Fourier series & Fourier Transforms

1. Orthogonal and orthonormal functions 
1.1. Definitions 
Two functions φ1 and φ2 
are orthogonal on an interval [a, b] if the 
integral of their product is null. That is:



∫ φ1(x) φ2(x) dx = 0 
on the interval [a, b]

These functions are orthonormal if:

∫ φ1(x) φ2(x) dx = 1 
on the interval [a, b]

If they are not, we can force them to be so by introducing the 
weight function w(x) as follows:


∫ w(x) φ1(x) φ2(x) dx = 1 
on the interval [a, b]


A set of functions {φn}, where n is 
an integer, is orthogonal on 
the intereval [a, b] if:


∫ φn(x) φm(x) dx = 0 
on the interval [a, b] for n ≠ m


A set of functions {φn}, where n is 
an integer, is orthonormal on 
the intereval [a, b] if:


∫ φn(x) φm(x) dx = 1 
on the interval [a, b] for n =  m

 That is the set {φn} is 
normalized to 1:


∫ φn(x) 2 dx = 1


The integral ∫ φn(x) 2 dx  
is written as: 
||φn(x)||2, and is called the 
norm of the function φn. This norm is 
different from 1 in the case of the set is not normalized to 1.

1.2. Examples 
1.2.1. Example 1 
The tho following functions are orthogonal on the interval [-2, + 2]
φ1(x) = x

φ2(x) = x2

Indeed:


∫ φ1(x) φ2(x) dx = 
∫ x. x2 dx = ∫ x3 dx = 
x4/4 [-2, + 2] = 0


The set (p/π)(1/2) {cos nπx/p, sin mπx/p} 
is orthonormal on the interval [- π + π]; 
The numbers "n" and "m" are integers, 
and "p" is real.

Indeed:

For orthogonality:

∫ cos nπx/p cos mπx/p  dx = 0 (E1)
∫ sin nπx/p sin mπx/p  dx = 0 (E2)
∫ cos nπx/p sin mπx/p  dx = 0 (E3)

For normality:

∫ cos2 nπx/p dx =  p  (E4)
∫ sin2 nπx/p dx =  p  (E5)

Proof:

We know:
cos (a + b) = cos a cos b - sin a  sin b (1)
cos (a - b) = cos a cos b + sin a  sin b (2)
sin (a + b) = sin a cos b + cos a  sin b (3)
sin (a - b) = sin a cos b - cos a  sin b (4)

Adding (1) and (2) gives:
cos a cos b = [cos (a + b) + cos (a - b)]/2

Subtracting (2) and (1) gives:
sin a sin b = [cos (a - b) - cos (a + b)]/2

Adding (3) and (4) gives:
sin a cos b = [sin (a + b) + sin (a - b)]/2

Setting a = b in (1) gives:
cos (2a) = cos2 a - sin2 a  = 
cos2 a - [1 - cos2 a] = 
2 cos2 a -  1. Then:
cos2 a  = [1 + cos (2a)]/2
And:
sin2 a = 1 - cos2 a  = 
1 - [1 + cos (2a)]/2 = [1 - cos (2a)]/2


1.
∫ cos nπx/p cos mπx/p dx = 
∫ [cos (π/p)(n + m)x + cos (π/p)(n - m)x]/2 dx = 
(1/2) ∫ [cos (π/p)(n + m)x] + (1/2) ∫ cos (π/p)(n - m)x] dx = 
(1/2)  (p/π(n + m)) [sin(π/p)(n + m)x] + 
(1/2)  p/π(n - m) [sin(π/p)(n - m)x]. 

Integration on [- p, + p] yields:

(1/2) (p/π) { [ sin π(n + m)]/(n + m) + [sin π(n + m)]/(n + m)  + 
[sin π(n - m)]/(n - m)  + [sin π(n - m)]/(n - m)} = 
(1/2) (p/ π) {2[sin π(n + m)]/(n + m) + 2[sin π(n - m)]/(n - m)} = 0
The equation (E1) is proved.

In the case of n = m, the first term is 0, and the second term 
to 2 as: (sin x)/x tends to 1, we have [sin π(n - m)]/(n - m)} = 
π [sin π(n - m)]/π(n - m)} tends to π and then:
∫ cos2 nπx/p  dx = p
The equation (E4) is proved. 
The set (1/p)1/2 { cos nπx/p, sin mπx/p } is normalozed to 1.

2.
∫ sin nπx/p sin mπx/p dx = 
∫ [cos (π/p)(n - m)x - (π/p)cos (n + m)x]/2 dx = 
(1/2)  (p/π(n - m)) [sin(π/p)(n - m)x] - 
(1/2)  p/π(n + m) [sin(π/p)(n + m)x]. 

Integration on [- p, + p] yields:
(1/2) (p/ π) {[sin π(n - m)]/(n - m) + 
[sin π(n - m)]/(n - m)  
- [sin π(n + m)]/(n + m)  - [sin π(n + m)]/(n + m)} = 
(1/2) (p/ π) {2[sin π(n - m)]/(n - m)} = 
(p/ π) {[ sin π(n - m)]/(n - m)} = 0
The equation (E2) is proved.

In the cas of n = m, since (sin x)/x tends to 1, we have:
∫ sin2 nπx/p  dx = p
The equation (E5) is then proved.

3.
∫ cos nπx/p sin mπx/p dx = 
(1/2) {∫ sin (n + m)πx/p + sin (n - m)πx/p dx} =  
(1/2) {- ( p/(n + m)π) cos (n + m)πx/p - 
( p/(n + m)π) cos (n - m)πx/p } = 
- (p/π)(1/2) { [cos (n + m)πx/p ]/(n + m) - 
 [cos (n - m)πx/p]/(n + m)}.

Integration on [- p, + p] yields:
∫ cos nπx/p sin mπx/p dx = 
- (p/π)(1/2) { [cos (n + m)π]/(n + m) - 
cos (n + m)π]/(n + m)  
-  [cos (n - m)π]/(n + m) - [cos (n - m)π]/(n + m)} = 0.
The equation (E3) is then proved.


2. Fourier series 
2.1.  Definition 
Is it possible to extend any function f defined on the interval [a, b] in 
series of the orthonormal set {φn}?; that is:
f(x) = ∑ cn φn (x) n: 0 → ∞. 
The response is yes. Here is the proof:
Integrating f(x) φm (x) from - a to + b yields:
∫ f(x) φm (x) dx  = ∫ ∑ cn φn (x) φm (x) dx = 
∑ cn ∫ φn (x) φm (x) dx 
Since the set is orthonormal, the integral of the product ∫ φn (x) φm (x) is 0 
for n ≠ m and 1 if n = m. Hence:
∑ cn ∫ φn (x) φm (x) dx = 
cn ∫ φn2 (x) dx = cn || φn2 (x)||. Therefore:

cn = ∫f(x)φn(x)dx/||φn(x)||2
n : 0 → ∞ and x : a → b. 

2.2. Particular case: Fourier series 
Let's take the orthonormal set (p/π)(1/2) {cos nπx/p, 
sin mπx/p} on the interval [- π + π].
Any function f defined on the interval [- p, + p] can be expended as:
f(x) = ∑ an cos nπx/p + bn sin mπx/p
n: 0 → ∞. 
with: an = ∫f(x) cos nπx/p dx/||cos2 nπx/p(x)|| = ∫f(x) cos nπx/p dx/(p), and 
bn = ∫f(x) sin nπx/p dx/||sin2 nπx/p(x)|| = ∫f(x) sin nπx/p dx/(p)

f(x) = ∑ an cos nπx/p + bn sin mπx/p
With the following coefficients:
an = (1/p) ∫f(x) cos nπx/p dx, and 
bn = (1/p) ∫f(x) sin nπx/p dx
n : 0 → ∞, and  x : - p → + p.

Particular cases:
If the function f is even, that is f(- x) = f(x), 
then the product f(x) cos nπx/p is also 
even, and f(x) sin nπx/p is odd; therefore:
an = (2/p) ∫f(x) cos nπx/p dx, and 
bn = (1/p) ∫f(x) sin nπx/p dx = 0
n : 0 → ∞, and  x : 0 → + p.
Hence:
f(x) = ∑ an cos nπx/p 

If the function f is odd, that is f(- x) = - f(x), 
then the product f(x) cos nπx/p is also 
odd, and f(x) sin nπx/p is even; therefore:
an = (1/p) ∫f(x) cos nπx/p dx = 0 , and 
bn = (2/p) ∫f(x) sin nπx/p dx
n : 0 → ∞, and  x : 0 → + p.
Hence:
f(x) = ∑ bn sin nπx/p 

To recap:

f is even:
f(x) = ∑ an cos nπx/p, with:
an = (2/p) ∫f(x) cos nπx/p dx
n : 0 → ∞, and  x : 0 → + p.
That is the cosine series.


f is odd:
f(x) = ∑ bn sin nπx/p, with:
bn = (2/p) ∫f(x) sin nπx/p dx
n : 0 → ∞, and  x : 0 → + p.
That is the sine series.


2. Complex Fourier series 
We know that cos x = [eix + e- ix]/2, and 
sin x = [eix - e- ix]/2i
Substituting these expressions in Fourier series of the function f 
defined on the interval [- p, + p] yields:

f(x) = ∑ an cos nπx/p + bn sin mπx/p = 
(1/2) ∑ an [einπx/p + e- inπx/p] 
- i bn [einπx/p - e- inπx/p] = 
(1/2) ∑ [an -  i bn] einπx/p + 
[an +  i bn] e- inπx/p] = 
∑ cn einπx/p + c- n e- inπx/p 

Therefore: 
f(x) = ∑ cn einπx/p + c-n e- inπx/p 
With:
cn  = [an -  i bn]/2, and its conjugate:
c- n  = [an +  i bn]/2

The expressions of cn and c- n are found as follows:

I = ∫ f(x) e -imπx/p dx = ∑ cn ∫ ei(n - m)πx/p dx + 
c- n ∫ e- i(n + m)πx/p dx 
x : - p → +p 
I = ∑ cn (p/i(n - m)π) [ei(n - m)π - e- i(n - m)π ] + 
c- n (ip/(n + m)π) [e- i(n + m)π - ei(n + m)π]  = 
∑ cn (2p/(n - m)π) [sin ((n - m)π)] - c- n ( - 2 p/(n + m)π) [sin((n + m)π)] = 
0 if n≠ m. 
If n = m , we have, according to (sin x)/x = 1 when x tends to 0: 
I = cn (2p) - c- n ( - p/n )π) [sin((2n)π)] = cn (2p) + 0 = 
2p cn
Therefore: 
cn = (1/2p) ∫ f(x) e- inπx/p dx


Similarly, 

J = ∫ f(x) e+ imπx/p dx = ∑ cn ∫ ei(n + m)πx/p dx + 
c- n ∫ e- i(n - m)πx/p dx 
x : - p → +p
J = ∑ cn (p/i(n + m)π) [ei(n + m)π - e- i(n + m)π ] + 
c- n (ip/(n - m)π) [e- i(n - m)π - ei(n - m)π]  = 
∑ cn (2p/(n + m)π [sin ((n + m)π)] - c- n ( - 2 p/(n - m)π) [sin((n - m)π)] = 
0 if n≠ m. 
If n = m , we have, according to (sin x)/x = 1 when x tends to 0: 
J = cn (2p/2n)π [sin ((2n)π)] - c- n ( - 2p) = 0 + 2p c- n
Therefore: 
c-n = (1/2p) ∫ f(x) e+ inπx/p dx


We recap as follows:

f is any function defined on the interval [- p, + p]
f(x) = ∑ cn einπx/p + c-n e- inπx/p 
With the coefficients:
cn = (1/2p) ∫ f(x) e - inπx/p dx, and
c-n = (1/2p) ∫ f(x) e + inπx/p dx
n : 0 → ∞, and  x : - p → + p.

Remark that, here, the number "n"  can be negative as 
well. So we can write:


f is any function defined on the interval [- p, + p]
f(x) = ∑ cn exp{ i(nπ/p)x }
With the coefficient:
cn = (1/2p) ∫ f(x) exp{ - i(nπ/p)x } dx, 
n : - ∞ → + ∞, and  x : - p → + p.


3. Fourier Transforms 
We can write:
S = ∑ f(n) = ∑ f(n) 1 = ∑ f(n) [(n + 1) - n ] = ∑ f(n) dn, 
therefore the sum S can be written as S = ∫ f(n) dn.

The above equations:
f(x) = ∑ cn exp{ i(nπ/p)x }
cn = (1/2p) ∫ f(x) exp{ - i(nπ/p)x } dx, 
n : - ∞ → + ∞, and  x : - p → + p.
can be then written as:
f(x) = ∫ c(n) exp{ i(nπ/p)x } dn
c(n) = (1/2p) ∫ f(x) exp{ - i(nπ/p)x } dx, 
n : - ∞ → + ∞, and  x : - p → + p.

let's change "n/p" into "x" and "πx" into "k"; with F(x) = p c(n), the 
above equations become then:
f(k) = ∫ F(x) exp{ ikx} dx
F(x) = (1/2π) ∫ f(k) exp{ - ikx } dk, 
x : - ∞ → + ∞, and  k : - pπ → + pπ.

The number "p" is real. It can take the value ∞. We have then:

f(k) = ∫ F(x) exp{ ikx} dx
F(x) = (1/2π) ∫ f(k) exp{ - ikx } dk, 
x : - ∞ → + ∞, and  k : - ∞ → + ∞

F(x) is the Fourier Transform of f(k). f(k) is the 
inverse transform  of F(x).