Nonlinear Equations - UFRJ

88 [CH. 7: NEWTON ITERATION
1
y = ψ(u)
3 − √ 7
5− √ 17
4
3− √ 7
5− √ 17
4
2
1 − √ 2/2
Figure 7.1: y = ψ(u)
Lemma 7.7. The function ψ(u) = 1 − 4u + 2u 2 is decreasing and
non-negative in [0, 1 − √ 2/2], and satisfies:
u
ψ(u) < 1 for u ∈ [0, (5 − √ 17)/4) (7.3)
u
ψ(u) ≤ 1 2
for u ∈ [0, (3 − √ 7)/2] . (7.4)
The proof of Lemma 7.7 is left to the reader (but see Figure 7.1).
Another useful result is:
Lemma 7.8. Let A be a n × n matrix. Assume ‖A − I‖ 2 < 1. Then
A has full rank and, for all y,
‖y‖
≤ ‖A −1 ‖y‖
y‖ 2 ≤
.
1 + ‖A − I‖ 2 1 − ‖A − I‖ 2
Proof. By hypothesis, ‖Ax‖ > 0 for all x ≠ 0 so that A has full rank.
Let y = Ax. By triangular inequality,
‖Ax‖ ≥ ‖x‖ − ‖(A − I)x‖ ≥ (1 − ‖(A − I)‖ 2 )‖x‖.
Also by triangular inequality,
‖Ax‖ ≤ ‖x‖ + ‖(A − I)x‖ ≤ (1 + ‖(A − I)‖ 2 )‖x‖.