Newton-interpolation
Vad?
Diskret mängd datapunkter
(x_0, y_0), (x_1, y_1), ..., (x_n, y_n)
(
x
0
,
y
0
)
,
(
x
1
,
y
1
)
,
.
.
.
,
(
x
n
,
y
n
)
(x_0, y_0), (x_1, y_1), ..., (x_n, y_n)
(
x
0
,
y
0
)
,
(
x
1
,
y
1
)
,
.
.
.
,
(
x
n
,
y
n
)
Kvadratiska polynom
Spline-interpolation
Newton-interpolation
Linjär interpolation
Polynominterpolation
f_{n-1}(x)=b_1+b_2(x-x_1)+b_3(x-x_1)(x-x_2)+ ...+b_n(x-x_1)(x-x_2)...(x-x_{n-1})
f
n
−
1
(
x
)
=
b
1
+
b
2
(
x
−
x
1
)
+
b
3
(
x
−
x
1
)
(
x
−
x
2
)
+
.
.
.
+
b
n
(
x
−
x
1
)
(
x
−
x
2
)
.
.
.
(
x
−
x
n
−
1
)
f_{n-1}(x)=b_1+b_2(x-x_1)+b_3(x-x_1)(x-x_2)+ ...+b_n(x-x_1)(x-x_2)...(x-x_{n-1})
f
n
−
1
(
x
)
=
b
1
+
b
2
(
x
−
x
1
)
+
b
3
(
x
−
x
1
)
(
x
−
x
2
)
+
.
.
.
+
b
n
(
x
−
x
1
)
(
x
−
x
2
)
.
.
.
(
x
−
x
n
−
1
)
b_1=f(x_1)
b
1
=
f
(
x
1
)
b_1=f(x_1)
b
1
=
f
(
x
1
)
b_3=f[x_3,x_2,x_1]
b
3
=
f
[
x
3
,
x
2
,
x
1
]
b_3=f[x_3,x_2,x_1]
b
3
=
f
[
x
3
,
x
2
,
x
1
]
b_2=f[x_2,x_1]
b
2
=
f
[
x
2
,
x
1
]
b_2=f[x_2,x_1]
b
2
=
f
[
x
2
,
x
1
]
b_n=f[x_n,x_{n-1},...,x_2,x_1]
b
n
=
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
,
x
1
]
b_n=f[x_n,x_{n-1},...,x_2,x_1]
b
n
=
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
,
x
1
]
Vad?
f[x_i,x_j]=\frac{f(x_i)-f(x_j)}{x_i-x_j}
f
[
x
i
,
x
j
]
=
f
(
x
i
)
−
f
(
x
j
)
x
i
−
x
j
f[x_i,x_j]=\frac{f(x_i)-f(x_j)}{x_i-x_j}
f
[
x
i
,
x
j
]
=
x
i
−
x
j
f
(
x
i
)
−
f
(
x
j
)
f[x_i,x_j,x_k]=\frac{f[x_i,x_j]-f[x_j,x_k]}{x_i-x_k}
f
[
x
i
,
x
j
,
x
k
]
=
f
[
x
i
,
x
j
]
−
f
[
x
j
,
x
k
]
x
i
−
x
k
f[x_i,x_j,x_k]=\frac{f[x_i,x_j]-f[x_j,x_k]}{x_i-x_k}
f
[
x
i
,
x
j
,
x
k
]
=
x
i
−
x
k
f
[
x
i
,
x
j
]
−
f
[
x
j
,
x
k
]
f[x_n,x_{n-1},...,x_2,x_1]=\frac{f[x_n,x_{n-1},...,x_2]-f[x_{n-1},x_{n-2},...,x_1]}{x_n-x_1}
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
,
x
1
]
=
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
]
−
f
[
x
n
−
1
,
x
n
−
2
,
.
.
.
,
x
1
]
x
n
−
x
1
f[x_n,x_{n-1},...,x_2,x_1]=\frac{f[x_n,x_{n-1},...,x_2]-f[x_{n-1},x_{n-2},...,x_1]}{x_n-x_1}
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
,
x
1
]
=
x
n
−
x
1
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
]
−
f
[
x
n
−
1
,
x
n
−
2
,
.
.
.
,
x
1
]
f_{n-1}(x)=f(x_1)+(x-x_1)f[x_2,x_1]+(x-x_1)(x-x_2)f[x_3,x_2,x_1]
f
n
−
1
(
x
)
=
f
(
x
1
)
+
(
x
−
x
1
)
f
[
x
2
,
x
1
]
+
(
x
−
x
1
)
(
x
−
x
2
)
f
[
x
3
,
x
2
,
x
1
]
f_{n-1}(x)=f(x_1)+(x-x_1)f[x_2,x_1]+(x-x_1)(x-x_2)f[x_3,x_2,x_1]
f
n
−
1
(
x
)
=
f
(
x
1
)
+
(
x
−
x
1
)
f
[
x
2
,
x
1
]
+
(
x
−
x
1
)
(
x
−
x
2
)
f
[
x
3
,
x
2
,
x
1
]
+...+(x-x_1)(x-x_2)...(x-x_{n-1})f[x_n,x_{n-1},...,x_2,x_1]
+
.
.
.
+
(
x
−
x
1
)
(
x
−
x
2
)
.
.
.
(
x
−
x
n
−
1
)
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
,
x
1
]
+...+(x-x_1)(x-x_2)...(x-x_{n-1})f[x_n,x_{n-1},...,x_2,x_1]
+
.
.
.
+
(
x
−
x
1
)
(
x
−
x
2
)
.
.
.
(
x
−
x
n
−
1
)
f
[
x
n
,
x
n
−
1
,
.
.
.
,
x
2
,
x
1
]
Image source: https://en.wikipedia.org/wiki/Newton_polynomial
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Hur?
f(x) = 2x^2
f
(
x
)
=
2
x
2
f(x) = 2x^2
f
(
x
)
=
2
x
2
f(x) = sin(x)
f
(
x
)
=
s
i
n
(
x
)
f(x) = sin(x)
f
(
x
)
=
s
i
n
(
x
)
f(x) = \frac{1}{1 + 25x^2}
f
(
x
)
=
1
1
+
2
5
x
2
f(x) = \frac{1}{1 + 25x^2}
f
(
x
)
=
1
+
2
5
x
2
1
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