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Thread: questions on fourier series. answers are provided. but I dont understand the process

  1. #1
    Junior Member
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    questions on fourier series. answers are provided. but I dont understand the process

    In the first picture, there are two circles i've indicated. My question is how to get the second one from the first circle. I can see that n changed from n≠ 0 to n=1. And I also don't understand why I have to change these n values.
    Moreover, in the second picture, I wasn't sure why that circled one has disappeared in the next equation.

    Thanks in advance!
    Attached Thumbnails Attached Thumbnails questions on fourier series. answers are provided. but I dont understand the process-maths1.jpg   questions on fourier series. answers are provided. but I dont understand the process-maths2.jpg  
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    \sum\limits_{n \ne 0} {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
    {{in}}} = \sum\limits_{ - \infty }^{ - 1} {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
    {{in}} + } \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
    {{in}}} = $


    $\displaystyle
    = - \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} e^{ - inx} }}
    {{in}} + } \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
    {{in}}} = \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} \left( {e^{inx} - e^{ - inx} } \right)}}
    {{in}}} $

    regarding the second exercise:
    $\displaystyle
    e^{ - i2\pi n}$ equals one for any integral n:

    $\displaystyle
    \int\limits_0^1 {\frac{{e^{ - i2\pi nx} }}
    {{i2\pi n}}} = \left. {\frac{{e^{ - i2\pi nx} }}
    {{4\pi ^2 n^2 }}} \right|_0^1 = \frac{{e^{ - i2\pi n} - 1}}
    {{4\pi ^2 n^2 }} = 0$
    Last edited by Peritus; Nov 2nd 2008 at 07:15 AM.
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