Metodos numericos
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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