VIVA INSTITUTE OF TECHNOLOGY, VIRAR
DEPT:MECHANICAL ENGINEERING
SECOND YEAR/SEM-III
SUBJECT:ENGINEERING MATHEMATICS-III
MODULE-III FOURIER SERIES LEC NO. 5
SUBJECT IN CHARGE
DR.J.C.JAIN
Problem2:Find the Fourier series of f(x) = 𝑥 in the interval (-𝜋, 𝜋).Hence
𝜋2 1 1 1
deduce that = 2 + 2 + 2 +--------
8 1 3 5
Solution: f(x)= 𝑥 -𝜋<x< 𝜋
i.E f(x)=-x -𝜋<x≤ 0
= x 0≤x< 𝜋
f(x)= 𝑥 is an even function.
Hence 𝑏𝑛 =0
The Fourier series of an even function with period 2𝜋 is given by
f(x)= 𝑎0 +σ∞𝑛=1 𝑎𝑛 𝑐𝑜𝑠𝑛𝑥…………………………(1)
𝜋 1
Now 𝑓 0 = 𝑎0 𝑥𝑑 𝑥
𝜋
𝜋
𝜋 1 1 𝑥2 𝜋
𝑥𝑑𝑥 0 = = =
𝜋 𝜋 2 0 2
𝜋 2
𝑓 0 = 𝑛𝑎 𝑥𝑑𝑥𝑛𝑠𝑜𝑐 𝑥
𝜋
2 𝜋
𝑥𝑑𝑥𝑛𝑠𝑜𝑐𝑥 0 =
𝜋
2 𝑥𝑛𝑛𝑖𝑠 𝜋 𝑥𝑛𝑠𝑜𝑐
= 𝑥 + (1) 2
𝜋 𝑛 𝑛 0
𝜋𝑛𝑠𝑜𝑐 2 1
= −
𝜋 𝑛2 𝑛2
2 𝑛
= 2 −1 −1
𝑛𝜋
𝜋 2 ∞ −1 𝑛 −1
f(x)= + σ𝑛=1 2 cosnx
2 𝜋 𝑛
𝜋 4 1 1 1
= - cosx+ 2 cos3x+ 2 𝑐𝑜𝑠5𝑥 + ⋯ … . …………………………(2)
2 𝜋 12 3 5
Put x=0 in (2) we get
𝜋 4 1 1 1 𝜋2 1 1 1
f(x)=0= - 2 + 2+ 2 + ⋯…. = = 2 + 2 + 2 +--------
2 𝜋 1 3 5 8 1 3 5
Problem3:Find the Fourier series of f(x) = sinax in the interval (-𝜋, 𝜋).
Solution: f(-x)=sina(-x) = -sinax
f(-x)=-f(x)
f(x) = sinax is an odd function.
𝑎0 = 0 𝑎𝑛𝑑𝑎𝑛 = 0
The Fourier series of an odd function with period 2𝜋 is given by
f(x)= σ∞
𝑛=1 𝑏𝑛 𝑠𝑖𝑛𝑛𝑥…………………………(1)
2 𝜋 2 𝜋
𝑏𝑛 = 0 𝑓 𝑥 𝑠𝑖𝑛𝑛𝑥𝑑𝑥 = 0 𝑠𝑖𝑛𝑎𝑥. 𝑠𝑖𝑛𝑛𝑥𝑑𝑥
𝜋 𝜋
1 𝜋
= 0 cos 𝑛 − 𝑎 𝑥 − cos 𝑛 + 𝑎 𝑥 dx
𝜋
1 sin 𝑛−𝑎 𝑥 sin 𝑛+𝑎 𝑥 𝜋
𝑏𝑛 = −
𝜋 𝑛−𝑎 𝑛+𝑎 0
1 sin 𝑛−𝑎 𝜋 sin 𝑛+𝑎 𝜋
= −
𝜋 𝑛−𝑎 𝑛+𝑎
1 𝑠𝑖𝑛𝑛𝜋.𝑐𝑜𝑠𝑎𝜋−𝑐𝑜𝑠𝑛𝜋.𝑠𝑖𝑛𝑎𝜋 𝑠𝑖𝑛𝑛𝜋.𝑐𝑜𝑠𝑎𝜋+𝑐𝑜𝑠𝑛𝜋.𝑠𝑖𝑛𝑎𝜋
= −
𝜋 𝑛−𝑎 𝑛+𝑎
1 − −1 𝑛 𝑠𝑖𝑛𝑎𝜋 −1 𝑛 𝑠𝑖𝑛𝑎𝜋
= −
𝜋 𝑛−𝑎 𝑛+𝑎
− −1 𝑛 𝑠𝑖𝑛𝑎𝜋 1 1
= +
𝜋 𝑛−𝑎 𝑛+𝑎
2𝑛 −1 𝑛 𝑠𝑖𝑛𝑎𝜋
𝑏𝑛 = 2 2 .
𝜋 𝑎 −𝑛
2𝑠𝑖𝑛𝑎𝜋 ∞ 𝑛 −1 𝑛 𝑠𝑖𝑛𝑎𝑥
Hence, f(x)= σ𝑛=1
𝜋 𝑎2 −𝑛2
Problem3:Find the Fourier series of f(x) = coshax in the interval (-𝜋, 𝜋).
Solution: f(-x)=cosha(-x)=coshax=f(x)
f(-x)=f(x)
f(x) = coshax is an even function.
Hence 𝑏𝑛 =0
The Fourier series of an even function with period 2𝜋 is given by
f(x)= 𝑎0 +σ∞
𝑛=1 𝑎𝑛 𝑐𝑜𝑠𝑛𝑥…………………………(1)
1 𝜋
Now 𝑎0 = 0 𝑓 𝑥 𝑑𝑥
𝜋
1 𝜋
= 0 𝑐𝑜𝑠ℎ𝑎𝑥𝑑𝑥
𝜋
1 𝜋 𝑒 𝑎𝑥 +𝑒 −𝑎𝑥
= 0 𝑑𝑥
𝜋 2
1 𝑒 𝑎𝑥 𝑒 −𝑎𝑥 𝜋
= +
2𝜋 𝑎 −𝑎 0
1
= 𝑒 𝑎𝜋 − 𝑒 −𝑎𝜋
2𝜋𝑎
𝑠𝑖𝑛ℎ𝑎𝜋
𝑎0 =
𝜋𝑎
2 𝜋
𝑎𝑛 = 0 𝑓 𝑥 𝑐𝑜𝑠𝑛𝑥𝑑𝑥
𝜋
2 𝜋
= 0 𝑐𝑜𝑠ℎ𝑎𝑥𝑐𝑜𝑠𝑛𝑥𝑑𝑥
𝜋
2 𝜋 𝑒 𝑎𝑥 +𝑒 −𝑎𝑥
= 0 𝑐𝑜𝑠𝑛𝑥𝑑𝑥
𝜋 2
1 𝜋 𝑎𝑥
= 0 (𝑒 𝑐𝑜𝑠𝑛𝑥 + 𝑒 −𝑎𝑥 𝑐𝑜𝑠𝑛𝑥)𝑑𝑥
𝜋
1 𝑒 𝑎𝑥 𝑒 −𝑎𝑥 𝜋
= 𝑎𝑐𝑜𝑠𝑛𝑥 + 𝑛𝑠𝑖𝑛𝑛𝑥 + −𝑎𝑐𝑜𝑠𝑛𝑥 + 𝑛𝑠𝑖𝑛𝑛𝑥
𝜋 𝑎2 +𝑛2 𝑎2 +𝑛2 0
𝑎𝑐𝑜𝑠𝑛𝜋 𝑎𝑐𝑜𝑠𝑛𝜋 2𝑎 −1 𝑛
= 2 2 (𝑒 𝑎𝜋 − 𝑒 −𝑎𝜋 )= 2 2 2sinha𝜋 = 2 2 sinha𝜋
𝜋(𝑎 +𝑛 ) 𝜋(𝑎 +𝑛 ) 𝜋(𝑎 +𝑛 )
sinha𝜋 2𝑎 −1 𝑛
Hence,f(x)= + sinha𝜋 σ∞
𝑛=1 (𝑎2 +𝑛2 ) 𝑐𝑜𝑠𝑛𝑥.
𝜋𝑎 𝜋
Problem4:Find the Fourier series of f(x) = 1-𝑥 2 in the interval (-1,1).
Solution: f(-x) = 1- −𝑥 2 =1-𝑥 2 = f(x)
∴f(x) = 1-𝑥 2 is an even function
Hence 𝑏𝑛 =0
The Fourier series of an even function with period 2𝑙 = 2 is given by
∞ 𝑛𝜋𝑥
f(x)= 𝑎0 +σ𝑛=1 𝑎𝑛 𝑐𝑜𝑠 …………………………(1)
𝑙
1 𝑙
Now 𝑎0 = 0 𝑓 𝑥 𝑑𝑥
𝑙 1
1 𝑥3 2
= 0 1−𝑥 2 𝑑𝑥 = 𝑥 − =
3 0 3
2 𝑙 𝑛𝜋𝑥 2 1 2 𝑛𝜋𝑥
𝑎𝑛 = 0 𝑓 𝑥 𝑐𝑜𝑠 𝑑𝑥 = 0 1−𝑥 𝑐𝑜𝑠 𝑑𝑥
𝑙 𝑙 1 1
𝑠𝑖𝑛𝑛𝜋𝑥 𝑐𝑜𝑠𝑛𝜋𝑥 𝑠𝑖𝑛𝑛𝜋𝑥 1
=2 1−𝑥 2 − −2𝑥 − 2 2 + (−2) − 3 3
𝑛𝜋 𝑛 𝜋 𝑛 𝜋 0
𝑐𝑜𝑠𝑛𝜋 −4 −1 𝑛
=2 −2 2 2 =
𝑛 𝜋 𝑛2 𝜋2
2 4 ∞ −1 𝑛
f(x)= - 2 σ𝑛=1 2 𝑐𝑜𝑠𝑛𝜋𝑥
3 𝜋 𝑛
1 1
Problem5:Find the Fourier series of f(x) = + 𝑥 - <x<0
2 2
1 1
= − 𝑥 0<x<
2 2
1 1 1
Solution: f(-x) = −𝑥 - <-x<0 or 0<x<
2 2 2
1 1 1
= +𝑥 0<-x< or - <x<0
2 2 2
f(-x) =f(x)
F(x) is an even function
Hence 𝑏𝑛 =0
The Fourier series of an even function with period 2𝑙 = 1 is given by
∞ 𝑛𝜋𝑥
f(x)= 𝑎0 +σ𝑛=1 𝑎𝑛 𝑐𝑜𝑠 …………………………(1)
𝑙
1 𝑙
Now 𝑎0 = 0 𝑓 𝑥 𝑑𝑥
𝑙
1/2
1 1/2 1 𝑥 𝑥2 1
= − 𝑥 𝑑𝑥 =2 − =
1/2 0 2 2 2 0 4
2 𝑙 𝑛𝜋𝑥 2 1/2 1 𝑛𝜋𝑥
𝑎𝑛 = 0 𝑓 𝑥 𝑐𝑜𝑠 𝑑𝑥 = − 𝑥 𝑐𝑜𝑠 𝑑𝑥
𝑙 𝑙 1/2 0 2 1/2
1 sin2nπx cos2nπx 1/2
=4 −x − (−1) − 2 2
2 2nπ 4n π 0
cosnπ 1 1
=4 − 2 2+ 2 2 = 2 2 1 − −1 𝑛
4n π 4n π n π
1 1 ∞ 1− −1 𝑛
f(x)= + σ 𝑐𝑜𝑠2𝑛𝜋𝑥.
4 𝜋2 𝑛=1 𝑛2
