Integrating Functions by Matrix Multiplication
47
Integrating Functions by Matrix Multiplication Thomas M. Everest University of Pittsburgh Undergraduate Mathematics Seminar, October 3, 2017 Everest Integrating Functions by Matrix Multiplication
Transcript of Integrating Functions by Matrix Multiplication
Integrating Functions by Matrix Multiplication
Thomas M. Everest
University of Pittsburgh
Undergraduate Mathematics Seminar,October 3, 2017
Everest Integrating Functions by Matrix Multiplication
Preliminaries
Linear Transformation
Suppose the V and W are vector spaces over the same field F.
T : V →W is a linear transformation if
1 T (v1 + v2) = Tv1 + Tv2, for all v1, v2 ∈ V ; and
2 T (kv) = kTv , for all k ∈ F and for all v ∈ V .
Everest Integrating Functions by Matrix Multiplication
Preliminaries
Linear Transformation
Suppose the V and W are vector spaces over the same field F.
T : V →W is a linear transformation if
1 T (v1 + v2) = Tv1 + Tv2, for all v1, v2 ∈ V ; and
2 T (kv) = kTv , for all k ∈ F and for all v ∈ V .
Everest Integrating Functions by Matrix Multiplication
Linear Transformation Example
Suppose that V = R4 and W = R3. Let T : V →W be definedby:
T
xyzw
=
x + 2ywz
for all v =
xyzw
∈ V
Everest Integrating Functions by Matrix Multiplication
Linear Transformation Example
T
x1y1z1w1
+
x2y2z2w2
= T
x1 + x2y1 + y2z1 + z2w1 + w2
=
(x1 + x2) + 2(y1 + y2)w1 + w2
z1 + z2
=
x1 + 2y1w1
z1
+
x2 + 2y2w2
z2
= T
x1y1z1w1
+ T
x2y2z2w2
Everest Integrating Functions by Matrix Multiplication
Linear Transformation Example
T
k
xyzw
= T
kxkykzkw
=
kx + 2kykwkz
=
k(x + 2y)kwkz
= k
x + 2ywz
= kT
xyzw
Everest Integrating Functions by Matrix Multiplication
Representing Linear Transformations with Matrices
Suppose that {v1, . . . , vn} is a basis for V and {w1, . . . ,wm} is abasis for W .
Write
Tv1 = a1,1w1 + · · ·+ am,1wm
Tv2 = a1,2w1 + · · ·+ am,2wm
...
Tvn = a1,nw1 + · · ·+ am,nwm
Everest Integrating Functions by Matrix Multiplication
Representing Linear Transformations with Matrices
Suppose that {v1, . . . , vn} is a basis for V and {w1, . . . ,wm} is abasis for W .
Write
Tv1 = a1,1w1 + · · ·+ am,1wm
Tv2 = a1,2w1 + · · ·+ am,2wm
...
Tvn = a1,nw1 + · · ·+ am,nwm
Everest Integrating Functions by Matrix Multiplication
Representing Linear Transformations with Matrices
Let
A =
a1,1 a1,2 · · · a1,na2,1 a2,2 · · · a2,n
......
. . ....
am,1 am,2 · · · am,n
m×n
Then, for any v ∈ V with coordinates
x1x2...xn
∈ Rn,
[Tv ] = A[v ]
Everest Integrating Functions by Matrix Multiplication
Representing Linear Transformations with Matrices
Let
A =
a1,1 a1,2 · · · a1,na2,1 a2,2 · · · a2,n
......
. . ....
am,1 am,2 · · · am,n
m×n
Then, for any v ∈ V with coordinates
x1x2...xn
∈ Rn,
[Tv ] = A[v ]
Everest Integrating Functions by Matrix Multiplication
Revisiting Linear Transformation Example
V = R4, W = R3, and T : V →W by T
xyzw
=
x + 2ywz
.
Let A =
1 2 0 00 0 0 10 0 1 0
.
Then, for any
xyzw
∈ R4, T
xyzw
=
1 2 0 00 0 0 10 0 1 0
xyzw
.
Everest Integrating Functions by Matrix Multiplication
Revisiting Linear Transformation Example
V = R4, W = R3, and T : V →W by T
xyzw
=
x + 2ywz
.
Let A =
1 2 0 00 0 0 10 0 1 0
.
Then, for any
xyzw
∈ R4, T
xyzw
=
1 2 0 00 0 0 10 0 1 0
xyzw
.
Everest Integrating Functions by Matrix Multiplication
Revisiting Linear Transformation Example
V = R4, W = R3, and T : V →W by T
xyzw
=
x + 2ywz
.
Let A =
1 2 0 00 0 0 10 0 1 0
.
Then, for any
xyzw
∈ R4, T
xyzw
=
1 2 0 00 0 0 10 0 1 0
xyzw
.
Everest Integrating Functions by Matrix Multiplication
Integration Example
Find
∫ (2ex + 3xex − 4x2ex
)dx .
∫ (2ex + 3xex − 4x2ex
)dx
= 2
(∫ex dx
)+ 3
(∫xex dx
)− 4
(∫x2ex dx
)
= + +
=
Everest Integrating Functions by Matrix Multiplication
Integration Example
Find
∫ (2ex + 3xex − 4x2ex
)dx .
∫ (2ex + 3xex − 4x2ex
)dx
= 2
(∫ex dx
)+ 3
(∫xex dx
)− 4
(∫x2ex dx
)
= + +
=
Everest Integrating Functions by Matrix Multiplication
Integration Example
Find
∫ (2ex + 3xex − 4x2ex
)dx .
∫ (2ex + 3xex − 4x2ex
)dx
= 2
(∫ex dx
)+ 3
(∫xex dx
)− 4
(∫x2ex dx
)
= + +
=
Everest Integrating Functions by Matrix Multiplication
Integration Example
Find
∫ (2ex + 3xex − 4x2ex
)dx .
∫ (2ex + 3xex − 4x2ex
)dx
= 2
(∫ex dx
)+ 3
(∫xex dx
)− 4
(∫x2ex dx
)
= + +
=
Everest Integrating Functions by Matrix Multiplication
Integration Example - New Approach
Find
∫ (2ex + 3xex − 4x2ex
)dx .
Let V = span{ex , xex , x2ex}.
Let D : V → V be the differentiation transformation. (Note that itis important that D(V ) ⊂ V ).
The matrix that represents this transformation is
A =
1 1 00 1 20 0 1
Everest Integrating Functions by Matrix Multiplication
Integration Example - New Approach
Find
∫ (2ex + 3xex − 4x2ex
)dx .
Let V = span{ex , xex , x2ex}.
Let D : V → V be the differentiation transformation. (Note that itis important that D(V ) ⊂ V ).
The matrix that represents this transformation is
A =
1 1 00 1 20 0 1
Everest Integrating Functions by Matrix Multiplication
Integration Example - New Approach
Find
∫ (2ex + 3xex − 4x2ex
)dx .
Let V = span{ex , xex , x2ex}.
Let D : V → V be the differentiation transformation. (Note that itis important that D(V ) ⊂ V ).
The matrix that represents this transformation is
A =
1 1 00 1 20 0 1
Everest Integrating Functions by Matrix Multiplication
The Inverse of Differentiation
Note that if D : V → V is the differentiation transformation, thenD−1 : V → V is the integration transformation (where +C = 0, sothat the transformation is linear).
FACTIf A represents the linear transformation D, then the inverse matrixA−1 represents the inverse transformation D−1.
In our case, if A =
1 1 00 1 20 0 1
, then A−1 =
1 −1 20 1 −20 0 1
.
Everest Integrating Functions by Matrix Multiplication
The Inverse of Differentiation
Note that if D : V → V is the differentiation transformation, thenD−1 : V → V is the integration transformation (where +C = 0, sothat the transformation is linear).
FACTIf A represents the linear transformation D, then the inverse matrixA−1 represents the inverse transformation D−1.
In our case, if A =
1 1 00 1 20 0 1
, then A−1 =
1 −1 20 1 −20 0 1
.
Everest Integrating Functions by Matrix Multiplication
The Inverse of Differentiation
Note that if D : V → V is the differentiation transformation, thenD−1 : V → V is the integration transformation (where +C = 0, sothat the transformation is linear).
FACTIf A represents the linear transformation D, then the inverse matrixA−1 represents the inverse transformation D−1.
In our case, if A =
1 1 00 1 20 0 1
, then A−1 =
1 −1 20 1 −20 0 1
.
Everest Integrating Functions by Matrix Multiplication
Finishing the Previous Example
Previously, we wanted to find
∫ (2ex + 3xex − 4x2ex
)dx .
Notice that 2ex + 3xex − 4x2ex is an element of V .
The coordinates of this vector under the given basis are
23−4
.
Therefore, to find our integral, we need to findD−1(2ex + 3xex − 4x2ex).
Everest Integrating Functions by Matrix Multiplication
Finishing the Previous Example
Previously, we wanted to find
∫ (2ex + 3xex − 4x2ex
)dx .
Notice that 2ex + 3xex − 4x2ex is an element of V .
The coordinates of this vector under the given basis are
23−4
.
Therefore, to find our integral, we need to findD−1(2ex + 3xex − 4x2ex).
Everest Integrating Functions by Matrix Multiplication
Finishing the Previous Example
Previously, we wanted to find
∫ (2ex + 3xex − 4x2ex
)dx .
Notice that 2ex + 3xex − 4x2ex is an element of V .
The coordinates of this vector under the given basis are
23−4
.
Therefore, to find our integral, we need to findD−1(2ex + 3xex − 4x2ex).
Everest Integrating Functions by Matrix Multiplication
The Calculation and the Interpretation
D−1(2ex + 3xex − 4x2ex)
=
1 −1 20 1 −20 0 1
23−4
=
−911−4
Therefore, we have that∫ (2ex + 3xex − 4x2ex
)dx = −9ex + 11xex − 4x2ex
Everest Integrating Functions by Matrix Multiplication
The Calculation and the Interpretation
D−1(2ex + 3xex − 4x2ex)
=
1 −1 20 1 −20 0 1
23−4
=
−911−4
Therefore, we have that∫ (2ex + 3xex − 4x2ex
)dx = −9ex + 11xex − 4x2ex
Everest Integrating Functions by Matrix Multiplication
The Extension
This technique now works quickly for any integral of this form.
Ex
Find
∫ (11xex +
2
3x2ex
)dx .
No need to restart!
Answer
D−1(11xex +2
3x2ex) =
1 −1 20 1 −20 0 1
01123
=
−29/329/32/3
.
Therefore,
∫ (11xex +
2
3x2ex
)dx = −29
3ex +
29
3xex +
2
3x2ex .
Everest Integrating Functions by Matrix Multiplication
The Extension
This technique now works quickly for any integral of this form.
Ex
Find
∫ (11xex +
2
3x2ex
)dx .
No need to restart!
Answer
D−1(11xex +2
3x2ex) =
1 −1 20 1 −20 0 1
01123
=
−29/329/32/3
.
Therefore,
∫ (11xex +
2
3x2ex
)dx = −29
3ex +
29
3xex +
2
3x2ex .
Everest Integrating Functions by Matrix Multiplication
The Extension
This technique now works quickly for any integral of this form.
Ex
Find
∫ (11xex +
2
3x2ex
)dx .
No need to restart!
Answer
D−1(11xex +2
3x2ex) =
1 −1 20 1 −20 0 1
01123
=
−29/329/32/3
.
Therefore,
∫ (11xex +
2
3x2ex
)dx = −29
3ex +
29
3xex +
2
3x2ex .
Everest Integrating Functions by Matrix Multiplication
The Extension
This technique now works quickly for any integral of this form.
Ex
Find
∫ (11xex +
2
3x2ex
)dx .
No need to restart!
Answer
D−1(11xex +2
3x2ex) =
1 −1 20 1 −20 0 1
01123
=
−29/329/32/3
.
Therefore,
∫ (11xex +
2
3x2ex
)dx = −29
3ex +
29
3xex +
2
3x2ex .
Everest Integrating Functions by Matrix Multiplication
The Problem
In order for this technique to work, we need D(V ) ⊂ V .
For example, we cannot use this technique to find∫ (ex
2+ xex
2)dx .
The problem is that D(ex2) = 2xex
2, D(xex
2) = (1 + 2x2)ex
2,
D(x2ex2) = (2x + 2x3)ex
2, . . . , etc.
Everest Integrating Functions by Matrix Multiplication
The Problem
In order for this technique to work, we need D(V ) ⊂ V .
For example, we cannot use this technique to find∫ (ex
2+ xex
2)dx .
The problem is that D(ex2) = 2xex
2, D(xex
2) = (1 + 2x2)ex
2,
D(x2ex2) = (2x + 2x3)ex
2, . . . , etc.
Everest Integrating Functions by Matrix Multiplication
The Problem
In order for this technique to work, we need D(V ) ⊂ V .
For example, we cannot use this technique to find∫ (ex
2+ xex
2)dx .
The problem is that D(ex2) = 2xex
2, D(xex
2) = (1 + 2x2)ex
2,
D(x2ex2) = (2x + 2x3)ex
2, . . . , etc.
Everest Integrating Functions by Matrix Multiplication
Other Examples That Do Work!
1 V = span{sin x , cos x , x sin x , x cos x}
2 V = span{ex sin x , ex cos x}
Matrices (where B is A−1 from before):
1 B =
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2 B =
[1/2 1/2−1/2 1/2
]
(Notice the pattern to each matrix - THIS IS THE WINNINGSTRATEGY!)
Everest Integrating Functions by Matrix Multiplication
Other Examples That Do Work!
1 V = span{sin x , cos x , x sin x , x cos x}2 V = span{ex sin x , ex cos x}
Matrices (where B is A−1 from before):
1 B =
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2 B =
[1/2 1/2−1/2 1/2
]
(Notice the pattern to each matrix - THIS IS THE WINNINGSTRATEGY!)
Everest Integrating Functions by Matrix Multiplication
Other Examples That Do Work!
1 V = span{sin x , cos x , x sin x , x cos x}2 V = span{ex sin x , ex cos x}
Matrices (where B is A−1 from before):
1 B =
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2 B =
[1/2 1/2−1/2 1/2
]
(Notice the pattern to each matrix - THIS IS THE WINNINGSTRATEGY!)
Everest Integrating Functions by Matrix Multiplication
Other Examples That Do Work!
1 V = span{sin x , cos x , x sin x , x cos x}2 V = span{ex sin x , ex cos x}
Matrices (where B is A−1 from before):
1 B =
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2 B =
[1/2 1/2−1/2 1/2
]
(Notice the pattern to each matrix - THIS IS THE WINNINGSTRATEGY!)
Everest Integrating Functions by Matrix Multiplication
Other Examples That Do Work!
1 V = span{sin x , cos x , x sin x , x cos x}2 V = span{ex sin x , ex cos x}
Matrices (where B is A−1 from before):
1 B =
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2 B =
[1/2 1/2−1/2 1/2
]
(Notice the pattern to each matrix - THIS IS THE WINNINGSTRATEGY!)
Everest Integrating Functions by Matrix Multiplication
1 Find
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx .
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2−13−2
=
2−4−2−3
Therefore,
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx =
2 sin x − 4 cos x − 2x sin x − 3x cos x .
2 Find
∫(2ex sin x − 4ex cos x) dx .[
1/2 1/2−1/2 1/2
] [2−4
]=
[−1−3
]Therefore,∫
(2ex sin x − 4ex cos x) dx = −ex sin x − 3ex cos x .
Everest Integrating Functions by Matrix Multiplication
1 Find
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx .
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2−13−2
=
2−4−2−3
Therefore,
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx =
2 sin x − 4 cos x − 2x sin x − 3x cos x .
2 Find
∫(2ex sin x − 4ex cos x) dx .[
1/2 1/2−1/2 1/2
] [2−4
]=
[−1−3
]Therefore,∫
(2ex sin x − 4ex cos x) dx = −ex sin x − 3ex cos x .
Everest Integrating Functions by Matrix Multiplication
1 Find
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx .
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2−13−2
=
2−4−2−3
Therefore,
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx =
2 sin x − 4 cos x − 2x sin x − 3x cos x .
2 Find
∫(2ex sin x − 4ex cos x) dx .[
1/2 1/2−1/2 1/2
] [2−4
]=
[−1−3
]Therefore,∫
(2ex sin x − 4ex cos x) dx = −ex sin x − 3ex cos x .
Everest Integrating Functions by Matrix Multiplication
1 Find
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx .
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2−13−2
=
2−4−2−3
Therefore,
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx =
2 sin x − 4 cos x − 2x sin x − 3x cos x .
2 Find
∫(2ex sin x − 4ex cos x) dx .
[1/2 1/2−1/2 1/2
] [2−4
]=
[−1−3
]Therefore,∫
(2ex sin x − 4ex cos x) dx = −ex sin x − 3ex cos x .
Everest Integrating Functions by Matrix Multiplication
1 Find
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx .
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2−13−2
=
2−4−2−3
Therefore,
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx =
2 sin x − 4 cos x − 2x sin x − 3x cos x .
2 Find
∫(2ex sin x − 4ex cos x) dx .[
1/2 1/2−1/2 1/2
] [2−4
]=
[−1−3
]
Therefore,∫(2ex sin x − 4ex cos x) dx = −ex sin x − 3ex cos x .
Everest Integrating Functions by Matrix Multiplication
1 Find
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx .
0 1 1 0−1 0 0 10 0 0 10 0 −1 0
2−13−2
=
2−4−2−3
Therefore,
∫(2 sin x − cos x + 3x sin x − 2x cos x) dx =
2 sin x − 4 cos x − 2x sin x − 3x cos x .
2 Find
∫(2ex sin x − 4ex cos x) dx .[
1/2 1/2−1/2 1/2
] [2−4
]=
[−1−3
]Therefore,∫
(2ex sin x − 4ex cos x) dx = −ex sin x − 3ex cos x .
Everest Integrating Functions by Matrix Multiplication
THE END
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Everest Integrating Functions by Matrix Multiplication
