01 polinomios lagrange
TRANSCRIPT
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Numerical MethodMétodos Numéricos
Esteban Velilla [email protected]
Jaime Valencia [email protected]
Departamento de Ingeniería eléctrica
Universidad de Antioquia
Interpolación - Interpolation
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Interpolation
Interpolation is the process of defining a
function that takes on specified values at
specified points.
18/04/2013 MN - GIMEL 2
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18/04/2013 MN - GIMEL 3
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LAGRANGE’S INTERPOLATION
El método de Lagrange permite calcular el polinomio de n-ésimo orden
(como máximo) que conecta a n+1 puntos:
1
1
1
1
)(
N
i j j ji
j
i
i
N
i
i N
x x
x x x L
x f x L x f
18/04/2013 4MN - GIMEL
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LAGRANGE’S INTERPOLATION
Si se tienen 2 puntos, entonces, el polinomio interpolante será una línea
recta, orden 1.
2211
2
1
1 x f x L x f x L x f x L x f i
i
i
21
22
11 1
1 )( x x
x x
x x
x x x L
j j j
j
12
12
21 2
2 )( x x
x x
x x
x x x L
j j j
j
2
12
11
21
21 x f
x x
x x x f
x x
x x x f
18/04/2013 5MN - GIMEL
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LAGRANGE’S INTERPOLATION
Si se tiene 3 puntos, entonces el polinomo será de orden 2:
332211
3
1
2 x f x L x f x L x f x L x f x L x f i
i
i
31
3
21
23
11 1
1 )( x x
x x
x x
x x
x x
x x x L
j j j
j
32
3
12
13
21 2
2 )( x x x x
x x x x
x x x x x L
j j j
j
23
2
13
13
31 3
3 )(
x x
x x
x x
x x
x x
x x x L
j j j
j
3
23
2
13
12
32
3
12
11
31
3
21
22 x f
x x
x x
x x
x x x f
x x
x x
x x
x x x f
x x
x x
x x
x x x f
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Encontrar el polinomio interpolante que pasa por los siguientes 4 puntos.
x = [0 1 2 3]
y = [-5 -6 -1 16];
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Ejemplo
1
1
1
1
)( N
i j j ji
j
i
i
N
i
i N
x x
x x x L
x f x L x f
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Ejemplo de Aplicación
Si se tienen 4 puntos: {(−2,−6), (−1, 0), (1, 0), (2, 6)}
Encontrar el polinomio interpolante de orden 3.
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P(x)= 1 · x^3 + 0 · x^2 − 1 · x + 0 = x^3 - x
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The interpolation meaning
>> x=0:1:3; , y=[-5 -6 -1 16]; , u = -.25:.01:3.25;
>> v = polyinterp(x,y,u);, plot(x,y,'o',u,v,'-')
18/04/2013 MN - GIMEL 9
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Vandermonde Matrix
18/04/2013 MN - GIMEL 10
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Example - Vander
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-5-6
-1
16