# On the Elementary Symmetric Polynomials and the Zeros of Legendre Polynomials

Authors: Maryam Salem Alatawi
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we seek to present some new identities for the elementary symmetric polynomials and use these identities to construct new explicit formulas for the Legendre polynomials. First, we shed light on the variable nature of elementary symmetric polynomials in terms of repetition and additive inverse by listing the results related to these. Sequentially, we have proven an important formula for the Legendre polynomials, in which the exponent moves one walk instead of twice as known. The importance of this formula appears throughout presenting Vieta’s formula for the Legendre polynomials in terms of their zeros and the results mentioned therein. We provide new identities for the elementary symmetric polynomials of the zeros of the Legendre polynomials. Finally, we propose the relationship between elementary symmetric polynomials for the zeros ofPnx and the zeros of Pn−1x.


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## Body

## 1. Introduction

In the last few decades, the importance of symmetric polynomials has emerged in many branches of pure and applied mathematics, such as representation theory [1], algebraic combinatorics [2], and numerical analysis [3–5]. The symmetric polynomials have several types, for instance, monomial, complete, and the elementary, see [6–9] and references therein. Throughout the current paper, we will focus on the elementary type. Let us first recall the fundamental concepts and well-known results that we will use subsequently, for further details see [6, 10] and references therein.For positive integern, the elementary symmetric polynomial (for short, ESP) of degree k denoted by eknx1,x2,…,xn is defined as the sum of all possible products of k variables of x1,x2,…,xn; that is,(1)eknx1,x2,…,xn=∑1≤ℓ1<ℓ2<⋯<ℓk≤n∏i=1kxℓi,with e0nx1,x2,…,xn=1 and eknx1,x2,…,xn=0 for k>n or k<0. It is easy to see that eknx1,x2,…,xn involves nk terms. For example, the ESPs for n=3 are given by the following:(2)e03x1,x2,x3=1,e13x1,x2,x3=x1+x2+x3,e23x1,x2,x3=x1x2+x1x3+x2x3,e33x1,x2,x3=x1x2x3.The generating function of the ESP,eknx, where x denotes the n− tuple of the variables x1,x2,…,xn is given by the following:(3)∏i=1n1+xiλ=∑k=0neknxλk.Notice that, from the generation function (3), we have the following recurrence relation:(4)eknx=ekn−1x1,x2,…,xn−1+xnek−1n−1x1,x2,…,xn−1,for all n≥1 and k=1,2,…,n with e0n−1⋅=1, see [10]. Moreover, we can deduce the following:(5)ekn−x=−1keknx.The orthogonal polynomials such as Hermite, Laguerre, and Jacobi have great importance in applied sciences, for instance, in quantum mechanics, engineering, and computational mathematics (e.g., [11–15] and references therein). The Legendre polynomials are a particular case of the Jacobi polynomials that can be applied to reveal Schoenberg’s representation of positive definite functions on the unit sphere, see [11]. Now, we shall go through some known results of the Legendre polynomials that we ought to use later on, see [16–24] for further details. The Legendre polynomials Pnxn=0∞ can be defined as the coefficients of βn in the following generating function, that is,(6)11−2xβ+β2=∑n=0∞Pnxβn,where β∈−1,1. Favard’s theorem ([19], Theorem 4.5.1) showed that the Legendre polynomials Pnxn=0∞ satisfies the following three-term recurrence relation:(7)Pnx=xPn−1x+1−1nxPn−1x−Pn−2x,for all n=2,3,4,…, with P0x=1 and P1x=x, see [25]. Also, the Legendre polynomials Pnx satisfies the following:(8)ddxPnx=nPn−1x+xddxPn−1x,for all n=1,2,…, see [22]. In addition to this, the Legendre polynomials have many explicit formulas, for instance the following representation:(9)Pnx=2−n∑k=0n/2−1knk2n−2knxn−2k,where ⋅ denotes the floor function, see [22]. The orthogonality property for a sequence of the Legendre polynomials Pnxn=0∞ on the closed interval −1,1 is given by the following:(10)∫−11PnxPmxdx=22n+1δnm,where δnm denotes the Kronecker delta which is equal to 1 if n=m, and 0 otherwise, see [26].The current paper is organized as follows: In Section2, we introduce new identities for the elementary symmetric polynomials. Section 3 is mainly devoted to the present new formulas for the Legendre polynomials in terms of their zeros. In Section 4, novel identities for the elementary symmetric polynomials of the zeros of the Legendre polynomials are presented. In Section 5, some illustrative examples are given.


## 2. New Identities for Elementary Symmetric Polynomials

In this section, we are mainly concerned with introducing the main results related to the ESPs that we will use subsequently.Lemma 1.
Fork∈ℤ+∪0 and for any even positive integer n, we have the following:(11)eknx1,x2,…,xn/2,−x1,−x2,…,−xn/2=0,ifk≡0mod2;−1k/2ek/2n/2x12,x22,…,xn/22,ifk≡0mod2.Proof:.
Following the generating function (3), we have as follows:(12)∑k=0neknx1,x2,…,xn/2,−x1,−x2,…,−xn/2λk=∏i=1n/21+xiλ∏i=1n/21−xiλ=∏i=1n/21−xi2λ2=∑k=0n/2−1kekn/2x12,x22,…,xn/22λ2k.
By comparing the coefficients on both sides, we complete the proof.
Moreover, we present the following result that studies the ESPs for another special case of the variablesx1,x2,…,xn.Lemma 2.
Fork∈ℤ+∪0 and for any odd positive integer n>1, we have(13)eknx1,x2,…,xn−1/2,0,−x1,−x2,…,−xn−1/2=0,ifk≡0mod2;−1k/2ek/2n+1/20,x12,x22,…,xn−1/22,ifk≡0mod2.Proof:.
Following the generating function (3), we have as follows:(14)∑k=0neknx1,x2,…,xn−1/2,0,−x1,−x2,…,−xn−1/2λk=∏i=1n−1/21+xiλ1+xn+1/2λ|xn+1/2=0∏i=1n−1/21−xiλ=∏i=1n+1/21−xi2λ2|xn+1/2=0=∑k=0n+1/2−1kekn+1/20,x12,x22,…,xn−1/22λ2k.
By comparing the coefficients on both sides, we obtain (13). This completes the proof.
The following results give a closed form of the ESPs for non-distinct variables. Notice that sinceekn⋅=0 for k>n, then the generating function of the ESP (3) is equivalent to the following:(15)∏i=1n1+xiλ=∑k=0∞eknxλk.Lemma 3.
Fork∈ℤ+∪0 and n a positive integer, we have as follows:(16)ek2nx1,x2,…,xn,x1,x2,…,xn=∑i=0keinx1,x2,…,xnek−inx1,x2,…,xn.Proof:.
Following the generating function (3), we have as follows:(17)∑k=02nek2nx1,x2,…,xn,x1,x2,…,xnλk=∏i=1n1+xiλ∏i=1n1+xiλ=∏i=1n1+xiλ2=∑k=0neknx1,x2,…,xnλk2.
By considering (15) and applying the Cauchy product ([27], Definition 9.4.6.), we deduce the following:(18)∑k=02nek2nx1,x2,…,xn,x1,x2,…,xnλk=∑k=02n∑i=0keinx1,x2,…,xnek−inx1,x2,…,xnλk.
This completes the proof.
Now, we seek to generalize Lemma3 in the following result.Theorem 1.
Fork∈ℤ+∪0 and n,ℓ positive integers, we have as follows:(19)ekn+ℓx1,x2,…,xn,y1,y2,…,yℓ=∑i=0keinx1,x2,…,xnek−iℓy1,y2,…,yℓ.Proof:.
Following the generating function (3), we have as follows:(20)∑k=0n+ℓekn+ℓx1,x2,…,xn,y1,y2,…,yℓλk=∏i=1n1+xiλ∏i=1ℓ1+yiλ=∑k=0neknx1,x2,…,xnλk∑k=0ℓeiℓy1,y2,…,yℓλi.
By considering (15) and applying the Cauchy product ([27], Definition 9.4.6.), we deduce the following:(21)∑k=0n+ℓekn+ℓx1,x2,…,xn,y1,y2,…,yℓλk=∑k=0n+ℓ∑i=0keinx1,x2,…,xnek−iℓy1,y2,…,yℓλk.
This completes the proof.


## 3. More Formulas for the Legendre Polynomials

Throughout the current section, we will introduce some results related to the Legendre polynomials. We start this section by giving a simple proof for an equivalent explicit formula for the Legendre polynomials.Lemma 4.
ConsiderPnn=0∞ to be a sequence of the Legendre polynomials. Then, the explicit formula (9) coincides with the following formula:(22)Pnx=2n∑k=0nnkn+k−12nxk.Proof:.
Letakn=2−nnk2n−2kn and bmn=2nnmn+m−1/2n. Then, we have the following:(23)bn−2k+1n=2nnn−2k+1n−kn,since n−kn=0 for all k≥1. Hence bn−2k+1n=0 for k=1,2,…,n/2−1. Now, it is sufficient to show that bn−2kn=akn for all k=0,1,…,n/2.(24)bn−2kn=2nnn−2k2n−2k−12n=2n−2k−12n−2k−3⋯−2k+1n−2k!2k!=2n−2k−12n−2k−3⋯3⋅1n−2k!−1−3⋯−2k−1−2k+12k!=2n−2k!2n−kn−k!1n−2k!2k!−1k2k!2kk!=2−n−1kn!n−k!k!2n−2k!n!n−2k!=2−nnk2n−2kn=akn.
This fact completes the proof.
It is easy to see that, according to formula (22), the Legendre polynomials take the following expression:(25)Pnx=αn,0+αn,1x+αn,2x2+⋯+αn,nxn,where(26)αn,k=2nnkn+k−12n,for all k=0,1,2,…,n.
Now, it is the time to give the following result which reformulates the Legendre polynomial explicit formula (22).Lemma 5.
ConsiderPnn=0∞ to be a sequence of the Legendre polynomials. Then, the explicit formula (22) is equivalent to the following:(27)Pnx=∑k=0k≡0mod2n−1k/22n−k−1!!k−1!!k!n−k!xn−k,where ⋅!! denotes the double factorial, with −1!!=1 and 0!!=0.Proof:.
The coefficient ofxn−k in equation (22) is given by the following:(28)αn,n−k=2nnn−k2n−k−12n=2n−k−12n−k−3⋯−k+1n−k!k!.
Assume thatα^n,k=2n−k−12n−k−3⋯−k+1, then we have two cases to consider based on k:(i)
Ifk≡0mod2, then α^n,k will vanish since it contains 0 as a factor; hence, so the coefficients in (27) equals to zero.(ii)
Ifk≡0mod2, then we have the following:(29)α^n,k=2n−k−1⋯3⋅1⋅−1⋅−3⋯−k+1=2n−k−1!!−1k/2k−1!!.
This fact completes the proof.
Notice that following Lemma5, we conclude that n/2 coefficients in (25) are equals to zero, where ⋅ denotes the ceiling function. Now, we state the following result, which constructs the Legendre polynomials in terms of their zeros.Theorem 2.
For any positive integern, consider x1,x2,…,xn to be the zeros of Legendre polynomials Pnx, then(30)Pnx=2−n+12n−1n∑k=0n−1n−ken−knx1,x2,…,xnxk.Proof: .
The leading coefficientαn,n in formula (25) is given by the following:(31)αn,n=2n2n−12n=2n2n−1/22n−2/22n−3/22n−4/2⋯2/21/2n!n−1!=2−n+12n−1n.
Now, we can rewrite formula (25) as follows:(32)Pnx=2−n+12n−1nqx,where qx be a monic polynomial of degree n. Since x1,x2,…,xn be the zeros of the Legendre polynomial of degree n, then the polynomial qx has n distinct linear factors. Therefore, we can write the polynomial qx as follows:(33)qx=∏i=1nx−xi.
By applying Vieta’s theorem [28], we deduce the following:(34)qx=∑k=0n−1n−ken−knxxk,which completes the proof.
On the other hand, for any Legendre polynomialPnx, there is a negative zero with the same absolute value for every positive zero. So , it is clear that e1nx1,x2,…,xn=0 for all n≥2.


## 4. On the Zeros of Legendre Polynomials

In this section, we present the results related to the zeros of the Legendre polynomials. Based on Lemma5 and Theorem 2, we state the following result that gives a closed-form for the ESPs of Legendre’s zeros.Corollary 1.
For any positive integersn and k. Consider x1,x2,…,xn to be the zeros of the Legendre polynomial Pnx, then(35)eknx1,x2,…,xn=−13k/22n−1nk2n−k−1!!k−1!!n−1!2n−1!,with e0nx1,x2,…,xn=1.The Legendre polynomialPnx is an even function when n is even; moreover, it has n zeros, half of the zeros are negative and the other half is positive. On the contrary, when n is odd, we find that the polynomial is an odd function and has zero at x=0 and half of the remaining zeros are negative and the other half positive. Based on the type of degree of the polynomial being even or odd and applying Lemmas 1 and 2, respectively, we conclude the following corollaries.Corollary 2.
For any positive even integern. The Legendre polynomial Pnx is given by the following:(36)Pnx=2−n+12n−1n∑k=0k≡0mod2n−13k/2ek/2n/2x12,x22,…,xn/22xn−k,for all positive zeros x1,x2,…,xn/2.Corollary 3.
For any positive odd integern. The Legendre polynomial Pnx is given by the following:(37)Pnx=2−n+12n−1n∑k=0k≡0mod2n−13k/2ek/2n+1/20,x12,x22,…,xn−1/22xn−k,for all positive zeros x1,x2,…,xn−1/2.It is clear thatCorollaries 2 and 3 reduce the computational cost, since in the case of odd k the coefficient becomes zero.Based on the orthogonality of the Legendre polynomials, we provide the following result.Lemma 6.
For any positive integern, we have the following:(38)∫−11Pnx2dx=2−n+3/22n−1n2∑k=0k≡0mod22n−1k/2ek/2nx12,x22,…,xn22n−k+1,for all zeros x1,x2,…,xn of Pnx.Proof:.
Following Theorem2, we deduce the following:(39)Pnx2=ϒn∑k=0n−1keknx1,x2,…,xnxn−k2,where ϒn=2−n+12n−1n2.
By considering (15) and applying the Cauchy product, we deduce the following:(40)Pnx2=ϒn∑k=02n−1k∑i=0keinx1,x2,…,xn×ek−inx1,x2,…,xnx2n−k.
By applying Lemma3, we obtain the following:(41)Pnx2=ϒn∑k=02n−1kek2nx1,x2,…,xn,x1,x2,…,xnx2n−k.
By integrating both sides on the interval−1,1, we conclude as follows:(42)∫−11Pnx2dx=ϒn∑k=02n−1k2n−k+1ek2nx1,x2,…,xn,x1,x2,…,xn+12n−k+1ek2nx1,x2,…,xn,x1,x2,…,xn.
Therefore, by considering the nature of Legendre’s zeros, we obtain (38) by expanding the previous summation on the right-hand side and then applying Lemma 1. This completes the proof.
Furthermore, we can deduce the following result.Corollary 4.
For any positive integern. Consider the Legendre polynomial Pnx with zeros x1,x2,…,xn, then(43)∑k=0k≡0mod22n−1k/2ek/2nx12,x22,…,xn22n−k+1=22n+12n−1n!n−1!2n−1!2.Similarly, following the orthogonality of the Legendre polynomials with distinct degrees, we deduce the following result.Corollary 5.
Letn,m be positive integers with n≠m. Consider the Legendre polynomial Pnx with zeros x1,x2,…,xn arranged in ascending order, namely, x1<x2<…<xn and similarly Pnx with zeros y1,y2,…,ym with y1<y2<⋯<ym, then(44)∑k=0k≡0mod2n+m−1k/2n+m−k+1ek/2n+m/2x12,x22,…,xn/22,y12,y22,…,ym/22=0.The following result gives a relation between the ESPs for the zeros of Legendre polynomialsPnx and Pn−1x.Theorem 3.
Letn be a positive integer. Consider the Legendre polynomial Pnx with zeros x1,x2,…,xn and Pn−1x with zeros y1,y2,…,yn−1, then(45)eknx1,x2,…,xn=n2n−k−12n−1n−kekn−1y1,y2,…,yn−1,for all k=0,1,2,…,n−1.Proof:.
Following the recurrence relation (8), we obtain the following:(46)2n−1n∑k=0n−1−1kn−keknxxn−k−1=n∑k=0n−1−1kekn−1yxn−k−1+∑k=0n−2−1kn−k−1ekn−1yxn−k−1,where y denotes the n−1− tuple y1,y2,…,yn−1. Based on the definition of ESPs, we can rewrite the last term in (46) as follows:(47)∑k=0n−2−1kn−k−1ekn−1yxn−k−1=∑k=0n−1−1kn−k−1ekn−1yxn−k−1.
Therefore, by comparing the coefficients ofxn−k−1, we obtain the following:(48)2n−1n−kneknx=nekn−1y+n−k−1ekn−1y,which completes the proof.
Following Theorem3, it is easy to construct the Legendre polynomial Pnx via the zeros of Pn−1x, looking forward the following corollary.Corollary 6.
Letn be a positive integer, with n>1. Assume that the Legendre polynomial Pn−1x has zeros y1,y2,…,yn−1, then(49)Pnx=2−n+1n2n−12n−1n∑k=1n−1n−kn+k−1ken−kn−1y1,y2,…,yn−1xk+2nn−12n.It is easy to see that ifn is odd, then the last term of (49) will vanish.


## 5. Numerical Examples

Throughout this section, we shall give some examples to illustrate the results are posted in Sections3 and 4. Let us start with the following example that illustrates the idea of Theorem 2.Example 1.
The zeros ofP3x are x1=3/5,x2=0 and x3=−3/5, then following (2), we have the following:(50)e03x1,x2,x3=1,e13x1,x2,x3=0,e23x1,x2,x3=−35,e33x1,x2,x3=0.
Therefore, by applying Theorem2, then we obtain as follows:(51)P3x=2−253−e33x1,x2,x3+e23x1,x2,x3x−e13x1,x2,x3x2+e03x1,x2,x3x3=1410−35x+x3.
Moreover, we deduce thatP3x=1/25x3−3x.The following example to illustrates the idea ofCorollaries 2 and 3.Example 2.
The zeros of Legendre polynomialP4x are x1=15+230/35,x2=15−230/35,x3=−15+230/35 and x4=−15−230/35, then following Corollary 2, we have the following:(52)P4x=358Λ4,0x4+Λ4,2x2+Λ4,4,where Λ4,1=Λ4,3=0. Moreover, we obtain as follows:(53)Λ4,0=e02x12,x22=1,Λ4,2=e12x12,x22=−3035,Λ4,4=e22x12,x22=105352.
Thus, we conclude thatP4x=1/835x4−30x2+3.Following Theorem3 and Corollary 6, we give the next example.Example 3.
The zeros of the Legendre polynomialP2x are x1=1/3 and x2=−1/3. Thus, following Corollary 6 we can construct the Legendre polynomial P3x as follows:(54)P3x=2−235533e2213,−13x−2e1213,−13x2+53e0213,−13x3=3253x3−x,therefore, P3x=1/25x3−3x.


## 6. Conclusion

Throughout this paper, we provided some results related to the elementary symmetric polynomials such as Lemmas1–3. The importance of the results mentioned in Section 2 appears by providing some new results related to zeros of the Legendre polynomials that have been presented in Section 3. Also, we constructed the Legendre polynomials via their zeros, and we presented the closed form of Vieta’s formula. By Theorem 3, we were able to construct Pnx via the zeros of Pn−1x for all n≥2, which reduce the computational costs for constructing the Legendre polynomials.



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