The Inverse of Power Series and the Partial Bell Polynomials

Miloud Mihoubi^{1} and Rachida Mahdid^{1} 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^{⟨−1⟩} of f around zero, three methods are used to compute the coefficients y_{n} for which the series f^{⟨−1⟩}(t) = ∑_{n≥0}y_{n} \frac{t^{n}}{n!} is the compositional inverse of the series f(t) = ∑_{n≥0}x_{n} \frac{t^{n}}{n!}. The first method

1This research is supported by PNR project 8/u160/3172.

is based on the solution on y = (y_{1}, y_{2}, …) in the equation

t = ∑_{k≥0} \frac{y_{k}}{k!} (∑_{j≥0}x_{j} \frac{t^{j}}{j!})^{k}.

If we set x_{n} = y_{n} = 0 if n ≤ 0, it was shown by Whittaker [20] that

y_{1} = −\frac{1}{x_{1}} , y_{n} = −\frac{(−1)^{n−1}}{n!x_{1}^{2n−1}} det(((1 + i − j)n + j − 1)x_{2+i−j})_{1≤i,j≤n−1}, n ≥ 2.

This can be reduced to solve the equation

t = ∑_{k≥1} \frac{y_{k}}{k!} (∑_{j≥1}x_{j} \frac{t^{j}}{j!})^{k} = ∑_{n≥1} \frac{t^{n}}{n!}∑^{n} ; k=1y_{k}B_{n,k}(x_{1}, x_{2}, …),

which is equivalent to solving the system

∑^{n} ; k=1y_{k}B_{n,k}(x_{1}, x_{2}, …) = δ_{n−1}, n ≥ 1,

where δ_{n} is the Kronicker’s symbol, i.e. δ_{0} = 1 and δ_{n} = 0 if n ≥ 1, and the polynomials B_{n,k}(x_{1}, x_{2}, …) are the (exponential) partial Bell polynomials defined by their generating function

∑^{∞} ; n=kB_{n,k}(x_{1}, x_{2}, …)\frac{t^{n}}{n!} = \frac{1}{k!}(∑^{∞} ; m=1x_{m} \frac{t^{m}}{m!})^{k}.

The second method is based on Lagrange’s inversion formula [1] for which we have

y_{n} = \frac{d^{n−1}}{dφ^{n−1}} (\frac{φ}{f(φ)})^{n}|_{φ=0}.

The third method is based on the n^{th} nested derivative of a function g defined in [8] by

D^{0}[g](φ) := 1, D^{n}[g](φ) := \frac{d}{dφ}(f(φ)D^{n−1}[f](φ)), n ≥ 1,

for which Dominici [8] showed that if f(t) = ∫_{α}^{t} \frac{1}{g(φ)}dφ, with g(α) ≠ 0, ±∞, we have

f^{⟨−1⟩}(t) = α + g(α)∑_{n≥1}D^{n}[g](α)\frac{t^{n}}{n!}.

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 r, s, d be integers with d ≥ 1 and x = (x_{1}, x_{2}, …) be a sequence of real numbers with x_{1} = 1. For r > max(s, −2s) we consider

H_{1}(t) = t(1 − ∑_{n≥1} \frac{((d − 1)n)!s}{rdn − s} \begin{matrix}( & {(r+1)dn−s \atop (d−1)n} & ) \\ ( & {(rd+1)n−s \atop n} & ) \end{matrix}B_{(rd+1)n−s,rdn−s}(x )\frac{t^{dn}}{n!} ),  (1)

and for r > max(−s, (d + 1)s) we consider

H_{2}(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s} \frac{B_{(r+1)n−s, rn−s}(x )}{\binom{(r+1)n−s}{rn−s}} \frac{t^{dn}}{n!} ).  (2)

We give below the explicit compositional inverses H_{1}^{⟨−1⟩} of H_{1} (Theorem 3) and H_{2}^{⟨−1⟩} of H_{2} (Theorem 16). Also, we present the explicit reciprocal power series \frac{t}{H_{1}(t)} of \frac{H_{1}(t)}{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 (f_{n}(φ)), in what follows we let (f_{n}(φ; α)) denote any sequence of binomial type such that

f_{n}(φ; α) := \frac{φ}{αn + φ}f_{n}(αn + φ),  (3)

see Comtet [7, pp. 133–175], Aigner [2, pp. 99–116] and Proposition 1 given in [12]. For example, the sequences (φ^{n}) and (φ(αn + φ)^{n−1}) are sequences of binomial type. Below, we use the following notation: B_{n,k}(x_{j}) or B_{n,k}(x ) with x = (x_{1}, x_{2}, …) for the

partial Bell polynomial B_{n,k}(x_{1}, x_{2}, …),

D_{z=0}^{k}f(z) for \frac{d^{k}f}{dz^{k}} (0), k ≥ 2, and D_{z=0}f(z) for \frac{df}{dz}(0),

\left[{n \atop k}\right] and \left[{n \atop k}\right]_{r} for the unsigned Stirling and r-Stirling numbers of the first kind, respectively,

\left\{{n \atop k}\right\} and \left\{{n \atop k}\right\}_{r} for the Stirling and r-Stirling numbers of the second kind, respectively.

2 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 = (x_{1}, x_{2}, …) be a sequence of real numbers with x_{1} = 1. Then, for r, s integers such that r > max(s, −2s), the reciprocal and compositional inverse of the power series

H(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s} \frac{B_{(r+1)n−s,rn−s}(x)}{\binom{(r+1)n−s}{n}} \frac{t^{n}}{n!})  (4)

are given by

H^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + s)n + s} \frac{B_{(r+s+1)n+s,(r+s)n+s}(x)}{\binom{(r+s+1)n+s}{n}} \frac{t^{n}}{n!}),  (5)

\frac{t}{H(t)} = 1 + ∑_{n≥1} \frac{s}{rn + s} \frac{B_{(r+1)n+s,rn+s}(x)}{\binom{(r+1)n+s}{n}} \frac{t^{n}}{n!}.  (6)

Proof. We consider only the case s ≥ 0 (for s < 0, we can proceed similarly). Let (f_{n}(φ)) be a sequence of binomial-type polynomials such that f_{n}(1) = x_{n+1}/(n + 1) with x_{2} ≠ 0 to ensure that Df_{1}(φ) ≠ 0. Then, by Proposition 1 given in [12], we get

f_{n}(k) = \binom{n + k}{k}^{−1}B_{n+k,k}(x ).  (7)

In [13, Theorem 1], we proved that

H(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s}f_{n}(rn − s)\frac{t^{n}}{n!}),

has inverse

H^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + s)n + s}f_{n}((r + s)n + s)\frac{t^{n}}{n!}),

and because for any binomial-type sequence of polynomials (f_{n}(φ)) we have

(1 + ∑_{n≥1} f_{n}(−s)\frac{t^{n}}{n!})^{−1} = 1 + ∑_{n≥1} f_{n}(s)\frac{t^{n}}{n!},

then, by replacing in the last identity f_{n}(s) by f_{n}(s; r) defined by (3) we get

\frac{t}{H(t)} = 1 + ∑_{n≥1} \frac{s}{rn + s}f_{n}(rn + s)\frac{t^{n}}{n!}.

Then, replace f_{n}(rn − s) and f_{n}((r + s)n + s) by their expressions obtained from the identity (7) by

f_{n}(rn − s) = \frac{B_{(r+1)n−s,rn−s}(x )}{\binom{(r+1)n−s}{n}} and f_{n}((r + s)n + s) = \frac{B_{(r+s+1)n+s,(r+s)n+s}(x )}{\binom{(r+s+1)n+s}{n}}.

The proposition remains true for the case x_{2} = 0 by continuity.

Lemma 2. Let x = (x_{1}, x_{2}, …) be a sequence of real numbers with x_{1} = 1, x_{n} = 0 if d ∤ n−1, and y = (y_{1}, y_{2}, …) with y_{j} = j!x_{d(j−1)+1}/(d(j − 1) + 1)!. Then, we have

\frac{B_{dn+k,k}(x)}{(dn + k)!} = \frac{B_{n+k,k}(y)}{(n + k)!} and B_{dn+k+l,k}(x) = 0 if 1 ≤ l ≤ d − 1.

Proof. Setting z = (z_{1}, z_{2}, …) with z_{j} = y_{j+1}/(j + 1). From the definition of partial Bell polynomials we have

∑_{n≥k} B_{n,k}(x )\frac{t^{n}}{n!} = \frac{t^{k}}{k!}(1 + ∑_{j≥1} z_{j} \frac{(t^{d})^{j}}{j!} )^{k} = \frac{t^{k}}{k!}∑^{k} ; l=0 \frac{(k)_{l}}{l!} (∑_{j≥1} z_{j} \frac{(t^{d})^{j}}{j!} )^{l}

i.e.

∑_{n≥k} B_{n,k}(x )\frac{t^{n}}{n!} = \frac{t^{k}}{k!}∑^{k} ; l=0(k)_{l} ∑_{n≥l} B_{n,l}(z )\frac{t^{dn}}{n!} = \frac{t^{k}}{k!} ∑_{n≥0} \frac{t^{dn}}{n!} ∑^{min(n,k)} ; l=0(k)_{l}B_{n,l}(z ).

Now, from the identity [3l] given in Comtet [7, pp. 136], we have

∑^{min(n,k)} ; l=0(k)_{l}B_{n,l}(z ) = \binom{n + k}{k}^{−1}B_{n+k,k}(y)

and the above expansion becomes

∑_{n≥k} B_{n,k}(x )\frac{t^{n}}{n!} = \frac{1}{k!} ∑_{n≥0} \frac{t^{dn+k}}{n!} \binom{n + k}{k}^{−1}B_{n+k,k}(y) = ∑_{n≥0} \frac{(dn + k)!}{(n + k)!}B_{n+k,k}(y)\frac{t^{dn+k}}{(dn + k)!}.

This gives B_{dn+k,k}(x ) = \frac{(dn+k)!}{(n+k)!}B_{n+k,k}(y) and B_{dn+k+l,k}(x ) = 0 if 1 ≤ l ≤ d − 1.

Theorem 3. Let y = (y_{1}, y_{2}, …) be a sequence of real numbers with y_{1} = 1, r, s, d be integers such that d ≥ 1, r > max(s, −2s) and let

U_{n}(r, s; d) = \frac{((d − 1)n)!s}{rdn + s} \begin{matrix}( & {(r+1)dn+s \atop (d−1)n} & ) \\ ( & {(rd+1)n+s \atop n} & ) \end{matrix}B_{(rd+1)n+s,rdn+s}(y), n ≥ 1, U_{0}(r, s; d) = 1. (8)

Then for

H_{1}(t) = t(1 + ∑_{n≥1} U_{n}(r, −s; d)\frac{t^{dn}}{n!} )  (9)

we have

H_{1}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} U_{n}(r + s, s; d)\frac{t^{dn}}{n!} ),  (10)

\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} U_{n}(r, s; d)\frac{t^{dn}}{n!} ,  (11)

Proof. In Proposition 1, choice x = (x_{1}, x_{2}, …) such that x_{n} = 0, if d ∤ n − 1 and x_{dn+1} = (dn + 1)!y_{n+1}/(n + 1)!, after that, apply Lemma 2.

Example 4. For y = (1!, 2!, …, (q + 1)!, 0, …) in Theorem 3, the identity

B_{n,k}(y) = \frac{n!}{k!}\binom{k}{n − k}_{q},  (12)

see [3], gives

H_{1}(t) = t(1 − s ∑_{n≥1} \frac{((d − 1)n)!}{rdn − s} \begin{matrix}( & {(r+1)dn−s \atop (d−1)n} & ) \\ ( & {(rd+1)n−s \atop n} & ) \end{matrix}\binom{rdn − s}{n}_{q} \frac{t^{dn}}{n!} ),
H_{1}^{⟨−1⟩}(t) = t(1 + s ∑_{n≥1} \frac{((d − 1)n)!}{(r + s)dn + s} \begin{matrix}( & {(r+s+1)dn+s \atop (d−1)n} & ) \\ ( & {((r+s)d+1)n+s \atop n} & ) \end{matrix}\binom{(r + s)dn + s}{n}_{q} \frac{t^{dn}}{n!} ),
\frac{t}{H_{1}(t)} = 1 + s ∑_{n≥1} \frac{((d − 1)n)!}{rdn + s}\binom{(r + 1)dn + s}{(d − 1)n}\binom{rdn + s}{n}_{q}t^{dn},

where \binom{k}{n}_{q} is the coefficients defined by (1 + φ + φ^{2} + ⋯ + φ^{q})^{k} = ∑_{n≥0} \binom{k}{n}_{q}φ^{n}. Example 5. For y = (0!, 1!, …) in Theorem 16, the identity B_{n,k}(y) = \left[{n \atop k}\right] gives

H_{1}(t) = t(1 − ∑_{n≥1} \frac{((d − 1)n)!s}{rdn − s} \begin{matrix}( & {(r+1)dn−s \atop (d−1)n} & ) \\ ( & {(rd+1)n−s \atop rdn−s} & ) \end{matrix}\left[{(rd + 1)n − s \atop rdn − s}\right]\frac{t^{dn}}{n!} ),
H_{1}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{((d − 1)n)!s}{(r + s)dn + s} \begin{matrix}( & {(r+s+1)dn+s \atop (d−1)n} & ) \\ ( & {((r+s)d+1)n+s \atop (r+s)dn+s} & ) \end{matrix}\left[{((r + s)d + 1)n + s \atop (r + s)dn + s}\right]\frac{t^{dn}}{n!} ),
\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} \frac{((d − 1)n)!s}{rdn + s} \begin{matrix}( & {(r+1)dn+s \atop (d−1)n} & ) \\ ( & {(rd+1)n+s \atop rdn+s} & ) \end{matrix}\left[{(rd + 1)n + s \atop rdn + s}\right]\frac{t^{dn}}{n!}

Example 6. For y = (1, 1, …) in Theorem 16, the identity B_{n,k}(y) = \left\{{n \atop k}\right\} gives

H_{1}(t) = t(1 − ∑_{n≥1} \frac{((d − 1)n)!s}{rdn − s} \begin{matrix}( & {(r+1)dn−s \atop (d−1)n} & ) \\ ( & {(rd+1)n−s \atop rdn−s} & ) \end{matrix}\left\{{(rd + 1)n − s \atop rdn − s}\right\}\frac{t^{dn}}{n!} ),
H_{1}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{((d − 1)n)!s}{(r + s)dn + s} \begin{matrix}( & {(r+s+1)dn+s \atop (d−1)n} & ) \\ ( & {((r+s)d+1)n+s \atop (r+s)dn+s} & ) \end{matrix}\left\{{((r + s)d + 1)n + s \atop (r + s)dn + s}\right\}\frac{t^{dn}}{n!} ),
\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} \frac{((d − 1)n)!s}{rdn + s} \begin{matrix}( & {(r+1)dn+s \atop (d−1)n} & ) \\ ( & {(rd+1)n+s \atop rdn+s} & ) \end{matrix}\left\{{(rd + 1)n + s \atop rdn + s}\right\}\frac{t^{dn}}{n!} ,

Proposition 7. Let r, s, d be integers with s ≠ 0, d ≥ 1 and (f_{n}(φ)) be binomial-type polynomials. Then, for r > max(s, −2s) we get

H_{1}(t) = t(1 − φ ∑_{n≥1}((d − 1)n)!\binom{(r + 1)dn − s}{(d − 1)n}\frac{f_{n}(αn − φ)}{αn − φ} \frac{t^{dn}}{n!} ),
H_{1}^{(−1)}(t) = t(1 + φ ∑_{n≥1}((d − 1)n)!\binom{(r + s + 1)dn + s}{(d − 1)n}\frac{f_{n}((α + sd)n + φ)}{(α + sd)n + φ} \frac{t^{dn}}{n!} ),
\frac{t}{H_{1}(t)} = 1 + φ ∑_{n≥1}((d − 1)n)!\binom{(r + 1)dn + s}{(d − 1)n}\frac{f_{n}(αn + φ)}{αn + φ} \frac{t^{dn}}{n!} .

Proof. For H_{1}, take y_{n} = f_{n}(φ; α) = \frac{φ}{αn+φ}f_{n}(αn + φ) in Theorem 3 and use Proposition 1 given in [12], after that, replace α by α − rd and φ by \frac{φ}{s}. Example 8. For f_{n}(φ) = φ^{n} in Proposition 7 we get

H_{1}(t) = t(1 − φ ∑_{n≥1}((d − 1)n)!\binom{(r + 1)dn − s}{(d − 1)n}(αn − φ)^{n−1} \frac{t^{dn}}{n!} ),
H_{1}^{(−1)}(t) = t(1 + φ ∑_{n≥1}((d − 1)n)!\binom{(r + s + 1)dn + s}{(d − 1)n}((α + sd)n + φ)^{n−1} \frac{t^{dn}}{n!} ),
\frac{t}{H_{1}(t)} = 1 + φ ∑_{n≥1}((d − 1)n)!\binom{(r + 1)dn + s}{(d − 1)n}(αn + φ)^{n−1} \frac{t^{dn}}{n!} ,

and for f_{n}(φ) = n!\binom{φ}{n} in Proposition 7 we get

H_{1}(t) = t(1 − φ ∑_{n≥1}((d − 1)n)!\binom{(r + 1)dn − s}{(d − 1)n}\binom{αn − φ}{n}t^{dn}),
H_{1}^{(−1)}(t) = t(1 + φ ∑_{n≥1}((d − 1)n)!\binom{(r + s + 1)dn + s}{(d − 1)n}\binom{(α + sd)n + φ}{n}t^{dn}),
\frac{t}{H_{1}(t)} = 1 + φ ∑_{n≥1}((d − 1)n)!\binom{(r + 1)dn + s}{(d − 1)n}\binom{αn + φ}{n}t^{dn}.

The following theorem generalizes Theorem 3. Theorem 9. Let x = (x_{1}, x_{2}, …) be a sequence of real numbers with x_{1} = 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

H_{1}(t) = t(1 − ∑_{n≥1} \frac{((d − 1)n)!v}{un − v} \binom{(u + 1)n − v}{un − v}^{−1}\binom{(r + 1)dn − s}{(d − 1)n}B_{(u+1)n−v,un−v}(x)\frac{t^{dn}}{n!} )

we have

H_{1}^{(−1)}(t) = t(1 + ∑_{n≥1} \frac{((d − 1)n)!v}{(u + sd)n + v} \begin{matrix}( & {(r+s+1)dn+s \atop (d−1)n} & ) \\ ( & {(u+sd+1)n+v \atop (u+sd)n+v} & ) \end{matrix}B_{(u+sd+1)n+v,(u+sd)n+v}(x)\frac{t^{dn}}{n!} ),
\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} \frac{((d − 1)n)!v}{un + v} \binom{(u + 1)n + v}{un + v}^{−1}\binom{(r + 1)dn + s}{(d − 1)n}B_{(u+1)n+v,un+v}(x)\frac{t^{dn}}{n!} .

Proof. Let (f_{n}(φ)) be a sequence of of binomial type of polynomials such that f_{n}(1) = x_{n+1}/(n + 1) with x_{2} ≠ 0 to ensure that Df_{1}(φ) ≠ 0. Then, f_{n}(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 x_{2} = 0 by continuity.

For u = rd and v = s in Theorem 9 we obtain Theorem 3.

Example 10. Set x = (x_{1}, x_{2}, …) with x_{n} = n!\binom{p+n−2}{p−1}, p ≥ 1, in Theorem 9. The identity [16, Example 13]

B_{n,k}(x) = \frac{n!}{k!}\binom{n + k(p − 1) − 1}{kp − 1}  (13)

implies for the power series

H_{1}(t) = t(1 − ∑_{n≥1} \frac{((d − 1)n)!v}{un − v} \binom{(r + 1)dn − s}{(d − 1)n}\binom{(up + 1)n − vp − 1}{upn − vp − 1}t^{dn})

we have

H_{1}^{(−1)}(t) = t(1 + ∑_{n≥1} \frac{((d − 1)n)!v}{(u + sd)n + v} \binom{(r + s + 1)dn + s}{(d − 1)n}\binom{((u + sd)p + 1)n + vp − 1}{(u + sd)pn + vp − 1}t^{dn}),

\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} \frac{((d − 1)n)!v}{un + v} \binom{(r + 1)dn + s}{(d − 1)n}\binom{(up + 1)n + vp − 1}{upn + vp − 1}t^{dn}.

Example 11. Set x = (x_{1}, x_{2}, …) with x_{n} = p\binom{j+p−1}{p−1}^{−1}\left[{p+q+j−1 \atop p+q}\right]_{q}, p ≥ 1, q ≥ 0, in Theorem 9. The identity [16, Example 13]

B_{n,k}(x) = \binom{kp}{k}\binom{n + (p − 1)k}{(p − 1)k}^{−1}\left[{n + (p + q − 1)k \atop (p + q)k}\right]_{kq}  (14)

implies for the power series

H_{1}(t) = t(1 − ∑_{n≥1} v \frac{((r + 1)dn − s)!}{((rd + 1)n − s)!} \frac{\binom{pA}{A}\left[{n+(p+q)A \atop (p+q)A}\right]Aq}{A\binom{n+A}{A}\binom{n+pA}{(p−1)A}} \frac{t^{dn}}{n!} ), A = un − v,

we have

H_{1}^{(−1)}(t) = t(1 + ∑_{n≥1} v \frac{((r + s + 1)dn + s)!}{(((r + s)d + 1)n + s)!} \frac{\binom{pB}{B}\left[{n+(p+q)B \atop (p+q)B}\right]Bq}{B\binom{n+B}{B}\binom{n+pB}{(p−1)B}} \frac{t^{dn}}{n!} ), B = (u + sd)n + v,

\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} v \frac{((r + 1)dn + s)!}{((rd + 1)n + s)} \frac{\binom{pC}{C}\left[{n+(p+q)C \atop (p+q)C}\right]Cq}{C\binom{n+C}{C}\binom{n+pC}{(p−1)C}} \frac{t^{dn}}{n!} , C = un + v,

Example 12. Set x = (x_{1}, x_{2}, …) with x_{n} = p\binom{j+p−1}{p−1}^{−1}\left\{{p+q+j−1 \atop p+q}\right\}_{q}, p ≥ 1, q ≥ 0, in Theorem 9. The identity [16, Example 13]

B_{n,k}(x) = \binom{kp}{k}\binom{n + (p − 1)k}{(p − 1)k}^{−1}\left\{{n + (p + q − 1)k \atop (p + q)k}\right\}_{kq}  (15)

implies for the power series

H_{1}(t) = t(1 − ∑_{n≥1} v \frac{((r + 1)dn − s)!}{((rd + 1)n − s)!} \frac{\binom{pA}{A}\left\{{n+(p+q)A \atop (p+q)A}\right\}Aq}{A\binom{n+A}{A}\binom{n+pA}{(p−1)A}} \frac{t^{dn}}{n!} ), A = un − v,

we have

H_{1}^{(−1)}(t) = t(1 + ∑_{n≥1} v \frac{((r + s + 1)dn + s)!}{(((r + s)d + 1)n + s)!} \frac{\binom{pB}{B}\left\{{n+(p+q)B \atop (p+q)B}\right\}Bq}{B\binom{n+B}{B}\binom{n+pB}{(p−1)B}} \frac{t^{dn}}{n!} ), B = (u + sd)n + v,

\frac{t}{H_{1}(t)} = 1 + ∑_{n≥1} v \frac{((r + 1)dn + s)!}{((rd + 1)n + s)} \frac{\binom{pC}{C}\left\{{n+(p+q)C \atop (p+q)C}\right\}Cq}{C\binom{n+C}{C}\binom{n+pC}{(p−1)C}} \frac{t^{dn}}{n!} , C = un + v,

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 = (x_{0}, x_{1}, …) with x_{0} = 1 and y = (y_{1}, y_{2}, …) be sequences of real numbers with y_{j} = jx_{j−1}. Then, the compositional inverse of

H(t) = \frac{t}{1 + ∑_{n≥1} x_{n} \frac{t^{n}}{n!}}  (16)

is given by

H^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{n!}{(2n + 1)!}B_{2n+1,n+1}(y)t^{n}).  (17)

Proof. Let (f_{n}(φ)) be a sequence of polynomials such that f_{n}(1) = x_{n} and assume that x_{1} ≠ 0 to ensure that Df_{1}(φ) ≠ 0. Then we get f_{n}(k) = \binom{n+k}{k}^{−1}B_{n+k,k}(y). In [13, Theorem 1], we proved that

H(t) = \frac{t}{1 + ∑_{n≥1} f_{n}(1)\frac{t^{n}}{n!}} = ∑_{n≥1} nf_{n−1}(−1)\frac{t^{n}}{n!}, H^{⟨−1⟩}(t) = ∑_{n≥1} f_{n−1}(n)\frac{t^{n}}{n!}.

Then, from the last identity, f_{n−1}(n) can be expressed by partial Bell polynomials as

f_{n−1}(n) = \binom{2n − 1}{n}^{−1}B_{2n−1,n}(y).

The lemma holds for the case x_{1} = 0 by continuity.

Proposition 14. Let x = (x_{1}, x_{2}, …), y = (y_{1}, y_{2}, …) be sequences of real numbers with x_{1} = 1, y_{j} = jx_{j−1} and d be a positive integer. Then, we have the pair of compositional inverse power series

H(t) = \frac{t}{1 + ∑_{n≥1} x_{n} \frac{t^{dn}}{n!}} ,  (18)
H^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{(dn)!}{((d + 1)n + 1)!}B_{(d+1)n+1,dn+1}(y)t^{dn}).  (19)

Proof. For d ≥ 2, choice in Lemma 13 x_{n} = 0 if d ∤ n. On using the notations of Lemma 2 we get y_{n} = nx_{n−1} = 0 if d ∤ n − 1 and

1 + ∑_{n≥1} x_{n} \frac{t^{dn}}{n!} = 1 + ∑_{n≥1} x_{dn} \frac{t^{dn}}{(dn)!}.

Let z = (z_{1}, z_{2}, …) with z_{j} = j!y_{d(j−1)+1}/(d(j − 1) + 1)!. Then, on using Lemma 2 we obtain: If d ∤ n − 1 then B_{2n+1,n+1}(y) = 0 and if d ∤ n − 1 we get

\frac{(dn)!}{(2dn + 1)!}B_{2dn+1,dn+1}(y) = \frac{(dn)!}{((d + 1)n + 1)!}B_{(d+1)n+1,dn+1}(z ).

Therefore, the pair of the power series given in Lemma 13 can be written as

H(t) = \frac{t}{1 + ∑_{n≥1} x_{dn}t^{dn}/(dn)!}, H^{⟨−1⟩}(t) = 1 + ∑_{n≥1} \frac{(dn)!}{((d + 1)n + 1)!}B_{(d+1)n+1,dn+1}(z )t^{dn}.

To finish this proof, replace n!x_{dn}/(dn)! by x_{n}.

Example 15. For x = (\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, …) in Proposition 14 we get

H(t) = \frac{t^{d+1}}{e^{t^{d}} − 1}, H^{⟨−1⟩}(t) = t ∑_{n≥0} \frac{1}{dn + 1}\binom{(d + 1)n + 1}{dn + 1}^{−1}\left\{{(d + 1)n + 1 \atop dn + 1}\right\}\frac{t^{dn}}{n!} ,

for x = (1!, 2!, 3!, …) in Proposition 14 we get

H(t) = t(1 − t^{d}), H^{⟨−1⟩}(t) = t ∑_{n≥0} \frac{1}{dn + 1}\binom{(d + 1)n}{dn}t^{dn}

and for x = (\frac{1!}{2}, \frac{2!}{3}, \frac{3!}{4}, …) in Proposition 14 we get

H(t) = −\frac{t^{d+1}}{ln(1 − t^{d})}, H^{⟨−1⟩}(t) = t ∑_{n≥0} \frac{1}{dn + 1}\binom{(d + 1)n + 1}{dn + 1}^{−1}\left[{(d + 1)n + 1 \atop dn + 1}\right]\frac{t^{dn}}{n!} .

Theorem 16. Let x = (x_{1}, x_{2}, …) be a sequence of real numbers with x_{1} = 1, r, s, d be integers such that r > max(s, −(d + 1)s), d ≥ 1 and let

V_{n}(r, s) = \frac{s}{rn + s} \frac{B_{(r+1)n+s, rn+s}(x)}{\binom{(r+1)n+s}{rn+s}}, n ≥ 1, V_{0}(r, s) = 1.  (20)

Then, we have the pair of compositional inverses

H_{2}(t) = t(1 + ∑_{n≥1} V_{n}(r, −s)\frac{t^{dn}}{n!} )  (21)

H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} V_{n}(r + sd, s)\frac{t^{dn}}{n!} ),  (22)

Proof. Let (f_{n}(φ)) be a sequence of binomial type such that f_{n}(1) = \frac{x_{n+1}}{n+1} . Necessarily f_{n}(φ) satisfies (7) . Then because

\frac{s}{rn + s}\binom{(r + 1)n + s}{rn + s}^{−1}B_{(r+1)n+s,rn+s}(x ) = f_{n}(s; r)

and

H(t) = t(1 + ∑_{n≥1} \frac{s}{rn + s} \frac{B_{(r+1)n+s, rn+s}(x )}{\binom{(r+1)n+s}{rn+s}} \frac{t^{dn}}{n!} )^{−1} = t(1 + ∑_{n≥1} f_{n}(s, r)\frac{t^{dn}}{n!} )^{−1}

we can state, on using Proposition 14 and Proposition 1 given [12], that

H^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + sd)n + s}f_{n}((r + sd)n + s)\frac{t^{dn}}{n!} )

and by the identity (7) we obtain

H^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + sd)n + s} \frac{B_{(r+sd+1)n+s, (r+sd)n+s}(x )}{\binom{(r+sd+1)n+s}{(r+sd)n+s}} \frac{t^{dn}}{n!} ).

It suffices to remark that from Proposition 1 we have

(1 + s ∑_{n≥1} \frac{B_{(r+1)n+s, rn+s}(x )}{(rn + s)\binom{(r+1)n+s}{rn+s}} \frac{t^{dn}}{n!} )^{−1} = 1 − s ∑_{n≥1} \frac{B_{(r+1)n−s, rn−s}(x )}{(rn − s)\binom{(r+1)n−s}{rn−s}} \frac{t^{dn}}{n!} .

Example 17. With x = (1!, 2!, …, (q + 1)!, 0, …) in Theorem 16, the identity (12) gives

H_{2}(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s}\binom{rn − s}{n}_{q}t^{dn}),

H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + sd)n + s}\binom{(r + sd)n + s}{n}_{q}t^{dn}).

With x = (1!, 2!, …) in Theorem 16 we obtain

H_{2}(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s}\binom{(r + 1)n − s − 1}{rn − s − 1}t^{dn}),
H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + sd)n + s}\binom{(r + sd + 1)n + s − 1}{(r + sd)n + s − 1}\frac{t^{dn}}{n!} ).

With x = (1, 1, …) in Theorem 16 we obtain

H_{2}(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s}\binom{(r + 1)n − s}{rn − s}^{−1}\left\{{(r + 1)n − s \atop rn − s}\right\}\frac{t^{dn}}{n!} ),
H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + sd)n + s}\binom{(r + sd + 1)n + s}{(r + sd)n + s}^{−1}\left\{{(r + sd + 1)n + s \atop (r + sd)n + s}\right\}\frac{t^{dn}}{n!} ).

With x = (0!, 1!, …) in Theorem 16 we obtain

H_{2}(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s}\binom{(r + 1)n − s}{rn − s}^{−1}\left[{(r + 1)n − s \atop rn − s}\right]\frac{t^{dn}}{n!} ),
H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{s}{(r + sd)n + s}\binom{(r + sd + 1)n + s}{(r + sd)n + s}^{−1}\left[{(r + sd + 1)n + s \atop (r + sd)n + s}\right]\frac{t^{dn}}{n!} ).

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 (f_{n}(φ)) be a sequence of binomial type. Then, we have the pair of compositional inverses

H_{2}(t) = t(1 − ∑_{n≥1} \frac{φ}{αn − φ}f_{n}(αn − φ)\frac{t^{dn}}{n!} ),
H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{φ}{(α + dφ)n + φ}f_{n}((α + dφ)n + φ)\frac{t^{dn}}{n!} ).

Proof. Set x_{n} = nf_{n−1}(φ; α) in Theorem 16 to get

H_{2}(t) = t(1 − ∑_{n≥1} \frac{sφ}{(α + rφ)n − sφ}f_{n}((α + rφ)n − sφ)\frac{t^{dn}}{n!} ),
H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{sφ}{(α + rφ + sdφ)n + sφ}f_{n}((α + rφ + sdφ)n + sφ)\frac{t^{dn}}{n!} )

and change sφ by φ a and α − rφ and sφ by φ.

Example 19. With f_{n}(φ) = φ^{n} in Proposition 18 we get

H_{2}(t) = t(1 − φ ∑_{n≥1}(αn − φ)^{n−1} \frac{t^{dn}}{n!} ),

H_{2}^{⟨−1⟩}(t) = t(1 + x ∑_{n≥1}((α + dφ)n + x)^{n−1} \frac{t^{dn}}{n!} )

With f_{n}(φ) = n!\binom{φ}{n} in Proposition 18 we get

H_{2}(t) = t(1 − ∑_{n≥1} \frac{φ}{αn − φ}\binom{αn − φ}{n}t^{dn}),

H_{2}^{⟨−1⟩}(t) = t(1 + ∑_{n≥1} \frac{φ}{(α + dφ)n + x}\binom{(α + dφ)n + φ}{n}t^{dn}).

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 = (x_{1}, x_{2}, …) 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

U_{n}(r, s; d) = s\frac{((r + 1)dn + s)!}{rdn + s}\binom{(rd + 1)n + s}{n}^{−1} \frac{B_{(rd+1)n+s−(f−1)[n/f],rdn+s}(x)}{((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

V_{n}(r, s) = (rn + s − 1)!n!s\frac{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 y_{1} = 1, y_{2} = ⋯ = y_{f} = 0 and n = mf + δ (0 ≤ δ ≤ f − 1) in Theorem 3 to get

\frac{B_{n+k,k}(y_{j})}{\binom{n+k}{k}} = \frac{B_{mf+δ+k,k}(y_{j})}{\binom{mf+δ+k}{k}}

= ∑_{i=0}^{mf+δ} \frac{k!}{(k − i)!}B_{mf+δ,i}(\frac{y_{j+1}}{j + 1})

= \frac{(mf + δ)!}{(m + δ)!} ∑_{i=0}^{m+δ} \frac{k!}{(k − i)!}B_{m+δ,i}(\frac{j!y_{j+f}}{(j + f)!})

= \frac{(mf + δ)!}{(m + δ)!}\binom{m + δ + k}{k}^{−1}B_{m+δ+k,k}(\frac{j!y_{j+f−1}}{(j + f − 1)!})

= \frac{n!k!}{(n + k − (f − 1)[n/f])!}B_{n+k−(f−1)[n/f],k}(\frac{j!y_{j+f−1}}{(j + f − 1)!}).

Then, for x_{n} = n!y_{n+f−1}/(n + f − 1)! we obtain

U_{n}(r, s; d) = \frac{((d − 1)n)!s}{rdn + s}\binom{(rd + 1)n + s}{n}^{−1}\binom{(r + 1)dn + s}{(d − 1)n}B_{(rd+1)n+s,rdn+s}(y_{j})

= s\frac{((r + 1)dn + s)!}{rdn + s}\binom{(rd + 1)n + s}{n}^{−1} \frac{B_{(rd+1)n+s−(f−1)[n/f],rdn+s}(x_{j})}{((rd + 1)n + s − (f − 1)[n/f])!},

V_{n}(r, s) = \frac{s}{rn + s} \frac{B_{(r+1)n+s, rn+s}(y_{j})}{\binom{(r+1)n+s}{rn+s}} = (rn + s − 1)!n!s\frac{B_{(r+1)n+s−(f−1)[n/f], rn+s}(x_{j})}{((r + 1)n + s − (f − 1)[n/f])!}.

Proposition 21. Let r, s be integers, a, x be real numbers and (f_{n}(φ)) be a binomial-type polynomials. Then, for r > max(s, −2s) the inverse power series given by (9) and (10) hold for

U_{n}(r, s; d) = \frac{((d − 1)n)!s}{rdn + s}\binom{(r + 1)dn + s}{(d − 1)n}D_{z=0}^{rdn+s}(e^{z}f_{n}((rdn + s)φ + z; α)),

and for r > max(s, −(d + 1)s) the inverse power series given by (21) and (22) hold for

V_{n}(r, s) = \frac{s}{rn + s}D_{z=0}^{rn+s}(e^{z}f_{n}((rn + s)φ + z; α)).

Proof. For y_{n} = nD_{z=0}(e^{z}f_{n−1}(φ + z; α)) in Theorems 3 and 16 and use the identity given in [15, Lemma 1] by B_{n,k}(jD_{z=0}(e^{z}f_{j−1}(φ + z; α))) = \binom{n}{k}D_{z=0}^{k}(e^{z}f_{n−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 = (x_{1}, x_{2}, …) be a sequence of real numbers with x_{1} = 1, s_{1}, …, s_{m}, r, s, m be integers and H_{r,s} be power series defined by

H_{r,s}(t) = t(1 − ∑_{n≥1} \frac{s}{rn − s} \frac{B_{(r+1)n−s,rn−s}(x)}{\binom{(r+1)n−s}{n}} \frac{t^{n}}{n!}), r > max(s, 0) := s^{+}.

Then, we have

t∏^{m} ; i=1 \frac{H_{r,s_{i}}(t)}{t} = H_{r,s}(t) with r > max(s^{+}, s_{1}^{+}, …, s_{m}^{+}), s = m ; ∑_{i=1}s_{i}.

Proof. Let (f_{n}(φ)) be a sequence of polynomials of binomial type such that f_{n}(1) = x_{n+1}/(n+ 1) and let (f(t; r))^{φ} be the exponential generating function of the sequence of binomial type (f_{n}(t; r)). Assume that x_{2} ≠ 0. By Proposition 1 given in [12] we get

∏^{m} ; i=1 \frac{H_{r,s_{i}}(t)}{t} = ∏^{m} ; i=1(∑_{n≥0} f_{n}(−s_{i}; r)\frac{t^{n}}{n!}) = ∏^{m} ; i=1(f(t; r))^{−s_{i}} = ∑_{n≥0} f_{n}(−s; r)\frac{t^{n}}{n!} = ∑_{n≥0} f_{n}(−s; r)\frac{t^{n}}{n!},

and this is exactly

1 − ∑_{n≥1} \frac{s}{rn − s} \frac{B_{(r+1)n−s,rn−s}(x )}{\binom{(r+1)n−s}{n}} \frac{t^{n}}{n!} = ∏^{m} ; i=1 \frac{H_{r,s_{i}}(t)}{t}.

The proposition remains true for the case x_{2} = 0 by continuity.

Corollary 23. Let y = (y_{1}, y_{2}, …) be a sequence of real numbers with y_{1} = 1, r, s, d be integers with d ≥ 1 and let U_{n}(r, s; d), V_{n}(r, s; d) be given by Theorem 3 and Theorem 16. Then, for r > max(s, −2s) we have

∑^{n} ; k=1B_{n,k}(U_{j}(r, −s; d))(dn + k)_{k−1} = U_{n}(r + s, s; d),

∑^{n} ; k=1B_{n,k}(U_{j}(r + s, s; d))(dn + k)_{k−1} = U_{n}(r, −s; d),

and for r > max(s, −(1 + d)s) we have

∑^{n} ; k=1B_{n,k}(V_{j}(r, −s))(dn + k)_{k−1} = V_{n}(r + sd, s),

∑^{n} ; k=1B_{n,k}(V_{j}(r + sd, s))(dn + k)_{k−1} = V_{n}(r, −s).

Proof. From Comtet [7, pp. 151], for h(t) = t(1 + ∑_{n≥1}a_{n} \frac{t^{dn}}{n!} ), we have

h^{⟨−1⟩}(t) = t(1 + ∑_{n≥1}b_{n} \frac{t^{dn}}{n!} ) with b_{n} = ∑^{n} ; k=1(−1)^{k}(dn + k)_{k−1}B_{n,k}(a_{1}, a_{2}, …).

Then, it suffices to combine with Theorem 3 by taking a_{n} = U_{n}(r, −s; d) and b_{n} = U_{n}(r + s, s; d) and combine with Theorem 16 by taking a_{n} = V_{n}(r, −s) and b_{n} = V_{n}(r + sd, s).

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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.

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