QUESTION BANK - ANSWERS SEMESTER: III MA 2211 - … UNIT... · Find the root mean square value of...

MAHALAKSHMI ENGINEERING COLLEGE, TRICHY

1 M.MARIA AROCKIA RAJ M.Sc.,M.Phil., AP/MATHEMATICS, MEC

MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI – 621213

QUESTION BANK - ANSWERS

SEMESTER: III

MA 2211 - TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS UNIT – I: FOURIER SERIES

PART-A

State Dirichlet's conditions on Fourier series

( ). ( ) is single valued ,finite and periodic function in ( , 2 )( ). ( ) is continuous or piece-wii f x c cii f x

Question :1 AUC A / M 2013, 2010 N / D 2011, N / D 2011

Answer :

se continuous with finite number of finite discontinuities in (c,c+2π)( ). ( ) has a finite number of maxima (or) minima in (c,c+2π).

What is meant by Harmonic analysis

iii f xQuestion : 2 AUC A / M 2013

Answ

0

The process of finding the Fourier series for a given function by numerical value is known as harmonic analisis. In harmonic analisis the Fourier coefficients , and of the functn na a b

er :

0

ion ( ) in (0, 2 ) given by

2 ,

cos2 ,

sin2

where 'n' is the number of terms appears in the table.

n

n

y f x

ya

n

y nxa

n

y nxb

n



2

20

0 0

0

,

Find the constant term in the expansion of cos as a Fourier series in the interval ,

2 2 1 cos 2cos2

1 1 sin 21 cos 2

x

xa xdx dx

x dx x

Question : 3 : AUC A / M 2012 N / D 2010

Answer :

0

0

12

1Constant term 2 2

Define Root Mean Square value of a function ( ), over the interval ( , )

The root mean square value of ( ) over the inte

x

a

f x a b

f x

Question : 4 : AUC A / M 2012 N / D 2012

Answer :

2

rval ( , )is defined as

( )R.M.S

Give the expression for the Fourier series co-efficient for .the function ( )over the interval ( 2, 2)

2 ( )sin

b

a

n

n

a b

f x dx

b a

b f x

nb f xl

Question : 5 : AUC A / M 2011

Answer :

2 2

0 0 0

20

22 20

1

2 ( )sin ( )sin2 2 2

Without finding the values of , and , for ( ) in the interval (0, )

find the value of ( )2

l

n n

n nn

x n x n xdx f x dx f x dxl

a a b f x x

a a b

Question : 6 : AUC A / M 2011

An

22 20

10

222 20

1 0

2 22 2 20

1 0

1 1We know that ( ) ..............(1)4 2

2(1) 2 ( ) ( )2

2( )2

nn

n nn

n nn

af x dx a

a a b f x dx

a a b x dx

swer :



2 5 52 2 4 50

1 0 0

2 2 2 2( ) 02 5 5 5

Find the coefficient of the Fourier series for the function sin in ( 2,2)

Since the function

n nn

n

a xa b x dx

b x x

Question : 7 : AUC N / D 2012

Answer :

3 32 2

0 0

( ) sin even 0

Find the root mean square value of the function ( ) in (0, )

( )3 3R.M.S

0 0 0 3

n

lb l

a

f x x x b

f x x l

x lf x dx x dxl

b a l l l


Answer :

Question : 9 : AUC N / D 2010A

2

5 522 22

0 0

Find the root mean square value of the function ( ) in (0, )

( )5 5R.M.S

0 0 0 5

Obtain the first term of the

lb l

a

f x x l

x lf x dx x dxl

b a l l l

nswer :

Question :10 : AUC N / D 20092

32

00 0 0

23

Fourier series for the function ( ) ,

2 2 2Since ( ) is even, ( )3

2 203 3

f x x x

xf x a f x dx x dx

Answer :



2

2

Find the Fourier series expansion for in the interval

Since ( ) is neither even nor odd the Fourier series expansionin the interval is given

x x x

f x x xx

PART B Question :1 AUC N / D 2012

Answer : .

0

1 1

0

2 32

0

2 3 2

by

( ) cos sin .....................(1)2

1 1Where ( ) , ( )cos

1 ( )sin

1 1 1Now ( ) ( )2 3

1 ( )2 3 2

n nn n

n

n

af x a nx b nx

a f x dx a f x nxdx

b f x nxdx

x xa f x dx x x dx

3

2 3 2 3 3 3

32 2

0

2

22

( )3

1 12 3 2 3 3 3

1 2 2 2 .....................(2)3 3 3

1 1( ) cos ( ) cos

1 sin cos sin( ) (1 2 ) (2)

n

a

a f x nxdx x x nxdx

nx nx nxx x xn n n

3

2

2

2

2

2

1 cos(1 2 ) since sinterms 0

1 (1 2 )cos

1 (1 2 )cos (1 2 )cos ( )

1 (1 2 )cos (1 2 )cos since cos( ) cos

1 cos 1 2 1 2

nxxn

x nxn

n nn

n nn

nn



2 2 2

2

22 3

23

4 4 4( 1) ( 1) ( 1) ............(3)

1 1( )sin ( )sin

1 cos sin cos( ) (1 2 ) (2)

1 2cos cos( )

1 2co

n n nn

n

an n n

b f x nxdx x x nxdx

nx nx nxx x xn n n

nx nxx xn n

2 23 3

2 23 3

2 2 2 2

s cos 2cos ( ) cos ( )( ) ( ( ) )

1 2cos cos 2cos cos( ) ( )

1 cos 1 cos( ) ( )

n n n nn n n n

n n n nn n n nn n

n n

2

02

1 1 1 1

2

21

)

1 ( 1) 2 22 ( 1) ( 1) ..................(4)

Use (2), (3), and (4) in (1) we get 2

4 23( ) cos sin ( 1) cos ( 1) sin2 2

( 1)4 cos3

nn n

n

n nn n

n n n n

n

n

bn n n

af x a nx b nx nx nxn n

nxn

1

0

1

00

( 1)2 sin

Find the half range Fourier cosine series expansion of ( ) sin in 0 .

( ) cos .......................(1)2

2 2 2( ) sin ( co

n

n

nn

nxn

f x x x x

af x a nx

a f x dx x xdx x

Question : 2 AUC N / D 2011

Answer :

00

0 0

0

s ) (1)( sin )

2 2 2cos sin cos cos 0

2 ( 1) 2, 2 .............................(2)

x x

x x x x x

a



0 0 0

0 0

1 20 0

1

2 2 2( )cos sin cos cos sin

2 2 1cos sin sin( 1) sin( 1)2

1 1sin( 1) sin( 1) I I ........................(*)

cosI sin( 1)

na f x nxdx x x nxdx x nx xdx

x nx xdx x n x n x dx

x n xdx x n xdx

x n xdx x

0 0

0

1

12

1 11 2

1 1

( 1) sin( 1)(1)( 1) ( 1

1 cos( 1)( 1)

1 cos( 1) 0 1( 1) ( 1)

Similarlly I 1( 1)

I I 1 1( 1) ( 1)

1 1 Since 1( 1) ( 1)

n

n

n n

n n n

n x n xn n

x n xn

nn n

n

n n

n n

1 1

1

1 12

11 22 2 2

1

1 11( 1) ( 1)

1 1 21 1( 1)( 1) 1

2 2 ( 1) 2 ( 1)( 1) ( 1) I I ...........( )1 1 1

Use (a) in (*) we get

1

n

n

n n

n nn

n

n n

n nn n n

an n n

a

2 2 2

0

10 0

0 0

2 ( 1) 2( 1) 2( 1) , 1 ....................(3)1 1 1

2 cos sin

2 2 1sin cos 2sin cos2

1 1 cos 2 sin 2sin 2 (1)2 4

n n n

n

n

a nn n n

a x nx xdx

a x x xdx x x x dx

x xx xdx x



0

1

0

1

01

2

2

1 1 1 1cos 2 cos 2 0 (1)2 2 2 2

1 ....................................(4)2

Use (2),(3) and (4) in (1) we get

( ) cos2

cos cos22 1 2( 1)cos cos2 2 1

nn

nn

n

n

x x

a

af x a nx

a a x a nx

x nxn

2

22

2

4

4 4 4

2

1 ( 1)1 cos 2 cos2 1

Find the Fourier series expansion for in the interval and hence prove that1 1 11 2 3 90

Since ( ) is even 0 and t

n

n

n

x nxn

x x

f x x b

Question : 3 AUC M / J 2013

Answer :

0

1

00 0

32

00 0 0

3

0

he Fourier series expansion in the interval is given by

( ) cos .......................(1)2

2 2Where ( ) , ( ) cos

2 2 2( )3

23

nn

n

xaf x a nx

a f x dx a f x nxdx

xa f x dx x dx

x

2

3

20

0

2

0

2 203 3

2 .........................(2)3

2 ( )cos

2 cos

n

a

a f x nxdx

x nxdx



2

02

22 2 0 0

0

2 2 2

2

2 sin cos sin(2 ) (2)

2 2 cos 4 cos since sinterms 0

4 4 4cos 0 ( 1) ( 1)

4 ( 1) .................................

n n

nn

nx nx nxx xn n n n n n

x nx x nxn n

nn n n

an

2

21

22 2 20

1

22

0 2

.(3)

Use (2) and (3) in (1) we get 2

43( ) ( 1) cos2

The Parseval's identity of ( ) in is given by

1 1( ) ( ).2 4 2

2 4( 1)Put , , 0 , ( ) , we get3

1

n

n

n nn

n

n n

f x nxn

f x x

af x dx a b

a a b f x xn

22

222

21

42

44

1

5 4 2

41

45

4 4 4

45 5

4

23 1 4( 1) 0)

2 4 2

41 1 16( 1)9 .

2 4 2

1 ( 1)8 .2 5 9

1 1 1 1810 9 1 2 3

1 1( ) 810 9 1

n

n

n

n

n

n

x dxn

x dxn

xn

x

4 4

5 4

4 4 4

4 4

4 4 4

1 12 3

2 1 1 1810 9 1 2 3

1 1 185 9 1 2 3



4 4

4 4 4

4 4 4

4 4 4 4 4 4

4

4 4 4

4

4 4 4

1 1 181 2 3 5 91 1 1 9 5 1 1 1 48 81 2 3 45 1 2 3 45

1 1 1 41 2 3 45 81 1 11 2 3 90

Find the Fourier series to represent t

Question : 4 : AUC M / J 2012

2

21

0

1

00 0

2

00

he function ( ) , and

1hence deduce (2 1) 8

Since ( ) is even, 0

( ) cos ..............(1)2

2 2( ) , ( )cos

2 2 2( )2

n

n

nn

n

f x x x

n

f x baf x a nx

a f x dx a f x nxdx

xa f x dx xdx

Answer :

0 0

0 0

2 2( ) cos cos

2 sin

na f x nxdx x nxdx

nxxn

0

2 0

2 2 2

2

21

cos(1)

2 cos

2 1 4cos cos0 ( 1) 1 , is odd

4 , is odd ..................................(3)

use (2) and (3) in (1) we get4( ) cos ...

2

n

n

n

nxn n

nxn

n nn n n

a nn

f x nxn

...........(1)



21

21

21

21

2

21

4 1( ) cos 2 1 .............(1)2 2 1

Put 0 in (1) we getL.H.S of (1) (0) 0

4 1R.H.S of (1)2 2 1

R.H.S L.H.S implies4 1 0

2 2 1

4 122 1

12 4 82 1

n

n

n

n

n

f x n xn

xf

n

n

n

n

Ques 2

0

1 1

2 2 32

00 0 0

Find the Fourier series expansion for ( ) ( ) in (0, 2 ).

( ) cos sin .............(1)2

1 1 1 ( )( ) ( )3( 1)

n nn n

f x x

af x a nx b nx

xa f x dx x dx

tion : 5 : AUC M / J 2012

Answer :

2

23 3 3

0

3 3

3 2

20

2

02

2

0

2

1 1( ) ( 2 ) ( 0)3 31 ( )

31 22

3 32 .........................(2)3

1 ( ) cos .

1 ( ) cos .

1 sin cos( ) 2( )( 1)

n

x

a

a f x nxdx

x nxdx

nx nxx xn n n

2

0

sin(2)( 1)( 1)

nxn n n



22 2

2 2 0 00

2 2

2

2

2

2

0

1 2( ) cos 2 ( ) cos since sin terms 0

2 2( 2 )cos 2 ( 0) cos 0

2 2

1 4 since cos 2 1

4 ..................................(3)

1 ( )sin

n

n

x nx x nxn n

nn n

n

nn

an

b f x nxdx

22

0

2

22

30

2 23 3

1 ( ) sin

1 cos sin cos( ) 2( )( 1) (2)( 1)( 1)

1 2cos cos( )

1 2cos 2 cos 2 2cos 0 cos0( 2 ) ( 0)

x nxdx


nx nxxn n

n nn n n n

2

0

22

3 3

2 2

3 3

0

1 1

since sin terms 0

1 2 1 2( )

1 2 2 0

0 ........................(4)

use (2),(3), and (4) in (1) we get

( ) cos sin2

n

n nn n

n n n n

n n n n

b

af x a nx b nx

2

21 1

2

21

243 cos 0sin

214 cos

3

n n

n

nx nxn

nxn



2

0

1

32

00 0 0

3

Find the half range Fourier cosine series of ( ) ( ) in (0, )

( ) cos .............(1)2

2 2 2 ( )( ) ( )3( 1)

2 ( )3

nn

f x x

af x a nx

xa f x dx x dx

x

Question : 6 : AUC M / J 2012

Answer :

3 3

0

3

3 2

20

0

2

0

2

2 ( ) ( 0)3

2 031 22

3 32 .........................(2)3

2 ( )cos .

2 ( ) cos .

2 sin cos sin( ) 2( )( 1) (2)( 1)( 1)

n

a

a f x nxdx

x nxdx

nx nx nxx xn n n n

2

0

22 2 0 0

0

2 2 2

0

2 2( )cos 4 ( ) cos since sin terms 0

4 4 40 ( 0) cos0 ..................................(3)

Use (2) and (3) in (1) we get

( )2

n

n n

x nx x nxn n

an n n

af x a

1

22 20

10

4

44

10

5 4

41

0

cos ..............(1)

1 1( )4 24

1 1 1694 2

1 185 1 9

nn

nn

n

n

nx

af x dx a

x dxn

xn



4 4

41

4 4 4

41

4 4

41

189 5

1 485 9 45

1 4 145 8 90

Find the Fourier series expansion of ( ) sin in .

Since ( ) sin odd×odd even , 0 and hence the Fourie

n

n

n

n

n

n

n

f x x x x

f x x x b


Answer :

0

1

0 00 0

0 0

0

rseries expansion is given by

( ) cos .......................(1)2

2 2 2( ) sin ( cos ) (1)( sin )

2 2 2cos sin cos cos 0

2 ( 1) 2, 2 ...................

nn

af x a nx

a f x dx x xdx x x x

x x x x x

a

0 0 0

0 0

1 20 0

1

..........(2)

2 2 2( ) cos sin cos cos sin

2 2 1cos sin sin( 1) sin( 1)2

1 1sin( 1) sin( 1) I I ........................(*)

I sin(

na f x nxdx x x nxdx x nx xdx

x nx xdx x n x n x dx

x n xdx x n xdx

x

0 0

0

1

12

cos( 1) sin( 1)1) (1)( 1) ( 1

1 cos( 1)( 1)

1 cos( 1) 0 1( 1) ( 1)

Similarlly I 1( 1)

n

n

n x n xn xdx xn n

x n xn

nn n

n



1 11 2

1 1 1 1

1

1 12

11 22 2

I I 1 1( 1) ( 1)

1 1 Since 1 1( 1) ( 1)

1 11( 1) ( 1)

1 1 21 1( 1)( 1) 1

2 2 ( 1)( 1) ( 1) I I1 1

n n

n n n n

n

n n

nn

n n

n n

n n

n nn n n

n n

2

2 2 2

0

1

2 ( 1) ...........( )1

Use (a) in (*) we get

1 2 ( 1) 2( 1) 2( 1) , 1 ....................(3)1 1 1

Use (2),(3) and (4) in (1) we get

( ) cos2

n

n n n

n n

nn

an

a a nn n n

af x a nx

01

2

22

22

2

0

1

cos cos22 1 2( 1)cos cos2 2 1

1 ( 1)1 cos 2 cos2 1

Find the Fourier series for ( ) 2 in 0 3.

Here 2 332

( ) cos2

nn

n

nn

n

n nn

a a x a nx

x nxn

x nxn

f x x x x

l

l

a n xf x a bl


Answer :

1

0

1 1

0

1 1

sin .

( ) cos sin .3 322 2

2 2( ) cos sin ...............(1)2 3 3

n

n nn n

n nn n

n xl

a n x n xf x a b

a n x n xf x a b



32 3 2 3

20

0 0 0

332

0

0

2 32

0 0

2

1 2 2( ) 2 23 3 3 2 32

2 2 27 29 (9 9) 03 3 3 3 3

0 .....................(2)

1 2 2 2( )sin (2 )sin3 3 3

2cos2 3(2 ) ( 223

3

l

l

n

x xa f x dx x x dx

xx

a

n x n xb f x dx x x dxl

n x

x x n

2 2

2sin32 )

49

n x

xn

3

3 3

03

23 3

0

3 3 3 3

3 3

2cos3( 2)

1681

2 3 2 81 2(2 ) cos 2cos3 2 3 16 3

2 3 81 813 cos 6 cos 6 0 (cos 0)3 2 8 8

2 9 813 2 8

n x

n

n x n xx xn n

n nn n n

n n

3 3

818n

2 32

0 0

22 2 3 3

3

3 ...................(3)

1 2 2 2( ) cos (2 ) cos3 3 3

2 2 2sin cos sin2 3 3 3(2 ) (2 2 ) ( 2)2 4 1633 9 81

n

l

n

n

bn

n x n xa f x dx x x dxl

n x n x n x

x x xn n n

3

0

3

3

2 2 2 20

0

2 2

2cos2 2 9 23(2 2 ) cos43 3 4 3

93 cos 2 cos 0 0

20 ...................(4)n

n xn xx

n n

nna



0

1 1

1

Use (2), (3), and (4) in (1) we get

( ) cos sin2

3 sin

, 0 1Find the Fourier series expansion of ( ) . Hence show that

2 , 1 2

1

n nn n

n

af x a n x b n x

n xn

x xf x

x x


2

2 2 2

0

1 1

0

1 1

2 1 2

00 0 1

11 2 2

0 1 0

1 1 ...... .1 3 5 8

Here 2 2 1

( ) cos sin2

( ) cos sin2

1 1( ) ( ) ( )1̀

(2 ) 22

n nn n

n nn n

l

l la n x n xf x a b

l laf x a n x b n x

a f x dx f x dx f x dxl

xxdx x dx x

Answer :

22

1

2

0

2 1 2

0 0 1

1 2

1

10

2

1 1 4 1 1 10 2(2) 2 2 2 02 2 2 2 2

0 .................(2)

1 1( ) cos cos ( ) cos 1

I I .

siI cos

l

n

x

l

a

n xa f x dx x n xdx l x n xdxl l

x n xdx x

n n xn

1

2 20

1

2 20

cos(1)

1 cos

n xn

n xn



2 2

2 2 2 2

2

21

1 cos cos 0

1 1( 1) 1 1 ( 1)

sinI (2 ) cos (2 )

n n

nn

n n

n xx n xdx xn

2

2 21

22 2 1

2 2

2 2 2 2

1 2 2 2

2 2

cos( 1)

1 cos

1 cos 2 cos

1 11 ( 1) 1 ( 1)

1 2I I 1 ( 1)

4 , if is odd.

0 , if is even

n n

nn

n xn

n xn

n nn

n n

al n

nn

n

2 1 2

0 0 1

1 2

0 1

1 2

1

1 2 20

.....................(3)

1 ( )sin ( )sin ( )sin

sin (2 )sin

I I

cos sinI sin (1)

l

nn xb f x dx f x n xdx f x n xdx

l l

x n xdx x n xdx

n x n xx n xdx xn n

1

0

1

0

2

2 2 21

1 1cos cos 0

( 1)

cos sinI (2 )sin (2 ) ( 1)

n

x n x nn n

n

n x n xx n xdx xn n

2

12

2

1

1 1(2 ) cos cos 2 0 lx n x nn n n



1 2

0

1 1

2 2odd od

( 1) 1I I

1 11 ( 1) 1 ( 1)

2 , if is odd....................(4)

0, if is even

( ) cos sin2

0 4 2cos sin2

n

n

n n

n nn n

n n

bn n

n n

nn

naf x a n x b n x

n xn n

d

2 21 1

2 21

2

4 1 2 1( ) cos(2 1) sin(2 1) ............(*)(2 1) (2 1)

Put 0 in the above series we get. . of * (0) 2

4 1R.H.S of (*) cos 0(2 1)

R.H.S of (*) . . of *4 1

n n

n

n x

f x n x n xn n

xL H S f

nL H S

21

2

21

0

1

00

cos 0 2(2 1)1

(2 1) 8

Obtain the half range Fourier cosine series for ( ) in (0, )

( ) cos ..........................(1)2

2 2( )

n

n

nn

n

n

f x x

af x a nx

a f x dx xd

Question :10 : AUC N / D 2012, 2010

Answer :

22

00 0

0 0

20

2 2 20

2 1 0 ......................(2)2

2 2( ) cos cos

2 sin cos(1)

2 2 2cos cos cos 0 ( 1) 1

n

n

xx a

a f x nxdx x nxdx

nx nxxn n

nx nn n n



2

2

0

1

2odd

21

2 1 ( 1)

4 if is odd...................(3)

=0,if is evenUse (2) and (3) in (1) we get

( ) cos2

4 cos2

4 1 cos(2 1)2 (2 1)

4 1Hence ( )2 (2

n

nn

n

n

n

nn

n

af x a nx

nxn

n xn

f xn

21

2

2 2 2

0

1 1

cos(2 1)1)

, 0Find the Fourier series expansion for ( ) .Hence show that

2 , 2

1 1 1 ...... .1 3 5 8

( ) cos sin .............2

n

n nn n

n x

x xf x

x x

af x a nx b nx

Question :11: AUC N / D 2010

Answer :

2 2

00 0

2

0

22 2

0

2 2 2

2 2 22 2

(1)

1 1( ) ( ) ( )

1 (2 )

1 22 2

1 (2 )0 2 (2 ) 2 ( )2 2 2

1 44 22 2 2

a f x dx f x dx f x dx

xdx x dx

x xx

2

0

1

......................(2)a



2

0

2

0

2

1 20

10 0

2 2 00

1 ( ) cos

1 ( )cos ( ) cos

1 1cos (2 ) cos I I ..........(*)

sin cosI cos (1)

cos 1 cos

na f x nxdx

f x nxdx f x nxdx

x nxdx x nxdx

nx nxx nxdx xn n n

nx nxn n

2

2 2

2

2

2

2

22 2

2

1 cos cos 0

1 1( 1) 1 1 ( 1) ...........................( )

I (2 )cos

sin cos(2 ) ( 1)

1 1cos cos 2 cos

1 1 ( 1) ............

n n

n

nn

an n

x nxdx

nx nxxn n

nx n nn n

n

2 2

2

2

2

0

........( )

Use (a) and (b) in * we get 1 1 11 ( 1) 1 ( 1)

1 2 1 ( 1)

4 if is odd................................(3)

0 if is even

1 ( )sin

1 (

n nn

n

n

b

an n

n

nn

n

b f x nxdx

f

2

0

2

0

)sin ( )sin

1 sin (2 )sin

x nxdx f x nxdx

x nxdx x nxdx



1 2

10

0 0

200

0

1 I I ..............................(A)

I sin .

cos sinsin (1)

cos sinsince 0

1 1cos cos 0

1 ( 1)n

x nxdx

nx nxx nxdx xn n n

nx nxxn n

x nx nn n

n

1

2

2

2

2

2

( 1)I .............................( )

I (2 )sin

cos sin(2 ) ( 1)

1 (2 ) cos

1 (2 2 ) cos 2 (2 ) cos

1 cos

( 1) ...................

n

n

cn

x nxdx

nx nxxn n

x nxn

n nn

nn

n

1 2

0

1

................( )

use (c) and (d) in (A) we get

1 1 ( 1) ( 1)I I

0 ....................................(4)

Use (2) ,(3) and (4) in (1) we get

( ) cos2

n n

n

n

nn

d

bn n

b

af x a nx

1

2 21,3,5 1 1

21

sin

4 4 1cos 0sin cos(2 1)2 2 (2 1)

4 1( ) cos(2 1)2 (2 1)

nn

n n n

n

b nx

nx nx n xn n

f x n xn



21

21

21

21

2

4 1( ) cos(2 1) .............(*)2 (2 1)

Put 0 we get L.H.S of (*) 04 1R.H.S of (*) cos0

2 (2 1)R.H.S of (*) L.H.S of (*)

4 1 02 (2 1)

4 1(2 1) 21

(2 1)

n

n

n

n

f x n xn

x

n

n

n

n

2

1

2

2 2 2

2

2 4 8

1 1 1Hence ......1 3 5 8

Find the Fourier series expansion for ( ) 1 in the interval ( 1,1)

Here 1Since ( ) is even the Fourier series expa

n

f x x

lf x


Answer :

0

1

00

11 32

0 00 0

1

0 0

nsion is

given by ( ) cos .............(1) , 2

2where ( ) ,

1 4 42 1 2 2 1 .....................(2)3 3 3 3

2 ( ) cos 2 ( ) cos 2

nn

l

l

n

af x a n x

a f x dxl

xa x dx x a

n xa f x dx f x n xdxl l

12

0

2

1 cos

sin2 1

x n xdx

n xxn

2 2 3 3

cos sin( 2 ) ( 2)n x n xxn n

1

0

12 2 2 2 2 20

2 2

4 4 4cos cos 0 ( 1)

4Hence ( 1) ...............(3)

n

nn

x n x nn n n

an



0

1

2 21 1

2

Use (2) and (3) in (1) we get

( ) cos .............(1)2

4 2 43 cos ( 1) cos2 3

Find the Fourier series expansion for ( ) in the

nn

nn

n n

af x a n x

a n x n xn

f x x x


0

1 1

0

0

interval

Since ( ) is neither even nor odd , the Fourier series expansion is

given by ( ) cos sin ..............(1) , 2

1where ( ) ,

1 ( )

n nn n

l

l

l x l

f xa n x n xf x a b

l l

a f x dxl

a f x dxl

Answer :

2 32

2 3 2 3 2 3 2 3 3

3

0

2

2

1 1( )2 3

1 1 22 3 2 3 2 3 2 3 3

2 .....................(2)3

1 1( )cos ( ) cos .

sin1 ( )

ll l

l l l

l l

nl l

x xx x dxl l

l l l l l l l l ll l

la

n x n xa f x dx x x dxl l l l

n xlx x nl

l

2 2 3 3

2 3

cos sin(1 2 ) ( 2)

n x n xl lx

n nl l

2

2 2

2 2

2

2 2 2 2

2

2 2

1 (1 2 ) cos

(1 2 ) cos (1 2 ) cos since cos( ) cos

( 1) 4 ( 1)1 2 1 2

4 ( 1) .................(3)

l

ll

l

n n

n

n

l n xxl n l

l l n l nnl ll ln n

lan



2

2

0

2 2

20

1 1( )sin ( )sin .

1 sin sin

1 sin

2 sin

cos sin2 (1)

l l

nl l

l l

l l

l

l

l

l

n x n xb f x dx x x dxl l l l

n x n xx dx x dxl l l

n xx dxl l

n xx dxl l

n x n xl lx n nl

l l

0

0

1 1

2 2

2 2

2 cos

2 2cos 0 ( 1)

2 ( 1) ..........................(4)

Use (2) , (3) and (4) in (1) we get

( ) cos sin2

4 ( 1) cos3

l

n

n

n

n nn n

n

n

l n xxl n l

ll nn n

lbn

a n x n xf x a bl l

l l n xn l

1 1

2 2

2 21 1

2

2 ( 1) sin

4 ( 1) 2 ( 1)cos sin3

Find the half range Fourier cosine series expansion for in the interval 0 and hence

deduce

n

n

n n

n n

l n xn l

l l n x l n xn l n l

x x

Question :14 : AUC M / J 2013

4

4 4

0

1

00 0

1 1that .....1 2 90

( ) cos .......................(1)2

2 2Where ( ) , ( ) cos

nn

n

af x a nx

a f x dx a f x nxdx

Answer :



32

00 0 0

233

0

20

0

2

0

2

2 2 2( )3

2 2 203 3 3

2 .........................(2)3

2 ( ) cos

2 cos

2 sin cos sin(2 ) (2)

n

xa f x dx x dx

x

a

a f x nxdx

x nxdx


02

22 2 0 0

0

2 2 2

2

2

21

2 2 cos 4 cos since sinterms 0

4 4 4cos 0 ( 1) ( 1)

4 ( 1) ..................................(3)

Use (2) and (3) in (1) we get 2

43( ) ( 1) cos2

n n

nn

n

n

x nx x nxn n

nn n n

an

f x nxn

2

21

2

21

22 20

10

22

0 2

22

222

0

( 1)4 cos3

( 1)( ) 4 cos3

By parseval's identity we have

1 1( ) ( ).4 2

2 4( 1)Put , , ( ) , we get3

231 1 4( 1)4 2

n

n

n

n

nn

n

n

n

nxn

f x nxn

af x dx a

a a f x xn

x dxn

2

1

0)n



42

44

10

5 4 2

410

45

4 4 40

45

4 4 4

4 4

4 4 4

4 4

4 4 4

41 1 16( 1)9 .

4 2

1 ( 1)8 .5 9

1 1 1 185 9 1 2 31 1 1 10 8

5 9 1 2 31 1 18

5 9 1 2 31 1 18

5 9 1 2 3

n

n

n

n

x dxn

xn

x

4 4

4 4 4

4 4 4

4 4 4 4 4 4

4

4 4 4

4

4 4 4

1 1 181 2 3 5 91 1 1 9 5 1 1 1 48 81 2 3 45 1 2 3 45

1 1 1 41 2 3 45 81 1 11 2 3 90

Find the half range Fourier co


2

2 2

0

1

0

1

00

sine series expansion for ( 1) in the interval 0 1 and hence1 1deduce that .....1 2 6

Here 1

( ) cos .........(1)2

( ) cos .........(1)2

2= ( ) 2

nn

nn

l

x x

la n xf x a

laf x a n x

a f x dx xl

Answer :

131

2

0 0

0

11 2

3

1 2 22 0 .................(2)3 3 3

xdx

a



12

0 0

2

2 ( ) cos 2 1 cos

sin2 1

l

nn xa f x dx x n xdx

l l

n xxn

2 2 3 3

cos sin2( 1) 2n x n xxn n

1

0

12 2 2 2 2 20

2 2

2 21,3,5....

2 21

4 4 4( 1) cos 0 cos 0

4 ................(3)

Use (2) and (3) in (1) we get2 43( ) cos2

1 4 1( ) cos ..............(*)3

Put 0 in

n

n

n

x n xn n n

an

f x n xn

f x n xn

x

2 21

2 2

2 21

2 21

2 2

21

(*) we get1 4 1R.H.S of (*) cos 03

(0) (1) (0 1) (1 1) 1L.H.S of (*) (0)2 2 2

R.H.S of (*) L.H.S of (*)1 4 1 13 24 1 1 1 3 2 1

2 3 6 61 1

6 4 24

n

n

n

n

nf ff

n

n

n


0

1

2

00 0

2

, 02Find the half range Fourier cosine series for ( )

( ),2

( ) sin ................(1)2

2 2( ) ( )

nn

ll l

l

lkx xf x

lk l x x l

a n xf x al

a f x dx kx dx k l x dxl l

013

Answer :



2 22 2

0 022

2 2 2 2 2 2 2 22

2

0

0

2

0

2 2( )2 2

2 28 2 2 8 8 2 2 8

2 28 2 2

2 ( )cos

2 ( ) cos

l l ll

ll

l

n

l

k k x xx dx l x dx lxl l

k l l l l k l l l lll l

k l kl klal

n xa f x dxl l

f xl

2

2

1 20

2

2

2

1 2 20

20

2

2 2

( ) cos

2 2cos ( ) cos I I

sin cosI cos (1)

cos sin

l

l

ll

l

l

l

n x n xdx f x dxl l

n x n x kkx dx k l x dxl l l l

n x n xn x l lx dx x n nl

l l

l n x l nxn l n

2

0

2 2 2

2 2 2 2

2 2 2

2 2 2 2

2

2

2 2

2

2

cos sin cos 0 02 2 2

cos sin2 2 2

I ( ) cos

sin cos( ) ( 1)

l

l

l

l

xl

l n l n ln n n

l n l n ln n n

n xl x dxl

n x n xl ll x n n

l l

l



2

2 2

2

2 2

2 2 2 2

2 2 2

2 2 2 2

2 2

1 2 2 2

( )sin cos

0 cos sin sin2 2 2

cos sin cos2 2 2

I I cos sin2 2 2

l

l

l n x l n xl xn l n l

l l l n l nnn n n

l l n l nnn n n

l n l nn n

2 2 2

2 2 2 2 cos sin2 2

l l l nnn n n

2

2 2

2 2 2 2

2 2 2 2 2 2 2 2

2 2

2 2 2 2

2

2 2

2

2 2 2 2

cos2

2 2cos 1 cos cos 1 ( 1)2 2

2 2cos if is even, 0 otherwise2

2 cos 12

2 2 4cos 1 cos 12 2

n

n

l nn

l n l l n lnn n n n

l n l nn n

l nn

k l n kl nal n n

2 2

.

4 cos 1 ..................(3)2

Use (2) and (3) in (1) we get the result

nkl na

n



2 2

2 20

0 0

0

0

sin cos

cos2 2 2sin sin

2 2cos cos cos 0

2 2( 1) 1 1 ( 1)

4 , if is odd

0 if is even

l

l

l l

l

n n

n

l n x l n xxn l n l

n xn x k n x k lk dx dx nl l l l l

l

k l n x k nl n l nk k

n nk n

b nn

,

1

odd 1

..................(2)

Use (2) in (1) we get

( ) sin

4 4 1 (2 1)sin sin(2 1)

, 02Obtain the sine series for ( )

( ),2

nn

n n

n xf x bl

k n x k n xn l n l

lx xf x

ll x x l


1

0

2

02

2

1 20

2

( ) sin

2 ( )sin

2 ( )sin ( )sin

2 2sin ( )sin I I

nn

l

n

ll

l

ll

l

n xf x bl

n xb f x dxl l

n x n xf x dx f x dxl l l

n x n xx dx l x dxl l l l

Answer :



2

2

1 2 20

20

2 2

2 20

2 2

2 20

2 2 2

2 2 2 2

cos sinI sin (1)

sin cos

sin cos

sin cos sin 02 2 2

l

l

l

l

n x n xn x l lx dx x n nl

l l

l n x l n xxn l n l

l n x l n xxn l n l

l n l n ln n n

2 2

2 2

2

2

2 2

2

2

2

2 2

2

0

sin cos2 2 2

I ( )sin

cos sin( ) ( 1)

( ) cos sin

0 0 cos2 2

l

l

l

l

l

l

l n l nn n

n xl x dxl

n x n xl ll x n n

l l

l n x l n xl xn l n l

l l n ln

2

2 2

2 2

2 2

2 2 2 2

1 2 2 2 2 2

2

2 2

2

1 2 2 2

2 2 2 2

sin2

cos sin2 2 2

I I sin cos cos sin2 2 2 2 2 2

2 sin .2

2 2 2I I sin2

4 4sin sin ........(2)2 2

Use (2) i

n

n

nn

l n l nn n

l n l n l n l nn n n n

l nn

l nbl l n

l n l nbn n

n (1) we get



2 21 1

2 21

2

0

1 1

0

4( ) sin sin sin2

4 1 sin sin2

Find the Fourier series for ( ) 2 in 0 2.

Here 2 21

( ) cos sin .2

nn n

n

n nn n

n x l n n xf x bl n l

l n n xn l

f x x x x

ll

a n x n xf x a bl l

a


Answer :

1 122 2 2 3

20

0 0 023

2

0

0

2 22

0 0

2

cos sin ..............(1)2

1 ( ) 2 22 3

8 12 8 443 3 3 3

4 .....................(2)3

1 ( )sin 2 sin

c2

n nn n

l

l

n

a n x b n x

x xa f x dx x x dxl

xx

a

n xb f x dx x x n xdxl l

x x

2

2 2 3 30

22

3 30

3 3 3 3

2

os sin cos(2 2 ) ( 2)

cos cos2 2

2cos 2 2cos 00 0

0 ...................(3)

(2 )cos

n

n

n x n x n xxn n n

n x n xx xn n

nn n

b

a x x n x

2

0

22

2 2 3 30

2

2 20

2 2 2 2

sin cos sin(2 ) (2 2 ) ( 2)

cos(2 2 )

cos 2 cos 0(2 4) (2 0)

dx

n x n x n xx x xn n n

n xxn

nn n



2 2 2 2 2 2 2 2

0

1 1

2 2 2 21 1 1

2 2 4 4 ...................(4)

Use (2), (3), and (4) in (1) we get

( ) cos sin2

2 4 2 4 1cos 0sin cos3 3

n

n nn n

n n n

an n n n

af x a n x b n x

n x n x n xn n

Question :19 : AUC M /

0

1

2 32

00 0 0

3 3 3 3 3

Find the half range cosine series of ( ) ( ) , when 0

( ) cos .............(1),2

2 2 2( )2 3

2 2 3 2 22 3 6 6

nn

f x x x x

af x a nx

x xa f x dx x x dx

J 2010

Answer :

2

2

0

0

2

0 0

2

3

Hence ..............(2)3

2 a ( ) cos

2 2( ) cos ( )cos

2 sin( )

n

a

f x nxdx

x x nxdx x x nxdx

nxx xn

2 3

cos sin( 2 ) ( 2)nx nxxn n

0

2 20

2 2

2

2 2( 2 ) cos ( 2 ) cos ( 0) cos0

2 2( 1) 1 ( 1)

4 , if is even..............(3)

0 , if is oddUse (2) and (3) in (1) we get the r

n n

x nx nn n

n n

nn

n

esult



0

The process of finding the Fourier series for a given function by numerical value is known as harmonic analisis. In harmonic analisis the Fourier coefficients a

Definition :HARMONIC ANALYSIS

0

1 1

, and of the function ( ) in (0, 2 ) given by

cos sin2 , 2 , 2

where 'n' is the number of terms appears in the table.

The term ( cos sin ) is called t

n n

n n

a b y f x

y y nx y nxa a b

n n n

a x b x

Note :1

2 2

he fundamental or first harmonic,the term ( cos 2 sin 2 ) is called the second harmonic and so on.

Find the Fourier series expansion of period 2 for the function ( )which is defined in

a x b x

y f x

Problem : 20 :

(0, 2 ) by means of the table of values given below.Find the series upto second harmonic.

x 0 3

23

4

3

53

2

y 1.0 1.4 1.9 1.7 1.5 1.2 1.0

0

1 2

Since the last value of is repitition of the firstTherefore only the first six values will be used, ( . ) 6

8.72 2 2.96

cos 1.1 0.32 2 0.37 2 06 6

yi e n

ya

n

y xa a

n

Solution :

1 2

01 1 2 2

.1

0.5196 0.17322 0.17 2 0.066 6

( ) cos sin cos 2 sin 2 .2

1.45 0.37cos 0.17sin 0.1cos 2 0.06sin 2

b b

ay f x a x b x a x b x

x x x x



The following table gives the variations of a periodic functionover a period T.

Problem : 21

x 0 6T

3T

2T

23T

56T

T

( )f x 1.98 1.3 1.05 1.03 0.88 2.5 1.98

x y cos x

cos 2x

sin x sin 2x cosy x cos 2y x siny x sin 2y x

0 1.0 1 1 0 0 1 1 0 0

3

1.4 0.5 0.5 0.866 0.866 0.7 0.7 1.212 1.212

23

1.9 0.5 0.5 0.866 0.866

0.95 0.95 1.65 1.645

1.7 1 1 0 0 1.7 1.7 0 0

43

1.5 0.5 0.5 0.866

0.866 0.75 0.75 1.299 1.299

53

1.2 0.5 0.5 0.866

0.866

0.6 0.6 1.039 1.039

8.7 1.1 0.3 0.5196 0.1732

y

cosy x

cos 2y x

siny x

sin 2y x



2Show that ( ) 0.75 0.37cos 1.004sin , where = .

Here the last value is the mere repetition of the first therefore we omit that value and consider the remaining 6 values.(i.e) 6.

Therefore

xf xT

n

Solution :

01 1

0

1

2 5 when takes the values 0, , , , , takes the 6 3 2 3 6

2 4 5values 0, , , , , .3 3 3 3

Let the Fourier series be of the form

( ) cos sin 0.75 0.37cos 1.004sin2

2 1.5

cos2 0.3

T T T T Tx

af x a b

ya

n

ya

n

1

7

sin2 1.00456

yb

n



y cos sin cosy siny

00 1.98 1.0 0 1.98 0

3

1.30 0.5 0.866 0.65 1.1258

23

1.05 0.5 0.866 0.525 0.9093

1.30 1 0 1.3 0

43

0.88 0.5 0.866 0.44 0.762

53

0.25 0.5 0.866 0.125 0.2165

4.6 1.12 3.013



Find the Fourier series as far as the second harmonic to representthe function given in the following data.

: 0 1 2 3 4 5( ) : 9 18 24 28 26 20

Solution:Here the length of the interval is 6( . ) , 2 6 (or)

xf x

i e l

Problem: 22

2 20

1 1

01 2 1 2

0

1

2

3

( ) cos sin .2

2 2cos cos sin sin (1)2 3 3 3 3

125Now , 2 2 41.666

cos cos ( 25)3 32 2 2 8.336 6

co2

n nn n

la n x n xf x a b

l la x x x xa a b b

ya

n

x xy ya

n

ya

1

2

2 2s cos ( 19)3 32 2 6.336 6

sin sin ( 3.4)3 32 2 2 1.136 6

2 2sin sin (20.8)3 32 2 2 6.96 6

x xy

n

x xy yb

n

x xy yb

n



x

3x

2

3x

y

cos3xy

sin3xy

2cos

3xy

2sin

3xy

0 0 0 9 9 0 9

0

1 3

23

18 9 15.7 -9 15.6

2 23

43

24 -12 20.9 -24 0

3 2 28 -28 0 28 0

4

43

83

26 -13 -22.6 -13 22.6

5 53

10

3

20 10 -17.4 -10 -17.4

-25 -3.4 -19 20.8

QUESTION BANK - ANSWERS SEMESTER: III MA 2211 - … UNIT... · Find the root mean square value of...

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Transcript of QUESTION BANK - ANSWERS SEMESTER: III MA 2211 - … UNIT... · Find the root mean square value of...