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Math Help - Fourier Series- Concerning one of the coefficients.

  1. #1
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    Fourier Series- Concerning one of the coefficients.

    Hello,

    I'm working on the following functi(on

    f(x)= 0 x\epsilon[-\pi,0)
    f(x)=1 x\epsilon[0,\pi)

    I was able to figure out the mean value by myself (which equals 1).

    I got lost when it was time to figure out the cosine coefficient.
    I was able to narrow it down to

    \frac{1}{n\pi} (sin(n\pi)-sin(0))

    after differentiating, but when I read my notes, it says the answer to that is 0.
    I thought the answer would be

    \frac{sin(n\pi)}{n\pi}

    Can someone explain this to me?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by DarkMalice View Post
    Hello,

    I'm working on the following functi(on

    f(x)= 0 x\epsilon[-\pi,0)
    f(x)=1 x\epsilon[0,\pi)

    I was able to figure out the mean value by myself (which equals 1).

    I got lost when it was time to figure out the cosine coefficient.
    I was able to narrow it down to

    \frac{1}{n\pi} (sin(n\pi)-sin(0))

    after differentiating, but when I read my notes, it says the answer to that is 0.
    I thought the answer would be

    \frac{sin(n\pi)}{n\pi}

    Can someone explain this to me?
    The sine function has the property that it vanishes at multiples of \pi. In other words, \sin n\pi = 0

    Problem solved!
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  3. #3
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    Quote Originally Posted by Opalg View Post
    The sine function has the property that it vanishes at multiples of \pi. In other words, \sin n\pi = 0

    Problem solved!
    Well duh! Wow for some reason i competely forgot pi equals 180 degrees. Associated it with 90 degrees and that's why I was messing up. Thank you! :P
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