Dr. Nirav Vyas fourierseries1.pdf

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.



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Periodic Function


The smallest positive number T , for which this relation holds, iscalled the period of f (x).



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Periodic Function



If T is the period of f (x) then



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Periodic Function




f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT )



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Periodic Function




f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT )

f (x) = f (x− T ) = f (x− 2T ) = . . . = f (x− nT )



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Periodic Function




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


P F


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Periodic Function




f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT )

f (x) = f (x− T ) = f (x− 2T ) = . . . = f (x− nT )


Eg. Sinx

, Cosx

, Secx

and Cosecx

are periodic functions with period2π


P F


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Periodic Function




f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT )

f (x) = f (x− T ) = f (x− 2T ) = . . . = f (x− nT )


Eg. Sinx

, Cosx

, Secx

and Cosecx

are periodic functions with period2π

tanx and cotx are periodic functions with period π.


E & O


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Even & Odd functions

A function f (x) is said to be even if f (−x) = f (x).




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Eg. x2 and cosx are even function.




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c

−c

f (x)dx = 2 c

0

f (x)dx ; if f (x) is an even function.




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c

−c

f (x)dx = 2 c

0


A function f (x) is said to be odd if f (−x) = −f (x).




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c

−c

f (x)dx = 2 c

0



Eg. x3 and sinx are odd function.




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c

−c

f (x)dx = 2 c

0



Eg. x3 and sinx are odd function.

c

−c

f (x)dx = 0 ; if f (x) is an odd function.


Some Important Formula


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Some Important Formula eax sin bx dx =

eax

a2 + b2(asinbx − bcosbx) + c




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eax


eax cos bx dx = e

ax

a2 + b2(a cosbx + b sinbx) + c




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eax


eax cos bx dx = e

ax


c+2π

c

sin nx dx = −

cos nx

n c+2π

c

= 0, n = 0




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eax


eax cos bx dx = e

ax


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




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eax


eax cos bx dx = e

ax


c+2π

c


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




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eax


eax cos bx dx = e

ax


c+2π

c


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




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S eax sin bx dx =

eax


eax cos bx dx = e

ax


c+2π

c


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




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c+2π

c

cosmxcosnxdx = 12

c+2π

c

2cosmxcosnxdx




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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




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c+2π

c

cosmxcosnxdx = 12

c+2π

c

2cosmxcosnxdx

= 12

c+2πc


= 12

sin (m + n)x

m + n +

sin (m − n)x

m − n c+2π

c

= 0, m = n

c+2π

c

sinmxsinnxdx




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c+2π

c

cosmxcosnxdx = 12

c+2π

c

2cosmxcosnxdx

= 12

c+2πc


= 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



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c+2π

c

cosmxcosnxdx = 12

c+2π

c

2cosmxcosnxdx

= 12

c+2πc


= 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



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c+2π

c

cosmxcosnxdx = 12

c+2π

c

2cosmxcosnxdx

= 12

c+2πc


= 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


c

cos2 nx dx = π

c+2π

c

sin2 nxdx




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c+2π

c

cosmxcosnxdx = 12

c+2π

c

2cosmxcosnxdx

= 12

c+2πc


= 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


c

cos2 nx dx = π

c+2π

c

sin2 nxdx = π




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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 + . . .




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v2 + u v3 − u

v4 + . . .

Eg. x3 e−2x dx




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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




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v2 + u v3 − u

v4 + . . .


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)




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v2 + u v3 − u

v4 + . . .


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




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v2 + u v3 − u

v4 + . . .


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




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v2 + u v3 − u

v4 + . . .


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

n + 12

π = (−1)n




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v2 + u v3 − u

v4 + . . .


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

n + 12

π = (−1)n and cos

n + 1

2

π = 0




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v2 + u v3 − u

v4 + . . .


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

n + 12

π = (−1)n and cos

n + 1

2

π = 0

where n is integer.


Fourier Series


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The Fourier series for the function f (x) in the intervalc < x < c + 2π is given by


Fourier Series


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f (x) = a0

2 +

∞n=1

an cos nx +∞n=1

bn sin nx


Fourier Series


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f (x) = a0

2 +

∞n=1

an cos nx +∞n=1

bn sin nx

where a0 = 1π

c+2πc

f (x) dx


Fourier Series


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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


Fourier Series


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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


Fourier Series


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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 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


Fourier Series

S i l C 1 If th i t l i < < d f ( ) i


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f (x) = a0

2 +

∞n=1

an cos nx +∞n=1

bn sin nx

where a0 = 1

π π

−π

f (x) dx = 0


Fourier Series

S i l C 1 If th i t l i < < d f ( ) i


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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


Fourier Series

Special Case 1: If the interval is c < x < c and f (x) is an


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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


Fourier Series

Special Case 1: If the interval is c < x < c and f (x) is an


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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



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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


Fourier Series



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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


Fourier Series



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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


Fourier Series



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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


Fourier Series



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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


Fourier Series



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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


bn = 1π

π−π

f (x) sin nx dx = 0


Fourier Series



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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


bn = 1π

π−π

f (x) sin nx dx = 0

because sin nx is an odd function , f (x)sin nx is an oddfunction


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 − . . .


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 − . . .


Example


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Sol. Step 1. The Fourier series of f (x) is given by


Example


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f (x) = a0

2 +

∞n=1

(an cos nx + bn sin nx) . . . (1)


Example


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f (x) = a0

2 +

∞n=1


where a0 = 1

π 2π

0

f (x) dx


Example


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f (x) = a0

2 +

∞n=1


where a0 = 1

π 2π

0

f (x) dx

an = 1

π

2π

0

f (x) cos nx dx



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Example


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Step 2. Now a0 = 1π

2π0

f (x) dx



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Example


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2π0

f (x) dx

a0 = 1

π

2π

0

(π − x)2

4 dx

= 14π

(π − x)3

(−3)

2π

0

= − 112π

(−π3 − π3)


Example


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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


Example


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Step 3. an = 1

π

2π

0

f (x) cosnxdx


Example


83/215

Step 3. an = 1

π

2π

0

f (x) cosnxdx

an = 1

π 2π

0

(π − x)2

4 cosnxdx


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


Example


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Step 4. bn = 1

π

2π

0

f (x) sin nx dx


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


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


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]


Example

Step 5 Substituting values of a0 a and b in (1) we get the


89/215


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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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


Example

Step 5 Substituting values of a0 an and bn in (1) we get the


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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 − . . .



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Example

S l St 1 Th F i i f f ( ) i i b


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Example

S l St 1 Th F i s i s f f ( ) is gi b


97/215


f (x) = a0

2 +

∞n=1



Example

Sol Step 1 The Fourier series of f (x) is given by


98/215


f (x) = a0

2 +

∞n=1


where a0 = 1

π 2π

0

f (x) dx



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Example



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f (x) = a0

2 +

∞n=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


Example



101/215


f (x) = a0

2 +

∞n=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


Example


102/215


2π

0

f (x) dx


Example


103/215


2π

0

f (x) dx

a0 = 1

π 2π

0

x dx


Example


104/215


2π

0

f (x) dx

a0 = 1

π 2π

0

x dx

= 1

4π

x2

2

2π

0



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Example


108/215

Step 3. an = 1π

2π

0

f (x) cosnxdx

an = 1

π 2π

0

xcosnxdx

x

sin nx

n

−−

cos nx

n

2π0



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Example


112/215

Step 4. bn = 1

π

2π

0

f (x) sin nx dx

bn = 1

π 2π

0

xsinnxdx

= −2

n


Example


113/215

Step 5. Substituting values of a0, an and bn in (1), we get therequired Fourier series of f (x) in the interval [0, 2π]



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Example


115/215


f (x) = 2π

2 + 0 +

∞

n=1−2

n sin nx

= π −∞n=1

sin nx

n


Example


116/215


f (x) = 2π

2 + 0 +

∞

n=1−2

n sin nx

= π −∞n=1

sin nx

n


Example


117/215

Ex. Determine the Fourier series expansion of thefunction f (x) = xsin x in the interval 0 ≤ x ≤ 2π.



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Example



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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


where a0 = 1

π 2π

0

f (x) dx


Example

Sol. Step 1. The Fourier series of f (x) is given by∞


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f (x) = a0

2 +

n=1


where a0 = 1

π 2π

0

f (x) dx

an = 1

π

2π

0

f (x) cos nx dx



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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



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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)


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



130/215


131/215

Example

Step 3. an = 1

π 2π

0

f (x) cosnxdx

2π


132/215

an =

1

π

2π

0

xsinxcosnxdx

= 1

2π 2π

0

x (2 sinxcosnx) dx


Example

Step 3. an = 1

π 2π

0

f (x) cosnxdx

1

2π


133/215

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


Example

Step 3. an = 1

π 2π

0

f (x) cosnxdx

1

2π


134/215

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


Example

an=

1

2π 2π0

x sin(

n+ 1)

x dx − 1

2π 2π0

x sin(

n −1)

x dx


135/215

= 2π0 ( + 1) 2π

0 ( 1)


Example

an=

1

2π 2π0

x sin(

n+ 1)

x dx − 1

2π 2π0

x sin(

n −1)

x dx


136/215

= 2π0 ( + 1) 2π

0 ( 1)

If n = 1


Example

an=

1

2π 2π0

x sin(

n+ 1)

x dx − 1

2π 2π0

x sin(

n −1)

x dx


137/215

2π0 ( + 1) 2π

0 ( 1)

If n = 1

∴ a1 = 1

2π

2π

0

x sin 2x dx


Example

an = 1

2π 2π0

x sin (n + 1)x dx − 1

2π 2π0

x sin (n − 1)x dx


138/215

2π

2π

If n = 1

∴ a1 = 1

2π

2π

0

x sin 2x dx

= 12π

−x

cos 2x

2

+ sin 2x

4

2π0


Example

an = 1

2π 2π0

x sin (n + 1)x dx − 1

2π 2π0

x sin (n − 1)x dx


139/215

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



140/215

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


141/215


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


142/215

∴ an = 1

2π

x

−

cos (n + 1)x

n + 1

−

−

sin (n + 1)x

(n + 1)2

2π0


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


143/215

∴ 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


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


144/215

∴ 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


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


145/215

∴ 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


Example

Step 4. bn =

1

π 2π0 f (x) sin nx dx


146/215


Example

Step 4. bn =

1



147/215

=

1

π

2π

0

xsinxsinnxdx



148/215

Example

Step 4. bn =

1


2


149/215

=

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



150/215


151/215

Example

bn =

1

2π 2π0 x (−cos (n + 1)x + cos (n − 1)x) dx

If 1


152/215

If n = 1

∴ b1 = 1

2π 2π

0

(−x cos 2x) dx + 1

2π 2π

0

x dx


Example

bn =

1


If 1


153/215

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


Example

bn =

1


If 1


154/215

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



155/215

Example

bn =

1


If n = 1


156/215

If n = 1



157/215

Example

bn =

1


If n = 1


158/215

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



159/215

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


160/215

bn(n > 1) in (1), we get the required Fourier series of f (x) inthe interval [0, 2π]


Example



161/215


f (x) =

−2

2 +

∞n=1

(an cos nx + bn sin nx)


Example



162/215


f (x) =

−2

2 +

∞n=1



Example

Ex. Find the Fourier series for the periodic function


163/215


f (x)= −π;−π < x


164/215


165/215


166/215

Example


f (x) = a02

+

∞n=1



167/215

where a0 = 1

π

π

−π

f (x) dx


Example


f (x) = a02

+

∞n=1



168/215

where a0 = 1

π

π

−π

f (x) dx

an = 1π

π−π

f (x) cosnxdx



169/215


170/215

Example

Step 2. Now a0 = 1

π π−π

f (x) dx


171/215


Example

Step 2. Now a0 = 1

π π−π

f (x) dx

= 1

0

f (x) dx +

π

f (x) dx


172/215

=

π

−π

f (x) dx +

0

f (x) dx


Example

Step 2. Now a0 = 1

π π−π

f (x) dx

= 1

0

f (x) dx +

π

f (x) dx


173/215

=

π

−π

f (x) dx +

0

f (x) dx

=

1

π 0

−π(−π) dx + π0 x dx



174/215

Example

Step 2. Now a0 = 1

π π−π

f (x) dx

= 1

0

f (x) dx +

π

f (x) dx


175/215

=π

−π

f (x) dx +

0

f (x) dx

=

1

π 0

−π(−π) dx + π0 x dx

= −

π

2



176/215

Example

Step 3. an = 1

π 0

−π

f (x) cosnxdx + π

0

f (x) cos nx dx

= 1

0

(−π) cosnxdx +

π

xcosnxdx


177/215

π

−π

( )

0


Example

Step 3. an = 1

π 0

−π

f (x) cosnxdx + π

0

f (x) cos nx dx

= 1

0

(−π) cosnxdx +

π

xcosnxdx


178/215

π

−π

( )

0

=

1

π−π

sin nxn

0−π +


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


179/215

π

−π

0

=

1

π−π

sin nxn

0−π +

xsin nx

n− (1)

−

cos nx

n2π

0


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


180/215

π

−π

0

=

1

π−π

sin nxn

0−π +

xsin nx

n− (1)

−

cos nx

n2π

0

=

(−1)n − 1

πn2


Example

Step 4. bn = 1

π

0

−π

f (x) sinnxdx + π

0

f (x) sin nx dx


181/215


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


182/215

π

−π

0


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


183/215

π

−π

0

=

1

π−π

−cosnxn

0−π +



184/215


185/215

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)


186/215



187/215

Example


f (x) = −π

+∞

(a cos nx + b sin nx)


188/215

f (x) =4

+n=1


= −π4

+∞n=1

(−1)

n

− 1πn2

cos nx +

∞n=1

1 − 2(−1)

n

n

sin nx


Example


f (x) = −π

+∞



189/215

f (x) 4

+n=1


= −π4

+∞n=1

(−1)

n

− 1πn2

cos nx +

∞n=1

1 − 2(−1)

n

n

sin nx



190/215

Example


f (x)= 2;−π < x < 0


191/215

f (x)= 2; π < x


192/215

= k; 0 < x < π


Example

Ex. Find the Fourier series for the periodic functionf (x)= −k;−π < x


193/215

= k; 0 < x < π

Hence deduce that 1 − 13 + 15 − 17 + . . . = π4



194/215

Example



195/215



196/215


197/215


198/215

Example


f (x) = a0

2 +

∞n=1


where a0 = 1 2π

f (x) dx


199/215

0π

0

f ( )

an = 1π

π0

f (x) cosnxdx

bn = 1

π

2π

0

f (x) sinnxdx



200/215

Example

Step 2. Now a0

=

1

π 2π0

f (

x)

dx


201/215


Example

Step 2. Now a0 = 1

π 2π0

f (x) dx

= 1

π

π

0

x dx +

2π

π

(2π − x) dx


202/215



203/215

Example

Step 2. Now a0 = 1

π 2π0

f (x) dx

= 1

π

π

0

x dx +

2π

π

(2π − x) dx


204/215

= 1

πx2

2π0

+

2πx − x2

22π

π

= π



205/215

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


206/215


Example

Step 3. an = 1π

π

0


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


207/215


Example

Step 3. an = 1π

π

0


π

f (x) cosnxdx

= 1

π π

0

xcosnxdx + 2π

π


1

π

x

sin nx

n

− (1)

−

cos nx

n2

π0

+1

(2 )sin nx

( 1)cos nx

2π


208/215

+π (2π − x) n − (−1)− n2 π


Example

Step 3. an = 1π

π

0


π

f (x) cosnxdx

= 1

π π

0

xcosnxdx + 2π

π


1

π

x

sin nx

n

− (1)

−

cos nx

n2

π0

+1

(2 )sin nx

( 1)cos nx

2π


209/215

+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


Example

Step 4. bn = 1

π 2π

0

f (x) sin nx dx


210/215



211/215

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


212/215

=π x n (1) n2 0



213/215


214/215

Example

Ex. Find the Fourier series of f (x)= −1;−π < x < −π2

= 0 ;−π2 < x < π

2


215/215

2 2

= 1 ; π

2 < x < π


Dr. Nirav Vyas fourierseries1.pdf

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