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Recursive sequences and girard-waring identities with applications in sequence transformation

  • Received: 01 January 2020 Revised: 01 May 2020
  • Primary: 05A15; Secondary: 05A05, 15B36, 15A06, 05A19, 11B83

  • We present here a generalized Girard-Waring identity constructed from recursive sequences. We also present the construction of Binet Girard-Waring identity and classical Girard-Waring identity by using the generalized Girard-Waring identity and divided differences. The application of the generalized Girard-Waring identity to the transformation of recursive sequences of numbers and polynomials is discussed.

    Citation: Tian-Xiao He, Peter J.-S. Shiue, Zihan Nie, Minghao Chen. Recursive sequences and girard-waring identities with applications in sequence transformation[J]. Electronic Research Archive, 2020, 28(2): 1049-1062. doi: 10.3934/era.2020057

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  • We present here a generalized Girard-Waring identity constructed from recursive sequences. We also present the construction of Binet Girard-Waring identity and classical Girard-Waring identity by using the generalized Girard-Waring identity and divided differences. The application of the generalized Girard-Waring identity to the transformation of recursive sequences of numbers and polynomials is discussed.



    Albert Girard published a class of identities in Amsterdam in 1629. Edward Waring published similar material in Cambridge in 1762-1782, which are referred as Girard-Waring identities A000330 [15]. These identities may be derived from the earlier work of Sir Isaac Newton. Surveys and some applications of these identities can be found in Comtet [2] (P. 198), Gould [3], Shapiro and one of the authors [5], and the first two authors [7]. We now give a different approach to derive Girard-Waring identities by using the Binet formula A097600 [15] of recursive sequences and divided differences. Meanwhile, this approach offers some formulas and identities that may have wider applications.

    This paper starts from an application of recursive sequences in the construction of a combinatorial identity referred to as generalized Girard-Waring identity from the Binet formula and the generating function of a recursive sequence. By using the generalized Girard-Waring identity, the Binet type Girard-Waring identity is derived, which yields the classical Girard-Waring identity by making use of divided differences. Many number and polynomial sequences can be defined, characterized, evaluated, and/or classified by linear recurrence relations with certain orders. A number sequence {an} is called sequence of order 2 if it satisfies the linear recurrence relation of order 2:

    an=pan1+qan2,n2, (1)

    for constants p,qR and q0 and initial conditions a0 and a1. Let α and β be two roots of of quadratic equation x2pxq=0. From He and Shiue [6], the general term of the sequence {an} can be presented by the following Binet formula.

    an={(a1βa0αβ)αn(a1αa0αβ)βn,ifαβ;na1αn1(n1)a0αn,ifα=β. (2)

    In the next section, from the above Binet formula we will construct a generalized Girard-Waring identity by using generating function of the recursive sequence shown in (1). Then the Binet type Girard-Waring identity will be derived accordingly. In Section 3, we present a way to construct classical Girard-Waring identity from the Binet type Girard-Waring identity by using the divided difference. Section 4 will give an application of the generalized Girard-Waring identity to the transformation of recursive sequences of numbers and polynomials.

    We now find the generating function of the sequence defined by (1).

    Proposition 1. Denote A(s)=n0ansn. Then

    A(s)=a0+(a1pa0)s1psqs2. (3)

    Furthermore, the Taylor expansion A165998 [15] of A(s) is

    A(s)=a0+n1(a1pn1+[n/2]j=11j(nj1j1)pn2j1qj(jpa0+(n2j)a1))sn.

    Proof. From the definition of A(s), we have

    A(s)=n0ansn=a0+a1s+n2ansn=a0+a1s+n2(pan1+qan2)sn=a0+a1s+psn1ansn+qs2n0ansn=a0+a1s+ps(A(s)a0)+qs2A(s).

    Hence, we obtain (3).

    We now give the Taylor series expansion of the right-hand side of (3) as follows:

    a0+(a1pa0)s1psqs2=(a0+(a1pa0)s)n0sn(p+qs)n
    =(a0+(a1pa0)s)n0nj=0(nj)pnjqjsn+j=(a0+(a1pa0)s)n0nj=0(njj)pn2jqjsn=a0n0nj=0(njj)pn2jqjsn+(a1pa0)n1nj=0(nj1j)pn2j1qjsn=a0+n1nj=0(a0(njj)pn2jqj+(a1pa0)(nj1j)pn2j1qj)sn=a0+n1nj=0(a0((njj)(nj1j))pn2jqj+a1(nj1j)pn2j1qj)sn=a0+n1(a1pn1+nj=1(a0(nj1j1)pn2jqj+a1(nj1j)pn2j1qj))sn=a0+n1(a1pn1+[n/2]j=1(nj1j1)pn2j1qj(pa0+n2jja1))sn=a0+n1(a1pn1+[n/2]j=11j(nj1j1)pn2j1qj(jpa0+(n2j)a1))sn,

    which completes the proof of the proposition.

    Corollary 1. Let (an) be the sequence defined by the recursive relation (1), and let α and β be two distinct roots of the characteristic polynomial of (1). Then we have the following generalized Girard-Waring identity:

    an=a1pn1+[n/2]j=11j(nj1j1)pn2j1qj(jpa0+(n2j)a1). (4)

    If a0=0 and a1=1, (4) implies the Binet type Girard-Waring identity

    an=αnβnαβ=[n/2]j=0(nj1j)pn2j1qj=[n/2]j=0(1)j(nj1j)(α+β)n2j1(αβ)j, (5)

    where p=α+β and q=αβ.

    Proof. From (4), we have

    an=a1pn1+[n/2]j=1(nj1j1)pn2j1qj(pa0+n2jja1), (6)

    or equivalently, (4). Hence, we obtain the following identity for all recursive sequences defined by (1)

    a1pn1+[n/2]j=11j(nj1j1)pn2j1qj(jpa0+(n2j)a1)
    =(a1βa0αβ)αn(a1αa0αβ)βn,

    where p=α+β and αβ=q, or equivalently,

    α=p+p2+4q2andβ=pp2+4q2.

    If a0=0 and a1=1, from (4) we have

    an=a1pn1+[n/2]j=1n2jj(nj1j1)pn2j1qj=a1pn1+[n/2]j=1(nj1j)pn2j1qj,

    which implies the first equation of (5) by noting (2). Substituting p=α+β and q=αβ into the first equation of (5), we obtain the second equation of (5) and complete the proof.

    Remark 1. He and Shapiro [5] used Riordan array approach to establish the Binet type Girard-Waring identity (5).

    We now prove the classical Girard-Waring identity

    αn+βn=0k[n/2](1)knnk(nkk)(α+β)n2k(αβ)k, (7)

    by using the first equation of (5). First, we need the following lemmas.

    Lemma 3.1. Let nN. Then

    jk=0nnk(nkk)(nj1+kjk)=(2nj1j). (8)

    Proof. From the Chu-Vandermonde formula and noting, the left-hand side of (8) can be written as

    LHS=jk=0nnk(nkk)(jki=0(n+nj1jki)(n+ki))=jk=0nnk(nkk)(ji=k(2nj1ji)(n+kik))=ji=0(ik=0nnk(nkk)(n+kik))(2nj1ji)=ji=0(2nj1ji)(ik=0(1)iknn2k+i(ik)(n2k+ii)),

    where on the first line

    (n+ki)=(1)i(nk)(nk1)(nki+1)i!=(1)i(nki).

    We split the inner sum of the rightmost hand of the above equation for the left-hand side of (8),

    ik=0(1)ik(ik)nn2k+i(n2k+ii),

    into two cases. For i=0, we have the above sum to be nn0+0=nn=1. For i>0, we have

    ik=0(1)ik(ik)nn2k+i(n2k+i)!(n2k)!i!=ik=0(1)ik(ik)n(n2k+i1)!(n2k)!i!=ik=0(1)ik(ik)n(n2k+i1)(n2k+i2)...(n2k+1)i!=0,

    where the last sum is zero because it is the finite difference of a polynomial with its degree one less than the order of the difference. Hence, the left-hand side of (8) becomes

    LHS=ji=0(2nj1ji)(ik=0(1)iknn2k+i(ik)(n2k+ii))=(2nj1j).

    Lemma 3.2. Let α and β be two roots of the characteristic polynomial of the recursive relation(1). Then

    0k[n/2](1)knnk(nkk)(α+β)n2k(αβ)k0k[n/2](1)k(nk1k)(α+β)n2k1(αβ)k=0kn(1)k(2nk1k)(α+β)2n2k1(αβ)k. (9)

    Proof. The left-hand side of (9) can be written as

    0k[n/2]0i[n/2](1)k+innk(nkk)(ni1i)(α+β)2n2(k+i)1(αβ)k+i=0k[n/2]0jn(1)jnnk(nkk)(n+kj1jk)(α+β)2n2j1(αβ)j
    =0jn(1)j(0k[n/2]nnk(nkk)(n+kj1jk))(α+β)2n2j1(αβ)j,

    where by using Lemma 3.1 the inner sum can be written as

    0kjnnk(nkk)(n+kj1jk)=(2nj1j).

    Thus we obtain (9).

    To prove Girard-Waring identity (7), we only need to mention that (9) implies

    0k[n/2](1)knnk(nkk)(α+β)n2k(αβ)k=0kn(1)k(2nk1k)(α+β)2n2k1(αβ)k/0k[n/2](1)k(nk1k)(α+β)n2k1(αβ)k=α2nβ2nαβ/αnβnαβ=αn+βn.

    There are some alternative forms of formula (7). As an example, we give the following one. If x+y+z=0, then (7) gives

    xn+yn=0k[n/2](1)knnk(nkk)(z)n2k(xy)k=(1)nzn+1k[n/2](1)nknnk(nkk)zn2k(xy)k,

    which implies

    xn+yn(1)nzn=1k[n/2](1)nknnk(nkk)zn2k(xy)k.

    Thus, when n is even, we have formula

    xn+ynzn=1k[n/2](1)nknnk(nkk)zn2k(xy)k, (10)

    while for odd n we have

    xn+yn+zn=1k[n/2](1)nknnk(nkk)zn2k(xy)k, (11)

    where x+y+z=0. Consequently, if n=3, then

    x3+y3+z3=3xyz, (12)

    which was shown in He and Shiue [7]. Saul and Andreescu [14] have shown that the cube vanishes if x=(y+z). Note that Euler used this to solve the general cubic. In [7], the following proposition as an application of (12) was presented.

    Proposition 2. Let x,yN. Then pxy(x+y)|(xp+yp(x+y)p) when p3 is a prime.

    As a source of Binet Girard-Waring identity, the generalized Girard-Waring identity (4) has many applications including a simple way in transferring recursive sequences of numbers and polynomials. For example, we consider Chebyshev polynomials of the first kind A028297 [15] defined by

    Tn(x)=2xTn1(x)Tn2(x) (13)

    for all n2 and T0(x)=1 and T1(x)=x. Then from Corollary 1, we have

    Tn(x)=x(2x)n1+[n/2]j=11j(nj1j1)(2x)n2j1(1)j(2xj+(n2j)x)=2n1xn+n[n/2]j=11j(nj1j1)(1)j2n2j1xn2j.

    Similarly, for Lucas numbers A000032 [15] defined by

    Ln=Ln1+Ln2

    for all n2 and L0=2 and L1=1, we have

    Ln=1+[n/2]j=11j(nj1j1)(2j+(n2j))=1+n[n/2]j=11j(nj1j1).

    From the expressions of Tn(x) and Ln, we may see that

    Tn(i2)=2n1(i2)n+n[n/2]j=11j(nj1j1)(1)j2n2j1(i2)n2j=(1)nin2+n[n/2]j=11j(nj1j1)(1)j21(1)n2jin2j=(1)nin2+n[n/2]j=11j(nj1j1)(1)nin2=i3n2Ln,

    or equivalently,

    Ln=2inT(i2),

    where i=1.

    In general, we have the following result for transferring a certain class of recursive sequences to the Chebyshev polynomial sequence of the first kind at certain points.

    Theorem 4.1. Let {an}n0 be a sequence defined by (1) with pa0=2a1, a00, and let {Tn(x)}n0 be the Chebyshev polynomial sequence of the first kind defined by (13). Then

    an=2a1pn1(2x0)nTn(x0), (14)

    where

    x0=±ip2q. (15)

    Namely, an shown in (14) can be expressed as

    an=(i)na0qn/2Tn(±ip2q). (16)

    Proof. From (14) we have

    Tn(x)=(2x)n2(1+n[n/2]j=11j(nj1j1)(1)j(2x)2j).

    If pa0=2a1, then from (4), we have

    an=a1pn1+[n/2]j=11j(nj1j1)pn2j1qjna1=a1pn1(1+n[n/2]j=11j(nj1j1)(qp2)j).

    Setting an=CnTn(x0) and using the latest forms of Tn(x) and an, we obtain

    a1pn1(1+n[n/2]j=11j(nj1j1)(qp2)j)=Cn(2x0)n2(1+n[n/2]j=11j(nj1j1)(1)j(2x0)2j),

    which implies

    a1pn1=Cn2(2x0)n (17)

    for the value x0 satisfying

    (2x0)2=qp2.

    Thus, we solve the last equation to get x0 shown in (15). Substituting x0 into (17) and solving for C yields

    Cn=2a1pn1(2x0)n=2a1pn1/(±ipq)n.

    By substituting x0 shown in (15) into (14) and noting 2a1=pa0, we obtain (16).

    Theorem 4.1 can be extended to recursive polynomial case.

    Corollary 2. Let {an(x)}n0 be a recursive polynomial sequence defined by

    an(x)=p(x)an1(x)+qan2(x)

    for n2, where p(x)R[x] and qR, with initial conditions a0(x) and a1(x) satisfying p(x)a0(x)=2a1(x). Then

    an(x)=(i)na0(x)qn/2Tn(±ip(x)2q),

    where Tn(x) is the nth Chebyshev polynomial of the first kind.

    Proof. The proof is similar as the proof of Theorem 4.1 and is omitted.

    Example 4.1. For instance, consider the Lucas polynomial sequence {Ln(x)} defined by

    Ln(x)=xLn1(x)+Ln2(x)

    for all n2 with the initial conditions L0(x)=2 and L1(x)=x. Thus

    Ln(x)=2(i)nTn(±ix2)

    for all n0. In addition, the Lucas numbers Ln=Ln(1) can be transferred to

    Ln=2(i)nTn(±i2)

    for all n0.

    Similarly, for the Pell-Lucas polynomials A122075 [15] Qn(x) defined by (see Horadam and Mahon [9])

    Qn(x)=2xQn1(x)+Qn2(x)

    for all n2 with initial conditions Q0(x)=2 and Q1(x)=2x, we have

    Qn(x)=2(i)nTn(±ix)

    for all n0. The first one of the above formulas is shown in Magnus, Oberhettinger, and Soni [12].

    For the Dickson polynomials of the first kind Dn(x) A000041 [15] defined by (see Lidl, Mullen, and Turnwald [11])

    Dn(x)=xDn1(x)aDn2(x)

    for all n2 with initial conditions D0(x)=2 and D1(x)=x, where aR, we have

    Dn(x)=(±1)n2an/2Tn(±x2a)

    for all n0.

    For the Viate polynomials of the second kind defined by (see Horadan [8])

    vn(x)=xvn1(x)vn2(x)

    for all n2 with the initial conditions v0(x)=2 and v1(x)=x, we have

    vn(x)=2(i)n(1)n/2Tn(±ix2i)=2(±1)nTn(±x2)

    for all n0. The first one of the above formulas can be seen in Jacobsthal [10] and Robbins [13].

    We now consider the recursive number or polynomial sequences defined by (1) with initial conditions a0=0 and a10, where pR[x] and qR. For instance,

    ˆUn+1=2xˆUnˆUn1 (18)

    for all n1, where initial conditions are ˆU0=0 and ˆU1=1. It is obvious that ˆUn+1=Un, the Chebyshev polynomials of the second kind A135929 [15]. By using (4), we have

    ˆUn(x)=ˆU1(2x)n1+[n/2]j=11j(nj1j1)(2x)n2j1(1)j(j(2x)ˆU0+(n2j)ˆU1)=(2x)n1+[n/2]j=1(nj1j)(1)j(2x)n2j1=(2x)n1[n/2]j=0(nj1j)(14x2)j. (19)

    From (1) and initial conditions a0=0 and a1=1, we obtain

    an=a1pn1+[n/2]j=11j(nj1j1)pn2j1qj(n2j)a1=a1pn1+[n/2]j=1(nj1j)pn2j1qja1=a1pn1[n/2]j=0(nj1j)p2jqj. (20)

    Comparing with the rightmost sides of (19) and (20), we have the following result.

    Theorem 4.2. Let {an}n0 be the sequence defined by (1) with a0=0 and a10, and let {Un(x)}n0 be the Chebyshev polynomial sequence of the second kind defined by (18). Then

    an=(i)n1a1q(n1)/2Un1(x0), (21)

    where

    x0=±ip2q. (22)

    Namely,

    an=(i)n1a1q(n1)/2Un1(±ip2q).

    Proof. Let an and ˆUn be the sequences shown in the rightmost of (20) and (19), respectively. Suppose an=CnˆUn(x0) for some x0. Then we may have

    p2jqj=(14x20)j

    for x0=±ip/(2q) and

    a1pn1=Cn(2x0)n1,

    which implies

    Cn=a1pn1(2x0)n1=a1pn1(±2ip2q)n1=a1q(n1)/2(±i)n1.

    Consequently, (21) follows.

    Example 4.2. Among all the homogeneous linear recurring sequences satisfying second order homogeneous linear recurrence relation (1) with a nonzero p and arbitrary initial conditions {a0,a1}, the Lucas sequence with respect to {p,q} is defined in one of the authors paper [4], which is the sequence satisfying (1) with initial conditions a0=0 and a1=1. The relationships among the recursive sequences and the Chebyshev polynomial sequence of the second kind at certain points and some nonlinear expressions are studied. Theorem 4.2 presents the relationships of the Chebyshev polynomial sequence of the second kind at some points with a more general class of recursive sequences defined by (1) with initial conditions a0=0 and a10. For instance, for the Fibonacci numbers Fn A000045 [15] with respect to {p,q}={1,1} and initial conditions a0=0 and a1=1, we have

    Fn=(i)n1Un1(±i2)

    (see Aharonov, Beardon, and Driver [1]). For the Pell numbers Pn A000129 [15] with respect to {p,q}={2,1} and initial conditions a0=0 and a1=1, we have

    Pn=(i)n1Un1(±i).

    For the Jacobsthal numbers (cf. [2]) Jn A001045 [15] with respect to {p,q}={1,2} and initial conditions a0=0 and a1=1, we have

    Jn=(2i)n1Un1(±i22).

    For the numbers An shown in the sequence of n coin flips that win on the last flip A198834 [15] defined by the recurrence relation (1) with respect to {p,q}={1,1} and initial conditions a0=0 and a1=2, we have {An}={0,2,2,4,6,10,16,} and

    An=2(i)n1Un1(±i2).

    For the numbers Bn shown in the sequence of the numerators of the fractions in a 'zero-transform' approximation A163271 [15] defined by the recurrence relation (1) with respect to {p,q}={2,1} and initial conditions a0=0 and a1=2, we have {Bn}={0,2,4,10,24,58,} and

    Bn=2(i)n1Un1(±i).

    Theorem 4.2 can be extended to recursive polynomial case as Chebyshev polynomials of the second kind.

    Corollary 3. Let {an(x)}n0 be the recursive polynomial sequence defined by

    an(x)=p(x)an1(x)+qan2(x)

    for n2, where p(x)R[x] and qR, with initial conditions a0(x)=0 and a1(x)0. Then

    an(x)=(i)n1a1(x)q(n1)/2Un1(±ip(x)2q).

    Proof. The result can be proved by using similar argument in the proof of Theorem 4.2.

    Example 4.3. For instance, for Pell polynomials A122075 [15] defined by (see Horadam and Mahon [9])

    Pn(x)=2xPn1(x)+Pn2(x)

    for all n2 with initial conditions P0(x)=0 and P1(x)=1, we have

    Pn(x)=(i)n1Un1(±ix)

    for all n0, where U1(x)=0. The above formulas are shown in Magnus, Oberhettinger, and Soni [12].

    Similarly, for Viate polynomials of the first kind defined by (see Horadan [8])

    Vn(x)=xVn1(x)Vn2(x)

    for all n2 with initial conditions V0(x)=0 and V1(x)=1, we have

    Vn(x)=(i)n1(1)(n1)/2Un1(±ix2i)=(±1)n1Un1(±x2).

    for all n0. The first one of the above formulas is shown in Horadan [8]. From the above formulas and the similar formulas of vn(x) shown above, one may obtain

    Vn(ix)=in1Fn(x)andvn(ix)=inLn(x),

    which are shown in Robins [13].

    Gegenbauer-Humbert polynomial sequences, denoted by {Pλ,y,C(x)}, is defined by the recurrence relation

    Pλ,y,Cn(x)=2λ+n1CnPλ,y,Cn1(x)y2λ+n2CnPλ,y,Cn2(x) (23)

    for all n2 with initial conditions Pλ,y,C0(x)=Cλ and Pλ,y,C1(x)=2λxCλ1. Particularly, for λ=1, we denote Pλ,y,Cn(x) by Py,Cn(x) and (23) becomes

    Py,Cn(x)=2xCPy,Cn1(x)yCPy,Cn2(x) (24)

    for all n2 and Py,C0(x)=C1 and Py,C1(x)=2xC2. From (4), we have the expression of Py,Cn(x) as

    Py,Cn(x)=Py,C1(x)(2xC)n1+[n/2]j=11j(nj1j1)(2xC)n2j1(yC)j(j2xCC1+(n2j)(2xC2))=(2xC2)(2xC)n1+[n/2]j=11j(nj1j1)(2xC)n2j1(yC)j(nj)(2xC2)=1C(2xC)n+[n/2]j=11C(njj)(2xC)n2j(yC)j=1C(2xC)n[n/2]j=0(njj)(yC4x2)j.

    Thus,

    P1,1n(x)=Un(x)=(2x)n[n/2]j=0(njj)(14x2)j (25)

    are the Chebyshev polynomials of the second kind with the first few elements as 1,2x,4x21,8x34x,, and

    P1,1n(x)=Pn(x)=(2x)n[n/2]j=0(njj)(14x2)j (26)

    are the Pell polynomials ([2]) with the first few elements as 1,2x,4x2+1,8x3+4x,.

    Similar to Theorem 4.2, we may establish the following result.

    Theorem 4.3. Let Py,Cn(x) and PyCn(x) be two Gegenbauer-Humbert polynomials defined by (24) with respect two difference complex parameter pairs (y,C) and (y,C), respectively. Then

    Py,Cn(x1)=αnPyCn(x2) (27)

    if the complex variables x1 and x2 satisfy

    x22=yCyCx21, (28)

    where

    αn=(CC)n+1(x1x2)n.

    Proof. Let Py,Cn(x1)=αnPyCn(x2). Then from the rightmost side of (25) for Py,Cn(x) and PyCn(x), respectively, we have

    1C(2x1C)n[n/2]j=0(njj)(yC4x21)j=αn1C(2x2C)n[n/2]j=0(njj)(yC4x22)j

    for

    yC4x21=yC4x22,

    or equivalently, (28), and αn satisfying

    1C(2x1C)n=αn1C(2x2C)n.

    Consequently, we obtain

    αn=Cn+1Cn+1(x1x2)n

    which completes the proof of the theorem.

    Example 4.4. The Chebyshev polynomials of the second kind P1,1n(x1) can be transferred to the Pell polynomials P1,1n(x2) by using

    P1,1n(x1)=(x1x2)nP1,1n(x2),

    where x22=x21. More precisely, noticing (26) and using x22=x21 we may write the righthand side of the above equation as

    (x1x2)nP1,1n(x2)=(x1x2)n(2x2)n[n/2]j=0(njj)(14x22)j=(2x1)n[n/2]j=0(njj)(14x21)j=P1,1n(x1),

    where the last equation comes from (25).

    The authors would like to express their sincere thanks to the editor and two referees for their comments.



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