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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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