Submit your draft for review!
primer encabezado(?)
segunda etc
tercera etc
primera fila
1
2
3
2ª fila
2
4
6
3ª fila
3
6
9
4ª fila
4
8
12
5ª fila
5
10
15
Newton's polynomial Example
edit
Another example:
The sequence
f
0
{\displaystyle f_{0}}
f
0
(
1
)
=
6
,
f
0
(
2
)
=
9
,
f
0
(
3
)
=
2
{\displaystyle f_{0}(1)=6,f_{0}(2)=9,f_{0}(3)=2}
f
0
(
4
)
=
5
{\displaystyle f_{0}(4)=5}
6
,
9
,
2
,
5
{\displaystyle 6,9,2,5}
x
0
=
1
{\displaystyle x_{0}=1}
x
3
=
4
{\displaystyle x_{3}=4}
You obtain the slope of order
1
{\displaystyle 1}
f
1
(
x
0
,
x
1
)
=
f
0
(
x
1
)
−
f
0
(
x
0
)
x
1
−
x
0
=
9
−
6
2
−
1
=
3
{\displaystyle f_{1}(x_{0},x_{1})={\frac {f_{0}(x_{1})-f_{0}(x_{0})}{x_{1}-x_{0}}}={\frac {9-6}{2-1}}=3}
f
1
(
x
1
,
x
2
)
=
f
0
(
x
2
)
−
f
0
(
x
1
)
x
2
−
x
1
=
2
−
9
3
−
2
=
−
7
{\displaystyle f_{1}(x_{1},x_{2})={\frac {f_{0}(x_{2})-f_{0}(x_{1})}{x_{2}-x_{1}}}={\frac {2-9}{3-2}}=-7}
f
1
(
x
2
,
x
3
)
=
f
0
(
x
3
)
−
f
0
(
x
2
)
x
3
−
x
2
=
5
−
2
4
−
3
=
3
{\displaystyle f_{1}(x_{2},x_{3})={\frac {f_{0}(x_{3})-f_{0}(x_{2})}{x_{3}-x_{2}}}={\frac {5-2}{4-3}}=3}
As we have the slopes of order
1
{\displaystyle 1}
f
2
(
x
0
,
x
1
,
x
2
)
=
f
1
(
x
1
,
x
2
)
−
f
1
(
x
0
,
x
1
)
x
2
−
x
0
=
−
7
−
3
3
−
1
=
−
5
{\displaystyle f_{2}(x_{0},x_{1},x_{2})={\frac {f_{1}(x_{1},x_{2})-f_{1}(x_{0},x_{1})}{x_{2}-x_{0}}}={\frac {-7-3}{3-1}}=-5}
f
2
(
x
1
,
x
2
,
x
3
)
=
f
1
(
x
2
,
x
3
)
−
f
1
(
x
1
,
x
2
)
x
3
−
x
1
=
3
−
(
−
7
)
4
−
2
=
5
{\displaystyle f_{2}(x_{1},x_{2},x_{3})={\frac {f_{1}(x_{2},x_{3})-f_{1}(x_{1},x_{2})}{x_{3}-x_{1}}}={\frac {3-(-7)}{4-2}}=5}
Finally, we define the slope of order
3
{\displaystyle 3}
f
3
(
x
0
,
x
1
,
x
2
,
x
3
)
=
f
2
(
x
1
,
x
2
,
x
3
)
−
f
2
(
x
0
,
x
1
,
x
2
)
x
3
−
x
0
=
5
−
(
−
5
)
4
−
1
=
10
3
{\displaystyle f_{3}(x_{0},x_{1},x_{2},x_{3})={\frac {f_{2}(x_{1},x_{2},x_{3})-f_{2}(x_{0},x_{1},x_{2})}{x_{3}-x_{0}}}={\frac {5-(-5)}{4-1}}={\frac {10}{3}}}
Once we have the slope, we can define the consequent polynomials:
p
0
(
x
)
=
6
{\displaystyle p_{0}(x)=6}
p
1
(
x
)
=
6
+
3
(
x
−
1
)
{\displaystyle p_{1}(x)=6+3(x-1)}
p
2
(
x
)
=
6
+
3
(
x
−
1
)
−
5
(
x
−
1
)
(
x
−
2
)
{\displaystyle p_{2}(x)=6+3(x-1)-5(x-1)(x-2)}
p
3
(
x
)
=
6
+
3
(
x
−
1
)
−
5
(
x
−
1
)
(
x
−
2
)
+
10
3
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
{\displaystyle p_{3}(x)=6+3(x-1)-5(x-1)(x-2)+{\frac {10}{3}}(x-1)(x-2)(x-3)}
