fourier series 2

Fourier Series 2

N. B. Vyas

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

Rajkot (Guj.)- INDIA

N. B. Vyas Fourier Series 2

Functions of any Period p = 2L

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

The Fourier series of f(x) is given by

f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ c+2L

c

f(x) dx

an =1

L

∫ c+2L

c

f(x) cos(nπxL

)dx

bn =1

L

∫ c+2L

c

f(x) sin(nπxL

)dx





f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]

where a0 =1

L

∫ c+2L

c

f(x) dx

an =1

L

∫ c+2L

c

f(x) cos(nπxL

)dx

bn =1

L

∫ c+2L

c

f(x) sin(nπxL

)dx





f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ c+2L

c

f(x) dx

an =1

L

∫ c+2L

c

f(x) cos(nπxL

)dx

bn =1

L

∫ c+2L

c

f(x) sin(nπxL

)dx



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

f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ 2L

0

f(x) dx

an =1

L

∫ 2L

0

f(x) cos(nπxL

)dx

bn =1

L

∫ 2L

0

f(x) sin(nπxL

)dx




f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]

where a0 =1

L

∫ 2L

0

f(x) dx

an =1

L

∫ 2L

0

f(x) cos(nπxL

)dx

bn =1

L

∫ 2L

0

f(x) sin(nπxL

)dx




f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ 2L

0

f(x) dx

an =1

L

∫ 2L

0

f(x) cos(nπxL

)dx

bn =1

L

∫ 2L

0

f(x) sin(nπxL

)dx



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

f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ L

−Lf(x) dx

an =1

L

∫ L

−Lf(x) cos

(nπxL

)dx

bn =1

L

∫ L

−Lf(x) sin

(nπxL

)dx




f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]

where a0 =1

L

∫ L

−Lf(x) dx

an =1

L

∫ L

−Lf(x) cos

(nπxL

)dx

bn =1

L

∫ L

−Lf(x) sin

(nπxL

)dx




f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ L

−Lf(x) dx

an =1

L

∫ L

−Lf(x) cos

(nπxL

)dx

bn =1

L

∫ L

−Lf(x) sin

(nπxL

)dx


Example

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

Hence deduce that 1− 1

22+

1

32− 1

42+ . . . =

π2

12


Example

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

f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ 2

0

f(x) dx

an =1

L

∫ 2

0

f(x) cos(nπxL

)dx

bn =1

L

∫ 2

0

f(x) sin(nπxL

)dx

Here p = 2L = 2⇒ L = 1


Example


f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]

where a0 =1

L

∫ 2

0

f(x) dx

an =1

L

∫ 2

0

f(x) cos(nπxL

)dx

bn =1

L

∫ 2

0

f(x) sin(nπxL

)dx

Here p = 2L = 2⇒ L = 1


Example


f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ 2

0

f(x) dx

an =1

L

∫ 2

0

f(x) cos(nπxL

)dx

bn =1

L

∫ 2

0

f(x) sin(nπxL

)dx

Here p = 2L = 2⇒ L = 1


Example

Step 2. Now a0 =1

1

∫ 2

0

f(x) dx

a0 =

∫ 2

0

(x)2 dx

=

[x3

3

]20

=8

3


Example

Step 3. an =1

1

∫ 2

0

f(x) cos(nπx

1

)dx

an =

∫ 2

0

x2 cos(nπx) dx

=

[x2(sin nπx

nπ

)− (2x)

(−cos nπx

n2π2

)+ 2

(−sin nπx

n3π3

)]20

=4

n2π2


Example

Step 4. bn =1

1

∫ 2

0

f(x) sin(nπx

1

)dx

bn =

∫ 2

0

x2 sin(nπx) dx

=

[x2(−cos nπx

nπ

)− (2x)

(−sin nπx

n2π2

)+ 2

(cos nπxn3π3

)]20

= − 4

nπ


Example

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

f(x) =8

6+∞∑n=1

(4

n2π2

)cos(nπx

1

)+∞∑n=1

(− 4

nπ

)sin(nπx

1

)=

4

3+

4

π2

∞∑n=1

cos(nπx)

n2− 4

π

∞∑n=1

sin(nπx)

n

Putting x = 1, we get

1 =4

3+

4

π2

[− 1

11+

1

22− 1

32+ . . .

]−1

3=

4

π2

[− 1

11+

1

22− 1

32+ . . .

]π2

12=

1

12− 1

22+

1

32− 1

42+ . . .


Example


f(x) =8

6+∞∑n=1

(4

n2π2

)cos(nπx

1

)+∞∑n=1

(− 4

nπ

)sin(nπx

1

)

=4

3+

4

π2

∞∑n=1

cos(nπx)

n2− 4

π

∞∑n=1

sin(nπx)

n


1 =4

3+

4

π2

[− 1

11+

1

22− 1

32+ . . .

]−1

3=

4

π2

[− 1

11+

1

22− 1

32+ . . .

]π2

12=

1

12− 1

22+

1

32− 1

42+ . . .


Example


f(x) =8

6+∞∑n=1

(4

n2π2

)cos(nπx

1

)+∞∑n=1

(− 4

nπ

)sin(nπx

1

)=

4

3+

4

π2

∞∑n=1

cos(nπx)

n2− 4

π

∞∑n=1

sin(nπx)

n


1 =4

3+

4

π2

[− 1

11+

1

22− 1

32+ . . .

]−1

3=

4

π2

[− 1

11+

1

22− 1

32+ . . .

]π2

12=

1

12− 1

22+

1

32− 1

42+ . . .


Example


f(x) =8

6+∞∑n=1

(4

n2π2

)cos(nπx

1

)+∞∑n=1

(− 4

nπ

)sin(nπx

1

)=

4

3+

4

π2

∞∑n=1

cos(nπx)

n2− 4

π

∞∑n=1

sin(nπx)

n


1 =4

3+

4

π2

[− 1

11+

1

22− 1

32+ . . .

]

−1

3=

4

π2

[− 1

11+

1

22− 1

32+ . . .

]π2

12=

1

12− 1

22+

1

32− 1

42+ . . .


Example


f(x) =8

6+∞∑n=1

(4

n2π2

)cos(nπx

1

)+∞∑n=1

(− 4

nπ

)sin(nπx

1

)=

4

3+

4

π2

∞∑n=1

cos(nπx)

n2− 4

π

∞∑n=1

sin(nπx)

n


1 =4

3+

4

π2

[− 1

11+

1

22− 1

32+ . . .

]−1

3=

4

π2

[− 1

11+

1

22− 1

32+ . . .

]

π2

12=

1

12− 1

22+

1

32− 1

42+ . . .


Example


f(x) =8

6+∞∑n=1

(4

n2π2

)cos(nπx

1

)+∞∑n=1

(− 4

nπ

)sin(nπx

1

)=

4

3+

4

π2

∞∑n=1

cos(nπx)

n2− 4

π

∞∑n=1

sin(nπx)

n


1 =4

3+

4

π2

[− 1

11+

1

22− 1

32+ . . .

]−1

3=

4

π2

[− 1

11+

1

22− 1

32+ . . .

]π2

12=

1

12− 1

22+

1

32− 1

42+ . . .


Example

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


Example


f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ 1

−1f(x) dx

an =1

L

∫ 1

−1f(x) cos

(nπxL

)dx

bn =1

L

∫ 1

−1f(x) sin

(nπxL

)dx

Here p = 2L = 2⇒ L = 1


Example


f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]

where a0 =1

L

∫ 1

−1f(x) dx

an =1

L

∫ 1

−1f(x) cos

(nπxL

)dx

bn =1

L

∫ 1

−1f(x) sin

(nπxL

)dx

Here p = 2L = 2⇒ L = 1


Example


f(x) =a02

+∞∑n=1

[an cos

(nπxL

)+ bn sin

(nπxL

)]where a0 =

1

L

∫ 1

−1f(x) dx

an =1

L

∫ 1

−1f(x) cos

(nπxL

)dx

bn =1

L

∫ 1

−1f(x) sin

(nπxL

)dx

Here p = 2L = 2⇒ L = 1


Example

Step 2. Now a0 =1

1

∫ 1

−1f(x) dx

a0 =

∫ 1

−12x dx

= 0


Example

Step 3. an =1

1

∫ 1

−1f(x) cos

(nπx1

)dx

an =

∫ 1

−12x cos(nπx) dx

=

[2x

(sin nπx

nπ

)− (2)

(−cos nπx

n2π2

)]1−1

=

[0 +

2(−1)n

n2π2− 0− 2(−1)n

n2π2

]= 0


Example

Step 3. an =1

1

∫ 1

−1f(x) cos

(nπx1

)dx

an =

∫ 1

−12x cos(nπx) dx

=

[2x

(sin nπx

nπ

)− (2)

(−cos nπx

n2π2

)]1−1

=

[0 +

2(−1)n

n2π2

− 0− 2(−1)n

n2π2

]= 0


Example

Step 3. an =1

1

∫ 1

−1f(x) cos

(nπx1

)dx

an =

∫ 1

−12x cos(nπx) dx

=

[2x

(sin nπx

nπ

)− (2)

(−cos nπx

n2π2

)]1−1

=

[0 +

2(−1)n

n2π2− 0− 2(−1)n

n2π2

]

= 0


Example

Step 3. an =1

1

∫ 1

−1f(x) cos

(nπx1

)dx

an =

∫ 1

−12x cos(nπx) dx

=

[2x

(sin nπx

nπ

)− (2)

(−cos nπx

n2π2

)]1−1

=

[0 +

2(−1)n

n2π2− 0− 2(−1)n

n2π2

]= 0


Example

Ex. Find the Fourier series of periodic function

f(x)= −1;−1 < x < 0= 1; 0 < x < 1

p = 2L = 2


Example

Ex. Find the Fourier series of periodic function

f(x)= 0;−2 < x < 0= 2; 0 < x < 2

p = 2L = 4


Fourier Half Range Series

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

If f(x) is represented on this interval by a Fourier series ofperiod 2L

Then such Fourier series are known as half range Fourierseries or half range expansions.

Types of Half Range Fourier series

1 Fourier Cosine Series

2 Fourier Sine Series


Fourier Cosine Series

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

f(x) =ao2

+∞∑n=1

an cos(nπx

l

)where a0 =

2

l

∫ l

0

f(x) dx

an =2

l

∫ l

0

f(x) cos(nπx

l

)dx





f(x) =ao2

+∞∑n=1

an cos(nπx

l

)

where a0 =2

l

∫ l

0

f(x) dx

an =2

l

∫ l

0

f(x) cos(nπx

l

)dx





f(x) =ao2

+∞∑n=1

an cos(nπx

l

)where a0 =

2

l

∫ l

0

f(x) dx

an =2

l

∫ l

0

f(x) cos(nπx

l

)dx


Fourier Sine Series


the Fourier sine series expansion of f(x) on the half rangeinterval [0, l] is given by

f(x) =∞∑n=1

bn sin(nπx

l

)bn =

2

l

∫ l

0

f(x) sin(nπx

l

)dx


Fourier Sine Series



f(x) =∞∑n=1

bn sin(nπx

l

)

bn =2

l

∫ l

0

f(x) sin(nπx

l

)dx


Fourier Sine Series



f(x) =∞∑n=1

bn sin(nπx

l

)bn =

2

l

∫ l

0

f(x) sin(nπx

l

)dx


Example

Ex. Find Fourier cosine and sine series of the functionf(x) = 1 for 0 ≤ x ≤ 2


Example

Sol. Here given interval is 0 ≤ x ≤ 2

∴ l = 2


Example

1 Fourier cosine series

Step 1. f(x) =a02

+∞∑n=1

an cos(nπx

l

)

where a0 =2

l

∫ l

0

f(x)dx

an =2

l

∫ l

0

f(x) cos(nπx

l

)


Example

1 Fourier cosine series

Step 1. f(x) =a02

+∞∑n=1

an cos(nπx

l

)where a0 =

2

l

∫ l

0

f(x)dx

an =2

l

∫ l

0

f(x) cos(nπx

l

)


Example

Step 2. a0 =2

2

∫ 2

0

f(x)dx

=

∫ 2

0

1dx = [x]20 = 2.


Example

Step 2. a0 =2

2

∫ 2

0

f(x)dx

=

∫ 2

0

1dx

= [x]20 = 2.


Example

Step 2. a0 =2

2

∫ 2

0

f(x)dx

=

∫ 2

0

1dx = [x]20

= 2.


Example

Step 2. a0 =2

2

∫ 2

0

f(x)dx

=

∫ 2

0

1dx = [x]20 = 2.


Example

Step 3. an =2

2

∫ 2

0

f(x)cos(nπx

2

)dx

=

∫ 2

0

(1) cos(nπx

2

)dx

=

sin(nπx

2

)(nπ

2

)2

0

=

(2

nπ

)(sin (nπ)− sin (0)) = 0


Example

Step 3. an =2

2

∫ 2

0

f(x)cos(nπx

2

)dx

=

∫ 2

0

(1) cos(nπx

2

)dx

=

sin(nπx

2

)(nπ

2

)2

0

=

(2

nπ

)(sin (nπ)− sin (0))

= 0


Example

Step 3. an =2

2

∫ 2

0

f(x)cos(nπx

2

)dx

=

∫ 2

0

(1) cos(nπx

2

)dx

=

sin(nπx

2

)(nπ

2

)2

0

=

(2

nπ

)(sin (nπ)− sin (0)) = 0


Example

∴ Fourier cosine series of f(x) is

f(x) =a02

+∞∑n=1

an cos(nπx

l

)=

2

2+ 0 = 1


Example


f(x) =a02

+∞∑n=1

an cos(nπx

l

)

=2

2+ 0 = 1


Example


f(x) =a02

+∞∑n=1

an cos(nπx

l

)=

2

2+ 0

= 1


Example


f(x) =a02

+∞∑n=1

an cos(nπx

l

)=

2

2+ 0 = 1


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