Fourier series 1

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Fourier series 1

  1. 1. Fourier Series N. B. Vyas Department of Mathematics,Atmiya Institute of Tech. and Science, Rajkot (Guj.) - INDIA N. B. Vyas Fourier Series
  2. 2. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. N. B. Vyas Fourier Series
  3. 3. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). N. B. Vyas Fourier Series
  4. 4. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). If T is the period of f (x) then N. B. Vyas Fourier Series
  5. 5. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). If T is the period of f (x) then f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT ) N. B. Vyas Fourier Series
  6. 6. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). If T is the period of f (x) then f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT ) f (x) = f (x − T ) = f (x − 2T ) = . . . = f (x − nT ) N. B. Vyas Fourier Series
  7. 7. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). If T is the period of f (x) then f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT ) f (x) = f (x − T ) = f (x − 2T ) = . . . = f (x − nT ) ∴ f (x) = f (x ± nT ), where n is a positive integer N. B. Vyas Fourier Series
  8. 8. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). If T is the period of f (x) then f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT ) f (x) = f (x − T ) = f (x − 2T ) = . . . = f (x − nT ) ∴ f (x) = f (x ± nT ), where n is a positive integerEg. Sinx, Cosx, Secx and Cosecx are periodic functions with period 2π N. B. Vyas Fourier Series
  9. 9. Periodic Function A function f (x) which satisfies the relation f (x) = f (x + T ) for all real x and some fixed T is called Periodic function. The smallest positive number T , for which this relation holds, is called the period of f (x). If T is the period of f (x) then f (x) = f (x + T ) = f (x + 2T ) = . . . = f (x + nT ) f (x) = f (x − T ) = f (x − 2T ) = . . . = f (x − nT ) ∴ f (x) = f (x ± nT ), where n is a positive integerEg. Sinx, Cosx, Secx and Cosecx are periodic functions with period 2π tanx and cotx are periodic functions with period π. N. B. Vyas Fourier Series
  10. 10. Even & Odd functions A function f (x) is said to be even if f (−x) = f (x). N. B. Vyas Fourier Series
  11. 11. Even & Odd functions A function f (x) is said to be even if f (−x) = f (x).Eg. x2 and cosx are even function. N. B. Vyas Fourier Series
  12. 12. Even & Odd functions A function f (x) is said to be even if f (−x) = f (x).Eg. x2 and cosx are even function. c c f (x)dx = 2 f (x)dx ; if f (x) is an even function. −c 0 N. B. Vyas Fourier Series
  13. 13. Even & Odd functions A function f (x) is said to be even if f (−x) = f (x).Eg. x2 and cosx are even function. c c f (x)dx = 2 f (x)dx ; if f (x) is an even function. −c 0 A function f (x) is said to be odd if f (−x) = −f (x). N. B. Vyas Fourier Series
  14. 14. Even & Odd functions A function f (x) is said to be even if f (−x) = f (x).Eg. x2 and cosx are even function. c c f (x)dx = 2 f (x)dx ; if f (x) is an even function. −c 0 A function f (x) is said to be odd if f (−x) = −f (x).Eg. x3 and sinx are odd function. N. B. Vyas Fourier Series
  15. 15. Even & Odd functions A function f (x) is said to be even if f (−x) = f (x).Eg. x2 and cosx are even function. c c f (x)dx = 2 f (x)dx ; if f (x) is an even function. −c 0 A function f (x) is said to be odd if f (−x) = −f (x).Eg. x3 and sinx are odd function. c f (x)dx = 0 ; if f (x) is an odd function. −c N. B. Vyas Fourier Series
  16. 16. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 N. B. Vyas Fourier Series
  17. 17. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 eax eax cos bx dx = 2 (a cosbx + b sinbx) + c a + b2 N. B. Vyas Fourier Series
  18. 18. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 eax eax cos bx dx = 2 (a cosbx + b sinbx) + c a + b2 c+2π cos nx c+2π sin nx dx = − = 0, n = 0 c n c N. B. Vyas Fourier Series
  19. 19. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 eax eax cos bx dx = 2 (a cosbx + b sinbx) + c a + b2 c+2π cos nx c+2π sin nx dx = − = 0, n = 0 c n c c+2π c+2π sin nx cos nx dx = = 0, n = 0 c n c N. B. Vyas Fourier Series
  20. 20. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 eax eax cos bx dx = 2 (a cosbx + b sinbx) + c a + b2 c+2π cos nx c+2π sin nx dx = − = 0, n = 0 c n c c+2π c+2π sin nx cos nx dx = = 0, n = 0 c n c c+2π 1 c+2π sin mx cos nx dx = 2sin mx cos nx dx c 2 c N. B. Vyas Fourier Series
  21. 21. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 eax eax cos bx dx = 2 (a cosbx + b sinbx) + c a + b2 c+2π cos nx c+2π sin nx dx = − = 0, n = 0 c n c c+2π c+2π sin nx cos nx dx = = 0, n = 0 c n c c+2π 1 c+2π sin mx cos nx dx = 2sin mx cos nx dx c 2 c 1 c+2π = [sin (m + n)x + sin (m − n)x] dx 2 c N. B. Vyas Fourier Series
  22. 22. Some Important Formula eax eax sin bx dx = (a sin bx − b cos bx) + c a2 + b 2 eax eax cos bx dx = 2 (a cosbx + b sinbx) + c a + b2 c+2π cos nx c+2π sin nx dx = − = 0, n = 0 c n c c+2π c+2π sin nx cos nx dx = = 0, n = 0 c n c c+2π 1 c+2π sin mx cos nx dx = 2sin mx cos nx dx c 2 c 1 c+2π = [sin (m + n)x + sin (m − n)x] dx 2 c c+2π 1 cos (m + n)x cos (m − n)x =− + = 0, n = 0 2 m+n m−n c N. B. Vyas Fourier Series
  23. 23. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c N. B. Vyas Fourier Series
  24. 24. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c N. B. Vyas Fourier Series
  25. 25. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx c N. B. Vyas Fourier Series
  26. 26. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx = 0 c c+2π sin nx cos nx dx c N. B. Vyas Fourier Series
  27. 27. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx = 0 c c+2π sin nx cos nx dx = 0, n = 0 c N. B. Vyas Fourier Series
  28. 28. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx = 0 c c+2π sin nx cos nx dx = 0, n = 0 c c+2π cos2 nx dx c N. B. Vyas Fourier Series
  29. 29. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx = 0 c c+2π sin nx cos nx dx = 0, n = 0 c c+2π cos2 nx dx = π c N. B. Vyas Fourier Series
  30. 30. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx = 0 c c+2π sin nx cos nx dx = 0, n = 0 c c+2π cos2 nx dx = π c c+2π sin2 nx dx c N. B. Vyas Fourier Series
  31. 31. Some Important Formula c+2π c+2π 1 cos mx cos nx dx = 2 2cos mx cos nx dx c c c+2π 1 = 2 [cos (m + n)x + cos (m − n)x] dx c c+2π 1 sin (m + n)x sin (m − n)x = 2 + = 0, m = n m+n m−n c c+2π sin mx sin nx dx = 0 c c+2π sin nx cos nx dx = 0, n = 0 c c+2π cos2 nx dx = π c c+2π sin2 nx dx = π c N. B. Vyas Fourier Series
  32. 32. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . . N. B. Vyas Fourier Series
  33. 33. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx N. B. Vyas Fourier Series
  34. 34. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 N. B. Vyas Fourier Series
  35. 35. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 1 = − e−2x (4x3 + 6x2 + 6x + 3) 8 N. B. Vyas Fourier Series
  36. 36. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 1 = − e−2x (4x3 + 6x2 + 6x + 3) 8 sin nπ = 0 N. B. Vyas Fourier Series
  37. 37. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 1 = − e−2x (4x3 + 6x2 + 6x + 3) 8 sin nπ = 0 and cos nπ = (−1)n N. B. Vyas Fourier Series
  38. 38. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 1 = − e−2x (4x3 + 6x2 + 6x + 3) 8 sin nπ = 0 and cos nπ = (−1)n sin n + 1 π = (−1)n 2 N. B. Vyas Fourier Series
  39. 39. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 1 = − e−2x (4x3 + 6x2 + 6x + 3) 8 sin nπ = 0 and cos nπ = (−1)n sin n + 1 π = (−1)n and cos n + 2 1 2 π=0 N. B. Vyas Fourier Series
  40. 40. Some Important Formula (Leibnitz’s Rule)To integrate the product of two functions, one of which is power of x . We apply the generalized rule of integration by parts u vdx = u v1 − u v2 + u v3 − u v4 + . . .Eg. x3 e−2x dx e−2x e−2x e−2x e−2x = x3 − 3x2 + 6x −6 −2 (−2)2 (−2)3 (−2)4 1 = − e−2x (4x3 + 6x2 + 6x + 3) 8 sin nπ = 0 and cos nπ = (−1)n sin n + 1 π = (−1)n and cos n + 2 1 2 π=0 where n is integer. N. B. Vyas Fourier Series
  41. 41. Fourier Series The Fourier series for the function f (x) in the interval c < x < c + 2π is given by N. B. Vyas Fourier Series
  42. 42. Fourier Series The Fourier series for the function f (x) in the interval c < x < c + 2π is given by ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 N. B. Vyas Fourier Series
  43. 43. Fourier Series The Fourier series for the function f (x) in the interval c < x < c + 2π is given by ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 c+2π 1 where a0 = f (x) dx π c N. B. Vyas Fourier Series
  44. 44. Fourier Series The Fourier series for the function f (x) in the interval c < x < c + 2π is given by ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 c+2π 1 where a0 = f (x) dx π c c+2π 1 an = f (x) cos nx dx π c N. B. Vyas Fourier Series
  45. 45. Fourier Series The Fourier series for the function f (x) in the interval c < x < c + 2π is given by ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 c+2π 1 where a0 = f (x) dx π c c+2π 1 an = f (x) cos nx dx π c c+2π 1 bn = f (x) sin nx dx π c N. B. Vyas Fourier Series
  46. 46. Fourier Series Corollary 1: If c = 0, the interval becomes 0 < x < 2π N. B. Vyas Fourier Series
  47. 47. Fourier Series Corollary 1: If c = 0, the interval becomes 0 < x < 2π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 N. B. Vyas Fourier Series
  48. 48. Fourier Series Corollary 1: If c = 0, the interval becomes 0 < x < 2π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 2π 1 where a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  49. 49. Fourier Series Corollary 1: If c = 0, the interval becomes 0 < x < 2π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  50. 50. Fourier Series Corollary 1: If c = 0, the interval becomes 0 < x < 2π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  51. 51. Fourier Series Corollary 2: If c = −π, the interval becomes −π << π N. B. Vyas Fourier Series
  52. 52. Fourier Series Corollary 2: If c = −π, the interval becomes −π << π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 N. B. Vyas Fourier Series
  53. 53. Fourier Series Corollary 2: If c = −π, the interval becomes −π << π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 π 1 where a0 = f (x) dx π −π N. B. Vyas Fourier Series
  54. 54. Fourier Series Corollary 2: If c = −π, the interval becomes −π << π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 π 1 where a0 = f (x) dx π −π π 1 an = f (x) cos nx dx π −π N. B. Vyas Fourier Series
  55. 55. Fourier Series Corollary 2: If c = −π, the interval becomes −π << π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 π 1 where a0 = f (x) dx π −π π 1 an = f (x) cos nx dx π −π π 1 bn = f (x) sin nx dx π −π N. B. Vyas Fourier Series
  56. 56. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π N. B. Vyas Fourier Series
  57. 57. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 N. B. Vyas Fourier Series
  58. 58. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 π 1 where a0 = f (x) dx = 0 π −π N. B. Vyas Fourier Series
  59. 59. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π where a0 = f (x) dx = 0 π −π 1 π an = f (x) cos nx dx = 0 π −π N. B. Vyas Fourier Series
  60. 60. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π where a0 = f (x) dx = 0 π −π 1 π an = f (x) cos nx dx = 0 π −π because cos nx is an even function , f (x)cos nx is an odd function N. B. Vyas Fourier Series
  61. 61. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π where a0 = f (x) dx = 0 π −π 1 π an = f (x) cos nx dx = 0 π −π because cos nx is an even function , f (x)cos nx is an odd function 1 π 2 π bn = f (x) sin nx dx = f (x) sin nx dx π −π π 0 N. B. Vyas Fourier Series
  62. 62. Fourier Series Special Case 1: If the interval is −c < x < c and f (x) is an odd function i.e. f (−x) = −f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π where a0 = f (x) dx = 0 π −π 1 π an = f (x) cos nx dx = 0 π −π because cos nx is an even function , f (x)cos nx is an odd function 1 π 2 π bn = f (x) sin nx dx = f (x) sin nx dx π −π π 0 because sin nx is an odd function , f (x)sin nx is an even function N. B. Vyas Fourier Series
  63. 63. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π N. B. Vyas Fourier Series
  64. 64. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 N. B. Vyas Fourier Series
  65. 65. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 π π 1 2 where a0 = f (x) dx = f (x) dx π −π π 0 N. B. Vyas Fourier Series
  66. 66. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π 2 π where a0 = f (x) dx = f (x) dx π −π π 0 1 π 2 π an = f (x) cos nx dx = f (x) cos nx dx π −π π 0 N. B. Vyas Fourier Series
  67. 67. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π 2 π where a0 = f (x) dx = f (x) dx π −π π 0 1 π 2 π an = f (x) cos nx dx = f (x) cos nx dx π −π π 0 because cos nx is an even function , f (x)cos nx is an even function N. B. Vyas Fourier Series
  68. 68. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π 2 π where a0 = f (x) dx = f (x) dx π −π π 0 1 π 2 π an = f (x) cos nx dx = f (x) cos nx dx π −π π 0 because cos nx is an even function , f (x)cos nx is an even function 1 π bn = f (x) sin nx dx = 0 π −π N. B. Vyas Fourier Series
  69. 69. Fourier Series Special Case 2: If the interval is −c < x < c and f (x) is an even function i.e. f (−x) = f (x). Let C = π ∞ ∞ a0 f (x) = + an cos nx + bn sin nx 2 n=1 n=1 1 π 2 π where a0 = f (x) dx = f (x) dx π −π π 0 1 π 2 π an = f (x) cos nx dx = f (x) cos nx dx π −π π 0 because cos nx is an even function , f (x)cos nx is an even function 1 π bn = f (x) sin nx dx = 0 π −π because sin nx is an odd function , f (x)sin nx is an odd function N. B. Vyas Fourier Series
  70. 70. Example 2 π−xEx. Obtain Fourier series of f (x) = in the 2 interval 0 ≤ x ≤ 2π. Hence deduce that π2 1 1 1 = 2 − 2 + 2 − ... 12 1 2 3 N. B. Vyas Fourier Series
  71. 71. Example 2 π−xEx. Obtain Fourier series of f (x) = in the 2 interval 0 ≤ x ≤ 2π. Hence deduce that π2 1 1 1 = 2 − 2 + 2 − ... 12 1 2 3 N. B. Vyas Fourier Series
  72. 72. ExampleSol. Step 1. The Fourier series of f (x) is given by N. B. Vyas Fourier Series
  73. 73. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 N. B. Vyas Fourier Series
  74. 74. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  75. 75. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  76. 76. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  77. 77. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  78. 78. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  79. 79. Example 1 2π Step 2. Now a0 = f (x) dx π 0 2π 1 (π − x)2 a0 = dx π 0 4 N. B. Vyas Fourier Series
  80. 80. Example 1 2π Step 2. Now a0 = f (x) dx π 0 2π 1 (π − x)2 a0 = dx π 0 4 2π 1 (π − x)3 1 = =− (−π 3 − π 3 ) 4π (−3) 0 12π N. B. Vyas Fourier Series
  81. 81. Example 1 2π Step 2. Now a0 = f (x) dx π 0 2π 1 (π − x)2 a0 = dx π 0 4 2π 1 (π − x)3 1 = =− (−π 3 − π 3 ) 4π (−3) 0 12π 2 π = 6 N. B. Vyas Fourier Series
  82. 82. Example 2π 1 Step 3. an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  83. 83. Example 1 2π Step 3. an = f (x) cos nx dx π 0 2π 1 (π − x)2 an = cos nx dx π 0 4 N. B. Vyas Fourier Series
  84. 84. Example 1 2π Step 3. an = f (x) cos nx dx π 0 2π 1 (π − x)2 an = cos nx dx π 0 4 1 = 2 n N. B. Vyas Fourier Series
  85. 85. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  86. 86. Example 1 2π Step 4. bn = f (x) sin nx dx π 0 2π 1 (π − x)2 bn = sin nx dx π 0 4 N. B. Vyas Fourier Series
  87. 87. Example 1 2π Step 4. bn = f (x) sin nx dx π 0 2π 1 (π − x)2 bn = sin nx dx π 0 4 =0 N. B. Vyas Fourier Series
  88. 88. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] N. B. Vyas Fourier Series
  89. 89. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] 2 ∞ π−x π2 1 = + cos nx 2 12 n=1 n2 N. B. Vyas Fourier Series
  90. 90. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] 2 ∞ π−x π2 1 = + cos nx 2 12 n=1 n2 π2 1 1 1 = + 2 cos x + 2 cos 2x + 2 cos 3x + . . . 12 1 2 3 N. B. Vyas Fourier Series
  91. 91. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] 2 ∞ π−x π2 1 = + cos nx 2 12 n=1 n2 π2 1 1 1 = + 2 cos x + 2 cos 2x + 2 cos 3x + . . . 12 1 2 3 Putting x = π, we get N. B. Vyas Fourier Series
  92. 92. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] 2 ∞ π−x π2 1 = + cos nx 2 12 n=1 n2 π2 1 1 1 = + 2 cos x + 2 cos 2x + 2 cos 3x + . . . 12 1 2 3 Putting x = π, we get π2 1 1 1 1 0= − 2 + 2 − 2 + 2 − ... 12 1 2 3 4 N. B. Vyas Fourier Series
  93. 93. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] 2 ∞ π−x π2 1 = + cos nx 2 12 n=1 n2 π2 1 1 1 = + 2 cos x + 2 cos 2x + 2 cos 3x + . . . 12 1 2 3 Putting x = π, we get π2 1 1 1 1 0= − 2 + 2 − 2 + 2 − ... 12 1 2 3 4 π2 1 1 1 1 = 2 − 2 + 2 − 2 + ... 12 1 2 3 4 N. B. Vyas Fourier Series
  94. 94. ExampleEx. Expand in a Fourier series the function f (x) = x in the interval 0 ≤ x ≤ 2π. N. B. Vyas Fourier Series
  95. 95. ExampleEx. Expand in a Fourier series the function f (x) = x in the interval 0 ≤ x ≤ 2π. N. B. Vyas Fourier Series
  96. 96. ExampleSol. Step 1. The Fourier series of f (x) is given by N. B. Vyas Fourier Series
  97. 97. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 N. B. Vyas Fourier Series
  98. 98. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  99. 99. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  100. 100. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  101. 101. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  102. 102. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  103. 103. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 2π 1 a0 = x dx π 0 N. B. Vyas Fourier Series
  104. 104. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 2π 1 a0 = x dx π 0 2 2π 1 x = 4π 2 0 N. B. Vyas Fourier Series
  105. 105. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 2π 1 a0 = x dx π 0 2 2π 1 x = 4π 2 0 = 2π N. B. Vyas Fourier Series
  106. 106. Example 2π 1 Step 3. an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  107. 107. Example 2π 1 Step 3. an = f (x) cos nx dx π 0 2π 1 an = x cos nx dx π 0 N. B. Vyas Fourier Series
  108. 108. Example 2π 1 Step 3. an = f (x) cos nx dx π 0 1 2π an = x cos nx dx π 0 2π sin nx cos nx x − − n n 0 N. B. Vyas Fourier Series
  109. 109. Example 2π 1 Step 3. an = f (x) cos nx dx π 0 1 2π an = x cos nx dx π 0 2π sin nx cos nx x − − n n 0 =0 N. B. Vyas Fourier Series
  110. 110. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  111. 111. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 2π 1 bn = x sin nx dx π 0 N. B. Vyas Fourier Series
  112. 112. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 2π 1 bn = x sin nx dx π 0 −2 = n N. B. Vyas Fourier Series
  113. 113. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] N. B. Vyas Fourier Series
  114. 114. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] ∞ 2π −2 f (x) = +0+ sin nx 2 n=1 n N. B. Vyas Fourier Series
  115. 115. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] ∞ 2π −2 f (x) = +0+ sin nx 2 n=1 n ∞ sin nx =π− n=1 n N. B. Vyas Fourier Series
  116. 116. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval [0, 2π] ∞ 2π −2 f (x) = +0+ sin nx 2 n=1 n ∞ sin nx =π− n=1 n N. B. Vyas Fourier Series
  117. 117. ExampleEx. Determine the Fourier series expansion of the function f (x) = xsin x in the interval 0 ≤ x ≤ 2π. N. B. Vyas Fourier Series
  118. 118. ExampleEx. Determine the Fourier series expansion of the function f (x) = xsin x in the interval 0 ≤ x ≤ 2π. N. B. Vyas Fourier Series
  119. 119. ExampleSol. Step 1. The Fourier series of f (x) is given by N. B. Vyas Fourier Series
  120. 120. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 N. B. Vyas Fourier Series
  121. 121. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  122. 122. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  123. 123. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  124. 124. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 2π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  125. 125. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  126. 126. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 2π 1 a0 = x sin x dx π 0 N. B. Vyas Fourier Series
  127. 127. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 1 2π a0 = x sin x dx π 0 1 = [−x cos x + sin x]2π 0 π N. B. Vyas Fourier Series
  128. 128. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 1 2π a0 = x sin x dx π 0 1 = [−x cos x + sin x]2π 0 π 1 = (−2 π cos 2π + sin 2π − 0 + sin 0) π N. B. Vyas Fourier Series
  129. 129. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 1 2π a0 = x sin x dx π 0 1 = [−x cos x + sin x]2π 0 π 1 = (−2 π cos 2π + sin 2π − 0 + sin 0) π −2π a0 = = −2 π N. B. Vyas Fourier Series
  130. 130. Example 2π 1 Step 3. an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  131. 131. Example 1 2π Step 3. an = f (x) cos nx dx π 0 1 2π an = x sin x cos nx dx π 0 N. B. Vyas Fourier Series
  132. 132. Example 1 2π Step 3. an = f (x) cos nx dx π 0 1 2π an = x sin x cos nx dx π 0 2π 1 = x (2 sin x cos nx) dx 2π 0 N. B. Vyas Fourier Series
  133. 133. Example 1 2π Step 3. an = f (x) cos nx dx π 0 1 2π an = x sin x cos nx dx π 0 2π 1 = x (2 sin x cos nx) dx 2π 0 2π 1 = x (sin (n + 1)x − sin (n − 1)x) dx 2π 0 N. B. Vyas Fourier Series
  134. 134. Example 1 2π Step 3. an = f (x) cos nx dx π 0 1 2π an = x sin x cos nx dx π 0 2π 1 = x (2 sin x cos nx) dx 2π 0 2π 1 = x (sin (n + 1)x − sin (n − 1)x) dx 2π 0 2π 2π 1 1 = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 N. B. Vyas Fourier Series
  135. 135. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 N. B. Vyas Fourier Series
  136. 136. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 N. B. Vyas Fourier Series
  137. 137. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 ∴ a1 = x sin 2x dx 2π 0 N. B. Vyas Fourier Series
  138. 138. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 ∴ a1 = x sin 2x dx 2π 0 2π 1 cos 2x sin 2x = −x + 2π 2 4 0 N. B. Vyas Fourier Series
  139. 139. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 ∴ a1 = x sin 2x dx 2π 0 2π 1 cos 2x sin 2x = −x + 2π 2 4 0 1 =− 2 N. B. Vyas Fourier Series
  140. 140. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 N. B. Vyas Fourier Series
  141. 141. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 N. B. Vyas Fourier Series
  142. 142. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 cos (n + 1)x sin (n + 1)x ∴ an = x − − − 2π n+1 (n + 1)2 0 N. B. Vyas Fourier Series
  143. 143. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 cos (n + 1)x sin (n + 1)x ∴ an = x − − − 2π n+1 (n + 1)2 0 2π 1 cos (n − 1)x sin (n − 1)x − x − − − 2π n−1 (n − 1)2 0 N. B. Vyas Fourier Series
  144. 144. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 cos (n + 1)x sin (n + 1)x ∴ an = x − − − 2π n+1 (n + 1)2 0 2π 1 cos (n − 1)x sin (n − 1)x − x − − − 2π n−1 (n − 1)2 0 1 2π 2π = − +0+ −0 2π n+1 n−1 N. B. Vyas Fourier Series
  145. 145. Example 2π 2π 1 1 an = x sin (n + 1)x dx − x sin (n − 1)x dx 2π 0 2π 0 If n = 1 2π 1 cos (n + 1)x sin (n + 1)x ∴ an = x − − − 2π n+1 (n + 1)2 0 2π 1 cos (n − 1)x sin (n − 1)x − x − − − 2π n−1 (n − 1)2 0 1 2π 2π = − +0+ −0 2π n+1 n−1 2 = 2 n −1 N. B. Vyas Fourier Series
  146. 146. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  147. 147. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 2π 1 = x sin x sin nx dx π 0 N. B. Vyas Fourier Series
  148. 148. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 2π 1 = x sin x sin nx dx π 0 2π 1 = x (2sin nx sin x) dx 2π 0 N. B. Vyas Fourier Series
  149. 149. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 2π 1 = x sin x sin nx dx π 0 2π 1 = x (2sin nx sin x) dx 2π 0 2π 1 = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 N. B. Vyas Fourier Series
  150. 150. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 N. B. Vyas Fourier Series
  151. 151. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 N. B. Vyas Fourier Series
  152. 152. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 2π 2π 1 1 ∴ b1 = (−x cos 2x) dx + x dx 2π 0 2π 0 N. B. Vyas Fourier Series
  153. 153. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 2π 2π 1 1 ∴ b1 = (−x cos 2x) dx + x dx 2π 0 2π 0 2π 1 sin 2x cos 2x x2 = x − − + 2π 2 4 2 0 N. B. Vyas Fourier Series
  154. 154. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 2π 2π 1 1 ∴ b1 = (−x cos 2x) dx + x dx 2π 0 2π 0 2π 1 sin 2x cos 2x x2 = x − − + 2π 2 4 2 0 b1 = π N. B. Vyas Fourier Series
  155. 155. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 N. B. Vyas Fourier Series
  156. 156. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 N. B. Vyas Fourier Series
  157. 157. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 2π 1 sin (n + 1)x cos (n + 1)x ∴ bn = x − − 2π n+1 (n + 1)2 0 N. B. Vyas Fourier Series
  158. 158. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 2π 1 sin (n + 1)x cos (n + 1)x ∴ bn = x − − 2π n+1 (n + 1)2 0 2π 1 sin (n − 1)x cos (n − 1)x + x + 2π n−1 (n − 1)2 0 N. B. Vyas Fourier Series
  159. 159. Example 2π 1 bn = x (−cos (n + 1)x + cos (n − 1)x) dx 2π 0 If n = 1 2π 1 sin (n + 1)x cos (n + 1)x ∴ bn = x − − 2π n+1 (n + 1)2 0 2π 1 sin (n − 1)x cos (n − 1)x + x + 2π n−1 (n − 1)2 0 =0 N. B. Vyas Fourier Series
  160. 160. Example Step 5. Substituting values of a0 , a1 , an (n > 1), b1 and bn (n > 1) in (1), we get the required Fourier series of f (x) in the interval [0, 2π] N. B. Vyas Fourier Series
  161. 161. Example Step 5. Substituting values of a0 , a1 , an (n > 1), b1 and bn (n > 1) in (1), we get the required Fourier series of f (x) in the interval [0, 2π] ∞ −2 f (x) = + (an cos nx + bn sin nx) 2 n=1 N. B. Vyas Fourier Series
  162. 162. Example Step 5. Substituting values of a0 , a1 , an (n > 1), b1 and bn (n > 1) in (1), we get the required Fourier series of f (x) in the interval [0, 2π] ∞ −2 f (x) = + (an cos nx + bn sin nx) 2 n=1 N. B. Vyas Fourier Series
  163. 163. ExampleEx. Find the Fourier series for the periodic function f (x)= −π; −π < x < 0 = x; 0 < x < π N. B. Vyas Fourier Series
  164. 164. ExampleEx. Find the Fourier series for the periodic function f (x)= −π; −π < x < 0 = x; 0 < x < π N. B. Vyas Fourier Series
  165. 165. ExampleSol. Step 1. The Fourier series of f (x) is given by N. B. Vyas Fourier Series
  166. 166. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 N. B. Vyas Fourier Series
  167. 167. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 π 1 where a0 = f (x) dx π −π N. B. Vyas Fourier Series
  168. 168. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 π 1 where a0 = f (x) dx π −π π 1 an = f (x) cos nx dx π −π N. B. Vyas Fourier Series
  169. 169. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 π 1 where a0 = f (x) dx π −π π 1 an = f (x) cos nx dx π −π π 1 bn = f (x) sin nx dx π −π N. B. Vyas Fourier Series
  170. 170. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 π 1 where a0 = f (x) dx π −π π 1 an = f (x) cos nx dx π −π π 1 bn = f (x) sin nx dx π −π N. B. Vyas Fourier Series
  171. 171. Example π 1 Step 2. Now a0 = f (x) dx π −π N. B. Vyas Fourier Series
  172. 172. Example π 1 Step 2. Now a0 = f (x) dx π −π 0 π 1 = f (x) dx + f (x) dx π −π 0 N. B. Vyas Fourier Series
  173. 173. Example π 1 Step 2. Now a0 = f (x) dx π −π 0 π 1 = f (x) dx + f (x) dx π −π 0 0 π 1 = (−π) dx + x dx π −π 0 N. B. Vyas Fourier Series
  174. 174. Example π 1 Step 2. Now a0 = f (x) dx π −π 0 π 1 = f (x) dx + f (x) dx π −π 0 0 π 1 = (−π) dx + x dx π −π 0 π =− 2 N. B. Vyas Fourier Series
  175. 175. Example π 1 Step 2. Now a0 = f (x) dx π −π 0 π 1 = f (x) dx + f (x) dx π −π 0 0 π 1 = (−π) dx + x dx π −π 0 π =− 2 N. B. Vyas Fourier Series
  176. 176. Example 0 π 1 Step 3. an = f (x) cos nx dx + f (x) cos nx dx π −π 0 N. B. Vyas Fourier Series
  177. 177. Example 1 0 π Step 3. an = f (x) cos nx dx + f (x) cos nx dx π −π 0 0 π 1 = (−π) cos nx dx + x cos nx dx π −π 0 N. B. Vyas Fourier Series
  178. 178. Example 1 0 π Step 3. an = f (x) cos nx dx + f (x) cos nx dx π −π 0 0 π 1 = (−π) cos nx dx + x cos nx dx π −π 0 0 1 sin nx = −π + π n −π N. B. Vyas Fourier Series
  179. 179. Example 1 0 π Step 3. an = f (x) cos nx dx + f (x) cos nx dx π −π 0 0 π 1 = (−π) cos nx dx + x cos nx dx π −π 0 0 π 1 sin nx sin nx cos nx = −π + x − (1) − π n −π n n2 0 N. B. Vyas Fourier Series
  180. 180. Example 1 0 π Step 3. an = f (x) cos nx dx + f (x) cos nx dx π −π 0 0 π 1 = (−π) cos nx dx + x cos nx dx π −π 0 0 π 1 sin nx sin nx cos nx = −π + x − (1) − π n −π n n2 0 n (−1) − 1 = πn2 N. B. Vyas Fourier Series
  181. 181. Example 0 π 1 Step 4. bn = f (x) sin nx dx + f (x) sin nx dx π −π 0 N. B. Vyas Fourier Series
  182. 182. Example 0 π 1 Step 4. bn = f (x) sin nx dx + f (x) sin nx dx π −π 0 0 π 1 = (−π) sin nx dx + x sin nx dx π −π 0 N. B. Vyas Fourier Series
  183. 183. Example 0 π 1 Step 4. bn = f (x) sin nx dx + f (x) sin nx dx π −π 0 0 π 1 = (−π) sin nx dx + x sin nx dx π −π 0 0 1 −cosnx = −π + π n −π N. B. Vyas Fourier Series
  184. 184. Example 0 π 1 Step 4. bn = f (x) sin nx dx + f (x) sin nx dx π −π 0 0 π 1 = (−π) sin nx dx + x sin nx dx π −π 0 0 π 1 −cosnx −cos nx −sin nx = −π + x − π n −π n n2 0 N. B. Vyas Fourier Series
  185. 185. Example 0 π 1 Step 4. bn = f (x) sin nx dx + f (x) sin nx dx π −π 0 0 π 1 = (−π) sin nx dx + x sin nx dx π −π 0 0 π 1 −cosnx −cos nx −sin nx = −π + x − π n −π n n2 0 n 1 − 2(−1) = n N. B. Vyas Fourier Series
  186. 186. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval (−π, π) N. B. Vyas Fourier Series
  187. 187. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval (−π, π) ∞ −π f (x) = + (an cos nx + bn sin nx) 4 n=1 N. B. Vyas Fourier Series
  188. 188. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval (−π, π) ∞ −π f (x) = + (an cos nx + bn sin nx) 4 n=1 ∞ ∞ −π (−1)n − 1 1 − 2(−1)n = + cos nx + sin nx 4 n=1 πn2 n=1 n N. B. Vyas Fourier Series
  189. 189. Example Step 5. Substituting values of a0 , an and bn in (1), we get the required Fourier series of f (x) in the interval (−π, π) ∞ −π f (x) = + (an cos nx + bn sin nx) 4 n=1 ∞ ∞ −π (−1)n − 1 1 − 2(−1)n = + cos nx + sin nx 4 n=1 πn2 n=1 n N. B. Vyas Fourier Series
  190. 190. ExampleEx. Find the Fourier series for the periodic function f (x)= 2; −π < x < 0 = 1; 0 < x < π N. B. Vyas Fourier Series
  191. 191. ExampleEx. Find the Fourier series for the periodic function f (x)= 2; −π < x < 0 = 1; 0 < x < π N. B. Vyas Fourier Series
  192. 192. ExampleEx. Find the Fourier series for the periodic function f (x)= −k; −π < x < 0 = k; 0 < x < π N. B. Vyas Fourier Series
  193. 193. ExampleEx. Find the Fourier series for the periodic function f (x)= −k; −π < x < 0 = k; 0 < x < π 1 1 1 π Hence deduce that 1 − + − + . . . = 3 5 7 4 N. B. Vyas Fourier Series
  194. 194. ExampleEx. Find the Fourier series of the function f (x)= x; 0 ≤ x < π = 2π ; x = π = 2π − x ; π < x < 2π 3π 2 1 1 1 Hence deduce that = 2 + 2 + 2 + ... 8 1 3 5 N. B. Vyas Fourier Series
  195. 195. ExampleSol. Step 1. The Fourier series of f (x) is given by N. B. Vyas Fourier Series
  196. 196. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 N. B. Vyas Fourier Series
  197. 197. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  198. 198. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 π 1 an = f (x) cos nx dx π 0 N. B. Vyas Fourier Series
  199. 199. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  200. 200. ExampleSol. Step 1. The Fourier series of f (x) is given by ∞ a0 f (x) = + (an cos nx + bn sin nx) . . . (1) 2 n=1 2π 1 where a0 = f (x) dx π 0 π 1 an = f (x) cos nx dx π 0 2π 1 bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  201. 201. Example 2π 1 Step 2. Now a0 = f (x) dx π 0 N. B. Vyas Fourier Series
  202. 202. Example 1 2π Step 2. Now a0 = f (x) dx π 0 π 2π 1 = x dx + (2π − x) dx π 0 π N. B. Vyas Fourier Series
  203. 203. Example 1 2π Step 2. Now a0 = f (x) dx π 0 π 2π 1 = x dx + (2π − x) dx π 0 π π 2π 1 x2 x2 = + 2πx − π 2 0 2 π N. B. Vyas Fourier Series
  204. 204. Example 1 2π Step 2. Now a0 = f (x) dx π 0 π 2π 1 = x dx + (2π − x) dx π 0 π π 2π 1 x2 x2 = + 2πx − π 2 0 2 π =π N. B. Vyas Fourier Series
  205. 205. Example π 2π 1 Step 3. an = f (x) cos nx dx + f (x) cos nx dx π 0 π N. B. Vyas Fourier Series
  206. 206. Example π 2π 1 Step 3. an = f (x) cos nx dx + f (x) cos nx dx π 0 π π 2π 1 = x cos nx dx + (2π − x) cos nx dx π 0 π N. B. Vyas Fourier Series
  207. 207. Example π 2π 1 Step 3. an = f (x) cos nx dx + f (x) cos nx dx π 0 π π 2π 1 = x cos nx dx + (2π − x) cos nx dx π 0 π π 1 sin nx cos nx = x − (1) − π n n2 0 N. B. Vyas Fourier Series
  208. 208. Example π 2π 1 Step 3. an = f (x) cos nx dx + f (x) cos nx dx π 0 π π 2π 1 = x cos nx dx + (2π − x) cos nx dx π 0 π π 1 sin nx cos nx = x − (1) − π n n2 0 2π 1 sin nx cos nx + (2π − x) − (−1) − π n n2 π N. B. Vyas Fourier Series
  209. 209. Example π 2π 1 Step 3. an = f (x) cos nx dx + f (x) cos nx dx π 0 π π 2π 1 = x cos nx dx + (2π − x) cos nx dx π 0 π π 1 sin nx cos nx = x − (1) − π n n2 0 2π 1 sin nx cos nx + (2π − x) − (−1) − π n n2 π 1 cos nπ 1 = 0+ 2 − 0+ 2 π n n 1 cos 2nπ cos nπ + 0− 2 − 0− π n n2 2 [(−1)n − 1] = πn2 N. B. Vyas Fourier Series
  210. 210. Example 2π 1 Step 4. bn = f (x) sin nx dx π 0 N. B. Vyas Fourier Series
  211. 211. Example 1 2π Step 4. bn = f (x) sin nx dx π 0 π 2π 1 = x sin nx dx + (2π − x) sin nx dx π 0 π N. B. Vyas Fourier Series
  212. 212. Example 1 2π Step 4. bn = f (x) sin nx dx π 0 π 2π 1 = x sin nx dx + (2π − x) sin nx dx π 0 π π 1 cos nx sin nx = x − − (1) − π n n2 0 N. B. Vyas Fourier Series
  213. 213. Example 1 2π Step 4. bn = f (x) sin nx dx π 0 π 2π 1 = x sin nx dx + (2π − x) sin nx dx π 0 π π 1 cos nx sin nx = x − − (1) − π n n2 0 2π 1 cos nx sin nx + (2π − x) − − (−1) − π n n2 π N. B. Vyas Fourier Series
  214. 214. Example 1 2π Step 4. bn = f (x) sin nx dx π 0 π 2π 1 = x sin nx dx + (2π − x) sin nx dx π 0 π π 1 cos nx sin nx = x − − (1) − π n n2 0 2π 1 cos nx sin nx + (2π − x) − − (−1) − π n n2 π =0 N. B. Vyas Fourier Series
  215. 215. ExampleEx. Find the Fourier series of f (x)= −1; −π < x < − π 2 = 0 ; −π < x < π 2 2 = 1; π < x < π 2 N. B. Vyas Fourier Series

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