## Univ-bejaia.dz

**132 **(2013), p. 139 -149.

**ALGEBRAIC AND TOPOLOGICAL STRUCTURES ON**
**THE SET OF MEAN FUNCTIONS AND**
**GENERALIZATION OF THE AGM MEAN**
Abstract. In this paper, we present new structures and results on theset

*MD *of mean functions on a given symmetric domain

*D *in R2. First,we construct on

*MD *a structure of abelian group in which the neutralelement is the

*arithmetic *mean; then we study some symmetries in thatgroup. Next, we construct on

*MD *a structure of metric space underwhich

*MD *is the closed ball with center the

*arithmetic *mean and radius1

*/*2. We show in particular that the

*geometric *and

*harmonic *means lieon the boundary of

*MD*. Finally, we give two theorems generalizing theconstruction of the AGM mean. Roughly speaking, those theorems showthat for any two given means

*M*1 and

*M*2, which satisfy some regular-ity conditions, there exists a unique mean

*M *satisfying the functionalequation

*M *(

*M*1

*, M*2) =

*M *.

Let

*D *be a nonempty symmetric domain in R2. A

*mean *function (or
simply a

*mean*) on

*D *is a function

*M *:

*D → *R satisfying the followingthree axioms:
(i)

*M *is symmetric, that is,

*M *(

*x, y*) =

*M *(

*y, x*) for all (

*x, y*)

*∈ D*.

(ii) For all (

*x, y*)

*∈ D*, we have min(

*x, y*)

*≤ M *(

*x, y*)

*≤ *max(

*x, y*).

(iii) For all (

*x, y*)

*∈ D*, we have

*M *(

*x, y*) =

*x *=

*⇒ x *=

*y*.

Note that because of (ii), the implication in (iii) is actually an equivalence.

Among the most known examples of mean functions, we cite:

*• *The arithmetic mean A deﬁned on R2 by: A(

*x, y*) =

*x*+

*y*.

The geometric mean G deﬁned on (0

*, *+

*∞*)2 by: G(

*x, y*) =

*• *The harmonic mean H deﬁned on (0

*, *+

*∞*)2 by: H(

*x, y*) = 2

*xy *.

*• *The Gauss arithmetic-geometric mean AGM deﬁned on (0

*, *+

*∞*)2 by thefollowing process:
Given positive real numbers

*x, y*, AGM(

*x, y*) is the common limit of the two
2010

*Mathematics Subject Classification. *Primary 20K99, 54E35; Secondary 39B22.

*Key words and phrases. *Means, abelian groups, metric spaces, symmetries.

sequences (

*xn*)

*n∈*N and (

*yn*)

*n∈*N deﬁned by

*x*0 =

*x , y*0 =

*y*

*xn*+1 =

*xn*+

*yn*
For a survey on mean functions, we refer to Chapter 8 of the book [Bor] in
which AGM takes the principal place. However, there are some diﬀerences
between that reference and the present paper. Indeed, in [Bor], only axiom
(ii) is taken to deﬁne a mean function; (iii) is added to obtain the so called

*strict *mean while (i) is not considered. In this paper, we shall see that the
three axioms (i), (ii) and (iii) are both necessary and suﬃcient to deﬁne a

*good *mean or a

*good *set of mean functions on a given domain. In particular,
axiom (iii), absent in [Bor], is necessary for the foundation of our algebraic
and topological structures (see Sections 2 and 3).

Given a nonempty symmetric domain

*D *in R2, we denote by

*MD *the
set of mean functions on

*D*. The purpose of this paper is on the one handto establish some algebraic and topological structures on

*MD *and to studysome of their properties and on the other hand to generalize in a natural
way the arithmetic-geometric mean AGM.

In the ﬁrst section, we deﬁne on

*MD *a structure of abelian group in
which the neutral element is the arithmetic mean. The study of this group
reveals that the arithmetic, geometric and harmonic means lie in a particular
class of mean functions that we call

*normal mean functions*. We then study
symmetries on

*MD *and we discover that the symmetry with respect toeach of the three means A, G and H oddly coincides with another type
of symmetry (with respect to the same means) which we call

*functional*
*symmetry*. The problem of describing the set of all means realizing that
In the second section, we deﬁne on

*MD *a structure of metric space which
turns out to be a closed ball with center A and radius 1

*/*2. We then use
the group structure to calculate the distance between two means on

*D*;this permits us in particular to establish a simple characterization of the
In the third section, we introduce the concept of

*functional middle *of
two mean functions on

*D *which generalizes in a natural way the arithmetic-geometric mean, so that the latter is the functional middle of the arithmetic
and geometric means. We establish two suﬃcient conditions for the existence
and uniqueness of the functional middle of two means. The ﬁrst one uses the
metric space structure of

*MD *by imposing on the two means in question
the condition that the distance between them is less than 1. The second
requires the two means in question to be continuous on

*D*. In the proof ofthe latter one, axiom (iii) plays a vital role.

2. An abelian group structure on

*MD*
Given a nonempty symmetric domain

*D *in R2, we denote by

*AD *the set
of asymmetric maps on

*D*, that is, maps

*f *:

*D → *R, satisfying

*f *(

*x, y*) =

*−f *(

*y, x*)
(

*∀*(

*x, y*)

*∈ D*)

*.*
It is clear that (

*AD, *+) (where + is the usual addition of maps from

*D *intoR) is an abelian group with neutral element the null map.

Now, consider

*φ *:

*MD → *R

*D *deﬁned by: (
log

*−M*(

*x,y*)

*−x*
*D , ∀*(

*x, y*)

*∈ D *:

*φ*(

*M *)(

*x, y*) := 0
The axioms (i)-(iii) ensure that

*−M*(

*x,y*)

*−x *(for

*x ̸*=

*y*) is well-deﬁned and

**Theorem 2.1. ***We have φ*(

*MD*) =

*AD. In addition, the map φ *:

*M →*

φ(

*M *)

*is a bijection from MD to AD and its inverse is given by*
*∀f ∈ AD , ∀*(

*x, y*)

*∈ D *:

*φ−*1(

*f*)(

*x, y*) =

*Proof. *Axiom (i) ensures that for all

*M ∈ MD*, we have

*φ*(

*M*)

*∈ AD*.

Next, if

*f *is an asymmetric map on

*D*, we easily verify that

*M *:

*D → *Rdeﬁned by

*M *(

*x, y*) :=

*x*+

*yef*(

*x,y*) (

*∀*(

*x, y*)

*∈ D*) is a mean on

*D *and

*φ*(

*M *) =

*f *.

Since obviously

*φ *is injective, the proof is ﬁnished.

We now transport, by

*φ*, the abelian group structure (

*AD, *+) onto

*MD*,
that is, we deﬁne on

*MD *the following composition law

*∗*:

*∀M*1

*, M*2

*∈ MD *:

*M*1

*∗ M*2 =

*φ−*1 (

*φ*(

*M*1) +

*φ*(

*M*2))

*.*
So (

*MD, ∗*) is an abelian group and

*φ *is a group isomorphism from (

*MD, ∗*)to (

*AD, *+). Furthermore, since the null map on

*D *is the neutral element of(

*AD, *+) and

*φ−*1(0) = A, the arithmetic mean A is the neutral element of(

*MD, ∗*).

By calculating explicitly

*M*1

*∗ M*2 (for

*M*1

*, M*2

*∈ MD*), we obtain:

**Proposition 2.2. ***The composition law ∗ on MD is defined by:*
{

*x*(

*M*1(

*x,y*)

*−y*)(

*M*2(

*x,y*)

*−y*)+

*y*(

*M*1(

*x,y*)

*−x*)(

*M*2(

*x,y*)

*−x*)

*if x ̸*=

*y*
(

*M*1(

*x,y*)

*−x*)(

*M*2(

*x,y*)

*−x*)+(

*M*1(

*x,y*)

*−y*)(

*M*2(

*x,y*)

*−y*)

*for M*1

*, M*2

*∈ MD and *(

*x, y*)

*∈ D.*
Now, it is easy to verify that the images of the geometric and harmonic
means under the isomorphism

*φ *are given by
(

*∀*(

*x, y*)

*∈ *(0

*, *+

*∞*)2)

*,*
*φ*(H)(

*x, y*) = log

*x − *log

*y*
(

*∀*(

*x, y*)

*∈ *(0

*, *+

*∞*)2)

*.*
From (2.2) and (2.3), we see that

*φ*(G) and

*φ*(H) (and trivially also

*φ*(A))
have a particular form: each can be written as

*h*(

*x*)

*− h*(

*y*), where

*h *is a realfunction of one variable.

To generalize, we deﬁne a

*normal mean *as a mean function

*M *:

*I*2

*→ *R
(

*I ⊂ *R) such that

*φ*(

*M *) has the form

*h*(

*x*)

*− h*(

*y*) for some map

*h *:

*I → *R.

Equivalently, a

*normal mean function *is a function

*M *:

*I*2

*→ *R (

*I ⊂ *R)which can be written as

*xP *(

*x*) +

*yP *(

*y*)

*P *(

*x*) +

*P *(

*y*)
where

*P *:

*I → *R is a positive function on

*I*.

**Study of some symmetries on the group **(

*MD, ∗*)

**. **We are now inter-

ested in the symmetric image of a given mean

*M*1 with respect to another

mean

*M*0 via the group structure (

*MD, ∗*). Denote by

*SM *the symmetry

with respect to

*M*0 in the group (

*MD, ∗*), deﬁned by

*∀M*1

*, M*2

*∈ MD *:

*SM *(

*M*
1) =

*M*2

*⇐⇒ M*1

*∗ M*2 =

*M*0

*∗ M*0

*.*
Using the group isomorphism

*φ*, we obtain by a simple calculation the ex-

**Proposition 2.3. ***For any M*0

*, M*1

*∈ MD,*
1

*− x*)(

*M*0

*− y*)2

*− y*(

*M*0

*− x*)2(

*M*1

*− y*)
(

*M*1

*− x*)(

*M*0

*− y*)2

*− *(

*M*0

*− x*)2(

*M*1

*− y*)

*where, for simplicity, we have written M*0

*for M*0(

*x, y*)

*, M*1

*for M*1(

*x, y*)

*andSM *(

*M*
As an application, we get the following immediate corollary:

**Corollary 2.4. ***For any M ∈ MD, we have:*
(1)

*S*A(

*M *) =

*x *+

*y − M *= 2A

*− M .*

(2)

*S*G(

*M *) =

*xy *= G2

*(when D ⊂ *(0

*, *+

*∞*)2

*).*
*(when D ⊂ *(0

*, *+

*∞*)2

*).*
(4)

*S*H =

*S*G

*◦ S*A

*◦ S*G

*.*
Now, we are going to deﬁne another symmetry on

*MD *(for

*D *of a certain
form), independent of the group structure (

*MD, ∗*). This new symmetry is
deﬁned by solving a functional equation but it curiously coincides, in many
cases, with the symmetry deﬁned above.

**Definition 2.5. **Let

*I *be a nonempty interval of R,

*D *=

*I*2 and

*M*0,

*M*1

and

*M*2 be three mean functions on

*D *such that

*M*1 and

*M*2 take their

values in

*I*. We say that

*M*2 is the

*functional symmetric mean *of

*M*1 with

respect to

*M*0 if the following functional equation is satisﬁed:

*M*0(

*M*1(

*x, y*)

*, M*2(

*x, y*)) =

*M*0(

*x, y*)
(

*∀*(

*x, y*)

*∈ D*)

*.*
Equivalently, we also say that

*M*0 is the

*functional middle *of

*M*1 and

*M*2.

According to axiom (iii), it is immediate that if the functional symmetric
mean exists then it is unique. This justiﬁes the following notation:

**Notation 2.6. **Given two mean functions

*M*0 and

*M*1 on

*D *=

*I*2 with values

in

*I *(where

*I *is an interval of R), we denote by

*σM *(

*M*
symmetric mean (if it exists) of

*M*1 with respect to

*M*0.

A simple calculation establishes the following:

**Proposition 2.7. ***Let M be a mean function on a suitable symmetric do-*
*σA*(

*M *) =

*x *+

*y − M,*
(

*for D ⊂ *(0

*, *+

*∞*)2)

*,*
(

*for D ⊂ *(0

*, *+

*∞*)2)

*.*
The remarkable phenomenon of the coincidence of the two symmetries
deﬁned on

*MD *in the particular cases of the means A, G and H leads tothe following question:

**Open question. **For which mean functions

*M *on

*D *= (0

*, *+

*∞*)2 the two

symmetries with respect to

*M *(in the sense of the group law introduced on

*MD *and in the functional sense) coincide?

**Example. **Using the deﬁnition of AGM (see Section 1), it is easy to show

that A and G are symmetric in the functional sense with respect to AGM.

Throughout this section, we ﬁx a nonempty symmetric domain

*D *in R2.

We suppose that

*D *contains at least one point (

*x*0

*, y*0) of R2 such that

*x*0

*̸*=

*y*0 (otherwise

*MD *reduces to a unique element). For all couples (

*M*1

*, M*2)
1(

*x, y*)

*− M*2(

*x, y*)

**Proposition 3.1. ***The map *d

*of M*2

*D into *[0

*, *+

*∞*]

*is a distance on MD.*

In addition, the metric space (

*MD, *d)

*is the closed ball with center *A

*(the*

arithmetic mean) and radius 1

*.*
*Proof. *First let us show that d(

*M*1

*, M*2) is ﬁnite for all

*M*1

*, M*2. For all(

*x, y*)

*∈ D*,

*x ̸*=

*y*, the numbers

*M*1(

*x, y*) and

*M*2(

*x, y*) lie in the interval[min(

*x, y*)

*, *max(

*x, y*)], so

*|M*1(

*x, y*)

*− M*2(

*x, y*)

*| ≤ *max(

*x, y*)

*− *min(

*x, y*) =

*|x − y|.*
1(

*x, y*)

*− M*2(

*x, y*)
that is, d(

*M*1

*, M*2)

*≤ *1. Further, since the three axioms of a distance aretrivially satisﬁed, d is a distance on

*MD*.

Now, given

*M ∈ MD*, let us show that d(

*M, *A)

*≤ *1. For all (

*x, y*)

*∈ D*,

*x ̸*=

*y*, the number

*M *(

*x, y*) lies in the closed interval with endpoints

*x *and

*y*, so

*|M*(

*x, y*)

*− *A(

*x, y*)

*| ≤ *max (

*x − *A(

*x, y*)

*, y − *A(

*x, y*))

*x − x *+

*y , y − x *+

*y*
*M *(

*x, y*)

*− *A(

*x, y*)
that is, d(

*M, *A)

*≤ *1, as required.

**Remark 3.2. **Given

*M*1

*, M*2

*∈ MD*, since the map (

*x, y*)

*→ M*1(

*x,y*)

*−M*2(

*x,y*)

is obviously asymmetric (on the set

*{*(

*x, y*)

*∈ D *:

*x ̸*=

*y}*), we also have
1(

*x, y*)

*− M*2(

*x, y*)
We now establish a practical formula for the distance between two mean

**Proposition 3.3. ***Let M*1

*and M*2

*be two mean functions on D. Set f*1 =

*φ*(

*M*1)

*and f*2 =

*φ*(

*M*2)

*. Then*
(

*x,y*)

*∈D *(

*ef*1 + 1)(

*ef*2 + 1)

*Proof. *Using (2.1), for all (

*x, y*)

*∈ D *we have

*M*1(

*x, y*) =

*φ−*1(

*f*1)(

*x, y*) =

*x*+

*yef*1(

*x,y*) and

*M*
2(

*x, y*) =

*φ−*1(

*f*2)(

*x, y*) =

*x*+

*yef*2(

*x,y*)
As an application, we get the following immediate corollary:

**Corollary 3.4. ***Let M be a mean function on D and f *:=

*φ*(

*M *)

*. Then,*

setting s := sup

*D f ∈ *[0

*, *+

*∞*]

*, we have*
*(We naturally suppose that es−*1 = 1

*when s *= +

*∞).*
*In particular, the mean M lies on the boundary of MD (that is, on the circlewith center *A

*and radius *1

*) if and only if *sup

**Examples: **The two means G and H lie on the boundary of

*MD*.

4. Construction of a functional middle of two means
Let

*I ⊂ *R (

*I ̸*=

*∅*) and let

*D *=

*I*2. The aim of this section is to
prove, under some

*regularly *conditions, the existence and uniqueness of the
functional middle of two given means

*M*1 and

*M*2 on

*D*; that is, the existenceand uniqueness of a new mean

*M *on

*D *satisfying the functional equation

*M *(

*M*1

*, M*2) =

*M.*
In this context, we obtain two results which only diﬀer in the condition
imposed on

*M*1 and

*M*2. The ﬁrst one requires d(

*M*1

*, M*2)

*̸*= 1 (where d isthe distance deﬁned in Section 3) while the second requires

*M*1 and

*M*2 tobe continuous on

*D *(by taking

*I *an interval of R). Notice further that ourway of establishing the existence of the functional middle is constructive
and generalizes the idea of the AGM mean. Our ﬁrst result is the following:

**Theorem 4.1. ***Let M*1

*and M*2

*be two mean functions on D *=

*I*2

*, with*

values in I and such that d(

*M*1

*, M*2)

*< *1

*. Then there exists a unique mean*

function M on D satisfying the functional equation
*M *(

*M*1

*, M*2) =

*M.*
*Moreover, for all *(

*x, y*)

*∈ D, M *(

*x, y*)

*is the common limit of the two realsequences *(

*xn*)

*and *(

*y*

*x*0 =

*x , y*0 =

*y,*

*xn*+1 =

*M*1(

*xn, yn*)

*Proof. *Let

*k *:= d(

*M*1

*, M*2). By hypothesis, we have

*k < *1. Let (

*xn*) and
(

*yn*) be as in the statement and let (

*u*
*un *:= min(

*xn, yn*) and

*vn *:= max(

*xn, yn*) (

*∀n ∈ *N)

*.*
*un*+1 = min(

*xn*+1

*, yn*+1) = min(

*M*1(

*xn, yn*)

*, M*2(

*xn, yn*))

*≥ *min(

*xn, yn*) =

*un*
(because

*M*1(

*xn, yn*)

*≥ *min(

*xn, yn*) and

*M*2(

*xn, yn*)

*≥ *min(

*xn, yn*)).

Similarly, for all

*n ∈ *N,

*vn*+1 = max(

*xn*+1

*, yn*+1) = max(

*M*1(

*xn, yn*)

*, M*2(

*xn, yn*))

*≤ *max(

*xn, yn*) =

*vn.*
*|vn*+1

*− un*+1

*| *=

*|*max(

*xn*+1

*, yn*+1)

*− *min(

*xn*+1

*, yn*+1)

*|*
=

*|M*1(

*xn, yn*)

*− M*2(

*xn, yn*)

*|≤ k|xn − yn|*
*|vn − un| ≤ kn|v*0

*− u*0

*|*
It follows (since

*k ∈ *[0

*, *1)) that (

*vn − un*) tends to 0 as

*n *tends to inﬁnity.

Thus the bounded monotonic sequences (

*un*) and (

*v*
*un ≤ xn ≤ vn *and

*un ≤ yn ≤ vn*
also converge to the same limit. Denote the
common limit of the four sequences by

*M *(

*x, y*).

Now we show that the map

*M *:

*D → *R just deﬁned is a mean function
on

*D *and satisﬁes

*M *(

*M*1

*, M*2) =

*M *. First we check the three axioms of amean function.

**(i) **Given (

*x, y*)

*∈ D*, on changing (

*x, y*) to (

*y, x*) in the deﬁnition of the

sequences (

*xn*) and (

*y*
, they remain unchanged except their ﬁrst terms
(since

*M*1 and

*M*2 are symmetric). So,

*M *(

*x, y*) =

*M *(

*y, x*)
(

*∀*(

*x, y*)

*∈ D*)

*.*
**(ii) **Given (

*x, y*)

*∈ D*, since the corresponding sequences (

*un*) and (

*v*
are respectively non-decreasing and non-increasing and since

*M *(

*x, y*) is their
common limit, we have

*u*0

*≤ M *(

*x, y*)

*≤ v*0, that is,
min(

*x, y*)

*≤ M *(

*x, y*)

*≤ *max(

*x, y*)

*.*
**(iii) **Fix (

*x, y*)

*∈ D*. Suppose that

*M *(

*x, y*) =

*x *and, towards a contradiction,

*x ̸*=

*y*. Since

*M*1 and

*M*2 are means, we have (by axiom (iii))

*M*1(

*x, y*)

*̸*=

*x *and

*M*2(

*x, y*)

*̸*=

*x.*
Then

*M *(

*x, y*) =

*x *= min(

*x, y*) =

*u*0. So (

*un*) is non-decreasing and con-
verges to

*u*0. It follows that (

*un*) is necessarily constant and in particular
min(

*M*1(

*x, y*)

*, M*2(

*x, y*)) =

*x,*
Then

*M *(

*x, y*) =

*x *= max(

*x, y*) =

*v*0. So (

*vn*) is non-increasing and con-
verges to

*v*0. It follows that (

*vn*) is constant and in particular

*v*
max(

*M*1(

*x, y*)

*, M*2(

*x, y*)) =

*x,*
which again contradicts (4.1), proving (iii).

To prove

*M *(

*M*1

*, M*2) =

*M *, note that changing in the deﬁnition of (

*xn*)

*n*
and (

*yn*) the couple (

*x, y*) of

*D *to (

*M*
1(

*x, y*)

*, M*2(

*x, y*)) just amounts to shift-
ing these sequences (namely we obtain (

*xn*+1) instead of (

*x*
instead of (

*yn*) ). Consequently, the common limit (which is

*M *(

*x, y*)) re-

*M *(

*M*1(

*x, y*)

*, M*2(

*x, y*)) =

*M *(

*x, y*)

*.*
It remains to show that

*M *is the unique mean satisfying the func-
tional equation

*M *(

*M*1

*, M*2) =

*M *. Let

*M ′ *be any mean function satis-fying

*M ′*(

*M*1

*, M*2) =

*M ′ *and ﬁx (

*x, y*)

*∈ D*. We associate to (

*x, y*) thesequence (

*xn, yn*)

*n∈*N in the statement of the theorem. Using the relation

*M ′*(

*M*1

*, M*2) =

*M ′*, we have

*M ′*(

*x, y*) =

*M ′*(

*x*1

*, y*1) =

*M ′*(

*x*2

*, y*2) =

*· · · *=

*M ′*(

*xn, yn*) =

*· · ·*
But since

*M ′ *is a mean, it follows that for all

*n ∈ *N,
min(

*xn, yn*)

*≤ M ′*(

*x, y*)

*≤ *max(

*xn, yn*)

*,*
*M ′*(

*x, y*) =

*M *(

*x, y*)

*,*
From Theorem 4.1, we derive the following corollary:

**Corollary 4.2. ***Let *M

*be a mean function on D *=

*I*2

*, with values in I.*

Then there exists a unique mean on D satisfying the functional equation
=

*M *(

*x, y*) (

*∀*(

*x, y*)

*∈ D*)

*.*
*In addition, for all *(

*x, y*)

*∈ D, M *(

*x, y*)

*is the common limit of the two realsequences *(

*xn*)

*and *(

*y*
*Proof. *Since the metric space (

*MD, *d) is the closed ball with center A andradius 1

*/*2 (see Proposition 3.1), we have d(M

*, *A)

*≤ *1

*/*2

*< *1. The corollarythen immediately follows from Theorem 4.1.

In the following theorem, we establish another suﬃcient condition for
the existence and uniqueness of the functional middle of two means.

**Theorem 4.3. ***Suppose that I is an interval of *R

*and let M*1

*and M*2

*be*

two mean functions on D =

*I*2

*with values in I. Suppose that M*1

*and M*2

*are continuous on D. Then there exists a unique mean function M on D*

satisfying the functional equation
*M *(

*M*1

*, M*2) =

*M.*
*In addition, for all *(

*x, y*)

*∈ D, M *(

*x, y*)

*is the common limit of the two realsequences *(

*xn*)

*and *(

*y*
*Proof. *Fix (

*x, y*)

*∈ D *and deﬁne (

*un*) and (

*v*
are convergent. Let

*u *=

*u*(

*x, y*) and

*v *=

*v*(

*x, y*) denote their respective limits (so

*u *and

*v *lie in [

*u*0

*, v*0] =[min(

*x, y*)

*, *max(

*x, y*)]

*⊂ I*).

Now, since

*M*1 and

*M*2 are symmetric on

*D*, we have, for all

*n ∈ *N,

*xn*+1 =

*M*1(

*un, vn*) and

*yn*+1 =

*M*2(

*un, vn*)

*.*
By continuity, the sequences (

*xn*) and (

*y*
respective limits are

*M*1(

*u, v*) and

*M*2(

*u, v*). Letting

*n → ∞ *in

*xn*+1 =

*M*1(

*xn, yn*), we obtain

*M*1(

*u, v*) =

*M*1 (

*M*1(

*u, v*)

*, M*2(

*u, v*))

*,*
*M*1(

*u, v*) =

*M*2(

*u, v*)

*.*
converge to the same limit. Denoting by

*M *(

*x, y*) this
common limit, we show as in the proof of Theorem 4.1 that

*M *is a mean
function on

*D *and that it is the unique mean on

*D *which satisﬁes thefunctional equation

*M *(

*M*1

*, M*2) =

*M *.

[Bor] J. M. Borwein and P. B. Borwein,

*Pi and the AGM, (A study in*
*Analytic Number Theory and Computational Complexity)*, Wiley., New
Department of Mathematics, University of B´

*E-mail address*:

[email protected]
Source: http://univ-bejaia.dz/staff/photo/pubs/1104-532-means.pdf

2013 Medicare High Performance 4 Tier Step Therapy Criteria ALZHEIMER'S DRUGS . 6 omeprazole/sodium bicarbonat . 18 ALZHEIMER'S DRUGS Affected Drugs STEP 1 DRUGS If the patient has tried a Step 1 drug, then authorization for a Step 2 drug may be given. Step 1 Drug(s): Donepezil Hcl, Galantamine Hydrobromide, Galantamine Hbr, Rivastigmine. Step 2 Drug(s): Exelon oral solution, Exelon

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