The Expanded Form of the Definition
edit
The usual definition of divided differences is equivalent to the
Expanded form
f
[
x
0
,
x
1
…
,
x
n
]
=
∑
j
=
0
n
f
(
x
j
)
∏
k
∈
{
0
,
…
,
n
}
∖
{
j
}
(
x
j
−
x
k
)
{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{\prod _{k\in \{0,\dots ,n\}\setminus \{j\}}(x_{j}-x_{k})}}}
(expanded
With help of a polynomial functions
q
{\displaystyle q}
q
(
ξ
)
=
(
ξ
−
x
0
)
⋯
(
ξ
−
x
n
)
{\displaystyle q(\xi )=(\xi -x_{0})\cdots (\xi -x_{n})}
f
[
x
0
,
…
,
x
n
]
=
∑
j
=
0
n
f
(
x
j
)
q
′
(
x
j
)
.
{\displaystyle f[x_{0},\dots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{q'(x_{j})}}\,.}
Since we will need the Expanded form (expanded
Proof of the expanded form
edit
For
n
=
1
{\displaystyle n=1}
expanded
f
[
x
0
,
x
1
]
=
f
[
x
1
]
−
f
[
x
0
]
x
1
−
x
0
=
f
(
x
0
)
(
x
0
−
x
1
)
+
f
(
x
1
)
(
x
1
−
x
0
)
.
{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}={\frac {f(x_{0})}{(x_{0}-x_{1})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})}}\,.}
We now assume (expanded
n
{\displaystyle n}
n
+
1
{\displaystyle n+1}
n
{\displaystyle n}
If the formula
f
[
x
0
,
x
1
,
…
,
x
n
]
=
∑
j
=
0
n
f
(
x
j
)
q
′
(
x
j
)
{\displaystyle f[x_{0},x_{1},\ldots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{q'(x_{j})}}}
q
(
ξ
)
=
∏
k
=
0
n
(
ξ
−
x
k
)
{\displaystyle q(\xi )=\prod _{k=0}^{n}(\xi -x_{k})}
q
1
(
ξ
)
=
∏
k
=
1
n
+
1
(
ξ
−
x
k
)
,
{\displaystyle q_{1}(\xi )=\prod _{k=1}^{n+1}(\xi -x_{k}),}
q
2
(
ξ
)
=
∏
k
=
0
n
(
ξ
−
x
k
)
{\displaystyle q_{2}(\xi )=\prod _{k=0}^{n}(\xi -x_{k})}
Q
(
ξ
)
=
∏
k
=
0
n
+
1
(
ξ
−
x
k
)
{\displaystyle Q(\xi )=\prod _{k=0}^{n+1}(\xi -x_{k})}
f
[
x
0
,
x
1
,
…
,
x
n
+
1
]
=
f
[
x
1
,
…
,
x
n
+
1
]
−
f
[
x
0
,
…
,
x
n
]
x
n
+
1
−
x
0
=
1
x
n
+
1
−
x
0
(
∑
j
=
0
n
f
(
x
j
+
1
)
q
1
′
(
x
j
+
1
)
−
∑
j
=
0
n
f
(
x
j
)
q
2
′
(
x
j
)
)
=
1
x
n
+
1
−
x
0
(
∑
k
=
1
n
+
1
f
(
x
k
)
q
1
′
(
x
k
)
−
∑
k
=
0
n
f
(
x
k
)
q
2
′
(
x
k
)
)
=
1
x
n
+
1
−
x
0
(
f
(
x
n
+
1
)
q
1
′
(
x
n
+
1
)
+
∑
k
=
1
n
f
(
x
k
)
(
1
q
1
′
(
x
k
)
−
1
q
2
′
(
x
k
)
)
−
f
(
x
0
)
q
2
′
(
x
0
)
)
.
{\displaystyle {\begin{aligned}f[x_{0},x_{1},\ldots ,x_{n+1}]&={\frac {f[x_{1},\ldots ,x_{n+1}]-f[x_{0},\ldots ,x_{n}]}{x_{n+1}-x_{0}}}\\&={\frac {1}{x_{n+1}-x_{0}}}\left(\sum _{j=0}^{n}{\frac {f(x_{j+1})}{q_{1}'(x_{j+1})}}-\sum _{j=0}^{n}{\frac {f(x_{j})}{q_{2}'(x_{j})}}\right)\\&={\frac {1}{x_{n+1}-x_{0}}}\left(\sum _{k=1}^{n+1}{\frac {f(x_{k})}{q_{1}'(x_{k})}}-\sum _{k=0}^{n}{\frac {f(x_{k})}{q_{2}'(x_{k})}}\right)\\&={\frac {1}{x_{n+1}-x_{0}}}\left({\frac {f(x_{n+1})}{q_{1}'(x_{n+1})}}+\sum _{k=1}^{n}f(x_{k})\left({\frac {1}{q_{1}'(x_{k})}}-{\frac {1}{q_{2}'(x_{k})}}\right)-{\frac {f(x_{0})}{q_{2}'(x_{0})}}\right)\,.\end{aligned}}}
We have,
(
x
n
+
1
−
x
0
)
q
1
′
(
x
n
+
1
)
=
(
x
n
+
1
−
x
0
)
∏
k
=
1
n
(
x
n
+
1
−
x
k
)
=
∏
k
=
0
n
(
x
n
+
1
−
x
k
)
=
Q
′
(
x
n
+
1
)
,
{\displaystyle (x_{n+1}-x_{0})q_{1}'(x_{n+1})=(x_{n+1}-x_{0})\prod _{k=1}^{n}(x_{n+1}-x_{k})=\prod _{k=0}^{n}(x_{n+1}-x_{k})=Q'(x_{n+1}),}
(
x
n
+
1
−
x
0
)
q
2
′
(
x
0
)
=
(
x
n
+
1
−
x
0
)
∏
k
=
1
n
(
x
0
−
x
k
)
=
−
∏
k
=
1
n
+
1
(
x
0
−
x
k
)
=
−
Q
′
(
x
0
)
{\displaystyle (x_{n+1}-x_{0})q_{2}'(x_{0})=(x_{n+1}-x_{0})\prod _{k=1}^{n}(x_{0}-x_{k})=-\prod _{k=1}^{n+1}(x_{0}-x_{k})=-Q'(x_{0})}
and
1
q
1
′
(
x
k
)
−
1
q
2
′
(
x
k
)
=
∏
j
=
1
,
j
≠
k
n
+
1
1
x
k
−
x
j
−
∏
j
=
0
,
j
≠
k
n
1
x
k
−
x
j
=
∏
j
=
0
,
j
≠
k
n
+
1
1
x
k
−
x
j
(
x
k
−
x
0
−
(
x
k
−
x
n
+
1
)
)
=
(
x
n
+
1
−
x
0
)
∏
j
=
0
,
j
≠
k
n
+
1
1
x
k
−
x
j
,
{\displaystyle {\begin{aligned}{\frac {1}{q_{1}'(x_{k})}}-{\frac {1}{q_{2}'(x_{k})}}&=\prod _{j=1,j\neq k}^{n+1}{\frac {1}{x_{k}-x_{j}}}-\prod _{j=0,j\neq k}^{n}{\frac {1}{x_{k}-x_{j}}}\\&=\prod _{j=0,j\neq k}^{n+1}{\frac {1}{x_{k}-x_{j}}}(x_{k}-x_{0}-(x_{k}-x_{n+1}))\\&=(x_{n+1}-x_{0})\prod _{j=0,j\neq k}^{n+1}{\frac {1}{x_{k}-x_{j}}}\,,\end{aligned}}}
which gives
f
[
x
0
,
x
1
,
…
,
x
n
+
1
]
=
∑
j
=
0
n
+
1
f
(
x
j
)
Q
′
(
x
j
+
1
)
=
∑
j
=
0
n
+
1
f
(
x
j
)
∏
k
∈
{
0
,
…
,
n
}
∖
{
j
}
(
x
j
−
x
k
)
.
{\displaystyle f[x_{0},x_{1},\ldots ,x_{n+1}]=\sum _{j=0}^{n+1}{\frac {f(x_{j})}{Q'(x_{j+1})}}=\sum _{j=0}^{n+1}{\frac {f(x_{j})}{\prod _{k\in \{0,\dots ,n\}\setminus \{j\}}(x_{j}-x_{k})}}\,.}
Hence, since the assertion holds for
n
=
1
{\displaystyle n=1}
n
+
1
{\displaystyle n+1}
n
{\displaystyle n}
Symmetry property of divided differences
edit
The divided differences have a number of special properties that can simplify work with them. One of the property is called the Symmetry Property which states that the Divided differences remain unaffected by permutations (rearrangement) of their variables.
Now we prove this symmetry property by showing that
f
[
x
0
,
x
1
…
,
x
n
]
=
f
[
x
1
,
x
0
…
,
x
n
]
etc.
{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=f[x_{1},x_{0}\dots ,x_{n}]\quad {\text{etc.}}}
When
n
=
1
{\displaystyle n=1}
f
[
x
1
,
x
0
]
=
f
[
x
1
]
−
f
[
x
0
]
x
1
−
x
0
=
f
(
x
0
)
(
x
0
−
x
1
)
+
f
(
x
1
)
(
x
1
−
x
0
)
=
f
[
x
0
]
−
f
[
x
1
]
x
0
−
x
1
=
f
[
x
0
,
x
1
]
.
{\displaystyle f[x_{1},x_{0}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}={\frac {f(x_{0})}{(x_{0}-x_{1})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})}}={\frac {f[x_{0}]-f[x_{1}]}{x_{0}-x_{1}}}=f[x_{0},x_{1}]\,.}
Hence
f
[
x
1
,
x
0
]
=
f
[
x
0
,
x
1
]
{\displaystyle f[x_{1},x_{0}]=f[x_{0},x_{1}]}
When
n
=
2
{\displaystyle n=2}
f
[
x
2
,
x
1
,
x
0
]
=
f
[
x
2
,
x
1
]
−
f
[
x
1
,
x
0
]
x
2
−
x
0
=
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
2
)
(
x
2
−
x
0
)
⋅
(
x
2
−
x
1
)
+
f
(
x
1
)
(
x
1
−
x
0
)
⋅
(
x
1
−
x
2
)
+
f
(
x
0
)
(
x
0
−
x
1
)
⋅
(
x
0
−
x
2
)
=
f
[
x
0
,
x
1
,
x
2
]
=
f
[
x
1
,
x
0
,
x
2
]
etc.
{\displaystyle {\begin{aligned}f[x_{2},x_{1},x_{0}]&={\frac {f[x_{2},x_{1}]-f[x_{1},x_{0}]}{x_{2}-x_{0}}}={\frac {{\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}-{\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}}{x_{2}-x_{0}}}\\&={\frac {f(x_{2})}{(x_{2}-x_{0})\cdot (x_{2}-x_{1})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})\cdot (x_{1}-x_{2})}}+{\frac {f(x_{0})}{(x_{0}-x_{1})\cdot (x_{0}-x_{2})}}\\&=f[x_{0},x_{1},x_{2}]=f[x_{1},x_{0},x_{2}]\quad {\text{etc.}}\end{aligned}}}
Hence
f
[
x
2
,
x
1
,
x
0
]
=
f
[
x
0
,
x
1
,
x
2
]
=
f
[
x
1
,
x
0
,
x
2
]
{\displaystyle f[x_{2},x_{1},x_{0}]=f[x_{0},x_{1},x_{2}]=f[x_{1},x_{0},x_{2}]}
Similarly, when
n
=
3
{\displaystyle n=3}
f
[
x
3
,
x
2
,
x
1
,
x
0
]
=
f
[
x
3
,
x
2
,
x
1
]
−
f
[
x
2
,
x
1
,
x
0
]
x
3
−
x
0
=
f
(
x
0
)
(
x
0
−
x
1
)
⋅
(
x
0
−
x
2
)
⋅
(
x
0
−
x
3
)
+
f
(
x
1
)
(
x
1
−
x
0
)
⋅
(
x
1
−
x
2
)
⋅
(
x
1
−
x
3
)
+
f
(
x
2
)
(
x
2
−
x
0
)
⋅
(
x
2
−
x
1
)
⋅
(
x
2
−
x
3
)
+
f
(
x
3
)
(
x
3
−
x
0
)
⋅
(
x
3
−
x
1
)
⋅
(
x
3
−
x
2
)
=
f
[
x
0
,
x
1
,
x
2
,
x
3
]
=
f
[
x
1
,
x
0
,
x
2
,
x
3
]
etc.
{\displaystyle {\begin{aligned}f[x_{3},x_{2},x_{1},x_{0}]&={\frac {f[x_{3},x_{2},x_{1}]-f[x_{2},x_{1},x_{0}]}{x_{3}-x_{0}}}\\&={\frac {f(x_{0})}{(x_{0}-x_{1})\cdot (x_{0}-x_{2})\cdot (x_{0}-x_{3})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})\cdot (x_{1}-x_{2})\cdot (x_{1}-x_{3})}}+{\frac {f(x_{2})}{(x_{2}-x_{0})\cdot (x_{2}-x_{1})\cdot (x_{2}-x_{3})}}+\\&\quad \quad {\frac {f(x_{3})}{(x_{3}-x_{0})\cdot (x_{3}-x_{1})\cdot (x_{3}-x_{2})}}\\&=f[x_{0},x_{1},x_{2},x_{3}]=f[x_{1},x_{0},x_{2},x_{3}]\quad {\text{etc.}}\end{aligned}}}
Hence
f
[
x
3
,
x
2
,
x
1
,
x
0
]
=
f
[
x
0
,
x
1
,
x
2
,
x
3
]
=
f
[
x
1
,
x
0
,
x
3
,
x
2
]
{\displaystyle f[x_{3},x_{2},x_{1},x_{0}]=f[x_{0},x_{1},x_{2},x_{3}]=f[x_{1},x_{0},x_{3},x_{2}]}
In general, we can use the Expanded Form (expanded
f
[
x
0
,
x
1
…
,
x
n
]
=
∑
j
=
0
n
f
(
x
j
)
∏
k
∈
{
0
,
…
,
n
}
∖
{
j
}
(
x
j
−
x
k
)
=
f
[
x
1
,
x
0
…
,
x
n
]
etc.
{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{\prod _{k\in \{0,\dots ,n\}\setminus \{j\}}(x_{j}-x_{k})}}=f[x_{1},x_{0}\dots ,x_{n}]\quad {\text{etc.}}}
Hence
f
[
x
0
,
x
1
…
,
x
n
]
=
f
[
x
1
,
x
0
…
,
x
n
]
{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=f[x_{1},x_{0}\dots ,x_{n}]}
n
t
h
{\displaystyle n^{th}}
Computing the divided differences in tabular form
edit
A difference table is again a convenient device for displaying differences, the standard diagonal form being used and thus the generation of the divided differences is outlined in Table below.
x
f
(
x
)
First Divided Difference
Second Divided Difference
Third Divided Difference
Fourth Divided Difference
x
0
f
[
x
0
]
f
[
x
0
,
x
1
]
=
f
[
x
1
]
−
f
[
x
0
]
x
1
−
x
0
x
1
f
[
x
1
]
f
[
x
0
,
x
1
,
x
2
]
=
f
[
x
1
,
x
2
]
−
f
[
x
0
,
x
1
]
x
2
−
x
0
f
[
x
1
,
x
2
]
=
f
[
x
2
]
−
f
[
x
1
]
x
2
−
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
x
2
f
[
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
,
x
4
]
=
f
[
x
1
,
x
2
,
x
3
,
x
4
]
−
f
[
x
0
,
x
1
,
x
2
,
x
3
]
x
4
−
x
0
f
[
x
2
,
x
3
]
=
f
[
x
3
]
−
f
[
x
2
]
x
3
−
x
2
f
[
x
1
,
x
2
,
x
3
,
x
4
]
=
f
[
x
2
,
x
3
,
x
4
]
−
f
[
x
1
,
x
2
,
x
3
]
x
4
−
x
1
x
3
f
[
x
3
]
f
[
x
2
,
x
3
,
x
4
]
=
f
[
x
3
,
x
4
]
−
f
[
x
2
,
x
3
]
x
4
−
x
2
f
[
x
3
,
x
4
]
=
f
[
x
4
]
−
f
[
x
3
]
x
4
−
x
3
x
4
f
[
x
4
]
{\displaystyle {\begin{matrix}x&f(x)&{\text{First Divided Difference}}&{\text{Second Divided Difference}}&{\text{Third Divided Difference}}&{\text{Fourth Divided Difference}}\\x_{0}&f[x_{0}]&&&\\&&f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}&\\x_{1}&f[x_{1}]&&f[x_{0},x_{1},x_{2}]={\frac {f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}}&\\&&f[x_{1},x_{2}]={\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}&&f[x_{0},x_{1},x_{2},x_{3}]={\frac {f[x_{1},x_{2},x_{3}]-f[x_{0},x_{1},x_{2}]}{x_{3}-x_{0}}}\\x_{2}&f[x_{2}]&&f[x_{1},x_{2},x_{3}]={\frac {f[x_{2},x_{3}]-f[x_{1},x_{2}]}{x_{3}-x_{1}}}&&f[x_{0},x_{1},x_{2},x_{3},x_{4}]={\frac {f[x_{1},x_{2},x_{3},x_{4}]-f[x_{0},x_{1},x_{2},x_{3}]}{x_{4}-x_{0}}}\\&&f[x_{2},x_{3}]={\frac {f[x_{3}]-f[x_{2}]}{x_{3}-x_{2}}}&&f[x_{1},x_{2},x_{3},x_{4}]={\frac {f[x_{2},x_{3},x_{4}]-f[x_{1},x_{2},x_{3}]}{x_{4}-x_{1}}}\\x_{3}&f[x_{3}]&&f[x_{2},x_{3},x_{4}]={\frac {f[x_{3},x_{4}]-f[x_{2},x_{3}]}{x_{4}-x_{2}}}&\\&&f[x_{3},x_{4}]={\frac {f[x_{4}]-f[x_{3}]}{x_{4}-x_{3}}}&\\x_{4}&f[x_{4}]&\\\end{matrix}}}
A Numerical Example 1
edit
For a function
f
{\displaystyle f}
x
1
=
2
f
[
x
1
]
=
2
x
0
=
1
f
[
x
0
]
=
−
6
x
2
=
4
f
[
x
2
]
=
12
{\displaystyle {\begin{matrix}x_{1}=2&f[x_{1}]=2&\\&\\x_{0}=1&f[x_{0}]=-6&\\&\\x_{2}=4&f[x_{2}]=12&\\\end{matrix}}}
find
f
[
x
0
,
x
1
,
x
2
]
{\displaystyle f[x_{0},x_{1},x_{2}]}
x
f
(
x
)
First Divided Difference
Second Divided Difference
x
1
=
2
f
[
x
1
]
=
2
f
[
x
1
,
x
0
]
=
(
−
6
)
−
2
1
−
2
=
8
x
0
=
1
f
[
x
0
]
=
−
6
f
[
x
1
,
x
0
,
x
2
]
=
6
−
8
4
−
2
=
−
1
f
[
x
0
,
x
2
]
=
12
−
(
−
6
)
4
−
1
=
6
x
2
=
4
f
[
x
2
]
=
12
{\displaystyle {\begin{matrix}x&f(x)&{\text{First Divided Difference}}&{\text{Second Divided Difference}}\\x_{1}=2&f[x_{1}]=2&&\\&&f[x_{1},x_{0}]={\frac {(-6)-2}{1-2}}=8&&\\x_{0}=1&f[x_{0}]=-6&&f[x_{1},x_{0},x_{2}]={\frac {6-8}{4-2}}=-1\\&&f[x_{0},x_{2}]={\frac {12-(-6)}{4-1}}=6&&\\x_{2}=4&f[x_{2}]=12&&\\\end{matrix}}}
f
[
x
1
,
x
0
,
x
2
]
=
−
1
{\displaystyle f[x_{1},x_{0},x_{2}]=-1}
f
[
x
1
,
x
0
,
x
2
]
=
f
[
x
0
,
x
1
,
x
2
]
{\displaystyle f[x_{1},x_{0},x_{2}]=f[x_{0},x_{1},x_{2}]}
f
[
x
0
,
x
1
,
x
2
]
=
−
1
{\displaystyle f[x_{0},x_{1},x_{2}]=-1}
A Numerical Example 2
edit
For a function
f
{\displaystyle f}
x
0
=
0.0
f
[
x
0
]
f
[
x
0
,
x
1
]
x
1
=
0.4
f
[
x
1
]
f
[
x
0
,
x
1
,
x
2
]
=
50
7
f
[
x
1
,
x
2
]
=
10
x
2
=
0.7
f
[
x
2
]
=
6
{\displaystyle {\begin{matrix}x_{0}=0.0&f[x_{0}]&&\\&&f[x_{0},x_{1}]&&\\x_{1}=0.4&f[x_{1}]&&&f[x_{0},x_{1},x_{2}]={\frac {50}{7}}&\\&&f[x_{1},x_{2}]=10&&\\x_{2}=0.7&f[x_{2}]=6&&\\\end{matrix}}}
Determine the missing entries in the table.
We have the formula
f
[
x
0
,
x
1
,
x
2
]
=
f
[
x
1
,
x
2
]
−
f
[
x
0
,
x
1
]
x
2
−
x
0
{\displaystyle f[x_{0},x_{1},x_{2}]={\frac {f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}}}
and substituting gives
50
7
=
(
10
−
f
[
x
0
,
x
1
]
)
0.7
.
{\displaystyle {\frac {50}{7}}={\frac {(10-f[x_{0},x_{1}])}{0.7}}\,.}
Thus,
f
[
x
0
,
x
1
]
=
−
0.7
⋅
(
50
7
)
+
10
=
5
{\displaystyle f[x_{0},x_{1}]=-0.7\cdot ({\frac {50}{7}})+{10}=5}
Using the formula
f
[
x
1
,
x
2
]
=
f
[
x
2
]
−
f
[
x
1
]
x
2
−
x
1
{\displaystyle f[x_{1},x_{2}]={\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}}
and substituting gives
10
=
(
6
−
f
[
x
1
]
)
0.3
.
{\displaystyle 10={\frac {(6-f[x_{1}])}{0.3}}\,.}
Thus,
f
[
x
1
]
=
6
−
3
=
3
.
{\displaystyle f[x_{1}]=6-3=3\,.}
Further,
f
[
x
0
,
x
1
]
=
f
[
x
1
]
−
f
[
x
0
]
x
1
−
x
0
.
{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}\,.}
So,
5
=
(
3
−
f
[
x
0
]
)
0.4
.
{\displaystyle 5={\frac {(3-f[x_{0}])}{0.4}}\,.}
Thus,
f
[
x
0
]
=
3
−
2
=
1
{\displaystyle f[x_{0}]=3-2=1}
Algorithm: Computing the Divided Differences
edit
Algorithm: Newton's Divided-Differences
edit
Given the points
(
x
i
,
f
(
x
i
)
)
,
i
=
0
,
1
,
.
.
.
,
n
.
{\displaystyle (x_{i},f(x_{i})),i=0,1,...,n.}
F
i
,
0
=
f
(
x
i
)
,
i
=
0
,
1
,
.
.
.
,
n
.
{\displaystyle F_{i},_{0}=f(x_{i}),i=0,1,...,n.}
i
=
1
:
n
.
{\displaystyle i=1:n.}
j
=
1
:
i
.
{\displaystyle j=1:i.}
F
i
,
j
=
F
i
,
j
−
1
−
F
i
−
1
,
j
x
i
−
x
i
−
j
{\displaystyle F_{i},_{j}={\frac {F_{i},_{j-1}-F_{i-1},_{j}}{x_{i}-x_{i-j}}}}
F
i
,
j
{\displaystyle F_{i},_{j}}
f
[
x
0
,
.
.
.
,
x
i
]
.
{\displaystyle f[x_{0},...,x_{i}].}
Relationship between Generalization of the Mean Value Theorem and the Derivatives
edit
Generalization of the Mean Value Theorem
edit
For any n + 1 pairwise distinct points x 0 , ..., x n n -times differentiable function f there exists an interior point
ξ
∈
(
min
{
x
0
,
…
,
x
n
}
,
max
{
x
0
,
…
,
x
n
}
)
{\displaystyle \xi \in (\min\{x_{0},\dots ,x_{n}\},\max\{x_{0},\dots ,x_{n}\})\,}
where the n th derivative of f equals
n
{\displaystyle n}
n
t
h
{\displaystyle n^{th}}
f
[
x
0
,
…
,
x
n
]
=
f
(
n
)
(
ξ
)
n
!
.
{\displaystyle f[x_{0},\dots ,x_{n}]={\frac {f^{(n)}(\xi )}{n!}}\,.}
This is called the Generalized Mean Value Theorem .
For
n
=
1
{\displaystyle n=1}
f
[
x
0
,
x
1
]
=
f
[
x
1
]
−
f
[
x
0
]
x
1
−
x
0
=
f
′
(
ξ
)
{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}=f'(\xi )}
for some
ξ
{\displaystyle \xi }
x
0
{\displaystyle x_{0}}
x
1
{\displaystyle x_{1}}
Mean Value Theorem .
We have extended MVT to higher order derivatives as
f
[
x
0
,
…
,
x
n
]
=
f
(
n
)
(
ξ
)
n
!
.
{\displaystyle f[x_{0},\dots ,x_{n}]={\frac {f^{(n)}(\xi )}{n!}}\,.}
What is the theorem telling us?
This theorem is telling us that the Newton's
n
t
h
{\displaystyle n^{th}}
n
t
h
{\displaystyle n^{th}}
f
{\displaystyle f}
A Numerical Example
edit
Let
f
(
x
)
=
cos
(
x
)
{\displaystyle f(x)=\cos(x)}
x
0
=
0.2
,
x
1
=
0.3
,
x
2
=
0.4
{\displaystyle x_{0}=0.2,x_{1}=0.3,x_{2}=0.4}
f
[
x
0
,
x
1
]
≈
f
′
(
ξ
)
{\displaystyle f[x_{0},x_{1}]\approx f'(\xi )}
f
[
x
0
,
x
1
,
x
2
]
≈
1
2
f
″
(
ξ
)
{\displaystyle f[x_{0},x_{1},x_{2}]\approx {\frac {1}{2}}f''(\xi )}
for some
ξ
{\displaystyle \xi }
x
0
,
x
1
{\displaystyle x_{0},x_{1}}
x
2
{\displaystyle x_{2}}
f
[
x
0
,
x
1
]
=
f
[
x
1
]
−
f
[
x
0
]
x
1
−
x
0
=
cos
(
0.3
)
−
cos
(
0.2
)
0.3
−
0.2
=
−
0.2473009
{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}={\frac {\cos(0.3)-\cos(0.2)}{0.3-0.2}}=-0.2473009}
If we chose
ξ
=
0.25
,
{\displaystyle \xi =0.25,}
0.25
∈
(
0.2
,
0.3
)
{\displaystyle 0.25\in (0.2,0.3)}
f
′
(
ξ
)
=
−
sin
(
0.25
)
=
−
0.24744040
{\displaystyle f'(\xi )=-\sin(0.25)=-0.24744040}
f
[
x
0
,
x
1
]
{\displaystyle f[x_{0},x_{1}]}
f
[
x
1
,
x
2
]
=
f
[
x
2
]
−
f
[
x
1
]
x
2
−
x
1
=
cos
(
0.4
)
−
cos
(
0.3
)
0.4
−
0.3
=
−
0.3427550
{\displaystyle f[x_{1},x_{2}]={\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}={\frac {\cos(0.4)-\cos(0.3)}{0.4-0.3}}=-0.3427550}
and
f
[
x
0
,
x
1
,
x
2
]
=
f
[
x
1
,
x
2
]
−
f
[
x
0
,
x
1
]
x
2
−
x
0
=
−
0.3427550
−
(
−
0.2473009
)
0.4
−
0.2
=
−
0.4772705
.
{\displaystyle f[x_{0},x_{1},x_{2}]={\frac {f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}}={\frac {-0.3427550-(-0.2473009)}{0.4-0.2}}=-0.4772705\,.}
Thus, by the Generalized Mean Value Theorem with
n
=
2
{\displaystyle n=2}
f
[
x
0
,
x
1
,
x
2
]
=
1
2
f
″
(
ξ
)
{\displaystyle f[x_{0},x_{1},x_{2}]={\frac {1}{2}}f''(\xi )}
for some
ξ
{\displaystyle \xi }
x
0
,
x
1
{\displaystyle x_{0},x_{1}}
x
2
{\displaystyle x_{2}}
f
″
(
ξ
)
=
−
cos
(
ξ
)
{\displaystyle f''(\xi )=-\cos(\xi )}
ξ
=
x
1
{\displaystyle \xi =x_{1}}
1
2
f
″
(
0.3
)
=
−
1
2
cos
(
0.3
)
=
−
0.4776682
{\displaystyle {\frac {1}{2}}f''(0.3)=-{\frac {1}{2}}\cos(0.3)=-0.4776682}
f
[
x
0
,
x
1
,
x
2
]
=
−
0.4772705
{\displaystyle f[x_{0},x_{1},x_{2}]=-0.4772705}
n
t
h
{\displaystyle n^{th}}
n
t
h
{\displaystyle n^{th}}
f
{\displaystyle f}
Reference
edit
Guide to Numerical Analysis by Peter R. Turner
Numerical Analysis by Richard L. Burden and J. Douglas Faires (EIGHT EDITION)
Elementary Numerical Analysis by Kendall Atkinson (Second Edition)
Applied Numerical Analysis by Gerald / Wheatley (Sixth Edition)
Theory and Problems of Numerical Analysis by Francis Scheid 
![{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{\prod _{k\in \{0,\dots ,n\}\setminus \{j\}}(x_{j}-x_{k})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0af78e21e5ccee96ecda3985867a2754f22e8fc8)
![{\displaystyle f[x_{0},\dots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{q'(x_{j})}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97116bef80032373a3071a7cf9f0901464439454)
![{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}={\frac {f(x_{0})}{(x_{0}-x_{1})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e52d5498d86c961d4f47b41b30ac345e80e57e4e)
![{\displaystyle f[x_{0},x_{1},\ldots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{q'(x_{j})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d2b174c7ff907416f5ebbeabafa7023022725bc)
![{\displaystyle {\begin{aligned}f[x_{0},x_{1},\ldots ,x_{n+1}]&={\frac {f[x_{1},\ldots ,x_{n+1}]-f[x_{0},\ldots ,x_{n}]}{x_{n+1}-x_{0}}}\\&={\frac {1}{x_{n+1}-x_{0}}}\left(\sum _{j=0}^{n}{\frac {f(x_{j+1})}{q_{1}'(x_{j+1})}}-\sum _{j=0}^{n}{\frac {f(x_{j})}{q_{2}'(x_{j})}}\right)\\&={\frac {1}{x_{n+1}-x_{0}}}\left(\sum _{k=1}^{n+1}{\frac {f(x_{k})}{q_{1}'(x_{k})}}-\sum _{k=0}^{n}{\frac {f(x_{k})}{q_{2}'(x_{k})}}\right)\\&={\frac {1}{x_{n+1}-x_{0}}}\left({\frac {f(x_{n+1})}{q_{1}'(x_{n+1})}}+\sum _{k=1}^{n}f(x_{k})\left({\frac {1}{q_{1}'(x_{k})}}-{\frac {1}{q_{2}'(x_{k})}}\right)-{\frac {f(x_{0})}{q_{2}'(x_{0})}}\right)\,.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3311f493e1896951098169231d3acf08ce91b643)
![{\displaystyle f[x_{0},x_{1},\ldots ,x_{n+1}]=\sum _{j=0}^{n+1}{\frac {f(x_{j})}{Q'(x_{j+1})}}=\sum _{j=0}^{n+1}{\frac {f(x_{j})}{\prod _{k\in \{0,\dots ,n\}\setminus \{j\}}(x_{j}-x_{k})}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/839c0cbd484b2d07c269aabeded396cacf319443)
![{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=f[x_{1},x_{0}\dots ,x_{n}]\quad {\text{etc.}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81b9da15b7e6065c19a762f2ba9eb658bb71cebd)
![{\displaystyle f[x_{1},x_{0}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}={\frac {f(x_{0})}{(x_{0}-x_{1})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})}}={\frac {f[x_{0}]-f[x_{1}]}{x_{0}-x_{1}}}=f[x_{0},x_{1}]\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fca0461f586461623dede0a8dfe7b369659cca4)
![{\displaystyle f[x_{1},x_{0}]=f[x_{0},x_{1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f2cb15eda72ead842e6fbe988fe39e4df88f172)
![{\displaystyle {\begin{aligned}f[x_{2},x_{1},x_{0}]&={\frac {f[x_{2},x_{1}]-f[x_{1},x_{0}]}{x_{2}-x_{0}}}={\frac {{\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}-{\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}}{x_{2}-x_{0}}}\\&={\frac {f(x_{2})}{(x_{2}-x_{0})\cdot (x_{2}-x_{1})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})\cdot (x_{1}-x_{2})}}+{\frac {f(x_{0})}{(x_{0}-x_{1})\cdot (x_{0}-x_{2})}}\\&=f[x_{0},x_{1},x_{2}]=f[x_{1},x_{0},x_{2}]\quad {\text{etc.}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81c205acf657a96bc6de20c3365461c683a14e7a)
![{\displaystyle f[x_{2},x_{1},x_{0}]=f[x_{0},x_{1},x_{2}]=f[x_{1},x_{0},x_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f5f5e7e01b26c69bc948d8786ff60988c5c383d)
![{\displaystyle {\begin{aligned}f[x_{3},x_{2},x_{1},x_{0}]&={\frac {f[x_{3},x_{2},x_{1}]-f[x_{2},x_{1},x_{0}]}{x_{3}-x_{0}}}\\&={\frac {f(x_{0})}{(x_{0}-x_{1})\cdot (x_{0}-x_{2})\cdot (x_{0}-x_{3})}}+{\frac {f(x_{1})}{(x_{1}-x_{0})\cdot (x_{1}-x_{2})\cdot (x_{1}-x_{3})}}+{\frac {f(x_{2})}{(x_{2}-x_{0})\cdot (x_{2}-x_{1})\cdot (x_{2}-x_{3})}}+\\&\quad \quad {\frac {f(x_{3})}{(x_{3}-x_{0})\cdot (x_{3}-x_{1})\cdot (x_{3}-x_{2})}}\\&=f[x_{0},x_{1},x_{2},x_{3}]=f[x_{1},x_{0},x_{2},x_{3}]\quad {\text{etc.}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f220e9a2598cdb10a45ecd71ec03d0b0a87b637)
![{\displaystyle f[x_{3},x_{2},x_{1},x_{0}]=f[x_{0},x_{1},x_{2},x_{3}]=f[x_{1},x_{0},x_{3},x_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0709fae835fb69f0dfc0fdf3bd2542744bcebf6)
![{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=\sum _{j=0}^{n}{\frac {f(x_{j})}{\prod _{k\in \{0,\dots ,n\}\setminus \{j\}}(x_{j}-x_{k})}}=f[x_{1},x_{0}\dots ,x_{n}]\quad {\text{etc.}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b4b81e2cfe982e3c30cc115c25e9e7d823259c0)
![{\displaystyle f[x_{0},x_{1}\dots ,x_{n}]=f[x_{1},x_{0}\dots ,x_{n}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c65a5eb3e67a185a562675401f43f46093c777d8)
![{\displaystyle {\begin{matrix}x&f(x)&{\text{First Divided Difference}}&{\text{Second Divided Difference}}&{\text{Third Divided Difference}}&{\text{Fourth Divided Difference}}\\x_{0}&f[x_{0}]&&&\\&&f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}&\\x_{1}&f[x_{1}]&&f[x_{0},x_{1},x_{2}]={\frac {f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}}&\\&&f[x_{1},x_{2}]={\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}&&f[x_{0},x_{1},x_{2},x_{3}]={\frac {f[x_{1},x_{2},x_{3}]-f[x_{0},x_{1},x_{2}]}{x_{3}-x_{0}}}\\x_{2}&f[x_{2}]&&f[x_{1},x_{2},x_{3}]={\frac {f[x_{2},x_{3}]-f[x_{1},x_{2}]}{x_{3}-x_{1}}}&&f[x_{0},x_{1},x_{2},x_{3},x_{4}]={\frac {f[x_{1},x_{2},x_{3},x_{4}]-f[x_{0},x_{1},x_{2},x_{3}]}{x_{4}-x_{0}}}\\&&f[x_{2},x_{3}]={\frac {f[x_{3}]-f[x_{2}]}{x_{3}-x_{2}}}&&f[x_{1},x_{2},x_{3},x_{4}]={\frac {f[x_{2},x_{3},x_{4}]-f[x_{1},x_{2},x_{3}]}{x_{4}-x_{1}}}\\x_{3}&f[x_{3}]&&f[x_{2},x_{3},x_{4}]={\frac {f[x_{3},x_{4}]-f[x_{2},x_{3}]}{x_{4}-x_{2}}}&\\&&f[x_{3},x_{4}]={\frac {f[x_{4}]-f[x_{3}]}{x_{4}-x_{3}}}&\\x_{4}&f[x_{4}]&\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/956090b962623b99ad1c892d003f49f15e24aee1)
![{\displaystyle {\begin{matrix}x_{1}=2&f[x_{1}]=2&\\&\\x_{0}=1&f[x_{0}]=-6&\\&\\x_{2}=4&f[x_{2}]=12&\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c17dffe7389f40623a27583d63ebeae248761b06)
![{\displaystyle f[x_{0},x_{1},x_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d46b58779485963bbee5306d1f8dcd8550e5a86b)
![{\displaystyle {\begin{matrix}x&f(x)&{\text{First Divided Difference}}&{\text{Second Divided Difference}}\\x_{1}=2&f[x_{1}]=2&&\\&&f[x_{1},x_{0}]={\frac {(-6)-2}{1-2}}=8&&\\x_{0}=1&f[x_{0}]=-6&&f[x_{1},x_{0},x_{2}]={\frac {6-8}{4-2}}=-1\\&&f[x_{0},x_{2}]={\frac {12-(-6)}{4-1}}=6&&\\x_{2}=4&f[x_{2}]=12&&\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22b8355206caf4085701e5f8afea543ea72e7069)
![{\displaystyle f[x_{1},x_{0},x_{2}]=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/509ddc844eba87ea0dab4e4554a3e0c62ee6f0e3)
![{\displaystyle f[x_{1},x_{0},x_{2}]=f[x_{0},x_{1},x_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f19d0abf1cb69c6afa340666832b6659b6e4d35)
![{\displaystyle f[x_{0},x_{1},x_{2}]=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7db9191a70d2a1785ae1ee58b83a04c5bd39b3e0)
![{\displaystyle {\begin{matrix}x_{0}=0.0&f[x_{0}]&&\\&&f[x_{0},x_{1}]&&\\x_{1}=0.4&f[x_{1}]&&&f[x_{0},x_{1},x_{2}]={\frac {50}{7}}&\\&&f[x_{1},x_{2}]=10&&\\x_{2}=0.7&f[x_{2}]=6&&\\\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e5e07e5beee6109470e8e6bd759998830a41bc3)
![{\displaystyle f[x_{0},x_{1},x_{2}]={\frac {f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/388ecbcbbcad736cc439d7c0f14acfef81e9a913)
![{\displaystyle {\frac {50}{7}}={\frac {(10-f[x_{0},x_{1}])}{0.7}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee0573a61033267e5fa26659f3e69329d8da156e)
![{\displaystyle f[x_{0},x_{1}]=-0.7\cdot ({\frac {50}{7}})+{10}=5}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b198093d8245afb070ddea96c78940a76d127e2)
![{\displaystyle f[x_{1},x_{2}]={\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/465fd4fa900214b1ad33d835dbe405e0f22f5d30)
![{\displaystyle 10={\frac {(6-f[x_{1}])}{0.3}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27e442dbbc52de4a4fbb8e41c125e10d68c54a86)
![{\displaystyle f[x_{1}]=6-3=3\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29a36de7b3e34827bd1fe4b2b2b616a2dddbb4dc)
![{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2faaf3b74f90174c80a4443af06e7be5d973654)
![{\displaystyle 5={\frac {(3-f[x_{0}])}{0.4}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/317f447c844d8a9d0dabdad4f20409d83ebc77f2)
![{\displaystyle f[x_{0}]=3-2=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5129ba2d30d2e12d68260b1ab1949e881c8694a6)
![{\displaystyle f[x_{0},...,x_{i}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c7645f748f0812cd50428942cef77f93ca78aaa)
![{\displaystyle f[x_{0},\dots ,x_{n}]={\frac {f^{(n)}(\xi )}{n!}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abb353a74c8e781757c67c94e8817bcbe923acad)
![{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}=f'(\xi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33e7fbc3247f56bc1d880cd7cf5cba5e88b5882b)
![{\displaystyle f[x_{0},x_{1}]\approx f'(\xi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2d698979afbdf6b70d4a82e3ebc51f63f1c5937)
![{\displaystyle f[x_{0},x_{1},x_{2}]\approx {\frac {1}{2}}f''(\xi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ffda46c9ff19ea35aaebbcb9cd2e3a9839cdd57)
![{\displaystyle f[x_{0},x_{1}]={\frac {f[x_{1}]-f[x_{0}]}{x_{1}-x_{0}}}={\frac {\cos(0.3)-\cos(0.2)}{0.3-0.2}}=-0.2473009}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63d12a17eb5450098ae31dafcec7f3364f7dda3c)
![{\displaystyle f[x_{0},x_{1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a31f8458da9ea4ec290a96134207b45c5d7a4f7c)
![{\displaystyle f[x_{1},x_{2}]={\frac {f[x_{2}]-f[x_{1}]}{x_{2}-x_{1}}}={\frac {\cos(0.4)-\cos(0.3)}{0.4-0.3}}=-0.3427550}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bd5c808de1e9aa1834d055245bbc18fb2cc8ae7)
![{\displaystyle f[x_{0},x_{1},x_{2}]={\frac {f[x_{1},x_{2}]-f[x_{0},x_{1}]}{x_{2}-x_{0}}}={\frac {-0.3427550-(-0.2473009)}{0.4-0.2}}=-0.4772705\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bef4de90aaa3b90e9b86de6b440a0b2d73000a89)
![{\displaystyle f[x_{0},x_{1},x_{2}]={\frac {1}{2}}f''(\xi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cb6cc45a74be8376485f99035d4bb57c785b9fb)
![{\displaystyle f[x_{0},x_{1},x_{2}]=-0.4772705}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab948176bad38fa9eb6e7e8fd2c631ffa437f9a8)