В послідовності
x
0
,
x
1
,
x
2
,
…
,
x
n
{\displaystyle x_{0},x_{1},x_{2},\ldots ,x_{n}}
(m +1) разів і назвемо її
z
0
,
z
1
,
…
,
z
(
n
+
1
)
(
m
+
1
)
−
1
{\displaystyle z_{0},z_{1},\ldots ,z_{(n+1)(m+1)-1}}
і будемо рахувати розділені різниці для них. Хоча деякі з них будуть невизначенностями
z
i
=
z
i
+
1
⟹
f
[
z
i
,
z
i
+
1
]
=
f
(
z
i
+
1
)
−
f
(
z
i
)
z
i
+
1
−
z
i
=
0
0
{\displaystyle z_{i}=z_{i+1}\implies f[z_{i},z_{i+1}]={\frac {f(z_{i+1})-f(z_{i})}{z_{i+1}-z_{i}}}={\frac {0}{0}}}
Ці невизначеності замінимо на
f
′
(
z
i
)
{\displaystyle f'(z_{i})}
Різниці вищих порядків (j > 2) зі співпадаючими точками замінимо на похідні вищих порядків за правилом:
f
(
j
)
(
x
i
)
j
!
.
{\displaystyle {\frac {f^{(j)}(x_{i})}{j!}}.}
Наблизимо функцію
f
(
x
)
=
x
8
+
1
{\displaystyle f(x)=x^{8}+1}
x
∈
{
−
1
,
0
,
1
}
{\displaystyle x\in \{-1,0,1\}}
x f (x )f ′(x )f ″(x )
−1
2
−8
56
0
1
0
0
1
2
8
56
Потроїмо точки
{
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
6
)
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
7
)
1
=
8
z
8
=
1
f
[
z
8
]
=
2
{\displaystyle {\begin{array}{llcclrrrrr}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_{6})}{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_{7})}{1}}=8&&&&&&&\\z_{8}=1&f[z_{8}]=2&&&&&&&&\\\end{array}}}
і побудуємо многочлен
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
=
x
8
+
1.
{\displaystyle {\begin{aligned}P(x)&=2-8(x+1)+28(x+1)^{2}\\&\quad -21(x+1)^{3}+15x(x+1)^{3}\\&\quad -10x^{2}(x+1)^{3}+4x^{3}(x+1)^{3}\\&\quad -1x^{3}(x+1)^{3}(x-1)+x^{3}(x+1)^{3}(x-1)^{2}\\&=x^{8}+1.\end{aligned}}}
взявши коефіцієнти з діагоналі (зверху), домноживши їх на
∏
i
=
0
k
−
1
(
x
−
z
i
)
{\displaystyle \prod _{i=0}^{k-1}(x-z_{i})}
,&(x_{1},y_{1}'),&\ldots ,&(x_{n},y_{n}'),\\[1ex]\vdots &\vdots &&\vdots \\[1.2ex](x_{0},y_{0}^{(m)}),&(x_{1},y_{1}^{(m)}),&\ldots ,&(x_{n},y_{n}^{(m)})\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90f592c4abb74f8754fc911d6dbd075aa773d751)
![{\displaystyle z_{i}=z_{i+1}\implies f[z_{i},z_{i+1}]={\frac {f(z_{i+1})-f(z_{i})}{z_{i+1}-z_{i}}}={\frac {0}{0}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc3588ca4962f1f296e2cdea2c000c7ec2a8e66f)
![{\displaystyle {\begin{array}{llcclrrrrr}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_{6})}{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_{7})}{1}}=8&&&&&&&\\z_{8}=1&f[z_{8}]=2&&&&&&&&\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edcde38a7c93e88fab4e386bb863ffdf554f758f)
