Your SlideShare is downloading. ×
  • Like
Fourier series 2
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Fourier series 2

  • 464 views
Published

 

Published in Career
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
464
On SlideShare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
34
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. Fourier Series 2 N. B. Vyas Department of Mathematics,Atmiya Institute of Tech. & Science, Rajkot (Guj.)- INDIA N. B. Vyas Fourier Series 2
  • 2. Functions of any Period p = 2L Let f (x) be a periodic function with an arbitrary period 2L defined in the interval c < x < c + 2L N. B. Vyas Fourier Series 2
  • 3. Functions of any Period p = 2L Let f (x) be a periodic function with an arbitrary period 2L defined in the interval c < x < c + 2L The Fourier series of f (x) is given by N. B. Vyas Fourier Series 2
  • 4. Functions of any Period p = 2L Let f (x) be a periodic function with an arbitrary period 2L defined in the interval c < x < c + 2L The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L N. B. Vyas Fourier Series 2
  • 5. Functions of any Period p = 2L Let f (x) be a periodic function with an arbitrary period 2L defined in the interval c < x < c + 2L The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L c+2L 1 where a0 = f (x) dx L c N. B. Vyas Fourier Series 2
  • 6. Functions of any Period p = 2L Let f (x) be a periodic function with an arbitrary period 2L defined in the interval c < x < c + 2L The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 c+2L where a0 = f (x) dx L c 1 c+2L nπx an = f (x) cos dx L c L N. B. Vyas Fourier Series 2
  • 7. Functions of any Period p = 2L Let f (x) be a periodic function with an arbitrary period 2L defined in the interval c < x < c + 2L The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 c+2L where a0 = f (x) dx L c 1 c+2L nπx an = f (x) cos dx L c L 1 c+2L nπx bn = f (x) sin dx L c L N. B. Vyas Fourier Series 2
  • 8. Functions of any Period p = 2L Corollary 1: If c = 0 the interval becomes 0 < x < 2L N. B. Vyas Fourier Series 2
  • 9. Functions of any Period p = 2L Corollary 1: If c = 0 the interval becomes 0 < x < 2L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L N. B. Vyas Fourier Series 2
  • 10. Functions of any Period p = 2L Corollary 1: If c = 0 the interval becomes 0 < x < 2L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 2L 1 where a0 = f (x) dx L 0 N. B. Vyas Fourier Series 2
  • 11. Functions of any Period p = 2L Corollary 1: If c = 0 the interval becomes 0 < x < 2L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 2L 1 where a0 = f (x) dx L 0 2L 1 nπx an = f (x) cos dx L 0 L N. B. Vyas Fourier Series 2
  • 12. Functions of any Period p = 2L Corollary 1: If c = 0 the interval becomes 0 < x < 2L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 2L 1 where a0 = f (x) dx L 0 2L 1 nπx an = f (x) cos dx L 0 L 2L 1 nπx bn = f (x) sin dx L 0 L N. B. Vyas Fourier Series 2
  • 13. Functions of any Period p = 2L Corollary 2: If c = −L the interval becomes −L < x < L N. B. Vyas Fourier Series 2
  • 14. Functions of any Period p = 2L Corollary 2: If c = −L the interval becomes −L < x < L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L N. B. Vyas Fourier Series 2
  • 15. Functions of any Period p = 2L Corollary 2: If c = −L the interval becomes −L < x < L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L L 1 where a0 = f (x) dx L −L N. B. Vyas Fourier Series 2
  • 16. Functions of any Period p = 2L Corollary 2: If c = −L the interval becomes −L < x < L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 L where a0 = f (x) dx L −L 1 L nπx an = f (x) cos dx L −L L N. B. Vyas Fourier Series 2
  • 17. Functions of any Period p = 2L Corollary 2: If c = −L the interval becomes −L < x < L ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 L where a0 = f (x) dx L −L 1 L nπx an = f (x) cos dx L −L L 1 L nπx bn = f (x) sin dx L −L L N. B. Vyas Fourier Series 2
  • 18. ExampleEx. The Fourier series of f (x) = x2 , 0 < x < 2 where f (x + 2) = f (x). N. B. Vyas Fourier Series 2
  • 19. ExampleEx. The Fourier series of f (x) = x2 , 0 < x < 2 where f (x + 2) = f (x). 1 1 1 π2 Hence deduce that 1 − 2 + 2 − 2 + . . . = 2 3 4 12 N. B. Vyas Fourier Series 2
  • 20. ExampleSol. Step 1: The Fourier series of f (x) is given by N. B. Vyas Fourier Series 2
  • 21. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L N. B. Vyas Fourier Series 2
  • 22. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 2 1 where a0 = f (x) dx L 0 N. B. Vyas Fourier Series 2
  • 23. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 2 where a0 = f (x) dx L 0 1 2 nπx an = f (x) cos dx L 0 L N. B. Vyas Fourier Series 2
  • 24. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 2 where a0 = f (x) dx L 0 1 2 nπx an = f (x) cos dx L 0 L 1 2 nπx bn = f (x) sin dx L 0 L N. B. Vyas Fourier Series 2
  • 25. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 2 where a0 = f (x) dx L 0 1 2 nπx an = f (x) cos dx L 0 L 1 2 nπx bn = f (x) sin dx L 0 L Here p = 2L = 2 ⇒ L = 1 N. B. Vyas Fourier Series 2
  • 26. Example 2 1 Step 2. Now a0 = f (x) dx 1 0 N. B. Vyas Fourier Series 2
  • 27. Example 2 1 Step 2. Now a0 = f (x) dx 1 0 2 a0 = (x)2 dx 0 N. B. Vyas Fourier Series 2
  • 28. Example 2 1 Step 2. Now a0 = f (x) dx 1 0 2 a0 = (x)2 dx 0 3 2 x = 3 0 N. B. Vyas Fourier Series 2
  • 29. Example 2 1 Step 2. Now a0 = f (x) dx 1 0 2 a0 = (x)2 dx 0 3 2 x = 3 0 8 = 3 N. B. Vyas Fourier Series 2
  • 30. Example 2 1 nπx Step 3. an = f (x) cos dx 1 0 1 N. B. Vyas Fourier Series 2
  • 31. Example 2 1 nπx Step 3. an = f (x) cos dx 1 0 1 2 an = x2 cos(nπx) dx 0 N. B. Vyas Fourier Series 2
  • 32. Example 2 1 nπx Step 3. an = f (x) cos dx 1 0 1 2 an = x2 cos(nπx) dx 0 2 sin nπx cos nπx sin nπx = x2 − (2x) − 2π2 +2 − 3 3 nπ n nπ 0 N. B. Vyas Fourier Series 2
  • 33. Example 2 1 nπx Step 3. an = f (x) cos dx 1 0 1 2 an = x2 cos(nπx) dx 0 2 sin nπx cos nπx sin nπx = x2 − (2x) − 2π2 +2 − 3 3 nπ n nπ 0 4 = 2 2 nπ N. B. Vyas Fourier Series 2
  • 34. Example 2 1 nπx Step 4. bn = f (x) sin dx 1 0 1 N. B. Vyas Fourier Series 2
  • 35. Example 2 1 nπx Step 4. bn = f (x) sin dx 1 0 1 2 bn = x2 sin(nπx) dx 0 N. B. Vyas Fourier Series 2
  • 36. Example 2 1 nπx Step 4. bn = f (x) sin dx 1 0 1 2 bn = x2 sin(nπx) dx 0 2 cos nπx sin nπx cos nπx = x2 − − (2x) − 2 2 +2 nπ nπ n3 π 3 0 N. B. Vyas Fourier Series 2
  • 37. Example 2 1 nπx Step 4. bn = f (x) sin dx 1 0 1 2 bn = x2 sin(nπx) dx 0 2 cos nπx sin nπx cos nπx = x2 − − (2x) − 2 2 +2 nπ nπ n3 π 3 0 4 =− nπ N. B. Vyas Fourier Series 2
  • 38. 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, 2) N. B. Vyas Fourier Series 2
  • 39. 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, 2) ∞ ∞ 8 4 nπx 4 nπx f (x) = + 2π2 cos + − sin 6 n=1 n 1 n=1 nπ 1 N. B. Vyas Fourier Series 2
  • 40. 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, 2) ∞ ∞ 8 4 nπx 4 nπx f (x) = + 2π2 cos + − sin 6 n=1 n 1 n=1 nπ 1 ∞ ∞ 4 4 cos(nπx) 4 sin(nπx) = + 2 − 3 π n=1 n2 π n=1 n N. B. Vyas Fourier Series 2
  • 41. 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, 2) ∞ ∞ 8 4 nπx 4 nπx f (x) = + 2π2 cos + − sin 6 n=1 n 1 n=1 nπ 1 ∞ ∞ 4 4 cos(nπx) 4 sin(nπx) = + 2 − 3 π n=1 n2 π n=1 n Putting x = 1, we get N. B. Vyas Fourier Series 2
  • 42. 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, 2) ∞ ∞ 8 4 nπx 4 nπx f (x) = + 2π2 cos + − sin 6 n=1 n 1 n=1 nπ 1 ∞ ∞ 4 4 cos(nπx) 4 sin(nπx) = + 2 − 3 π n=1 n2 π n=1 n Putting x = 1, we get 4 4 1 1 1 1 = + 2 − 1 + 2 − 2 + ... 3 π 1 2 3 N. B. Vyas Fourier Series 2
  • 43. 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, 2) ∞ ∞ 8 4 nπx 4 nπx f (x) = + 2π2 cos + − sin 6 n=1 n 1 n=1 nπ 1 ∞ ∞ 4 4 cos(nπx) 4 sin(nπx) = + 2 − 3 π n=1 n2 π n=1 n Putting x = 1, we get 4 4 1 1 1 1 = + 2 − 1 + 2 − 2 + ... 3 π 1 2 3 1 4 1 1 1 − = 2 − 1 + 2 − 2 + ... 3 π 1 2 3 N. B. Vyas Fourier Series 2
  • 44. 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, 2) ∞ ∞ 8 4 nπx 4 nπx f (x) = + 2π2 cos + − sin 6 n=1 n 1 n=1 nπ 1 ∞ ∞ 4 4 cos(nπx) 4 sin(nπx) = + 2 − 3 π n=1 n2 π n=1 n Putting x = 1, we get 4 4 1 1 1 1 = + 2 − 1 + 2 − 2 + ... 3 π 1 2 3 1 4 1 1 1 − = 2 − 1 + 2 − 2 + ... 3 π 1 2 3 2 π 1 1 1 1 = 2 − 2 + 2 − 2 + ... 12 1 2 3 4 N. B. Vyas Fourier Series 2
  • 45. ExampleEx. Find the Fourier series of f (x) = 2x in −1 < x < 1 where p = 2L = 2. N. B. Vyas Fourier Series 2
  • 46. ExampleSol. Step 1: The Fourier series of f (x) is given by N. B. Vyas Fourier Series 2
  • 47. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L N. B. Vyas Fourier Series 2
  • 48. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 1 where a0 = f (x) dx L −1 N. B. Vyas Fourier Series 2
  • 49. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 1 where a0 = f (x) dx L −1 1 1 nπx an = f (x) cos dx L −1 L N. B. Vyas Fourier Series 2
  • 50. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 1 where a0 = f (x) dx L −1 1 1 nπx an = f (x) cos dx L −1 L 1 1 nπx bn = f (x) sin dx L −1 L N. B. Vyas Fourier Series 2
  • 51. ExampleSol. Step 1: The Fourier series of f (x) is given by ∞ a0 nπx nπx f (x) = + an cos + bn sin 2 n=1 L L 1 1 where a0 = f (x) dx L −1 1 1 nπx an = f (x) cos dx L −1 L 1 1 nπx bn = f (x) sin dx L −1 L Here p = 2L = 2 ⇒ L = 1 N. B. Vyas Fourier Series 2
  • 52. Example 1 1 Step 2. Now a0 = f (x) dx 1 −1 N. B. Vyas Fourier Series 2
  • 53. Example 1 1 Step 2. Now a0 = f (x) dx 1 −1 1 a0 = 2x dx −1 N. B. Vyas Fourier Series 2
  • 54. Example 1 1 Step 2. Now a0 = f (x) dx 1 −1 1 a0 = 2x dx −1 =0 N. B. Vyas Fourier Series 2
  • 55. Example 1 1 nπx Step 3. an = f (x) cos dx 1 −1 1 N. B. Vyas Fourier Series 2
  • 56. Example 1 1 nπx Step 3. an = f (x) cos dx 1 −1 1 1 an = 2x cos(nπx) dx −1 N. B. Vyas Fourier Series 2
  • 57. Example 1 1 nπx Step 3. an = f (x) cos dx 1 −1 1 1 an = 2x cos(nπx) dx −1 1 sin nπx cos nπx = 2x − (2) − 2 2 nπ nπ −1 N. B. Vyas Fourier Series 2
  • 58. Example 1 1 nπx Step 3. an = f (x) cos dx 1 −1 1 1 an = 2x cos(nπx) dx −1 1 sin nπx cos nπx = 2x − (2) − 2 2 nπ nπ −1 2(−1)n = 0+ 2 2 nπ N. B. Vyas Fourier Series 2
  • 59. Example 1 1 nπx Step 3. an = f (x) cos dx 1 −1 1 1 an = 2x cos(nπx) dx −1 1 sin nπx cos nπx = 2x − (2) − 2 2 nπ nπ −1 n n 2(−1) 2(−1) = 0+ 2 2 −0− 2 2 nπ nπ N. B. Vyas Fourier Series 2
  • 60. Example 1 1 nπx Step 3. an = f (x) cos dx 1 −1 1 1 an = 2x cos(nπx) dx −1 1 sin nπx cos nπx = 2x − (2) − 2 2 nπ nπ −1 n n 2(−1) 2(−1) = 0+ 2 2 −0− 2 2 nπ nπ =0 N. B. Vyas Fourier Series 2
  • 61. ExampleEx. Find the Fourier series of periodic function f (x)= −1; −1 < x < 0 = 1; 0 < x < 1 p = 2L = 2 N. B. Vyas Fourier Series 2
  • 62. ExampleEx. Find the Fourier series of periodic function f (x)= 0; −2 < x < 0 = 2; 0 < x < 2 p = 2L = 4 N. B. Vyas Fourier Series 2
  • 63. Fourier Half Range Series A function f (x) defined only on the interval of the form 0 < x < L. N. B. Vyas Fourier Series 2
  • 64. Fourier Half Range Series A function f (x) defined only on the interval of the form 0 < x < L. If f (x) is represented on this interval by a Fourier series of period 2L N. B. Vyas Fourier Series 2
  • 65. Fourier Half Range Series A function f (x) defined only on the interval of the form 0 < x < L. If f (x) is represented on this interval by a Fourier series of period 2L Then such Fourier series are known as half range Fourier series or half range expansions. N. B. Vyas Fourier Series 2
  • 66. Fourier Half Range Series A function f (x) defined only on the interval of the form 0 < x < L. If f (x) is represented on this interval by a Fourier series of period 2L Then such Fourier series are known as half range Fourier series or half range expansions. Types of Half Range Fourier series N. B. Vyas Fourier Series 2
  • 67. Fourier Half Range Series A function f (x) defined only on the interval of the form 0 < x < L. If f (x) is represented on this interval by a Fourier series of period 2L Then such Fourier series are known as half range Fourier series or half range expansions. Types of Half Range Fourier series 1 Fourier Cosine Series N. B. Vyas Fourier Series 2
  • 68. Fourier Half Range Series A function f (x) defined only on the interval of the form 0 < x < L. If f (x) is represented on this interval by a Fourier series of period 2L Then such Fourier series are known as half range Fourier series or half range expansions. Types of Half Range Fourier series 1 Fourier Cosine Series 2 Fourier Sine Series N. B. Vyas Fourier Series 2
  • 69. Fourier Cosine Series Let f (x) be piecewise continuous on [o, l]. N. B. Vyas Fourier Series 2
  • 70. Fourier Cosine Series Let f (x) be piecewise continuous on [o, l]. the Fourier cosine series expansion of f (x) on the half range interval [0, l] is given by N. B. Vyas Fourier Series 2
  • 71. Fourier Cosine Series Let f (x) be piecewise continuous on [o, l]. the Fourier cosine series expansion of f (x) on the half range interval [0, l] is given by ∞ ao nπx f (x) = + an cos 2 n=1 l N. B. Vyas Fourier Series 2
  • 72. Fourier Cosine Series Let f (x) be piecewise continuous on [o, l]. the Fourier cosine series expansion of f (x) on the half range interval [0, l] is given by ∞ ao nπx f (x) = + an cos 2 n=1 l l 2 where a0 = f (x) dx l 0 N. B. Vyas Fourier Series 2
  • 73. Fourier Cosine Series Let f (x) be piecewise continuous on [o, l]. the Fourier cosine series expansion of f (x) on the half range interval [0, l] is given by ∞ ao nπx f (x) = + an cos 2 n=1 l l 2 where a0 = f (x) dx l 0 l 2 nπx an = f (x) cos dx l 0 l N. B. Vyas Fourier Series 2
  • 74. Fourier Sine Series Let f (x) be piecewise continuous on [o, l]. N. B. Vyas Fourier Series 2
  • 75. Fourier Sine Series Let f (x) be piecewise continuous on [o, l]. the Fourier sine series expansion of f (x) on the half range interval [0, l] is given by N. B. Vyas Fourier Series 2
  • 76. Fourier Sine Series Let f (x) be piecewise continuous on [o, l]. the Fourier sine series expansion of f (x) on the half range interval [0, l] is given by ∞ nπx f (x) = bn sin n=1 l N. B. Vyas Fourier Series 2
  • 77. Fourier Sine Series Let f (x) be piecewise continuous on [o, l]. the Fourier sine series expansion of f (x) on the half range interval [0, l] is given by ∞ nπx f (x) = bn sin n=1 l l 2 nπx bn = f (x) sin dx l 0 l N. B. Vyas Fourier Series 2
  • 78. ExampleEx. Find Fourier cosine and sine series of the function f (x) = 1 for 0 ≤ x ≤ 2 N. B. Vyas Fourier Series 2
  • 79. ExampleSol. Here given interval is 0 ≤ x ≤ 2 N. B. Vyas Fourier Series 2
  • 80. ExampleSol. Here given interval is 0 ≤ x ≤ 2 ∴l=2 N. B. Vyas Fourier Series 2
  • 81. Example 1 Fourier cosine series ∞ a0 nπx Step 1. f (x) = + an cos 2 n=1 l N. B. Vyas Fourier Series 2
  • 82. Example 1 Fourier cosine series ∞ a0 nπx Step 1. f (x) = + an cos 2 n=1 l l 2 where a0 = f (x)dx l 0 N. B. Vyas Fourier Series 2
  • 83. Example 1 Fourier cosine series ∞ a0 nπx Step 1. f (x) = + an cos 2 n=1 l 2 l where a0 = f (x)dx l 0 2 l nπx an = f (x) cos l 0 l N. B. Vyas Fourier Series 2
  • 84. Example 2 2 Step 2. a0 = f (x)dx 2 0 N. B. Vyas Fourier Series 2
  • 85. Example 2 2 Step 2. a0 = f (x)dx 2 0 2 = 1dx 0 N. B. Vyas Fourier Series 2
  • 86. Example 2 2 Step 2. a0 = f (x)dx 2 0 2 = 1dx = [x]2 0 0 N. B. Vyas Fourier Series 2
  • 87. Example 2 2 Step 2. a0 = f (x)dx 2 0 2 = 1dx = [x]2 = 2. 0 0 N. B. Vyas Fourier Series 2
  • 88. Example 2 2 nπx Step 3. an = f (x)cos dx 2 0 2 N. B. Vyas Fourier Series 2
  • 89. Example 2 2 nπx Step 3. an = f (x)cos dx 2 0 2 2 nπx = (1) cos dx 0 2 N. B. Vyas Fourier Series 2
  • 90. Example 2 2 nπx Step 3. an = f (x)cos dx 2 0 2 2 nπx = (1) cos dx 0 2 nπx 2   sin =  2  nπ  2 0 N. B. Vyas Fourier Series 2
  • 91. Example 2 2 nπx Step 3. an = f (x)cos dx 2 0 2 2 nπx = (1) cos dx 0 2 nπx 2   sin =  2  nπ  2 0 2 = (sin (nπ) − sin (0)) nπ N. B. Vyas Fourier Series 2
  • 92. Example 2 2 nπx Step 3. an = f (x)cos dx 2 0 2 2 nπx = (1) cos dx 0 2 nπx 2   sin =  2  nπ  2 0 2 = (sin (nπ) − sin (0)) = 0 nπ N. B. Vyas Fourier Series 2
  • 93. Example ∴ Fourier cosine series of f (x) is N. B. Vyas Fourier Series 2
  • 94. Example ∴ Fourier cosine series of f (x) is ∞ a0 nπx f (x) = + an cos 2 n=1 l N. B. Vyas Fourier Series 2
  • 95. Example ∴ Fourier cosine series of f (x) is ∞ a0 nπx f (x) = + an cos 2 n=1 l 2 = +0 2 N. B. Vyas Fourier Series 2
  • 96. Example ∴ Fourier cosine series of f (x) is ∞ a0 nπx f (x) = + an cos 2 n=1 l 2 = +0 =1 2 N. B. Vyas Fourier Series 2