A Generalization of the logarithmic and the Gauss means

Vol. 17, No.1, April 2009, pp 117-124 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

117

A Generalization of the logarithmic and the Gauss means

Mih´aly Bencze ⁸

ABSTRACT. In this paper we present a generalization of the classical logarithm and Gauss means, and we give some interesting applications.

MAIN RESULTS

1. THE LOGARITHMIC MEAN

The logarithmic mean of two positive numbers aand bis the numberL(a, b) defined as

L(a, b) =

_a−b

lna−lnb if a6=b

a if a=b

see [1]. G. Hardy, J.E. Littlewood and G. P´olya have discovered many applications of logarithmic mean. Their book Inequalities has had quite a few succesors, and yet new properties of these means continue to be discovered.

Definition 1.1. Ifa_k>0 (k= 1,2, ..., n) anda_i6=a_j (i, j∈ {1,2, ..., n}, i6=j) n≥2, then

L(a₁, a₂, ..., a_n) = 1

−(n−1)Pⁿ

k=1

u_klna_k

_n−1¹ (1.1)

denote the logarithmic mean of positive numbersa₁, a₂, ..., a_n,where

u_k= 1

(a₁−a_k)...(a_k−1−a_k) (a_k+1−a_k)...(a_n−a_k) for all k∈ {1,2, ..., n}.

Ifn= 2, thenu₁= _a⁻¹

2−a1, u₂= _a ¹

2−a1 and

8Received: 25.03.2006

2000Mathematics Subject Classification. 26D15.

Key words and phrases. Logarithmic mean, Gauss mean etc.

L(a1, a2) = _u1lna1+u¹ _2lna2 = _ln^a_a²₂⁻₋^a_lna¹ ₁, therefore we reobtain the classical logarithmic mean.

Remark 1.2. By a simple calculation we get:

Xn k=1

u_k = 0 (1.2)

Theorem 1.3. We have the following relation:

1

L(a1, a2, ..., an) =





(n−1) Z∞ 0

dx Qn k=1

(x+a_k)







1 n−1

(1.3)

Proof. We have the following descomposition:

1 Qn k=1

(x+a_k)

=− Xn k=1

u_k x+ak

,

therefore Z∞ 0

dx Qn k=1

(x+a_k)

=− Xn k=1

u_k Z∞

0

dx

x+a_k =− Xn k=1

u_kln (x+a_k)|^∞0 =

=−ln Yn k=1

(x+a_k)^uk|^∞0 = Xn k=1

u_klna_k Because Pⁿ

k=1

u_k= 0,then we obtain

1

L(a₁, a₂, ..., a_n) =





(n−1) Z∞ 0

dx Qn k=1

(x+a_k)







1 n−1

= −(n−1) Xn k=1

u_klna_k

!_n¹₋₁

which finish the proof.

In following we denote A(a₁, a₂, ..., a_n) = _n¹ Pⁿ

k=1

a_k the arithmetic and G(a₁, a₂, ..., a_n) = ⁿ

sQn k=1

a_k the geometric mean.

Remark 1.3. The mean L can also be expressed in terms of a divided difference, of order n-1, of the logarithmic function

Theorem 1.4. We have the following inequalities:

G(a1, a2, ..., an)≤L(a1, a2, ..., an)≤A(a1, a2, ..., an) (1.4) Proof. Using theAM−GM inequality we obtain

Yn k=1

(x+a_k)≤(x+A)ⁿ From Huygen’s inequality we get

Yn k=1

(x+a_k)≥(x+G)ⁿ, therefore

1

(n−1)Aⁿ⁻¹ = Z∞

0

dx (x+A)ⁿ ≤

Z∞ 0

dx Qn k=1

(x+a_k)

= 1

(n−1)Lⁿ⁻¹ ≤

≤ Z∞ 0

dx

(x+G)ⁿ = 1 (n−1)Gⁿ⁻¹ which finish the proof.

Corollary 1.4.1. Ifx_k∈R (k= 1,2, ..., n) are different, then we have the following inequalities:

e^{A(x1,x2,...,xn)}≤ 1

(n−1) _n

P

k=1

−xk

(e^x1−e^x2)...(e^xk−1−e^xk)(e^xk+1−e^xk)...(e^xn−e^x1)

_n¹

−1 ≤

≤A(e^x1, e^x2, ..., e^xn) (1.5) Proof. In Theorem 1.4 we take

a_k=e^xk (k= 1,2, ..., n) Corollary 1.4.2. Ift≥0, then

tanht≤t≤sinht (1.6) Proof. In (1.5) we taken= 2, and after elementary calculus we get

tanhx₁−x₂

2 ≤ x₁−x₂

2 ≤sinhx₁−x₂ 2 , and in these we denote t= ^x1−₂^x2 ≥0.

These inequalities are fundamental inequalities in analysis, therefore Corollary 1.4.1 offer a lot of generalizations of these.

Corollary 1.4.3. Ifa_k>0 (k= 1,2, ..., n) are different, then we have the following inequalities:

G(a₁, a₂, ..., a_n)≤ 1

−(n−1) Pⁿ

k=1

ln_ak^a1...ln^ak_ak⁻¹ln^ak+1_ak ...ln^an_aklnak

L(a1,ak)...L(ak−1,ak)L(ak+1,ak)...L(an,ak)

_n−1¹ ≤

≤A(a₁, a₂, ..., a_n) (1.7)

Proof. In Theorem 1.4 we consider the substitutions a_i−a_j = L(a_i, a_j)

ln^a_aⁱ

j

(i, j ∈ {1,2, ..., n}, i6=j) (1.8) Using the inequalitiesG(a_i, a_j)≤L(a_i, a_j)≤A(a_i, a_j) we obtain from corollary 1.4.3 a cathegory of new inequalities.

Remark 1.5. If in Corollary 1.4.3 we replace a_k by a^t_k (k= 1,2, ..., n), respectively, then we obtain a lot of new inequalities.

By example, if we consider n= 2, then we obtain

Gt(a1, a2) =G a^t₁, a^t₂

≤ a^t₂−a^t₁

t(lna₂−lna₁) ≤A a^t₁, a^t₂

=At(a1, a2) (1.9) or

G_t(a₁, a₂) =tG_t(a₁, a₂)a₂−a₁

a^t₂−a^t₁ ≤L(a₁, a₂)≤tA_t(a₁, a₂)a₂−a₁ a^t₂−a^t₁ =

=A_t(a₁, a₂) (1.10) with properties: G_−t(a₁, a₂) =G_t(a₁, a₂), A_−t(a₁, a₂) =A_t(a₁, a₂),

G₀(a₁, a₂) =A₀(a₁, a₂) =L(a₁, a₂), G₁(a₁, a₂) =G(a₁, a₂), A1(a1, a2) =A(a1, a2).

For fixed a₁ and a₂, G_t(a₁, a₂)is a decreasing function of |t|,and A_t(a₁, a₂) is an increasing function of |t|,therefore G_t(a₁, a₂)≤L(a₁, a₂)≤A_t(a₁, a₂) is a very fundamental inequality (see [3]).

In same way the idea of Corollary 1.4.3 and of remark 1.5 can be continued with hard calculus, and we introduced in same way the new means

G_t(a₁, a₂, ..., a_n) and A_t(a₁, a₂, ..., a_n) for which we obtain the fundamental inequalities G_t(a₁, a₂, ..., a_n)≤L(a₁, a₂, ..., a_n)≤A_t(a₁, a₂, ..., a_n) which offer for all t∈R a lot of new refinements for the inequalities proved in Theorem 1.4.

Remark 1.6. If in Remark 1.5 we choose t= 2^−m (m∈N), then we obtain G₂−m(a1, a2)≤L(a1, a2)≤A₂−m(a1, a2) (1.11) After elementary calculus we get

G₂−(m+1)(a₁, a₂) Ym k=1

A

a²₁^−k, a²₂^−k

≤L(a₁, a₂)≤

≤A₂−(m+1)(a₁, a₂) Ym k=1

A

a²₁^−k, a²₂^−k

If we letm→ ∞ in two formulas above, then L(a1, a2) =

Y∞ k=1

A

a²₁^−k, a²₂^−k

(1.12) (See [3])

Using these inductively for

G₂^m(a₁, a₂, ..., a_n)≤L(a₁, a₂, ..., a_n)≤A₂^m(a₁, a₂, ..., a_n) we obtained a product formula for L(a₁, a₂, ..., a_n).

2. THE GAUSS MEAN

Given positive numbersaand b, inductively define two sequences as a0=a, b0=b, an+1=A(an, bn), bn+1=G(an, bn).

Then (a_n)_n≥0 is decreasing, and (b_n)_n≥0 is increasing. Alla_n and b_n are betweenaand b. So both sequence converge. By induction we obtain a_n+1−bn+1 ≤ ¹₂(a_n−b_n),and hence the sequences converge to a common limit, denoted byAG(a, b) which is called the Gauss arithmetic-geometric mean.

Gauss showed that 1

AG(a, b) = 2 π

Z∞ 0

p dx

(x²+a²) (x²+b²) (2.1) and G(a, b)≤AG(a, b)≤A(a, b).

Definition 2.1. Ifa_k>0 (k= 1,2, ..., n),then the Gauss mean of a₁, a₂, ..., a_n is defined in following way

1

AG(a1, a2, ..., an) = 2 π

Z∞ 0







1 Qn k=1

x²+a²_k







1 n

dx. (2.2)

IfQ(a₁, a₂, ..., a_n) = s

1 n

Pn k=1

a²_k,then we obtain the following:

Theorem 2.2. We have the following inequalities:

G(a1, a2, ..., an)≤AG(a1, a2, ..., an)≤Q(a1, a2, ..., an) (2.3) Proof. Using the AM-GM inequality we have

Yn k=1

x²+a²_kn¹

≤x²+Q² From Huygen’s inequality we obtain

Yn k=1

x²+a²_kn¹

≥x²+G²

therefore 1

Q(a₁, a₂, ..., a_n) = 2 π

Z∞ 0

dx

x²+Q² ≤ 2 π

Z∞ 0

dx

x²+a²_k¹_n =AG(a₁, a₂, ..., a_n)≤

≤ 2 π

Z∞ 0

dx

x²+G² = 1

G(a1, a2, ..., an) Definition 2.3. Ifa_k>0 (k= 1,2, ..., n),then

1

B_α(a₁, a₂, ..., a_n) =cα





(n−1) Z∞

0

dx Qn k=1

x^α+a^α_kα¹







1 n−1

(2.4)

when c_α is a constant, depend only ofα∈R.This mean generalize the logarithmic and the Gauss means too.

Theorem 2.4. Ifa_k>0 (k= 1,2, ..., n) andα∈(−∞,0]∪[1,+∞),then L(a₁, a₂, ..., a_n)≤c_α2^{n(α(n−1)α−1)}B_α(a₁, a₂, ..., a_n) and if α∈(0,1) then holds the reverse inequality.

Proof. We have the inequalities

(x^α+a^α_k)^α1 ≥ x+a_k 2^1−1α (k= 1,2, ..., n),therefore

1

B_α(a₁, a₂, ..., a_n) =c_α





(n−1) Z∞ 0

dx Qn k=1

x^α+a^α_kα¹







1 n−1

≤

≤cα2

n(α−1) (n−1)α





(n−1) Z∞ 0

dx Qn k=1

(x+a_k)







1 n−1

Ifα= 1 and c₁ = 1,then B₁(a₁, a₂, ..., a_n) =L(a₁, a₂, ..., a_n) and ifα= 2, c₂ = ²_π, n= 2, thenB₂(a₁, a₂) =AG(a₁, a₂).

REFERENCES

[1] Hardy, G., Littlewood, J.E. and P´olya, G.,Inequalities, Cambridge University Press, Second edition, 1952.

[2] Bullen,P.S., Mitrinovic, D.S. and Vasic, P.M., Means and their inequalities,D. Reidel, 1998.

[3] Bhatia, R.,The logarithmic mean, Resonance, Vol. 13, june 2008, pp.

583-594.

Str. H˘armanului 6 505600 S˘acele-N´egyfalu, Jud. Bra¸sov, Romania

E-mail: benczemihaly@yahoo.com