0.1 Error analysis for Iterative Methods Definition 1 Suppose {pn ...

0.1 Error analysis for Iterative Methods
Definition 1 Suppose {p n } ∞ n=0 is a sequence that converges
to p, with p n ≠ p for all n. If there exist constants
λ and α such that
lim
n→∞
|p n+1 − p|
|p n − p| α = λ,
then {p n } ∞ n=0 converges to p of order α, with
asymptotic error constant λ.
An iterative technique of the form p n = g(p n−1 ) is said
to be of order α if the sequence {p n } ∞ n=0 converges to p =
g(p) of order α.
• If α = 1(and λ < 1), the sequence is linearly convergent.
• If α = 2, the seqiuence is quadratically convergent.
Theorem 1 Let g ∈ C[a, b] be such that g([a, b]) ⊆ [a, b].
Suppose, in addition, that g ′ is continuous on (a, b) and
that a positive constant k < 1 exists with
|g ′ (x)| ≤ k, for all x ∈ (a, b)
. If g ′ (p) ≠ 0, where p is the unique fixed point of g(x),
then for any p 0 in [a, b], the sequence
p n = g(p n−1 ), for all n ≥ 1
converges linearly to p.
Proof:
By the condition, the function g(x) has a unique fixed
point p and for any p 0 ∈ [a, b] , the sequence converges
to p. since
p n+1 − p = g(p n ) − g(p) = g ′ (ξ n )(p n − p).
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where ξ n lies between p n and p.
Since p n → p , then ξ n → p. since g ′ is continuous on
(a, b), so
lim
n→∞ g′ (ξ n ) = g ′ (p)
Thus,
|p n+1 − p|
lim = |g ′ (p)|
n→∞ |p n − p|
Hence, if g ′ (p) ≠ 0, the sequence {p n } converges linearly
to p with asymptotic error constant |g ′ (p)|.
So, if we want to get the higher order, the condition
g ′ (p) = 0 needs to be occurred.
Theorem 2 Assume p is a solution of x = g(x), and
suppose that g(x) is continuously differentiable in some
neighboring interval about p with |g ′ (p)| < 1. Then p n =
g(p n−1 ) converges to p provided p 0 is chosen sufficiently
close to p
Proof:
Since lim x→p g ′ (x) = g ′ (p) and |g ′ (p)| < 1 for some
neighborhood of p, then we can pick up an interval I =
[p − ɛ, p + ɛ] with
Max x∈I |g ′ (x)| ≤ λ < 1
Let x ∈ I. Then |x − p| ≤ ɛ. So
|g(x) − p| = |g(x) − g(p)| = |g ′ (ξ)(x − p)| ≤ λ|x − p| ≤ ɛ
Thus we have that g(I) ⊆ I and |g ′ (x)| ≤ λ < 1 for all
x ∈ I, hence p n converges to p for any p 0 ∈ I.
Example x 2 − 3 = 0
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• g(x) = x 2 + x − 3
• g(x) = 3/x
• g(x) = 1 2 (x + 3 x )
Theorem 3 Assume that α is a root of x = g(x), and
that g(x) is p times continuously differentiable for all x
near to α, for some p ≥ 2. furthermore, assume
g ′ (α) = . . . = g (p−1) (α) = 0
Then if the initial guess x 0 is chosen sufficient close to
α, the iteration
x n+1 = g(x n ), n ≥ 0
will have order of convergence p, and
lim
n→∞
|x n+1 − α|
|(x n − α)| = |g(p) (α)|
p |p!|
3