The Real And Complex Number Systems

so, by induction, we find
dp n1 , p n n dp 1 , p 0 n dx, fx.
Use the triangel inequality we find, for m n,
m1
dp m , p n dp k1 , p k
kn
m1
dx, fx k
kn
dx, fx n m
1
dx, fx
n
1 . *
Since n 0asn , we know that p n is a Cauchy sequence. And since S is
complete, we have p n p S. The uniqueness is from the inequality,
dfx, fy dx, y.
From (*), we know that (let m )
dp, p n dx, fx n
1 .
This inequality, which is useful in numberical work, provides an estimate for the
distance from p n to the fixed point p. An example is given in (b)
(b) Take fx 1 x 2/x, S 1, . Prove that f is contraction of S with
2
contraction constant 1/2 and fixed point p 2 . Form the sequence p n starting wth
x p 0 1 and show that p n 2 2 n .
Proof: First, fx fy 1 x 2/x 1 y 2/y 1 yx
x y 2
2 2 2 xy , then we
have
|fx fy| 1 y x
x y 2
2 xy
1 2 x y 1 xy
2
1 2 |x y| since 1 2 xy 1.
So, f is a contraction of S with contraction constant 1/2. By Fixed Point Theorem, we
know that there is a unique p such that fp p. Thatis,
1
2 p 2 p p p 2 .( 2 is not our choice since S 1, .)
By (a), it is easy to know that
p n 2 2 n .
Remark: Here is a modefied Fixed Point Theorem: Let f be function defined on a
complete metric space S. If there exists a N such that df N x f N y dx, y for all
x, y S, where 0 1. Then f has a unique fixed point p S.
Proof: Since f N is a contraction defined on a complete metric space, with the
contraction constant , with 0 1, by Fixed Point Theorem, we know that there
exists a unique point p S, such that