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8/20/2019 Dr. Nirav Vyas fourierseries1.pdf
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Fourier Series
N. B. Vyas
Department of Mathematics,Atmiya Institute of Tech. and Science,
Rajkot (Guj.) - INDIA
N. B. Vyas Fourier Series
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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.
N. B. Vyas Fourier Series
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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).
N. B. Vyas Fourier Series
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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
N. B. Vyas Fourier Series
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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 )
N. B. Vyas Fourier Series
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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 )
N. B. Vyas Fourier Series
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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
N. B. Vyas Fourier Series
P F
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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
N. B. Vyas Fourier Series
P F
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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
E & O
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Even & Odd functions
A function f (x) is said to be even if f (−x) = f (x).
N. B. Vyas Fourier Series
Even & Odd functions
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Even & Odd functions
A function f (x) is said to be even if f (−x) = f (x).
Eg. x2 and cosx are even function.
N. B. Vyas Fourier Series
Even & Odd functions
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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
−c
f (x)dx = 2 c
0
f (x)dx ; if f (x) is an even function.
N. B. Vyas Fourier Series
Even & Odd functions
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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
−c
f (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).
N. B. Vyas Fourier Series
Even & Odd functions
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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
−c
f (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.
N. B. Vyas Fourier Series
Even & Odd functions
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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
−c
f (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
−c
f (x)dx = 0 ; if f (x) is an odd function.
N. B. Vyas Fourier Series
Some Important Formula
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Some Important Formula eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
N. B. Vyas Fourier Series
Some Important Formula
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Some Important Formula eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
eax cos bx dx = e
ax
a2 + b2(a cosbx + b sinbx) + c
N. B. Vyas Fourier Series
Some Important Formula
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Some Important Formula eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
eax cos bx dx = e
ax
a2 + b2(a cosbx + b sinbx) + c
c+2π
c
sin nx dx = −
cos nx
n c+2π
c
= 0, n = 0
N. B. Vyas Fourier Series
Some Important Formula
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Some Important Formula eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
eax cos bx dx = e
ax
a2 + b2(a cosbx + b sinbx) + c
c+2π
c
sin nx dx = −cos nx
n
c+2π
c
= 0, n = 0
c+2πc
cos nx dx =
sin nxn
c+2πc
= 0, n = 0
N. B. Vyas Fourier Series
Some Important Formula
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Some Important Formula eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
eax cos bx dx = e
ax
a2 + b2(a cosbx + b sinbx) + c
c+2π
c
sin nx dx = −cos nx
n
c+2π
c
= 0, n = 0
c+2πc
cos nx dx =
sin nxn
c+2πc
= 0, n = 0
c+2π
c
sinmxcosnxdx = 1
2 c+2π
c
2sinmxcosnxdx
N. B. Vyas Fourier Series
Some Important Formula
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Some Important Formula eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
eax cos bx dx = e
ax
a2 + b2(a cosbx + b sinbx) + c
c+2π
c
sin nx dx = −cos nx
n
c+2π
c
= 0, n = 0
c+2πc
cos nx dx =
sin nxn
c+2πc
= 0, n = 0
c+2π
c
sinmxcosnxdx = 1
2 c+2π
c
2sinmxcosnxdx
= 12
c+2π
c
[sin (m + n)x + sin (m − n)x] dx
N. B. Vyas Fourier Series
Some Important Formula
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S eax sin bx dx =
eax
a2 + b2(asinbx − bcosbx) + c
eax cos bx dx = e
ax
a2 + b2(a cosbx + b sinbx) + c
c+2π
c
sin nx dx = −cos nx
n
c+2π
c
= 0, n = 0
c+2πc
cos nx dx =
sin nxn
c+2πc
= 0, n = 0
c+2π
c
sinmxcosnxdx = 1
2 c+2π
c
2sinmxcosnxdx
= 12
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 = 0
N. B. Vyas Fourier Series
Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
N. B. Vyas Fourier Series
Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
= 12
c+2
π
c
[cos (m + n)x + cos (m − n)x] dx
N. B. Vyas Fourier Series
Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
= 12
c+2πc
[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 = n
c+2π
c
sinmxsinnxdx
N. B. Vyas Fourier Series
Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
= 12
c+2πc
[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 = n
c+2π
c
sinmxsinnxdx = 0
c+2π
c
sinnxcosnxdx
N. B. Vyas Fourier Series
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Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
= 12
c+2πc
[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 = n
c+2π
c
sinmxsinnxdx = 0
c+2π
c
sinnxcosnxdx = 0, n = 0 c+2π
c
cos2 nx dx
N. B. Vyas Fourier Series
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Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
= 12
c+2πc
[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 = n
c+2π
c
sinmxsinnxdx = 0
c+2π
c
sinnxcosnxdx = 0, n = 0 c+2π
c
cos2 nx dx = π
c+2π
c
sin2 nxdx
N. B. Vyas Fourier Series
Some Important Formula
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c+2π
c
cosmxcosnxdx = 12
c+2π
c
2cosmxcosnxdx
= 12
c+2πc
[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 = n
c+2π
c
sinmxsinnxdx = 0
c+2π
c
sinnxcosnxdx = 0, n = 0 c+2π
c
cos2 nx dx = π
c+2π
c
sin2 nxdx = π
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = u v1 − u
v2 + u v3 − u
v4 + . . .
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = u v1 − u
v2 + u v3 − u
v4 + . . .
Eg. x3 e−2x dx
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
8 e−2x (4x3 + 6x2 + 6x + 3)
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
8
e−2x (4x3 + 6x2 + 6x + 3)
sin nπ = 0
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
8
e−2x (4x3 + 6x2 + 6x + 3)
sin nπ = 0 and cos nπ = (−1)n
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
8
e−2x (4x3 + 6x2 + 6x + 3)
sin nπ = 0 and cos nπ = (−1)n
sin
n + 12
π = (−1)n
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
8
e−2x (4x3 + 6x2 + 6x + 3)
sin nπ = 0 and cos nπ = (−1)n
sin
n + 12
π = (−1)n and cos
n + 1
2
π = 0
N. B. Vyas Fourier Series
Some Important Formula
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(Leibnitz’s Rule)To integrate the product of twofunctions, one of which is power of x . We apply the
generalized rule of integration by parts uvdx = 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
8
e−2x (4x3 + 6x2 + 6x + 3)
sin nπ = 0 and cos nπ = (−1)n
sin
n + 12
π = (−1)n and cos
n + 1
2
π = 0
where n is integer.
N. B. Vyas Fourier Series
Fourier Series
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The Fourier series for the function f (x) in the intervalc < x < c + 2π is given by
N. B. Vyas Fourier Series
Fourier Series
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The Fourier series for the function f (x) in the intervalc < x < c + 2π is given by
f (x) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
N. B. Vyas Fourier Series
Fourier Series
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The Fourier series for the function f (x) in the intervalc < x < c + 2π is given by
f (x) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1π
c+2πc
f (x) dx
N. B. Vyas Fourier Series
Fourier Series
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The Fourier series for the function f (x) in the intervalc < x < c + 2π is given by
f (x) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1π
c+2πc
f (x) dx
an = 1
π
c+2π
c
f (x) cosnxdx
N. B. Vyas Fourier Series
Fourier Series
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The Fourier series for the function f (x) in the intervalc < x < c + 2π is given by
f (x) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1π
c+2πc
f (x) dx
an = 1
π
c+2π
c
f (x) cosnxdx
bn = 1π
c+2πc
f (x) sin nx dx
N. B. Vyas Fourier Series
Fourier Series
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Corollary 1: If c = 0, the interval becomes 0 < x
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Corollary 1: If c = 0, the interval becomes 0 < x
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Fourier Series
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Corollary 1: If c = 0, the interval becomes 0 < x
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Corollary 1: If c = 0, the interval becomes 0 < x
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Corollary 2: If c = −π, the interval becomes −π
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Corollary 2: If c = −π, the interval becomes −π
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Corollary 2: If c = −π, the interval becomes −π
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Corollary 2: If c = −π, the interval becomes −π
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Corollary 2: If c = −π, the interval becomes −π
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Fourier Series
S i l C 1 If th i t l i d f ( ) i
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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) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
N. B. Vyas Fourier Series
Fourier Series
S i l C 1 If th i t l i < < d f ( ) i
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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) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1
π π
−π
f (x) dx = 0
N. B. Vyas Fourier Series
Fourier Series
S i l C 1 If th i t l i < < d f ( ) i
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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) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1
π π
−π
f (x) dx = 0
an = 1
π
π
−π
f (x) cosnxdx = 0
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is c < x < c and f (x) is an
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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) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1
π π
−π
f (x) dx = 0
an = 1
π
π
−π
f (x) cosnxdx = 0
because cos nx is an even function , f (x)cos nx is an oddfunction
N. B. Vyas Fourier Series
Fourier Series
Special Case 1: If the interval is c < x < c and f (x) is an
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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) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
where a0 = 1
π π
−π
f (x) dx = 0
an = 1
π
π
−π
f (x) cosnxdx = 0
because cos nx is an even function , f (x)cos nx is an oddfunction
bn = 1π
π−π
f (x) sin nx dx = 2π
π0
f (x) sin nx dx
N. B. Vyas Fourier Series
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Fourier Series
Special Case 2: If the interval is −c < x < c and f (x) is an
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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
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f (x) is an
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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) = a0
2 +
∞n=1
an cos nx +∞n=1
bn sin nx
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f (x) is an
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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
−π
f (x) dx = 2
π π
0
f (x) dx
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f (x) is an
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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
−π
f (x) dx = 2
π π
0
f (x) dx
an = 1π
π
−π
f (x) cosnxdx = 2π
π
0
f (x) cosnxdx
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f (x) is an
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Spec a Case t e te va s c < x < c a d f (x) s aeven 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
−π
f (x) dx = 2
π π
0
f (x) dx
an = 1π
π
−π
f (x) cosnxdx = 2π
π
0
f (x) cosnxdx
because cos nx is an even function , f (x)cos 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 an
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p f ( )even 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
−π
f (x) dx = 2
π π
0
f (x) dx
an = 1π
π
−π
f (x) cosnxdx = 2π
π
0
f (x) cosnxdx
because cos nx is an even function , f (x)cos nx is an evenfunction
bn = 1π
π−π
f (x) sin nx dx = 0
N. B. Vyas Fourier Series
Fourier Series
Special Case 2: If the interval is −c < x < c and f (x) is an
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p f ( )even 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
−π
f (x) dx = 2
π π
0
f (x) dx
an = 1π
π
−π
f (x) cosnxdx = 2π
π
0
f (x) cosnxdx
because cos nx is an even function , f (x)cos nx is an evenfunction
bn = 1π
π−π
f (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
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Ex. Obtain Fourier series of f (x) =
π − x
2
2in the
interval 0 ≤ x ≤ 2π. Hence deduce thatπ2
12 =
1
12 −
1
22 +
1
32 − . . .
N. B. Vyas Fourier Series
Example
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Ex. Obtain Fourier series of f (x) =
π − x
2
2in the
interval 0 ≤ x ≤ 2π. Hence deduce thatπ2
12 =
1
12 −
1
22 +
1
32 − . . .
N. B. Vyas Fourier Series
Example
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Sol. Step 1. The Fourier series of f (x) is given by
N. B. Vyas Fourier Series
Example
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞n=1
(an cos nx + bn sin nx) . . . (1)
N. B. Vyas Fourier Series
Example
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞n=1
(an cos nx + bn sin nx) . . . (1)
where a0 = 1
π 2π
0
f (x) dx
N. B. Vyas Fourier Series
Example
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞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
N. B. Vyas Fourier Series
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Example
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Step 2. Now a0 = 1π
2π0
f (x) dx
N. B. Vyas Fourier Series
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Example
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Step 2. Now a0 = 1π
2π0
f (x) dx
a0 = 1
π
2π
0
(π − x)2
4 dx
= 14π
(π − x)3
(−3)
2π
0
= − 112π
(−π3 − π3)
N. B. Vyas Fourier Series
Example
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Step 2. Now a0 = 1π
2π0
f (x) dx
a0 = 1
π
2π
0
(π − x)2
4 dx
= 14π
(π − x)3
(−3)
2π
0
= − 112π
(−π3 − π3)
= π2
6
N. B. Vyas Fourier Series
Example
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Step 3. an = 1
π
2π
0
f (x) cosnxdx
N. B. Vyas Fourier Series
Example
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Step 3. an = 1
π
2π
0
f (x) cosnxdx
an = 1
π 2π
0
(π − x)2
4 cosnxdx
N. B. Vyas Fourier Series
Example
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Step 3. an = 1
π
2π
0
f (x) cosnxdx
an = 1
π 2π
0
(π − x)2
4
cosnxdx
= 1
n2
N. B. Vyas Fourier Series
Example
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Step 4. bn = 1
π
2π
0
f (x) sin nx dx
N. B. Vyas Fourier Series
Example
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Step 4. bn = 1
π
2π
0
f (x) sin nx dx
bn =
1
π 2π0
(π − x)2
4 sin nx dx
N. B. Vyas Fourier Series
Example
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Step 4. bn = 1
π
2π
0
f (x) sin nx dx
bn =
1
π 2π0
(π − x)2
4 sin nx dx= 0
N. B. Vyas Fourier Series
Example
Step 5 Substituting values of a a and b in (1) we get the
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Step 5. Substituting values of a0, an and bn in (1), we get the
required Fourier series of f (x) in the interval [0, 2π]
N. B. Vyas Fourier Series
Example
Step 5 Substituting values of a0 a and b in (1) we get the
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Step 5. Substituting values of a0, an and bn in (1), we get the
required Fourier series of f (x) in the interval [0, 2π]π − x
2
2=
π2
12 +
∞n=1
1
n2 cos nx
N B Vyas Fourier Series
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Example
Step 5 Substituting values of a0 a and b in (1) we get the
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Step 5. Substituting values of a0, an and bn in (1), we get the
required Fourier series of f (x) in the interval [0, 2π]π − x
2
2=
π2
12 +
∞n=1
1
n2 cos nx
=
π2
12 +
1
12 cos x +
1
22 cos 2x +
1
32 cos 3x + . . .Putting x = π, we get
N B Vyas Fourier Series
Example
Step 5 Substituting values of a0 an and bn in (1) we get the
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Step 5. Substituting values of a0, an and bn in (1), we get the
required Fourier series of f (x) in the interval [0, 2π]π − x
2
2=
π2
12 +
∞n=1
1
n2 cos nx
=
π2
12 +
1
12 cos x +
1
22 cos 2x +
1
32 cos 3x + . . .Putting x = π, we get
0 = π2
12 −
1
12 +
1
22 −
1
32 +
1
42 − . . .
N B Vyas Fourier Series
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Example
S l St 1 Th F i i f f ( ) i i b
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Sol. Step 1. The Fourier series of f (x) is given by
N B Vyas Fourier Series
Example
S l St 1 Th F i s i s f f ( ) is gi b
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞n=1
(an cos nx + bn sin nx) . . . (1)
N B Vyas Fourier Series
Example
Sol Step 1 The Fourier series of f (x) is given by
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞n=1
(an cos nx + bn sin nx) . . . (1)
where a0 = 1
π 2π
0
f (x) dx
N B Vyas Fourier Series
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Example
Sol Step 1 The Fourier series of f (x) is given by
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞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) sinnxdx
N. B. Vyas Fourier Series
Example
Sol Step 1 The Fourier series of f (x) is given by
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Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞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) sinnxdx
N. B. Vyas Fourier Series
Example
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Step 2. Now a0 = 1π
2π
0
f (x) dx
N. B. Vyas Fourier Series
Example
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Step 2. Now a0 = 1π
2π
0
f (x) dx
a0 = 1
π 2π
0
x dx
N. B. Vyas Fourier Series
Example
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Step 2. Now a0 = 1π
2π
0
f (x) dx
a0 = 1
π 2π
0
x dx
= 1
4π
x2
2
2π
0
N. B. Vyas Fourier Series
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Example
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Step 3. an = 1π
2π
0
f (x) cosnxdx
an = 1
π 2π
0
xcosnxdx
x
sin nx
n
−−
cos nx
n
2π0
N. B. Vyas Fourier Series
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Example
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Step 4. bn = 1
π
2π
0
f (x) sin nx dx
bn = 1
π 2π
0
xsinnxdx
= −2
n
N. B. Vyas Fourier Series
Example
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Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f (x) in the interval [0, 2π]
N. B. Vyas Fourier Series
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Example
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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=1−2
n sin nx
= π −∞n=1
sin nx
n
N. B. Vyas Fourier Series
Example
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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=1−2
n sin nx
= π −∞n=1
sin nx
n
N. B. Vyas Fourier Series
Example
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Ex. Determine the Fourier series expansion of thefunction f (x) = xsin x in the interval 0 ≤ x ≤ 2π.
N. B. Vyas Fourier Series
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Example
Sol. Step 1. The Fourier series of f (x) is given by
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N. B. Vyas Fourier Series
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Example
Sol. Step 1. The Fourier series of f (x) is given by∞
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f (x) = a0
2 +
∞n=1
(an cos nx + bn sin nx) . . . (1)
where a0 = 1
π 2π
0
f (x) dx
N. B. Vyas Fourier Series
Example
Sol. Step 1. The Fourier series of f (x) is given by∞
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f (x) = a0
2 +
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
N. B. Vyas Fourier Series
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Example
Step 2. Now a0 = 1
2π
0
f (x) dx
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π0
a0 = 1
π
2π
0
xsinxdx
N. B. Vyas Fourier Series
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Example
Step 2. Now a0 = 1
2π
0
f (x) dx
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π0
a0 = 1
π
2π
0
xsinxdx
= 1
π [−xcosx + sin x]2π
0
= 1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
N. B. Vyas Fourier Series
Example
Step 2. Now a0 = 1
π 2π
0
f (x) dx
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π0
a0 = 1
π
2π
0
xsinxdx
= 1
π
[−xcosx + sin x]2π0
= 1
π(−2 π cos 2π + sin 2π − 0 + sin 0)
a0 = −2π
π = −2
N. B. Vyas Fourier Series
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Example
Step 3. an = 1
π 2π
0
f (x) cosnxdx
2π
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an =
1
π
2π
0
xsinxcosnxdx
= 1
2π 2π
0
x (2 sinxcosnx) dx
N. B. Vyas Fourier Series
Example
Step 3. an = 1
π 2π
0
f (x) cosnxdx
1
2π
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an =
1
π
2π
0
xsinxcosnxdx
= 1
2π 2π
0
x (2 sinxcosnx) dx
= 1
2π
2π
0
x (sin (n + 1)x − sin (n − 1)x) dx
N. B. Vyas Fourier Series
Example
Step 3. an = 1
π 2π
0
f (x) cosnxdx
1
2π
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an =
1
π
2π
0
xsinxcosnxdx
= 1
2π 2π
0
x (2 sinxcosnx) dx
= 1
2π
2π
0
x (sin (n + 1)x − sin (n − 1)x) dx
= 1
2π 2π
0
xsin (n + 1)x dx − 1
2π 2π
0
xsin (n − 1)x dx
N. B. Vyas Fourier Series
Example
an=
1
2π 2π0
x sin(
n+ 1)
x dx − 1
2π 2π0
x sin(
n −1)
x dx
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= 2π0 ( + 1) 2π
0 ( 1)
N. B. Vyas Fourier Series
Example
an=
1
2π 2π0
x sin(
n+ 1)
x dx − 1
2π 2π0
x sin(
n −1)
x dx
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= 2π0 ( + 1) 2π
0 ( 1)
If n = 1
N. B. Vyas Fourier Series
Example
an=
1
2π 2π0
x sin(
n+ 1)
x dx − 1
2π 2π0
x sin(
n −1)
x dx
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2π0 ( + 1) 2π
0 ( 1)
If n = 1
∴ a1 = 1
2π
2π
0
x sin 2x dx
N. B. Vyas Fourier Series
Example
an = 1
2π 2π0
x sin (n + 1)x dx − 1
2π 2π0
x sin (n − 1)x dx
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2π
2π
If n = 1
∴ a1 = 1
2π
2π
0
x sin 2x dx
= 12π
−x
cos 2x
2
+ sin 2x
4
2π0
N. B. Vyas Fourier Series
Example
an = 1
2π 2π0
x sin (n + 1)x dx − 1
2π 2π0
x sin (n − 1)x dx
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2π
2π
If n = 1
∴ a1 = 1
2π
2π
0
x sin 2x dx
= 12π
−x
cos 2x
2
+ sin 2x
4
2π0
= −1
2
N. B. Vyas Fourier Series
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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
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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
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∴ an = 1
2π
x
−
cos (n + 1)x
n + 1
−
−
sin (n + 1)x
(n + 1)2
2π0
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 = 12
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∴ 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)22π
0
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 = 12
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∴ 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)22π
0
= 1
2π
−
2π
n + 1 + 0 +
2π
n − 1 − 0
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 = 12π
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∴ 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)22π
0
= 1
2π
−
2π
n + 1 + 0 +
2π
n − 1 − 0
=
2
n2 − 1
N. B. Vyas Fourier Series
Example
Step 4. bn =
1
π 2π0 f (x) sin nx dx
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N. B. Vyas Fourier Series
Example
Step 4. bn =
1
π 2π0 f (x) sin nx dx
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=
1
π
2π
0
xsinxsinnxdx
N. B. Vyas Fourier Series
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Example
Step 4. bn =
1
π 2π0 f (x) sin nx dx
2
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=
1
π
2π
0
xsinxsinnxdx
=
1
2π 2π0
x (2sinnxsinx) dx
= 1
2π
2π
0
x (−cos (n + 1)x + cos (n − 1)x) dx
N. B. Vyas Fourier Series
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Example
bn =
1
2π 2π0 x (−cos (n + 1)x + cos (n − 1)x) dx
If 1
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If n = 1
∴ b1 = 1
2π 2π
0
(−x cos 2x) dx + 1
2π 2π
0
x dx
N. B. Vyas Fourier Series
Example
bn =
1
2π 2π0 x (−cos (n + 1)x + cos (n − 1)x) dx
If 1
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If n = 1
∴ b1 = 1
2π 2π
0
(−x cos 2x) dx + 1
2π 2π
0
x dx
= 12π
x
−
sin 2x
2
−
cos 2x
4 +
x2
2
2π
0
N. B. Vyas Fourier Series
Example
bn =
1
2π 2π0 x (−cos (n + 1)x + cos (n − 1)x) dx
If 1
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If n = 1
∴ b1 = 1
2π 2π
0
(−x cos 2x) dx + 1
2π 2π
0
x dx
= 12π
x
−
sin 2x
2
−
cos 2x
4 +
x2
2
2π
0
b1 = π
N. B. Vyas Fourier Series
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Example
bn =
1
2π 2π0 x (−cos (n + 1)x + cos (n − 1)x) dx
If n = 1
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If n = 1
N. B. Vyas Fourier Series
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Example
bn =
1
2π 2π0 x (−cos (n + 1)x + cos (n − 1)x) dx
If n = 1
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If n = 1
∴ bn = 1
2π x−sin (n + 1)x
n + 1 − cos (n + 1)x
(n + 1)2 2π
0
+ 1
2π
x
sin (n − 1)x
n − 1
+
cos (n − 1)x
(n − 1)2
2π
0
N. B. Vyas Fourier Series
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Example
Step 5. Substituting values of a0, a1, an(n > 1), b1 andb (n > 1) in (1) we get the required Fourier series of f (x) in
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bn(n > 1) in (1), we get the required Fourier series of f (x) inthe interval [0, 2π]
N. B. Vyas Fourier Series
Example
Step 5. Substituting values of a0, a1, an(n > 1), b1 andb (n > 1) in (1) we get the required Fourier series of f (x) in
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bn(n > 1) in (1), we get the required Fourier series of f (x) inthe interval [0, 2π]
f (x) =
−2
2 +
∞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 andb (n > 1) in (1) we get the required Fourier series of f (x) in
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bn(n > 1) in (1), we get the required Fourier series of f (x) inthe interval [0, 2π]
f (x) =
−2
2 +
∞n=1
(an cos nx + bn sin nx)
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic function
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Ex. Find the Fourier series for the periodic function
f (x)= −π;−π < x
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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)
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where a0 = 1
π
π
−π
f (x) 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)
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where a0 = 1
π
π
−π
f (x) dx
an = 1π
π−π
f (x) cosnxdx
N. B. Vyas Fourier Series
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Example
Step 2. Now a0 = 1
π π−π
f (x) dx
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N. B. Vyas Fourier Series
Example
Step 2. Now a0 = 1
π π−π
f (x) dx
= 1
0
f (x) dx +
π
f (x) dx
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=
π
−π
f (x) dx +
0
f (x) dx
N. B. Vyas Fourier Series
Example
Step 2. Now a0 = 1
π π−π
f (x) dx
= 1
0
f (x) dx +
π
f (x) dx
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=
π
−π
f (x) dx +
0
f (x) dx
=
1
π 0
−π(−π) dx + π0 x dx
N. B. Vyas Fourier Series
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Example
Step 2. Now a0 = 1
π π−π
f (x) dx
= 1
0
f (x) dx +
π
f (x) dx
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=π
−π
f (x) dx +
0
f (x) dx
=
1
π 0
−π(−π) dx + π0 x dx
= −
π
2
N. B. Vyas Fourier Series
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Example
Step 3. an = 1
π 0
−π
f (x) cosnxdx + π
0
f (x) cos nx dx
= 1
0
(−π) cosnxdx +
π
xcosnxdx
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π
−π
( )
0
N. B. Vyas Fourier Series
Example
Step 3. an = 1
π 0
−π
f (x) cosnxdx + π
0
f (x) cos nx dx
= 1
0
(−π) cosnxdx +
π
xcosnxdx
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π
−π
( )
0
=
1
π−π
sin nxn
0−π +
N. B. Vyas Fourier Series
Example
Step 3. an = 1
π 0
−π
f (x) cosnxdx + π
0
f (x) cos nx dx
= 1
π
0
(−π) cosnxdx +
π
xcosnxdx
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π
−π
0
=
1
π−π
sin nxn
0−π +
xsin nx
n− (1)
−
cos nx
n2π
0
N. B. Vyas Fourier Series
Example
Step 3. an = 1
π 0
−π
f (x) cosnxdx + π
0
f (x) cos nx dx
= 1
π
0
(−π) cosnxdx +
π
xcosnxdx
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π
−π
0
=
1
π−π
sin nxn
0−π +
xsin nx
n− (1)
−
cos nx
n2π
0
=
(−1)n − 1
πn2
N. B. Vyas Fourier Series
Example
Step 4. bn = 1
π
0
−π
f (x) sinnxdx + π
0
f (x) sin nx dx
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N. B. Vyas Fourier Series
Example
Step 4. bn = 1
π
0
−π
f (x) sinnxdx + π
0
f (x) sin nx dx=
1
π
0
(−π) sinnxdx +
π
0
xsinnxdx
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π
−π
0
N. B. Vyas Fourier Series
Example
Step 4. bn = 1
π
0
−π
f (x) sinnxdx + π
0
f (x) sin nx dx=
1
π
0
(−π) sinnxdx +
π
0
xsinnxdx
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π
−π
0
=
1
π−π
−cosnxn
0−π +
N. B. Vyas Fourier Series
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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)
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N. B. Vyas Fourier Series
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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) = −π
+∞
(a cos nx + b sin nx)
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f (x) =4
+n=1
(an cos nx + bn sin nx)
= −π4
+∞n=1
(−1)
n
− 1πn2
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) = −π
+∞
(an cos nx + bn sin nx)
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f (x) 4
+n=1
(an cos nx + bn sin nx)
= −π4
+∞n=1
(−1)
n
− 1πn2
cos nx +
∞n=1
1 − 2(−1)
n
n
sin nx
N. B. Vyas Fourier Series
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Example
Ex. Find the Fourier series for the periodic function
f (x)= 2;−π < x < 0
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f (x)= 2; π < x
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= k; 0 < x < π
N. B. Vyas Fourier Series
Example
Ex. Find the Fourier series for the periodic functionf (x)= −k;−π < x
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= k; 0 < x < π
Hence deduce that 1 − 13 + 15 − 17 + . . . = π4
N. B. Vyas Fourier Series
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Example
Sol. Step 1. The Fourier series of f (x) is given by
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N. B. Vyas Fourier Series
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Example
Sol. Step 1. The Fourier series of f (x) is given by
f (x) = a0
2 +
∞n=1
(an cos nx + bn sin nx) . . . (1)
where a0 = 1 2π
f (x) dx
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0π
0
f ( )
an = 1π
π0
f (x) cosnxdx
bn = 1
π
2π
0
f (x) sinnxdx
N. B. Vyas Fourier Series
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Example
Step 2. Now a0
=
1
π 2π0
f (
x)
dx
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N. B. Vyas Fourier Series
Example
Step 2. Now a0 = 1
π 2π0
f (x) dx
= 1
π
π
0
x dx +
2π
π
(2π − x) dx
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N. B. Vyas Fourier Series
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Example
Step 2. Now a0 = 1
π 2π0
f (x) dx
= 1
π
π
0
x dx +
2π
π
(2π − x) dx
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= 1
πx2
2π0
+
2πx − x2
22π
π
= π
N. B. Vyas Fourier Series
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Example
Step 3. an = 1π
π
0
f (x) cos nx dx + 2π
π
f (x) cosnxdx
= 1
π π
0
xcosnxdx + 2π
π
(2π − x) cosnxdx
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N. B. Vyas Fourier Series
Example
Step 3. an = 1π
π
0
f (x) cos nx dx + 2π
π
f (x) cosnxdx
= 1
π π
0
xcosnxdx + 2π
π
(2π − x) cosnxdx=
1
π
x
sin nx
n
− (1)
−
cos nx
n2
π0
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N. B. Vyas Fourier Series
Example
Step 3. an = 1π
π
0
f (x) cos nx dx + 2π
π
f (x) cosnxdx
= 1
π π
0
xcosnxdx + 2π
π
(2π − x) cosnxdx=
1
π
x
sin nx
n
− (1)
−
cos nx
n2
π0
+1
(2 )sin nx
( 1)cos nx
2π
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+π (2π − x) n − (−1)− n2 π
N. B. Vyas Fourier Series
Example
Step 3. an = 1π
π
0
f (x) cos nx dx + 2π
π
f (x) cosnxdx
= 1
π π
0
xcosnxdx + 2π
π
(2π − x) cosnxdx=
1
π
x
sin nx
n
− (1)
−
cos nx
n2
π0
+1
(2 )sin nx
( 1)cos nx
2π
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+π (2π − x) n − (−1)− n2 π
= 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
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N. B. Vyas Fourier Series
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Example
Step 4. bn = 1
π 2π
0
f (x) sin nx dx
= 1π
π
0
xsinnxdx +
2π
π
(2π − x) sinnxdx
= 1
x −cos nx
− (1) −sin nx
π
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=π x n (1) n2 0
N. B. Vyas Fourier Series
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Example
Ex. Find the Fourier series of f (x)= −1;−π < x < −π2
= 0 ;−π2 < x < π
2
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2 2
= 1 ; π
2 < x < π
N. B. Vyas Fourier Series