SOLUTION OF EQUATIONS FOR ITERATIVOS METHODS

Jeannie Castaño 2053298

A iterative method is a method that progressively it calculates approaches to the solution of a problem. To difference of the direct methods, in which the process should be finished to have the answer, in the iterative methods you can suspend the process to the i finish of an iteration and an approach is obtained to the solution.

For example the method of Newton-Raphson

)(

)('1 xf

xfxx iii

JACOBI

2

2

1 `

0

2 20 0

1 00

2 21 1

2 11

2 2

2

2 23 2

2

:

( 1) 1 4 1 21,2

2

( ) 2

( ) 2 1

3

0

2 3 3 23 2,2

2 1 6 1

2 2,2 2,2 22,2 2,011

2 1 4,4 1

2 2,011 2,02,011

2

) 2

1

(

i i ii i i

i i

Example

x

f x x xx x x

f x x

x

i

x xx x

x

x xx x

x

x xx x

f x x

x

x

11 22,00004

2.2,011 1

Etcetera

The following steps should be continued:

1. First the equation de recurrence is determined. Of the equation i and the incognito i clears. In notation matricial is written:

where x is the vector of incognito 2. It takes an approach for the solutions and

to this it is designated for

3. You itera in the cycle that changes the approach

The Jacobi method de is the iterative method to solve system of equations simple ma and it is applied alone to square systems, that is to say to systems with as many equations as incognito.

x c Bx

1i ix c Bx

1

1

0.20 0.00 0.40

0.00 0.25 0.00

0.20 0.00 0.40

0.00 0.25 0.00

0.20 0.00(1.00) 0.40(2.00) 0.60

0.00 0.25(1.00) 0.00(2.00) 0.25

0.20 0.00( 0.60) 0.40(0.25) 0.1

i

i

i

x x y

y x y

x x

y y

x x

y y

x

2

2

3

3

4

4

0

0.00 0.25( 0.60) 0.00(0.25) 0.15

0.20 0.00(0.10) 0.40( 0.15) 0.26

0.00 0.25(0.10) 0.00( 0.15) 0.025

0.20 0.00(0.26) 0.40(0.025) 0.190

0.00 0.25(0.26) 0.00(0.025) 0.065

i

i

i

i

i

i

x

y y

x x

y y

x x

y y

x

5

5

6

6

0.20 0.00(0.19) 0.40(0.065) 0.174

0.00 0.25(0.19) 0.00(0.065) 0.0475

0.20 0.00(0.174) 0.40(0.0475) 0.181

0.00 0.25(0.174) 0.00(0.0475) 0.0435

i

i

i

x

y y

x x

y y

Leaving of x=1, y=2 applies two iterations of the Jacobi method to solve the system:

5 2 1

4 0

x y

x y

This Di it is used as unemployment approach in the iterations until Di it is smaller than certain given value.

Being Di: 1 1max ,i i i iDi x x y y

i xi yi Xi+1 Yi+1 Di

0 1.000 2.000 -0.600 0.250 1.750

1 -0.600 0.250 0.100 -0.150 0.700

2 0.100 -0.150 0.260 0.025 0.175

3 0.260 0.025 0.190 0.065 0.070

4 0.190 0.065 0.174 0.047 0.017

5 0.174 0.047 0.181 0.043 0.007

6 0.181 0.043 0.182 0.045 0.001

Given a square system of n linear equations with unknown x:

Where:

GAUSS-SEIDEL

Ax B

11 12 1

21 22 2

1 2

n

n

n n nn

a a a

a a aA

a a a

1 1

2 2

n n

x b

x bX B

x b

Then A can be decomposed into a lower triangular component L*, and a strictly upper triangular component U:

The system of linear equations may be rewritten as:

11 11 1

21 22 2

1 2

0 0 0

0 0 0* * ,

0 0 0

n

n

n n nn

a a a

a a aA L U L U

a a a

* *L x b U x

GAUSS-SEIDEL

The Gauss–Seidel method is an iterative technique that solves the left hand side of this expression for x, using previous value for x on the right hand side. Analytically, this may be written as:

However, by taking advantage of the triangular form of L*, the elements of x(k+1) can be computed sequentially using forward substitution:

1 1* *k kx L b U x

1 11, 1,2,3.....k k k

i ij j ij jj i j iii

x b a x a x i na

GAUSS-SEIDEL

5 10

8 2 11

4 3

x y z

y x z

x y z

10

5

y zx

11 2 2

8

xy

3

4

y xz

1ra iteration

To solve the following exercise for the method of Gauss-Seidel with an initial value of (0,0,0). Use three iterations

10 0 02

5x

11 2(2) 0 7

8 8y

73 2 3384 32

z

2da iteration

7 3310 638 32

5 32x

63 3311 2( ) 25932 32

8 256y

259 633 1013256 32

4 1024z

3ra iteration

259 101310 10263256 1024

5 5120x

10263 101311 2( )

5120 1024 0,9975348

y

102633 0,997534

5120 1,0017394

z

To solve the previous exercise for the gauss method for relaxation. .Use two iterations5 10

8 2 11

4 3

x y z

y x z

x y z

(1 )new previousi i ix x x

1,05

10 0 02

5x

11 2(2) 0 7

8 8y

73 2 3384 32

z

Replacing with the initial values (0,0,0)

GAUSS-SEIDEL RELAXATION

1ra iteration

1,05(2) (1 1,05)(0)

2,1

11 2(2,1) 0 17

8 2017

1,05( ) (1 1,05)(0)20

357

400357

3 2,1 16834004 16001683

1,05( ) (1 1,05)(0)1600

1,10446875

n

n

n

n

n

n

x

x

y

y

y

z

x

y

2da iteration357

10 1,10446875400 1,95760625

5

1,05(1,95760625) (1 1,05)(2,1)

1,90486563

11 2(1,90486563) 1,104468751,025436953

8357

1,05(1,025436953) (1 1,05)( )400

1,032083801

3 1,032083801 1,90486563

4

n

n

n

n

x

x

x

y

y

y

z

0,9681954573

1,05(0,9681954573) (1 1,05)(1,10446875)

0,9613817926

n

n

z

z

BIBLIOGRAFIAhttp://es.wikipedia.org/wiki/M%C3%A9todo_iterativo

http://search.conduit.com/Results.aspx?q=METODOS+ITERATIVOS&meta=all&hl=es&gl=co&SearchSourceOrigin=13&SelfSearch=1&ctid=CT2247187

http://en.wikipedia.org/wiki/Gauss%E2%80%93Seidel_method