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Fourier cosine series of cos(x) from x=0 to Pi

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier cosine series for f(x) = cos x, 0 < x < Pi

    EDIT: I believe we are talking about a half-range extension..


    2. Relevant equations
    Fourier cosine series:
    f(x) = a0/2 + Sum(n = 1 to infinity) (an * cos (nx))

    where a0 = 2/L * integral (x = 0, x = L) (f(x) dx)

    and an = 2/L * integral (x = 0, x = L) (f(x) * cos (n*Pi*x/L) dx)

    Sorry if thats hard to understand...


    3. The attempt at a solution

    I graphed the function and extended the period to - Pi, i.e. -Pi < x < Pi

    I found a0 = 2/Pi * integral (x = 0, x = L) (cos x dx) = 0
    an = 2/Pi * integral (x = 0, x = L) (cos x * cos (nx) dx)
    and evaluated the integral using IBP...

    an = 2/Pi * (ncos(x)*sin (nx) - sin(x)cos(nx))*1/(n2-1)| x = 0, x = Pi
    which gives me an = 0!

    So the Fourier cosine series would be:
    f(x) = 0

    I would really appreciate any clarification as getting 0 for the series doesnt sound right..

    Thank you in advance
     
    Last edited: May 17, 2009
  2. jcsd
  3. May 17, 2009 #2

    tiny-tim

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    Hi alias99! :smile:
    Nooo … no need for IBP here (and it doesn't help anyway :redface:) …

    learn your https://www.physicsforums.com/library.php?do=view_item&itemid=18"

    cosx cos(nx) = … ? :smile:
     
    Last edited by a moderator: Apr 24, 2017
  4. May 17, 2009 #3
    Hi tiny-tim! :smile: (reminds me of the Nudie juices..hmm)
    Thanks for your response ;)

    You're right, I do need to learn my identities :tongue:

    I can now evaluate the integral much easier (Lol... funny thing is I did manage to get a result by IBP.. it looked like it would have continued indefinetely but I got the original integral on the RHS, took it over to the LHS and made it the subject)

    so now..

    cosx cos(nx) = 1/2 (cos (x-nx) + cos (x+nx)) :smile: which can be integrated fairly easily

    I still get 0! hehe.. I guess it is 0 after all. probably has something to do with cos(x) having an area under the curve of 0 between x = -Pi and x = Pi (or even x = 0 to Pi)

    I'm probably going to state that the Fourier cosine series of cos (x) with a period of Pi is cos(x) .. or is it 0.. lol :tongue:

    I'd appreciate just a word of confirmation, if you dont mind :)
     
  5. May 18, 2009 #4

    tiny-tim

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    Hi alias99! :smile:
    Well, it seemed a strange question …
    … if f is 2π-periodic, then yes obviously the only term in its Fourier cosine series is the first one, cosx.
     
  6. May 18, 2009 #5
    Great! Thanks alot tiny-tim :biggrin:
     
  7. Jan 15, 2011 #6
    I also just tried to do this and it looks like this is pretty trivial if you try to get the exponential Fourier series instead? Because cosx can be written as e^jwt+e^-jwt
     
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