RIORDAN MATRICES IN THE RECIPROCATION OF QUADRATIC POLYNOMIALS Ana Luz´on a
a,∗
and Manuel A. Mor´onb
Departamento de Matem´atica Aplicada a los RR. NN. E.T.S. Ingenieros de Montes.
Universidad Polit´ecnica de Madrid. 28040-Madrid, SPAIN. email
[email protected] b
Departamento de Geometr´ıa y Topolog´ıa. Facultad de Matem´aticas. Universidad Complutense de Madrid. 28040- Madrid, SPAIN. ma
[email protected]
∗
Corresponding author: email:
[email protected]. Telephone number: 34913366399. Fax number: 34915439557.
ABSTRACT: We iterate contractive one-degree polynomials with coefficients in the ring K[[x]] of formal power series to calculate the reciprocal in K[[x]] of a quadratic polynomial. Doing this we meet with the structure of Riordan array. We interpret certain changes of variable as a Riordan array. We finish the paper using our techniques to find new ways to get known formulas for the sum of powers of natural numbers involving Stirling and Eulerian numbers. MSC: Primary: 41A80, 41A65. Secondary: 05E20, 05A99 Keywords: Banach’s Fixed Point Theorem, reciprocal of a quadratic polynomial, Riordan matrices, changes of variables.
1
1. Introduction
In this paper we deal with two types of arithmetical triangles. Those of the first type are remainders that appear in an iterative process, those of the second are changes of variables. We will recognize these triangles as elements of the Riordan group. Unlike the classical way of approaching the problem of the generalization of the structure of the Pascal triangle, see [22], [19], [20], we run into this structure. It appears naturally as a reminder associated to an iterative process to calculate the reciprocal of certain power series. To carry out the iterations we will use the following well-known result: Banach’s Fixed Point Theorem (BFPT) Let (X, d) be a complete metric space and f : X → X contractive. Then f has a unique fixed point x0 and f n (x) → x0 for every x ∈ X. In the above statement f n = f ◦ f ◦ · · · ◦ f . Recall that a map is contractive, concretely c-contractive, if there is a real number c ∈ [0, 1) such that d(f (x), f (y)) ≤ cd(x, y). We recommend, for example, [7] for the description of some of the applications of this result. f In [13] we construct the Riordan group from an iterative process to calculate using a g Generalized Banach Fixed Point Theorem. There we show that the structure of Riordan arrays, the reciprocation operation in the ring K[[x]] and some fixed point problems are intrinsically related. To consolidate the above idea, in this paper we expose some elements of the Riordan group as remainders of an iterative process. Actually, we show that these remainders are generalizations of the Pascal triangle, T (1 | 1−x) according to our notation. In fact we want to state that the arithmetical pattern used to construct the Pascal triangle is, intrinsically, in the reciprocation of any quadratic polynomial. So, this paper describes a natural framework where Riordan arrays appear as a consequence and not as the main objective. The other main idea is that the successive approximation method gives rise to such pattern of behavior. Due to the way we approached the Riordan group in [13], the natural notation is ours. There we find a reparametrization of a Riordan array in terms of two formal power series. Our approach allowed us, for example, to give a new algorithm to construct the Riordan arrays avoiding the A, and the Z sequences (see [14] for definitions). In fact, we can obtain with the T (f | g) notation very nice expressions for these sequences: 2
Let f , g ∈ K[[x]] with f (0), g(0) 6= 0. Denote by A and Z the A-sequence and Z-sequence of the Riordan array T (f | g). Then T −1 (1 | g) = T (1 | A). More generally: T −1 (f | g) = ³ ¯ ´ ¯A . T fg(0) (A − xZ) (0) Some words more about the notation: Between the different notations for Riordan arrays, see different authors: [2], [8], [11], [14], [21], [24], [27] and [13]. We prefer our own notation in [13] because, for our propose, is more friendly. In the next table we show one of the standard and the T (f | g) notations for elements of the Riordan group: Name Identity Pascal Remainders Fundamental equality
(d(t),th(t))
T(f |g)
(1, t) 1 t , 1−t 1−t −ct 1 −c , Q(t) a + bt a + bt f (t) t 1 , = (f (t), t) g(t) , g(t) g(t)
T (1 | 1)
Appel subgroup element
(d(t),t)
Associated subgroup element
(1,th(t))
Bell subgroup element
(d(t),td(t))
t g(t)
T (1 | 1 − t) 1 T Q | a+bt −c T (f |g) = T (f |1)T (1|g) T (d | 1) 1 1 T | h h 1 T 1| d
In the above table we suppose that f (0), g(0), h(0), d(0) 6= 0. In Section 2 we develop, in a significant example, the theory that we will describe in the following sections. In Section 3 we consider the polynomial function induced by a certain polynomial P (S) ∈ K[[x]][S] of degree one which is contractive for a suitable complete metric in K[[x]] and whose unique fixed point is just the reciprocal
1 Q
of a quadratic polynomial Q(x) = a + bx + cx2 ,
with a 6= 0. Later we compare the iterations at S = 0 of P with the Taylor polynomials of 1 . Q
In this way we define the remainder. We identify this remainder with an element of the
Riordan group. We define the family of polynomials associated to a Riordan array (which, in some sense, is to consider those arrays by rows not by columns as usual) and studying those we find that doing linear changes of variables in the polynomial we arrive to what we will call the Pascal triangle associated to the series
1 . Q
Finally we get the Pascal triangle as
product of certain matrices. There are similar results in literature. See [15], [16] and [17]. We ³ ´ 1 a+bx show a factorization of the Pascal Triangle in terms of the remainder T Q | −c and the corresponding two changes of variables. 3
³ The remainders T
1 Q
|
a+bx −c
´ has been used in the literature. When Q(1) = 0 and ac < 0,
ab < 0 the corresponding remainder is related to the description of a probabilistic modeling of a certain movement of a particle in the plane, see [4]. Actually all our remainders and linear changes of variables are 7-matrices as called in [4]. Moreover the used iteration process generates all 7-matrices if we allow to start at any series as initial condition. We will not treat it here. On the contrary we always start to iterate at S = 0. In Section 4 we focus on the study of the change of variables obtained in the Section 3. Actually, we show that, in general, if we multiply any element of Riordan group, T (f | g) by T (1 | a + bx) we are doing a change of variables in the associated family of polynomials of T (f | g). We note that the set of treated changes of variables has a representation as a subgroup of the Riordan group. Inside this subgroup we find different elements of order 2 and some conjugation relations. In Section 5 we go back to our motivating example. Using the family of polynomials associated in this case, we get some known formulas of the sums of powers of natural numbers. From now on, we consider N = {0, 1, 2, · · · } the natural numbers in the field K of characteristic 0. 2. Motivation: BFPT and the arithmetic-geometric series There is an obvious way to sum the geometric series
P∞ k=0
xk using the BFPT. See [1] for a
proof without words. It is natural to wonder if we can sum the arithmetic-geometric series P∞ k−1 using the BFPT. It is easy to see that there are not any one-degree polynomial k=1 kx P f (t) = g(x)t + h(x) and any point x0 such that the partial sum nk=0 (k + 1)xk = f n+1 (x0 ). Since f (x0 ) = g(x)x0 + h(x) = 1, f 2 (x0 ) = f (f (x0 )) = f (1) = g(x) + h(x) = 1 + 2x and f 3 (x0 ) = f (f (f (x0 ))) = f (1 + 2x) = g(x)(1 + 2x) + h(x) we obtain that x0 = − 31 and f (x) = 32 xt + 1 + 12 x and from here f 4 (− 13 ) 6= 1 + 2x + 3x2 + 4x3 . In view of this, we are going to iterate a polynomial whose fixed point is the sum of the arithmetic-geometric series, that P 1 1 k−1 2 = (1−x) is ∞ 2 . Since the equality t = (1−x)2 can be converted to t = 1 + (2x − x )t, k=1 kx we consider the polynomial f (t) = 1 + (2x − x2 )t and we do the first iterations in t = 0: f (0) = 1 f 2 (0) = 1 + 2x − x2 4
f 3 (0) = 1 + 2x + 3x2 − 4x3 + x4 f 4 (0) = 1 + 2x + 3x2 + 4x3 − 11x4 + 6x5 − x6 f 5 (0) = 1 + 2x + 3x2 + 4x3 + 5x4 − 26x5 + 23x6 − 8x7 + x8 f 6 (0) = 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 − 57x6 + 72x7 − 39x8 + 10x9 − x10 We can observe that in each iteration the partial sum appears plus a remainder. We want to control the difference with the partial sum. For it, we began to write the coefficients of the remainder, that is:
−1
−4 1 −11 6 −1 A1 = −8 1 −26 23 −57 72 −39 10 −1 −120 201 −150 59 −12 1 .. .. .. .. .. .. . . . . . . . . .
Observing the above triangle we can see some resemblances with the Pascal Triangle: the rule of construction is similar, we can obtain some numerical identities from its rows and columns, we get some known sequences, etc. In fact: Proposition 1.
(i) The rule of contruction is: an,k = 2an−1,k −an−1,k−1 for any n, k ≥
1, a0,k = 0 if k ≥ 1 and an auxiliary column an,0 = n + 1 for n ∈ N. (ii) The first column are eulerian numbers except for the sign. (iii) For every ai,j , the sum of all elements to the right in its row and all elements above Pn Pi−1 in its column is zero. k=j+1 aik = 0. k=1 akj + ¡ ¢ P k n+j−1−k n+j−2k (iv) The general term is an,j = n + j − 1 + j−1 2 . k=1 (−1) n+j−2k (v) The sum of the elements in any row are triangular numbers with negative sign. ¡ ¢ n−k P 4 . (vi) Using the general term we obtain: n = nk=1 (−1)k−1 2n−k k−1 Returning to the iterations, we can write the (n+2)-iteration as µ f
n+2
(0) = Tn+1
1 (1 − x)2 5
¶ + xn+2 pn (x)
where Tn (S) is the n-degree Taylor polynomial of S(x) at x = 0 and we call (pn (x))n∈N the family of polynomials associated to this triangle. In this case: p0 (x) = −1 p1 (x) = −4 + x p2 (x) = −11 + 6x − x2 p3 (x) = −26 + 23x − 8x2 + x3 p4 (x) = −57 + 72x − 39x2 + 10x3 − x4 In general and using the recurrence of the elements of the above triangle we get
pn+1 (x) = (2 − x)pn (x) − (n + 2) and pn (x) = −
n X
(k + 1)(2 − x)n−k
k=0
In view of the last expression we do the changes t = 2 − x and qn (t) = −pn (2 − t). If we consider the coefficients as above, for the new family of polynomials, we obtain:
1
2 3 A2 = 4 5 6 .. .
1 2 1 3 2 1 4 3 2 1 5 .. .
4 .. .
3 .. .
2 .. .
1 .. . . . .
Working with this family and its derivatives we obtain the next expressions:
qn (t) =
n X
k
(n + 1 − k)t ,
and
k=0
qn(k) (1)
µ ¶ n+2 = k! k+2
Once again, in view of the last expression it is reasonable to expand qn (t) at t = 1,
qn (t) =
¶ n µ X n+2 k=0
k+2 6
(t − 1)k
Now we can do the change s = t − 1 and consider the family rn (s) = qn (s + 1). If we put the coefficients as above we get the Pascal 1 3 6 A3 = 10 15 21 .. .
triangle with the two first columns deleted: 1 4
1
10
5
1
20 15
6
1
35 35 21 7 .. .. .. .. . . . .
1 .. . . . .
µ
¶ ¯ 1 ¯2x − 1 . We can see that the three above matrices are Riordan arrays. In fact A1 = T 2 (1 − x) ¶ µ ¯ 1 ¯ 1 . After the secAfter the first change of variable it is transformed into A2 = T 2 µ ¶ (1 − x) ¯ 1 ¯1 − x . We also have the following ond change of variable it becomes A3 = T (1 − x)2 equalities relating them: ¶ µ ¶ µ ¯ ¯ 1 1 ¯ ¯ 2x − 1 = T 1 T (1 | 2x − 1) T (1 − x)2 (1 − x)2 and
µ T
¯ 1 ¯1 (1 − x)2
¶
µ =T
¶ ¯ 1 ¯1 − x T (1 | 1 + x) (1 − x)2
Note that this means that the matrices T (1 | 2x − 1) and T (1 | 1 + x) can be interpreted as the changes of variables in the family of polynomials made before. This example motivates the next section where we will do the analogous development for any quadratic polynomial. 3. An iterative process to calculate the reciprocal of quadratic polynomials In this section we want to show that the development described in the previous one is no more than a particular example of a general phenomenon about quadratic polynomials. For any quadratic polynomial Q we find a contractive first degree polynomial function PQ 1 in K[[x]] whose unique fixed point is just . We iterate this function. So, in each iteration Q 7
1 appears plus a remainder. The study of these remainders and the Q identification of them as Riordan arrays is the first main aim of this section. The second
the partial sum of
main aim is the proof, by means of changes of variables, that the structure of Pascal triangle is in each remainder. This is the reason why we define the Pascal triangle associated to a power series. Along this section we use [13] for notation and basic results. In particular as supposed there K is always a field with characteristic zero and K[[x]] is the ring of formal power series over P K. Denote by ω(f ) the order of f = n≥0 fn xn . Recall that ω(f ) is the smallest nonnegative integer number p such that fp 6= 0 if any exist. Otherwise, that is if f = 0, we write ω(f ) = ∞. See [18] for details and for main properties of ultrametrics. We recall here some facts in [13] that we need: Proposition 2. The map d : K[[x]]×K[[x]] → R+ defined by d(f, g) = ultrametric on K[[x]]. Moreover d(f, g) ≤
1 2k+1
1 2ω(f −g)
is a complete
if and only if Tk (f )=Tk (g). Finally the sum
and product of series are continuous if we consider the corresponding product topology in K[[x]] × K[[x]] Proposition 3. Let f, h ∈ K[[x]] with f (0) = 0. Then the first degree polynomial map P : K[[x]] → K[[x]] defined by P (S) = f S + h is 12 -contractive independently on f and h. In fact d(P (S1 ), P (S2 )) = consequently
f g
1 d(S1 , S2 ). 2ω(f )
Moreover the unique fixed point of P is just
h 1−f
and
³X ´ h = fn h 1−f n≥0
P P Corollary 4. Let f, g ∈ K[[x]] with g(0) 6= 0. If f = n≥0 fn xn and g = n≥0 gn xn and P = n≥0 dn xn , then dn = − gg10 dn−1 − gg02 dn−2 · · · − ggn0 d0 + fgn0 , for n ≥ 1, d0 = fg00 .
Note that in the above corollary we iterate the function P : K[[x]] → K[[x]] defined by P (S) = ( g0g−g )S + 0
f g0
which is at least 12 -contractive and whose unique fixed point is fg .
Another of the results in [13] is the following: Algorithm for T (f | g) P P f = n≥0 fn xn , g = n≥0 gn xn with g0 6= 0, T (f | g) = (ci,j ) with i, j ≥ 1. 8
f0 f1 c11 c12 c13 c14 c15 · · · f2 c21 c22 c23 c24 c25 · · · f3 .. .
c31 .. .
c32 .. .
c33 .. .
c34 .. .
c35 · · · .. . ···
fn .. .
cn1 .. .
cn2 .. .
cn3 .. .
cn4 .. .
cn5 · · · .. .. . .
with ci,j = 0 if j > i and the following rules for i ≥ j: If j > 1 ci,j
g1 g2 gi−1 ci−1,j−1 1 = − ci−1,j − ci−2,j · · · − c1,j + = g0 g0 g0 g0 g0
à ci−1,j−1 −
i−1 X
! gk ci−k,j
k=1
and if j = 1 ci,1 = −
g1 g2 gi−1 fi−1 1 ci−1,1 − ci−2,1 · · · − c1,1 + = g0 g0 g0 g0 g0
with the agreement
P0 k=1
= 0. Note that c11 =
à fi−1 −
i−1 X
! gk ci−k,1
k=1
f0 . g0
If Q(x) = a + bx + cx2 with a 6= 0 and from Proposition 3, we have that the first degree polynomial function in K[[x]] defined by PQ (S) = (( −b )x + ( −c )x2 )S + a1 is contractive and its a a unique fixed point is
1 . Q
So we iterate this function and we compare these iterations with the ³ ´ corresponding Taylor polynomial of Q1 , Tn Q1 . In this process a lower triangle matrix (an,k ), depending on Q, appears. This matrix describes the remainder as we did in the motivating example. The following theorem shows, in particular, an algorithm to get the entries of the matrix (an,k ). Studying this matrix, and using our Algorithm for T (f | g) above we realize that (an,k ) is a Riordan array. Theorem 5. Let Q(x) = a + bx + cx2 be a polynomial with a 6= 0. Consider the one degree P −c 1 1 2 k polynomial function in the ring K[[x]], PQ (S) = (( −b )x + ( )x )S + . If = k≥0 dk x a a a Q ³ ´ P k−1 ) with then PQn+1 (0) = Tn Q1 + xn+1 ( ∞ k=1 an,k x (1) d0 = 1/a, d1 = (−b/a)d0 and dn = (−b/a)dn−1 + (−c/a)dn−2 , n ≥ 2. 9
(2) an,k = 0, ∀k > n. If we call an,0 = dn for n ∈ N, we have an,k = (−b/a)an−1,k + (−c/a)an−1,k−1 for n, k ≥ 1. Proof. First, since PQ is at least
1 -contractive 2
in the expression PQn+1 (0), that is Tn
³ ´
³ ´ 1 Q
then Tn
is just the n-degree polynomial
is obtained from PQn+1 (0) eliminating all powers, in
1 Q
the unknown, greater than n. The part (1) is well-known and it was proved again in the Corollary 4. We are going to use induction in order to prove (2). Note first that PQ (0) =
1 a
= T0 ( Q1 ). So, a0,k = 0 for k > 0,
now 1 1 1 1 PQ2 (0) = ((−b/a)x + (−c/a)x2 ) + = + (−b/a) x + x2 (−c/a2 ) a a a a consequently a1,1 = −c/a2 = (−b/a)a0,1 + (−c/a)a0,0 , because a0,1 = 0 and a0,0 = d0 =
1 a
and
a1,k = 0 for k > 1. Suppose that the result is true for m − 1 ∈ N then PQm+1 (0)
2
m−1 X
= ((−b/a)x + (−c/a)x )(
k
m
dk x + x (
1/a+
(−b/a)dk x
k=0
k+1
m−2 X
(−c/a)dk x
+
k+2
+(−c/a)dm−1 x
am−1,k xk−1 )) +
k=1
k=0
m−1 X
∞ X
m+1
+x
m+1
1 = a
∞ X ( ((−b/a)+(−c/a)x)am−1,k xk−1 ) = k=1
k=0
m X ((−b/a)dk−1 + (−c/a)dk−2 )xk + 1/a + (−b/a)d0 x + k=2
+x
m+1
∞ ∞ X X k−1 ((−b/a)am−1,1 + (−c/a)dm−1 + (−b/a)am−1,k x + (−c/a)am−1,k xk ) = k=2
= Tm (1/Q) + x
m+1
k=1
∞ X ( ((−b/a)am−1,k + (−c/a)am−1,k−1 )xk−1 ) k=1
Consequently am,k = (−b/a)am−1,k +(−c/a)am−1,k−1 . Now if k > m then am,k = (−b/a)am−1,k + (−c/a)am−1,k−1 . By induction hypothesis we have am,k = 0
¤
Let C 1 = (an,k )n,k∈N , Note that it is a lower triangular matrix. Let us expand some few term Q
of C 1 avoiding, a priori, the null entries: Q
10
1 a −b a2 b2 a3
−c a2
−
−b3 a4
c a2 2bc a3
+
.. .
c2 a3
2bc a3 −3cb2 a4
+
.. .
c2 a3
1 −xc Q a+bx
1 Q
−3bc2 a4
−c3 a4
.. .
.. .
...
1 x 2 c2 Q (a+bx)2
1 −x3 c3 Q (a+bx)3
...
Let R 1 = (bn,k )n,k∈N with bn,k = an+1,k+1 , that is, C 1 without the first column and the first Q
Q
row. As an easy and direct application of our algorithm for T (f | g) quoted above, we obtain the structure of these matrices in the following: Corollary 6. Let Q = a + bx + cx2 with a, c 6= 0. The matrices R 1 and C 1 are the Q
following Riordan arrays: µ ¶ 1 a + bx | R1 = T Q Q −c
µ and
C1 = T Q
1 a + bx a + bx | Q −c −c
Q
¶
One the principal tools in this paper is: Definition 7. Consider T (f | g) and suppose that (bi,j )i,j∈N is the associated matrix to T (f | g). Then the family of polynomials associated to T (f | g) , we denote it by (pn )n∈N , is pn (x) =
n X
bn,j xj ,
with n ∈ N
j=0
Let us consider the sequence of polynomials associated to R 1 defined by p0 (x) = b0,0 , pn (x) = Q Pn k k=0 bn,k x , n ∈ N where bn,k = an+1,k+1 are those described in Proposition 5. With this notation we can rewrite the main formula in Proposition 5 as: PQn+2 (0) = Tn+1 (1/Q) + xn+2 pn (x) or pn (x) =
PQn+2 (0) − Tn+1 ( Q1 ) xn+2
So using Proposition 5 again, we obtain easily Proposition 8. µ pn+1 (x) =
b + cx −a
¶ pn (x) + (−c/a)dn+1 , for n ≥ 0 11
Consequently pn (x) = (−c/a)
à n X
µ dn−k
k=0
b + cx −a
¶k !
In view of the above proposition it is natural to consider the change of variable, supposing c 6= 0, t =
b+cx . −a
Define qn (t) = (−a/c)pn ( at+b ). Consequently −c
(1)
qn (t) =
n X
dn−k tk
k=0
If we consider the matrix M = (mn,k )n,k∈N where the entries in the row n are the coefficients of the polynomial qn in increasing power order we have: ³ ´ Proposition 9. M = T Q1 | 1 . ³
´ | 1 is a lower triangular Toeplitz matrix whose columns are, ³ ´ 1 1 beginning at the main diagonal, the coefficients of Q , that is, dn . If now we read T Q | 1 Proof. We know that T
1 Q
by rows, avoiding the a priori null entries, we get for the first row d0 , the second: d1 , d0 , the third d2 , d1 , d0 and so on, that is, the matrix M .
¤
Related to the polynomials qn defined in (1) we obtain the next expression for the number (k)
(k)
qn (1), where qn means the kth-derivative of qn : qn(k) (1)
=
n X j=k
j! dn−j (j − k)!
Now, expanding qn (t) at t = 1, qn (t) =
n µ ¶ n X X j
k
k=0 j=k
dn−j (t − 1)k
Once more, in view of the above expression, it is natural to consider the change of variable s = t − 1. We define the next family of polynomials: rn (s) = qn (s + 1). Consequently n X n µ ¶ X j rn (s) = dn−k sk k k=0 j=k If we define the matrix L = (ln,k )n,k∈N where the n-th row is constituted by means of the coefficients of rn , as before, we get: ³ ´ Proposition 10. L = T Q1 | 1 − x . 12
Proof. We write the matrix L d0 d0 + d1
d0
d0 + d1 + d2
2d0 + d1
d0
d0 + d1 + d2 + d3 3d0 + 2d1 + d2 3d0 + d1 ↓
↓
1 1 Q 1−x
³ that is T
1 Q
..
.
↓ x2
1 x Q (1−x)2
1 Q (1−x)3
& ···
1 xn−1 Q (1−x)n
···
´ |1−x
¤
Remark 11. The above proposition reflects the known ubiquity of the Pascal triangle. More concretely the rule of construction of the Pascal triangle is in any of the remainders described before, up some linear changes of variables. P Given a power series f = n≥0 an xn , if we call T (f | 1 − x) the Pascal triangle associated to power series f, then the Pascal triangle associated to the power series f ≡ 1 is just the classical Pascal triangle. Moreover the rule of construction of T (f | 1 − x) is just the same as that of Pascal triangle. Note that T (f | 1 − x) = T (f | 1)T (1 | 1 − x). So, the last equality says that, up the changes of variables, to construct these remainders one only has to know 1 and the rule of construction of Pascal triangle. the coefficients of Q If we look again at Proposition 8 and to the first change of variable, we can say that to ³ ´ make this change of variable is just the same thing as to multiply the matrix T Q1 | a+bx by −c ¡ c+bx ¢ T 1 | −a because µ ¶ µ ¶ µ ¶ 1 1 a + bx c + bx T |1 =T | T 1| Q Q −c −a ¡ ¢ ¡ ¢ note that T −1 1 | a+bx = T 1 | c+bx . −c −a In a similar way the last change of variable, after Proposition 9, is the same as to multiply ³ ´ the matrix T Q1 | 1 by the Pascal triangle T (1 | 1 − x) because µ T
1 |1−x Q
¶
µ =T
¶ 1 | 1 T (1 | 1 − x) Q
We can summarize all above in the next factorization theorem: 13
Theorem 12. µ ¶ ¶ µ ¶ µ 1 a + bx a + bx 1 −1 T | | 1 − x T (1 | 1 − x)T 1 | =T Q −c Q −c or equivalently µ T
1 |1−x Q
¶
µ =T
1 a + bx | Q −c
¶
µ T
−1
a + bx 1| −c
¶ T (1 | 1 − x)
or equivalently µ ¶ µ ¶ µ ¶ a + bx 1 1 a + bx −1 T (1 | 1 − x) = T 1 | T | T |1−x −c Q −c Q Remark 13. Note that the global change of variable in the sequence of associated polynomials given by s=
a + b + cx −a
−a as + a + b )pn ( ) c −c ³ ´ corresponds to multiply the remainder by the matrix T 1 | a+(a+b)x . −c rn (s) = (
4. The group of generalized linear change of variables In this section we are going to relate the change of variables represented by T (1 | a + bx), and some mild generalizations, with some results and problems in the related literature. First, to reinforce the idea of change of variable, we can generalize the procedure described in the previous section to a general arithmetical triangle T (f | g): Proposition 14. Let pn (x) =
n X
cn,k xk be the associated polynomials to T (f |g) with gen-
k=0
n
1X eral term (cn,k )n,k∈N . Then the associated polynomials to T (f | ag+bx) are qn (t) = cn,k tk a k=0 with t =
x−b . a
Proof. First we observe that T (f | ag + bx) = T (f |g)T (1|a + bx). On the other hand, it is µ ¶ µ ¶k−m k b 1 − . easy to find the general term, (lk,m )n,k∈N , of T (1|a+bx), that is: lk,m = m+1 m a a 14
We define the associated polynomials to T (f | ag + bx) as pˆn (x) = P αn,m = nk=m cn,k lk,m . Then pˆn (x) =
n X
αn,m x =
m=0
=
n X n X
m
n X k X k=0 m=0
m
cn,k lk,m x =
m=0 k=m
à cn,k
n X k X
Pn m=0
αn,m xm where
cn,k lk,m xm =
k=0 m=0
µ ¶ µ ¶k−m ! b k − xm = m+1 a a m 1
µ ¶k k µ ¶ n n X 1 X k 1 1X x−b k−m m k cn,k k+1 (−b) x = cn,k k+1 (x − b) = cn,k a m a a k=0 a m=0 k=0 k=0
n X
¤ Consider now the sets LCV= {T (1 | α + βx) with α, β ∈ K and α 6= 0}, we call it the set of linear changes of variable, and GLCV= {T (λ | α + βx) with λ, α, β ∈ K and λ, α 6= 0}, we call it the set of generalized linear changes of variables. For this set we have Proposition 15. GLCV is a subgroup of the Riordan group and, of course, LCV is a subgroup of GLCV. Proof. T (1 | 1) ∈ GLCV and T (λ1 | α1 + β1 x)T −1 (λ2 | α2 + β2 x) = T ( λλ12 |
α1 +(β1 −β2 )x ) α2
¤
In order to compare with some results in the literature we have to say that the group LCV contains all the Pascal-like triangle denoted by Pb in [2]. In our notation Pb = T (1 | 1 − bx) It is clear that if the field K is algebraically closed, remember that we supposed always of characteristic zero, then GLCV contains elements of any finite order. In fact if ωn is a primitive n − th root of unity and λ is a n − th root of unity then T (λ | ωn + αx) has order n for any α ∈ K. For reasons explained by Shapiro in [23] and Cameron and Nkwanta in [2], researcher in combinatorics focus on Riordan matrices of order 2. We realize that, for combinatorial interest it is common to concentrate on elements with nonnegative entries. Following [2] we consider M = T (−1 | −1) so R = T (f | g) has pseudo order 2 if and only if RM = T (−f | −g) has order 2. Note that M has order 2. Consider f = T (1 | −1) = −M . Obviously M f is not conjugated to M in also the order 2 matrix M f = RM R−1 but it is the Riordan group. That is, there is not a Riordan matrix R with M not a negative answer to question Q8 of Shapiro in [23] because there all elements of order 2 15
considered had 1 as the first entry in the main diagonal (because of the definition of Riordan matrix considered). Our result, related to the group GLCV is the following Proposition 16. let α ∈ K, with α 6= 0. Then: f in the (i) The elements T (1|αx − 1) have order 2 and all of them are conjugated to M group LCV and then in the Riordan group. (ii) The elements T (−1|αx − 1) have order 2 and all of them are conjugated to M in the group GLCV and then in the Riordan group. Proof. The proof of the whole proposition follows from the equality, easy to check, T (1 | αx − 1) = T (1 | αx − 2)T (1 | −1)T −1 (1 | αx − 2) We found this by using elementary operations in certain related matices. In fact this equality proves all in (i) now T (−1 | αx − 1) = T (−1 | 1)T (1 | αx − 1) but T (−1 | 1) = −T (1 | 1) and then commutes with any other. Then (ii) follows multiplying by the left in (i) by T (−1 | 1) . f = M. Note also that T (−1 | 1)M ¤ For elements of order 2 in the Riordan group we have some results in the next proposition but caution: Abusing on the language, order 2 in the next proposition means that the square is the identity, that is we allow the identity to have order 2. We do it for avoid some unnecessary restrictions. Proposition 17.
(i) T (f | g) is of order 2 if and only if T (−f | g) is of order 2.
(ii) T (f | g) is of order 2 if and only if T (f n | g) is of order 2 ∀ n ∈ Z. (iii) T (f | g) is of order 2 if and only if T (f g n | g) is of order 2 ∀ n ∈ Z. Proof. In our notation an element T (f | g) has order 2 if and only if T 2 (f | g) = T (1 | 1) but T 2 (f | g) = T (f f ( xg ) | gg( xg )). So T (f | g) has order 2 if and only if gg( xg ) = 1 and f f ( xg ) = 1. Let us prove only the part (iii). Since gg( xg ) = 1 and f f ( xg ) = 1, then f g n f ( xg )g n ( xg ) = 1 because of the commutativity of the product of the series.
¤
The above and the below propositions allow us to get more elements of order 2 in the associated subgroup. Moreover these new elements of order 2 are also conjugated in this subgroup of the Riordan group, to the matrix M as asked by Shapiro in [23]. 16
Proposition 18. Let α ∈ K, with α 6= 0. Then the elements, of the associated subgroup, T (αx − 1|αx − 1) have order 2 and all of them are conjugated, inside the associated subgroup, to M = T (−1 | −1). Proof. Recall that T (1 | αx − 1) = T (1 | αx − 2)T (1 | −1)T −1 (1 | αx − 2) then T (αx − 1 | αx − 1) = T (αx − 2 | αx − 2)T (−1 | −1)T −1 (αx − 2 | αx − 2) ¤ Remark 19. Recently some positive answers to some problems posed by Shapiro in [23] has been given in [5] and [6]. They are related to involutions in the Riordan group and to the problem of conjugation with the Matrix M . After some e-mail messages with the authors of [5] we realize that a good choice of an invertible matrix B in their Theorem 2.5 for Riordan involutions D = (g(x), f (x)) conjugated to the matrix M is, using their notation, ´ ´ ³ ³ (x)) , ±(1 − f (x)) , because the original one proposed in [5] is not always B = exp Φ(x,xf 2 invertible. We choose this corresponding B in the special case in Proposition 18. 5. A special remainder: Sums of powers of natural numbers As we mentioned in the introduction one of the applications of Riordan arrays is to use them to find numerical or combinatorial identities, see for example [20], [22], [24], [25] and [27], among many others. The usual way to do that is by means of the action of a Riordan array to a particular power series or using the inverses of elements in the Riordan group. We hope that our way to use the change of variables could help, in the future, to get some others identities. But now, we want to show how the use of the family of polynomials associated to a T (f | g) can be used to get, in a different way, some equalities. To do that we choose our arithmetic-geometric series motivating example. 17
So, when we try to sum
P∞
k=0 (k + 1)x
k
using the corresponding iteration process, we consider ³ ´ 1 Q1 (x) = (1 − x)2 , and the remainder, in this case, is R 1 = T (1−x) 2 | 2x − 1 . Recall that
R
1 Q1
Q1
−1
−4 1 −11 6 −1 = −26 23 −8 1 −57 72 −39 10 −1 −120 201 −150 59 −12 1 .. .. .. .. .. .. . . . . . . . . .
We are going to obtain two different formulas for the sums of powers of natural numbers. Although both of them are known, what is new is our way to obtain that from the family of ´ ³ 1 | 2x − 1 . polynomials associated to T (1−x) 2 After our first change of the variable in the associated sequence of polynomials t =
b+cx , −a
we
obtain the sequence of polynomials qn (t) = (−a/c)pn ( at+b ) that was described as −c qn (t) =
n X
dn−k tk
k=0
For the particular case of Q1 this sequence is qn (t) = (n + 1) + nt + (n − 1)t2 + · · · + 2tn−1 + tn with the recurrence qn+1 (t) = tqn (t) + (n + 2) ¡ ¢ (k) In this case, qn (1) = k! n+2 for n, k ∈ N. Where f (k) represents the k-th derivative of f . k+2 For this sequence of polynomials we can obtain the following formula for the sequence of the derivatives, we leave the proof to the reader. Proposition 20. 0 (t) qn+1
= (n + 1)qn (t) − 2
n−1 X
qk (t)
k=0
Let us denote by (Qp )p∈N the sequence of the upper factorial polynomials. That is, Qp (x) = x(x + 1) · · · (x + p − 1), p ∈ N, Q0 (x) ≡ 1. We have our own proof of the following fact: 18
Proposition 21.
n X
Qp (k) =
k=1
Qp+1 (n) p+1
Proof. Let p ∈ N. Derivating p − 2 times in the expression in the previous proposition and valuating at t = 1 we have n+p−3 (p−1) qn+p−1 (1)
= (n + p −
(p−2) 1)qn+p−2 (1)
X
−2
(p−2)
qk
(1)
k=0
so
µ ¶ µ ¶ µ ¶ n+p−3 X n+p+1 n+p k+2 (p − 1)! = (n + p − 1)(p − 2)! −2 (p − 2)! p+1 p p k=0
Avoiding the null terms and changing the variable, l = k − p + 3, in the sum we have ¢ P ¡ (n+p)!n 2 nl=1 l+p−1 = 2 n!(p+1)! that is no more than another way to write the stated result. ¤ p We now are going to find a formula for the sum of powers of natural numbers, using the Stirling numbers, that is equivalent to that in [9] page 275 (see also [12]). Following [26], the numbers s(n, k) = (−1)n−k c(n, k) are known as the Stirling numbers of the first kind and c(n, k) are called the signless Stirling numbers of the first kind. Where the numbers c(n, k) satisfy the recurrence c(n, k) = (n − 1)c(n − 1, k) + c(n − 1, k − 1) n, k ≥ 1 with the initial conditions c(n, k) = 0 if n ≤ 0 or k ≤ 0 except for c(0, 0) = 1. On the other hand, the Stirling numbers of the second kind satisfy the following basic recurrence: S(n, k) = kS(n − 1, k) + S(n − 1, k − 1), n, k ≥ 1 with the initial condition S(0, 0) = 1. Thus the matrix s = (s(n, k))n,k∈N and the matrix S = (S(n, k))n,k∈N are mutually inverses (see [9] or [26]). Let I be the identity matrix then, ((−1)n−k c(n, k))n,k∈N · (S(n, k))n,k∈N = I Consequently the inverse of the matrix C = (c(n, k))n,k∈N is S = ((−1)n+k S(n, k))n,k∈N . Corollary 22.
Pn k=1
kp =
¡
Pp
¢
n+j p+j S(p, j)j! n−1 j=1 (−1)
where S(p, j) are the Stirling num-
bers of the second kind. Proof. In [26] we can find that Hence
Pp k=0 n X k=1
c(p, k)xk = Qp (x).
Qp (k) =
p X j=0
19
c(p, j)
n X k=1
kj
From above proposition we have p X
c(p, j)
j=0
n X
kj =
k=1
Qp+1 (n) p+1
So consider the p + 1 equalities m X
c(m, j)
j=0
n X
kj =
k=1
Qm+1 (n) , m+1
m = 0, 1 · · · p
Then using the inverse matrix S of C we obtain the announced formula.
¤
We are going to finish describing a very singular way to obtain a formula for the sum of powers using L’Hopital rule where Eulerian numbers appear without invoking them. Apart from the L’Hopital rule the main tools to obtain this formula are some equalities described in [10]. For each p ≥ 1, consider the sequence of polynomials qn,p (t) = (n + 1)p + np t + (n − 1)p t2 + · · · + 2p tn−1 + 1p tn Note that for p = 1 we have the above sequence qn (t). Proposition 23. For t 6= 1, P p g(t) + (−1)p+1 pk=1 h p−k itn+k qn−1,p (t) = (1 − t)p+1 p
where g(t) is a p-degree polynomial and h p−k i are just the Eulerian numbers. Proof. Since qn,p (t) = (n + 1)p + tqn−1,p (t) and n+1 X qn,p (t) − qn−1,p (t) = (k p − (k − 1)p )tn+1−k k=1
Then, if t 6= 1 qn−1,p (t) =
np −
So if t 6= 1, qn−1,p (t) =
(1 − t)p np −
Pn
k=1 (k
Pn
p
− (k − 1)p )tn+1−k 1−t
− 1)p − k p )tn+1−k (1 − t)p (1 − t)p+1
k=1 ((k
20
Expanding the numerator in powers of t, for our purpose we are only interested in the coefficients from tp+1 to tn+p , and after some computations we find that the numerator is: µ ¶ µ ¶ n−p p+1 p p−k X X X X p+1−j p + 1 p n+1−k p+1−j p + 1 g(t)+ (−1) (p+k−j) t + (−1) (p+1−k−j)p tn+k j j k=1 j=0 k=1 j=0 Where g(t) is a p-degree polynomial. All the coefficients in the middle term are zero (see [3] or [10] page 244). Finally, as one can see in [9] page 255, the coefficients in the third term of p
the expression are, up the factor (−1)p+1 , the Eulerian numbers h p−k i. Corollary 24.
Pn k=1
Proof. Since qn−1,p (1) =
kp = Pn k=1
Pp
¡
Pp−k
k=1
¢
j p+1 j=0 (−1) j
(p + 1 − k − j)p
¤
¡n+k¢ . p+1
k p , we only need to apply the L’Hopital rule p + 1-times to the
formula in the previuous proposition.
¤
Acknowledgment: The authors thank the referee(s) for his/her suggestions and comments which improved an earlier version of this paper. The first author was partially supported by DGES grant MAT 2005-05730-C02-02. The second author was partially supported by DGES grant MTM-2006-0825.
References [1] The Viewpoints 2000 Group. Proof without Words: Geometric Series. Mathematics Magazine. Vol.74 No 4 (Oct 2001) 320. [2] N. T. Cameron. and A. Nkwanta. On some (pseudo) involutions in the Riordan Group. Journal of Integer Sequences. Vol. 8 (2005) Article 05.3.7. [3] T. Carroll, P Hilton and J. Pedersen. Eulerian Numbers, Pseudo-Eulerian Coefficients, and Weighted Sums in Pascal’s Triangle. Niew Archief voor Wiskunde. 9 no 1 (March 1991) 41-64. [4] G.-S. Cheon, S-G. Hwang, S-H Rim, S-Z. Song. Matrices determined by a linear recurrence relation among entries. Linear Algebra Appl. 373 (2003) 89-99. [5] G.-S. Cheon, H. Kim. Simple proofs of open problems about the structure of involutions in the Riordan group. Linear Algebra Appl. In press.
(2007) doi:10.1016/jlaa.2007.08.027.
[6] G.-S. Cheon, H. Kim, L.W. Shapiro. Riordan group involutions. Linear Algebra Appl. In press. (2007) doi:10.1016/jlaa.2007.09.003. [7] J. Dugundji and A. Granas. Fixed Point Theory. Monografie Matematyczne. Tom 61. PWN- Polish Scientific Publishers. Volume I (1982.) 21
[8] G. P. Egorychev and E. V. Zima. Decomposition and group theoretic characterization of pairs of inverse relations of the Riordan type. Acta Applicandae Mathematicae. 85 (2005.) 93-109 [9] R. Graham, D. Knuth and O. Patashnik. Concrete Mathematics. Addison-Wesley. 1989. [10] P. Hilton, D. Holton and J. Pedersen. Mathematical Reflections:from a room with many windows. Springer. 2002. [11] I.C. Huang. Inverse Relations and Schauder Bases. Journal of Combinatorial Theory. S. A 97 2002 (203224). [12] D. Knuth. Johann Faulhaber and sums of powers. Math. Comp. 61 (1993) 277-294. [13] A. Luz´on and M. A. Mor´on. Ultrametrics, Banach’s fixed point theorem and the Riordan group. Discr. Appl. Math. (In press.) (2007) doi:10.1016/j.dam.2007.10.026. [14] D. Merlini, D. G. Rogers, R. Sprugnoli, M.C. Verri. On some alternative characterizations of Riordan arrays. Canadian J. Math. 49(2) (1997) 301-320. [15] A. Nkwanta and N. Knox. A note on Riordan matrices. Contemporary Mathematics 252. (1999). [16] A. Nkwanta. A Riordan matrix approach to unifying a selected class of combinatorial arrays. Congressus numeratium 160. (2003) 33-45. [17] P. Peart and L. Woodson. Triple factorization of some riordan arrays. Fibonacci Quart. 31. (1993) 121-128. [18] Alain M. Robert. A course in p-adic analysis. Graduate Texts in Mathematics. 198 Springer N.Y. 2000. [19] D.G. Rogers. Pascal triangles, Catalan numbers and renewal arrays. Discrete Math. 22 (1978) 301-310. [20] L. W. Shapiro, S. Getu, W.J. Woan and L. Woodson. The Riordan group. Discrete Appl. Math. 34 (1991) 229-239. [21] L. W. Shapiro. Bijections and the Riordan group. Theoretical Computer Science. 307 (2003) 403-413. [22] L. W. Shapiro. A Catalan triangle. Discrete Math. 14 (1976) 83-90. [23] L. W. Shapiro. Some open question about random walks, involutions, limiting distributions, and generating functions. Advan. Appl. Math. 27 (2001) 585-596. [24] R. Sprugnoli. Riordan arrays and combinatorial sums. Discrete Math. 132 (1994) 267-290. [25] R.
Sprugnoli.
Riordan
arrays
proofs
of
identities
in
Gould’s
book.
http://www.dsi.unifi.it/ resp/GouldBK.pdf (2006). [26] Richard P. Stanley. Enumerative Combinatorics. Vol. I
Cambridge University Press. 1997.
[27] X. Zhao, S. Ding and T. Wang. Some summation rules related to Riordan arrays. Discrete Math. 281 (2004) 295-307.
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