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Axioms 2018, 7(3), 56; doi:10.3390/axioms7030056

Article
Some Identities for Euler and Bernoulli Polynomials and Their Zeros
1
Department of Mathematics, Kwangwoon University, Seoul 139-701, Korea
2
Department of Mathematics, Hannam University, Daejeon 306-791, Korea
*
Author to whom correspondence should be addressed.
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 [1,2,3,4,5,6,7,8,9,10,11,12,13,14]). The Appell polynomials can be defined by considering the following generating function:
A ( t ) e x t = A 0 ( x ) + A 1 ( x ) t 1 ! + A 2 ( x ) t 2 2 ! + + A n ( x ) t n n ! + = n = 0 A n ( x ) t n n ! , ( see [ 5 , 7 , 8 ] ) ,
where
A ( t ) = A 0 + A 1 t 1 ! + A 2 t 2 2 ! + + A n t n n ! + , A 0 0 .
Alternatively, the sequence A n ( x ) is Appell sequence for ( g ( t ) , t ) if and only if
1 g ( t ) e x t = n = 0 A n ( x ) t n n ! , ( see [ 5 , 7 , 8 ] ) ,
where
g ( t ) = n = 0 g n t n n ! , g 0 0 .
Differentiating generating Equation (1) with respect to x and equating coefficients of t n n ! , we have
d d x A n ( x ) = n A n 1 ( x ) , n = 0 , 1 , 2 , 3 , .
The typical examples of Appell polynomials are the Bernoulli and Euler polynomials (see [1,2,3,4,5,6,7,8,9,10,11,12,13,14]). It is well known that the Bernoulli polynomials are defined by the generating function to be
t e t 1 e x t = n = 0 B n ( x ) t n n ! .
When x = 0 , B n = B n ( 0 ) are called the Bernoulli numbers. The Euler polynomials are given by the generating function to be
2 e t + 1 e x t = n = 0 E n ( x ) t n n ! .
When x = 0 , E n = E n ( 0 ) are called the Euler numbers.
The Bernoulli polynomials B n ( r ) ( x ) of order r are defined by the following generating function
t e t 1 r e x t = n = 0 B n ( r ) ( x ) t n n ! , ( | t | < 2 π ) .
The Frobenius–Euler polynomials of order r , denoted by H n ( r ) ( u , x ) , are defined as
1 u e t u r e x t = n = 0 H n ( r ) ( u , x ) t n n ! .
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 e t + 1 e ( x + i y ) t = n = 0 E n ( x + i y ) t n n ! .
On the other hand, we observe that
e ( x + i y ) t = e x t e i y t = e x t ( cos y t + i sin y t ) .
From Equations (6) and (7), we have
n = 0 E n ( x + i y ) t n n ! = 2 e t + 1 e ( x + i y ) t = 2 e t + 1 e x t ( cos y t + i sin y t ) ,
and
n = 0 E n ( x i y ) t n n ! = 2 e t + 1 e ( x i y ) t = 2 e t + 1 e x t ( cos y t i sin y t ) .
Thus, by (8) and (9), we can derive
2 e t + 1 e x t cos y t = n = 0 E n ( x + i y ) + E n ( x i y ) 2 t n n ! ,
and
2 e t + 1 e x t sin y t = n = 0 E n ( x + i y ) + E n ( x i y ) 2 i t n n ! .
It follows that we define the following cosine–Euler polynomials and sine–Euler polynomials.
Definition 1.
The cosine–Euler polynomials E n ( C ) ( x , y ) and sine–Euler polynomials E n ( S ) ( x , y ) are defined by means of the generating functions
n = 0 E n ( C ) ( x , y ) t n n ! = 2 e t + 1 e x t cos y t ,
and
n = 0 E n ( S ) ( x , y ) t n n ! = 2 e t + 1 e x t sin y t ,
respectively.
Note that E n ( C ) ( x , 0 ) = E n ( x ) , E n ( S ) ( x , 0 ) = 0 , ( n 0 ) . The cosine–Euler and sine–Euler polynomials can be determined explicitly. A few of them are
E 0 ( C ) ( x , y ) = 1 , E 1 ( C ) ( x , y ) = 1 2 + x , E 2 ( C ) ( x , y ) = x + x 2 y 2 , E 3 ( C ) ( x , y ) = 1 4 3 x 2 2 + x 3 + 3 y 2 2 3 x y 2 , E 4 ( C ) ( x , y ) = x 2 x 3 + x 4 + 6 x y 2 6 x 2 y 2 + y 4 ,
and
E 0 ( S ) ( x , y ) = 0 , E 1 ( S ) ( x , y ) = y , E 2 ( S ) ( x , y ) = y + 2 x y , E 3 ( S ) ( x , y ) = 3 x y + 3 x 2 y y 3 , E 4 ( S ) ( x , y ) = y 6 x 2 y + 4 x 3 y + 2 y 3 4 x y 3 .
By (10)–(13), we have
E n ( C ) ( x , y ) = E n ( x + i y ) + E n ( x i y ) 2 , E n ( S ) ( x , y ) = E n ( x + i y ) E n ( x i y ) 2 i .
Clearly, we can get the following explicit representations of E n ( x + i y )
E n ( x + i y ) = k = 0 n n k ( x + i y ) n k E k , E n ( x + i y ) = k = 0 n n k ( i y ) n k E k ( x ) .
Let
e x t cos y t = k = 0 C k ( x , y ) t k k ! , e x t sin y t = k = 0 S k ( x , y ) t k k ! .
Then, by Taylor expansions of e x t cos y t and e x t sin y t , we get
e x t cos y t = k = 0 m = 0 [ k 2 ] k 2 m ( 1 ) m x k 2 m y 2 m t k k !
and
e x t sin y t = k = 0 m = 0 [ k 1 2 ] k 2 m + 1 ( 1 ) m x k 2 m 1 y 2 m + 1 t k k ! ,
where [ ] denotes taking the integer part. By (14)–(16), we get
C k ( x , y ) = m = 0 [ k 2 ] k 2 m ( 1 ) m x k 2 m y 2 m ,
and
S k ( x , y ) = m = 0 [ k 1 2 ] k 2 m + 1 ( 1 ) m x k 2 m 1 y 2 m + 1 , ( k 0 ) .
The two polynomials can be determined explicitly. A few of them are
C 0 ( x , y ) = 1 , C 1 ( x , y ) = x , C 2 ( x , y ) = x 2 y 2 , C 3 ( x , y ) = x 3 3 x y 2 , C 4 ( x , y ) = x 4 6 x 2 y 2 + y 4 , C 5 ( x , y ) = x 5 10 x 3 y 2 + 5 x y 4 , C 6 ( x , y ) = x 6 15 x 4 y 2 + 15 x 2 y 4 y 6 ,
and
S 0 ( x , y ) = 0 , S 1 ( x , y ) = y , S 2 ( x , y ) = 2 x y , S 3 ( x , y ) = 3 x 2 y y 3 , S 4 ( x , y ) = 4 x 3 y 4 x y 3 , S 5 ( x , y ) = 5 x 4 y 10 x 2 y 3 + y 5 , S 6 ( x , y ) = 6 x 5 y 20 x 3 y 3 + 6 x y 5 .
Now, we observe that
2 e t + 1 e x t cos y t = l = 0 E l t l l ! m = 0 C m ( x , y ) t m m ! = n = 0 l = 0 n n l E l C n l ( x , y ) t n n ! .
Therefore, we obtain the following theorem:
Theorem 1.
For n 0 , we have
E n ( C ) ( x , y ) = l = 0 n n l E l C n l ( x , y )
and
E n ( S ) ( x , y ) = l = 0 n n l E l S n l ( x , y ) .
From (12), we have
2 e x t cos y t = l = 0 E l ( C ) ( x , y ) t l l ! e t + 1 = n = 0 l = 0 n n l E l ( C ) ( x , y ) + E n ( C ) ( x , y ) t n n ! .
By (14) and (18), we get
C n ( x , y ) = 1 2 l = 0 n n l E l ( C ) ( x , y ) + E n ( C ) ( x , y ) .
Therefore, we obtain the following theorem:
Theorem 2.
For n 0 , we have
C n ( x , y ) = 1 2 l = 0 n n l E l ( C ) ( x , y ) + E n ( C ) ( x , y ) ,
and
S n ( x , y ) = 1 2 l = 0 n n l E l ( S ) ( x , y ) + E n ( S ) ( x , y ) .
From (12), we note that
n = 0 E n ( C ) ( 1 x , y ) t n n ! = 2 e t + 1 e ( 1 x ) t cos y t = 2 e t + 1 e x t cos ( y t ) = l = 0 ( 1 ) l E l t l l ! m = 0 ( 1 ) m C m , ( x , y ) t m m ! = n = 0 l = 0 n n l E l C n l ( x , y ) ( 1 ) n n ! t n .
Therefore, we obtain the following theorem:
Theorem 3.
For n 0 , we have
E n ( C ) ( 1 x , y ) = ( 1 ) n l = 0 n n l E l C n l ( x , y ) = ( 1 ) n E n ( C ) ( x , y ) ,
and
E n ( S ) ( 1 x , y ) = ( 1 ) n + 1 E n ( S ) ( x , y ) = ( 1 ) n + 1 l = 0 n n l E l S n l ( x , y ) .
Now, we observe that
n = 0 E n ( C ) ( x + 1 , y ) t n n ! = 2 e t + 1 e ( x + 1 ) t cos y t = 2 e t + 1 e x t ( e t 1 + 1 ) cos y t = 2 e x t cos y t 2 e t + 1 e x t cos y t = n = 0 2 C n ( x , y ) E n ( C ) ( x , y ) t n n ! .
By comparing the coefficients on the both sides, we get
E n ( C ) ( x + 1 , q ) + E n ( C ) ( x , y ) = 2 C n ( x , y ) , ( n 0 ) .
Therefore, we obtain the following theorem:
Theorem 4.
For n 0 , we have
E n ( C ) ( x + 1 , y ) + E n ( C ) ( x , y ) = 2 C n ( x , y ) ,
and
E n ( S ) ( x + 1 , y ) + E n ( S ) ( x , y ) = 2 S n ( x , y ) .
From (14) and (15), we have
k = 0 C k ( 0 , y ) t k k ! = m = 0 ( 1 ) m y 2 m t 2 m ( 2 m ) ! .
Therefore, by Theorem 4 and (23), we obtain the following corollary:
Corollary 1.
For n 0 , we have
E 2 n ( C ) ( 1 , y ) + E 2 n ( C ) ( 0 , y ) = 2 ( 1 ) n y 2 n ,
and
E 2 n + 1 ( S ) ( 1 , y ) + E 2 n + 1 ( S ) ( 0 , y ) = 2 ( 1 ) n y 2 n + 1 .
By (12), we get
n = 0 E n ( C ) ( x + r , y ) t n n ! = 2 e x t e t + 1 cos y t e r t = l = 0 E l ( C ) ( x , y ) t l l ! k = 0 r k t k k ! = n = 0 k = 0 n n k E k ( C ) ( x , y ) r n k t n n ! .
Therefore, by comparing the coefficients on the both sides, we obtain the following theorem:
Theorem 5.
For n 0 , r N , we have
E n ( C ) ( x + r , y ) = k = 0 n n k E k ( C ) ( x , y ) r n k ,
and
E n ( S ) ( x + r , y ) = k = 0 n n k E k ( S ) ( x , y ) r n k .
Taking r = 1 in Theorem 5, we obtain the following corollary:
Corollary 2.
For n 0 , we have
2 C n ( x , y ) = E n ( C ) ( x , y ) + k = 0 n n k E k ( C ) ( x , y ) ,
and
2 S n ( x , y ) = E n ( S ) ( x , y ) + k = 0 n n k E k ( S ) ( x , y ) .
From Corollary 2, we note that
E n ( C ) ( 0 , y ) + k = 0 n n k E k ( C ) ( 0 , y ) = 0 , if   n = 2 m + 1 , 2 ( 1 ) m y 2 m , if   n = 2 m ,
and
E n ( S ) ( 0 , y ) + k = 0 n n k E k ( S ) ( 0 , y ) = 2 ( 1 ) m y 2 m + 1 , if   n = 2 m + 1 , 0 , if   n = 2 m .
By (12), we get
n = 1 x E n ( C ) ( x , y ) t n n ! = x 2 e t + 1 e x t cos y t = 2 e t + 1 t e x t cos y t = n = 1 n E n 1 ( C ) ( x , y ) t n n ! .
Comparing the coefficients on the both sides of (27), we have
x E n ( C ) ( x , y ) = n E n 1 ( C ) ( x , y ) .
Similarly, for n 1 , we have
x E n ( S ) ( x , y ) = n E n 1 ( S ) ( x , y ) , y E n ( C ) ( x , y ) = n E n 1 ( S ) ( x , y ) , y E n ( S ) ( x , y ) = n E n 1 ( C ) ( x , y ) .
Now, we consider the Bernoulli polynomials that are given by the generating function to be
t e t 1 e ( x + i y ) t = n = 0 B n ( x + i y ) t n n ! .
We also have
n = 0 B n ( x + i y ) t n n ! = t e t 1 e ( x + i y ) t = t e t 1 e x t ( cos y t + i sin y t ) ,
and
n = 0 B n ( x i y ) t n n ! = t e t 1 e ( x i y ) t = t e t 1 e x t ( cos y t i sin y t ) .
Thus, by (28) and (29), we can derive
t e t 1 e x t cos y t = n = 0 B n ( x + i y ) + B n ( x i y ) 2 t n n ! ,
and
t e t 1 e x t sin y t = n = 0 B n ( x + i y ) + B n ( x i y ) 2 i t n n ! .
It follows that we define the following cosine–Bernoulli and sine–Bernoulli polynomials.
Definition 2.
The cosine–Bernoulli polynomials B n ( C ) ( x , y ) and sine–Bernoulli polynomials B n ( S ) ( x , y ) are defined by means of the generating functions
n = 0 B n ( C ) ( x , y ) t n n ! = t e t 1 e x t cos y t ,
and
n = 0 B n ( S ) ( x , y ) t n n ! = t e t 1 e x t sin y t ,
respectively.
By (30), (31), (32), and (33), we have
B n ( C ) ( x , y ) = B n ( x + i y ) + B n ( x i y ) 2 , B n ( S ) ( x , y ) = B n ( x + i y ) B n ( x i y ) 2 i .
Note that B n ( C ) ( x , 0 ) = B n ( x ) are the Bernoulli polynomials. The cosine–Bernoulli and sine–Bernoulli polynomials can be determined explicitly. A few of them are
B 0 ( C ) ( x , y ) = 1 , B 1 ( C ) ( x , y ) = 1 2 + x , B 2 ( C ) ( x , y ) = 1 6 x + x 2 y 2 , B 3 ( C ) ( x , y ) = x 2 3 x 2 2 + x 3 + 3 y 2 2 3 x y 2 , B 4 ( C ) ( x , y ) = 1 30 + x 2 2 x 3 + x 4 y 2 + 6 x y 2 6 x 2 y 2 + y 4 ,
and
B 0 ( S ) ( x , y ) = 0 , B 1 ( S ) ( x , y ) = y , B 2 ( S ) ( x , y ) = y + 2 x y , B 3 ( S ) ( x , y ) = y 2 3 x y + 3 x 2 y y 3 , B 4 ( S ) ( x , y ) = 2 x y 6 x 2 y + 4 x 3 y + 2 y 3 4 x y 3 .
From (32), we have
n = 0 B n ( C ) ( x , y ) t n n ! = t e t 1 e x t cos y t , = l = 0 B n t l l ! m = 0 C m ( x , y ) t m m ! = n = 0 l = 0 n n l B l C n l ( x , y ) t n n ! .
Comparing the coefficients on the both sides of (34), we obtain the following theorem:
Theorem 6.
For n 0 , we have
B n ( C ) ( x , y ) = l = 0 n n l B l C n l ( x , y ) ,
and
B n ( S ) ( x , y ) = l = 0 n n l B l S n l ( x , y ) .
By replacing x by 1 x in (32), we get
n = 0 B n ( C ) ( 1 x , y ) t n n ! = t e t 1 e ( 1 x ) t cos y t = t 1 e t e x t cos y t = n = 0 ( 1 ) n B n ( x , y ) t n n ! .
Therefore, we obtain the following theorem:
Theorem 7.
For n 0 , we have
B n ( C ) ( 1 x , y ) = ( 1 ) n B n ( C ) ( x , y ) ,
and
B n ( S ) ( 1 x , y ) = ( 1 ) n + 1 B n ( S ) ( x , y ) .
Now, we observe that
n = 0 B n ( C ) ( x + 1 , q ) t n n ! = t e t 1 e ( x + 1 ) t cos y t = t e x t cos y t + t e t 1 e x t cos y t = n = 1 n C n 1 ( x , y ) t n n ! + n = 0 B n ( C ) ( x , y ) t n n ! = n = 0 n C n 1 ( x , y ) + B n ( C ) ( x , y ) t n n ! .
Thus, by (36), we get
B n ( C ) ( x + 1 , y ) = n C n 1 ( x , y ) + B n ( C ) ( x , y ) , ( n 1 ) .
Therefore, by (37), we obtain the following theorem:
Theorem 8.
For n 1 , we have
B n ( C ) ( x + 1 , y ) B n ( C ) ( x , y ) = n C n 1 ( x , y ) ,
and
B n ( S ) ( x + 1 , y ) B n ( S ) ( x , y ) = n S n 1 ( x , y ) .
Now, we define the new type polynomials that are given by the generating functions to be
2 e t + 1 cos y t = n = 0 E n ( C ) ( y ) t n n ! ,
and
2 e t + 1 sin y t = n = 0 E n ( S ) ( y ) t n n ! ,
respectively.
Note that E n ( C ) ( 0 ) = E n , E n ( S ) ( 0 ) = 0 , E n ( C ) ( 0 , y ) = E n ( C ) ( y ) , E n ( S ) ( 0 , y ) = E n ( S ) ( y ) , ( n 0 ) . The new type polynomials can be determined explicitly. A few of them are
E 0 ( C ) ( y ) = 1 , E 1 ( C ) ( x , y ) = 1 2 , E 2 ( C ) ( x , y ) = y 2 , E 3 ( C ) ( y ) = 1 4 + 3 y 2 2 , E 4 ( C ) ( y ) = y 4 , E 5 ( C ) ( y ) = 1 2 5 y 2 2 5 y 4 2 , E 6 ( C ) ( y ) = y 6 ,
and
E 0 ( S ) ( x , y ) = 0 , E 1 ( S ) ( x , y ) = y , E 2 ( S ) ( x , y ) = y , E 3 ( S ) ( x , y ) = y 3 , E 4 ( S ) ( x , y ) = y + 2 y 3 E 5 ( S ) ( x , y ) = y 5 , E 6 ( S ) ( x , y ) = 3 y 5 y 3 3 y 5 .
From (38) and (39), we derive the following equations:
2 e t + 1 cos y t = k = 0 m = 0 [ k 2 ] k 2 m ( 1 ) m E k 2 m y 2 m t k k !
and
2 e t + 1 sin y t = k = 0 m = 0 [ k 1 2 ] k 2 m + 1 ( 1 ) m E k 2 m 1 y 2 m + 1 t k k ! .
By (38)–(41), we get
E n ( C ) ( y ) = m = 0 [ n 2 ] n 2 m ( 1 ) m y 2 m E n 2 m ,
and
E n ( S ) ( y ) = m = 0 [ n 1 2 ] n 2 m + 1 ( 1 ) m y 2 m + 1 E n 2 m 1 , ( k 0 ) .
From (12), (13), (38) and (39), we derive the following theorem:
Theorem 9.
For n 0 , we have
E n ( C ) ( x , y ) = k = 0 n n k x n k E k ( C ) ( y ) ,
and
E n ( S ) ( x , y ) = k = 0 n n k x n k E k ( S ) ( y ) .
Now, we define the new type polynomials that are given by the generating functions to be
t e t 1 cos y t = n = 0 B n ( C ) ( y ) t n n ! ,
and
t e t 1 sin y t = n = 0 B n ( S ) ( y ) t n n ! ,
respectively.
Note that B n ( C ) ( 0 ) = B n , B n ( S ) ( 0 ) = 0 , B n ( C ) ( 0 , y ) = B n ( C ) ( y ) , B n ( S ) ( 0 , y ) = B n ( S ) ( y ) , ( n 0 ) . The new type polynomials can be determined explicitly. A few of them are
B 0 ( C ) ( x , y ) = 1 , B 1 ( C ) ( x , y ) = 1 2 , B 2 ( C ) ( x , y ) = 1 6 y 2 , B 3 ( C ) ( x , y ) = 3 y 2 2 , B 4 ( C ) ( x , y ) = 1 30 y 2 + y 4 , B 5 ( C ) ( x , y ) = 5 y 4 2 ,
and
B 0 ( S ) ( x , y ) = 0 , B 1 ( S ) ( x , y ) = y , B 2 ( S ) ( x , y ) = y , B 3 ( S ) ( x , y ) = y 2 y 3 , B 4 ( S ) ( x , y ) = 2 y 3 , B 5 ( S ) ( x , y ) = y 6 5 y 3 3 + y 5 .
From (44) and (45), we derive the following equations:
t e t 1 cos y t = k = 0 m = 0 [ k 2 ] k 2 m ( 1 ) m B k 2 m y 2 m t k k !
and
t e t 1 sin y t = k = 0 m = 0 [ k 1 2 ] k 2 m + 1 ( 1 ) m B k 2 m 1 y 2 m + 1 t k k ! .
By (44)–(47), we get
B n ( C ) ( y ) = m = 0 [ n 2 ] n 2 m ( 1 ) m y 2 m B n 2 m ,
and
B n ( S ) ( y ) = m = 0 [ n 1 2 ] n 2 m + 1 ( 1 ) m y 2 m + 1 B n 2 m 1 , ( k 0 ) .
From (32), (33), (44) and (45), we derive the following theorem:
Theorem 10.
For n 0 , we have
B n ( C ) ( x , y ) = k = 0 n n k x n k B k ( C ) ( y ) ,
and
B n ( S ) ( x , y ) = k = 0 n n k x n k B k ( S ) ( y ) .
We remember that the classical Stirling numbers of the first kind S 1 ( n , k ) and S 2 ( n , k ) are defined by the relations (see [12])
( x ) n = k = 0 n S 1 ( n , k ) x k and x n = k = 0 n S 2 ( n , k ) ( x ) k ,
respectively. Here, ( x ) n = x ( x 1 ) ( x n + 1 ) denotes the falling factorial polynomial of order n. The numbers S 2 ( n , m ) also admit a representation in terms of a generating function
( e t 1 ) m = m ! n = m S 2 ( n , m ) t n n ! .
By (12), (51) and by using Cauchy product, we get
n = 0 E n ( C ) ( x , y ) t n n ! = 2 e t + 1 ( 1 ( 1 e t ) ) x cos y t = 2 e t + 1 cos y t l = 0 x + l 1 l ( 1 e t ) l = l = 0 < x > l ( e t 1 ) l l ! 2 e t + 1 e l t cos y t = l = 0 < x > l n = 0 S 2 ( n , l ) t n n ! n = 0 E n ( C ) ( l , y ) t n n ! = n = 0 l = 0 i = l n n i S 2 ( i , l ) E n i ( C ) ( l , y ) < x > l t n 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:
Theorem 11.
For n Z + , we have
E n ( C ) ( x , y ) = l = 0 i = l n n i S 2 ( i , l ) E n i ( C ) ( l , y ) < x > l , E n ( S ) ( x , y ) = l = 0 i = l n n i S 2 ( i , l ) E n i ( S ) ( l , y ) < x > l .
By (12), (38), (50), (51) and by using Cauchy product, we have
n = 0 E n ( C ) ( x , y ) t n n ! = 2 e t + 1 ( ( e t 1 ) + 1 ) x cos ( y t ) = 2 e t + 1 cos ( y t ) l = 0 x l ( e t 1 ) l = l = 0 ( x ) l ( e t 1 ) l l ! 2 e t + 1 cos ( y t ) = l = 0 ( x ) l n = 0 S 2 ( n , l ) t n n ! n = 0 E n ( C ) ( y ) t n n ! = n = 0 l = 0 i = l n n i ( x ) l S 2 ( i , l ) E n i ( C ) ( y ) t n n ! .
By comparing the coefficients on both sides of (53), we have the following theorem:
Theorem 12.
For n Z + , we have
E n ( C ) ( x , y ) = l = 0 i = l n n i ( x ) l S 2 ( i , l ) E n i ( C ) ( y ) , E n ( S ) ( x , y ) = l = 0 i = l n n i ( x ) l S 2 ( i , l ) E n i ( S ) ( y ) .
By (4), (12), (38), (50), (51) and by using Cauchy product, we have
n = 0 E n ( C ) ( x , y ) t n n ! = 2 e t + 1 e x t cos ( y t ) = ( e t 1 ) r r ! r ! t r t e t 1 r e x t n = 0 E n ( C ) ( y ) t n n ! = ( e t 1 ) r r ! n = 0 B n ( r ) ( x ) t n n ! n = 0 E n ( C ) ( y ) t n n ! r ! t r = n = 0 l = 0 n n l l + r r S 2 ( l + r , r ) i = 0 n l n l i B i ( r ) ( x ) E n l i ( C ) ( y ) t n n ! .
By comparing the coefficients on both sides, we have the following theorem:
Theorem 13.
For n Z + and r N , we have
E n ( C ) ( x , y ) = l = 0 n n l l + r r S 2 ( l + r , r ) i = 0 n l n l i E n l i ( C ) ( y ) B i ( r ) ( x ) , E n ( S ) ( x , y ) = l = 0 n n l l + r r S 2 ( l + r , r ) i = 0 n l n l i E n l i ( S ) ( y ) B i ( r ) ( x ) .
By (5), (12), (38), (50), (51) and by using the Cauchy product, we get
n = 0 E n ( C ) ( x , y ) t n n ! = 2 e t + 1 e x t cos ( y t ) = ( e t u ) r ( 1 u ) r 1 u e t u r e x t 2 e t + 1 cos ( y t ) = n = 0 H n ( r ) ( u , x ) t n n ! i = 0 r r i e i t ( u ) r i 1 ( 1 u ) r 2 e t + 1 cos ( y t ) = 1 ( 1 u ) r i = 0 r r i ( u ) r i n = 0 H n ( r ) ( u , x ) t n n ! n = 0 E n ( C ) ( i , y ) t n n ! = n = 0 1 ( 1 u ) r i = 0 r r i ( u ) r i l = 0 n n l H l ( r ) ( u , x ) E n l ( C ) ( i , y ) t n n ! .
By comparing the coefficients on both sides, we have the following theorem:
Theorem 14.
For n Z + and r N , we have
E n ( C ) ( x , y ) = 1 ( 1 u ) r i = 0 r l = 0 n r i n l ( u ) r i H l ( r ) ( u , x ) E n l ( C ) ( i , y ) , E n ( S ) ( x , y ) = 1 ( 1 u ) r i = 0 r l = 0 n r i n l ( u ) r i H l ( r ) ( u , x ) E n l ( S ) ( i , y ) .
By Theorems 12–14, we have the following corollary.
Corollary 3.
For n Z + and r N , we have
l = 0 i = l n n i ( x ) l S 2 ( i , l ) E n i ( C ) ( y ) = 1 ( 1 u ) r i = 0 r l = 0 n r i n l ( u ) r i H l ( r ) ( u , x ) E n l ( C ) ( i , y ) = l = 0 n n l l + r r S 2 ( l + r , r ) i = 0 n l n l i E n l i ( C ) ( y ) B i ( r ) ( x ) .

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 polynomials. Using a computer, a realistic study for the cosine–Euler polynomials E n ( C ) ( x , y ) and sine–Euler polynomials E n ( S ) ( x , y ) is very interesting. It is the aim of this paper to observe an interesting phenomenon of “scattering” of the zeros of the the cosine–Euler polynomials E n ( C ) ( x , y ) and sine–Euler polynomials E n ( S ) ( 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 E n ( C ) ( 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 .
We plot the zeros of the sine–Euler polynomials E n ( S ) ( x , y ) (Figure 2).
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 .
We observe that E n ( C ) ( x , a ) , x C has R e ( x ) = 1 2 reflection symmetry in addition to the usual I m ( x ) = 0 reflection symmetry analytic complex functions, where a R ( Figure 1 and Figure 2).
Since
n = 0 E n ( C ) ( 1 x , y ) ( 1 ) n t n n ! = 2 e t + 1 e ( 1 x ) ( t ) cos y t = 2 e t + 1 e x t cos y t = n = 0 E n ( C ) ( x , y ) t n n ! ,
we obtain
E n ( C ) ( x , y ) = ( 1 ) n E n ( C ) ( 1 x , y ) , E n ( C ) ( x , y ) = ( 1 ) n E n ( C ) ( 1 x , y ) , E n ( S ) ( x , y ) = ( 1 ) n E n ( S ) ( 1 x , y ) , E n ( S ) ( x , y ) = ( 1 ) n + 1 E n ( S ) ( 1 x , y ) .
Hence, we have the following theorem:
Theorem 15.
If n 1 ( mod 2 ) , then
E n ( C ) ( 1 / 2 , y ) = 0 , B n ( C ) ( 1 / 2 , y ) = 0 , f o r   n N .
If n 0 ( mod 2 ) , then
E n ( S ) ( 1 / 2 , y ) = 0 , B n ( S ) ( 1 / 2 , y ) = 0 , f o r   n N .
Our numerical results for numbers of real and complex zeros of the cosine–Euler polynomials E n ( C ) ( x , y ) = 0 are displayed (Table 1).
Our numerical results for numbers of real and complex zeros of the sine–Euler polynomials E n ( S ) ( x , y ) = 0 are displayed (Table 2).
Stacks of zeros of the cosine–Euler polynomials E n ( C ) ( 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 of the cosine–Euler polynomials E n ( C ) ( 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 sine–Euler polynomials E n ( S ) ( x , y ) for 1 n 40 from a 3D structure are presented (Figure 5).
In Figure 5 (left), we choose x = 3 . In Figure 3 (right), we choose x = 1 . The plot of real zeros of the sine–Euler polynomials E n ( S ) ( x , y ) for 1 n 40 structure are presented (Figure 6).
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 E n ( C ) ( x , y ) . We also hope to verify a remarkable regular structure of the complex roots of the cosine–Euler polynomials E n ( C ) ( x , y ) . Next, we calculated an approximate solution satisfying E n ( C ) ( x , y ) = 0 , x R . The results are given in Table 3.
Next, we calculated an approximate solution satisfying E n ( S ) ( x , y ) = 0 , y R . The results are given in Table 4.

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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Figure 1. Zeros of E n ( C ) ( x , y ) .
Figure 1. Zeros of E n ( C ) ( x , y ) .
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Figure 2. Zeros of E n ( S ) ( x , y ) .
Figure 2. Zeros of E n ( S ) ( x , y ) .
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Figure 3. Stacks of zeros of E n ( C ) ( x , y ) , 1 n 40 .
Figure 3. Stacks of zeros of E n ( C ) ( x , y ) , 1 n 40 .
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Figure 4. Real zeros of E n ( C ) ( x , y ) .
Figure 4. Real zeros of E n ( C ) ( x , y ) .
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Figure 5. Stacks of zeros of E n ( S ) ( x , y ) , 1 n 40 .
Figure 5. Stacks of zeros of E n ( S ) ( x , y ) , 1 n 40 .
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Figure 6. Real zeros of E n ( S ) ( x , y ) .
Figure 6. Real zeros of E n ( S ) ( x , y ) .
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Table 1. Numbers of real and complex zeros of E n ( C ) ( x , y ) .
Table 1. Numbers of real and complex zeros of E n ( C ) ( x , y ) .
Degree n y = 3 y = 2
Real ZerosComplex ZerosReal ZerosComplex Zeros
11010
22020
33030
44040
55050
66060
77070
88080
99090
10100100
Table 2. Numbers of real and complex zeros of E n ( S ) ( x , y ) .
Table 2. Numbers of real and complex zeros of E n ( S ) ( x , y ) .
Degree n x = 3 x = 1
Real ZerosComplex ZerosReal ZerosComplex Zeros
11010
21010
33030
43030
55050
65014
77070
87016
99090
109018
Table 3. Approximate solutions of E n ( C ) ( x , 3 ) = 0 , x R .
Table 3. Approximate solutions of E n ( C ) ( x , 3 ) = 0 , x R .
Degree nx
1 0.50000
2 −2.5414,3.5414
3 −4.7678,0.50000,5.7678
4 −6.8305,−0.82832,1.8283,7.8305
5 −8.8303,−1.8336,0.50000,2.8336,9.8303
6 −10.799,−2.7017,−0.40666,1.4067,3.7017,11.799
7−12.751,−3.4960,−1.1389,0.50000,2.1389,4.4960,13.751
Table 4. Approximate solutionsof E n ( S ) ( 3 , y ) = 0 , y R .
Table 4. Approximate solutionsof E n ( S ) ( 3 , y ) = 0 , y R .
Degree ny
10.00000
20.00000
3 −6.0000,0,6.0000
4 −3.3912,0,3.3912
5 −10.687,−2.4038,0,2.4038,10.687
6 −5.9045,−1.8630,0,1.8630,5.9045
7−15.241,−4.1727,−1.5184,0,1.5184,4.1727,15.241

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