Consider the system
A
x
=
b
{\displaystyle Ax=b\,\!}
x
0
{\displaystyle \,\!x_{0}}
r
0
=
b
−
A
x
0
{\displaystyle \,\!r_{0}=b-Ax_{0}}
v
1
=
r
0
ν
{\displaystyle \,\!v_{1}={\frac {r_{0}}{\nu }}}
V
k
=
(
v
1
v
2
⋯
v
m
)
{\displaystyle \,\!V_{k}=(v_{1}\,v_{2}\,\cdots v_{m})}
A
V
k
=
V
k
+
1
H
k
{\displaystyle \,\!AV_{k}=V_{k+1}H_{k}}
where
H
k
{\displaystyle \,\!H_{k}}
(
k
+
1
)
×
k
{\displaystyle \,\!(k+1)\times k}
x
{\displaystyle \displaystyle x}
x
(
c
)
=
x
0
+
V
k
c
{\displaystyle \,\!x(c)=x_{0}+V_{k}c}
choosing
c
k
{\displaystyle \,\!c_{k}}
‖
r
(
c
)
‖
=
‖
b
−
A
x
(
c
)
‖
{\displaystyle \,\!\|r(c)\|=\|b-Ax(c)\|}
‖
⋅
‖
{\displaystyle \,\!\|\cdot \|}
Show that
c
k
{\displaystyle \,\!c_{k}}
‖
ν
e
1
−
H
k
c
‖
{\displaystyle \|\nu e_{1}-H_{k}c\|}
We wish to show that
‖
b
−
A
x
(
c
)
‖
=
‖
ν
e
1
−
H
k
c
‖
{\displaystyle \,\!\|b-Ax(c)\|=\|\nu e_{1}-H_{k}c\|}
‖
b
−
A
x
(
c
)
‖
=
‖
b
−
A
(
x
0
+
V
k
c
)
‖
=
‖
b
−
A
x
0
−
A
V
k
c
‖
=
‖
r
0
−
A
V
k
c
‖
=
‖
r
0
−
V
k
+
1
H
k
c
‖
=
‖
ν
v
1
−
V
k
+
1
H
k
c
‖
=
‖
V
k
+
1
(
ν
e
1
−
H
k
c
)
⏟
h
c
‖
=
(
V
k
+
1
h
c
,
V
k
+
1
h
c
)
1
2
=
(
(
V
k
+
1
h
c
)
T
V
k
+
1
h
c
)
1
2
=
(
h
c
T
V
k
+
1
T
V
k
+
1
h
c
)
1
2
=
(
h
c
T
h
c
)
1
2
=
‖
h
c
‖
=
‖
ν
e
1
−
H
k
c
‖
{\displaystyle {\begin{aligned}\|b-Ax(c)\|&=\|b-A(x_{0}+V_{k}c)\|\\&=\|b-Ax_{0}-AV_{k}c\|\\&=\|r_{0}-AV_{k}c\|\\&=\|r_{0}-V_{k+1}H_{k}c\|\\&=\|\nu v_{1}-V_{k+1}H_{k}c\|\\&=\|V_{k+1}\underbrace {(\nu e_{1}-H_{k}c)} _{h_{c}}\|\\&=(V_{k+1}h_{c},V_{k+1}h_{c})^{\frac {1}{2}}\\&=((V_{k+1}h_{c})^{T}V_{k+1}h_{c})^{\frac {1}{2}}\\&=(h_{c}^{T}V_{k+1}^{T}V_{k+1}h_{c})^{\frac {1}{2}}\\&=(h_{c}^{T}h_{c})^{\frac {1}{2}}\\&=\|h_{c}\|\\&=\|\nu e_{1}-H_{k}c\|\end{aligned}}}
Suppose we choose to solve the least squares problem in part a for
c
k
{\displaystyle c_{k}\!\,}
We would like to transform
H
k
{\displaystyle H_{k}\!\,}
(
k
+
1
)
×
k
{\displaystyle (k+1)\times k\!\,}
The cost is on the order of
4
k
2
+
1
2
k
2
=
4.5
k
2
{\displaystyle 4k^{2}+{\frac {1}{2}}k^{2}=4.5k^{2}}
edit
We need
k
{\displaystyle k\!\,}
k
{\displaystyle k\!\,}
H
k
{\displaystyle H_{k}\!\,}
R
{\displaystyle R\!\,}
Each successive Given's Rotations multiply on an upper Hessenberg matrix requires four fewer multiplies because each previous subdiagonal entry has been zeroed out by a Given's Rotation multiply.
Hence the cost of
k
{\displaystyle k\!\,}
4
k
+
4
(
k
−
1
)
+
…
+
4
⋅
1
<
4
k
⋅
k
=
4
k
2
{\displaystyle 4k+4(k-1)+\ldots +4\cdot 1<4k\cdot k=4k^{2}}
R
{\displaystyle R\!\,}
(
k
+
1
)
×
k
{\displaystyle (k+1)\times k\!\,}
k
{\displaystyle k\!\,}
1
+
2
+
…
+
k
=
k
(
k
+
1
)
2
=
k
2
2
+
k
2
{\displaystyle 1+2+\ldots +k={\frac {k(k+1)}{2}}={\frac {k^{2}}{2}}+{\frac {k}{2}}}
We wish to approximate the integral
I
(
f
)
≡
∫
a
b
f
(
x
)
d
x
{\displaystyle I(f)\equiv \int _{a}^{b}f(x)dx}
The composite trapezoidal rule is
Q
T
,
n
(
f
)
=
h
2
(
f
(
a
)
+
f
(
b
)
)
+
h
∑
k
=
1
n
−
1
f
(
x
k
)
{\displaystyle Q_{T,n}(f)={\frac {h}{2}}(f(a)+f(b))+h\sum _{k=1}^{n-1}f(x_{k})\!\,}
The error is
I
(
f
)
−
Q
T
,
n
(
f
)
=
−
(
b
−
a
)
f
′
′
(
ξ
)
12
h
2
{\displaystyle I(f)-Q_{T,n}(f)=-{\frac {(b-a)f^{\prime \prime }(\xi )}{12}}h^{2}}
where
ξ
∈
[
a
,
b
]
{\displaystyle \xi \in [a,b]\!\,}
Derivation of Composite Trapezoidal Error
edit
The local error, the error on one interval, is
−
h
3
12
f
′
′
(
η
)
{\displaystyle -{\frac {h^{3}}{12}}f^{\prime \prime }(\eta )\!\,}
Observe that
n
min
η
∈
[
a
,
b
]
f
′
′
(
η
i
)
≤
∑
i
=
1
n
f
′
′
(
η
i
)
≤
n
max
η
∈
[
a
,
b
]
f
′
′
(
η
i
)
{\displaystyle n\min _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\leq \sum _{i=1}^{n}f^{\prime \prime }(\eta _{i})\leq n\max _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\!\,}
which implies
min
η
∈
[
a
,
b
]
f
′
′
(
η
i
)
≤
∑
i
=
1
n
f
′
′
(
η
i
)
n
≤
max
η
∈
[
a
,
b
]
f
′
′
(
η
i
)
{\displaystyle \min _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\leq \sum _{i=1}^{n}{\frac {f^{\prime \prime }(\eta _{i})}{n}}\leq \max _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\!\,}
Hence, the Intermediate Value Theorem implies there exists a
ξ
∈
[
a
,
b
]
{\displaystyle \xi \in [a,b]\!\,}
f
′
′
(
ξ
)
=
∑
i
=
1
n
f
′
′
(
η
i
)
n
{\displaystyle f^{\prime \prime }(\xi )=\sum _{i=1}^{n}{\frac {f^{\prime \prime }(\eta _{i})}{n}}\!\,}
Multiplying both sides of the equation by
n
{\displaystyle n\!\,}
n
f
′
′
(
ξ
)
=
∑
i
=
1
n
f
′
′
(
η
i
)
{\displaystyle nf^{\prime \prime }(\xi )=\sum _{i=1}^{n}f^{\prime \prime }(\eta _{i})\!\,}
Using this relationship, we have
I
(
f
)
−
Q
T
,
n
(
f
)
=
∑
i
=
1
n
I
i
(
f
)
−
Q
i
(
f
)
=
∑
i
=
1
n
[
−
h
3
12
f
′
′
(
η
i
)
]
=
−
h
3
12
f
′
′
(
ξ
)
n
=
−
h
3
12
f
′
′
(
ξ
)
(
b
−
a
h
)
=
−
h
2
12
f
′
′
(
ξ
)
(
b
−
a
)
{\displaystyle {\begin{aligned}I(f)-Q_{T,n}(f)&=\sum _{i=1}^{n}I_{i}(f)-Q_{i}(f)\\&=\sum _{i=1}^{n}\left[-{\frac {h^{3}}{12}}f^{\prime \prime }(\eta _{i})\right]\\&=-{\frac {h^{3}}{12}}f^{\prime \prime }(\xi )n\\&=-{\frac {h^{3}}{12}}f^{\prime \prime }(\xi )({\frac {b-a}{h}})\\&=-{\frac {h^{2}}{12}}f^{\prime \prime }(\xi )(b-a)\end{aligned}}}
Derivation of Local Error
edit
The error in polynomial interpolation can be found by using the following theorem:
Assume
f
n
+
1
{\displaystyle f^{n+1}\!\,}
[
a
,
b
]
{\displaystyle [a,b]\!\,}
{
x
0
,
x
1
,
…
,
x
n
|
x
∈
[
a
,
b
]
}
.
{\displaystyle \{x_{0},x_{1},\ldots ,x_{n}|x\in [a,b]\}.\!\,}
P
n
{\displaystyle P_{n}\!\,}
f
{\displaystyle f\!\,}
{
x
j
}
j
=
0
n
{\displaystyle \{x_{j}\}_{j=0}^{n}\!\,}
ξ
x
{\displaystyle \xi _{x}\!\,}
ξ
{\displaystyle \xi \!\,}
x
{\displaystyle x\!\,}
f
(
x
)
−
p
n
(
x
)
=
(
x
−
x
0
)
(
x
−
x
1
)
.
.
.
(
x
−
x
n
)
(
n
+
1
)
!
f
(
n
+
1
)
(
ξ
x
)
{\displaystyle f(x)-p_{n}(x)={\frac {(x-x_{0})(x-x_{1})...(x-x_{n})}{(n+1)!}}f^{(n+1)}(\xi _{x})\!\,}
ξ
x
{\displaystyle \xi _{x}\!\,}
(
m
,
M
)
{\displaystyle (m,M)\!\,}
m
=
min
{
x
0
,
x
1
,
…
,
x
n
,
X
}
{\displaystyle m=\min\{x_{0},x_{1},\ldots ,x_{n},X\}\!\,}
M
=
max
{
x
0
,
x
1
,
…
,
x
n
,
X
}
{\displaystyle M=\max\{x_{0},x_{1},\ldots ,x_{n},X\}\!\,}
Applying the theorem yields,
f
(
x
)
−
p
1
(
x
)
=
(
x
−
a
)
(
x
−
b
)
f
′
′
(
ξ
x
)
2
{\displaystyle f(x)-p_{1}(x)=(x-a)(x-b){\frac {f^{\prime \prime }(\xi _{x})}{2}}}
Hence,
E
(
f
)
=
∫
a
b
f
(
x
)
−
p
1
(
x
)
d
x
=
∫
a
b
(
x
−
a
)
⏟
>
0
(
x
−
b
)
⏟
<
0
⏟
w
(
x
)
f
′
′
(
ξ
x
)
2
d
x
{\displaystyle {\begin{aligned}E(f)&=\int _{a}^{b}f(x)-p_{1}(x)dx\\&=\int _{a}^{b}\underbrace {\underbrace {(x-a)} _{>0}\underbrace {(x-b)} _{<0}} _{w(x)}{\frac {f^{\prime \prime }(\xi _{x})}{2}}dx\end{aligned}}}
a
{\displaystyle a\!\,}
x
−
a
{\displaystyle x-a\!\,}
b
{\displaystyle b\!\,}
x
−
b
{\displaystyle x-b\!\,}
w
(
x
)
<
0
{\displaystyle w(x)<0\!\,}
mean value theorem of integration
ζ
∈
[
a
,
b
]
{\displaystyle \zeta \in [a,b]\!\,}
E
(
f
)
=
f
′
′
(
ζ
)
2
∫
a
b
(
x
−
a
)
(
x
−
b
)
d
x
{\displaystyle E(f)={\frac {f^{\prime \prime }(\zeta )}{2}}\int _{a}^{b}(x-a)(x-b)dx\!\,}
f
′
′
(
ζ
)
{\displaystyle f^{\prime \prime }(\zeta )}
x
{\displaystyle x\!\,}
Integrating
∫
a
b
(
x
−
a
)
(
x
−
b
)
d
x
{\displaystyle \int _{a}^{b}(x-a)(x-b)dx\!\,}
E
(
f
)
=
f
′
′
(
ζ
)
2
(
−
(
b
−
a
)
6
)
=
−
f
′
′
(
ζ
)
(
b
−
a
)
12
{\displaystyle {\begin{aligned}E(f)&={\frac {f^{\prime \prime }(\zeta )}{2}}\left(-{\frac {(b-a)}{6}}\right)\\&=-{\frac {f^{\prime \prime }(\zeta )(b-a)}{12}}\!\,\end{aligned}}}
Use the error formula to derive a new quadrature rule obtained by performing one step of extrapolation on the composite trapezoidal rule. What is this rule, and how does its error depend on
h
{\displaystyle h\!\,}
f
{\displaystyle f\!\,}
STOER AND BUESCH pg 162
INTRODUCTION TO APPLIED NUMERICAL ANALYSIS by RICHARD HAMMING pg 178
The error for the composite trapezoidal rule at
n
{\displaystyle n\!\,}
[
a
,
b
]
{\displaystyle [a,b]\!\,}
E
n
=
I
(
f
)
−
Q
T
,
n
=
−
(
b
−
a
)
h
2
12
+
O
(
h
3
)
{\displaystyle E_{n}=I(f)-Q_{T,n}={\frac {-(b-a)h^{2}}{12}}+{\mathcal {O}}(h^{3})\!\,}
2
n
{\displaystyle 2n\!\,}
2
n
{\displaystyle 2n\!\,}
[
a
,
b
]
{\displaystyle [a,b]\!\,}
E
2
n
=
I
(
f
)
−
Q
T
,
2
n
=
2
⋅
−
(
b
−
a
)
(
h
2
)
2
12
+
O
(
h
3
)
=
−
(
b
−
a
)
h
2
24
+
O
(
h
3
)
{\displaystyle E_{2n}=I(f)-Q_{T,2n}=2\cdot {\frac {-(b-a)({\frac {h}{2}})^{2}}{12}}+{\mathcal {O}}(h^{3})={\frac {-(b-a)h^{2}}{24}}+{\mathcal {O}}(h^{3})\!\,}
h
2
{\displaystyle h^{2}\!\,}
E
n
{\displaystyle E_{n}\!\,}
E
2
n
{\displaystyle E_{2n}\!\,}
h
3
{\displaystyle h^{3}\!\,}
E
n
−
2
E
2
n
=
−
(
b
−
a
)
h
2
12
−
2
⋅
−
(
b
−
a
)
h
2
24
+
O
(
h
3
)
=
O
(
h
3
)
{\displaystyle {\begin{aligned}E_{n}-2E_{2n}&={\frac {-(b-a)h^{2}}{12}}-2\cdot {\frac {-(b-a)h^{2}}{24}}+{\mathcal {O}}(h^{3})\\&={\mathcal {O}}(h^{3})\end{aligned}}}
E
n
{\displaystyle E_{n}\!\,}
E
2
n
{\displaystyle E_{2n}\!\,}
I
(
f
)
−
Q
T
,
n
−
2
I
(
f
)
−
2
Q
T
,
2
n
=
O
(
h
3
)
{\displaystyle I(f)-Q_{T,n}-2I(f)-2Q_{T,2n}={\mathcal {O}}(h^{3})\!\,}
I
(
f
)
=
2
Q
T
,
2
n
−
Q
T
,
n
+
O
(
h
3
)
{\displaystyle I(f)=2Q_{T,2n}-Q_{T,n}+{\mathcal {O}}(h^{3})\!\,}
Q
^
{\displaystyle {\hat {Q}}\!\,}
Q
^
=
2
Q
T
,
2
n
−
Q
T
,
n
{\displaystyle {\hat {Q}}=2Q_{T,2n}-Q_{T,n}\!\,}
O
(
h
3
)
{\displaystyle {\mathcal {O}}(h^{3})\!\,}
Consider the shifted QR iteration for computing a 2 x 2 matrix
A
{\displaystyle A\!\,}
A
0
=
A
{\displaystyle A_{0}=A\!\,}
A
k
−
σ
k
I
=
Q
k
R
k
A
k
+
1
=
R
k
Q
k
+
σ
k
I
{\displaystyle A_{k}-\sigma _{k}I=Q_{k}R_{k}\quad \quad \quad A_{k+1}=R_{k}Q_{k}+\sigma _{k}I}
σ
k
{\displaystyle \sigma _{k}\!\,}
If
A
k
=
[
a
11
a
12
a
21
a
22
]
,
{\displaystyle A_{k}={\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}},}
Q
k
{\displaystyle Q_{k}\!\,}
Let
A
~
k
{\displaystyle {\tilde {A}}_{k}\!\,}
σ
k
{\displaystyle \sigma _{k}\!\,}
A
k
{\displaystyle A_{k}\!\,}
A
~
k
=
A
k
−
σ
k
I
=
[
a
11
−
σ
k
a
12
a
21
a
22
−
σ
k
]
,
{\displaystyle {\tilde {A}}_{k}=A_{k}-\sigma _{k}I={\begin{bmatrix}a_{11}-\sigma _{k}&a_{12}\\a_{21}&a_{22}-\sigma _{k}\end{bmatrix}},}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
A
~
k
{\displaystyle {\tilde {A}}_{k}\!\,}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
A
~
k
{\displaystyle {\tilde {A}}_{k}\!\,}
A
~
k
{\displaystyle {\tilde {A}}_{k}\!\,}
Q
k
T
A
~
k
=
[
∗
∗
0
∗
]
=
R
k
{\displaystyle Q_{k}^{T}{\tilde {A}}_{k}={\begin{bmatrix}*&*\\0&*\end{bmatrix}}=R_{k}}
∗
{\displaystyle *\!\,}
R
k
{\displaystyle R_{k}\!\,}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
Q
k
T
A
~
k
=
R
k
Q
k
Q
k
T
A
~
k
=
Q
k
R
k
A
~
k
=
Q
k
R
k
{\displaystyle {\begin{aligned}Q_{k}^{T}{\tilde {A}}_{k}&=R_{k}\\Q_{k}Q_{k}^{T}{\tilde {A}}_{k}&=Q_{k}R_{k}\\{\tilde {A}}_{k}&=Q_{k}R_{k}\end{aligned}}}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
Q
k
T
=
[
c
s
−
s
c
]
{\displaystyle Q_{k}^{T}={\begin{bmatrix}c&s\\-s&c\end{bmatrix}}}
c
=
a
11
−
σ
k
r
s
=
a
21
r
r
=
(
a
11
−
σ
k
)
2
+
a
21
2
{\displaystyle {\begin{aligned}c&={\frac {a_{11}-\sigma _{k}}{r}}\\s&={\frac {a_{21}}{r}}\\r&={\sqrt {(a_{11}-\sigma _{k})^{2}+a_{21}^{2}}}\end{aligned}}}
Q
k
T
{\displaystyle Q_{k}^{T}\!\,}
Q
k
=
[
c
−
s
s
c
]
{\displaystyle Q_{k}={\begin{bmatrix}c&-s\\s&c\end{bmatrix}}}
From hypothesis and part (a),
A
k
+
1
=
R
k
Q
k
+
σ
k
I
=
[
∗
∗
0
r
22
]
[
c
−
s
s
c
]
+
[
σ
k
0
0
σ
k
]
{\displaystyle {\begin{aligned}A_{k+1}&=R_{k}Q_{k}+\sigma _{k}I\\&={\begin{bmatrix}*&*\\0&r_{22}\end{bmatrix}}{\begin{bmatrix}c&-s\\s&c\end{bmatrix}}+{\begin{bmatrix}\sigma _{k}&0\\0&\sigma _{k}\end{bmatrix}}\end{aligned}}}
A
k
+
1
(
2
,
1
)
{\displaystyle A_{k+1}(2,1)\!\,}
A
k
{\displaystyle A_{k}\!\,}
A
k
+
1
(
2
,
1
)
{\displaystyle A_{k+1}(2,1)\!\,}
R
k
{\displaystyle R_{k}\!\,}
Q
k
{\displaystyle Q_{k}\!\,}
σ
k
I
{\displaystyle \sigma _{k}I\!\,}
A
k
+
1
(
2
,
1
)
=
(
0
⋅
c
+
r
22
⋅
s
)
+
0
=
r
22
s
{\displaystyle {\begin{aligned}A_{k+1}(2,1)&=(0\cdot c+r_{22}\cdot s)+0\\&=r_{22}s\end{aligned}}}
r
22
{\displaystyle r_{22}\!\,}
R
k
{\displaystyle R_{k}\!\,}
Q
k
{\displaystyle Q_{k}\!\,}
A
k
−
σ
k
I
=
Q
k
R
k
Q
k
T
(
A
k
−
σ
k
I
)
=
Q
k
T
Q
k
R
k
Q
k
T
(
A
k
−
σ
k
I
)
=
R
k
R
k
=
Q
k
T
(
A
k
−
σ
k
I
)
=
[
c
s
−
s
c
]
[
a
11
−
σ
k
a
12
δ
a
22
−
σ
k
]
{\displaystyle {\begin{aligned}A_{k}-\sigma _{k}I&=Q_{k}R_{k}\\Q_{k}^{T}(A_{k}-\sigma _{k}I)&=Q_{k}^{T}Q_{k}R_{k}\\Q_{k}^{T}(A_{k}-\sigma _{k}I)&=R_{k}\\R_{k}&=Q_{k}^{T}(A_{k}-\sigma _{k}I)\\&={\begin{bmatrix}c&s\\-s&c\end{bmatrix}}{\begin{bmatrix}a_{11}-\sigma _{k}&a_{12}\\\delta &a_{22}-\sigma _{k}\end{bmatrix}}\end{aligned}}}
R
k
{\displaystyle R_{k}\!\,}
r
2
2
{\displaystyle r_{2}2\!\,}
r
22
=
−
s
⋅
a
12
+
c
⋅
(
a
22
−
σ
k
)
{\displaystyle r_{22}=-s\cdot a_{12}+c\cdot (a_{22}-\sigma _{k})\!\,}
r
22
{\displaystyle r_{22}\!\,}
A
k
+
1
(
2
,
1
)
{\displaystyle A_{k+1}(2,1)\!\,}
A
k
+
1
(
2
,
1
)
=
r
22
⋅
s
=
(
−
s
a
12
+
c
(
a
22
−
σ
k
)
)
s
=
−
s
2
a
12
+
c
s
(
a
22
−
σ
k
)
=
−
δ
2
a
12
(
a
11
−
σ
k
)
2
+
δ
2
+
(
a
11
−
σ
k
)
(
a
22
−
σ
k
)
δ
(
a
11
−
σ
k
)
2
+
δ
2
=
−
δ
2
a
12
+
δ
(
a
11
−
σ
k
)
(
a
22
−
σ
k
)
(
a
11
−
σ
k
)
2
+
δ
2
≈
−
δ
2
a
12
+
δ
(
a
11
−
σ
k
)
(
a
22
−
σ
k
)
(
a
11
−
σ
k
)
2
{\displaystyle {\begin{aligned}A_{k+1}(2,1)&=r_{22}\cdot s\\&=(-sa_{12}+c(a_{22}-\sigma _{k}))s\\&=-s^{2}a_{12}+cs(a_{22}-\sigma _{k})\\&={\frac {-\delta ^{2}a_{12}}{(a_{11}-\sigma _{k})^{2}+\delta ^{2}}}+{\frac {(a_{11}-\sigma _{k})(a_{22}-\sigma _{k})\delta }{(a_{11}-\sigma _{k})^{2}+\delta ^{2}}}\\&={\frac {-\delta ^{2}a_{12}+\delta (a_{11}-\sigma _{k})(a_{22}-\sigma _{k})}{(a_{11}-\sigma _{k})^{2}+\delta ^{2}}}\\&\approx {\frac {-\delta ^{2}a_{12}+\delta (a_{11}-\sigma _{k})(a_{22}-\sigma _{k})}{(a_{11}-\sigma _{k})^{2}}}\end{aligned}}}
δ
{\displaystyle \delta \!\,}
σ
k
=
a
22
{\displaystyle \sigma _{k}=a_{22}\!\,}
A
k
+
1
(
2
,
1
)
≈
−
δ
2
a
12
+
δ
(
a
11
−
σ
k
)
(
a
22
−
σ
k
)
(
a
11
−
σ
k
)
2
≈
−
δ
2
a
12
+
δ
(
a
11
−
a
22
)
(
a
22
−
a
22
)
(
a
11
−
a
22
)
2
=
−
δ
2
a
12
(
a
11
−
a
22
)
2
{\displaystyle {\begin{aligned}A_{k+1}(2,1)&\approx {\frac {-\delta ^{2}a_{12}+\delta (a_{11}-\sigma _{k})(a_{22}-\sigma _{k})}{(a_{11}-\sigma _{k})^{2}}}\\&\approx {\frac {-\delta ^{2}a_{12}+\delta (a_{11}-a_{22})(a_{22}-a_{22})}{(a_{11}-a_{22})^{2}}}\\&={\frac {-\delta ^{2}a_{12}}{(a_{11}-a_{22})^{2}}}\end{aligned}}}
|
A
k
+
1
(
2
,
1
)
|
≈
|
δ
2
a
12
(
a
11
−
a
22
)
2
|
≤
|
δ
2
|
{\displaystyle {\begin{aligned}|A_{k+1}(2,1)|&\approx \left|{\frac {\delta ^{2}a_{12}}{(a_{11}-a_{22})^{2}}}\right|\\&\leq |\delta ^{2}|\end{aligned}}}
|
a
12
|
≤
(
a
11
−
a
22
)
2
{\displaystyle |a_{12}|\leq (a_{11}-a_{22})^{2}\!\,}
A
k
+
1
(
2
,
1
)
{\displaystyle A_{k+1}(2,1)\!\,}
δ
2
{\displaystyle \delta ^{2}\!\,}
δ
{\displaystyle \delta \!\,}
![{\displaystyle \xi \in [a,b]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/893a28f648b829efe4f91fb8ce73756768f9854b)
![{\displaystyle n\min _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\leq \sum _{i=1}^{n}f^{\prime \prime }(\eta _{i})\leq n\max _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47e5ba850937e6dcaffbd2b634b55e0c4a9f0084)
![{\displaystyle \min _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\leq \sum _{i=1}^{n}{\frac {f^{\prime \prime }(\eta _{i})}{n}}\leq \max _{\eta \in [a,b]}f^{\prime \prime }(\eta _{i})\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1be923bf10a3cb91d1ab74608b7330352df39603)
![{\displaystyle {\begin{aligned}I(f)-Q_{T,n}(f)&=\sum _{i=1}^{n}I_{i}(f)-Q_{i}(f)\\&=\sum _{i=1}^{n}\left[-{\frac {h^{3}}{12}}f^{\prime \prime }(\eta _{i})\right]\\&=-{\frac {h^{3}}{12}}f^{\prime \prime }(\xi )n\\&=-{\frac {h^{3}}{12}}f^{\prime \prime }(\xi )({\frac {b-a}{h}})\\&=-{\frac {h^{2}}{12}}f^{\prime \prime }(\xi )(b-a)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c0627e36f3e9fdfbbafcc9650718dafd7e8f3b8)
![{\displaystyle [a,b]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ef1354dbfeea71ab05cf297ffc40786b7813474)
![{\displaystyle \{x_{0},x_{1},\ldots ,x_{n}|x\in [a,b]\}.\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8b021354de5ba795361031d5f5b376074691d4f)
![{\displaystyle \zeta \in [a,b]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d52ae9847d4c106da77b203b56bc0ccb5908fa5)
