Ratkaisu
Etsitään siis pisteiden (x
0
,f(x
0
))=(
-
2,2), (x
1
,f(x
1
))=(
-
1,1), (x
2
,f(x
2
))=(1,1) ja (x
3
,f(x
3
))=(3,2) kautta kulkevaa 3. asteen polynomia P
3
(x) muodossa
P
3
(x)=a
0
+(x
-
x
0
)a
1
+(x
-
x
0
)(x
-
x
1
)a
2
+(x
-
x
0
)(x
-
x
1
)(x
-
x
2
)a
3
,
jolle pätee
P
3
(x
0
) = a
0
=f(x
0
):=f[x
0
],
P
3
(x
1
) = a
0
+a
1
(x
1
-
x
0
)=f(x
0
)+a
1
(x
1
-
x
0
)=f(x
1
)
Ž
a
1
=
f(x
1
)
-
f(x
0
)
x
1
-
x
0
:=f[x
0
,x
1
].
Edelleen
P
3
(x
2
)
=
a
0
+a
1
(x
2
-
x
0
)+a
2
(x
2
-
x
0
)(x
2
-
x
1
)
=
f(x
0
)+
f(x
1
)
-
f(x
0
)
x
1
-
x
0
(x
2
-
x
0
)+a
2
(x
2
-
x
0
)(x
2
-
x
1
)=f(x
2
)
Ž
a
2
=
f(x
2
)
-
f(x
1
)
x
2
-
x
1
-
f(x
1
)
-
f(x
0
)
x
1
-
x
0
x
2
-
x
0
=
f[x
1
,x
2
]
-
f[x
0
,x
1
]
x
2
-
x
0
:=f[x
0
,x
1
,x
2
].
Vastaavasti
P
3
(x
3
)
=
a
0
+a
1
(x
3
-
x
0
)+a
2
(x
3
-
x
0
)(x
3
-
x
1
)
+
a
3
(x
3
-
x
0
)(x
3
-
x
1
)(x
3
-
x
2
)=f(x
3
)
Ž
a
3
=
f[x
1
,x
2
,x
3
]
-
f[x
0
,x
1
,x
2
]
x
3
-
x
0
:=f[x
0
,x
1
,x
2
,x
3
].
Yleisesti pätee
f[x
0
,
¼
,x
n
]=
f[x
1
,
¼
,x
n
]
-
f[x
0
,
¼
,x
n
-
1
]
x
n
-
x
0
ja Newtonin
jaettujen erotusten
avulla muodostettu inteprolaatiopolynomi on muotoa
P
n
(x)=f[x
0
]+
n
å
k=1
f[x
0
,x
1
,
¼
,x
k
](x
-
x
0
)(x
-
x
1
)
¼
(x
-
x
k
-
1
).
Polynomin kertoimet a
k
=f[x
0
,x
1
,
¼
,x
k
] on tapana laskea jaettujen erotusten kaavio avulla
x
0
f[x
0
]
x
1
f[x
1
]
f[x
0
,x
1
]=
f[x
1
]
-
f[x
0
]
x
1
-
x
0
x
2
f[x
2
]
f[x
1
,x
2
]=
f[x
2
]
-
f[x
1
]
x
2
-
x
1
f[x
0
,x
1
,x
2
]=
f[x
1
,x
2
]
-
f[x
0
,x
1
]
x
2
-
x
0
x
3
f[x
3
]
f[x
2
,x
3
]=
f[x
3
]
-
f[x
2
]
x
3
-
x
2
f[x
1
,x
2
,x
3
]=
f[x
2
,x
3
]
-
f[x
1
,x
2
]
x
3
-
x
1
f[x
0
,x
1
,x
2
,x
3
]
=
f[x
1
,x
2
,x
3
]
-
f[x
0
,x
1
,x
2
]
x
3
-
x
0
Esimerkin tapauksessa kaavio saadaan muotoon
i
x
i
f[x
i
]
f[x
i
,x
i+1
]
f[x
i
,x
i+1
,x
i+2
]
f[x
i
,x
i+1
,x
i+2
,x
i+3
]
0
-2.0
2.0
1
-1.0
1.0
-1.0
2
1.0
1.0
0.0
1/3
3
3.0
2.0
1/2
1/8
-1/24
Poimimalla polynomin P
3
(x) kertoimet saadaan
P
3
(x)
=
2.0
-
1.0(x+2.0)+1/3(x+2.0)(x+1.0)
-
1/24(x+2.0)(x+1.0)(x
-
1.0)
=
3
4
+
1
24
x+
1
4
x
2
-
1
24
x
3
.
