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Therefore, since a n− 1
b n− 1 , it follows immediately from (9.20) that
b n b n− 1 b n− 1 = b n− 1 .
1
2 ( a n− 1 + a n− 1 )= a n− 1
a n
and
Also,
2 a n− 1 b n− 1 2
1
a n − b n =
a n− 1 b n− 1 a n− 1 + b n− 1
1
2
1
2 ( a n− 1
=
b n− 1 ) .
Therefore, a n − b n (1 / 2) n ( a − b ), so a n − b n 0. Since the a n 's are
a decreasing sequence bounded below by the increasing sequence of b n 's, it
follows immediately that the two seque nces conv erge to the same limit, so
M ( a, b ) exists. If b n− 1 1, then a n− 1 + b n− 1 2, so
16 a n− 1 b n− 1 2
a n b n
8
1
=
b n− 1 2 a n 1 + b n 1 2
4
16 a n− 1
1
= a n 1
2
b n− 1
.
8
Inequality 9.22 follows easily by induction.
The limit M ( a, b ) is called the arithmetic-geometric mean of a and b .
Since
M ( ca, cb )= cM ( a, b ) ,
we can always rescale a and b to make b
1. Also, since M ( b, a )= M ( a, b )
(because a 1 and b 1 are symmetric in a, b ), we may always arrange that a
b .
By Inequality (9.21), a n
b n < 1 for su ciently large n .Thenumbers a n + m
and b n + m give approximations to M ( a, b ). Inequality (9.22) predicts that
the number of decimal places of accuracy doubles with each iteration. This
phenomenon occurs in the above example.
The reasons we are interested in the arithmetic-geometric mean are the
following two propositions.
PROPOSITION 9.24
Let a, b be positive realnum bers. D efine
I ( a, b )= π/ 2
0
a 2 cos 2 θ + b 2 sin 2 θ
.
 
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