Cryptography Reference
In-Depth Information
Example
9
.1
Let
a
=
√
2and
b
=1. Then
a
1
=1
.
207106781186547524400844362
...
b
1
=1
.
189207115002721066717499970
...
a
2
=1
.
198156948094634295559172166
...
b
2
=1
.
198123521493120122606585571
...
a
3
=1
.
198140234793877209082878869
...
b
3
=1
.
198140234677307205798383788
...
a
4
=1
.
198140234735592207440631328
...
b
4
=1
.
198140234735592207439213655
....
The sequences are converging very quickly to the limit
a
∞
=
b
∞
=1
.
198140234735592207439922492
....
The rapid convergence is explained by the following.
PROPOSITION 9.23
Suppose
a
≥
b>
0
.Then
b
n−
1
≤ b
n
≤ a
n
≤ a
n−
1
and
1
2
(
a
n−
1
− b
n−
1
)
.
0
≤ a
n
− b
n
≤
(9.21)
T herefore
M
(
a, b
) = lim
n
a
n
= lim
n
b
n
→∞
→∞
exists. M oreover, if
b ≥
1
then
8
a
n
−
b
n
8
2
m
a
n
+
m
−
b
n
+
m
≤
(9.22)
for all
m, n
≥
0
.
PROOF
The fact that
a
n
≥
b
n
for all
n
is the arithmetic-geometric mean
inequality, or the fact that
a
n
− b
n
=
1
2
(
√
a
n−
1
−
b
n−
1
)
2
≥
0
.
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