Nonlinear Equations - UFRJ

[SEC. 9.3: APPROXIMATE ZEROS 129
and
Df(X) = Df(Z) + ∑ k≥2
1
k − 1! Dk f(Z)(X − Z) k−1 .
Combining the two equations, above, we obtain:
f(X) − Df(X)(X − Z) = ∑ k≥2
k − 1
D k f(Z)(X − Z) k .
k!
Using Lemma 7.6 with d = 2, the rightmost term in (9.2) is
bounded above by
‖f(X) − Df(X)(X − Z)‖ ≤ ∑ (k − 1)γ k−1 ‖X − Z‖ k
k≥2
(9.3)
γ‖X − Z‖ 2
=
(1 − γ‖X − Z‖) 2 .
Combining Lemma 9.4 and (9.3) in (9.2), we deduce that
‖N(f, X) − Z‖ ≤
γ‖X − Z‖2
ψ(γ‖X − Z‖) .
By induction, u i ≤ γ‖X i −Z i ‖. When u 0 ≤ (3− √ 7)/2, we obtain
as in Lemma 7.10 that
d proj (X i , Z)
d proj (X 0 , Z) ≤ ‖X i − Z‖
‖X 0 − Z‖ ≤ u i
≤ 2 −2i +1 .
u 0
We have seen in Lemma 7.10 that the bound above fails for i = 1
when u 0 > (3 − √ 7)/2.
The same comments as the ones for theorem 7.5 are in order. We
actually proved stronger theorems, see exercises.
Exercise 9.1. Show that the projective distance in P n satisfies the
triangle inequality. Same question in the multi-projective case.
Exercise 9.2. Restate and prove Theorem 7.11 in the context of
pseudo-Newton iteration.
Exercise 9.3. Restate and prove Theorem 7.12 in the context of
pseudo-Newton iteration.