Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 4, Article 101, 2005
ON THE HOMOGENEOUS FUNCTIONS WITH TWO PARAMETERS AND ITS
MONOTONICITY
ZHEN-HANG YANG
Z HEJIANG E LECTRIC P OWER VOCATIONAL T ECHNICAL C OLLEGE
H ANGZHOU , Z HEJIANG , C HINA , 311600
yzhkm@163.com
Received 18 May, 2005; accepted 04 August, 2005
Communicated by P.S. Bullen
A BSTRACT. Suppose f (x, y) is a positive homogeneous function defined on U(j R+ × R+ ),
1
h p p i p−q
,b )
call Hf (a, b; p, q) = ff (a
homogeneous function with two parameters. If f (x, y) is
q
q
(a ,b )
2nd differentiable, then the monotonicity in parameters p and q of Hf (a, b; p, q) depend on the
signs of I1 = (ln f )xy , for variable a and b depend on the sign of I2a = [(ln f )x ln(y/x)]y and
I2b = [(ln f )y ln(x/y)]x respectively. As applications of these results, a serial of inequalities
for arithmetic mean, geometric mean, exponential mean, logarithmic mean, power-Exponential
mean and exponential-geometric mean are deduced.
Key words and phrases: Homogeneous function with two parameters, f -mean with two-parameter, Monotonicity, Estimate
for lower and upper bounds.
2000 Mathematics Subject Classification. Primary 26B35, 26E60; Secondary 26A48, 26D07.
1. I NTRODUCTION
The so-called two-parameter mean or extended mean values between two unequal positive
numbers a and b were defined first by K.B. Stolarsky in [10] as
 1
q(ap −bp ) p−q


p 6= q, pq 6= 0

p(aq −bq )


1

p p p


a −b

p 6= 0, q = 0

 p(ln a−ln b)
1
.
(1.1)
E(a, b; p, q) =
q
aq −bq

p
=
0,
q
=
6
0

q(ln a−ln b)


p


a ln a−bp ln b
1

exp
−
p = q 6= 0

ap −bp
p


 √
ab
p=q=0
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
Thanks for Mr. Zhang Zhihua.
155-05
2
Z HEN -H ANG YANG
The monotonicity of E(a, b; p, q) has been researched by E. B. Leach and M. C. Sholander in
[4], and others also in [9, 8, 7, 6, 5, 11, 14, 15, 17] using different ideas and simpler methods.
As the generalized power-mean, C. Gini obtained a similar two-parameter type mean in [1].
That is:

1
ap +bp p−q


p 6= q

 aq +bq
p
p ln b
(1.2)
G(a, b; p, q) =
exp( a lnapa+b
) p = q 6= 0 .
+bp



 √
ab
p=q=0
Recently, the sufficient and necessary conditions comparing the two-parameter mean with
the Gini mean were put forward by using the so-called concept of “strong inequalities” ([3]).
1
p p p−q
,b )
From the above two-parameter type means, we find that their forms are both ff (a
,
(aq ,bq )
where f (x, y) is a homogeneous function of x and y.
The main aim of this paper is to establish the concept of “two-parameter homogeneous func1
p p p−q
f (a ,b )
tions”, and study the monotonicity of functions in the form of f (aq ,bq )
. As applications
of the main results, we will deduce three inequality chains which contain the arithmetic, geometric, exponential, logarithmic, power-exponential and exponential-geometric means, prove
an upper bound for the Stolarsky mean in [12], and present two estimated expressions for the
exponential mean.
2. BASIC C ONCEPTS AND M AIN R ESULTS
Definition 2.1. Assume that f : U(j R+ × R+ ) → R+ is a homogeneous function of variable
x and y, and is continuous and exists first order partial derivative, (a, b) ∈ R+ × R+ with a 6= b,
(p, q) ∈ R × R. If (1, 1) ∈
/ U, then we define
1
f (ap , bp ) p−q
(2.1)
Hf (a, b; p, q) =
(p 6= q, pq 6= 0),
f (aq , bq )
(2.2)
Hf (a, b; p, p) = limHf (a, b; p, q) = Gf,p (a, b)
(p = q 6= 0),
q→p
where
(2.3)
xfx (x, y) ln x + yfy (x, y) ln y
,
Gf,p (a, b) = Gf (a , b ), Gf (x, y) = exp
f (x, y)
1
p
p
p
fx (x, y) and fy (x, y) denote partial derivative to 1st and 2nd variable of f (x, y) respectively.
If (1, 1) ∈ U, then define further
1
f (ap , bp ) p
(2.4)
Hf (a, b; p, 0) =
(p 6= 0, q = 0),
f (1, 1)
1
f (aq , bq ) q
(2.5)
(p = 0, q 6= 0),
Hf (a, b; 0, q) =
f (1, 1)
(2.6)
fx (1,1)
fy (1,1)
Hf (a, b; 0, 0) = limHf (a, b; p, 0) = a f (1,1) b f (1,1)
p→0
(p = q = 0).
From Lemma 3.1, Hf (a, b; p, q) is still a homogeneous function of positive numbers a and b.
We call it a homogeneous function for positive numbers a and b with two parameters p and q,
and call it a two-parameter homogeneous function for short. To avoid confusion, we also denote
it by Hf (p, q) or Hf (a, b) or Hf .
J. Inequal. Pure and Appl. Math., 6(4) Art. 101, 2005
http://jipam.vu.edu.au/
H OMOGENEOUS F UNCTIONS WITH T WO PARAMETERS
3
If f (x, y) is a positive 1-order homogeneous mean function defined on R+ × R+ , then call
Hf (a, b; p, q) the two-parameter f -mean of positive numbers a and b.
Remark 2.1. If f (x, y) is a positive 1-order homogeneous function defined on R+ × R+ , and
is continuous and exists 1st order partial derivative, and satisfies f (x, y) = f (y, x), then
√
Gf,0 (a, b) = Hf (a, b; 0, 0) = ab.
In fact, by (2.3), we have
fx (1, 1) ln a + fy (1, 1) ln b
Gf,0 (a, b) = exp
= Hf (a, b; 0, 0).
f (1, 1)
Since f (x, y) is a positive 1-order homogeneous function, from (3.1) of Lemma 3.2, we obtain
1 · fx (1, 1) 1 · fy (1, 1)
+
= 1.
f (1, 1)
f (1, 1)
(2.7)
If f (x, y) = f (y, x), then fx (x, y) = fy (y, x), so we have
fx (1, 1) = fy (1, 1).
(2.8)
By (2.7) and (2.8), we get
fx (1, 1)
fy (1, 1)
1
=
= ,
f (1, 1)
f (1, 1)
2
√
thereby Gf,0 = ab.
√ Thus it can be seen that despite the form of f (x, y) we always have Hf (a, b; 0, 0) = Gf,0 (a, b) =
ab, so long as f (x, y) is a positive 1-order homogeneous symmetric function defined on
R+ × R+ .
Example 2.1. In Definition 2.1, let f (x, y) = L(x, y)
(1.1), i.e.
 1

q(ap −bp ) p−q


q
q
p(a −b )




1

p p

p

 L (a , b )
1
(2.9)
HL (a, b; p, q) =
q (aq , bq )
L






GL,p (a, b)





G(a, b)
=
x−y
ln x−ln y
(x, y > 0, x 6= y), we get
p 6= q, pq 6= 0
p 6= 0, q = 0
p = 0, q 6= 0
,
p = q 6= 0
p=q=0
where
1
GL,p (a, b) = Ep (a, b) = E p (ap , bp ) = Ep ,
1
a a−b
√
a
E(a, b) = e−1
,
G(a,
b)
=
ab.
bb
Remark 2.2. That
−1
E(a, b) = e
aa
bb
1
a−b
(a, b > 0 with a 6= b)
is called the exponential mean of unequal positive numbers a and b, and is also called the
identical mean and denoted by I(a, b). To avoid confusion, we adopt our terms and notations in
what follows.
J. Inequal. Pure and Appl. Math., 6(4) Art. 101, 2005
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4
Z HEN -H ANG YANG
Example 2.2. In Definition 2.1, let f (x, y) = A(x, y) = x+y
(x, y > 0, x 6= y), we get (1.2),
2
i.e.
 p p 1
a +b
p−q

p 6= q

 aq +bq
(2.10)
HA (a, b; p, q) =
GA,p (a, b) p = q 6= 0 ,



G(a, b)
p=q=0
1
a
b
where GA,p (a, b) = Zp (a, b) = Z p (ap , bp ) = Zp . Z(a, b) = a a+b b a+b is called the powerexponential mean between positive numbers a and b.
1
−1 xx x−y
Example 2.3. In Definition 2.1, let f (x, y) = E(x, y) = e
(x, y > 0, x 6= y), then
yy
 1

E(ap ,bp ) p−q

p 6= q

 E(aq ,bq )
(2.11)
HE (a, b; p, q) =
GE,p (a, b)




G(a, b)
p = q 6= 0 ,
p=q=0
G2
1
where GE,p (a, b) = Yp (a, b) = Y p (ap , bp ) = Yp . Y (a, b) = Ee1− L2 is called the exponentialgeometric mean between positive numbers a and b, where E = E(a, b), L = L(a, b), G =
G(a, b).
Example 2.4. In Definition 2.1, let f (x, y) = D(x, y) = |x − y| (x, y > 0, x 6= y), then
 1
 ap −bp p−q
p 6= q, pq 6= 0
aq −bq
(2.12)
HD (a, b; p, q) =
,
 G (a, b) p = q 6= 0
D,p
1
1
1
where GD,p (a, b) = GD,p = e p E p (ap , bp ) = e p Ep .
In order to avoid confusion, we rename HL (a, b; p, q) (or E(a, b; p, q)) and HA (a, b; p, q) (or
G(a, b; p, q)) as the two-parameter logarithmic mean and two-parameter arithmetic mean respectively. In the same way, we call HE (a, b; p, q) in Example 2.3 the two-parameter exponential mean.
In Example 2.4, since D(x, y) = |x−y| is not a certain mean between positive numbers x and
y, but one absolute value function of difference of two positive numbers, we call HD (a, b; p, q)
a two-parameter homogeneous function of difference.
It is obvious that the conception of two-parameter homogeneous functions has greatly developed the extension of the concept of two-parameter means.
For monotonicity of two-parameter homogeneous functions Hf (a, b; p, q), we have the following main results.
Theorem 2.3. Let f (x, y) be a positive n−order homogenous function defined on U(j R+ ×
R+ ), and be second order differentiable. If I1 = (ln f )xy > (<)0, then Hf (p, q) is strictly
increasing (decreasing) in both p and q on (−∞, 0) ∪ (0, +∞).
Corollary 2.4.
(1) HL (p, q), HA (p, q), HE (p, q) are strictly increasing both p and q on (−∞, +∞),
(2) HD (p, q) is strictly decreasing both p and q on (−∞, 0) ∪ (0, +∞).
Theorem 2.5. Let f (x, y) be a positive 1-order homogeneous function defined on U(j R+ ×
R+ ), and be second order differentiable.
J. Inequal. Pure and Appl. Math., 6(4) Art. 101, 2005
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H OMOGENEOUS F UNCTIONS WITH T WO PARAMETERS
5
(1) If I2a = [(ln f )x ln(y/x)]y > (<)0, then Hf (a, b) is strictly increasing (decreasing) in
a.
(2) If I2b = [(ln f )y ln(x/y)]x > (<)0, then Hf (a, b) is strictly increasing (decreasing) in
b.
Corollary 2.6. HL (a, b), HD (a, b) is strictly increasing in both a and b.
3. L EMMAS AND P ROOFS OF THE M AIN R ESULTS
For proving the main results in this article, we need some properties of homogeneous functions in [16]. For convenience, we quote them as follows.
Lemma 3.1. Let f (x, y), g(x, y) be n, m-order homogenous functions over Ω respectively, then
f · g, f /g (g 6= 0) are n + m, n − m-order homogenous functions over Ω respectively.
If for a certain p with (xp , y p ) ∈ Ω, and f p (x, y) exists, then f (xp , y p ), f p (x, y) are both
np-order homogeneous functions over Ω.
Lemma 3.2. Let f (x, y) be a n-order homogeneous function over Ω, and fx , fy both exist, then
fx , fy are both (n − 1)-order homogeneous functions over Ω, furthermore we have
xfx + yfy = nf.
(3.1)
In particular, when n = 1 and f (x, y) is second order differentiable over Ω, then
xfx + yfy = f,
xfxx + yfxy = 0,
xfxy + yfyy = 0.
(3.2)
(3.3)
(3.4)
Lemma 3.3. Let f (x, y) be a positive n−order homogenous function defined on U(j R+ ×R+ ),
and be second order differentiable. Set
T (t) = ln f (at , bt ), where x = at , y = bt , a, b > 0,
then
T 00 (t) = −xyI1 (ln b − ln a)2 , where I1 =
∂ 2 ln f (x, y)
= (ln f )xy .
∂x∂y
Proof. Since f (x, y) is a positive n-order homogeneous function, from (3.1), we can obtain
x(ln f )x + y(ln f )y = n or x(ln f )x = n − y(ln f )y , y(ln f )y = n − x(ln f )x , so
at fx (at , bt ) ln a + bt fy (at , bt ) ln b
T (t) =
f (at , bt )
xfx (x, y) ln a + yfy (x, y) ln b
=
f (x, y)
= x(ln f )x ln a + y(ln f )y ln b.
0
(3.5)
(3.6)
(3.7)
Hence
∂T 0 (t) dx ∂T 0 (t) dy
+
∂x dt
∂y dt
= [y(ln f )y (ln b − ln a) + n ln a]x at ln a
T 00 (t) =
+ [x(ln f )x (ln a − ln b) + n ln b]y bt ln b
= y(ln f )yx (ln b − ln a)x ln a + x(ln f )xy (ln a − ln b)y ln b
= −xy(ln f )xy (ln b − ln a)2
= −xyI1 (ln b − ln a)2 .
J. Inequal. Pure and Appl. Math., 6(4) Art. 101, 2005
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6
Z HEN -H ANG YANG
Lemma 3.4. Let f (x, y) be a positive 1-order homogeneous function defined on U(j R+ ×R+ ),
and be second order differentiable. Set
S(t) =
txfx (x, y)
, where x = at , y = bt a, b > 0,
f (x, y)
then
S 0 (t) = xyI2a , where I2a = [(ln f )x ln(y/x)]y .
Proof.
xfx (x, y)
d xfx (x, y)
S (t) =
+t
f (x, y)
dt f (x, y)
∂(x(ln f )x ) dx ∂(x(ln f )x ) dy
= x(ln f )x + t
+
∂x
dt
∂y
dt
∂(x(ln f )x ) t
∂(x(ln f )x ) t
= x(ln f )x + t
a ln a +
b ln b
∂x
∂y
= x(ln f )x + t [x(x(ln f )x )x ln a + y(x(ln f )x )y ln b] .
0
(x,y)
By Lemma 3.1, that x(ln f )x = xff x(x,y)
is a 0-order homogeneous function, from (3.1) of
Lemma 3.2, we obtain x [x(ln f )x ]x + y [x(ln f )x ]y = 0 or x [x(ln f )x ]x = −y [x(ln f )x ]y ,
hence
S 0 (t) = x(ln f )x + ty [x(ln f )x ]y (ln b − ln a)
= x(ln f )x + txy(ln f )xy (ln b − ln a)
= x(ln f )x + xy(ln f )xy (ln bt − ln at )
= x(ln f )x + xy(ln f )xy (ln y − ln x)
= xy y −1 (ln f )x + (ln f )xy ln(y/x)
= xy [(ln f )x ln(y/x)]y = xyI2a .
Based on the above lemmas, then next we will go on proving the main results in this paper.
Proof of Theorem 2.3. Since Hf (p, q) is symmetric with respect to p and q, we only need to
prove the monotonicity for p of ln Hf .
1) When p 6= q,
1
f (ap , bp )
T (p) − T (q)
ln
=
,
q
q
p − q f (a , b )
p−q
∂ ln Hf
(p − q)T 0 (p) − T (p) + T (q)
=
.
∂p
(p − q)2
ln Hf =
Set g(p) = (p − q)T 0 (p) − T (p) + T (q), then g(q) = 0, g 0 (p) = (p − q)T 00 (p), and then exist
ξ = q + θ(p − q) with θ ∈ (0, 1) by Mean-value Theorem, such that
g(p) − g(q)
∂ ln Hf
g 0 (ξ)
(ξ − q)T 00 (ξ)
=
=
=
= (1 − θ)T 00 (ξ).
∂p
(p − q)2
p−q
p−q
By Lemma 3.3, T 00 (ξ) = −xyI1 (ln b − ln a)2 , x = aξ , y = bξ . Obviously, when I1 < (>)0, we
∂ ln H
get ∂p f > (<)0.
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H OMOGENEOUS F UNCTIONS WITH T WO PARAMETERS
7
2) When p = q, from (2.2) and (3.6),
1
ln Hf = ln Gfp (ap , bp ) =
xfx (x, y) ln x + yfy (x, y) ln y
= T 0 (p),
f (x, y)
∂ ln Hf
= T 00 (p) = −xyI1 (ln b − ln a)2 .
∂p
∂ ln H
when I1 < (>)0, we get ∂p f > (<)0.
Combining 1) with 2), the proof is completed.
Proof of Corollary 2.4. It follows from Theorem 2.3 that the monotonicity of Hf (p, q) depends
on the sign of I1 = (ln f )xy .
1) For f (x, y) = L(x, y),
1
1
−
2
(x − y)
xy(ln x − ln y)2
1
√ 2
2
=
(
xy)
−
L
(x,
y)
.
xy(x − y)2
√
By the well-known inequality L(x, y) > xy ([13]), we have I1 < 0.
2) For f (x, y) = A(x, y),
1
I1 = (ln f )xy = −
< 0.
(x + y)2
I1 = (ln f )xy =
3) For f (x, y) = E(x, y),
1
[2(x − y) − (x + y)(ln x − ln y)]
(x − y)3
2(ln x − ln y)
x+y
=
L(x, y) −
.
(x − y)3
2
I1 = (ln f )xy =
By the well-known inequality L(x, y) <
4) For f (x, y) = D(x, y),
x+y
2
([13]), we have I1 < 0.
I1 = (ln f )xy =
1
> 0.
(x − y)2
Applying mechanically Theorem 2.3, we immediately obtain Corollary 2.4.
Proof of Theorem 2.5.
1) Since
p−1
∂ ln Hf
1
pa fx (ap , bp ) qaq−1 fx (aq , bq )
S(p) − S(q)
=
−
=
,
p
p
q
q
∂a
p−q
f (a , b )
f (a , b )
a(p − q)
by the Mean-value Theorem, there exists ξ = q + θ(p − q) with θ ∈ (0, 1), such that
∂ ln Hf
S(p) − S(q)
=
= a−1 S 0 (ξ).
∂a
a(p − q)
From Lemma 3.4, S 0 (ξ) = xyI2a , where x = aξ , y = bξ . Obviously ,if I2a > 0, then
∂ ln Hf
∂ ln H
> 0, so Hf (a, b) is strictly increasing in a; If I2a < 0, then ∂a f < 0, so Hf (a, b) is
∂a
strictly decreasing in a.
2) It can be proved in the same way.
J. Inequal. Pure and Appl. Math., 6(4) Art. 101, 2005
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8
Z HEN -H ANG YANG
Proof of Corollary 2.6.
1) For f (x, y) = L(x, y),
I2a = [(ln f )x ln(y/x)]y =
x/y − 1 − ln(x/y)
.
(x − y)2
By the well-known inequality ln x < x − 1 (x > 0, x 6= 1), we have I2a > 0.
2) For f (x, y) = D(x, y),
I2a = [(ln f )x ln(y/x)]y =
x/y − 1 − ln(x/y)
> 0.
(x − y)2
Since HL (a, b), HD (a, b) are both symmetric with respect to a and b, applying mechanically
Theorem 2.5, we immediately obtain Corollary 2.6.
4. S OME A PPLICATIONS
As direct applications of theorems and lemmas in this paper, we will present several examples
as follows.
Example 4.1 (a G-A inequality chain). By 1) of Corollary 2.4, for f (x, y) = A(x, y), L(x, y) and
E(x, y), Hf (p, q) are strictly increasing in both p and q. So there are
1
Hf (a, b; 0, 0) < Hf (a, b; 1, 0) < Hf a, b; 1,
2
< Hf (a, b; 1, 1) < Hf (a, b; 1, 2).
(4.1)
From it we can obtain the following inequalities respectively, that are
√
√
ab < L(a, b) <
(4.2)
√
(4.4)
E(a2 ,b2 )
E(a,b)
< E(a, b) <
a+b
;
2
a+b
ab <
<
2
= Z(a, b), then (4.4) can be written into that
√
(4.5)
√ !2
b
2
a+b
a2 + b 2
√
< Z(a, b) <
;
√
a+b
a+ b
2

√
E(a, b)
E(a2 , b2 )
ab < E(a, b) <  √ √  < Y (a, b) <
.
E(a,
b)
E
a, b
(4.3)
Notice
a+
2
√ √ G2
ab < E(a, b) < Z
a, b < E exp 1 − 2 < Z(a, b).
L
2
The inequality (4.2) was proved by [13], which shows that can be inserted L, A+G
and E
2
between G and A, so we call (4.2) the G-A inequality chain. (4.3) and (4.4) are the same in
form completely, so we call (4.1) the G-A inequality chain for homogeneous functions.
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H OMOGENEOUS F UNCTIONS WITH T WO PARAMETERS
Remark 4.1. That
E(a2 ,b2 )
E(a,b)
9
= Z(a, b) is a new identical equation for mean. In fact,
−1
E(a, b)Z(a, b) = e
bb
aa
1
b−a
b
b
b
a
b b+a a b+a
a
a
= e−1 b b+a + b−a a b+a − b−a
2b2
−2a2
= e−1 b b2 −a2 a b2 −a2
! 21 2
b −a
2 b2
(b
)
= e−1
= E(a2 , b2 ).
(a2 )a2
It shows that Z(a, b) is not only one “geometric mean”, but also one ratio of one exponential mean to another. Thus inequalities involving Z(a, b) may be translated into inequalities
involving exponential mean.
Example 4.2 (An estimation for upper bound of Stolarsky mean). From 2) of Corollary 2.4,
we can prove expediently an estimation for the upper bound of the Stolarsky mean presented by
[12]:
1
p
p−1
p
1
b
−
a
Sp (a, b) < p 1−p (a + b) with p > 2, where Sp (a, b) =
.
p(b − a)
In fact, from 2) of Corollary 2.4, when p, q ∈ (−∞, 0) ∪ (0, +∞), HD (p, q) is strictly
decreasing in both p and q, so when p > 2, we have HD (a, b; 1, p) < HD (a, b; 1, 2).
Notice
p
1
1
a − bp p−1
(4.6)
HD (a, b; p, 1) =
= p p−1 Sp (a, b) (p > 0),
a−b
1
1
1
thus when p > 2, we obtain p p−1 Sp (a, b) < 2 2−1 S2 (a, b) = a + b, i.e. Sp (a, b) < p 1−p (a + b).
Example 4.3 (Reversed inequalities and estimations for exponential mean). By 1) of Corollary 2.4, HL (p, q) is strictly increasing in both p and q, so when p1 ∈ (0, 1), p2 ∈ (1, +∞), we
have
HL (a, b; p1 , 1) < HL (a, b; 1, 1) < HL (a, b; p2 , 1),
i.e.
Sp1 (a, b) < E(a, b) < Sp2 (a, b).
(4.7)
On the other hand, By 2) of Corollary 2.4, when p, q ∈ (−∞, 0) ∪ (0, +∞), HD (p, q) is
strictly monotone decreasing in both p and q. So when p1 ∈ (0, 1), p2 ∈ (1, +∞), we have
(4.8)
HD (a, b; p1 , 1) > HD (a, b; 1, 1) > HD (a, b; p2 , 1).
From (4.6), (4.8) can be written into
1
1
p2p2 −1 Sp2 (a, b) < eE(a, b) < p1p1 −1 Sp1 (a, b)
or
(4.9)
1 p21−1
1 1
p2 Sp2 (a, b) < E(a, b) < p1p1 −1 Sp1 (a, b).
e
e
J. Inequal. Pure and Appl. Math., 6(4) Art. 101, 2005
http://jipam.vu.edu.au/
10
Z HEN -H ANG YANG
Combining (4.7) with (4.9), we have
(4.10)
1 1
Sp1 (a, b) < E(a, b) < p1p1 −1 Sp1 (a, b), where p1 ∈ (0, 1),
e
1 p21−1
Sp2 (a, b) < E(a, b) < Sp2 (a, b), where p2 ∈ (1, +∞).
p
e 2
1
In particular, when p1 = , p2 = 2, by (4.10), (4.11), we get
2
√ !2
√ !2
√
√
4
a+ b
a+ b
(4.12)
< E(a, b) <
,
2
2
e
2 a+b
a+b
(4.13)
< E(a, b) <
.
e
2
2
The inequalities (4.12) and (4.13) may be denoted simply by
4A+G
A+G
(4.14)
<E<
,
2
e 2
2
(4.15)
A < E < A.
e
The inequalities (4.14) and (4.15) make certain a bound of error that exponential mean E are
estimated by A or A+G
.
2
(4.11)
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