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Journal of Integer Sequences, Vol. 15 (2012),
Article 12.3.7
23 11
The Inverse of Power Series and
the Partial Bell Polynomials
Miloud Mihoubi1 and Rachida Mahdid1
Faculty of Mathematics
University of Science and Technology Houari Boumediene (USTHB)
PB 32
El Alia 16111 Algiers
Algeria
miloudmihoubi@gmail.com
mmihoubi@usthb.dz
rachida-mahdid@hotmail.fr
Abstract
Using the Bell polynomials, in this paper we give the explicit compositional inverses
and/or the reciprocals of some power series. We illustrate the obtained results by some
examples on Stirling numbers.
1
Introduction
The applications of the partial Bell polynomials have attracted the attention of several authors. Comtet [7] studied these polynomials, and Riordan [17] used them in combinatorial
analysis and Roman [18] in umbral calculus. Recently, more applications of these polynomials have appeared in different frameworks, including integration [6], inverse relations [13],
congruences [14], Dyck paths [11] and Blissard problem [10], all of which motivate us to
apply these polynomials to determine the explicit compositional inverses and/or reciprocals
of some power series.
Indeed, recall that the study of the existence of the compositional inverses of power
series is a well-known result of complex analysis; see Forsyth [9] and Stanley [19, Proposition
5.4.1]. Under some conditions on a function f, to find the compositional inverse f h−1i of
f around zero, three methods are used to compute the coefficients yn for which the series
1
This research is supported by PNR project 8/u160/3172.
1
f h−1i (t) =
P
n≥0
n
yn tn! is the compositional inverse of the series f (t) =
is based on the solution on y = (y1 , y2 , . . .) in the equation
X yk X tj k
t=
xj
.
k! j≥0 j!
k≥0
P
n
xn tn! . The first method
n≥0
If we set xn = yn = 0 if n ≤ 0, it was shown by Whittaker [20] that
1
(−1)n−1
y1 = − , yn = −
det ((1 + i − j)n + j − 1)x2+i−j
, n ≥ 2.
x1
n!x12n−1
1≤i,j≤n−1
This can be reduced to solve the equation
n
X yk X tj k X tn X
xj
=
yk Bn,k (x1 , x2 , . . .),
t=
k!
j!
n!
j≥1
n≥1
k=1
k≥1
which is equivalent to solving the system
n
X
yk Bn,k (x1 , x2 , . . .) = δn−1 ,
n ≥ 1,
k=1
where δn is the Kronicker’s symbol, i.e. δ0 = 1 and δn = 0 if n ≥ 1, and the polynomials
Bn,k (x1 , x2 , . . .) are the (exponential) partial Bell polynomials defined by their generating
function
∞
k
∞
X
1 X tm
tn
.
xm
Bn,k (x1 , x2 , . . .) =
n!
k! m=1 m!
n=k
The second method is based on Lagrange’s inversion formula [1] for which we have
n ϕ
dn−1
.
yn =
dϕn−1 f (ϕ) ϕ=0
The third method is based on the nth nested derivative of a function g defined in [8] by
d
0
n
n−1
D [g](ϕ) := 1, D [g](ϕ) :=
f (ϕ)D [f ](ϕ) , n ≥ 1,
dϕ
Rt 1
dϕ, with g(α) 6= 0, ±∞, we have
for which Dominici [8] showed that if f (t) = α g(ϕ)
f h−1i (t) = α + g(α)
X
tn
Dn [g](α) .
n!
n≥1
For our contribution, we show that the partial Bell polynomials define two families of power
series for which we can obtain explicit compositional inverses. We also give the reciprocals
of power series connected to these families. For the applications, we give some examples on
power series whose coefficients are related to Stirling and r-Stirling numbers. Indeed, let
2
r, s, d be integers with d ≥ 1 and x = (x1 , x2 , . . .) be a sequence of real numbers with x1 = 1.
For r > max(s, −2s) we consider
X ((d − 1)n)!s (r+1)dn−s
tdn
(d−1)n
H1 (t) = t 1 −
,
(1)
B(rd+1)n−s,rdn−s (x )
(rd+1)n−s
rdn − s
n!
n≥1
n
and for r > max(−s, (d + 1)s) we consider
X s B(r+1)n−s, rn−s (x ) tdn H2 (t) = t 1 −
.
(r+1)n−s
rn − s
n!
n≥1
rn−s
h−1i
(2)
h−1i
We give below the explicit compositional inverses H1
of H1 (Theorem 3) and H2
of H2
H1 (t)
t
(Theorem 16). Also, we present the explicit reciprocal power series H1 (t) of t .
The mathematical tools used are based on the connection between the partial Bell polynomials and the polynomials of binomial type. For a given real number α and for any sequence
of binomial type (fn (ϕ)), in what follows we let (fn (ϕ; α)) denote any sequence of binomial
type such that
ϕ
fn (αn + ϕ),
(3)
fn (ϕ; α) :=
αn + ϕ
see Comtet [7, pp. 133–175], Aigner [2, pp. 99–116] and Proposition 1 given in [12].
For example, the sequences (ϕn ) and (ϕ(αn + ϕ)n−1 ) are sequences of binomial type.
Below, we use the following notation: Bn,k (xj ) or Bn,k (x ) with x = (x1 , x2 , . . .) for the
partial Bell polynomial Bn,k (x1 , x2 , . . .),
dk f
(0),
dz k
k
Dz=0
f (z) for
n
k
and
n
k
2
n
and
k r
n
k ≥ 2, and Dz=0 f (z) for
df
(0),
dz
for the unsigned Stirling and r-Stirling numbers of the first kind, respectively,
k r
for the Stirling and r-Stirling numbers of the second kind, respectively.
The first family of power series and their inverses
We give in this section the reciprocal and/or the compositional inverse of a power series
given from a large family of power series which have coefficients can be expressed in terms
of partial Bell polynomials.
Proposition 1. Let x = (x1 , x2 , . . .) be a sequence of real numbers with x1 = 1. Then, for
r, s integers such that r > max(s, −2s), the reciprocal and compositional inverse of the power
series
X s B(r+1)n−s,rn−s (x) tn (4)
H(t) = t 1 −
(r+1)n−s
rn
−
s
n!
n≥1
n
3
are given by
H
h−1i
X
(t) = t 1 +
n≥1
B(r+s+1)n+s,(r+s)n+s (x) tn
s
,
(r+s+1)n+s
(r + s)n + s
n!
n
t
s B(r+1)n+s,rn+s (x) tn
=1+
.
(r+1)n+s
H(t)
rn + s
n!
n≥1
n
X
(5)
(6)
Proof. We consider only the case s ≥ 0 (for s < 0, we can proceed similarly). Let (fn (ϕ))
be a sequence of binomial-type polynomials such that fn (1) = xn+1 /(n + 1) with x2 6= 0 to
ensure that Df1 (ϕ) 6= 0. Then, by Proposition 1 given in [12], we get
−1
n+k
Bn+k,k (x ).
(7)
fn (k) =
k
In [13, Theorem 1], we proved that
X
H(t) = t 1 −
n≥1
s
tn
fn (rn − s)
,
rn − s
n!
has inverse
H
h−1i
X
(t) = t 1 +
n≥1
s
tn
,
fn ((r + s)n + s)
(r + s)n + s
n!
and because for any binomial-type sequence of polynomials (fn (ϕ)) we have
−1
X
X
tn
tn
=1+
fn (s) ,
1+
fn (−s)
n!
n!
n≥1
n≥1
then, by replacing in the last identity fn (s) by fn (s; r) defined by (3) we get
X s
tn
t
=1+
fn (rn + s) .
H(t)
rn + s
n!
n≥1
Then, replace fn (rn − s) and fn ((r + s)n + s) by their expressions obtained from the identity
(7) by
fn (rn − s) =
B(r+s+1)n+s,(r+s)n+s (x )
B(r+1)n−s,rn−s (x )
and fn ((r + s)n + s) =
.
(r+1)n−s
(r+s+1)n+s
n
n
The proposition remains true for the case x2 = 0 by continuity.
Lemma 2. Let x = (x1 , x2 , . . .) be a sequence of real numbers with x1 = 1, xn = 0 if d ∤ n−1,
and y = (y1 , y2 , . . .) with yj = j!xd(j−1)+1 /(d(j − 1) + 1)!. Then, we have
Bdn+k,k (x)
Bn+k,k (y)
=
and Bdn+k+l,k (x) = 0 if 1 ≤ l ≤ d − 1.
(dn + k)!
(n + k)!
4
Proof. Setting z = (z1 , z2 , . . .) with zj = yj+1 /(j + 1).
From the definition of partial Bell polynomials we have
l
k
X (td )j k tk X
tn
(k)l X (td )j
tk
Bn,k (x ) =
1+
zj
zj
=
n!
k!
j!
k!
l!
j!
j≥1
j≥1
n≥k
l=0
X
i.e.
min(n,k)
k
X
tdn
tk X
tk X tdn X
tn
(k)l
Bn,l (z )
=
(k)l Bn,l (z ).
Bn,k (x ) =
n!
k! l=0
n!
k! n≥0 n! l=0
n≥l
n≥k
X
Now, from the identity [3l] given in Comtet [7, pp. 136], we have
min(n,k)
X
l=0
−1
n+k
Bn+k,k (y )
(k)l Bn,l (z ) =
k
and the above expansion becomes
−1
X
X (dn + k)!
tn
1 X tdn+k n + k
tdn+k
Bn,k (x ) =
Bn+k,k (y )
.
Bn+k,k (y ) =
n!
k! n≥0 n!
(n + k)!
(dn + k)!
k
n≥0
n≥k
This gives Bdn+k,k (x ) =
(dn+k)!
Bn+k,k (y )
(n+k)!
and Bdn+k+l,k (x ) = 0 if 1 ≤ l ≤ d − 1.
Theorem 3. Let y = (y1 , y2 , . . .) be a sequence of real numbers with y1 = 1, r, s, d be integers
such that d ≥ 1, r > max(s, −2s) and let
(r+1)dn+s
((d − 1)n)!s (d−1)n
Un (r, s; d) =
B(rd+1)n+s,rdn+s (y), n ≥ 1, U0 (r, s; d) = 1. (8)
(rd+1)n+s
rdn + s
n
Then for
X
tdn
H1 (t) = t 1 +
Un (r, −s; d)
n!
n≥1
(9)
we have
h−1i
H1 (t)
X
tdn
=t 1+
Un (r + s, s; d)
,
n!
n≥1
X
t
tdn
=1+
Un (r, s; d) ,
H1 (t)
n!
n≥1
(10)
(11)
Proof. In Proposition 1, choice x = (x1 , x2 , . . .) such that xn = 0, if d ∤ n − 1 and xdn+1 =
(dn + 1)!yn+1 /(n + 1)!, after that, apply Lemma 2.
5
Example 4. For y = (1!, 2!, . . . , (q + 1)!, 0, . . .) in Theorem 3, the identity
n!
k
Bn,k (y) =
,
k! n − k q
(12)
see [3], gives
X ((d − 1)n)! (r+1)dn−s
rdn − s tdn
(d−1)n
,
H1 (t) = t 1 − s
(rd+1)n−s
n
rdn
−
s
n!
q
n≥1
n
(r+s+1)dn+s X ((d − 1)n)!
(r + s)dn + s tdn
(d−1)n
h−1i
,
H1 (t) = t 1 + s
(r + s)dn + s ((r+s)d+1)n+s
n
q n!
n≥1
n
X ((d − 1)n)! (r + 1)dn + srdn + s
t
tdn ,
=1+s
n
(d
−
1)n
H1 (t)
rdn
+
s
q
n≥1
where
k
n q
is the coefficients defined by (1 + ϕ + ϕ2 + · · · + ϕq )k =
k
ϕn .
n q
P
n≥0
n
gives
n
gives
Example 5. For y = (0!, 1!, . . .) in Theorem 16, the identity Bn,k (y) =
k
X ((d − 1)n)!s (r+1)dn−s
(rd + 1)n − s tdn
(d−1)n
H1 (t) = t 1 −
,
(rd+1)n−s
rdn − s
rdn − s
n!
n≥1
rdn−s
(r+s+1)dn+s X
((r + s)d + 1)n + s tdn
((d
−
1)n)!s
(d−1)n
h−1i
,
H1 (t) = t 1 +
((r+s)d+1)n+s
(r
+
s)dn
+
s
(r
+
s)dn
+
s
n!
n≥1
(r+s)dn+s
(r+1)dn+s X ((d − 1)n)!s (d−1)n
(rd + 1)n + s tdn
t
=1+
(rd+1)n+s
rdn + s
H1 (t)
rdn + s
n!
n≥1
rdn+s
Example 6. For y = (1, 1, . . .) in Theorem 16, the identity Bn,k (y) =
k
X ((d − 1)n)!s (r+1)dn−s
(rd + 1)n − s tdn
(d−1)n
,
H1 (t) = t 1 −
(rd+1)n−s
rdn
−
s
rdn
−
s
n!
n≥1
rdn−s
(r+s+1)dn+s X
((r + s)d + 1)n + s tdn
((d
−
1)n)!s
(d−1)n
h−1i
,
H1 (t) = t 1 +
((r+s)d+1)n+s
(r
+
s)dn
+
s
(r
+
s)dn
+
s
n!
n≥1
(r+s)dn+s
(r+1)dn+s X ((d − 1)n)!s (d−1)n
t
(rd + 1)n + s tdn
=1+
,
(rd+1)n+s
H1 (t)
rdn + s
n!
rdn + s
n≥1
rdn+s
Proposition 7. Let r, s, d be integers with s 6= 0, d ≥ 1 and (fn (ϕ)) be binomial-type
6
polynomials. Then, for r > max(s, −2s) we get
X
(r + 1)dn − s fn (αn − ϕ) tdn
H1 (t) = t 1 − ϕ
((d − 1)n)!
,
(d − 1)n
αn − ϕ n!
n≥1
X
(r + s + 1)dn + s fn ((α + sd)n + ϕ) tdn
(−1)
,
H1 (t) = t 1 + ϕ
((d − 1)n)!
(α
+
sd)n
+
ϕ
n!
(d
−
1)n
n≥1
X
t
(r + 1)dn + s fn (αn + ϕ) tdn
=1+ϕ
((d − 1)n)!
.
H1 (t)
(d − 1)n
αn + ϕ n!
n≥1
ϕ
fn (αn + ϕ) in Theorem 3 and use Proposition 1
Proof. For H1 , take yn = fn (ϕ; α) = αn+ϕ
given in [12], after that, replace α by α − rd and ϕ by ϕs .
Example 8. For fn (ϕ) = ϕn in Proposition 7 we get
dn
X
(r + 1)dn − s
n−1 t
,
H1 (t) = t 1 − ϕ
((d − 1)n)!
(αn − ϕ)
n!
(d − 1)n
n≥1
dn
X
(r + s + 1)dn + s
(−1)
n−1 t
,
H1 (t) = t 1 + ϕ
((d − 1)n)!
((α + sd)n + ϕ)
n!
(d
−
1)n
n≥1
X
t
tdn
(r + 1)dn + s
(αn + ϕ)n−1 ,
=1+ϕ
((d − 1)n)!
(d − 1)n
H1 (t)
n!
n≥1
and for fn (ϕ) = n! ϕn in Proposition 7 we get
X
(r + 1)dn − s αn − ϕ dn
,
H1 (t) = t 1 − ϕ
((d − 1)n)!
t
(d − 1)n
n
n≥1
X
(r + s + 1)dn + s (α + sd)n + ϕ dn
(−1)
,
H1 (t) = t 1 + ϕ
((d − 1)n)!
t
(d − 1)n
n
n≥1
X
t
(r + 1)dn + s αn + ϕ dn
=1+ϕ
((d − 1)n)!
t .
(d
−
1)n
n
H1 (t)
n≥1
The following theorem generalizes Theorem 3.
Theorem 9. Let x = (x1 , x2 , . . .) be a sequence of real numbers with x1 = 1 and r, s, u, v, d
be integers such that d ≥ 1, r > max(s, −2s) and u > max(|v|, −sd − v). Then for
X ((d − 1)n)!v (u + 1)n − v −1 (r + 1)dn − s
tdn
B(u+1)n−v,un−v (x)
H1 (t) = t 1 −
un
−
v
n!
un
−
v
(d
−
1)n
n≥1
we have
X ((d − 1)n)!v (r+s+1)dn+s
tdn
(d−1)n
=t 1+
,
B(u+sd+1)n+v,(u+sd)n+v (x)
(u + sd)n + v (u+sd+1)n+v
n!
n≥1
(u+sd)n+v
X ((d − 1)n)!v (u + 1)n + v −1 (r + 1)dn + s
tdn
t
=1+
B(u+1)n+v,un+v (x) .
H1 (t)
un + v
n!
un + v
(d − 1)n
n≥1
(−1)
H1 (t)
7
Proof. Let (fn (ϕ)) be a sequence of of binomial type of polynomials such that fn (1) =
xn+1 /(n + 1) with x2 6= 0 to ensure that Df1 (ϕ) 6= 0. Then, fn (ϕ) satisfies (7). Set α = u
and ϕ = v in Proposition 7, after that, use the identity (7) in the three power series of
Proposition 7, respectively, for k = un − v, k = (u + sd)n + v and k = un + v.
The theorem remains true for the case x2 = 0 by continuity.
For u = rd and v = s in Theorem 9 we obtain Theorem 3.
Example 10. Set x = (x1 , x2 , . . .) with xn = n! p+n−2
, p ≥ 1, in Theorem 9. The identity
p−1
[16, Example 13]
n! n + k(p − 1) − 1
(13)
Bn,k (x) =
k!
kp − 1
implies for the power series
X ((d − 1)n)!v (r + 1)dn − s(up + 1)n − vp − 1 H1 (t) = t 1 −
tdn
un
−
v
(d
−
1)n
upn
−
vp
−
1
n≥1
we have
X ((d − 1)n)!v (r + s + 1)dn + s((u + sd)p + 1)n + vp − 1 tdn ,
=t 1+
(u
+
sd)n
+
v
(d
−
1)n
(u
+
sd)pn
+
vp
−
1
n≥1
X ((d − 1)n)!v (r + 1)dn + s(up + 1)n + vp − 1
t
=1+
tdn .
H1 (t)
un + v
(d − 1)n
upn + vp − 1
n≥1
(−1)
H1 (t)
Example 11. Set x = (x1 , x2 , . . .) with xn = p
Theorem 9. The identity [16, Example 13]
Bn,k (x) =
kp
k
n + (p − 1)k
(p − 1)k
j+p−1 −1 p+q+j−1
,
p+q
p−1
q
p ≥ 1, q ≥ 0, in
−1 n + (p + q − 1)k
(p + q)k
kq
(14)
implies for the power series
pA n+(p+q)A
X ((r + 1)dn − s)! A
(p+q)A Aq tdn
n+pA , A = un − v,
H1 (t) = t 1 −
v
((rd + 1)n − s)! A n+A
n!
(p−1)A
A
n≥1
we have
pB n+(p+q)B
dn
X
(p+q)B
B
((r
+
s
+
1)dn
+
s)!
Bq t
(−1)
n+pB H1 (t) = t 1 +
v
, B = (u + sd)n + v,
n+B
(((r
+
s)d
+
1)n
+
s)!
n!
B
(p−1)B
B
n≥1
pC n+(p+q)C
X ((r + 1)dn + s)! C
(p+q)C Cq tdn
t
n+pC =1+
v
, C = un + v,
n+C
H1 (t)
((rd
+
1)n
+
s)
n!
C
(p−1)C
C
n≥1
8
−1 p+q+j−1
, p ≥ 1, q ≥ 0, in
Example 12. Set x = (x1 , x2 , . . .) with xn = p j+p−1
p+q
p−1
q
Theorem 9. The identity [16, Example 13]
−1 n + (p + q − 1)k
kp n + (p − 1)k
(15)
Bn,k (x) =
(p + q)k
(p − 1)k
k
kq
implies for the power series
we have
n+(p+q)A
pA
dn
X ((r + 1)dn − s)! A
(p+q)A
Aq t
n+pA H1 (t) = t 1 −
v
, A = un − v,
n+A
((rd
+
1)n
−
s)!
n!
A
(p−1)A
A
n≥1
n+(p+q)B pB
dn
X
(p+q)B
B
((r
+
s
+
1)dn
+
s)!
Bq t
(−1)
n+pB , B = (u + sd)n + v,
H1 (t) = t 1 +
v
n+B
(((r
+
s)d
+
1)n
+
s)!
n!
B
(p−1)B
B
n≥1
n+(p+q)C pC
dn
X ((r + 1)dn + s)! C
(p+q)C
t
Cq t
n+pC =1+
v
, C = un + v,
H1 (t)
((rd + 1)n + s) C n+C
n!
(p−1)C
C
n≥1
3
The second family of power series and their inverses
We give in this section the compositional inverse of a power series given from a second family
of power series which have coefficients can be expressed in terms of partial Bell polynomials.
Lemma 13. Let x = (x0 , x1 , . . .) with x0 = 1 and y = (y1 , y2 , . . .) be sequences of real
numbers with yj = jxj−1 . Then, the compositional inverse of
H(t) =
t
1+
is given by
H
h−1i
X
(t) = t 1 +
n≥1
P
n≥1
n
xn tn!
n!
n
B2n+1,n+1 (y)t .
(2n + 1)!
(16)
(17)
Proof. Let (fn (ϕ)) be a sequence of polynomials such that fn (1) = xn and assume that
−1
Bn+k,k (y). In [13, Theorem
x1 6= 0 to ensure that Df1 (ϕ) 6= 0. Then we get fn (k) = n+k
k
1], we proved that
X
X
t
tn
tn
h−1i
P
H(t) =
,
H
(t)
=
f
(n)
.
=
nf
(−1)
n
n−1
n−1
n!
n!
1 + n≥1 fn (1) tn!
n≥1
n≥1
Then, from the last identity, fn−1 (n) can be expressed by partial Bell polynomials as
−1
2n − 1
B2n−1,n (y).
fn−1 (n) =
n
The lemma holds for the case x1 = 0 by continuity.
9
Proposition 14. Let x = (x1 , x2 , . . .), y = (y1 , y2 , . . .) be sequences of real numbers with
x1 = 1, yj = jxj−1 and d be a positive integer. Then, we have the pair of compositional
inverse power series
H(t) =
t
dn ,
1 + n≥1 xn tn!
X
(dn)!
dn
h−1i
.
B(d+1)n+1,dn+1 (y)t
H
(t) = t 1 +
((d
+
1)n
+
1)!
n≥1
P
(18)
(19)
Proof. For d ≥ 2, choice in Lemma 13 xn = 0 if d ∤ n. On using the notations of Lemma 2
we get yn = nxn−1 = 0 if d ∤ n − 1 and
1+
X
xn
n≥1
X
tdn
tdn
=1+
xdn
.
n!
(dn)!
n≥1
Let z = (z1 , z2 , . . .) with zj = j!yd(j−1)+1 /(d(j − 1) + 1)!.
Then, on using Lemma 2 we obtain:
If d ∤ n − 1 then B2n+1,n+1 (y ) = 0 and if d ∤ n − 1 we get
(dn)!
(dn)!
B2dn+1,dn+1 (y ) =
B(d+1)n+1,dn+1 (z ).
(2dn + 1)!
((d + 1)n + 1)!
Therefore, the pair of the power series given in Lemma 13 can be written as
H(t) =
1+
X
t
(dn)!
h−1i
B(d+1)n+1,dn+1 (z )tdn .
,
H
(t)
=
1
+
dn /(dn)!
((d
+
1)n
+
1)!
x
t
n≥1 dn
n≥1
P
To finish this proof, replace n!xdn /(dn)! by xn .
Example 15. For x = ( 12 , 31 , 14 , . . .) in Proposition 14 we get
X 1 (d + 1)n + 1−1 (d + 1)n + 1 tdn
td+1
h−1i
, H
(t) = t
H(t) = td
,
dn
+
1
dn
+
1
dn
+
1
n!
e −1
n≥0
for x = (1!, 2!, 3!, . . .) in Proposition 14 we get
d
H(t) = t(1 − t ), H
h−1i
(t) = t
X
n≥0
(d + 1)n dn
1
t
dn + 1
dn
and for x = ( 1!2 , 2!3 , 3!4 , . . .) in Proposition 14 we get
X 1 (d + 1)n + 1−1 (d + 1)n + 1 tdn
td+1
h−1i
, H
(t) = t
.
H(t) = −
ln(1 − td )
dn
+
1
n!
dn
+
1
dn
+
1
n≥0
10
Theorem 16. Let x = (x1 , x2 , . . .) be a sequence of real numbers with x1 = 1, r, s, d be
integers such that r > max(s, −(d + 1)s), d ≥ 1 and let
Vn (r, s) =
s B(r+1)n+s, rn+s (x)
, n ≥ 1, V0 (r, s) = 1.
(r+1)n+s
rn + s
rn+s
Then, we have the pair of compositional inverses
X
tdn
H2 (t) = t 1 +
Vn (r, −s)
n!
n≥1
X
tdn
h−1i
H2 (t) = t 1 +
Vn (r + sd, s)
,
n!
n≥1
(20)
(21)
(22)
n+1
. Necessarily fn (ϕ)
Proof. Let (fn (ϕ)) be a sequence of binomial type such that fn (1) = xn+1
satisfies (7) . Then because
−1
(r + 1)n + s
s
B(r+1)n+s,rn+s (x ) = fn (s; r)
rn + s
rn + s
and
X
H(t) = t 1 +
n≥1
s B(r+1)n+s, rn+s (x ) tdn
(r+1)n+s
rn + s
n!
rn+s
−1
−1
X
tdn
=t 1+
fn (s, r)
n!
n≥1
we can state, on using Proposition 14 and Proposition 1 given [12], that
X
tdn
s
h−1i
fn ((r + sd)n + s)
H
(t) = t 1 +
(r
+
sd)n
+
s
n!
n≥1
and by the identity (7) we obtain
X
h−1i
H
(t) = t 1 +
n≥1
B(r+sd+1)n+s, (r+sd)n+s (x ) tdn
s
.
(r+sd+1)n+s
(r + sd)n + s
n!
(r+sd)n+s
It suffices to remark that from Proposition 1 we have
X B(r+1)n−s, rn−s (x ) tdn
X B(r+1)n+s, rn+s (x ) tdn −1
=
1
−
s
1+s
.
(r+1)n+s n!
(r+1)n−s n!
(rn
+
s)
(rn
−
s)
n≥1
n≥1
rn+s
rn−s
Example 17. With x = (1!, 2!, . . . , (q + 1)!, 0, . . .) in Theorem 16, the identity (12) gives
X s rn − s
dn
H2 (t) = t 1 −
,
t
n
rn
−
s
q
n≥1
X
s
(r + sd)n + s dn
h−1i
H2 (t) = t 1 +
.
t
(r + sd)n + s
n
q
n≥1
11
With x = (1!, 2!, . . .) in Theorem 16 we obtain
X s (r + 1)n − s − 1 H2 (t) = t 1 −
tdn ,
rn
−
s
−
1
rn
−
s
n≥1
X
(r + sd + 1)n + s − 1 tdn
s
h−1i
.
H2 (t) = t 1 +
(r
+
sd)n
+
s
(r
+
sd)n
+
s
−
1
n!
n≥1
With x = (1, 1, . . .) in Theorem 16 we obtain
−1 s
(r + 1)n − s
(r + 1)n − s tdn
,
rn − s
rn − s
rn − s
n!
n≥1
−1 X
s
(r + sd + 1)n + s tdn
(r + sd + 1)n + s
h−1i
H2 (t) = t 1 +
.
(r + sd)n + s
(r + sd)n + s
(r + sd)n + s
n!
n≥1
X
H2 (t) = t 1 −
With x = (0!, 1!, . . .) in Theorem 16 we obtain
−1 (r + 1)n − s tdn
s
(r + 1)n − s
,
rn
−
s
rn
−
s
rn
−
s
n!
n≥1
−1 X
(r + sd + 1)n + s tdn
(r + sd + 1)n + s
s
h−1i
.
H2 (t) = t 1 +
(r
+
sd)n
+
s
(r
+
sd)n
+
s
(r
+
sd)n
+
s
n!
n≥1
X
H2 (t) = t 1 −
Other examples can be derived by using the identities (13), (14) and (15).
Proposition 18. Let r, s, d be integers with r > max(s, −(d + 1)s), d ≥ 1 and (fn (ϕ)) be a
sequence of binomial type. Then, we have the pair of compositional inverses
X ϕ
tdn
,
fn (αn − ϕ)
H2 (t) = t 1 −
αn − ϕ
n!
n≥1
X
tdn
ϕ
h−1i
fn ((α + dϕ)n + ϕ)
.
H2 (t) = t 1 +
(α
+
dϕ)n
+
ϕ
n!
n≥1
Proof. Set xn = nfn−1 (ϕ; α) in Theorem 16 to get
X
sϕ
tdn
H2 (t) = t 1 −
fn ((α + rϕ)n − sϕ)
,
(α
+
rϕ)n
−
sϕ
n!
n≥1
X
sϕ
tdn
h−1i
H2 (t) = t 1 +
fn ((α + rϕ + sdϕ)n + sϕ)
(α + rϕ + sdϕ)n + sϕ
n!
n≥1
and change sϕ by ϕ a and α − rϕ and sϕ by ϕ.
12
Example 19. With fn (ϕ) = ϕn in Proposition 18 we get
dn
X
n−1 t
H2 (t) = t 1 − ϕ
(αn − ϕ)
,
n!
n≥1
dn
X
h−1i
n−1 t
H2 (t) = t 1 + x
((α + dϕ)n + x)
n!
n≥1
With fn (ϕ) = n! ϕn in Proposition 18 we get
X ϕ αn − ϕ tdn ,
H2 (t) = t 1 −
n
αn
−
ϕ
n≥1
X
ϕ
(α + dϕ)n + ϕ dn
h−1i
.
t
H2 (t) = t 1 +
n
(α + dϕ)n + x
n≥1
4
Consequences and complementary results
We give in this section some properties and complementary remarks on the compositional
inverse by taking particular cases of Theorems (3) and (16).
Corollary 20. Let x = (x1 , x2 , . . .) be a sequence of real numbers and r, s, d, f be integers
with d ≥ 1 and f ≥ 2. Then, for r > max(s, −2s) the inverse power series given by (9) and
(10) hold for
−1
B(rd+1)n+s−(f −1)[n/f ],rdn+s (x)
((r + 1)dn + s)! (rd + 1)n + s
Un (r, s; d) = s
,
n
rdn + s
((rd + 1)n + s − (f − 1)[n/f ])!
and for r > max(s, −(d + 1)s) the inverse power series given by (21) and (22) hold for
Vn (r, s) = (rn + s − 1)!n!s
B(r+1)n+s−(f −1)[n/f ], rn+s (x)
,
((r + 1)n + s − (f − 1)[n/f ])!
where [ϕ] is the largest integer ≤ ϕ.
Proof. Set y1 = 1, y2 = · · · = yf = 0 and n = mf + δ (0 ≤ δ ≤ f − 1) in Theorem 3 to get
Bmf +δ+k,k (yj )
Bn+k,k (yj )
=
n+k
mf +δ+k
k
k
mf +δ
=
=
=
=
k!
yj+1
Bmf +δ,i
(k − i)!
j+1
i=0
m+δ
j!yj+f
(mf + δ)! X k!
Bm+δ,i
(m + δ)! i=0 (k − i)!
(j + f )!
−1
(mf + δ)! m + δ + k
j!yj+f −1
Bm+δ+k,k
(m + δ)!
k
(j + f − 1)!
n!k!
j!yj+f −1
.
Bn+k−(f −1)[n/f ],k
(n + k − (f − 1)[n/f ])!
(j + f − 1)!
X
13
Then, for xn = n!yn+f −1 /(n + f − 1)! we obtain
−1 ((d − 1)n)!s (rd + 1)n + s
(r + 1)dn + s
Un (r, s; d) =
B(rd+1)n+s,rdn+s (yj )
rdn + s
n
(d − 1)n
−1
B(rd+1)n+s−(f −1)[n/f ],rdn+s (xj )
((r + 1)dn + s)! (rd + 1)n + s
,
=s
rdn + s
n
((rd + 1)n + s − (f − 1)[n/f ])!
B(r+1)n+s−(f −1)[n/f ], rn+s (xj )
s B(r+1)n+s, rn+s (yj )
= (rn + s − 1)!n!s
Vn (r, s) =
.
(r+1)n+s
rn + s
((r
+
1)n
+
s
−
(f
−
1)[n/f
])!
rn+s
Proposition 21. Let r, s be integers, a, x be real numbers and (fn (ϕ)) be a binomial-type
polynomials. Then, for r > max(s, −2s) the inverse power series given by (9) and (10) hold
for
((d − 1)n)!s (r + 1)dn + s
rdn+s z
Un (r, s; d) =
Dz=0
(e fn ((rdn + s)ϕ + z; α)),
rdn + s
(d − 1)n
and for r > max(s, −(d + 1)s) the inverse power series given by (21) and (22) hold for
s
Vn (r, s) =
Drn+s (ez fn ((rn + s)ϕ + z; α)).
rn + s z=0
Proof. For yn = nDz=0 (ez fn−1 (ϕ + z; α)) in Theorems 3 and 16 and use the identity given
k
in [15, Lemma 1] by Bn,k (jDz=0 (ez fj−1 (ϕ + z; α))) = nk Dz=0
(ez fn−k (kϕ + z; α)).
Theorems 3 and 16 remain true when one use a finite product of power series as follows:
Proposition 22. Let x = (x1 , x2 , . . .) be a sequence of real numbers with x1 = 1, s1 , . . . , sm ,
r, s, m be integers and Hr,s be power series defined by
X s B(r+1)n−s,rn−s (x) tn Hr,s (t) = t 1 −
, r > max(s, 0) := s+ .
(r+1)n−s
rn
−
s
n!
n≥1
n
Then, we have
m
m
Y
X
Hr,si (t)
+ +
+
t
= Hr,s (t) with r > max(s , s1 , . . . , sm ), s =
si .
t
i=1
i=1
Proof. Let (fn (ϕ)) be a sequence of polynomials of binomial type such that fn (1) = xn+1 /(n+
1) and let (f (t; r))ϕ be the exponential generating function of the sequence of binomial type
(fn (t; r)). Assume that x2 6= 0. By Proposition 1 given in [12] we get
Y
m m
m
X
Y
tn
tn X
tn
Hr,si (t) Y X
−si
=
fn (−s; r) ,
=
fn (−s; r) =
fn (−si ; r)
(f (t; r)) =
t
n!
n! n≥0
n!
i=1 n≥0
i=1
i=1
n≥0
and this is exactly
m
1−
X
n≥1
s B(r+1)n−s,rn−s (x ) tn Y Hr,si (t)
=
.
(r+1)n−s
rn − s
n!
t
i=1
n
The proposition remains true for the case x2 = 0 by continuity.
14
Corollary 23. Let y = (y1 , y2 , . . .) be a sequence of real numbers with y1 = 1, r, s, d be
integers with d ≥ 1 and let Un (r, s; d), Vn (r, s; d) be given by Theorem 3 and Theorem 16.
Then, for r > max(s, −2s) we have
n
X
Bn,k (Uj (r, −s; d))(dn + k)k−1 = Un (r + s, s; d),
k=1
n
X
Bn,k (Uj (r + s, s; d))(dn + k)k−1 = Un (r, −s; d),
k=1
and for r > max(s, −(1 + d)s) we have
n
X
Bn,k (Vj (r, −s))(dn + k)k−1 = Vn (r + sd, s),
k=1
n
X
Bn,k (Vj (r + sd, s))(dn + k)k−1 = Vn (r, −s).
k=1
Proof. From Comtet [7, pp. 151], for h(t) = t(1 +
P
dn
an tn! ), we have
n≥1
h−1i
h
n
X tdn X
with bn =
bn
(t) = t 1 +
(−1)k (dn + k)k−1 Bn,k (a1 , a2 , . . .).
n!
n≥1
k=1
Then, it suffices to combine with Theorem 3 by taking an = Un (r, −s; d) and bn = Un (r +
s, s; d) and combine with Theorem 16 by taking an = Vn (r, −s) and bn = Vn (r + sd, s).
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15
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2010 Mathematics Subject Classification: Primary 11B73; Secondary 30B10, 70H03, 05A10.
Keywords: Partial Bell polynomials; binomial-type polynomials; reciprocals and compositional inverses; Stirling numbers; binomial coefficients.
Received April 28 2011; revised version received March 15 2012. Published in Journal of
Integer Sequences, March 25 2012.
Return to Journal of Integer Sequences home page.
16