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Fourier coefficient

  1. May 14, 2005 #1
    I'm supposed to show
    [tex]
    \hat f\left( n \right) = - \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}
    [/tex]
    where [tex]\hat f\left( n \right)[/tex] is the Fourier coefficient and [tex]f(x)[/tex] is a [tex]2\pi[/tex]-periodic and Riemann integrable on [tex][\pi,-\pi][/tex].

    This is what I've done:
    I use the formulae for the Fourier coefficient of a [tex]2\pi[/tex]-periodic function
    [tex]
    \hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right)e^{ - inx} dx}
    [/tex]
    and a simple change of variable
    [tex]
    \hat f\left( n \right) = \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)e^{ - in\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)} dx} = \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)e^{ - inx} e^{ - i\pi } dx} = - \frac{1}{{2\pi }}\int\limits_{ - \pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}}^{\pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}
    [/tex].

    Everything seems to agree except for the limits of the integral.
    1) If I have done some mistake, I'd appreciate it someon would point it out.
    2) If I haven't done any mistakes, what's the reasoning behind this? Don't the limits matter as longs as they are [tex]2\pi[/tex] apart?
     
  2. jcsd
  3. May 14, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    You'll have to prove that

    [tex] \int_{-\pi+\frac{\pi}{n}}^{-\pi} f\left(x+\frac{\pi}{n}\right) \ dx =-\int_{\pi}^{\pi+\frac{\pi}{n}} f\left(x+\frac{\pi}{n}\right) \ dx [/tex]

    Daniel.
     
  4. May 14, 2005 #3
    Ok, here's my reasoning.
    The integrand

    [tex]
    f(x+\frac{\pi}{n})e^{-inx}
    [/tex]

    is [tex]2\pi[/tex]-periodic. Hence a change from [tex]x+\frac{\pi}{n}[/tex] to [tex]x+\frac{\pi}{n}+2\pi[/tex] leaves the integrand unchanged and I get

    [tex]
    \int\limits_{ - \pi }^{ - \pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx} = \int\limits_\pi ^{\pi + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} {f\left( {x + {\pi \mathord{\left/
    {\vphantom {\pi n}} \right.
    \kern-\nulldelimiterspace} n}} \right)e^{ - inx} dx}
    [/tex]

    which allows me to change the limits of the original integral to [tex]-\pi[/tex] and [tex]\pi[/tex] without changing its value.
    Am I right?
     
    Last edited: May 15, 2005
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