Nonlinear Equations - UFRJ

[SEC. 7.3: ESTIMATES FROM DATA AT A POINT 99
t 0 = 0 t 1
t 2
ζ 1 ζ 2
Figure 7.4: y = h βγ (t).
Proof. By differentiating (7.9), one obtains
( 1
h ′ βγ(t) = h βγ (t) + 1
)
1
+
t − ζ 1 t − ζ 2 γ −1 − t
and hence the Newton operator is
1
N(h βγ , t) = t −
1
t−ζ 1
+ 1
t−ζ 2
+ 1 .
γ −1 −t
A tedious calculation shows that N(h βγ , t) is a rational function
of degree 2. Hence, it is defined by 5 coefficients, or by 5 values.
In order to solve the recurrence for t i , we change coordinates using
a fractional linear transformation. As the Newton operator will have
two attracting fixed points (ζ 1 and ζ 2 ), we will map those points to 0
and ∞ respectively. For convenience, we will map t 0 = 0 into y 0 = 1.
Therefore, we set
S(t) = ζ 2t − ζ 1 ζ 2
ζ 1 t − ζ 1 ζ 2
and S −1 (y) = −ζ 1ζ 2 y + ζ 1 ζ 2
−ζ 1 y + ζ 2