EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 6, No. 4, 2013, 405-412
ISSN 1307-5543 – www.ejpam.com

A Note on the Generalized Bernoulli and Euler Polynomials

Bao Quoc Ta

Department of Mathematics, Åbo Akademi University, FIN-20500 Åbo, Finland

Abstract. In this paper we use probabilistic methods to derive some results on the generalized Bernoulli
and generalized Euler polynomials. Our approach is based on the properties of Appell polynomials as-
sociated with uniformly distributed and Bernoulli distributed random variables and their sums.

2010 Mathematics Subject Classifications: 11B68, 26C05, 60E05, 62E15
Key Words and Phrases: Appell polynomials, Generalized Bernoulli polynomials, Generalized Euler
polynomials.

1. Introduction

We start with recalling the definition and basic properties of Appell polynomials. Let ξ be
a random variable with some exponential moments, i.e., E(eλ|ξ|) < ∞ for some λ > 0. The
Appell polynomials Q(ξ)n , n= 0, 1,2 . . . associated with ξ are defined via the expansion

eux

E(euξ)
=
∞
∑

n=1

un

n!
Q(ξ)n (x). (1)

Clearly, in case ξ≡ 0 it holds

Q(0)n (x) = xn, n= 0, 1,2, . . . . (2)

Notice also that Q(ξ)0 (x) = 1 for all x .
The Appell polynomials have the following properties (see, e.g., Salminen [6])

(i) Mean value property:
E(Q(ξ)n (ξ+ x)) = xn. (3)

(ii) Recursive differential equation:

d

d x
Q(ξ)n (x) = nQ(ξ)n−1(x). (4)

Email address: tbao@abo.fi

http://www.ejpam.com 405 c© 2013 EJPAM All rights reserved.



B. Ta / Eur. J. Pure Appl. Math, 6 (2013), 405-412 406

(iii) If ξ1 and ξ2 are independent random variables then

Q(ξ1+ξ2)
n (x + y) =

n
∑

k=0

�

n

k

�

Q(ξ1)
k (x)Q(ξ2)

n−k(y).

Choosing here ξ2 = 0 and x = 0 gives

Q(ξ1)
n (y) =

n
∑

k=0

�

n

k

�

Q(ξ1)
k (0)yn−k.

Bernoulli and Euler polynomials are, in fact, Appell polynomials as seen in the following
examples.

Example 1 (Bernoulli polynomials). Let θ be uniformly distributed random variable on [0,1],
i.e., θ ∼ U[0,1]. Then

eux

E(euθ )
=

ueux

eu− 1
=
∞
∑

n=0

un

n!
Bn(x).

The polynomials x → Bn(x), n = 0,1, . . . are called the Bernoulli polynomials (see also [5, p
809]). Using (3) and (4) we may find the explicit expressions: B0(x) = 1, B1(x) = x − 1

2
,

B2(x) = x2− x + 1
6
, . . .

Example 2. [Euler polynomials] Let η be a random variable such that
P(η= 0) = P(η= 1) = 1/2, i.e., η∼ Ber(1/2). Then

eux

E(euθ )
=

2eux

eu+ 1
=
∞
∑

n=0

un

n!
En(x).

The polynomials x → En(x), n = 0, 1, . . . are called the Euler polynomials(see [5, p 809]) and
we have, e.g., E0(x) = 1, E1(x) = x − 1

2
, E2(x) = x2− x , . . .

In the next section we will define the generalized Bernoulli and the generalized Euler
polynomials. We also give new probabilistic proofs for some of their properties via Appell
polynomials. In the third section we derive a new identity between the generalized Bernoulli
and the generalized Euler polynomials which extends the results in Cheon [1] and Srivastava
and Pintér [7].

2. Generalized Bernoulli and Generalized Euler Polynomials

Recall, e.g., from Luke [4, p 18] and Erdélyi [3, p 253] (see also Comtet [2, p 227]) that
for a real or complex number m, the generalized Bernoulli polynomials B(m)n , n = 0, 1, . . . are
defined via

umeux

(eu− 1)m
=
∞
∑

n=0

un

n!
B(m)n (x). (5)



B. Ta / Eur. J. Pure Appl. Math, 6 (2013), 405-412 407

From (5) it immediately follows

B(0)n (x) =xn, (6)

B(m+l)
n (x + y) =

n
∑

i=0

�

n

i

�

B(m)i (x)B(l)n−i(y), (7)

B(m)n (x + y) =
n
∑

i=0

�

n

i

�

B(m)i (x)yn−i , (8)

B(m)n (x + 1)− B(m)n (x) =nB(m−1)
n−1 (x). (9)

In case m is an integer, we may use a probabilistic approach via Appell polynomials. In-
deed, setting θ (m) :=

∑m
i=1 θi and θ (0) := 0, where {θi} is an i.i.d sequence of random vari-

ables such that θi ∼ U[0,1], it holds

E(euθ (m)) =
� eu− 1

u

�m
.

Consequently, the Appell polynomials Q(θ
(m))

n associated with θ (m) are the generalized Bernoulli
polynomials B(m)n .

We exploit the mean value property (3) to give a proof of formula (8) as follows: From
(6), (3), and (7) we obtain

E
�

B(m)n (x + θ1)
�

=
n
∑

i=0

�

n

i

�

B(m−1)
i (0)E(Bn−i(x + θ1)) = B(m−1)

n (x). (10)

On the other hand, also from (7)

E
�

B(m)n (x + θ1)
�

=
n
∑

i=0

�

n

i

�

B(m)n−i(0)E(x + θ1)
i

=
n
∑

i=0

�

n

i

�

B(m)n−i(0)
1

i+ 1
[(x + 1)i+1− x i+1]

=
1

n+ 1
(B(m)n+1(x + 1)− B(m)n+1(x)). (11)

Combining (10) and (11) gives (8).

Remark 1.

(i) From (10), by induction, for any positive integer l ≤ m, we obtain

E
�

B(m)n (x +
l
∑

i=1

θi)
�

= B(m−l)
n (x) (12)

which coincides with the mean value property (3) in case m= l.



B. Ta / Eur. J. Pure Appl. Math, 6 (2013), 405-412 408

(ii) For non-integer m, there does not exist a random variable θ (m) such that
� eu−1

u

�m is the
moment generating function of θ (m). This follows, e.g., from the fact the uniform distribu-
tion is not infinitely divisible. Hence, we can connect the generalized Bernoulli polynomials
with Appell polynomials only in case m is an integer.

We can generalize the Euler polynomials similarly as the Bernoulli polynomials. The gen-
eralized Euler polynomials are defined via (see [3])

2meux

(eu+ 1)m
=
∞
∑

n=0

un

n!
E(m)n (x), (13)

and it holds

E(0)n (x) =xn, (14)

E(k+l)
n (x + y) =

n
∑

i=0

�

n

i

�

E(k)i (x)E
(l)
n−i(y), (15)

E(m)n (x + y) =
n
∑

i=0

�

n

i

�

E(m)i (x)yn−i , (16)

E(m)n (x + 1) + E(m)n (x) =2E(m−1)
n (x). (17)

In case m is an integer, let η j , i = 1 . . . m be an i.i.d sequence of random variables such

that η j ∼ Ber(1/2). The Appell polynomials Q(η
(m))

n associated with the random variable
η(m) :=

∑m
j=1η j are the generalized Euler polynomials E(m)n (x).

Formula (17) is proved similarly as formula (8). It is seen that a formula analogous (12)
is valid for the generalized Euler polynomials, i.e.,

E
�

E(m)n (x +
l
∑

j=1

η j)
�

= E(m−l)
n (x). (18)

We also note similarly as for the generalized Bernoulli polynomials that the generalized
Euler polynomials can be connected with the Appell polynomials only if m is an integer.

3. Relationships Between the Generalized Bernoulli and the Generalized Euler
Polynomials

In this section we will generalize results in Cheon [1] and in Srivastava and Pintér [7].
Let us introduce the polynomials Q((m)+(l))n obtained from the expansion, for m, l ∈ C

� u

eu− 1

�m� 2

eu+ 1

�l
eux =

∞
∑

n=0

un

n!
Q((m)+(l))n (x). (19)



B. Ta / Eur. J. Pure Appl. Math, 6 (2013), 405-412 409

It holds

Q((m)+(l))n (x + y) =
n
∑

k=0

�

n

k

�

B(m)k (x)E(l)n−k(y), (20)

and

Q((m)+(l))n (x) =
n
∑

k=0

�

n

k

�

Q((m)+(l))k (0)xn−k. (21)

Furthermore, since

� u

eu− 1

�m� 2

eu+ 1

�l
=
h� u

eu− 1

�m−1� 2

eu+ 1

�li u

eu− 1

=
h� u

eu− 1

�m� 2

eu+ 1

�l−1i 2

eu+ 1
,

we have

Q((m)+(l))n (x) =
n
∑

k=0

�

n

k

�

Q((m−1)+(l))
k (0)Bn−k(x)

=
n
∑

k=0

�

n

k

�

Q((m)+(l−1))
k (0)En−k(x). (22)

In case m, l are integers, let us consider θ (m) :=
∑m

i=1 θi and η(l) :=
∑l

j=1η j , where
θi ∼ U[0,1], i = 1, . . . , m and η j ∼ Ber(1/2), j = 1, . . . , l are independent. Then it is seen
that Q((m)+(l))n , n= 0, 1 . . . , are the Appell polynomials associated with θ (m)+η(l).

Lemma 1. The following decomposition holds for all m, l ∈ C

Q((m)+(l))n (x) =Q((m)+(l−1))
n (x)−

n

2
Q((m−1)+(l))

n−1 (x). (23)

Proof. In the first equality of (22), substitute x + θ1 instead of x , take expectations, use
the mean value property (3) and apply (21) to obtain

E
�

Q((m)+(l))n (x + θ1)
�

=Q((m−1)+(l))
n (x).

Calculating similarly as in (11) we get

E
�

Q((m)+(l))n (x + θ1)
�

=
1

n+ 1

h

Q((m)+(l))n+1 (x + 1)−Q((m)+(l))n+1 (x)
i

.

Consequently
Q((m)+(l))n (x + 1)−Q((m)+(l))n (x) = nQ((m−1)+(l))

n−1 (x). (24)

Moreover, also by (22)

E
�

Q((m)+(l))n (x +η1)
�

=Q((m)+(l−1))
n (x), (25)



B. Ta / Eur. J. Pure Appl. Math, 6 (2013), 405-412 410

and from (21) we have

E
�

Q((m)+(l))n (x +η1)
�

=
n
∑

k=0

�

n

k

�

Q((m)+(l))k (0)E
�

(x +η1)
n−k�

=
1

2

h

Q((m)+(l))n (x + 1) +Q((m)+(l))n (x)
i

. (26)

Combining (25) and (26) implies

Q((m)+(l))n (x + 1) +Q((m)+(l))n (x) = 2Q((m)+(l−1))
n (x). (27)

Subtracting (27) and (24) completes the proof.

Remark 2. Formulas (24) and (27) generalize formulas (8) and (17) respectively.

Our main formula which connects the generalized Bernoulli and the generalized Euler
polynomials is given in the next theorem.

Theorem 1. For all m, l ∈ C, it holds

n
∑

k=0

�

n

k

�

B(m)k (x)E(l−1)
n−k (y) =

n
∑

k=0

�

n

k

�

h

B(m)k (x) +
k

2
B(m−1)

k−1 (x)
i

E(l)n−k(y). (28)

Proof. Using (20) it is seen that (23) can be developed as follows

n
∑

k=0

�

n

k

�

B(m)k (x)E(l)n−k(y)

=Q(m)+(l)n (x + y)

=Q(m)+(l−1)
n (x + y)−

n

2
Q(m−1)+(l)

n (x + y)

=
n
∑

k=0

�

n

k

�

B(m)k (x)E(l−1)
n−k (y)−

n

2

n−1
∑

k=0

�

n− 1

k

�

B(m−1)
k (x)E(l)n−k−1(y) (29)

=
n
∑

k=0

�

n

k

�

B(m)k (x)E(l−1)
n−k (y)−

n
∑

k=0

�

n

k

�

k

2
B(m−1)

k−1 (x)E(l)n−k(y), (30)

from which (28) readily follows.

Corollaries 1 and 2 below can be found in Srivastava and Pintér [7] as Theorem 1 and
Theorem 2, respectively.

Corollary 1. For all m ∈ C, it holds

B(m)n (x + y) =
n
∑

k=0

�

n

k

�

h

B(m)k (x) +
k

2
B(m−1)

k−1 (x)
i

En−k(y). (31)



B. Ta / Eur. J. Pure Appl. Math, 6 (2013), 405-412 411

Proof. This follows from (28) by putting l = 1 and using (14) and (7).

Corollary 2. For all l ∈ C, it holds

E(l)n (x + y) =
n
∑

k=0

�

n

k

�

2

k+ 1

h

E(l−1)
k+1 (y)− E(l)k+1(y)

i

Bn−k(x). (32)

Proof. Notice that in the step from (29) to (30) in the proof of Theorem 1 we have the
identity

n
∑

k=0

�

n

k

�

k

2
B(m−1)

k−1 (x)E(l)n−k(y) =
n

2

n−1
∑

k=0

�

n− 1

k

�

B(m−1)
k (x)E(l)n−k−1(y).

Hence,
n
∑

k=0

�

n

k

�

k

2
B(0)k−1(x)E

(l)
n−k(y) =

n

2
E(l)n−1(x + y).

From this and (28) with m= 1 we now have

E(l)n−1(x + y) =
2

n

n
∑

k=0

�

n

k

�

h

E(l−1)
k (y)− E(l)k (y)

i

Bn−k(x).

Changing n to n+ 1 and noting that E(l−1)
0 (y) = E(l)0 (y) = 1 we have

E(l)n (x + y) =
2

n+ 1

n+1
∑

k=1

�

n+ 1

k

�

h

E(l−1)
k (y)− E(l)k (y)

i

Bn+1−k(x),

which implies (32).

We recall also the following formula due to Cheon [1]

Bn(y) =
n
∑

k=0,k 6=1

�

n

k

�

Bk(0)En−k(y), (33)

which is now obtained from (31) by taking x = 0 and m = 1. The next result is derived in
Srivastava and Pintér [7]. We conclude this paper by giving a new proof for this.

Proposition 1. Formula (33) is equivalent to

2nBn(x/2) =
n
∑

k=0

�

n

k

�

Bk(0)En−k(x). (34)

Proof. From (23) putting m= 1, l = 1, we have

Q(θ
(1)+η(1))

n (x) = Bn(x)−
n

2
En−1(x).



REFERENCES 412

On the other hand
eux

E(eu(θ (1)+η(1)))
=

2ueux

e2u− 1
=
∞
∑

n=0

un

n!
2nBn(x/2),

and, hence,
Q(θ

(1)+η(1))
n (x) = 2nBn(x/2).

So we obtain
Bn(x)−

n

2
En−1(x) = 2nBn(x/2),

which implies the equivalence of (33) and (34).

ACKNOWLEDGEMENTS The author would like to thank Professor Paavo Salminen for valu-
able comments and discussions which improved this paper. This work was financially sup-
ported by The Finnish Doctoral Programme in Stochastics and Statistics.

References

[1] Gi-S. Cheon. A note on the Bernoulli and Euler polynomials. Applied Mathematics Letters,
16(3):365–368, 2003.

[2] L. Comtet. Advanced combinatorics. D. Reidel Publishing Co., Dordrecht, enlarged edition,
1974.

[3] A. Erdélyi, W. Magnus, F. Oberhettinger, and F.G. Tricomi. Higher transcendental functions,
volume 3. Robert E. Krieger Publishing Co. Inc., Melbourne, Fla., 1981.

[4] Y. L. Luke. The special functions and their approximations, volume 1. Academic Press, New
York, 1969.

[5] M. M. Abramowitz and I. Stegun. Handbook of mathematical functions. Dover publica-
tions, Inc., New York, 9th edition, 1970.

[6] P. Salminen. Optimal stopping, Appell polynomials, and Wiener-Hopf factorization.
Stochastics, 83(4-6):611–622, 2011.

[7] H.M. Srivastava and Á. Pintér. Remarks on some relationships between the Bernoulli and
Euler polynomials. Applied Mathematics Letters, 17(4):375–380, 2004.