Dati
k
+
1
{\displaystyle k+1}
(
x
0
,
y
0
)
,
…
,
(
x
k
,
y
k
)
.
{\displaystyle (x_{0},y_{0}),\ldots ,(x_{k},y_{k}).}
Definiamo le differenze divise come:
[
y
ν
]
:=
y
ν
,
ν
∈
{
0
,
…
,
k
}
,
{\displaystyle [y_{\nu }]:=y_{\nu },\qquad \nu \in \{0,\ldots ,k\},}
[
y
ν
,
…
,
y
ν
+
j
]
:=
[
y
ν
+
1
,
…
,
y
ν
+
j
]
−
[
y
ν
,
…
,
y
ν
+
j
−
1
]
x
ν
+
j
−
x
ν
,
ν
∈
{
0
,
…
,
k
−
j
}
,
j
∈
{
1
,
…
,
k
}
.
{\displaystyle [y_{\nu },\ldots ,y_{\nu +j}]:={\frac {[y_{\nu +1},\ldots ,y_{\nu +j}]-[y_{\nu },\ldots ,y_{\nu +j-1}]}{x_{\nu +j}-x_{\nu }}},\qquad \nu \in \{0,\ldots ,k-j\},\ j\in \{1,\ldots ,k\}.}
Definiamo le differenze divise all'indietro come:
[
y
ν
]
:=
y
ν
,
ν
∈
{
0
,
…
,
k
}
{\displaystyle [y_{\nu }]:=y_{\nu },\qquad \nu \in \{0,\ldots ,k\}}
[
y
ν
,
…
,
y
ν
−
j
]
:=
[
y
ν
,
…
,
y
ν
−
j
+
1
]
−
[
y
ν
−
1
,
…
,
y
ν
−
j
]
x
ν
−
x
ν
−
j
,
ν
∈
{
j
,
…
,
k
}
,
j
∈
{
1
,
…
,
k
}
.
{\displaystyle [y_{\nu },\ldots ,y_{\nu -j}]:={\frac {[y_{\nu },\ldots ,y_{\nu -j+1}]-[y_{\nu -1},\ldots ,y_{\nu -j}]}{x_{\nu }-x_{\nu -j}}},\qquad \nu \in \{j,\ldots ,k\},\ j\in \{1,\ldots ,k\}.}
dove
j
{\displaystyle j}
ordine della differenza divisa.
Se i punti
{
x
0
,
x
1
,
…
,
x
k
}
{\textstyle \{x_{0},x_{1},\dots ,x_{k}\}}
f
{\displaystyle f}
(
x
0
,
f
(
x
0
)
)
,
…
,
(
x
k
,
f
(
x
k
)
)
,
{\displaystyle (x_{0},f(x_{0})),\ldots ,(x_{k},f(x_{k})),}
si può trovare la notazione
f
[
x
ν
]
:=
f
(
x
ν
)
,
ν
∈
{
0
,
…
,
k
}
,
{\displaystyle f[x_{\nu }]:=f(x_{\nu }),\qquad \nu \in \{0,\ldots ,k\},}
f
[
x
ν
,
…
,
x
ν
+
j
]
:=
f
[
x
ν
+
1
,
…
,
x
ν
+
j
]
−
f
[
x
ν
,
…
,
x
ν
+
j
−
1
]
x
ν
+
j
−
x
ν
,
ν
∈
{
0
,
…
,
k
−
j
}
,
j
∈
{
1
,
…
,
k
}
.
{\displaystyle f[x_{\nu },\ldots ,x_{\nu +j}]:={\frac {f[x_{\nu +1},\ldots ,x_{\nu +j}]-f[x_{\nu },\ldots ,x_{\nu +j-1}]}{x_{\nu +j}-x_{\nu }}},\qquad \nu \in \{0,\ldots ,k-j\},\ j\in \{1,\ldots ,k\}.}
Altre scritture equivalenti sono:
[
x
0
,
…
,
x
n
]
f
;
{\displaystyle [x_{0},\ldots ,x_{n}]f;}
f
n
[
x
0
,
…
,
x
n
]
;
{\displaystyle f_{n}[x_{0},\ldots ,x_{n}];}
[
x
0
,
…
,
x
n
;
f
]
;
{\displaystyle [x_{0},\ldots ,x_{n};f];}
D
[
x
0
,
…
,
x
n
]
f
.
{\displaystyle D[x_{0},\ldots ,x_{n}]f.}
Quando due argomenti risultano coincidenti possiamo ugualmente dare un significato alla corrispondente differenza divisa di ordine
1
{\displaystyle 1}
f
′
(
x
)
{\displaystyle f'(x)}
[ 1]
f
[
x
0
,
x
0
]
=
lim
x
→
x
0
f
[
x
0
,
x
]
=
lim
x
→
x
0
f
(
x
)
−
f
(
x
0
)
x
−
x
0
=
f
′
(
x
0
)
.
{\displaystyle f[x_{0},x_{0}]=\lim _{x\to x_{0}}f[x_{0},x]=\lim _{x\to x_{0}}{\frac {f(x)-f(x_{0})}{x-x_{0}}}=f'(x_{0}).}
Più in generale, definiamo
f
[
x
0
,
x
0
,
…
,
x
0
⏟
k
+
1
]
=
f
(
k
)
(
x
0
)
k
!
,
{\displaystyle f[\underbrace {x_{0},x_{0},\dots ,x_{0}} _{k+1}]={\frac {f^{(k)}(x_{0})}{k!}},}
la cui esistenza è dimostrabile[ 2]
Differenze divise per
ν
=
0
{\displaystyle \nu =0}
j
{\displaystyle j}
[
y
0
]
=
y
0
;
[
y
0
,
y
1
]
=
y
1
−
y
0
x
1
−
x
0
;
[
y
0
,
y
1
,
y
2
]
=
[
y
1
,
y
2
]
−
[
y
0
,
y
1
]
x
2
−
x
0
=
y
2
−
y
1
x
2
−
x
1
−
y
1
−
y
0
x
1
−
x
0
x
2
−
x
0
=
y
2
−
y
1
(
x
2
−
x
1
)
(
x
2
−
x
0
)
−
y
1
−
y
0
(
x
1
−
x
0
)
(
x
2
−
x
0
)
;
[
y
0
,
y
1
,
y
2
,
y
3
]
=
[
y
1
,
y
2
,
y
3
]
−
[
y
0
,
y
1
,
y
2
]
x
3
−
x
0
.
{\displaystyle {\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0};\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}};\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {[y_{1},y_{2}]-[y_{0},y_{1}]}{x_{2}-x_{0}}}={\frac {{\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}-{\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}}{x_{2}-x_{0}}}={\frac {y_{2}-y_{1}}{(x_{2}-x_{1})(x_{2}-x_{0})}}-{\frac {y_{1}-y_{0}}{(x_{1}-x_{0})(x_{2}-x_{0})}};\\{\mathopen {[}}y_{0},y_{1},y_{2},y_{3}]&={\frac {[y_{1},y_{2},y_{3}]-[y_{0},y_{1},y_{2}]}{x_{3}-x_{0}}}.\end{aligned}}}
Per evidenziare il processo ricorsivo, le differenze divise possono essere messe in forma tabellare
x
0
y
0
=
[
y
0
]
[
y
0
,
y
1
]
x
1
y
1
=
[
y
1
]
[
y
0
,
y
1
,
y
2
]
[
y
1
,
y
2
]
[
y
0
,
y
1
,
y
2
,
y
3
]
x
2
y
2
=
[
y
2
]
[
y
1
,
y
2
,
y
3
]
[
y
2
,
y
3
]
x
3
y
3
=
[
y
3
]
{\displaystyle {\begin{matrix}x_{0}&y_{0}=[y_{0}]&&&\\&&[y_{0},y_{1}]&&\\x_{1}&y_{1}=[y_{1}]&&[y_{0},y_{1},y_{2}]&\\&&[y_{1},y_{2}]&&[y_{0},y_{1},y_{2},y_{3}]\\x_{2}&y_{2}=[y_{2}]&&[y_{1},y_{2},y_{3}]&\\&&[y_{2},y_{3}]&&\\x_{3}&y_{3}=[y_{3}]&&&\\\end{matrix}}}
Data una funzione
f
{\displaystyle f}
(
x
0
,
f
(
x
0
)
)
,
(
x
1
,
f
(
x
1
)
)
{\displaystyle (x_{0},f(x_{0})),(x_{1},f(x_{1}))}
1
{\displaystyle 1}
A
1
=
f
[
x
0
,
x
1
]
=
y
1
−
y
0
x
1
−
x
0
=
f
(
x
1
)
−
f
(
x
0
)
x
1
−
x
0
=
Δ
f
Δ
x
{\displaystyle A_{1}=f[x_{0},x_{1}]={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}={\frac {f(x_{1})-f(x_{0})}{x_{1}-x_{0}}}={\frac {\Delta f}{\Delta x}}}
è il rapporto incrementale costruito su due punti per la quantità
h
=
x
1
−
x
0
{\displaystyle h=x_{1}-x_{0}}
Per induzione matematica, non è difficile dimostrare che
f
[
x
0
,
x
1
,
…
,
x
n
]
=
∑
k
=
0
n
f
(
x
k
)
∏
j
=
0
,
j
≠
k
n
(
x
k
−
x
j
)
.
{\displaystyle f[x_{0},x_{1},\dots ,x_{n}]=\sum _{k=0}^{n}{\frac {f(x_{k})}{\prod _{j=0,\ j\neq k}^{n}{(x_{k}-x_{j})}}}.}
Questa espressione ci permette di affermare che
f
[
x
0
,
x
1
,
…
,
x
n
]
{\displaystyle f[x_{0},x_{1},\dots ,x_{n}]}
invariante a permutazione
f
[
x
0
,
x
1
,
…
,
x
n
]
=
f
[
x
i
0
,
x
i
1
,
…
,
x
i
n
]
,
{\displaystyle f[x_{0},x_{1},\dots ,x_{n}]=f[x_{i_{0}},x_{i_{1}},\dots ,x_{i_{n}}],}
dove
(
i
0
,
i
1
,
…
,
i
n
)
{\textstyle (i_{0},i_{1},\dots ,i_{n})}
permutazione di
(
0
,
1
,
…
,
n
)
{\textstyle (0,1,\dots ,n)}
[ 1]
1 2 Monegato, Giovanni., Metodi e algoritmi per il calcolo numerico ISBN 9788879922654 , OCLC 956017867 . URL consultato il 29 aprile 2019 . ↑ Isaacson, Eugene., Analysis of numerical methods 252, ISBN 0486680290 , OCLC 30032279 . URL consultato il 29 aprile 2019 .
![{\displaystyle [y_{\nu }]:=y_{\nu },\qquad \nu \in \{0,\ldots ,k\},}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/a715cb3e1f9666b594c6be04b6ebee8e80f7ee48)
![{\displaystyle [y_{\nu },\ldots ,y_{\nu +j}]:={\frac {[y_{\nu +1},\ldots ,y_{\nu +j}]-[y_{\nu },\ldots ,y_{\nu +j-1}]}{x_{\nu +j}-x_{\nu }}},\qquad \nu \in \{0,\ldots ,k-j\},\ j\in \{1,\ldots ,k\}.}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/a3fbf429b65bad77486a9d90314d4af6e5edf0c1)
![{\displaystyle [y_{\nu }]:=y_{\nu },\qquad \nu \in \{0,\ldots ,k\}}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/f0b559f584afa31efe09d04b4be0e091807e66c6)
![{\displaystyle [y_{\nu },\ldots ,y_{\nu -j}]:={\frac {[y_{\nu },\ldots ,y_{\nu -j+1}]-[y_{\nu -1},\ldots ,y_{\nu -j}]}{x_{\nu }-x_{\nu -j}}},\qquad \nu \in \{j,\ldots ,k\},\ j\in \{1,\ldots ,k\}.}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/a9ca4b8abdebfbfaf509c340f8aa07f1508e8c8b)
![{\displaystyle f[x_{\nu }]:=f(x_{\nu }),\qquad \nu \in \{0,\ldots ,k\},}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/4e07542916cc075fb44964d3ca138cd144f3ab0a)
![{\displaystyle f[x_{\nu },\ldots ,x_{\nu +j}]:={\frac {f[x_{\nu +1},\ldots ,x_{\nu +j}]-f[x_{\nu },\ldots ,x_{\nu +j-1}]}{x_{\nu +j}-x_{\nu }}},\qquad \nu \in \{0,\ldots ,k-j\},\ j\in \{1,\ldots ,k\}.}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/424ee9e0456399b61b1ed8c66e90ca761fa7dc3e)
![{\displaystyle [x_{0},\ldots ,x_{n}]f;}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/f7ff10dbb6cfb92ba8bf42b2a1c51ee17fb82e58)
![{\displaystyle f_{n}[x_{0},\ldots ,x_{n}];}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/33f106376c754c889895ed56eecfcd021724eefb)
![{\displaystyle [x_{0},\ldots ,x_{n};f];}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/ca7e8489bcddcc6ff0370c8a70f61ed7cf64c426)
![{\displaystyle D[x_{0},\ldots ,x_{n}]f.}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/22c86cfbfca09860822cab7c1573654ae8187897)
![{\displaystyle f[x_{0},x_{0}]=\lim _{x\to x_{0}}f[x_{0},x]=\lim _{x\to x_{0}}{\frac {f(x)-f(x_{0})}{x-x_{0}}}=f'(x_{0}).}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/2c7cd2df9234af2b96200b2a77be954eb5f8f0c8)
![{\displaystyle f[\underbrace {x_{0},x_{0},\dots ,x_{0}} _{k+1}]={\frac {f^{(k)}(x_{0})}{k!}},}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/dfa3d4425549aacf2af9aada4f349d86bde20b41)
![{\displaystyle {\begin{aligned}{\mathopen {[}}y_{0}]&=y_{0};\\{\mathopen {[}}y_{0},y_{1}]&={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}};\\{\mathopen {[}}y_{0},y_{1},y_{2}]&={\frac {[y_{1},y_{2}]-[y_{0},y_{1}]}{x_{2}-x_{0}}}={\frac {{\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}-{\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}}{x_{2}-x_{0}}}={\frac {y_{2}-y_{1}}{(x_{2}-x_{1})(x_{2}-x_{0})}}-{\frac {y_{1}-y_{0}}{(x_{1}-x_{0})(x_{2}-x_{0})}};\\{\mathopen {[}}y_{0},y_{1},y_{2},y_{3}]&={\frac {[y_{1},y_{2},y_{3}]-[y_{0},y_{1},y_{2}]}{x_{3}-x_{0}}}.\end{aligned}}}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/642b1f928c20d2fb5b81453f58e681a3a9ef00de)
![{\displaystyle {\begin{matrix}x_{0}&y_{0}=[y_{0}]&&&\\&&[y_{0},y_{1}]&&\\x_{1}&y_{1}=[y_{1}]&&[y_{0},y_{1},y_{2}]&\\&&[y_{1},y_{2}]&&[y_{0},y_{1},y_{2},y_{3}]\\x_{2}&y_{2}=[y_{2}]&&[y_{1},y_{2},y_{3}]&\\&&[y_{2},y_{3}]&&\\x_{3}&y_{3}=[y_{3}]&&&\\\end{matrix}}}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/d60ef92b00038923edcedaecccbadd25373b07c1)
![{\displaystyle A_{1}=f[x_{0},x_{1}]={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}={\frac {f(x_{1})-f(x_{0})}{x_{1}-x_{0}}}={\frac {\Delta f}{\Delta x}}}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/375706add51292ed4fa69ee4fc2764e47e204485)
![{\displaystyle f[x_{0},x_{1},\dots ,x_{n}]=\sum _{k=0}^{n}{\frac {f(x_{k})}{\prod _{j=0,\ j\neq k}^{n}{(x_{k}-x_{j})}}}.}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/c5587eba42123b548060c650861413545c2f350f)
![{\displaystyle f[x_{0},x_{1},\dots ,x_{n}]}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/1272da335f766e083aad2a751b88c53423974d5b)
![{\displaystyle f[x_{0},x_{1},\dots ,x_{n}]=f[x_{i_{0}},x_{i_{1}},\dots ,x_{i_{n}}],}](/%20https://reactormag.com/five-https-wikimedia.org/api/rest_v1/media/math/render/svg/b69706751f8c0dfbe9ee02935aff04d420641d7b)
