En el análisis numérico , la interpolación de Hermite , nombrada así en honor a Charles Hermite , es un método de interpolación de puntos de datos como una función polinómica . El polinomio de Hermite generado está estrechamente relacionado con el polinomio de Newton , en tanto que ambos se derivan del cálculo de diferencias divididas.
Consiste en buscar un polinomio por pedazos
H
n
(
x
)
{\displaystyle H_{n}(x)}
[
x
(
i
−
1
)
,
x
i
]
,
1
≤
i
≤
n
{\displaystyle [x_{(i-1)},x_{i}],1\leq i\leq n}
f
′
(
x
)
{\displaystyle f'(x)}
{
x
0
,
.
.
.
,
x
n
}
{\displaystyle \{x_{0},...,x_{n}\}}
f
(
x
)
{\displaystyle f(x)}
función que se quiere interpolar.
La función
H
n
(
x
)
{\displaystyle H_{n}(x)}
n
{\displaystyle n}
sistemas lineales de ecuaciones de tamaño
4
×
4
{\displaystyle 4\times 4}
La desventaja de la interpolación de Hermite es que requiere de la disponibilidad de los
{
f
′
(
x
i
)
,
0
≤
i
≤
n
}
{\displaystyle \{f'(x_{i}),0\leq i\leq n\}}
Considerada la función
f
(
x
)
=
x
8
+
1
{\displaystyle f(x)=x^{8}+1}
x
∈
{
−
1
,
0
,
1
}
{\displaystyle x\in \{-1,0,1\}}
x ƒ (x )ƒ '(x )ƒ ''(x )
−1
2
−8
56
0
1
0
0
1
2
8
56
Puesto que tenemos dos derivadas para trabajar, construiremos el conjunto
{
z
i
}
=
{
−
1
,
−
1
,
−
1
,
0
,
0
,
0
,
1
,
1
,
1
}
{\displaystyle \{z_{i}\}=\{-1,-1,-1,0,0,0,1,1,1\}}
z
0
=
−
1
f
[
z
0
]
=
2
f
′
(
z
0
)
1
=
−
8
z
1
=
−
1
f
[
z
1
]
=
2
f
″
(
z
1
)
2
=
28
f
′
(
z
1
)
1
=
−
8
f
[
z
3
,
z
2
,
z
1
,
z
0
]
=
−
21
z
2
=
−
1
f
[
z
2
]
=
2
f
[
z
3
,
z
2
,
z
1
]
=
7
15
f
[
z
3
,
z
2
]
=
−
1
f
[
z
4
,
z
3
,
z
2
,
z
1
]
=
−
6
−
10
z
3
=
0
f
[
z
3
]
=
1
f
[
z
4
,
z
3
,
z
2
]
=
1
5
4
f
′
(
z
3
)
1
=
0
f
[
z
5
,
z
4
,
z
3
,
z
2
]
=
−
1
−
2
−
1
z
4
=
0
f
[
z
4
]
=
1
f
″
(
z
4
)
2
=
0
1
2
1
f
′
(
z
4
)
1
=
0
f
[
z
6
,
z
5
,
z
4
,
z
3
]
=
1
2
1
z
5
=
0
f
[
z
5
]
=
1
f
[
z
6
,
z
5
,
z
4
]
=
1
5
4
f
[
z
6
,
z
5
]
=
1
f
[
z
7
,
z
6
,
z
5
,
z
4
]
=
6
10
z
6
=
1
f
[
z
6
]
=
2
f
[
z
7
,
z
6
,
z
5
]
=
7
15
f
′
(
z
7
)
1
=
8
f
[
z
8
,
z
7
,
z
6
,
z
5
]
=
21
z
7
=
1
f
[
z
7
]
=
2
f
″
(
z
7
)
2
=
28
f
′
(
z
8
)
1
=
8
z
8
=
1
f
[
z
8
]
=
2
{\displaystyle {\begin{matrix}z_{0}=-1&f[z_{0}]=2&&&&&&&&\\&&{\frac {f'(z_{0})}{1}}=-8&&&&&&&\\z_{1}=-1&f[z_{1}]=2&&{\frac {f''(z_{1})}{2}}=28&&&&&&\\&&{\frac {f'(z_{1})}{1}}=-8&&f[z_{3},z_{2},z_{1},z_{0}]=-21&&&&&\\z_{2}=-1&f[z_{2}]=2&&f[z_{3},z_{2},z_{1}]=7&&15&&&&\\&&f[z_{3},z_{2}]=-1&&f[z_{4},z_{3},z_{2},z_{1}]=-6&&-10&&&\\z_{3}=0&f[z_{3}]=1&&f[z_{4},z_{3},z_{2}]=1&&5&&4&&\\&&{\frac {f'(z_{3})}{1}}=0&&f[z_{5},z_{4},z_{3},z_{2}]=-1&&-2&&-1&\\z_{4}=0&f[z_{4}]=1&&{\frac {f''(z_{4})}{2}}=0&&1&&2&&1\\&&{\frac {f'(z_{4})}{1}}=0&&f[z_{6},z_{5},z_{4},z_{3}]=1&&2&&1&\\z_{5}=0&f[z_{5}]=1&&f[z_{6},z_{5},z_{4}]=1&&5&&4&&\\&&f[z_{6},z_{5}]=1&&f[z_{7},z_{6},z_{5},z_{4}]=6&&10&&&\\z_{6}=1&f[z_{6}]=2&&f[z_{7},z_{6},z_{5}]=7&&15&&&&\\&&{\frac {f'(z_{7})}{1}}=8&&f[z_{8},z_{7},z_{6},z_{5}]=21&&&&&\\z_{7}=1&f[z_{7}]=2&&{\frac {f''(z_{7})}{2}}=28&&&&&&\\&&{\frac {f'(z_{8})}{1}}=8&&&&&&&\\z_{8}=1&f[z_{8}]=2&&&&&&&&\\\end{matrix}}}
y el polinomio generado es
P
(
x
)
=
2
−
8
(
x
+
1
)
+
28
(
x
+
1
)
2
−
21
(
x
+
1
)
3
+
15
x
(
x
+
1
)
3
−
10
x
2
(
x
+
1
)
3
+
4
x
3
(
x
+
1
)
3
−
1
x
3
(
x
+
1
)
3
(
x
−
1
)
+
x
3
(
x
+
1
)
3
(
x
−
1
)
2
=
2
−
8
+
28
−
21
−
8
x
+
56
x
−
63
x
+
15
x
+
28
x
2
−
63
x
2
+
45
x
2
−
10
x
2
−
21
x
3
+
45
x
3
−
30
x
3
+
4
x
3
+
x
3
+
x
3
+
15
x
4
−
30
x
4
+
12
x
4
+
2
x
4
+
x
4
−
10
x
5
+
12
x
5
−
2
x
5
+
4
x
5
−
2
x
5
−
2
x
5
−
x
6
+
x
6
−
x
7
+
x
7
+
x
8
=
x
8
+
1.
{\displaystyle {\begin{aligned}P(x)&=2-8(x+1)+28(x+1)^{2}-21(x+1)^{3}+15x(x+1)^{3}-10x^{2}(x+1)^{3}\\&\quad {}+4x^{3}(x+1)^{3}-1x^{3}(x+1)^{3}(x-1)+x^{3}(x+1)^{3}(x-1)^{2}\\&=2-8+28-21-8x+56x-63x+15x+28x^{2}-63x^{2}+45x^{2}-10x^{2}-21x^{3}\\&\quad {}+45x^{3}-30x^{3}+4x^{3}+x^{3}+x^{3}+15x^{4}-30x^{4}+12x^{4}+2x^{4}+x^{4}\\&\quad {}-10x^{5}+12x^{5}-2x^{5}+4x^{5}-2x^{5}-2x^{5}-x^{6}+x^{6}-x^{7}+x^{7}+x^{8}\\&=x^{8}+1.\end{aligned}}}
mediante la adopción de los coeficientes de la diagonal de la tabla de diferencia dividida, y multiplicando el k -ésimo coeficiente por
∏
i
=
0
k
−
1
(
x
−
z
i
)
{\displaystyle \prod _{i=0}^{k-1}(x-z_{i})}
![{\displaystyle [x_{(i-1)},x_{i}],1\leq i\leq n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92e26f088e33ec1505fa89a81d71b7e9f7f12a90)
![{\displaystyle {\begin{matrix}z_{0}=-1&f[z_{0}]=2&&&&&&&&\\&&{\frac {f'(z_{0})}{1}}=-8&&&&&&&\\z_{1}=-1&f[z_{1}]=2&&{\frac {f''(z_{1})}{2}}=28&&&&&&\\&&{\frac {f'(z_{1})}{1}}=-8&&f[z_{3},z_{2},z_{1},z_{0}]=-21&&&&&\\z_{2}=-1&f[z_{2}]=2&&f[z_{3},z_{2},z_{1}]=7&&15&&&&\\&&f[z_{3},z_{2}]=-1&&f[z_{4},z_{3},z_{2},z_{1}]=-6&&-10&&&\\z_{3}=0&f[z_{3}]=1&&f[z_{4},z_{3},z_{2}]=1&&5&&4&&\\&&{\frac {f'(z_{3})}{1}}=0&&f[z_{5},z_{4},z_{3},z_{2}]=-1&&-2&&-1&\\z_{4}=0&f[z_{4}]=1&&{\frac {f''(z_{4})}{2}}=0&&1&&2&&1\\&&{\frac {f'(z_{4})}{1}}=0&&f[z_{6},z_{5},z_{4},z_{3}]=1&&2&&1&\\z_{5}=0&f[z_{5}]=1&&f[z_{6},z_{5},z_{4}]=1&&5&&4&&\\&&f[z_{6},z_{5}]=1&&f[z_{7},z_{6},z_{5},z_{4}]=6&&10&&&\\z_{6}=1&f[z_{6}]=2&&f[z_{7},z_{6},z_{5}]=7&&15&&&&\\&&{\frac {f'(z_{7})}{1}}=8&&f[z_{8},z_{7},z_{6},z_{5}]=21&&&&&\\z_{7}=1&f[z_{7}]=2&&{\frac {f''(z_{7})}{2}}=28&&&&&&\\&&{\frac {f'(z_{8})}{1}}=8&&&&&&&\\z_{8}=1&f[z_{8}]=2&&&&&&&&\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/751e0b3ad317c2c01256fc99f6b192f1cf99a4ae)
Datos: Q44451