Nonlinear Equations - UFRJ

100 [CH. 7: NEWTON ITERATION
Let us look at the sequence y i = S(t i ). By construction y 0 = 1, and
subsequent values are given by the recurrence
It is an exercise to check that
y i+1 = S(N(h βγ , S −1 (y i ))).
y i+1 = qy 2 i , (7.11)
Therefore we have y i = q 2i −1 , and equation (7.10) holds.
Proposition 7.17. Under the conditions of Proposition 7.16, 0 is
an approximate zero of the second kind for h βγ if and only if
α = βγ ≤ 13 − 3√ 17
.
4
Proof. Using the closed form for t i , we get:
with
t i+1 − t i = 1 − q2i+1 −1
1 − ηq 2i+1 −1 − 1 − q2i −1
1 − ηq 2i −1
In the particular case i = 0,
Hence
C i =
= q 2i −1 (1 − η)(1 − q 2i )
(1 − ηq 2i+1 −1
)(1 − ηq 2i −1
)
t 1 − t 0 = 1 − q
1 − ηq = β
t i+1 − t i
β
= C i q 2i −1
(1 − η)(1 − ηq)(1 − q 2i )
(1 − q)(1 − ηq 2i+1 −1
)(1 − ηq 2i −1
) .
Thus, C 0 = 1. The reader shall verify in Exercise 7.6 that C i is a
non-increasing sequence. Its limit is non-zero.
From the above, it is clear that 0 is an approximate zero of the
second kind if and only if q ≤ 1/2. Now, if we clear denominators