Nonlinear Equations - UFRJ

[SEC. 7.3: ESTIMATES FROM DATA AT A POINT 101
and rearrange terms in (1 + α − √ ∆)/(1 + α + √ ∆) = 1/2, we obtain
the second degree polynomial
2α 2 − 13α + 2 = 0.
This has solutions (13 ± √ √ 17)/2. When 0 ≤ α ≤ α 0 = (13 −
17)/2, the polynomial values are positive and hence q ≤ 1/2.
Proof of Theorem 7.15. Let β = β(f, x 0 ) and γ = γ(f, x 0 ). Let h βγ
and the sequence t i be as in Proposition 7.16. By construction, ‖x 1 −
x 0 ‖ = β = t 1 − t 0 . We use the following notations:
Those will be compared to
β i = β(f, x i ) and γ i = γ(f, x i ).
ˆβ i = β(h βγ , t i )) and ˆγ i = γ(h βγ , t i )).
Induction hypothesis: β i ≤ ˆβ i and for all l ≥ 2,
‖Df(x i ) −1 D l f(x i )‖ ≤ − h(l) βγ (t i)
h ′ βγ (t i) .
The initial case when i = 0 holds by construction.
assume that the hypothesis holds for i. We will estimate
So let us
β i+1 ≤ ‖Df(x i+1 ) −1 Df(x i )‖‖Df(x i ) −1 f(x i+1 )‖ (7.12)
and
γ i+1 ≤ ‖Df(x i+1 ) −1 Df(x i )‖ ‖Df(x i) −1 D k f(x i+1 )‖
. (7.13)
k!
By construction, f(x i ) + Df(x i )(x i+1 − x i ) = 0. The Taylor expansion
of f at x i is therefore
Df(x i ) −1 f(x i+1 ) = ∑ k≥2
Df(x i ) −1 D k f(x i )(x i+1 − x i ) k
k!