A Determinantal Expression and a Recurrence Relation for the Euler Polynomials

A DETERMINANTAL EXPRESSION AND A RECURRENCE RELATION FOR THE EULER POLYNOMIALS

FENG QI

Institute of Mathematics, Henan Polytechnic University, Jiaozuo 454010, Henan Province, China; College of Mathematics, Inner Mongolia University for

Nationalities, Tongliao 028043, Inner Mongolia Autonomous Region, China; Department of Mathematics, College of Science, Tianjin Polytechnic University,

Tianjin 300387, China

BAI-NI GUO

School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo 454010, Henan Province, China

Abstract. In the paper, by a very simple approach, the author establishes an expression in terms of a lower Hessenberg determinant for the Euler polyno-mials. By the determinantal expression, the author finds a recurrence relation for the Euler polynomials. By the way, the author derives the corresponding expression and recurrence relation for the Euler numbers.

1. _{Main results}

It is known that a matrix H = (hij)n×n is called a lower (or an upper, respec-tively) Hessenberg matrix if hij = 0 for all pairs (i, j) such that i+ 1 < j (or

j + 1 < i, respectively). Correspondingly, we can define a lower (or an upper, respectively) Hessenberg determinant.

It is general knowledge that the Bernoulli numbers and polynomials Bk and

Bk(u) can be generated respectively by

z ez₋₁ =

∞ X

k=0

Bk

zk k! = 1−

z

2 +

∞ X

k=1

B2k

z2k

(2k)!

E-mail addresses:[email protected], [email protected], [email protected], [email protected].

2010 Mathematics Subject Classification. Primary 11B68, Secondary 11C20, 11Y55, 15A15, 26A06, 33B10, 65F40.

Key words and phrases. determinantal expression; recurrence relation; Euler polynomial; Euler number; Hessenberg determinant.

This paper was typeset usingAMS-LA_TEX.

and

zeuz

ez₋₁ =

∞ X

k=0

Bk(u)

zk

k!

for |z| < 2π. Because the function x

ex₋₁ −1 +

x

2 is odd in x ∈ R, all of the Bernoulli numbersB2k+1 fork ∈Nequal 0. The first six Bernoulli numbers B2k for− ≤k≤5 are

B0= 1, B2= 1

6, B4=− 1

30, B6= 1

42, B8=− 1

30, B10= 5 66. In [4, Section 21.5] and [5, p. 1], it was listed that

Bk=k!

1 0 0 · · · 0 0 1

1

2! 1 0 · · · 0 0 0

1 3!

1

2! 1 · · · 0 0 0

..

. ... ... . .. ... ... ... 1

(k−1)! 1 (k−2)!

1

(k−3)! · · · 1 0 0 1

k!

1 (k−1)!

1 (k−2)! · · ·

1

2! 1 0 1

(k+1)! 1 k!

1 (k−1)! · · ·

1 3!

1 2! 0

.

In [7, Theorem 1.2], the Bernoulli polynomialsBk(u) fork∈Nwere expressed by

a lower Hessenberg determinant

Bk(u) = (−1)k

1

`+ 1 <sub>`+ 1</sub>

m

(1−u)`−m+1−(−u)`−m+1 ₁

≤`≤k,0≤m≤k−1 and, consequently, the Bernoulli numbersBkfork∈Nwere represented by a lower

Hessenberg determinant

Bk = (−1)k

1

`+ 1 <sub>`+ 1</sub>

m

₁

≤`≤k,0≤m≤k−1

.

It is common knowledge that the Euler numbersEk and the Euler polynomials

Ek(x) can be generated respectively by

(1) 2e

t/2

et_{+ 1} =

∞ X

k=0

Ek

k! _t

2 k

and

(2) 2e

xt

et_{+ 1} =

∞ X

k=0

Ek(x)

k! t k

which converge uniformly with respect to t ∈ (−π, π). By these definitions, it is easy to see that

(3) Ek= 2kEk

1 2

.

Since the generating function 2e_ett/₊₁2 of the Euler numbers Ek is even on (−π, π), thenE2k−1= 0 for allk∈N. The first nine Euler numbersE2k for 0≤k≤8 are

1, −1, 5, −61, 1385, −50521, 2702765, −199360981

and the first six Euler polynomialsEk(x) for 0≤k≤5 are

1, x−1 2, x

2_{−x, x}3₋3 2x

2₊1 4, x

4_−2x3_{+x, x}5₋5 2x

4₊5 2x

At the website [9], the Euler numbersE2kwere represented by a lower Hessenberg determinant

E2k= (−1)k(2k)!

1

2! 1 0 · · · 0 0

1 4!

1

2! 1 · · · 0 0

..

. ... ... . .. ... ... 1

(2k−2)! 1 (2k−4)!

1

(2k−6)! · · · 1 2! 1 1

(2k)! 1 (2k−2)!

1

(2k−4)! · · · 1 4! 1 2!

, k∈_N.

In [8, Theorem 1.1], the Euler numbersE2k fork∈Nwere represented by a lower

Hessenberg and sparse determinant

E2k = (−1)k

_i

j−1

cos

(i−j+ 1)π 2 _(2k) ×(2k) .

There have been many explicit expressions for the Euler numbers Bk and the Euler polynomials Ek(x). See, for example, the recently-published papers [2, 3, 8] and plenty of references therein.

In this paper, by a very simple approach, we will establish a new expression in terms of a lower Hessenberg determinant for the Euler polynomialsEk(x). By the new determinantal expression, we will find a recurrence relation for the Euler poly-nomials. By the way, we will derive the corresponding expression and recurrence relation for the Euler numbersEk.

Our main results can be summarized up as two theorems below.

Theorem 1. Fork≥0, the Euler polynomials Ek(x)can be expressed as

(4) Ek(x) = (−1)k 2k

1 2 0 0 · · · 0 0 0

x 1₀ 2 0 · · · 0 0 0

x2 2₀ 2₁ 2 · · · 0 0 0

x3 3

0 3 1 3 2

· · · 0 0 0

..

. ... ... ... . .. ... ... ...

xk−2 k−2 0

k−2 1

k−2 2

· · · k_k−₋2₃

2 0

xk−1 k−₀1 k−₁1 k−₂1 · · · k_k−₋1₃ k−1 k−2

2

xk k

0 k 1 k 2

· · · k k−3

k

k−2

k

k−1

and, consequently, the Euler numbers Ek can be expressed as

(5) Ek = (−1)k

1 2 0 0 · · · 0 0 0

1 2

1 0

2 0 · · · 0 0 0

1 22 2 0 2 1

2 · · · 0 0 0

1 23 3 0 3 1 3 2

· · · 0 0 0

..

. ... ... ... . .. ... ... ... 1

2k−2 k−2

0

k−2 1

k−2 2

· · · k_k−₋2₃

2 0

1 2k−1

k−1 0

k−1 1

k−1 2

· · · k_k−₋1₃ k−1 k−2

2 1 2k k 0 k 1 k 2

· · · k k−3

k

k−2

k

k−1

.

Theorem 2. The Euler polynomials and numbersEk(x)andEk satisfy the recur-rence relations

(6) Ek(x) =−

1 2

k−1

X

`=0

_k

`

E`(x) +xk

and, consequently,

(7) Ek=−2k−1

k−1

X

`=0 1 2`

_k

`

E`+ 1.

2. _{Proofs of Theorems 1 and 2}

Now we start out to prove our main results.

Proof of Theorem 1. In [6, Section 2.2, p. 849], [7, p. 94], and [8, Lemma 2.1], Exercise 5) in [1, p. 40] was reformulated as the following conclusion. Let u(x) and v(x) 6= 0 be two differentiable functions. LetU(n+1)×1(x) be an (n+ 1)×1 matrix whose elementsuk,1(x) = u(k−1)(x) for 1≤k≤n+ 1, let V(n+1)×n(x) be an (n+ 1)×nmatrix whose elements

vi,j(x) =

 

 _i−1

j−1

v(i−j)_{(x), i−j ≥0}

0, i−j <0

for 1≤i≤n+ 1 and 1≤j ≤n, and let|W(n+1)×(n+1)(x)|denote the determinant of the (n+ 1)×(n+ 1) matrix

W(n+1)×(n+1)(x) = U(n+1)×1(x) V(n+1)×n(x)

.

Then thenth derivative of the ratio u(x)_v(x) can be computed by

(8) d

n

dxn

_u(x)

v(x)

= (−1)n

W(n+1)×(n+1)(x)

vn+1_(x) .

Letu(t) =ext andv(t) =et+ 1. Then, by virtue of the formula (8),

dk dtk

_ext

et_{+ 1}

= (−1)

k

×

ext _et_{+ 1 0 · · · 0 0 0}

xext 1

0

et _et_{+ 1 · · · 0 0 0}

x2ext 2₀et 2₁et · · · 0 0 0

x3ext 3₀et 3₁et · · · 0 0 0 ..

. ... ... . .. ... ... ...

xk−2ext k−₀2et k−₁2et · · · k_k−₋2₃

et et+ 1 0

xk−1_ext k−1 0

et k−1 1

et _{· · ·} k−1 k−3

et k−1 k−2

et _et_{+ 1}

xk_ext k 0

et k

1

et _{· · ·} k k−3

et k

k−2

et k

k−1

et →(−1) k 2k+1

1 2 0 0 · · · 0 0 0

x 1₀

2 0 · · · 0 0 0

x2 2₀ 2₁ 2 · · · 0 0 0

x3 3

0 3 1 3 2

· · · 0 0 0

..

. ... ... ... . .. ... ... ...

xk−2 k−₀2 k−₁2 k−₂2 · · · k_k−₋2₃

2 0

xk−1 k−1 0

k−1 1

k−1 2

· · · k_k−₋1₃ k−1 k−2

2

xk k

0 k 1 k 2

· · · k k−3

k

k−2

k

k−1

ast→0 fork≥0. By the equation (2), the formula (4) is thus proved.

Considering the relation (3) and taking x = 1₂ in (4) lead to the formula (5)

readily. The proof of Theorem 1 is complete.

Proof of Theorem 2. Let

Dk+1(x) =

1 2 0 0 · · · 0 0 0

x 1₀ 2 0 · · · 0 0 0

x2 2

0

2

1

2 · · · 0 0 0

x3 3

0 3 1 3 2

· · · 0 0 0

..

. ... ... ... . .. ... ... ...

xk−2 k−2 0

k−2 1

k−2 2

· · · k_k−₋2₃

2 0

xk−1 k−₀1 k−₁1 k−₂1 · · · k_k−₋1₃ k−1 k−2

2

xk k

0 k 1 k 2

· · · _k−k₃ k k−2

k

k−1

for k ≥ 0. Expanding Dk+1(x) according to the last column consecutively and inductively reveals

Dk+1(x) =

_k

k−1

1 2 0 0 · · · 0 0

x 1₀ 2 0 · · · 0 0

x2 2

0

2

1

2 · · · 0 0

x3 3

0 3 1 3 2

· · · 0 0

..

. ... ... ... . .. ... ...

xk−2 k−2 0

k−2 1

k−2 2

· · · k−2 k−3

2

xk−1 k−₀1 k−₁1 k−₂1 · · · k_k−₋1₃ k−1 k−2

−2

1 2 0 0 · · · 0 0

x 1₀

2 0 · · · 0 0

x2 2

0

2

1

2 · · · 0 0

x3 3

0 3 1 3 2

· · · 0 0

..

. ... ... ... . .. ... ...

xk−2 k−2 0

k−2 1

k−2 2

· · · k−2 k−3

2

xk k

0 k 1 k 2

· · · _k−k₃ k k−2

= k k−1

1 2 0 0 · · · 0 0

x 1₀

2 0 · · · 0 0

x2 2₀ 2₁ 2 · · · 0 0

x3 3

0 3 1 3 2

· · · 0 0

..

. ... ... ... . .. ... ...

xk−2 k−₀2 k−₁2 k−₂2 · · · k_k−₋2₃ 2

xk−1 k−₀1 k−₁1 k−₂1 · · · k_k−₋1₃ k−1 k−2

−2 _k

k−2

1 2 0 0 · · · 0 0

x 1₀

2 0 · · · 0 0

x2 2

0

2

1

2 · · · 0 0

x3 3₀ 3₁ 3₂ · · · 0 0

..

. ... ... ... . .. ... ...

xk−3 k−3 0

k−3 1

k−3 2

· · · k_k−₋3₄ 2

xk−2 k−2 0

k−2 1

k−2 2

· · · k−2 k−4

k−2 k−3

+ 22

_k

k−3

1 2 0 0 · · · 0 0

x 1₀ 2 0 · · · 0 0

x2 2

0

2

1

2 · · · 0 0

x3 3

0 3 1 3 2

· · · 0 0

..

. ... ... ... . .. ... ...

xk−4 k−4 0

k−4 1

k−4 2

· · · k_k−₋4₅ 2

xk−3 k−₀3 k−₁3 k−₂3 · · · k_k−₋3₅ k−3 k−4

−23

1 2 0 0 · · · 0 0

x 1₀

2 0 · · · 0 0

x2 2

0

2

1

2 · · · 0 0

x3 3₀ 3₁ 3₂ · · · 0 0

..

. ... ... ... . .. ... ...

xk−4 k−4 0

k−4 1

k−4 2

· · · k_k−₋4₅ 2

xk k

0 k 1 k 2

· · · k k−5

k

k−4

= m X `=0 (−1)`₂`

_k

k−`−1

×

1 2 0 · · · 0 0

x 1₀

2 · · · 0 0

x2 2

0

2

1

· · · 0 0

x3 3

0

3

1

· · · 0 0

..

. ... ... . ... ...

xk−m−2 k−m−2 0

k−m−2 1

· · · k−m−2 k−m−3

2

xk k

0

k

1

· · · _k−mk₋₃ k k−m−2

= k−2

X

`=0 (−1)`₂`

_k

k−`−1

Dk−`(x) + (−1)k−12k−1

1 2

xk k

0

= k−2

X

`=0 (−1)`2`

_k

k−`−1

Dk−`(x) + (−1)k−12k−1

_k

0

−2xk

= k−1

X

`=0 (−1)`2`

_k

k−`−1

Dk−`(x) + (−1)k2kxk

= k−1

X

`=0 (−1)`₂`

_k

k−`−1

₂k−`−1 (−1)k−`−1

₍₋₁₎k−`−1

2k−`−1 Dk−`(x)

+ (−1)k₂k_xk

= (−1)k−1₂k−1 k−1

X

`=0

_k

k−`−1

Ek−`−1(x) + (−1)k2kxk.

Accordingly, it follows that

Ek(x) = (−1)k

2k Dk+1(x) =− 1 2

k−1

X

`=0

_k

k−`−1

Ek−`−1(x) +xk

=−1 2

k−1

X

`=0

k `

E`(x) +xk.

The relation (6) is thus proved.

Takingx= 1₂ in (6) and making use of the relation (3) result in the relation (7).

The proof of Theorem 2 is complete.

Remark 1. The expression (5) can also be obtained by applying the formula (8) to the functionsu(t) =et/2_{andv(t) =e}t_{+ 1 and considering the generating function}

2et/2

et₊₁ in the equation (1).

References

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2015:219, 8 pages; Available online athttp://dx.doi.org/10.1186/s13660-015-0738-9. [9] WikiPedia, Euler number, From the Free Encyclopedia; Available online at https://en.

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URL:https://qifeng618.wordpress.com