CSVTU (Mechanical and Civil) III sem last 6 years examination papers solution (Mathematics)

DIMAT ME SOLVED PAPERS
MATHEMATICS-III
MECHANICAL
SUBJECT CODE 300311(14)

UNIT 01
FOURIER SERIES
Nov-Dec 2006
2


nx
n cos
1. Prove that 2
x =
+ 4( 1
- )
, - < x < , hence show that:
2
3
=
n
n 1
2
1
1
1
1

-
+
-
+ ....... =
.
2
2
2
2
1
2
3
4
12
Ans: Here
2
f (x) = x is an even function in
- < x < .

So, b = 0 .
n



3
3
2
2
2
2 x
2

2


2
a =
f (x)dx = x dx =
=

- 0 =

0


3
3

3
0
0
0
2
2


2
a =
f (x)cosnxdx = x cosnxdx
n


0
0

2
sin nx
cos nx
sin nx


2
a = x
+ 2x
- 2

n

2
3

n
n
n
0
2
( 1
- )n

( 1
- )n


a = 0 - 0 + 2
- 0 - 0 + 0 = 4

n
2
2

n

n
a



Then
0
f (x) =
+ (a cos nx)
2
n
n 1
=
2

n
2

( 1
- )

( 1
- )n



f (x) =
+ 4
cos nx =
+ 4
cos nx (Proved).
2
2
3
=
n

=
n
n
3
1
n 1



Putting x = 0 we get
2

( 1
- )n


0 =
+ 4

2
3
=
n
n 1

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DIMAT ME SOLVED PAPERS

n
2
( 1
- )



4
-
=

2
=
n
n

3
1

n
2
( 1
- )



-
=

2
=
n
n

12
1
2
1
1
1
1




-
+
-
+ ....... =
(Proved).
2
2
2
2
1
2
3
4
12
x
0 x 1
2. If f (x) =
. Show that in the interval (0, 2)

(2 - x) 1 x 2

4 cos x
cos 3 x
cos 5 x

f (x) =
-
+
+
+ ........

and deduce that
2
2
2

2
1
3
5

2
1
1
1

+
+
+ ....... =

2
2
2
1
3
5
8
x
0 x 1
Ans: f (x) =


(2 - x) 1 x 2

Given function is of period T = 2l = 2 l = 1 .
a

n x
n x

Its Fourier series is
0
f (x) =
+ a cos
+ b sin

2
n
n


n 1
=
l
l
2
2
2
1
1

Where a = f (x)dx = f (x)dx = f (x)dx
0
l
1
0
0
0
1
2
1
2

a = f (x)dx + f (x)dx = xdx + (2 - x)dx
0
0
1
0
1
1
2
2
2

x

x

a = + 2x -

0
2



2


0
1
1
1

a = - 0 + 4 - 2 - 2 + = -------- (1)
0
2
2
2
2
1
n x

a = f (x)cos
dx = f (x)cos n xdx
n
l
l
0
0

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DIMAT ME SOLVED PAPERS
1
2

a = xcos n xdx + (2 - x)cos n xdx
n
0
1
1
2
sin n x cos n x

sin n x
cos n x

a = x
+
+ (2 - x)
-

n

2
2
2
2

n
n


n
n

0
1

( 1
- )n
1


1
( 1
- )n

a = 0 +
- 0 -
+ 0 -
- 0 +

n
2
2
2
2
2
2
2
2

n
n

n
n
2{( 1
- )n - }
1

a =
-------------
(2)
n
2
n
2
2
1
n x

b = f (x)sin
dx = f (x)sin n xdx
n
l
l
0
0
1
2

b = xsin n xdx + (2 - x)sin n xdx
n
0
1
1
2

cos n x
sin n x

cos n x
sin n x

b = -x
+
+ -(2 - x)
-

n

2
2
2
2

n
n


n
n

0
1
( 1
- )n


( 1
- )n


b = -
+ 0 + 0 - 0 + 0 - 0 +
+ 0 = 0 ---------- 93)
n

n


n

a

n x
n x

So, Fourier Series is
0
f (x) =
+ a cos
+ b sin

2
n
n


n 1
=
l
l
a


0
f (x) =
+ (a cos n x)
2
n
n 1
=




{( 1
- )n - }
1
2

f (x) =
+


cos n x
2
2



=
n
n 1


4 cos x
cos 3 x
cos 5 x



f (x) =
-

+
+
+ ......... (Ans)
2
2
2
2

1
3
5



Now, putting x = 1 we get

---

DIMAT ME SOLVED PAPERS

4 1
-
-1
-1



=
-

+
+
+ .........
2
2
2
2
1
3
5


4 1
1
1




=

+
+
+ .........
2
2
2
2
1
3
5

2
1
1
1




+
+
+ ....... =
(Ans)
2
2
2
1
3
5
8
3. The following table gives the variation of periodic current over a period:
t(sec)
:
0
T / 6 T / 3 T / 2
2T / 3
5T / 6
T
(
A amp) : 1.98 1.30
1.05
1.30
-0.88
-0.25 1.98
Show that there is a direct current part of 0.75 amp. in the variable current and
obtain the amplitude of the first harmonic.
Ans: Let the Fourier series to represent A in (0, T) be
a

2 t
2 t

0
A =
+ a cos
+ b sin

2
n
n


n 1
=
T
T
a
2 t
2 t
4 t
4 t

0
A =
+ a cos
+ b sin
+ a cos
+ b sin
+ .....
1
1
2
2
2
T
T
T
T

To evaluate all the coefficients, let us form the table
2 t
2 t
2 t
2 t
2 t
t
cos
sin
A
A cos
Asin
T
T
T
T
T
0
0
1
0
1.98
1.98
0
T / 6
/ 3
0.5
0.866
1.30
0.65
1.126

T / 3
2 / 3
0
- .5
0.866
1.05
0
- .525
0.909

T / 2

1
-
0.866
1.30
1
- .3
0
2T / 3 4 / 3
0
- .5
0
- .866
0
- .88
0.44
0.762
5T / 6
5 / 3
0.5
0
- .866
0
- .25
0
- .125
0.217
Sum =
4.5
1.12
3.014


1
1

Now, a = 2 x A = (4.5) = 1.5
0
6
3

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DIMAT ME SOLVED PAPERS
1
2 t
1

a = 2 x
Acos
=
(1.12) = 0.373
1
6
T
3
1
2 t
1

b = 2 x Asin
=
(3.014) = 1.005
1
6
T
3
a
1.5

Thus the direct current part in the variable current
0
=
=
= 0.75
2
2


Amplitude of the first harmonic
2
2
2
2
= a + b = (0.373) + (1.005) = 1.072 (Ans)
1
1
Examinations--Nov-Dec 2007
4. Obtain the Fourier series for ( )
x
f x
e-
=
in the interval 0 < x < 2 .

2
1

Ans: a =
f (x)dx
0
0
2
1


- x
=
e dx
0
1
2
- x
1


=
x (-e
=
1- e-


0
) (
2
)


2
1


a =

n
f (x)cosnxdx
0
2
1


- x
=
e cosnxdx
0
2
1
- x
e


=

(- cos nx + n sin nx)
2
1+ n
0
2
-
-2
1 e
1

1
1- e



=

( 1
- + 0) -
( 1
- + 0) =
x

2
2
2
1+ n
1+ n

1+ n
2
1


b =

n
f (x)sin nxdx
0
2
1



- x
=
e sin nxdx
0

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DIMAT ME SOLVED PAPERS
2
1
- x
e


=

(- sin nx - n cos nx)
2
1+ n
0
- x
-2
1 e
1

2
- e


=

( 0
- - 2 ) -
( 0
- - 0) =

2
2
2
1+ n
1+ n

1+ n
a


0
f (x) =
+ (a cos nx + b sin nx)
2
n
n
n=0

2
-
2
1
-

-

-
1 1 e
2e
=
(
2
1- e
) + x
cos nx -
sin nx
2
2
2

+
+
=

n
n
n
1
1
0


2
-
-2
1- e
1- e cos x
cos 2x
cos 3x



-
sin x
sin 2x
sin 3x
2
=
+

+
+
+ ..... - 2e

+
+
+ .....
2
2
2
2
2
2
2

1+1
1+ 2
1+ 3

1+1
1+ 2
1+ 3

(Ans)
5. Find the Fourier series expansion for f(x), if:






-x

- < x < 0

f (x) =

x
0 < x <
2
1
1
1


Deduce that
+
+
+ ....... =
.
2
2
2
1
3
5
8
-x

- < x < 0

Ans: f (x) =

x
0 < x <



2

2
2
x




f (x)dx = xdx =


=
- 0 =


2


2

2
0
0

0
0
0




2
2

2
2


x
x







2
f (x)dx = -xdx + xdx = - +


= 0 +
+
- 0 =




2
2




2
2

-
-
0

-
0




f (x)dx = 2 f (x)dx
-
0


So, the given function is an even function. So its Fourier series is given by

---

DIMAT ME SOLVED PAPERS
a



0
f (x) =
+ a cos nx
2
n
n 1
=



2

2
2
2
2 x
2



a =
f (x)dx = xdx =


=

- 0 =
0


2


2

0
0

0


2
2
2 sin nx
cos nx


a =
=
xcosnxdx = x
+

n
f (x)cosnxdx



2

n
n

0
0
0
2
cos
1
2 ( 1
- )n
n
-1


=
0 +
- 0 -
=


2
2
2

n
n

n

a


0
f (x) =
+ a cos nx
2
n
n 1
=


2 ( 1
- )n -1


=
+
cos nx
2
2
=
n
n 1


4
cos 3x
cos 5x

=
-
cos x +
+
+ ........ (Ans)------------- (1)
2

9
25


By putting x = 0 in (1) we get

4 1
1
1


0 =
-

+
+
+ ........
2
2
2
2
1
3
5

4 1
1
1





+
+
+ ........ =

2
2
2
1
3
5

2
2
1
1
1


+
+
+ ........ =
(Proved)
2
2
2
1
3
5

8
6. If f (x) = cos x , expand f (x) as a Fourier series in the interval (
- , ) .

Sol. Q f (x) = cos x
f (-x) = cos(-x) = cos x = f (x)


f (x) is an even function of x .

---

DIMAT ME SOLVED PAPERS
The Fourier series of f (x) in given by
a


0
f (x) =
+ a cos nx
2
n
n 1
=
[ Q
b = 0 , when f (x) in an even function ]
n

2
Where a =
f (x)dx
0
0
2
=
cos x dx
0
/ 2
2



=
cos xdx + (- cos x)dx
0
/ 2



Q cos x is + v ,
e when0 < x <


2





and - v ,
e when
< x < .


2


2

2

/ 2
=
[sin x]
-
[sin x]
0
/ 2


2
2
=
[(1- 0)] -
[(0 -1)]


2
2
=
+



4
a =

0

2
and a =
cos x.cosnx dx
n
0
/ 2
2
2
=
cos .xcosnx dx + (-cos x)cosnx dx


0
/ 2
/2
1
1
=
[cos(n +1)x + cos(n -1)x]dx - [cos(n +1)x + cos(n -1)x]dx


0
/ 2

---

DIMAT ME SOLVED PAPERS
/2

1 sin(n +1)x
sin(n -1)x
1 sin(n +1)x
sin(n -1)x
=
+


-
+


,

n +1
n -1


n +1
n -1

0
/2
For
n 1






sin(n +1)
sin(n -1)
sin(n +1)
sin(n -1)
1

1


2
2
=

+

2
2
+

+


n +1
n -1



n +1
n -1







n
n
cos
cos
2


2
2
=

-


n +1
n -1





n
2
- cos
2


2
=


2

n -1




n
4 cos

2
a = -
, for n 1
n
2
(n -1)
To find a
1
2
Q a =
cos x.cos x dx
1
0
/ 2
2
2
=
cos .xcos x dx + (-cos x)cos x dx


0
/2
/ 2
2
2

2
2
=
cos x dx - cos x dx


0
/2
2 1
1
=
. .
-
(1+ cos2x)dx
2 2
/2

1
1
sin 2x
=
-
x +



2

2
/2

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DIMAT ME SOLVED PAPERS
1
1


=
-
( + 0) -


+ 0
2

2

1
1
=
-

2
2
a = 0
1
The Fourier series for f (x) = cos x is
2

4
n

cos x =
+ 0.cos x + -
cos
cos nx
2


-
=
n
n
(
1)
2
2

2
4 1
1

cos x =
-
- cos 2x + 0 -
cos 4x -....



3
15

2
4 1
1

cos x =
+
cos 2x +
cos 4x + ....



3
15

Ans.


YEAR 2008

7. Explain conditions for a Fourier expansion.
Ans: A function f (x) defined in the interval [
- , ] can be expressed in Fourier series
1

f (x) =
a + a cos nx + b sin nx where a , a ,b are real coefficients if the
0
2
n
n
0
n
n
n 1
=
following
(i) Dirichlet's condition for convergence are satisfied in the interval [
- , ] .
(ii) f (x) and its integral are finite and single valued.
(iii) f (x) has finite number of discontinuities.
f (x) has finite number of maxima and minima.
8. Find a Fourier series to represent
2
x - x from x =
- to x = .
Ans: Here
2
f (x) = x - x



3
2
2
1
1
x
x
2
a =
f (x)dx = (-x + x)dx = - +
0



3
2
-
-
-

---