𝑋
is closed and hence complete (Exercise 1.107). Therefore,
𝑓
has a fixed point. That
is, there exists
𝑥
0
∈
𝑋
such that
𝑥
0
=
𝑓
(
𝑥
0
) =
1
2
(
𝑥
0
+
2
𝑥
0
)
Rearranging
2
𝑥
2
0
=
𝑥
2
0
+ 2 =
⇒
𝑥
2
0
= 2
so that
𝑥
0
=
√
2.
Letting
𝑥
0
= 2
𝑥
1
=
1
2
(2 + 1) =
3
2
Using the error bounds in Corollary 2.5.1,
𝜌
(
𝑥
𝑛
,
√
2)
≤
𝛽
𝑛
1
−
𝛽
𝜌
(
𝑥
0
, 𝑥
1
)
=
(1
/
2)
𝑛
1
/
2
1
/
2
=
1
2
𝑛
=
1
1024
<
0
.
001
when
𝑛
= 10.
Therefore, we conclude that 10 iterations are ample to reduce the
error below 0
.
001. Actually, with experience, we can refine this
a priori
estimate. In
Example 1.64, we calculated the first five terms of the sequence to be
(2
,
1
.
5
,
1
.
416666666666667
,
1
.
41421568627451
,
1
.
41421356237469)
We observe that
𝜌
(
𝑥
3
, 𝑥
4
) = 1
.
41421568627451
−
1
.
41421356237469) = 0
.
0000212389982
so that using the second inequality of Corollary 2.5.1
𝜌
(
𝑥
4
,
√
2)
≤
1
/
2
1
/
2
0
.
0000212389982
<
0
.
001
𝑥
4
= 1
.
41421356237469 is the desired approximation.
2.121
Choose any
𝑥
0
∈
𝑆
. Define the sequence
𝑥
𝑛
=
𝑓
(
𝑥
𝑛
) =
𝑓
𝑛
(
𝑥
0
). Then (
𝑥
𝑛
) is a
Cauchy sequence in
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 Fall '10
 Dr.DuMond
 Macroeconomics, All rights reserved, Metric space, Fixed point, ???? ????

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