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Math Help - quick fourier series question

  1. #1
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    quick fourier series question

    (2/(n^2))(cos(nPi)-cos(-nPi))

    can be written as (2/(n^2))*2 for Even n
    and (2/(n^2))*(-2) for n ODD

    so this means it can be written (4/(n^2))(-1)^n


    How? I cant get my head around why. if i sub in an even n and odd n say 2 and 1 i do not get 2 or -2! help
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by sheep99 View Post
    (2/(n^2))(cos(nPi)-cos(-nPi))

    can be written as (2/(n^2))*2 for Even n
    and (2/(n^2))*(-2) for n ODD

    so this means it can be written (4/(n^2))(-1)^n


    How? I cant get my head around why. if i sub in an even n and odd n say 2 and 1 i do not get 2 or -2! help
    Are you sure that it's \cos(n\pi)-\cos(-n\pi) and not \cos(n\pi)+\cos(-n\pi)?

    Because \cos(n\pi)-\cos(-n\pi) = 0 for all n because \cos(n\pi)=\cos(-n\pi).
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  3. #3
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    its how its written in my mark scheme, your totally right when i use + works out beautifully, the lecturer has just wasted about 10mins of my life. thanks!
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