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