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Fourier Series

  1. Oct 31, 2005 #1
    Find the Fourier Series in terms of [itex] \phi_{n} = \sin{nx} [/itex] of the step function

    f(x) = 0 for [itex] 0 \leq x \leq \frac{1}{2} \pi [/itex]
    = 1 for [itex] \frac{1}{2} \pi < x \leq \pi [/itex]

    Fourier Series is [itex] \Sigma c_{n} \phi_{n} [/itex]
    and for hte interval for x between 0 and 1/2 pi
    [tex] c_{n} = \frac{\int_{0}^{\frac{1}{2}\pi} 0 dx}{\int_{0}^{\frac{1}{2}\pi} (\sin{nx})^2 dx [/tex]
    because rho is assumed to be 1
    the numerator is a constant and assumed to be 1
    the denominator
    [tex] \int_{0}^{\frac{1}{2}\pi} (\sin{nx})^2 dx = \int_{0}^{\frac{1}{2} \pi} (\frac{e^{ix}-e^{-ix}}{2i})^2 = \frac{-1}{4} [\frac{e^{2ixn}}{2in}- \frac{e^{-2inx}}{2in} -2x]_{0}^{\frac{1}{2} \pi} = \frac{\sin{n \pi} - \pi}{-4} = \frac{\pi}{4}[/tex]

    P.s. Not done typing this yet but can you tell me if im going in the right direction?
    Last edited: Oct 31, 2005
  2. jcsd
  3. Oct 31, 2005 #2
    i don't think it's that complicated. just use integration stuff you learned in 1st-year calculus to deal with that sine integral. that's all you need to know. i think there's some identity that will change that sin^2 into cos something. i can't remember it off the top of my head but i think that how i learned to reduce an even power of a trig function.

    & what is rho? i don';t think that appears in anything you've written there.
  4. Oct 31, 2005 #3
    rho is supposed to be the weight

    what is the integral of (Sin nx)^2??
    can anyone help!
  5. Oct 31, 2005 #4
    aha I found the identity in my 2nd edition of edwards/penney's calculus text, in the section on trig integrals:
    [tex]sin^{2}n\theta = \frac{1 - cos(2n\theta)}{2}[/tex]
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