A NOTE ON NEWTON’S INEQUALITY
SLAVKO SIMIC
Mathematical Institute SANU, Kneza Mihaila 36
11000 Belgrade, Serbia.
Newton’s Inequality
Slavko Simic
vol. 10, iss. 2, art. 44, 2009
EMail: ssimic@turing.mi.sanu.ac.rs
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Received:
25 March, 2009
Accepted:
12 April, 2009
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
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Key words:
Symmetric functions, Weighted means.
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Abstract:
We present a generalization of Newton’s inequality, i.e., an inequality of mixed
form connecting symmetric functions and weighted means. Two open problems
are also stated.
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1
Introduction
3
Newton’s Inequality
Slavko Simic
vol. 10, iss. 2, art. 44, 2009
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1.
Introduction
A well-known theorem of Newton [1] states the following:
Theorem 1.1. If all the zeros of a polynomial
(1.1)
Pn (x) = e0 xn + e1 xn−1 + · · · + ek xn−k + · · · + en ,
e0 = 1,
Newton’s Inequality
are real, then its coefficients satisfy
Slavko Simic
(n)
ek−1 ek+1 ≤ Ak e2k ,
(1.2)
(n)
where Ak :=
k = 1, 2, . . . , n − 1;
n−k
k
.
k+1 n+1−k
For a sequence a = {ai }ni=1 of real numbers, by putting
(1.3)
n
n
Y
X
Pn (x) =
(x + ai ) =
ek xn−k ,
i=1
k=0
we see that the coefficient ek = ek (a) represents the kth elementary symmetric function of a, i.e. the sum of all the products, k at a time, of different ai ∈ a.
There are several generalizations of Newton’s inequality [2], [3]. In this article
we give another one. For this purpose define the sequences a0i := a/{ai }, i =
1, 2, . . . , n, and by ek (a0i ) denote the k-th elementary symmetric function over a0i .
We have:
Theorem 1.2. Let c =
fying
(1.4)
{ci }ni=1
be a weight sequence of non-negative numbers satisn
X
i=1
ci = 1,
vol. 10, iss. 2, art. 44, 2009
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and, for an arbitrary sequence a = {ai }ni=1 of real numbers, define
(c)
Ek
(1.5)
:=
n
X
ci ek (a0i ),
(c)
E0 = 1,
i=1
or equivalently,
(1.6)
(c)
Ek = ek − ek−1 f1 + ek−2 f22 − · · · + (−1)r ek−r frr + · · · + (−1)k fkk ,
Newton’s Inequality
Slavko Simic
where
vol. 10, iss. 2, art. 44, 2009
n
X
fs :=
(1.7)
! 1s
ci asi
.
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i=1
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Then
(1.8)
(c)
(c)
(n−1)
Ek−1 Ek+1 ≤ Ak
(c)
Ek
2
,
k = 1, 2, . . . , n − 1.
Proof. We shall give an easy proof supposing that the sequence c consists of arbitrary positive rational numbers. Since a and c are independent of each other, the
truthfulness of the above theorem follows by the continuity principle.
Therefore, let p = {pk }nk=1 be an arbitrary sequence of positive integers and put
(1.9)
pi
ci = Pn
1
pk
,
i = 1, 2, . . . , n; p ∈ N.
Now, for a given real sequence a, consider the polynomial Q(x) defined by
(1.10)
n
Y
Q(x) :=
(x + ai )pi .
i=1
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Since all its zeros are real, by the well-known Gauss theorem, the zeros of Q0 (x),
0
Q (x) = Q(x)
(1.11)
n
X
i=1
pi
,
x + ai
are also real.
In particular, the same is valid for the polynomial R(x) defined by
n
n
Y
X
R(x) :=
(x + ai )
(1.12)
i=1
i=1
Newton’s Inequality
ci
.
(x + ai )
Slavko Simic
vol. 10, iss. 2, art. 44, 2009
Since
(c)
(c)
(c)
the result follows by simple application of Theorem 1.1.
P
Remark 1. Since ni=1 ek (a0i ) = (n − k)ek (a), putting ci = 1/n, i = 1, 2, . . . , n
in (1.5) and (1.8), we obtain the assertion from Theorem 1.1. Hence our result
represents a generalization of Newton’s theorem.
P
(c)
1/s
Also, denoting by fs (a) = fs := ( ni=1 ci asi ) , s > 0, the classical weighted
mean (with weights c) of order s, and using the identity
ek (a0i ) = ek (a) − ai ek−1 (a0i ),
(1.14)
(c)
an equivalent form of Ek arises, i.e.,
(1.15)
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R(x) = xn−1 + E1 xn−2 + · · · + Ek xn−1−k + · · · + En−1 ,
(1.13)
(c)
Ek = ek − ek−1 f1 + ek−2 f22 − · · · + (−1)r ek−r frr + · · · + (−1)k fkk .
Putting this in (1.8), we obtain a mixed inequality connecting elementary symmetric functions with weighted means of integer order.
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Problem 1.1. An interesting fact is that non-negativity of c is not a necessary condition for (1.8) to hold. We shall illustrate this point by an example. For k = 1, n = 3,
we have
(c)
(c)
(c)
(E1 )2 − 4E0 E2
= (c1 (a2 + a3 ) + c2 (a1 + a3 ) + c3 (a1 + a2 ))2 − 4(c1 a2 a3 + c2 a1 a3 + c3 a1 a2 )
= (1 − c2 )2 (a1 − a2 )2 + 2(c1 − c2 c3 )(a1 − a2 )(a3 − a1 ) + (1 − c3 )2 (a3 − a1 )2 ,
and this quadratic form is positive semi-definite whenever c1 c2 c3 ≥ 0.
Hence, in this case the inequality (1.8) is valid for all real sequences a with c
satisfying
(1.16)
c1 + c2 + c3 = 1,
Problem 1.2. There is an interesting application of Theorem 1.2 to the well known
Turan’s problem. Under what conditions does the sequence of polynomials {Qn (x)}
satisfy Turan’s inequality
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Qn−1 (x)Qn+1 (x) ≤ (Qn (x)) ,
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for each x ∈ [a, b] and n ∈ [n1 , n2 ]?
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This problem is solved for many classes of polynomials [4]. We shall consider
here the following question [5].
An arbitrary sequence {di }, i = 1, 2, . . . of real numbers generates a sequence of
polynomials {Pn (x)}, n = 0, 1, 2, . . . defined by
(1.18)
Slavko Simic
vol. 10, iss. 2, art. 44, 2009
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c1 c2 c3 ≥ 0.
Therefore there remains the seemingly difficult problem of finding true bounds
for the sequence c satisfying (1.4), such that the inequality (1.8) holds for an arbitrary
real sequence a.
(1.17)
Newton’s Inequality
Pn (x) := xn + d1 xn−1 + d2 xn−2 + · · · + dn−1 x + dn ,
P0 (x) := 1.
Denote also by An the set of zeros of Pn (x).
Now, if for some m > 1 the set Am consists of real numbers only, then from
Theorem 1.2, it follows that
(1.19)
Pn−1 (a)Pn+1 (a) ≤ Pn2 (a),
for each a ∈ Am and n ∈ [1, m − 1].
Is it possible to establish some simple conditions such that the Turan inequality
(1.20)
Pn−1 (x)Pn+1 (x) ≤ Pn2 (x),
Newton’s Inequality
Slavko Simic
vol. 10, iss. 2, art. 44, 2009
holds for each x ∈ [min a, max a]a∈Am and n ∈ [1, m − 1].
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Camb. Univ.
Press, Cambridge, 1978.
[2] J.N. WHITELEY, A generalization of a theorem of Newton, Proc. Amer. Math.
Soc., 13 (1962), 144–151.
[3] K.V. MENON, Inequalities for symmetric functions, Duke Math. J., 35 (1968),
37–45.
[4] S. SIMIC, Turan’s inequality for Appell polynomials, J. Ineq. Appl., 1 (2006),
1–7.
Newton’s Inequality
Slavko Simic
vol. 10, iss. 2, art. 44, 2009
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[5] S. SIMIC, A log-concave sequence, Siam Problems and Solutions, [ONLINE:
http//:www.siam.org].
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