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Fourier series

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier series defined in the interval (-π,π) and sketch its sum over several periods.
    i) f(x) = 0 (-π < x < 1/2π) f(x) = 1 (1/2π < x < π)

    2. Relevant equations

    ao/2 + ∑(ancos(nx) + bnsin(nx))
    a0= 1/π∫f(x)dx
    an = 1/π ∫f(x)cos(nx) dx
    bn = 1/π ∫f(x) sin(nx)

    3. The attempt at a solution

    I've worked out the a0 by splitting the limits and integrating individually for -π < x < 1/2π and 1/2π < x < π. When I did this i got a0 = ½

    for an = 1/π ∫(upper limit π, lower limit 1/2π) cos(nx) dx
    = 1/π[1/n sin(nx)]
    = 1/π ((1/n. sin(πn) - 1/n.sin(πn/2))
    = 1/nπ(0- sin(πn/2)

    Here is where i get stuck as sin(πn/2) is 0 for even values of n and alternates between 1, -1 for odd values.
    Can i leave this written in sin form of the Fourier series as every other example i've changed the value of sin/cos to either 0 or (-1)n.

    Thanks for any help!
     
  2. jcsd
  3. Dec 9, 2015 #2
    Hi struggles:

    You have already noticed that the integrals for the coefficients can be limited to {π/2,π}. That's a good start.

    I think it might be helpful to next separate the function f(x) into two parts:
    f(x) = S(x) + A(x)
    where S(x) is symmetric, S(x) = S(-x)
    and
    A(x) is anti-symmertric, A(x) = -A(-x).​
    Then think about the coefficients for A(x) and S(x) separately.

    You may also want to think about the general form of the integrals and try a substitution
    y = nx, dx = dy/n,​
    including the lower and upper bounds of the integral.

    Hope this helps.

    Regards,
    Buzz
     
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