2 Fourier series(F.S.) Dirichlet’s conditions - General Fourier series - Odd and even functions - Half range sine series - Half range cosine series - Complex form of Fourier Series - Parseval’s identity - Harmonic Analysis. 2.0.1 Periodic function A function f (x) is said to be periodic if and only if f (x + p)= f (x) for some p for x. The smallest value of p is called period of the function. Examples : 1. sin(x +2π) = sin x, 2π is period for f (x) = sin x 2. cos(x +2π) = cos x, 2π is period for f (x) = cos x 3. tan(x + π) = tan x, π is period for f (x) = tan x 4. sin x + 2π n = sin nx, 2π n is period for f (x) = sin nx 5. cos x + 2π n = cos nx, 2π n is period for f (x) = cos nx 6. tan x + π n = tan nx, π n is period for f (x) = tan nx 2.0.2 Continuity of a function A function f (x) is said to be continuous in the interval [a, b], if it is continuous at every point of the interval. 2.0.3 Left Hand Limit The left hand limit of f (x) at x = a is defined as x approaches a from left and denoted by f (a-) and is defined by f (a-) = lim h→0 f (a - h) 89
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2 Fourier series(F.S.) - RMD Engineering CollegeMA6351 Transforms and Partial Di erential Equations by K A Niranjan Kumar 95 a n = 1 ˇ Z2ˇ 0 2ˇx x2 cosnxdx = 1 ˇ 2ˇx x2 sinnx
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2 Fourier series(F.S.)
Dirichlet’s conditions - General Fourier series - Odd and evenfunctions - Half range sine series - Half range cosine series -Complex form of Fourier Series - Parseval’s identity - HarmonicAnalysis.
2.0.1 Periodic function
A function f(x) is said to be periodic if and only if f(x + p) = f(x) forsome p for x. The smallest value of p is called period of the function.
Examples :
1. sin(x+ 2π) = sin x, 2π is period for f(x) = sin x
2. cos(x+ 2π) = cos x, 2π is period for f(x) = cos x
3. tan(x+ π) = tan x, π is period for f(x) = tan x
4. sin
(x+
2π
n
)= sinnx,
2π
nis period for f(x) = sinnx
5. cos
(x+
2π
n
)= cosnx,
2π
nis period for f(x) = cosnx
6. tan(x+
π
n
)= tannx,
π
nis period for f(x) = tannx
2.0.2 Continuity of a function
A function f(x) is said to be continuous in the interval [a, b], if it iscontinuous at every point of the interval.
2.0.3 Left Hand Limit
The left hand limit of f(x) at x = a is defined as x approaches a fromleft and denoted by f(a−) and is defined by
f(a−) = limh→0
f(a− h)
89
90 Unit II - FOURIER SERIES (F.S.)
2.0.4 Right Hand Limit
The right hand limit of f(x) at x = a is defined as x approaches a fromright and denoted by f(a+) and is defined by
f(a+) = limh→0
f(a+ h)
Note : A function f(x) is said to be continuous at x = a is
f(a−) = f(a) = f(a+)
2.1 Dirichlet’s conditions
If a function f(x) is defined in c ≤ x ≤ c + 2`, it can be expanded asFourier series of the form
f(x) =a02+
∞∑n=1
(an cos
nπx
`+ bn sin
nπx
`
)where a0, an, bn are Fourier constant coefficients, provided:
(i) f(x) is periodic, single valued and finite in (c, c+ 2`).
(ii) f(x) is continuous (or) piecewise continuous with finite number ofdiscontinuities in (c, c+ 2`).
(iii) f(x) has at the most a finite number of maxima or minima in(c, c+ 2`).
2.2 General Fourier series
In (c, c+ 2`), Fourier series is
f(x) =a02+
∞∑n=1
(an cos
nπx
`+ bn sin
nπx
`
)where a0, an, bn are Fourier coefficients which can be found by Euler’sformulae.
2.2.1 Euler’s formula
In (c, c+ 2`), Fourier series is
f(x) =a02+
∞∑n=1
(an cos
nπx
`+ bn sin
nπx
`
)
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 91
where
a0 =1
`
∫ c+2`
c
f(x) dx
an =1
`
∫ c+2`
c
f(x)cos(nπx
`
)dx
bn =1
`
∫ c+2`
c
f(x)sin(nπx
`
)dx
Case(i): If c = 0, (c, c+ 2`) becomes (0, 2`).∴ Fourier series is
f(x) =a02+
∞∑n=1
(an cos
nπx
`+ bn sin
nπx
`
)where
a0 =1
`
∫ 2`
0
f(x) dx
an =1
`
∫ 2`
0
f(x)cos(nπx
`
)dx
bn =1
`
∫ 2`
0
f(x)sin(nπx
`
)dx
Case(ii): If c = −`, (c, c+ 2`) becomes (−`, `).∴ Fourier series is
f(x) =a02+
∞∑n=1
(an cos
nπx
`+ bn sin
nπx
`
)where
a0 =1
`
∫ `
−`
f(x) dx
an =1
`
∫ `
−`
f(x)cos(nπx
`
)dx
bn =1
`
∫ `
−`
f(x)sin(nπx
`
)dx
Case(iii): If c = 0 and ` = π, (c, c+ 2`) becomes (0, 2π).∴ Fourier series is
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx)
where
92 Unit II - FOURIER SERIES (F.S.)
a0 =1
π
∫ π
0
f(x) dx
an =1
π
∫ π
0
f(x)cosnx dx
bn =1
π
∫ π
0
f(x)sinnx dx
Case(iv): If c = −π and ` = π, (c, c+ 2`) becomes (−π, π).∴ Fourier series is
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx)
where
a0 =1
π
∫ π
−π
f(x) dx
an =1
π
∫ π
−π
f(x)cosnx dx
bn =1
π
∫ π
−π
f(x)sinnx dx
While finding series(Fourier/Cosine/Sine/Complex/Harmonic):Analyse what series, what interval and f (x ) with related formulae.Intervals : (0, 2π), (−π, π), (0, π), (0, 2`), (−`, `), (0, `)
(* denotes class work problems)
�� � Worked Examples
2.2.2 Examples under (0, 2π)
Example 2.1. Find the Fourier series of f(x) = x2 in (0, 2π) and with
period 2π.
{a0 =
8
3π2, an =
4
n2, bn =
−4π
n
}Solution : Given f(x) = x2 defined in the interval (0, 2π).∴ The Fourier series of f(x) is given by
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx) (1)
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 93
where a0 =1
π
2π∫0
f (x) dx
an =1
π
2π∫0
f (x) cosnxdx
bn =1
π
2π∫0
f (x) sinnxdx
Now, a0 =1
π
2π∫0
x2dx =1
π
[x3
3
]=
1
π
[8π3
3− 0
]=
8π2
3
an =1
π
2π∫0
x2 cosnxdx
=1
π
[x2(sinnx
n
)− 2x
(− cosnx
n2
)+ 2
(− sinnx
n3
)]2π0
=1
π
[x2
sinnx
n+ 2x
cosnx
n2− 2
sinnx
n3
]2π0
=1
π
[(0 +
4π
n2− 0
)− (0 + 0− 0)
][∵ cos 2nπ = 1]
∴ an =4
n2
bn =1
π
2π∫0
x2 sinnxdx
=1
π
[x2(− cosnx
n
)− 2x
(− sinnx
n2
)+ 2
(cosnxn3
)]2π0
=1
π
[−x2cosnx
n+ 2x
sinnx
n2+ 2
cosnx
n3
]2π0
=1
π
[(−4π2
n+ 0 +
2
n3
)−(0 + 0 +
2
n3
)]=1
π
(−4π2
n
)∴ bn =
−4π
n
Sub. the value of a0, an, bn in (1)
94 Unit II - FOURIER SERIES (F.S.)
f(x) =1
2
(8π2
3
)+
∞∑n=1
(4
n2cosnx− 4π
nsinnx
)=4π2
3+ 4
∞∑n=1
(cosnx
n2− π sinnx
n
)Example 2.2. * Find the Fourier series of f(x) = x in (0, 2π) with f(x+
2π) = f(x).
{Ans : a0 = 2π, an = 0, bn =
−2
n
}Example 2.3. Find the Fourier series of f(x) = (π − x)2 in (0, 2π) of
periodicity 2π.
{a0 =
2
3π2, an =
4
n2, bn = 0
}Example 2.4. * Obtain the Fourier series expansion for the functionf(x) = x(2π − x) in 0 ≤ x ≤ 2π, with period 2π. Show that112 +
122 +
132 + · · · = π2
6 {a0 =
4
3π2, an = − 4
n2, bn = 0
}
Solution: Given f(x) = x(2π − x)
= 2πx− x2
The Fourier series of f(x) is given by
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx) (1)
where a0 =1
π
2π∫0
f (x) dx,an =1
π
2π∫0
f (x) cosnxdx, bn =1
π
2π∫0
f (x) sinnxdx
Now, a0 =1
π
2π∫0
(2πx− x2
)dx
=1
π
[2πx2
2− x3
3
]2π0
=1
π
[πx2 − x3
3
]2π0
=1
π
[(4π3 − 8π2
3
)− 0
]=
1
π
(4π3
3
)a0 =
4π2
3
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 95
an =1
π
2π∫0
(2πx− x2
)cosnxdx
=1
π
[(2πx− x2
)(sinnxn
)− (2π − 2x)
(− cosnx
n2
)+(−2)
(− sinnx
n3
)]2π0
=1
π
[(2πx− x2
) sinnxn
+ (2π − 2x)cosnx
n2+
2 sinnx
n3
]2π0
=1
π
[(0 + (−2π)
1
n2+ 0
)−(0 + 2π
1
n2 + 0
)]=
1
π
[−2π
n2− 2π
n2
]=
1
π
(−4π
n2
)an =
−4
n2
bn =1
π
2π∫0
(2πx− x2
)sinnxdx
=1
π
[(2πx− x2
)(− cosnx
n
)− (2π − 2x)
(− sinnx
n2
)+(−2)
(cosnxn3
)]2π0
=1
π
[−(2πx− x2
) cosnxn
+ (2π − 2x)sinnx
n2− 2
cosnx
n3
]2π0
=1
π
[(0 + 0− 2
n3
)−(0 + 0− 2
n3
)]bn = 0
Sub. the value of a0, an and bn in (1),
f (x) =1
2
(4π2
3
)+
∞∑n=1
(−4
n2
)cosnx+ 0
f (x) =2π2
3− 4
∞∑n=1
cosnx
n2
96 Unit II - FOURIER SERIES (F.S.)
Deduction :
Since f (x) = x (2π − x)
∴ x (2π − x) =2π2
3− 4
∞∑n=1
cosnx
n2
Put x = 0,
0 =2π2
3− 4
∞∑n=1
1
n2
4∞∑n=1
1
n2=
2π2
3
i.e.,∞∑n=1
1
n2=
2π2
12
i.e.,1
12+
1
22+
1
32+ · · · = π2
6.
Example 2.5. Find the Fourier series of periodicity 2π for
f(x) =
{x, (0, π)
2π − x, (π, 2π)
Solution: Given f(x) =
{x, (0, π)
2π − x, (π, 2π)Since the function f(x) is defined in the interval (0, 2π).∴ The Fourier series of f(x) is given by
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx) (1)
where a0 =1
π
2π∫0
f (x) dx,an =1
π
2π∫0
f (x) cosnxdx, bn =1
π
2π∫0
f (x) sinnxdx
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 97
Now, a0 =1
π
π∫0
xdx+
2π∫π
(2π − x) dx
=
1
π
[(x2
2
)π
0
+
(2πx− x2
2
)2π
π
]
=1
π
[(π2
2− 0
)+
((4π2 − 2π2
)−(2π2 − π2
2
))]=
1
π
[π2
2+ 2π2 − 3π2
2
]=
1
π
(π2 + 4π2 − 3π2
2
)=
1
π
(2π2
2
)a0 = π
an =1
π
π∫0
x cosnxdx+
2π∫π
(2π − x) cosnxdx
=
1
π
[[x
(sinnx
n
)− 1 ·
(− cosnx
n2
)]π0
+
[(2π − x)
(sinnx
n
)− (−1)
(− cosnx
n2
)]2ππ
]
=1
π
[[x sinnx
n+
cosnx
n2
]π0
+
[(2π − x)
sinnx
n− cosnx
n2
]2ππ
]
=1
π
[[(0 +
(−1)n
n2
)−(0 +
1
n2
)]+
(0− 1
n2
)−(0− (−1)n
n
)]=
1
π
[(−1)n
n2− 1
n2− 1
n2+
(−1)n
n2
]=
1
π
[2 (−1)n
n2− 2
n2
]an =
2
n2π[(−1)n − 1]
98 Unit II - FOURIER SERIES (F.S.)
bn =1
π
π∫0
x sinnxdx+
2π∫π
(2π − x) sinnxdx
=
1
π
[[x
(− cosnx
n
)− 1
(− sinnx
n2
)]π0
+
[(2π − x)
(− cosnx
n
)− (−1)
(− sinnx
n2
)]2ππ
]
=1
π
[[−x cosnx
n+
sinnx
n2
]π0
+
[− (2π − x)
cosnx
n− sinnx
n2
]2ππ
]
=1
π
[[(−π (−1)n
n+ 0
)− (0 + 0)
]+
[(0− 0)−
(−π (−1)n
n− 0
)]]=
1
π
[−π (−1)n
n+π (−1)n
n
]bn = 0
Sub. the value of a0, an and bn in (1)
f (x) =π
2+
∞∑n=1
(2
n2π((−1)n − 1) cosnx+ 0
)f (x) =
π
2+
2
π
∞∑n=1
((−1)n − 1
n2
)cosnx
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 99
For deduction partsContinuous point Discontinuous point
Example 2.6. Find the Fourier series for the function f(x) =1
2(π − x)
in(0, 2π) with period 2π and deduce 1− 1
3+
1
5− · · · = π
4.{
a0 = 0, an = 0, bn =1
n&x =
π
2
}Example 2.7. Find the Fourier series for the function
f(x) =
{x, (0, π)
2π − x, (π, 2π)and deduce
1
12+
1
32+
1
52+ · · · = π2
8.{
a0 = π, an =2
πn2[(−1)n − 1] , bn = 0&x = 0
}Example 2.8. * Find the Fourier series for f(x) = x (2π − x) in (0, 2π)
and deduce∞∑n=1
1
n2=π2
6.
{a0 =
4π
3
2
, an =−4
n2, bn = 0&x = 0
}
100 Unit II - FOURIER SERIES (F.S.)
Example 2.9. Find the Fourier series for f(x) = x sinx in (0, 2π) and
deduce1
(1)(3)− 1
(3)(4)+
1
(5)(7)− · · · = π − 2
4.{
a0=−2, an=−2
1−n2(n 6= 1) , a1=
−1
2, bn=0 (n 6= 1) , b1=π&x=
π
2
}Solution : Given f(x) = x sinx in (0, 2π).∴ The Fourier series of f(x) is given by
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx) (1)
where a0 =1
π
2π∫0
f (x) dx, an =1
π
2π∫0
f (x) cosnxdx, bn =1
π
2π∫0
f (x) sinnxdx
Now, a0 =1
π
2π∫0
x sinxdx
=1
π[x (− cosx)− 1 (− sinx)]2π0
=1
π[−x cosx+ sinx]2π0
=1
π[(−2π + 0)− (0 + 0)]
a0 = −2
an =1
π
2π∫0
x sinx cosnxdx
=1
π
2π∫0
x cosnx sinxdx
=1
π
2π∫0
x1
2(sin (n+ 1)x− sin (n− 1)x) dx
=1
2π
2π∫0
x (sin (n+ 1)x− sin (n− 1)x) dx
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 101
=1
2π
[x
(− cos (n+ 1)x
n+ 1+
cos (n− 1)x
n− 1
)−1
(− sin (n+ 1)x
(n+ 1)2+
sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[x
(cos (n− 1)x
n− 1− cos (n+ 1)x
n+ 1
)+sin (n+ 1)x
(n+ 1)2− sin (n− 1)x
(n− 1)2
]2π0
=1
2π
[[2π
(1
n− 1− 1
n+ 1
)+ 0− 0
]− [0 + 0− 0]
]=
1
n− 1− 1
n+ 1=n+ 1− (n− 1)
n2 − 1
an =2
n2 − 1, n 6= 1
a1 =1
π
2π∫0
x sinx cosxdx (∵ an = 1π
2π∫0
x sinx cosnxdx)
=1
2π
2π∫0
x2 sin x cosxdx
=1
2π
2π∫0
x sin 2xdx
=1
2π
[x
(− cos 2x
2
)− 1 ·
(− sin 2x
4
)]2π0
=1
2π
[−x cos 2x
2+
sin 2x
4
]2π0
=1
2π
[(−2π
2+ 0
)− (0 + 0)
]a1 =
−1
2
102 Unit II - FOURIER SERIES (F.S.)
bn =1
π
2π∫0
x sinx sinnxdx
=1
π
2π∫0
x sinnx sinxdx
=1
π
2π∫0
x1
2(cos (n− 1)x− cos (n+ 1)x)dx
=1
π
2π∫0
x (cos (n− 1)x− cos (n+ 1)x)dx
=1
2π
[x
(sin (n− 1)x
n− 1− sin (n+ 1)x
n+ 1
)−1
(− cos (n− 1)x
(n− 1)2+
cos (n+ 1)x
(n+ 1)2
)]2π0
=1
2π
[x
(sin (n+ 1)x
n+ 1− sin (n− 1)x
n− 1
)+
(cos (n− 1)x
(n− 1)2− cos (n+ 1)x
(n+ 1)2
)]2π0
=1
2π
[(0 +
1
(n − 1)2− 1
(n + 1)2
)−(0 +
1
(n − 1)2− 1
(n + 1)2
)]bn = 0, n 6= 1
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 103
b1 =1
π
2π∫0
x sinx sinxdx =1
π
2π∫0
x sin2 xdx
=1
π
2π∫0
x1
2(1− cos 2x) dx
=1
2π
2π∫0
(x− x cos 2x) dx
=1
2π
[x2
2−[x
(sin 2x
2
)− 1
(− cos 2x
4
)]]2π0
=1
2π
[x2
2− x sin 2x
2− cos 2x
4
]2π0
=1
2π
[(4π2
2− 0− 1
4
)−(0− 0− 1
4
)]=
1
2π
(2π2)
b1 = π
From (1),
f (x) =a02+
∞∑n=1
(an cosnx+ bn sinnx)
=a02+
∞∑n=1
an cosnx+∞∑n=1
bn sinnx
=a02+ a1 cosx+
∞∑n=2
an cosnx+ b1 sinx+∞∑n=2
bn sinnx
=1
2(−2)− 1
2cosx+
∞∑n=2
(2
n2 − 1
)cosnx+ π sinx+ 0
i.e., f (x) = −1− 12 cosx+ 2
∞∑n=2
(cos nxn2−1
)+ π sinx
Example 2.10. * Find the Fourier series for f(x) = x cosx in (0, 2π).{a0 = 0, an = 0 (n 6= 1) , a1 = π, bn =
2n
1− n2(n 6= 1) , b1 =
−1
2
}
104 Unit II - FOURIER SERIES (F.S.)
Example 2.11. Obtain Fourier series for f(x) = eax in (0, 2π).
Solution : Given f(x) = eax in (0, 2π).The Fourier series of f(x) is given by
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx) (1)
where
a0 =1
π
2π∫0
f (x) dx, an =1
π
2π∫0
f (x) cosnxdx, bn =1
π
2π∫0
f (x) sinnxdx
Now, a0 =1
π
2π∫0
eaxdx =1
π
[eax
a
]2π0
=1
aπ[eax]2π0 =
1
aπ
[ea2π − e0
]∴ a0 =
1
aπ
(e2aπ − 1
)an =
1
π
2π∫0
eax cosnxdx =1
π
[eax
a2 + n2(a cosnx+ n sinnx)
]2π0
=1
π
[e2aπ
a2 + n2(a+ 0)− 1
a2 + n2(a+ 0)
]=1
π
[ae2aπ
a2 + n2− a
a2 + n2
]=
a
π (a2 + n2)
[e2aπ − 1
]∴ an =
a(e2aπ − 1
)π (a2 + n2)
bn =1
π
2π∫0
eax sinnxdx =1
π
[eax
a2 + n2(a sinnx− cosnx)
]2π0
=1
π
[e2aπ
a2 + n2(0− n)− 1
a2 + n2(0− n)
]=1
π
[−ne2aπ
a2 + n2+
n
a2 + n2
]=
−nπ (a2 + n2)
[e2aπ − 1
]∴ bn =
−n(e2aπ − 1
)π (a2 + n2)
Sub. the value of a0, an, bn in (1)
f(x) =1
2
(e2aπ − 1
aπ
)+
∞∑n=1
[a(e2aπ − 1
)π (a2 + n2)
cosnx−n(e2aπ − 1
)π (a2 + n2)
sinnx
]
=e2aπ − 1
2aπ+
∞∑n=1
(e2aπ − 1
π (a2 + n2)
)[a cosnx− n sinnx]
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 105
=e2aπ − 1
2aπ+e2aπ − 1
π
[ ∞∑n=1
(1
a2 + n2
)(a cosnx− n sinnx)
]
=e2aπ − 1
π
[1
2a+
∞∑n=1
1
a2 + n2(a cosnx− n sinnx)
]Example 2.12. * Find the Fourier series for f(x) = e−x in (0, 2π) and
deduce∞∑n=2
(−1)n
1 + n2, further derive a series for cosec hπ.{a0 =
1− e−2π
π, an =
1− e−2π
π (1 + n2), bn =
n
π
1− e−2π
1 + n2&x = π
}Note : General Fourier series is used in full range intervals as 2π or 2`
length. i.e., intervals of the form (0, 2π), (0, 2`), (−π, π), (−`, `).In intervals (−π, π) & (−`, `), use even function or odd function orneither even nor odd function(or none) ideas.
2.3 Odd and even functions
(1) If
f(−x)=
{f(x)⇒Gn. f(x) is an even fn.[symmetric about y-axis]
−f(x)⇒Gn. f(x) is an odd fn.[symmetric about origin]
Example 2.28. Find the Fourier series for f(x) = (`− x)2 in (0, 2`) and
deduce∞∑n=1
1
n2=π2
6.
{a0 =
2`2
3, an =
4`2
n2π2, bn = 0&x = 0
}
Example 2.29. Find the Fourier series for f(x) =
{x, 0 ≤ x ≤ 3
6− x, 3 ≤ x ≤ 6{a0 = 3, an =
−12
n2π2(n = odd), bn = 0
}Example 2.30. * Find the Fourier series for
f(x)=
x
`, 0<x<`
2`−x`
, `<x <2`.
{a0=1, an=
[ −4
n2π2, (n=odd)
0, (n=even)
], bn=0
}
2.7.2 Examples under (−`, `)
Example 2.31. Find the Fourier series for f(x) = e−x in (−1, 1).{a0 = 2 sinh 1, an =
2 (−1)n sinh 1
1 + n2π2, bn =
2nπ (−1)n sinh 1
1 + n2π2
}
Example 2.32. * Find the Fourier series for f(x)=
0, −2<x<−1k, −1<x<10, 1<x<2
.
128 Unit II - FOURIER SERIES (F.S.)a0=k, an=2k sin
(nπ2
)nπ
, bn=0 (∵ even function)
2.7.3 Examples under (0, `)
Example 2.33. Find the Fourier sine series for
f(x)=
x,
[0,`
2
]`−x,
[`
2, `
] .
a0 = 0
(∵ cosineseries
), an = 0
(∵ cosineseries
), bn =
4` sin(nπ
2
)n2π2
Example 2.34. Find the Fourier cosine series for
f(x) =
{1, 0 ≤ x ≤ a/2
−1, a/2 ≤ x ≤ a.a0 = 0, an =
4 sin(nπ
2
)nπ
, bn = 0 (∵ cosine series)
Example 2.35. * Find the sine series for f(x) = x− x2 in 0 < x < 1.{a0 = an = 0
(∵ cosineseries
), bn =
[ 8
n3π3, n = odd
0, n = even
}
Example 2.36. * Find the Half range cosine series for f(x) = kx(`− x)
in (0, `).
a0 = k`2
3, an =
0, when n is odd−4k`2
n2π2, when n is even
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 129
2.8 Complex form of Fourier Series
Interval Fourier Series Complex Fourier coefficient
(0, 2π) f(x) =∞∑
n=−∞Cne
inx Cn =1
2π
∫ 2π
0
f(x) e−inx dx
(−π, π) f(x) =∞∑
n=−∞Cne
inx Cn =1
2π
∫ π
−π
f(x) e−inx dx
(0, 2`) f(x) =∞∑
n=−∞Cne
inπx/` Cn =1
2`
∫ 2`
0
f(x) e−inπx/` dx
(−`, `) f(x) =∞∑
n=−∞Cne
inπx/` Cn =1
2`
∫ `
−`
f(x) e−inπx/` dx
(0, π) f(x) =∞∑
n=−∞Cne
inx Cn =1
π
∫ π
0
f(x) e−inx dx
(0, `) f(x) =∞∑
n=−∞Cne
inπx/` Cn =1
`
∫ `
0
f(x) e−inπx/` dx
Example 2.37. Derive complex form for f(x) = eax in (0, 2π).{cn =
(a+ in)(e2aπ − 1
)2π (a2 + n2)
}Example 2.38. Find the complex form of the series for the function
f(x) = x in (−`, `).{cn =
−` (−1)n
inπ
}
130 Unit II - FOURIER SERIES (F.S.)
2.9 Harmonic Analysis
So far, we found Fourier series for a function f(x) given by the formulain one (or) more interval. Now, there is a process of finding a Fourier seriesfor the function f(x) given by a Table (or) by numerical values (or) by theGraph is known as Harmonic Analysis.We know that Fourier series for f(x) in (0, 2π) is
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx)
(or)
f(x) =a02+
∞∑n=1
an cosnx+∞∑n=1
bn sinnx (1)
where a0 =1
π
∫ 2π
0
f(x) dx
=2
2π
∫ 2π
0
f(x) dx
= 2
[1
(2π − 0)
∫ 2π
0
f(x) dx
][∵
1
b− a
∫ b
a
f(x)dx = Mean value off(x) in (a, b)
]∴ a0 = 2 [ Meanvalueoff(x) in (0, 2π)] = 2
[∑f (x)
n
]=
2
n
[∑f (x)
]Now, an =
1
π
∫ 2π
0
f(x)cosnx dx
=2
2π
∫ 2π
0
f(x) cosnxdx
= 2
[1
(2π − 0)
∫ 2π
0
f(x)cosnx dx
][∵
1
b− a
∫ b
a
f(x) cosnxdx = Mean value off(x) cosnx in (a, b)
]∴ an = 2 [ Mean value off(x) cosnx in (0, 2π)] =
2
n
[∑f (x) cosnx
]lly, bn = 2 [ Mean value off(x) sinnx in (0, 2π)] =
2
n
[∑f (x) sinnx
]Note: In equation (1),
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 131
1. The terma02
is called constant term / direct part / direct current part
of Fourier series.
2. In (0, 2π), (i) (a1 cosx+ b1 sinx) is called fundamental (or) firstharmonic. (ii) (a2 cos 2x+ b2 sin 2x) is called octave (or) secondharmonic of F.S.
(i) Fourier series upto first harmonic is
f(x) =a02+ (a1 cosx+ b1 sinx)
(ii) Fourier series upto second harmonic is
f(x) =a02+ (a1 cosx+ b1 sinx) + (a2 cos 2x+ b2 sin 2x)
4.(i) Fourier series upto 2 coefficients in Fourier cosine series is
f(x) =a02+ (a1 cosx+ a2 cos 2x)
(ii) Fourier series upto 2 coefficients in Fourier sine series is
f(x) = b1 sinx+ b2 sin 2x
5. In (0, 2`), Fourier series upto 2nd harmonic is
f(x) =a02+ a1 cos
(πx`
)+ b1 sin
(πx`
)+ a2 cos
(2πx
`
)+ b2 sin
(2πx
`
)6. Amplitude of the nth harmonic = An =
√a2n + b2n
2.9.1 Types of Harmonic Table Data
1. π form (Radian form) →180◦ (Degree form)2. θ◦ form (Degree form)
3. T form
(Use θ◦ =
2πx
T
)4. ` form → 2` = Number of data
⇒ ` =Number of data
2
2.9.2 Examples under π form(Radian form)
Example 2.39. The table of values of the function y = f(x) is given