First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 2. SOLUTIONS OF EQUATIONS: ROOT-FINDING 86
If this limit were zero, then we would have
|p n+1 − p|
lim
n→∞ |p n − p|
which means the denominator is growing at a larger rate than the numerator. We could then
ask if
|p n+1 − p|
lim
= nonzero constant
n→∞ |p n − p|
α
for some α>1.
Theorem 46. Assume p is a solution of g(x) =x where g ∈ C α (I) for some interval I that
contains p, and for some α ≥ 2. Furthermore assume
=0,
g ′ (p) =g ′′ (p) =... = g (α−1) (p) =0, and g (α) (p) ≠0.
Then if the initial guess p 0 is sufficiently close to p, the fixed-point iteration p n = g(p n−1 ),n≥
1, will have order of convergence of α, and
Proof. From Taylor’s theorem,
lim
n→∞
p n+1 − p
(p n − p) = g(α) (p)
.
α α!
p n+1 = g(p n )=g(p)+(p n − p)g ′ (p)+... + (p n − p) α−1
g (α−1) (p)+ (p n − p) α
g (α) (ξ n )
(α − 1)!
α!
where ξ n isanumberbetweenp n and p, and all numbers are in I. From the hypothesis, this
simplifies as
p n+1 = p + (p n − p) α
g (α) (ξ n ) ⇒ p n+1 − p
α!
(p n − p) = g(α) (ξ n )
.
α α!
From Theorem 44, ifp 0 is chosen sufficiently close to p, then lim n→∞ p n = p. The order of
convergence is α with
|p n+1 − p|
lim
n→∞ |p n − p| = lim |g (α) (ξ n )|
α n→∞ α!
= |g(α) (p)|
α!
≠0.