Für die Newton-Cotes-Formel folgt dann:
∫
a
b
f
(
x
)
d
x
≈
∫
a
b
p
n
(
x
)
d
x
=
(
b
−
a
)
∑
i
=
0
n
w
i
f
(
x
i
)
{\displaystyle \int \limits _{a}^{b}f(x)\,dx\approx \int \limits _{a}^{b}p_{n}(x)\,dx=(b-a)\sum _{i=0}^{n}w_{i}f(x_{i})}
mit den Gewichten
w
i
=
1
b
−
a
∫
a
b
l
i
(
x
)
d
x
{\displaystyle w_{i}={\frac {1}{b-a}}\int \limits _{a}^{b}l_{i}(x)\,dx}
Die Gewichte sind symmetrisch, das heißt
w
n
−
i
=
w
i
{\displaystyle w_{n-i}=w_{i}}
l
i
(
x
)
=
∏
0
≤
j
≤
n
j
≠
i
x
−
x
j
x
i
−
x
j
=
(
x
−
x
0
)
⋯
(
x
−
x
i
−
1
)
(
x
−
x
i
+
1
)
⋯
(
x
−
x
n
)
(
x
i
−
x
0
)
⋯
(
x
i
−
x
i
−
1
)
(
x
i
−
x
i
+
1
)
⋯
(
x
i
−
x
n
)
{\displaystyle l_{i}(x)=\prod _{\begin{smallmatrix}0\leq j\leq n\\j\neq i\end{smallmatrix}}{\frac {x-x_{j}}{x_{i}-x_{j}}}={\frac {(x-x_{0})\cdots (x-x_{i-1})(x-x_{i+1})\cdots (x-x_{n})}{(x_{i}-x_{0})\cdots (x_{i}-x_{i-1})(x_{i}-x_{i+1})\cdots (x_{i}-x_{n})}}}
Wegen der speziellen Wahl der Stützstellen integrieren diese Formeln bei ungeradem
n
{\displaystyle n}
n
{\displaystyle n}
n
{\displaystyle n}
n
+
1
{\displaystyle n+1}
n
{\displaystyle n}
n
{\displaystyle n}
Genauigkeitsgrad .
Speziell gilt für
f
(
x
)
=
1
{\displaystyle f(x)=1}
∫
a
b
f
(
x
)
d
x
=
∫
a
b
1
d
x
=
b
−
a
=
(
b
−
a
)
∑
i
=
0
n
w
i
⋅
1
=
(
b
−
a
)
∑
i
=
0
n
w
i
{\displaystyle \int \limits _{a}^{b}f(x)\,dx=\int \limits _{a}^{b}1\,dx=b-a=(b-a)\sum _{i=0}^{n}w_{i}\cdot 1=(b-a)\sum _{i=0}^{n}w_{i}}
∑
i
=
0
n
w
i
=
1
{\displaystyle \sum _{i=0}^{n}w_{i}=1}
Falls
∑
i
=
0
n
|
w
i
|
>
∑
i
=
0
n
w
i
=
1
{\displaystyle \sum _{i=0}^{n}|w_{i}|>\sum _{i=0}^{n}w_{i}=1}
Vorzeichen der Fall ist, besteht die Gefahr, dass sich die Rundungsfehler aufschaukeln oder Auslöschung eintritt. Daher sind aus numerischen Gründen Formeln mit positiven Gewichten zu bevorzugen. Da für großes
n
{\displaystyle n}
Interpolationspolynom
p
n
(
x
)
{\displaystyle p_{n}(x)}
n
{\displaystyle n}
E
(
f
)
=
∫
a
b
f
(
x
)
d
x
−
∫
a
b
p
n
(
x
)
d
x
{\displaystyle E(f)=\int \limits _{a}^{b}f(x)\,dx-\int \limits _{a}^{b}p_{n}(x)\,dx}
ist der Fehler (Verfahrensfehler), der bei der Anwendung der Newton-Cotes-Formel gemacht wird. Dieser hat bei der speziellen Wahl der Stützstellen für
(
p
+
1
)
{\displaystyle (p+1)}
[
a
,
b
]
{\displaystyle [a,b]}
stetig differenzierbar reellwertige Funktionen
f
(
x
)
{\displaystyle f(x)}
E
(
f
)
=
K
⋅
f
(
p
+
1
)
(
ξ
)
{\displaystyle E(f)=K\cdot f^{(p+1)}(\xi )}
wobei
K
{\displaystyle K}
f
(
x
)
{\displaystyle f(x)}
ξ
∈
[
a
,
b
]
{\displaystyle \xi \in [a,b]}
E
(
f
)
{\displaystyle E(f)}
Integral exakt ausrechnen, im Widerspruch zu der Tatsache, dass es unendlich viele Integrale gibt, die man nicht exakt berechnen kann. Der Fehler ist Null für alle Funktionen, deren
(
p
+
1
)
{\displaystyle (p+1)}
p
{\displaystyle p}
p
{\displaystyle p}
p
+
1
{\displaystyle p+1}
(polynomiale) Ordnung der Newton-Cotes-Formel bezeichnet.
Mit Hilfe des Verfahrensfehlers erhält man die Fehlerabschätzung:
|
E
(
f
)
|
≤
|
K
|
⋅
max
a
≤
ξ
≤
b
|
f
(
p
+
1
)
(
ξ
)
|
{\displaystyle |E(f)|\leq |K|\cdot \max _{a\leq \xi \leq b}\left|f^{(p+1)}(\xi )\right|}
Der exakte Fehler ist immer kleiner oder gleich dieser Fehlerabschätzung, wie auch die unten angegebenen Beispiele zeigen.
Die angegebenen Stützstellen
t
i
{\displaystyle t_{i}}
[
0
,
1
]
{\displaystyle [0,1]}
t
0
=
0
,
t
i
=
i
n
,
t
n
=
1
{\displaystyle t_{0}=0,t_{i}={\frac {i}{n}},t_{n}=1}
[
a
,
b
]
{\displaystyle [a,b]}
x
i
=
a
+
t
i
⋅
(
b
−
a
)
{\displaystyle x_{i}=a+t_{i}\cdot (b-a)}
n
{\displaystyle n}
Name
Stützstellen
t
i
{\displaystyle t_{i}}
Gewichte
w
i
{\displaystyle w_{i}}
E
(
f
)
{\displaystyle E(f)}
1
Trapezregel
0
1
{\displaystyle 0\quad 1}
1
2
1
2
{\displaystyle {\frac {1}{2}}\quad {\frac {1}{2}}}
−
(
b
−
a
)
3
12
f
″
(
ξ
)
{\displaystyle -{\frac {(b-a)^{3}}{12}}f''(\xi )}
2
Simpson-Regel Keplersche Fassregel
0
1
2
1
{\displaystyle 0\quad {\frac {1}{2}}\quad 1}
1
6
4
6
1
6
{\displaystyle {\frac {1}{6}}\quad {\frac {4}{6}}\quad {\frac {1}{6}}}
−
(
b
−
a
2
)
5
90
f
(
4
)
(
ξ
)
{\displaystyle -{\frac {\left({\frac {b-a}{2}}\right)^{5}}{90}}f^{(4)}(\xi )}
3
3/8-Regel
0
1
3
2
3
1
{\displaystyle 0\quad {\frac {1}{3}}\quad {\frac {2}{3}}\quad 1}
1
8
3
8
3
8
1
8
{\displaystyle {\frac {1}{8}}\quad {\frac {3}{8}}\quad {\frac {3}{8}}\quad {\frac {1}{8}}}
−
3
(
b
−
a
3
)
5
80
f
(
4
)
(
ξ
)
{\displaystyle -{\frac {3\left({\frac {b-a}{3}}\right)^{5}}{80}}f^{(4)}(\xi )}
4
Milne -Regel Boole -Regel
0
1
4
2
4
3
4
1
{\displaystyle 0\quad {\frac {1}{4}}\quad {\frac {2}{4}}\quad {\frac {3}{4}}\quad 1}
7
90
32
90
12
90
32
90
7
90
{\displaystyle {\frac {7}{90}}\quad {\frac {32}{90}}\quad {\frac {12}{90}}\quad {\frac {32}{90}}\quad {\frac {7}{90}}}
−
8
(
b
−
a
4
)
7
945
f
(
6
)
(
ξ
)
{\displaystyle -{\frac {8\left({\frac {b-a}{4}}\right)^{7}}{945}}f^{(6)}(\xi )}
5
6-Punkt-Regel
0
1
5
2
5
3
5
4
5
1
{\displaystyle 0\quad {\frac {1}{5}}\quad {\frac {2}{5}}\quad {\frac {3}{5}}\quad {\frac {4}{5}}\quad 1}
19
288
75
288
50
288
50
288
75
288
19
288
{\displaystyle {\frac {19}{288}}\quad {\frac {75}{288}}\quad {\frac {50}{288}}\quad {\frac {50}{288}}\quad {\frac {75}{288}}\quad {\frac {19}{288}}}
−
275
(
b
−
a
5
)
7
12
096
f
(
6
)
(
ξ
)
{\displaystyle -{\frac {275\left({\frac {b-a}{5}}\right)^{7}}{12\,096}}f^{(6)}(\xi )}
6
Weddle-Regel (nach Thomas Weddle , 1817–1853)[ 1]
0
1
6
2
6
3
6
4
6
5
6
1
{\displaystyle 0\quad {\frac {1}{6}}\quad {\frac {2}{6}}\quad {\frac {3}{6}}\quad {\frac {4}{6}}\quad {\frac {5}{6}}\quad 1}
41
840
216
840
27
840
272
840
27
840
216
840
41
840
{\displaystyle {\frac {41}{840}}\quad {\frac {216}{840}}\quad {\frac {27}{840}}\quad {\frac {272}{840}}\quad {\frac {27}{840}}\quad {\frac {216}{840}}\quad {\frac {41}{840}}}
−
9
(
b
−
a
6
)
9
1400
f
(
8
)
(
ξ
)
{\displaystyle -{\frac {9\left({\frac {b-a}{6}}\right)^{9}}{1400}}f^{(8)}(\xi )}
Die gekürzten Werte aller Gewichte bis
n
=
10
{\displaystyle n=10}
[ 2]
n
Gewichte
1
{
1
2
,
1
2
}
{\displaystyle \{{\tfrac {1}{2}},{\tfrac {1}{2}}\}}
2
{
1
6
,
2
3
,
1
6
}
{\displaystyle \{{\tfrac {1}{6}},{\tfrac {2}{3}},{\tfrac {1}{6}}\}}
3
{
1
8
,
3
8
,
3
8
,
1
8
}
{\displaystyle \{{\tfrac {1}{8}},{\tfrac {3}{8}},{\tfrac {3}{8}},{\tfrac {1}{8}}\}}
4
{
7
90
,
16
45
,
2
15
,
16
45
,
7
90
}
{\displaystyle \{{\tfrac {7}{90}},{\tfrac {16}{45}},{\tfrac {2}{15}},{\tfrac {16}{45}},{\tfrac {7}{90}}\}}
5
{
19
288
,
25
96
,
25
144
,
25
144
,
25
96
,
19
288
}
{\displaystyle \{{\tfrac {19}{288}},{\tfrac {25}{96}},{\tfrac {25}{144}},{\tfrac {25}{144}},{\tfrac {25}{96}},{\tfrac {19}{288}}\}}
6
{
41
840
,
9
35
,
9
280
,
34
105
,
9
280
,
9
35
,
41
840
}
{\displaystyle \{{\tfrac {41}{840}},{\tfrac {9}{35}},{\tfrac {9}{280}},{\tfrac {34}{105}},{\tfrac {9}{280}},{\tfrac {9}{35}},{\tfrac {41}{840}}\}}
7
{
751
17280
,
3577
17280
,
49
640
,
2989
17280
,
2989
17280
,
49
640
,
3577
17280
,
751
17280
}
{\displaystyle \{{\tfrac {751}{17280}},{\tfrac {3577}{17280}},{\tfrac {49}{640}},{\tfrac {2989}{17280}},{\tfrac {2989}{17280}},{\tfrac {49}{640}},{\tfrac {3577}{17280}},{\tfrac {751}{17280}}\}}
8
{
989
28350
,
2944
14175
,
−
464
14175
,
5248
14175
,
−
454
2835
,
5248
14175
,
−
464
14175
,
2944
14175
,
989
28350
}
{\displaystyle \{{\tfrac {989}{28350}},{\tfrac {2944}{14175}},-{\tfrac {464}{14175}},{\tfrac {5248}{14175}},-{\tfrac {454}{2835}},{\tfrac {5248}{14175}},-{\tfrac {464}{14175}},{\tfrac {2944}{14175}},{\tfrac {989}{28350}}\}}
9
{
2857
89600
,
15741
89600
,
27
2240
,
1209
5600
,
2889
44800
,
2889
44800
,
1209
5600
,
27
2240
,
15741
89600
,
2857
89600
}
{\displaystyle \{{\tfrac {2857}{89600}},{\tfrac {15741}{89600}},{\tfrac {27}{2240}},{\tfrac {1209}{5600}},{\tfrac {2889}{44800}},{\tfrac {2889}{44800}},{\tfrac {1209}{5600}},{\tfrac {27}{2240}},{\tfrac {15741}{89600}},{\tfrac {2857}{89600}}\}}
10
{
16067
598752
,
26575
149688
,
−
16175
199584
,
5675
12474
,
−
4825
11088
,
17807
24948
,
−
4825
11088
,
5675
12474
,
−
16175
199584
,
26575
149688
,
16067
598752
}
{\displaystyle \{{\tfrac {16067}{598752}},{\tfrac {26575}{149688}},-{\tfrac {16175}{199584}},{\tfrac {5675}{12474}},-{\tfrac {4825}{11088}},{\tfrac {17807}{24948}},-{\tfrac {4825}{11088}},{\tfrac {5675}{12474}},-{\tfrac {16175}{199584}},{\tfrac {26575}{149688}},{\tfrac {16067}{598752}}\}}
Für
n
=
8
{\displaystyle n=8}
w
i
<
0
{\displaystyle w_{i}<0}
i
=
2
,
4
,
6
{\displaystyle i=2,4,6}
∑
i
=
0
n
|
w
i
|
=
1
,
45
…
{\displaystyle \textstyle \sum _{i=0}^{n}|w_{i}|=1{,}45\dots }
n
=
10
{\displaystyle n=10}
∑
i
=
0
n
|
w
i
|
=
3,064
794
…
{\displaystyle \sum _{i=0}^{n}|w_{i}|=3{,}064794\dots }
Beispiel :
∫
1
3
1
x
d
x
=
ln
(
3
)
−
ln
(
1
)
=
ln
(
3
)
=
1,098
612
…
{\displaystyle \int \limits _{1}^{3}{\frac {1}{x}}\,dx=\ln(3)-\ln(1)=\ln(3)=1{,}098612\dots }
Näherung mit Simpson-Regel (
n
=
2
{\displaystyle n=2}
h
=
b
−
a
n
=
2
2
=
1
{\displaystyle h={\frac {b-a}{n}}={\frac {2}{2}}=1}
x
0
=
a
=
1
{\displaystyle x_{0}=a=1}
∫
1
3
p
2
(
x
)
d
x
=
2
⋅
(
1
6
f
(
1
)
+
4
6
f
(
2
)
+
1
6
f
(
3
)
)
=
2
⋅
(
1
6
⋅
1
+
4
6
⋅
1
2
+
1
6
⋅
1
3
)
=
10
9
=
1
,
1
¯
{\displaystyle \int \limits _{1}^{3}p_{2}(x)\,dx=2\cdot \left({\frac {1}{6}}f(1)+{\frac {4}{6}}f(2)+{\frac {1}{6}}f(3)\right)=2\cdot \left({\frac {1}{6}}\cdot 1+{\frac {4}{6}}\cdot {\frac {1}{2}}+{\frac {1}{6}}\cdot {\frac {1}{3}}\right)={\frac {10}{9}}=1{,}{\overline {1}}}
Verfahrensfehler : Mit
f
(
4
)
(
ξ
)
=
4
!
ξ
5
{\displaystyle f^{(4)}(\xi )={\frac {4!}{\xi ^{5}}}}
E
(
f
)
=
−
1
90
⋅
(
2
2
)
5
⋅
4
!
ξ
5
=
−
4
15
⋅
1
ξ
5
{\displaystyle E(f)=-{\frac {1}{90}}\cdot \left({\frac {2}{2}}\right)^{5}\cdot {\frac {4!}{\xi ^{5}}}=-{\frac {4}{15}}\cdot {\frac {1}{\xi ^{5}}}}
ξ
∈
[
1
,
3
]
{\displaystyle \xi \in [1,3]}
Fehlerabschätzung :
|
E
(
f
)
|
≤
4
15
⋅
max
1
≤
ξ
≤
3
|
1
ξ
5
|
=
4
15
⋅
1
1
=
0
,
2
6
¯
{\displaystyle |E(f)|\leq {\frac {4}{15}}\cdot \max _{1\leq \xi \leq 3}\left|{\frac {1}{\xi ^{5}}}\right|={\frac {4}{15}}\cdot {\frac {1}{1}}=0{,}2{\overline {6}}}
Exakter Fehler :
a
˘
|
E
(
f
)
|
=
|
∫
1
3
1
x
d
x
−
∫
1
3
p
2
(
x
)
d
x
|
=
|
1,098
612
…
−
1
,
1
¯
|
=
0,012
498
…
<
0
,
2
6
¯
{\displaystyle {\breve {a}}|E(f)|=\left|\int \limits _{1}^{3}{\frac {1}{x}}\,dx-\int \limits _{1}^{3}p_{2}(x)\,dx\right|=\left|1{,}098612\ldots -1{,}{\overline {1}}\right|=0{,}012498\ldots <0{,}2{\overline {6}}}
Die Stützstellen
t
i
{\displaystyle t_{i}}
[
0
,
1
]
{\displaystyle [0,1]}
t
0
=
1
n
+
2
,
t
i
=
i
+
1
n
+
2
,
t
n
=
n
+
1
n
+
2
{\displaystyle t_{0}={\tfrac {1}{n+2}},t_{i}={\tfrac {i+1}{n+2}},t_{n}={\tfrac {n+1}{n+2}}}
Intervall
[
a
,
b
]
{\displaystyle [a,b]}
x
i
=
a
+
t
i
⋅
(
b
−
a
)
{\displaystyle x_{i}=a+t_{i}\cdot (b-a)}
n
{\displaystyle n}
Name
Stützstellen
t
i
{\displaystyle t_{i}}
Gewichte
w
i
{\displaystyle w_{i}}
E
(
f
)
{\displaystyle E(f)}
0
Rechteckregel Mittelpunktsregel Tangententrapezregel
1
2
{\displaystyle {\frac {1}{2}}}
1
{\displaystyle 1\quad }
(
b
−
a
)
3
24
f
″
(
ξ
)
{\displaystyle {\frac {(b-a)^{3}}{24}}f''(\xi )}
1
1
3
2
3
{\displaystyle {\frac {1}{3}}\quad {\frac {2}{3}}}
1
2
1
2
{\displaystyle {\frac {1}{2}}\quad {\frac {1}{2}}}
3
(
b
−
a
3
)
3
4
f
″
(
ξ
)
{\displaystyle {\frac {3\left({\frac {b-a}{3}}\right)^{3}}{4}}f''(\xi )}
2
1
4
2
4
3
4
{\displaystyle {\frac {1}{4}}\quad {\frac {2}{4}}\quad {\frac {3}{4}}}
2
3
−
1
3
2
3
{\displaystyle {\frac {2}{3}}\quad -{\frac {1}{3}}\quad {\frac {2}{3}}}
14
(
b
−
a
4
)
5
45
f
(
4
)
(
ξ
)
{\displaystyle {\frac {14\left({\frac {b-a}{4}}\right)^{5}}{45}}f^{(4)}(\xi )}
3
1
5
2
5
3
5
4
5
{\displaystyle {\frac {1}{5}}\quad {\frac {2}{5}}\quad {\frac {3}{5}}\quad {\frac {4}{5}}}
11
24
1
24
1
24
11
24
{\displaystyle {\frac {11}{24}}\quad {\frac {1}{24}}\quad {\frac {1}{24}}\quad {\frac {11}{24}}}
95
(
b
−
a
5
)
5
144
f
(
4
)
(
ξ
)
{\displaystyle {\frac {95\left({\frac {b-a}{5}}\right)^{5}}{144}}f^{(4)}(\xi )}
4
1
6
2
6
3
6
4
6
5
6
{\displaystyle {\frac {1}{6}}\quad {\frac {2}{6}}\quad {\frac {3}{6}}\quad {\frac {4}{6}}\quad {\frac {5}{6}}}
11
20
−
14
20
26
20
−
14
20
11
20
{\displaystyle {\frac {11}{20}}\quad -{\frac {14}{20}}\quad {\frac {26}{20}}\quad -{\frac {14}{20}}\quad {\frac {11}{20}}}
41
(
b
−
a
6
)
7
140
f
(
6
)
(
ξ
)
{\displaystyle {\frac {41\left({\frac {b-a}{6}}\right)^{7}}{140}}f^{(6)}(\xi )}
5
1
7
2
7
3
7
4
7
5
7
6
7
{\displaystyle {\frac {1}{7}}\quad {\frac {2}{7}}\quad {\frac {3}{7}}\quad {\frac {4}{7}}\quad {\frac {5}{7}}\quad {\frac {6}{7}}}
611
1440
−
453
1440
562
1440
562
1440
−
453
1440
611
1440
{\displaystyle {\frac {611}{1440}}\quad -{\frac {453}{1440}}\quad {\frac {562}{1440}}\quad {\frac {562}{1440}}\quad -{\frac {453}{1440}}\quad {\frac {611}{1440}}}
5257
(
b
−
a
7
)
7
8640
f
(
6
)
(
ξ
)
{\displaystyle {\frac {5257\left({\frac {b-a}{7}}\right)^{7}}{8640}}f^{(6)}(\xi )}
6
1
8
2
8
3
8
4
8
5
8
6
8
7
8
{\displaystyle {\frac {1}{8}}\quad {\frac {2}{8}}\quad {\frac {3}{8}}\quad {\frac {4}{8}}\quad {\frac {5}{8}}\quad {\frac {6}{8}}\quad {\frac {7}{8}}}
460
945
−
954
945
2196
945
−
2459
945
2196
945
−
954
945
460
945
{\displaystyle {\frac {460}{945}}\quad -{\frac {954}{945}}\quad {\frac {2196}{945}}\quad -{\frac {2459}{945}}\quad {\frac {2196}{945}}\quad -{\frac {954}{945}}\quad {\frac {460}{945}}}
3956
(
b
−
a
8
)
9
14
175
f
(
8
)
(
ξ
)
{\displaystyle {\frac {3956\left({\frac {b-a}{8}}\right)^{9}}{14\,175}}f^{(8)}(\xi )}
Für
n
=
5
{\displaystyle n=5}
∑
i
=
0
n
|
w
i
|
=
3252
1440
=
2,258
333
…
{\displaystyle \textstyle \sum _{i=0}^{n}|w_{i}|={\frac {3252}{1440}}=2{,}258333\dots }
n
=
6
{\displaystyle n=6}
∑
i
=
0
n
|
w
i
|
=
9679
945
=
10
,
24
…
{\displaystyle \textstyle \sum _{i=0}^{n}|w_{i}|={\frac {9679}{945}}=10{,}24\dots }
Von diesen Formeln ist nur die Rechteckregel empfehlenswert. Die Formel für
n
=
1
{\displaystyle n=1}
Beispiel :
∫
1
3
1
x
d
x
=
ln
(
3
)
−
ln
(
1
)
=
ln
(
3
)
=
1,098
612
…
{\displaystyle \int \limits _{1}^{3}{\frac {1}{x}}\,dx=\ln(3)-\ln(1)=\ln(3)=1{,}098612\dots }
Näherung mit der Formel für
n
=
2
{\displaystyle n=2}
h
=
b
−
a
n
+
2
=
2
4
=
1
2
{\displaystyle h={\frac {b-a}{n+2}}={\frac {2}{4}}={\frac {1}{2}}}
x
0
=
a
+
h
=
3
2
{\displaystyle x_{0}=a+h={\frac {3}{2}}}
∫
1
3
p
2
(
x
)
d
x
=
2
⋅
(
2
3
f
(
3
2
)
−
1
3
f
(
4
2
)
+
2
3
f
(
5
2
)
)
=
2
⋅
(
2
3
⋅
2
3
−
1
3
⋅
2
4
+
2
3
⋅
2
5
)
=
49
45
=
1
,
0
8
¯
{\displaystyle \int \limits _{1}^{3}p_{2}(x)\,dx=2\cdot \left({\frac {2}{3}}f\!\left({\frac {3}{2}}\right)-{\frac {1}{3}}f\!\left({\frac {4}{2}}\right)+{\frac {2}{3}}f\!\left({\frac {5}{2}}\right)\right)=2\cdot \left({\frac {2}{3}}\cdot {\frac {2}{3}}-{\frac {1}{3}}\cdot {\frac {2}{4}}+{\frac {2}{3}}\cdot {\frac {2}{5}}\right)={\frac {49}{45}}=1{,}0{\overline {8}}}
Verfahrensfehler : Mit
f
(
4
)
(
ξ
)
=
4
!
ξ
5
{\displaystyle f^{(4)}(\xi )={\frac {4!}{\xi ^{5}}}}
E
(
f
)
=
14
45
⋅
(
2
4
)
5
⋅
4
!
ξ
5
=
7
30
⋅
1
ξ
5
{\displaystyle E(f)={\frac {14}{45}}\cdot \left({\frac {2}{4}}\right)^{5}\cdot {\frac {4!}{\xi ^{5}}}={\frac {7}{30}}\cdot {\frac {1}{\xi ^{5}}}}
ξ
∈
[
1
,
3
]
{\displaystyle \xi \in [1,3]}
Fehlerabschätzung :
|
E
(
f
)
|
≤
7
30
⋅
max
1
≤
ξ
≤
3
|
1
ξ
5
|
=
7
30
⋅
1
1
=
0
,
2
3
¯
{\displaystyle |E(f)|\leq {\frac {7}{30}}\cdot \max _{1\leq \xi \leq 3}\left|{\frac {1}{\xi ^{5}}}\right|={\frac {7}{30}}\cdot {\frac {1}{1}}=0{,}2{\overline {3}}}
Exakter Fehler :
|
E
(
f
)
|
=
|
∫
1
3
1
x
d
x
−
∫
1
3
p
2
(
x
)
d
x
|
=
|
1,098
612
…
−
1
,
0
8
¯
|
=
0,009
723
…
<
0
,
2
3
¯
{\displaystyle |E(f)|=\left|\int \limits _{1}^{3}{\frac {1}{x}}\,dx-\int \limits _{1}^{3}p_{2}(x)\,dx\right|=\left|1{,}098612\ldots -1{,}0{\overline {8}}\right|=0{,}009723\ldots <0{,}2{\overline {3}}}
Diese Formeln sind nach Colin Maclaurin benannt. Die Stützstellen
t
i
{\displaystyle t_{i}}
[
0
,
1
]
{\displaystyle [0,1]}
t
0
=
1
2
n
+
2
,
t
i
=
2
i
+
1
2
n
+
2
,
t
n
=
2
n
+
1
2
n
+
2
{\displaystyle t_{0}={\tfrac {1}{2n+2}},t_{i}={\tfrac {2i+1}{2n+2}},t_{n}={\tfrac {2n+1}{2n+2}}}
Intervall
[
a
,
b
]
{\displaystyle [a,b]}
x
i
=
a
+
t
i
⋅
(
b
−
a
)
{\displaystyle x_{i}=a+t_{i}\cdot (b-a)}
n
{\displaystyle n}
Name
Stützstellen
t
i
{\displaystyle t_{i}}
Gewichte
w
i
{\displaystyle w_{i}}
E
(
f
)
{\displaystyle E(f)}
0
Rechteckregel Mittelpunktsregel Tangententrapezregel
1
2
{\displaystyle {\frac {1}{2}}}
1
{\displaystyle 1\quad }
(
b
−
a
)
3
24
f
″
(
ξ
)
{\displaystyle {\frac {(b-a)^{3}}{24}}f''(\xi )}
1
1
4
3
4
{\displaystyle {\frac {1}{4}}\quad {\frac {3}{4}}}
1
2
1
2
{\displaystyle {\frac {1}{2}}\quad {\frac {1}{2}}}
(
b
−
a
2
)
3
12
f
″
(
ξ
)
{\displaystyle {\frac {\left({\frac {b-a}{2}}\right)^{3}}{12}}f''(\xi )}
2
1
6
1
2
5
6
{\displaystyle {\frac {1}{6}}\quad {\frac {1}{2}}\quad {\frac {5}{6}}}
3
8
2
8
3
8
{\displaystyle {\frac {3}{8}}\quad {\frac {2}{8}}\quad {\frac {3}{8}}}
21
(
b
−
a
3
)
5
640
f
(
4
)
(
ξ
)
{\displaystyle {\frac {21\left({\frac {b-a}{3}}\right)^{5}}{640}}f^{(4)}(\xi )}
3
1
8
3
8
5
8
7
8
{\displaystyle {\frac {1}{8}}\quad {\frac {3}{8}}\quad {\frac {5}{8}}\quad {\frac {7}{8}}}
13
48
11
48
11
48
13
48
{\displaystyle {\frac {13}{48}}\quad {\frac {11}{48}}\quad {\frac {11}{48}}\quad {\frac {13}{48}}}
103
(
b
−
a
4
)
5
1440
f
(
4
)
(
ξ
)
{\displaystyle {\frac {103\left({\frac {b-a}{4}}\right)^{5}}{1440}}f^{(4)}(\xi )}
4
1
10
3
10
5
10
7
10
9
10
{\displaystyle {\frac {1}{10}}\quad {\frac {3}{10}}\quad {\frac {5}{10}}\quad {\frac {7}{10}}\quad {\frac {9}{10}}}
275
1152
100
1152
402
1152
100
1152
275
1152
{\displaystyle {\frac {275}{1152}}\quad {\frac {100}{1152}}\quad {\frac {402}{1152}}\quad {\frac {100}{1152}}\quad {\frac {275}{1152}}}
5575
(
b
−
a
5
)
7
193
536
f
(
6
)
(
ξ
)
{\displaystyle {\frac {5575\left({\frac {b-a}{5}}\right)^{7}}{193\,536}}f^{(6)}(\xi )}
Für
n
=
6
{\displaystyle n=6}
∑
i
=
0
n
|
w
i
|
=
1,363
…
{\displaystyle \sum _{i=0}^{n}|w_{i}|=1{,}363\dots }
n
=
8
{\displaystyle n=8}
∑
i
=
0
n
|
w
i
|
=
3,433
…
{\displaystyle \sum _{i=0}^{n}|w_{i}|=3{,}433\dots }
Beispiel :
∫
1
3
1
x
d
x
=
ln
(
3
)
−
ln
(
1
)
=
ln
(
3
)
=
1,098
612
…
{\displaystyle \int \limits _{1}^{3}{\frac {1}{x}}\,dx=\ln(3)-\ln(1)=\ln(3)=1{,}098612\dots }
Näherung mit der Formel für
n
=
2
{\displaystyle n=2}
h
=
b
−
a
n
+
1
=
2
3
{\displaystyle h={\frac {b-a}{n+1}}={\frac {2}{3}}}
x
0
=
a
+
h
2
=
4
3
{\displaystyle x_{0}=a+{\frac {h}{2}}={\frac {4}{3}}}
∫
1
3
p
2
(
x
)
d
x
=
2
⋅
(
3
8
f
(
4
3
)
+
2
8
f
(
6
3
)
+
3
8
f
(
8
3
)
)
=
2
⋅
(
3
8
⋅
3
4
+
2
8
⋅
3
6
+
3
8
⋅
3
8
)
=
105
96
=
1,093
75
{\displaystyle \int \limits _{1}^{3}p_{2}(x)\,dx=2\cdot \left({\frac {3}{8}}f\!\left({\frac {4}{3}}\right)+{\frac {2}{8}}f\!\left({\frac {6}{3}}\right)+{\frac {3}{8}}f\!\left({\frac {8}{3}}\right)\right)=2\cdot \left({\frac {3}{8}}\cdot {\frac {3}{4}}+{\frac {2}{8}}\cdot {\frac {3}{6}}+{\frac {3}{8}}\cdot {\frac {3}{8}}\right)={\frac {105}{96}}=1{,}09375}
Verfahrensfehler : Mit
f
(
4
)
(
ξ
)
=
4
!
ξ
5
{\displaystyle f^{(4)}(\xi )={\frac {4!}{\xi ^{5}}}}
E
(
f
)
=
21
640
⋅
(
2
3
)
5
⋅
4
!
ξ
5
=
14
135
⋅
1
ξ
5
{\displaystyle E(f)={\frac {21}{640}}\cdot \left({\frac {2}{3}}\right)^{5}\cdot {\frac {4!}{\xi ^{5}}}={\frac {14}{135}}\cdot {\frac {1}{\xi ^{5}}}}
ξ
∈
[
1
,
3
]
{\displaystyle \xi \in [1,3]}
Fehlerabschätzung :
|
E
(
f
)
|
≤
14
135
⋅
max
1
≤
ξ
≤
3
|
1
ξ
5
|
=
14
135
⋅
1
1
=
0
,
1
037
¯
{\displaystyle |E(f)|\leq {\frac {14}{135}}\cdot \max _{1\leq \xi \leq 3}\left|{\frac {1}{\xi ^{5}}}\right|={\frac {14}{135}}\cdot {\frac {1}{1}}=0{,}1{\overline {037}}}
Exakter Fehler :
|
E
(
f
)
|
=
|
∫
1
3
1
x
d
x
−
∫
1
3
p
2
(
x
)
d
x
|
=
|
1,098
612
…
−
1,093
75
|
=
0,000
486
…
<
0
,
1
037
¯
{\displaystyle |E(f)|=\left|\int \limits _{1}^{3}{\frac {1}{x}}\,dx-\int \limits _{1}^{3}p_{2}(x)\,dx\right|=|1{,}098612\ldots -1{,}09375|=0{,}000486\ldots <0{,}1{\overline {037}}}
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle \xi \in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cc05d1fb45b90e25c99bc6a57473d508d3e9c23)
![{\displaystyle [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
![{\displaystyle \xi \in [1,3]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b658b27a1b1babd01aa16dc7426ceddb92d85fb)
