Introduction to Fixed Point Iteration Method and its ... · Introduction Fixed Point Iteration...

Table of contentsIntroduction

Fixed Point Iteration MethodCondition for Convergence

ApplicationAppendix

Introduction to Fixed Point Iteration Method andits application

Damodar Rajbhandari

St. Xavier’s College

Nepal, 2016

Damodar Rajbhandari Fixed point iteration method



ApplicationAppendix

Table of contents

1 Introduction

2 Fixed Point Iteration Method

3 Condition for Convergence

4 Application

5 Appendix




ApplicationAppendix

Introduction

What is fixed point?Does anybody knows, What is it?




ApplicationAppendix

Introduction

What is fixed point?→If a function φ(x) intersected by y = x then, the x-co-ordinate of an intersection is known as fixed point.

Formal definition

”A fixed point of a function is an element of function’s domainthat is mapped to itself by the function.”




ApplicationAppendix

Introduction

What is fixed point?→If a function φ(x) intersected by y = x then, the x-co-ordinate of an intersection is known as fixed point.

Formal definition

”A fixed point of a function is an element of function’s domainthat is mapped to itself by the function.”




ApplicationAppendix

Lets see an example 1 [See its matlab code in Appendix Section];




ApplicationAppendix

In the previous figure, what are the fixed points?




ApplicationAppendix

Fixed Point’s response on your answer:

”Finally, you knew me!”




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Fixed Point’s response on your answer:

”Finally, you knew me!”




ApplicationAppendix

What is ”Fixed Point Iteration Method”?

In Numerical analysis, It is a method of computing fixedpoints by doing no. of iteration to the functions.

So, What?

You’ll get roots for your desire equation.




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So, What?





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So, What?





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So, What?





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What is the reason behind changing my equation into x = φ(x)form?

→ You’re in the right place! This presentation is for you!Hehe, Thank Me!




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→ You’re in the right place! This presentation is for you!Hehe, Thank Me!




ApplicationAppendix

Fixed Point Iteration Method

Let me answer your question first!




ApplicationAppendix


Suppose, We have a function like f (x) = x − φ(x).(Not mandatory!)

If we want to find a root of this equation then, we haveto do like this;

f (x) = 0

So, the function becomes x = φ(x)




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f (x) = 0





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f (x) = 0





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f (x) = 0





ApplicationAppendix

Let me ask you a question!

In the previous example 1, y = x is a line. Lets say, y = φ(x) then,the function φ(x) = x becomes y = x . Is this an equation of a linethat will intersect the equation φ(x)?

ANSWER: No! It is a fixed point that will satisfy f (x) = 0.




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What is the primary algorithm for this method?

1. First, Convert your function f (x) into x = φ(x) form.

2. Iterate the value of x to the function φ(x) until you get thedesire precision such that the precision value should berepeated after you again iterate it.




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ApplicationAppendix

Can you show me in geometrical perspective?

CLICK THE ”ICON” TO GLORIFY YOURSELF!


https://youtu.be/XhSsu5P1NdY



ApplicationAppendix

Can you show me in geometrical perspective?



https://youtu.be/XhSsu5P1NdY



ApplicationAppendix

Condition for Convergence

We can make lots of functions in the form of x = φ(x) by using thef (x). Then, Can every functions will give me the root of the f (x)?

→ May be! But, the rate of convergence would be slow.




ApplicationAppendix


We can make lots of functions in the form of x = φ(x) by using thef (x). Then, Can every functions will give me the root of the f (x)?→ May be! But, the rate of convergence would be slow.




ApplicationAppendix

So, How can I decide whether the function φ(x) would convergesfast or not?

→ You have to need a test called ”Condition for Convergence”Lets talk about it!




ApplicationAppendix

So, How can I decide whether the function φ(x) would convergesfast or not?→ You have to need a test called ”Condition for Convergence”

Lets talk about it!




ApplicationAppendix

So, How can I decide whether the function φ(x) would convergesfast or not?→ You have to need a test called ”Condition for Convergence”Lets talk about it!




ApplicationAppendix


”If α be the root of the f (x) which is equivalent to x = φ(x) andα is contained in the interval I of φ(x) and |φ′(x)| < 1 ∀ x ∈ I .Then, if x0 is any point in I so that, the sequence defined by

xn = φ(xn−1) | n ≥ 1will converges to the unique fixed point x in I . Thus, that uniquefixed point x of φ(x) will be the root of f (x).”




ApplicationAppendix

Basic Preliminaries

1. Visualization of xn = φ(xn−1)

2. Mean Value Theorem




ApplicationAppendix

Visualization of xn = φ(xn−1)




ApplicationAppendix

Mean Value Theorem

Lagrange’s MVT

If φ(x) is a real valued function continuous on the closed interval[x , xn−1], and differentiable in the open interval (x , xn−1) then,there exists a point ε ∈ (x , xn−1) such that

φ(xn−1)− φ(x) = φ′(ε)(xn−1 − x)




ApplicationAppendix

Proof

By Uniqueness Theorem, ”A unique fixed point exists in I”.

Since, φ(x) maps I into itself. The sequence {x}∞n=0 is defined forall n ≥ 0 and xn ∈ I for all n.If x is fixed point then, it should follow x = φ(x). Also, fromabove;

xn = φ(xn−1)Now, lets write as ;|xn − x | = |φ(xn−1)− φ(x)|or , |xn − x | = |φ′(ε)(xn−1 − x)| Applying Lagrange MVT in rightpart.




ApplicationAppendix

Proof

By Uniqueness Theorem, ”A unique fixed point exists in I”.Since, φ(x) maps I into itself. The sequence {x}∞n=0 is defined forall n ≥ 0 and xn ∈ I for all n.

If x is fixed point then, it should follow x = φ(x). Also, fromabove;





ApplicationAppendix

Proof

By Uniqueness Theorem, ”A unique fixed point exists in I”.Since, φ(x) maps I into itself. The sequence {x}∞n=0 is defined forall n ≥ 0 and xn ∈ I for all n.If x is fixed point then, it should follow x = φ(x). Also, fromabove;

xn = φ(xn−1)

Now, lets write as ;|xn − x | = |φ(xn−1)− φ(x)|or , |xn − x | = |φ′(ε)(xn−1 − x)| Applying Lagrange MVT in rightpart.




ApplicationAppendix

Proof


xn = φ(xn−1)Now, lets write as ;|xn − x | = |φ(xn−1)− φ(x)|

or , |xn − x | = |φ′(ε)(xn−1 − x)| Applying Lagrange MVT in rightpart.




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Proof






ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.

Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |. . .∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |

Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |. . .∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |

⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |. . .∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |

⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |. . .∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |

. . .∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |. . .

∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

or , |xn − x | ≤ |φ′(ε)||xn−1 − x | Applying Cauchy-Schwarz’sinequality in right part.Since, ε ∈ (x , xn−1) so, assume |φ′(ε)| = k .∴ |xn − x | ≤ k |xn−1 − x |Applying the inequality of the hypothesis inductively gives,|xn − x | ≤ k |xn−1 − x |⇒ |xn − x | ≤ k .k |xn−2 − x | = k2|xn−2 − x |⇒ |xn − x | ≤ k2.k|xn−3 − x | = k3|xn−3 − x |. . .∴ |xn − x | ≤ kn|xn−n − x | = kn|x0 − x |




ApplicationAppendix

Proof

Suppose, k < 1 and taking limn→∞

in both part.

limn→∞

|xn − x | ≤ limn→∞

kn|x0 − x |⇒ lim

n→∞|xn − x | < 0 Since, lim

n→∞kn → 0

Proof.

So, {x}∞n=0 converges to x.




ApplicationAppendix

Proof


in both part.

limn→∞

|xn − x | ≤ limn→∞

kn|x0 − x |

⇒ limn→∞

|xn − x | < 0 Since, limn→∞

kn → 0

Proof.





ApplicationAppendix

Proof


in both part.

limn→∞

|xn − x | ≤ limn→∞



n→∞kn → 0

Proof.





ApplicationAppendix

Proof


in both part.

limn→∞

|xn − x | ≤ limn→∞



n→∞kn → 0

Proof.





ApplicationAppendix

Application

1. Find the roots of the equation f (x) = x3 + x − 1? Correctupto two decimal places.

→ Given, equation f (x) = x3 + x − 1. To find the root of thisequation, we say f (x) = 0 i.e.

x3 + x − 1 = 0By testing on f (a).f (b) < 0. We estimated, the root shouldlies on [0, 1].

There are many ways to change the equation to the fixed pointform x = φ(x) using simple algebraic manipulation.

My possible choices for φ(x) are;x = φ1(x) = 1− x3

x = φ2(x) = (1− x)1/3




ApplicationAppendix

Application

1. Find the roots of the equation f (x) = x3 + x − 1? Correctupto two decimal places.→ Given, equation f (x) = x3 + x − 1. To find the root of thisequation, we say f (x) = 0 i.e.




x = φ2(x) = (1− x)1/3




ApplicationAppendix

Application





x = φ2(x) = (1− x)1/3




ApplicationAppendix

Application





x = φ2(x) = (1− x)1/3




ApplicationAppendix

Application

Previous theorem will suggests us, which form of φ(x) willconverges reliably and rapidly! So that, we can be able to rejectthe one which converges slowly.

Taking the derivative of above forms with respect to x . We getφ′1(x) = −3x2

φ′2(x) = −13(1−x)

23




ApplicationAppendix

Application

Previous theorem will suggests us, which form of φ(x) willconverges reliably and rapidly! So that, we can be able to rejectthe one which converges slowly.

Taking the derivative of above forms with respect to x . We getφ′1(x) = −3x2

φ′2(x) = −13(1−x)

23




ApplicationAppendix

Application

Test on the value of x = 0.8. On putting in above equation andtaking absolute value of it. We get

|φ′1(0.8)| = | − 1.92| > 1|φ′2(0.8)| = | − 0.97| < 1

Since,|φ′1(0.8)| > 1 So,we reject φ1(x) and we choose φ2(x) foriteration.




ApplicationAppendix

Application

Test on the value of x = 0.8. On putting in above equation andtaking absolute value of it. We get|φ′1(0.8)| = | − 1.92| > 1|φ′2(0.8)| = | − 0.97| < 1





ApplicationAppendix

Application

Test on the value of x = 0.8. On putting in above equation andtaking absolute value of it. We get|φ′1(0.8)| = | − 1.92| > 1|φ′2(0.8)| = | − 0.97| < 1





ApplicationAppendix

Application

Taking initial approximation of x as 0.5. We can see in theattached video, how it converges!



https://youtu.be/-bSkR0o0-GU



ApplicationAppendix

Application

From the ”Fixed Point Iteration” program in Matlab[See the codeof it in Appendix Section], we can get the result as,

Hence, the root of the given equation corrected upto 2 decimalplaces is 0.68




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Appendix

Matlab Code for the figure(example 1)




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Appendix

Matlab Code for ”Fixed Point Iteration Method”




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You can find the MATLAB files of;1. Plotting code at Click Me!2. Program code at Click Me!


https://dx.doi.org/10.6084/m9.figshare.4285463.v1

https://dx.doi.org/10.6084/m9.figshare.4285592.v1

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