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Math Help - 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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    <br />
\sum\limits_{n \ne 0} {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}<br />
{{in}}}  = \sum\limits_{ - \infty }^{ - 1} {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}<br />
{{in}} + } \sum\limits_1^\infty  {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}<br />
{{in}}}  =


    <br />
 =  - \sum\limits_1^\infty  {\frac{{\left( { - 1} \right)^{n + 1} e^{ - inx} }}<br />
{{in}} + } \sum\limits_1^\infty  {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}<br />
{{in}}}  = \sum\limits_1^\infty  {\frac{{\left( { - 1} \right)^{n + 1} \left( {e^{inx}  - e^{ - inx} } \right)}}<br />
{{in}}}

    regarding the second exercise:
    <br />
e^{ - i2\pi n} equals one for any integral n:

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