The Pascal Function and Its Applications to Bernoulli Numbers and Bernoulli and Euler Numbers and Euler Polynomials

Tian-Xiao He ∗ Jeff H.-C. Liao † and Peter J.-S. Shiue ‡ Dedicated to Professor L. C. Hsu on the occasion of his 95th birthday

Abstract A Pascal matrix function is introduced by Call and Velleman in [3]. In this paper, we will use the function to give a unified approach in the study of Bernoulli numbers and Bernoulli poly- nomials. Many well-known and new properties of the Bernoulli numbers and polynomials can be established by using the Pascal matrix function. The approach is also applied to the study of Euler numbers and Euler polynomials.

AMS Subject Classification: 05A15, 65B10, 33C45, 39A70, 41A80.

Key Words and Phrases: Pascal matrix, Pascal matrix func- tion, , Bernoulli , Euler number, Euler polynomial.

∗Department of , Illinois Wesleyan University, Bloomington, Illinois 61702 †Institute of Mathematics, Academia Sinica, Taipei, Taiwan ‡Department of Mathematical Sciences, University of Nevada, Las Vegas, Las Vegas, Nevada, 89154-4020. This work is partially supported by the Tzu (?) Tze foundation, Taipei, Taiwan, and the Institute of Mathematics, Academia Sinica, Taipei, Taiwan. The author would also thank to the Institute of Mathematics, Academia Sinica for the hospitality.

1 2 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

1 Introduction

A large literature scatters widely in books and journals on Bernoulli numbers Bn, and Bn(x). They can be studied by means of the binomial expression connecting them,

n X n B (x) = B xn−k, n ≥ 0. (1) n k k k=0 The study brings consistent attention of researchers working in combi- natorics, number theory, etc. In this paper, we will establish a matrix form of the above binomial expression using the following generalized Pascal matrix P [x] introduced by Call and Velleman in [3]:

n  P [x] := xn−k , (2) k n,k≥0 which we also call the Pascal matrix function and P [1] is clearly the clas- sical Pascal matrix. Many well-known and new properties of Bernoulli numbers and Bernoulli polynomials can be obtained through this ma- T trix approach. In fact, by denoting B(x) = (B0(x),B1(x),...) and T B = B(0) = (B1,B2,...) , we may write (1) as

B(x) = P [x]B(0). (3) P [x] is a homomorphic mapping from R to the infinite lower triangular matrices with real entries due to Theorem 2 shown [3]:

Theorem 1.1 [3] (Homomorphism Theorem) For any x, y ∈ R,

P [x + y] = P [x]P [y]. (4)

By noting

ik ii − j  = , (5) k j j k − j one immediately find the (i, j) entry of the right-hand side of (4) can be written as

i X i k xi−k yk−j, (6) k j k=j Pascal Matrix Function 3

while the (i, j) entry of the left-hand side of (4) is

i−j i i i X i − j X ii − j  (x+y)i−j = ykxi−j−k = xi−kyk−j, j j k j k − j k=0 k=j which is exactly (6) completing the proof of Theorem 1.1. In [4], iden- tity (5) is used to rederive several known properties and relationships involving the Bernoulli and Euler polynomials. In this paper, a uni- fied approach by means of P [x] is presented in the study of Bernoulli polynomials and numbers and Euler polynomials and numbers. For any square matrix A, the exponential of A is defined as the following matrix in a form:

1 1 X 1 eA = I + A + A2 + A3 + ··· = Ak. 2! 3! k! k=0 We say the series converges for every A if each entry of eA converges. From [13] and the definition of eA, we have

(α+β)A αA βA e = e e for any α, β ∈ R, (eA)−1 = e−A, tA tA tA Dte = Ae = e A. (7)

i i−j For n ∈ N ∪ 0, denote Pn[x] = ( j x )0≤i,j≤n. [3] shows there exists xHn a unique Hn = (hi,j)0≤i,j≤n such that Pn[x] = e . In fact, from (7), one may have

H = D exHn = D P [x] . (8) n x x=0 x n x=0

Denote H = (hi,j)0≤i,j. Then (4) of Theorem 1.1 can be proved in one line:

P [x + y] = e(x+y)H = exH eyH = P [x]P [y].

It is easy to see that the entry hi,j of H and Hn, n = 0, 1,..., are

 i if i = j + 1, h = (H) = , (9) i,j i,j 0 otherwise. 4 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

and the entry of Hk are

 i! if i = j + k, (Hk) = j! (10) i,j 0 otherwise.

k Hence, Hn = 0 if k ≥ n. In next section, we will define new matrix functions from Pn[x], which and the matrix relationship (3) between the Bernoulli numbers and the Bernoulli polynomials through P [x] will be used to re-build some well-known properties of the Bernoulli numbers and the Bernoulli polynomials in Section 3. Finally, in Section 4, a similar approach is used to study the Euler numbers and Euler polynomials and their relationships with Bernoulli numbers and Bernoulli polynomials.

2 Matrix functions Ln[x] and L¯n[x] related to Pn[x]

We now present the of P [x]. From [1] (on Page 240), there R 1 holds expression of the linear mapping Ln : Pn[x] 7→ 0 Pn[x]dx from matrix functions to matrices as

n−1 Z 1 X Hk L = P [x]dx = n , (11) n n (k + 1)! 0 k=0

xHn which is simply from the fact Pn[x] = e and the last formula in (7). Notice that L is a nonsingular lower with all main diagonal entries equal to 1. We now extend matrices (11) to matrix functions

n−1 Z x Z x X Hk L [x] := P [t]dt = etHn dt = n xk+1, (12) n n (k + 1)! 0 0 k=0 R x which can be considered a mapping Ln[x]: Pn[t] 7→ 0 Pn[t]dt associ- ated with any x ∈ R. It is obvious Ln[1] ≡ Ln defined in (11). From (10) and noting k = i − j, the entries of Ln[x] can be given as

 1 ixi−j+1, if i ≥ j ≥ 0, (L [x]) = i−j+1 j (13) n ij 0, otherwise. Pascal Matrix Function 5

Particularly,

 1 i, if i ≥ j, (L ) = i−j+1 j (14) n ij 0, otherwise.

We now give some properties of Ln[x] and its relationships with Pn[x].

Proposition 2.1 Let matrices Hn, Pn[x], and Ln[x] be defined as above. Then there holds

HnLn[x] = Ln[x]Hn = Pn[x] − In. (15)

Proof. By transferring the indexes, we have

n X Hk H L [x] = L [x]H = n xk n n n n k! k=1 n X Hk = n xk − I = P [x] − I . k! n n n k=0

Inspired by [1], we define

¯ −1 Ln = Dn(−1)LnDn(−1) , and ¯ −1 Ln[x] = Dn(−1)Ln[x]Dn(−1) , (16)

n−1 ¯ ¯ where Dn(−1) = diag(1, −1, 1, −1,..., (−1) ). Hence, Ln = Ln[1]. −1 Note. Since Dn(−1)HnDn(−1) = −Hn, we have

n−1 k −1 X Dn(−1)H Dn(−1) D (−1)P [x]D (−1) = n xk n n n k! k=0 n−1 k X (Dn(−1)HnDn(−1)) = xk k! k=0 n−1 k n−1 k X (−Hn) X (Hn) = xk = (−x)k = P [−x]. k! k! n k=0 k=0 ¯ The relationship between Ln[x] and Ln[x] can be shown below. 6 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

¯ Proposition 2.2 Let Ln[x] and Ln[x] be defined as before. Then there hold

¯ ¯ Ln[x] = −Ln[−x],Ln[x] = −Ln[−x], and (17) ¯ ¯ Ln = −Ln[−1],Ln = −Ln[−1]. (18)

Proof. From (16),

Z x  Z x Z −x ¯ −1 Ln[x] = Dn(−1) Pn[t]dt Dn(−1) = Pn[−t]dt = − Pn[t]dt, 0 0 0 which implies (17), and (18) follows as well.

¯ Using the above relationship between Ln[x] and Ln[x] and Proposi- tion 2.1, we immediately have

¯ Proposition 2.3 Let Hn, Pn[x], and Ln[x] be defined as before. Then

n−1 k X (−Hn) L¯ [x] = xk+1, and (19) n (k + 1)! k=0 ¯ ¯ −HnLn[x] = −Ln[x]Hn = P [−x] − In, (20) n−1 k n−1 k X (−Hn) X H (−x)k+1 = − n xk+1. (21) (k + 1)! (k + 1)! k=0 k=0

Proof. From (16) and (17)

¯ −1 Ln[x] = Dn(−1)Ln[x]Dn(−1) n−1 k X (Dn(−1)HnDn(−1)) = xk+1 (k + 1)! k=0 n−1 k X (−Hn) = xk+1, (k + 1)! k=0 i.e., formula (19). Again, (17) and (15) yield Pascal Matrix Function 7

¯ −HnLn[x] = HnLn[−x] = Pn[−x] − In and

¯ −Ln[x]Hn = Ln[−x]Hn = Pn[−x] − In. Substituting x = −x into (12) and noting (17) and (19), we have

n−1 k n−1 k X (−Hn) X H xk+1 = L¯ [x] = −L [−x] = − n (−x)k+1, (k + 1)! n n (k + 1)! k=0 k=0

which implies (21).

¯ Corollary 2.4 Let Hn, Pn[x], Ln[x], and Ln[x] be defined as before. Then

¯ Ln[x] = Pn[x]Ln[x], or equivalently

Ln[x]Dn(−1) = Pn[x]Dn(−1)Ln[x]. (22)

Proof. For k = 1, 2, . . . , n, we have

¯ k k−1 k−1 k Pn[x]Ln[x]Hn = Pn[x](−Pn[−x]+In)Hn = (Pn[x]−In)Hn = Ln[x]Hn,

k which imply (22) from the structure of Hn.

¯ Proposition 2.5 Let Pn[x], Ln[x], and Ln[x] be defined as before. Then

Ln[y] − Ln[x] = Pn[x]Ln[y − x] = Ln[y − x]Pn[x], ¯ ¯ Ln[y] − Ln[x] = Pn[−y]Ln[y − x] = Ln[y − x]Pn[−y]. (23) 8 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Proof. The first formula of (8) can be proved as follows, and the second formula is from the first formula and the relationship between Ln[x] and ¯ Ln[x] shown in (17). In fact,

Z y Z y−x Z y−x Ln[y] − Ln[x] = Pn[t]dt = Pn[t + x]dt = Pn[x] Pn[t]dt, x 0 0 (24) which completes the proof.

¯ Proposition 2.6 Let Hn, Ln[x], and Ln[x] be defined as before. Then for a 6= 0,

1 1 D (a)L [x]D = L [ax], n n n a a n 1 1 D (a)L¯ [x] = L¯ [ax]. (25) n n a a n

Proof. For a 6= 0, the left-hand side of the first equation of (25) can be changed to

  n−1 k 1 X (aHn) D (a)L [x]D = xk+1 n n n a (k + 1)! k=0 n−1 1 X Hk = n (ax)k+1, a (k + 1)! k=0 which proves the proposition.

¯ As n → ∞, matrix functions of nth order, Ln[x] and Ln[x], can be extended to the infinite lower triangular matrix functions L[x]: P [x] 7→ R x ¯ R −x R x 0 P [t]dt and L[x]: P [x] 7→ − 0 P [t]dt(= 0 P [−t]dt), respectively. The properties shown in Propositions 2.1-2.6 can be extended to the case of infinite series associated with certain convergence conditions accordingly. Pascal Matrix Function 9

3 Pascal matrix function applied to Bernoulli numbers and Bernoulli polynomials

Since

P [−x]P [x] = P [x]P [−x] = P [0] = I, (26) there exists

P [x]−1 = P [−x]. Thus, (3) implies the following inverse relationship between B(x) and B:

T T Theorem 3.1 Let B(x) = (B0(x),B1(x),...) and B = (B0,B1,...) , and let P [x] be defined as (2). Then there holds a pair of inverse rela- tionship

B(x) = P [x]B(0) B = B(0) = P [−x]B(x). (27) And the latter can be presented as

   0    B0 0 0 ··· 0 0 ··· B0(x) 1 1  B1   0 (−x) 1 ··· 0 0 ···   B1(x)   .   ......   .   .   ......   .   .  =  ......   .  ,    n n n n−1 n    Bn 0 (−x) 1 (−x) ··· n 0 ··· Bn(x)  .   ......   .  ...... (28) which implies

n X n B = (−x)n−kB (x) (29) n k k k=0 for n = 0, 1,....

The proof of the theorem is straightforward from (26) and is omit- ted. Similarly, we have 10 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Proposition 3.2 Let B(x) and P [x] be defined as above. Then there hold

n X n B (x + y) = (P [x]B(y)) = xn−kB (y), n n k k k=0 n X n B (y) = P [−x]B (x + y) = (−x)n−kB (x + y). (30) n n n k k k=0

We now consider recursive relations of Bn(x).

Proposition 3.3 Let B(x), B, and P [x] be defined as above, and let η(x) = (0, 1, x, x2,...)T . Then there hold

B(x + 1) − B(x) = P [x](B(1) − B(0)) = P [x]η(0), (31) (P [1] − I)B(x) = P [x]η(0), (32)

which imply

n−1 Bn(x + 1) − Bn(x) = nx , and (33) n−1 X n B (x) = nxn−1, (34) k k k=0 respectively.

T Proof. It is well-known that Bn(1) − Bn(0) = (δn,1) for n = 0, 1,..., where δ is the Kronecker symbol. Hence, by using η(x) = (0, 1, x, . . .)T , we may write

B(1) − B(0) = η(0). Hence,

B(x+1)−B(x) = P [x](B(1)−B(0)) = P [x]η(0) = (0, 1, 0,...)T , (35) which implies (33): Pascal Matrix Function 11

n−1 Bn(x + 1) − Bn(x) = nx .

On the other hand, we may write Bn(x + 1) − Bn(x) as

B(x+1)−B(x) = P [x+1]B−P [x]B = P [1]P [x]B−P [x]B = (P [1]−I)B(x).

Thus (32) follows from (31). Since the n + 1st row of the matrix on the rightmost side is

n n  n  T B (x), B (x),..., B (x), 0,... , 0 0 1 1 n − 1 n−1 by comparing the components on the two sides of (32), we may have (34).

Substituting x = 0 into (32) and (34), we immediately have

(P [1] − I)B = η(0), n−1 X n B = δ , n = 0, 1, 2,.... k k 1,n k=0

From expressions (9) and (10), one may obtain the differential for- mulas for Bn(x) readily.

Proposition 3.4 Let B(x) and P [x] be defined as above. Then there hold

n B(k)(x) = k! B (x). (36) n k n−k Particularly, for k = 1

0 Bn(x) = nBn−1(x). 12 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Proof. From (10), we have

k k k xH k k DxB(x) = (DxP [x])B = (Dxe )B = H P [x]B = H B(x). Comparing the nth components on both side vectors of the above equa- tion, we obtain (36).

From the previous section, we have seen that the Ln : R 1 R x ¯ Pn 7→ 0 Pn[t]dt and its extension Ln[x]: Pn 7→ 0 Pn[t]dt as well as Ln ¯ and Ln[x] have closed relationships with Pn[x], and, hence, they can be used to derive properties of Bernoulli numbers and Bernoulli polynomials accordingly. Denote bn(x) = (B0(x),B1(x),...,Bn−1(x)) and bn = (B0,B1,...,Bn−1). From the definition of Ln shown in (11) and Theorem 3.1 (see also [1]),

Z 1 Z 1 Lnbn = Pn[t]dtbn = bn(t)dt = e0, (37) 0 0 n where e0 = (1, 0,..., 0), the first element of the standard basis of R , R 1 and the last step comes from 0 Bk(x)dx = δk,0. In [1], a of ˆ mappings, {Ln}n≥0, from e0 to bn for any n ≥ 0 is defined as

n−1 X Bk Lˆ = Hk. (38) n k! n k=0 k Thus, by using k!ek = Hne0, we have (see also [1])

n−1 ˆ X Lne0 = Bkek = bn, (39) k=0 n where ek is the kth element of the standard basis of R . From the pair ˆ of relations (37) and (39), we may sayL n and Ln are inverse each other ˆ ˆ in the sense of (LnLn)e0 = e0 and (LnLn)bn = bn. In addition, denote n−1 T ξn(x) = (1, x, . . . , x ) , there hold

ˆ ˆ bn(x) = Pn[x]bn = P [x]Lne0 = LnPn[x]eo n−1 n−1 X xk X = Lˆ Hke = Lˆ xke = Lˆ ξ (x) (40) n k! 0 n k n n k=0 k=0 Pascal Matrix Function 13

and

Lnbn(x) = LnPn[x]bn = Pn[x]Lnbn n−1 n−1 X xk X = Hke = xke = ξ (x). (41) k! n 0 k n k=0 k=0 (40) is given in [1] as (42). From the pair relations of (40) and (41), ˆ we may also sayL n and Ln are inverse each other in the sense of ˆ ˆ (LnLn)ξn(x) = ξn(x) and (LnLn)bn(x) = bn(x). By using (10), we ˆ also have that the entries of Ln are

   B i if n − 1 ≥ i ≥ j ≥ 0, ˆ i−j j Ln = (42) i,j 0 otherwise. From (40) and (14) we immediately obtain

Proposition 3.5 Let Bn(x) be the nth Bernoulli polynomial. Then

n X 1 n xn = B (x). n − k + 1 k k k=0

Theorem 3.6 For any n ≥ 1, there holds

¯ bn(x) = HnLn[x]bn(t) + bn = −HnLn[−x]bn(t) + bn Z x = Hn bn(t)dt + bn (43) 0 and then Z x Bn(x) = n Bn−1(t)dt + Bn. (44) 0 Proof. From (27) of Theorem 3.1 and (15) of Proposition 2.1, we have

Z x bn(x) − bn = (Pn[x] − In)bn = HnLn[x]bn = Hn Pn[t]bndt, 0 which implies (43) and (44). 14 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Theorem 3.7 For any n ≥ 0, there holds

Z x+1 bn(t)dt = ξn(x) (45) x and then

Z x+1 n Bn(t)dt = x . (46) x Proof. From Proposition 2.5, we may have

Z x+1 Z x+1 Z x bn(t)dt = bn(t)dt − bn(t)dt x 0 0 = (Ln[x + 1] − Ln[x])bn = Pn[x]Lnbn

= LnPn[x]bn = Lnbn(x) = ξn(x), where the last step is from (41), which completes the proof.

Theorem 3.8 For any n ≥ 0, there holds Z y bn(y) − bn(x) = H bn(t)dt (47) x and then Z y 1 Bn(t)dt = (Bn+1(y) − Bn+1(x)). (48) x n + 1 Proof. From (27) of Theorem 3.1 and (15) of Proposition 2.1, we have

bn(y) − bn(x) = (Pn[y] − Pn[x])bn

= [(Pn[y] − In) − (Pn[x] − In)] bn

= (HnLn[y] − HnLn[x]) bn Z y Z x  = H Pn[t]dt − Pn[t]dt bn 0 0 Z y Z y = Hn Pn[t]dtbn = Hn bn(t)dt, x x completing the proof of the theorem. Pascal Matrix Function 15

Theorem 3.6 can be considered as a special case of Theorem 3.8. Remark 3.1 It should be noted that the Pascal matrix function P [x] we used in the study of properties of Bernoulli number set B and Bernoulli polynomial set B(x) is a type of globe approach. Some properties based on some internal relationships of B and B(x) may not be obtained by using our approach. For instance, the matrix form of the well-known property

 x x + 1 B (x) = 2n−1 B + B (49) n n 2 n 2 is

1  x x + 1 B(x) = D(2) B + B , 2 2 2 or equivalently, 1 P [x]B = P [x](I + P )D(2)B. 2 However, P [x] and (1/2)P [x](I+P )D(2) are different because (1/2)(I+ P )D(2) 6= I.

4 Pascal matrix function and Euler num- bers and Euler polynomials

Euler polynomials En(x) can be presented in terms of Euler numbers as

n    n−k X n 1 Ek E (x) = x − , n = 0, 1,..., (50) n k 2 2k k=0 n T where En = 2 En(1/2). Denote E(x) = (E0(x),E1(x),...) and E = T (E0,E1,...) . By making use of the Pascal matrix, we may write (50) as a matrix form

 1 1  1 1 E(x) = P x − D E = P x − E , (51) 2 2 2 2 where 16 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

D[t] = diag[1, t, t2,...]. 2 T Since En(1) + En(0) = 2δ0,n, by denoting ξ(x) = (1, x, x ,...) and using the homomorphism of mapping P , we have

Proposition 4.1 Let En(x), E, and P [x] be defined as before. Then there hold

E(x) = P [x]E(0), (52) E[x + 1] + E[x] = 2P [x]ξ(0), (53) (P [1] + I)E(x) = 2P [x]ξ(0), (54) E[x + y] = P [x]E(y). (55)

In addition, (53), (54), and (55) imply

n En(x + 1) + En(x) = 2x , (56) n X n E (x) + E (x) = 2xn, and (57) k k n k=0 n X n E (x + y) = xn−kE (y) (58) n k k k=0 respectively.

Proof. (52) is from (51) due to

 1 1  1 1 E(x) = P x − E = P [x]P − E = P [x]E(0). 2 2 2 2

Because of

E(x + 1) + E(x) = P [x](E(1) + E(0)), we obtain (53). On the other hand,

E(x + 1) + E(x) = P [x + 1]E(0) + P [x]E(0) = (P [1] + I)P [x]E(0). Pascal Matrix Function 17

Thus, (54) follows from (53). From (52), we can find (55)

E(x + y) = P [x + y]E(0) = P [x]P [y]E(0) = P [x]E(y). (56), (57), and (58) are from the comparison of components of two side vectors of (53), (54), and (55), respectively.

We now discuss the relationships between Bernoulli polynomials Bn(x) and Euler polynomials En(x). Denote en(x) = (E0(x),E1(x),..., T En−1(x)) . From (54), we have

(Pn[1] + In)en(x) = 2Pnξ(0) = 2ξn(x), or equivalently,

−1 en(x) = 2(Pn + In) ξn(x), (59) which and formulas (38) and (40) will be used rederive some relation- ships between Bernoulli polynomials and Euler polynomials shown in [4] and [11].

Theorem 4.2 Let bn(x) and en(x) be defined as above. Then

ˆ Dn(2)bn(x) = Lnen(2x). (60) Hence,

n −n X Bn(x) = 2 Bn−kEk(2x) (61) k=0 for all n ≥ 0.

Proof. From the definition of Pn[x], we have

Dn(2)Pn[x] = Pn[2x]Dn(2), 2 n−1 where Dn(2) is the n×n diag(1, 2, 2 ,..., 2 ). Thus, using (27) yields

Dn(2)bn(x) = Dn(2)Pn[x]bn = Pn[2x]Dn(2)bn ˆ = Pn[2x]Dn(2)Lne0 (62) 18 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Substituting y = 2 and x = 1 into (23) of Proposition 2.5, we obtain

Ln[2] = Pn[1]Ln[2 − 1] + Ln[1] = (Pn + In)Ln, where Ln[2] can be written as 2D(2)LnD(1/2) due to (25) of Proposi- tion 2.6. Thus,

2D(2)LnD(1/2) = (Pn + In)Ln, or equivalently,

2D(2)Ln = (Pn + In)LnD(2). ˆ Applying the both sides operators of the last equation to Lne0 yields

ˆ ˆ 2D(2)(LnLn)e0 = (Pn + In)LnD(2)Lne0, which implies

ˆ ˆ −1 D(2)Lne0 = 2Ln(Pn + In) D(2)e0 ˆ −1 = 2Ln(Pn + In) e0 ˆ −1 = 2Ln(Pn + In) ξn(0) ˆ = Lnen(0), where the last step is due to (59). By combining the above equation and (62), we finally have

ˆ ˆ ˆ D(2)bn(x) = Pn[2x]Dn(2)Lne0 = LnPn[2x]en(0) = Lnen(2x), where (52) is applied in the last step, which completes the proof of the theorem with using (42).

Theorem 4.2 can be used to derive many relationships between Bernoulli polynomials and Euler polynomials. For instance, we have

Corollary 4.3 Let bn(x) and en(x) be defined as above. Then  x + 1 x H e (x) = D (2) b − B . (63) n n n n 2 2 Pascal Matrix Function 19

Hence,

2n+1  x + 1 x E (x) = B − B (64) n n + 1 n+1 2 n+1 2 for n ≥ 0. Furthermore, 2  x E (x) = B (x) − 2n+1B . (65) n n + 1 n+1 n+1 2

Proof. From (60), there holds x e (x) = L D (2)b . n n n n 2 Hence, noting (15) of Proposition 2.1 we have

x H e (x) = H L D (2)b n n n n n n 2 x = (P − I )D (2)b n n n n 2 1 x x = D (2)P b − D (2)b n n 2 n 2 n n 2  x + 1 x = D (2) b − b . n n 2 n 2

Combining (49) and (64), we may prove (65).

We can also derive the following known results (see, for example, [9]) by using our simple unified approach.

Corollary 4.4 Let bn(x) and en(x) be defined as above. Then

Z x+1/2 en(2x) = 2Dn(2) bn(t)dt. (66) x Hence,

Z x+1/2 n+1 En(2x) = 2 Bn(t)dt (67) x for n ≥ 0. Particularly, Euler numbers 20 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Z 3/4 2n+1 En = 2 Bn(t)dt (68) 1/4 for n ≥ 0.

Proof. From (25) of Proposition 2.6, we have

1 1 L = 2D (2)L D . n n n 2 n 2

Thus, by using (60) and Ln[x + 1/2] − Ln[x] = Pn[x]Ln[1/2], we have

1 e (2x) = L D (2)b (x) = 2D (2)L P [x]b n n n n n n 2 n   1  = 2D (2) L x + − L [x] b n n 2 n n Z x+1/2 = 2Dn(2) Pn[t]dtbn x Z x+1/2 = 2Dn(2) bn(t)dt, x

n which implies (67) and (68) after inputing En = 2 En(1/2).

Note. In Sun and Pan’s work [10, 12], they have presented numerous new identities by using the finite difference and differentiation. But we do not know whether our matrix method can be applied or not. Further investigations are needed.

5 An extension of Bernoulli polynomials and their application in numerical anal- ysis

0 The properties of Bn(x) = nBn−1(x) and Bn(0) = Bn(1) for n ≥ 2 make important rule in the applications of Bernoulli polynomials. For Pascal Matrix Function 21

R 1 instance, from those properties we immediately have 0 Bn(x)dx = 0 for every n ≥ 1, which can be used to construct numerical integra- tion formula as follows (see a survey in [6]). Due to the property 0 Bn(x) = nBn−1(x), we may say that Bernoulli polynomials Bn(x) are “deformations” of standard polynomials xn. We start from the Daroux formula. The Darboux formula, first given in 1876, is an expansion formula for an with Taylor formula as one of its special cases. Let f(z) be normal analytic on the line connecting points a and z, and let φ(x) be a polynomial of degree n. Then there holds

d Pn k k (n−k) (k) dx k=1(−1) (z − a) φ (x)f (a + x(z − a)) = −(z − a)φ(n)(x)f 0(a + x(z − a)) +(−1)n(z − a)n+1φ(x)f (n+1)(a + x(z − a)), which can be proved using the . By taking the integral in terms of x from 0 to 1 and noting that φ(n)(x) = φ(n)(0), we obtain

φ(n)(0)[f(z) − f(a)] n X = (−1)k−1(z − a)k φ(n−k)(1)f (k)(z) − φ(n−k)(0)f (k)(a) k=1 Z 1 +(−1)n(z − a)n+1 φ(x)f (n+1)(a + x(z − a))dx. (69) 0 Equation (69) is the Darboux formula. Now we will consider several of its special cases. Let φ(x) = (x−1)n in equation (69). Then φ(n)(0) = n!, φ(n−k)(1) = 0, and φ(n−k)(0) = (−1)kn!/k!, for 1 ≤ k ≤ n. This is the Taylor formula with a Cauchy integral form remainder. Without a loss of generality, let the highest power term of φ(x) be xn and F (x) be the of f(x)(F (x) is assumed to be n order continuously differentiable; then the following integral quadrature formula will be obtained by setting a = 0 and z = 1 in equation (69)

1 n k−1 Z X (−1) x=1 F (x)dx = φ(n−k)(x)F (k−1)(x) n! x=0 0 k=1 (−1)n Z 1 + φ(x)F (n)(x)dx. (70) n! 0 22 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

Particularly, let φ(x) = Bn(x), the nth Bernoulli polynomial, then (70) becomes

Z 1 F (x)dx 0 n F (1) + F (0) X (−1)k−1 = + B (0) F (k−1)(1) − F (k−1)(0) 2 k! k k=2 n Z 1 (−1) (n) + Bn(x)F (x)dx (71) n! 0 for n ≥ 2, which is a numerical integration formula with the remainder associated with Bernoulli polynomials. For n = 1, the term in (71) vanishes, and its right-hand side can be shown as identical as its left-hand side by using for the integral term. For n = 0, the right-hand side of (71) reduces to the same integral as its left-hand side. In this sense, (71) holds for all integers n ≥ 0. To prove (71) from (70), we need use the fact Bn(1) = Bn(0) for n ≥ 2 and (36), i.e., n! Bn−k = B (x). n k! k In fact, equation (70) can easily be verified directly by applying n times integral by parts to its integral form remainder. One may rewrite the above formula into a more general form (a suitable transformation of variable is needed).

b n k−1 Z X (−1) x=b F (x)dx = φ(n−k)(x)F (k−1)(x) + R (72) n! x=a n a k=1 where remainder Rn is

n Z b (−1) (n) Rn = φ(x)F (x)dx. (73) n! a Equations (72) and (73) can be understood as the integral form of the Darboux formula. However, the applications of the Darboux formula to numerical integration were only given attention to after 1940 (see [8]). Pascal Matrix Function 23

The formula shown in equation (72) has two features: (1) Except for the remainder, the right-hand side of the equation consists of the values of the integrand and its at end points of the integral interval. Hence, the formula is a boundary type quadrature formula (BTQF). (2) By choosing a suitable weight function φ(t) in the remainder (73), we can make Rn ≡ Rn(F ) have the smallest possible estimate in various norms. In addition, setting φ(t) as the nth order generalized Bernoulli poly- ˜ nomial Bn(x) defined on interval [a, b] in equation (72), we obtain the ˜ Euler-Maclaurin type formula. Here, Bn(x), x ∈ [a, b], n ∈ N0, are defined as

x − a B˜ (x) = (b − a)nB . (74) n n b − a

˜ ˜ ˜ Hence, for [a, b] = [0, 1] Bn(x) = Bn(x). Denote B(x) := (B0(x), ˜ T ˜ B1(x),...) for x ∈ [a, b]. Particularly, for [a, b] = [0, 1] B(x) = B(x) defined before. From (3), we have

x − a B˜(x) = P B. (75) b − a

Using (75), we may transfer the properties of Bn(x) to their general- ˜ ization Bn(x). For instance,

x − a B˜0 (x) = n(b − a)n−1B = nB˜ (x), n ≥ 0, (76) n n−1 b − a n−1 and

˜ ˜ Bn(b) = Bn(a), n ≥ 2. (77)

From Theorem 3.1 and Proposition 3.3, we have

˜ ˜ Proposition 5.1 Let Bn(x), B(x), B(x), and B be defined as before. Then there hold 24 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

x − a a − x B˜(x) = P B,B = P B˜(x), (78) b − a b − a ˜ ˜ n−1 Bn(x + b − a) − Bn(x) = n(b − a)(x − a) , (79) n−1 X n (b − a)n−kB˜ (x) = n(b − a)(x − a)n−1. (80) k n k=0

Proof. The correction of the formulas (78)-(80) may be proved from (75) and (76) straightforwardly.

˜ The analogies of some other properties of Bn(x) for Bn(x) can be ˜ derived similarly. Substituting φ(x) = Bn(x) into (72) and (73) and ap- plying (75) and (76) yield the following Euler-Maclaurin type formula.

˜ Proposition 5.2 Let Bn(x) be defined as before. Then for any analytic function F (x) defined on [a, b] there holds

Z b F (x)dx a n F (1) + F (0) X (−1)k−1 = (b − a) + B˜ (a) F (k−1)(b) − F (k−1)(a) 2 k! k k=2 n Z b (−1) ˜ (n) + Bn(x)F (x)dx (81) n! a for all n ≥ 2.

Proof. Denote g(x) = F (a + x(b − a). Then from (71) and noting (k) k (k) ˜ ˜ g (x) = (b − a) F (a + x(b − a)) and Bk(a) = Bk(b) for n ≥ 2, we have Pascal Matrix Function 25

Z b Z 1 F (x)dx = (b − a) g(x)dx a 0 " n g(1) + g(0) X (−1)k−1 = (b − a) + B (0) g(k−1)(1) − g(k−1)(0) 2 k! k k=2 n Z 1  (−1) (n) + Bn(x)g (x)dx n! 0 n F (b) + F (a) X (−1)k−1 = (b − a) + B˜ (a) F (k−1)(b) − F (k−1)(a) 2 k! k k=2 n Z 1 n+1 (−1) (n) +(b − a) Bn(x)F (a + x(b − a))dx, n! 0 which implies (81).

˜ Proposition 5.3 Let Bn(x) be defined as before. Denote the remain- der of Euler-Maclaurin formula (81) by Rn(F ), i.e.,

n Z b (−1) ˜ (n) Rn(F ) := Bn(x)F (x)dx. n! a

Then for any analytic function F (x) defined on [a, b] with L2([a, b]) 1/2 (n) R b (n) 2  norm kF (x)k2 := a F (x) dx =: Mn, Euler-Maclaurin for- mula (81) has error bound s (b − a)(−1)n−1B |R (F )| ≤ M (b − a)n 2n , (82) n n (2n)! where B2n is the 2nth Bernoulli number.

Proof. Using the Cauchy-Schwarz’s inequality and formula (see [7])

Z 1 n−1 m!n! Bn(x)Bm(x)dx = (−1) Bn+m, 0 (m + n)! we have 26 T. X. He, J. H.-C. Liao, and P. J.-S. Shiue

1 Z b ˜ (n) |Rn(F )| ≤ Bn(x)F (x) dx n! a 1/2 Z b   2 ! Mn 2n x − a = (b − a) Bn dx n! a b − a M Z 1 1/2 = n (b − a)2n+1 |B (x)|2 dx n! n √ 0 M b − a  (n!)2 1/2 = n (b − a)n (−1)n−1B , n! (2n)! 2n

n−1 which implies (82). Here, (−1) B2n > 0 for n = 1, 2,....

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