Post on 05-Dec-2014
description
Fourier Series
N. B. Vyas
Department of Mathematics,Atmiya Institute of Tech. and Science,
Rajkot (Guj.) - INDIA
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Periodic Function
A function f(x) which satisfies the relation f(x) = f(x+ T ) forall real x and some fixed T is called Periodic function.
The smallest positive number T , for which this relation holds, iscalled the period of f(x).
If T is the period of f(x) then
f(x) = f(x+ T ) = f(x+ 2T ) = . . . = f(x+ nT )
f(x) = f(x− T ) = f(x− 2T ) = . . . = f(x− nT )
∴ f(x) = f(x± nT ), where n is a positive integer
Eg. Sinx, Cosx, Secx and Cosecx are periodic functions with period2π
tanx and cotx are periodic functions with period π.
N. B. Vyas Fourier Series
Even & Odd functions
A function f(x) is said to be even if f(−x) = f(x).
Eg. x2 and cosx are even function.∫ c
−cf(x)dx = 2
∫ c
0
f(x)dx ; if f(x) is an even function.
A function f(x) is said to be odd if f(−x) = −f(x).Eg. x3 and sinx are odd function.∫ c
−cf(x)dx = 0 ; if f(x) is an odd function.
N. B. Vyas Fourier Series
Even & Odd functions
A function f(x) is said to be even if f(−x) = f(x).
Eg. x2 and cosx are even function.
∫ c
−cf(x)dx = 2
∫ c
0
f(x)dx ; if f(x) is an even function.
A function f(x) is said to be odd if f(−x) = −f(x).Eg. x3 and sinx are odd function.∫ c
−cf(x)dx = 0 ; if f(x) is an odd function.
N. B. Vyas Fourier Series
Even & Odd functions
A function f(x) is said to be even if f(−x) = f(x).
Eg. x2 and cosx are even function.∫ c
−cf(x)dx = 2
∫ c
0
f(x)dx ; if f(x) is an even function.
A function f(x) is said to be odd if f(−x) = −f(x).Eg. x3 and sinx are odd function.∫ c
−cf(x)dx = 0 ; if f(x) is an odd function.
N. B. Vyas Fourier Series
Even & Odd functions
A function f(x) is said to be even if f(−x) = f(x).
Eg. x2 and cosx are even function.∫ c
−cf(x)dx = 2
∫ c
0
f(x)dx ; if f(x) is an even function.
A function f(x) is said to be odd if f(−x) = −f(x).
Eg. x3 and sinx are odd function.∫ c
−cf(x)dx = 0 ; if f(x) is an odd function.
N. B. Vyas Fourier Series
Even & Odd functions
A function f(x) is said to be even if f(−x) = f(x).
Eg. x2 and cosx are even function.∫ c
−cf(x)dx = 2
∫ c
0
f(x)dx ; if f(x) is an even function.
A function f(x) is said to be odd if f(−x) = −f(x).Eg. x3 and sinx are odd function.
∫ c
−cf(x)dx = 0 ; if f(x) is an odd function.
N. B. Vyas Fourier Series
Even & Odd functions
A function f(x) is said to be even if f(−x) = f(x).
Eg. x2 and cosx are even function.∫ c
−cf(x)dx = 2
∫ c
0
f(x)dx ; if f(x) is an even function.
A function f(x) is said to be odd if f(−x) = −f(x).Eg. x3 and sinx are odd function.∫ c
−cf(x)dx = 0 ; if f(x) is an odd function.
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c
∫eax cos bx dx =
eax
a2 + b2(a cosbx+ b sinbx) + c∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c∫
eax cos bx dx =eax
a2 + b2(a cosbx+ b sinbx) + c
∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c∫
eax cos bx dx =eax
a2 + b2(a cosbx+ b sinbx) + c∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0
∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c∫
eax cos bx dx =eax
a2 + b2(a cosbx+ b sinbx) + c∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0
∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c∫
eax cos bx dx =eax
a2 + b2(a cosbx+ b sinbx) + c∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c∫
eax cos bx dx =eax
a2 + b2(a cosbx+ b sinbx) + c∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫eax sin bx dx =
eax
a2 + b2(a sin bx− b cos bx) + c∫
eax cos bx dx =eax
a2 + b2(a cosbx+ b sinbx) + c∫ c+2π
c
sin nx dx = −[cos nx
n
]c+2π
c= 0, n 6= 0∫ c+2π
c
cos nx dx =
[sin nx
n
]c+2π
c
= 0, n 6= 0∫ c+2π
c
sinmx cos nx dx =1
2
∫ c+2π
c
2sinmx cos nx dx
=1
2
∫ c+2π
c
[sin (m+ n)x+ sin (m− n)x] dx
= −1
2
[cos (m+ n)x
m+ n+cos (m− n)x
m− n
]c+2π
c
= 0, n 6= 0
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx
= 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx
= 0, n 6= 0∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0
∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx
= π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx = π
∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx
= π
N. B. Vyas Fourier Series
Some Important Formula∫ c+2π
c
cosmx cos nx dx = 12
∫ c+2π
c
2cosmx cos nx dx
= 12
∫ c+2π
c
[cos (m+ n)x+ cos (m− n)x] dx
= 12
[sin (m+ n)x
m+ n+sin (m− n)x
m− n
]c+2π
c
= 0,m 6= n∫ c+2π
c
sinmx sin nx dx = 0∫ c+2π
c
sin nx cos nx dx = 0, n 6= 0∫ c+2π
c
cos2 nx dx = π∫ c+2π
c
sin2 nx dx = π
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)
= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0
and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n
and cos(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Some Important Formula
(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply thegeneralized rule of integration by parts∫u vdx = u v1 − u′ v2 + u′′ v3 − u′′′ v4 + . . .
Eg.
∫x3 e−2x dx
= x3(e−2x
−2
)− 3x2
(e−2x
(−2)2
)+ 6x
(e−2x
(−2)3
)− 6
(e−2x
(−2)4
)= −1
8e−2x (4x3 + 6x2 + 6x+ 3)
sin nπ = 0 and cos nπ = (−1)n
sin(n+ 1
2
)π = (−1)n and cos
(n+ 1
2
)π = 0
where n is integer.
N. B. Vyas Fourier Series
Fourier Series
The Fourier series for the function f(x) in the intervalc < x < c+ 2π is given by
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ c+2π
c
f(x) dx
an =1
π
∫ c+2π
c
f(x) cos nx dx
bn =1
π
∫ c+2π
c
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
The Fourier series for the function f(x) in the intervalc < x < c+ 2π is given by
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ c+2π
c
f(x) dx
an =1
π
∫ c+2π
c
f(x) cos nx dx
bn =1
π
∫ c+2π
c
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
The Fourier series for the function f(x) in the intervalc < x < c+ 2π is given by
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ c+2π
c
f(x) dx
an =1
π
∫ c+2π
c
f(x) cos nx dx
bn =1
π
∫ c+2π
c
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
The Fourier series for the function f(x) in the intervalc < x < c+ 2π is given by
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ c+2π
c
f(x) dx
an =1
π
∫ c+2π
c
f(x) cos nx dx
bn =1
π
∫ c+2π
c
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
The Fourier series for the function f(x) in the intervalc < x < c+ 2π is given by
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ c+2π
c
f(x) dx
an =1
π
∫ c+2π
c
f(x) cos nx dx
bn =1
π
∫ c+2π
c
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 1: If c = 0, the interval becomes 0 < x < 2π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 1: If c = 0, the interval becomes 0 < x < 2π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 1: If c = 0, the interval becomes 0 < x < 2π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 1: If c = 0, the interval becomes 0 < x < 2π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 1: If c = 0, the interval becomes 0 < x < 2π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 2: If c = −π, the interval becomes −π << π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 2: If c = −π, the interval becomes −π << π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 2: If c = −π, the interval becomes −π << π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 2: If c = −π, the interval becomes −π << π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Corollary 2: If c = −π, the interval becomes −π << π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is −c < x < c and f(x) is anodd function i.e. f(−x) = −f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx = 0
an =1
π
∫ π
−πf(x) cos nx dx = 0
because cos nx is an even function , f(x)cos nx is an oddfunction
bn =1
π
∫ π
−πf(x) sin nx dx =
2
π
∫ π
0
f(x) sin nx dx
because sin nx is an odd function , f(x)sin nx is an evenfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f(x) is aneven function i.e. f(−x) = f(x). Let C = π
f(x) =a02
+∞∑n=1
an cos nx+∞∑n=1
bn sin nx
where a0 =1
π
∫ π
−πf(x) dx =
2
π
∫ π
0
f(x) dx
an =1
π
∫ π
−πf(x) cos nx dx =
2
π
∫ π
0
f(x) cos nx dx
because cos nx is an even function , f(x)cos nx is an evenfunction
bn =1
π
∫ π
−πf(x) sin nx dx = 0
because sin nx is an odd function , f(x)sin nx is an oddfunction
N. B. Vyas Fourier Series
Example
Ex. Obtain Fourier series of f(x) =
(π − x2
)2
in the
interval 0 ≤ x ≤ 2π. Hence deduce thatπ2
12=
1
12− 1
22+
1
32− . . .
N. B. Vyas Fourier Series
Example
Ex. Obtain Fourier series of f(x) =
(π − x2
)2
in the
interval 0 ≤ x ≤ 2π. Hence deduce thatπ2
12=
1
12− 1
22+
1
32− . . .
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
(π − x)2
4dx
=1
4π
[(π − x)3
(−3)
]2π0
= − 1
12π(−π3 − π3)
=π2
6
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
(π − x)2
4dx
=1
4π
[(π − x)3
(−3)
]2π0
= − 1
12π(−π3 − π3)
=π2
6
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
(π − x)2
4dx
=1
4π
[(π − x)3
(−3)
]2π0
= − 1
12π(−π3 − π3)
=π2
6
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
(π − x)2
4dx
=1
4π
[(π − x)3
(−3)
]2π0
= − 1
12π(−π3 − π3)
=π2
6
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
(π − x)2
4cos nx dx
=1
n2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
(π − x)2
4cos nx dx
=1
n2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
(π − x)2
4cos nx dx
=1
n2
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
bn =1
π
∫ 2π
0
(π − x)2
4sin nx dx
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
bn =1
π
∫ 2π
0
(π − x)2
4sin nx dx
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
bn =1
π
∫ 2π
0
(π − x)2
4sin nx dx
= 0
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π]
(π − x2
)2
=π2
12+∞∑n=1
1
n2cos nx
=π2
12+
1
12cos x+
1
22cos 2x+
1
32cos 3x+ . . .
Putting x = π, we get
0 =π2
12− 1
12+
1
22− 1
32+
1
42− . . .
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π](π − x2
)2
=π2
12+∞∑n=1
1
n2cos nx
=π2
12+
1
12cos x+
1
22cos 2x+
1
32cos 3x+ . . .
Putting x = π, we get
0 =π2
12− 1
12+
1
22− 1
32+
1
42− . . .
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π](π − x2
)2
=π2
12+∞∑n=1
1
n2cos nx
=π2
12+
1
12cos x+
1
22cos 2x+
1
32cos 3x+ . . .
Putting x = π, we get
0 =π2
12− 1
12+
1
22− 1
32+
1
42− . . .
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π](π − x2
)2
=π2
12+∞∑n=1
1
n2cos nx
=π2
12+
1
12cos x+
1
22cos 2x+
1
32cos 3x+ . . .
Putting x = π, we get
0 =π2
12− 1
12+
1
22− 1
32+
1
42− . . .
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π](π − x2
)2
=π2
12+∞∑n=1
1
n2cos nx
=π2
12+
1
12cos x+
1
22cos 2x+
1
32cos 3x+ . . .
Putting x = π, we get
0 =π2
12− 1
12+
1
22− 1
32+
1
42− . . .
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π](π − x2
)2
=π2
12+∞∑n=1
1
n2cos nx
=π2
12+
1
12cos x+
1
22cos 2x+
1
32cos 3x+ . . .
Putting x = π, we get
0 =π2
12− 1
12+
1
22− 1
32+
1
42− . . .
π2
12=
1
12− 1
22+
1
32− 1
42+ . . .
N. B. Vyas Fourier Series
Example
Ex. Expand in a Fourier series the function f(x) = xin the interval 0 ≤ x ≤ 2π.
N. B. Vyas Fourier Series
Example
Ex. Expand in a Fourier series the function f(x) = xin the interval 0 ≤ x ≤ 2π.
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x dx
=1
4π
[x2
2
]2π0
= 2π
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x dx
=1
4π
[x2
2
]2π0
= 2π
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x dx
=1
4π
[x2
2
]2π0
= 2π
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x dx
=1
4π
[x2
2
]2π0
= 2π
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x cos nx dx[x
(sin nx
n
)−(−cos nx
n
)]2π0
= 0
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x cos nx dx
[x
(sin nx
n
)−(−cos nx
n
)]2π0
= 0
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x cos nx dx[x
(sin nx
n
)−(−cos nx
n
)]2π0
= 0
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x cos nx dx[x
(sin nx
n
)−(−cos nx
n
)]2π0
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
bn =1
π
∫ 2π
0
x sin nx dx
=−2n
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
bn =1
π
∫ 2π
0
x sin nx dx
=−2n
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
bn =1
π
∫ 2π
0
x sin nx dx
=−2n
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π]
f(x) =2π
2+ 0 +
∞∑n=1
−2nsin nx
= π −∞∑n=1
sin nx
n
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π]
f(x) =2π
2+ 0 +
∞∑n=1
−2nsin nx
= π −∞∑n=1
sin nx
n
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π]
f(x) =2π
2+ 0 +
∞∑n=1
−2nsin nx
= π −∞∑n=1
sin nx
n
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval [0, 2π]
f(x) =2π
2+ 0 +
∞∑n=1
−2nsin nx
= π −∞∑n=1
sin nx
n
N. B. Vyas Fourier Series
Example
Ex. Determine the Fourier series expansion of thefunction f(x) = xsin x in the interval 0 ≤ x ≤ 2π.
N. B. Vyas Fourier Series
Example
Ex. Determine the Fourier series expansion of thefunction f(x) = xsin x in the interval 0 ≤ x ≤ 2π.
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ 2π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x sin x dx
=1
π[−x cos x+ sin x]2π0
=1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
a0 =−2ππ
= −2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x sin x dx
=1
π[−x cos x+ sin x]2π0
=1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
a0 =−2ππ
= −2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x sin x dx
=1
π[−x cos x+ sin x]2π0
=1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
a0 =−2ππ
= −2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x sin x dx
=1
π[−x cos x+ sin x]2π0
=1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
a0 =−2ππ
= −2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
a0 =1
π
∫ 2π
0
x sin x dx
=1
π[−x cos x+ sin x]2π0
=1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
a0 =−2ππ
= −2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x sin x cos nx dx
=1
2π
∫ 2π
0
x (2 sin x cos nx) dx
=1
2π
∫ 2π
0
x (sin (n+ 1)x − sin (n− 1)x) dx
=1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x sin x cos nx dx
=1
2π
∫ 2π
0
x (2 sin x cos nx) dx
=1
2π
∫ 2π
0
x (sin (n+ 1)x − sin (n− 1)x) dx
=1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x sin x cos nx dx
=1
2π
∫ 2π
0
x (2 sin x cos nx) dx
=1
2π
∫ 2π
0
x (sin (n+ 1)x − sin (n− 1)x) dx
=1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x sin x cos nx dx
=1
2π
∫ 2π
0
x (2 sin x cos nx) dx
=1
2π
∫ 2π
0
x (sin (n+ 1)x − sin (n− 1)x) dx
=1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 2π
0
f(x) cos nx dx
an =1
π
∫ 2π
0
x sin x cos nx dx
=1
2π
∫ 2π
0
x (2 sin x cos nx) dx
=1
2π
∫ 2π
0
x (sin (n+ 1)x − sin (n− 1)x) dx
=1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n = 1
∴ a1 =1
2π
∫ 2π
0
x sin 2x dx
=1
2π
[−x(cos 2x
2
)+sin 2x
4
]2π0
= −1
2
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n = 1
∴ a1 =1
2π
∫ 2π
0
x sin 2x dx
=1
2π
[−x(cos 2x
2
)+sin 2x
4
]2π0
= −1
2
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n = 1
∴ a1 =1
2π
∫ 2π
0
x sin 2x dx
=1
2π
[−x(cos 2x
2
)+sin 2x
4
]2π0
= −1
2
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n = 1
∴ a1 =1
2π
∫ 2π
0
x sin 2x dx
=1
2π
[−x(cos 2x
2
)+sin 2x
4
]2π0
= −1
2
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n = 1
∴ a1 =1
2π
∫ 2π
0
x sin 2x dx
=1
2π
[−x(cos 2x
2
)+sin 2x
4
]2π0
= −1
2
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n 6= 1
∴ an =1
2π
[x
(−cos (n+ 1)x
n+ 1
)−(−sin (n+ 1)x
(n+ 1)2
)]2π0
− 1
2π
[x
(−cos (n− 1)x
n− 1
)−(−sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[− 2π
n+ 1+ 0 +
2π
n− 1− 0
]=
2
n2 − 1
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n 6= 1
∴ an =1
2π
[x
(−cos (n+ 1)x
n+ 1
)−(−sin (n+ 1)x
(n+ 1)2
)]2π0
− 1
2π
[x
(−cos (n− 1)x
n− 1
)−(−sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[− 2π
n+ 1+ 0 +
2π
n− 1− 0
]=
2
n2 − 1
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n 6= 1
∴ an =1
2π
[x
(−cos (n+ 1)x
n+ 1
)−(−sin (n+ 1)x
(n+ 1)2
)]2π0
− 1
2π
[x
(−cos (n− 1)x
n− 1
)−(−sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[− 2π
n+ 1+ 0 +
2π
n− 1− 0
]=
2
n2 − 1
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n 6= 1
∴ an =1
2π
[x
(−cos (n+ 1)x
n+ 1
)−(−sin (n+ 1)x
(n+ 1)2
)]2π0
− 1
2π
[x
(−cos (n− 1)x
n− 1
)−(−sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[− 2π
n+ 1+ 0 +
2π
n− 1− 0
]=
2
n2 − 1
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n 6= 1
∴ an =1
2π
[x
(−cos (n+ 1)x
n+ 1
)−(−sin (n+ 1)x
(n+ 1)2
)]2π0
− 1
2π
[x
(−cos (n− 1)x
n− 1
)−(−sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[− 2π
n+ 1+ 0 +
2π
n− 1− 0
]
=2
n2 − 1
N. B. Vyas Fourier Series
Example
an =1
2π
∫ 2π
0
x sin (n+ 1)x dx− 1
2π
∫ 2π
0
x sin (n− 1)x dx
If n 6= 1
∴ an =1
2π
[x
(−cos (n+ 1)x
n+ 1
)−(−sin (n+ 1)x
(n+ 1)2
)]2π0
− 1
2π
[x
(−cos (n− 1)x
n− 1
)−(−sin (n− 1)x
(n− 1)2
)]2π0
=1
2π
[− 2π
n+ 1+ 0 +
2π
n− 1− 0
]=
2
n2 − 1
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
∫ 2π
0
x sin x sin nx dx
=1
2π
∫ 2π
0
x (2sin nx sin x) dx
=1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
∫ 2π
0
x sin x sin nx dx
=1
2π
∫ 2π
0
x (2sin nx sin x) dx
=1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
∫ 2π
0
x sin x sin nx dx
=1
2π
∫ 2π
0
x (2sin nx sin x) dx
=1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
∫ 2π
0
x sin x sin nx dx
=1
2π
∫ 2π
0
x (2sin nx sin x) dx
=1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n = 1
∴ b1 =1
2π
∫ 2π
0
(−x cos 2x) dx+ 1
2π
∫ 2π
0
x dx
=1
2π
[x
(−sin 2x
2
)− cos 2x
4+x2
2
]2π0
b1 = π
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n = 1
∴ b1 =1
2π
∫ 2π
0
(−x cos 2x) dx+ 1
2π
∫ 2π
0
x dx
=1
2π
[x
(−sin 2x
2
)− cos 2x
4+x2
2
]2π0
b1 = π
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n = 1
∴ b1 =1
2π
∫ 2π
0
(−x cos 2x) dx+ 1
2π
∫ 2π
0
x dx
=1
2π
[x
(−sin 2x
2
)− cos 2x
4+x2
2
]2π0
b1 = π
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n = 1
∴ b1 =1
2π
∫ 2π
0
(−x cos 2x) dx+ 1
2π
∫ 2π
0
x dx
=1
2π
[x
(−sin 2x
2
)− cos 2x
4+x2
2
]2π0
b1 = π
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n = 1
∴ b1 =1
2π
∫ 2π
0
(−x cos 2x) dx+ 1
2π
∫ 2π
0
x dx
=1
2π
[x
(−sin 2x
2
)− cos 2x
4+x2
2
]2π0
b1 = π
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n 6= 1
∴ bn =1
2π
[x
(−sin (n+ 1)x
n+ 1
)−(cos (n+ 1)x
(n+ 1)2
)]2π0
+1
2π
[x
(sin (n− 1)x
n− 1
)+
(cos (n− 1)x
(n− 1)2
)]2π0
= 0
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n 6= 1
∴ bn =1
2π
[x
(−sin (n+ 1)x
n+ 1
)−(cos (n+ 1)x
(n+ 1)2
)]2π0
+1
2π
[x
(sin (n− 1)x
n− 1
)+
(cos (n− 1)x
(n− 1)2
)]2π0
= 0
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n 6= 1
∴ bn =1
2π
[x
(−sin (n+ 1)x
n+ 1
)−(cos (n+ 1)x
(n+ 1)2
)]2π0
+1
2π
[x
(sin (n− 1)x
n− 1
)+
(cos (n− 1)x
(n− 1)2
)]2π0
= 0
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n 6= 1
∴ bn =1
2π
[x
(−sin (n+ 1)x
n+ 1
)−(cos (n+ 1)x
(n+ 1)2
)]2π0
+1
2π
[x
(sin (n− 1)x
n− 1
)+
(cos (n− 1)x
(n− 1)2
)]2π0
= 0
N. B. Vyas Fourier Series
Example
bn =1
2π
∫ 2π
0
x (−cos (n+ 1)x + cos (n− 1)x) dx
If n 6= 1
∴ bn =1
2π
[x
(−sin (n+ 1)x
n+ 1
)−(cos (n+ 1)x
(n+ 1)2
)]2π0
+1
2π
[x
(sin (n− 1)x
n− 1
)+
(cos (n− 1)x
(n− 1)2
)]2π0
= 0
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, a1, an(n > 1), b1 andbn(n > 1) in (1), we get the required Fourier series of f(x) inthe interval [0, 2π]
f(x) =−22
+∞∑n=1
(an cos nx+ bn sin nx)
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, a1, an(n > 1), b1 andbn(n > 1) in (1), we get the required Fourier series of f(x) inthe interval [0, 2π]
f(x) =−22
+∞∑n=1
(an cos nx+ bn sin nx)
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, a1, an(n > 1), b1 andbn(n > 1) in (1), we get the required Fourier series of f(x) inthe interval [0, 2π]
f(x) =−22
+∞∑n=1
(an cos nx+ bn sin nx)
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
f(x)= −π;−π < x < 0= x; 0 < x < π
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
f(x)= −π;−π < x < 0= x; 0 < x < π
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ π
−πf(x) dx
an =1
π
∫ π
−πf(x) cos nx dx
bn =1
π
∫ π
−πf(x) sin nx dx
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ π
−πf(x) dx
=1
π
[∫ 0
−πf(x) dx+
∫ π
0
f(x) dx
]=
1
π
[∫ 0
−π(−π) dx+
∫ π
0
x dx
]= −π
2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ π
−πf(x) dx
=1
π
[∫ 0
−πf(x) dx+
∫ π
0
f(x) dx
]
=1
π
[∫ 0
−π(−π) dx+
∫ π
0
x dx
]= −π
2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ π
−πf(x) dx
=1
π
[∫ 0
−πf(x) dx+
∫ π
0
f(x) dx
]=
1
π
[∫ 0
−π(−π) dx+
∫ π
0
x dx
]
= −π2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ π
−πf(x) dx
=1
π
[∫ 0
−πf(x) dx+
∫ π
0
f(x) dx
]=
1
π
[∫ 0
−π(−π) dx+
∫ π
0
x dx
]= −π
2
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ π
−πf(x) dx
=1
π
[∫ 0
−πf(x) dx+
∫ π
0
f(x) dx
]=
1
π
[∫ 0
−π(−π) dx+
∫ π
0
x dx
]= −π
2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 0
−πf(x) cos nx dx+
∫ π
0
f(x) cos nx dx
=1
π
[∫ 0
−π(−π) cos nx dx+
∫ π
0
x cos nx dx
]=
1
π
[−π[sin nx
n
]0−π
+
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
]
=(−1)n − 1
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 0
−πf(x) cos nx dx+
∫ π
0
f(x) cos nx dx
=1
π
[∫ 0
−π(−π) cos nx dx+
∫ π
0
x cos nx dx
]
=1
π
[−π[sin nx
n
]0−π
+
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
]
=(−1)n − 1
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 0
−πf(x) cos nx dx+
∫ π
0
f(x) cos nx dx
=1
π
[∫ 0
−π(−π) cos nx dx+
∫ π
0
x cos nx dx
]=
1
π
[−π[sin nx
n
]0−π
+
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
]
=(−1)n − 1
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 0
−πf(x) cos nx dx+
∫ π
0
f(x) cos nx dx
=1
π
[∫ 0
−π(−π) cos nx dx+
∫ π
0
x cos nx dx
]=
1
π
[−π[sin nx
n
]0−π
+
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
]
=(−1)n − 1
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
∫ 0
−πf(x) cos nx dx+
∫ π
0
f(x) cos nx dx
=1
π
[∫ 0
−π(−π) cos nx dx+
∫ π
0
x cos nx dx
]=
1
π
[−π[sin nx
n
]0−π
+
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
]
=(−1)n − 1
πn2
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
[∫ 0
−πf(x) sin nx dx+
∫ π
0
f(x) sin nx dx
]
=1
π
[∫ 0
−π(−π) sin nx dx+
∫ π
0
x sin nx dx
]=
1
π
[−π[−cosnx
n
]0−π
+
[x
(−cos nx
n
)−(−sin nxn2
)]π0
]
=1− 2(−1)n
n
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
[∫ 0
−πf(x) sin nx dx+
∫ π
0
f(x) sin nx dx
]=
1
π
[∫ 0
−π(−π) sin nx dx+
∫ π
0
x sin nx dx
]
=1
π
[−π[−cosnx
n
]0−π
+
[x
(−cos nx
n
)−(−sin nxn2
)]π0
]
=1− 2(−1)n
n
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
[∫ 0
−πf(x) sin nx dx+
∫ π
0
f(x) sin nx dx
]=
1
π
[∫ 0
−π(−π) sin nx dx+
∫ π
0
x sin nx dx
]=
1
π
[−π[−cosnx
n
]0−π
+
[x
(−cos nx
n
)−(−sin nxn2
)]π0
]
=1− 2(−1)n
n
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
[∫ 0
−πf(x) sin nx dx+
∫ π
0
f(x) sin nx dx
]=
1
π
[∫ 0
−π(−π) sin nx dx+
∫ π
0
x sin nx dx
]=
1
π
[−π[−cosnx
n
]0−π
+
[x
(−cos nx
n
)−(−sin nxn2
)]π0
]
=1− 2(−1)n
n
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
[∫ 0
−πf(x) sin nx dx+
∫ π
0
f(x) sin nx dx
]=
1
π
[∫ 0
−π(−π) sin nx dx+
∫ π
0
x sin nx dx
]=
1
π
[−π[−cosnx
n
]0−π
+
[x
(−cos nx
n
)−(−sin nxn2
)]π0
]
=1− 2(−1)n
n
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (−π, π)
f(x) =−π4
+∞∑n=1
(an cos nx+ bn sin nx)
=−π4
+∞∑n=1
((−1)n − 1
πn2
)cos nx+
∞∑n=1
(1− 2(−1)n
n
)sin nx
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (−π, π)
f(x) =−π4
+∞∑n=1
(an cos nx+ bn sin nx)
=−π4
+∞∑n=1
((−1)n − 1
πn2
)cos nx+
∞∑n=1
(1− 2(−1)n
n
)sin nx
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (−π, π)
f(x) =−π4
+∞∑n=1
(an cos nx+ bn sin nx)
=−π4
+∞∑n=1
((−1)n − 1
πn2
)cos nx+
∞∑n=1
(1− 2(−1)n
n
)sin nx
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f(x) in the interval (−π, π)
f(x) =−π4
+∞∑n=1
(an cos nx+ bn sin nx)
=−π4
+∞∑n=1
((−1)n − 1
πn2
)cos nx+
∞∑n=1
(1− 2(−1)n
n
)sin nx
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
f(x)= 2;−π < x < 0= 1; 0 < x < π
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
f(x)= 2;−π < x < 0= 1; 0 < x < π
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
f(x)= −k;−π < x < 0= k; 0 < x < π
Hence deduce that 1− 1
3+
1
5− 1
7+ . . . =
π
4
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
f(x)= −k;−π < x < 0= k; 0 < x < π
Hence deduce that 1− 1
3+
1
5− 1
7+ . . . =
π
4
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series of the function
f(x)= x; 0 ≤ x < π= 2π ;x = π= 2π − x ; π < x < 2π
Hence deduce that3π2
8=
1
12+
1
32+
1
52+ . . .
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f(x) is given by
f(x) =a02
+∞∑n=1
(an cos nx+ bn sin nx) . . . (1)
where a0 =1
π
∫ 2π
0
f(x) dx
an =1
π
∫ π
0
f(x) cos nx dx
bn =1
π
∫ 2π
0
f(x) sin nx dx
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
=1
π
[∫ π
0
x dx+
∫ 2π
π
(2π − x) dx]
=1
π
[(x2
2
)π0
+
(2πx− x2
2
)2π
π
]= π
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
=1
π
[∫ π
0
x dx+
∫ 2π
π
(2π − x) dx]
=1
π
[(x2
2
)π0
+
(2πx− x2
2
)2π
π
]= π
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
=1
π
[∫ π
0
x dx+
∫ 2π
π
(2π − x) dx]
=1
π
[(x2
2
)π0
+
(2πx− x2
2
)2π
π
]
= π
N. B. Vyas Fourier Series
Example
Step 2. Now a0 =1
π
∫ 2π
0
f(x) dx
=1
π
[∫ π
0
x dx+
∫ 2π
π
(2π − x) dx]
=1
π
[(x2
2
)π0
+
(2πx− x2
2
)2π
π
]= π
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
[∫ π
0
f(x) cos nx dx+
∫ 2π
π
f(x) cos nx dx
]
=1
π
[∫ π
0
x cos nx dx+
∫ 2π
π
(2π − x) cos nx dx]
=1
π
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
+1
π
[(2π − x)
(sin nx
n
)− (−1)
(−cos nx
n2
)]2ππ
=1
π
[(0 +
cos nπ
n2
)−(0 +
1
n2
)]+1
π
[(0− cos 2nπ
n2
)−(0− cos nπ
n2
)]=
2 [(−1)n − 1]
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
[∫ π
0
f(x) cos nx dx+
∫ 2π
π
f(x) cos nx dx
]=
1
π
[∫ π
0
x cos nx dx+
∫ 2π
π
(2π − x) cos nx dx]
=1
π
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
+1
π
[(2π − x)
(sin nx
n
)− (−1)
(−cos nx
n2
)]2ππ
=1
π
[(0 +
cos nπ
n2
)−(0 +
1
n2
)]+1
π
[(0− cos 2nπ
n2
)−(0− cos nπ
n2
)]=
2 [(−1)n − 1]
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
[∫ π
0
f(x) cos nx dx+
∫ 2π
π
f(x) cos nx dx
]=
1
π
[∫ π
0
x cos nx dx+
∫ 2π
π
(2π − x) cos nx dx]
=1
π
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
+1
π
[(2π − x)
(sin nx
n
)− (−1)
(−cos nx
n2
)]2ππ
=1
π
[(0 +
cos nπ
n2
)−(0 +
1
n2
)]+1
π
[(0− cos 2nπ
n2
)−(0− cos nπ
n2
)]=
2 [(−1)n − 1]
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
[∫ π
0
f(x) cos nx dx+
∫ 2π
π
f(x) cos nx dx
]=
1
π
[∫ π
0
x cos nx dx+
∫ 2π
π
(2π − x) cos nx dx]
=1
π
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
+1
π
[(2π − x)
(sin nx
n
)− (−1)
(−cos nx
n2
)]2ππ
=1
π
[(0 +
cos nπ
n2
)−(0 +
1
n2
)]+1
π
[(0− cos 2nπ
n2
)−(0− cos nπ
n2
)]=
2 [(−1)n − 1]
πn2
N. B. Vyas Fourier Series
Example
Step 3. an =1
π
[∫ π
0
f(x) cos nx dx+
∫ 2π
π
f(x) cos nx dx
]=
1
π
[∫ π
0
x cos nx dx+
∫ 2π
π
(2π − x) cos nx dx]
=1
π
[x
(sin nx
n
)− (1)
(−cos nx
n2
)]π0
+1
π
[(2π − x)
(sin nx
n
)− (−1)
(−cos nx
n2
)]2ππ
=1
π
[(0 +
cos nπ
n2
)−(0 +
1
n2
)]+1
π
[(0− cos 2nπ
n2
)−(0− cos nπ
n2
)]=
2 [(−1)n − 1]
πn2
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
[∫ π
0
x sin nx dx+
∫ 2π
π
(2π − x) sin nx dx]
=1
π
[x(−cos nx
n
)− (1)
(−sin nx
n2
)]π0
+1
π
[(2π − x)
(−cos nx
n
)− (−1)
(−sin nx
n2
)]2ππ
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
[∫ π
0
x sin nx dx+
∫ 2π
π
(2π − x) sin nx dx]
=1
π
[x(−cos nx
n
)− (1)
(−sin nx
n2
)]π0
+1
π
[(2π − x)
(−cos nx
n
)− (−1)
(−sin nx
n2
)]2ππ
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
[∫ π
0
x sin nx dx+
∫ 2π
π
(2π − x) sin nx dx]
=1
π
[x(−cos nx
n
)− (1)
(−sin nx
n2
)]π0
+1
π
[(2π − x)
(−cos nx
n
)− (−1)
(−sin nx
n2
)]2ππ
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
[∫ π
0
x sin nx dx+
∫ 2π
π
(2π − x) sin nx dx]
=1
π
[x(−cos nx
n
)− (1)
(−sin nx
n2
)]π0
+1
π
[(2π − x)
(−cos nx
n
)− (−1)
(−sin nx
n2
)]2ππ
= 0
N. B. Vyas Fourier Series
Example
Step 4. bn =1
π
∫ 2π
0
f(x) sin nx dx
=1
π
[∫ π
0
x sin nx dx+
∫ 2π
π
(2π − x) sin nx dx]
=1
π
[x(−cos nx
n
)− (1)
(−sin nx
n2
)]π0
+1
π
[(2π − x)
(−cos nx
n
)− (−1)
(−sin nx
n2
)]2ππ
= 0
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series of
f(x)= −1;−π < x < −π2
= 0 ;−π2 < x < π
2
= 1 ; π2 < x < π
N. B. Vyas Fourier Series