# FOURIER SERIES FOR DISCONTINUOUS FUNCTIONS

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مواضيع المحاضرة: FOURIER SERIES FOR DISCONTINUOUS FUNCTIONS
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### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

1

### FOURIER SERIES FOR DISCONTINUOUS FUNCTIONS

Let the function f(x) be defined by
f(x)= f1(x). c< x< x0
= f2(x).x0< x < c+ 2π .
Where x0 is the point of discontinuity in the interval (c .c+ 2π )
In such cases also , we obtain the Fouri er series for f(x) in the usual way The values of
a0 .an .bn are evaluated by
a0= 1
π[ ∫ f1(x)dx + ∫ f2(x)dx ] ; 2+2π
x0
x0
c
an= 1
π[ ∫ f1(x)cos n xdx + ∫ f2(x)cos nx dx ] ; 2+2π
x0
x0
c
bn= 1
π[ ∫ f1(x)sin nx dx + ∫ f2(x)cos nx dx ]
2+2π
x0
x0
c

### If is the point of finite discontinuity, then the sum of the Fourier series

= 1
2[ limh→0f( x0− h)+ limh→0f( x0+ h)]
= 1
2[ limh→0f( x0− 0)+ limh→0f( x0+ 0)]= 1
2(FB + FC )
Remarks
1. It may be seen from the graph, that at a point of finite discontinuity x= x0 there is a finite
jump equal to BC in the value of the function f(x) at x= x0
2. A given function f(x) may be defined by different formulae in different regions Such types of
functions are quite common in Fourier Series.
3. At a point of discontinuity the sum of the series is equal to the mean of the limits on the right
and left
FUNCTION DEFINED IN TWO OR MORE SUB - RANGES
Example 4. Find the Fourier series to represent the function f (x) given by:
f (x)= [−k for − π < x< 0
k for 0< x< π Hence show that ∶

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

2

1− 1
3+ 1
5− 1
7+ ⋯ = π
4
Solution. f(x){−k .for − π < x<
k .for 0< x< π ⋯ (1)
a0= 1
π∫ f(x)dx = 1
π[∫ −kdx ∫ kdx
π
0
0
−π
π
−π
]= 1
π[−kx ]0
−π+ [kx ]π
0]
= 1
π k[0− π+ π− 0]= 0
an= 1
π∫ f(x)cos nx dx = 1
π[∫ −kcos nx dx + ∫ kcos nx dx
π
0
0
−π
π
−π
]
= 1
π k[− {sin nx
n } 0
−π− {sin nx
n }π
0 ]= 1
πk[−0+ 0]= 0
bn= 1
π∫ f(x)sin nx dx = 1
π[∫ −ksin nx dx + ∫ ksin nx dx
π
0
0
−π
π
−π
]
= 1
π k[− {cos nx
n } 0
−π− {cos nx
n }π
0 ]
= 1
π k [1
n− (−1)n
n − (−1)n
n + 1
n]= 1
πk[2
n− 2(−1)2
n ]
If n is even bn= 0
If n is odd bn= 4k

f(x)= a0+ ∑ an

n=1
cos nx + ∑ bn

n=1
sin nx
f(x)= ∑ bn

n=1
sin nx = b1sin x+ b2sin2 x+ b3sin3 x+ b4sin4 x+ ⋯
Thus required Fourier sin series is
f (x)= 4k
π sin x+ 4k
3π sin 3x + 4k
5π sin 5x + ⋯
⇒ f (x)= 4k
π [sin x+ 1
3sin 3x + 1
5sin 5x + ⋯ ] ⋯ (2)

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

3

### Putting x= π

2 in (2). we get
k= 4k
π [sin π
2+ 1
3sin 3π
2 + 1
5sin 5π
2 + ⋯
1= 4
π[1+ 1
3(−1)+ 1
5(1)+ 1
7(−1)+ ⋯ ]
= 4
π [1− 1
3+ 1
5− 1
7+ ⋯ ]⇒ 4
π = 1− 1
3+ 1
5− 1
7+ ⋯
Example 5. find the Fourier series of the function
f(x)=
{

−1 for − π < x< − π
2
0 for − π
2< x< π
2
+1 for π
2 < x< π

### Solution : Let f(x)= a0

2+ a1cos x+ a2cos 2x + ⋯ + b1sin x+ b2sin 2x + ⋯ (1)
a0= 1
π∫ f(x)dx = 1
π∫ (−1)dx + 1
π∫ 0 dx + 1
π ∫ 1 dx
π
π2
π/2
−π/2
−π2
−π
π
−π

= 1
π[−x]−π−π/2+ 1
π[x]π/2π = 1
π[π
2− π+ π− π
2]= 0
an= 1
π∫ f(x)cos nx dx
π
−π

= 1
π∫ (−1)cos nx dx + 1
π∫ (0)cos nx dx + 1
π ∫ (1)cos nx dx
π
π/2
π2
−π2
−π/2
−π

π[sin nx
n ]
−π
−π/2
+ 1
π [sin nx
n ]
π/2
π

π[
sin nπ
2
n ]+ 1
π [
sin nπ
2
n ]= 0
bn= 1
π∫ f(x)sin nx dx
π
−π

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

4

= 1
π∫ (−1)sin nx dx + 1
π∫ (0)sin nx dx + 1
π ∫ (1)sin nx dx
π
π/2
π/2
−π/2
−π/2
−π

= 1
π[cos nx
n ]−π
−π/2
− 1
π [cos nx
n ]π/2
π

= 1
nπ[cos nπ
2 − cos nπ]− 1
nπ[cos nπ− cos nπ
2 ] = 2
nπ[cos nπ
2 − cos nπ]
b1= 2
π. b2= − 2
π. b3= 2

Putting the values of a0.an. bn in (1). we get
f(x)= 1
π[2sin x− 2sin 2x + 2
3sin 3x + ⋯ ]
DISCONTINUOUS FUNCTION
At a point of discontinuity, Fourier series gives the value f(x) of as the arithmetic mean of left and
right limits
At the point of discontinuity, x= c
Example 6. Find the Fourier series for if f(x)= {−π − π < x< 0
x. 0< x< π
Deduce that 1
12+ 1
32+ 1
52+ ⋯ = π2
8
Solution Let f(x)= a0
2+ a1cos x+ a2cos 2x + ⋯ + ancos nx + ⋯ + b1sin x+
b2sin 2x + ⋯ + bnsin nx + ⋯ ⋯ (1)
a0= 1
π ∫ f(x)dx
π
−π

### Then a0= 1

π[ ∫ (−π)dx + ∫ x dx ]= 1
π [−π(x)−π0 + (x2/2)0π π
−π
0
−π ]= 1
π(−π2+ π2
2)= − π
2;
an= 1
π ∫ f(x)cos nx dx
π
−π

an= 1
π [∫ (−π)cos nx dx
0
−π
+ ∫ xcos nx dx
π
0
]= 1
π[ −π(sin nx
n )
−π
0
+ (sin nx
n + cos nx
n2 )
−π
0
]

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

5

= 1
π[0+ 1
n2cos nπ− 1
n2]= 1
πn2 (cos nπ− 1)
bn= 1
π ∫ f(x)sin nx dx
π
−π

= 1
π [∫ (−π)sin nx dx
0
−π
+ ∫ xsin nx dx
π
0
]
= 1
π[ (πsin nx
n )
−π
0
+ (−xcos nx
n + sin nx
n2 )
0
π
]
= 1
π [π
n (1− cos nπ)− π
ncos nπ]= 1
n(1− 2cos nπ)
f(x)= − π
4− 2
π[cos x+ cos 3x
32 + cos 5x
52 + ⋯ ] + 3sin x− sin2x
2 + sin3x
3 − sin4x
4 ⋯ (2)
Putting x= 0 in (2). we get f(0)= − π
4− 2
π[1+ 1
32+ 1
52+ ⋯ ∞ ] ⋯ (3)
Now f(x) discontinuous at x= 0
But f (0− 0)= −π and f (0+ 0)= 0
∴ f(0)= 1
2[ f(0− 0)+ f(0+ 0)]= − π
2
From (3), − π
2= − π
4− 2
π[1
12+ 1
32+ 1
52+ ⋯ ]⇒ π2
8 = 1
12+ 1
32+ 1
52+ ⋯ proved
Example 7. Obtain Fourier Series of the function
f(x)= { x. − π < x< 0
−x. 0< x< π
And hence show that 1
12+ 1
32+ 1
52+ ⋯ ⋯ ⋯ = π2
8
Solution. We have, f(x)= { x. − π < x< 0
−x. 0< x< π
Fourier Series
Here f(x) is an even function so bn= 0
a0= 2
π ∫ f(x)dx
π
0
= 2
π ∫ −xdx
π
0
= − 2
π [π2
2]
0
π
= − 2
π [π2
2]= −π

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

6

an= 2
π ∫ f(x)cos nx dx
π
0
= 2
π ∫ −xcos nxdx
π
0
= − 2
π [xsin nx
n + cos nx
n2 ]
0
π

= − 2
π[(−1)n
n2 − 1
n2]= 2
πn2[1− (−1)n]= {
0 . n is even
4
πn2. n is odd
Fourier series : f(x)= − π
4+ 4
π[cos x
12 + cos 3x
32 + cos 5x
52 + ⋯ ]
At the point of discontinuity
f(0)= 1
2[f (0− 0)+ f(0+ 0)= − 1
2(0− 0)= 0
Putting x= 0 in the above, we get
0= − π
2+ π
4[1
12+ 2
32+ 1
52+ ⋯ ]⇒ 1
12+ 2
32+ 1
52+ ⋯ ⋯ π2
8 Ans
Example 8. Find the Fourier series of the function defined as
f(x)= { x+ π. for 0≤ x≤ π
−x− π. for − π ≤ x< π and f (x+ 2π )= f(x)
Solution a0= 1
π ∫ f(x)dx π
−π = 1
π∫ f(x)dx 0
−π + 1
π ∫ f(x)dx π
0
= 1
π ∫ (−x− π)dx +
0
−π
1
π ∫ (x+ π)dx
π
0
= 1
π[− x2
2 − πx]
−π
0
+ 1
π [x2
2 + πx]
−π
0

= 1
π [x2
2 + πx]+ [x2
2 + π2]= π[1
2− 1]+ π[1
2+ 1]= π
an= 1
π ∫ f(x)cos nxdx = 1
π ∫ f(x)cos nxdx + 1
π ∫ f(x)cos nx dx
π
0
0
−π
0
−π

= 1
π ∫ (−x− π)cos nxdx + 1
π ∫ (x+ π)cos nx dx
π
0
0
−π

= 1
π[(−x− π)sin nx
n − (−1){− cos nx
n2 }]
−π
0
+ 1
π[(x+ π)sin nx
n − (−1){− cos nx
n2 }]
0
π

= 1
π[− 1
n2+ (−1)n
n2 ]+ 1
π[− (−1)n
n2 + 1
n2]= 2
n2π[(−1)n− 1]= −4
n2π . if n is odd
= 0.if n is even

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

7

bn= 1
π ∫ f(x)sin nxdx = 1
π ∫ f(x)sin nxdx + 1
π ∫ f(x)sin nx dx
π
0
0
−π
π
−π

= 1
π ∫ (−x− π)sin nxdx + 1
π ∫ (x+ π)sin nx dx
π
0
0
−π

= 1
π[(−x− π)[− cos nx
n ]− (−1){− sin nx
n2 }]
−π
0
+ 1
π[(x+ π)[− cos nx
n ]− (−1){− sin nx
n2 }]
0
π

= 1
π[π
n]+ 1
π[− 2π
n (−1)n+ π
n]= 1
n[(1)− 2(−1)n+ (1)]= 2
n[1− (−1)n]
= 4
n . if n is odd
= 0. if n is even
Fourier series is : f(x)= a0
2+ a1cos x+ a2cos 2x + ⋯ + b1sin x+ b2sin 2x + ⋯
f(x)= π
2− π
4 [cos x
12 + cos 3x
32 + ⋯ ]+ [sinx
1 + sin3x
3 ]
Problems
1- Find the Fourier series of the function
f(x)= {−1 for − π < x< 0
1 for 0< x< π
Where f(x+ 2π )= f(x) Ans 4
π[1
1sin x+ 1
3sin 3x + 1
5sin 5x + 1
7sin 7x + ⋯ ]
2. Find the Fourier series of the function
f(x)= {
0 for − π ≤ x≤ 0
1 for 0 < x< π
2
0 for π
2≤ x≤ π

Ans 1
4+ 1
π [cos x
1 + cos 2x
2 + cos 3x
3 + ⋯ + sinx
1 + sin2x
2 + sin3x
3 ]
3. Obtain a Fourier series to represent the following periodic function
f(x)= { 0 when 0 < x< π
1. when π < x< 2π Ans 1
2− 2
π[sin x= 1
3sin 3x = 1
5sin 5x + ⋯ ]
4. Find the Fourier expansion the function defined in single period by the relations

### Lecture N o.2-10 by Hussein J. AbdulHussein

Advanced Calculus II Al Muthanna University, College of Science

8

### f(x)= {1 for 0< x< π

2 for π < x< 2π
And from it deduce that π
4= 1− 1
3+ 1
5− 1
7+ ⋯ Ans ∙3
2− 2
π[sin x+ 1
3sin 3x + 1
5sin 5x + ⋯ ]
5. Find a Fourier series to represent the function
f(x)= {
0 for − π < x≤ 0
1
4πx for 0 < x< 2π and hence deduce that π2
8 = 1
12+ 2
32+ 1
52+ ⋯
Ans π2
16 + ∑ [[(−1)n− 1]
4n 2

n=1
cos nx − (−1)nπ
4n sin nx + ⋯ ]
6. Find the Fourier series for f(x). If
Fourier series
f(x)= {
−π for − π < x≤ 0
x for 0 < x< π −π
2 for x= 0

### Deduce that 1

12+ 2
32+ 1
52+ ⋯ = π2
8
Ans − π
4− 4
π[cos x+ cos 3x
32 + cos 5x
52 + ⋯ ]+ 3sin x− 1
2sin 2x + 3
2sin 3x − 1
4sin 4x + ⋯
7. Obtain a Fourier series to represent the function
f(x)=| x |for − π < x< 0
And hence deduce π2
8 = 1
12+ 2
32+ 1
52+ ⋯ Ans π
4− 4
π[cosx + 1
32cos 3x + 1
52 cos 5x + ⋯ ]
8. Example as a Fourier series , the function f(x) defined as
f(x)=
{

2
π
2 for − π
2 < x< π
2
π− x for π
2 < x< π

### 9. Obtain a Fourier s eries to represent the function

f(x)=| sin x |for − π < x< π ( Hint } = sin x for 0 Lecture N o.2-10 by Hussein J. AbdulHussein Advanced Calculus II Al Muthanna University, College of Science

9

Ans π
4− 4
π[1
3cos2x + 1
15 cos 4x + 1
35 cos 6+ ⋯ ]
10. An alternating current after passing through a rectifier has the form
i{Isin θ for 0< θ< π
0 for π < θ< 2π ,Find the Fourier series of the function
Ans ∙I
π− 2I
π [cos θ
3 + cos4 θ
15 + ⋯ ]+ I
2sin θ
11. If f(x)= { 0 for − π < x< 0
sin x for 0 < x< π
Prove that f(x)= 1
π+ sin x
2 − 2
π ∑ cos 2mx
4m2−1
∞m=1 Hence show that 1
1∙3− 1
3∙5+ 1
5∙7⋯ ∞ = 1
4(π− 2)

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