Partial Bell Polynomials and Inverse Relations

Miloud Mihoubi^{1} 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 B_{n,k} are 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}  (1)

and are given explicitly by the formula

B_{n,k}(x_{1}, x_{2}, …) = ∑_{π(n,k)} \frac{n!}{k_{1}!k_{2}! ⋯}(\frac{x_{1}}{1!} )^{k_{1}}(\frac{x_{2}}{2!} )^{k_{2}} ⋯,

where π(n, k) is the set of all nonnegative integers (k_{1}, k_{2}, …) such that

k_{1} + k_{2} + k_{3} + ⋯ = k and k_{1} + 2k_{2} + 3k_{3} + ⋯ = n,

1Research supported by LAID3 Laboratory of USTHB University.

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^{⟨−1⟩}, the following inverse relations hold

\begin{matrix}y_{n} = ∑_{j=1}^{n} D_{x=a}^{j}F (x)B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = ∑_{j=1}^{n} D_{x=f(a)}^{j}F^{⟨−1⟩}(x)B_{n,j}(y_{1}, y_{2}, …). \end{matrix}

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 {f_{n}(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} = ∑_{n≥1} nf_{n−1}(x)\frac{t^{n}}{n!}

is given by

h^{⟨−1⟩}(t) = ∑_{n≥1} f_{n−1}(−nx)\frac{t^{n}}{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 = h^{⟨−1⟩}(t) = ∑_{n≥1} D_{z=0}^{n−1}(f(z)^{−nx})\frac{t^{n}}{n!} = ∑_{n≥1} f_{n−1}(−nx)\frac{t^{n}}{n!}.

Corollary 2. Let {f_{n}(x)} be a binomial-type sequence of polynomials and let a be a real number. Then the compositional inverse function of

h(t; a) = ∑_{n≥1} \frac{nx}{a(n − 1) + x}f_{n−1}(a(n − 1) + x)\frac{t^{n}}{n!}

is given by

h^{⟨−1⟩}(t; a) = − ∑_{n≥1} \frac{nx}{a(n − 1) − nx}f_{n−1}(a(n − 1) − nx)\frac{t^{n}}{n!}.

Proof. This result follows by replacing {f_{n}(x)} in Theorem 1, by the binomial-type sequence of polynomials {f_{n}(x; a)}, where

f_{n}(x; a) = \frac{x}{an + x}f_{n}(an + x),  (2)

see Mihoubi [5, 6, 7].

Theorem 3. Let {f_{n}(x)} be a binomial-type sequence of polynomials and a be a real number. Then the following inverse relations hold

\begin{matrix}y_{n} = ∑_{j=1}^{n} \frac{xj}{a(j−1)+x}f_{j−1}(a(j − 1) + x)B_{n,j}(x_{1}, x_{2}, …) \\ x_{n} = − ∑_{j=1}^{n} \frac{xj}{a(j−1)−jx}f_{j−1}(a(j − 1) − jx)B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (3)

Proof. For any function F having power formal series with compositional inverse F^{⟨−1⟩}, Chou, Hsu and Shiue [2, Remark 1] have proved that

\begin{matrix}y_{n} = ∑_{j=1}^{n} D_{x=a}^{j}F (x)B_{n,j}(x_{1}, x_{2}, …) \\ x_{n} = ∑_{j=1}^{n} D_{x=f(a)}^{j}F^{⟨−1⟩}(x)B_{n,j}(y_{1}, y_{2}, …). \end{matrix}

To prove (3), it suffices to take

F (t) := ∑_{n=1}^{∞} \frac{nx}{a(n − 1) + x}f_{n−1}(a(n − 1) + x)\frac{t^{n}}{n!}

and then use Corollary 2.

Now, let f_{n}(x) in Theorem 3 be one of the next binomial-type sequence of polynomials

f_{n}(x) = x^{n},

f_{n}(x) = (x)_{(n)} := x(x − 1) ⋯ (x − n + 1), n ≥ 1, with (x)_{(o)} = 1, f_{n}(x) = (x)^{(n)} := x(x + 1) ⋯ (x + n − 1), n ≥ 1, with (x)^{(o)} = 1,

f_{n}(x) = n!\binom{x}{n}_{q} := ∑_{j=o}^{n} B_{n,j}(\binom{1}{1}_{q}, …, i!\binom{1}{i}_{q}, …)(x)_{(j)},

f_{n}(x) = B_{n}(x) := ∑_{j=o}^{n} S(n, k)x^{k},

where B_{n}(.), S(n, k) and \binom{k}{n}_{q} are, respectively, the single variable Bell polynomials, the

Stirling numbers of second kind and the coefficients defined by

(1 + x + x^{2} + ⋯ + x^{q})^{k} = ∑_{n≥0} \binom{k}{n}_{q}x^{n},

see Belbachir, Bouroubi and Khelladi [4]. We deduce the following results:

Corollary 4. Let a and x be real numbers. Then the following inverse relations hold: For f_{n}(x) = x^{n}, we get

\begin{matrix}y_{n} = ∑_{j=1}^{n} xj(a(j − 1) + x)^{j−2}B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = − ∑_{j=1}^{n} xj(a(j − 1) − jx)^{j−2}B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (4)

For f_{n}(x) = (x)_{(n)}, we get

\begin{matrix}y_{n} = ∑_{j=1}^{n} \frac{xj}{a(j−1)+x}(a(j − 1) + x)_{(j−1)}B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = − ∑_{j=1}^{n} \frac{xj}{a(j−1)−jx}(a(j − 1) − jx)_{(j−1)}B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (5)

For f_{n}(x) = (x)^{(n)}, we get

\begin{matrix}y_{n} = ∑_{j=1}^{n} \frac{xj}{a(j−1)+x}(a(j − 1) + x)^{(j−1)}B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = − ∑_{j=1}^{n} \frac{xj}{a(j−1)−jx}(a(j − 1) − jx)^{(j−1)}B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (6)

For f_{n}(x) = n!\binom{x}{n}_{q}, we get

\begin{matrix}y_{n} = ∑_{j=1}^{n} \frac{x(j−1)!}{a(j−1)+x} \binom{a(j−1)+x}{j−1}_{q}B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = − ∑_{j=1}^{n} \frac{xj!}{a(j−1)−jx} \binom{a(j−1)−jx}{j−1}_{q}B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (7)

For f_{n}(x) = B_{n}(x), we get

\begin{matrix}y_{n} = ∑_{j=1}^{n} \frac{xj}{a(j−1)+x}B_{j−1}(a(j − 1) + x)B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = − ∑_{j=1}^{n} \frac{xj}{a(j−1)−jx}B_{j−1}(a(j − 1) − jx)B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (8)

Example 5. For a = 0, x = 1 in (5), we obtain

\begin{matrix}y_{n} = ∑_{j=0}^{n} \binom{n}{j}x_{j}x_{n−j} with x_{0} = \frac{1}{2}, \\ x_{n} = ∑_{j=0}^{n−1}(−1)^{j} \frac{(2j)!}{(j)!}B_{n,j+1}(y_{1}, y_{2}, …). \end{matrix}  (9)

For x_{n} = 2^{(n−2)/2}, x_{n} = \frac{1}{2} or x_{n} = \frac{1}{2^{n+1}} in (9), we get

∑_{j=0}^{n−1} \frac{(2j)!}{(j)!}(−4)^{n−j}S(n, j + 1) = (−1)^{n}2^{n+1}.

For x_{1} = 1, x_{n} = 0, n ≥ 2, in (9), we get

∑_{j=n−[n/2]}^{n+1}(−1)^{j}\binom{n + 1}{j}\binom{2j}{n} = 0, n ≥ 0.

Take x = a = x_{1} = 1, x_{2} = 2 and x_{n} = 0, n ≥ 3, in (5) and from the identity of Ceralosi [1]

∑_{j=1}^{n} j!B_{n,j}(1!, 2!, 0, 0, …) = n!F_{n}, n ≥ 1,

we obtain

∑_{j=1}^{n}(−1)^{j}j!B_{n,j}(1!F_{1}, 2!F_{2}, …) = 0, n ≥ 3,

where F_{n}, n = 0, 1, 2, ⋯, are the Fibonacci numbers. Take x = 1, a = x_{1} = 0, x_{2} = 2 and x_{n} = n!, n ≥ 3, in (5), from the identity of Ceralosi [1]

∑_{j=1}^{n} j!B_{n,j}(0, 2!, 3!, …) = n!F_{n−2}, n ≥ 2,

we obtain

∑_{j=1}^{n}(−1)^{j−1}j!B_{n,j}(0, 2!F_{0}, 3!F_{1}, …) = n!, n ≥ 2.

Theorem 6. Let r, s be nonnegative integers, rs ≠ 0, and let {u_{n}} be a sequence of real numbers with u_{1} = 1. Then

y_{n} = s ∑_{j=1}^{n} \frac{j}{U_{j}} \binom{U_{j}+j−1}{U_{j}}^{−1}B_{U_{j}+j−1,U_{j}}(1, u_{2}, u_{3}, …)B_{n,j}(x_{1}, x_{2}, …),
x_{1} = y_{1} and for n ≥ 2 we have
x_{n} = y_{n} − s ∑_{j=2}^{n} \frac{j}{V_{j}} \binom{V_{j}+j−1}{V_{j}}^{−1}B_{V_{j}+j−1,V_{j}}(1, u_{2}, u_{3}, …)B_{n,j}(y_{1}, y_{2}, …),

(10)

where U_{j} = (r + 2s)(j − 1) + s and V_{j} = (r + s)(j − 1) − s.

Proof. Let n, r, s be nonnegative integers, nr(nr + s) ≥ 1, z_{n}(r) := \frac{B_{(r+1)n,nr}(1,u_{2},u_{3},...)}{nr\binom{(r+1)n}{nr}}, and

consider the binomial-type sequence of polynomials {f_{n}(x)} defined by

f_{n}(x) := ∑_{j=1}^{n} B_{n,j}(z_{1}(r), z_{2}(r), …)x^{j} with f_{o}(x) = 1,

see Roman [8]. Then from the identity

∑_{j=1}^{n} B_{n,j}(z_{1}(r), z_{2}(r), …)s^{j} = \frac{s}{nr + s}\binom{(r + 1)n + s}{nr + s}^{−1}B_{(r+1)n+s,nr+s}(1, u_{2}, u_{3}, …),

see Mihoubi [5, 6, 7], we get

f_{n}(s) = \frac{s}{nr + s}\binom{(r + 1)n + s}{nr + s}^{−1}B_{(r+1)n+s,nr+s}(1, u_{2}, u_{3}, …).  (11)

To obtain (10), we set a = 0, x = s in (3) and use the expression of f_{n}(s) given by (11), with r + 2s instead of r.

Example 7. From the well-known identity B_{n,k}(1!, 2!, …, i!, …) = \binom{n−1}{k−1} \frac{n!}{k!}, we get

y_{n} = s ∑_{j=1}^{n} j!\frac{((r+2s+1)(j−1)+s−1)!}{((r+2s)(j−1)+s)!}B_{n,j}(x_{1}, x_{2}, …),
x_{1} = y_{1} and for n ≥ 2 we have
x_{n} = y_{n} − s ∑_{j=2}^{n} j!\frac{((r+s+1)(j−1)+s−1)!}{((r+s)(j−1)+s)!}B_{n,j}(y_{1}, y_{2}, …).

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 u_{n} = (−1)^{n−1}(n − 1)!, u_{n} = (n − 1)! and u_{n} = 1 for all n ≥ 1, respectively.

Corollary 8. Let u, r, s be nonnegative integers, a, α be real numbers, αurs ≠ 0, and {f_{n}(x)} be a binomial-type sequence of polynomials. Then

y_{n} = s ∑_{j=1}^{n} \frac{j}{α^{T_{j} −u(j−1)}T_{j}} D_{z=0}^{T_{j}}(e^{αz}f_{j−1}(T_{j}x + z; a))B_{n,j}(x_{1}, x_{2}, …),
x_{1} = y_{1} and for n ≥ 2 we have
x_{n} = y_{n} − s ∑_{j=2}^{n} \frac{j}{α^{R_{j} −u(j−1)}R_{j}} D_{z=0}^{R_{j}}(e^{αz}f_{j−1}(R_{j}x + z; a))B_{n,j}(y_{1}, y_{2}, …),

where T_{j} = (u + r + 2s)(j − 1) + s and R_{j} = (u + r + s)(j − 1) − s.

Proof. Set in Theorem 6

u_{n} = \frac{n}{(u(n − 1) + 1)α}D_{z=0}^{u(n−1)+1}(e^{αz}f_{n−1}((u(n − 1) + 1)x + z; a))

and use the first identity of Mihoubi [6, Theorem 2].

Corollary 9. Let u, r, s be nonnegative integers, urs ≠ 0, a be real number and let {f_{n}(x)} be a binomial-type sequence of polynomials. Then

y_{n} = s ∑_{j=1}^{n} \frac{j!}{α^{T_{j} −u(j−1)}(T_{j}+j−1)!T_{j}} D_{z=0}^{T_{j}}f_{T_{j}+j−1}(T_{j}x + z; a)B_{n,j}(x_{1}, x_{2}, …),
x_{1} = y_{1} and for n ≥ 2 we have
x_{n} = y_{n} − s ∑_{j=2}^{n} \frac{j!}{α^{R_{j} −u(j−1)}(R_{j}+j−1)!R_{j}} D_{z=0}^{R_{j}}f_{R_{j}+j−1}(R_{j}x + z; a)B_{n,j}(y_{1}, y_{2}, …),

where T_{j} = (u + r + 2s)(j − 1) + s and R_{j} = (u + r + s)(j − 1) − s.

Proof. Set in Theorem 6

u_{n} = \frac{n!D_{z=0}^{u(n−1)+1}f_{((u+1)(n−1)+1)}((u(n − 1) + 1)x + z; a)}{((u + 1)(n − 1) + 1)!(u(n − 1) + 1)α}

and use the second identity of Mihoubi [6, Theorem 2].

Theorem 10. Let d be an integer ≥ 1. The inverse relations hold

\begin{matrix}y_{n} = ∑^{n} ; j=1(−1)^{j}(dn + j)_{(j−1)}B_{n,j}(x_{1}, x_{2}, …), \\ x_{n} = ∑^{n} ; j=1(−1)^{j}(dn + j)_{(j−1)}B_{n,j}(y_{1}, y_{2}, …). \end{matrix}  (12)

Proof. Let

f(t) = t(1 + ∑_{n≥1}x_{n} \frac{t^{dn}}{n!} ) and f^{⟨−1⟩}(t) = t(1 + ∑_{n≥1}y_{n} \frac{t^{dn}}{n!} ).

The proof now follows from Comtet [3, Theorem F, p. 151]. Example 11. Take d = 1 and x_{n} = n!, n ≥ 1, in Theorem 10, we get

f(t) = \frac{t}{1 − t} and f^{⟨−1⟩}(t) = \frac{t}{1 + t},

i.e. y_{n} = (−1)^{n}n!, and the relations (12) give

∑^{n} ; j=1(−1)^{n−j}\binom{n}{j}\binom{n + j}{n + 1} = n.

Take d = 2 and x_{n} = n!, n ≥ 1, we get

f(t) = \frac{t}{1 − t^{2}} and f^{⟨−1⟩}(t) = \frac{1}{2t}(−1 + \sqrt{1 + 4t^{2}}),

i.e. y_{n} = (−1)^{n} \frac{(2n)!}{(n + 1)!}, n ≥ 1, and the relations (12) give

∑^{n} ; j=1(−1)^{n−j}\binom{n}{j}\binom{2n + j}{2n + 1} = \frac{n}{n + 1}\binom{2n}{n} = nC_{n},

where C_{n}, n = 0, 1, 2, ⋯, are the Catalan numbers. Theorem 12. The following inverse relations hold

\begin{matrix}y_{n} = \frac{1}{nr} \binom{(r+1)n}{nr}^{−1}B_{(r+1)n,nr}(1, x_{1}, x_{2}, …), r ≥ 1, \\ x_{n} = (n + 1) ∑_{j=1}^{n} B_{n, j}(y_{1}, y_{2}, …)(−1)^{j−1}(nr − 1)^{j−1}. \end{matrix}

Proof. From Mihoubi [7, Theorem 1] we have

x_{1}^{k} ∑_{j=1}^{n} B_{n,j}(y_{1}, y_{2}, …)(k − nr)^{j−1} = \frac{x_{1}^{nr}}{k} \binom{n + k}{k}^{−1}B_{n+k,k}(x_{1}, x_{2}, x_{3}, …),

with y_{n} = \frac{1}{nr} \binom{(r+1)n}{nr}^{−1}B_{(r+1)n,nr}(1, x_{1}, x_{2}, …), nrk ≥ 1. It just suffices to set k = 1, x_{1} = 1, and replace x_{n} by x_{n−1}.

3 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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