Metodos numericos

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INSTITUTO TECNOLÓGICO DE MINATITLÁN

NOMBRE DE LOS INTEGRANTES DEL EQUIPO
ALEJANDRA ALVARADO GARCÍA
GABRIEL HUGO BORROMEO ALCÁNTARA

NOMBRE DEL MAESTRO
JOSE ANTONIO MOLINA CARRILLO

NOMBRE DE LA MATERIA
METODOS NUMERICOS

TRABAJO
UNIDAD III

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)...
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