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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
Same point At extremes At middle
Substitute thecontinuous pointin f(x) directlyExample:(1)
f(x) =x
2in (0, 2π) at x = πisf(x = π) =
π
2(2)f(x) = x3 − x
in (0, 2π) at x =π
2isf(x =
π
2
)=(π
2
)3−(π2
)
f(x) = Average offunction at extremesExample:
(1)f(x) =x
2in (0, 2π) at
x = 0 isf(x = 0)
=f(x = 0) + f(x = 2π)
2
=02 +
2π2
2=π
2(2)f(x) = cos xin (0, 2π) at x = 2π is
f(x = 2π)
=cos 0 + cos 2π
2
=1 + 1
2= 1
f(x) = Average of
LHL & RHLExample:(1)f(x) ={
0, 0 < x < ππ, π < x < 2π
in (0, 2π) isf(x = π)
=f(π−) + f(π+)
2wheref(π−) = lim
h→0f(π − h)
= 0
&
f(π+) = limh→0
f(π + h)
= π
∴ f(x = π) =0 + π
2
=π
2
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]
(2) If f (−x) 6=
{f(x)
−f(x)
}⇒ Gn. f(x) is neither even nor odd fn.
Note: (even function)× (even function) = (even function)
(odd function)× (odd function) = (even function)
(even function)× (odd function) = (odd function)
Examples :Even fns:
±a, x2, cosx, |x| , |cosx| , |sinx| , x sinx, sin2 x,√1−cosx, e±x2
Odd fns:±x, x3, x3 − x, x cosx, x2 sinx, · · ·
Neither Even nor odd fns:e±ax,±x2 ± x, · · ·
Note :
a∫−a
f(x)dx =
a∫
0
f(x)dx, if f(x) is even
0, if f(x) is odd
106 Unit II - FOURIER SERIES (F.S.)
If f(x) =
{f1(x), in (−π, 0)f2(x), in (0, π)
such that
(a) f1(−x) = f2(x), then given f(x) is said to be an even function in(−π, π).
(b) f1(−x) = −f2(x), then given f(x) is said to be an odd functionin (−π, π).
2.3.1 Fourier series formula for the intervals (−π, π), (−`, `)
Interval (−`, `) (−π, π)
If f(x) is even,⇒ bn = 0
f(x) =a02
+∞∑n=1
an cosnπx
`
where
a0 =1
`
∫ `
−`
f(x)dx (or)
=2
`
∫ `
0
f(x)dx
an =1
`
∫ `
−`
f(x) cosnπx
`dx (or)
=2
`
∫ `
0
f(x) cosnπx
`dx
f(x) =a02
+∞∑n=1
an cosnx
where
a0 =1
π
∫ π
−π
f(x)dx (or)
=2
π
∫ π
0
f(x)dx
an =1
π
∫ π
−π
f(x) cosnxdx (or)
=2
π
∫ π
0
f(x) cosnxdx
If f(x) is odd,⇒a0 = an = 0
f(x) =∞∑n=1
bn sinnπx
`
where
bn =1
`
∫ `
−`
f(x) sinnπx
`dx (or)
=2
`
∫ `
0
f(x) sinnπx
`dx
f(x) =∞∑n=1
bn sinnx
where
bn =1
π
∫ π
−π
f(x) sinnxdx (or)
=2
π
∫ π
0
f(x) sinnxdx
If f(x) is none(neither odd noreven)
f(x) =a02
+∞∑n=1
an cosnπx
`
+∞∑n=1
bn sinnπx
`
where
a0 =1
`
∫ `
−`
f(x)dx
an =1
`
∫ `
−`
f(x) cosnπx
`dx
bn =1
`
∫ `
−`
f(x) sinnπx
`dx
f(x) =a02
+∞∑n=1
an cosnx
+∞∑n=1
bn sinnx
where
a0 =1
π
∫ π
−π
f(x)dx
an =1
π
∫ π
−π
f(x) cosnxdx
bn =1
π
∫ π
−π
f(x) sinnxdx
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 107
2.3.2 Root Mean Square Value
Root mean square value (RMS value) of f(x) in (a, b) is
y =
√1
(b− a)
∫ b
a
[f (x)]2 dx
Eg: Find R.M.S. for f(x) = x in (0, π) is
y =
√1
(π − 0)
∫ π
0
[f (x)]2 dx =
√π2
3=
π√3
2.4 Parseval’s Identity
If the power of n in Fourier series expansion < the power of n indeduction part, then use appropriate Parseval’s Identity.
Interval Simplified Parseval’s Identity for deduction
(−`, `) 1
`
∫ `
−`
[f(x)]2 dx =a202+
∞∑n=1
(a2n + b2n
)(−π, π) 1
π
∫ π
−π
[f(x)]2 dx =a202+
∞∑n=1
(a2n + b2n
)(0, 2`)
1
`
∫ 2`
0
[f(x)]2 dx =a202+
∞∑n=1
(a2n + b2n
)(0, 2π)
1
π
∫ 2π
0
[f(x)]2 dx =a202+
∞∑n=1
(a2n + b2n
)
2.4.1 Examples under (−π, π)
Example 2.13. Find the Fourier series for the function f (x) = x+ x2 in(−π, π) and deduce that 1
12 +122 +
132 + · · · = π2
6 ·
Solution: Given f (x) = x+ x2 in (−π, π).Take f1(x) = x. Clearly f1(x) = x is an odd function in (−π, π).
108 Unit II - FOURIER SERIES (F.S.)
The Fourier series for the function of f(x) in (−π, π) is given by
f (x) =∞∑n=1
bn sinnx (1)
where bn =1
π
π∫−π
f (x) sinnxdx
i.e., bn =2
π
π∫0
f (x) sinnxdx [∵ f(x) is odd function]
bn =2
π
π∫0
x sinnxdx
=2
π
[x
(− cosnx
n
)− (1)
(− sinnx
n2
)]π0
=2
π
[−x(cosnx
n
)+ (1)
(sinnx
n2
)]π0
=2
π
[−π(−1)n
n+ 0− (0 + 0)
]bn =
−2(−1)n
n
Fourier series of f1(x) = x is given by f1 (x) = −2∞∑n=1
(−1)n
n sinnx
Take f2(x) = x2. Clearly f2(x) = x2 is an even function in (−π, π).The Fourier series for f(x) in (−π, π) is given by
f (x) =a02+
∞∑n=1
an cosnx (2)
where a0 =1
π
π∫−π
f (x) dx, an =1
π
π∫−π
f (x) cosnxdx
i.e., a0 =2
π
π∫0
f (x) dx, an =2
π
π∫0
f (x) cosnxdx [∵ f(x) is even function]
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 109
a0 =2
π
π∫0
x2dx =2
π
[x3
3
]π0
=2
π
[π3
3− 0
]
a0 =2π2
3·
an =2
π
π∫0
x2 cosnxdx
=2
π
[x2(sinnx
n
)− 2x
(− cosnx
n2
)+ 2
(− sinnx
n3
)]π0
=2
π
[x2 sinnx
n+
2x cosnx
n2− 2 sinnx
n3
]π0
=2
π
[(0 +
2π (−1)n
n2− 0
)− (0 + 0− 0)
]=
2
π
[2π (−1)n
n2
]an =
4 (−1)n
n2
Fourier series of f2(x) = x2 is given by
f2 (x) =π2
3+ 4
∞∑n=1
(−1)n
n2cosnx
Since f (x) = x+ x2
The Fourier series of f (x) is
f (x) = −2∞∑n=1
(−1)n
nsinnx+
π2
3+ 4
∞∑n=1
(−1)n
n2cosnx
f (x) =π2
3− 2
∞∑n=1
(−1)n
nsinnx+ 4
∞∑n=1
(−1)n
n2cosnx.
Deduction : Put x = π. Here x = π is a point of discontinuity whichis one end of the given interval (−π, π).
Sum of the Fourier series of f(x) isf (−π) + f (π)
2
110 Unit II - FOURIER SERIES (F.S.)
∴f (−π) + f (π)
2=π2
3− 0 + 4
∞∑n=1
(−1)n
n2(−1)n
−π + π2 + π + π2
2=π2
3+ 4
∞∑n=1
(−1)2n
n2
π2 =π2
3+ 4
∞∑n=1
1
n2
π2 − π2
3= 4
∞∑n=1
1
n2
2π2
3= 4
∞∑n=1
1
n2
π2
6=
∞∑n=1
1
n2
1
12+
1
22+
1
32+ · · · = π2
6·
Example 2.14. Find the Fourier series for the function f(x) = x2 in[−π, π]with period 2π and deduce
(i)1
12+
1
22+
1
32+ ... =
π2
6(ii)
1
12− 1
22+
1
32− ... =
π2
12
(iii)1
12+
1
32+
1
52+ ... =
π2
8(iv)
1
14+
1
24+
1
34+ ... =
π4
90{a0 =
2
3π2, an =
4 (−1)n
n2, bn = 0(∵ even)
}{
(i)x = π(or − π) (ii)x = 0 (iii) add i and ii(iv) Use Parseval′s identity
}Solution: Given f(x) = x2 in [−π, π].We know that, the Fourier series of f(x) is given by (Refer above example),
f (x) =π2
3+ 4
∞∑n=1
(−1)n
n2cosnx (1)
Deduction:(i) Put x = π. [Here x = π is a point of discontinuity which is one end ofthe given interval (−π, π)]
∴ Sum of the Fourier series of f(x) isf (−π) + f (π)
2
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 111
i.e.,f (−π) + f (π)
2=π2
3+ 4
∞∑n=1
(−1)n
n2(−1)n
π2 + π2
2=π2
3+ 4
∞∑n=1
(−1)2n
n2
2π2
2=π2
3+ 4
∞∑n=1
1
n2
π2 − π2
3= 4
∞∑n=1
1
n2
2π2
3= 4
∞∑n=1
1
n2
π2
6=
∞∑n=1
1
n2
1
12+
1
22+
1
32+ · · · = π2
6(2)
(ii) Put x = 0
0 =π2
3+ 4
∞∑n=1
(−1)n
n2
−π2
3= 4
[−1
12+
1
22− 1
32+ · · ·
]−π2
3= −4
[1
12− 1
22+
1
32− · · ·
]π2
12=
1
12− 1
22+
1
32− · · ·
1
12− 1
22+
1
32− · · · = π2
12(3)
(iii) (1) + (2) ⇒
2
12+ 0 +
2
32+ 0 + · · · = π2
6+π2
12
2
(1
12+
1
32+
1
52+ · · ·
)=
3π2
12
1
12+
1
32+
1
52+ · · · = π2
8·
112 Unit II - FOURIER SERIES (F.S.)
Example 2.15. * Find the Fourier series for f(x) = x in(−π, π)with
period 2π.
{a0 = 0(∵ odd), an = 0(∵ odd), bn =
−2
n(−1)n
}
Example 2.16. * Find the Fourier series for f(x) = x2 − x in(−π, π).{a0 =
2
3π2, an =
4
n2(−1)n, bn =
2
n(−1)n
}
Example 2.17. Find the Fourier series for
f(x) =
{x− 1, −π < x < 0x+ 1, 0 < x < π
and deduce 1− 1
3+
1
5− · · · = π
4.
Solution : Given f(x) =
{x− 1, −π < x < 0x+ 1, 0 < x < π
f(−x) ={
−x− 1, −π < −x < 0−x+ 1, 0 < −x < π
=
{−x− 1, 0 < x < π−x+ 1, −π < x < 0
=
{−x+ 1, −π < x < 0−x− 1, 0 < x < π
=−{x− 1, −π < x < 0x+ 1, 0 < x < π
f(−x) =− f(x)∴ f(x) is an odd function in −π < x < π.∴ The Fourier series for the odd function f(x) is given by
f (x) =∞∑n=1
bn sinnx (1)
where bn =2
π
∫ π
0
f (x) sinnxdx
=2
π
∫ π
0
(x+ 1) sinnxdx
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 113
=2
π
[(x+ 1)
(− cosnx
n
)− 1
(− sinnx
n2
)]π0
=2
π
[− (x+ 1)
cosnx
n+
sinnx
n2
]π0
=2
π
[(− (π + 1)
(−1)n
n+ 0
)−(−1
n+ 0
)]=2
π
[1
n− (π + 1)
(−1)n
n
]∴ bn =
2
nπ[1− (π + 1) (−1)n]
From (1), the Fourier series of f(x) is
f(x) =∞∑n=1
2
nπ[1− (π + 1) (−1)n] sinnx
=2
π
∞∑n=1
[1− (π + 1) (−1)n]sinnx
n
Example 2.18. * Find the Fourier series for
f(x) =
{π + x, −π < x < 0π − x, 0 < x < π
and deduce∞∑n=1
1
(2n− 1)2=π2
8.{
a0 = π, an =4
πn2(n = odd), bn = 0(∵ even) & x = 0
}
Example 2.19. Expand f(x) = | cosx| in a Fourier series for in theinterval (−π, π).
Solution : Given f(x) =|cosx|f(−x) = |cos(−x)| = |cosx|f(−x) =f(x)
∴ f(x) is an even function in −π < x < π.∴ The Fourier series for the even function f(x) is given by
f (x) =a02+
∞∑n=1
an cosnx (1)
where a0 =2
π
∫ π
0
f (x)dx, an =2
π
∫ π
0
f (x) cosnxdx
114 Unit II - FOURIER SERIES (F.S.)
Now, a0 =2
π
∫ π
0
|cosx|dx
=2
π
∫ π
2
0
|cosx| dx+∫ π
π
2
|cosx| dx
=2
π
∫ π
2
0
cosxdx+
∫ π
π
2
(− cosx)dx
=2
π
∫ π
2
0
cosxdx−∫ π
π
2
cosxdx
=2
π
(sinx)π20 − (sinx)ππ
2
=2
π[(1− 0)− (0− 1)] =
2
π[1 + 1]
∴ a0 =4
π
an =2
π
∫ π
0
|cosx| cosnxdx
=2
π
∫ π
2
0
|cosx| cosnxdx+∫ π
π
2
|cosx| cosnxdx
=2
π
∫ π
2
0
cosx cosnxdx+
∫ π
π
2
(− cosx) cosnxdx
=2
π
∫ π
2
0
cosnx cosxdx−∫ π
π
2
cosnx cosxdx
=2
π
∫ π
2
0
1
2[cos (n+ 1)x+ cos (n− 1)x] dx
−∫ π
π
2
1
2[cos (n+ 1)x+ cos (n− 1)x] dx
[∵ cosA cosB =
1
2cos (A+B) + cos (A−B)
]
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 115
=1
π
∫ π
2
0
[cos (n+ 1)x+ cos (n− 1)x] dx
−∫ π
π
2
[cos (n+ 1)x+ cos (n− 1)x] dx
=1
π
(sin (n+ 1)x
n+ 1+
sin (n− 1)x
n− 1
)π2
0
−(sin (n+ 1)x
n+ 1+
sin (n− 1)x
n− 1
)π
π
2
=1
π
sin (n+ 1)
π
2n+ 1
+sin (n− 1)
π
2n− 1
− 0
−
0−sin (n+ 1)
π
2n+ 1
+sin (n− 1)
π
2n− 1
=1
π
2 sin (n+ 1)π
2n+ 1
+2 sin (n− 1)
π
2n− 1
=2
π
sin(nπ
2+π
2
)n+ 1
+sin(nπ
2− π
2
)n− 1
=2
π
sin nπ2 cosπ
2+ cosn
π
2sin
π
2n+ 1
+sin
nπ
2cos
π
2− cosn
π
2sin
π
2n− 1
=2
π
cos nπ2n+ 1
−cos
nπ
2n− 1
=2 cos
nπ
2π
[1
n+ 1− 1
n− 1
]=
2 cosnπ
2π
[−1− (+1)
n2 − 1
]
∴ an =−4 cos
nπ
2π (n2 − 1)
,n 6= 1
116 Unit II - FOURIER SERIES (F.S.)
Now, a1 =2
π
∫ π
0
|cosx| cosxdx
∵ an =2
π
∫ π
0
|cosx| cosnxdx
∴ a1 =2
π
∫ π
0
|cosx| cosxdx
=2
π
∫ π
2
0
cosx cosxdx+
∫ π
π
2
− cosx cosxdx
=2
π
∫ π
2
0
cos2 xdx−∫ π
π
2
cos2 xdx
=2
π
∫ π
2
0
1
2(1 + cos 2x) dx−
∫ π
π
2
1
2(1 + cos 2x) dx
=1
π
∫ π
2
0
(1 + cos 2x) dx−∫ π
π
2
1
2(1 + cos 2x) dx
=1
π
(x+ sin 2x
2
)π2
0
−(x+
sin 2x
2
)π
π
2
=1
π
[[(π2− 0)− 0]−[(π + 0)−
(π2+ 0)]]
=1
π
[π2− π
2
]∴ a1 =0From (1), the Fourier series of f(x) is
f(x) =a02+ a1 cosx+
∞∑n=2
an cosnx
=1
2
(4
π
)+ 0 +
∞∑n=2
−4 cosnπ
2π (n2 − 1)
cosnx
=2
π− 4
π
∞∑n=1
cosnπ
2n2 − 1
cosnx.
Example 2.20. * Find the Fourier series forf(x) = | sinx| in the interval (−π, π).a0 = 4
π, an =
−4
π (n2 − 1), (n = even)
0, (n = odd), bn = 0(∵ even)
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 117
Example 2.21. Find the Fourier series for f(x) =√1− cosx in
(−π, π).
Solution : Given f(x) =√1− cosx
f(−x) =√1− cos(−x) =
√1− cosx
f(−x) =f(x)∴ f(x) =
√1− cosx is an even function in −π < x < π.
∴ The Fourier series for the even function f(x) is given by
f (x) =a02+
∞∑n=1
an cosnx (1)
where a0 =2
π
∫ π
0
f (x)dxan =2
π
∫ π
0
f (x) cosnxdx
Now, a0 =2
π
∫ π
0
√1− cosxdx
∵√1− cosx =
√2 sin2
x
2
=√2 sin
x
2
=2
π
∫ π
0
√2 sin
x
2dx =
2√2
π
∫ π
0
sinx
2dx
=2√2
π
− cosx
21
2
π
0
=−4
√2
π
[cos
x
2
]π0
=−4
√2
π[0− 1]
[∵ cos
π
2= 0, cos 0 = 1
]∴ a0 =
4√2
π
118 Unit II - FOURIER SERIES (F.S.)
an =2
π
∫ π
0
√1− cosx cosnxdx
=2
π
∫ π
0
√2 sin
x
2cosnxdx
=2√2
π
∫ π
0
cosnx sinx
2dx
=2√2
π
∫ π
0
1
2
(sin
(n+
1
2
)x− sin
(n− 1
2
)x
)dx
=
√2
π
∫ π
0
(sin
(n+
1
2
)x− sin
(n− 1
2
)x
)dx
=
√2
π
−cos
(n+
1
2
)x
n+1
2
+
cos
(n− 1
2
)x
n− 1
2
π
0
=
√2
π
− cos
(n+
1
2
)π
n+1
2
+
cos
(n− 1
2
)π
n− 1
2
−
−1
n+1
2
+1
n− 1
2
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 119
=
√2
π
cos(nπ − π
2
)n− 1
2
−cos(nπ +
π
2
)n+
1
2
+1
n+1
2
− 1
n− 1
2
=
√2
π
cosnπ cos
π
2+ sinnπ sin
π
2
n− 1
2
−cosnπ cos
π
2− sinnπ sin
π
2
n+1
2
+
n− 1
2−(n+
1
2
)n2 − 1
4
=
√2
π
0− 0 +
−1
2− 1
2
n2 − 1
4
[∵ cos
π
2= 0, sinnπ = 0
]
=
√2
π
−1
n2 − 1
4
an =
−√2
π
(n2 − 1
4
)Sub. the value of a0 and an in (1)
f(x) =1
2
(4√2
π
)+
∞∑n=1
−√2
π
(n2 − 1
4
) cosnx
=2√2
π−
√2
π
∞∑n=1
cosnx
n2 − 1
4
Example 2.22. Find the Fourier series for the function f(x) = x cosx inthe interval (−π, π).
Solution : Given f(x) =x cosx
f(−x) =(−x) cos(−x) = −x cosxf(−x) =− f(x)
∴ f(x) = x cosx is an odd function in −π < x < π.∴ The Fourier series for the odd function f(x) is given by
120 Unit II - FOURIER SERIES (F.S.)
f (x) =∞∑n=1
bn sinnx (1)
where bn =2
π
∫ π
0
f (x) sinnxdx
=2
π
∫ π
0
x cosx sinnxdx
=2
π
∫ π
0
x sinnx cosxdx
=2
π
∫ π
0
x1
2(sin (n+ 1)x+ sin (n− 1)x) dx
=1
π
∫ π
0
x (sin (n+ 1)x+ sin (n− 1)x) dx
=1
π
{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
)}π
0
=1
π
{−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
)}π
0
=1
π
[[−π
((−1)n+1
n+ 1+
(−1)n−1
n− 1
)+ 0
]− [0 + 0]
]
=−
((−1)n+1
n+ 1+
(−1)n−1
n− 1
)= −
((−1) (−1)n
n+ 1+
(−1)n
(−1) (n− 1)
)=−
(− (−1)n
n+ 1− (−1)n
n− 1
)= (−1)n
(1
n+ 1+
1
n− 1
)=(−1)n
(n− 1 + n+ 1
n2 − 1
)∴ bn =
2n (−1)n
n2 − 1, n 6= 1
Now, b1 =2
π
∫ π
0
x cosx sinxdx
[∵ bn =
2
π
∫ π
0
x cosx sinxdx
]=1
π
∫ π
0
x2 sin x cosxdx
=1
π
∫ π
0
x sin 2xdx
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 121
=1
π
[x
(− cos 2x
2
)− 1
(− sin 2x
4
)]π0
=1
π
[x
(− cos 2x
2
)− 1
(− sin 2x
4
)]π0
=1
π
[−x cos 2x
2+
sin 2x
4
]π0
=1
π
[(−π2+ 0)− (0 + 0)
]∴ b1 =−1
2From (1), the Fourier series of f(x) is
f(x) =b1 sinx+∞∑n=2
bn sinnx
=−1
2sinx+
∞∑n=2
2n (−1)n
n2 − 1sinnx
=−1
2sinx+ 2
∞∑n=2
n (−1)n
n2 − 1sinnx
Example 2.23. Obtain Fourier series for f (x) =
{0, −π < x < 0
sinx, 0 < x < π
and deduce that1
1.3+
1
3.5+
1
5.7+ · · · = 1
2.
Solution :
Given f(x) =
{0, −π < x < 0
sinx, 0 < x < π
f(−x) ={
0, 0 ≤ x ≤ πsin(−x), −π ≤ x ≤ 0
=
{0, 0 ≤ x ≤ π
− sinx, −π ≤ x ≤ 0
=
{− sinx, −π ≤ x ≤ 0
0, 0 ≤ x ≤ π
f(−x) 6={
f(x)−f(x)
∴ f(x) is neither even function nor odd function.The Fourier series for[ neither odd nor even function(none function)] f(x)
is given by
f(x) =a02+
∞∑n=1
(an cosnx+ bn sinnx) (1)
122 Unit II - FOURIER SERIES (F.S.)
where
a0 =1
π
∫ π
−π
f(x)dx, an =1
π
∫ π
−π
f(x) cosnxdx, bn =1
π
∫ π
−π
f(x) sinnxdx
Now, a0 =1
π
[∫ 0
−π
0dx+
∫ π
0
sinxdx
]=
1
π
∫ π
0
sinxdx
=1
π[− cosx]π0 = −1
π[cosx]π0 = −1
π[−1− 1]
∴ a0 =2
π
an =1
π
[∫ 0
−π
0 cosnxdx+
∫ π
0
sinx cosnxdx
]=1
π
∫ π
0
cosnx sinxdx
=1
π
∫ π
0
1
2[sin (n+ 1)x− sin (n− 1)x] dx
=1
2π
∫ π
0
[sin (n+ 1)x− sin (n− 1)x] dx
=1
2π
[− cos (n+ 1)x
n+ 1+
cos (n− 1)x
n− 1
]π0
=1
2π
[(− (−1)n+1
n+ 1+
− (−1)n−1
n− 1
)−(
−1
n+ 1− 1
n− 1
)]
=1
2π
[(− (−1) (−1)n
n+ 1+
(−1)n
(−1) (n− 1)
)+
(1
n+ 1− 1
n− 1
)]=
1
2π
[(− (−1) (−1)n
n+ 1+
(−1)n
(−1) (n− 1)
)+
(1
n+ 1− 1
n− 1
)]=
1
2π
[((−1)n
n+ 1− (−1)n
n− 1
)+
(1
n+ 1− 1
n− 1
)]=
1
2π
[(−1)n
(1
n+ 1− 1
n− 1
)+
(1
n+ 1− 1
n− 1
)]=
1
2π
[1
n+ 1− 1
n− 1
]((−1)n + 1)
=1
2π
[n− 1− (n+ 1)
n2 − 1
]((−1)n + 1)
=1
2π
(−2 ((−1)n + 1)
n2 − 1
)∴ an =−1
π
((−1)n + 1
n2 − 1
), n 6= 1
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 123
a1 =1
π
∫ π
−π
f (x) cos xdx =1
π
[∫ 0
−π
0 cos xdx+
∫ π
0
sinx cosxdx
]=
1
2π
∫ π
0
2 sin x cosxdx =1
2π
∫ π
0
sin 2xdx
=1
2π
[− cos 2x
2
]π0
=−1
4π[cos 2x]π0 =
−1
4π[1− 1]
∴ a1 =0
bn =1
π
[∫ 0
−π
0 sinnxdx+
∫ π
0
sinx sinnxdx
]=1
π
∫ π
0
sinnx sinxdx
=1
π
∫ π
0
1
2[cos (n− 1)x− cos (n+ 1)x] dx
=1
2π
∫ π
0
[cos (n− 1)x− cos (n+ 1)x] dx
=1
2π
[sin (n− 1)x
n− 1− sin (n+ 1)x
n+ 1
]π0
, n 6= 1
=1
2π[(0− 0)− (0− 0)] , n 6= 1
∴ bn =0, n 6= 1
b1 =1
π
∫ π
−π
f (x) sin xdx =1
π
[∫ 0
π
0 sin xdx+
∫ π
0
sinx sinxdx
]=1
π
∫ π
0
sin2 xdx
=1
π
∫ π
0
1
2(1− cos 2x) dx =
1
2π
∫ π
0
(1− cos 2x) dx
=1
2π
[x− sin 2x
2
]π0
=1
2π[(π − 0)− (0− 0)]
∴ b1 =1
2From (1),
f(x) =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
π
)+ 0 +
∞∑n=2
−1
n
((−1)n + 1
n2 − 1
)cosnx+
1
2sinx+ 0
124 Unit II - FOURIER SERIES (F.S.)
∴ f(x) =1
π− 1
π
∞∑n=2
((−1)n + 1
n2 − 1
)cosnx+
sinx
2(2)
This is required Fourier series of the given f(x).For deduction partLet x = 0 is point of discontinuity at middle of the interval.
∴ f(x) value is calculated by as follows:
f(x = 0) =f(0−) + f(0+)
2
=limh→0
f(0− h) + limh→0
f(0 + h)
2
=limh→0
f(−h) + limh→0
f(+h)
2
=limh→0
0 + limh→0
sinh
2=
0 + 0
2=0
∴ (2) ⇒ 0 =1
π− 1
π
∞∑n=2
((−1)n + 1
n2 − 1
).1 + 0
−1
π=−1
π
∞∑n=2
((−1)n + 1
n2 − 1
)1 =
∞∑n=2
((−1)n + 1
n2 − 1
)=
2
3+ 0 +
2
15+ 0 +
2
35+ · · ·
1
2=1
3+
1
15+
1
35+ · · ·
1
2=
1
1.3+
1
3.5+
1
5.7+ · · ·
1
1.3+
1
3.5+
1
5.7+· · · =1
2
Example 2.24. * Find the Fourier series for f(x) = e−x in (−π, π) and
hence deduce 2∞∑n=2
(−1)n
(n2 + 1)=
π
sinh π.{
a0 =2 sinh π
π, an =
(−1)n
π (1 + n2)(2 sinh π), bn =
2n (−1)n sinh π
π (1 + n2)&x = 0
}
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 125
2.5 Half range Fourier series in (0, `)
Half range cosine series(or) cosine series
Half range sine series(or) sine series
Don’t check for even or odd function, just subtitute thefollowing formulae
bn = 0 a0 = an = 0
f(x) =a02+
∞∑n=1
an cosnπx
`
where
a0 =2
`
∫ `
0
f(x)dx
an =2
`
∫ `
0
f(x) cosnπx
`dx
f(x) =∞∑n=1
an sinnπx
`
where
bn =2
`
∫ `
0
f(x) sinnπx
`dx
Note :Half range Fourier series in (0, π), substitute ` = π in above formulae.
2.6 Parseval’s Identity
(If the power of n in Fourier series expansion < the power of n in deductionpart, then use appropriate Parseval’s Identity)
(0, `)2
`
∫ `
0
[f(x)]2 dx =a202+
∞∑n=1
(a2n + b2n
)(0, π)
2
π
∫ π
0
[f(x)]2 dx =a202+
∞∑n=1
(a2n + b2n
)Example 2.25. Find Half range Fourier sine series for f(x) = x in (0, π)
and deduce1
12+
1
22+
1
32+ ... =
π2
6.{
a0 = 0
(∵ sineseries
), an = 0
(∵ sineseries
), bn =
−2 (−1)n
n&(Par′s id
)}
126 Unit II - FOURIER SERIES (F.S.)
Example 2.26. * Find Half range Fourier cosine series for f(x) = x in
(0, π) and deduce1
14+
1
34+
1
54+ ... =
π4
96.{
a0 = π, an =
{ −4
πn2, (n = odd)
0, (n = even), bn = 0
(∵ cosseries
)&(
Par′s id)}
Example 2.27. Expand x(π − x) in half range sine series in the interval(0, π).
Solution: Given f(x) = x(π − x) in (0, π)
= πx− x2
The sine series of f(x) is given by
f (x) =∞∑n=1
bn sinnx→ (1)
where bn =2
π
π∫0
f (x) sinnxdx
=2
π
π∫0
(πx− x2
)sinnxdx
=2
π
[(πx− x2
)(− cosnx
n
)− (π − 2x)
(− sinnx
n2
)+(−2)
(cosnxn3
)]π0
=2
π
[−(πx− x2
) cosnxn
+ (π − 2x)sinnx
n2−2
cosnx
n3
]π0
=2
π
[(0 + 0− 2 (−1)n
n3
)−(0 + 0− 2
n3
)]=
2
π
[2
n3− 2 (−1)n
n3
]bn =
4
πn3[1− (−1)n]
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 127
Sub. the value of bn in (1)
f (x) =∞∑n=1
4
πn3(1− (−1)n) sinnx
=4
π
∞∑n=1
(1− (−1)n
n3
)sinnx
=4
π
[2
13sinx+ 0 +
2
33sin 3x+ 0 + · · ·
]=
8
π
[sinx
13+
sin 3x
33+
sin 5x
53+ · · ·
]
2.7 Change of Interval {[0, 2`] , [−`, `] , [0, `]}
*Note : Just change π as ` and nx asnπx
`in previous π format formulas.
2.7.1 Examples under (0, 2`)
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
below:x 0 π/3 2π/3 π 4π/3 5π/3 2π
f(x) 1.0 1.4 1.9 1.7 1.5 1.2 1.0. Find the Fourier
series upto 2nd harmonic to represent y = f(x) in terms of x in (0, 2π).
132 Unit II - FOURIER SERIES (F.S.)
Solution : Letf(x) =
a02+ a1 cosx+ a2 cos 2x+ b1 sinx+ b2 sin 2x (1)
be the F.S. upto second harmonic.Since the first and last values of y are same in the given table, leave the
first column (or) last column of the table. Hence only the first six columnvalues will be used.
x y cosx cos 2x sinx sin 2x y cosx y cos 2x y sinx y sin 2x
0 1.0 1 1 0 0 1 1 0 0π
31.4 0.5 -0.5 0.866 0.866 0.7 -0.7 1.2124 1.2124
2π
31.9 -0.5 -0.5 0.866 -0.866 -0.95 -0.95 1.6454 -1.6454
π 1.7 -1 1 0 0 -1.7 1.7 0 04π
31.5 -0.5 -0.5 -0.866 0.866 -0.75 -0.75 -1.299 1.299
5π
31.2 0.5 -0.5 -0.866 -0.866 0.6 -0.6 -1.0392 -1.0392
Sum 8.7 - - - - -1.1 -0.3 0.5196 -0.1732
a0 =2× 1
6
[∑y]= 2× 1
6[8.7] = 2.9 ⇒ a0
2=
2.9
2= 1.45
a1 =2× 1
6
[∑y cosx
]= 2× 1
6[−1.1] = −0.37
a2 =2× 1
6
[∑y cos 2x
]= 2× 1
6[−0.3] = −0.1
b1 =2× 1
6
[∑y sinx
]= 2× 1
6[0.5196] = 0.17
b2 =2× 1
6
[∑y sin 2x
]= 2× 1
6[−0.1732] = −0.06
Hence the required Fourier Series upto second harmonic for the data is
(1) ⇒ y = f(x) = 1.45− 0.37 cos x− 0.1 cos 2x+ 0.17 sin x− 0.06 sin 2x
Example 2.40. * Determine upto second harmonic of Fourier series for
the following data:x 0 π/3 2π/3 π 4π/3 5π/3 2π
f(x) 1.98 1.3 1.05 1.3 –0.88 –0.25 1.98{a0 = 1.5, a1 = 0.3733, b1 = 1.00453, a2 = 0.89, b2 = −0.109693}
2.9.3 Examples under θ◦ form(Degree form)
Example 2.41. Find an emprical form of the functionf(x) = a0+ a1 cosx+ b1 sinx with period 2π.
xo: 0 60 120 180 240 300 360
y = f(x) 40 31 –13 20 3.7 –21 40
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 133
{a0 = 20.233, a1 = 9.883, b1 = 10.18993}
Example 2.42. Compute the first two harmonic of the Fourier series forf(x) from the following data. Expand f(x) as series of sine.
θ : x 0 30 60 90 120 150 180
T : f(x) 0 5224 8097 7850 5499 2626 0
{b1 = 7849.712, b2 = 1499.91067}
2.9.4 Examples under T form: (θ = 2πx/T )
Example 2.43. The following table gives the vibration of periodic currentover a period. Find Fourier series upto 1st harmonic.
T(sec) 0 T/6 T/3 T/2 2T/3 5T/6 T
I(Amp) 1.98 1.3 1.05 1.3 –0.88 –0.25 1.98
{a0 = 1.5, a1 = 0.373, b1 = 1.0045}
2.9.5 Problems under ` form: (2` = Number of data)
Example 2.44. Find the first harmonic of the Fourier series for f(x) for
the data.x 0 1 2 3 4 5
f(x) 9 18 24 28 26 20{a0 = 41.67, a1 = −8.33, b1 = −1.16}
134 Unit II - FOURIER SERIES (F.S.)
Exceptional Problems in (0, 2`) format
Example 2.45. Find the Fourier series for
f(x) =
{sinx, 0 ≤ x ≤ π/4cosx, π/4 ≤ x ≤ π/2{
a0 =−8√2π, an =
8
π (16n2 − 1)
[(−1)n√
2− 1
], bn = 0
}Example 2.46. * Find the Fourier series for the function
f(x) =
{x, 0 < x < `/2
`− x, `/2 < x < `and the series
∞∑n=1
1
(2n− 1)4.
2.10 Assignment I[Fourier series]
1. Find the Fourier series of f(x) =1
2(π − x) in the interval (0, 2π).
Hence deduce that 1− 1
3+
1
5− 1
7+ · · ·∞ =
π
4.
2. Find the Fourier series expansion of f(x) = ex in the interval (0, 2`).
3. Obtain the Fourier series for the function
f (x) =
{1− x, −π < x < 01 + x, 0 < x < π
. Hence deduce that
1
12+
1
32+
1
52+ ...∞ =
π2
8.
4. Find the Fourier series for the function given by
f (x) =
1 +
2x
`, −` ≤ x ≤ 0
1− 2x
`, 0 ≤ x ≤ `
. Hence deduce that
∞∑n=1
1
(2n− 1)2=π2
8.
5. Express f(x) as a Fourier sine series where
f(x) =
1
4− x, 0 < x <
1
2
x− 3
4,1
2< x < 1
.
6. Find the half range Fourier cosine series for the functionf (x) = x (π − x) in 0 < x < π. Deduce that1
14+
1
24+
1
34+ · · ·∞ =
π4
90.
7. Find the complex form of the Fourier series of f (x) = eax,−π < x < π.
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 135
8. Obtain a Fourier series upto the second harmonics from the datax: 0 π/3 2π/3 π 4π/3 5π/3 2π
f(x): 0.8 0.6 0.4 0.7 0.9 1.1 0.8
9. The following table gives the vibration of periodic current over aperiod. Find the Fourier upto 2nd harmonic.
T(sec): 0 T/6 T/6 T/2 2T/3 5T/6 T
I(Amp): 1.98 1.3 1.05 1.3 -0.88 -0.25 1.98
10. Find the Fourier series as for as the second harmonic to represent the
function given in the following tablex: 0 1 2 3 4 5
f(x): 9 18 24 28 26 20
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