Cryptography Reference
In-Depth Information
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
dθ
a
2
cos
2
θ
+
b
2
sin
2
θ
.
Search WWH ::
Custom Search