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Journal of Integer Sequences, Vol. 13 (2010),
Article 10.4.5
23 11
Partial Bell Polynomials and Inverse
Relations
Miloud Mihoubi1
USTHB
Faculty of Mathematics
P.B. 32 El Alia
16111 Algiers
Algeria
miloudmihoubi@hotmail.com
Abstract
Chou, Hsu and Shiue gave some applications of Faà di Bruno’s formula for the
characterization of inverse relations. In this paper, we use partial Bell polynomials and
binomial-type sequence of polynomials to develop complementary inverse relations.
1
Introduction
Recall that the (exponential) partial Bell polynomials Bn,k are defined by their generating
function
∞
∞
X
1 X tm k
tn
xm
(1)
Bn,k (x1 , x2 , · · · ) =
n!
k!
m!
m=1
n=k
and are given explicitly by the formula
Bn,k (x1 , x2 , . . .) =
X
π(n,k)
x 1 k1 x 2 k2
n!
··· ,
k1 !k2 ! · · · 1!
2!
where π(n, k) is the set of all nonnegative integers (k1 , k2 , . . .) such that
k1 + k2 + k3 + · · · = k and k1 + 2k2 + 3k3 + · · · = n,
1
Research supported by LAID3 Laboratory of USTHB University.
1
Comtet [3] has studied the partial and complete Bell polynomials and has given their basic
properties. Riordan [6] has shown the applications of the Bell polynomials in combinatorial
analysis and Roman [7] in umbral calculus. Chou, Hsu and Shiue [2] have used these polynomials to characterize some inverse relations. They have proved that, for any function F
having power formal series with compositional inverse F h−1i , the following inverse relations
hold
P
j
yn = nj=1 Dx=a
F (x)Bn,j (x1 , x2 , . . .),
xn =
Pn
j
h−1i
Dx=f
(x)Bn,j (y1 , y2 , . . .).
(a) F
j=1
In this paper, we link their results to those of Mihoubi [5, 6, 7] on partial Bell polynomials
and binomial-type sequence of polynomials.
2
Bell polynomials and inverse relations
Using the compositional inverse function with binomial-type sequence of polynomials, we
determine some inverse relations and the connections with the partial Bell polynomials.
Theorem 1. Let {fn (x)} be a binomial-type sequence of polynomials with exponential generating function (f (t))x . Then the compositional inverse function of
h(t) = t(f (t))x =
X
nfn−1 (x)
n≥1
is given by
hh−1i (t) =
X
fn−1 (−nx)
n≥1
tn
n!
tn
.
n!
Proof. To obtain the compositional inverse function of h it suffices to solve the equation
z = tf (z)−x . The Lagrange inversion formula ensures that the last equation has an unique
solution defined around zero by
z = hh−1i (t) =
X
n−1
Dz=0
(f (z)−nx )
n≥1
tn
tn X
=
fn−1 (−nx) .
n! n≥1
n!
Corollary 2. Let {fn (x)} be a binomial-type sequence of polynomials and let a be a real
number. Then the compositional inverse function of
h(t; a) =
X
n≥1
tn
nx
fn−1 (a(n − 1) + x)
a(n − 1) + x
n!
is given by
h−1i
h
(t; a) = −
X
n≥1
tn
nx
fn−1 (a(n − 1) − nx) .
a(n − 1) − nx
n!
2
Proof. This result follows by replacing {fn (x)} in Theorem 1, by the binomial-type sequence
of polynomials {fn (x; a)}, where
fn (x; a) =
x
fn (an + x),
an + x
(2)
see Mihoubi [5, 6, 7].
Theorem 3. Let {fn (x)} be a binomial-type sequence of polynomials and a be a real number.
Then the following inverse relations hold
P
xj
fj−1 (a(j − 1) + x)Bn,j (x1 , x2 , . . .)
yn = nj=1 a(j−1)+x
(3)
P
xj
xn = − nj=1 a(j−1)−jx
fj−1 (a(j − 1) − jx)Bn,j (y1 , y2 , . . .).
Proof. For any function F having power formal series with compositional inverse F h−1i ,
Chou, Hsu and Shiue [2, Remark 1] have proved that
P
j
yn = nj=1 Dx=a
F (x)Bn,j (x1 , x2 , . . .)
xn =
Pn
j=1
j
h−1i
Dx=f
(x)Bn,j (y1 , y2 , . . .).
(a) F
To prove (3), it suffices to take
F (t) :=
∞
X
n=1
nx
tn
fn−1 (a(n − 1) + x)
a(n − 1) + x
n!
and then use Corollary 2.
Now, let fn (x) in Theorem 3 be one of the next binomial-type sequence of polynomials
fn (x) = xn ,
fn (x) = (x)(n) := x(x − 1) · · · (x − n + 1), n ≥ 1, with (x)(o) = 1,
fn (x) = (x)(n):= x(x + 1) · · · (x + n − 1), n ≥ 1, with (x)(o) = 1,
P
x
fn (x) = n!
:= nj=o Bn,j ( 11 q , . . . , i! 1i q , . . .)(x)(j) ,
n q
P
fn (x) = Bn (x) := nj=o S(n, k)xk ,
k
where Bn (.), S(n, k) and
are, respectively, the single variable Bell polynomials, the
n q
Stirling numbers of second kind and the coefficients defined by
X k 2
q k
(1 + x + x + · · · + x ) =
xn ,
n
q
n≥0
see Belbachir, Bouroubi and Khelladi [4]. We deduce the following results:
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Corollary 4. Let a and x be real numbers. Then the following inverse relations hold:
For fn (x) = xn , we get
P
yn = nj=1 xj(a(j − 1) + x)j−2 Bn,j (x1 , x2 , . . .),
Pn
xn = −
j=1
For fn (x) = (x)(n) , we get
P
yn = nj=1
xn = −
xj(a(j − 1) − jx)
xj
(a(j
a(j−1)+x
xj
j=1 a(j−1)−jx (a(j
xj
(a(j
a(j−1)+x
xj
j=1 a(j−1)−jx (a(j
xn = −
x
For fn (x) = n!
, we get
n q
P
yn = nj=1
xn = −
For fn (x) = Bn (x), we get
P
yn = nj=1
xn = −
Example 5.
(5)
− 1) − jx)(j−1) Bn,j (y1 , y2 , . . .).
− 1) + x)(j−1) Bn,j (x1 , x2 , . . .),
Pn
(4)
Bn,j (y1 , y2 , . . .).
− 1) + x)(j−1) Bn,j (x1 , x2 , . . .),
Pn
For fn (x) = (x)(n) , we get
P
yn = nj=1
j−2
− 1) − jx)
(j−1)
Bn,j (y1 , y2 , . . .).
x(j−1)! a(j−1)+x
Bn,j (x1 , x2 , . . .),
j−1
a(j−1)+x
q
Pn
(6)
(7)
a(j−1)−jx
xj!
Bn,j (y1 , y2 , . . .).
j=1 a(j−1)−jx
j−1
q
xj
B (a(j
a(j−1)+x j−1
− 1) + x)Bn,j (x1 , x2 , . . .),
Pn
xj
j=1 a(j−1)−jx Bj−1 (a(j
(8)
− 1) − jx)Bn,j (y1 , y2 , . . .).
For a = 0, x = 1 in (5), we obtain
For xn = 2(n−2)/2 , xn =
yn =
Pn
xn =
Pn−1
j=0
1
or xn =
2
n−1
X
(2j)!
j=0
(j)!
n
j
xj xn−j
1
with x0 = ,
2
j (2j)!
j=0 (−1) (j)! Bn,j+1 (y1 , y2 , . . .).
1
2n+1
in (9), we get
(−4)n−j S(n, j + 1) = (−1)n 2n+1 .
For x1 = 1, xn = 0, n ≥ 2, in (9), we get
n+1
X
2j
j n+1
(−1)
= 0, n ≥ 0.
j
n
j=n−[n/2]
4
(9)
Take x = a = x1 = 1, x2 = 2 and xn = 0, n ≥ 3, in (5) and from the identity of Ceralosi [1]
n
X
j=1
we obtain
n
X
j=1
j!Bn,j (1!, 2!, 0, 0, . . .) = n!Fn , n ≥ 1,
(−1)j j!Bn,j (1!F1 , 2!F2 , . . .) = 0, n ≥ 3,
where Fn , n = 0, 1, 2, · · · , are the Fibonacci numbers.
Take x = 1, a = x1 = 0, x2 = 2 and xn = n!, n ≥ 3, in (5), from the identity of Ceralosi [1]
n
X
j=1
we obtain
n
X
j=1
j!Bn,j (0, 2!, 3!, . . .) = n!Fn−2 , n ≥ 2,
(−1)j−1 j!Bn,j (0, 2!F0 , 3!F1 , . . .) = n!, n ≥ 2.
Theorem 6. Let r, s be nonnegative integers, rs 6= 0, and let {un } be a sequence of real
numbers with u1 = 1. Then
−1
P
yn = s nj=1 Ujj Uj +j−1
BUj +j−1,Uj (1, u2 , u3 , . . .)Bn,j (x1 , x2 , . . .),
Uj
x1 = y1 and for n ≥ 2 we have
xn = yn − s
(10)
j Vj +j−1 −1
BVj +j−1,Vj (1, u2 , u3 , . . .)Bn,j (y1 , y2 , . . .),
j=2 Vj
Vj
Pn
where Uj = (r + 2s)(j − 1) + s and Vj = (r + s)(j − 1) − s.
Proof. Let n, r, s be nonnegative integers, nr(nr + s) ≥ 1, zn (r) :=
B(r+1)n,nr (1,u2 ,u3 ,...)
nr((r+1)n
nr )
, and
consider the binomial-type sequence of polynomials {fn (x)} defined by
fn (x) :=
n
X
Bn,j (z1 (r), z2 (r), . . .)xj with fo (x) = 1,
j=1
see Roman [8]. Then from the identity
−1
(r + 1)n + s
s
B(r+1)n+s,nr+s (1, u2 , u3 , . . .),
j=1 Bn,j (z1 (r), z2 (r), . . .)s =
nr + s
nr + s
Pn
j
see Mihoubi [5, 6, 7], we get
−1
s
(r + 1)n + s
fn (s) =
B(r+1)n+s,nr+s (1, u2 , u3 , . . .).
nr + s
nr + s
(11)
To obtain (10), we set a = 0, x = s in (3) and use the expression of fn (s) given by (11),
with r + 2s instead of r.
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Example 7.
From the well-known identity Bn,k (1!, 2!, . . . , i!, . . .) =
yn = s
Pn
j=1
n−1 n!
,
k−1 k!
j! ((r+2s+1)(j−1)+s−1)!
Bn,j (x1 , x2 , . . .),
((r+2s)(j−1)+s)!
we get
x1 = y1 and for n ≥ 2 we have
xn = yn − s
Pn
j=2
j! ((r+s+1)(j−1)+s−1)!
Bn,j (y1 , y2 , . . .).
((r+s)(j−1)+s)!
Similar relations can be obtained for the Stirling numbers of the first kind, the unsigned
Stirling numbers of the first kind and the Stirling numbers of the second kind by setting
un = (−1)n−1 (n − 1)!, un = (n − 1)! and un = 1 for all n ≥ 1, respectively.
Corollary 8. Let u, r, s be nonnegative integers, a, α be real numbers, αurs 6= 0, and {fn (x)}
be a binomial-type sequence of polynomials. Then
P
Tj
j
(eαz fj−1 (Tj x + z; a))Bn,j (x1 , x2 , . . .),
Dz=0
yn = s nj=1 Tj −u(j−1)
α
Tj
x1 = y1 and for n ≥ 2 we have
xn = yn − s
Rj
j
αz
j=2 αRj −u(j−1) Rj Dz=0 (e fj−1 (Rj x
Pn
+ z; a))Bn,j (y1 , y2 , . . .),
where Tj = (u + r + 2s)(j − 1) + s and Rj = (u + r + s)(j − 1) − s.
Proof. Set in Theorem 6
un =
n
u(n−1)+1 αz
D
(e fn−1 ((u(n − 1) + 1)x + z; a))
(u(n − 1) + 1)α z=0
and use the first identity of Mihoubi [6, Theorem 2].
Corollary 9. Let u, r, s be nonnegative integers, urs 6= 0, a be real number and let {fn (x)}
be a binomial-type sequence of polynomials. Then
P
Tj
yn = s nj=1 Tj −u(j−1)j!
Dz=0
fTj +j−1 (Tj x + z; a)Bn,j (x1 , x2 , . . .),
α
(Tj +j−1)!Tj
x1 = y1 and for n ≥ 2 we have
xn = yn − s
Rj
j!
j=2 αRj −u(j−1) (Rj +j−1)!Rj Dz=0 fRj +j−1 (Rj x
Pn
+ z; a)Bn,j (y1 , y2 , . . .),
where Tj = (u + r + 2s)(j − 1) + s and Rj = (u + r + s)(j − 1) − s.
Proof. Set in Theorem 6
u(n−1)+1
n!Dz=0
f((u+1)(n−1)+1) ((u(n − 1) + 1)x + z; a)
un =
((u + 1)(n − 1) + 1)!(u(n − 1) + 1)α
and use the second identity of Mihoubi [6, Theorem 2].
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Theorem 10. Let d be an integer ≥ 1. The inverse relations hold
yn =
xn =
n
P
(−1)j (dn + j)(j−1) Bn,j (x1 , x2 , . . .),
j=1
n
P
(−1)j (dn + j)(j−1) Bn,j (y1 , y2 , . . .).
j=1
Proof. Let
f (t) = t 1 +
X tdn X tdn xn
and f h−1i (t) = t 1 +
yn
.
n!
n!
n≥1
n≥1
The proof now follows from Comtet [3, Theorem F, p. 151].
Take d = 1 and xn = n!, n ≥ 1, in Theorem 10, we get
Example 11.
f (t) =
t
t
and f h−1i (t) =
,
1−t
1+t
i.e. yn = (−1)n n!, and the relations (12) give
n
X
n+j
n−j n
= n.
(−1)
j
n
+
1
j=1
Take d = 2 and xn = n!, n ≥ 1, we get
f (t) =
i.e. yn = (−1)n
√
t
1
h−1i
2 ,
and
f
(t)
=
−1
+
1
+
4t
1 − t2
2t
(2n)!
, n ≥ 1, and the relations (12) give
(n + 1)!
n
X
n
2n + j
2n
n−j n
=
(−1)
= nCn ,
j
2n + 1
n+1 n
j=1
where Cn , n = 0, 1, 2, · · · , are the Catalan numbers.
Theorem 12. The following inverse relations hold
1 (r+1)n −1
B(r+1)n,nr (1, x1 , x2 , . . .), r ≥ 1,
yn = nr
nr
P
xn = (n + 1) nj=1 Bn, j (y1 , y2 , . . .)(−1)j−1 (nr − 1)j−1 .
Proof. From Mihoubi [7, Theorem 1] we have
xk1
n
X
j=1
with yn =
Bn,j (y1 , y2 , . . .)(k − nr)
j−1
−1
n+k
xnr
1
=
Bn+k,k (x1 , x2 , x3 , . . .),
k
k
1 (r+1)n −1
B(r+1)n,nr (1, x1 , x2 , . . .),
nr
nr
nrk ≥ 1.
It just suffices to set k = 1, x1 = 1, and replace xn by xn−1 .
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(12)
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Acknowledgments
The author thanks the anonymous referee for his/her careful reading and valuable suggestions. He also thanks Professor Benaissa Larbi for his English corrections.
References
[1] M. Cerasoli, Two identities between Bell polynomials and Fibonacci numbers. Boll. Un.
Mat. Ital. A, (5) 18 (1981), 387–394.
[2] W. S. Chou, L. C. Hsu, P. J. S. Shiue, Application of Faà di Bruno’s formula in characterization of inverse relations. J. Comput. Appl. Math 190 (2006), 151–169.
[3] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, 1974.
[4] H. Belbachir, S. Bouroubi, A. Khelladi, Connection between ordinary multinomials, generalized Fibonacci numbers, partial Bell partition polynomials and convolution powers
of discrete uniform distribution. Ann. Math. Inform. 35 (2008), 21–30.
[5] M. Mihoubi, Bell polynomials and binomial type sequences. Discrete Math. 308 (2008),
2450–2459.
[6] M. Mihoubi, The role of binomial type sequences in determination identities for Bell polynomials. To appear, Ars Combin. Preprint available at
http://arxiv.org/abs/0806.3468v1.
[7] M. Mihoubi, Some congruences for the partial Bell polynomials. J. Integer Seq. 12
(2009), Article 09.4.1.
[8] S. Roman, The Umbral Calculus, 1984.
2010 Mathematics Subject Classification: Primary 05A10, 05A99; Secondary 11B73, 11B75.
Keywords: inverse relations, partial Bell polynomials, binomial-type sequence of polynomials.
Received December 5 2009; revised version received April 4 2010. Published in Journal of
Integer Sequences, April 5 2010.
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