𝑁)𝑘−1
[𝑘/2]
{
(−1)𝑗 (𝑘 + 2𝑗)!
𝑘𝜋
)
sin
(
∑
{
2 𝑗=0 (2𝑗)! (𝑘 − 2𝑗)!
{
⋅
∀𝑘 = 0, 1, 2, . . . ,
(51)
(
𝑁𝜋√𝑡
[(𝑘−1)/2]
∑
𝑗=0
(𝑘) 󵄨󵄨
(𝑘) 󵄨󵄨
󵄨󵄨
󵄨󵄨
sin√𝑥
󵄨
) 󵄨󵄨󵄨 = (𝑁𝜋)2𝑘 (
) 󵄨󵄨󵄨
. (52)
󵄨
√𝑥 𝑥 󵄨󵄨󵄨𝑥=(𝑁𝜋)2
𝑡 󵄨󵄨󵄨𝑡=1
Finally, from (47) (49) we get that
(𝑁)
𝑎𝑘−1
=
𝑘𝜋
1
− cos ( )
2
(2𝑁𝜋)2𝑗
×
which implies that
sin (𝑁𝜋√𝑡)
(−1)𝑗 (𝑘 + 2𝑗 + 1)!
1 }
⋅
.
(2𝑗 + 1)! (𝑘 − 2𝑗 − 1)! (2𝑥)2𝑗+1 }
}
(56)
From Lemmas 6 and 7, and (53), we get that
and, hence,
𝑑𝑘 𝜙 (𝑁2 𝜋2 𝑡)
𝑘𝜋
1
+ cos (𝑥 −
)
2
(2𝑥)2𝑗
(𝑘) 󵄨󵄨
2𝑁2 𝜋2𝑘 (−1)𝑁+𝑘+1 sin√𝑥
)
(
√𝑥 𝑥
𝑘!
󵄨󵄨
󵄨󵄨
,
󵄨󵄨
󵄨󵄨𝑥=(𝑁𝜋)2
𝑘 ≥ 1.
(53)
⋅
(57)
(−1)𝑗 (𝑘 + 2𝑗 + 1)!
(2𝑗 + 1)! (𝑘 − 2𝑗 − 1)!
}
1
.
2𝑗+1 }
(2𝑁𝜋)
}
This completes the proof of Theorem 4.
4. Examples
The direct formula for 𝑎𝑘(𝑁) can be found from Theorem 4.
However, we would like to use Theorem 3 to present some
6
Journal of Applied Mathematics
concrete examples to show how to calculate 𝑎𝑘(𝑁) for small 𝑘.
The difficult part of the recursion formula (14) is for 𝑗 ≥ 1 to
calculate the sum
∞
From (58)–(63), we get that
∞
𝑏1 = −𝑎1(𝑁) = 𝜆 ∑ [𝐴 𝑟 + 𝐵𝑟 ] = −3𝜆2 ,
𝑟=1
𝑟 ≠ 𝑁
𝑗
1 𝑗 ∞
1
1
=
(
[
)
+
]
𝑏𝑗 = ∑
∑
2
2 𝑗
2𝑁 𝑟=1 𝑁 − 𝑟 𝑁 + 𝑟
𝑟=1 (𝑁 − 𝑟 )
1
𝑟 ≠ 𝑁
𝑗
𝑟 ≠ 𝑁
∞
∞
𝑏2 = 𝜆2 ∑ {[𝐴2𝑟 + 𝐵𝑟2 ] + 2𝜆 [𝐴 𝑟 + 𝐵𝑟 ]}
(58)
𝑟=1
𝑟 ≠ 𝑁
𝑗
= 𝜆 ∑ [𝐴 𝑟 + 𝐵𝑟 ] ,
= 𝜆2 ⋅ {[2𝜁 (2) − 5𝜆2 ] − 6𝜆2 }
𝑟=1
𝑟 ≠ 𝑁
where we denote 𝐴 𝑟 = 1/(𝑁 − 𝑟), 𝐵𝑟 = 1/(𝑁 + 𝑟), and
𝜆 = 1/2𝑁.
It follows from 𝐴 𝑟 𝐵𝑟 = 𝜆[𝐴 𝑟 + 𝐵𝑟 ] that
2
[𝐴 𝑟 + 𝐵𝑟 ] = [𝐴2𝑟 + 𝐵𝑟2 ] + 2𝜆 [𝐴 𝑟 + 𝐵𝑟 ] ,
= − 11𝜆4 + 2𝜁(2) 𝜆2 ,
∞
𝑏3 = 𝜆3 ∑ {[𝐴3𝑟 + 𝐵𝑟3 ] + 3𝜆 [𝐴2𝑟 + 𝐵𝑟2 ] + 6𝜆2 [𝐴 𝑟 + 𝐵𝑟 ]}
𝑟=1
𝑟 ≠ 𝑁
= 𝜆3 ⋅ {−9𝜆3 + 3𝜆 [2𝜁(2) − 5𝜆2 ] − 18𝜆3 }
3
[𝐴 𝑟 + 𝐵𝑟 ] = [𝐴3𝑟 + 𝐵𝑟3 ] + 3𝜆 [𝐴2𝑟 + 𝐵𝑟2 ] + 6𝜆2 [𝐴 𝑟 + 𝐵𝑟 ] .
(59)
= − 42𝜆6 + 6𝜁(2)𝜆4 ,
∞
𝑏4 = 𝜆4 ∑ {[𝐴4𝑟 + 𝐵𝑟4 ] + 4𝜆 [𝐴3𝑟 + 𝐵𝑟3 ]
Generally, we can use induction on 𝑗 to prove that if for 𝑗 ≥ 2
we have gotten some positive integers 𝑐1 , 𝑐2 , . . . , 𝑐𝑗−1 such that
𝑗
[𝐴 𝑟 + 𝐵𝑟 ] =
[𝐴𝑗𝑟
+
𝐵𝑟𝑗 ]
𝑐1 𝜆 [𝐴𝑗−1
𝑟
+
+
𝐵𝑟𝑗−1 ]
+ ⋅ ⋅ ⋅ + 𝑐𝑗−1 𝜆𝑗−1 [𝐴 𝑟 + 𝐵𝑟 ] ,
𝑟=1
𝑟 ≠ 𝑁
+ 10𝜆2 [𝐴2𝑟 + 𝐵𝑟2 ]
+20𝜆3 [𝐴 𝑟 + 𝐵𝑟 ]}
(60)
= − 163𝜆8 + 20𝜁(2) 𝜆6 + 2𝜁(4) 𝜆4 .
then the expression for [𝐴 𝑟 + 𝐵𝑟 ]𝑗+1 is
(64)
From formula (14), we get that
𝑗+1
[𝐴 𝑟 + 𝐵𝑟 ]
=
[𝐴𝑗+1
𝑟
+
𝐵𝑟𝑗+1 ]
+ (1 +
𝑐1 ) 𝜆 [𝐴𝑗𝑟
+
𝑎2(𝑁) = −
𝐵𝑟𝑗 ]
𝑗−1
+ (1 + 𝑐1 + 𝑐2 ) 𝜆2 [𝐴𝑗−1
𝑟 + 𝐵𝑟 ]
(61)
+ ⋅ ⋅ ⋅ + (1 + 𝑐1 + 𝑐2 + ⋅ ⋅ ⋅ + 𝑐𝑗−1 ) 𝜆𝑗−1 [𝐴2𝑟 + 𝐵𝑟2 ]
𝑟=1
𝑟 ≠ 𝑁
1≤𝑟<𝑁
= 2𝜁 (𝑖) −
1
= − {[−42𝜆6 + 6𝜁(2) 𝜆4 ] + 3𝜆2 [−11𝜆4 + 2𝜁(2) 𝜆2 ]
3
1
1
+
]
(𝑁 − 𝑟)𝑖 (𝑁 + 𝑟)𝑖
1
1
+∑[
+
]
𝑖
(𝑁 + 𝑟)𝑖
𝑟>𝑁 (𝑁 − 𝑟)
−3𝜆2 [10𝜆4 − 𝜁(2) 𝜆2 ]}
(62)
1+2
= 2𝜁 (𝑖) − (1 + 2𝑖 ) 𝜆𝑖 .
(2𝑁)𝑖
∑ [𝐴𝑖𝑟 + 𝐵𝑟𝑖 ] = −
𝑟=1
𝑟 ≠ 𝑁
35
5𝜁(2)
−
,
64𝑁6 16𝑁4
1
1
= 126𝜆8 − 21𝜁(2) 𝜆6 − 𝜁(4) 𝜆4 + 𝜁2 (2) 𝜆4
2
2
=
𝑖
1+2
= − (1 + 2𝑖 ) 𝜆𝑖 .
(2𝑁)𝑖
= 35𝜆6 − 5𝜁(2) 𝜆4 =
1
𝑎4(𝑁) = − {𝑎0(𝑁) 𝑏4 + 𝑎1(𝑁) 𝑏3 + 𝑎2(𝑁) 𝑏2 + 𝑎3(𝑁) 𝑏1 }
4
𝑖
Similarly, if 𝑖 ≥ 1 is an odd integer, then we have
∞
5
𝜁(2)
−
,
4
8𝑁
4𝑁2
1
𝑎3(𝑁) = − {𝑎0(𝑁) 𝑏3 + 𝑎1(𝑁) 𝑏2 + 𝑎2(𝑁) 𝑏1 }
3
Note that if 𝑖 ≥ 2 is an even integer, then we have
∑ [𝐴𝑖𝑟 + 𝐵𝑟𝑖 ] = ∑ [
2
1
{[−11𝜆4 + 2𝜁(2)𝜆2 ] − (3𝜆2 ) }
2
= 10𝜆4 − 𝜁(2) 𝜆2 =
+ 2 (1 + 𝑐1 + 𝑐2 + ⋅ ⋅ ⋅ + 𝑐𝑗−1 ) 𝜆𝑗 [𝐴 𝑟 + 𝐵𝑟 ] .
∞
= −
1 (𝑁)
{𝑎 𝑏 + 𝑎1(𝑁) 𝑏1 }
2 0 2
63
21𝜁(2) 3𝜁(4)
−
+
.
8
128𝑁
64𝑁6
64𝑁4
(63)
(65)
Journal of Applied Mathematics
7
For 𝑘 = 2, 3, 4, using 𝑎1(𝑁) , 𝑎2(𝑁) , and 𝑎3(𝑁) in formula (12),
respectively, we will get identities (6) and (8). Moreover, for
𝑛 ≥ 5, we have
𝐸 (2𝑛, 5) =
21
63
𝜁(2𝑛) − 𝜁(2) 𝜁 (2𝑛 − 2)
128
64
3
+ 𝜁(4) 𝜁 (2𝑛 − 4) .
64
(66)
Acknowledgment
This work is supported by the National Natural Science
Foundation of China (1127208).
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Optimization
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