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EJERCICIOS. DIFERENCIAS FINITAS
INTEGRANTES:
Narvaez Salas Edson
Valverde Flores Alma
Velazquez Lopez Mario Jordi
Vidals Ramos Iván Miguel
Villaseñor Leon Israel

1. Dados los siguientesdatos interpolar:
X | Y |
1 | 0.7651977 |
1.3 | 0.620086 |
1.6 | 0.4554022 |
1.9 | 0.2818186 |
2.2 | 0.1103623 |
Xk=1.5
Xk=1.1
Xk=2

X | Y | Δyi | Δ 2yi | Δ 3yi | Δ 4yi |1 | 0.7651977 | -0.1451117 | -0.0195721 | 0.0106723 | 0.0003548 |
1.3 | 0.620086 | -0.1646838 | -0.0088998 | 0.0110271 |   |
1.6 | 0.455402 | -0.1735836 | 0.0021273 |   |   |
1.9 | 0.281819 |-0.1714563 |   |   |   |
2.2 | 0.110362 |   |   |   |   |

Xk=1.5(Avanzada de Newton)

Yk= y0 +ka0+k(k-1) (b0)+ k(k-1) (k-2) (c0) , K= XK-X0 ,K=1.5-1.3= 2
2!3! h 0.3 3

Yk= (0.620086)+(2/3)( -0.1646838)+ (2/3)(2/3-1)( -0.0088998)+ (2/3)(2/3-1)(2/3-2)( 0.0110271)2! 3!

Yk=0.5118302


Xk=1.1(Avanzada de Newton)
K= XK-X0 K=1.1-1= 1
h 0.3 3Yk= (0.7651977)+(1/3)( -0.1451117)+ (1/3)(1/3-1)( -0.0195721 )+ (1/3)(1/3-1)(1/3-2)( 0.0106723)+ (1/3)(1/3-1)(1/3-2)(1/3-3)( 0.0003548)2! 3! 4!
Yk= 0.7196459

Xk=2 (Retrasada de Newton)
K= X0-Xk K=2.2-2= 2
h0.3 3

Yk= (0.110362)+(2/3)( -0.1714563)+ (2/3)(2/3-1)( 0.0021273 )+ (2/3)(2/3-1)(2/3-2)( 0.0110271)+ (2/3)(2/3-1)(2/3-2)(2/3-3)( 0.0003548)2! 3! 4!

Yk=0.0036439

2. La siguiente tabla proporciona las presiones de vapor en...
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