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