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Research Article

On the Elementary Symmetric Polynomials and the Zeros of

Legendre Polynomials

Maryam Salem Alatawi

Department of Mathematics, Faculty of Science, University of Tabuk, Tabuk 71491, Saudi Arabia

Correspondence should be addressed to Maryam Salem Alatawi; msoalatawi@ut.edu.sa

Received 28 February 2022; Accepted 22 April 2022; Published 28 May 2022

Academic Editor: Barbara Martinucci

Copyright ©2022 Maryam Salem Alatawi. is is an open access article distributed under the Creative Commons Attribution

License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is

properly cited.

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. e

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 of Pn(x)and

the zeros of Pn−1(x).

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].

e symmetric polynomials have several types, for instance,

monomial, complete, and the elementary, see [6–9] and

references therein. roughout the current paper, we will

focus on the elementary type. Let us ﬁrst recall the funda-

mental concepts and well-known results that we will use

subsequently, for further details see [6, 10] and references

therein.

For positive integer n, the elementary symmetric poly-

nomial (for short, ESP) of degree kdenoted by

e(n)

k(x1, x2,. . . , xn)is deﬁned as the sum of all possible

products of kvariables of x1, x2,. . . , xn; that is,

e(n)

kx1, x2,. . . , xn

�

1≤ℓ1<ℓ2<···<ℓk≤n

k

i�1

xℓi,(1)

with e(n)

0(x1, x2,. . . , xn) � 1 and e(n)

k(x1, x2,. . . , xn) � 0 for

k>nor k<0. It is easy to see that e(n)

k(x1, x2,. . . , xn)in-

volves n

k

terms. For example, the ESPs for n�3 are given

by the following:

e(3)

0x1, x2, x3

�1,

e(3)

1x1, x2, x3

�x1+x2+x3,

e(3)

2x1, x2, x3

�x1x2+x1x3+x2x3,

e(3)

3x1, x2, x3

�x1x2x3.

(2)

e generating function of the ESP, e(n)

k(x), where x

denotes the n−tuple of the variables (x1, x2,. . . , xn)is given

by the following:

n

i�1

1+xiλ

�

n

k�0

e(n)

k(x)λk.(3)

Notice that, from the generation function (3), we have

the following recurrence relation:

Hindawi

Journal of Mathematics

Volume 2022, Article ID 4139728, 9 pages

https://doi.org/10.1155/2022/4139728

e(n)

k(x) � e(n−1)

kx1, x2,. . . , xn−1

+xne(n−1)

k−1x1, x2,. . . , xn−1

,(4)

for all n≥1 and k�1,2,..., n with e(n−1)

0(·) � 1, see [10].

Moreover, we can deduce the following:

e(n)

k(− x) �(− 1)ke(n)

k(x).(5)

e orthogonal polynomials such as Hermite, Laguerre,

and Jacobi have great importance in applied sciences, for

instance, in quantum mechanics, engineering, and com-

putational mathematics (e.g., [11–15] and references

therein). e 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. e Legendre

polynomials Pn(x)

∞

n�0can be deﬁned as the coeﬃcients of

βnin the following generating function, that is,

1

����������

1−2xβ+β2

�

∞

n�0Pn(x)βn,(6)

where β∈(− 1,1). Favard’s theorem ([19], eorem 4.5.1)

showed that the Legendre polynomials Pn(x)

∞

n�0satisﬁes

the following three-term recurrence relation:

Pn(x) � xPn−1(x) + 1−1

n

xPn−1(x) − Pn−2(x)

,(7)

for all n�2,3,4,..., with P0(x) � 1 and P1(x) � x, see [25].

Also, the Legendre polynomials Pn(x)satisﬁes the following:

d

dxPn(x)

�nPn−1(x) + xd

dxPn−1(x)

,(8)

for all n�1,2,. . ., see [22]. In addition to this, the Legendre

polynomials have many explicit formulas, for instance the

following representation:

Pn(x) � 2−n

⌊n/2 ⌋

k�0(− 1)kn

k

2n−2k

n

xn−2k,(9)

where ⌊ · ⌋ denotes the ﬂoor function, see [22]. e or-

thogonality property for a sequence of the Legendre poly-

nomials Pn(x)

∞

n�0on the closed interval [− 1,1]is given by

the following:

1

−1

Pn(x)Pm(x)dx�2

2n+1δnm,(10)

where δnm denotes the Kronecker delta which is equal to 1 if

n�m, and 0 otherwise, see [26].

e current paper is organized as follows: In Section 2,

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 poly-

nomials are presented. In Section 5, some illustrative ex-

amples 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. For k∈Z+∪0

{ } and for any even positive integer

n, we have the following:

e(n)

kx1, x2,. . . , xn/2,−x1,−x2,. . . ,−xn/2

�0,if k≡0(mod2);

(− 1)k/2e(n/2)

k/2 x2

1, x2

2,. . . , x2

n/2

,if k≡0(mod2).

⎧⎨

⎩(11)

Proof: Following the generating function (3), we have as

follows:

n

k�0

e(n)

kx1, x2,. . . , xn/2,−x1,−x2,. . . ,−xn/2

λk

�

n/2

i�1

1+xiλ

n/2

i�1

1−xiλ

�

n/2

i�1

1−x2

iλ2

�

n/2

k�0(− 1)ke(n/2)

kx2

1, x2

2,. . . , x2

n/2

λ2k.

(12)

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 variables

x1, x2,. . . , xn.□

Lemma 2. For k∈Z+∪0

{ } and for any odd positive integer

n>1, we have

e(n)

kx1, x2,..., x(n−1)/2,0,−x1,−x2,...,−x(n−1)/2

�0,if k≡0(mod2);

(− 1)k/2e((n+1)/2)

k/2 0, x2

1, x2

2,..., x2

(n−1)/2

,if k≡0(mod2).

⎧

⎨

⎩(13)

Proof: Following the generating function (3), we have as

follows:

n

k�0

e(n)

kx1, x2,. . . , x(n−1)/2,0,−x1,−x2,. . . ,−x(n−1)/2

λk

�

(n−1)/2

i�1

1+xiλ

1+x(n+1)/2λ

|x(n+1)/2�0

(n−1)/2

i�1

1−xiλ

�

(n+1)/2

i�1

1−x2

iλ2

|x(n+1)/2�0

�

(n+1)/2

k�0(− 1)ke(n+1)/2

k0, x2

1, x2

2,. . . , x2

(n−1)/2

λ2k.

(14)

2Journal of Mathematics

By comparing the coeﬃcients on both sides, we obtain

(13). is completes the proof.

e following results give a closed form of the ESPs for

non-distinct variables. Notice that since e(n)

k(·) � 0 for k>n,

then the generating function of the ESP (3) is equivalent to

the following:

n

i�1

1+xiλ

�

∞

k�0

e(n)

k(x)λk.(15)

□

Lemma 3. For k∈Z+∪0

{ } and na positive integer, we have

as follows:

e(2n)

kx1, x2,. . . , xn, x1, x2,. . . , xn

�

k

i�0

e(n)

ix1, x2,. . . , xn

e(n)

k−ix1, x2,. . . , xn

.(16)

Proof: Following the generating function (3), we have as

follows:

2n

k�0

e(2n)

kx1, x2,. . . , xn, x1, x2,. . . , xn

λk

�

n

i�1

1+xiλ

n

i�1

1+xiλ

�

n

i�1

1+xiλ

⎡

⎣⎤

⎦2

�

n

k�0

e(n)

kx1, x2,. . . , xn

λk

⎡

⎣⎤

⎦2

.

(17)

By considering (15) and applying the Cauchy product

([27], Deﬁnition 9.4.6.), we deduce the following:

2n

k�0

e(2n)

kx1, x2,. . . , xn, x1, x2,. . . , xn

λk

�

2n

k�0

k

i�0

e(n)

ix1, x2,. . . , xn

e(n)

k−ix1, x2,. . . , xn

⎡

⎣⎤

⎦λk.

(18)

is completes the proof.

Now, we seek to generalize Lemma 3 in the following

result. □

Theorem 1. For k∈Z+∪0

{ } and n, ℓ positive integers, we

have as follows:

e(n+ℓ)

kx1, x2,. . . , xn, y1, y2,. . . , yℓ

�

k

i�0

e(n)

ix1, x2,. . . , xn

e(ℓ)

k−iy1, y2,. . . , yℓ

.(19)

Proof: Following the generating function (3), we have as

follows:

n+ℓ

k�0

e(n+ℓ)

kx1, x2,. . . , xn, y1, y2,. . . , yℓ

λk

�

n

i�1

1+xiλ

ℓ

i�1

1+yiλ

�

n

k�0

e(n)

kx1, x2,. . . , xn

λk

⎛

⎝⎞

⎠

ℓ

k�0

e(ℓ)

iy1, y2,. . . , yℓ

λi

⎛

⎝⎞

⎠.

(20)

By considering (15) and applying the Cauchy product

([27], Deﬁnition 9.4.6.), we deduce the following:

n+ℓ

k�0

e(n+ℓ)

kx1, x2,. . . , xn, y1, y2,. . . , yℓ

λk

�

n+ℓ

k�0

k

i�0

e(n)

ix1, x2,. . . , xn

e(ℓ)

k−iy1, y2,. . . , yℓ

⎡

⎣⎤

⎦λk.

(21)

is completes the proof. □

3. More Formulas for the Legendre Polynomials

roughout 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. Consider Pn

∞

n�0to be a sequence of the Legendre

polynomials. en, the explicit formula (9) coincides with the

following formula:

Pn(x) � 2n

n

k�0

n

k

⎛

⎝⎞

⎠n+k−1

2

n

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠xk.(22)

Proof: Let an

k�2−nn

k

2n−2k

n

and

bn

m�2nn

m

(n+m−1)/2

n

. en, we have the

following:

bn

n−2k+1�2nn

n−2k+1

n−k

n

,(23)

since n−k

n

�0 for all k≥1. Hence bn

n−2k+1�0 for

k�1,2,...,⌊n/2 ⌋ − 1. Now, it is suﬃcient to show that

bn

n−2k�an

kfor all k�0,1,. . . ,⌊n/2 ⌋.

Journal of Mathematics 3

bn

n−2k�2nn

n−2k

⎛

⎝⎞

⎠2n−2k−1

2

n

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

�(2n−2k−1)(2n−2k−3)· ·· (− 2k+1)

(n−2k)!(2k)!

�(2n−2k−1)(2n−2k−3)· ·· 3·1

(n−2k)!(− 1)(− 3)· ·· (− 2k−1)(− 2k+1)

(2k)!

�(2n−2k)!

2n−k(n−k)!

1

(n−2k)!(2k)!(− 1)k(2k)!

2kk!

�2−n(− 1)kn!

(n−k)!k!(2n−2k)!

n!(n−2k)!

�2−nn

k

⎛

⎝⎞

⎠2n−2k

n

⎛

⎝⎞

⎠�an

k.

(24)

is fact completes the proof.

It is easy to see that, according to formula (22), the

Legendre polynomials take the following expression:

Pn(x) � αn,0+αn,1x+αn,2x2+ · ·· + αn,nxn,(25)

where

αn,k �2nn

k

⎛

⎝⎞

⎠n+k−1

2

n

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠,(26)

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. Consider Pn

∞

n�0to be a sequence of the Legendre

polynomials. en, the explicit formula (22) is equivalent to

the following:

Pn(x) �

n

k�0

k≡0(mod2)

(− 1)k/2(2n−k−1)!!(k−1)!!

k!(n−k)!xn−k,(27)

where (·)!! denotes the double factorial, with (− 1)!! �1and

(0)!! �0.

Proof: e coeﬃcient of xn−kin equation (22) is given by the

following:

αn,n−k�2nn

n−k

⎛

⎝⎞

⎠2n−k−1

2

n

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

�(2n−k−1)(2n−k−3)· ·· (− k+1)

(n−k)!k!.

(28)

Assume that

αn,k � (2n−k−1)(2n−k−3)· ·· (− k+1),

then we have two cases to consider based on k:

(i) If k≡0(mod2), then

αn,k will vanish since it contains

0 as a factor; hence, so the coeﬃcients in (27) equals

to zero.

(ii) If k≡0(mod2), then we have the following:

αn,k �(2n−k−1)· ··(3) · (1) · (− 1) · (− 3)·· ·(− k+1)

�(2n−k−1)!!(− 1)k/2(k−1)!!.

(29)

is fact completes the proof.

Notice that following Lemma 5, we conclude that ⌈n/2 ⌉

coeﬃcients in (25) are equals to zero, where ⌈ · ⌉denotes the

ceiling function. Now, we state the following result, which

4Journal of Mathematics

constructs the Legendre polynomials in terms of their

zeros. □

Theorem 2. For any positive integer n, consider

x1, x2,. . . , xnto be the zeros of Legendre polynomials Pn(x),

then

Pn(x) � 2−n+12n−1

n

n

k�0(− 1)n−ke(n)

n−kx1, x2,. . . , xn

xk.

(30)

Proof: . e leading coeﬃcient αn,n in formula (25) is given

by the following:

αn,n �2n

2n−1

2

n

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

�2n((2n−1)/2)((2n−2)/2)((2n−3)/2)((2n−4)/2)·· · (2/2)(1/2)

n!(n−1)!

�2−n+12n−1

n

⎛

⎝⎞

⎠.

(31)

Now, we can rewrite formula (25) as follows:

Pn(x) � 2−n+12n−1

n

q(x),(32)

where q(x)be a monic polynomial of degree n. Since

x1, x2,. . . , xnbe the zeros of the Legendre polynomial of

degree n, then the polynomial q(x)has ndistinct linear

factors. erefore, we can write the polynomial q(x)as

follows:

q(x) �

n

i�1

x−xi

.(33)

By applying Vieta’s theorem [28], we deduce the

following:

q(x) �

n

k�0(− 1)n−ke(n)

n−k(x)xk,(34)

which completes the proof.

On the other hand, for any Legendre polynomial Pn(x),

there is a negative zero with the same absolute value for every

positive zero. So , it is clear that e(n)

1(x1, 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 Lemma 5 and eorem

2, we state the following result that gives a closed-form for

the ESPs of Legendre’s zeros.

Corollary 1. For any positive integers nand k. Consider

x1, x2,. . . , xnto be the zeros of the Legendre polynomial

Pn(x), then

e(n)

kx1, x2,. . . , xn

�(− 1)⌈3k/2 ⌉2n−1n

k

⎛

⎝⎞

⎠(2n−k−1)!!(k−1)!!(n−1)!

(2n−1)!,

(35)

with e(n)

0(x1, x2,. . . , xn) � 1.

e Legendre polynomial Pn(x)is an even function

when nis even; moreover, it has nzeros, half of the zeros are

negative and the other half is positive. On the contrary, when

nis 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 integer n. e Legendre

polynomial Pn(x)is given by the following:

Pn(x) � 2−n+12n−1

n

n

k�0

k≡0(mod2)

(− 1)3k/2e(n/2)

k/2 x2

1, x2

2,..., x2

n/2

xn−k,

(36)

for all positive zeros x1, x2,. . . , xn/2.

Corollary 3. For any positive odd integer n. e Legendre

polynomial Pn(x)is given by the following:

Pn(x) � 2−n+12n−1

n

n

k�0

k≡0(mod2)(− 1)3k/2e((n+1)/2)

k/2 0, x2

1, x2

2,. . . , x2

(n−1)/2

xn−k,

(37)

for all positive zeros x1, x2,. . . , x(n−1)/2.

It is clear that Corollaries 2 and 3 reduce the compu-

tational cost, since in the case of odd kthe coeﬃcient be-

comes zero.

Based on the orthogonality of the Legendre polynomials,

we provide the following result.

Lemma 6. For any positive integer n, we have the following:

1

−1

Pn(x)

2dx

�2−n+(3/2)2n−1

n

⎛

⎝⎞

⎠

⎡

⎢

⎢

⎢

⎣⎤

⎥

⎥

⎥

⎦2

2n

k�0

k≡0(mod2)

(− 1)k/2e(n)

k/2 x2

1, x2

2,. . . , x2

n

2n−k+1,

(38)

Journal of Mathematics 5

for all zeros x1, x2,. . . , xnof Pn(x).

Proof: Following eorem 2, we deduce the following:

Pn(x)

2�Υn

n

k�0(− 1)ke(n)

kx1, x2,. . . , xn

xn−k

⎡

⎣⎤

⎦2

,(39)

where Υn�2−n+12n−1

n

2

.

By considering (15) and applying the Cauchy product,

we deduce the following:

Pn(x)

2�Υn

2n

k�0(− 1)k

k

i�0

e(n)

ix1, x2,. . . , xn

×e(n)

k−ix1, x2,. . . , xn

⎡

⎣⎤

⎦x2n−k.(40)

By applying Lemma 3, we obtain the following:

Pn(x)

2�Υn

2n

k�0(− 1)ke(2n)

kx1, x2,. . . , xn, x1, x2,. . . , xn

x2n−k.

(41)

By integrating both sides on the interval [− 1,1], we

conclude as follows:

1

−1

Pn(x)

2dx�Υn

2n

k�0

(− 1)k

2n−k+1e(2n)

kx1, x2,. . . , xn, x1, x2,. . . , xn

+1

2n−k+1e(2n)

kx1, x2,. . . , xn, x1, x2,. . . , xn

.

(42)

erefore, 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. is completes

the proof.

Furthermore, we can deduce the following result. □

Corollary 4. For any positive integer n. Consider the Leg-

endre polynomial Pn(x)with zeros x1, x2,. . . , xn, then

2n

k�0

k≡0(mod2)

(− 1)k/2e(n)

k/2 x2

1, x2

2,..., x2

n

2n−k+1�2

2n+12n−1n!(n−1)!

(2n−1)!

2

.

(43)

Similarly, following the orthogonality of the Legendre

polynomials with distinct degrees, we deduce the following

result.

Corollary 5. Let n, m be positive integers with n≠m. Con-

sider the Legendre polynomial Pn(x)with zeros x1, x2,. . . , xn

arranged in ascending order, namely, x1<x2<... <xnand

similarly Pn(x)with zeros y1, y2,. . . , ymwith

y1<y2<·· · <ym, then

n+m

k�0

k≡0(mod2)

(− 1)k/2

n+m−k+1e(⌈(n+m)/2⌉)

k/2 x2

1, x2

2,. . . , x2

⌈n/2 ⌉, y2

1, y2

2,. . . , y2

⌈m/2 ⌉

�0.

(44)

e following result gives a relation between the ESPs for

the zeros of Legendre polynomials Pn(x)and Pn−1(x).

Theorem 3. Let n be a positive integer. Consider the Leg-

endre polynomial Pn(x)with zeros x1, x2,. . . , xnand

Pn−1(x)with zeros y1, y2,. . . , yn−1, then

e(n)

kx1, x2,. . . , xn

�n(2n−k−1)

(2n−1)(n−k)e(n−1)

ky1, y2,..., yn−1

,

(45)

for all k�0,1,2,. . . , n −1.

Proof: Following the recurrence relation (8), we obtain the

following:

2n−1

n

n−1

k�0(− 1)k(n−k)e(n)

k(x)xn−k−1

�n

n−1

k�0(− 1)ke(n−1)

k(y)xn−k−1+

n−2

k�0(− 1)k(n−k−1)e(n−1)

k(y)xn−k−1,

(46)

6Journal of Mathematics

where ydenotes the (n−1)− tuple (y1, y2,. . . , yn−1). Based

on the deﬁnition of ESPs, we can rewrite the last term in (46)

as follows:

n−2

k�0(− 1)k(n−k−1)e(n−1)

k(y)xn−k−1�

n−1

k�0(− 1)k(n−k−1)e(n−1)

k(y)xn−k−1.

(47)

erefore, by comparing the coeﬃcients of xn−k−1, we

obtain the following:

(2n−1)(n−k)

ne(n)

k(x) � ne(n−1)

k(y) +(n−k−1)e(n−1)

k(y),

(48)

which completes the proof.

Following eorem 3, it is easy to construct the Legendre

polynomial Pn(x)via the zeros of Pn−1(x), looking forward

the following corollary. □

Corollary 6. Let nbe a positive integer, with n>1. Assume

that the Legendre polynomial Pn−1(x)has zeros

y1, y2,. . . , yn−1, then

Pn(x) � 2−n+1n

2n−1

2n−1

n

⎛

⎝⎞

⎠

n

k�1(− 1)n−kn+k−1

k

e(n−1)

n−ky1, y2,. . . , yn−1

xk+2n

n−1

2

n

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠.(49)

It is easy to see that if nis odd, then the last term of (49)

will vanish.

5. Numerical Examples

roughout this section, we shall give some examples to

illustrate the results are posted in Sections 3 and 4. Let us

start with the following example that illustrates the idea of

eorem 2.

Example 1. e zeros of P3(x)are x1����

3/5

√, x2�0 and

x3� − ���

3/5

√, then following (2), we have the following:

e(3)

0x1, x2, x3

�1,

e(3)

1x1, x2, x3

�0,

e(3)

2x1, x2, x3

� − 3

5,

e(3)

3x1, x2, x3

�0.

(50)

erefore, by applying eorem 2, then we obtain as

follows:

P3(x) � 2−25

3

⎛

⎝⎞

⎠−e(3)

3x1, x2, x3

+e(3)

2x1, x2, x3

x−e(3)

1x1, x2, x3

x2+e(3)

0x1, x2, x3

x3

�1

4

(10) − 3

5x+x3

.

(51)

Moreover, we deduce that P3(x) � (1/2)(5x3−3x).

e following example to illustrates the idea of Corol-

laries 2 and 3.

Example 2. e zeros of Legendre polynomial P4(x)are

x1��������������

(15 +2��

30

√)/35

, x2��������������

(15 −2��

30

√)/35

, x3�

−�������������

(15 +2��

30

√)/35

and x4� − �������������

(15 −2��

30

√)/35

, then fol-

lowing Corollary 2, we have the following:

P4(x) � 35

8Λ4,0x4+Λ4,2x2+Λ4,4

,(52)

where Λ4,1�Λ4,3�0. Moreover, we obtain as follows:

Λ4,0�e(2)

0x2

1, x2

2

�1,

Λ4,2�e(2)

1x2

1, x2

2

� − 30

35,

Λ4,4�e(2)

2x2

1, x2

2

�105

(35)2.

(53)

us, we conclude that P4(x) � (1/8)(35x4−30x2+3).

Journal of Mathematics 7

Following eorem 3 and Corollary 6, we give the next

example.

Example 3. e zeros of the Legendre polynomial P2(x)are

x1�1/ �

3

√and x2� − (1/ �

3

√). us, following Corollary 6

we can construct the Legendre polynomial P3(x)as follows:

P3(x) � 2−23

5

5

3

⎛

⎝⎞

⎠3e(2)

2

1

�

3

√,−1

�

3

√

x−2e(2)

1

1

�

3

√,−1

�

3

√

x2+5

3e(2)

0

1

�

3

√,−1

�

3

√

x3

�3

2

5

3x3−x

,

(54)

therefore, P3(x) � (1/2)(5x3−3x).

6. Conclusion

roughout this paper, we provided some results related to

the elementary symmetric polynomials such as Lemmas 1–3.

e importance of the results mentioned in Section 2 ap-

pears 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 eorem 3, we were able to construct Pn(x)via the zeros

of Pn−1(x)for all n≥2, which reduce the computational

costs for constructing the Legendre polynomials.

Data Availability

No data were used to support this study.

Conflicts of Interest

e authors declare that they have no conﬂicts of interest.

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