CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF ...
CHAPTER 101 FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD 2π EXERCISE 362 Page 1066
1. Determine the Fourier series for the periodic function: − 2, when − π 〈 x 〈 0 f(x) = + 2, when 0 〈 x 〈 π which is periodic outside this range of period 2π.
The periodic function is shown in the diagram below.
∞
f(x) = a0 + ∑ ( an cos nx + bn sin nx )
The Fourier series is given by:
(1)
n =1
a0 =
1 2π
π
∫ π f ( x) d x = −
1 2π
{∫
}
π
0 −π
−2 d x + ∫ 2 d x = 0
=
{
1 2π
{[−2 x]
+ [2 x] 0
π
0 −π
}
1 {[(0) − (2π )] + [(2π ) − (0)]=} 0 2π
}
π 1 π 1 0 an =∫ f ( x) cos nx d x =∫ −2 cos nx d x + ∫ 2 cos nx d x
π
π
−π
=
−π
0
0 π 1 2 2 sin nx sin nx − + n = 0 π n −π 0
{
}
π 1 π 1 0 bn =∫ f ( x) sin nx d x = ∫ −2sin nx d x + ∫ 2sin nx d x
π
π
−π
−π
0
= 0 π 1 2 2 2 cos nx cos nx + − = n π n {[ cos 0 − cos n(−π ) ] − [ cos nπ − cos 0]} π n −π 0
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= bn
When n is even,
When n is odd, = bn Hence,
b1 =
8
π
,
2 ]} 0 {[1 − 1] − [1 − 1= πn 2 2 8 − 1]} = ( 4) [1 − −1] − [ −1= { πn πn πn
b3 =
8 , 3π
b5 =
8 and so on 5π
Substituting onto equation (1) gives: f(x) = 0 + 0 +
i.e.
f(x) =
8
π
sin x +
8 8 sin 3 x + sin 5 x + ... 3π 5π
8 1 1 sin x + sin 3 x + sin 5 x + ... 3 5 π
2. For the Fourier series in Problem 1, deduce a series for
When x =
π 2
, f(x) = 2, hence
2=
8
π
sin
π 2
+
π 4
at the point where x =
π 2
8 3π 8 5π sin sin + + ... 3π 2 5π 2
8 1 1 1 1 − + − + ... π 3 5 7
i.e.
2=
and
2π 1 1 1 =1 − + − + ... 8 3 5 7
π
1 1 1 =1 − + − + ... 4 3 5 7
i.e.
3. For the waveform shown below determine (a) the Fourier series for the function and (b) the sum of the Fourier series at the points of discontinuity.
∞
(a) The Fourier series is given by:
f(x) = a0 + ∑ ( an cos nx + bn sin nx )
(1)
n =1
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a0 =
= an
π
1 2π
x) d x ∫ π f (=
1
π
1 2π
−
f ( x) cos nx= dx π∫π −
{∫
−π /2
0d x + ∫
−π
π {∫ 1
−π /2 −π
π /2 −π /2
}
π
1d x + ∫ = 0d x π /2
(0) cos nx d x + ∫
π /2 −π /2
1 2π
] } {[ x= π /2
−π /2
1 π 1 π 1 − = {π } −= 2π 2 2 2π 2
}
π
1cos nx d x + ∫ (0) cos nx d x π2
π /2 1 π 1 1 π = ∫ cos nx d x = sin nx = sin n − sin −n −π /2 π n 2 −π /2 π n 2 π /2
When n = 1, = a1
1
π
1) (1 − −=
2
π
1 When n = 2, a2 = ( 0 − 0 ) =0 =a4 =a6 and so on for all even values of n 2π 1 2 When n = 3, a1 = ( −1 − 1) =− 3π 3π
When n = 5, = a5 a7 = −
Hence,
= bn
1
π
π
∫π −
1 2 −1) (1 −= 5π 5π 2 , 7π
a9 =
dx f ( x) sin nx=
π {∫ 1
2 9π
−π /2 −π
and so on
(0) sin nx d x + ∫
π /2 −π /2
1sin nx d x + ∫
π
π /2
}
(0) sin nx d x
π /2 π /2 1 1 1 π π sin nx d x cos nx = = − − = cos n − cos −n ∫−π /2 π n π n 2 2 −π
Whatever value of n is chosen, bn = 0 Substituting onto equation (1) gives:
i.e.
f(x) =
1 2 2 2 2 cos 3 x + cos 5 x − cos 7 x + ... + 0 + cox − 2 π 3π 5π 7π
f(x) =
1 2 1 1 + cos x − cos 3 x + cos 5 x + ... 2 π 3 5
(b) The sum of the Fourier series at the points of discontinuity (i.e. at π/2, π, 3π/2, ...) is: 1+ 0 1 = 2 2
4. For Problem 3, draw graphs of the first three partial sums of the Fourier series and show that as the series is added together term by term the result approximates more and more closely to the 1509
© 2014, John Bird
function it represents.
In the diagram below graphs of
π
cos x,
2 2 1 cos 3x and cos 5x and f(x) = are shown 3π 5π 2
2 2 cos 3x + cos 5x is also shown 5π π 3π 1 2 2 2 Finally, a graph of f(x) = + cos x – cos 3x + cos 5x is sketched. If further harmonics 2 π 3π 5π
A graph of
2
2
cos x –
were added then the waveform would approach that shown in Problem 3
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5. Find the term representing the third harmonic for the periodic function of period 2π given by: 0, when − π 〈 x 〈 0 f(x) = 1, when 0 〈 x 〈 π The periodic function is shown in the diagram below.
= an
bn
1
π
π
∫π −
1 π 1 = ∫ f ( x) sin nx d x
π
π
1 sin nx f ( x= ) cos nx d x = 1cos nx d x = 0 ∫ π 0 π n 0 π
1
π∫
−π
π 0
sin nx d x
π 1 cos nx 1 1 = − − {cos nπ − cos 0} =− {1 cos nπ } = π n 0 πn πn
The third harmonic is when n = 3, i.e.
= b3
1 1 2 −1} 3π } {1 −= {1 − cos = 3π 3π 3π ∞
Since the Fourier series is given by: f(x) = a0 + ∑ ( an cos nx + bn sin nx ) , n =1
2 sin 3 x 3π
the 3rd harmonic term is:
6. Determine the Fourier series for the periodic function of period 2π defined by: 0, when − π 〈 t 〈 0 π f(t) = 1, when 0 〈 t 〈 2 π − 1, when 2 〈 t 〈 π The function has a period of 2π 1511
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The periodic function is shown in the diagram below
a0 =
= an
1 2π
1
π
1 2π
∫π −
f (t ) d t =
π
f (= t ) cos nt d t π∫π −
{∫
0 −π
0dt + ∫
π {∫
π /2
1
0
π /2 0
π
}
{
}
1 π /2 π [t ] 0 + [ −t ] π /2 2π 1 π π = − 0 − ( 0 ) + ( −π ) − = 2π 2 2
1d t + ∫ = −1d t π /2
cos nt d t + ∫
π π /2
}
− cos nx d x
π /2 π nπ 1 sin nt sin nt 1 nπ = − = − 0 − sin nπ − sin sin π n 0 2 2 n π /2 nπ
an = 0
When n is even,
1 π π 1 2 [1 − 0] − [0 − 1]}= { sin − 0 − sin π − sin = π 2 2 π π
When n = 1,
a= 1
When n = 3,
1 a3 = 3π
When n = 5,
= a5
It follows that
a7 = −
bn =
1
1 5π
π
f (= t ) sin nt d t π∫π −
3π 3π 1 2 − 0 − sin 3π − sin = {[ −1 − 0] − [ 0 − −1]} = − sin 2 2 3π 3π 5π 5π 1 2 − 0 − sin 5π − sin = [1 − 0] − [0 − 1= ]} { sin 2 2 5π 5π
2 , 7π
π {∫ 1
a9 = π /2 0
2 9π
and so on
sin nt d t + ∫
π π /2
− sin nt d t
}
= π /2 π 1 cos nt nπ nπ cos nt 1 + − cos 0 + cos nπ − cos − = − cos π n 0 2 2 n π /2 π n
= When n is odd,
= bn
1 nπ 1 − cos π n 2
nπ + cos nπ − cos 2
nπ 1 + cos nπ = 1 − 2 cos 2 π n
1 1} 0 {1 − 0 −= πn
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When n is even,
b2=
1 4 2 = {1 − 2(−1) + 1}= 2π 2π π
b= 4
1 {1 − 2(1) + 1=} 0 4π
b6 =
1 4 2 = {1 − 2(−1) + 1}= 6π 6π 3π
b8 = 0 ,
Similarly,
b10 =
2 , and so on 5π
∞
Substituting into f(t) = a0 + ∑ ( an cos nt + bn sin nt ) n =1
gives:
f(x) = 0 +
2
π
cos t −
2 2 2 cos 3t + cos 5t − cos 7t + ... 3π 5π 7π 2
+
i.e. f(x) =
π
sin 2t +
2 2 sin 6t + sin10t + ... 3π 5π
2 1 1 1 1 cos t − cos 3t + cos 5t − ... + sin 2t + sin 6t + sin10t + ... 3 5 3 5 π
7. Show that the Fourier series for the periodic function of period 2π defined by: θ , when − π 〈 θ 〈 0 f(θ) = sin θ , when 0 〈 θ 〈 π is given by: f(θ) =
2 1 cos 2θ cos 4θ cos 6θ − − − ... − π 2 (3) (3)(5) (5)(7)
The periodic function is shown in the diagram below
a0 =
1 2π
π
θ ∫ π f (θ ) d= −
1 2π
{∫
0 −π
π
}
0 d θ + ∫ sin θ d = θ 0
=
1 2π
{[− cosθ ]=} π 0
{
}
1 ( − cos π ) − ( − cos 0 ) 2π
1 1 1 π) −1) (1 − cos= (1 −= 2π 2π π
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π {∫ 1
an
=
0 −π
π
0 cos nθ d θ + ∫ sin θ cos nθ d θ 0
}
1 π1 1 θ − nθ ) ∫ 0 sin (θ + nθ ) + sin (= π 2 2π
{∫ sin θ (1 + n) + sin θ (1 − n) dθ } from 6, page 716 π
0
of the textbook 1 = 2π
π
cos θ (1 + n ) θ (1 − n ) 1 cos π (1 + n ) cos π (1 − n ) 1 1 − cos = − − − −− − 1+ n 1 − n 0 2π 1+ n 1− n 1 + n 1 − n
When n is odd, an =
1 2π
1 1 1 1 − 1 + n − 1 − n + 1 + n + 1 − n = 0
When n = 2,
a= 2
1 cos 3π cos(−π ) 1 1 1 1 1 1 4 2 − + + = − − 1 + − 1= − = − 2π 3 −1 3 −1 2π 3 3 2π 3 3π
When n = 4,
a= 4
1 2π =
When n = 6,
=
π {∫ 1
bn
=
0 −π
1 1 1 1 − + − 5 3 5 3
1 3 − 5 + 3 − 5 1 4 2 − = − = 2π (3)(5) 2π (3)(5) π (3)(5)
1 2π
a= 6
cos 5π cos(−3π ) 1 1 1 − + + = − 5 −3 5 −3 2π
cos 7π cos(−5π ) 1 1 1 1 1 1 1 − + + = − − + − −5 7 7 −5 2π 7 5 7 5
1 5 − 7 + 5 − 7 1 4 2 − = − = 2π (5)(7) 2π (5)(7) π (5)(7) π
0sin nθ d θ + ∫ sin θ sin nθ d θ 0
}
1 π 1 1 − ∫ 0 − cos (θ + nθ ) − cos (θ − nθ ) = π 2 2π
π
sin θ (1 + n) sin θ (1 − n) = 0 from 9, page − 1 − n 0 1 + n 716 of the textbook
∞
Substituting into f(θ) = a0 + ∑ ( an cos nθ + bn sin nθ ) n =1
gives:
f(θ) =
i.e.
f(θ) =
1
π
−
2 2 2 cos 2θ − cos 4θ − cos 6θ − ... + 0 3π (3)(5)π (5)(7)π
2 1 cos 2θ cos 4θ cos 6θ − − − ... − π 2 (3) (3)(5) (5)(7)
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1. Determine the Fourier series for the periodic function: − 2, when − π 〈 x 〈 0 f(x) = + 2, when 0 〈 x 〈 π which is periodic outside this range of period 2π.
The periodic function is shown in the diagram below.
∞
f(x) = a0 + ∑ ( an cos nx + bn sin nx )
The Fourier series is given by:
(1)
n =1
a0 =
1 2π
π
∫ π f ( x) d x = −
1 2π
{∫
}
π
0 −π
−2 d x + ∫ 2 d x = 0
=
{
1 2π
{[−2 x]
+ [2 x] 0
π
0 −π
}
1 {[(0) − (2π )] + [(2π ) − (0)]=} 0 2π
}
π 1 π 1 0 an =∫ f ( x) cos nx d x =∫ −2 cos nx d x + ∫ 2 cos nx d x
π
π
−π
=
−π
0
0 π 1 2 2 sin nx sin nx − + n = 0 π n −π 0
{
}
π 1 π 1 0 bn =∫ f ( x) sin nx d x = ∫ −2sin nx d x + ∫ 2sin nx d x
π
π
−π
−π
0
= 0 π 1 2 2 2 cos nx cos nx + − = n π n {[ cos 0 − cos n(−π ) ] − [ cos nπ − cos 0]} π n −π 0
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= bn
When n is even,
When n is odd, = bn Hence,
b1 =
8
π
,
2 ]} 0 {[1 − 1] − [1 − 1= πn 2 2 8 − 1]} = ( 4) [1 − −1] − [ −1= { πn πn πn
b3 =
8 , 3π
b5 =
8 and so on 5π
Substituting onto equation (1) gives: f(x) = 0 + 0 +
i.e.
f(x) =
8
π
sin x +
8 8 sin 3 x + sin 5 x + ... 3π 5π
8 1 1 sin x + sin 3 x + sin 5 x + ... 3 5 π
2. For the Fourier series in Problem 1, deduce a series for
When x =
π 2
, f(x) = 2, hence
2=
8
π
sin
π 2
+
π 4
at the point where x =
π 2
8 3π 8 5π sin sin + + ... 3π 2 5π 2
8 1 1 1 1 − + − + ... π 3 5 7
i.e.
2=
and
2π 1 1 1 =1 − + − + ... 8 3 5 7
π
1 1 1 =1 − + − + ... 4 3 5 7
i.e.
3. For the waveform shown below determine (a) the Fourier series for the function and (b) the sum of the Fourier series at the points of discontinuity.
∞
(a) The Fourier series is given by:
f(x) = a0 + ∑ ( an cos nx + bn sin nx )
(1)
n =1
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a0 =
= an
π
1 2π
x) d x ∫ π f (=
1
π
1 2π
−
f ( x) cos nx= dx π∫π −
{∫
−π /2
0d x + ∫
−π
π {∫ 1
−π /2 −π
π /2 −π /2
}
π
1d x + ∫ = 0d x π /2
(0) cos nx d x + ∫
π /2 −π /2
1 2π
] } {[ x= π /2
−π /2
1 π 1 π 1 − = {π } −= 2π 2 2 2π 2
}
π
1cos nx d x + ∫ (0) cos nx d x π2
π /2 1 π 1 1 π = ∫ cos nx d x = sin nx = sin n − sin −n −π /2 π n 2 −π /2 π n 2 π /2
When n = 1, = a1
1
π
1) (1 − −=
2
π
1 When n = 2, a2 = ( 0 − 0 ) =0 =a4 =a6 and so on for all even values of n 2π 1 2 When n = 3, a1 = ( −1 − 1) =− 3π 3π
When n = 5, = a5 a7 = −
Hence,
= bn
1
π
π
∫π −
1 2 −1) (1 −= 5π 5π 2 , 7π
a9 =
dx f ( x) sin nx=
π {∫ 1
2 9π
−π /2 −π
and so on
(0) sin nx d x + ∫
π /2 −π /2
1sin nx d x + ∫
π
π /2
}
(0) sin nx d x
π /2 π /2 1 1 1 π π sin nx d x cos nx = = − − = cos n − cos −n ∫−π /2 π n π n 2 2 −π
Whatever value of n is chosen, bn = 0 Substituting onto equation (1) gives:
i.e.
f(x) =
1 2 2 2 2 cos 3 x + cos 5 x − cos 7 x + ... + 0 + cox − 2 π 3π 5π 7π
f(x) =
1 2 1 1 + cos x − cos 3 x + cos 5 x + ... 2 π 3 5
(b) The sum of the Fourier series at the points of discontinuity (i.e. at π/2, π, 3π/2, ...) is: 1+ 0 1 = 2 2
4. For Problem 3, draw graphs of the first three partial sums of the Fourier series and show that as the series is added together term by term the result approximates more and more closely to the 1509
© 2014, John Bird
function it represents.
In the diagram below graphs of
π
cos x,
2 2 1 cos 3x and cos 5x and f(x) = are shown 3π 5π 2
2 2 cos 3x + cos 5x is also shown 5π π 3π 1 2 2 2 Finally, a graph of f(x) = + cos x – cos 3x + cos 5x is sketched. If further harmonics 2 π 3π 5π
A graph of
2
2
cos x –
were added then the waveform would approach that shown in Problem 3
1510
© 2014, John Bird
5. Find the term representing the third harmonic for the periodic function of period 2π given by: 0, when − π 〈 x 〈 0 f(x) = 1, when 0 〈 x 〈 π The periodic function is shown in the diagram below.
= an
bn
1
π
π
∫π −
1 π 1 = ∫ f ( x) sin nx d x
π
π
1 sin nx f ( x= ) cos nx d x = 1cos nx d x = 0 ∫ π 0 π n 0 π
1
π∫
−π
π 0
sin nx d x
π 1 cos nx 1 1 = − − {cos nπ − cos 0} =− {1 cos nπ } = π n 0 πn πn
The third harmonic is when n = 3, i.e.
= b3
1 1 2 −1} 3π } {1 −= {1 − cos = 3π 3π 3π ∞
Since the Fourier series is given by: f(x) = a0 + ∑ ( an cos nx + bn sin nx ) , n =1
2 sin 3 x 3π
the 3rd harmonic term is:
6. Determine the Fourier series for the periodic function of period 2π defined by: 0, when − π 〈 t 〈 0 π f(t) = 1, when 0 〈 t 〈 2 π − 1, when 2 〈 t 〈 π The function has a period of 2π 1511
© 2014, John Bird
The periodic function is shown in the diagram below
a0 =
= an
1 2π
1
π
1 2π
∫π −
f (t ) d t =
π
f (= t ) cos nt d t π∫π −
{∫
0 −π
0dt + ∫
π {∫
π /2
1
0
π /2 0
π
}
{
}
1 π /2 π [t ] 0 + [ −t ] π /2 2π 1 π π = − 0 − ( 0 ) + ( −π ) − = 2π 2 2
1d t + ∫ = −1d t π /2
cos nt d t + ∫
π π /2
}
− cos nx d x
π /2 π nπ 1 sin nt sin nt 1 nπ = − = − 0 − sin nπ − sin sin π n 0 2 2 n π /2 nπ
an = 0
When n is even,
1 π π 1 2 [1 − 0] − [0 − 1]}= { sin − 0 − sin π − sin = π 2 2 π π
When n = 1,
a= 1
When n = 3,
1 a3 = 3π
When n = 5,
= a5
It follows that
a7 = −
bn =
1
1 5π
π
f (= t ) sin nt d t π∫π −
3π 3π 1 2 − 0 − sin 3π − sin = {[ −1 − 0] − [ 0 − −1]} = − sin 2 2 3π 3π 5π 5π 1 2 − 0 − sin 5π − sin = [1 − 0] − [0 − 1= ]} { sin 2 2 5π 5π
2 , 7π
π {∫ 1
a9 = π /2 0
2 9π
and so on
sin nt d t + ∫
π π /2
− sin nt d t
}
= π /2 π 1 cos nt nπ nπ cos nt 1 + − cos 0 + cos nπ − cos − = − cos π n 0 2 2 n π /2 π n
= When n is odd,
= bn
1 nπ 1 − cos π n 2
nπ + cos nπ − cos 2
nπ 1 + cos nπ = 1 − 2 cos 2 π n
1 1} 0 {1 − 0 −= πn
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When n is even,
b2=
1 4 2 = {1 − 2(−1) + 1}= 2π 2π π
b= 4
1 {1 − 2(1) + 1=} 0 4π
b6 =
1 4 2 = {1 − 2(−1) + 1}= 6π 6π 3π
b8 = 0 ,
Similarly,
b10 =
2 , and so on 5π
∞
Substituting into f(t) = a0 + ∑ ( an cos nt + bn sin nt ) n =1
gives:
f(x) = 0 +
2
π
cos t −
2 2 2 cos 3t + cos 5t − cos 7t + ... 3π 5π 7π 2
+
i.e. f(x) =
π
sin 2t +
2 2 sin 6t + sin10t + ... 3π 5π
2 1 1 1 1 cos t − cos 3t + cos 5t − ... + sin 2t + sin 6t + sin10t + ... 3 5 3 5 π
7. Show that the Fourier series for the periodic function of period 2π defined by: θ , when − π 〈 θ 〈 0 f(θ) = sin θ , when 0 〈 θ 〈 π is given by: f(θ) =
2 1 cos 2θ cos 4θ cos 6θ − − − ... − π 2 (3) (3)(5) (5)(7)
The periodic function is shown in the diagram below
a0 =
1 2π
π
θ ∫ π f (θ ) d= −
1 2π
{∫
0 −π
π
}
0 d θ + ∫ sin θ d = θ 0
=
1 2π
{[− cosθ ]=} π 0
{
}
1 ( − cos π ) − ( − cos 0 ) 2π
1 1 1 π) −1) (1 − cos= (1 −= 2π 2π π
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π {∫ 1
an
=
0 −π
π
0 cos nθ d θ + ∫ sin θ cos nθ d θ 0
}
1 π1 1 θ − nθ ) ∫ 0 sin (θ + nθ ) + sin (= π 2 2π
{∫ sin θ (1 + n) + sin θ (1 − n) dθ } from 6, page 716 π
0
of the textbook 1 = 2π
π
cos θ (1 + n ) θ (1 − n ) 1 cos π (1 + n ) cos π (1 − n ) 1 1 − cos = − − − −− − 1+ n 1 − n 0 2π 1+ n 1− n 1 + n 1 − n
When n is odd, an =
1 2π
1 1 1 1 − 1 + n − 1 − n + 1 + n + 1 − n = 0
When n = 2,
a= 2
1 cos 3π cos(−π ) 1 1 1 1 1 1 4 2 − + + = − − 1 + − 1= − = − 2π 3 −1 3 −1 2π 3 3 2π 3 3π
When n = 4,
a= 4
1 2π =
When n = 6,
=
π {∫ 1
bn
=
0 −π
1 1 1 1 − + − 5 3 5 3
1 3 − 5 + 3 − 5 1 4 2 − = − = 2π (3)(5) 2π (3)(5) π (3)(5)
1 2π
a= 6
cos 5π cos(−3π ) 1 1 1 − + + = − 5 −3 5 −3 2π
cos 7π cos(−5π ) 1 1 1 1 1 1 1 − + + = − − + − −5 7 7 −5 2π 7 5 7 5
1 5 − 7 + 5 − 7 1 4 2 − = − = 2π (5)(7) 2π (5)(7) π (5)(7) π
0sin nθ d θ + ∫ sin θ sin nθ d θ 0
}
1 π 1 1 − ∫ 0 − cos (θ + nθ ) − cos (θ − nθ ) = π 2 2π
π
sin θ (1 + n) sin θ (1 − n) = 0 from 9, page − 1 − n 0 1 + n 716 of the textbook
∞
Substituting into f(θ) = a0 + ∑ ( an cos nθ + bn sin nθ ) n =1
gives:
f(θ) =
i.e.
f(θ) =
1
π
−
2 2 2 cos 2θ − cos 4θ − cos 6θ − ... + 0 3π (3)(5)π (5)(7)π
2 1 cos 2θ cos 4θ cos 6θ − − − ... − π 2 (3) (3)(5) (5)(7)
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