# On Certain Matrices of Bernoulli Numbers

Authors: Ruiming Zhang; Li-Chen Chen
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

In this work we compute the determinant and inverse matrices for a certain symmetric matrix of Rayleigh sums. As a special case we also obtain the determinants and inverses for the matrices of the Bernoulli numbers and related numbers.


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## Body

## 1. Introduction

The sequence of Bernoulli numbersBn is one of the most important sequences in mathematics. It has deep connections to number theory, for instance, the Bernoulli numbers are used to express the values of ζ(2n), where ζ(s) is the Riemann zeta function and n is a positive integer [1, 2]. The Bernoulli numbers are also very important in analysis, for example, they appear in the Euler-Maclaurin formula [1], which is very important in mathematics and physics. The Bernoulli numbers are also very important in asymptotics of q-special functions; for example, in [3] we proved a complete asymptotic expansion of q-Gamma function Γq(z) on the complex plane in terms of Bernoulli polynomials and Bernoulli polynomials. The applications of Bernoulli numbers in applied mathematics are just too many to list all of them; just to name a few, for example, see [4–6]. The Rayleigh sums σν(n) generalize ζ(2n) and it is known that σ1/2(n) is a rational multiple of B2n [7]. In this work we first derive the inverse and determinant of a certain symmetric matrix defined by σν(n) and then specialize the result to the matrices defined by Bernoulli numbers Bn and related numbers Sn.But we have to emphasize that the present work demonstrated a method to compute inverses of certain Hankel matrices, not just determinants. In fact there are many known methods to compute determinants; for example, see [1, 8–11].


## 2. Preliminaries

Forν>-1 the Bessel function of first kind is defined by [1, 7, 11, 12]:
(1)Jν(z)=1Γ(ν+1)∑k=0∞‍(-1)kk!(ν+1)k(z2)2k+ν,
where
(2)1Γ(a)=a∏j=1∞‍(1+aj)(1+1j)-a,(a)n=Γ(a+n)Γ(a),n∈Z,a∈C.
As a special case we have
(3)J1/2(z)=2πzsinz.
It is known that the even entire function Jν(z)z-ν has infinitely many zeros, all of which are real. Let
(4)0<jν,1<jν,2<⋯
be all its positive zeros; then the Rayleigh sum is defined by [7]
(5)σν(n)=∑k=1∞‍1jν,k2n,n∈N.
Clearly [1],
(6)σ1/2(n)=∑n=1∞‍1(nπ)2k=(-1)n+122n-1B2n(2n)!,
where the Bernoulli numbers Bn are defined by [1, 2, 12]
(7)zez-1=∑n=0∞‍Bnznn!,|z|<2π.
The related numbers {Sn}n=1∞ are defined by [2, 13]
(8)Sn=2(2π)n∑k=-∞∞‍1(4k+1)n
for n=2,3,… and S1=1; it is known that
(9)(-1)n+1B2n(2n)!=S2n22n(22n-1),n∈N.


## 3. Main Results

Theorem 1.
Given a nonnegative integern, one has
(10)det⁡(σν(j+k+1))j,k=1n=2(n+1)(2n+1)((ν+1)/2)n+1∏k=0n‍(ν+1)2k2,{(σν(j+k+1))j,k=1n}-1=(∑m=0n‍((-1)j+k(2m+ν+1)(m+j)!(m+k)!×(ν+1)m-j(ν+1)m-k)-1∑m=0n‍×(ν+1)m-j(ν+1)m-k)-1hhhhhh.h×(ν+1)m+j(ν+1)m+khhhhhh.h×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!hhhhhhhhhhh∑m=0n‍×(ν+1)m-j(ν+1)m-k)-1))j,k=0n,
for ν>-1.Corollary 2.
For any nonnegative integern, one has
(11)det⁡(B2j+2k+2(2j+2k+2)!)j,k=1n=1(3/4)n+1∏k=0n‍(3/2)2k2,{(B2(j+k+1)(2j+2k+2)!)j,k=0n}-1=(∑m=0n‍((m+34)(m+j)!(m+k)!(32)m+j(32)m+k(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1hhhhhhhhhh×((2j)!(2k)!(m-j)!(m-k)!(32)m-jhhhhhhhhhhhhhl∑m=0n‍(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1))j,k=0n,
or, equivalently,
(12)det⁡(S2j+2k+24j+k+1-1)j,k=0n=4(n+1)2(3/4)n+1∏k=0n‍(3/2)2k2,(13){(S2j+2k+24j+k+1-1)j,k=0n}-1=(∑m=0n‍((m+34)(m+j)!(m+k)!(32)m+j(32)m+k(2j)!(2k)!(m-j)!(m-k)!(32)m-j(-4)j+k(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1)hhhhhhhhlh×((-4)j+k(2j)!(2k)!(m-j)!(m-k)!(32)m-jhhhhhhhhllhhhh∑m=0n‍(2j)!(2k)!(m-j)!(m-k)!(32)m-j(-4)j+k(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-j(32)m-k)-1))j,k=0n.


## 4. Proofs

Given a probability measuredμ(x) on R such that ∫R‍x2ndμ(x)<∞ for all n∈R, we define the inner product for dμ(x) square integrable functions f(x) and g(x) by
(14)(f,g)=∫-∞∞‍f(x)g(x)dμ(x).
For each n∈N∪{0}, let Gn=(mj,k)j,k=0n with mj,k=(uj,uk) for j,k=0,1,…,n where {uk(x)}k=0∞ is a sequence of polynomials with u0(x)=1 such that, for each n, {uk(x)}k=0n are linearly independent. Then there is a unique orthonormal system {pk(x)}k=0∞ [1, 10, 11]:
(15)pn(x)=1det⁡Gndet⁡Gn-1×det⁡(m0,0m0,1m0,2⋯m0,nm1,0m1,1m1,2⋯m1,n⋮⋮⋮⋱⋮mn-1,0mn-1,1mn-1,2⋯mn-1,nu0(x)u1(x)u2(x)⋯un(x)),
with positive leading coefficient in un(x). Clearly we have pn(x)=∑j=0n‍an,juj(x) for some real numbers aj,k for j,k=0,1,…,n and aj,k=0 for k>j.Lemma 3.
For each nonnegative integern, let Gn=(mj,k)j,k=0n and An=(aj,k)j,k=0n. Then
(16)det⁡Gn=∏j=0n‍aj,j-2,Gn-1=AnTAn.Proof.
From (15) and pn(x)=∑j=0n‍an,juj(x) it is clear that
(17)an,n=det⁡Gn-1det⁡Gn,det⁡Gn=∏j=0n‍aj,j-2.
For each n, since both {pk(x)}k=0n and {uk(x)}k=0n are a basis for the same set of polynomials, An must be invertible for each n∈N∪{0}. We denote An-1=(sj,k)j,k=0n; then uj(x)=∑l=0n‍sj,lpl(x) for j=0,1,…,n. Clearly, sj,l=0 for l>j. Thus,
(18)mj,k=(uj(x),uk(x))=∑m=0n‍sj,msk,m,
for j,k=0,1,…,n, which is
(19)Gn=An-1(An-1)T=An-1(AnT)-1,
and hence Gn-1=AnTAn.


### 4.1. Proof of Theorem1

The normalized even order Lommel polynomials are defined by [11]
(20)hn(x)=2n+ν+12h2n,ν+1(x)=∑k=0n‍(-1)n-k2n+ν+1(n+k)!(ν+1)n+k22k+1(2k)!(n-k)!(ν+1)n-k(x2)k,
for n∈N and h0(x)=ν+1/2. They satisfy the orthogonal relation
(21)∑k=1∞‍1jν,k2hm(1jν,k)hn(1jν,k)=δm,n.
For n=0,1,…, it is clear that the nth moment with respect to the measure of orthogonality is
(22)mn=∑k=1∞‍1jν,k2+2n=σν(n+1).
Let un(x)=x2n for n=0,1,…; then
(23)an,k=(-1)n-k2n+ν+1(n+k)!(ν+1)n+k22k+1(2k)!(n-k)!(ν+1)n-k.
By Lemma 3, the matrix (σν(j+k+1))j,k=0n has determinant
(24)∏k=0n‍24k+2(ν+1+2k)(ν+1)2k2=2(n+1)(2n+1)((ν+1)/2)n+1∏k=0n‍(ν+1)2k2
and its inverse (γj,k)j,k=0n has elements
(25)γj,k=∑m=0n‍((-1)j+k(2m+ν+1)(m+j)!(m+k)!×(ν+1)m-j(ν+1)m-k)-1hhhh×(ν+1)m+j(ν+1)m+khhhhh×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!hhhhhhhh×(ν+1)m-j(ν+1)m-k)-1).


### 4.2. Proof of Corollary2

From (24), (25), and (6), we get
(26)det⁡((-1)j+k22j+2k+1B2(j+k+1)(2j+2k+2)!)j,k=0n=2(n+1)(2n+1)(3/4)n+1∏k=0n‍(3/2)2k2,{((-1)j+k22j+2k+1B2(j+k+1)(2j+2k+2)!)j,k=0n}-1=(∑m=0n‍((-1)j+k(2m+32)(m+j)!(m+k)!(32)m+j(32)m+k∑m=0n‍×(32)m-k)-1∑m=0n‍×(32)m-k)-1hhhhhhhhhh×(32)m+j(32)m+khhhhhhhhhh×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!(32)m-j∑m=0n‍×(32)m-khhhhhhhhhhhhh∑m=0n‍×(32)m-k)-1))j,k=0n.
They are simplified to
(27)det⁡(B2(j+k+1)(2j+2k+2)!)=1(3/4)n+1∏k=0n‍(3/2)2k2,{(B2(j+k+1)(2j+2k+2)!)j,k=0n}-1=(∑m=0n‍((m+34)(m+j)!(m+k)!(32)m+j(32)m+k(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1hhhhhhhhhh×((2j)!(2k)!(m-j)!(m-k)!(32)m-jhhhhhhhhhhhhhl∑m=0n‍((m+34)(m+j)!(m+k)!(32)m+j(32)m+k)(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1))j,k=0n.
By (9) we get
(28)det⁡(S2j+2k+22(4j+k+1-1))j,k=0n=2(n+1)(2n+1)(3/4)n+1∏k=0n‍(3/2)2k2,{(S2j+2k+22(4j+k+1-1))j,k=0n}-1=(∑m=0n‍((-1)j+k(2m+32)(m+j)!(m+k)!(32)m+j(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1hhhhhhhhh×(32)m+khhhhhhhhhl×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!(32)m-jhhhhhhhhhlhhh(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1))j,k=0n,
which are simplified to (12) and (13), respectively.


## 4.1. Proof of Theorem1

The normalized even order Lommel polynomials are defined by [11]
(20)hn(x)=2n+ν+12h2n,ν+1(x)=∑k=0n‍(-1)n-k2n+ν+1(n+k)!(ν+1)n+k22k+1(2k)!(n-k)!(ν+1)n-k(x2)k,
for n∈N and h0(x)=ν+1/2. They satisfy the orthogonal relation
(21)∑k=1∞‍1jν,k2hm(1jν,k)hn(1jν,k)=δm,n.
For n=0,1,…, it is clear that the nth moment with respect to the measure of orthogonality is
(22)mn=∑k=1∞‍1jν,k2+2n=σν(n+1).
Let un(x)=x2n for n=0,1,…; then
(23)an,k=(-1)n-k2n+ν+1(n+k)!(ν+1)n+k22k+1(2k)!(n-k)!(ν+1)n-k.
By Lemma 3, the matrix (σν(j+k+1))j,k=0n has determinant
(24)∏k=0n‍24k+2(ν+1+2k)(ν+1)2k2=2(n+1)(2n+1)((ν+1)/2)n+1∏k=0n‍(ν+1)2k2
and its inverse (γj,k)j,k=0n has elements
(25)γj,k=∑m=0n‍((-1)j+k(2m+ν+1)(m+j)!(m+k)!×(ν+1)m-j(ν+1)m-k)-1hhhh×(ν+1)m+j(ν+1)m+khhhhh×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!hhhhhhhh×(ν+1)m-j(ν+1)m-k)-1).


## 4.2. Proof of Corollary2

From (24), (25), and (6), we get
(26)det⁡((-1)j+k22j+2k+1B2(j+k+1)(2j+2k+2)!)j,k=0n=2(n+1)(2n+1)(3/4)n+1∏k=0n‍(3/2)2k2,{((-1)j+k22j+2k+1B2(j+k+1)(2j+2k+2)!)j,k=0n}-1=(∑m=0n‍((-1)j+k(2m+32)(m+j)!(m+k)!(32)m+j(32)m+k∑m=0n‍×(32)m-k)-1∑m=0n‍×(32)m-k)-1hhhhhhhhhh×(32)m+j(32)m+khhhhhhhhhh×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!(32)m-j∑m=0n‍×(32)m-khhhhhhhhhhhhh∑m=0n‍×(32)m-k)-1))j,k=0n.
They are simplified to
(27)det⁡(B2(j+k+1)(2j+2k+2)!)=1(3/4)n+1∏k=0n‍(3/2)2k2,{(B2(j+k+1)(2j+2k+2)!)j,k=0n}-1=(∑m=0n‍((m+34)(m+j)!(m+k)!(32)m+j(32)m+k(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1hhhhhhhhhh×((2j)!(2k)!(m-j)!(m-k)!(32)m-jhhhhhhhhhhhhhl∑m=0n‍((m+34)(m+j)!(m+k)!(32)m+j(32)m+k)(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1))j,k=0n.
By (9) we get
(28)det⁡(S2j+2k+22(4j+k+1-1))j,k=0n=2(n+1)(2n+1)(3/4)n+1∏k=0n‍(3/2)2k2,{(S2j+2k+22(4j+k+1-1))j,k=0n}-1=(∑m=0n‍((-1)j+k(2m+32)(m+j)!(m+k)!(32)m+j(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1hhhhhhhhh×(32)m+khhhhhhhhhl×(4j+k+1(2j)!(2k)!(m-j)!(m-k)!(32)m-jhhhhhhhhhlhhh(2j)!(2k)!(m-j)!(m-k)!(32)m-j×(32)m-k)-1))j,k=0n,
which are simplified to (12) and (13), respectively.



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