Nonlinear Equations - UFRJ

[SEC. 10.2: PROOF OF THEOREM 10.5 147
By (10.9) and taking µ ≥ √ 2, D ≥ 2, β(F ti+1 , X i ) ≤ (1 − ɛ 1 )α/2.
Therefore,
d proj (N(F ti+1 , X i ), z ti+1 ) ≤ 2 1 − ɛ 1 1 − α
1
D 3/2 µ ψ(α) α2 r 1 (α)
1 − (1 − ɛ 1 )α/2
using (10.13).
Putting all together,
L(f t , z t ; t i , t i+1 ) ≥
2
D 3/2√ n ×
×
(ɛ 2−a 0) √ 1−ɛ 2 1
(1+ɛ 1)
− a 0 r 0 (a 0 ) − (1 − ɛ 1 )
1−α
ψ(α)(1−(1−ɛ 1)α/2) α2 r 1 (α)
1 + ɛ 1 + π(1 − ɛ 1 )αr 0 (α)/ √ D
The final bound was obtained numerically, assuming D ≥ 2. We
check computationally that
2
(ɛ 2−a 0) √ 1−ɛ 2 1
(1+ɛ 1)
− a 0 r 0 (a 0 ) − (1 − ɛ 1 )
1−α
ψ(α)(1−(1−ɛ 1)α/2) α2 r 1 (α)
1 + ɛ 1 + π(1 − ɛ 1 )αr 0 (α)/ √ 2
≥ C −1
(10.18)
Proof of Theorem 10.5. Suppose the algorithm terminates. We claim
that for each t i , X i is a (β, µ, a 0 )-certified approximate zero of F ti ,
and that its associated zero is z ti . This is true by hypothesis when
i = 0. Therefore, assume this is true up to a certain i.
Recall that β(F, X) scales as ‖X‖. In particular,
β(F ti+1 , X i+1 ) = β(F t i+1
, N(F ti+1 , X i ))
‖N(F ti+1 , X i )‖
By (10.9) again, β(F ti+1 , X i ) ≤ (1 − ɛ 1 )α/2.
We apply (10.14) to obtain that
D 3/2
2 β(F s, X i+1 )µ(f s , [N(F s , X i )]) ≤ a 0 .
≤ β(F t i+1
, N(F ti+1 , X i ))
.
1 − β(F ti+1 , X i )
From (10.11), X i is an approximate zero of the second kind for
F s , s ∈ [t i , t i+1 ]. Since both α(F s , X i ) and β(F s , X i ) are bounded