problem on fourier series

 

Find the Fourier series representing

                                                   f (x) = x, 0 < x < 2π

and sketch its graph from x = −4π to x = −4π

Solution:   Let f (x) =a₀/2+a₁ cos x + a₂ cos 2x +b₁ sin x +b₂ sin 2x .....

Hence   a₀ =(1/ π)  ^2π ∫₀ x dx =(1/ π)[x^2/2]₀^2π  = 2π

a_n = (1/ π) ^2π ∫₀ f (x) cos nx dx

  =(1/ π)[x (sin nx/n) -1 (- cos nx/n²) ]₀^2π    =(1/ π)[cos2n π/n²  - 1/n²]

    =(1/n² π) [1-1]=0

b_n= (1/ π) ^2π ∫₀ f (x) sin nx dx =(1/ π) ^2π ∫₀ x sin nx dx

=(1/ π)[x(- cos nx/n²)-1 (-sin nx/n²)]₀^2π     =(1/ π)[- 2πcos2nπ/n]= -2/n

 Substituting the values of a0 an, bn , in (1), we get

X=π-2[sin x +1/2 sin 2x +1/3  sin 3x +………]

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Example: Find the Fourier sine series for the function

                                            f (x) = e^ax     for −π ≤ x ≤ π

                      where a is constant.

 Ans:  ∫ e^ax sin bx dx = (e^ax/ a² +b²) [ a sin bx -bcos nx]

 b_n =∫ e^ax sin nx dx

 = (2/ π) [ (e^ax/ a² +n²) (a sin nx-ncos nx) ]₀^π    

   = (2/ π) [ (e^{a π}    / a² +n²) (a sin n π -ncos n π) + n/ a² +n²]

      = (2/ π) ( n/ a² +n²) [-(-1)ⁿ  e^{a π} +1]

= {n/ (a² +n²) π} [1-(-1)ⁿ e^{a π} ]

b₁ = 2(1 + e^{a π})/ (a² +1²) π            b₂ = 2.2(1 - e^{a π})/ (a² +2²) π    

e^ax = (2/ π)[{ (1 + e^{a π})/ (a² +1²) π} sin x + { 2(1 - e^{a π})/ (a² +2²) π} + ……….]

 

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Expand f(x) = e^x in a cosine series over (0, 1).

Ans:   f(x) = e^x    and c =1

        a₀ = (2/ c) ^c ∫₀ f (x) dx =(2/ 1) ¹ ∫₀  e^x dx =2(e-1)

        a_n = (2/ c) ^c ∫₀ f (x) cos(nπx/c) dx ==(2/ 1) ¹ ∫₀  e^x cos(nπx/1) dx

              = 2[ (e^x/ π²n² +1) (nπ sin nπx + cos nπx) ]₀¹   

              = 2[ (e^x/ π²n² +1) (nπ sin nπx + cos nπx) - 1/ (π²n² +1)]

               =2/ (π²n² +1) [ (-1)ⁿ e -1]

    f(x)= a₀/2 + a₁ cos πx + a₂ cos 2πx + a₃ cos 3πx +…………..

         e^x  = e  -1 + 2[ (-e-1/ π²+1) cos πx +(-e-1/4 π²+1) cos 2πx +(-e-1/ 9π²+1) cos 3πx + …...

By using the sine series for f(x) = 1 in 0 < x < π show that

      π²/8  =1 + 1/3² + 1/5² + 1/7²  + 1/9²+ ……

Ans : sine series is f(x) = Σ b_n sin nx  

b_n = (2/ π)  ^π ∫₀ f(x) sin nx dx

     = (2/ π)  ^π ∫₀ (1) sin nx dx =(2/ π)(- cos nx/n)₀^π

    = (-2/ nπ) [cos nπ -1]     = (-2/ nπ) [(-1)ⁿ -1]

  =     -4/nπ            if n is odd

0                              if n is even

       Then, the sine series is

                 1=(4/π )sin x  + (4/3π )sin 3x + (4/5π )sin 5x + (4/7π )sin 7x  +………

       ^c ∫₀ [f (x)]² dx = c/2[b₁² + b₂² + b₃² + b₄² + b₅² +….]

             ^c ∫₀ [1]² dx = π/2[(4/π )²+ (4/3π )²+ (4/5π )² + (4/7π )²+………]

            [x]₀^π = (π/2) (16/ π²)[1 + 1/3²+ 1/5² + 1/7²+………]

                  π = (π/2) (16/ π²)[1 + 1/3²+ 1/5² + 1/7²+………]

               π²/8 =1+ 1/3²+ 1/5² + 1/7²+……

 

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