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Fourier Series of Secant

  1. Feb 15, 2013 #1
    So I know that sec(x) has period 2Pi, and it's even so I don't need to figure out coefficients for bn.

    Let's take the limits of the integral to go from -3/2 Pi to 1/2 Pi. How do I integrate sec(x) sin(nx) dx?! Am I on the right path?

    PS: I know that this doesn't satisfy the Dirichlet Theorem, but the textbook I'm using still says to compute it.
     
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  3. Feb 16, 2013 #2

    haruspex

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    Seems to me that integral is zero for n even and indeterminate for n odd.
     
  4. Feb 16, 2013 #3

    rbj

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    well since [itex]\sec(x) = 1/\cos(x) [/itex] then i think you can express the cosine functions, both on the top and on the bottom as exponentials with imaginary argument. i think you can get to an integral that looks like

    [tex] c_n = \frac{1}{2 \pi} \int_{-\pi}^{+\pi} \frac{2}{e^{i (n+1) x}+e^{i (n-1) x}} \ dx [/tex]

    i think you can find an antiderivative of that. might still come out as indeterminate. but that's the integral you have to solve.
     
    Last edited: Feb 16, 2013
  5. Feb 16, 2013 #4

    vela

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    I evaluated the first few odd n integrals using Mathematica, and it looks like ##a_{2n+1} = (-1)^n 2\pi##.
     
  6. Feb 16, 2013 #5

    haruspex

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    The reason I thought it indeterminate is that integrating over a 2π range effectively cancels a +∞ with a -∞. E.g. with n=1, it's the integral of tan, which is log cos. An interval that spans π/2 will appear to give a sensible answer, but in reality it's undefined.
     
  7. Feb 16, 2013 #6

    vela

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    mathskier indicated the wrong integrand. It should be sec x cos(nx), which is the integrand I used.
     
  8. Feb 16, 2013 #7

    haruspex

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    That certainly makes a difference. It's not hard to show that the integral of sec x cos(nx) over a 2pi range is the negative of that of sec x cos((n-2)x), for n > 2.
     
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