Download - Fourier series 2
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Fourier Series 2
N. B. Vyas
Department of Mathematics,Atmiya Institute of Tech. & Science,
Rajkot (Guj.)- INDIA
N. B. Vyas Fourier Series 2
![Page 2: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/2.jpg)
Functions of any Period p = 2L
Let f(x) be a periodic function with an arbitrary period 2Ldefined in the interval c < x < c+ 2L
The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ c+2L
c
f(x) dx
an =1
L
∫ c+2L
c
f(x) cos(nπxL
)dx
bn =1
L
∫ c+2L
c
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 3: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/3.jpg)
Functions of any Period p = 2L
Let f(x) be a periodic function with an arbitrary period 2Ldefined in the interval c < x < c+ 2L
The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ c+2L
c
f(x) dx
an =1
L
∫ c+2L
c
f(x) cos(nπxL
)dx
bn =1
L
∫ c+2L
c
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 4: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/4.jpg)
Functions of any Period p = 2L
Let f(x) be a periodic function with an arbitrary period 2Ldefined in the interval c < x < c+ 2L
The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]
where a0 =1
L
∫ c+2L
c
f(x) dx
an =1
L
∫ c+2L
c
f(x) cos(nπxL
)dx
bn =1
L
∫ c+2L
c
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 5: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/5.jpg)
Functions of any Period p = 2L
Let f(x) be a periodic function with an arbitrary period 2Ldefined in the interval c < x < c+ 2L
The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ c+2L
c
f(x) dx
an =1
L
∫ c+2L
c
f(x) cos(nπxL
)dx
bn =1
L
∫ c+2L
c
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 6: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/6.jpg)
Functions of any Period p = 2L
Let f(x) be a periodic function with an arbitrary period 2Ldefined in the interval c < x < c+ 2L
The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ c+2L
c
f(x) dx
an =1
L
∫ c+2L
c
f(x) cos(nπxL
)dx
bn =1
L
∫ c+2L
c
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 7: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/7.jpg)
Functions of any Period p = 2L
Let f(x) be a periodic function with an arbitrary period 2Ldefined in the interval c < x < c+ 2L
The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ c+2L
c
f(x) dx
an =1
L
∫ c+2L
c
f(x) cos(nπxL
)dx
bn =1
L
∫ c+2L
c
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 8: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/8.jpg)
Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2L
0
f(x) dx
an =1
L
∫ 2L
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2L
0
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 9: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/9.jpg)
Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]
where a0 =1
L
∫ 2L
0
f(x) dx
an =1
L
∫ 2L
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2L
0
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 10: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/10.jpg)
Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2L
0
f(x) dx
an =1
L
∫ 2L
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2L
0
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 11: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/11.jpg)
Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2L
0
f(x) dx
an =1
L
∫ 2L
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2L
0
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 12: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/12.jpg)
Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2L
0
f(x) dx
an =1
L
∫ 2L
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2L
0
f(x) sin(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 13: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/13.jpg)
Functions of any Period p = 2L
Corollary 2: If c = −L the interval becomes −L < x < L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ L
−Lf(x) dx
an =1
L
∫ L
−Lf(x) cos
(nπxL
)dx
bn =1
L
∫ L
−Lf(x) sin
(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 14: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/14.jpg)
Functions of any Period p = 2L
Corollary 2: If c = −L the interval becomes −L < x < L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]
where a0 =1
L
∫ L
−Lf(x) dx
an =1
L
∫ L
−Lf(x) cos
(nπxL
)dx
bn =1
L
∫ L
−Lf(x) sin
(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 15: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/15.jpg)
Functions of any Period p = 2L
Corollary 2: If c = −L the interval becomes −L < x < L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ L
−Lf(x) dx
an =1
L
∫ L
−Lf(x) cos
(nπxL
)dx
bn =1
L
∫ L
−Lf(x) sin
(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 16: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/16.jpg)
Functions of any Period p = 2L
Corollary 2: If c = −L the interval becomes −L < x < L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ L
−Lf(x) dx
an =1
L
∫ L
−Lf(x) cos
(nπxL
)dx
bn =1
L
∫ L
−Lf(x) sin
(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 17: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/17.jpg)
Functions of any Period p = 2L
Corollary 2: If c = −L the interval becomes −L < x < L
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ L
−Lf(x) dx
an =1
L
∫ L
−Lf(x) cos
(nπxL
)dx
bn =1
L
∫ L
−Lf(x) sin
(nπxL
)dx
N. B. Vyas Fourier Series 2
![Page 18: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/18.jpg)
Example
Ex. The Fourier series of f(x) = x2, 0 < x < 2 wheref(x+ 2) = f(x).
Hence deduce that 1− 1
22+
1
32− 1
42+ . . . =
π2
12
N. B. Vyas Fourier Series 2
![Page 19: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/19.jpg)
Example
Ex. The Fourier series of f(x) = x2, 0 < x < 2 wheref(x+ 2) = f(x).
Hence deduce that 1− 1
22+
1
32− 1
42+ . . . =
π2
12
N. B. Vyas Fourier Series 2
![Page 20: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/20.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2
0
f(x) dx
an =1
L
∫ 2
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2
0
f(x) sin(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 21: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/21.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]
where a0 =1
L
∫ 2
0
f(x) dx
an =1
L
∫ 2
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2
0
f(x) sin(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 22: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/22.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2
0
f(x) dx
an =1
L
∫ 2
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2
0
f(x) sin(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 23: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/23.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2
0
f(x) dx
an =1
L
∫ 2
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2
0
f(x) sin(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 24: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/24.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2
0
f(x) dx
an =1
L
∫ 2
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2
0
f(x) sin(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 25: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/25.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 2
0
f(x) dx
an =1
L
∫ 2
0
f(x) cos(nπxL
)dx
bn =1
L
∫ 2
0
f(x) sin(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 26: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/26.jpg)
Example
Step 2. Now a0 =1
1
∫ 2
0
f(x) dx
a0 =
∫ 2
0
(x)2 dx
=
[x3
3
]20
=8
3
N. B. Vyas Fourier Series 2
![Page 27: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/27.jpg)
Example
Step 2. Now a0 =1
1
∫ 2
0
f(x) dx
a0 =
∫ 2
0
(x)2 dx
=
[x3
3
]20
=8
3
N. B. Vyas Fourier Series 2
![Page 28: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/28.jpg)
Example
Step 2. Now a0 =1
1
∫ 2
0
f(x) dx
a0 =
∫ 2
0
(x)2 dx
=
[x3
3
]20
=8
3
N. B. Vyas Fourier Series 2
![Page 29: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/29.jpg)
Example
Step 2. Now a0 =1
1
∫ 2
0
f(x) dx
a0 =
∫ 2
0
(x)2 dx
=
[x3
3
]20
=8
3
N. B. Vyas Fourier Series 2
![Page 30: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/30.jpg)
Example
Step 3. an =1
1
∫ 2
0
f(x) cos(nπx
1
)dx
an =
∫ 2
0
x2 cos(nπx) dx
=
[x2(sin nπx
nπ
)− (2x)
(−cos nπx
n2π2
)+ 2
(−sin nπx
n3π3
)]20
=4
n2π2
N. B. Vyas Fourier Series 2
![Page 31: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/31.jpg)
Example
Step 3. an =1
1
∫ 2
0
f(x) cos(nπx
1
)dx
an =
∫ 2
0
x2 cos(nπx) dx
=
[x2(sin nπx
nπ
)− (2x)
(−cos nπx
n2π2
)+ 2
(−sin nπx
n3π3
)]20
=4
n2π2
N. B. Vyas Fourier Series 2
![Page 32: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/32.jpg)
Example
Step 3. an =1
1
∫ 2
0
f(x) cos(nπx
1
)dx
an =
∫ 2
0
x2 cos(nπx) dx
=
[x2(sin nπx
nπ
)− (2x)
(−cos nπx
n2π2
)+ 2
(−sin nπx
n3π3
)]20
=4
n2π2
N. B. Vyas Fourier Series 2
![Page 33: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/33.jpg)
Example
Step 3. an =1
1
∫ 2
0
f(x) cos(nπx
1
)dx
an =
∫ 2
0
x2 cos(nπx) dx
=
[x2(sin nπx
nπ
)− (2x)
(−cos nπx
n2π2
)+ 2
(−sin nπx
n3π3
)]20
=4
n2π2
N. B. Vyas Fourier Series 2
![Page 34: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/34.jpg)
Example
Step 4. bn =1
1
∫ 2
0
f(x) sin(nπx
1
)dx
bn =
∫ 2
0
x2 sin(nπx) dx
=
[x2(−cos nπx
nπ
)− (2x)
(−sin nπx
n2π2
)+ 2
(cos nπxn3π3
)]20
= − 4
nπ
N. B. Vyas Fourier Series 2
![Page 35: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/35.jpg)
Example
Step 4. bn =1
1
∫ 2
0
f(x) sin(nπx
1
)dx
bn =
∫ 2
0
x2 sin(nπx) dx
=
[x2(−cos nπx
nπ
)− (2x)
(−sin nπx
n2π2
)+ 2
(cos nπxn3π3
)]20
= − 4
nπ
N. B. Vyas Fourier Series 2
![Page 36: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/36.jpg)
Example
Step 4. bn =1
1
∫ 2
0
f(x) sin(nπx
1
)dx
bn =
∫ 2
0
x2 sin(nπx) dx
=
[x2(−cos nπx
nπ
)− (2x)
(−sin nπx
n2π2
)+ 2
(cos nπxn3π3
)]20
= − 4
nπ
N. B. Vyas Fourier Series 2
![Page 37: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/37.jpg)
Example
Step 4. bn =1
1
∫ 2
0
f(x) sin(nπx
1
)dx
bn =
∫ 2
0
x2 sin(nπx) dx
=
[x2(−cos nπx
nπ
)− (2x)
(−sin nπx
n2π2
)+ 2
(cos nπxn3π3
)]20
= − 4
nπ
N. B. Vyas Fourier Series 2
![Page 38: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/38.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)=
4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 39: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/39.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)
=4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 40: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/40.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)=
4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 41: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/41.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)=
4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 42: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/42.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)=
4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]
−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 43: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/43.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)=
4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 44: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/44.jpg)
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (0, 2)
f(x) =8
6+∞∑n=1
(4
n2π2
)cos(nπx
1
)+∞∑n=1
(− 4
nπ
)sin(nπx
1
)=
4
3+
4
π2
∞∑n=1
cos(nπx)
n2− 4
π
∞∑n=1
sin(nπx)
n
Putting x = 1, we get
1 =4
3+
4
π2
[− 1
11+
1
22− 1
32+ . . .
]−1
3=
4
π2
[− 1
11+
1
22− 1
32+ . . .
]π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series 2
![Page 45: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/45.jpg)
Example
Ex. Find the Fourier series of f(x) = 2x in −1 < x < 1where p = 2L = 2.
N. B. Vyas Fourier Series 2
![Page 46: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/46.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 1
−1f(x) dx
an =1
L
∫ 1
−1f(x) cos
(nπxL
)dx
bn =1
L
∫ 1
−1f(x) sin
(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 47: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/47.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]
where a0 =1
L
∫ 1
−1f(x) dx
an =1
L
∫ 1
−1f(x) cos
(nπxL
)dx
bn =1
L
∫ 1
−1f(x) sin
(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 48: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/48.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 1
−1f(x) dx
an =1
L
∫ 1
−1f(x) cos
(nπxL
)dx
bn =1
L
∫ 1
−1f(x) sin
(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 49: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/49.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 1
−1f(x) dx
an =1
L
∫ 1
−1f(x) cos
(nπxL
)dx
bn =1
L
∫ 1
−1f(x) sin
(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 50: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/50.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 1
−1f(x) dx
an =1
L
∫ 1
−1f(x) cos
(nπxL
)dx
bn =1
L
∫ 1
−1f(x) sin
(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 51: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/51.jpg)
Example
Sol. Step 1: The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
[an cos
(nπxL
)+ bn sin
(nπxL
)]where a0 =
1
L
∫ 1
−1f(x) dx
an =1
L
∫ 1
−1f(x) cos
(nπxL
)dx
bn =1
L
∫ 1
−1f(x) sin
(nπxL
)dx
Here p = 2L = 2⇒ L = 1
N. B. Vyas Fourier Series 2
![Page 52: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/52.jpg)
Example
Step 2. Now a0 =1
1
∫ 1
−1f(x) dx
a0 =
∫ 1
−12x dx
= 0
N. B. Vyas Fourier Series 2
![Page 53: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/53.jpg)
Example
Step 2. Now a0 =1
1
∫ 1
−1f(x) dx
a0 =
∫ 1
−12x dx
= 0
N. B. Vyas Fourier Series 2
![Page 54: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/54.jpg)
Example
Step 2. Now a0 =1
1
∫ 1
−1f(x) dx
a0 =
∫ 1
−12x dx
= 0
N. B. Vyas Fourier Series 2
![Page 55: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/55.jpg)
Example
Step 3. an =1
1
∫ 1
−1f(x) cos
(nπx1
)dx
an =
∫ 1
−12x cos(nπx) dx
=
[2x
(sin nπx
nπ
)− (2)
(−cos nπx
n2π2
)]1−1
=
[0 +
2(−1)n
n2π2− 0− 2(−1)n
n2π2
]= 0
N. B. Vyas Fourier Series 2
![Page 56: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/56.jpg)
Example
Step 3. an =1
1
∫ 1
−1f(x) cos
(nπx1
)dx
an =
∫ 1
−12x cos(nπx) dx
=
[2x
(sin nπx
nπ
)− (2)
(−cos nπx
n2π2
)]1−1
=
[0 +
2(−1)n
n2π2− 0− 2(−1)n
n2π2
]= 0
N. B. Vyas Fourier Series 2
![Page 57: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/57.jpg)
Example
Step 3. an =1
1
∫ 1
−1f(x) cos
(nπx1
)dx
an =
∫ 1
−12x cos(nπx) dx
=
[2x
(sin nπx
nπ
)− (2)
(−cos nπx
n2π2
)]1−1
=
[0 +
2(−1)n
n2π2− 0− 2(−1)n
n2π2
]= 0
N. B. Vyas Fourier Series 2
![Page 58: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/58.jpg)
Example
Step 3. an =1
1
∫ 1
−1f(x) cos
(nπx1
)dx
an =
∫ 1
−12x cos(nπx) dx
=
[2x
(sin nπx
nπ
)− (2)
(−cos nπx
n2π2
)]1−1
=
[0 +
2(−1)n
n2π2
− 0− 2(−1)n
n2π2
]= 0
N. B. Vyas Fourier Series 2
![Page 59: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/59.jpg)
Example
Step 3. an =1
1
∫ 1
−1f(x) cos
(nπx1
)dx
an =
∫ 1
−12x cos(nπx) dx
=
[2x
(sin nπx
nπ
)− (2)
(−cos nπx
n2π2
)]1−1
=
[0 +
2(−1)n
n2π2− 0− 2(−1)n
n2π2
]
= 0
N. B. Vyas Fourier Series 2
![Page 60: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/60.jpg)
Example
Step 3. an =1
1
∫ 1
−1f(x) cos
(nπx1
)dx
an =
∫ 1
−12x cos(nπx) dx
=
[2x
(sin nπx
nπ
)− (2)
(−cos nπx
n2π2
)]1−1
=
[0 +
2(−1)n
n2π2− 0− 2(−1)n
n2π2
]= 0
N. B. Vyas Fourier Series 2
![Page 61: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/61.jpg)
Example
Ex. Find the Fourier series of periodic function
f(x)= −1;−1 < x < 0= 1; 0 < x < 1
p = 2L = 2
N. B. Vyas Fourier Series 2
![Page 62: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/62.jpg)
Example
Ex. Find the Fourier series of periodic function
f(x)= 0;−2 < x < 0= 2; 0 < x < 2
p = 2L = 4
N. B. Vyas Fourier Series 2
![Page 63: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/63.jpg)
Fourier Half Range Series
A function f(x) defined only on the interval of the form0 < x < L.
If f(x) is represented on this interval by a Fourier series ofperiod 2L
Then such Fourier series are known as half range Fourierseries or half range expansions.
Types of Half Range Fourier series
1 Fourier Cosine Series
2 Fourier Sine Series
N. B. Vyas Fourier Series 2
![Page 64: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/64.jpg)
Fourier Half Range Series
A function f(x) defined only on the interval of the form0 < x < L.
If f(x) is represented on this interval by a Fourier series ofperiod 2L
Then such Fourier series are known as half range Fourierseries or half range expansions.
Types of Half Range Fourier series
1 Fourier Cosine Series
2 Fourier Sine Series
N. B. Vyas Fourier Series 2
![Page 65: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/65.jpg)
Fourier Half Range Series
A function f(x) defined only on the interval of the form0 < x < L.
If f(x) is represented on this interval by a Fourier series ofperiod 2L
Then such Fourier series are known as half range Fourierseries or half range expansions.
Types of Half Range Fourier series
1 Fourier Cosine Series
2 Fourier Sine Series
N. B. Vyas Fourier Series 2
![Page 66: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/66.jpg)
Fourier Half Range Series
A function f(x) defined only on the interval of the form0 < x < L.
If f(x) is represented on this interval by a Fourier series ofperiod 2L
Then such Fourier series are known as half range Fourierseries or half range expansions.
Types of Half Range Fourier series
1 Fourier Cosine Series
2 Fourier Sine Series
N. B. Vyas Fourier Series 2
![Page 67: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/67.jpg)
Fourier Half Range Series
A function f(x) defined only on the interval of the form0 < x < L.
If f(x) is represented on this interval by a Fourier series ofperiod 2L
Then such Fourier series are known as half range Fourierseries or half range expansions.
Types of Half Range Fourier series
1 Fourier Cosine Series
2 Fourier Sine Series
N. B. Vyas Fourier Series 2
![Page 68: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/68.jpg)
Fourier Half Range Series
A function f(x) defined only on the interval of the form0 < x < L.
If f(x) is represented on this interval by a Fourier series ofperiod 2L
Then such Fourier series are known as half range Fourierseries or half range expansions.
Types of Half Range Fourier series
1 Fourier Cosine Series
2 Fourier Sine Series
N. B. Vyas Fourier Series 2
![Page 69: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/69.jpg)
Fourier Cosine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier cosine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =ao2
+∞∑n=1
an cos(nπx
l
)where a0 =
2
l
∫ l
0
f(x) dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 70: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/70.jpg)
Fourier Cosine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier cosine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =ao2
+∞∑n=1
an cos(nπx
l
)where a0 =
2
l
∫ l
0
f(x) dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 71: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/71.jpg)
Fourier Cosine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier cosine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =ao2
+∞∑n=1
an cos(nπx
l
)
where a0 =2
l
∫ l
0
f(x) dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 72: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/72.jpg)
Fourier Cosine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier cosine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =ao2
+∞∑n=1
an cos(nπx
l
)where a0 =
2
l
∫ l
0
f(x) dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 73: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/73.jpg)
Fourier Cosine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier cosine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =ao2
+∞∑n=1
an cos(nπx
l
)where a0 =
2
l
∫ l
0
f(x) dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 74: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/74.jpg)
Fourier Sine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier sine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =∞∑n=1
bn sin(nπx
l
)bn =
2
l
∫ l
0
f(x) sin(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 75: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/75.jpg)
Fourier Sine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier sine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =∞∑n=1
bn sin(nπx
l
)bn =
2
l
∫ l
0
f(x) sin(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 76: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/76.jpg)
Fourier Sine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier sine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =∞∑n=1
bn sin(nπx
l
)
bn =2
l
∫ l
0
f(x) sin(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 77: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/77.jpg)
Fourier Sine Series
Let f(x) be piecewise continuous on [o, l].
the Fourier sine series expansion of f(x) on the half rangeinterval [0, l] is given by
f(x) =∞∑n=1
bn sin(nπx
l
)bn =
2
l
∫ l
0
f(x) sin(nπx
l
)dx
N. B. Vyas Fourier Series 2
![Page 78: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/78.jpg)
Example
Ex. Find Fourier cosine and sine series of the functionf(x) = 1 for 0 ≤ x ≤ 2
N. B. Vyas Fourier Series 2
![Page 79: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/79.jpg)
Example
Sol. Here given interval is 0 ≤ x ≤ 2
∴ l = 2
N. B. Vyas Fourier Series 2
![Page 80: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/80.jpg)
Example
Sol. Here given interval is 0 ≤ x ≤ 2
∴ l = 2
N. B. Vyas Fourier Series 2
![Page 81: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/81.jpg)
Example
1 Fourier cosine series
Step 1. f(x) =a02
+∞∑n=1
an cos(nπx
l
)
where a0 =2
l
∫ l
0
f(x)dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)
N. B. Vyas Fourier Series 2
![Page 82: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/82.jpg)
Example
1 Fourier cosine series
Step 1. f(x) =a02
+∞∑n=1
an cos(nπx
l
)where a0 =
2
l
∫ l
0
f(x)dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)
N. B. Vyas Fourier Series 2
![Page 83: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/83.jpg)
Example
1 Fourier cosine series
Step 1. f(x) =a02
+∞∑n=1
an cos(nπx
l
)where a0 =
2
l
∫ l
0
f(x)dx
an =2
l
∫ l
0
f(x) cos(nπx
l
)
N. B. Vyas Fourier Series 2
![Page 84: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/84.jpg)
Example
Step 2. a0 =2
2
∫ 2
0
f(x)dx
=
∫ 2
0
1dx = [x]20 = 2.
N. B. Vyas Fourier Series 2
![Page 85: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/85.jpg)
Example
Step 2. a0 =2
2
∫ 2
0
f(x)dx
=
∫ 2
0
1dx
= [x]20 = 2.
N. B. Vyas Fourier Series 2
![Page 86: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/86.jpg)
Example
Step 2. a0 =2
2
∫ 2
0
f(x)dx
=
∫ 2
0
1dx = [x]20
= 2.
N. B. Vyas Fourier Series 2
![Page 87: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/87.jpg)
Example
Step 2. a0 =2
2
∫ 2
0
f(x)dx
=
∫ 2
0
1dx = [x]20 = 2.
N. B. Vyas Fourier Series 2
![Page 88: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/88.jpg)
Example
Step 3. an =2
2
∫ 2
0
f(x)cos(nπx
2
)dx
=
∫ 2
0
(1) cos(nπx
2
)dx
=
sin(nπx
2
)(nπ
2
)2
0
=
(2
nπ
)(sin (nπ)− sin (0)) = 0
N. B. Vyas Fourier Series 2
![Page 89: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/89.jpg)
Example
Step 3. an =2
2
∫ 2
0
f(x)cos(nπx
2
)dx
=
∫ 2
0
(1) cos(nπx
2
)dx
=
sin(nπx
2
)(nπ
2
)2
0
=
(2
nπ
)(sin (nπ)− sin (0)) = 0
N. B. Vyas Fourier Series 2
![Page 90: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/90.jpg)
Example
Step 3. an =2
2
∫ 2
0
f(x)cos(nπx
2
)dx
=
∫ 2
0
(1) cos(nπx
2
)dx
=
sin(nπx
2
)(nπ
2
)2
0
=
(2
nπ
)(sin (nπ)− sin (0)) = 0
N. B. Vyas Fourier Series 2
![Page 91: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/91.jpg)
Example
Step 3. an =2
2
∫ 2
0
f(x)cos(nπx
2
)dx
=
∫ 2
0
(1) cos(nπx
2
)dx
=
sin(nπx
2
)(nπ
2
)2
0
=
(2
nπ
)(sin (nπ)− sin (0))
= 0
N. B. Vyas Fourier Series 2
![Page 92: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/92.jpg)
Example
Step 3. an =2
2
∫ 2
0
f(x)cos(nπx
2
)dx
=
∫ 2
0
(1) cos(nπx
2
)dx
=
sin(nπx
2
)(nπ
2
)2
0
=
(2
nπ
)(sin (nπ)− sin (0)) = 0
N. B. Vyas Fourier Series 2
![Page 93: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/93.jpg)
Example
∴ Fourier cosine series of f(x) is
f(x) =a02
+∞∑n=1
an cos(nπx
l
)=
2
2+ 0 = 1
N. B. Vyas Fourier Series 2
![Page 94: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/94.jpg)
Example
∴ Fourier cosine series of f(x) is
f(x) =a02
+∞∑n=1
an cos(nπx
l
)
=2
2+ 0 = 1
N. B. Vyas Fourier Series 2
![Page 95: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/95.jpg)
Example
∴ Fourier cosine series of f(x) is
f(x) =a02
+∞∑n=1
an cos(nπx
l
)=
2
2+ 0
= 1
N. B. Vyas Fourier Series 2
![Page 96: Fourier series 2](https://reader033.vdocument.in/reader033/viewer/2022061202/547b282bb4795968098b4c2f/html5/thumbnails/96.jpg)
Example
∴ Fourier cosine series of f(x) is
f(x) =a02
+∞∑n=1
an cos(nπx
l
)=
2
2+ 0 = 1
N. B. Vyas Fourier Series 2