dr. nirav vyas fourierseries2.pdf
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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
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Functions of any Period p = 2L
Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2L
N. B. Vyas Fourier Series 2
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Functions of any Period p = 2L
Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2LThe Fourier series of f (x ) is given by
N. B. Vyas Fourier Series 2
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Functions of any Period p = 2L
Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2LThe Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
c +2 L
c
f (x ) dx
N. B. Vyas Fourier Series 2
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Functions of any Period p = 2L
Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2LThe Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
c +2 L
c
f (x ) dx
a n = 1
L c +2 L
c
f (x ) cosnπx
L dx
N. B. Vyas Fourier Series 2
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Functions of any Period p = 2L
Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2LThe Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
c +2 L
c
f (x ) dx
a n = 1
L c +2 L
c
f (x ) cosnπx
L dx
bn = 1L
c +2 L
c
f (x ) sinnπx
L dx
N. B. Vyas Fourier Series 2
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Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
N. B. Vyas Fourier Series 2
f d
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Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
N. B. Vyas Fourier Series 2
F i f P i d 2L
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Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L 2 L
0f (x) dx
N. B. Vyas Fourier Series 2
F ti f P i d 2L
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Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L 2 L
0f (x) dx
a n = 1L
2 L
0f (x ) cos
nπxL
dx
N. B. Vyas Fourier Series 2
F ti f P i d 2L
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Functions of any Period p = 2L
Corollary 1: If c = 0 the interval becomes 0 < x < 2L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
2 L
0f (x) dx
a n = 1L
2 L
0f (x ) cos
nπxL
dx
bn = 1L 2 L
0f (x ) sin nπxL dx
N. B. Vyas Fourier Series 2
Functions of any Period 2L
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Functions of any Period p = 2L
Corollary 2: If c = − L the interval becomes − L < x < L
N. B. Vyas Fourier Series 2
Functions of any Period p = 2L
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Functions of any Period p = 2L
Corollary 2: If c = − L the interval becomes − L < x < L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
N. B. Vyas Fourier Series 2
Functions of any Period p = 2L
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Functions of any Period p = 2L
Corollary 2: If c = − L the interval becomes − L < x < L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
L
− Lf (x) dx
N. B. Vyas Fourier Series 2
Functions of any Period p = 2L
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Functions of any Period p = 2L
Corollary 2: If c = − L the interval becomes − L < x < L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
L
− Lf (x) dx
a n = 1L
L
− L
f (x) cosnπx
L dx
N. B. Vyas Fourier Series 2
Functions of any Period p = 2L
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Functions of any Period p = 2L
Corollary 2: If c = − L the interval becomes − L < x < L
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1L
L
− Lf (x) dx
a n = 1L
L
− L
f (x) cosnπx
L dx
bn = 1L L
− Lf (x ) sin nπxL dx
N. B. Vyas Fourier Series 2
Example
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Example
Ex. The Fourier series of f (x ) = x2, 0 < x < 2 wheref (x + 2) = f (x ).
N. B. Vyas Fourier Series 2
Example
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Example
Ex. The Fourier series of f (x ) = x2, 0 < x < 2 wheref (x + 2) = f (x ).
Hence deduce that 1 −122
+ 132
−142
+ . . . = π2
12
N. B. Vyas Fourier Series 2
Example
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p
Sol. Step 1: The Fourier series of f (x ) is given by
N. B. Vyas Fourier Series 2
Example
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p
Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
N. B. Vyas Fourier Series 2
Example
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p
Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1
L 2
0
f (x ) dx
N. B. Vyas Fourier Series 2
Example
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p
Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1
L 2
0
f (x ) dx
a n = 1L
2
0f (x ) cos
nπxL
dx
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1
L 2
0
f (x ) dx
a n = 1L
2
0f (x ) cos
nπxL
dx
bn = 1
L 2
0f (x ) sin
nπx
L dx
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin
nπxL
where a0 = 1
L 2
0
f (x ) dx
a n = 1L
2
0f (x ) cos
nπxL
dx
bn = 1
L 2
0f (x ) sin
nπx
L dx
Here p = 2L = 2 ⇒ L = 1
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
2
0f (x ) dx
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
2
0f (x ) dx
a 0 =
2
0
(x)2 dx
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
2
0f (x ) dx
a 0 =
2
0
(x)2 dx
=x 3
3
2
0
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
2
0f (x ) dx
a 0 =
2
0
(x)2 dx
=x 3
3
2
0
= 8
3
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11
2
0f (x ) cos
nπx1
dx
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11
2
0f (x ) cos
nπx1
dx
a n =
2
0x 2 cos (nπx ) dx
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11
2
0f (x ) cos
nπx1
dx
a n =
2
0x 2 cos (nπx ) dx
= x 2sin nπx
nπ− (2x) −
cos nπxn 2 π 2
+ 2 −sin nπx
n 3 π 3
2
0
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11
2
0f (x ) cos
nπx1
dx
a n = 2
0x 2 cos (nπx ) dx
= x 2sin nπx
nπ− (2x) −
cos nπxn 2 π 2
+ 2 −sin nπx
n 3 π 3
2
0
= 4
n2
π2
N. B. Vyas Fourier Series 2
Example
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Step 4. bn = 11
2
0f (x ) sin
nπx1
dx
N. B. Vyas Fourier Series 2
Example
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Step 4. bn = 11
2
0f (x ) sin
nπx1
dx
bn = 2
0x 2 sin (nπx ) dx
N. B. Vyas Fourier Series 2
Example
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Step 4. bn = 11
2
0f (x ) sin
nπx1
dx
bn = 2
0x 2 sin (nπx ) dx
= x 2 −cos nπx
nπ− (2x ) −
sin nπxn 2 π 2
+ 2cos nπx
n 3 π 3
2
0
N. B. Vyas Fourier Series 2
Example
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Step 4. bn = 11
2
0f (x ) sin
nπx1
dx
bn = 2
0x 2 sin (nπx ) dx
= x 2 −cos nπx
nπ− (2x ) −
sin nπxn 2 π 2
+ 2cos nπx
n 3 π 3
2
0
= −4
nπ
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
f (x ) = 86
+∞
n =1
4n 2 π 2
cosnπx
1+
∞
n =1
−4
nπsin
nπx1
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
f (x ) = 86
+∞
n =1
4n 2 π 2
cosnπx
1+
∞
n =1
−4
nπsin
nπx1
= 4
3 +
4
π2
∞
n =1
cos (nπx )
n2
−4
π
∞
n =1
sin (nπx )
n
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
f (x ) = 86
+∞
n =1
4n 2 π 2
cosnπx
1+
∞
n =1
−4
nπsin
nπx1
= 4
3 +
4
π2
∞
n =1
cos (nπx )
n2
−4
π
∞
n =1
sin (nπx )
nPutting x = 1, we get
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
f (x ) = 86
+∞
n =1
4n 2 π 2
cosnπx
1+
∞
n =1
− 4nπ
sinnπx
1
= 4
3 +
4
π2
∞
n =1
cos (nπx )
n2
−4
π
∞
n =1
sin (nπx )
nPutting x = 1, we get
1 = 43
+ 4π 2
−111
+ 122
−132
+ . . .
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
f (x ) = 86
+∞
n =1
4n 2 π 2
cosnπx
1+
∞
n =1
− 4nπ
sinnπx
1
= 4
3 +
4
π2
∞
n =1
cos (nπx )
n2
−4
π
∞
n =1
sin (nπx )
nPutting x = 1, we get
1 = 43
+ 4π 2
−111
+ 122
−132
+ . . .
− 13
= 4π 2
− 111
+ 122
− 132
+ . . .
N. B. Vyas Fourier Series 2
Example
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Step 5. Substituting values of a0 , a n and bn in (1), we get therequired Fourier series of f (x ) in the interval (0 , 2)
f (x ) = 86
+∞
n =1
4n 2 π 2
cosnπx
1+
∞
n =1
− 4nπ
sinnπx
1
= 4
3 +
4
π2
∞
n =1
cos (nπx )
n2
−4
π
∞
n =1
sin (nπx )
nPutting x = 1, we get
1 = 43
+ 4π 2
−111
+ 122
−132
+ . . .
− 13
= 4π 2
− 111
+ 122
− 132
+ . . .
π 2
12 =
112
−122
+ 132
−142
+ . . .
N. B. Vyas Fourier Series 2
Example
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Ex. Find the Fourier series of f (x ) = 2 x in − 1 < x < 1
where p = 2L = 2.
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin nπx
L
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin nπx
L
where a0 = 1L
1
−
1
f (x) dx
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cosnπx
L+ bn sin nπx
L
where a0 = 1L
1
−
1
f (x) dx
a n = 1L
1
− 1f (x) cos
nπxL
dx
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cos nπxL
+ bn sin nπxL
where a0 = 1L
1
−
1
f (x) dx
a n = 1L
1
− 1f (x) cos
nπxL
dx
bn = 1
L 1
− 1
f (x ) sinnπx
L dx
N. B. Vyas Fourier Series 2
Example
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Sol. Step 1: The Fourier series of f (x ) is given by
f (x ) = a02
+∞
n =1
a n cos nπxL
+ bn sin nπxL
where a0 = 1L
1
− 1f (x) dx
a n = 1L
1
− 1f (x) cos
nπxL
dx
bn = 1
L 1
− 1
f (x ) sinnπx
L dx
Here p = 2L = 2 ⇒ L = 1
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
1
− 1f (x ) dx
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
1
− 1f (x ) dx
a 0 = 1
− 12x dx
N. B. Vyas Fourier Series 2
Example
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Step 2. Now a0 = 11
1
− 1f (x ) dx
a 0 = 1
− 12x dx
= 0
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11 1
− 1f (x) cos nπx1
dx
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11 1
− 1f (x) cos nπx1 dx
a n = 1
− 12xcos (nπx ) dx
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11 1
− 1f (x) cos nπx1 dx
a n = 1
− 12xcos (nπx ) dx
= 2x sin nπxnπ
− (2) − cos nπxn 2 π 2
1
− 1
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11 1
− 1f (x) cos nπx1 dx
a n = 1
− 12xcos (nπx ) dx
= 2x sin nπxnπ
− (2) − cos nπxn 2 π 2
1
− 1
= 0 + 2(− 1)n
n 2 π 2
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11 1
− 1f (x) cos nπx1 dx
a n = 1
− 12xcos (nπx ) dx
= 2x sin nπxnπ
− (2) − cos nπxn 2 π 2
1
− 1
= 0 + 2(− 1)n
n 2 π 2 − 0 −
2(− 1)n
n 2 π 2
N. B. Vyas Fourier Series 2
Example
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Step 3. an = 11 1
− 1f (x) cos nπx1 dx
a n = 1
− 12xcos (nπx ) dx
= 2x sin nπxnπ
− (2) − cos nπxn 2 π 2
1
− 1
= 0 + 2(− 1)n
n 2 π 2 − 0 −
2(− 1)n
n 2 π 2
= 0
N. B. Vyas Fourier Series 2
Example
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Ex. Find the Fourier series of periodic functionf (x )= − 1; − 1 < x < 0
= 1; 0 < x < 1 p = 2L = 2
N. B. Vyas Fourier Series 2
Example
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Ex. Find the Fourier series of periodic functionf (x )= 0; − 2 < x < 0
= 2; 0 < x < 2 p = 2L = 4
N. B. Vyas Fourier Series 2
Fourier Half Range Series
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A function f (x) dened only on the interval of the form0 < x < L .
N. B. Vyas Fourier Series 2
Fourier Half Range Series
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A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2L
N. B. Vyas Fourier Series 2
Fourier Half Range Series
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A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2L
Then such Fourier series are known as half range Fourier
series or half range expansions .
N. B. Vyas Fourier Series 2
Fourier Half Range Series
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A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2LThen such Fourier series are known as half range Fourierseries or half range expansions .Types of Half Range Fourier series
N. B. Vyas Fourier Series 2
Fourier Half Range Series
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A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2LThen such Fourier series are known as half range Fourierseries or half range expansions .Types of Half Range Fourier series
1 Fourier Cosine Series
N. B. Vyas Fourier Series 2
Fourier Half Range Series
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A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2LThen such Fourier series are known as half range Fourierseries or half range expansions .Types of Half Range Fourier series
1 Fourier Cosine Series2 Fourier Sine Series
N. B. Vyas Fourier Series 2
Fourier Cosine Series
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Let f (x) be piecewise continuous on [o, l ].
N. B. Vyas Fourier Series 2
Fourier Cosine Series
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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
N. B. Vyas Fourier Series 2
Fourier Cosine Series
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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 ) = ao
2 +
∞
n =1
a n cosnπx
l
N. B. Vyas Fourier Series 2
Fourier Cosine Series
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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 ) = ao
2 +
∞
n =1
a n cosnπx
l
where a0 = 2l
l
0f (x) dx
N. B. Vyas Fourier Series 2
Fourier Cosine Series
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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 ) = ao
2 +
∞
n =1
a n cosnπx
l
where a0 = 2l
l
0f (x) dx
a n =
2l
l
0f
(x
)cos
nπx
ldx
N. B. Vyas Fourier Series 2
Fourier Sine Series
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Let f (x) be piecewise continuous on [o, l ].
N. B. Vyas Fourier Series 2
Fourier Sine Series
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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
N. B. Vyas Fourier Series 2
Fourier Sine Series
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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 =1bn sin
nπxl
N. B. Vyas Fourier Series 2
Fourier Sine Series
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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 =1bn sin
nπxl
bn = 2l
l
0f (x ) sin
nπxl
dx
N. B. Vyas Fourier Series 2
Example
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Ex. Find Fourier cosine and sine series of the function
f (x ) = 1 for 0 ≤ x ≤ 2
N. B. Vyas Fourier Series 2
Example
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Sol. Here given interval is 0 ≤ x ≤ 2
N. B. Vyas Fourier Series 2
Example
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Sol. Here given interval is 0 ≤ x ≤ 2
∴ l = 2
N. B. Vyas Fourier Series 2
Example
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1
Fourier cosine series
Step 1. f (x ) = a02
+∞
n =1
a n cosnπx
l
N. B. Vyas Fourier Series 2
Example
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1
Fourier cosine series
Step 1. f (x ) = a02
+∞
n =1
a n cosnπx
l
where a0 = 2l l
0f (x )dx
N. B. Vyas Fourier Series 2
Example
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1
Fourier cosine series
Step 1. f (x ) = a02
+∞
n =1
a n cosnπx
l
where a0 = 2l l
0f (x )dx
a n = 2
l l
0f (x ) cos
nπx
l
N. B. Vyas Fourier Series 2
Example
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Step 2. a0 = 22
2
0f (x )dx
N B Vyas Fourier Series 2
Example
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Step 2. a0 = 22
2
0f (x )dx
= 2
01dx
N B Vyas Fourier Series 2
Example
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Step 2. a0 = 22
2
0f (x )dx
= 2
01dx = [x ]20
N B Vyas Fourier Series 2
Example
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Step 2. a0 = 22
2
0f (x )dx
= 2
01dx = [x ]20 = 2 .
N B Vyas Fourier Series 2
Example
2 2 nπx
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Step 3. an = 2
2 0f (x )cos
nπx
2dx
N B Vyas Fourier Series 2
Example
2 2 nπx
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Step 3. an = 2
2 0f (x )cos
nπx
2dx
= 2
0(1) cos
nπx2
dx
N B Vyas Fourier Series 2
Example
2 2 nπx
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Step 3. an = 2
2 0f (x )cos
π
2dx
= 2
0(1) cos
nπx2
dx
=sin
nπx2
nπ2
2
0
N B Vyas Fourier Series 2
Example
S 3 2 2
f ( )nπx
d
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Step 3. an =2 0
f (x )cos2
dx
= 2
0(1) cos
nπx2
dx
=sin
nπx2
nπ2
2
0
= 2nπ (sin (nπ ) − sin (0))
N B Vyas Fourier Series 2
Example
S 3 2 2
f ( )nπx
d
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Step 3. an =2 0
f (x )cos2
dx
= 2
0(1) cos
nπx2
dx
=sin
nπx2
nπ2
2
0
= 2nπ (sin (nπ ) − sin (0)) = 0
N B Vyas Fourier Series 2
Example
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∴ Fourier cosine series of f (x ) is
N B Vyas Fourier Series 2
Example
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∴ Fourier cosine series of f (x ) is
f (x ) = a02
+∞
n =1
a n cosnπx
l
N B Vyas Fourier Series 2
Example
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∴ Fourier cosine series of f (x ) is
f (x ) = a02
+∞
n =1
a n cosnπx
l
= 22
+ 0
N B Vyas Fourier Series 2
Example
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∴ Fourier cosine series of f (x ) is
f (x ) = a02
+∞
n =1
a n cosnπx
l
= 22
+ 0 = 1
N B V Fourier Series 2