Trabajo metodos numericos
-
Upload
carolina-castillo -
Category
Documents
-
view
6 -
download
2
description
Transcript of Trabajo metodos numericos
TALLER
1. Realice un mapa conceptual de los métodos iterativos empleados en la solución de ecuaciones diferenciales de valor inicial.
2. Plantee y solucione dos ejercicios sobre diferenciación numérica explicando paso a paso el procedimiento utilizado.
f ( x )=sin (x ) en x1=π4 y h=0.01
La fórmula para derivación numérica es: f ' (x1 )=f (x1+h )−f (x1 )
h
Así:
f ' (2 )=f ( π4 +0.01)−f ( π4 )
0.01=sin( π4 +0.01)−sin ( π4 )
0.01=0.714142−√2
20.01
=0.7035
f ( x )=x3 en x1=4 y h=0.001
f ' (4 )= f(4+0.001 )−f (4 )
0.001=4.001
3−43
0.001=64.048012−64
0.001=0.048012001
0.001=48.012001
3. Solucione el siguiente ejercicio utilizando la Regla del Trapecio, la Regla de Simpson 1/3 y la Regla de Simpson 3/8. Compare los resultados y haga un pequeño análisis.
Métodos iteracivos para E.D.O de V.I
Método de Euler Método de Taylor Método de Euler modificado
Método de Runge.Kutta de
cuarto orden
Método predictor-corrector
∫0
1x
2 x+1dx
Solución:
- Regla del trapecio:
∆ x=b−an
=1−04
=14=0,25
Veamos
∫a
b
f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]
Reemplazando
∫0
1x
2 x+1dx=1−0
2(4 )[ f (0 )+2 f (0,25 )+2 f (0,5 )+2 f (0,75 )+ f (1 ) ]
¿ 18 [0+2 0,25
2 (0,25 )+1+2 0,52 (0,5 )+1
+2 0,752 (0,75 )+1
+ 12 (1 )+1 ]
¿ 18 [0+13 + 1
2+ 35+13 ]
¿ 18 [ 5330 ]= 53
240=0,221
- Regla Simpson 1/3
Veamos
∫a
b
f ( x )dx=b−a3n [f (x0 )+3 f ( x1 )+3 f (x2 )+2 f ( x3 )+…+2 f (xn−2 )+4 f (xn−1 )+f (xn ) ]
Reemplazando
∫0
1x
2 x+1dx=1−0
3(4 ) [ f (x0 )+4 f (x1 )+2 f (x2 )+4 f (x3 )+ f (xn) ]
¿ 112
[ f (0 )+4 f (0,25 )+2 f (0,5)+4 f (0,75 )+ f (1 ) ]
¿ 112 [0+4 16+2 14 +4 3
10+ 13¿]
¿ 112 [ 2710 ]= 9
40=0,225
- Regla Simpson 3/8 Veamos
∫0
1x
2 x+1dx=
3(1−0)8(4) [f (x0 )+3 f (x1 )+3 f (x2 )+2 f (x3 )+ f (xn) ]
¿ 332
[ f (0 )+3 f (0,25 )+3 f (0,5)+2 f (0,75 )+f (1 ) ]
¿ 332 [0+3 16 +3 1
4+2 310
+ 13¿]
¿ 332 [ 13160 ]=131640=0,205
El valor real de esa integral es 0,225; luego podría decirse que los tres métodos son aproximados al valor real pero en este caso el mejor método el que más se aproxima es la Regla Simpson 1/3.
4. Solucione el siguiente ejercicio utilizando la regla del trapecio, la regla de Simpson 1/3 y la regla de Simpson 3/8. Compare los resultados y haga un pequeño análisis (dividiendo en 4 intervalos).
∫0
43√x exdx
REGLA DEL TRAPECIO:
∫a
b
f ( x )dx ≈ b−a2n
[ f ( x0 )+2 f (x1 )+2 f ( x2 )+…+2 f (xn−1 )+f (xn )]
x0=0 , x1=1 , x2=2 , x3=3 , x4=4
∫0
43√x exdx ≈ 4−0
2 (4 )[ 3√0e0+2 ( 3√1e1 )+2 ( 3√2e2)+2 ( 3√3e3 )+ 3√4 e4 ]=1
2[0+5.436563+18.619254+57.936713+86.669160 ]=84.330845
REGLA DE SIMPSON 1/3 Caso general
∫a
b
f ( x )dx ≈ h3 [ f (x0 )+4 f ( x1 )+2 f (x2 )+4 f (x3 )+2 f (x4 )+…+ f (xn) ]
h=b−an
h=4−04
=1
∫0
43√x exdx=1
3[ 3√0e0+4 ( 3√1e1)+2 ( 3√2e2 )+4 ( 3√3e3 )+ 3√4e4 ]=1
3[0+10.873127+18.619254+115.873428+86.669160 ]=232.034969
3=77.34498966
REGLA DE SIMPSON 3/8
∫a
b
f ( x )dx=3h8
¿ con h=b−a3
=43
x0=0 , x1=43, x2=
83, x3=4
∫0
43√x exdx=12
24 [ 3√0e0+3( 3√ 43 e4 /3)+3( 3√ 83 e83 )+ ( 3√4 e4 )]=12 [0+12.526415+59.8727837+86.6691608 ]=159.0683595
2=79.53417975
El valor real aproximado de esta integral es: 76.9621
Por lo tanto El método que más se aproxima es el de la regla de Simpson 1/3.
5. Solucione el siguiente ejercicio utilizando la integración de Romberg. Usando segmentos de longitud 1, 1/3 y 1/6.
∫0
1
ex2
dx
Solución: Para n=1Usamos la fórmula del trapecio, esto es;
∫a
b
f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]
I 1=∫0
1
ex2
dx=1−02(1) [ f (x0 )+ f (x1 ) ]
¿ 12
[ f (0 )+ f (1 ) ]
¿ 12
[e (0 )2+e(1)2 ]
¿ 12
[1+e ]
¿1,86
Para n=3
∆ x=b−an
=1−03
=13
Usamos la fórmula del trapecio, esto es;
∫a
b
f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]
I 2=∫0
1
ex2
dx=1−02(3) [ f (x0 )+2 f (x1 )+2 f (x2 )+f (x3 ) ]
¿ 16
[ f (0 )+2 f (1/3 )+2 f (2/3 )+ f (1 ) ]
¿ 16
[ e(0)2+2e(1/3)2+2e(2/3)2+e (1)2 ]
¿ 16
[1+2e 19+2e 49+e]¿1,512
Para n=6
∆ x=b−an
=1−06
=16
Usamos la fórmula del trapecio, esto es;
∫a
b
f ( x )dx=b−a2n [f (x0 )+2 f (x1 )+2 f (x2 )+…+2 f (xn−1)+ f (xn ) ]
I 3=∫0
1
ex2
dx=1−02(6) [f (x0 )+2 f (x1 )+2 f (x2 )+2 f (x3 )+2 f (x4 )+2 f ( x5 )+ f (x6)]
¿ 16
[ f (0 )+2 f (1/6 )+2 f (1/3 )+2 f (1/2 )+2 f (2 /3 )+2 f (5 /6 )+ f (1 ) ]
¿ 16
[ e(0)2+2e(1/6)2+2e(1/3)2+2e(1/2)2+2e(2/3 )2+2e(5 /6)2+e (1)2 ]
¿ 16
[1+2e 136+2e 19+2e14+2e 49+2e 2536+e]¿1,475
NIVELI
NIVELII
NIVELIII
I 1 I 4=43I 2−
13I 1
I 5=43I3−
13I2
1615I 5−
115I 4
I 2
I 3
NIVELI
NIVELII
NIVELIII
1,86I 4=
43
(1,512 )−13
(1,86 )=1,396
I 5=43
(1,475 )−13
(1,512 )=1,4626
1615
(1,4626 )− 115
(1,396 )=1,467041,512
1,475
6. Solucione el siguiente ejercicio utilizando la integración de Romberg. Usando segmentos de longitud 1, ½, 1/4, 1/8
∫1
2
ex ln ( x )dx
Formulas del trapecio: Cuando n=1 h2[ f (a )+ f (b )]
Cuando n=2,3 ,…,∞ h2[ f (a )+2∑
j=1
n−1
f (a+ jh )+ f (b )] y
h=limite superior−limite inferior
n
Segmentos: 1, ½, ¼, 1/8.
Cuando h=1 n=1
I 1=12
[e1ln (1 )+e2 ln (2 ) ]=12
[0+5.121703 ]=2.560851
Cuando h=1/2 n=2
I 2=14 [e1 ln (1 )+2(e1+
12 ln(1+12 ))+e2 ln (2 )]= 14 [0+3.634337+5.121703 ]=2.189010
Cuando h=1/4 n=4
I 3=18¿
Cuando h=1/8 n=8
I 4=116
¿
Nivel 1 Nivel 2 Nivel 3 Nivel 4I 1=2.560851 I 5=2.065063 I 8=2.0625862 I=2.062585184I 2=2.189010 I 6=2.062741 I 9=2.0625852I 3=2.0943085 I 7=2.062595I 4=2.070524
43I 2−
13I1=2.065063
43I 3−
13I2=2.062741
43I 4−
13I 3=2.062595
1615I 6−
115I 5=2.0625862
1615I 7−
115I 6=2.0625852
6463I 9−
163I 8=2.062585184
7. Usar el Método de Euler mejorado para aproximar y (2,3) tomando h=0.1 dada la ecuación diferencial
y '=2 x+ y−3y (2 )=1
x1=2+0.1=2.1
u1=1+0.1 (2 (2 )+1−3 )=1.2
y1=1+0.1( 12 ) [(2 (2 )+1−3 )+(2 (2.1 )+(1.2 )−3 ) ]=1.22
x2=2.1+0.1=2.2
u2=1.22+0.1 (2 (2.1 )+1.22−3 )=1.462
y2=1.22+0.1( 12 )[(2 (2.1 )+1.22−3 )+(2 (2.2 )+(1.462 )−3 ) ]=1.4841
x3=2.2+0.1=2.3
u3=1.4841+0.1 (2 (2.2 )+1.4841−3 )=1.77251
y3=1.4841+0.1 (12 )[(2 (2.2 )+1.4841−3 )+(2 (2.3 )+(1.77251 )−3 )]=1.7969305
x4=2.3+0.1=2.4
u4=1.797+0.1 (2 (2.3 )+1.797−3 )=2.1367
y4=1.797+0.1( 12 ) [(2 (2.3 )+1.797−3 )+(2 (2.4 )+(2.1367 )−3 )]=2.1637
x5=2.4+0.1=2.5
u5=2.1637+0.1 (2 (2.4 )+2.1637−3 )=2.56
y5=2.1637+0.1( 12 )[ (2 (2.4 )+2.1637−3 )+ (2 (2.5 )+ (2.56 )−3 )]=2.59
x6=2.5+0.1=2.6
u6=2.59+0.1 (2 (2.5 )+2.59−3 )=3.5
y6=2.59+0.1( 12 )[ (2 (2.5 )+2.59−3 )+(2 (2.6 )+(3.5 )−3 )]=3.1
x7=2.6+0.1=2.7
u7=3.1+0.1 (2 (2.6 )+3.1−3 )=3.63
y7=3.1+0.1(12 )[ (2 (2.6 )+3.1−3 )+ (2 (2.7 )+(3.63 )−3 )]=3.66
x8=2.7+0.1=2.8
u8=3.66+0.1 (2 (2.7 )+3.66−3 )=4.26
y8=3.66+0.1( 12 )[ (2 (2.7 )+3.66−3 )+(2 (2.8 )+(4.26 )−3 )]=4.3
x9=2.8+0.1=2.9
u9=4.3+0.1 (2 (2.8 )+4.3−3 )=5
y9=4.3+0.1( 12 ) [(2 (2.8 )+4.3−3 )+(2 (2.9 )+(5 )−3 )]=5.035
x10=2.9+0.1=3
u10=5.035+0.1 (2 (2.9 )+5.035−3 )=5.8185
y10=5.035+0.1(12 )[(2 (2.9 )+5.035−3 )+(2 (3 )+ (5.8185 )−3 )]=5.8676
8. Utilizar el método de Runge-Kutta para aproximar y (0.5 )
Dada la siguiente ecuación diferencial:
y '=2 xy
y (0 )=1
El método de Runge-Kutta de segundo orden se presenta con las siguientes expresiones:
k 1=f (xn , y n) ; k2=f (xn+h , yn+hk1 )
yn+1= yn+h2(k1+k2)
h=0.25 , x0=0 , y0=1k 1=f (x0 , y0 )=f (0,1 )=0
Para esta iteración x1=0.25k 2=f (x0+h , y0+hk1 )=f (0+0.25 ,1+0.25 (0 ) )= f (0.25 ,1 )=0.5
y1= y (0.25 )= y0+h2
(k 1+k2 )=1+ 0.252
(0+0.5 )=1716
=1.0625
La segunda iteración es:
k 1=f (x1 , y1 )=f (0.25 ,1.0625 )=1732
=0.53125
k 2=f (x1+h , y1+hk1 )=f (0.25+0.25 ,1.0625+0.25 (0.53125 ))=f (0.5 ,1.1953125 )=1.1953125
y2= y (0.5 )= y1+h2
(k1+k 2 )=1.0625+ 0.252
(0.53125+1.1953125 )=1.278320313
9. Hallar la solución aproximada que proporciona el Método de Adams-Bashorth de segundo, tercer y cuarto orden para la ecuación: h=0.1, x en [1,2]
y '=4−2 xy2
y (1 )=1- Segundo orden
Por el método de Euler
x0=1
y0=1
f 0=( 4−2 (1 )12 )=2
x1=1.1
y1=1+0.1( 4−2 (1 )12 )=1.2
f 1=( 4−2 (1.1 )1.22 )=1.25
Por método de Adam’s Bashorth
x3=0.3
y3=1.2+0.12
(3 (1.25 )−2 )=1.3
x4=0.4
y4=1.3+0.12
(3 (1.25 )−2 )=1.3875
x5=0.5
y5=1.3875+0.12
(3 (1.25 )−2 )=1.475
x6=0.6
y6=1.475+0.12
(3 (1.25 )−2 )=1.5625
x7=0.7
y7=1.5625+0.12
(3 (1.25 )−2 )=1.65
x8=0.8
y8=1.65+0.12
(3 (1.25 )−2 )=1.7375
x9=0.9
y9=1.7375+0.12
(3 (1.25 )−2 )=1.825
x10=1
y10=1.825+0.12
(3 (1.25 )−2 )=1.9
- Tercer orden
x2=1.2
y2=1.2+0.1( 4−2 (1.1 )(1.2)2 )=1.325
f 2=( 4−2 (1.2 )1.3252 )=0.911
Por método de Adam’s Bashorth
x3=1.3
y3=1.325+0.112
[23 (0.911)−16 (1.25 )+5 (2 ) ]=1.416275
x4=1.4
y4=1.416275+0.112
[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.50755
x5=1.5
y5=1.50755+0.112
[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.598825
x6=1.6
y6=1.5 98825+0.112
[23 (0.911)−16 (1.25 )+5 (2 ) ]=1.6901
x7=1.7
y7=1.6901+0.112
[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.781375
x8=1.8
y8=1.781375+0.112
[23 (0.911)−16 (1.25 )+5 (2 ) ]=1.87265
x9=1.9
y9=1.87265+0.112
[23 (0.911 )−16 (1.25 )+5 (2 ) ]=1.963925
x10=2
y10=1.963925+0.112
[23 (0.911 )−16 (1.25 )+5 (2 ) ]=2.0552
- Cuarto orden
x3=1.3
y3=1.325+0.1( 4−2 (1.2 )(1.325)2 )=1.42
f 3=( 4−2 (1.3 )1.422 )=0.695
Por método de Adam’s Bashorth
x4=0.4
y4=1.42+0.124
[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.473
x5=0.5
y5=1.473+0.124
[55 (0.695 )−59 (0.911)+37 (1.25 )−9 (2 ) ]=1.526
x6=0.6
y6=1.526+0.124
[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.58
x7=0.7
y7=1.58+0.124
[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.633
x8=0.8
y8=1.633+0.124
[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.686
x9=0.9
y9=1.686+0.124
[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.74
x10=1
y10=1.74+0.124
[55 (0.695 )−59 (0.911 )+37 (1.25 )−9 (2 ) ]=1.793