UNIDAD 1. INTERPOLACIÓN
UNIDAD 1 INTERPOLACIÓN 1. Introducción. La idea básica de la interpolación es hallar un polinomio o función que cumpla con pasar por todos los puntos de datos
, , , , … , ,
, y poder estimar los valores
entre ellos por medio de un polinomio.
2. Interpolación de Lagrange.
Para un conjunto de puntos, el polinomio de Lagrange es:
donde
, , ∑∏= =≠
( − ) = = ( − ) ≠
Coordenadas del punto interpolado.
Número de puntos empleados en el proceso. El grado del polinomio es
usual es emplear entre tres y cinco puntos.
−1
y lo más
Sirven para enumerar los puntos conocidos y los términos en la sumatoria y en el
producto.
Sumatoria que representa la suma de los términos colocados a su derecha. Indica multiplicación.
La expresión anterior se puede reescribir como
donde
= − = − ≠
es llamado polinomios fundamentales fundamentales de Lagrange.
M. GUERRERO RODÍGUEZ
1
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UNIDAD 1. INTERPOLACIÓN Actividad 1. Escriba los polinomios de Lagrange para dos puntos, tres puntos y cuatro
puntos. Ejemplo 1. Dados los puntos (9.0, 2.1972), (9.5, 2.2513), (11.0, 2.3979), obtener un
polinomio cuadrático,
valores de . Calcular el valor de Solución.
, al introducir el vector u de los valores de , el vector v de de los , cuando
9. 2
.
La respuesta se presenta en Excel. x = u
y =v
9 9.5 11
2.1972 2.2513 2.3979
La gráfica de los valores anteriores, se muestra a continuación, nótese que esta gráfica no es totalmente una línea recta.
2.45 2.4 2.35 y
2.3 2.25 2.2 2.15 8.5
9
9.5
10
10.5
x
Un polinomio de segundo orden tiene la forma:
11
11.5
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UNIDAD 1. INTERPOLACIÓN
9, 2 . 1 972 972 → 2. 1 972 9 9 9 81 9.11,5,22.2.3513 979 →→ 2.2.2513 513979 9.5 9.5 9.5 90.25 3979 1111 11 11 121 La solución del sistema de ecuaciones anterior se presenta enseguida, el cual es resuelto en Excel.
1 = 1 1
9 9.5 11
81 90.25 121
→
=
1
9
81
1
9.5
90.25
1
11
121
104.5
-132
28.5
-20.5 26.66 26.66666 66667 67 -6.1666666 -6.16666667 7 1 -1.33333333 -1.33333333 0.333 0.333333 33333 33
2.1972 = 2.2513 2.3979 =
2.1972
→
2.2513 2.3979 0.77595
→ 0.20501667 -0.00523333
El polinomio de segundo orden queda de la forma siguiente:
..−. La gráfica del polinomio, en Geogebra, se muestra enseguida:
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UNIDAD 1. INTERPOLACIÓN
9. 2 9.2 0.775950.20501667 05016679.2 −0.00523333 05233339.2
Evaluando el polinomio para
p(9.2) =
:
2.219154
Ejemplo 2. Dados los valores de la siguiente tabla
1
2
1.54
1.5
3
5
1.42 0.66
Encontrar el polinomio de interpolación de Lagrange. Solución.
El diagrama de dispersión de estos puntos se presenta a continuación en Geogebra
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UNIDAD 1. INTERPOLACIÓN Polinomios fundamentales de Lagrange.
−−−−−− 1 −− 2211 −− 3311 −− 55 − 18 −10 31−30 31−30 −−−−−− 2 −− 1122 −− 3322 −− 55 13 − 9 23−15 23−15 −−−−−− 3 −− 1133 −− 2233 −− 55 − 14 − 8 17−10 17−10 −−−−−− 5 −− 1155 −− 2255 −− 33 241 − 6 11−6 11−6 El polinomio de interpolación de Lagrange es
−. . −..
Sustituyendo en la ecuación anterior cada uno de los polinomios fundamentales de Lagrange y los valores de (proporcionados en la tabla), resulta que el polinomio tiene la forma siguiente:
La gráfica del polinomio anterior, en Geogebra, es:
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UNIDAD 1. INTERPOLACIÓN
A continuación se presenta la solución en wxMaxima, se obtiene el diagrama de dispersión, el polinomio de Lagrange y su representación gráfica.
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UNIDAD 1. INTERPOLACIÓN Diagrama de dispersión.
Gráfica del polinomio.
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UNIDAD 1. INTERPOLACIÓN Actividad 2. Resolver el ejemplo 1 haciendo uso de una interpolación de Lagrange.
r esolver un problema de física, para lo cual requiere conocer Actividad 3. Un ingeniero debe resolver la densidad del agua a 43.7 0 C con la mayor precisión posible y recurre a una tabla de datos de un texto:
t, 0C
Densidad, g/cm3
Volumen de 1 g/cm3
0
0.99987
1.00013
2
0.99997
1.00003
4
1.00000
1.00000
6
0.99997
1.00003
10
0.99973
1.00027
20
0.99823
1.00177
50
0.98807
1.01207
75
0.97489
1.02576
100
0.95838
1.04343
Empleé: a) Interpolación gráfica. b) Un polinomio de segundo grado:
c) Un polinomio de Lagrange de tres puntos.
3. Método de diferencias divididas (Polinomio de Interpolación de Newton). N ewton). Definición.
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UNIDAD 1. INTERPOLACIÓN el cual se conoce como polinomio de interpolación de Newton.
+, + +[ [, , … , +] = ( − )
Al agregar un nuevo par de interpolación
, el nuevo polinomio es
Ejemplo 3. Dada la siguiente tabla, obtener el polinomio de interpolación de Newton.
2
4
6
8
4
8
14
16
Solución.
El ejemplo es resuelto en Excel. x
f 2
4
2,4 = 4
8 2,4,6 =
8
0.25
3 2,4,6,8 2,4,6,8 = 14 4,6,8 = -0.5 6,8 = 1
4,6 = 6
2
-0.125
16
Polinomio de interpolación interpolación de N ewton.
= 4 + 2 − 2 + 0.25 − 2 − 4 − 0. 0.125 − 2 − 4 − 6
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UNIDAD 1. INTERPOLACIÓN -5
92.375
0
8
10
8
11
-1.625
20
-392
A continuación se presenta un programa en wxMaxima para resolver este ejemplo.
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UNIDAD 1. INTERPOLACIÓN
4. Interpolación con splines cúbicos. Una función spline está formada por varios polinomios, cada uno definido en un intervalo y que se unen bajo ciertas condiciones de continuidad. Se ha observado que las splines cúbicas son las más adecuadas. Dada una partición
{ < <⋯< }
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UNIDAD 1. INTERPOLACIÓN b) Condiciones de continuidad (en nodos interiores).
+ ++, 0, 1 , … , −2 ′′ +++,, 0, 0,11,,2…,…, ,−2−2 + + + 4−2
c) Condiciones de suavidad (en nodos interiores).
Nótese que el número de ecuaciones es
, lo cual nos indica que para determinar el spline
de forma única es necesario tener dos condiciones adicionales. Las condiciones más habituales,
impuestas sobre los extremos del intervalo, son:
0, 0, spline cúbico naturalral, , , spline cúbico sujeto
Ejemplo 4. Construir un spline cúbico sujeto que se ajuste a los datos de la siguiente tabla
0
1
2
3
0
1/2
2
3/2
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UNIDAD 1. INTERPOLACIÓN
Modelos matemáticos del spline cúbico para la tabla de datos pr oporcionada.
si ∈[ [ ] ∈ 0, 0 , 1 sisi ∈[ ∈∈ [[1,2,12,,23]]
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UNIDAD 1. INTERPOLACIÓN De los intervalos se observa que las posibles discontinuidades de la derivada están en
2
.
1
y en
Las siguientes ecuaciones permiten hacer que la primera derivada sea continua:
3 2 3 2 12 4 12 4 [ ] 6 2 si ∈ 0, 0 , 1 [ ] si ∈[ ∈ 1, 1 , 2 ´´ 662 2 si ∈ [2,2,3]
Ecuaciones de la segunda derivada de
.
También, se observa que las posibles discontinuidades de la segunda derivada están en
2
.
Para que la segunda derivada sea continua se obtienen las ecuaciones siguientes:
6 2 6 2 → 3 3
1
y en
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UNIDAD 1. INTERPOLACIÓN
3 2 3 2 12 4 12 4 3 6 6 0.2 27 6 −1 3
Enseguida se tiene el sistema de ecuaciones en forma matricial. Matriz de coeficientes.
=
0
0
0
1
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
8
4
2
1
0
0
0
0
0
0
0
0
0
0
0
0
8
4
2
1
0
0
0
0
0
0
0
0
2277
9
3
1
3
2
1
0
-3
-2
-1
0
0
0
0
0
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UNIDAD 1. INTERPOLACIÓN Vector de términos independientes. 0 0.5 0.5 2 2 =
1.5 0 0 0 0 0.2 -1
Coeficientes del polinomio. a1 =
0.48
b1 =
- 0. 18
c1 =
0.2
d1 =
0
a2 =
- 1. 04
b2=
4.38
c2 =
- 4. 36
d2 =
1.52
a3 =
0.68
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UNIDAD 1. INTERPOLACIÓN x
s0( x )
s1(x )
0
0
0.2
0.03664
0.4
0.08192
0.6
0.15888
0.8
0.29056
1
0.5
s2(x )
0.5
1.2
0.79808
1.4
1.14704
1.6
1.49696
1.8
1.79792
2
2
22
2 2 06704
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UNIDAD 1. INTERPOLACIÓN
, , , , … , , ++ −− , ++ −− −
Para el siguiente conjunto de datos
, entre dos puntos consecutivos del
conjunto de datos se tiene una recta cuya pendiente es
Y que pasa por el conjunto inicial par de puntos es
, entonces la ecuación de la recta que interpola entre ese
Nótese que la interpolación lineal se hace por pedazos y no entrega un solo polinomio para todo el
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UNIDAD 1. INTERPOLACIÓN Actividad 4. Empleando interpolación lineal calcule el valor del logaritmo de 4.7 . Use los datos
siguientes: Núm.
Logaritmo10
3
0.477121
4.5
0.653212
5
0.698970
a) Interpolar entre 3 y 5. b) Interpolar entre 4.5 y 5. Calcular el error relativo porcentual para ambos incisos, basado en el valor verdadero,