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Math Help - complex foorier question

  1. #1
    MHF Contributor
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    complex foorier question

    a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt\\
    \frac{e^{in\pi}}{2\pi}\int_{-\pi}^{\pi}\overline{f(t)}\overline{e^{int}}dt\\
    =\frac{e^{in\pi}}{2\pi}\int_{\pi}^{-\pi}e^{-i(-n)t}=e^{in\pi}\overline{a_{-n}}
    why there is \overline{a_{-n}}??
    by the first formula it supposed to be only a_{-n}


    i understand how they removed the conjugat of exponenet
    but i cant see how they removed it from the f part
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  2. #2
    Senior Member
    Joined
    Mar 2009
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    The formula you gave for a_n was
     a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} \! f(t)e^{-int} \, dt, but they tell you what the value of
    \frac{e^{in\pi}}{2\pi}\int_{-\pi}^{\pi} \! \overline{f(t)}\overline{e^{int}} \,dt. Now you have to convert the original formula of a_n to look like the integral you know the value of, i.e.,
    a_{-n} = \frac{1}{2\pi}\int_{-\pi}^{\pi} \! f(t)e^{-i(-n)t} \, dt
    \overline{a_{-n}} = \frac{1}{2\pi}\int_{-\pi}^{\pi} \! \overline{f(t)e^{int}} \, dt
    e^{in\pi}\overline{a_{-n}} = \frac{e^{in\pi}}{2\pi}\int_{-\pi}^{\pi} \! \overline{f(t)e^{int}} \, dt.
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