Binomial sequences Andrzej Nowicki Toru´n30.09.2017
Contents
1 Introduction 1
2 Notations and preliminary facts 2
3 Initial properties of principal sequences 4
4 Principal power series 6
5 Properties of binomial sequences 10
6 Linear operators of type zero 11
7 Examples of binomial sequences 17 7.1 Successive powers of x ...... 17 7.2 Lower and upper factorials ...... 18 7.3 Abel polynomials ...... 19 7.4 Laguerre polynomials ...... 23 7.5 Other examples ...... 23
1 Introduction
Throughout this article K is a field of characteristic zero, K[x] is the ring of poly- nomials in one variable x over K, and K[x, y] is the ring of polynomials in two variables x, y over K. Moreover, K[x][[t]] is the ring of formal power series in one variable t over K[x]. Let F = (F (x)) be a nonzero sequence of polynomials in K[x]. We say that F n n>0 is a binomial sequence if
n X n F (x + y) = F (x) F (y) n k k n−k k=0 for all n > 0. The equalities are in the ring K[x, y]. The assumption that F is nonzero means that there exists a nonnegative integer n such that Fn(x) 6= 0. We will say that a binomial sequence F = (F (x)) is strict if every polynomial F (x) is nonzero. n n>0 n The well known binomial theorem can be stated by saying that (xn) is a strict n>0 binomial sequence. Several other such strict sequences exist. The sequence of lower factorials x(n) , defined by x(0) = 1 and x(n) = x(x−1)(x−2) ··· (x−n+1) for n n>0 > 1, is a strict binomial sequence. The same property has the sequence of upper factorials Andrzej Nowicki, 2017, Binomial sequences 2
x(n) , defined by x(0) = 1 and x(n) = x(x + 1)(x + 2) ··· (x + n − 1) for n 1. n>0 > The sequence of Abel’s polynomials (A (x)) , defined by A (x) = 1 and A (x) = n n>0 0 n n−1 x(x − n) for n > 1, is a strict binomial sequence (see Subsection 7.3). Many interesting results concerning binomial sequences can be find for example in [13], [3], [20], [21], [23], [5], [15], and others. There exists a full description of all strict binomial sequences. The important role of such description play results of I. M. Sheffer [24], on linear operators of type zero, published in 1939. Later, in 1957, H. L. Krall [13], applying these results, proved that F = (F (x)) is a strict binomial sequence if and only if there exists a formal power n n>0 ∞ P n series H(t) = ant , belonging to K[[t]] with a1 6= 0 and without the constant term, n=1 such that ∞ X Fn(x) tn = exH(t). n! n=0 In Section 6 we present his proof and some basic properties of linear operators of type zero. Several other proofs and applications of this result can be find; see for example: [21], [20] and [12]. We have here the assumption that F is strict. In the known proofs this assumption is very important. In this case every polynomial Fn(x) is nonzero and moreover, deg Fn(x) = n for all n > 0. However there exist non-strict binomial sequences. We have the obvious example F = (1, 0, 0,... ). The sequence (F (x)) defined by n n>0 (2m)! F (x) = xm and F (x) = 0 for all m 0, 2m m! 2m+1 > is also a non-strict binomial sequence. Some other such examples are in Section 7. In this article we present a description of all binomial sequences. We prove (see Theorem 5.5) that if in the above mentioned result of Krall [13] we omit the assumption a1 6= 0, then this result is also valid for non-strict binomial sequences.
2 Notations and preliminary facts
We denote by N the set {1, 2, 3,... }, of all natural numbers, and by N0 the set {0, 1, 2,... } = N ∪ {0}, of all nonnegative integers. D E If i1, . . . , is ∈ N0, where s > 1, then we denote by i1, . . . , is the generalized Newton integer (i + ··· + i )! 1 s . i1! ··· is! D E D ED E i1+i2 In particular, hi1i = 1, hi1, i2i = , i1, i2, i3 = i1 + i2, i3 i1, i2 . i1
Let F = (F ) be a nonzero sequence of polynomials belonging to K[x]. Let us n n>0 repeat that F is a binomial sequence if X Fn(x + y) = hi, jiFi(x)Fj(y) for all n > 0. i+j=n Andrzej Nowicki, 2017, Binomial sequences 3
We shall say that F is a principal sequence, if X Fn(x + y) = Fi(x)Fj(y) for all n > 0. i+j=n Here the sums range over all pairs of nonnegative integers (i, j) such that i + j = n. We say that a binomial sequence F is strict if all the polynomials Fn are nonzero. Moreover, we say that a principal sequence F is strict if all the polynomials Fn are nonzero. Proposition 2.1. Let F = (F ) and P = (P ) be nonzero sequences of polyno- n n>0 n n>0 mials from K[x] such that 1 P = F for n 0. n n! n > The sequence F is binomial if and only if the sequence P is principal. Moreover, F is a strict binomial sequence if and only if P is a strict principal sequence. Proof. Assume that F is binomial. Then we have
1 1 P P 1 1 Pn(x + y) = n! Fn(x + y) = n! hi, jiFi(x)Fj(y) = i! Fi(x) j! Fj(y) i+j=n i+j=n P = Pi(x)Pj(y). i+j=n
Hence, it is clear that P is principal. The opposite implication is also clear. Thus, if we have a result for principal sequences, then by the above proposition we obtain a similar result for binomial sequences. Let R be a commutative ring with identity. We shall denote by R<
∞ ! X X n f ∗ g = hi, jiaibj t . n=0 i+j=n
∞ P n The ring R<
∞ ! X X n (f ∗ g) ∗ h = f ∗ (g ∗ h) = hi, j, kiaibjck t . n=0 i+j+k=n If R is a domain containing Q, then R<
Proposition 2.2. If Q ⊂ R, then the rings R<
∞ ! ∞ X X f(n) σ f(n)tn = tn n! n=0 n=0 is an isomorphism of rings.
Proof. It is clear that σ is a bijection, σ(1) = 1 and σ(F + G) = σ(F ) + σ(G) for ∞ ∞ ∞ F,G ∈ R<
∞ ∞ ! P 1 n P 1 P n σ(FG) = n! (f ∗ g)(n)t = n! hi, jif(i)g(j) t n=0 n=0 i+j=n
∞ ! ∞ ∞ P P f(i) g(j) n P f(n) n P g(n) n = i! j! t = n! t n! t = σ(F )σ(G). n=0 i+j=n n=0 n=0
This completes the proof.
3 Initial properties of principal sequences
Proposition 3.1. If P = (P ) is a principal sequence, then P = 1. n n>0 0
Proof. Suppose P0 = 0. Then P1(x) = P1(x + 0) = P0(x)P1(0) + P1(x)P0(x) = 0 + 0 = 0. Let n > 2 and assume that P0 = P1 = ··· = Pn−1 = 0. Then
n−1 n−1 X X Pn(x) = Pn(x + 0) = P0(x)Pn(0) + Pn(x)P0(0) + Pk(x)Pn−k(0) = 0 + 0 + 0 = 0. k=1 k=1
Hence, by induction, Pn = 0. Thus, if P0 = 0 then P is the zero sequence; but it is a contradiction because by definition every principal sequence is nonzero. n n−1 Therefore, P0 6= 0. Let P0 = pnx +pn−1x +···+p0, where n > 0, p0, . . . , pn ∈ K, and pn 6= 0. Since P0(x + x) = P0(x)P0(x), we have the equality
n n n−1 n−1 2 2n 2 2 pnx + 2 pn−1x + ··· + 2p1x + p0 = pnx + ··· + p0.
2 If n > 1, then pn = 0 but this contradicts the assumption pn 6= 0. Thus, n = 0 and 2 2 P0 = p0 ∈ K r {0}. Moreover, p0 = p0, because P0(0) = P0(0 + 0) = P0(0) . Hence, P0 = p0 = 1.
Proposition 3.2. If P = (P ) is a principal sequence, then P (0) = 0 for all n 1. n n>0 n >
Proof. We already know from Proposition 3.1 that P0 = 1. This implies that P1(0) = P1(0 + 0) = P1(0) + P1(0), so P1(0) = 0. Let n > 1 and assume that P1(0) = P2(0) = ··· = Pn(0) = 0. Then X Pn+1(0) = Pn+1(0 + 0) = Pi(0)Pj(0) = Pn+1(0) + Pn+1(0) i+j=n+1 Andrzej Nowicki, 2017, Binomial sequences 5 and so, Pn+1(0) = 0.
Assume that P = (P ) is an arbitrary principal sequence. We do not assume that n n>0 P is strict. There exist many non-strict such sequences. For example P = (1, 0, 0,... ) is a non-strict principal sequence. Next such examples we may obtain by the following proposition.
Proposition 3.3. Let (P ) be a principal sequence and let s be a positive integer. n n>0 Let (R ) be a sequence of polynomials defined by n n>0
Rms = Pm for m > 0, and R = 0 when s n. Then (R ) is a non-strict principal sequence. n - n n>0 P Proof. It is obvious that Rn(x + y) = 0 = Gi(x)Gj(y) in the case when i+j=n P P s - n. If n = sm with m ∈ N0, then Ri(x)Rj(y) = Rsi(x)Rsj(y) = i+j=sm si+sj=bm P Pi(x)Pj(y) = Pm(x + y) = Rsm(x + y). i+j=m Note also the following general property of principal sequences.
Proposition 3.4. Let (P (x)) be a principal sequence of polynomials from K[x] and n n>0 let 0 6= a ∈ K. Let n Rn(x) = a Pn(x) for n > 0. Then (R (x)) is a principal sequence. n n>0 Proof. We have
n n P Rn(x + y) = a Pn(x + y) = a Pi(x)Pj(y) i+j=n P i j P = (a Pi(x)) (a Pj(y)) = Ri(x)Rj(y). i+j=n i+j=n
This completes the proof.
In the next proposition we characterize strict principal sequences.
Proposition 3.5. Let (P (x)) be a strict principal sequence. Then: n n>0 (1) P1(x) = ax for some 0 6= a ∈ K; (2) deg Pn(x) = n for all n > 0; 1 n n (3) the initial monomial of each Pn(x) is equal to n! a x .
m m−1 Proof. Let P1(x) = amx +am−1x +···+a0, where m > 0 and a0, . . . , am ∈ K with am 6= 0. Since P1(x + y) = P0(x)P1(y) + P1(x)P0(y) and P0(x) = P0(y) = 1, we m have (putting y = x) P1(2x) = 2P1(x) and so, (2 − 2)am = 0. Hence, m = 1 because am 6= 0. We know also that F1(0) = 0 (see Proposition 3.2). Therefore,
P1(x) = ax for some 0 6= a ∈ K. Andrzej Nowicki, 2017, Binomial sequences 6
Now let s > 2 and assume that the initial monomial of every Pk(x), for k = 1, . . . , s−1, 1 k k is equal to k! a x . Look at the equality
s−1 X Ps(2x) − 2Ps(x) = Pk(x)Ps−k(x). k=1 On the right side we have a polynomial and its initial monomial is equal to
s−1 s−1 X 1 1 1 X s 2s − 2 akxk as−kxs−k = asxs = asxs 6= 0. k! (s − k)! s! k s! k=1 k=1
m m−1 This implies that Ps(x) 6= 0. Let Ps(x) = amx + am−1x + ··· + a0, where m > 0 and a0, . . . , am ∈ K with am 6= 0. Then the initial monomial of Fs(2x) − 2Fs(x) is m m equal to (2 − 2) amx . Hence, 2s − 2 (2m − 2) a xm = asxs m s!
1 s and hence, m = s and am = s! a . Therefore, deg Ps(x) = s and the initial monomial 1 s s of Ps(x) equals s! a x . This completes the proof.
Colorary 3.6. A principal sequence (P ) is strict if and only if P 6= 0. n n>0 1
4 Principal power series
In this section K[x][[t]] is the ring of formal power series over K[x] in one variable t. Every element of this ring is of the form
∞ X n P (x) = Pn(x)t , n=0 where (P (x)) is a sequence of polynomials belonging to K[x]. We shall say that n n>0 the series P (x) is principal if (P (x)) is a principal sequence. n n>0
∞ P n Proposition 4.1. Let P (x) = Pn(x)t ∈ K[x][[t]]. The series P (x) is principal if n=0 and only if in the ring K[x, y][[t]] it satisfies the equality
P (x + y) = P (x)P (y).
Proof. Assume that the series P (x) is principal. Then (P (x)) is a principal n n>0 sequence, and then
∞ ∞ ! P n P P n P (x + y) = Pn(x + y)t = Pi(x)Pj(y) t n=0 n=0 i+j=n
∞ ∞ P n P n = Pn(x)t Pn(y)t = P (x)P (y). n=0 n=0 Andrzej Nowicki, 2017, Binomial sequences 7
Thus if P (x) is principal, then it is clear that P (x + y) = P (x)P (y). The opposite implication is also clear. ∞ P n Let F = Fnt be a formal power series belonging to K[X][[t]], and let G = n=0 ∞ P n Gnt ∈ K[x][[t]] be a formal power series without the constant term. Consider the n=1 substitution ∞ ∞ ! X X j n F (G) = Fn Gjt t . n=0 j=1 Since G has no the constant term, F (G) is a formal power series belonging to K[x][[t]]. ∞ P 1 n Let us use this substitution for the power series F = e = n! t and G = xH(t) where n=0 ∞ P n H(t) = ant ∈ K[[t]]. Denote this substitution by P (x). Thus, we have n=1
xH(t) 1 2 3 P (x) = e = P0(x) + P1(x)t + P2(x)t + P3(x)t + ··· , where each Pj(x) is a polynomial belonging to K[x]. Initial examples:
P0(x) = 1,
P1(x) = a1x, 1 2 2 P2(x) = 2 a1x + a2x, 1 3 3 2 P3(x) = 6 a1x + a1a2x + a3x, 1 4 4 1 2 3 2 1 2 2 P4(x) = 24 a1x + 2 a1a2x + a1a3x + 2 a2x + a4x, 1 5 5 1 3 4 1 2 2 3 2 P5(x) = 120 a1x + 6 a1a2x + 2 (a1a3 + a1a2) x + (a1a4 + a2a3) x + a5x. Proposition 4.2. Let H(t) ∈ K[[t]] be a formal power series without the constant term, and let P (x) = exH(t). Then P (x) is a formal power series belonging to K[x][[t]] and this series is principal. ∞ ∞ P n P n 0 Moreover, if P (x) = Pn(x)t and H(t) = ant , then an = Pn(0) for all n > 1, n=0 n=1 0 where each Pn(x) is the derivative of Pn(x).
Proof. Since H(t) is without the constant term, the substitution exH(t) is well defined and it is really an element of K[x][[t]]. Moreover,
P (x + y) = e(x+y)H(t) = exH(t)+yH(t) = exH(t)eyH(t) = P (x)P (y).
Hence, by Proposition 4.1, the series P (x) is principal. d Now we use the derivation dx of the ring K[x][[t]], and we have
∞ X 0 n 0 xH(t)0 xH(t) Pn(x)t = P (x) = e = H(t)e . n=1
∞ ∞ P 0 n 0 P n 0 Hence, Pn(0)t = H(t)e = H(t) = ant and hence, an = Pn(0) for all n > 1. n=1 n=1 Andrzej Nowicki, 2017, Binomial sequences 8
Now we shall prove that every principal power series is of the above form exH(t), where H(t) ∈ K[[t]] is a power series without the constant term. Before our proof, let us recall some well known properties of formal power series.
Assume that R is a commutative ring with identity containing the field Q, of rational numbers, and let R[[t]] be the ring of formal power series over R. Denote by M the ideal tR[[t]], and let 1+M = {1 + f; f ∈ M}. Note that M is the set of all power series from R[[t]] without the constant terms, and 1 + M is the set of all power series from R[[t]] with constant terms equal to 1. We have two classical functions Log : 1 + M → M and Exp : M → 1 + M, defined by
∞ 1 2 1 3 1 4 P (−1)n+1 n Log (1 + ξ) = ξ − 2 ξ + 3 ξ − 4 ξ + ... = n ξ , n=1 ∞ 1 2 1 3 P 1 n ξ Exp (ξ) = 1 + ξ + 2! ξ + 3! ξ + ... = n! ξ = e , n=0 for all ξ ∈ M. It is well known that Log (Exp (ξ)) = ξ and Exp (Log (1 + ξ)) = 1 + ξ, for all ξ ∈ M. As a consequence of these facts we obtain
Lemma 4.3. With the above notations: (1) if ξ, ν ∈ M and eξ = eν, then ξ = ν; (2) for every ξ ∈ M there exists a unique ν ∈ M such that eν = 1 + ξ.
Now let R be the polynomial ring K[x], where K is a field of characteristic zero.
Lemma 4.4. Let F (x) be a polynomial from K[x] such that
(x + y)F (x + y) = xF (x) + yF (y).
Then F (x) ∈ K.
Proof. Suppose that F (x) 6∈ K. Let deg F (x) = n > 1, and let F (x) = n n−1 anx + an−1x + ··· + a0, where a0, . . . , an ∈ K with an 6= 0. Putting y = x, we have n n 2xF (2x) = 2xF (x) and so, F (2x) = F (x). This implies that 2 an = an, so 2 = 1 and we have the contradiction: 0 = n > 1.
Now we are ready to prove the following main result of this section.
Theorem 4.5. Let P = (P (x)) be a nonzero sequence of polynomials from K[x]. n n>0 Then P is a principal sequence if and only if there exists a formal power series H(t), belonging to K[[t]] and without the constant term, such that
∞ X n xH(t) Pn(x)t = e . n=0
∞ P n Proof. Put P (x) = Pn(x)t . We already know (see Proposition 4.2) that if n=0 P (x) = exH(t) where H(t) ∈ K[[t]] is without the constant term, then P is principal. Now assume that P is principal. Since P is nonzero, we know by Proposition 3.1 that P0(x) = 1. Thus, by Lemma 4.3(2), there exists a formal power series B(x) ∈ Andrzej Nowicki, 2017, Binomial sequences 9
∞ B(x) P n K[x][[t]], without the constant term, such that P (x) = e . Put B(x) = Bn(x)t , n=1 where each Bn(x) is a polynomial from K[x]. Observe that, by Proposition 3.2, we have P (0) = 1. Hence, 1 = P (0) = eB(0) and hence, by Lemma 4.3(1), we have the equality B(0) = 0. Therefore, Bn(0) = 0 for all n > 1. This implies that for every n > 1 there ∞ P n exists a polynomial An(x) ∈ K[x] such that Bn(x) = xAn(x). Put A(x) = An(x)t . n=1 Then B(x) = xA(x), and we have P (x) = exA(x), where A(x) is a power series from K[x][[t]] without the constant term. Since P is principal, we know, by Proposition 4.1, that P (x + y) = P (x)P (y). Hence e(x+y)A(x+y) = P (x + y) = P (x)P (y) = exA(x)eyA(y) = exA(x)+yA(y) and hence, (x + y)A(x + y) = xA(x) + yA(y) (see Lemma 4.3(1)), that is,
∞ ∞ X n X n (x + y)An(x + y) t = xAn(x) + yAn(y) t . n=1 n=1
So (x+y)An(x+y) = xAn(x)+yAn(y) for all n > 1 and so, by Lemma 4.4, every An(x) ∞ P n xH(t) is a constant an belonging to K. Consequently A(x) = ant . Thus, P (x) = e , n=1 ∞ P n where H(t) = ant . This completes the proof. n=1
The next propositions are consequences of the above theorem.
Proposition 4.6. If A(x),B(x) ∈ K[x][[t]] are principal power series, then the product A(x)B(x) is a principal power series.
Proof. It follows from Theorem 4.5 that A(x) = exH1(t) and B(x) = exH2(t), where H1(t),H2(2) are some formal power series from K[[t]] without the constant terms. Then xH(t) A(x)B(x) = e , where H(t) = H1(t) + H2(t) is a formal power series from K[[t]] without the constant term. Hence, again by Theorem 4.5, the power series A(x)B(x) is principal.
∞ P n Proposition 4.7. Let P (x) = Pn(x)t ∈ K[x][[t]] be a principal power series. n=0 Then P (x) is invertible in K[x][[t]], and the inverse P (x)−1 is a principal power series. Moreover, ∞ −1 X n P (x) = Pn(−x)t . n=0 Proof. It follows from Theorem 4.5 that P (x) = exH(t), where H(t) is a formal power series from K[[t]] without the constant term. Then P (x)P (−x) = exH(t)e−xH(t) = e0 = 1, and hence P (x)−1 = P (−x) = ex(−H(t)), and, again by Theorem 4.5, the series −1 P (x) is principal. Thus, the set of all principal power series from K[x][[t]] is a subgroup of the multi- plicative group of the ring K[x][[t]]. Andrzej Nowicki, 2017, Binomial sequences 10
5 Properties of binomial sequences
In the previous sections we proved several essential properties of principal sequences.
Let us recall (see Proposition 2.1) that a sequence of polynomials (Pn(x))n 0 is principal > if and only if n!Pn(x) is a binomial sequence. The following propositions are n>0 immediate consequences of Proposition 2.1 and the suitable propositions from Section 3. Proposition 5.1. Let F = Fn(x) be a binomial sequence. Then: n>0 (1) F0(x) = 1; (2) Fn(0) = 0 for all n > 1. (3) Let s be a positive integer, and let G = (G (x)) be a sequence defined by n n>0 (ms)! G (x) = F (x) for m 0, ms m! m > and Gn(x) = 0 when s - n. Then G is a binomial sequence. (4) Let 0 6= a ∈ K. Let G (x) = anF (x) for n 0. Then (G (x)) is a binomial n n > n n>0 sequence.
Proposition 5.2. If F = (F (x)) is a strict binomial sequence, then n n>0 (1) F1(x) = ax for some 0 6= a ∈ K; (2) deg Fn(x) = n for all n > 0; n n (3) the initial monomial of each Fn(x) equals a x .
Proof. Use Propositions 3.5 and 2.1.
Colorary 5.3. A binomial sequence (F ) is strict if and only if F 6= 0. n n>0 1 Proposition 5.4. Let H(t) ∈ K[[t]] be a formal power series without the constant term, and let ∞ X Fn(x) exH(t) = tn. n! n=0 ∞ Then (F (x)) is a binomial sequence. Moreover, if H(t) = P a tn, then n!a = n n>0 n n n=1 0 0 Fn(0) for all n > 1, where each Fn(x) is the derivative of Fn(x).
The following theorem is the main result of this article. It is an extension of Krall’s result [13] mentioned in Introduction.
Theorem 5.5. Let F = (F (x)) be a nonzero sequence of polynomials from K[x]. n n>0 Then F is a binomial sequence if and only if there exists a formal power series H(t), belonging to K[x][[t]] and without the constant term, such that
∞ X Fn(x) tn = exH(t). n! n=0 Andrzej Nowicki, 2017, Binomial sequences 11
Proof. Use Theorem 4.5 and 2.1. Let us recall (see Section 2) that we denote by K[x]<
Proposition 5.6. If F,G ∈ K[x]<
∞ P n Proposition 5.7. Let F = Fn(x)t be a formal power series belonging to K[x]<< n=0 t>>. If F is binomial, then F is invertible in K[x]<
∞ −1 X n F = Fn(−x)t . n=0 Proposition 5.8. The set of all binomial series from K[x]<
It follows from Theorem 5.5 that every binomial sequence (F (x)) is uniquely n n>0 ∞ P Fn(x) n xH(t) determined by the formula n! t = e , where H(t) ∈ K[[t]] is a formal power n=0 series without the constant term. Thus for every sequence H = (h1, h2,... ) of elements of K we obtain the unique nonzero binomial sequence F = (F (x)) defined by the n n>0 ∞ ∞ P Fn(x) n xH(t) P n formula n! t = e , where H(t) = hnt . In this case we shall say that F is n=0 n=1 the binomial sequence determined by H(t).
∞ Proposition 5.9. Let H(t) = P h tn ∈ K[[t]], and let (F ) be the binomial se- n n n>0 n=1 quence determined by H(t). Let 0 6= a ∈ K. Then (anF ) is the binomial sequence n n>0 ∞ P n n determined by H(at) = hna t . n=1
Proof. This proposition follows from Theorem 5.5 and Proposition 5.1(4).
6 Linear operators of type zero
In this section we consider strict binomial sequences. We recall some important results of I. M. Sheffer [24] and H. L. Krall [13], mentioned in Introduction. Throughout d this section we denote by d the ordinary derivative dx . Assume that F is a polynomial belonging to K[x]. We know that dn(F ) = 0 for all n > deg F . Moreover, dn(xn) = n! and
n m m−n d (x ) = m(m − 1) ··· (m − n + 1)x for n 6 m. Andrzej Nowicki, 2017, Binomial sequences 12
Proposition 6.1. Let J : K[x] → K[x] be a K-linear map. Then there exists a unique sequence (L (x)) , of polynomials from K[x], such that n n>0
∞ X n J(F ) = Ln(x)d (F ) n=0 for every F ∈ K[x].
n Proof. Put Fn = J(x ) for all n ∈ N0. We define the Ln(x) recurrently by the relation n n X n−k Fn = J(x ) = Lk(x) · n(n − 1) ··· (n − k + 1)x , k=0 for n > 0. That is,
L0 = F0,
L1 = F1 − xL0, 1 2 L2 = 2 (F2 − x L0 − 2xL1) , 1 3 2 L3 = 6 (F3 − x L0 − 3x L1 − 6xL2) , 1 4 3 2 L4 = 24 (F4 − x L0 − 4x L1 − 12x L2 − 24xL3) , and so on. Then, for every m ∈ N0, we have the equality
∞ m X n m J(x ) = Ln(x)d (x ). n=0
∞ P n But the mappings J and d are K-linear, hence J(F ) = Ln(x)d (F ), for all F ∈ K[x]. n=0 It is obvious that such sequence (L (x)) is unique. n n>0 Thus, for every K-linear mapping J : K[x] → K[x] we have the unique sequence (L (x)) associated with J. In this case the mapping J is said to be an operator of n n>0 type zero ([24], [13]) if its associated sequence is of the form: Ln(x) = cn ∈ K for all n > 0 with c0 = 0 and c1 6= 0, that is, if
2 3 J(F ) = c1d(F ) + c2d (F ) + c3d (F ) + ··· for all F ∈ K[x], where cn ∈ K for n > 1 and c1 6= 0. There are many interesting papers on operators of type zero, their generalizations and applications ([1], [25]). Now we present some properties of operators of type zero.
Proposition 6.2 ([24]). Let J be an operator of type zero. If F ∈ K[x] is a nonzero polynomial of degree n > 1, then J(F ) is a nonzero polynomial of degree n − 1.
2 n Proof. Put J = c1d + c2d + ··· , with c1 6= 0, and let F = anx + ··· + a1x + a0, n−1 where a0, . . . , an ∈ K, an 6= 0. Then d(F ) = nanx + ··· is a nonzero polynomial of degree n − 1, and the degrees of all the polynomials d2(F ), d3(F ),... are smaller than ( n − 1. Since c1 6= 0, the polynomial J(F ) = c1d(F ) + c2d F ) + ··· is nonzero, and its degree is equal to n − 1. Andrzej Nowicki, 2017, Binomial sequences 13
Proposition 6.3. Let J be an operator of type zero, and let G ∈ K[x] be a nonzero polynomial of degree n − 1 > 0. Then there exists a unique polynomial F ∈ K[x] of degree n such that J(F ) = G and F (0) = 0.
2 n−1 Proof. Put J = c1d + c2d + ··· , with c1 6= 0, and G = g0 + g1x + ··· + gn−1x , where g0, . . . , gn−1 ∈ K, gn−1 6= 0. We shall construct a polynomial
2 n F = f1x + f2x + ··· + fnx with f1, . . . , fn ∈ K and fn 6= 0, such that J(F ) = G. If 1 6 j 6 m, the we use the notation: w(m, j) = m(m − 1) ··· (m − j + 1).
n j P k−j Observe that, for all j −1, . . . , n, we have d (F ) = w(k, j)fkx . If G = J(F ), then k=j we have the following equalities:
n n n P j P P k−j G = cjd (F ) = cj w(k, j)fkx j=1 j=1 k=j
0 1 2 n−1 = c1 w(1, 1)f1x + w(2, 1)f2x + w(3, 1)f3x + ··· + w(n, 1)fnx
0 1 2 n−2 + c2 w(2, 2)f2x + w(3, 2)f3x + w(4, 2)f4x + ··· + w(n, 2)fnx
0 1 2 n−3 + c3 w(3, 3)f3x + w(4, 3)f4x + w(5, 3)f5x + ··· + w(n, 3)fnx . . 0 1 + cn−1 w(n − 1, n − 1)fn−1x + w(n, n − 1)fnx
0 + cn w(n, n)fnx .
n−1 Comparing the coefficients of x , we have gn−1 = c1w(n, 1)fn = nc1fn. But nc1 6= 0, 1 so fn = gn−1. Thus, if J(F ) = G, then the coefficient fn uniquely determined. Now nc1 n−2 compare the coefficients of x . We have gn−2 = (n − 1)c1fn−1 + c2w(n, 2)fn. But fn is already constructed and (n − 1)c1 6= 0, so the coefficient fn−1 is also uniquely determined.
Repeating this procedure we obtain the coefficients fn, fn−1, . . . , f2. In the final step, we compare the coefficients of x0 and we obtain the equality
g0 = c1f1 + (2!)c2f2 + ··· + (n!)cnfn.
But the coefficients f2, f3, . . . , fn are already uniquely determined and c1 6= 0, so the coefficient f1 is also uniquely determined. This completes the proof.
As a consequence of Proposition 6.3 we obtain
Proposition 6.4 ([24]). If J is an operator of type zero, then there exists a unique sequence (B (x)) , of nonzero polynomials from K[x], such that: n n>0 (1) B0(x) = 1; (2) Bn(0) = 0 for n > 1; (3) J(Bn(x)) = Bn−1(x) for n > 0 where B−1(x) = 0. Andrzej Nowicki, 2017, Binomial sequences 14
Proof. Put B0(x) = 1.Then of course J(B0(x)) = 0 = B−1(x). Let n > 0 and assume that the polynomials B0(x),B1(x),...,Bn(x) are already defined. Then, by Proposition 6.3, there exists a unique nonzero polynomial Bn+1(x) ∈ K[x] such that Bn+1(0) = 0 and J (Bn+1(x)) = Bn(x). Thus, by induction, we obtain a uniquely determined sequence (B (x)) satisfying the given conditions. n n>0 The polynomial sequence (B (x)) from the above proposition is said to be the n n>0 basic sequence of J (see [24], [13]). We will prove that this sequence is principal.
2 Let J = c1d + c2d + ... be a fixed operator of type zero. Let us recall that c1 6= 0 and cn ∈ K for n > 1. Denote by M(t) the formal power series from K[[t]], defined by
1 2 3 M(t) = c1t + c2t + c3t + ··· .
Since M(t) is without the constant term and c1 6= 0. There exists a unique formal power series 1 2 3 H(t) = s1t + s2t s3t + · · · ∈ K[[t]]
−1 such that s1 = c1 6= 0 and H M(t) = M H(t) = t. Consider the formal power series A(x) = exH(t). This series belongs to K[x][[t]]. Put
xH(t) 2 A(x) = e = A0(x) + A1(x)t + A2(x)t + ··· , where An ∈ K[x] for all n > 0. It is clear that A0(x) = 1, An(0) = 0 for n > 1. Moreover, each An(x) is nonzero and deg An(x) = n.
Lemma 6.5 ([24]). If J and A are as above, then J An(x) = An−1(x) for n > 1.
d Proof. Let us extend the derivative d = dx : K[x] → K[x] to the derivative d : K[x][[t]] → K[x][[t]] putting d(t) = 0. Then
∞ ! ∞ X n X n d fn(x)t = d fn(x) t n=0 n=0 and, for every k > 0, we have
∞ ! ∞ k X n X k n d fn(x)t = d fn(x) t . n=0 n=0 Extend also the operator J : K[x] → K[x] to the K[[t]]-linear mapping J : K[x][[t]] → K[x][[t]] defined by ∞ X n J(ϕ) = cnd (ϕ), n=1 for ϕ ∈ K[x][[t]. Since for every F ∈ K[x] there exists an m such that dm(F ) = 0, the ∞ ∞ P p P p extended operator J is well defined. Observe that J Ap(x)t = J Ap(x) t . p=0 p=0 Andrzej Nowicki, 2017, Binomial sequences 15
In fact,
∞ ∞ ∞ ∞ ∞ P p P n P p P P n p J Ap(x)t = cnd Ap(x)t = cn d Ap(x) t p=0 n=1 p=0 n=1 p=0 ∞ ∞ ∞ ∞ P P n p P P n p = cnd Ap(x) t = cnd Ap(x) t n=1 p=0 p=0 n=1 ∞ P p = J Ap(x) t . p=0
xH(t) xH(t) k xH(t) k xH(t) Observe also that d e = H(t)e and d e = H(t) e for all k > 0. Hence,
∞ ∞ ∞ P p xH(t) P n xH(t) P n xH(t) J Ap(x)t = J e = cnd e = cnH(t) e p=0 n=1 n=1 ∞ P n xH(t) xH(t) xH(t) = cnH(t) e = M(H)e = te n=1 ∞ ∞ P p P p = t Ap(x)t = Ap−1(x)t . p=0 p=1
∞ ∞ P p P p Hence, we proved that J Ap(x) t = Ap−1(x)t and this implies that J An(x) = p=1 p=1 An−1(x) for all n > 1. This completes the proof.
Theorem 6.6 ([24]). If (B (x)) is the basic sequence of an operator J = P c dn n n>0 n n=1 of type zero. then ∞ X n xH(t) Bn(x)t = e , n=0 where H(t) ∈ K[[t]] is the formal power series (without the constant term) such that ∞ P n M(H) = H(M) = t, where M(t) = cnt . n=1
∞ xH(t) P n Proof. Put e = An(x)t . It is clear that A0(x) = 1 and An(0) = 0 for n=0 n > 1. Moreover we know, by Lemma 6.5, that J (An(x)) = Jn−1(x) for all n > 0. Hence, by Proposition 6.4, the sequence (A (x)) is the basic sequence of J. Thus, n n>0 ∞ P n xH(t) Bn(x) = An(x) for n > 0, and we have the equality Bn(x)t = e . n=0
Theorem 6.7 ([24], [13]). The basic sequence of every operator of type zero is a strict principal sequence.
Proof. This is an immediate consequence of Theorem 6.6 and Proposition 4.2.
Now we shall prove that every strict principal sequence is the basic sequence of an operator of type zero. For this aim, first we prove two lemmas. Let us recall that K is a field of characteristic zero. Andrzej Nowicki, 2017, Binomial sequences 16
Lemma 6.8. Let F (x),G(x) be two polynomials from K[x] such that F (x + y) − F (x) − F (y) = G(x + y) − G(x) − G(y). Then F (x) = G(x) + px for some p ∈ K. n n−1 n n−1 Proof. Let F (x) = anx +an−1x +···+a1x+a0 and G(x) = bnx +bn−1x + ··· + b1x + b0, where a0, . . . , an, b0, . . . , bn ∈ K. We do not assume that an 6= 0 and bn 6= 0. Putting y = x, we have the equality F (2x) − 2F (x) = G(2x) − 2G(x), that is, n n n−1 n−1 2 (2 − 2)anx + (2 − 2)an−1x + ··· + 4a2x + a0 n n n−1 n−1 2 = (2 − 2)bnx + (2 − 2)bn−1x + ··· + 4b2x + b0.
Observe that we have not the monomials a1x and b1x. This equality implies that aj = bj for j = 2, 3, . . . , n and a0 = b0. Thus, F (x) = G(x)+px where p = a1 −b1 ∈ K. Lemma 6.9. Let (P ) be a strict principal sequence. Then there exists a sequence n n>0 (c ) , of elements of K, such that c 6= 0, and for every n 1, n n>1 1 > Vn Pj = Pj−1 for j = 1, 2, . . . , n,
2 n where Vn = c1d + c2d + ··· + cnd . Proof. ([13]). We define the sequence (c ) recurrently by the following way. n n>1 We know (see Proposition 3.5) that P1 = ax for some 0 6= a ∈ K, and the initial 1 n coefficient of each polynomial Pn, for n > 1, is equal to n! a . 1 Let c1 = a and V1 = c1d. Then 1 a V P = d(ax) = = 1 = P . 1 1 a a 0 Thus, c1 is determined. Let n > 2 and assume that the elements c1, . . . , cn−1 are 2 n−1 already determined. Consider the operator Vn−1 = c1d + c2d + ··· + cn−1d . We already know that Vn−1 (Pj) = Pj−1 for j = 1, 2, . . . , n − 1. Since Vn−1 is an operator of type zero, there exists the basic sequence (B ) of V (see Proposition 6.4). It m m>0 n−1 follows from Proposition 6.3 that then Bj = Pj for all j = 0, 1, . . . , n − 1. Moreover, we know from Theorem 6.7 that (B ) is a principal sequence. Hence, m m>0 n−1 n−1 P P Pn(x + y) − Pn(x) − Pn(y) = Pk(x)Pn−k(y) = Bk(x)Bn−k(y) k=1 k=1
= Bn(x + y) − Bn(x) − Bn(y) and hence, by Lemma 6.9, Pn = Bn+px for some p ∈ K. Moreover, since B1 = P1 = ax, 1 n the initial coefficient of Bn is equal to n! a (see Proposition 3.5). We define p c = − . n an+1 n n Let Vn = c1d + ··· + cnd = Vn−1 + cnd . Then it is clear that Vn(Pj) = Pj−1 for all j = 1, 2, . . . , n − 1. We shall show that it is also true for j = n, that is, that Vn(Pn) = Pn−1. In fact, n n Vn(Pn) = Vn−1(Pn) + cnd (Pn) = Vn−1(Bn + px) + cnd (Bn + px) n p n = Vn−1(Bn) + pVn−1(x) + cnd (Bn) = Bn−1 + pc1 − an+1 a p p = Bn−1 + a − a = Bn−1 = Pn−1. This completes the proof. Andrzej Nowicki, 2017, Binomial sequences 17
Theorem 6.10 ([13]). Every strict principal sequence is the basic sequence of an operator of type zero.
Proof. Let P = (P ) be a strict principal sequence. Let (c ) be the n n>0 n n>1 sequence of elements from K, defined in Lemma 6.9. It follows from this lemma that ∞ P n P is the basic sequence of the operator cnd . n=1 Now, by Proposition 2.1 and the above facts, we obtain
Theorem 6.11 ([13]). A sequence (F ) , of polynomials from K[x], is a strict n n>0 polynomial sequence if and only if Fn is the basic sequence of an operator of type n! n>0 zero.
We will say that (c ) is a strict sequence, if c ∈ K for all n 1 and c 6= 0. n n>1 n > 1 Given an arbitrary strict sequence C = (cn)n>1, we obtain a unique strict binomial Fn sequence (Fn) such that is the basic sequence of the operator n>0 n! n>0
2 3 J = c1d + c2d + c3d + ··· .
d We call it the C-sequence. Recall that d is the ordinary derivative dx . Every polynomial Fn(x) is here nonzero, and its degree equals n. Moreover, every strict binomial sequence is a C-sequence for some strict sequence C.
7 Examples of binomial sequences
7.1 Successive powers of x It is well known that (xn) is a strict binomial sequence of polynomials. It is n>0 the first classical example of binomial sequences. It is not difficult to verify that it is the C-sequence for C = (1, 0, 0,... ), and it is the binomial sequence determined by H(t) = t. The binomial sequence (axn) , where 0 6= a ∈ K, is determined by n>0 H(t) = at.
Example 7.1. Let F (x) = (2n)! xn and F (x) = 0 for all n 0. Then (F (x)) 2n n! 2n+1 > n n>0 is the binomial sequence determined by H(t) = t2. This sequence is non-strict. Let 0 6= a ∈ K and let s be a positive integer. Let F = (F (x)) , where n n>0 (ms)! F (x) = anxn for m 0, ms m! > and Fn(x) = 0 when s - n. Then F is the binomial sequence determined by H(t) = s (at) . If s > 2, then this sequence is non=strict. Andrzej Nowicki, 2017, Binomial sequences 18
7.2 Lower and upper factorials Let a ∈ K. Consider the polynomial sequence (W (x)) defined by n n>0 ( 1, for n = 0, Wn(x) = x x + a x + 2a ··· x + (n − 1)a , for n > 1.
2 3 2 2 In particular, W1(x) = x, W2(x) = x + ax, W3(x) = x + 3ax + 2a x, and
Wn+1(x) = (x + na)Wn(x) for all n > 0. Proposition 7.2. The sequence Wn(x) is binomial. n>0
Proof. We shall show, by induction, that for all n > 0, X Wn(x + y) = hi, jiWi(x)Wj(y) i+j=n
It is obvious for n 6 1. Assume that it is true for some n > 1. Then Wn+1(x + y) = n P n = (x + y + na)Wn(x + y) = (x + y + na) k Wk(x)Wn−k(y) k=0 n n P n P n = k (x + ka)Wk(x)Wn−k(y) + k y + (n − k)a Wk(x)Wn−k(y) k=0 k=0 n n P n P n = k Wk+1(x)Wn−k(y) + k Wk(x)Wn+1−k(y) k=0 k=0 n−1 n P n P n = Wn+1(x) + Wn+1(y) + k Wk+1(x)Wn−k(y) + k Wk(x)Wn+1−k(y) k=0 k=1 n n P n P n = Wn+1(x) + Wn+1(y) + k−1 Wk(x)Wn+1−k(y) + k Wk(x)Wn+1−k(y) k=1 k=1 n P n n = Wn+1(x) + Wn+1(y) + k−1 + k Wk(x)Wn+1−k(y) k=1 n P n+1 = Wn+1(x) + Wn+1(y) + k Wk(x)Wn+1−k(y) k=1 n+1 P n+1 = k Wk(x)Wn+1−k(y). k=0
This completes the proof. Note that (W (x)) is the binomial sequence determined by n n>0 ∞ X an−1 H(t) = tn. n n=1 Two special cases of such sequences Wn(x) are well known. For a = −1 we n 0 > have the sequence x(n) of lower factorials, defined by x(0) = 1 and n>0
x(n) = x(x − 1)(x − 2) ··· (x − n + 1) for n > 1. Andrzej Nowicki, 2017, Binomial sequences 19
2 3 2 In particular, x(1) = x, x(2) = x − x, x(3) = x − 3x + 2x, and x(n+1) = (x − n)x(n). For a = 1 we have the sequence x(n) of upper factorials, defined by x(0) = 1 n>0 and (n) x = x(x + 1)(x + 2) ··· (x + n − 1) for n > 1. In particular, x(1) = x, x(2) = x2 + x, x(3) = x3 + 3x2 + 2x, and x(n+1) = (x + n)x(n). (n) It follows from Proposition 7.2 that x(n) and x are strict binomial n>0 n>0 sequences. Moreover,
1 Proposition 7.3. The sequence x(n) is the C-sequence for C = , and it is n>0 n! n>1 ∞ P (−1)n−1 n the binomial sequence determined by H(t) = n t . n=1 (n) (−1)n+1 The sequence x n 0 is the C-sequence for C = n! , and it is the bino- > n>1 ∞ P 1 n mial sequence determined by H(t) = n t . n=1
7.3 Abel polynomials
Now we examine the sequence (A (x)) of Abel polynomials, defined by n n>0 n−1 An(x) = x(x − an) , where a is an element of K. The first few polynomials:
2 2 A0(x) = 1,A1(x) = x, A2(x) = x(x − 2a),A3(x) = x(x − 6ax + 9a ). We will show that this sequence is binomial. We will prove this fact trough a series of lemmas below. Let 1 B (x) = A (x) for n 0. n n! n > Lemma 7.4. For every n > 1 and all 0 6 k 6 n − 1, 1 B(k)(x) = (x − ka)(x − na)n−1−k. n (n − k)!
(k) Here Bn (x) is the k-th derivative of Bn(x). Proof. By induction on k. It is obvious for k = 0. Assume that it is true for some k > 0. Then 0 0 (k+1) (k) 1 n−1−k Bn (x) = Bn (x) = (n−k)! (x − ka)(x − na)
1 n−1−k n−2−k = (n−k)! (x − na) + (n − 1 − k)(x − ka)(x − na)
1 n−2−k = (n−k)! (x − na) (x − na) + (n − 1 − k)(x − ka)
1 n−2−k = (n−k)! (x − na) (n − k) x − a(1 + k)
1 n−1−(k+1) = (n−(k+1))! (x − na) x − a(1 + k) .
This completes the proof of this lemma. Andrzej Nowicki, 2017, Binomial sequences 20
n X n Lemma 7.5. pan−pzn−1 = n(z + a)n−1. p p=1
n d P n n−p p n Proof. Use the derivative dz for the equality p a z = (z + a) − 1. p=1
n X n Lemma 7.6. (z + pa)an−pzp−1 = (z + a + an)(z + a)n−1 − an. p p=1
Proof. By Lemma 7.5, we have
n n n P n n−p p−1 P n n−p p P n n−p p−1 p a (z + pa)z = p a z + a p pa z p=1 p=1 p=1 = (z + a)n − an + an(z + a)n−1 = (z + a + an)(z + n + 1)(z + a)n−1 − 1.
This completes the proof.
n−1 X n n−1−k Lemma 7.7. ak x − (k + 1)a x − (n + 1)a = x(x − an)n−1 − an. k k=0 Proof. Using Lemma 7.6 for z = x − (n + 1)a, we obtain that the left side of the n−1 n P n k n−k−1 P n n−p p−1 above equality is equal to k a z + (n − k)a z = p a z + pa z k=0 p=1 n−1 n n−1 n = (z + a + an)(z + a) − a = x(x − an) − a .
Proposition 7.8. (A (x)) is a strict binomial sequence. It is the C-sequence for n n>0 1 1 1 C = 1, a, a2, a3, a4,... . 2! 3! 4!
1 n−1 2 Proof. Put c1 = 1 and cn = (n−1)! a for all n > 2, and let J = c1d + c2d + ··· . We need to show that J (Bn+1(x)) = Bn(x), that is, that
(1) (2) (n+1) Bn(x) = c1Bn+1(x) + c2Bn+1(x) + ··· + cn+1Bn+1 (x) for all n > 0. For n = 0 and n = 1 it is obvious. Assume that n > 2. Then, by the previous lemmas, we have
P (k) J (Bn+1(x)) = Bn+1(x) k=1 n n−k P ak−1 1 n = (k−1)!(n−(k−1))! (x − ka) x − (n + 1)a + n! a k=1 n−1 n−(k+1) P ak 1 n = k!(n−k)! x − (k + 1)a x − (n + 1)a + n! a k=0 n−1 n−(k+1) 1 P n k 1 n = n! k a x − (k + 1)a x − (n + 1)a + n! a k=0
1 n−1 n n 1 n−1 = n! x(x − an) − a + a = n! x(x − an) = Bn(x). Andrzej Nowicki, 2017, Binomial sequences 21
This completes the proof.
Thus, we already know that (A (x)) is a binomial sequence. It is not difficult n n>0 to check that this sequence is determined by
∞ X (−na)n−1 H(t) = tn. n! n=1 The fact that (A (x)) is a binomial sequence means that in the polynomial ring n n>0 n P n K[x, y] we have the equalities An(x + y) = k Ak(x)An−k(y) for all n > 0. Hence, k=0 for a ∈ K and n > 0, the following identity holds n X n k−1 n−k−1 (1) (x + y)(x + y − na)n−1 = x x − ka y y − (n − k)a . k k=0 Now we present a second proof of the above identity (1). In 1826, Abel deduced an identity which is
n X n (2) (x + y)n = x(x − ka)k−1(y + ka)n−k, k k=0 for a ∈ K. Many authors offered different proofs of this identity ([9], [7], [22], [8]). In 2004, M. Lipnowski [14] and G. Zheng [27] presented elegant and short proofs in Solutions of Problem 310 of Mathematical Olympiads’ Correspondence Program. There are many applications of the Abel identity ([7], [22], [11], [19], [26]).
Proposition 7.9. The identity (1) follows from the Abel identity.
Proof. Substitute in (2) the element −a to the places of a, and next substitute y + na to the places of y. Then we get
n X n k−1 n−k n (3) x x + ka y + (n − k)a = x + y + na . k k=0
n Call Un(x, y, a) the left hand side of (3). Then Un(x, y, a) = (x + y + na) , and looking at Un−1(x, y + a, a), we obtain the identity
n−1 X n − 1 k−1 n−1−k n−1 (4) x x + ka y + (n − k)a = x + y + na . k k=0
n n−k−1 P n k−1 Put P = k x x + ka) y y + (n − k)a . Then we have k=0
n k−1 n−k n P n (x + y + na) = k x x + ka y + (n − k)a k=0 n k−1 n−1−k P n = k x x + ka y + (n − k)a y + (n − k)a k=0 = P + Q, Andrzej Nowicki, 2017, Binomial sequences 22
n k−1 n−1−k P n where Q = k x x+ka (n−k)a y +(n−k)a . Using (4) and the identity k=0 n n−1 n−1 n (n − k) k = n k we get Q = na(x + y + na) . Hence, P = (x + y + na) − Q = (x + y + na)n − na(x + y + na)n−1 = (x + y + na)n−1(x + y), and hence,
n X n n−k−1 (x + y)(x + y + na)n−1 = x x + ka)k−1y y + (n − k)a . k k=0
Now, putting −a instead of a, we obtain (1). This completes the proof.
Note also the following proposition.
Proposition 7.10. The Abel identity follows from the identity (1).
Proof. Substitute in (1) the element −a to the places of a, and next substitute y + na to the places of y. Then we get
n X n k−1 n−k−1 (5) (x + y + na)(x + y)n−1 = x x − ka (y + na) y + ka . k k=0 We prove the Abel identity (2) by induction. When n = 0, then it is obvious. Assume that for n > 1,
n−1 X n − 1 (6) (x + y)n−1 = x(x − ka)k−1(y + ka)n−1−k. k k=0
n−1 n Then, by (5), (6) and the identity n k = (n − k) k , we have (x + y)n = (x + y + na)(x + y)n−1 − na(x + y)n−1 n k−1 n−k−1 P n = k x x − ka (y + ka) + (n − k)a y + ka k=0 n−1 P n−1 k−1 n−1−k −na k x(x − ka) (y + ka) k=0 n k−1 n−k P n = k x x − ka y + ka k=0 n−1 P n−1 k−1 n−1−k +na k x(x − ka) (y + ka) k=0 n−1 P n−1 k−1 n−1−k −na k x(x − ka) (y + ka) k=0 n k−1 n−k P n = k x x − ka y + ka . k=0
This completes the proof. Andrzej Nowicki, 2017, Binomial sequences 23
7.4 Laguerre polynomials Let (L (x)) be the sequence of polynomials from K[x] defined by L (x) = 1 and n n>0 0 n X n!n − 1 L (x) = xk, for n 1. n k! k − 1 > k=1 They are called the Laguerre1 polynomials ([4], [6], [20], [12]). The first few polynomials:
L1(x) = x, L2(x) = (x + 2)x, 2 L3(x) = (x + 6x + 6)x, 3 2 L4(x) = (x + 12x + 36x + 24)x, 4 3 2 L5(x) = (x + 20x + 120x + 240x + 120)x, 5 4 3 2 L6(x) = (x + 30x + 300x + 1200x + 1800x + 720)x, 6 5 4 3 2 L7(x) = (x + 42x + 630x + 4200x + 12600x + 15120x + 5040)x. Proposition 7.11. (L (x)) is the a strict binomial sequence. This sequence is n n>0 determined by ∞ X H(t) = tn. n=1
7.5 Other examples 1 1 1 Example 7.12 ([13]). Consider the strict sequence C = 1, 0, − 3! , 0, 5! , 0, − 7! , 0,... . The initial terms of the C-sequence (Fn(x)) are: 3 F0(x) = 1, F3(x) = x + x, 4 2 F1(x) = x, F4(x) = x + 4x , 2 5 3 F2(x) = x , F5(x) = x + 10x + 9x. Example 7.13. Initial terms of the C-sequence for C = (1, 1, 0, 0, 0,... ):
F0(x) = 1, F1(x) = x, F2(x) = x(x − 2), 2 F3(x) = x(x − 6x + 12), 3 2 F4(x) = x(x − 12x + 60x − 120), 4 3 2 F5(x) = x(x − 20x + 180x − 840x + 1680), 5 4 3 2 F6(x) = x(x − 30x + 420x − 3360x + 15120x − 30240), 6 5 4 3 2 F7(x) = x(x − 42x + 840x − 10080x + 75600x − 332640x + 665280). Example 7.14. Initial terms of the C-sequence for C = (1, 0, 1, 0, 0, 0,... ):
F0(x) = 1, F1(x) = x, 2 F2(x) = x , 2 F3(x) = (x − 6)x, 2 2 F4(x) = (x − 24)x , 4 2 F5(x) = (x − 60x + 360)x, 2 2 2 F6(x) = (x − 120x + 2520)x , 6 4 2 F7(x) = (x − 210x + 10080x − 60480)x.
1Edmond Nicolas Laguerre (1834-1886), a French mathematician. Andrzej Nowicki, 2017, Binomial sequences 24
Example 7.15. Initial terms of the C-sequence for C = (1, 0, 0, 1, 0, 0, 0,... ):
3 F0(x) = 1, F4(x) = (x − 24)x, 3 2 F1(x) = x, F5(x) = (x − 120)x , 2 3 3 F2(x) = x , F6(x) = (x − 360)x , 3 6 3 F3(x) = x , F7(x) = (x − 840x + 20160)x.
In [16], we find a description of C-sequences for J = ad − bdp+1 where a, b ∈ R, a 6= 0 and p > 1.
Example 7.16. Initial terms of the C-sequence for C = (1, 1, 1, 0, 0, 0, 0,... ):
F0(x) = 1, F1(x) = x, F2(x) = (x − 2)x, 2 F3(x) = (x − 6x + 6)x, 2 2 F4(x) = (x − 6) x, 4 3 2 F5(x) = (x − 20x + 120x − 120x − 480)x, 5 4 3 2 F6(x) = (x − 30x + 300x − 840x − 2520x + 10080)x, 6 5 4 3 2 F7(x) = (x − 42x + 630x − 3360x − 5040x + 90720x − 151200)x.
Example 7.17. Let (F (x)) be the binomial sequence determined by H(t) = t − n n>0 1 5 n 120 t . Then Fn(x) = x for 0 6 n 6 4 and
4 4 4 F5(x) = (x − 1)x, F8(x) = (x − 56)x , 4 2 4 5 F6(x) = (x − 6)x , F9(x) = (x − 126)x , 4 3 8 4 2 F7(x) = (x − 21)x , F10(x) = (x − 252x + 126)x .
The next example is a generalization of the previous example.
Example 7.18. Let (F (x)) be the binomial sequence determined by H(t) = t − n n>0 1 s+1 n (s+1)! t with s > 1. Then Fn(x) = x for 0 6 n 6 s and
s s+k k Fs+k(x) = x − s+1 x , for k = 1, 2, . . . , s + 1.
In the next examples we present initial terms of two non-strict binomial sequences.
Example 7.19. Let (F (x)) be the binomial sequence determined by H(t) = 1 t2 + n n>0 2 1 3 6 t . Then
3 F0(x) = 1, F7(x) = 105x , 3 F1(x) = 0, F8(x) = 35(3x + 8)x , 3 F2(x) = x, F9(x) = 140(9x + 2)x , 4 F3(x) = x, F10(x) = 315(3x + 20)x , 2 4 F4(x) = 3x , F11(x) = 1925(9x + 8)x , 2 2 4 F5(x) = 10x , F12(x) = 385(27x + 360x + 40)x , 2 5 F6(x) = 5(3x + 2)x , F13(x) = 30030(9x + 20)x . Andrzej Nowicki, 2017, Binomial sequences 25
Example 7.20. Let (F (x)) be the binomial sequence determined by H(t) = 1 t3 + n n>0 6 1 4 24 t . Then
4 F0(x) = 1, F13(x) = a13x , 4 F1(x) = 0, F14(x) = a14x , 4 F2(x) = 0, F15(x) = a15(8x + 15)x , 4 F3(x) = x, F16(x) = a16(32x + 3)x , 5 F4(x) = 3x, F17(x) = a17x , 5 F5(x) = 0, F18(x) = a18(8x + 45)x , 2 5 F6(x) = 10x , F19(x) = a19(32x + 15)x , 2 5 F7(x) = 35x , F20(x) = a20(80x + 3)x , 2 6 F8(x) = 35x , F21(x) = a21(8x + 105)x , 3 6 F9(x) = 280x , F22(x) = a22(32x + 45)x , 3 6 F10(x) = 2100x , F23(x) = a23(16x + 3)x , 3 2 6 F11(x) = 5775x , F24(x) = a24(128x + 3360x + 63)x , 3 7 F12(x) = 1925(8x + 3)x , F25(x) = a25(32x + 105)x , where a13, a14, . . . , a25 are some positive integers.
References
[1] L. Aceto and I. Cacao, A matrix approach to Sheffer polynomials, J. Math. Anal. Appl., 446 (2017) 87-100.
[2] P. Berthelot, Cohomologie cristalline des sch´emasde caract´eristique p > 0, Lecture Notes in Mathematics, Vol. 407, Springer-Verlag, Berlin, 1974.
[3] L. Brand, Binomial expansions in factorial powers, American Mathematical Monthly 67(10) (1960) 953-957.
[4] J. W. Brown, On zero types sets of Laguerre polynomials, Duke Math. J. 35 (1968) 821-823.
[5] A. di Bucchianico, Probabilistic and Analytical Aspects of the Umbral Calculus, Amsterdam, CWI, 1997.
[6] L. Carlitz, Some generating functions for Laguerre polynomials, Duke Math. J. 35 (1968) 825-827.
[7] L. Comtet, Advanced Combinatorics, D. Reidel Publ. Co., Dordrechet/Boston, 1974.
[8] S. B. Ekhad and J. E. Majewicz, A short WZ-style proof of Abel’s identitity, The Electronic Journal of Combinatorics, 3(2) (1996).
[9] D. Foata, Enumerating k-trees, Discr Math. , 1 (1971) 181-186.
[10] R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics. A Foundation for Computer Science, Adison-Wesley, 1989. Andrzej Nowicki, 2017, Binomial sequences 26
[11] F. Huang and B. Liu, The Abel-type polynomials identities, The Electronic Journal of Combinatorics, 17 (2010) 1-7.
[12] V. V. Kisil, Polynomial sequences of binomial type and path integrals, Leeds-Math- Pure, Preprint 2001.
[13] H. L. Krall, Polynomials with the binomial property, American Mathematical Monthly 64(5) (1957) 342-343.
[14] M. Lipnowski, A solution of Problem 310, Mathematical Olympiads’ Correspon- dence Program, Canada, 2004.
[15] M. Mihoubi, Bell polynomials and binomial type sequences, Discrete Mathematics 308 (2008) 2450-2459.
[16] W. Mlotkowski and A. Romanowicz, A family of sequences of binomial type, Preprint 2015.
[17] A. Nowicki, Arithmetic functions (in Polish), Podr´o˙zepo Imperium Liczb, vol.5, Second Edition, Wydawnictwo OWSIiZ, Toru´n,Olsztyn, 2012.
[18] A. Nowicki, Factorials and binomial coefficiens (in Polish), Podr´o˙zepo Imperium Liczb, vol.11, Second Edition, Wydawnictwo OWSIiZ, Toru´n,Olsztyn, 2013.
[19] P. Petrullo, Outcomes of the Abel identity, Preprint, 2011.
[20] S. Roman and G.-C. Rota, The umbral calculus, Adv. in Math. 27 (1978) 95-188.
[21] G.-C. Rota, D. Kahaner, and A. Odlyzko, Finite operator calculus, J. Math. Anal. Appl. 42 (1973) 685-760.
[22] G.-C. Rota, J. Shen, B. D. Taylor, All polynomials of binomial type are represented by Abel polynomials, Ann. Sc. Norm. Super. Pisa Cl. Sci., 25 (1997), 731-738.
[23] J. Schneider, Polynomial sequences of binomial-type arising in graph theory, Preprint 2014.
[24] I. M. Sheffer, Some properties of polynomial sets of type zero, Duke. Math. J. 5 (1939) 590-622.
[25] A. K. Shukla and S. J. Rapeli, An extension of Sheffer polynomials, Proyecciones Journal of Mathematics 30(2) (2011) 265-275.
[26] S. Sykora, An Abel’s identity and its corollaries, Preprint 2014.
[27] G. Zheng, A solution of Problem 310, Mathematical Olympiads’ Correspondence Program, Canada, 2004.
Nicolaus Copernicus University, Faculty of Mathematics and Computer Science, 87-100 Toru´n,Poland, (e-mail: [email protected]).