A NOTE ON NEWTON’S INEQUALITY

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Newton’s Inequality Slavko Simic vol. 10, iss. 2, art. 44, 2009

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A NOTE ON NEWTON’S INEQUALITY

SLAVKO SIMIC

Mathematical Institute SANU, Kneza Mihaila 36 11000 Belgrade, Serbia.

EMail:ssimic@turing.mi.sanu.ac.rs

Received: 25 March, 2009

Accepted: 12 April, 2009

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26E60.

Key words: Symmetric functions, Weighted means.

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

A well-known theorem of Newton [1] states the following:

Theorem 1.1. If all the zeros of a polynomial

(1.1) P_n(x) =e₀xⁿ+e₁xⁿ⁻¹+· · ·+e_kx^n−k+· · ·+e_n, e₀ = 1, are real, then its coefficients satisfy

(1.2) ek−1ek+1 ≤A⁽ⁿ⁾_k e²_k, k= 1,2, . . . , n−1;

whereA⁽ⁿ⁾_k := _k+1^k _n+1−k^n−k .

For a sequencea={a_i}ⁿ_i=1of real numbers, by putting

(1.3) Pn(x) =

n

Y

i=1

(x+ai) =

n

X

k=0

ekx^n−k,

we see that the coefficiente_k=e_k(a)represents thekth elementary symmetric func- tion ofa, i.e. the sum of all the products,kat a time, of differenta_i ∈a.

There are several generalizations of Newton’s inequality [2], [3]. In this article we give another one. For this purpose define the sequences a⁰_i := a/{a_i}, i = 1,2, . . . , n, and by e_k(a⁰_i) denote the k-th elementary symmetric function over a⁰_i. We have:

Theorem 1.2. Letc={ci}ⁿ_i=1 be a weight sequence of non-negative numbers satis- fying

(1.4)

n

Xci = 1,

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and, for an arbitrary sequencea={a_i}ⁿ_i=1of real numbers, define

(1.5) E_k^(c):=

n

X

i=1

c_ie_k(a⁰_i), E₀^(c) = 1,

or equivalently,

(1.6) E_k^(c) =e_k−e_k−1f₁+e_k−2f₂²− · · ·+ (−1)^re_k−rf_r^r+· · ·+ (−1)^kf_k^k, where

(1.7) fs :=

n

X

i=1

cia^s_i

!¹_s .

Then

(1.8) E_k−1^(c) E_k+1^(c) ≤A⁽ⁿ⁻¹⁾_k E_k^(c)2

, k = 1,2, . . . , n−1.

Proof. We shall give an easy proof supposing that the sequencec consists of arbi- trary positive rational numbers. Since a and c are independent of each other, the truthfulness of the above theorem follows by the continuity principle.

Therefore, letp={p_k}ⁿ_k=1 be an arbitrary sequence of positive integers and put

(1.9) ci = p_i

Pn

1 p_k, i= 1,2, . . . , n; p∈N.

Now, for a given real sequencea, consider the polynomialQ(x)defined by

(1.10) Q(x) :=

n

Y

i=1

(x+ai)^pⁱ.

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Since all its zeros are real, by the well-known Gauss theorem, the zeros ofQ⁰(x),

(1.11) Q⁰(x) =Q(x)

n

X

i=1

p_i x+a_i,

are also real.

In particular, the same is valid for the polynomialR(x)defined by

(1.12) R(x) :=

n

Y

i=1

(x+a_i)

n

X

i=1

c_i (x+a_i).

Since

(1.13) R(x) =xⁿ⁻¹+E₁^(c)xⁿ⁻² +· · ·+E_k^(c)x^n−1−k+· · ·+E_n−1^(c) , the result follows by simple application of Theorem 1.1.

Remark 1. Since Pn

i=1e_k(a⁰_i) = (n −k)e_k(a), putting c_i = 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.

Also, denoting byfs^(c)(a) =f_s := (Pn

i=1c_ia^s_i)^1/s, s >0, the classical weighted mean (with weightsc) of orders, and using the identity

(1.14) e_k(a⁰_i) = e_k(a)−a_iek−1(a⁰_i), an equivalent form ofE_k^(c)arises, i.e.,

(1.15) E_k^(c) =e_k−ek−1f₁+ek−2f₂²− · · ·+ (−1)^rek−rf_r^r+· · ·+ (−1)^kf_k^k. 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 ofcis not a necessary condi- tion for (1.8) to hold. We shall illustrate this point by an example. Fork = 1, n = 3, we have

(E₁^(c))² −4E₀^(c)E₂^(c)

= (c1(a2+a3) +c2(a1+a3) +c3(a1+a2))²−4(c1a2a3+c2a1a3+c3a1a2)

= (1−c₂)²(a₁−a₂)²+ 2(c₁−c₂c₃)(a₁−a₂)(a₃−a₁) + (1−c₃)²(a₃−a₁)², and this quadratic form is positive semi-definite wheneverc₁c₂c₃ ≥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, c₁c₂c₃ ≥0.

Therefore there remains the seemingly difficult problem of finding true bounds for the sequencecsatisfying (1.4), such that the inequality (1.8) holds for an arbitrary real sequencea.

Problem 1.2. There is an interesting application of Theorem1.2to the well known Turan’s problem. Under what conditions does the sequence of polynomials{Q_n(x)}

satisfy Turan’s inequality

(1.17) Qn−1(x)Q_n+1(x)≤(Q_n(x))², for eachx∈[a, b]andn∈[n₁, n₂]?

This problem is solved for many classes of polynomials [4]. We shall consider here the following question [5].

An arbitrary sequence{d_i}, i= 1,2, . . . of real numbers generates a sequence of polynomials{Pn(x)}, n= 0,1,2, . . . defined by

(1.18) Pn(x) := xⁿ+d1xⁿ⁻¹+d2xⁿ⁻²+· · ·+dn−1x+dn, P0(x) := 1.

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Denote also byA_nthe set of zeros ofP_n(x).

Now, if for some m > 1 the set A_m consists of real numbers only, then from Theorem1.2, it follows that

(1.19) Pn−1(a)P_n+1(a)≤P_n²(a), for eacha∈A_mandn∈[1, m−1].

Is it possible to establish some simple conditions such that the Turan inequality (1.20) Pn−1(x)P_n+1(x)≤P_n²(x),

holds for eachx∈[mina,maxa]a∈Am andn∈[1, m−1].

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References

[1] G.H. HARDY, J.E. LITTLEWOODANDG. 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.

[5] S. SIMIC, A log-concave sequence, Siam Problems and Solutions, [ONLINE:

http//:www.siam.org].