Nonlinear Equations - UFRJ

106 [CH. 7: NEWTON ITERATION
Proof. In order to estimate the higher derivatives, we expand:
1
l! Df(x)−1 D l f(y) = ∑ ( ) k + l Df(x) −1 D k+l f(x)(y − x) k
l
k + l
k≥0
and by Lemma 7.6 for d = l + 1,
1
l! ‖Df(x)−1 D l γ(f, x)l−1
f(y)‖ ≤
(1 − u) l+1 .
Combining with Lemma 7.9,
1
l! ‖Df(y)−1 D l γ(f, x)l−1
f(y)‖ ≤
(1 − u) l−1 ψ(u) .
Taking the l − 1-th power,
γ(f, y) ≤
γ(f, x)
(1 − u)ψ(u) .
Proof of Theorem 7.19. We have necessarily α < 3 − 2 √ 2 or r is
undefined. Then (Theorem 7.18) there is a zero ζ of f with ‖x 0 −ζ‖ ≤
rβ(f, x 0 ). Then, Lemma 7.20 implies that ‖x 0 − ζ‖γ(f, ζ) ≤ u 0 . Now
apply Theorem 7.12.
Exercise 7.6. The objective of this exercise is to show that C i is
non-increasing.
1. Show the following trivial lemma: If 0 ≤ s < a ≤ b, then
a−s
b−s ≤ a b .
2. Deduce that q ≤ η.
3. Prove that C i+1 /C i ≤ 1.
Exercise 7.7. Show that
ζ 1 γ(ζ 1 ) = 1 + α − √ ∆
3 − α + √ ∆
(
ψ
1
1+α− √ ∆
4
).