Vol. 11 No 1 - Pi Mu Epsilon

24 ANATOLYSHTEKHMAN
·'
ARITHMETIC-GEOMETRIC MEANS 25
The main purpose of our paper is to introduce the AG and to give a partial
solution to an open problem posed by M.K. Vamanurthy and M. Vuorinen [7] p. 165.
We will use some definitions from [3], [7], and [2].
DEFINITION 1. Let x, y > 0. The logarithmic mean (L) is defined as,
x-y
L(x,y) = ----=----
logx - logy
since
we obtain
ix- xoi, iY- xoi < 8 and ic- xi < ix- Yi
ic- xi+ ix- xoi ~ ix- Yi + 8 ~ ix- xoi + ixo - Yi + 8 ~ 38
for x # y and L(x, y) = x for x = y.
2. AG and L are continuous. Since L(x, y) is a piecewise defined function
and AG ( x, y) is defined recursively we would like to take a moment to prove their
continuity. This will support our conjecture below.
THEOREM 1. A(x, y) is a continuous function on [0, oo) 2 .
Proof Define a function f by
1
f( X, y, Z) = ---;::::====::::::;;:=
y' x 2 cos 2 z + y 2 sin 2 z
f is a continuous function on ~ 2 \ {(0, 0)} x ~. Gauss proved [2] p.484 that for x, y > 0
AG(x, y) = ~ 1" f(x, y, z)dz.
For any (x, y) # (0, 0) and any sequence (xn, Yn) ---+ (x, y) the sequence f(x n, Yn, z)
uniformly converges to f(x, y, z) Hence by the interchange theorem (see e.g., [4],
Theorem II, p. 461) it follows that the AG(x, y) is continuous at (x, y). To prove
that the AG(x, y) is continuous at (0, 0) let us pick c > 0 and let 8 = c. Assume that
y'x 2 + y 2 < 8. We have
iAG(x, y)- AG(O, O)i = iAG(x, y)i
< max{ixi , iYi}
max{fx2,Vij2}
< y'x2 +y2 = c
That concludes that the AG is continuous function for any x, y ~ 0.
THEOREM 2. L is a continuous function on (0, oo) 2 .
Proof L is piecewise defined so we shall have two cases:
Case 1 - when the point (xo, Yo) is on the diagonal {(x, x) : x > 0}. We must
show that for every c > 0 there is a corresponding number 8 > 0 such that whenever
x,y > 0
y'(x - xo) 2 + (y- xo) 2 < 8 =====> iL(x, y) - L(xo, xo) i < c
If x = y then the above implication holds with 8 = c. If x # y then by the mean
value theorem, [6]
iL(x, y) - L(xo, xo)i =
where cis between x andy.
I x-y I I x - y I
1 og x - 1 og y c x - y
- xo = 1 ( ) - xo = ic - xo l
ic - xoi = ic - x + x - xoi ~ ic - x i + ix - xoi (2)
D
(1)
Therefore ic- x 0 i ~ 38. Hence by 1 and 2 it is clear that if we take 8 < ~ the
implication
y'(x- x 0 )2 + (y- xo) 2 < 8 =====> iL(x, y)- L(xo, xo)i < c
holds for all x, y > 0.
Case 2 - when the point is not on the diagonal. In this case L(x, y) is continuous
as a composition of continuous functions such as xjy, x-y, and log(x). That concludes
the proof. D
3. Is AGt ~ L for some t E (0, 1). It is well known that A(x, y) ~ G(x, y).
This can be generalized to a weighted mean inequality
whenever w 1 , w 2 ~ 0 and w 1 + w2 = 1. To read more about weighted and modified
inequalities we suggest [1]. In 1991 B.C. Carlson and M. Vuorinen proved that
AG(x, y) ~ L(x, y) (3)
More recently inequalities similar to 3 have been studied involving means "modified"
in the following way:
DEFINITION 2. For real numbers t # 0, we define
Mt (X' y) = M (X ' y ) t '
where M can be any of the means: G, AG, or L.
For example,
1
t t
AGt = AG ( x , y ) t ,
Lt (X' y) = L (X ' y ) t •
AGt has many properties of AG. In particular since the function xt is continuous for
x > 0 so composed with Theorems 2 and 3 we obtain the following result:
COROLLARY 3. Ift > 0 then AGt(x, y) and Lt(x, y) are continuous functions on
(0, oo) 2 .
However, if we modify AG in the inequality 3, its behavior is not entirely clear:
PROBLEM 4. Does there exists a t E (0, 1 ) so that AGt dominates L, i.e. so that
AGt(x, y) ~ L(x, y)for all x, y > 0 ?
THEOREM 5. For positive x andy both AGt(x, y) and Lt(x, y) are continuous,
strictly increasing functions oft.
Proof See Theorem 1.2 of [7]. D
THEOREM 6. AGt(x, y) < L(x, y) for all t E (0, ~) and x, y E ~+.
t
t
t
t
1
1