Dr. Nirav Vyas fourierseries2.pdf

8/20/2019 Dr. Nirav Vyas fourierseries2.pdf

1/96

Fourier Series 2

N. B. Vyas

Department of Mathematics,Atmiya Institute of Tech. & Science,

Rajkot (Guj.)- INDIA

N. B. Vyas Fourier Series 2


2/96

Functions of any Period p = 2L

Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2L



3/96


Let f (x) be a periodic function with an arbitrary period 2 Ldened in the interval c < x < c + 2LThe Fourier series of f (x ) is given by



4/96


5/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

c +2 L

c

f (x ) dx



6/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

c +2 L

c

f (x ) dx

a n = 1

L c +2 L

c

f (x ) cosnπx

L dx



7/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

c +2 L

c

f (x ) dx

a n = 1

L c +2 L

c

f (x ) cosnπx

L dx

bn = 1L

c +2 L

c

f (x ) sinnπx

L dx



8/96


Corollary 1: If c = 0 the interval becomes 0 < x < 2L


f d


9/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL


F i f P i d 2L


10/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L 2 L

0f (x) dx


F ti f P i d 2L


11/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L 2 L

0f (x) dx

a n = 1L

2 L

0f (x ) cos

nπxL

dx


F ti f P i d 2L


12/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

2 L

0f (x) dx

a n = 1L

2 L

0f (x ) cos

nπxL

dx

bn = 1L 2 L

0f (x ) sin nπxL dx


Functions of any Period 2L


13/96


Corollary 2: If c = − L the interval becomes − L < x < L




14/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL




15/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

L

− Lf (x) dx




16/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

L

− Lf (x) dx

a n = 1L

L

− L

f (x) cosnπx

L dx




17/96



f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1L

L

− Lf (x) dx

a n = 1L

L

− L

f (x) cosnπx

L dx

bn = 1L L

− Lf (x ) sin nπxL dx


Example


18/96

Example

Ex. The Fourier series of f (x ) = x2, 0 < x < 2 wheref (x + 2) = f (x ).


Example


19/96

Example

Ex. The Fourier series of f (x ) = x2, 0 < x < 2 wheref (x + 2) = f (x ).

Hence deduce that 1 −122

+ 132

−142

+ . . . = π2

12


Example


20/96

p

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


Example


21/96

p


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL


Example


22/96

p


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1

L 2

0

f (x ) dx


Example


23/96

p


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1

L 2

0

f (x ) dx

a n = 1L

2

0f (x ) cos

nπxL

dx


Example


24/96


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1

L 2

0

f (x ) dx

a n = 1L

2

0f (x ) cos

nπxL

dx

bn = 1

L 2

0f (x ) sin

nπx

L dx


Example


25/96


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin

nπxL

where a0 = 1

L 2

0

f (x ) dx

a n = 1L

2

0f (x ) cos

nπxL

dx

bn = 1

L 2

0f (x ) sin

nπx

L dx

Here p = 2L = 2 ⇒ L = 1


Example


26/96

Step 2. Now a0 = 11

2

0f (x ) dx


Example


27/96

Step 2. Now a0 = 11

2

0f (x ) dx

a 0 =

2

0

(x)2 dx


Example


28/96

Step 2. Now a0 = 11

2

0f (x ) dx

a 0 =

2

0

(x)2 dx

=x 3

3

2

0


Example


29/96

Step 2. Now a0 = 11

2

0f (x ) dx

a 0 =

2

0

(x)2 dx

=x 3

3

2

0

= 8

3


Example


30/96

Step 3. an = 11

2

0f (x ) cos

nπx1

dx


Example


31/96

Step 3. an = 11

2

0f (x ) cos

nπx1

dx

a n =

2

0x 2 cos (nπx ) dx


Example


32/96

Step 3. an = 11

2

0f (x ) cos

nπx1

dx

a n =

2

0x 2 cos (nπx ) dx

= x 2sin nπx

nπ− (2x) −

cos nπxn 2 π 2

+ 2 −sin nπx

n 3 π 3

2

0


Example


33/96

Step 3. an = 11

2

0f (x ) cos

nπx1

dx

a n = 2

0x 2 cos (nπx ) dx

= x 2sin nπx

nπ− (2x) −

cos nπxn 2 π 2

+ 2 −sin nπx

n 3 π 3

2

0

= 4

n2

π2


Example


34/96

Step 4. bn = 11

2

0f (x ) sin

nπx1

dx


Example


35/96

Step 4. bn = 11

2

0f (x ) sin

nπx1

dx

bn = 2

0x 2 sin (nπx ) dx


Example


36/96

Step 4. bn = 11

2

0f (x ) sin

nπx1

dx

bn = 2

0x 2 sin (nπx ) dx

= x 2 −cos nπx

nπ− (2x ) −

sin nπxn 2 π 2

+ 2cos nπx

n 3 π 3

2

0


Example


37/96

Step 4. bn = 11

2

0f (x ) sin

nπx1

dx

bn = 2

0x 2 sin (nπx ) dx

= x 2 −cos nπx

nπ− (2x ) −

sin nπxn 2 π 2

+ 2cos nπx

n 3 π 3

2

0

= −4

nπ


Example


38/96

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


Example


39/96


f (x ) = 86

+∞

n =1

4n 2 π 2

cosnπx

1+

∞

n =1

−4

nπsin

nπx1


Example


40/96


f (x ) = 86

+∞

n =1

4n 2 π 2

cosnπx

1+

∞

n =1

−4

nπsin

nπx1

= 4

3 +

4

π2

∞

n =1

cos (nπx )

n2

−4

π

∞

n =1

sin (nπx )

n


Example


41/96


f (x ) = 86

+∞

n =1

4n 2 π 2

cosnπx

1+

∞

n =1

−4

nπsin

nπx1

= 4

3 +

4

π2

∞

n =1

cos (nπx )

n2

−4

π

∞

n =1

sin (nπx )

nPutting x = 1, we get


Example


42/96


f (x ) = 86

+∞

n =1

4n 2 π 2

cosnπx

1+

∞

n =1

− 4nπ

sinnπx

1

= 4

3 +

4

π2

∞

n =1

cos (nπx )

n2

−4

π

∞

n =1

sin (nπx )


1 = 43

+ 4π 2

−111

+ 122

−132

+ . . .


Example


43/96


f (x ) = 86

+∞

n =1

4n 2 π 2

cosnπx

1+

∞

n =1

− 4nπ

sinnπx

1

= 4

3 +

4

π2

∞

n =1

cos (nπx )

n2

−4

π

∞

n =1

sin (nπx )


1 = 43

+ 4π 2

−111

+ 122

−132

+ . . .

− 13

= 4π 2

− 111

+ 122

− 132

+ . . .


Example


44/96


f (x ) = 86

+∞

n =1

4n 2 π 2

cosnπx

1+

∞

n =1

− 4nπ

sinnπx

1

= 4

3 +

4

π2

∞

n =1

cos (nπx )

n2

−4

π

∞

n =1

sin (nπx )


1 = 43

+ 4π 2

−111

+ 122

−132

+ . . .

− 13

= 4π 2

− 111

+ 122

− 132

+ . . .

π 2

12 =

112

−122

+ 132

−142

+ . . .


Example


45/96

Ex. Find the Fourier series of f (x ) = 2 x in − 1 < x < 1

where p = 2L = 2.


Example


46/96



Example


47/96


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin nπx

L


Example


48/96


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin nπx

L

where a0 = 1L

1

−

1

f (x) dx


Example


49/96


f (x ) = a02

+∞

n =1

a n cosnπx

L+ bn sin nπx

L

where a0 = 1L

1

−

1

f (x) dx

a n = 1L

1

− 1f (x) cos

nπxL

dx


Example


50/96


f (x ) = a02

+∞

n =1

a n cos nπxL

+ bn sin nπxL

where a0 = 1L

1

−

1

f (x) dx

a n = 1L

1

− 1f (x) cos

nπxL

dx

bn = 1

L 1

− 1

f (x ) sinnπx

L dx


Example


51/96


f (x ) = a02

+∞

n =1

a n cos nπxL

+ bn sin nπxL

where a0 = 1L

1

− 1f (x) dx

a n = 1L

1

− 1f (x) cos

nπxL

dx

bn = 1

L 1

− 1

f (x ) sinnπx

L dx

Here p = 2L = 2 ⇒ L = 1


Example


52/96

Step 2. Now a0 = 11

1

− 1f (x ) dx


Example


53/96

Step 2. Now a0 = 11

1

− 1f (x ) dx

a 0 = 1

− 12x dx


Example


54/96

Step 2. Now a0 = 11

1

− 1f (x ) dx

a 0 = 1

− 12x dx

= 0


Example


55/96

Step 3. an = 11 1

− 1f (x) cos nπx1

dx


Example


56/96

Step 3. an = 11 1

− 1f (x) cos nπx1 dx

a n = 1

− 12xcos (nπx ) dx


Example


57/96

Step 3. an = 11 1


a n = 1


= 2x sin nπxnπ

− (2) − cos nπxn 2 π 2

1

− 1


Example


58/96

Step 3. an = 11 1


a n = 1


= 2x sin nπxnπ

− (2) − cos nπxn 2 π 2

1

− 1

= 0 + 2(− 1)n

n 2 π 2


Example


59/96

Step 3. an = 11 1


a n = 1


= 2x sin nπxnπ

− (2) − cos nπxn 2 π 2

1

− 1

= 0 + 2(− 1)n

n 2 π 2 − 0 −

2(− 1)n

n 2 π 2


Example


60/96

Step 3. an = 11 1


a n = 1


= 2x sin nπxnπ

− (2) − cos nπxn 2 π 2

1

− 1

= 0 + 2(− 1)n

n 2 π 2 − 0 −

2(− 1)n

n 2 π 2

= 0


Example


61/96

Ex. Find the Fourier series of periodic functionf (x )= − 1; − 1 < x < 0

= 1; 0 < x < 1 p = 2L = 2


Example


62/96

Ex. Find the Fourier series of periodic functionf (x )= 0; − 2 < x < 0

= 2; 0 < x < 2 p = 2L = 4


Fourier Half Range Series


63/96

A function f (x) dened only on the interval of the form0 < x < L .




64/96

A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2L




65/96

A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2L

Then such Fourier series are known as half range Fourier

series or half range expansions .




66/96

A function f (x) dened only on the interval of the form0 < x < L .If f (x ) is represented on this interval by a Fourier series of period 2LThen such Fourier series are known as half range Fourierseries or half range expansions .Types of Half Range Fourier series




67/96


1 Fourier Cosine Series




68/96


1 Fourier Cosine Series2 Fourier Sine Series


Fourier Cosine Series


69/96

Let f (x) be piecewise continuous on [o, l ].




70/96

Let f (x) be piecewise continuous on [o, l ].the Fourier cosine series expansion of f (x ) on the half rangeinterval [0 , l] is given by




71/96


f (x ) = ao

2 +

∞

n =1

a n cosnπx

l




72/96


f (x ) = ao

2 +

∞

n =1

a n cosnπx

l

where a0 = 2l

l

0f (x) dx




73/96


f (x ) = ao

2 +

∞

n =1

a n cosnπx

l

where a0 = 2l

l

0f (x) dx

a n =

2l

l

0f

(x

)cos

nπx

ldx


Fourier Sine Series


74/96

Let f (x) be piecewise continuous on [o, l ].


Fourier Sine Series


75/96

Let f (x) be piecewise continuous on [o, l ].the Fourier sine series expansion of f (x) on the half rangeinterval [0 , l] is given by


Fourier Sine Series


76/96


f (x ) =

∞

n =1bn sin

nπxl


Fourier Sine Series


77/96


f (x ) =

∞

n =1bn sin

nπxl

bn = 2l

l

0f (x ) sin

nπxl

dx


Example


78/96

Ex. Find Fourier cosine and sine series of the function

f (x ) = 1 for 0 ≤ x ≤ 2


Example


79/96

Sol. Here given interval is 0 ≤ x ≤ 2


Example


80/96

Sol. Here given interval is 0 ≤ x ≤ 2

∴ l = 2


Example


81/96

1

Fourier cosine series

Step 1. f (x ) = a02

+∞

n =1

a n cosnπx

l


Example


82/96

1


Step 1. f (x ) = a02

+∞

n =1

a n cosnπx

l

where a0 = 2l l

0f (x )dx


Example


83/96

1


Step 1. f (x ) = a02

+∞

n =1

a n cosnπx

l

where a0 = 2l l

0f (x )dx

a n = 2

l l

0f (x ) cos

nπx

l


Example


84/96

Step 2. a0 = 22

2

0f (x )dx

N B Vyas Fourier Series 2

Example


85/96

Step 2. a0 = 22

2

0f (x )dx

= 2

01dx


Example


86/96

Step 2. a0 = 22

2

0f (x )dx

= 2

01dx = [x ]20


Example


87/96

Step 2. a0 = 22

2

0f (x )dx

= 2

01dx = [x ]20 = 2 .


Example

2 2 nπx


88/96

Step 3. an = 2

2 0f (x )cos

nπx

2dx


Example

2 2 nπx


89/96

Step 3. an = 2

2 0f (x )cos

nπx

2dx

= 2

0(1) cos

nπx2

dx


Example

2 2 nπx


90/96

Step 3. an = 2

2 0f (x )cos

π

2dx

= 2

0(1) cos

nπx2

dx

=sin

nπx2

nπ2

2

0


Example

S 3 2 2

f ( )nπx

d


91/96

Step 3. an =2 0

f (x )cos2

dx

= 2

0(1) cos

nπx2

dx

=sin

nπx2

nπ2

2

0

= 2nπ (sin (nπ ) − sin (0))


Example

S 3 2 2

f ( )nπx

d


92/96

Step 3. an =2 0

f (x )cos2

dx

= 2

0(1) cos

nπx2

dx

=sin

nπx2

nπ2

2

0

= 2nπ (sin (nπ ) − sin (0)) = 0


Example


93/96

∴ Fourier cosine series of f (x ) is


Example


94/96


f (x ) = a02

+∞

n =1

a n cosnπx

l


Example


95/96


f (x ) = a02

+∞

n =1

a n cosnπx

l

= 22

+ 0


Example


96/96


f (x ) = a02

+∞

n =1

a n cosnπx

l

= 22

+ 0 = 1

N B V Fourier Series 2

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Dr. Nirav Vyas fourierseries2.pdf