Journal of Inequalities in Pure and
Applied Mathematics
A POLYNOMIAL INEQUALITY GENERALISING AN
INTEGER INEQUALITY
volume 3, issue 4, article 52,
2002.
ROGER B. EGGLETON AND WILLIAM P. GALVIN
Department of Mathematics
Illinois State University
Normal, IL 61790-4520,
USA.
EMail: roger@math.ilstu.edu
School of Mathematical and Physical Sciences
University of Newcastle
Callaghan, NSW 2308
Australia.
EMail: mmwpg@cc.newcastle.edu.au
Received 1 May, 2002;
accepted 7 June, 2002.
Communicated by: H. Gauchman
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
043-02
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Abstract
For any a := (a1 , a2 , . . . , an ) ∈ (R+ )n , we establish inequalities between the two
homogeneous polynomials ∆Pa (x, t) := (x + a1 t)(x + a2 t) · · · (x + an t) − xn and
Sa (x, y) := a1 xn−1 + a2 xn−2 y + · · · + an yn−1 in the positive orthant x, y, t ∈ R+ .
Conditions for ∆Pa (x, t) ≤ tSa (x, y) yield a new proof and broad generalization
of the number theoretic inequality that for base b ≥ 2 the sum of all nonempty
products of digits of any m ∈ Z+ never exceeds m, and equality holds exactly
when all auxiliary digits are b − 1. Links with an inequality of Bernoulli are also
noted. When n ≥ 2 and a is strictly positive, the surface ∆Pa (x, t) = tSa (x, y)
lies between the planes y = x + t max{ai : 1 ≤ i ≤ n − 1} and y = x + t min{ai :
1 ≤ i ≤ n − 1}. For fixed t ∈ R+ , we explicitly determine functions α, β, γ, δ
of a such that this surface is y = x + αt + βt2 x−1 + O(x−2 ) as x → ∞, and
y = γt + δx + O(x2 ) as x → 0 + .
2000 Mathematics Subject Classification: Primary 26D15, 26C99.
Key words: Polynomial inequality, sums of products of digits, Bernoulli inequality.
The first author gratefully acknowledges the hospitality of the School of Mathematical
and Physical Sciences, University of Newcastle, while this paper was being written.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Polynomial Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3
Conditions for Equality to Hold . . . . . . . . . . . . . . . . . . . . . . . . . 13
References
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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1.
Introduction
For any finite sequence of real numbers a, let Πa be the product of all terms in
a, and let T (a), the total sum of products of a, be the sum of all products Πx as
x runs through the nonempty subsequences x ⊆ a. Thus
T (a) := Σ{Πx : x ⊆ a, x 6= ω},
where ω is the empty sequence. As usual we observe the convention that Πω =
1. There is a rather surprising inequality which T (a) must satisfy in the case of
integer sequences. In particular, for given integers b ≥ 2 and m ≥ 0, let a be
the sequence of digits in the base b representation of m. Then
T (a) ≤ m
holds for every such integer m and base b, as shown in [2]. Moreover the inequality is sharp: T (a) = m holds precisely when the auxiliary digits of m, if
any, are all b − 1. (The leading digit of n is the most significant digit; the less
significant digits, if any, are its auxiliary digits.) For example
T (3, 7, 7) = 255 ≤ 377(b) ,
where 377(b) is the base b representation of m = 255, 313, 377, 447, . . . when
b = 8, 9, 10, 11, . . . . We also note in passing that if a is the base b digit sequence
of m then T (a) is odd precisely when at least one of the digits of m is odd.
Our main purpose in this paper is to show that the integer inequality just
described is an instance of a much more general inequality between polynomials. We shall establish the polynomial inequality and investigate some of its
properties.
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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2.
Polynomial Inequality
Let a be any nonempty finite sequence of real numbers, say
a := (a1 , a2 , . . . , an ) ∈ Rn , with n ≥ 1.
With a we associate two homogeneous polynomials in two real variables, the
product polynomial
Pa (x, t) := (x + a1 t)(x + a2 t) · · · (x + an t) =
n
Y
(x + ar t),
r=1
and the sum polynomial
Sa (x, y) := a1 x
n−1
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
+ a2 x
n−2
y + · · · + an y
n−1
=
n
X
ar xn−r y r−1 .
r=1
Here we shall study these polynomials when a ∈ (R+ )n , where R+ := {x ∈
R : x ≥ 0}. It turns out that it is natural to compare t times the sum polynomial
with the first difference of the product polynomial,
∆Pa (x, t) := Pa (x, t) − Pa (x, 0) = Pa (x, t) − xn .
Note that tSa (x, y) and ∆Pa (x, t) are both homogeneous of degree n.
With a we also associate two bounds when n ≥ 2:
M (a) := max{ar : 1 ≤ r ≤ n − 1}
and m(a) := min{ar : 1 ≤ r ≤ n − 1}.
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n
Theorem 2.1. For any finite nonnegative sequence a ∈ (R+ ) with n ≥ 1, the
inequality
0 ≤ ∆Pa (x, t) ≤ tSa (x, y)
holds for all x, y, t ∈ R+ , provided y ≥ x + tM (a) if n ≥ 2. The reverse
inequality
∆Pa (x, t) ≥ tSa (x, y) ≥ 0
holds for all x, y, t ∈ R+ , provided y ≤ x + tm(a) if n ≥ 2.
A Polynomial Inequality
Generalising an Integer
Inequality
Proof. An easy induction on n establishes the identity
n
n
r−1
Y
X
Y
n
n−r
Pa (x, t) =
(x + ar t) = x +
ar x t (x + as t).
r=1
r=1
s=1
For x, t ∈ R+ we have x + as t ≥ 0 for each s, so
0≤
Roger B. Eggleton and
William P. Galvin
r−1
Y
(x + as t) ≤ y r−1
s=1
holds trivially if r = 1, and for r ≥ 2 it certainly holds if
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y ≥ max{x + as t : 1 ≤ s ≤ r − 1} = x + t · max{as : 1 ≤ s ≤ r − 1}.
Because each ar ∈ R+ , it follows for x, t ∈ R+ that
0 ≤ ∆Pa (x, t) = Pa (x, t) − xn
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=t
≤t
n
X
r=1
n
X
ar x
n−r
r−1
Y
(x + as t)
s=1
ar xn−r y r−1 = tSa (x, y)
r=1
holds trivially if n = 1, and for n ≥ 2 it holds if y ≥ x + tM (a). An entirely
similar argument establishes the reverse inequality in the theorem.
Let us define
Σ(a) :=
n
X
A Polynomial Inequality
Generalising an Integer
Inequality
ar .
Roger B. Eggleton and
William P. Galvin
r=1
+ n
If a ∈ (R ) and n ≥ 2 then
0 ≤ m(a) ≤ M (a) ≤ Σ(a).
Title Page
Note that Sa (1, 1) = Σ(a). This constant plays a natural role in bounding our
polynomial inequalities away from zero. Specifically, we have
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+ n
Corollary 2.2. Let a ∈ (R ) be a finite nonnegative sequence with n ≥ 3 and
M (a) > m(a). Then for all strictly positive x, y, t ∈ R+ the inequality
0 < tΣ(a)xn−1 < ∆Pa (x, t) < tSa (x, y)
holds provided y ≥ x + tM (a), and the reverse inequality
∆Pa (x, t) > tSa (x, y) ≥ tΣ(a)z
n−1
holds provided y ≤ x + tm(a), with z := min{x, y}.
>0
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Proof. We sharpen the details of the proof of Theorem 2.1. The condition
M (a) > m(a) ensures that M (a) > 0, so if x, t are strictly positive reals
then x + as t > x for at least one s ≤ n − 1, and
r−1
Y
(x + as t) > xr−1
s=1
holds for some r ≤ n. Then
A Polynomial Inequality
Generalising an Integer
Inequality
∆Pa (x, t) = Pa (x, t) − xn
=
n
X
ar xn−r t
r=1
>
n
X
r−1
Y
(x + as t)
Roger B. Eggleton and
William P. Galvin
s=1
ar xn−1 t = tΣ(a)xn−1 > 0.
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r=1
Contents
If y ≥ x + tM (a), then M (a) > m(a) ensures that
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r−1
Y
(x + as t) < y r−1
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holds for at least one r ≤ n, so
n
∆Pa (x, t) = Pa (x, t) − x < t
n
X
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ar x
n−r r−1
y
= tSa (x, y).
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For the second inequality, if 0 < y ≤ x + tm(a) then M (a) > m(a) ensures
that
r−1
Y
(x + as t) > y r−1
s=1
holds for at least one r ≤ n, so
∆Pa (x, t) = Pa (x, t) − xn
n
X
>t
ar xn−r y r−1
A Polynomial Inequality
Generalising an Integer
Inequality
r=1
= tSa (x, y)
n
X
≥t
ar z n−1 = tΣ(a)z n−1 > 0,
Roger B. Eggleton and
William P. Galvin
r=1
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where z := min{x, y}.
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n
Corollary 2.3. For any real c and given finite sequence a ∈ R , if n = 1 or if
n ≥ 2 and M (a) = m(a) = c, then
∆Pa (x, t) = tSa (x, x + ct)
is an identity for all x, t ∈ R.
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+ n
+
Proof. First suppose a ∈ (R ) and c, x, t ∈ R . If n = 1 both inequalities
in Theorem 2.1 hold, so ∆Pa (x, t) = tSa (x, x + ct). The same result holds if
n ≥ 2 when M (a) = m(a) = c and y = x + ct. Since we have a degree n
polynomial equality which holds for more than n values of x and more than n
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values of t, it must in fact be a polynomial identity, and therefore holds for all
x, t ∈ R and a ∈ Rn with M (a) = m(a).
We shall now show that the integer inequality proved in [2], and the conditions under which it is an equality, are directly deducible from the above results.
Thus Theorem 2.1 provides a new proof of the results in [2] as well as placing
them in a much more general context.
Corollary 2.4. For any integers b ≥ 2 and m ≥ 0, let a ∈ (Z+ )n be the
sequence of base b digits of m. Then the total sum of products of these digits
satisfies T (a) ≤ m, with equality precisely when every auxiliary digit of m is
b − 1.
Proof. Assume that the base b digits of m are arranged in a in order of increasing significance, so an is the leading digit. Then Sa (1, b) = m. Furthermore
M (a) ≤ b − 1. Put x = 1, t = 1 and y = b. Then y ≥ x + tM (a), so the first
inequality in Theorem 2.1 yields
T (a) = Pa (1, 1) − 1 = ∆Pa (1, 1) ≤ Sa (1, b) = m,
as required. Now consider when equality holds. By Corollary 2.2, the strict
inequality T (a) < m holds if n ≥ 3 and the auxiliary digits are not all equal,
so suppose n ≥ 2 and all auxiliary digits are equal to M (a). Corollary 2.3
shows that T (a) = m∗ , where m∗ = Sa (1, M (a) + 1) is the integer with base
M (a) + 1 digit sequence a if we permit the slightly nonstandard possibility that
the leading digit may exceed M (a). Thus m∗ = m if M (a) = b − 1, and
m∗ < m if M (a) < b − 1. If n = 1, Corollary 2.3 confirms the already obvious
T (a) = m.
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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We now note some examples of Theorem 2.1.
Example 2.1. With t = 1, a = (a, b, c, d) ∈ (R+ )4 , and the change of variables
x ← t, y ← x with x, t ∈ R+ , we have
(t + a)(t + b)(t + c)(t + d) − t4 ≤ at3 + bt2 x + ctx2 + dx3
when x ≥ t + max{a, b, c}. The reverse inequality holds when x ≤ t +
min{a, b, c}.
Example 2.2. With t = 1, a = (d, c, b, a) ∈ (R+ )4 , and the change of variables
x ← t, y ← x with x, t ∈ R+ , we have
A Polynomial Inequality
Generalising an Integer
Inequality
(t + a)(t + b)(t + c)(t + d) − t4 ≤ ax3 + btx2 + ct2 x + dt3
Roger B. Eggleton and
William P. Galvin
when x ≥ t + max{b, c, d}. The reverse inequality holds when x ≤ t +
min{b, c, d}.
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Example 2.3. In Example 2.2, let t = 1 and replace (a, b, c, d) in that example
with (a, bt, ct2 , dt3 ), where a, b, c, d, t are strictly positive. Then
(1 + a)(1 + bt)(1 + ct2 )(1 + dt3 ) − 1 ≤ ax3 + btx2 + ct2 x + dt3
when x ≥ 1 + max{bt, ct2 , dt3 }.
Example 2.4. Replace (a, b, c, d) in Example 2.2 by (a, bt−1 , ct−2 , dt−3 ), so
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(t + a)(t2 + b)(t3 + c)(t4 + d) − t10 ≤ t6 (ax3 + bx2 + cx + d)
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when x ≥ t + max{bt−1 , ct−2 , dt−3 }.
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Example 2.5. Evidently
∆Pa (1, 1) ≥ Sa (1, 1) = Σ(a)
holds for any a ∈ (R+ )n with n ≥ 1, and holds with strict inequality if n ≥ 2
and a has at least two strictly positive terms. However, it is interesting to note
that it also holds with strict inequality for any a ∈ (−1, 0)n with n ≥ 2, a result
which goes back to Jacques [= James = Jakob] Bernoulli (1654-1705) in the
case where the sequence a is constant (see [1, Theorem 58]). Our focus in the
present paper is on cases in which a ∈ (R+ )n .
The reverse of a given finite sequence a := (a1 , a2 , . . . , an ) ∈ Rn with n ≥ 1
is the sequence aR := (an , . . . , a2 , a1 ) ∈ Rn . Then
PaR (x, t) = Pa (x, t) and SaR (x, y) = Sa (y, x).
Let max(a) := max{ar : 1 ≤ r ≤ n} and min(a) := min{ar : 1 ≤ r ≤ n}. If
n ≥ 2 we have
R
max{M (a), M (a )} = max(a) and
R
min{m(a), m(a )} = min(a).
With these observations, combining the principles used in Examples 2.1 and 2.2
readily yields
+ n
Corollary 2.5. For any finite nonnegative sequence a ∈ (R ) with n ≥ 1, the
inequality
0 ≤ ∆Pa (t, 1) ≤ min{Sa (t, x), Sa (x, t)}
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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holds for all x, t ∈ R+ , provided x ≥ t + max(a) if n ≥ 2. The reverse
inequality
∆Pa (t, 1) ≥ max{Sa (t, x), Sa (x, t)} ≥ 0
holds for all x, t ∈ R+ , provided x ≤ t + min(a) if n ≥ 2.
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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3.
Conditions for Equality to Hold
When does the inequality studied in Theorem 2.1 become an equality? To reduce this to a problem in two variables, let us examine the t = 1 cross-section.
Suppose n ≥ 2 and a ∈ (R+ )n is strictly positive, that is, ar > 0 for 1 ≤ r ≤ n.
We have from Theorem 2.1:
(
≤ Sa (x, y) when y ≥ x + M (a),
∆Pa (x, 1)
≥ Sa (x, y) when y ≤ x + m(a).
If x, y are strictly positive, then
∂
Sa (x, y) > 0,
∂y
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
and continuity of Sa (x, y) as a function of y ensures the following result:
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Lemma 3.1. For any strictly positive x ∈ R+ and any strictly positive sequence
a ∈ (R+ )n with n ≥ 2, there is a unique y0 > 0 such that

 < Sa (x, y) if y > y0 ,
∆Pa (x, 1) = Sa (x, y0 )

> Sa (x, y) if 0 < y < y0 .
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Furthermore
x + m(a) ≤ y0 ≤ x + M (a).
In what follows we shall determine y0 more explicitly. It is convenient to
introduce some notation. Let Σk (a) be the kth elementary symmetric function
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of the sequence a, defined to be the sum of products Πx as x runs through all
the k-term subsequences x ⊆ a. Thus
Σk (a) := Σ{Πx : x ⊆ a, |x| = k}.
n
In particular Σ1 (a) = Σ(a) = Σnr=1 ar and Σ2 (a) = Σn−1
r=1 Σs=r+1 ar as . Again let
n X
r−1
Wk (a) :=
ar .
k−1
r=1
We call Wk (a) the kth binomially-weighted sum of the sequence a. Note that
W1 (a) = Σ1 (a).
Lemma 3.2. For any finite strictly positive sequence a ∈ (R+ )n and any positive integer k ≤ n, we have
min(a)k
Σk (a)
max(a)k
≤
≤
,
max(a)
Wk (a)
min(a)
with strict inequalities when a is not constant.
Proof. Let a∗ ∈ (R+ )n be the constant sequence with every term equal to
max(a). Then
n
∗
max(a)k ,
Σk (a) ≤ Σk (a ) =
k
and the inequality is strict when a is not constant. Also
n n X
X
r−1
r−1
n
Wk (a) =
ar ≥
min(a) =
min(a) > 0,
k−1
k−1
k
r=1
r=1
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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so
Σk (a)
max(a)k
≤
,
Wk (a)
min(a)
with strict inequality when a is not constant. An entirely similar argument establishes the other inequality in the lemma.
For any real c > 0, if c ∈ (R+ )n is the constant sequence with every
term equal to c, then Lemma 3.2 shows that Σk (c)/Wk (c) = ck−1 . Hence
1
(Σk (a)/Wk (a)) k−1 is a measure of central tendency for the terms of the sequence a ∈ (R+ )n , for each integer k in the interval 2 ≤ k ≤ n. The case k = 2
enters into the asymptotic behaviour of y0 , as we now show.
Theorem 3.3. For strictly positive x, y ∈ R+ and any strictly positive sequence
a ∈ (R+ )n with n ≥ 2, the equality ∆Pa (x, 1) = Sa (x, y) holds for large x
when
y = x + α + O(x−1 )
(x → ∞),
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
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where
Σ2 (a)
α :=
.
W2 (a)
Proof. Let y0 = x + f0 (x), so ∆Pa (x, 1) = Sa (x, x + f0 (x)). Then m(a) ≤
f0 (x) ≤ M (a) by Lemma 3.1, so O(f0 (x)) = O(1) as x → ∞. Hence
Sa (x, x + f0 (x)) =
=
n
X
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ar xn−r (x + f0 (x))r−1
r=1
n
X
r=1
!
ar
x
n−1
+
n
X
r=1
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!
(r − 1)ar
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f0 (x)x
n−2
+ O(x
n−3
)
J. Ineq. Pure and Appl. Math. 3(4) Art. 52, 2002
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= Σ1 (a)xn−1 + W2 (a)f0 (x)xn−2 + O(xn−3 ).
Also
∆Pa (x, 1) = (x + a1 )(x + a2 ) · · · (x + an ) − xn
= Σ1 (a)xn−1 + Σ2 (a)xn−2 + O(xn−3 ).
But these two expressions are equal, so for large x it follows that
f0 (x) =
Σ2 (a)
+ O(x−1 ).
W2 (a)
A Polynomial Inequality
Generalising an Integer
Inequality
Roger B. Eggleton and
William P. Galvin
By Theorem 3.3, if we put y0 = x + α + f1 (x) then O(f1 (x)) = O(x−1 ) as
x → ∞. Explicit expansion of ∆Pa (x, 1) and Sa (x, x + α + f1 (x)) as far as
terms in xn−3 yields
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Corollary 3.4. For any finite strictly positive sequence a ∈ (R+ )n with n ≥ 3,
the equality ∆Pa (x, 1) = Sa (x, y) holds for large x, y ∈ R+ when
y = x + α + βx−1 + O(x−2 )
(x → ∞),
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where
Σ2 (a)
α :=
W2 (a)
and
Σ3 (a) − α2 W3 (a)
β :=
.
W2 (a)
From Lemma 3.1 we immediately deduce
Corollary 3.5. If M (a) = m(a) = c, then α = c and β = 0.
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Next we shall consider y0 when x is small but positive. It will be convenient to use G(a) to denote the geometric mean of {ar : 1 ≤ r ≤ n − 1}, so
1
G(a) := (a1 a2 · · · an−1 ) n−1 . For any finite strictly positive sequence a ∈ (R+ )n
−1
−1
−1
we define a−1 := (a−1
1 , a2 , . . . , an ), so Σ1 (a ) is the sum of reciprocals of
the terms of a. Of course, Σ1 (a−1 ) = Σn−1 (a)/Σn (a). This sum enters into the
small scale behaviour of y0 , as we now show.
Theorem 3.6. For strictly positive x, y ∈ R+ and any strictly positive sequence
a ∈ (R+ )n with n ≥ 2, the equality ∆Pa (x, 1) = Sa (x, y) holds for small x
when
y = γ + δx + O(x2 )
(x → 0+),
where
−1
γ := G(a) and
δ :=
γan Σ1 (a ) − an−1
.
(n − 1)an
Proof. For 0 < x < M (a) let y0 = g0 (x), so ∆Pa (x, 1) = Sa (x, g0 (x)).
Lemma 3.1 ensures that m(a) < g0 (x) < 2M (a), so O(g0 (x)) = O(1) as
x → 0 + . Then
Sa (x, g0 (x)) =
n
X
an−r+1 xr−1 g0 (x)n−r = an g0 (x)n−1 + O(x)
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and
∆Pa (x, 1) = (a1 a2 · · · an ) + O(x),
so equality of these expressions implies that
g0 (x) = G(a) + O(x).
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Now let y0 = G(a) + g1 (x), so O(g1 (x)) = O(x) as x → 0 + . Then
Sa (x, G(a) + g1 (x))
n
X
=
an−r+1 xr−1 (G(a) + g1 (x))n−r
r=1
= an G(a)n−1 + (n − 1)an G(a)n−2 g1 (x) + an−1 xG(a)n−2 + O(x2 )
and
(
∆Pa (x, 1) = (a1 a2 · · · an ) 1 +
n
X
!
a−1
r
)
x + O(x2 ) .
r=1
Equality of these two expressions implies that
g1 (x) =
(an G(a)Σ1 (a−1 ) − an−1 ) x
+ O(x2 ),
(n − 1)an
and the theorem follows.
A Polynomial Inequality
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Roger B. Eggleton and
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From Lemma 3.1 we deduce
Corollary 3.7. If M (a) = m(a) = c, then γ = c and δ = 1.
Let us now consider the geometry underlying Theorems 3.3 and 3.6. The
positive quadrant x, y ∈ R+ is divided into an “S-region”, where
∆Pa (x, 1) < Sa (x, y),
and a “∆P -region”, where
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∆Pa (x, 1) > Sa (x, y).
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The boundary between these two regions is
E1 (a) := {(x, y) ∈ (R+ )2 : ∆Pa (x, 1) = Sa (x, y)}.
On this boundary curve the polynomials ∆Pa (x, 1) and Sa (x, y) are equal, so
we call E1 (a) the equipoise curve for a. Lemma 3.1 ensures that E1 (a) lies
in the strip between the parallel lines y = x + M (a) and y = x + m(a). By
Theorem 3.3 the equipoise curve is asymptotic to y = x + α, and by Theorem
3.6 it cuts the y-axis at y = G(a), with slope δ.
When n = 2, we have M (a) = m(a) = α = G(a) = a1 and E1 (a) is
the line y = x + a1 . When n ≥ 3, as x → ∞ the equipoise curve approaches
the asymptote from the S-region side if β > 0, and from the ∆P -region side if
β < 0.
It appears likely that the equipoise curve never crosses the asymptote, though
we were not able to demonstrate this in general. The condition for such a crossing to occur is a polynomial of degree n − 3 in x, so such crossings are possible
only when n ≥ 4. However it seems unlikely that there are ever any solutions
with x > 0. When n = 3, it is clear that E1 (a) must be entirely on one side of
the asymptote unless α2 = a1 a2 . In the latter case, β = 0 and E1 (a) actually
coincides with the asymptote; this behaviour is demonstrated by E1 (1, 4, 4) for
example.
Throughout the preceding discussion in this section we have been comparing
∆Pa (x, 1) with Sa (x, y) in the positive x, y-quadrant. A simple observation
enables us to deduce the corresponding information comparing ∆Pa (x, t) with
tSa (x, y) in the positive x, y, t-orthant. For any t ∈ R+ and a ∈ (R+ )n , let
ta := (ta1 , ta2 , . . . , tan ) ∈ (R+ )n . Then
Pta (x, 1) = Pa (x, t) and Sta (x, y) = tSa (x, y),
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so all the relevant facts about ∆Pa (x, t) = tSa (x, y) follow from our earlier
results in this section by replacing a by ta. In particular, the equipoise surface
E2 (a) := {(x, y, t) ∈ (R+ )3 : ∆Pa (x, t) = tSa (x, y)}
lies in the region between the planes y = x + tM (a) and y = x + tm(a), which
coincide if M (a) = m(a), and otherwise intersect in the line y = x, t = 0. For
any fixed t > 0 the equipoise surface satisfies
y = x + αt + βt2 x−1 + O(x−2 )
(x → ∞)
and
y = γt + δx + O(x2 )
(x → 0+).
However, the device of replacing a by ta does not provide any information
about the comparison of the product and sum polynomials for a general finite
sequence a ∈ Rn . As hinted at by Bernoulli’s Inequality, mentioned in Example
2.5, there is potentially much of interest in this more general case.
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Roger B. Eggleton and
William P. Galvin
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, 2nd ed.,
Cambridge, 1959.
[2] A.Y. ÖZBAN, W. P. GALVIN AND R.B. EGGLETON, An inequality for
the digits of an integer, Austral. Math. Soc. Gazette, 28 (2001), 9–10.
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William P. Galvin
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