# Metodos numericos

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INSTITUTO TECNOLÓGICO DE MINATITLÁN

NOMBRE DE LOS INTEGRANTES DEL EQUIPO
GABRIEL HUGO BORROMEO ALCÁNTARA

NOMBRE DEL MAESTRO
JOSE ANTONIO MOLINA CARRILLO

NOMBRE DE LA MATERIA
METODOS NUMERICOS

TRABAJO

INTERPOLACIÓN LINEAL

* Calcule el PI Y PT cuando la presión del agua es de 6.44 kg/cm

Formulax=x1+x2-x1y2-x1(x-x1)

x=8+10-87-56.44-5

X=9.44 PI
y=y1+y2-y1x2-x1(x-x1)
y=4+7-410-8(9.44-8)
Y=6.16 PT/PSI
Agua kg/cm2 | PT/PSI | PI |
5 | 4 | 8 |
6.44 | 6.16 | 9.44 |
7 | 7 | 10 |

Problemas langrage interpolación

y=x-x2x-x3x-x4x-x5x1-x2x1-x3x1-x4x1-x5
y=(x-x2)(x-x3)(x-x4)(x-x5)(x2-x2)(x2-x3)(x2-x4)(x2-x5)y=(x-x2)(x-x3)(x-x4)(x-x5)(x3-x2)(x3-x3)(x3-x4)(x3-x5)
y=(x-x2)(x-x3)(x-x4)(x-x5)(x4-x2)(x4-x3)(x4-x4)(x4-x5)
y=(x-x2)(x-x3)(x-x4)(x-x5)(x5-x2)(x5-x3)(x5-x4)(x5-x5)

kg/cm2 | Pr/ PSI | PI |
0 | 4 | 0 |
1.95 | 7 | 10 |
3.7 | 9 | 30 |
5.38 | 11 | X
50 |
7.7 | 12.84 | 68.47 |
8 | 13 | 70 |

y=68.47-1068.47-3068.47-5068.47-700-100-300-500-70=-653507.984610500000=0y=68.47-068.47-3068.47-5068.47-7010-010-3010-5010-70=-765276.0682-4800001.95=3.108934027
y=68.47-068.47-1068.47-5068.47-7030-030-1030-5030-70=-1163132.0954800003.7=8.965809899
y=68.47-068.47-1068.47-3068.47-7050-050-1050-3050-70=-2376410.891-80000005.38=1.598136
y=68.47-068.47-1068.47-3068.47-5070-070-1070-3070-50=2844608.533600008=6.772877381
y=0+3.108934027+8.965809899+1.598136 +6.772877381=20.44577463y=68.47-1068.47-3068.47-5068.47-700-100-300-500-70=653507.984610500004=2.489554227
y=68.47-068.47-3068.47-5068.47-7010-010-3010-5010-70=-765276.0682-4800007=1.112359328
y=68.47-068.47-1068.47-5068.47-7030-030-1030-5030-70=-1163132.0954800009=-21.800872678
y=68.47-068.47-1068.47-3068.47-7050-050-1050-3050-70-2376410.891-800000011=3.267584975y=68.47-068.47-1068.47-3068.47-5070-070-1070-3070-50=2844608.5336000013=11.00592574

y=2.489554227 +1.112359328-21.800872678+3.267584975+11.0092574=3.9254484

Problema 3
X | F(x) | Interpolar |
2 | 0.69314718 | 0.648636716 |
3 | 1.098612289 | 0.716703787 |
4 | 1.386294361 | 0.84718956 |
5 | 1.609437912 | 0.92429024 |
6 | 1.791759469 | 0.648636716 |
7 | 1.945910149 | 0.716703787 |
Formula
f1x-f(x0)x-x0=f(x1)-f(x0)x-x0

f1x=f(x0)+f(x1)-f(x0)x1-x0(x1-x0)* Calcule el logaritmo de 3=x usando interpolación lineal.
| x0 | x1 |
A | 1 | 7 |
B | 1 | 6 |
C | 1 | 5 |
D | 1 | 4 |
X1 X
1 2 3 4 5 6 7

fx=0+1.945919149-07-13-1=0.648636716
fx=0+1.791759469-06-13-1=0.716703787
fx=0+1.609437912-05-13-1=0.84718956
fx=0+1.386294361-04-13-1=0.92429024

1.098612289100 = 59.04145826
0.648636716 x

1.098612289 100 = 65.3729006
0.716703787 x

1.098612289 100 = 77.11451697
0.84718956 x

1.098612289 100 = 84.12396705
0.92419624 x

INTEPORLACION CUBICA DE NEWTON

x | F(X) | X0 | num | x |
In 1 | 0 | 1 | 9 |X1 |
In2 | 0.69314718 | 1 | 8 | X2 |
In3 | 1.098612289 | 1 | 7 | X3 |
In5 | 1.609437912 | 1 | 6 | X4 |
In6 | 1.791759469 | 1 | 5 | X5 |
In7 | 1.945910149 | 1 | 3 | X6 |
In8 | 2.079441542 | 1 | 2 | X7 |
In9 | 2.197224577 | | | |

a=2.197224577-(0)9-1 = 0.274653072

b=2.079441542-(2.197224577)8-9= 0.117783035

c=(1.945910149) -2.0794415427-8 =0.133531393d=(1.791759469) -1.9459101496-7 =0.15415068

e=(1.609437912) -1.7917594695-6 =0.182321557

f=(1.098612289) -1.6094379123-5 =0.255412811

g=(0.69314718) -1.0986122892-3 =0.405465109

h=(0.117783035) -0.2746530728-1 =-0022410005

i=(0.133531393) -0.1177830357-9 =-0.007874179

j=(0.15415068) -0.1335313936-8 =0.010309643
k=(0.182321557) -0.154150685-7 =0.014085438

l=(0.255412811)...