gauss seidel - metodos numericos
DESCRIPTION
metodos numericos- metodo iterativo- gauss seidel- ejercicio de explicacionTRANSCRIPT
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LATEX Ingenieria Mecatronica 1
UNIVERSIDAD DE LAS FUERZAS ARMADAS - ESPE
DEPARTAMENTO DE CIENCIAS EXACTAS
METODOS NUMERICOS
TRABAJO DE INVESTIGACION
Nombre: Andres Enrquez
CURSO : H - 203
Profesor: Ing. Patricio Pugarn
Fecha: Martes 30 de Junio de 2015
SANGOLQUI - ECUADOR
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LATEX Ingenieria Mecatronica 2
1. INVESTIGACION
1.1. GAUSS SEIDEL
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LATEX Ingenieria Mecatronica 3
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LATEX Ingenieria Mecatronica 4
EJERCICIOResuelva por el metodo de Gauss-Seidel
4x1 x2 + 2x3 + 0x4 = 33x1 + 6x2 + x3 x4 = 1x1 + 2x2 + 4x3 + 0x4 = 14x1 + 3x2 2x3 + 10x4 = 0
A =
4 1 2 03 6 1 11 2 4 04 3 2 10
b =
3110
N =
4 1 2 00 6 1 10 0 4 00 0 0 10
P =
0 0 0 03 0 0 01 2 0 04 3 2 0
N1 =
4 1 2 0 1 0 0 00 6 1 1 0 1 0 00 0 4 0 0 0 1 00 0 0 10 0 0 0 1
1 14
12
0 14
0 0 00 1 1
61 0 1/6 0 0
0 0 1 0 0 0 14
00 0 0 1 0 0 0 1
10
16F4
1 1
412
0 14
0 0 00 1 1
60 0 1
60 1
60
0 0 1 0 0 0 14
00 0 0 1 0 0 0 1
10
16F3 + F2
1 14
0 0 14
0 18
00 1 0 0 0 1
6 1
24160
0 0 1 0 0 0 14
00 0 0 1 0 0 0 1
10
14F2 + F1
1 14
0 0 14
124
1396
1240
0 1 0 0 0 16
124
160
0 0 1 0 0 0 14
00 0 0 1 0 0 0 1
10
; N1 =
14
124
1396
1240
0 16
124
160
0 0 14
00 0 0 1
10
xk+1 = N
1 b + N1 P xk
N1 b =
14
124
1396
1240
0 16
124
160
0 0 14
00 0 0 1
10
3110
=
89/965/241/4
0
N1 P xk =
14
124
1396
1240
0 16
124
160
0 0 14
00 0 0 1
10
.
0 0 0 03 0 0 01 2 0 04 3 2 0
xkykzkwk
xk+1yk+1zk+1wk+1
=
0,9271 0,0063xk 0,2583yk 0,0083zk0,2083 0,5250xk 0,0333kk 0,0333zk0,2500 0,2500xk 0,5000yk 0
0 0,400xk 0,300yk 0,200zk
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LATEX Ingenieria Mecatronica 5
k = 0;
x1 =y1 =z1 =w1 =
0,92710,20830,2500
0
k = 1;
x2 =y2 =z2 =w2 =
0,97300,48820,58590,4833
k = 2;
x3 =y3 =z3 =w3 =
0,79000,33830,24920,3599
k = 3;
x4 =y4 =z4 =w4 =
0,83270,22600,27840,2644
k = 4;
x5 =y5 =z5 =w5 =
0,86120,24570,34520,3210
k = 5;
x6 =y6 =z6 =w6 =
0,85530,26350,34250,3398
k = 6;
x7 =y7 =z7 =w7 =
0,85050,26090,33210,3316
k = 7;
x8 =y8 =z8 =w8 =
0,85150,25800,33220,3284
k = 8;
x9 =y9 =z9 =w9 =
0,85230,25840,33390,3296