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Fourier Series of n*(pi/2)

  1. Feb 28, 2012 #1
    ok i know that

    sin n*(pi/2)
    = 1 if n=1,5,9,13...
    = -1 if n=3,7,11,15...
    = 0 if n is even

    cos n*(pi/2)
    = 0 if n is odd
    = -1 if n=0,4,8,12
    = 1 if n=2,6,10,14...

    is there a simpler way of expressing this?
    for example simple way to express cos(n*pi)=cos(-n*pi)=(-1)^n

    is there a similar way to express cos n*(pi/2) and sin n*(pi/2)
    thanks
     
  2. jcsd
  3. Feb 28, 2012 #2

    tiny-tim

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    Hi Deathfish! :smile:
    standard trick …

    sin((2n + 1)π/2) = (-1)n

    and i suppose

    cos((2n)π/2) = (-1)n :wink:
     
  4. Feb 28, 2012 #3
    and how do you use the term in Fourier series?

    lets say you encounter the term .. sin n*(pi/2)

    just replace with sin((2n + 1)π/2) ? don't know how to use this expression properly

    any simple example will be helpful.
     
  5. Feb 28, 2012 #4

    tiny-tim

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    i'm not seeing what the difficulty is :confused:

    you just replace the sin, or cos, with (-1)something :smile:
     
  6. Feb 28, 2012 #5
    ok what is the (something)
     
  7. Feb 28, 2012 #6

    tiny-tim

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    sin((2n + 1)π/2) = (-1)n



    cos((2n)π/2) = (-1)n :wink:
     
  8. Feb 28, 2012 #7
    ok is it because the 'n' is arbitrary you can just replace sin n*(pi/2) with sin((2n + 1)π/2) ?
     
  9. Feb 28, 2012 #8

    tiny-tim

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    abritrary and odd :wink:

    yes :smile:

    (though of course, it's a different n …

    any odd n is 2m + 1, then we rename m as n :wink:)​
     
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