Problem 2:Continuation of proof of trapezoidal error
edit
P. 27-1
Continue the proof of Trapezoidal Rule Error to steps 4a and 4b and determine
P
6
(
t
)
{\displaystyle {\boldsymbol {P_{6}(t)}}}
P
7
(
t
)
{\displaystyle {\boldsymbol {P_{7}(t)}}}
From steps 3a and 3b we get the expression
E
=
[
P
2
(
t
)
g
(
1
)
(
t
)
+
P
4
(
t
)
g
(
3
)
(
t
)
]
−
1
+
1
−
∫
−
1
+
1
P
5
(
t
)
g
(
5
)
(
t
)
d
t
⏟
D
{\displaystyle \displaystyle {\boldsymbol {E=[P_{2}(t)g^{(1)}(t)+P_{4}(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace {\int _{-1}^{+1}P_{5}(t)g^{(5)}(t)dt} _{D}}}}
(Summary p 26-3)
P
4
(
t
)
=
−
t
4
24
+
t
2
12
−
7
360
=
c
1
(
t
4
4
!
)
+
c
3
(
t
2
2
!
)
+
c
5
{\displaystyle \displaystyle P_{4}(t)={\frac {-t^{4}}{24}}+{\frac {t^{2}}{12}}-{\frac {7}{360}}=c_{1}({\frac {t^{4}}{4!}})+c_{3}({\frac {t^{2}}{2!}})+c_{5}}
P
5
(
t
)
=
−
t
5
120
+
t
3
36
−
7
t
360
=
c
1
(
t
5
5
!
)
+
c
3
(
t
3
3
!
)
+
c
5
(
t
)
{\displaystyle P_{5}(t)={\frac {-t^{5}}{120}}+{\frac {t^{3}}{36}}-{\frac {7t}{360}}=c_{1}({\frac {t^{5}}{5!}})+c_{3}({\frac {t^{3}}{3!}})+c_{5}(t)}
(Summary p 26-3)
∫
−
1
+
1
P
5
(
t
)
⏟
v
′
g
(
5
)
(
t
)
⏟
u
d
t
=
[
g
(
5
)
(
t
)
P
6
(
t
)
]
−
1
+
1
−
∫
−
1
+
1
[
P
6
(
t
)
g
(
6
)
(
t
)
⏟
E
]
d
t
{\displaystyle \displaystyle \int _{-1}^{+1}\underbrace {P_{5}(t)} _{v'}\underbrace {g^{(5)}(t)} _{u}dt=[g^{(5)}(t)P_{6}(t)]_{-1}^{+1}-\underbrace {\int _{-1}^{+1}[{P_{6}(t)g^{(6)}(t)}} _{E}]dt}
where,
P
6
(
t
)
=
∫
P
5
(
t
)
=
c
1
(
t
6
6
!
)
+
c
3
(
t
4
4
!
)
+
c
5
(
t
2
2
!
)
+
c
7
{\displaystyle P_{6}(t)=\int P_{5}(t)=c_{1}({\frac {t^{6}}{6!}})+c_{3}({\frac {t^{4}}{4!}})+c_{5}({\frac {t^{2}}{2!}})+c_{7}}
P
6
(
t
)
=
−
t
6
720
+
t
4
144
−
7
t
2
720
+
α
{\displaystyle P_{6}(t)={\frac {-t^{6}}{720}}+{\frac {t^{4}}{144}}-{\frac {7t^{2}}{720}}+\alpha }
∫
−
1
+
1
[
P
6
(
t
)
⏟
v
′
g
(
6
)
(
t
)
⏟
u
]
d
t
=
D
=
[
g
(
6
)
(
t
)
P
7
(
t
)
]
−
1
+
1
−
∫
−
1
+
1
P
7
(
t
)
g
(
7
)
(
t
)
d
t
{\displaystyle \displaystyle \int _{-1}^{+1}[{\underbrace {P_{6}(t)} _{v'}\underbrace {g^{(6)}(t)} _{u}}]dt=D=[g^{(6)}(t)P_{7}(t)]_{-1}^{+1}-\int _{-1}^{+1}P_{7}(t)g^{(7)}(t)dt}
where,
P
7
(
t
)
=
∫
P
6
(
t
)
==
c
1
(
t
7
7
!
)
+
c
3
(
t
5
5
!
)
+
c
5
(
t
3
3
!
)
+
c
7
(
t
)
+
c
8
{\displaystyle P_{7}(t)=\int P_{6}(t)==c_{1}({\frac {t^{7}}{7!}})+c_{3}({\frac {t^{5}}{5!}})+c_{5}({\frac {t^{3}}{3!}})+c_{7}(t)+c_{8}}
P
7
(
t
)
=
−
t
7
5040
+
t
5
720
−
7
t
3
2160
+
α
t
+
β
{\displaystyle P_{7}(t)={\frac {-t^{7}}{5040}}+{\frac {t^{5}}{720}}-{\frac {7t^{3}}{2160}}+\alpha t+\beta }
P
7
(
t
)
{\displaystyle \displaystyle {P_{7}(t)}}
P
7
(
+
1
)
=
P
7
(
−
1
)
=
P
7
(
0
)
=
0
{\displaystyle \displaystyle {P_{7}(+1)=P_{7}(-1)=P_{7}(0)=0}}
P
7
(
t
=
0
)
=
0
⇒
β
=
c
8
=
0
{\displaystyle \displaystyle {P_{7}(t=0)=0\Rightarrow \beta =c_{8}=0}}
P
(
−
1
)
=
0
=
−
(
−
1
)
7
7
!
+
1
6
(
(
−
1
)
5
5
!
)
−
7
360
(
(
−
1
)
3
3
!
)
−
α
⇒
α
=
c
7
=
31
15120
{\displaystyle P(-1)=0=-{\frac {(-1)^{7}}{7!}}+{\frac {1}{6}}({\frac {(-1)^{5}}{5!}})-{\frac {7}{360}}({\frac {(-1)^{3}}{3!}})-\alpha \Rightarrow \alpha =c_{7}={\frac {31}{15120}}}
P
6
(
t
)
=
−
t
6
600
+
t
4
108
−
7
t
2
720
+
31
15120
{\displaystyle \displaystyle P_{6}(t)={\frac {-t^{6}}{600}}+{\frac {t^{4}}{108}}-{\frac {7t^{2}}{720}}+{\frac {31}{15120}}}
P
7
(
t
)
=
∫
P
6
(
t
)
=
−
t
7
4200
+
t
5
540
−
7
t
3
2160
+
31
15120
t
{\displaystyle P_{7}(t)=\int P_{6}(t)={\frac {-t^{7}}{4200}}+{\frac {t^{5}}{540}}-{\frac {7t^{3}}{2160}}+{\frac {31}{15120}}t}
(Summary p 26-3)
--Egm6341.s10.team2.niki 21:38, 23 March 2010 (UTC)
Problem 7: Understanding the derivation of the proof of Trapezoidal Error
edit
P. 28-2
Redo steps in the proof of the Trapezoidal Rule error by trying to cancel terms with odd order derivatives of "g"
We begin with equation (5) on P. 21-1 which is the result of transformation of variables on equation (1) P. 21-1
(Prob 4 HW4 ) (P. 21-1 )
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
∫
−
1
1
g
k
(
t
)
d
t
−
[
g
k
(
−
1
)
+
g
k
(
+
1
)
]
]
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\int _{-1}^{1}g_{k}(t)dt-[g_{k}(-1)+g_{k}({+1})]]}
(5) on P.21-1
From Prob 5 HW4 , we can express the above equation as:
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
∫
−
1
+
1
(
−
t
)
g
(
1
)
(
t
)
d
t
]
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\int _{-1}^{+1}(-t)g^{(1)}(t)dt]}
(1)
Integrating the term withing the square brackets by "Integration by parts" Prob 7 HW4 we can rewrite (1) as follows
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
[
P
2
(
t
)
g
(
1
)
(
t
)
]
−
1
+
1
−
∫
−
1
+
1
P
2
(
t
)
g
(
2
)
(
t
)
d
t
]
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[[P_{2}(t)g^{(1)}(t)]_{-1}^{+1}-\int _{-1}^{+1}P_{2}(t)g^{(2)}(t)dt]}
(2)
In order to eliminate terms with even powers of
h
{\displaystyle \displaystyle {h}}
g
(
t
)
{\displaystyle \displaystyle {g(t)}}
P
2
(
t
)
{\displaystyle \displaystyle {P_{2}(t)}}
We have
P
2
(
t
)
{\displaystyle \displaystyle {P_{2}(t)}}
P. 21-3
P
2
(
t
)
=
c
1
(
t
2
2
!
)
+
c
3
{\displaystyle \displaystyle P_{2}(t)=c_{1}({\frac {t^{2}}{2!}})+c_{3}}
(1)p21-3
c
2
=
0
{\displaystyle \displaystyle c_{2}=0}
(2)p21-3
Setting
P
2
(
−
1
)
=
0
{\displaystyle \displaystyle {P_{2}(-1)=0}}
c
3
=
1
/
2
{\displaystyle \displaystyle {c_{3}=1/2}}
P
2
(
t
)
=
c
1
(
t
2
2
!
)
+
c
3
{\displaystyle \displaystyle P_{2}(t)=c_{1}({\frac {t^{2}}{2!}})+c_{3}}
(3)
So following this method, the next term to be eliminated will have P4(t)
P
4
(
t
)
=
c
1
(
t
4
4
!
)
+
c
3
(
t
2
2
!
)
+
c
4
(
t
)
+
c
5
{\displaystyle \displaystyle P_{4}(t)=c_{1}({\frac {t^{4}}{4!}})+c_{3}({\frac {t^{2}}{2!}})+c_{4}(t)+c_{5}}
(4)
Setting
P
4
(
−
1
)
=
0
a
n
d
P
4
(
1
)
=
0
{\displaystyle \displaystyle {P_{4}(-1)=0andP_{4}(1)=0}}
c
4
=
0
a
n
d
c
4
=
−
5
/
24
{\displaystyle \displaystyle {c_{4}=0andc_{4}=-5/24}}
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
P
2
g
(
1
)
]
−
1
+
1
−
[
P
3
g
(
2
)
]
−
1
+
1
+
[
P
4
g
(
3
)
]
−
1
+
1
−
[
P
5
g
(
4
)
]
−
1
+
1
+
[
P
6
g
(
5
)
]
−
1
+
1
−
[
P
7
g
(
6
)
]
−
1
+
1
−
∫
−
1
+
1
P
7
g
(
7
)
d
t
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}{[P_{2}g^{(1)}]_{-1}^{+1}-[P_{3}g^{(2)}]_{-1}^{+1}}+[P_{4}g^{(3)}]_{-1}^{+1}-[P_{5}g^{(4)}]_{-1}^{+1}+[P_{6}g^{(5)}]_{-1}^{+1}-[P_{7}g^{(6)}]_{-1}^{+1}-\int _{-1}^{+1}P_{7}g^{(7)}dt}
(5)
E
n
1
=
h
2
∑
k
=
0
n
−
1
−
[
P
3
g
(
2
)
]
−
1
+
1
+
−
[
P
5
g
(
4
)
]
−
1
+
1
+
−
[
P
7
g
(
6
)
]
−
1
+
1
−
∫
−
1
+
1
P
7
g
(
7
)
d
t
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}{-[P_{3}g^{(2)}]_{-1}^{+1}}+-[P_{5}g^{(4)}]_{-1}^{+1}+-[P_{7}g^{(6)}]_{-1}^{+1}-\int _{-1}^{+1}P_{7}g^{(7)}dt}
(6)
E
n
1
=
h
2
∑
k
=
0
n
−
1
(
∑
i
=
1
m
−
[
P
(
2
i
+
1
)
g
(
2
i
)
]
−
1
+
1
)
−
∫
−
1
+
1
P
(
2
m
+
1
)
g
(
2
m
+
1
)
d
t
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}(\sum _{i=1}^{m}-[P_{(2i+1)}g^{(2i)}]_{-1}^{+1})-\int _{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt}
(7)
manipulating the terms yields,
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
∑
i
=
1
m
−
[
(
P
(
2
i
+
1
)
(
+
1
)
g
(
2
i
)
(
+
1
)
)
−
(
P
(
2
i
+
1
)
(
−
1
)
g
(
2
i
)
(
−
1
)
)
]
−
∫
−
1
+
1
P
(
2
m
+
1
)
g
(
2
m
+
1
)
d
t
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\sum _{i=1}^{m}-[(P_{(2i+1)}(+1)g^{(2i)}(+1))-(P_{(2i+1)}(-1)g^{(2i)}(-1))]-\int _{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt}
(8)
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
∑
i
=
1
m
[
(
P
(
2
i
+
1
)
(
−
1
)
g
(
2
i
)
(
−
1
)
)
−
(
P
(
2
i
+
1
)
(
+
1
)
g
(
2
i
)
(
+
1
)
)
]
−
∫
−
1
+
1
P
(
2
m
+
1
)
g
(
2
m
+
1
)
d
t
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\sum _{i=1}^{m}[(P_{(2i+1)}(-1)g^{(2i)}(-1))-(P_{(2i+1)}(+1)g^{(2i)}(+1))]-\int _{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt}
(9)
Transforming g(t) back to f(x) we get [see [prob 6 HW4 ]
E
n
1
=
h
2
∑
k
=
0
n
−
1
[
∑
i
=
1
m
[
(
P
(
2
i
+
1
)
(
−
1
)
(
h
2
i
2
)
f
(
2
i
)
(
x
k
)
)
−
(
P
(
2
i
+
1
)
(
+
1
)
(
h
2
i
2
)
f
(
2
i
)
(
x
k
+
1
)
)
]
−
∫
x
k
x
k
+
1
P
(
2
m
+
1
)
(
x
)
(
h
2
m
+
1
2
)
f
(
2
m
+
1
)
(
x
)
)
d
t
{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\sum _{i=1}^{m}[(P_{(2i+1)}(-1)({\frac {h^{2i}}{2}})f^{(2i)}(x_{k}))-(P_{(2i+1)}(+1)({\frac {h^{2i}}{2}})f^{(2i)}(x_{k+1}))]-\int _{x_{k}}^{x_{k+1}}P_{(2m+1)}(x)({\frac {h^{2m+1}}{2}})f^{(2m+1)}(x))dt}
(10)
To see the difference in the two approaches we must compare the equations from the two methods. From (1) P. 27-1 we have
E
n
1
=
∑
r
=
1
l
h
2
r
d
2
r
[
f
(
2
r
−
1
)
(
b
)
−
f
(
2
r
−
1
)
(
a
)
]
−
h
(
2
l
)
2
(
2
l
)
∑
k
=
0
n
−
1
∫
x
k
x
k
+
1
P
2
l
(
t
k
(
x
)
)
f
(
2
l
)
(
x
)
d
x
{\displaystyle \displaystyle E_{n}^{1}=\sum _{r=1}^{l}h^{2r}d_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-{\frac {h^{(2l)}}{2^{(2l)}}}\sum _{k=0}^{n-1}\int _{x_{k}}^{x_{k+1}}P_{2l}(t_{k}(x))f^{(2l)}(x)dx}
(11)
It is seen that the first term in eqn 10 is a summation as against the term of eqn (11) which is dependent only on the end points.
![{\displaystyle \displaystyle {\boldsymbol {E=[P_{2}(t)g^{(1)}(t)+P_{4}(t)g^{(3)}(t)]_{-1}^{+1}-\underbrace {\int _{-1}^{+1}P_{5}(t)g^{(5)}(t)dt} _{D}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f05afbde2f04704b76067518a639169b2a054390)
![{\displaystyle \displaystyle \int _{-1}^{+1}\underbrace {P_{5}(t)} _{v'}\underbrace {g^{(5)}(t)} _{u}dt=[g^{(5)}(t)P_{6}(t)]_{-1}^{+1}-\underbrace {\int _{-1}^{+1}[{P_{6}(t)g^{(6)}(t)}} _{E}]dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15ab1895cdc848658ce1640bbc292a68be63d20e)
![{\displaystyle \displaystyle \int _{-1}^{+1}[{\underbrace {P_{6}(t)} _{v'}\underbrace {g^{(6)}(t)} _{u}}]dt=D=[g^{(6)}(t)P_{7}(t)]_{-1}^{+1}-\int _{-1}^{+1}P_{7}(t)g^{(7)}(t)dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b40dab8864516a657a793cc5948dbf7e1c2cb4bc)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\int _{-1}^{1}g_{k}(t)dt-[g_{k}(-1)+g_{k}({+1})]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e148dfc1c782ebec210d1d746b18f5421d22a25)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\int _{-1}^{+1}(-t)g^{(1)}(t)dt]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23ba69610a19e3c0b3f1c8ffc15fb07f922d0f58)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[[P_{2}(t)g^{(1)}(t)]_{-1}^{+1}-\int _{-1}^{+1}P_{2}(t)g^{(2)}(t)dt]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71743df11bc628a2c222acf156b3be5c29407da9)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}{[P_{2}g^{(1)}]_{-1}^{+1}-[P_{3}g^{(2)}]_{-1}^{+1}}+[P_{4}g^{(3)}]_{-1}^{+1}-[P_{5}g^{(4)}]_{-1}^{+1}+[P_{6}g^{(5)}]_{-1}^{+1}-[P_{7}g^{(6)}]_{-1}^{+1}-\int _{-1}^{+1}P_{7}g^{(7)}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3968df3b7795984f9a47cafd402118ef6bc0463d)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}{-[P_{3}g^{(2)}]_{-1}^{+1}}+-[P_{5}g^{(4)}]_{-1}^{+1}+-[P_{7}g^{(6)}]_{-1}^{+1}-\int _{-1}^{+1}P_{7}g^{(7)}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f8fd6e40e5669cd2a71a7589b851751090750ac)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}(\sum _{i=1}^{m}-[P_{(2i+1)}g^{(2i)}]_{-1}^{+1})-\int _{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce9ca8e8f3c0dce7b271ea5247ba04ad8b317546)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\sum _{i=1}^{m}-[(P_{(2i+1)}(+1)g^{(2i)}(+1))-(P_{(2i+1)}(-1)g^{(2i)}(-1))]-\int _{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f77cf155d7f535da2d074188d0bc69bb73430a0)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\sum _{i=1}^{m}[(P_{(2i+1)}(-1)g^{(2i)}(-1))-(P_{(2i+1)}(+1)g^{(2i)}(+1))]-\int _{-1}^{+1}P_{(2m+1)}g^{(2m+1)}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb0d2006fcc402bd3a28eb7006ee2005df1835d4)
![{\displaystyle \displaystyle E_{n}^{1}={\frac {h}{2}}\sum _{k=0}^{n-1}[\sum _{i=1}^{m}[(P_{(2i+1)}(-1)({\frac {h^{2i}}{2}})f^{(2i)}(x_{k}))-(P_{(2i+1)}(+1)({\frac {h^{2i}}{2}})f^{(2i)}(x_{k+1}))]-\int _{x_{k}}^{x_{k+1}}P_{(2m+1)}(x)({\frac {h^{2m+1}}{2}})f^{(2m+1)}(x))dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/473c518ad274ae0164d0beb4e45994bb9f54e9f3)
![{\displaystyle \displaystyle E_{n}^{1}=\sum _{r=1}^{l}h^{2r}d_{2r}[f^{(2r-1)}(b)-f^{(2r-1)}(a)]-{\frac {h^{(2l)}}{2^{(2l)}}}\sum _{k=0}^{n-1}\int _{x_{k}}^{x_{k+1}}P_{2l}(t_{k}(x))f^{(2l)}(x)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4b3b0d0ff6756bfcea6898c0fc42c20bf5ecafb)