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Fourier series

  1. Apr 25, 2004 #1
    Hi there!!

    Can anyone please help me with this one..?

    Find the Fourier Sine series of f(x)=sin(x/2) for interval (0,pi)

    thanx a lot :)
     
  2. jcsd
  3. Apr 25, 2004 #2
    well, sin(x/2) is already a periodical function with period pi, so even if you construct a furier series of this one you would get the same: sin(x/2)
     
  4. Apr 26, 2004 #3
    get the standard Fourier series for sin(X) and replace X=x/2, and thats it!
     
  5. Apr 26, 2004 #4
    Wouldn't it have a period of 4pi?
     
  6. Apr 26, 2004 #5
    Well hmm...not really :rolleyes:

    I did get almoust correct answer BUT...
    you see b_n is zero but a_n is not...and this is where i stuck ...i cant get correct a_n
    by the way...can you tell me is this true?

    cos(npi) + cos(0pi) = (-1)^(n+1)

    thanx again :smile:
     
  7. Apr 26, 2004 #6
    (Assuming n is an integer). It's not true. cos(npi) is either 1 or -1, and cos(0pi) = cos(0) = 1, so the LHS is either 1 + 1 = 2 or -1 + 1 = 0, while the RHS is either 1 or -1... So you can never get equality.
     
  8. Apr 26, 2004 #7
    aha

    Yes i just got it..
    it is when you use integr. in series
    so then it would be ((-1)^n +1)

    thanx

    :smile:
     
  9. Apr 26, 2004 #8
    couldn't you just use the regular Taylor series?
     
  10. Apr 27, 2004 #9
    hm

    Not really :frown:
    the task...as so on my exam..they strictly want it with fourier ...
    bcs we do also taylor..and in task they mention which one they want..
     
  11. Apr 27, 2004 #10

    matt grime

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    THe answers would be different for a start as well.
     
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