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Aug 14, 2018 - 1. (x, y) = y,. E. (S). 2. (x, y) = −y + 2xy,. E. (S). 3. (x, y) = −3xy + 3x2y − y3,. E. (S). 4. (x, y) = y − 6x2y + 4x3y + 2y3 − 4xy3. By (10)–(13), we have.
axioms Article

Some Identities for Euler and Bernoulli Polynomials and Their Zeros Taekyun Kim 1 and Cheon Seoung Ryoo 2, * 1 2

*

Department of Mathematics, Kwangwoon University, Seoul 139-701, Korea; [email protected] Department of Mathematics, Hannam University, Daejeon 306-791, Korea Correspondence: [email protected]; Tel.: +82-42-629-7525

Received: 30 June 2018; Accepted: 11 August 2018; Published: 14 August 2018

Abstract: In this paper, we study some special polynomials which are related to Euler and Bernoulli polynomials. In addition, we give some identities for these polynomials. Finally, we investigate the zeros of these polynomials by using the computer. Keywords: Appell sequence; Appell numbers and polynomials; Bernoulli and Euler polynomials; cosine–Bernoulli and cosine–Euler polynomials; sine–Bernoulli and sine–Euler polynomials MSC: 11B68; 11S40; 11S80

1. Introduction Many mathematicians have studied in the area of the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, and tangent numbers and polynomials. The class of Appell polynomial sequences is one of the important classes of polynomial sequences. The Appell polynomial sequences arise in numerous problems of applied mathematics, mathematical physics and several other mathematical branches (see [1–14]). The Appell polynomials can be defined by considering the following generating function: A(t)e xt = A0 ( x ) + A1 ( x ) ∞

=



An ( x )

n =0

where A ( t ) = A0 + A1

t t2 tn + A2 ( x ) + · · · + A n ( x ) + · · · 1! 2! n!

(1)

tn , (see [5,7,8]), n!

t t2 tn + A2 + · · · + An + · · · , A0 6= 0. 1! 2! n!

Alternatively, the sequence An ( x ) is Appell sequence for ( g(t), t) if and only if ∞ 1 xt tn e = ∑ An ( x ) , (see [5,7,8]), g(t) n! n =0

where



g(t) =



n =0

gn

tn , g0 6= 0. n!

Differentiating generating Equation (1) with respect to x and equating coefficients of

tn n! ,

we have

d An ( x ) = nAn−1 ( x ), n = 0, 1, 2, 3, · · · . dx

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The typical examples of Appell polynomials are the Bernoulli and Euler polynomials (see [1–14]). It is well known that the Bernoulli polynomials are defined by the generating function to be ∞ t tn xt e = B ( x ) . n ∑ et − 1 n! n =0

(2)

When x = 0, Bn = Bn (0) are called the Bernoulli numbers. The Euler polynomials are given by the generating function to be ∞ 2 tn xt e = E ( x ) . (3) n ∑ et + 1 n! n =0 When x = 0, En = En (0) are called the Euler numbers. (r ) The Bernoulli polynomials Bn ( x ) of order r are defined by the following generating function 

t et − 1

r



∑ Bn

e xt =

(r )

(x)

n =0

tn , n!

(|t| < 2π ).

(4)

(r )

The Frobenius–Euler polynomials of order r, denoted by Hn (u, x ), are defined as 

1−u et − u

r

e xt =



∑ Hn

n =0

(r )

(u, x )

tn . n!

(5)

The values at x = 0 are called Frobenius–Euler numbers of order r; when r = 1, the polynomials or numbers are called ordinary Frobenius–Euler polynomials or numbers. In this paper, we study some special polynomials which are related to Euler and Bernoulli polynomials. In addition, we give some identities for these polynomials. Finally, we investigate the zeros of these polynomials by using the computer. 2. Cosine–Bernoulli, Sine–Bernoulli, Cosine–Euler and Sine–Euler Polynomials In this section, we define the cosine–Bernoulli, sine–Bernoulli, cosine–Euler and sine–Euler polynomials. Now, we consider the Euler polynomials that are given by the generating function to be ∞ 2 tn ( x +iy)t e = E ( x + iy ) . n ∑ et + 1 n! n =0

(6)

On the other hand, we observe that e( x+iy)t = e xt eiyt = e xt (cos yt + i sin yt).

(7)

From Equations (6) and (7), we have ∞

tn

∑ En (x + iy) n!

=

2 2 e( x+iy)t = t e xt (cos yt + i sin yt), et + 1 e +1

(8)

=

2 2 e( x−iy)t = t e xt (cos yt − i sin yt). et + 1 e +1

(9)

n =0

and



tn

∑ En (x − iy) n!

n =0

Thus, by (8) and (9), we can derive ∞ 2 xt e cos yt = ∑ et + 1 n =0



En ( x + iy) + En ( x − iy) 2



tn , n!

(10)

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and

∞ 2 xt e sin yt = ∑ et + 1 n =0



En ( x + iy) + En ( x − iy) 2i



tn . n!

(11)

It follows that we define the following cosine–Euler polynomials and sine–Euler polynomials. (C )

(S)

Definition 1. The cosine–Euler polynomials En ( x, y) and sine–Euler polynomials En ( x, y) are defined by means of the generating functions ∞

∑ En

(C )

( x, y)

n =0

and



∑ En

(S)

( x, y)

n =0

2 tn = t e xt cos yt, n! e +1

(12)

tn 2 = t e xt sin yt, n! e +1

(13)

respectively. (C )

(S)

Note that En ( x, 0) = En ( x ), En ( x, 0) = 0, (n ≥ 0). The cosine–Euler and sine–Euler polynomials can be determined explicitly. A few of them are (C )

E0 ( x, y) = 1,

1 (C ) E1 ( x, y) = − + x, 2

(C )

E2 ( x, y) = − x + x2 − y2 , (C )

E3 ( x, y) =

1 3x2 3y2 − + x3 + − 3xy2 , 4 2 2

(C )

E4 ( x, y) = x − 2x3 + x4 + 6xy2 − 6x2 y2 + y4 , and

(S)

(S)

E0 ( x, y) = 0,

E1 ( x, y) = y,

(S)

E2 ( x, y) = −y + 2xy, (S)

E3 ( x, y) = −3xy + 3x2 y − y3 , (S)

E4 ( x, y) = y − 6x2 y + 4x3 y + 2y3 − 4xy3 . By (10)–(13), we have En ( x + iy) + En ( x − iy) , 2 En ( x + iy) − En ( x − iy) (S) En ( x, y) = . 2i (C )

En ( x, y) =

Clearly, we can get the following explicit representations of En ( x + iy) n

  n En ( x + iy) = ∑ ( x + iy)n−k Ek , k k =0 n   n En ( x + iy) = ∑ (iy)n−k Ek ( x ). k k =0 Let e xt cos yt =



tk

∑ Ck (x, y) k! ,

k =0

e xt sin yt =



tk

∑ Sk (x, y) k! .

k =0

(14)

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Then, by Taylor expansions of e xt cos yt and e xt sin yt, we get e xt cos yt =









[ k−2 1 ]  m =0

k =0

where [

  k tk (−1)m x k−2m−1 y2m+1  , 2m + 1 k!

(16)



m =0

∑∑

e xt sin yt =

(15)

∑∑

k =0

and

  tk k (−1)m x k−2m y2m  2m k!

[ 2k ]

] denotes taking the integer part. By (14)–(16), we get [ 2k ]

Ck ( x, y) =





m =0

and

[ k−2 1 ] 

Sk ( x, y) =



m =0

 k (−1)m x k−2m y2m , 2m

 k (−1)m x k−2m−1 y2m+1 , (k ≥ 0). 2m + 1

The two polynomials can be determined explicitly. A few of them are C0 ( x, y) = 1,

C2 ( x, y) = x2 − y2 ,

C1 ( x, y) = x,

C3 ( x, y) = x3 − 3xy2 ,

C4 ( x, y) = x4 − 6x2 y2 + y4 ,

C5 ( x, y) = x5 − 10x3 y2 + 5xy4 ,

C6 ( x, y) = x6 − 15x4 y2 + 15x2 y4 − y6 ,

and S0 ( x, y) = 0,

S1 ( x, y) = y, 2

3

S3 ( x, y) = 3x y − y ,

S2 ( x, y) = 2xy,

S4 ( x, y) = 4x3 y − 4xy3 ,

S5 ( x, y) = 5x4 y − 10x2 y3 + y5 ,

S6 ( x, y) = 6x5 y − 20x3 y3 + 6xy5 .

Now, we observe that ! ! ∞ tl tm ∑ El l! ∑ Cm (x, y) m! m =0 l =0 ! ∞ n   n tn = ∑ ∑ El Cn−l ( x, y) . l n! n =0 l =0

2 e xt cos yt = et + 1



Therefore, we obtain the following theorem: Theorem 1. For n ≥ 0, we have (C ) En ( x, y)

and (S) En ( x, y)

n

  n =∑ El Cn−l ( x, y) l l =0 n

  n =∑ El Sn−l ( x, y). l l =0

(17)

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From (12), we have ∞

xt

2e cos yt =



l =0 ∞

=



n =0

tl (C ) El ( x, y) l!

! et + 1



!   n (C ) tn (C ) ∑ l El (x, y) + En (x, y) n! . l =0 n

(18)

By (14) and (18), we get 1 Cn ( x, y) = 2

!   n (C ) (C ) ∑ l El (x, y) + En (x, y) . l =0 n

(19)

Therefore, we obtain the following theorem: Theorem 2. For n ≥ 0, we have 1 Cn ( x, y) = 2

!   n (C ) (C ) ∑ l El (x, y) + En (x, y) , l =0

1 Sn ( x, y) = 2

!   n (S) (S) ∑ l El (x, y) + En (x, y) . l =0

and

n

n

From (12), we note that ∞

∑ En

(C )

(1 − x, y)

n =0

tn 2 = t e(1− x)t cos yt n! e +1 2 e− xt cos (−yt) +1 ! ! ∞ ∞ tl tm l m = ∑ (−1) El ∑ (−1) Cm, (x, y) m! l! m =0 l =0 ! ∞ n   n (−1)n n = ∑ ∑ El Cn−l ( x, y) t . l n! n =0 l =0

=

e−t

Therefore, we obtain the following theorem: Theorem 3. For n ≥ 0, we have (C )

En (1 − x, y) = (−1)n

n

  n ∑ l El Cn−l (x, y) l =0 (C )

= (−1)n En ( x, y), and

(S)

(S)

En (1 − x, y) = (−1)n+1 En ( x, y) n   n = (−1)n+1 ∑ El Sn−l ( x, y). l l =0

(20)

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Now, we observe that ∞

∑ En

(C )

( x + 1, y)

n =0

tn 2 = t e( x+1)t cos yt n! e +1 2 e xt (et − 1 + 1) cos yt et + 1 2 e xt cos yt = 2e xt cos yt − t e +1  tn ∞  (C ) = ∑ 2Cn ( x, y) − En ( x, y) . n! n =0

=

(21)

By comparing the coefficients on the both sides, we get (C )

(C )

En ( x + 1, q) + En ( x, y) = 2Cn ( x, y), (n ≥ 0).

(22)

Therefore, we obtain the following theorem: Theorem 4. For n ≥ 0, we have (C )

(C )

(S)

(S)

En ( x + 1, y) + En ( x, y) = 2Cn ( x, y), and En ( x + 1, y) + En ( x, y) = 2Sn ( x, y). From (14) and (15), we have ∞

tk

∑ Ck (0, y) k!



=

t2m

∑ (−1)m y2m (2m)! .

(23)

m =0

k =0

Therefore, by Theorem 4 and (23), we obtain the following corollary: Corollary 1. For n ≥ 0, we have (C )

(C )

E2n (1, y) + E2n (0, y) = 2(−1)n y2n , and (S)

(S)

E2n+1 (1, y) + E2n+1 (0, y) = 2(−1)n y2n+1 . By (12), we get ∞



 2e xt = cos yt ert n! et + 1 ! ! ∞ ∞ k tl (C ) kt = ∑ El ( x, y) ∑ r k! l! l =0 k =0 ! ∞ n   n (C ) tn = ∑ ∑ Ek ( x, y)r n−k . k n! n =0 k =0

t (C ) En ( x + r, y)

n =0

n



(24)

Therefore, by comparing the coefficients on the both sides, we obtain the following theorem: Theorem 5. For n ≥ 0, r ∈ N, we have

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n

(C )

En ( x + r, y) = and (S) En ( x + r, y)

  n (C ) ∑ k Ek (x, y)rn−k , k =0 n

  n (S) = ∑ Ek ( x, y)r n−k . k k =0

Taking r = 1 in Theorem 5, we obtain the following corollary: Corollary 2. For n ≥ 0, we have 2Cn ( x, y) =

(C ) En ( x, y) +

2Sn ( x, y) =

(S) En ( x, y) +

and

n

  n (C ) ∑ k Ek (x, y), k =0 n

  n (S) ∑ k Ek (x, y). k =0

From Corollary 2, we note that (C ) En (0, y) +

and (S) En (0, y) +

By (12), we get

(   n (C ) 0, ∑ k Ek (0, y) = 2(−1)m y2m , k =0 n

(   n (S) 2(−1)m y2m+1 , ∑ k Ek (0, y) = 0, k =0 n



∂ (C ) tn ∂ ∑ ∂x En (x, y) n! = ∂x n =1



if n = 2m + 1, if n = 2m,

if n = 2m + 1, if n = 2m.

2 e xt cos yt et + 1

et

Comparing the coefficients on the both sides of (27), we have ∂ (C ) (C ) En ( x, y) = nEn−1 ( x, y). ∂x Similarly, for n ≥ 1, we have ∂ (S) (S) En ( x, y) = nEn−1 ( x, y), ∂x ∂ (C ) (S) En ( x, y) = −nEn−1 ( x, y), ∂y ∂ (S) (C ) En ( x, y) = nEn−1 ( x, y). ∂y

(26)



2 te xt cos yt +1  tn ∞  (C ) . = ∑ nEn−1 ( x, y) n! n =1

=

(25)

(27)

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Now, we consider the Bernoulli polynomials that are given by the generating function to be ∞ t tn ( x +iy)t e = B ( x + iy ) . n ∑ et − 1 n! n =0

We also have ∞

tn

∑ Bn (x + iy) n!

=

t t e( x+iy)t = t e xt (cos yt + i sin yt), et − 1 e −1

(28)

=

t t e( x−iy)t = t e xt (cos yt − i sin yt). et − 1 e −1

(29)

n =0

and



tn

∑ Bn (x − iy) n!

n =0

Thus, by (28) and (29), we can derive

and

∞ t e xt cos yt = ∑ t e −1 n =0



∞ t e xt sin yt = ∑ t e −1 n =0



Bn ( x + iy) + Bn ( x − iy) 2



Bn ( x + iy) + Bn ( x − iy) 2i



tn , n!

(30)

tn . n!

(31)

It follows that we define the following cosine–Bernoulli and sine–Bernoulli polynomials. (C )

(S)

Definition 2. The cosine–Bernoulli polynomials Bn ( x, y) and sine–Bernoulli polynomials Bn ( x, y) are defined by means of the generating functions ∞

∑ Bn

(C )

( x, y)

n =0

and



∑ Bn

(S)

( x, y)

n =0

tn t = t e xt cos yt, n! e −1

tn t = t e xt sin yt, n! e −1

(32)

(33)

respectively. By (30), (31), (32), and (33), we have Bn ( x + iy) + Bn ( x − iy) , 2 Bn ( x + iy) − Bn ( x − iy) (S) Bn ( x, y) = . 2i (C )

Bn ( x, y) =

(C )

Note that Bn ( x, 0) = Bn ( x ) are the Bernoulli polynomials. The cosine–Bernoulli and sine–Bernoulli polynomials can be determined explicitly. A few of them are (C )

B0 ( x, y) = 1,

1 (C ) B1 ( x, y) = − + x, 2

1 − x + x 2 − y2 , 6 x 3x2 3y2 (C ) B3 ( x, y) = − + x3 + − 3xy2 , 2 2 2 1 (C ) B4 ( x, y) = − + x2 − 2x3 + x4 − y2 + 6xy2 − 6x2 y2 + y4 , 30 (C )

B2 ( x, y) =

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(S)

(S)

(S)

B0 ( x, y) = 0, B1 ( x, y) = y, B2 ( x, y) = −y + 2xy, y (S) B3 ( x, y) = − 3xy + 3x2 y − y3 , 2 (S)

B4 ( x, y) = 2xy − 6x2 y + 4x3 y + 2y3 − 4xy3 . From (32), we have ∞

∑ Bn

(C )

( x, y)

n =0

tn t = t e xt cos yt, n! e −1 ! ! ∞ ∞ tm tl = ∑ Bn ∑ Cm (x, y) m! l! m =0 l =0 ! n   ∞ n tn Bl Cn−l ( x, y) = ∑ ∑ . l n! n =0 l =0

(34)

Comparing the coefficients on the both sides of (34), we obtain the following theorem: Theorem 6. For n ≥ 0, we have (C )

Bn ( x, y) = and (S)

Bn ( x, y) =

n

  n ∑ l Bl Cn−l (x, y), l =0 n

  n ∑ l Bl Sn−l (x, y). l =0

By replacing x by 1 − x in (32), we get ∞

∑ Bn

(C )

(1 − x, y)

n =0

t tn = t e(1− x)t cos yt n! e −1

=

t e− xt cos yt 1 − e−t

=

∑ (−1)n Bn (x, y) n! .



tn

n =0

Therefore, we obtain the following theorem: Theorem 7. For n ≥ 0, we have (C )

(C )

Bn (1 − x, y) = (−1)n Bn ( x, y), and (S)

(35)

(S)

Bn (1 − x, y) = (−1)n+1 Bn ( x, y).

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Now, we observe that ∞

∑ Bn

(C )

( x + 1, q)

n =0

tn t = t e( x+1)t cos yt n! e −1 t e xt cos yt et − 1 ∞ ∞ tn tn (C ) = ∑ nCn−1 ( x, y) + ∑ Bn ( x, y) n! n=0 n! n =1  tn ∞  (C ) = ∑ nCn−1 ( x, y) + Bn ( x, y) . n! n =0

= te xt cos yt +

(36)

Thus, by (36), we get (C )

(C )

(37)

Bn ( x + 1, y) = nCn−1 ( x, y) + Bn ( x, y), (n ≥ 1). Therefore, by (37), we obtain the following theorem: Theorem 8. For n ≥ 1, we have (C )

(C )

(S)

(S)

Bn ( x + 1, y) − Bn ( x, y) = nCn−1 ( x, y), and Bn ( x + 1, y) − Bn ( x, y) = nSn−1 ( x, y). Now, we define the new type polynomials that are given by the generating functions to be

et

∞ tn 2 (C ) cos yt = ∑ En (y) , +1 n! n =0

(38)

et

∞ 2 tn (S) sin yt = ∑ En (y) , +1 n! n =0

(39)

and

respectively. (C )

(S)

(C )

(C )

(S)

(S)

Note that En (0) = En , En (0) = 0, En (0, y) = En (y), En (0, y) = En (y), (n ≥ 0). The new type polynomials can be determined explicitly. A few of them are 1 1 3y2 (C ) (C ) (C ) E1 ( x, y) = − , E2 ( x, y) = −y2 , E3 (y) = + , 2 4 2 1 5y2 5y4 (C ) (C ) (C ) − , E6 (y) = −y6 , E4 (y) = y4 , E5 (y) = − − 2 2 2 (C )

E0 (y) = 1,

and

(S)

E0 ( x, y) = 0, (S)

(S)

(S)

E3 ( x, y) = −y3 ,

E5 ( x, y) = y5 ,

E6 ( x, y) = −3y − 5y3 − 3y5 .

E1 ( x, y) = y,

E4 ( x, y) = y + 2y3

(S)

E2 ( x, y) = −y,

(S)

(S)

From (38) and (39), we derive the following equations:   k  [2]  ∞ k 2 k m 2m  t ∑ (− 1 ) E y cos yt = ∑ k − 2m et + 1 2m k! k =0 m =0

(40)

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and ∞

et



[ k−2 1 ] 

2 sin yt = ∑  ∑ +1 m =0 k =0



k tk (−1)m Ek−2m−1 y2m+1  . 2m + 1 k! 

(41)

By (38)–(41), we get (C ) En (y)

[ n2 ]

 n (−1)m y2m En−2m , 2m

(42)

 n (−1)m y2m+1 En−2m−1 , (k ≥ 0). 2m + 1

(43)

=





m =0

and (S) En (y)

1  [ n− 2 ]



=

m =0

From (12), (13), (38) and (39), we derive the following theorem: Theorem 9. For n ≥ 0, we have n

(C )

En ( x, y) = and

  n (C ) ∑ k xn−k Ek (y), k =0 n

(S)

En ( x, y) =

  n (S) ∑ k xn−k Ek (y). k =0

Now, we define the new type polynomials that are given by the generating functions to be

and

∞ tn t (C ) cos yt = B ( y ) , n ∑ et − 1 n! n =0

(44)

∞ t tn (S) sin yt = B ( y ) , n ∑ et − 1 n! n =0

(45)

respectively. (C )

(S)

(C )

(C )

(S)

(S)

Note that Bn (0) = Bn , Bn (0) = 0, Bn (0, y) = Bn (y), Bn (0, y) = Bn (y), (n ≥ 0). The new type polynomials can be determined explicitly. A few of them are 1 (C ) B1 ( x, y) = − , 2

(C )

B0 ( x, y) = 1, (C )

B3 ( x, y) = and

(S)

3y2 , 2

B0 ( x, y) = 0, (S)

B3 ( x, y) =

(C )

B2 ( x, y) =

(C )

1 − y2 + y4 , 30

B1 ( x, y) = y,

B2 ( x, y) = −y,

B4 ( x, y) = − (S)

y − y3 , 2

(S)

1 − y2 , 6 (C )

B5 ( x, y) =

5y4 , 2

(S)

B4 ( x, y) = 2y3 ,

y 5y3 (S) B5 ( x, y) = − − + y5 . 6 3

From (44) and (45), we derive the following equations:

and

  k  [2]  ∞ k k t m 2m  t ∑ cos yt = (− 1 ) B y ∑ k − 2m et − 1 2m k! k =0 m =0

(46)

 k −1   [ 2 ] ∞ k k t m 2m+1  t ∑ (− 1 ) B y sin yt = . ∑ k − 2m − 1 et − 1 2m + 1 k! m =0 k =0

(47)

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By (44)–(47), we get (C ) Bn (y)

[ n2 ]

 n (−1)m y2m Bn−2m , 2m

(48)

 n (−1)m y2m+1 Bn−2m−1 , (k ≥ 0). 2m + 1

(49)

=





m =0

and (S) Bn (y)

1  [ n− 2 ]

=



m =0

From (32), (33), (44) and (45), we derive the following theorem: Theorem 10. For n ≥ 0, we have n

(C )

Bn ( x, y) = and

  n (C ) ∑ k xn−k Bk (y), k =0 n

(S)

Bn ( x, y) =

  n (S) ∑ k xn−k Bk (y). k =0

We remember that the classical Stirling numbers of the first kind S1 (n, k ) and S2 (n, k) are defined by the relations (see [12])

( x )n =

n

n

k =0

k =0

∑ S1 (n, k)xk and xn = ∑ S2 (n, k)(x)k ,

(50)

respectively. Here, ( x )n = x ( x − 1) · · · ( x − n + 1) denotes the falling factorial polynomial of order n. The numbers S2 (n, m) also admit a representation in terms of a generating function

(et − 1)m = m!





S2 (n, m)

n=m

tn . n!

(51)

By (12), (51) and by using Cauchy product, we get ∞



n =0

t (C ) En ( x, y)

n

n!



=

2 et + 1



(1 − (1 − e−t ))− x cos yt ∞

 x+l−1 (1 − e − t ) l l l =0   ∞ t 2 ( e − 1) l e−lt cos yt = ∑ < x >l t+1 l! e l =0 

=

2 et + 1



cos yt ∑



=



tn



∑ < x >l ∑ S2 (n, l ) n! ∑ En n =0

l =0 ∞

=





n =0



n =0

(C )

(−l, y)

tn n!

!   n tn (C ) ∑ ∑ i S2 (i, l )En−i (−l, y) < x >l n! , l =0 i = l n

where < x >l = x ( x + 1) · · · ( x + l − 1)(l ≥ 1) with < x >0 = 1. By comparing the coefficients on both sides of (52), we have the following theorem:

(52)

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Theorem 11. For n ∈ Z+ , we have ∞

n

  n (C ) ∑ ∑ i S2 (i, l )En−i (−l, y) < x >l , l =0 i = l ∞ n   n (S) (S) En ( x, y) = ∑ ∑ S2 (i, l ) En−i (−l, y) < x >l . i l =0 i = l (C )

En ( x, y) =

By (12), (38), (50), (51) and by using Cauchy product, we have ∞



t (C ) En ( x, y)

n

n!

n =0



=

2 t e +1



((et − 1) + 1) x cos(yt)

∞   2 x = t ( e t − 1) l cos(yt) ∑ l e +1 l =0   ∞ ( e t − 1) l 2 = ∑ ( x )l cos(yt) l! et + 1 l =0 ∞

=



n =0

l =0 ∞

=

tn



∑ (x)l ∑ S2 (n, l ) n! ∑ En ∑

n =0



(C )

(y)

n =0

tn n!

!   n tn (C ) ∑ ∑ i (x)l S2 (i, l )En−i (y) n! . l =0 i = l n

By comparing the coefficients on both sides of (53), we have the following theorem: Theorem 12. For n ∈ Z+ , we have ∞

n

  n (C ) = ∑∑ ( x )l S2 (i, l ) En−i (y), i l =0 i = l ∞ n   n (S) (S) En ( x, y) = ∑ ∑ ( x )l S2 (i, l ) En−i (y). i l =0 i = l (C ) En ( x, y)

By (4), (12), (38), (50), (51) and by using Cauchy product, we have ∞

∑ En

(C )

( x, y)

n =0



2 et + 1

tn n!



e xt cos(yt)  r ∞ (et − 1)r r! t tn (C ) xt = e E ( y ) n ∑ r! tr e t − 1 n! n =0 ! ! ∞ ∞ ( e t − 1)r tn tn r! (r ) (C ) = ∑ Bn (x) n! ∑ En (y) n! tr r! n =0 n =0 !  ∞ n (n) n−l  n − l tn ( r ) ( C ) = ∑ ∑ l +l r S2 (l + r, r ) ∑ Bi ( x ) En−l −i (y) . i n! n =0 l =0 ( r ) i =0

=

By comparing the coefficients on both sides, we have the following theorem:

(53)

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Theorem 13. For n ∈ Z+ and r ∈ N, we have n

(nl)

n−l 

 n − l (C ) (r ) = ∑ l +r S2 (l + r, r ) ∑ En−l −i (y)Bi ( x ), i ( ) i =0 l =0 r  n (n) n−l  n − l (S) (r ) (S) En−l −i (y)Bi ( x ). En ( x, y) = ∑ l +l r S2 (l + r, r ) ∑ i ) i =0 l =0 (

(C ) En ( x, y)

r

By (5), (12), (38), (50), (51) and by using the Cauchy product, we get ∞

  tn 2 = e xt cos(yt) ∑ t+1 n! e n =0     2 (et − u)r 1 − u r xt e cos(yt) = (1 − u )r e t − u et + 1     ∞ 1 tn r r it 2 (r ) e (−u)r−i = ∑ Hn (u, x ) ∑ cos(yt) n! i=0 i (1 − u )r e t + 1 n =0 r   ∞ 1 r t n ∞ (C ) tn (r ) = (−u)r−i ∑ Hn (u, x ) ∑ En (i, y) ∑ r (1 − u ) i =0 i n! n=0 n! n =0 !     ∞ r n r n 1 tn ( r ) ( C ) r −i = ∑ (− u ) H . ( u, x ) E ( i, y ) ∑ ∑ l n − l (1 − u ) r i =0 i l n! n =0 l =0 (C )

En ( x, y)

By comparing the coefficients on both sides, we have the following theorem: Theorem 14. For n ∈ Z+ and r ∈ N, we have r

n

   r n (r ) (C ) ∑ ∑ i l (−u)r−i Hl (u, x)En−l (i, y), i =0 l =0 r n    r n 1 (r ) (S) (S) En ( x, y) = ∑ i l (−u)r−i Hl (u, x)En−l (i, y). (1 − u)r i∑ =0 l =0 (C )

En ( x, y) =

1 (1 − u )r

By Theorems 12–14, we have the following corollary. Corollary 3. For n ∈ Z+ and r ∈ N, we have ∞

n

  n (C ) ∑ ∑ i (x)l S2 (i, l )En−i (y) l =0 i = l r n    r n 1 (r ) (C ) = ∑ i l (−u)r−i Hl (u, x)En−l (i, y) (1 − u)r i∑ =0 l =0  n (n) n−l  n − l (C ) (r ) l = ∑ l +r S2 (l + r, r ) ∑ En−l −i (y)Bi ( x ). i i =0 l =0 ( r ) 3. Distribution of Zeros of the Cosine–Euler and Sine–Euler Polynomials This section aims to demonstrate the benefit of using numerical investigation to support theoretical prediction and to discover a new interesting pattern of the zeros of the cosine–Euler and sine–Euler (C )

polynomials. Using a computer, a realistic study for the cosine–Euler polynomials En ( x, y) and (S)

sine–Euler polynomials En ( x, y) is very interesting. It is the aim of this paper to observe an interesting (C )

phenomenon of “scattering” of the zeros of the the cosine–Euler polynomials En ( x, y) and sine–Euler

Axioms 2018, 7, 56

15 of 19

(S)

polynomials En ( x, y) in a complex plane. We investigate the beautiful zeros of the cosine–Euler and sine–Euler polynomials by using a computer. We plot the zeros of the cosine–Euler polynomials (C )

En ( x, y) (Figure 1). In Figure 1 (top-left), we choose n = 30 and y = −3. In Figure 1 (top-right), we choose n = 30 and y = 0. In Figure 1 (bottom-left), we choose n = 30 and y = 1/2. In Figure 1 (bottom-right), we choose n = 30 and y = 3. (S)

We plot the zeros of the sine–Euler polynomials En ( x, y) (Figure 2).

Im(x)

10

10

5

5

0

Im(x)

-5

0

-5

-10

-50

-10 0

-50

50

Re(x)

Im(x)

0

50

Re(x)

10

10

5

5

0

Im(x)

-5

0

-5

-10

-50

-10 0

-50

50

Re(x)

0

50

Re(x)

(C )

Figure 1. Zeros of En ( x, y).

In Figure 2 (top-left), we choose n = 30 and x = −3. In Figure 2 (top-right), we choose n = 30 and x = −1. In Figure 2 (bottom-left), we choose n = 30 and x = 1. In Figure 2 (bottom-right), we choose n = 30 and x = 3. (C ) We observe that En ( x, a), x ∈ C has Re( x ) = 21 reflection symmetry in addition to the usual Im( x ) = 0 reflection symmetry analytic complex functions, where a ∈ R( Figures 1 and 2). Since ∞ (−1)n tn 2 (C ) ∑ En (1 − x, −y) n! = e−t + 1 e(1−x)(−t) cos yt n =0

=

∞ 2 tn (C ) xt e cos yt = En ( x, y) , ∑ t e +1 n! n =0

we obtain (C )

(C )

En ( x, y) = (−1)n En (1 − x, y),

(S)

(S)

En ( x, y) = (−1)n+1 En (1 − x, y).

En ( x, y) = (−1)n En (1 − x, −y), En ( x, y) = (−1)n En (1 − x, −y), Hence, we have the following theorem:

(C )

(S)

(C )

(S)

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Theorem 15. If n ≡ 1 (mod 2), then (C )

Bn (1/2, y) = 0, for n ∈ N.

(S)

Bn (1/2, y) = 0, for n ∈ N.

En (1/2, y) = 0,

(C )

If n ≡ 0 (mod 2), then En (1/2, y) = 0,

(S)

Our numerical results for numbers of real and complex zeros of the cosine–Euler polynomials (C ) En ( x, y)

Im(x)

= 0 are displayed (Table 1).

10

10

5

5

0

0

Im(x)

-5

-5

-10 -30

-20

-10

0

10

20

-10 -30

30

-20

-10

0

Re(x)

Im(x)

10

20

30

10

20

30

Re(x)

10

10

5

5

0

0

Im(x)

-5

-5

-10 -30

-20

-10

0

10

20

-10 -30

30

-20

-10

0

Re(x)

Re(x)

(S)

Figure 2. Zeros of En ( x, y). (C )

Table 1. Numbers of real and complex zeros of En ( x, y).

Degree n 1 2 3 4 5 6 7 8 9 10

y = −3

y=2

Real Zeros

Complex Zeros

Real Zeros

Complex Zeros

1 2 3 4 5 6 7 8 9 10

0 0 0 0 0 0 0 0 0 0

1 2 3 4 5 6 7 8 9 10

0 0 0 0 0 0 0 0 0 0

Our numerical results for numbers of real and complex zeros of the sine–Euler polynomials (S) En ( x, y)

= 0 are displayed (Table 2).

Axioms 2018, 7, 56

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(C )

Stacks of zeros of the cosine–Euler polynomials En ( x, y) for 1 ≤ n ≤ 40 from a 3D structure are presented (Figure 3). In Figure 3 (left), we choose y = −3. In Figure 3 (right), we choose y = 1/2. The plot of real zeros (C )

of the cosine–Euler polynomials En ( x, y) for 1 ≤ n ≤ 40 structure are presented (Figure 4). In Figure 4 (left), we choose y = −3. In Figure 4 (right), we choose y = 1/2. Stacks of zeros of the (S)

sine–Euler polynomials En ( x, y) for 1 ≤ n ≤ 40 from a 3D structure are presented (Figure 5). (S)

Table 2. Numbers of real and complex zeros of En ( x, y).

Degree n 1 2 3 4 5 6 7 8 9 10

x = −3

x=1

Real Zeros

Complex Zeros

Real Zeros

Complex Zeros

1 1 3 3 5 5 7 7 9 9

0 0 0 0 0 0 0 0 0 0

1 1 3 3 5 1 7 1 9 1

0 0 0 0 0 4 0 6 0 8

(C )

Figure 3. Stacks of zeros of En ( x, y), 1 ≤ n ≤ 40.

(C )

Figure 4. Real zeros of En ( x, y).

Axioms 2018, 7, 56

18 of 19

(S)

Figure 5. Stacks of zeros of En ( x, y), 1 ≤ n ≤ 40.

In Figure 5 (left), we choose x = −3. In Figure 3 (right), we choose x = 1. The plot of real zeros of (S)

the sine–Euler polynomials En ( x, y) for 1 ≤ n ≤ 40 structure are presented (Figure 6).

(S)

Figure 6. Real zeros of En ( x, y).

In Figure 6 (left), we choose x = −3. In Figure 6 (right), we choose x = 1. We observe a remarkable regular structure of the complex roots of the cosine–Euler polynomials (C )

En ( x, y). We also hope to verify a remarkable regular structure of the complex roots of the (C )

cosine–Euler polynomials En ( x, y). (C ) En ( x, y)

Next, we calculated an approximate solution satisfying

= 0, x ∈ R. The results are given in Table 3. (C )

Table 3. Approximate solutions of En ( x, −3) = 0, x ∈ R. Degree n 1 2 3 4 5 6 7

x

−12.751,

−6.8305, −8.8303, −10.799, −3.4960,

−2.5414, −4.7678, −0.82832, −1.8336, −2.7017, −1.1389,

0.50000 3.5414 0.50000, 1.8283, 0.50000, −0.40666, 0.50000,

5.7678 7.8305 2.8336, 1.4067, 2.1389, (S)

9.8303 3.7017, 4.4960,

11.799 13.751

Next, we calculated an approximate solution satisfying En ( x, y) = 0, y ∈ R. The results are given in Table 4.

Axioms 2018, 7, 56

19 of 19

(S)

Table 4. Approximate solutionsof En (−3, y) = 0, y ∈ R. Degree n 1 2 3 4 5 6 7

y

−15.241,

−10.687, −5.9045, −4.1727,

0.00000 0.00000 −6.0000, 0, −3.3912, 0, −2.4038, 0, −1.8630, 0, −1.5184, 0,

6.0000 3.3912 2.4038, 1.8630, 1.5184,

10.687 5.9045 4.1727,

15.241

Author Contributions: T.K. and C.S.R. wrote and checked the results of the paper; C.S.R. conducted numerical experiments of this paper; T.K. completed the revision of the article. Funding: This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No. 2017R1A2B4006092). Acknowledgments: The authors would like to thank the referees for their valuable comments. Conflicts of Interest: The authors declare no conflict of interest.

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