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Fixed Point Method
Adrian Down
January 31, 2006
1 Review: Sufficient condition for a fixed point
1.1 Condition for the existence of a fixed point
Theorem. Given g(x) continuous in [a, b] and g(x) in [a, b] ∀x ∈ [a, b], then ∃ at least one fixed point α such that g(α) = α.
Note. g(x) must lie in the square with sides [a, b] for iteration to be possible. If xν ∈ [a, b], then xν+1 = g(xν ) ∈ [a, b], and we can perform the next iteration.
1.2 Uniqueness of the fixed point
Theorem. If, in addition to the hypothesis above, g is differentiable and |g′(x)| < λ for some λ such that 0 < λ < 1 , ∀x ∈ (a, b), then the fixed point α ∈ [a, b] is unique.
Proof. Suppose α and β are distinct fixed points ∈ (a, b) such that a < α < β < b. By the mean value theorem, ∃ζ ∈ (α, β) such that
g′(ζ) =
g(β) − g(α) β − α
Because β and α are fixed points by assumption, g(α) = α and g(β) = β. Hence,
g′(ζ) =
β − α β − α
This contradicts the hypothesis that |g′(x)| < λ < 1, ∀x ∈ (a, b). Hence two unique fixed points cannot exist.
2 Fixed point iteration method
2.1 General method
Theorem. Assume g(x) satisfies the hypothesis of both theorems above. Choose any x 0 ∈ [a, b]. Define the sequence { xn } by
xν = g(xν− 1 ) ∀ν ≥ 1
Then xν converges to the unique fixed point x = α ∈ [a, b].
Proof.
|xν − α| = |g(xν− 1 ) − g(α)|
By the mean value theorem,
|xν − α| = |g′(ζν− 1 )||xν− 1 − α| ≤ λ|xν− 1 − α|
for some ζν− 1 ∈ (xν− 1 , α). Continuing iteratively,
|xν − α| ≤ λ|xν− 1 − α| ≤ λ^2 |xν− 2 − α| ≤... ≤ λν^ |x 0 − α| ≤ λν^ |b − a| −−→ ν→ 0
because 0 < λ < 1.
2.2 Rate of convergence
We would like to know how many iterations are required to obtain m signif- icant figures with this iterative method. Using the previous proof,
|xν − α| ≤ λν^ |b − a| ≤ 10 −m
To compare powers of 10, use the log 10 function. Reformulate log λ to make it positive,
|xν − α| ≤ 10 log^ |b−a|−ν^ log^
(^1) λ ≤ 10 −m
Comparing powers of 10,
log |b − a| − ν log
λ
≤ −m
⇒ ν log
λ
≥ log |b − a| + m
3.3 First-order convergence
Definition (First-order convergence). An approximation that achieves the same number of significant figures per iteration is called first order conver- gence.
We want to find the convergence error of a partial sum of terms in the continued fraction approximation.
|g′(x)| =
(1 + x)^2
≤ 1 ∈ x > 0
From the previous error estimate, the error is determined by
λˆ = |g′(α)| = 1 (1 + α)^2
= g^2 (α) = α^2 =. 38197
Thus we achieve about 1 significant figure for ever 2 iterations.
ν xν ν = |xν − α| 0 1
- 6666
6 .61904. 7 .61764.
Table 1: Calculated values of the continued fraction
From the calculated values in 3.3, ^76 ≈ .386. This is as we expect, since ˆλ = .38197.
3.4 Higher order convergence
Suppose g′(α) = 0.
|xν − α| |xν− 1 − α|
as ν → ∞. It appears that the continued fraction method converges faster than first order in this case. Hence the error estimate must include higher order terms. Assume g′(α) = 0, and g′′(α) ∈ (a, b) has |g′′(x)| ≤ 2 M ∀x ∈ (a, b). Then by the Taylor series approximation, modulo some remainders,
g(x) − g(α) =
g′′(ζ)(x − α)^2
for some ζ between x and a. To find the error, consider the term
xν − α = g(xν− 1 ) − g(α) =
g′′(ζν− 1 )(xν− 1 − α)^2
for ζν− 1 between α and xν− 1. Substituting the assumed bound for the second derivative,
|xν − α| ≤ M |xν− 1 − α|^2 ≤ M
M |xν− 2 − α|^2
Collecting terms,
M
M |xν− 2 − α|^2
= M 1+2|xν− 2 − α|^2 ·^2 ≤ M 1+^
M |xν− 3 − α|^2
= M 1+2+
2 |xν− 3 − α|^2 ·^2 ·^2 ≤ M 1+2+...+
ν− 1 |x 0 − α|^2
ν
Recognize the exponent of M as a geometric series,
∑^ ν−^1
n=
2 n^ = 2ν^ − 1
⇒ |xν − α| ≤ (M |x 0 − α|)^2
ν (^) − 1 |x 0 − α|
Convergence occurs if M |x 0 − α| < 1, which will certainly occur if M |b − a| <
