2.4 error analysis for iterative methodszxu2/acms40390f12/lec-2.4.pdfΒ Β· newtonβs method as fixed...
TRANSCRIPT
2.4 Error Analysis for Iterative Methods
1
Definition β’ Order of Convergence Suppose {ππ}π=0
β is a sequence that converges to π with ππ β π for all π. If positive constants π and πΌ exist with
limπββ
|ππ+1 β π|
|ππ β π|πΌ= π
then {ππ}π=0β is said to converges to π of order πΆ with asymptotic
error constant π. An iterative technique ππ = π(ππβ1) is said to be of order πΆ if the sequence {ππ}π=0
β converges to the solution π = π(π) of order πΆ. β’ Special cases
1. If πΌ = 1 (and π < 1), the sequence is linearly convergent 2. If πΌ = 2, the sequence is quadratically convergent 3. If πΌ < 1, the sequence is sub-linearly convergent (undesirable, very slow) 4. If πΌ = 1 and π = 0 or 1 < πΌ < 2, the sequence is super-linearly convergent
β’ Remark: High order (πΌ) βΉ faster convergence (more desirable) π is less important than the order (πΌ)
2
Linear vs. Quadratic
3
Suppose we have two sequences converging to 0 with:
limπββ
|ππ+1|
|ππ|= 0.9, lim
πββ
|ππ+1|
|ππ|2= 0.9
Roughly we have: ππ β 0.9 ππβ1 β β― β 0.9π π0 ,
ππ β 0.9|ππβ1|2 β β― β 0.92πβ1 π0 , Assume π0 = π0 = 1
Fixed Point Convergence β’ Theorem
Let π β πΆ[π, π] be such that π π₯ β [π, π] for all π₯ β π, π . Suppose πβ² is continuous on (π, π) and that 0 < π < 1 exists with |πβ²(π₯)| β€ π for all π₯ β π, π .
If πβ²(π) β 0, then for all number π0 in [π, π], the sequence ππ = π(ππβ1) converges only linearly to the unique fixed point π in π, π .
β’ Proof: ππ+1 β π = π ππ β π π = πβ² ππ ππ β π , ππ β (ππ, π)
Since {ππ}π=0β converges to π, {ππ}π=0
β converges to π.
Since πβ² is continuous, limπββ
πβ²(ππ) = πβ²(π)
limπββ
|ππ+1βπ|
|ππβπ|= lim
πββπβ² ππ = |πβ²(π)| βΉ linear convergence
4
Speed up Convergence of Fixed Point Iteration
β’ If we look for faster convergence methods, we must have πβ² π = 0
β’ Theorem Let π be a solution of π₯ = π π₯ . Suppose πβ² π = 0 and πβ²β² is continuous with πβ²β² π₯ < π on an open interval πΌ containing π. Then there exists a πΏ > 0 such that for π0 β π β πΏ, π + πΏ , the sequence defined by ππ+1 =π ππ , when π β₯ 0, converges at least quadratically to π. For sufficiently large π
ππ+1 β π <π
2|ππ β π|2
Remark:
Look for quadratically convergent fixed point methods which π π = π and πβ² π = 0.
5
Newtonβs Method as Fixed Point Problem
Solve π π₯ = 0 by fixed point method. We write the problem as an equivalent fixed point problem:
π π₯ = π₯ β π π₯ solve: π₯ = π(π₯) π π₯ = π₯ β πΌπ π₯ solve π₯ = π π₯ , πΌ is a constant
π π₯ = π₯ β π π₯ π π₯ solve π₯ = π π₯ , π π₯ is differentiable
Newtonβs method is derived by the last form:
Find differentiable π π₯ with πβ² π = 0 when π π = 0.
πβ² π₯ =π
ππ₯π₯ β π π₯ π π₯ = 1 β πβ²π β ππβ²
Use πβ² π = 0 when π π = 0 πβ² π = 1 β πβ² π β 0 β π π πβ² π = 0
π π = 1/πβ²(π)
This gives Newtonβs method
ππ+1 = π ππ = ππ βπ(ππ)
πβ²(ππ)
6
Multiple Roots
β’ How to modify Newtonβs method when πβ² π = 0. Here π is the root of π π₯ = 0.
β’ Definition: Multiplicity of a Root A solution π of π π₯ = 0 is a zero of multiplicity π of π if for π₯ β π, we can write π π₯ = π₯ β π ππ π₯ , where limπ₯βπ π(π₯) β 0.
β’ Theorem π β πΆ1[π, π] has a simple zero at π in (π, π) if and only if π π = 0, but πβ²(π) β 0.
β’ Theorem The function π β πΆπ[π, π] has a zero of multiplicity π at point π in (π, π) if and only if 0 = π π = πβ² π = πβ²β² π = β― = π πβ1 π , but π π (π) β 0
7
Newtonβs Method for Zeroes of Higher Multiplicity (π > 1)
Define the new function π π₯ =π(π₯)
πβ²(π₯).
Write π π₯ = π₯ β π ππ(π₯), hence
π π₯ =π(π₯)
πβ²(π₯)= π₯ β π
π π₯
ππ π₯ + π₯ β π πβ² π₯
Note that π is a simple zero of π π₯ .
β’ Apply Newtonβs method to π π₯ to give:
π₯ = π π₯ = π₯ βπ π₯
πβ² π₯
= π₯ βπ π₯ πβ² π₯
πβ² π₯ 2 β π π₯ πβ²β² π₯
β’ Quadratic convergence: ππ+1 = ππ βπ ππ πβ² ππ
πβ² ππ2βπ ππ πβ²β² ππ
8
Drawbacks:
β’ Compute πβ²β²(π₯) is expensive
β’ Iteration formula is more complicated β more expensive to compute
β’ Roundoff errors in denominator β both πβ²(π₯) and π(π₯) approach zero.
9