![Page 1: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/1.jpg)
Solutions of Equations in One Variable
Fixed-Point Iteration I
Numerical Analysis (9th Edition)
R L Burden & J D Faires
Beamer Presentation Slidesprepared byJohn Carroll
Dublin City University
c© 2011 Brooks/Cole, Cengage Learning
![Page 2: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/2.jpg)
Theoretical Basis Example Formulation I Formulation II
Outline
1 Introduction & Theoretical Framework
2 Motivating the Algorithm: An Example
3 Fixed-Point Formulation I
4 Fixed-Point Formulation II
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 2 / 59
![Page 3: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/3.jpg)
Theoretical Basis Example Formulation I Formulation II
Outline
1 Introduction & Theoretical Framework
2 Motivating the Algorithm: An Example
3 Fixed-Point Formulation I
4 Fixed-Point Formulation II
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 3 / 59
![Page 4: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/4.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
Prime Objective
In what follows, it is important not to lose sight of our prime
objective:
Given a function f (x) where a ≤ x ≤ b, find values p such that
f (p) = 0
Given such a function, f (x), we now construct an auxiliary function
g(x) such that
p = g(p)
whenever f (p) = 0 (this construction is not unique).
The problem of finding p such that p = g(p) is known as the fixed
point problem.
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 4 / 59
![Page 5: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/5.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
A Fixed Point
If g is defined on [a, b] and g(p) = p for some p ∈ [a, b], then the
function g is said to have the fixed point p in [a, b].
Note
The fixed-point problem turns out to be quite simple both
theoretically and geometrically.
The function g(x) will have a fixed point in the interval [a, b]whenever the graph of g(x) intersects the line y = x .
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 5 / 59
![Page 6: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/6.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
The Equation f (x) = x − cos(x) = 0
If we write this equation in the form:
x = cos(x)
then g(x) = cos(x).
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 6 / 59
![Page 7: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/7.jpg)
Theoretical Basis Example Formulation I Formulation II
Single Nonlinear Equation f (x) = x − cos(x) = 0
y
x
1
1
−1
g(x) = cos(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 7 / 59
![Page 8: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/8.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
x = cos(x)
y
x1
1
x
g(x) = cos(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 8 / 59
![Page 9: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/9.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
p = cos(p) p ≈ 0.739
y
x1
1
0.739
0.739
x
g(x) = cos(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 9 / 59
![Page 10: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/10.jpg)
Theoretical Basis Example Formulation I Formulation II
Existence of a Fixed Point
If g ∈ C[a, b] and g(x) ∈ [a, b] for all x ∈ [a, b] then the function g has
a fixed point in [a, b].
Proof
If g(a) = a or g(b) = b, the existence of a fixed point is obvious.
Suppose not; then it must be true that g(a) > a and g(b) < b.
Define h(x) = g(x) − x ; h is continuous on [a, b] and, moreover,
h(a) = g(a) − a > 0, h(b) = g(b) − b < 0.
The Intermediate Value Theorem IVT implies that there exists
p ∈ (a, b) for which h(p) = 0.
Thus g(p) − p = 0 and p is a fixed point of g.
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 10 / 59
![Page 11: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/11.jpg)
Theoretical Basis Example Formulation I Formulation II
g(x) is Defined on [a, b]
y
x
a b
g(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 11 / 59
![Page 12: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/12.jpg)
Theoretical Basis Example Formulation I Formulation II
g(x) ∈ [a, b] for all x ∈ [a, b]
y
x
a b
a
b
g(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 12 / 59
![Page 13: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/13.jpg)
Theoretical Basis Example Formulation I Formulation II
g(x) has a Fixed Point in [a, b]
y
x
a b
a
b
g(x)
x
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 13 / 59
![Page 14: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/14.jpg)
Theoretical Basis Example Formulation I Formulation II
g(x) has a Fixed Point in [a, b]
y
x
y 5 x
y 5 g(x)
p 5 g(p)
a p b
a
b
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 14 / 59
![Page 15: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/15.jpg)
Theoretical Basis Example Formulation I Formulation II
Illustration
Consider the function g(x) = 3−x on 0 ≤ x ≤ 1. g(x) is
continuous and since
g′(x) = −3−x log 3 < 0 on [0, 1]
g(x) is decreasing on [0, 1].
Hence
g(1) =1
3≤ g(x) ≤ 1 = g(0)
i.e. g(x) ∈ [0, 1] for all x ∈ [0, 1] and therefore, by the preceding
result, g(x) must have a fixed point in [0, 1].
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 15 / 59
![Page 16: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/16.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
g(x) = 3−x
x
y
1
1
y 5 x
y 5 32x
s1, ad
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 16 / 59
![Page 17: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/17.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
An Important Observation
It is fairly obvious that, on any given interval I = [a, b], g(x) may
have many fixed points (or none at all).
In order to ensure that g(x) has a unique fixed point in I, we must
make an additional assumption that g(x) does not vary too rapidly.
Thus we have to establish a uniqueness result.
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 17 / 59
![Page 18: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/18.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
Uniqueness Result
Let g ∈ C[a, b] and g(x) ∈ [a, b] for all x ∈ [a, b]. Further if g′(x) exists
on (a, b) and
|g′(x)| ≤ k < 1, ∀ x ∈ [a, b],
then the function g has a unique fixed point p in [a, b].
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 18 / 59
![Page 19: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/19.jpg)
Theoretical Basis Example Formulation I Formulation II
g′(x) is Defined on [a, b]
y
x
a b
g‘(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 19 / 59
![Page 20: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/20.jpg)
Theoretical Basis Example Formulation I Formulation II
−1 ≤ g′(x) ≤ 1 for all x ∈ [a, b]
y
x
a b
1
−1
g‘(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 20 / 59
![Page 21: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/21.jpg)
Theoretical Basis Example Formulation I Formulation II
Unique Fixed Point: |g′(x)| ≤ 1 for all x ∈ [a, b]
y
x
a b
1
−1
g‘(x)
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 21 / 59
![Page 22: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/22.jpg)
Theoretical Basis Example Formulation I Formulation II
Functional (Fixed-Point) Iteration
Proof of Uniqueness Result
Assuming the hypothesis of the theorem, suppose that p and q
are both fixed points in [a, b] with p 6= q.
By the Mean Value Theorem MVT Illustration , a number ξ exists
between p and q and hence in [a, b] with
|p − q| = |g(p) − g(q)| =∣
∣g′(ξ)∣
∣ |p − q|
≤ k |p − q|
< |p − q|
which is a contradiction.
This contradiction must come from the only supposition, p 6= q.
Hence, p = q and the fixed point in [a, b] is unique.
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 22 / 59
![Page 23: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/23.jpg)
Theoretical Basis Example Formulation I Formulation II
Outline
1 Introduction & Theoretical Framework
2 Motivating the Algorithm: An Example
3 Fixed-Point Formulation I
4 Fixed-Point Formulation II
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 23 / 59
![Page 24: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/24.jpg)
Theoretical Basis Example Formulation I Formulation II
A Single Nonlinear Equaton
Model Problem
Consider the quadratic equation:
x2 − x − 1 = 0
Positive Root
The positive root of this equations is:
x =1 +
√5
2≈ 1.618034
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 24 / 59
![Page 25: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/25.jpg)
Theoretical Basis Example Formulation I Formulation II
Single Nonlinear Equation f (x) = x2− x − 1 = 0
y
x
1−
−1−
1−1 1.5
y = x2 − x − 1
We can convert this equation into a fixed-point problem.
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 25 / 59
![Page 26: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/26.jpg)
Theoretical Basis Example Formulation I Formulation II
Outline
1 Introduction & Theoretical Framework
2 Motivating the Algorithm: An Example
3 Fixed-Point Formulation I
4 Fixed-Point Formulation II
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 26 / 59
![Page 27: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/27.jpg)
Theoretical Basis Example Formulation I Formulation II
Single Nonlinear Equation f (x) = x2 − x − 1 = 0
One Possible Formulation for g(x)
Transpose the equation f (x) = 0 for variable x :
x2 − x − 1 = 0
⇒ x2 = x + 1
⇒ x = ±√
x + 1
g(x) =√
x + 1
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 27 / 59
![Page 28: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/28.jpg)
Theoretical Basis Example Formulation I Formulation II
xn+1 = g (xn) =√
xn + 1 with x0 = 0
y
x
g(x) =√
x + 1
y = x
x0
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 28 / 59
![Page 29: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/29.jpg)
Theoretical Basis Example Formulation I Formulation II
Fixed Point: g(x) =√
x + 1 x0 = 0
n pn pn+1 |pn+1 − pn|1 0.000000000 1.000000000 1.000000000
2 1.000000000 1.414213562 0.414213562
3 1.414213562 1.553773974 0.139560412
4 1.553773974 1.598053182 0.044279208
5 1.598053182 1.611847754 0.013794572
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 29 / 59
![Page 30: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/30.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 30 / 59
![Page 31: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/31.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 31 / 59
![Page 32: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/32.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 32 / 59
![Page 33: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/33.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 33 / 59
![Page 34: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/34.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 34 / 59
![Page 35: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/35.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 35 / 59
![Page 36: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/36.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 36 / 59
![Page 37: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/37.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x
g(x) =√
x + 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 37 / 59
![Page 38: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/38.jpg)
Theoretical Basis Example Formulation I Formulation II
xn+1 = g (xn) =√
xn + 1 with x0 = 0
Rate of Convergence
We require that |g′(x)| ≤ k < 1. Since
g(x) =√
x + 1 and g′(x) =1
2√
x + 1> 0 for x ≥ 0
we find that
g′(x) =1
2√
x + 1< 1 for all x > −3
4
Note
g′(p) ≈ 0.30902
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 38 / 59
![Page 39: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/39.jpg)
Theoretical Basis Example Formulation I Formulation II
Fixed Point: g(x) =√
x + 1 p0 = 0
n pn−1 pn |pn − pn−1| en/en−1
1 0.0000000 1.0000000 1.0000000 —
2 1.0000000 1.4142136 0.4142136 0.41421
3 1.4142136 1.5537740 0.1395604 0.33693
4 1.5537740 1.5980532 0.0442792 0.31728
5 1.5980532 1.6118478 0.0137946 0.31154...
......
......
12 1.6180286 1.6180323 0.0000037 0.30902
13 1.6180323 1.6180335 0.0000012 0.30902
14 1.6180335 1.6180338 0.0000004 0.30902
15 1.6180338 1.6180339 0.0000001 0.30902
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 39 / 59
![Page 40: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/40.jpg)
Theoretical Basis Example Formulation I Formulation II
Outline
1 Introduction & Theoretical Framework
2 Motivating the Algorithm: An Example
3 Fixed-Point Formulation I
4 Fixed-Point Formulation II
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 40 / 59
![Page 41: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/41.jpg)
Theoretical Basis Example Formulation I Formulation II
Single Nonlinear Equation f (x) = x2− x − 1 = 0
A Second Formulation for g(x)
Transpose the equation f (x) = 0 for variable x :
x2− x − 1 = 0
⇒ x2 = x + 1
⇒ x = 1 +1
x
g(x) = 1 +1
x
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 41 / 59
![Page 42: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/42.jpg)
Theoretical Basis Example Formulation I Formulation II
xn+1 = g (xn) = 1xn
+ 1 with x0 = 1
y
x1
g(x) = 1
x+ 1
y = x
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 42 / 59
![Page 43: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/43.jpg)
Theoretical Basis Example Formulation I Formulation II
Fixed Point: g(x) =1
x+ 1 x0 = 1
n pn pn+1 |pn+1 − pn|
1 1.000000000 2.000000000 1.000000000
2 2.000000000 1.500000000 0.500000000
3 1.500000000 1.666666667 0.166666667
4 1.666666667 1.600000000 0.066666667
5 1.600000000 1.625000000 0.025000000
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 43 / 59
![Page 44: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/44.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 44 / 59
![Page 45: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/45.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1 x1
x2
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 45 / 59
![Page 46: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/46.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1 x1
x2x2
x3
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 46 / 59
![Page 47: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/47.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1 x1
x2x2
x3
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 47 / 59
![Page 48: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/48.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1 x1
x2x2
x3
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 48 / 59
![Page 49: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/49.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1 x1
x2x2
x3
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 49 / 59
![Page 50: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/50.jpg)
Theoretical Basis Example Formulation I Formulation II
y
x1
g(x) = 1
x+ 1
y = x
x0
x1 x1
x2x2
x3
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 50 / 59
![Page 51: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/51.jpg)
Theoretical Basis Example Formulation I Formulation II
xn+1 = g (xn) = 1xn
+ 1 with x0 = 1
Rate of Convergence
We require that |g′(x)| ≤ k < 1. Since
g(x) =1
x+ 1 and g′(x) = − 1
x2< 0 for x
we find that
g′(x) =1
2√
x + 1> −1 for all x > 1
Note
g′(p) ≈ −0.38197
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 51 / 59
![Page 52: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/52.jpg)
Theoretical Basis Example Formulation I Formulation II
Fixed Point: g(x) =1
x+ 1 p0 = 1
n pn−1 pn |pn − pn−1| en/en−1
1 1.0000000 2.0000000 1.0000000 —
2 2.0000000 1.5000000 0.5000000 0.50000
3 1.5000000 1.6666667 0.1666667 0.33333
4 1.6666667 1.6000000 0.0666667 0.40000
5 1.6000000 1.6250000 0.0250000 0.37500...
......
......
12 1.6180556 1.6180258 0.0000298 0.38197
13 1.6180258 1.6180371 0.0000114 0.38196
14 1.6180371 1.6180328 0.0000043 0.38197
15 1.6180328 1.6180344 0.0000017 0.38197
Numerical Analysis (Chapter 2) Fixed-Point Iteration I R L Burden & J D Faires 52 / 59
![Page 53: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/53.jpg)
Questions?
![Page 54: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/54.jpg)
Reference Material
![Page 55: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/55.jpg)
Intermediate Value Theorem
If f ∈ C[a, b] and K is any number between f (a) and f (b), then there
exists a number c ∈ (a, b) for which f (c) = K .
x
y
f (a)
f (b)
y 5 f (x)
K
(a, f (a))
(b, f (b))
a bc
(The diagram shows one of 3 possibilities for this function and interval.)Return to Existence Theorem
![Page 56: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/56.jpg)
Mean Value Theorem: Illustration (1/3)
Assume that f ∈ C[a, b] and f is differentiable on (a, b).
y
x
f(x)
a b
f(a)
f(b)
![Page 57: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/57.jpg)
Mean Value Theorem: Illustration (2/3)
Measure the slope of the line joining a, f (a)] and [b, f (b)].
y
x
f(x)
a b
f(a)
f(b)slope =
f(b)−f(a)b−a
![Page 58: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/58.jpg)
Mean Value Theorem: Illustration (3/3)
Then a number c exists such that
f′(c) =
f (b) − f (a)
b − a
y
x
c
f(x)
a b
f(a)
f(b)
slope = f ′(c)
slope = f(b)−f(a)b−a
![Page 59: Solutions of Equations in One Variable Fixed-Point Iteration ITheoretical Basis Example Formulation I Formulation II Outline 1 Introduction & Theoretical Framework 2 Motivating the](https://reader034.vdocuments.mx/reader034/viewer/2022042909/5f3b19791a9e70474a43eb4b/html5/thumbnails/59.jpg)
Mean Value Theorem
If f ∈ C[a, b] and f is differentiable on (a, b), then a number c exists
such that
f′(c) =
f (b) − f (a)
b − a
y
xa bc
Slope f 9(c)
Parallel lines
Slopeb 2 a
f (b) 2 f (a)
y 5 f (x)
Return to Fixed-Point Uniqueness Theorem