MAT 4725 Numerical Analysis Section 7.3 Iterative Techniques .

MAT 4725Numerical Analysis

Section 7.3

Iterative Techniques

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Induction – P(1)Basic algebraic proofsSee me ….

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Focus on theoretical issues Jacobi Iterative Method

The need of iterative methods

4 2( ) 2 3 0f x x x x

The need of iterative methods

Ax b

Finding inverse… o Extremely expensive (time and storage)o More or less o Difficult/Impossible to implement in real

time

3( )O n

Chapter 7 Iterative Techniques in Matrix Algebra

Section 2.2 Fixed-Point Iteration

1 1

To solve To find a fixed point

( ) 0 for some ( )

( ), until n n n n

f x g x

x g x x x TOL

Chapter 7 Iterative Techniques in Matrix Algebra

Section 7.3 Iterative Techniques

( 1) ( )

To solve To find a fixed point

for some ( )

( ), until n n

G xAx b

x G x some conditions related to

the distance of 2 vectors

are satisfied

Section 2.2 Example 1

Show that if p is a fixed point of

then p is a solution of

4 3( )

2

x xg x

4 2( ) 2 3 0f x x x x

Idea

(0)

( )

( ) ( 1)

( ) ( 1)

Rewrite into

Given initial approximation ,

we can find by

Stop if

k

k k

k k

Ax b x Tx c

x

x

x Tx c

x x TOL

Example 1

(0) (1) (2)

Write the system

1 2 4 11

2 5 2 3

4 1 1 8

into the form of .

If 0, find , .

x Tx c

x x x

Example 1

Is there a pattern between and ?

0 2 41 2 4

2 22 5 2 , 0

5 54 1 1

4 1 0

A T

A T

Jacobi Iterative Method

11 12 1 1 1

2 2

1 2

n

n n nn n n

Ax b

a a a x b

x b

a a a x b

Jacobi Iterative Method:Matrix Form

11 12 1

2122

1,

1 , 1

Rewrite

0 0 00 0 0

0 00 0 0 0

00 0 0

00 0 0 0 0

It can be shown that the above process is the same as

n

n n

n n nnn

A D L U

a a a

aa

a

a aa

Convergence

( ) ( 1)k kx Tx c

Lemma

1

1 2

0

If ( ) 1, then ( ) exists and

( ) j

j

T I T

I T I T T T

c.f.

Theorem

( ) ( 1)

1( ) (0)

0

( )

If <1, then , 1

converges to the unique solution of

( ) ( ) : , 1

( )

[( ) is unique]

n n

nn n j

j

n

T x Tx c n

x Tx c

i P n x T x T c n

ii x x

iii x

Theorem

( ) ( 1)

1( ) (0)

0

( )

If <1, then , 1

converges to the unique solution of

( ) ( ) : , 1

( )

[( ) is unique]

n n

nn n j

j

n

T x Tx c n

x Tx c

i P n x T x T c n

ii x x

iii x

HW

HW

Corollary( ) ( 1)

( ) (0)

( ) (1) (0)

If 1, and , 1

then

(i)

(ii) 1

n n

nn

n

n

T x Tx c n

x x T x x

Tx x x x

T

HW

Homework

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