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
[1] | Tian-Xiao He, Peter J.-S. Shiue, Zihan Nie, Minghao Chen . Recursive sequences and girard-waring identities with applications in sequence transformation. 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=pan−1+qan−2,n≥2, | (1) |
for constants
an={(a1−βa0α−β)αn−(a1−αa0α−β)βn,ifα≠β;na1αn−1−(n−1)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)=a0+(a1−pa0)s1−ps−qs2. | (3) |
Furthermore, the Taylor expansion A165998 [15] of
A(s)=a0+∑n≥1(a1pn−1+[n/2]∑j=11j(n−j−1j−1)pn−2j−1qj(jpa0+(n−2j)a1))sn. |
Proof. From the definition of
A(s)=∑n≥0ansn=a0+a1s+∑n≥2ansn=a0+a1s+∑n≥2(pan−1+qan−2)sn=a0+a1s+ps∑n≥1ansn+qs2∑n≥0ansn=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+(a1−pa0)s1−ps−qs2=(a0+(a1−pa0)s)∑n≥0sn(p+qs)n |
=(a0+(a1−pa0)s)∑n≥0n∑j=0(nj)pn−jqjsn+j=(a0+(a1−pa0)s)∑n≥0n∑j=0(n−jj)pn−2jqjsn=a0∑n≥0n∑j=0(n−jj)pn−2jqjsn+(a1−pa0)∑n≥1n∑j=0(n−j−1j)pn−2j−1qjsn=a0+∑n≥1n∑j=0(a0(n−jj)pn−2jqj+(a1−pa0)(n−j−1j)pn−2j−1qj)sn=a0+∑n≥1n∑j=0(a0((n−jj)−(n−j−1j))pn−2jqj+a1(n−j−1j)pn−2j−1qj)sn=a0+∑n≥1(a1pn−1+n∑j=1(a0(n−j−1j−1)pn−2jqj+a1(n−j−1j)pn−2j−1qj))sn=a0+∑n≥1(a1pn−1+[n/2]∑j=1(n−j−1j−1)pn−2j−1qj(pa0+n−2jja1))sn=a0+∑n≥1(a1pn−1+[n/2]∑j=11j(n−j−1j−1)pn−2j−1qj(jpa0+(n−2j)a1))sn, |
which completes the proof of the proposition.
Corollary 1. Let
an=a1pn−1+[n/2]∑j=11j(n−j−1j−1)pn−2j−1qj(jpa0+(n−2j)a1). | (4) |
If
an=αn−βnα−β=[n/2]∑j=0(n−j−1j)pn−2j−1qj=[n/2]∑j=0(−1)j(n−j−1j)(α+β)n−2j−1(αβ)j, | (5) |
where
Proof. From (4), we have
an=a1pn−1+[n/2]∑j=1(n−j−1j−1)pn−2j−1qj(pa0+n−2jja1), | (6) |
or equivalently, (4). Hence, we obtain the following identity for all recursive sequences defined by (1)
a1pn−1+[n/2]∑j=11j(n−j−1j−1)pn−2j−1qj(jpa0+(n−2j)a1) |
=(a1−βa0α−β)αn−(a1−αa0α−β)βn, |
where
α=p+√p2+4q2andβ=p−√p2+4q2. |
If
an=a1pn−1+[n/2]∑j=1n−2jj(n−j−1j−1)pn−2j−1qj=a1pn−1+[n/2]∑j=1(n−j−1j)pn−2j−1qj, |
which implies the first equation of (5) by noting (2). Substituting
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=∑0≤k≤[n/2](−1)knn−k(n−kk)(α+β)n−2k(αβ)k, | (7) |
by using the first equation of (5). First, we need the following lemmas.
Lemma 3.1. Let
j∑k=0nn−k(n−kk)(n−j−1+kj−k)=(2n−j−1j). | (8) |
Proof. From the Chu-Vandermonde formula and noting, the left-hand side of (8) can be written as
LHS=j∑k=0nn−k(n−kk)(j−k∑i=0(n+n−j−1j−k−i)(−n+ki))=j∑k=0nn−k(n−kk)(j∑i=k(2n−j−1j−i)(−n+ki−k))=j∑i=0(i∑k=0nn−k(n−kk)(−n+ki−k))(2n−j−1j−i)=j∑i=0(2n−j−1j−i)(i∑k=0(−1)i−knn−2k+i(ik)(n−2k+ii)), |
where on the first line
(−n+ki)=(−1)i(n−k)(n−k−1)⋯(n−k−i+1)i!=(−1)i(n−ki). |
We split the inner sum of the rightmost hand of the above equation for the left-hand side of (8),
i∑k=0(−1)i−k(ik)nn−2k+i(n−2k+ii), |
into two cases. For
i∑k=0(−1)i−k(ik)nn−2k+i(n−2k+i)!(n−2k)!i!=i∑k=0(−1)i−k(ik)n(n−2k+i−1)!(n−2k)!i!=i∑k=0(−1)i−k(ik)n(n−2k+i−1)(n−2k+i−2)...(n−2k+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=j∑i=0(2n−j−1j−i)(i∑k=0(−1)i−knn−2k+i(ik)(n−2k+ii))=(2n−j−1j). |
Lemma 3.2. Let
∑0≤k≤[n/2](−1)knn−k(n−kk)(α+β)n−2k(αβ)k⋅∑0≤k≤[n/2](−1)k(n−k−1k)(α+β)n−2k−1(αβ)k=∑0≤k≤n(−1)k(2n−k−1k)(α+β)2n−2k−1(αβ)k. | (9) |
Proof. The left-hand side of (9) can be written as
∑0≤k≤[n/2]∑0≤i≤[n/2](−1)k+inn−k(n−kk)(n−i−1i)(α+β)2n−2(k+i)−1(αβ)k+i=∑0≤k≤[n/2]∑0≤j≤n(−1)jnn−k(n−kk)(n+k−j−1j−k)(α+β)2n−2j−1(αβ)j |
=∑0≤j≤n(−1)j(∑0≤k≤[n/2]nn−k(n−kk)(n+k−j−1j−k))(α+β)2n−2j−1(αβ)j, |
where by using Lemma 3.1 the inner sum can be written as
∑0≤k≤jnn−k(n−kk)(n+k−j−1j−k)=(2n−j−1j). |
Thus we obtain (9).
To prove Girard-Waring identity (7), we only need to mention that (9) implies
∑0≤k≤[n/2](−1)knn−k(n−kk)(α+β)n−2k(αβ)k=∑0≤k≤n(−1)k(2n−k−1k)(α+β)2n−2k−1(αβ)k/∑0≤k≤[n/2](−1)k(n−k−1k)(α+β)n−2k−1(αβ)k=α2n−β2nα−β/αn−βnα−β=αn+βn. |
There are some alternative forms of formula (7). As an example, we give the following one. If
xn+yn=∑0≤k≤[n/2](−1)knn−k(n−kk)(−z)n−2k(xy)k=(−1)nzn+∑1≤k≤[n/2](−1)n−knn−k(n−kk)zn−2k(xy)k, |
which implies
xn+yn−(−1)nzn=∑1≤k≤[n/2](−1)n−knn−k(n−kk)zn−2k(xy)k. |
Thus, when
xn+yn−zn=∑1≤k≤[n/2](−1)n−knn−k(n−kk)zn−2k(xy)k, | (10) |
while for odd
xn+yn+zn=∑1≤k≤[n/2](−1)n−knn−k(n−kk)zn−2k(xy)k, | (11) |
where
x3+y3+z3=3xyz, | (12) |
which was shown in He and Shiue [7]. Saul and Andreescu [14] have shown that the cube vanishes if
Proposition 2. Let
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)=2xTn−1(x)−Tn−2(x) | (13) |
for all
Tn(x)=x(2x)n−1+[n/2]∑j=11j(n−j−1j−1)(2x)n−2j−1(−1)j(2xj+(n−2j)x)=2n−1xn+n[n/2]∑j=11j(n−j−1j−1)(−1)j2n−2j−1xn−2j. |
Similarly, for Lucas numbers A000032 [15] defined by
Ln=Ln−1+Ln−2 |
for all
Ln=1+[n/2]∑j=11j(n−j−1j−1)(2j+(n−2j))=1+n[n/2]∑j=11j(n−j−1j−1). |
From the expressions of
Tn(−i2)=2n−1(−i2)n+n[n/2]∑j=11j(n−j−1j−1)(−1)j2n−2j−1(−i2)n−2j=(−1)nin2+n[n/2]∑j=11j(n−j−1j−1)(−1)j2−1(−1)n−2jin−2j=(−1)nin2+n[n/2]∑j=11j(n−j−1j−1)(−1)nin2=i3n2Ln, |
or equivalently,
Ln=2inT(−i2), |
where
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=2a1pn−1(2x0)nTn(x0), | (14) |
where
x0=±ip2√q. | (15) |
Namely,
an=(∓i)na0qn/2Tn(±ip2√q). | (16) |
Proof. From (14) we have
Tn(x)=(2x)n2(1+n[n/2]∑j=11j(n−j−1j−1)(−1)j(2x)−2j). |
If
an=a1pn−1+[n/2]∑j=11j(n−j−1j−1)pn−2j−1qjna1=a1pn−1(1+n[n/2]∑j=11j(n−j−1j−1)(qp2)j). |
Setting
a1pn−1(1+n[n/2]∑j=11j(n−j−1j−1)(qp2)j)=Cn(2x0)n2(1+n[n/2]∑j=11j(n−j−1j−1)(−1)j(2x0)−2j), |
which implies
a1pn−1=Cn2(2x0)n | (17) |
for the value
−(2x0)−2=qp2. |
Thus, we solve the last equation to get
Cn=2a1pn−1(2x0)n=2a1pn−1/(±ip√q)n. |
By substituting
Theorem 4.1 can be extended to recursive polynomial case.
Corollary 2. Let
an(x)=p(x)an−1(x)+qan−2(x) |
for
an(x)=(∓i)na0(x)qn/2Tn(±ip(x)2√q), |
where
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)=xLn−1(x)+Ln−2(x) |
for all
Ln(x)=2(∓i)nTn(±ix2) |
for all
Ln=2(∓i)nTn(±i2) |
for all
Similarly, for the Pell-Lucas polynomials A122075 [15]
Qn(x)=2xQn−1(x)+Qn−2(x) |
for all
Qn(x)=2(∓i)nTn(±ix) |
for all
For the Dickson polynomials of the first kind
Dn(x)=xDn−1(x)−aDn−2(x) |
for all
Dn(x)=(±1)n2an/2Tn(±x2√a) |
for all
For the Viate polynomials of the second kind defined by (see Horadan [8])
vn(x)=xvn−1(x)−vn−2(x) |
for all
vn(x)=2(∓i)n(−1)n/2Tn(±ix2i)=2(±1)nTn(±x2) |
for all
We now consider the recursive number or polynomial sequences defined by (1) with initial conditions
ˆUn+1=2xˆUn−ˆUn−1 | (18) |
for all
ˆUn(x)=ˆU1(2x)n−1+[n/2]∑j=11j(n−j−1j−1)(2x)n−2j−1(−1)j(j(2x)ˆU0+(n−2j)ˆU1)=(2x)n−1+[n/2]∑j=1(n−j−1j)(−1)j(2x)n−2j−1=(2x)n−1[n/2]∑j=0(n−j−1j)(−14x2)j. | (19) |
From (1) and initial conditions
an=a1pn−1+[n/2]∑j=11j(n−j−1j−1)pn−2j−1qj(n−2j)a1=a1pn−1+[n/2]∑j=1(n−j−1j)pn−2j−1qja1=a1pn−1[n/2]∑j=0(n−j−1j)p−2jqj. | (20) |
Comparing with the rightmost sides of (19) and (20), we have the following result.
Theorem 4.2. Let
an=(∓i)n−1a1q(n−1)/2Un−1(x0), | (21) |
where
x0=±ip2√q. | (22) |
Namely,
an=(∓i)n−1a1q(n−1)/2Un−1(±ip2√q). |
Proof. Let
p−2jqj=(−14x20)j |
for
a1pn−1=Cn(2x0)n−1, |
which implies
Cn=a1pn−1(2x0)n−1=a1pn−1(±2ip2√q)n−1=a1q(n−1)/2(±i)n−1. |
Consequently, (21) follows.
Example 4.2. Among all the homogeneous linear recurring sequences satisfying second order homogeneous linear recurrence relation (1) with a nonzero
Fn=(∓i)n−1Un−1(±i2) |
(see Aharonov, Beardon, and Driver [1]). For the Pell numbers
Pn=(∓i)n−1Un−1(±i). |
For the Jacobsthal numbers (cf. [2])
Jn=(∓√2i)n−1Un−1(±i2√2). |
For the numbers
An=2(∓i)n−1Un−1(±i2). |
For the numbers
Bn=2(∓i)n−1Un−1(±i). |
Theorem 4.2 can be extended to recursive polynomial case as Chebyshev polynomials of the second kind.
Corollary 3. Let
an(x)=p(x)an−1(x)+qan−2(x) |
for
an(x)=(∓i)n−1a1(x)q(n−1)/2Un−1(±ip(x)2√q). |
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)=2xPn−1(x)+Pn−2(x) |
for all
Pn(x)=(∓i)n−1Un−1(±ix) |
for all
Similarly, for Viate polynomials of the first kind defined by (see Horadan [8])
Vn(x)=xVn−1(x)−Vn−2(x) |
for all
Vn(x)=(∓i)n−1(−1)(n−1)/2Un−1(±ix2i)=(±1)n−1Un−1(±x2). |
for all
Vn(ix)=in−1Fn(x)andvn(ix)=inLn(x), |
which are shown in Robins [13].
Gegenbauer-Humbert polynomial sequences, denoted by
Pλ,y,Cn(x)=2λ+n−1CnPλ,y,Cn−1(x)−y2λ+n−2CnPλ,y,Cn−2(x) | (23) |
for all
Py,Cn(x)=2xCPy,Cn−1(x)−yCPy,Cn−2(x) | (24) |
for all
Py,Cn(x)=Py,C1(x)(2xC)n−1+[n/2]∑j=11j(n−j−1j−1)(2xC)n−2j−1(−yC)j(j2xCC−1+(n−2j)(2xC−2))=(2xC−2)(2xC)n−1+[n/2]∑j=11j(n−j−1j−1)(2xC)n−2j−1(−yC)j(n−j)(2xC−2)=1C(2xC)n+[n/2]∑j=11C(n−jj)(2xC)n−2j(−yC)j=1C(2xC)n[n/2]∑j=0(n−jj)(−yC4x2)j. |
Thus,
P1,1n(x)=Un(x)=(2x)n[n/2]∑j=0(n−jj)(−14x2)j | (25) |
are the Chebyshev polynomials of the second kind with the first few elements as
P−1,1n(x)=Pn(x)=(2x)n[n/2]∑j=0(n−jj)(14x2)j | (26) |
are the Pell polynomials ([2]) with the first few elements as
Similar to Theorem 4.2, we may establish the following result.
Theorem 4.3. Let
Py,Cn(x1)=αnPy′C′n(x2) | (27) |
if the complex variables
x22=y′C′yCx21, | (28) |
where
αn=(C′C)n+1(x1x2)n. |
Proof. Let
1C(2x1C)n[n/2]∑j=0(n−jj)(−yC4x21)j=αn1C′(2x2C′)n[n/2]∑j=0(n−jj)(−y′C′4x22)j |
for
−yC4x21=−y′C′4x22, |
or equivalently, (28), and
1C(2x1C)n=αn1C′(2x2C′)n. |
Consequently, we obtain
αn=C′n+1Cn+1(x1x2)n |
which completes the proof of the theorem.
Example 4.4. The Chebyshev polynomials of the second kind
P1,1n(x1)=(x1x2)nP−1,1n(x2), |
where
(x1x2)nP−1,1n(x2)=(x1x2)n(2x2)n[n/2]∑j=0(n−jj)(14x22)j=(2x1)n[n/2]∑j=0(n−jj)(−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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1. | Tian-Xiao He, Peter J.-S. Shiue, Identities for linear recursive sequences of order $ 2 $, 2021, 29, 2688-1594, 3489, 10.3934/era.2021049 | |
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