Identities involving Frobenius-Euler polynomials arising from non-linear differential equations

0.12 MB·English

by  T. Kim

IDENTITIES INVOLVING FROBENIUS-EULER POLYNOMIALS
ARISING FROM NON-LINEAR DIFFERENTIAL EQUATIONS
2
1 TAEKYUNKIM
0
2
Abstract. In this paper we consider non-linear differential equations which

n are closely related to the generating functions of Frobenius-Euler polynomi-
a als. Fromournon-lineardifferential equations, wederivesomenew identities
J between thesumsofproductsofFrobenius-EulerpolynomialsandFrobenius-

4 Eulerpolynomialsofhigherorder.
2

]
T
N
1. Introduction
.
h
t Let u ∈ C with u 6= 1. Then the Frobenius-Euler polynomials are defined by
a
generating function as follows:
m
[ (1) Fu(t,x)= e1t−−uuext = ∞ Hn(x|u)tnn!, (see [1,2]).
1 n=0
X
v
In the special case, x = 0, H (0 | u) = H (u) are called the n-th Frobenius-Euler
8 n n
numbers (see [2]).
8
0 By (1), we get
5
n
1. (2) Hn(x|u)= nl xn−lHl(u) for n∈Z+ =N∪{0}.
0 l=0(cid:18) (cid:19)
X
2
Thus, by (1) and (2), we get the recurrence relation for H (u) as follows:
1 n
:
v 1−u if n=0,
n
i (3) H0(u)=1, H(u)+1 −Hn(u)=
X ( 0 if n>0.
(cid:0) (cid:1)
ar with the usual convention about replacing H(u)n by Hn(u) (see [2,10,12]).
The Bernoulli and Euler polynomials can be defined by
t ∞ tn 2 ∞ tn
ext = B (x) , ext = E (x) .
et−1 n n! et+1 n n!
n=0 n=0
X X
Inthespecialcase,x=0,B (0)=B arethen-thBernoullinumbersandE (0)=
n n n
E are the n-th Euler numbers.
n
The formula for a product of two Bernoulli polynomials are given by
(4)
∞
m n B B (x) m!n!
B (x)B (x)= n+ m 2r m+n−2r +(−1)m+1 B ,
m n 2r 2r m+n−2r (m+n)! m+n
r=0 (cid:18) (cid:19) (cid:18) (cid:19) !
X
where m+n≥2 and m = m! = m(m−1)···(m−n+1) (see [1,3]).
n n!(m−n)! n!
1
(cid:0) (cid:1)
2 TAEKYUNKIM
From (1), we note that H (x | −1) = E (x). In [10], Nielson also obtained
n n
similar formulas for E (x)E (x) and E (x)B (x).
n m m n
In view point of (4), Carlitz have considered the following identities for the
Frobenius-Euler polynomials as follows:
(5)
(1−α)(1−β)
H (x|α)H (x|β)=H (x|αβ)
m n m+n 1−αβ
m n
α(1−β) m β(1−β) n
+ H (α)H (x|αβ)+ H (β)H (x|αβ),
1−αβ r r m+n−r 1−αβ s s m+n−s
r=0(cid:18) (cid:19) s=0(cid:18) (cid:19)
X X
where α,β ∈C with α6=1, β 6=1 and αβ 6=1 (see [2]).
In particular, if α6=1 and αβ =1, then
m
m B (x)
H (x|α)H (x|α−1)=−(1−α) H (α) m+n−r+1
m n r
r m+n−r+1
r=1(cid:18) (cid:19)
X
n
n B (x)
−(1−α−1) H (α−1) m+n−s+1
s
s m+n−s+1
s=1(cid:18) (cid:19)
X
m!n!
+(−1)n+1 (1−α)H (α).
(m+n+1)! m+n+1
Forr ∈N,then-thFrobenius-Eulerpolynomialsoforderraredefinedbygenerating
function as follows:
Fr(t,x)=F (t,x)×F (t,x)×···×F (t,x)
u u u u
r−times
1−u 1−u 1−u
=| × {z ×···× } ext
(6) et−u et−u et−u
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
r−times
=|∞ H(r)(x|u)tn {zfor u∈C with u6=}1.
n n!
n=0
X
Inthespecialcase,x=0,H(r)(0|u)=H(r)(u)arecalledthen-thFrobenius-Euler
n n
numbers of order r (see [1-15]).
In this paper we derive non-linear differential equations from (1) and we study
the solutions of non-linear differential equations. Finally, we give some new and
interesting identities and formulae for the Frobenius-Euler polynomials of higher
order by using our non-linear differential equations.
2. Computation of sums of the products of Frobenius-Euler numbers
and polynomials
In this section we assume that
1
F =F(t)= , and FN(t,x)=F ×···×F ext for N ∈N.
(7) et−u
N−times
Thus, by (7), we get | {z }
dF(t) −et 1 u
(8) F(1) = = =− + =−F +uF2.
dt (et−u)2 et−u (et−u)2
IDENTITIES INVOLVING FROBENIUS-EULER POLYNOMIALS 3
By (8), we get
F(1)(t,x)=F(1)(t)etx =−F(t,x)+uF2(t,x),
(9)
and F(1)+F =uF2.
Let us consider the derivative of (8) with respect to t as follows:
′ ′′ ′
(10) 2uFF =F +F .
Thus, by (10) and (8), we get
(11) 2!u2F3−2uF2 =F′′ +F′.
From (11), we note that
(12) 2!u2F3 =F(2)+3F′ +2F, where F(2) = d2F.
dt2
Thus, by the derivative of (12) with respect to t, we get
(13) 2!u23F2F′ =F(3)+3F(2)+2F(1), and F(1) =uF2−F.
By (13), we see that
(14) 3!u3F4F =F(3)+6F(2)+11F(1)+6F.
Thus, from (14), we have
3!u4F4(t,x)=F(3)(t,x)+6F(2)(t,x)+11F(1)(t,x)+6F(t,x).
Continuing this process, we set
N−1
(15) (N −1)!uN−1FN = a (N)F(k),
k
k=0
X
where F(k) = dkF and N ∈N.
dtk
Now we try to find the coefficient a (N) in (15). From the derivative of (15)
k
with respect to t, we have
N−1 N
(16) N!uN−1FN−1F(1) = a (N)F(k+1) = a (N)F(k).
k k−1
k=0 k=1
X X
By (8), we easily get
(17) N!uN−1FN−1F(1) =N!uN−1FN−1(uF2−F)=N!uNFN+1−N!uN−1FN.
From (16) and (17), we can derive the following equation (18):
N
N!uNFN+1 =N(N −1)!uN−1FN + a (N)F(k)
k−1
k=1
(18) X
N−1 N
=N a (N)F(k)+ a (N)F(k).
k k−1
k=0 k=1
X X
In (15), replacing N by N +1, we have
N
(19) N!uNFN+1 = a (N +1)F(k).
k
k=0
X
4 TAEKYUNKIM
By (18) and (19), we get
N
a (N +1)F(k) =N!uNFN+1
k
k=0
(20) X
N−1 N
=N a (N)F(k)+ a (N)F(k).
k k−1
k=0 k=1
X X
By comparing coefficients on the both sides of (20), we obtain the following equa-
tions:
(21) Na0(N)=a0(N +1), aN(N +1)=aN−1(N).
For 1≤k ≤n−1, we have
(22) ak(N +1)=Nak(N)+ak−1(N),
where a (N)=0 for k ≥N or k <0. From (21), we note that
k
(23) a0(N +1)=Na0(N)=N(N −1)a0(N −1)=···=N(N −1)···2a0(2).
By (8) and (15), we get
1
(24) F +F′ =uF2 = a (2)F(k) =a (2)F +a (2)F(1).
k 0 1
k=0
X
By comparing coefficients on the both sides of (24), we get
(25) a0(2)=1, and a1(2)=1.
From(23)and(25),we havea (N)=(N−1)!. By the secondtermof(21), we see
0
that
(26) aN(N +1)=aN−1(N)=aN−2(N −1)=···=a1(2)=1.
Finally, we derive the value of a (N) in (15) from (22).
k
Let us consider the following two variable function with variables s, t:
tN
(27) g(t,s)= ak(N) sk, where |t|<1.
N!
N≥10≤k≤N−1
X X
By (22) and (27), we get
tN
a (N +1) sk
k+1 N!
N≥10≤k≤N−1
X X
tN tN
(28) = Nak+1(N +1)N!sk+ ak(N)N!sk
N≥10≤k≤N−1 N≥10≤k≤N−1
X X X X
tN
= Na (N) sk+g(t,s).
k+1 N!
N≥10≤k≤N−1
X X
IDENTITIES INVOLVING FROBENIUS-EULER POLYNOMIALS 5
It is not difficult to show that
(29)
tN 1 tN
Na (N) sk = Na (N) sk+1
k+1 N! s k+1 N!
N≥10≤k≤N−1 N≥10≤k≤N−1
X X X X
1 tN 1 tNsk a (N)tN
= a (N) sk = a (N) − 0
k k
s (N −1)! s (N −1)! (N −1)!
!
N≥11≤k≤N N≥1 0≤k≤N
X X X X
1 tN t tN−1sk 1
= a (N) sk−tN = a (N) −
k k
s (N −1)! s (N −1)! 1−t
! !
N≥1 0≤k≤N N≥10≤k≤N
X X X X
t ′ 1
= g (t,s)− .
s 1−t
(cid:16) (cid:17)
From (28) and (29), we can derive the following equation:
tNsk t ′ 1
(30) ak+1(N +1) N! = s g (t,s)− 1−t +g(t,s).
NX≥10≤kX≤N−1 (cid:16) (cid:17)
(31)
tN−1
The left hand side of (13)= a (N) sk
k+1 (N −1)!
N≥21≤k≤N−2
X X
tN−1sk−1 1 tN−1
= a (N) = a (N) sk
k k
(N −1)! s (N −1)!
!
N≥21≤k≤N−1 N≥21≤k≤N−1
X X X X
1 tN−1 tN−1
= a (N) sk−a (N)
s k (N −1)! 0 (N −1)!
!
NX≥2(cid:16)0≤kX≤N−1 (cid:17)
1 tN−1 t
= a (N) sk−
k
s (N −1)! 1−t
!
N≥20≤k≤N−1
X X
= 1 a (N) tN−1 sk−a (1)− t = 1 g′(t,s)− 1 .
s k (N −1)! 0 1−t s 1−t
!
NX≥10≤kX≤N−1 (cid:16) (cid:17)
By (30) and (31), we get
t ′ 1 1 ′ 1
(32) g(t,s)+ g (t,s)− = g (t,s)− .
s 1−t s 1−t
(cid:16) (cid:17) (cid:16) (cid:17)
Thus, by (32), we easily see that
t−1 ′ 1−t t−1 ′ 1
(33) 0=g(t,s)+ g (t,s)+ =g(t,s)+ g (t,s)+ .
s s(1−t) s s
By (33), we get
t−1 ′ 1
(34) g(t,s)+ g (t,s)=− .
s s
To solve (34), we consider the solution of the following homogeneous differential
equation:
t−1 ′
(35) 0=g(t,s)+ g (t,s).
s
6 TAEKYUNKIM
Thus, by (35), we get
t−1 ′
(36) −g(t,s)= g (t,s).
s
By (33), we get
′
g (t,s) s
(37) = .
g(t,s) 1−t
From (37), we have the following equation:
(38) logg(t,s)=−slog(1−t)+C.
By (38), we see that
(39) g(t,s)=e−slog(1−t)λ where λ=eC.
By using the variant of constant, we set
(40) λ=λ(t,s).
From (39) and (40), we note that
g′(t,s)= dg(t,s) =λ′(t,s)e−slog(1−t)+ λ(t,s)e−slog(1−t)s
dt 1−t
(41)
=λ′(t,s)e−slog(1−t)+ g(t,s)s,
1−t
where λ′(t,s)= dλ(t,s).
dt
By multiply t−1 on both sides in (41), we get
s
(42) t−1g′(t,s)+g(t,s)=λ′t−1e−slog(1−t).
s s
From (34) and (42), we get
(43) −1 =λ′t−1e−slog(1−t).
s s
Thus, by (43), we get
(44) λ′ =λ′(t,s)=(1−t)s−1.
If we take indefinite integral on both sides of (44), we get
(45) λ= λ′dt= (1−t)s−1dt=−1(1−t)s+C ,
s 1
Z Z
where C is constant.
1
By (39) and (45), we easily see that
1
(46) g(t,s)=e−slog(1−t) − (1−t)s+C .
s 1
Let us take t=0 in (46). Then, by (27) a(cid:0)nd (46), we get (cid:1)
1 1
(47) 0=− +C , C = .
s 1 1 s
IDENTITIES INVOLVING FROBENIUS-EULER POLYNOMIALS 7
Thus, by (46) and (47), we have
1 1 1
g(t,s)=e−slog(1−t) − (1−t)s = (1−t)−s 1−(1−t)s
s s s
(48)
(1−t)−s−(cid:16)1 1 (cid:17) (cid:0) (cid:1)
= = e−slog(1−t)−1 .
s s
From (48) and Taylor expansion, we(cid:0)can derive the f(cid:1)ollowing equation (49):
1 sn n sn−1 ∞ tl n
g(t,s)= −log(1−t) =
s n! n! l
!
n≥1 n≥1 l=1
X (cid:0) (cid:1) X X
sn−1 ∞ tl1 ∞ tln
(49) = ×···×
n! l l
1 n !
nX≥1 lX1=1 lXn=1
sn−1 1
= tN.
n! l l ···l
1 2 n!
nX≥1 NX≥n l1+··X·+ln=N
Thus, by (49), we get
sk 1
g(t,s)= tN
(k+1)! l l ···l
1 2 k+1!
(50) kX≥0 NX≥k+1 l1+···+Xlk+1=N
N! 1 tN
= sk.
(k+1)! l l ···l N!
1 2 k+1!
NX≥1 0≤kX≤N−1 l1+···+Xlk+1=N
From (27) and (50), we can derive the following equation (51):
N! 1
a (N)= .
(51) k
(k+1)! l l ···l
1 2 k+1
l1+···+Xlk+1=N
Therefore, by (15) and (51), we obtain the following theorem.
Theorem 1. For u ∈ C with u 6= 1, and N ∈ N, let us consider the following
non-linear differential equation with respect to t:
N−1
N 1 1
(52) FN(t)= F(k)(t),
uN−1 (k+1)! l l ···l
1 2 k+1
Xk=0 l1+···+Xlk+1=N
where F(k)(t) = dkF(t) and FN(t) = F(t)×···×F(t). Then F(t) = 1 is a
dtk et−u
N−times
solution of (52).
| {z }
Let us define F(k)(t,x)=F(k)(t)etx. Then we obtain the following corollary.
Corollary 2. For N ∈N, we set
N
N 1 1
(53) FN(t,x)= F(k)(t,x).
uN−1 (k+1)! l l ···l
1 2 k+1
Xk=0 l1+···+Xlk+1=N
Then etx is a solution of (53).
et−u
8 TAEKYUNKIM
From (1) and (6), we note that
1−u ∞ tn
= H (u) ,
et−u n n!
n=0
X
and
(54)
1−u 1−u 1−u ∞ tn
× ×···× = H(N)(u) ,
et−u et−u et−u n n!
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) nX=0
N−times
| {z }
where H(N)(u) are called the n-th Frobenius-Euler numbers of order N.
n
By (7) and (54), we get
1 1 1
FN(t)= × ×···×
et−u et−u et−u
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
N−times
1 1−u 1−u 1−u
=| {z× ×···×}
(1−u)N et−u et−u et−u
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
N−times
(55)
= 1 |∞ H(N)(u)tl, {z }
(1−u)N l l!
l=0
X
and
1−u 1 1 ∞ tl
F(t)= = H (u) .
et−u 1−u 1−u l l!
(cid:16) (cid:17)(cid:16) (cid:17) Xl=0
From (55), we note that
dkF(t) ∞ tl
(56) F(k)(t)= dtk = Hl+k(u)l!.
l=0
X
Therefore, by (52), (55) and (56), we obtain the following theorem.
Theorem 3. For N ∈N, n∈Z , we have
+
1−u N−1N−1 1 H (u)
H(N)(u)=N n+k .
n u (k+1)! l l ···l
1 2 k+1
(cid:16) (cid:17) kX=0 l1+···+Xlk+1=N
IDENTITIES INVOLVING FROBENIUS-EULER POLYNOMIALS 9
From (55), we can derive the following equation:
∞ tn 1−u 1−u 1−u
H(N)(u) = × ×···×
n n! et−u et−u et−u
nX=0 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
N−times
| ∞ tl {z ∞ } tlN
= H (u) 1 ×···× H (u)
l1 l ! lN l !
1 ! N !
(57) lX1=0 lXN=0
∞ H (u)H (u)···H (u)n! tn
= l1 l2 lN
l !l !···l ! n!
nX=0 l1+··X·+lN=0 1 1 N !
∞ n tn
= H (u)···H (u) .
l ,··· ,l ! l1 lN n!
nX=0 l1+··X·+lN=0(cid:18)1 N (cid:19) !
Therefore, by (57), we obtain the following corollary.
Corollary 4. For N ∈N, n∈Z , we have
+
n
H (u)H (u)···H (u)
l ,··· ,l ! l1 l2 lN
l1+··X·+lN=n(cid:18)1 N (cid:19)
1−u N−1N−1 1 H (u)
=N n+k .
u (k+1)! l l ···l
1 2 k+1
(cid:16) (cid:17) kX=0 l1+···+Xlk+1=N
By (53), we obtain the following corollary.
Corollary 5. For N ∈N, n∈Z , we have
+
H(N)(x|u)
n
1−u N−1N−1 1 1 n n
=N H (u)xn−m.
u (k+1)! l l ···l m m+k
(cid:16) (cid:17) kX=0 l1+···+Xlk+1=N 1 2 k+1 mX=0(cid:18) (cid:19)
From (6), we note that
∞ tn 1−u 1−u 1−u
H(N)(x|u) = × ×···× ext
n n! et−u et−u et−u
nX=0 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
N−times
(58) =| ∞ H(N)(u)tn {z ∞ xmtm }
n n! m!
! !
n=0 m=0
X X
∞ n n tn
= xn−lH(N)(u) .
l l n!
n=0 l=0(cid:18) (cid:19) !
X X
By comparing coefficients on both sides of (58), we get
n
n
(59) H(N)(x|u)= xn−lH(N)(u).
n l l
l=0(cid:18) (cid:19)
X
10 TAEKYUNKIM
By the definition of notation, we get
∞ tl ∞ xm
F(k)(t,x)=F(k)(t)etx = H (u) tm
l+k l! m!
! !
l=0 m=0
X X
∞ n n tn
= H (u)xn−l .
l l+k n!
n=0 l=0(cid:18) (cid:19) !
X X
From (6), we note that
(60)
∞ tn 1−u 1−u
H(N)(x|u) = ×···× ext
n n! et−u et−u
nX=0 (cid:16) (cid:17) (cid:16) (cid:17)
N−times
=| ∞ H (u){Hzl1(u)tl1 ×}···× ∞ HlN(u)tlN ∞ xmtm
l1 l ! l ! m!
lX1=0 1 ! lXN=0 N !mX=0
∞ H (u)H (u)···H (u) tn
= l1 l2 lN xmn!
l !l !···l !m! n!
nX=0 l1+···+XlN+m=n 1 1 N !
∞ n tn
= H (u)···H (u)xm .
l ,··· ,l ,m l1 lN n!
nX=0 l1+···+XlN+m=n(cid:18)1 N (cid:19) !
By comparing coefficients on both sides of (58), we get
n
H(N)(x|u)= H (u)···H (u)xm.
n l ,··· ,l ,m l1 lN
l1+···+XlN+m=n(cid:18)1 N (cid:19)
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