Inner Products and Fourier Series The Inner Product

If 0 and 1 are continuous functions on the interval c+ß ,d, the standard inner product of 0 and 1 is the following quantity: Ø0 ß 1Ù œ ( 0 aBb 1aBb .B ,

+

The standard inner product of two functions is analogous to the dot product of two vectors: Ø0 ß 1Ù

v†w

is like

EXAMPLE 1 Find the inner product of the functions 0 aBb œ B and 1aBb œ B#  " on the interval c!ß #d. SOLUTION

We have: Ø0 ß 1Ù œ ( 0 aBb 1aBb .B œ ( BˆB#  "‰ .B œ ' #

#

!

è

!

PROPERTIES OF THE STANDARD INNER PRODUCT 1. Ø0 ß 0 Ù € !, and Ø0 ß 0 Ù œ ! only if 0 is the constant zero function. 2. Ø0 ß 1Ù œ Ø1ß 0 Ù 3. Ø0 ß 1  2Ù œ Ø0 ß 1Ù  Ø0 ß 2Ù 4. Ø0 ß -1Ù œ -Ø0 ß 1Ù for any scalar - .

When discussing the inner product of functions, we use much of the same terminology as we do for the dot product of vectors. For example: The norm of a function 0 is the quantity l0 l œ ÈØ0 ß 0 Ù œ Ë( 0 aBb# .B. ,

•

+

•

We refer to a function 0 as a unit vector if l0 l œ " (i.e. Ø0 ß 0 Ù œ ").

•

Two functions 0 and 1 are orthogonal if Ø0 ß 1Ù œ !.

•

Functions 0" ß á ß 08 are orthonormal if each 03 is a unit vector and each pair 03 ß 04 are orthogonal: l03 l œ "

and

Ø03 ß 04 Ù œ ! for 3 Á 4

EXAMPLE 2 Let W œ SpaneBß B# f on the interval c!ß $d. Find an orthonormal basis for W . SOLUTION

We use the Gram-Schmidt process. First, the norm of the function B is: lBl œ ÈØBß BÙ œ Ë( B# .B œ $ $

!

" Since B has length $, the function B is a unit vector. This reduces the basis to: $ " W œ Spanœ Bß B#  $ " Next we find the inner product of B# with B: $ # ¤B ß

Therefore, the function B# 

$ " " #( B¥ œ ( B# Œ B .B œ $ $ % !

#( " * " # Œ B œ B  B is orthogonal to B. This reduces the basis % $ % $

to: " * W œ Spanœ Bß B#  B $ % * Finally, we must make B#  B into a unit vector. We have: % $ * * * * * $ # #  Bß B#  B B  B œ B œ Ë( ŒB#  B .B œ Ê ¾ ¾ ˤ ¥ % % % % % & ! #

* * $ % & * & % Since B#  B has length Ê , the function Ê ŒB#  B œ Ê Œ B#  B is a unit % % & * $ % $ * vector. Therefore: " & % #  $ Bß Ê $ Œ * B  BŸ is an orthonormal basis for W .

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An orthonormal basis of functions works just like an orthonormal basis of vectors. In particular, recall that we can decompose any vector v using an orthonormal basis u" ß á ß u8 : v œ av † u" bu"  av † u# bu#  â  av † u8 bu8

An analogous result holds for functions:

FOURIER DECOMPOSITION

Let Z be a vector space of functions on the interval c+ß ,d, and suppose that 0" ß á ß 08 is an orthonormal basis for Z . Then for any function 0 − Z :

0 aBb œ Ø0 ß 0" Ù 0" aBb  Ø0 ß 0# Ù 0# aBb  â  Ø0 ß 08 Ù 08 aBb

EXAMPLE 3 Let: 0" aBb œ 1,

0# aBb œ È$ a#B  "b,

and

0$ aBb œ 'È& ˆB#  B  "' ‰

on the interval c!ß "d. Given that e0" ß 0# ß 0$ f is an orthonormal basis for T$ , express the function 0 aBb œ B# as a linear combination of 0" ß 0# ß 0$ . SOLUTION

We have:

Ø0 ß 0" Ù œ €B# ß "¡ œ ( B# .B œ "

!

" $

Ø0 ß 0# Ù œ €B# ß È$ a#B  "b¡ œ ( B# È$ a#B  "b .B œ "

!

" È # $

Ø0 ß 0$ Ù œ €B# ß 'È& ˆB#  B  "' ‰¡ œ ( B# 'È& ˆB#  B  "' ‰ .B œ "

!

" 'È &

Therefore: 0 aB b œ

" " " 0" a B b  0# a B b  0$ a B b $ #È $ 'È &

è

Fourier Series

Fourier series are a special kind of Fourier decomposition on the interval c1ß 1d, using the following inner product: Ø0 ß 1Ù œ

" 1 ( 0 aBb 1aBb .B 1 1

THEOREM (ORTHONORMAL TRIG FUNCTIONS) With respect to the above inner product, the following functions are all orthonormal on the interval c1ß 1d:

1. The constant function

" , È#

2. The functions cos Bß cos #Bß cos $Bß á , and 3. The functions sin Bß sin #Bß sin $Bß á .

This theorem is really just saying something about the values of certain integrals. For example, sin $B and sin #B are orthogonal because: €sin $Bß sin #B¡ œ

" 1 ( sin $B sin #B .B œ !. 1 1

Similarly, sin $B is a unit vector because: Øsin $Bß sin $BÙ œ NOTE 1. 2. 3.

" 1 # ( sin $B .B œ " 1 1

The sorts of integrals that come up here can be evaluated using the following obscure trig identities:

cosaB  Cb  cosaB  Cb # sinaB  Cb  sinaB  Cb cos B sin C œ # cosaB  Cb  cosaB  Cb cos B cos C œ # sin B sin C œ

For example, Øsin $Bß sin #BÙ œ (

1

sin $B sin #B .B œ 1

" 1 ( acos B  cos &Bb .B œ !, # 1

and lsin $Bl œ ÈØsin $Bß sin $BÙ œ Ë(

" 1 sin $B sin $B .B œ Ë ( a"  cos 'Bb .B œ È1 # 1 1 1

EXAMPLE 4 Given that cos% B − Spane"ß cos Bß cos #Bß cos $Bß cos %Bf, express cos% B as a linear combination of ", cos B, cos #B, cos $B, and cos %B. SOLUTION

Using a calculator or computer algebra system, we have: ¢cos% Bß

" È# £

œ

" 1 cos% B $ .B œ ( È È 1 1 # % #

¢cos% Bß cos B£ œ

" 1 & ( cos B .B œ ! 1 1

¢cos% Bß cos #B£ œ

" 1 " % ( cos B cos #B .B œ 1 1 #

¢cos% Bß cos $B£ œ

" 1 % ( cos B cos $B .B œ ! 1 1

¢cos% Bß cos %B£ œ

" 1 " % ( cos B cos %B .B œ 1 1 )

Therefore: $ " " " $ " " cos% B œ Œ È   Œ  cos #B  Œ  cos %B œ  cos #B  cos %B # ) ) # ) % # È#

è

EXAMPLE 5 On the interval c1ß 1d, it is possible to express B$ as an infinite linear combination of the functions sin 8B: B$ œ " +8 sin 8B _

8œ"

Assuming this fact, find the values of the coefficients +8 . SOLUTION

We have:

" 1 " $a8# B#  #b sin 8B Ba8# B#  'b cos 8B  ¢B ß sin 8B£ œ ( B$ sin 8B .B œ ” • 8% 8$ 1 1 1 1 1

$

Now, sin 81 œ ! for any integer value of 8, while cos 81 œ a"b8 . Therefore, this integral evaluates to " 1a8# 1#  'b a"b8 a1ba8# 1#  'ba"b8 a"b8 a#ba8# 1#  'b œ Œ   œ 1 8$ 8$ 8$

Therefore: B œ "Œ _

$

8œ"

_ # a"b8 a#ba8# 1#  'b "# 8 #1 " sin 8B œ a " b  $  sin 8B  Œ $ 8 8 8 8œ"

è

In the last example, it's not at all obvious that B$ can be expressed as a Fourier sine series. But assuming that it can, then we know that we've found the coefficients. It turns out that any continuous function on the interval c1ß 1d can be expressed as a Fourier series: FOURIER SERIES

Any continuous function 0 on the interval c1ß 1d can be expressed as a Fourier series: 0 aBb œ +  " ,8 cos 8B  " -8 sin 8B _

_

8œ"

8œ"

This series converges for all values of B in c1ß 1d.

The above theorem also holds for functions with finitely many jump discontinuities, except that the series may not converge to the correct value at the jump point. EXAMPLE 6 Let 0 aBb be the following step function: 0 aB b œ œ

!

if 1 Ÿ B  !

"

if !  B Ÿ 1

Find the Fourier series for 0 aBb on the interval c1ß 1d. SOLUTION

¢0 ß

We have: " È# £

œ

" 1 0 aB b " 1 " ( È .B œ Œ È#  œ È 1 1 1 # #

" 1 " 1 " sin 8B ¢0 ß cos 8B£ œ ( 0 aBb cos 8B .B œ ( cos 8B .B œ ” • œ ! 1 1 1 ! 1 8 ! 1

¢0 ß sin 8B£ œ

" 1 " 1 " cos 8B 1 “ ( 0 aBb sin 8B .B œ ( sin 8B . B œ ’ 1 1 1 ! 1 8 !

The last expression is equal to ! when 8 is even, and is

# when 8 is odd. Therefore: 81

" " # # # 0 aBb œ Œ È  È  Œ  sin B  Œ  sin $B  Œ  sin &B  â 1 # # $1 &1 œ

" # # #  sin B  sin $B  sin &B  â # 1 $1 &1

è

There are other advantages to having an orthonormal basis. For example, recall that: a+" u"  â  +8 u8 b † a," u"  â  ,8 u8 b œ +" ,"  â  +8 ,8

for any orthonormal basis u" ß á ß u8 . Similarly:

l+" u"  â  +8 u8 l œ È+" #  â  +8 #

These rules work for functions as well: EXAMPLE 7 Let:

0 aBb œ "  # cos B  $ cos #B  & cos $B

and

Find the inner product Ø0 ß 1Ù on the interval c1ß 1d

1aBb œ #  cos B  $ cos #B  cos $B

In terms of our orthonormal basis:

SOLUTION

" 0 aBb œ ˆÈ#‰ È

and so:

 # cos B  $ cos #B  & cos $B # " 1aBb œ Š#È#‹ È  cos B  $ cos #B  cos $B #

Ø0 ß 1Ù œ ˆÈ#‰ˆ#È#‰  a#ba"b  a$ba$b  a&ba"b œ "'

EXAMPLE 8 Compute (

1

1

è

asin #B  # sin %B  $ sin 'Bb# .B.

This integral is related to the norm of sin #B  # sin %B  $ sin 'B:

SOLUTION

(

1

1

asin #B  # sin %B  $ sin 'Bb# .B œ 1lsin #B  # sin %B  $ sin 'Bl# œ 1ˆ"#  ##  $# ‰ œ "%1

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