TPDE_UNIT II-FOURIER SERIES_PPT.pptx

1. TRANSFORMS AND PARTIAL DIFFERENTIAL UNIT II: FOURIER SERIES

2. Periodic Functions

3. 0  T f   

4.            1 1 0 sin cos 2 n n n n n b n a a f   

5. Definition of a Fourier series A Fourier series may be defined as an expansion of a function in a series of sines and cosines such as The coefficients are related to the periodic function f(x) Henceforth we assume f satisfies the following (Dirichlet) conditions: (1) f(x) is a periodic function; (2) f(x) has only a finite number of finite discontinuities; (3) f(x) has only a finite number of extrem values, maxima and minima in the interval [0,2p].            1 1 0 sin cos 2 n n n n nx b nx a a x f

6. 0 0 p 2p cos  cos 2 cos 3

7. 0 0 p 2p sin  sin 2 sin 3

8. A Fourier series is a convenient representation of a periodic function. A Fourier series consists of a sum of sines and cosine terms. Sines and cosines are the most fundamental periodic functions.

9. Fourier series   p p ,    p 2 , 0            1 1 0 sin cos 2 n n n n nx b nx a a x f

10.  dx x f a   p p p 1 0    , 2 , 1 cos 1    n dx nx x f an p p p    , 2 , 1 sin 1    n dx nx x f bn p p p FOURIER COEFFICIENTS FOR (-π,π)

11. FOURIER COEFFICIENTS FOR (0,2π)  dx x f a   p p 2 0 0 1    , 2 , 1 cos 1 2 0    n dx nx x f an p p    , 2 , 1 sin 1 2 0    n dx nx x f bn p p

12. Fourier series ) , ( l l  ) 2 , 0 ( l                        1 1 0 sin cos 2 n n n n l x n b l x n a a x f p p

13.  dx x f l a l l   1 0 FOURIER COEFFICIENTS FOR ) , ( l l           l l n dx l x n x f l a p cos ) ( 1          l l n dx l x n x f l b p sin ) ( 1

14. FOURIER COEFFICIENTS FOR ) 2 , 0 ( l  dx x f l a l   2 0 0 1         l n dx l x n x f l a 2 0 cos ) ( 1 p         l n dx l x n x f l b 2 0 sin ) ( 1 p

15. Interval Parsevals Identity ) , ( l l              1 2 2 2 0 2 2 ) ( 1 n n n l l b a a dx x f l            1 2 2 2 0 2 2 0 2 ) ( 1 n n n l b a a dx x f l             1 2 2 2 0 2 2 ) ( 1 n n n b a a dx x f p p p             1 2 2 2 0 2 2 ) ( 1 n n n b a a dx x f p p p   p p ,    p 2 , 0 ) 2 , 0 ( l

16. Even and Odd Functions Even Functions  f( The value of the function would be the same when we walk equal distances along the X-axis in opposite directions. Mathematically speaking -

17. Odd Functions  f( The value of the function would change its sign but with the same magnitude when we walk equal distances along the X-axis in opposite directions. Mathematically speaking -

18. Even functions can solely be represented by cosine waves because, cosine waves are even functions. A sum of even functions is another even function.

19. Odd functions can solely be represented by sine waves because, sine waves are odd functions. A sum of odd functions is another odd function.

20. The Fourier series of an even function   x f is expressed in terms of a cosine series. The Fourier series of an odd function   x f is expressed in terms of a sine series.       1 sin n n nx b x f        1 0 cos 2 n n nx a a x f

21. 1. Find the Fourier series expansion of f(x) = x2, 0 < x < 2π. Hence deduce that 8 .... .......... 5 1 3 1 1 1 ) ( 12 .... .......... 3 1 2 1 1 1 ) ( 6 .... .......... 3 1 2 1 1 1 ) ( 2 2 2 2 2 2 2 2 2 2 2 2 p p p             iii ii i

22. . Fourier series is       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f     p p p p 2 0 2 2 0 0 1 ) ( 1 dx x dx x f a p p 2 0 3 3 1        x         0 3 8 1 3 p p 3 8 2 p 

23.   p p 2 0 cos ) ( 1 dx nx x f an   p p 2 0 2 cos 1 dx nx x p p 2 0 3 2 2 sin ) 2 ( cos ) 2 ( sin ) ( 1                              n nx n nx x n nx x                    0 0 0 0 ) 1 ( ) 4 ( 0 1 2 n p p 2 4 n 

24.   p p 2 0 sin ) ( 1 dx nx x f bn   p p 2 0 2 sin 1 dx nx x p p 2 0 3 2 2 cos ) 2 ( sin ) 2 ( cos ) ( 1                              n nx n nx x n nx x                          3 3 2 2 0 0 2 0 4 1 n n n p p n p 4  

25.       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f                     1 2 2 sin 4 cos 4 3 8 2 1 n nx n nx n p p                      .. .......... 3 3 sin 2 2 sin 1 sin 4 ..... .......... 3 3 cos 2 2 cos 1 cos 4 3 4 ) ( 2 2 2 2 x x x x x x x f p p Put x = 0 in the above series we get ) 0 ( 4 .. .......... 3 1 2 1 1 1 4 3 4 ) 0 ( 2 2 2 2 p p             f --------------- (1)

26. But x = 0 is the point of discontinuity. So we have 2 2 2 2 ) 4 ( ) 0 ( 2 ) 2 ( ) 0 ( ) 0 ( p p p      f f f Hence equation (1) becomes            .. .......... 3 1 2 1 1 1 4 3 4 2 2 2 2 2 2 p p            .. .......... 3 1 2 1 1 1 4 3 4 2 2 2 2 2 2 p p           .. .......... 3 1 2 1 1 1 4 3 2 2 2 2 2 p ) 2 ( ...... .......... 3 1 2 1 1 1 6 2 2 2 2            p

27. Now, put x = π (which is point of continuity) in the above series we get ) 0 ( 4 .. .......... 3 1 2 1 1 1 4 3 4 2 2 2 2 2 p p p                          .. .......... 3 1 2 1 1 1 4 3 4 2 2 2 2 2 p p             .. .......... 3 1 2 1 1 1 4 3 2 2 2 2 p ) 3 ( ...... .......... 3 1 2 1 1 1 12 2 2 2 2            p

28. Adding (2) and (3), we get            .. .......... 5 1 3 1 1 1 2 12 6 2 2 2 2 2 p p           .. .......... 5 1 3 1 1 1 2 12 3 2 2 2 2 p ...... .......... 5 1 3 1 1 1 8 2 2 2 2     p

29. 2. Expand in Fourier series of f(x) = x sinx for 0 < x < 2π and deduce the result 4 2 .......... 7 . 5 1 5 . 3 1 3 . 1 1      p Sol. Fourier series is       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f   p p 2 0 0 ) ( 1 dx x f a

30.   p p 2 0 sin 1 dx x x   p p 2 0 ) sin )( 1 ( ) cos ( 1 x x x       ) 0 0 ( ) 0 2 ( 1      p p 2  

31.   p p 2 0 cos ) ( 1 dx nx x f an   p p 2 0 cos sin 1 dx nx x x   p p 2 0 ) sin cos 2 ( 2 1 dx x nx x   1 , ) 1 sin( ) 1 sin( 2 1 2 0       n dx x n x n x p p

32.       p p p p 2 0 2 0 ) 1 sin( 2 1 ) 1 sin( 2 1 dx x n x dx x n x p p 2 0 2 ) 1 ( ) 1 sin( ) 1 ( 1 ) 1 cos( ) ( 2 1                             n x n n x n x p p 2 0 2 ) 1 ( ) 1 sin( ) 1 ( 1 ) 1 cos( ) ( 2 1                             n x n n x n x 1 ) 1 ( 1 ) 1 ( 2 2 2 2       n n Note:

33.                                            0 0 0 1 ) 1 ( 2 2 1 0 0 0 1 ) 1 ( 2 2 1 2 2 2 2 n n n n p p p p 1 1 1 1      n n ) 1 )( 1 ( ) 1 ( ) 1 (        n n n n an 1 , 1 2 2    n n an

34. When n = 1, we have   p p 2 0 1 cos ) ( 1 dx x x f a   p p 2 0 cos sin 1 dx x x x   p p 2 0 2 sin 2 1 dx x x p p 2 0 4 2 sin ) 1 ( 2 2 cos 2 1                       x x x                        ) 0 0 ( 0 2 1 2 2 1 p p 2 1  

35.     p p p p 2 0 2 0 sin sin 1 sin ) ( 1 dx nx x x dx nx x f bn   p p 2 0 ) sin sin 2 ( 2 1 dx x nx x   1 , ) 1 cos( ) 1 cos( 2 1 2 0       n dx x n x n x p p       p p p p 2 0 2 0 ) 1 cos( 2 1 ) 1 cos( 2 1 dx x n x dx x n x

36. p p p p 2 0 2 2 0 2 ) 1 ( ) 1 cos( ) 1 ( 1 ) 1 sin( ) ( 2 1 ) 1 ( ) 1 cos( ) 1 ( 1 ) 1 sin( ) ( 2 1                                                       n x n n x n x n x n n x n x                                                     2 2 2 2 2 2 2 2 ) 1 ( 1 0 ) 1 ( ) 1 ( 0 2 1 ) 1 ( 1 0 ) 1 ( ) 1 ( 0 2 1 n n n n n n p p                                                 2 2 2 2 ) 1 ( 1 0 ) 1 ( 1 0 2 1 ) 1 ( 1 0 ) 1 ( 1 0 2 1 n n n n p p 1 , 0   n bn

37. When n = 1, we have     p p p p 2 0 2 0 1 sin sin 1 sin ) ( 1 dx x x x dx x x f b   p p 2 0 2 sin 1 dx x x          p p 2 0 2 2 cos 1 1 dx x x p p 2 0 2 4 2 cos ) 1 ( 2 2 sin 2 2 1                             x x x x                               2 1 0 0 2 1 0 2 2 1 2 p p p 

38. ) sin cos ( 2 ) ( 1 0       n n n nx b nx a a x f            2 1 2 1 0 sin sin cos cos 2 n n n n nx b x b nx a x a a 0 sin cos ) 1 )( 1 ( 2 cos 2 1 2 2 2            x nx n n x n p                ........ .......... 6 . 4 5 cos 5 . 3 4 cos 4 . 2 3 cos 3 . 1 2 cos 2 sin cos 2 1 1 sin x x x x x x x x p

39. Put x = in the above series we get                    ........ .......... 0 7 . 5 1 0 5 . 3 1 0 3 . 1 1 2 ) 1 ( 0 1 ) 1 ( 2 p p              ........ .......... 7 . 5 1 5 . 3 1 3 . 1 1 2 1 2 p p              ........ .......... 7 . 5 1 5 . 3 1 3 . 1 1 2 2 2 2p p              ........ .......... 7 . 5 1 5 . 3 1 3 . 1 1 2 2 2 p ....... .......... 7 . 5 1 5 . 3 1 3 . 1 1 4 2      p

40. 3. Find the Fourier series of         p p p 2 , 2 0 , 1 ) ( x x x f Hence evaluate the value of the series ...... .......... 5 1 3 1 1 1 2 2 2    Sol. Fourier series is       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f

41.       p p p p p p p 2 0 2 0 0 ) 2 ( 1 ) 1 ( 1 ) ( 1 dx dx dx x f a     p p p p p 2 0 2 1 x x       ) 2 ( 2 ) 0 ( 1 p p p p p     3 2 1   

42.       p p p p p p p 2 0 2 0 cos ) 2 ( 1 cos ) 1 ( 1 cos ) ( 1 dx nx dx nx dx nx x f an p p p p p 2 0 sin 2 sin 1               n nx n nx ) 0 0 ( 2 ) 0 0 ( 1     p p 0 

43.       p p p p p p p 2 0 2 0 sin ) 2 ( 1 sin ) 1 ( 1 sin ) ( 1 dx nx dx nx dx nx x f bn p p p p p 2 0 cos 2 cos 1               n nx n nx ] ) 1 ( 1 [ 2 ] 1 ) 1 [( 1 n n n n        p p   n n n ) 1 ( 2 2 1 ) 1 ( 1        p p n n 1 ) 1 (   

44.       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f               1 sin 1 ) 1 ( cos . 0 2 3 n n nx n nx p            ..... .......... .......... 3 3 sin 2 2 sin 1 sin 2 2 3 x x x p

45. When we put x = 0, , π, 2π we will not get the given series. So, using Parseval’s identity for Fourier series we have        p p 2 0 1 2 2 2 0 2 ) ( 2 )] ( [ 1 n n n b a a dx x f                      p p p p p p 0 2 1 2 2 2 2 2 2 1 ) 1 ( 0 2 ) 3 ( ) 2 ( 1 ) 1 ( 1 n n n dx dx                    ........ .......... 0 5 4 0 3 4 0 1 4 1 2 9 4 1 2 2 2 2 2 0 p p p p p p x x                   ........ .......... 5 1 3 1 1 1 4 2 9 2 4 0 1 2 2 2 2 p p p p p p

46.            ........ .......... 5 1 3 1 1 1 4 2 9 5 2 2 2 2 p           ........ .......... 5 1 3 1 1 1 4 2 1 2 2 2 2 p ....... .......... 5 1 3 1 1 1 8 2 2 2 2     p

47. 4.Find the Fourier series expansion for the function p p p 2 2 0 , ) ( period with x in x x f     ... 5 1 3 1 1    of sum the deduce Hence Solution:       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f   p p 2 0 0 ) ( 1 dx x f a

48.    p p p 2 0 ) ( 1 dx x p p p 2 0 2 2 ) ( 1          x   2 2 2 1 p p p    0 

49.    p p p 2 0 cos ) ( 1 nxdx x   p p 2 0 cos ) ( 1 nxdx x f an                                           p p p p 2 0 2 2 0 cos 1 sin 1 n nx n nx x   0 1 1 1 2    p n

50.   p p 2 0 sin ) ( 1 nxdx x f bn    p p p 2 0 sin ) ( 1 nxdx x                                             p p p p 2 0 2 2 0 sin 1 cos 1 n nx n nx x n 2               0 1 1 p p p n

51. Hence       1 sin 2 ) ( n n nx x x f p 2 ... 5 1 3 1 1 p     x putting by obtained be can series required The      1 sin 2 2 n n nx p p            ... 2 4 sin 4 1 2 3 sin 3 1 2 2 sin 2 1 2 sin 2 2 p p p p p           ... 5 1 3 1 1 2 2 p 4 ... 5 1 3 1 1 p          

52.          p p p 2 0 2 2 1 dx x p p p 2 0 3 3 ) ( 4 1          x   6 12 1 2 3 3 p p p p      6 2 p 

53. 4.Find the Fourier series expansion for the function 6 1 Pr . 2 2 0 , 2 ) ( 2 2 1 2 p p p p               n that ove Hence period with x in x x f n Solution:       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f   p p 2 0 0 ) ( 1 dx x f a

54.   p p 2 0 cos ) ( 1 nxdx x f an          p p p 2 0 2 cos 2 1 nxdx x                                                                p p p p p p 2 0 3 2 0 2 2 0 2 sin 1 2 cos 2 sin 4 1 n nx n nx x n nx x                       0 cos 2 0 4 1 2 0 2 p p p nx x n   p p p     2 4 2 n 2 1 n 

55.   p p 2 0 sin ) ( 1 nxdx x f bn          p p p 2 0 2 sin 2 1 nxdx x                                                                p p p p p p 2 0 3 2 0 2 2 0 2 cos 1 2 sin 2 cos 4 1 n nx n nx x n nx x               1 1 2 0 1 4 1 3 2 2 n n p p p =0

56. Hence 0  x putting      1 2 2 2 1 12 4 n n p p             1 2 2 2 cos 12 2 n n nx x p p 6 12 4 1 2 2 2 1 2 p p p       n n

57. 2 1 1 4 1 2 0 int cos 1 exp . 5 1 2         n n that deduce hence and x erval the in x of ansion fourier the Obtain p Solution:       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f   p p 2 0 0 ) ( 1 dx x f a

58. 2 1 1 4 1 2 0 int cos 1 exp . 5 1 2         n n that deduce hence and x erval the in x of ansion fourier the Obtain p Solution:       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f   p p 2 0 0 ) ( 1 dx x f a

59.    p p 2 0 0 cos 1 1 dx x a   p p 2 0 2 sin 2 1 dx x p p 2 0 2 1 2 cos 2           x   p p 2 4 1 1 2 2     

60.   p p 2 0 cos ) ( 1 nxdx x f an    p p 2 0 cos cos 1 1 dx nx x   p p 2 0 cos 2 sin 2 1 nxdx x

61. dx x n x n                        p p 2 0 2 1 sin 2 1 sin 2 1 2 p p 2 0 2 1 2 1 cos 2 1 2 1 cos 2 2                                n x n n x n 1 2 2 1 cos 1 2 2 1 cos                   p p n n Since

62.                2 4 1 2 1 2 1 2 n n n p         2 4 1 4 2 n p                                                n n n n 2 1 1 2 1 1 2 1 1 2 1 1 2 2 p               n n 2 1 1 2 1 1 2 2 2 p

63.   p p 2 0 sin ) ( 1 nxdx x f bn    p p 2 0 sin cos 1 1 dx nx x   p p 2 0 sin 2 sin 2 1 nxdx x dx x n x n                        p p 2 0 2 1 cos 2 1 cos 2 1 2

64. 0 2 1 2 1 sin 2 1 2 1 sin 2 2 2 0                                p p n x n n x n         1 2 cos 4 1 2 4 2 2 ) ( n nx n x f Hence p p

65.   2 1 1 4 1 0 ) 0 ( 1 4 2 4 2 2 ) 0 ( 0 1 2 1 2              n n n f n f x putting  p p

66. 6.Find the Fourier series for   p 2 , 0 ) ( in e x f ax  Solution:          1 1 0 sin cos 2 ) ( n n n n nx b nx a a x f   p p 2 0 0 ) ( 1 dx x f a   p p 2 0 1 dx eax p p 2 0 1        a eax   1 1 2   p p a e a

67.           p p p p a a a e e e a 1   p p p a a ea sinh 2  2 sinh x x e e x    

68.   p p 2 0 cos ) ( 1 nxdx x f an   p p 2 0 cos 1 nxdx eax   p p 2 0 2 2 sin cos 1          nx n nx a n a eax   0 2 sin 1 2 cos sin cos cos : 2 2       p p n n bx b bx a b a e bxdx e note ax ax 

69.     a ae n a a    p p 2 2 2 1              p p p p a a a e e e n a a 2 2   p p p a n a aea sinh 2 2 2  

70.   p p 2 0 sin ) ( 1 nxdx x f bn   p p 2 0 sin 1 nxdx eax   p p 2 0 2 2 cos sin 1          nx n nx a n a eax   bx b bx a b a e bxdx e note ax ax cos sin sin : 2 2    

71.     n ne n a a     p p 2 2 2 1     1 2 2 2     p p a e n a n   p p p a n a nea sinh 2 2 2   

72. 7.Find the fourier series expansion for the function f(x) = 1 + x + x2 in (–π, π). Deduce 6 .... .......... 3 1 2 1 1 1 2 2 2 2 p     Sol. The given function is neither an even nor an odd function.       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f         p p p p p p dx x x dx x f a ) 1 ( 1 ) ( 1 2 0

73. p p p           3 2 1 3 2 x x x                            3 2 3 2 1 3 2 3 2 p p p p p p p 3 2 2 3 2 2 1 2 3 p p p p          

74.         p p p p p p dx nx x x dx nx x f an cos ) 1 ( 1 cos ) ( 1 2 p p p                                  3 2 2 sin ) 2 ( cos ) 2 1 ( sin ) 1 ( 1 n nx n nx x n nx x x                             0 ) 1 ( ) 2 1 ( 0 0 ) 1 ( ) 2 1 ( 0 1 2 2 n n n n p p p

75.   p p p 2 1 2 1 ) 1 ( 2      n n 2 2 ) 1 ( 4 ) 4 ( ) 1 ( n n n n     p p n n n n n n ) 1 ( cos ) ( cos 0 sin ) 1 ( cos        p p p p

76.         p p p p p p dx nx x x dx nx x f bn sin ) 1 ( 1 sin ) ( 1 2 p p p                                  3 2 2 cos ) 2 ( sin ) 2 1 ( cos ) 1 ( 1 n nx n nx x n nx x x                                   3 2 3 2 ) 1 ( 2 0 ) 1 ( ) 1 ( ) 1 ( 2 0 ) 1 ( ) 1 ( 1 n n n n n n n n p p p p p

77.   2 2 1 1 ) 1 ( p p p p p         n n n n n n n n 1 ) 1 ( 2 ) 1 ( 2 ) 2 ( ) 1 (          p p                               1 1 2 2 1 0 sin ) 1 ( 2 cos ) 1 ( 4 3 2 2 2 1 ) sin cos ( 2 ) ( n n n n n n nx n nx n nx b nx a a x f p

78.                        ....... 3 3 sin 2 2 sin 1 sin 2 ...... 3 3 cos 2 2 cos 1 cos 4 3 1 2 2 2 2 x x x x x x p (i.e.)                       .. .......... 3 3 sin 2 2 sin 1 sin 2 .. .......... 3 3 cos 2 2 cos 1 cos 4 3 1 ) ( 2 2 2 2 x x x x x x x f p Put x = π in the above series we get ) 1 ..( )......... 0 ( 2 .. .......... 3 1 2 1 1 1 4 3 1 ) ( 2 2 2 2               p p f

79. But x = π is the point of discontinuity. So we have 2 2 2 2 1 2 2 2 2 ) 1 ( ) 1 ( 2 ) ( ) ( ) ( p p p p p p p p p              f f f Hence equation (1) becomes               .. .......... 3 1 2 1 1 1 4 3 1 1 2 2 2 2 2 p p            .. .......... 3 1 2 1 1 1 4 3 2 2 2 2 2 p p           .. .......... 3 1 2 1 1 1 4 3 2 2 2 2 2 p ...... .......... 3 1 2 1 1 1 6 2 2 2 2     p

80. . 8. Find the Fourier series expansion of (π – x)2 in –π < x < π. Sol. Fourier series is       1 0 ) sin cos ( 2 ) ( n n n nx b nx a a x f        p p p p p p p dx x dx x f a 2 0 ) ( 1 ) ( 1 p p p p           3 ) ( 1 3 x   3 8 0 3 1 p p    3 8 2 p 

81.        p p p p p p p dx nx x dx nx x f an cos ) ( 1 cos ) ( 1 2 p p p p p                                  3 2 2 sin ) 2 ( cos )] 1 )( ( 2 [ sin ) ( 1 n nx n nx x n nx x                      0 ) 1 ( ) 4 ( 0 0 0 0 1 2 n n p p 2 ) 1 ( 4 n n  

82.        p p p p p p p dx nx x dx nx x f bn sin ) ( 1 sin ) ( 1 2 p p p p p                                  3 2 2 cos ) 2 ( sin )] 1 )( ( 2 [ cos ) ( 1 n nx n nx x n nx x                             3 2 3 ) 1 ( 2 0 ) 1 ( ) 4 ( ) 1 ( 2 0 0 1 n n n n n n p p n n ) 1 ( 4   p                             1 2 2 1 0 sin ) 1 ( 4 cos ) 1 ( 4 3 8 2 1 ) sin cos ( 2 ) ( n n n n n n nx n nx n nx b nx a a x f p p

83. 9.Obtain the fourier series expansion for 8 ....... 5 1 3 1 1 0 , 2 1 0 , 2 1 ) ( 2 2 2 p p p p p                  that deduce hence and x x x x x f Solution: ) 0 , ( ) ( ) , 0 ( 2 1 ) ( 2 1 ) ( ) , 0 ( ) ( ) 0 , ( 2 1 ) ( 2 1 ) ( p p p p p p p p                 in x f in x x x f and in x f in x x x f

84. 0  n b t coefficien fourier the Hence   p p 0 0 ) ( 2 dx x f a dx x          p p p 0 2 1 2 p p p 0 2 2         x x 0         1 0 cos 2 n n nx a a x f

85.            p p p p p 0 0 cos 2 1 2 cos ) ( 2 nxdx x nxdx x f an p p p p 0 2 cos 2 sin 2 1 2                                    n nx n nx x p p p p p 0 2 2 2 cos 2 2                n n n   n n ) 1 ( 1 4 2 2    p       odd is n when n even is n when an , 8 , 0 2 2 p

86. The required Fourier Series        5 , 3 , 1 2 2 cos 8 n nx n x f p             ... 5 5 cos 3 3 cos 1 cos 8 ., . 2 2 2 2 x x x x f e i p Putting x=0 , we get,             ... 5 1 3 1 1 1 8 0 2 2 2 2 p f           ... 5 1 3 1 1 1 8 1 2 2 2 2 p           ... 5 1 3 1 1 1 8 2 2 2 2 p

87. function odd an is x f ax x f Solution ) ( sin ) ( :     p p p     x in ax x f function the for period of ansion series fourier the find eger an nor zero neither is a If , sin ) ( 2 exp , int . 10       1 0 sin ) ( 0 , 0 n n n nx b x f a a Hence     p p p p 0 0 sin sin 2 sin ) ( 2 nxdx ax nxdx x f bn

88.                p p 0 cos cos 1 xdx a n xdx a n     p p 0 sin sin 1             a n x a n a n x a n                 a n a n a n a n p p p sin sin 1             a n a n a n a n a n a n p p p p p p p p p sin cos cos sin sin cos cos sin 1

89.                  a n a a n a n n p p p sin ) 1 ( sin ) 1 ( 1                   a n a a n a n n p p p sin ) 1 ( sin ) 1 ( 1 1 1 eger an not is a a note int , sin :   p               a n a n a n 1 1 sin ) 1 ( 1 p p     2 2 1 sin 2 ) 1 ( a n a n b n n     p p

90. 8 ..... 5 1 3 1 1 , ) ( . 11 2 2 2 p p p         deduce and x x x f function the represent to series fourier the Obtain function even an is x f x x x f x x f Solution ) ( ) ( ) ( :              1 0 cos 2 n n nx a a x f   p p 0 0 ) ( 2 dx x f a

91.   p p 0 0 2 dx x a   p p 0 0 2 xdx a p p p         0 2 2 2 x   p p 0 cos ) ( 2 nxdx x f an

92.   p p 0 cos 2 nxdx x   p p 0 cos 2 nxdx x p p 0 2 cos 1 sin 2                      n nx n nx x         2 2 1 cos 2 n n np p   1 ) 1 ( 2 2    n n p        odd is n if n even is n if an , 4 , 0 2 p

93.        odd n nx n x f cos 4 2 ) ( 2 p p            ...... 5 5 cos 3 3 cos cos 4 2 ) ( 2 2 x x x x f p p             ...... 5 1 3 1 1 4 2 0 0 2 2 p p x put ...... 5 1 3 1 1 8 2 2 2     p

94. p p     x x x f function the represent to series fourier the Obtain , cos ) ( . 11 function even an is x f x x x f x x f Solution ) ( cos ) cos( ) ( cos ) ( :              1 0 cos 2 n n nx a a x f   p p 0 0 ) ( 2 dx x f a

95.      p p p p p 2 2 0 cos 2 cos 2 xdx xdx   p p 0 0 cos 2 dx x a       p p p p 2 2 0 sin sin 2 x x      1 1 2   p p 4 

96.   p p 0 cos ) ( 2 nxdx x f an   p p 0 cos cos 2 nxdx x      p p p p 2 2 0 cos cos cos cos 2 nxdx x nxdx x                       p p p p 2 2 0 1 cos 1 cos 2 1 1 cos 1 cos 2 1 2 dx x n x n dx x n x n                                           p p p p 2 2 0 1 1 sin 1 1 sin 1 1 sin 1 1 sin 1 n x n n x n n x n n x n

97.                               1 2 1 sin 1 2 1 sin 1 2 1 sin 1 2 1 sin 1 n n n n n n n n p p p p p                     1 2 1 sin 2 1 2 1 sin 2 1 n n n n p p p                  p   p 2 sin cos 2 sin : note

98.                     1 2 1 sin 2 1 2 1 sin 2 1 n n n n p p p               1 2 cos 1 2 cos 2 n n n n p p p           1 1 1 1 2 cos 2 n n n p p

99.            1 1 1 2 cos 2 2 n n n n p p          1 2 2 cos 2 2 n n p p    1 Pr 1 2 cos 4 2     n ovided n n p p

100.    p p 0 1 cos cos 2 1 xdx x a n When              p p p p 2 2 0 cos cos cos cos 2 xdx x xdx x             p p p p 2 2 2 0 2 cos cos 2 dx xdx                           p p p p 2 2 0 2 2 cos 1 2 2 cos 1 2 dx x dx x

101.                         p p p p 2 2 0 4 2 sin 2 4 2 sin 2 2 x x x x          4 2 4 2 p p p p 0 1  a   nx n n x f n cos 1 2 cos 4 2 ) ( 2 2         p p p

102. •11.Expand f(x) = x – x2 as a Fourier series in –l < x < l and using this series •find the root square mean value of f(x) in the interval. Sol. Fourier series is             1 0 sin cos 2 ) ( n n n l x n b l x n a a x f p p        l l l l dx x x l dx x f l a ) ( 1 ) ( 1 2 0 l l x x l          3 2 1 3 2                       3 2 3 2 1 3 2 3 2 l l l l l 3 2 3 2 1 2 3 l l l            

103.        l l l l n dx l x n x x l dx l x n x f l a p p cos ) ( 1 cos ) ( 1 2 l l l n l x n l n l x n x l n l x n x x l                                                          3 3 3 2 2 2 2 sin ) 2 ( cos ) 2 1 ( sin ) ( 1 p p p p p p                                               0 ) 1 ( ) 2 1 ( 0 0 ) 1 ( ) 2 1 ( 0 1 2 2 2 2 2 2 p p n l l n l l l n n   l l n l l n 2 1 2 1 ) 1 ( 2 2 2      p   l n l n 4 ) 1 ( 2 2    p 2 2 1 2 ) 1 ( 4 p n l n  

104.        l l l l n dx l x n x x l dx l x n x f l b p p sin ) ( 1 sin ) ( 1 2 l l l n l x n l n l x n x l n l x n x x l                                                          3 3 3 2 2 2 2 cos ) 2 ( sin ) 2 1 ( cos ) ( 1 p p p p p p                                                    3 3 3 2 3 3 3 2 ) 1 ( 2 0 ) 1 ( ) ( ) 1 ( 2 0 ) 1 ( ) ( 1 p p p p n l n l l l n l n l l l l n n n n

105.   2 2 ) 1 ( l l l l n l l n       p   l n n 2 ) 1 ( 1 p    p n l n 1 ) 1 ( 2                1 0 sin cos 2 ) ( n n n l x n b l x n a a x f p p                            1 1 2 2 1 2 2 sin ) 1 ( 2 cos ) 1 ( 4 3 2 2 1 n n n l x n n l l x n n l l p p p p

106. RMS value of f(x) in (–l, l ) is       1 2 2 2 0 2 ) ( 2 1 4 n n n b a a y                          1 2 2 2 2 2 4 4 2 2 4 2 2 ) 1 ( 4 ) 1 ( 16 2 1 3 2 4 1 n n n n l n l l p p             1 2 2 2 4 4 4 4 2 2 8 9 .) . ( n n l n l l y e i p p

107. 12.Find the Fourier series expansion of            l x l x l l x x x f 2 , 2 0 , ) ( Hence deduce the value of     1 4 ) 1 2 ( 1 n n Sol. Let 2 2 l L l L    then the given function becomes          L x L x L L x x x f 2 , 2 0 , ) ( Fourier series is             1 0 sin cos 2 ) ( n n n L x n b L x n a a x f p p

108.        L L L L dx x L L dx x L dx x f L a 2 0 2 0 0 ) 2 ( 1 ) ( 1 ) ( 1 L L L x L L x L 2 2 0 2 2 ) 2 ( 1 2 1                                  2 0 1 0 2 1 2 2 L L L L L L L    2 2

109.   L n dx L x n x f L a 2 0 cos ) ( 1 p      L L L dx L x n x L L dx L x n x L 2 0 cos ) 2 ( 1 cos 1 p p L L L L n L x n L n L x n x L L L n L x n L n L x n x L 2 2 2 2 0 2 2 2 cos ) 1 ( sin ) 2 ( 1 cos ) 1 ( sin ) ( 1                                                                                 p p p p p p p p                                               2 2 2 2 2 2 2 2 2 2 2 2 ) 1 ( 0 0 1 0 ) 1 ( 0 1 p p p p n L n L L n L n L L n n     1 ) 1 ( 2 ) 1 ( 1 1 ) 1 ( 1 2 2 2 2 2          n n n n L n L L p p

110.   L n dx L x n x f L b 2 0 sin ) ( 1 p      L L L dx L x n x L L dx L x n x L 2 0 sin ) 2 ( 1 sin 1 p p L L L L n L x n L n L x n x L L L n L x n L n L x n x L 2 2 2 2 0 2 2 2 sin ) 1 ( cos ) 2 ( 1 sin ) 1 ( cos ) ( 1                                                                                   p p p p p p p p                                         0 ) 1 ( 0 0 1 0 0 0 ) 1 ( 1 2 2 p p n L L n L L n n 0 

111.             1 0 sin cos 2 ) ( n n n L x n b L x n a a x f p p                 1 2 2 0 cos ] 1 ) 1 [( 2 2 n n L x n n L L p p                1 2 2 2 cos ] 1 ) 1 [( 4 n n l x n n l l p p

112. Using Parseval’s identity for Fourier series we have        L n n n b a a dx x f L 2 0 1 2 2 2 0 2 ) ( 2 )] ( [ 1                       L L L n n n L L dx x L L dx x L 0 2 1 4 4 2 2 2 2 2 0 1 ) 1 ( 4 2 ) 2 ( 1 ) ( 1 p                              ........ .......... 0 5 4 0 3 4 0 1 4 4 2 3 ) 2 ( 1 3 1 4 4 4 4 2 2 2 3 0 3 p L L x L L x L L L L                            ........ .......... 5 1 3 1 1 1 16 2 3 0 1 0 3 1 4 4 4 4 2 2 3 3 p L L L L L L

113.            ........ .......... 5 1 3 1 1 1 16 2 3 2 4 4 4 4 2 2 2 p L L L           ........ .......... 5 1 3 1 1 1 16 6 4 4 4 4 2 2 p L L ....... .......... 5 1 3 1 1 1 96 4 4 4 4     p      1 4 4 ) 1 2 ( 1 96 .) . ( n n e i p

114. . 13. Find the Fourier series expansion of          l x l l x x l x f 2 , 0 0 , ) ( Hence deduce the value of the series .... ...... 7 1 5 1 3 1 1     ... .......... 5 1 3 1 1 1 2 2 2    and Sol. Fourier series is             1 0 sin cos 2 ) ( n n n l x n b l x n a a x f p p        l l l l dx l dx x l l dx x f l a 2 0 2 0 0 ) 0 ( 1 ) ( 1 ) ( 1

115. l x l l 0 2 2 ) ( 1            2 0 2 1 l l    2 l  0 cos ) ( 1 cos ) ( 1 0 2 0       l l n dx l x n x l l dx l x n x f l a p p l l n l x n l n l x n x l l 0 2 2 2 cos ) 1 ( sin ) ( 1                                          p p p p                        2 2 2 2 2 2 0 ) 1 ( 0 1 p p n l n l l n

116.   1 ) 1 ( 1 1 2 2 2     n n l l p   1 ) 1 ( 1 2 2     n n l p 0 sin ) ( 1 sin ) ( 1 0 2 0       l l n dx l x n x l l dx l x n x f l b p p l l n l x n l n l x n x l l 0 2 2 2 sin ) 1 ( cos ) ( 1                                           p p p p                  0 } 0 0 { 1 2 p n l l p n l 

117.             1 0 sin cos 2 ) ( n n n l x n b l x n a a x f p p                  1 2 2 1 sin cos ] 1 ) 1 [( 4 n n l x n n l l x n n l l p p p p 1 ( ....... .......... 3 sin 3 1 2 sin 2 1 sin 1 1 ....... .......... 5 cos 5 1 3 cos 3 1 cos 1 1 2 4 ) ( .) . ( 2 2 2 2                              l x l x l x l l x l x l x l l x f e i p p p p p p p p Put 2 l x  (which is point of continuity) in equation (1), we get

118.                ....... .......... 2 5 sin 5 1 4 sin 4 1 2 3 sin 3 1 sin 2 1 2 sin 1 1 ) 0 ( 2 4 2 2 p p p p p p p l l l l l                ....... .......... 7 1 0 5 1 0 3 1 0 1 4 2 p l l l             ....... .......... 7 1 5 1 3 1 1 4 2 p l l l            ....... .......... 7 1 5 1 3 1 1 4 p l l ....... .......... 7 1 5 1 3 1 1 4      p

119. Put x = l in equation (1) we get             ...... .......... 5 1 3 1 1 1 2 4 ) ( 2 2 2 2 p l l l f But x = l is the point of discontinuity. So we have f(x) = l – x f(l–) = l – l =0 f(x) = 0 f(l) = 0 0 2 ) 0 ( ) 0 ( 2 ) ( ) ( ) (        l f l f l f           .. .......... 5 1 3 1 1 1 2 4 0 2 2 2 2 p l l            .. .......... 5 1 3 1 1 1 2 4 2 2 2 2 p l l ...... .......... 5 1 3 1 1 1 8 2 2 2 2    p

120. 14.Find the Fourier series for the function            2 0 , 1 0 2 , 1 ) ( x x x x x f Deduce that 8 ) 1 2 ( 1 2 1 2 p      n n Solution: f(– x) = 1 – x in (–2, 0) = f(x) in (0, 2) and f(– x) = 1 + x in (0, 2) = f(x) in (–2, 0) Hence f(x) is an even function.       1 0 2 cos 2 ) ( n n x n a a x f p

121.      2 0 2 0 0 ) 1 ( ) ( 2 2 dx x dx x f a 2 0 2 2         x x   0 ) 0 ( ) 2 2 (          2 0 2 0 2 cos ) 1 ( 2 cos ) ( 2 2 dx x n x dx x n x f an p p

122. 2 0 2 2 4 2 cos ) 1 ( 2 2 sin ) 1 (                                          p p p p n x n n x n x                        2 2 2 2 4 0 ) 1 ( 4 0 p p n n n            odd is n when n even is n when a n n n , 8 , 0 ) 1 ( 1 4 2 2 2 2 p p

123.      1 2 2 2 cos 8 2 0 ) ( n x n n x f p p           .... .......... .......... 2 5 cos 5 1 2 3 cos 3 1 2 cos 1 1 8 ) ( 2 2 2 2 x x x x f p p p p Put x = 0 in the above series we get          .. .......... 5 1 3 1 1 1 8 ) 0 ( 2 2 2 2 p f

124. But x = 0 is the point of discontinuity. So we have 1 2 2 2 ) 1 ( ) 1 ( 2 ) 0 ( ) 0 ( ) 0 (         f f f Hence equation (1) becomes          .. .......... 5 1 3 1 1 1 8 1 2 2 2 2 p ...... .......... 5 1 3 1 1 1 8 2 2 2 2    p      1 2 2 ) 1 2 ( 1 8 .) . ( n n e i p

125. •15.Find the Fourier series of periodicity 3 for f(x) = 2x – x2 in 0 < x < 3. Sol. Fourier series is             1 0 3 2 sin 3 2 cos 2 ) ( n n n x n b x n a a x f p p      3 0 2 3 0 0 ) 2 ( 3 2 ) ( ) 2 / 3 ( 1 dx x x dx x f a 3 0 3 2 3 2 2 3 2         x x                 ) 0 0 ( 3 27 9 3 2 0 

126.      3 0 2 3 0 3 2 cos ) 2 ( 3 2 cos ) ( ) 2 / 3 ( 1 dx x n x x dx nx x f an p 3 0 3 3 2 2 2 27 8 3 2 sin ) 2 ( 9 4 3 2 cos ) 2 2 ( 3 2 3 2 sin ) 2 ( 3 2                                                         p p p p p p n x n n x n x n x n x x                                        0 4 9 ) 2 ( 0 0 4 9 ) 4 ( 0 3 2 2 2 2 2 p p n n         2 2 4 54 3 2 p n 2 2 9 p n  

127.      3 0 2 3 0 3 2 sin ) 2 ( 3 2 sin ) ( ) 2 / 3 ( 1 dx x n x x dx nx x f bn p 3 0 3 3 2 2 2 27 8 3 2 cos ) 2 ( 9 4 3 2 sin ) 2 2 ( 3 2 3 2 cos ) 2 ( 3 2                                                         p p p p p p n x n n x n x n x n x x                                             3 3 3 3 8 27 2 0 0 8 27 2 0 2 3 ) 3 ( 3 2 p p p n n n p n 3 

128.             1 0 3 2 sin 3 2 cos 2 ) ( n n n x n b x n a a x f p p             1 2 2 3 2 sin 3 3 2 cos 9 n x n n x n n p p p p

129. HALF RANGE FOURIER SERIES COSINE SERIES        1 0 cos 2 n n nx a a x f   p p 0 0 ) ( 2 dx x f a   p p 0 cos ) ( 2 nxdx x f an FOR THE INTERVAL   p , 0 SINE SERIES       1 sin n n nx b x f   p p 0 sin ) ( 2 nxdx x f bn

130. HALF RANGE FOURIER SERIES FOR THE INTERVAL   l , 0 COSINE SERIES SINE SERIES              1 0 cos 2 n n l x n a a x f p   l dx x f l a 0 0 ) ( 2         l n dx l x n x f l a 0 cos ) ( 2 p             1 sin n n l x n b x f p         l n dx l x n x f l b 0 sin ) ( 2 p

131. •16. Find the half range sine series for f(x) = 2 in 0 < x < p. Solution:     p p p p 0 0 sin 2 2 sin ) ( 2 dx nx dx nx x f bn p p 0 cos 4        n nx   1 ) 1 ( 4     n np       odd is n when n even is n when bn , 8 , 0 p Half range sine series is         1 1 sin 8 sin ) ( n n n nx n nx b x f p

132. . 17. Expand f(x) = cos x, 0 < x < π in a Fourier sine series. Sol. Fourier sine series is     1 sin ) ( n n nx b x f     p p p p 0 0 sin cos 2 sin ) ( 2 dx nx x dx nx x f bn   p p 0 cos sin 2 1 dx x nx 2SinACosB = Sin(A+B) + Sin(A–B)

133. 1 , ] ) 1 sin( ) 1 [sin( 1 0       n dx x n x n p p p p 0 1 ) 1 cos( 1 ) 1 cos( 1                           n x n n x n 1 1 ) 1 ( ) 1 cos( ) 1 ( ) 1 cos(         n n n n p p                                1 1 1 1 1 ) 1 ( 1 ) 1 ( 1 1 1 n n n n n n p

134.                               1 1 1 1 1 1 1 1 ) 1 ( 1 n n n n n p                            1 1 1 1 1 1 1 1 ) 1 ( 1 n n n n n p                        1 2 1 2 ) 1 ( 1 2 2 n n n n n p   1 , 1 ) 1 ( ) 1 ( 2 2      n n n b n n p

135. When n = 1, we have     p p p p 0 0 1 sin cos 2 sin ) ( 2 dx x x dx x x f b   p p 0 2 sin 1 dx x p p 0 2 2 cos 1        x 0 ) 1 1 ( 2 1     p          2 1 1 sin sin sin ) ( n n n n nx b x b nx b x f        2 2 sin ) 1 ( ] 1 ) 1 ( [ 2 n n nx n n p

136. 18.Find the half range cosine series for the function f(x) = x (π – x) in 0 < x < π. Deduce that 90 .. .......... 3 1 2 1 1 1 4 4 4 4 p     Sol. Half range fourier cosine series is      1 0 cos 2 ) ( n n nx a a x f      p p p p p 0 0 0 ) ( 2 ) ( 2 dx x x dx x f a p p p 0 3 2 3 2 2         x x

137.                   ) 0 0 ( 3 2 2 3 3 p p p        6 2 3 p p 3 2 p       p p p p p 0 0 cos ) ( 2 cos ) ( 2 dx nx x x dx nx x f an p p p p 0 3 2 2 sin ) 2 ( cos ) 2 ( sin ) ( 2                                 n nx n nx x n nx x x

138.                           0 ) 1 )( ( 0 0 ) 1 )( ( 0 2 2 2 n n n p p p   1 ) 1 ( 2 2     n n p p   odd is n when even is n when n a n n n , 0 , 4 1 ) 1 ( 2 2 2       

139.                    even n n n nx n nx a a x f cos 4 3 2 1 cos 2 ) ( 2 2 1 0 p            ..... .......... 6 6 cos 4 4 cos 2 2 cos 4 6 2 2 2 2 x x x p Parseval’s identity for half range fourier cosine series is       1 2 2 0 0 2 2 )] ( [ 2 n n a a dx x f p p

140.                even n n dx x x 4 2 2 0 2 2 16 3 2 1 ] [ 2 p p p p               .... .......... 6 1 4 1 2 1 16 18 ) 2 ( 2 4 4 4 4 0 3 4 2 2 p p p p p dx x x x                    .... .......... 3 1 2 1 1 1 2 16 18 4 2 5 3 2 4 4 4 4 4 0 4 5 3 2 p p p p p x x x

141.                             .... .......... 3 1 2 1 1 1 18 0 2 5 3 2 4 4 4 4 5 5 5 p p p p p ......... .......... 3 1 2 1 1 1 18 30 2 4 4 4 4 5            p p p ... .......... 3 1 2 1 1 1 18 15 4 4 4 4 4      p p ... .......... 3 1 2 1 1 1 90 .) . ( 4 4 4 4     p e i

142. 20.Find the half range sine series of f(x) = x cos x in (0, π). Sol. Fourier sine series is     1 sin ) ( n n nx b x f     p p p p 0 0 sin cos 2 sin ) ( 2 dx nx x x dx nx x f bn   p p 0 ) cos sin 2 ( 1 dx x nx x 1 , ] ) 1 sin( ) 1 [sin( 1 0       n dx x n x n x p p

143. 1 , ) 1 sin( 1 ) 1 sin( 1 0 0        n dx x n x dx x n x p p p p p p p p 0 2 0 2 ) 1 ( ) 1 sin( ) 1 ( 1 ) 1 cos( 1 ) 1 ( ) 1 sin( ) 1 ( 1 ) 1 cos( 1                                                         n x n n x n x n x n n x n x bn                                            0 0 0 1 ) 1 ( 1 0 0 0 1 ) 1 ( 1 1 1 n n n n p p p p 1 ) 1 ( 1 ) 1 ( 2        n n n n            1 1 1 1 ) 1 ( n n n

144.           ) 1 )( 1 ( 2 ) 1 ( n n n n 1 , 1 ) 1 ( 2 .) . ( 2     n n n b e i n n When n = 1, we have     p p p p 0 0 1 sin cos 2 sin ) ( 2 dx x x x dx x x f b   p p 0 2 sin 1 dx x x

145. p p 0 4 2 sin ) 1 ( 2 2 cos 1                       x x x 2 1 } 0 0 { 0 2 1 1                          p p          2 1 1 sin sin sin ) ( n n n n nx b x b nx b x f         2 2 sin 1 ) 1 ( 2 sin 2 1 n n nx n n x

146. 21.Obtain the sine series for              l x l in x l l x in x x f 2 2 0 ) ( Sol. Fourier sine series is     1 sin ) ( n n l x n b x f p   l n dx l x n x f l b 0 sin ) ( 2 p      l l l dx l x n x l l dx l x n x l 2 / 2 / 0 sin ) ( 2 sin 2 p p

147. l l l l n l x n l n l x n x l l l n l x n l n l x n x l 2 / 2 2 2 2 / 0 2 2 2 sin ) 1 ( cos ) ( 2 sin ) 1 ( cos ) ( 2                                                                                   p p p p p p p p                                                                                                                             2 2 2 2 2 2 2 sin . 2 cos . 2 } 0 0 { 2 0 0 2 sin . 2 cos . 2 2 p p p p p p p p n n l n n l l l n n l n n l l l            2 2 2 2 sin . 2 2 p p n n l l 2 sin 4 2 2 p p n n l bn      1 sin ) ( n n l x n b x f p     1 2 2 sin 2 sin 4 n l x n n n l p p p

148. 22.Find the half range cosine series for the function f(x) = x in 0 < x < l. Hence deduce the value of the series     1 4 ) 1 2 ( 1 n n Sol. Half range Fourier cosine series is      1 0 cos 2 ) ( n n l x n a a x f p   l dx x f l a 0 0 ) ( 2 l l x l dx x l 0 2 0 2 2 2          l l l          0 2 2 2

149.     l l n dx l x n x l dx l x n x f l a 0 0 cos 2 cos ) ( 2 p p l l n l x n l n l x n x l 0 2 2 2 cos ) 1 ( sin ) ( 2                                        p p p p                        2 2 2 2 2 2 0 ) 1 ( 0 2 p p n l n l l n

150.        even is n when odd is n when n l an , 0 , 4 2 2 p   1 ) 1 ( 2 2 2    n n l p      1 0 cos 2 ) ( n n l x n a a x f p       1 2 2 cos 4 2 n l x n n l l p p

151. Using Parseval’s identity for half range Fourier cosine series we have       l n n a a dx x f l 0 1 2 2 0 2 2 )] ( [ 2             l n n l l dx x l 0 1 4 4 2 2 2 16 2 ) ( 2 p                  ........ .......... 5 1 3 1 1 1 16 2 3 2 4 4 4 4 2 2 0 3 p l l x l l            ........ .......... 5 1 3 1 1 1 16 2 3 2 4 4 4 4 2 2 2 p l l l

152.           ........ .......... 5 1 3 1 1 1 16 6 4 4 4 4 2 2 p l l ....... .......... 5 1 3 1 1 1 96 4 4 4 4     p      1 4 4 ) 1 2 ( 1 96 .) . ( n n e i p

153. 23.Prove that           .... .......... 5 sin 5 1 3 sin 3 1 sin 4 1 l x l x l x p p p p in the interval 0 < x < l Sol. Since RHS contains sine series and given 0 < x < l, we have to find half range Fourie sine series for f(x) = 1     1 sin ) ( n n l x n b x f p     l l n dx l x n l dx l x n x f l b 0 0 sin ) 1 ( 2 sin ) ( 2 p p

154.                                      p p p p n l n l l l n l x n l n l ) 1 ( 2 cos 2 0   1 ) 1 ( 2 1     n np            1 1 1 sin ] 1 ) 1 [( 2 sin ) ( n n n n l x n n l x n b x f p p p              . .......... 0 5 sin 5 2 0 3 sin 3 2 0 sin 1 2 2 l x l x l x p p p p           ... .......... 5 sin 5 1 3 sin 3 1 sin 4 1 l x l x l x p p p p

155. 24.Obtain the half range cosine series for f(x) = (x – 2)2 in the interval 0 < x < 2. Sol. Half range cosine series is      1 0 2 cos 2 ) ( n n x n a a x f p      2 0 2 2 0 0 ) 2 ( ) ( 2 2 dx x dx x f a 2 0 3 3 ) 2 (         x 3 8 3 8 0               

156.      2 0 2 2 0 2 cos ) 2 ( 2 cos ) ( 2 2 dx x n x dx x n x f an p p 2 0 3 3 2 2 2 8 2 sin ) 2 ( 4 2 cos )] 2 ( 2 [ 2 2 sin ) 2 (                                                        p p p p p p n x n n x n x n x n x                     0 16 0 0 0 0 2 2 p n 2 2 16 n p                  .... 2 3 cos 3 1 2 2 cos 2 1 2 cos 1 1 16 3 4 ) ( .) . ( 2 cos 16 6 8 ) ( 2 2 2 2 1 2 2 x x x x f e i x n n x f n p p p p p p

157. COMPLEX OR EXPONENTIAL FORM OF FOURIER SERIES dx e x f c e c x f x In dx e x f c e c x f x In inx n n inx n inx n n inx n                       p p p p p p p p 2 0 ) ( 2 1 , ) ( 2 0 ) ( 2 1 , ) (

158. dx e x f l c e c x f l x In dx e x f l c e c x f l x l In l x in l n n l x in n l x in l l n n l x in n p p p p                       2 0 ) ( 2 1 , ) ( 2 0 ) ( 2 1 , ) (

159. 25.Find the complex form of the Fourier series of x e x f   ) ( in –1 < x < 1 Sol. The complex form of Fourier series of f(x) is given by       n x n i n e C x f p ) ( 2l = 2 l = 1 dx e x f C x n i n     1 1 ) ( ) 1 ( 2 1 p dx e e x n i x      1 1 2 1 p

160. dx e x n i      1 1 ) 1 ( 2 1 p 1 1 ) 1 ( ) 1 ( 2 1             p p n i e x n i   ) 1 ( ) 1 ( ) 1 ( 2 1 p p p n i n i e e n i          p p p p n i n i e e e e n n i 1 1 2 2 ) 1 ( 2 ) 1 (       

161.   ) sin (cos ) sin (cos ) 1 ( 2 ) 1 ( 1 1 2 2 p p p p p p n i n e n i n e n n i           n n n e e n n i C ) 1 ( ) 1 ( ) 1 ( 2 ) 1 ( 1 1 2 2         p p   1 1 2 2 ) 1 ( 2 ) 1 ( ) 1 (       e e n n i n p p 1 sinh 2 ) 1 ( 2 ) 1 ( ) 1 ( 2 2 p p n n i n     2 2 1 ) 1 ( 1 sinh ) 1 ( p p n n i C n n               n x n i n e n n i x f p p p 2 2 1 ) 1 ( 1 sinh ) 1 ( ) (

162. 26.Find the complex form of the Fourier series of p p     x in ax x f cos ) ( Solution:       n x n i n e C x f p ) ( dx e x f c inx n     p p p ) ( 2 1 dx e ax inx     p p p cos 2 1

163.   p p p             ax a ax in a n i e inx sin cos 2 1 2 2 2   bx b bx a b a e bx e note ax ax sin cos cos : 2 2                       p p p p p p p a a a in n a e a a a in n a e in in sin cos sin cos 2 1 2 2 2 2         p p p p p p p in in in in e e a a e e a in n a         sin cos 2 1 2 2

164.    p p p p p n a a n a in n a cos sin 2 sin cos 2 2 1 2 2       p p p n a a n a cos sin 2 2 1 2 2   0 sin  p n      2 2 sin 1 n a a a n    p p             n inx n e n a a a x f 2 2 sin 1 ) ( p p

165. HARMONIC ANALYSIS The process of finding the Fourier series for a function given by numerical value is known as harmonic analysis. In harmonic analysis the Fourier coefficients a0, an and bn of the function y = f(x) in (0, 2π) are given by a0 = 2 [mean value of y in (0, 2π)] an = 2 [mean value of y cosnx in (0, 2π)] bn = 2 [mean value of y sinnx in (0, 2π)]

166. 27. Find the Fourier series expansion up to third harmonic from the following data: x : 0 1 2 3 4 5 f(x) : 9 18 24 28 26 20 Sol. Here the length of the interval is 6 (i.e.) 2l = 6  l = 3             1 0 3 sin 3 cos 2 ) ( n n n x n b x n a a x f p p 3 3 sin 2 sin sin 3 cos 2 cos cos 2 ) ( .) . ( 3 3 sin 3 2 sin 3 sin 3 3 cos 3 2 cos 3 cos 2 ) ( 3 2 1 3 2 1 0 3 2 1 3 2 1 0 x where b b b a a a a x f e i x b x b x b x a x a x a a x f p        p p p p p p                Fourier series is

167. x y θ=πx/3 ycosθ ycos2θ ycos3θ ysinθ ysin2θ ysin3θ 0 9 0 9 9 9 0 0 0 1 18 π/3 9 –9 –18 15.588 15.588 0 2 24 2π/3 –12 –12 24 20.784 –20.784 0 3 28 π –28 28 –28 0 0 0 4 26 4π/3 –13 –13 26 –22.516 22.516 0 5 20 5π/3 10 –10 –20 –17.32 –17.32 0 Total 125 –25 –7 –7 –3.464 0 0

168. 667 . 41 6 125 2 2 ] [ 2 0                    n y y of value mean a 333 . 8 6 25 2 cos 2 ] cos [ 2 1                     n y y of value mean a   333 . 2 6 7 2 2 cos 2 ] 2 cos [ 2 2                     n y y of value mean a   333 . 2 6 7 2 3 cos 2 ] 3 cos [ 2 3                     n y y of value mean a   Here n = 6

169. 155 . 1 6 464 . 3 2 sin 2 ] sin [ 2 1                     n y y of value mean b   0 6 0 2 2 sin 2 ] 2 sin [ 2 2                    n y y of value mean b   0 6 0 2 3 sin 2 ] 3 sin [ 2 3                    n y y of value mean b   3 sin 155 . 1 3 cos 333 . 2 2 cos 333 . 2 cos 333 . 8 833 . 20 ) ( .) . ( 3 sin 0 2 sin 0 sin 155 . 1 3 cos 333 . 2 2 cos 333 . 2 cos 333 . 8 2 667 . 41 ) ( x where x f e i x f p                         

170. 28.Find the Fourier series expansion up to second harmonic from the following data: 10 11 17 20 15 12 10 : ) ( 2 3 5 3 4 3 2 3 0 : x f x p p p p p p Sol. Since the last value of y is a repetition of the first, only the first six values will be used. Fourier series is       1 0 ) sin cos ( 2 ) ( n n n x n b x n a a x f x b x b x a x a a x f e i 2 sin sin 2 cos cos 2 ) ( .) . ( 2 1 2 1 0     

171. x y ycosx ycos2x ysinx ysin2x 0 10 10 10 0 0 π/3 12 6 –6 10.392 10.392 2π/3 15 –7.5 –7.5 12.99 –12.99 π 20 –20 20 0 0 4π/3 17 –8.5 –8.5 –14.722 14.722 5π/3 11 5.5 –5.5 –9.526 –9.526 Total 85 –14.5 2.5 –0.866 2.598

172. Here n = 6 333 . 28 6 85 2 2 ] [ 2 0                    n y y of value mean a 833 . 4 6 5 . 14 2 cos 2 ] cos [ 2 1                     n x y x y of value mean a 833 . 0 6 5 . 2 2 2 cos 2 ] 2 cos [ 2 2                    n x y x y of value mean a

173. 289 . 0 6 866 . 0 2 sin 2 ] sin [ 2 1                     n x y x y of value mean b 866 . 0 6 598 . 2 2 2 sin 2 ] 2 sin [ 2 2                    n x y x y of value mean b x x x x x f e i x x x x x f 2 sin 866 . 0 sin 289 . 0 2 cos 833 . 0 cos 833 . 4 1665 . 14 ) ( .) . ( 2 sin 866 . 0 sin 289 . 0 2 cos 833 . 0 cos 833 . 4 2 333 . 28 ) (           

174. 29.Find the Fourier series expansion up to first harmonic from the following data: 7 . 15 4 . 12 9 4 . 6 3 . 5 6 3 . 8 6 . 11 15 6 . 17 7 . 18 18 : ) ( 11 10 9 8 7 6 5 4 3 2 1 0 : x f x Sol. Here the length of the interval is 12 (i.e.) 2l = 12  l = 6 Fourier series is             1 0 6 sin 6 cos 2 ) ( n n n x n b x n a a x f p p 6 sin cos 2 ) ( .) . ( 6 sin 6 cos 2 ) ( 1 1 0 1 1 0 x where b a a x f e i x b x a a x f p    p p       

175. x y θ=πx/6 cosθ ycosθ ysinθ 0 18 0 1 18 0 1 18.7 π/6 0.866 16.1942 9.35 2 17.6 2π/6 0.5 8.8 15.2416 3 15 3π/6 0 0 15 4 11.6 4π/6 –0.5 –5.8 10.0456 5 8.3 5π/6 –0.866 –7.1878 4.15 6 6 π –1 –6 0 7 5.3 7π/6 –0.866 –4.5898 –2.65 8 6.4 8π/6 –0.5 –3.2 –5.5424 9 9 9π/6 0 0 –9 10 12.4 10π/6 0.5 6.2 –10.7384 11 15.7 11π/6 0.866 13.5962 –7.85 Total 144 36.0128 18.0064

176. 24 12 144 2 2 ] [ 2 0                    n y y of value mean a 002 . 6 12 0128 . 36 2 cos 2 ] cos [ 2 1                    n y y of value mean a   001 . 3 12 0064 . 18 2 sin 2 ] sin [ 2 1                    n y y of value mean b   Here n = 12 6 sin 001 . 3 cos 002 . 6 12 ) ( .) . ( sin 001 . 3 cos 002 . 6 2 24 ) ( x where x f e i x f p             

177. 1. Write down the form of the Fourier series of an odd function in (– l, l) and the associated Euler’s formula for the Fourier coefficients. l x n b x f n n p sin ) ( 1     dx l x n x f l b l n p sin ) ( 2 0   Sol.

178. 2. If f(x) = 3x – 4x3 defined in the interval (– 2, 2) then find the value of a1 in the Fourier series expansion. Sol. Since f(x) is an odd function, an = 0.  0 1  a      1 0 cos 2 ) ( n n nx a a x f 3 2 0 3 2 3 2 2 ) ( 2 2 3 0 3 0 2 0 0 p p p p p p p p p                     x dx x dx x f a 3 2 2 0 p  a 3. Obtain the first term of the Fourier series for the function f(x) = x2, – π < x < π Sol. f(x) = x2 is an even function. Fourier series is Hence the first term of the Fourier series is

179. dx x x a 2 2 cos 2 2 1 4 0 2 p   0 1 0 1 0 cos ) 1 ( sin cos 2 2 4 0 2 4 0                                         p p p p p p p x x x dx x x 4.If f(x) = 2x in the interval (0, 4) then find the value of a2 in the Fourier series expansion. Sol.

180. 5. Define root mean square value of a function Sol. The root mean square value of f(x) over the interval (a, b) is defined as         b a b a dx x f a b y a b dx x f y S M R 2 2 2 )] ( [ 1 )] ( [    l dx x f l y S M R 0 2 2 )] ( [ 1 5 0 5 1 5 1 1 ] [ 1 4 5 0 5 0 4 0 2 2 l l l x l dx x l dx x l l l l                     5 2 l y   6.Find the root mean square value of f(x) = x2 in (0, l) Sol.

181. 7.Find the root mean square value of a function f(x) in (0, 2π) Sol. The root mean square value of f(x) over the interval (0, 2π) is defined as       p p p p 2 0 2 2 2 0 2 )] ( [ 2 1 )] ( [ 2 1 dx x f y dx x f y S M R    1 0 2 2 )] ( [ 1 1 dx x f y S M R 3 1 3 1 0 3 ) 1 ( ) 1 ( 1 0 3 1 0 2                            x dx x 3 1   y 8.Find the root mean square value of f(x) = 1 – x in 0 < x < 1 Sol.

182.    p p 2 0 2 2 )] ( [ 2 1 dx x f y S M R 3 3 2 2 1 3 3 2 1 3 ) ( 2 1 ) ( 2 1 2 3 3 3 2 0 3 2 0 2 p p p p p p p p p p p p                                              x dx x 3 p   y 9.Find the root mean square value of f(x) = π – x in 0 < x < 2π Sol.

183. 10.Write the sufficient conditions for a function f(x) to satisfy for the existence of a Fourier series. Sol. i) f(x) is defined and single valued except possibly at a finite number of points in (–π, π) ii) f(x) is periodic with period 2π iii) f(x) and ) (x f  are piecewise continuous in (–π, π) Then the Fourier series of f(x) converges to f(x) a) if x is a point of continuity 2 ) 0 ( ) 0 ( )    x f x f b if x is a point of discontinuity.

184. 11.Find the sum of the Fourier series for         2 1 , 2 1 0 , ) ( x x x x f at x = 1 Sol. Here x = 1 is a point of discontinuity 2 3 2 2 1 2 ) 0 1 ( ) 0 1 ( ) 1 (        f f f 12.State the Parseval’s identity for Fourier series. Sol. The Parseval’s identity for Fourier series in the interval (c, c + 2l) is ) ( 2 )] ( [ 1 2 1 2 2 0 2 2 n n n l c c b a a dx x f l         The Parseval’s identity for Fourier series in the interval (c, c + 2π) is ) ( 2 )] ( [ 1 2 1 2 2 0 2 2 n n n c c b a a dx x f         p p

185. 12.Write the formula for finding Fourier coefficients. Sol.    l c c dx x f l a 2 0 ) ( 1    l c c n dx l x n x f l a 2 cos ) ( 1 p    l c c n dx l x n x f l b 2 sin ) ( 1 p 13.Define RMS value of a function. Sol. The RMS value of a function f(x) in (a,b) is defined by       b a b a dx x f a b y dx x f a b y 2 2 2 )] ( [ 1 )] ( [ 1

186. 14.State the Parseval’s identity for Fourier series. Sol. The Parseval’s identity for Fourier series in the interval (c, c + 2l) is ) ( 2 )] ( [ 1 2 1 2 2 0 2 2 n n n l c c b a a dx x f l         The Parseval’s identity for Fourier series in the interval (c, c + 2π) is ) ( 2 )] ( [ 1 2 1 2 2 0 2 2 n n n c c b a a dx x f         p p 15.Find the mean square value of the function f(x) = x in the interval (0, l). Sol. Mean square value is 3 0 3 1 3 1 1 )] ( [ 1 2 3 0 3 0 2 0 2 2 l l l x l dx x l dx x f l y l l l                    

187. 16.Find the value of an in the cosine series expansion of f(x) = 10 in the interval (0,10). Sol.   0 0 sin 20 10 10 sin 2 10 cos ) 10 ( 10 2 10 cos ) ( 10 2 10 0 10 0 10 0                   p p p p p p n n n x n dx x n dx x n x f an . 17. What is the constant term a0 and the coefficient an in the Fourier series expansion of f(x) = x – x3 in (–p,p). Sol. Since the interval is (–p,p), let us verify whether the function is odd or even f(– x) = (– x) – (–x)3 = – x + x3 = – (x – x3) = – f(x) The given function is an odd function. Hence a0 = 0 and an = 0.

188. . 18. Find the constant term in the Fourier series corresponding to f(x) = cos2 x expanded in the interval (–p,p). Sol. Since the interval is (–p,p), let us verify whether the function is odd or even. f(–x) = cos2 (–x)= cos2x = f(x). Hence the function is even.   1 ) 0 ( ) 0 ( 1 2 2 sin 1 2 2 cos 1 2 cos 2 ) ( 2 0 0 0 2 0 0                    p p p p p p p p p p x x dx x dx x dx x f a Hence the constant term in the Fourier expansion is 2 1 2 0  a

189. 19. Find the constant term in the Fourier expansion of f(x) = x2 – 2 in –2 < x < 2 Sol. f(–x) = (–x)2 – 2 = x2 – 2 = f(x), which is an even function 3 4 ) 0 ( 4 3 8 2 3 ) 2 ( ) ( 2 2 2 0 3 2 0 2 2 0 0                               x x dx x dx x f a Hence the constant term in the Fourier expansion is 3 2 2 3 / 4 2 0     a

190.             1 2 2 sin 2 cos 4 ) 1 ( 3 n n nx n nx n p ... .......... 3 1 2 1 1 1 2 2 2    20. If the Fourier series of the function f(x) = x + x2, in the interval (–p,p) is , then find the value of the infinite series Sol. Given              1 2 2 sin 2 cos 4 ) 1 ( 3 ) ( n n nx n nx n x f p Put x = π in the above series we get               1 2 2 0 ) 1 ( 4 ) 1 ( 3 ) ( n n n n f p p --------------- (1) But x = π is the point of discontinuity. So we have

191. 2 2 2 2 2 2 2 ) ( ) ( 2 ) ( ) ( ) ( p p p p p p p p p           f f f Hence equation (1) becomes ..... .......... 3 1 2 1 1 1 6 ..... .......... 3 1 2 1 1 1 4 3 2 ) 1 ( 4 3 0 ) 1 ( 4 ) 1 ( 3 2 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2                                   p p p p p p n n n n n n n

192. 21. Does f(x) = tan x posses a Fourier series? Justify your answer. Sol. For a function f(x) to have Fourier series expansion it must satisfy all the three criteria in Dirichlet’s conditions. But f(x) = tan x has value ∞ at 2 x p  and so it is a discontinuous point and moreover it is an infinite discontinuity. So it does not have a Fourier series expansion.

TPDE_UNIT II-FOURIER SERIES_PPT.pptx

TPDE_UNIT II-FOURIER SERIES_PPT.pptx

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