Multinomial convolution polynomials

Jiang Zeng

Département de Mathématique, Université Louis Pasteur, 7, rue René Descartes, 67084 Strasbourg Cédex, France

October 26, 1994

Abstract

In [9] Knuth shows how to derive the convolution formulas of Hagen, Rothe and Abel from Vandermonde’s convolution or binomial theorem for integer exponents. In the present paper, we shall first present a short and elementary proof of the multi-extension of the above convolution formulas, due to Raney and Mohanty. In the second part we shall present a multi-version of Knuth’s approach to convolution polynomials and derive another short proof of the above formulas.

1 Introduction

Recall that a family of polynomials {F_{n}(x)}_{n≥0} is said to be of convolution type if F_{n}(x) has degree ≤ n and satisfies the convolution condition :

F_{n}(x + y) = ∑_{k=0}^{n} F_{k}(x)F_{n−k}(y).

The two typical families of convolution polynomials are the binomial theorem for integers

\frac{(x + y)^{n}}{n!} = ∑_{k=0}^{n} \frac{x^{k}}{k!} \frac{y^{n−k}}{(n − k)!}

and Vandermonde’s convolution

\binom{x + y}{n} = ∑_{k=0}^{n} \binom{x}{k}\binom{y}{n − k}.

Some important extensions of the above identities have been given by Abel, Rothe and Hagen in the last centry. More precisely, for a, b ∈ C and n ∈ N let

A_{n}(a, b) = \frac{a}{a + bn}\binom{a + bn}{n}.

In 1891 Hagen [7, 4, 6] proved the following identity :

∑_{k=0}^{n}(p + qk)A_{k}(a, b)A_{n−k}(c, b) = \frac{p(a + c) + aqn}{a + c}A_{n}(a + c, b).

In the case q = 0, this formula reduces to Rothe’s convolution formula [17, 5, 16] dated 1793:

∑_{k=0}^{n} A_{k}(a, b)A_{n−k}(c, b) = A_{n}(a + c, b).

Note that Hagen’s formula implies also Abel’s identity [1] :

(x + y)^{n} = ∑_{k=0}^{n} \binom{n}{k}x(x − kz)^{k−1}(y + kz)^{n−k},

where x, y, z ∈ C and n ∈ N.

In a refreshing paper about convolution polynomials, Knuth [9] shows how to derive these seemingly non trivial formulas from the basic binomial theorem for integers and Vandermonde’s convolution.

Since the multi-extensions of the above identities are also known, it is natural to give a multi-extension of Knuth’s results in [9]. To express them more concisely, we shall use the vector notation as follows. Throughout this paper m will be a fixed natural number. For a = (a_{1}, ⋯, a_{m}) ∈ N^{m} and b = (b_{1}, ⋯, b_{m}) ∈ N^{m} set |a| = ∑_{i=1}^{m} a_{i}, a! = a_{1}! ⋯ a_{m}!, a + b = (a_{1} + b_{1}, ⋯, a_{m} + b_{m}) and a · b = ∑_{i=1}^{m} a_{i}b_{i}. We order the elements of N^{m} by lexicographic order, i.e., a < a if a_{1} < b_{1} or there is a i > 1 such that a_{1} = b_{1}, …, a_{i−1} = b_{i−1} but a_{i} < b_{i}. Also for 1 ≤ i ≤ m let e_{i} = (δ_{i1}, δ_{i2}, ⋯, δ_{im}). Finally, for a complex number x and n ∈ Z^{m}, we define

\binom{x}{n} = \{ \begin{matrix}x(x − 1) ⋯ (x − |n| + 1)/n!, & if n = (n_{1}, ⋯, n_{m}) ∈ N^{m}; \\ 0, & if some n_{i} is negative. \end{matrix}

We now define the multi-analogue of A_{n}(a, b) by

A_{n}(x, b) = \frac{x}{x + b · n}\binom{x + b · n}{n}.

The multi-extension of Rothe’s convolution formula can then be stated as follows :

∑_{k} A_{k}(a, b)A_{n−k}(c, b) = A_{n}(a + c, b).  (1.1)

Remark: The author first learned this identity in July 1988 at Oberwolfach, where Louck [11] communicated it as a conjecture. It was shortly proved independently by Strehl, Paule and the author by using different methods and reported at the 20th session of le sminaire Lotharingien de Combinatoire held at Alghero, Italie, in September 1988. Shortly later, we noted that formula (1.1) was already established by Raney [15, Th. 2.2, Th. 2.3] in 1960. We refer the reader to the recent paper of Strehl [18] for a good account of the various aspects of this formula.

Curiously enough Raney’s derivation of (1.1) was entirely different from the three formentioned ones. Indeed, Strehl’s proof is of combinatorial nature [18], Paule’s proof is based on the one variable Lagrange inversion formula [14] and the author’s proof is inductive and will be produced below. Note that Chu [2] quoted Raney’s identity explicitly in his paper to derive a combinatorial interpretation of the generalized Catalan numbers.

A multi-extension of Hagen’s identity has also been given by Mohanty [12], by applying the multivariable Lagrange inversion formula, in the following form :

∑_{k}(p + q · k)A_{k}(a, b)A_{n−k}(c, b) = \frac{p(a + c) + a(q · n)}{a + c}A_{n}(a + c, b), (1.2)

where q = (q_{1}, ⋯, q_{m}) ∈ C^{m} and a, c, p ∈ C.

The aim of this paper is two folds. First we present a very short and elementary proof of (1.1) and (1.2) from scratch, which is similar in spirit to Good’s short proof of Dyson’s conjecture [4]. In the the second part of this paper we shall present a natural multi-extension of Knuth’s approach [9] to ordinary convolution polynomials. As a consequence of this generalization we see that (1.2) and (1.1) are respectively equivalent to the following trivial convolution identities :

\frac{(x + y)^{|n|}}{n!} = ∑_{0≤k≤n} \frac{x^{|k|}}{k!} \frac{y^{|n−k|}}{(n − k)!}  (1.3)

and

\binom{x + y}{n} = ∑_{0≤k≤n} \binom{x}{k}\binom{y}{n − k}.  (1.4)

As remarked by Knuth [9], in the special case that each polynomial F_{n}(x) has degree exactly n, the polynomials n!F_{n}(x) are said to be of binomial type [13]. We note that in the later case Joni [10] has given another generalization of binomial polynomials.

2 A short proof of formula (1.2)

Notice that formula (1.2) is eqivalent to (1.1) and the following identity :

∑_{k}(q · k)A_{k}(a, b)A_{n−k}(c, b) = \frac{a(q · n)}{a + c}A_{n}(a + c, b)  (2.1)

A further look shows that formula (1.3) is actually a consequence of (2.1). Indeed, exchanging a with c, and k with n − k in (2.1) yields

∑_{k}(q · (n − k))A_{k}(a, b)A_{n−k}(c, b) = \frac{c(q · n)}{a + c}A_{n}(a + c, b)  (2.2)

Summing up the two above identities side by side and taking q = n, we get (1.1). Therefore it suffices to prove (2.1). However it seems useful to first give an independent proof of (1.1) to illustrate the method. In the sequel we assume a to belong to N. Let

S_{n}(a, c, b) = ∑_{k} A_{k}(a, b)A_{n−k}(c, b).

Then, first

S_{n}(0, c, b) = A_{n}(c, b),  S_{0}(a, c, b) = 1.  (2.3)

Next, by the definition of A_{n}(a, b) we have

A_{n}(a, b) = \frac{1}{a − 1 + b · n} (a − 1 + ∑_{i=1}^{m} \frac{b_{i}n_{i}}{a + b · n}) \binom{a + b · n}{n} (2.4)
= \frac{1}{a − 1 + b · n} [(a − 1)\binom{a + b · n}{n} + ∑_{i=1}^{m} b_{i}\binom{a − 1 + b · n}{n − e_{i}}]
= A_{n}(a − 1, b) + ∑_{i=1}^{m} A_{n−e_{i}}(a + b_{i} − 1, b).

The last line is due to \binom{x}{n} = \binom{x−1}{n} + ∑_{i=1}^{m} \binom{x−1}{n−e_{i}}. So that

S_{n}(a, c, b) = ∑_{k} (A_{k}(a − 1, b) + ∑_{i=1}^{m} A_{k−e_{i}}(a + b_{i} − 1, b)) A_{n−k}(c, b)
= S_{n}(a − 1, c, b) + ∑_{i=1}^{m} S_{n−e_{i}}(a + b_{i} − 1, c, b). (2.5)

Eqs. (2.3)-(2.5) show that S_{n}(a, c, b) and A_{n}(a, c, b) satisfy the same recurrence and the initial conditions. Therefore they are equal.

Now we turn to (2.1). Note that (2.1) is equivalent to the following m identities:

∑_{k} k_{j}A_{k}(a, b)A_{n−k}(c, b) = \frac{a n_{j}}{a + c}A_{n}(a + c, b),  (1 ≤ j ≤ m). (2.6)

For 1 ≤ j ≤ m, let T_{n}^{(j)}(a, c, b) and S_{n}^{(j)}(a, c, b) be respectively the left-hand and right-hand sides of (2.6). By definition, we first have

S_{n}^{(j)}(a, c, b) = (a − 1)\binom{a − 1 + c + b · n}{n − e_{j}} + \binom{a − 1 + c + b · n}{n − e_{j}}. (2.7)

Now writing the first \binom{a−1+c+b·n}{n−e_{j}} as

\binom{a − 2 + c + b · n}{n − e_{j}} + ∑_{i=1}^{m} \binom{a − 2 + c + b · n}{n − e_{i}}

and the second one as

(\frac{b · (n − e_{j})}{a − 1 + c + b · n} + \frac{a − 1 + c + b_{j}}{a − 1 + c + b · n}) \binom{a − 1 + c + b · n}{n − e_{j}}

= ∑_{i=1}^{m} b_{i}\binom{a − 2 + c + b · n}{n − e_{j} − e_{i}} + A_{n−e_{j}}(a − 1 + c + b_{j}, b),

we see that

S_{n}^{(j)}(a, c, b) = S_{n}^{(j)}(a − 1, c, b) + ∑_{i=1}^{m} S_{n−e_{i}}^{(j)}(a − 1 + b_{i}, c, b) (2.8)
+ A_{n−e_{j}}(a − 1 + c + b_{j}, b).

Also we have the boundary conditions :

S_{n}^{(j)}(0, c, b) = 0 and S_{0}^{(j)}(a, c, b) = 0.  (2.9)

On the other hand, by (2.4) we verify easily that T_{n}^{(j)}(a, c, b) satisfies the same recurrence relation (2.8) and boundary condition (2.9). Thus completes proof of (2.6).

3 Multi-convolution polynomials

A family of polynomials {F_{n}(x)}_{n≥0} forms a multinomial convolution family if F_{n}(x) has degree ≤ |n| and if the convolution condition

F_{n}(x + y) = ∑_{0≤k≤n} F_{k}(x) F_{n−k}(y)  (3.1)

holds for all x and y and for all n ≥ 0. In the case that m = 1 we recover the monomial convolution family studied by Knuth [9]. We should note that this definition is different from the one of “higher dimentional polynomials of binomial type” studied by Joni [10]. Many such families are known, and they appear frequently in applications. For example, we can let F_{n}(x) = x^{|n|}/n!, the condition (1.7) is equivalent to the binomial theorem for integer exponents. Or we can let F_{n}(x) be the multinomial coefficient \binom{x}{n}; the corresponding identity (1.8) may be called multi-Vandermonde’s convolution.

Knuth showed that convolution polynomials arise as coefficients when a power series of one variable is raised to the power x. Now we show that multi-convolution polynomials arise as coefficients when a power series of several variables is raised to the power x. Let z^{n} = z_{1}^{n_{1}} … z_{m}^{n_{m}}. For a formal series F (z) = ∑_{n≥0} F_{n}z^{n} let [z^{n}]F (z) be the coefficient of z^{n} in F (z).

Theorem 1 Let

F (z) = 1 + ∑_{n≠0} F_{n}z^{n}  (3.2)

be any power series with F (0) = 1. Then the polynomials

F_{n}(x) = [z^{n}]F (z)^{x}  (3.3)

form a multi-convolution family. Conversely, every convolution family arises in this way or is identically zero.

Proof: It is easy to verify that the coefficient of z^{n} in F (z)^{x} is indeed a polynomial in x of degree ≤ |n| because

F (z)^{x} = 1 + ∑_{k≥1} \binom{x}{k}(∑_{n≠0} F_{n}z^{n})^{k}

This construction produces a convolution family because of the rule for forming coefficients of the product F (z)^{x+y} = F (z)^{x}F (z)^{y}. Conversely, suppose the polynomials

F_{n}(x) = (f_{n0} + f_{n1}x + … + f_{n|n|}x^{|n|})/n!

form a convolution family. The condition F_{0}(x) = F_{0}(x)^{2} can hold only if F_{0}(x) = 0 or F_{0}(x) = 1. In the former case it is easy to prove by induction that F_{n}(x) = 0 for all n because F_{n}(x) = ∑_{0<k<n} F_{k}(x) F_{n−k}(0). Otherwise, the condition

F_{n}(2 x) = 2F_{n}(x) + ∑_{0<k<n} F_{k}(x) F_{n−k}(x)  (3.4)

for n > 0 implies that F_{n}(0) = f_{n0} = 0 for n > 0 by induction. For k > 1, by equating coefficients of x^{k} on both sides of (3.4) we find that the coefficient f_{nk} is forced to have certain values based on the coefficients of F_{i}(x) with i < n, because 2^{k}f_{nk} occurs on the left and 2f_{nk} on the right. The coefficient f_{n1} can, however, be choosen freely. Any such choice must make

F_{n}(x) = [z^{n}] exp(x ∑_{n≠0} f_{n1}z^{n}),

by induction on n.

Examples: The first example (1.3), that is F_{n}(x) = x^{|n|}/n!, comes from the power series F (z) = e^{z_{1}+z_{2}+...+z_{m}}; the second example (1.4), that is F_{n}(x) = \binom{x}{n}, comes from F (z) = 1 + z_{1} + z_{2} + … + z_{m}.

As in the case of one-variable convolution polynomials [9], every multiconvolution family {F_{n}(x)} satisfies another general convolution formula in addition to the one we began with. In deed, we have the following result.

Lemma 2 (A derived convolution). If {F_{n}(x)} is a set of multi-convolution polynomilas, then

∑_{k=0}^{n} k_{i}F_{k}(x)F_{n−k}(y) = xn_{i}F_{n}(x + y) (1 ≤ i ≤ m).

or equivalently for all q ∈ C^{m} we have

(x + y) ∑_{k=0}^{n}(q · k)F_{k}(x)F_{n−k}(y) = x(q · n)F_{n}(x + y).

Proof: The alternative convolution formula is proved by differentiating the basic identity F (z)^{x} = ∑_{n≥0} F_{n}(x)z^{n} with respect to z_{i} (1 ≤ i ≤ m) and

xz_{i} \frac{∂}{∂z_{i}} F (z)F (z)^{x−1} = ∑_{n} n_{i}F_{n}(x)z^{n}.

Now observe that ∑ k_{i}F_{k}(x)F_{n−k}(y) is the coefficient of z^{n} in the product xz_{i}(∂F (z)/∂z_{i})F (z)^{x+y−1} while n_{i}F_{n}(x + y) is the coefficient of z^{n} in (x + y)z_{i}(∂F (z)/∂z_{i})F (z)^{x+y−1}.

Lemma 3 If F_{n}(x) = [z^{n}]F (z)^{x}, then

\frac{x}{x + n_{i}} F_{n}(x + n_{i}) = [z^{n}]G_{i}(z)^{x}, 1 ≤ i ≤ m

where G_{i}(z) = F (z_{1}, …, z_{i}G_{i}(z), …, z_{m}).

Proof: Since F_{n}(x) is a polynomial of degree ≤ |n|, so it is sufficient to prove the above identity for integer values k of x. Let w(z) = z_{i}/F (z), then w(z_{1}, …, z_{i}G_{i}(z), …, z_{m}) = z_{i}. By Lagrange’s inversion formula (see [3, p. 159]), we have

[z_{i}^{n_{i}}](z_{i}^{k}G_{i}^{k}(z)) = \frac{k}{n_{i}} [z_{i}^{n_{i}−k}]F (z)^{n_{i}}, 1 ≤ i ≤ m.

Consequently,

[z_{i}^{n_{i}}](G_{i}^{k}(z)) = \frac{k}{n_{i} + k} [z_{i}^{n_{i}}]F (z)^{n_{i}+k}, 1 ≤ i ≤ m.

It follows that for 1 ≤ i ≤ m we have

[z^{n}]G_{i}(z)^{x} = \frac{k}{n_{i} + k} [z^{n}]F (z)^{n_{i}+k} = \frac{k}{n_{i} + k} F_{n}(n_{i} + k).

This completes the proof.

Theorem 4 Let F_{n}(x) be any family of polynomials in x such that F_{n}(x) has degree ≤ |n|. If

F_{n}(2 x) = ∑_{k=0}^{n} F_{k}(x) F_{n−k}(x)  (3.5)

holds for all n and x, then the following identities hold for all n, x, y and t

\frac{(x + y) F_{n}(x + y + t · n)}{x + y + t · n}

= ∑_{k=0}^{n} \frac{x F_{k}(x + t · k)}{x + t · k} · \frac{y F_{n−k}(y + t · (n − k))}{y + t · (n − k)}

\frac{(q · n) F_{n}(x + y + t · n)}{x + y + t · n}

= ∑_{k=0}^{n} \frac{(q · k) F_{k}(x + t · k)}{x + t · k} · \frac{y F_{n−k}(y + t · (n − k))}{y + t · (n − k)}.

Proof: The proof of Theorem 1 shows that the multinomial convolution condition could be weakened to (3.4), or (3.5). Hence (3.5) implies that polynomials F_{n}(x) form a set of convolution polynomials. Now, for 1 ≤ i ≤ m, the process of going from F_{n}(x) to xF_{n}(x + n_{i})/(x + n_{i}) of Lemma 3 can be iterated : another replacement gives xF_{n}(x + 2n_{i})/(x + 2n_{i}), and after t_{i} iterations we discover that the polynomials xF_{n}(x+t_{i}n_{i})/(x+t_{i}n_{i}) also form a convolution family. This holds for all positive integers t_{i} (1 ≤ i ≤ m), and the convolution condition is expressible as a set of polynomial relations in t_{i} (1 ≤ i ≤ m); therefore xF_{n}(x + t_{i}n_{i})/(x + t_{i}n_{i}) is a convolution family for all complex numbers t_{i} (1 ≤ i ≤ m). Thus by applying the above process we obtain succesfully a sequence of convolution polynomials families

\frac{x}{x + t_{1}n_{1} + … + t_{i}n_{i}} F_{n}(x + t_{1}n_{1} + … + t_{i}n_{i})

for i = 1, 2, …, m and the last family corresponds to the first identity. The second one follows by combining the first one with the derived convolution formula of Lemma 2.

Example 1: Let F_{n}(x) = \binom{x}{n}, then it follows from (1.4) and Theorem 4 that

\frac{x + y}{x + y + t · n}\binom{x + y + t · n}{n}

= ∑_{k=0}^{n} \frac{x}{x + t · k}\binom{x + t · k}{k}\frac{y}{y + t · (n − k)}\binom{y + t · (n − k)}{n − k},

\frac{(q · n)}{x + y + t · n}\binom{x + y + t · n}{n}

= ∑_{k=0}^{n} \frac{q · k}{x + t · k}\binom{x + t · k}{k}\frac{y}{y + t · (n − k)}\binom{y + t · (n − k)}{n − k}.

Combining the above two identies yields (1.2).

Example 2: Let F_{n}(x) = x^{n}/n!, then it follows from (1.3) and Theorem 4 that

\frac{(x + y)(x + y + t · n)^{|n|−1}}{n!}

= ∑_{k=0}^{n} \frac{x(x + t · k)^{|k|−1}}{k!} \frac{y(y + t · (n − k))^{|n−k|−1}}{(n − k)!} ,

\frac{(q · n)(x + y + t · n)^{|n|−1}}{n!}

= ∑_{k=0}^{n} \frac{(q · k)(x + t · k)^{|k|−1}}{k!} \frac{y(y + t · (n − k))^{|n−k|−1}}{(n − k)!} .

Combining the above two identies we obtain

∑_{k}(p + q · k)B_{k}(x, t)B_{n−k}(y, t) = \frac{p(x + y) + x(q · n)}{x + y}B_{n}(x + y, t), (3.6)

where B_{n}(x, t) = x(x+t·n)^{|n|−1}/n!. If we take q = t and p = 1 in the above formula we obtain a multi-extension of Abel’s identity :

(x + y + n · t)^{|n|} = ∑_{k=0}^{n} \binom{n}{k}(x + k · t)^{|k|}y(y + (n − k) · t)^{|n−k|−1},

where \binom{n}{k} = ∏_{i=1}^{m} \binom{n_{i}}{k_{i}}, or equivalently

(x + y)^{|n|} = ∑_{k=0}^{n} \binom{n}{k}(x − k · t)^{|n−k|}y(y + k · t)^{|k|−1}.

We can also derive (3.6) from the first example as follows. If we substitute a, c and b by xa, xc and xb = (xb_{1}, ⋯, xb_{m}) respectively in (1.2), we get a polynomial identity in x. Equating the coefficients of x^{|n|} in both sides of this identity yields (3.6).

Note that formula (3.6) generalizes also Hurwitz’s multinomial extension of Abel’s identity [8]. To see this, setting n = 1 = (1, …, 1), q = 0 and p = 1, then (3.6) reduces to

∑_{k∈{0,1}^{m}} x(x + k · t)^{|k|−1}y(y + (1 − k) · t)^{|1−k|−1} = (x + y)(x + y + |t|)^{m−1}.

Exemple 3 Consider the set [n] = {1} × [n_{1}] + {2} × [n_{2}] + … + {m} × [n_{m}]. For each element α = (i, j) of [n] we call i the color of α. A partition π of [n] is called colored partition (cf. [19]) if each block of π is only composed of elements with different colors. Let

F (z) = exp(e_{1}(z) + e_{2}(z) + … + e_{m}(z)),

where e_{k}(z) is the k-th (1 ≤ k ≤ m) elementary symmetric function of z_{1}, z_{2}, …, z_{m}. Then the corresponding family of convolution polynomials is

F_{n}(x) = [z^{n}]e^{x(e_{1}(z)+e_{2}(z)+...+e_{m}(z))}
= \frac{\left\{{n \atop 0}\right\} + \left\{{n \atop 1}\right\}x + … + \left\{{n \atop |n|}\right\}x^{|n|}}{n!} ,

where \left\{{n \atop k}\right\} is the number of colored partitions in k blocks of [n] (see [19]). We

note that if n = (1, 1, …, 1), the number \left\{{n \atop k}\right\} is nothing but Stirling number of second kind S(m, k), that is, the number of partitions in k blocks of [m].

We have shown how to create new multinomial convolution families from given ones exactly as in the monomial case. The addiional constructions mentioned by Knuth [9] can also be extended. For example, if F_{n}(x) and G_{n}(x) are convolution families, then so is the family H_{n}(x) defined by

H_{n}(x) = ∑_{k=0}^{n} F_{k}(x)G_{n−k}(x).

as did by Knuth [9] in the case m = 1.

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