PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 97(111) (2015), 23–31 DOI: 10.2298/PIM141129002M

HOPF ALGEBRA OF PROJECTION FUNCTIONS

Žarko Mijajlović and Aleksandar Pejović

Abstract. We study Hopf algebras over projection functions of the complex vector C^{X} appropriate for computing inversion formulas from discrete mathematics. Using calculus of projection functions introduced in this way, we derived various inversion formulas, including Gould’s inversion formula and its generalizations.

1. Introduction

The main idea of the paper is explained by the following example. Let C be the set of complex numbers, Z the set of integers, N the set of nonnegative integers and C^{Z} the complex vector space. We remind that a function π_{n} : C^{Z} → C, n ∈ Z, is a projection if π_{n}(f) = f(n), f ∈ C^{Z}. Let A be the subspace of the vector space of linear functionals of C^{Z} where A is generated by projections π_{n}, n ∈ Z. We introduce an associative and commutative algebra A = (A, ·) over A defining multiplication of projections by π_{m} · π_{n} = π_{m+n} . Obviously, the power (π_{1})^{n} is equal to π_{n}. Hence, if π denotes π_{1}, then π^{n} = π_{n}. By projection calculus we shall mean calculation in the algebra A. It appears that the algebra A is very appropriate for computing various inversion formulas from discrete mathematics.

Now we proceed to our example. Let F = {f ∈ C^{Z} : ⋀ _{n<0} f(n) = 0}. Obviously F is a subspace of C^{Z} and we may identify F and C^{N}. Also, f ∈ F if and only if for all n < 0, π^{n}(f) = π_{n}(f) = 0. We shall prove by projection calculus the following well known inversion formula

(1.1) g_{n} = ∑_{k} ( {n \atop k}) f_{n−2k} ⇔ f_{n} = ∑_{2k⩽n} (−1)^{k} \frac{n}{n − k} ( {n − k \atop k} ) g_{n−2k} , f, g ∈ F.

For this purpose let us introduce the functional θ = π + π^{−1} . Then

(1.2)  θ^{n} = ∑_{k} ( {n \atop k}) π^{n−2k} .

2010 Mathematics Subject Classification: 03B35; 03C13; 03G27.

Key words and phrases: Hopf algebra, projection, inverse formula.

Partially supported by Serbian Ministry of Science, Grant III 44006.

Communicated by Gradimir Milovanović.

If f ∈ F and g ∈ F is defined by g_{n} = g(n) = θ^{n}(f), n ∈ N, then by (1.2) we have g_{n} = ∑ _{k} \binom{n}{k}f_{n−2k} . For the proof of equivalence (1.1), we express π^{n} by a polynomial of θ using Tchebychev polynomials. Let T_{n}(x) be a Tchebychev polynomial of the first kind and C_{n}(x) = \frac{1}{2}T_{n}(2x). Then C_{n}(x) is also called Tchebychev polynomial of the first kind and it is well known C_{n}(x) satisfies the following identities (see for example [1, 13]):

(1.3)  C_{n}(x + x^{−1} ) = x^{n} + x^{−n} , n ⩾ 0,
(1.4)  C_{n}(x) = ∑_{2k⩽n} (−1)^{k} \frac{n}{n − k} ( {n − k \atop k} ) x^{n−2k} , n > 0.

Hence, we have

(1.5)  π^{n} = C_{n}(θ) − π^{−n} , n ∈ Z.

Therefore, if f ∈ F and n > 0 then, as π^{−n} (f) = 0, we have f_{n} = π^{n}(f) = C_{n}(θ)(f) − π^{n}(f) = C_{n}(θ)(f). Hence

f_{n} = ∑_{2k⩽n} (−1)^{k} \frac{n}{n − k} ( {n − k \atop k} ) θ^{n−2k} (f) = ∑_{2k⩽n} (−1)^{k} \frac{n}{n − k} ( {n − k \atop k} ) g_{n−2k} .

Thus we proved direction (⇒) of equivalence (1.1). The other direction follows from the following observation. The equalities g_{m} = ∑ _{k} \binom{m}{k}f_{m−2k} , m = 0, 1, …, n, can be written as P ·F = G, where P is a regular triangular matrix, G = [g_{0}, g_{1}, …, g_{n}] and F = [f_{0}, f_{1}, …, f_{n}]. Then the righthand side of the equivalence (1.1) is written as F = Q · G where Q = P^{−1} and entries of Q are exactly coefficients appearing in expansion of f_{n} by g_{n−2k} . As from F = Q · G follows P · F = G, it also follows the direction (⇐) in (1.1).

2. Hopf algebra of projection functions

We show that the algebra A discussed in the previous section and similar algebras naturally bear the structure of Hopf algebra. Even if the next definitions and analysis can be applied to an arbitrary field K, we shall assume K = C. Let I ⊆ C, I ≠ ∅ and A the subspace of the vector space of linear functionals of the complex space C^{I}, generated by projection functions π_{i}, i ∈ I. Suppose that I is a subgroup of the additive group of C or of the multiplicative part C^{∗} of C. Assuming the usual notation for Hopf algebras and related notions (see for example [4] or [18]), it is easy to see that in both of the following two cases we obtain a Hopf algebra.

Additive case, I is a subgroup of (C, +, 0). Hopf algebra H_{I} = (A, ▽, 1, △, ε) over the complex field C is defined as follows: ▽(π_{i} ⊗ π_{j}) = π_{i+j} , in multiplicative notation π_{i} · π_{j} = π_{i+j} , 1(z) = π_{0}, z ∈ C, △(π_{i}) = π_{i} ⊗ π_{i} and ε(π_{i}) = 0, i, j ∈ I. The map a: π_{i} ↦ π_{−i} , i ∈ I, is the antipod.

Multiplicative case, I is a subgroup of (C^{∗}, ·, 1). Hopf algebra H_{I} = (A, ▽, 1, △, ε) over the complex field C is defined taking: ▽(π_{i} ⊗ π_{j}) = π_{ij} , in multiplicative notation π_{i} · π_{j} = π_{ij} , 1(z) = π_{1}, z ∈ C, △(π_{i}) = π_{i} ⊗ π_{i} and ε(π_{i}) = 1, i, j ∈ I. The map a: π_{i} ↦ π_{i^{−1}} , i ∈ I, is the antipod.

Obviously, in both cases H_{I} is commutative and in fact H_{I} is a Hopf subalgebra of the dual Hopf algebra of the group Hopf algebra C[I]. If I = (Z, +, 0) we obtain algebra A presented in the previous section.

Even if the additive and the multiplicative cases are similarly defined, they may produce examples of quite different nature. In an additive case if we take π = π_{1}, we may write π^{i} instead of π_{i} and if I is the set of real numbers, then the ring A_{I} = (A, +, ·, 0, 1) is an integral domain and it is isomorphic to the ring of polynomials over C in the variable π, see [5] and [14]. If I is the additive group of integers, then A_{I} = ℤ[π, π^{−1} ] is the ring of Laurent polynomials in the indeterminate π. On the other hand, in the multiplicative case, if I = ⟨ε⟩, ε^{n} = 1, then A_{I} has divisors of zero. For example, if ε is a primitive root of x^{3} − 1 and a = 1 + π_{ε} + π_{ε^{2}} , b = 1 + ε^{2}π_{ε} + επ_{ε^{2}} , then ab = 0.

Let S ⊆ I and S′ = I∖S. Then we can identify the space C^{S} with the subspace F_{S} = {f ∈ C^{Z} : ⋀ _{x∈S′} f(x) = 0} of C^{I}. In the example from the previous section, obviously S is the set of nonnegative integers. For deriving inversion formulas for functions f: S → C, we shall use their replicas in F_{S}. This derivation is related but not the same as that one in the umbral calculus, see for example [3]. We also note that Tchebychev polynomials and their variants have been already the subject of investigation in context of Hopf algebras and from the purely algebraic point of view, see for example [2] and [6].

3. Linear functional θ = π + π^{−r}

Let us suppose notation and definitions as previously introduced. Here we shall consider linear functionals of C^{Z} of the form

θ = π + π^{−r} , r is a nonnegative integer,

in the ring A_{I} = ℤ[π, π^{−1} ]. Then for m = r + 1

(3.1)  θ^{n} = ∑_{k} ( {n \atop k}) π^{n−mk} , n = 1, 2, . . ..

We shall prove that π^{n} can be expressed as stated in the following theorem.

Theorem 3.1. There are polynomials P_{n}(x) and Q_{n}(x) with integer coefficients such that

(3.2)  π^{n} = S_{n}(θ) − Q_{n}(π^{−1} ), Q_{n}(0) = 0.

Our main aim is to find explicitly polynomials S_{n}(x) and Q_{n}(x). For this purpose, we shall need some properties of symmetric functions related to the polynomial p(x) = x^{m} +ax^{m−1} +b, a, b ∈ C. Let λ_{1}, λ_{2}, …, λ_{m} be the roots of p(x) and s_{n}(a, b) = λ_{1}^{n} + λ_{2}^{n} + ⋯ + λ_{m}^{n} the n^{th} power sum of the roots. Using Girard-Waring formula for symmetric functions, Gould (see [9] or [12]) derived the formula

(3.3)  s_{n}(a, b) = ∑_{0⩽k⩽n/m} (−1)^{n−rk} \frac{n}{n − rk} ( {n − rk \atop k} ) a^{n−mk} b^{k}, m = r + 1.

Using this formula, the following proposition is easily deduced.

Proposition 3.1. Let m = r + 1 and u_{n} = s_{n}(−a, b), where s_{n}(a, b) is defined

by (3.3). Then for all positive integers n

(3.4)  u_{n} = ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) a^{n−mk} b^{k}.

If 1 ⩽ n ⩽ r, then u_{n} = a^{n} and also u_{m} = a^{m} − mb. The sequence u_{n} with these initial conditions is the unique solution of the difference equation

(3.5)  v_{n+1} − av_{n} + bv_{n−r} = 0.

Proof. Identity (3.4) immediately follows from (3.3). For the second part of the proposition, we can write (3.4) as

(3.6)  u_{n} = a^{n} + ∑_{1⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) a^{n−mk} b^{k}.

If n ⩽ r, then the sum in (3.6) is empty, hence u_{n} = a^{n}. Similarly,

u_{m} = a^{m} − mb + ∑_{2⩽k⩽n/m} (−1)^{k} \frac{m}{m − rk} ( {m − rk \atop k} ) a^{m−mk} b^{k} = a^{m} − mb.

The sequence u_{n} satisfies (3.5) since u_{n} is a linear combination of the roots of the characteristic equation of (3.5). The order of this recurrence is m and values for u_{1}, u_{2}, …, u_{m} are determined, hence the uniqueness follows. □

Now we proceed to the proof of identity (3.2). For this purpose, we shall deliver recurrence relations for the polynomials S_{n}(θ) and R_{n}(π).

Lemma 3.1. Let r be a nonnegative integer, m = r + 1, θ = π + π^{−r} and 1 ⩽ l ⩽ r. Then for the polynomials in (3.2) we can take:

(a) S_{l}(θ) = θ^{l} and S_{m} = θ^{m} − m.

(b) Q_{l}(t) = t^{rl} ((1+t^{−m} )^{l}−t^{−ml} ), Q_{m}(t) = t^{rm} ((1+t^{−m} )^{m} −t^{−m^{2}} −mt^{m−m^{2}} ).

Proof. By identity (3.1), after short calculation we have

π^{l} = θ^{l} − ∑_{k⩾1} ( {l \atop k}) π^{l−mk} = θ^{l} − π^{−rl} ((1 + π^{m})^{l} − π^{ml}).

Hence, S_{l}(θ) = θ^{l} and Q_{l}(π^{−1} ) = π^{−rl} ((1 + π^{m})^{l} − π^{ml}) for 1 ⩽ l ⩽ r. Taking t = π^{−1} , we have Q_{l}(t) = t^{rl} ((1 + t^{−m} )^{l} − t^{−ml} ).

According to identity (3.1) we also have

π^{m} = θ^{m} − m − ∑_{k⩾2} ( {m \atop k}) π^{m−mk} ,

hence

π^{m} = (θ^{m} − m) − π^{−rm} ((1 + π^{m})^{m} − π^{m^{2}} − mπ^{m^{2} −m} ).

Therefore S_{m} = θ^{n} − m and Q_{m}(π^{−1} ) = π^{−rm} ((1 + π^{m})^{m} − π^{m^{2}} − mπ^{m^{2} −m} ). Taking t = π^{−1} , we have Q_{m}(t) = t^{rm} ((1 + t^{−m} )^{m} − t^{−m^{2}} − mt^{−m^{2} +m} ). □

Now we shall deliver the recursive relations for polynomials S_{n}(t) and Q_{n}(t) appearing in (3.2). In the following we shall take m = r + 1.

Assuming identity (3.2), we have θπ^{n} = θS_{n}(θ) − θR_{n}(π^{−1} ), hence

\begin{aligned}(π + π^{−r} )π^{n} & = θS_{n}(θ) − (π + π^{−r} )Q_{n}(π^{−1} ), \\ π^{n+1} & = θS_{n}(θ) − π^{n−r} − (π + π^{−r} )R_{n}(π^{−1} ). \end{aligned} ; (3.7)

Assuming recurrence (3.2) for n − r, i.e., π^{n−r} = S_{n−r} (θ) − Q_{n−r} (π^{−1} ), we obtain

(3.8)  π^{n+1} = θS_{n}(θ) − S_{n−r} (θ) − ((π + π^{−r} )Q_{n}(π^{−1} ) − Q_{n−r} (π^{−1} )).

With regard to (3.2) we have π^{n+1} = S_{n+1} (θ) − Q_{n+1} (π^{−1} ) and comparing with (3.8), we have

\begin{aligned}S_{n+1} (θ) & = θS_{n}(θ) − S_{n−r} (θ), \\ Q_{n+1} (π^{−1} ) & = (π + π^{−r} )Q_{n}(π^{−1} ) − Q_{n−r} (π^{−1} ). \end{aligned} ; (3.9)

Taking the substitution t = π^{−1} and using (3.9) it is easy to deduce that Q_{n}(t) satisfies the recurrence

(3.10)  tQ_{n+1} (t) = (1 + t^{r+1} )Q_{n}(t) − tQ_{n−r} (t).

Lemma 3.2. We have Q_{n}(0) = 0 for all n > 0.

Proof. Let m = r + 1. First assume 1 ⩽ n ⩽ r. With regard to Lema 3.1 we have

Q_{n}(t) = t^{rn} ((1 + t^{−m} )^{n} − t^{−mn} )

= t^{rn} ( 1 + ( {n \atop 1}) t^{−m} + ⋯+ ( {n \atop n − 1} ) t^{−m(n−1)} ) .

As rn − m(n − 1) = m − n > 0, it follows that t | Q_{n}(t). Further,

Q_{m}(t) = t^{rm} ((1 + t^{−m} )^{m} − t^{−m^{2}} − mt^{−m^{2} +m} )

= t^{rm} ( 1 + ( {m \atop 1}) t^{−m} + ⋯+ ( {m \atop m − 2} ) t^{−m(m−2)} ) .

As −m^{2} + 2m + rm = m > 0, it follows that t | Q_{n}(t).

Therefore, we proved that t | Q_{n}(t) for n ⩽ m, i.e., Q_{n}(0) = 0. For n > m we use the recurrence (3.10). We see immediately that t | Q_{n}(t). □

Corollary 3.1. The constant term of Q_{n}(t) is equal to 0.

Proof of Theorem 3.1. The proof immediately follows by induction from recurrence relations 3.9, derivations 3.7, 3.8 and the previous corollary. □

Now we deliver the explicit forms of the polynomials S_{n}(x) and Q_{n}(t).

Proposition 3.2. Let r be a positive integer, m = r+1 and assume a sequence of polynomials S_{n}(x) ∈ ℤ[x] satisfies:

(a) S_{n+1} (x) = xS_{n}(x) − S_{n−r} (x)

(b) S_{l}(x) = x^{l}, 1 ⩽ l ⩽ r, and S_{m}(x) = x^{m} − m.

Then for all positive integers n

(3.11)  S_{n}(x) = ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) x^{n−mk} .

Proof. The characteristic equation of the recurrence (a) is

(3.12)  λ^{m} − xλ^{m−1} + 1 = 0.

It is easy to see that equation (3.12) has no multiple roots, hence if λ_{1}, λ_{2}, …, λ_{m} are roots of (3.12) then

S_{n}(x) = c_{1}λ_{1}^{n} + c_{2}λ_{2}^{n} + ⋯+ c_{m}λ_{m}^{n},

for some unique constants c_{1}, c_{2}, …, c_{m}. As any linear combination of the n-th powers of λ_{1}, λ_{2}, …, λ_{m} satisfies recurrence (3.11), due to the initial conditions (b) and the uniqueness of the constants c_{1}, c_{2}, …, c_{m}, according to Proposition 3.1 we have c_{i} = 1, i ⩽ m, and

{□ \atop (3.13)} S_{n}(x) = s_{n}(−x, 1) = ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) x^{n−mk} .

We note that for the most applications the explicit form 3.13 of the polynomials S_{n}(t) and the recurrence 3.2 are sufficient. However, if one wants to find the polynomials Q_{n}(t), it is possible and convenient to introduce new polynomials R_{n}(t) and h_{n}(t) with integer coefficients related to Q_{n}(t) in the following way

\begin{aligned}Q_{n}(t^{−1} ) & = t^{−rn} R_{n}(t), n ∈ N, \\ R_{n}(t) & = h_{n}(t^{m}), n ∈ N. \end{aligned} ; (3.14)

We also note that the polynomials S_{n}(x) are related to the so called incomplete polynomials, see [17], and to the orthogonal polynomials on the radial rays in the complex plane which were introduced by Milovanović in [15] and studied in details in [16].

It is easy to prove the following proposition.

Proposition 3.3. Let r be a nonnegative integer, m = r + 1, θ = π + π^{−r} and 1 ⩽ l ⩽ r. Then

(a) R_{l}(π) = (1 + π^{m})^{l} − π^{ml}, R_{m}(π) = (1 + π^{m})^{m} − π^{m^{2}} − mπ^{m^{2} −m} .

(b) deg(R_{n}(π)) < rn, n ∈ N.

(c) h_{n+1} (t) = (1 + t)h_{n}(x) − t^{r}h_{n−r} (x)

(d) h_{l}(x) = (1 + t)^{l} − t^{l}, 1 ⩽ l ⩽ r, and h_{m}(x) = (1 + t)^{m} − t^{m} − mt^{m−1} .

From the next theorem and relations 3.14 immediately follows the explicit form of the polynomial Q_{n}(t).

Theorem 3.2. Assume that a sequence of polynomials h_{n}(t) ∈ ℤ[t] satisfies conditions (c) and (d) in the previous proposition. Then for all positive integers n

(3.15)  h_{n}(t) = f_{n}(t) + g_{n}(t),

where

f_{n}(t) = (t + 1)^{n} − t^{n}, g_{n}(t) = ∑_{1⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) (1 + t)^{n−mk} t^{rk} .

Proof. The characteristic equation of the recurrence (a) is

λ^{m} − (1 + t)λ^{m−1} + t^{r} = 0.

The roots λ_{1}, λ_{2}, …, λ_{m} of this equation are distinct and its one root is t, so we can take λ_{r+1} = λ_{m} = t. Let u_{n} = λ_{1}^{n} + λ_{2}^{n} + ⋯ + λ_{r}^{n} and s_{n} = u_{n} + λ_{m}^{n} = u_{n} + t^{n}. From Proposition 3.1, s_{n} = (1 + t)^{n} for 1 ⩽ n ⩽ r and s_{m} = (1 + t)^{m} − mt^{m−1} . Hence, u_{n} = (1 + t)^{n} − t^{n} for 1 ⩽ n ⩽ r and u_{m} = (1 + t)^{m} − t^{m} − mt^{m−1} . Also, u_{n} obviously satisfies the recurrence u_{n+1} (t) = (1+t)u_{n}(t)−t^{r}u_{n−r} (t). Therefore, due to the the same initial conditions for u_{n} and h_{n}, by induction we have immediately u_{n} = h_{n} for all n, i.e., h_{n}(t) = s_{n} − t^{n}. From Proposition 3.1, we have

s_{n} = ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) (1 + t)^{n−mk} t^{rk} .

wherefrom we immediately deliver (3.15). □

4. Gould inversion formula

As an application of the projection calculus and a the operator θ introduced in the previous section, we prove the Gould inversion formula, see [7] or [8]. This formula is a generalization of inversion formula 1.1 and it states

g_{n} = ∑_{k} ( {n \atop k}) f_{n−mk} ⇔ f_{n} = ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) g_{n−mk} ,

where f, g ∈ F, r is a nonnegative integer and m = r + 1. For the proof we use the same technique as in the case r = 1 where we used projection calculus, Tchebychev polynomials and their crucial properties 1.3 and 1.4. By use of this technique, the proof of the Gould formula directly follows from Theorem 3.1 and the explicit form 3.13 of the polynomial S_{n}(x).

Using linear functional identities (3.1) and (3.1) and explicit forms of the polynomials S_{n}(θ) and Q_{n}(π^{−1} ), we can generalize the Gould inversion formula. Namely, we can find new inversion formulas for

(4.1)  g_{n} = ∑_{k} ( {n \atop k}) f_{n−mk+l} , 0 ⩽ l < m.

It is particularly simple to deliver the inversion formula for the case m = 2. To see this, assume m = 2 and let us introduce the new linear functional σ_{n} = πθ^{n}. Then σ_{n} = ∑ _{k} ( {n \atop k}) π^{n−2k+1} . By (1.5), i.e., inversion formula π^{n} = C_{n}(θ) − π^{−n} ,

we have π^{n+1} = πC_{n}(θ) − π^{1−n} and so

π^{n+1} = ∑_{k} (−1)^{k} \frac{n}{n − k} ( {n − k \atop k} ) σ_{n−2k+1} − π^{1−n} .

Hence we obtain the following inversion formula for (4.1), case m = 2, l = 1:

(4.2)  f_{n+1} = ∑ (−1)^{k} \frac{n}{n − k} ( {n − k \atop k} ) g_{n−2k+1} − f_{1−n} .

We note that for a given sequence g ∈ F, the sequence f is not uniquely determined by (4.2) as indices in g_{i} are shifted by one:

g_{0} = f_{1}, g_{1} = f_{2} + f_{0}, g_{2} = f_{3} + 2f_{1}, …

while

f_{1} = g_{0}, f_{2} = g_{1} − f_{0}, f_{3} = g_{2} − 2g_{0}, …

where f_{0} is arbitrary. Also note that f_{1−n} vanishes for n > 1. In a similar manner we can obtain the inversion formula for (4.1) in the general case. So assume (4.1) and let us introduce the linear functional σ_{n} = π^{l}θ^{n}. From Theorem 3.1 and Proposition 3.2 we have

π^{n+l} = π^{l}S_{n}(θ) − π^{l}Q_{n}(π^{−1} )

= ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) σ_{n−mk+l} − π^{l}Q_{n}(π^{−1} ).

By (3.14), Lemma 3.2 and as Q_{n}(0) = 0

Q_{n}(t^{−1} ) = t^{−rn} h_{n}(t^{m}) = c_{n}t^{−λ_{n}} + c′_{n}t^{−λ_{n} −m} + c′′_{n}t^{−λ_{n} −2m} + ⋯

where 1 ⩽ λ_{n} ⩽ m. Hence π^{l}Q_{n}(π^{−1} ) = c_{n}π^{l−λ_{n}} + H_{n}(π^{−1} ) for some polynomial H_{n}(t) with integer coefficients. Therefore we obtain the inversion formula for (4.1), 0 ⩽ l < m:

f_{n+l} = ∑_{0⩽k⩽n/m} (−1)^{k} \frac{n}{n − rk} ( {n − rk \atop k} ) g_{n−mk+l} − c_{n}f_{l−λ_{n}} .

We see that f_{0}, f_{1}, …, f_{l} can be chosen arbitrarily. The coefficient c_{n} can be obtained from the representation of the polynomial h_{n} given by Theorem 3.2. Here we shall find the power λ_{n}:

Proposition 4.1. We have λ_{n} = m − ρ_{m}(n) where ρ_{m}(n) is the remainder of division of n by m (remainder function).

Proof. Note that m = r + 1. As Q_{n}(t) = t^{rn} h_{n}(t^{−m} ) and Q_{n}(0) = 0, for powers t^{rn−lm} of terms in Q_{n}(t) we have rn − lm > 0, so l < rn/m. First assume m | n. Then for the smallest power t^{λ_{n}} = t^{rn−l̄n} in Q_{n}(t) we have l̄ = rn/m − 1, so λ_{n} = rn − l̄m = m = m − ρ_{m}(n). Assume m ∤ n. Then for l̄= [rn/m] we have l̄= [n − n/m] = n − 1 − [n/m], so λ_{n} = rn − l̄m = m − ρ_{m}(n). □

In the similar way we can deliver various inversion formulas such as appearing in [11] by studying associated functionals in the Hopf algebra A. For example, for delivering the inverse formula for g_{n} = f_{n} + f_{n−1} + f_{n−2} , see [10], one may use the functional θ = π + π^{−1} + π^{−2} .

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State University of Novi Pazar (Received 29 11 2014) Novi Pazar Serbia

 zarkom@matf.bg.ac.rs

Institute of Mathematics Serbian Academy of Science and Arts Belgrade Serbia

 pejovica@mi.sanu.ac.rs