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Fourier series 2

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-06-14at42751AM.png


    3. The attempt at a solution

    I don't see how if n = 0 then the answer is 1/2. By my reckoning

    1/pi * sin nx/n =

    1/pi * 0 = 0

    I also don't see where the 2 comes from when sin(npi/2) first shows up.
     
    Last edited: Jun 14, 2012
  2. jcsd
  3. Jun 14, 2012 #2
    If n=0, you should calculate the integral differently. Also lim x->0 sin(x)/x is not 0, so you get the correct result even if you just do take the limit.

    The upper limit substitution is always zero. You get pi/2 from the lower limit.

    It comes from the fact that then they are equal... Work it out. Plug in values n=1,3,5,... and see how sine behaves, then see what the 2 does in the denominator.
     
  4. Jun 14, 2012 #3
    I still don't get this.

    Also lim x->0 sin(x)/x is not 0 - That's correct, but 1/pi * sin 0/0 = 1/pi * 1/0 = undefined.

    Thanks, I understand now.

    I was able to get this before you answered.
     
  5. Jun 14, 2012 #4
    No it's not.
     
  6. Jun 14, 2012 #5
    Do the n's cancel? If not, then 1/pi * sin nx/n must be 1/pi * 1/0

    If yes, then 1/pi * sin pi/2 = 1/pi
     
  7. Jun 14, 2012 #6
    Look at the graph of sinx/x. It is a continuous and differentiable function at x=0.

    http://www.wolframalpha.com/input/?i=graph++sinx/x

    n's don't cancel. In integrals, the points that you are evaluating is the limit of the function at those values.
     
  8. Jun 14, 2012 #7
    <deleted>
     
  9. Jun 14, 2012 #8
    even if sin x/x is continuous and differentiable I still don't see how they got 1/pi * pi/2
     
  10. Jun 14, 2012 #9
    You are trying to find,

    [tex]\frac{1}{\pi }\int_{\pi/2}^{\pi} cos(nx) dx[/tex]

    At n=0,

    [tex]\frac{1}{\pi }\int_{\pi/2}^{\pi} cos(0)dx = \frac{1}{\pi }\int_{\pi/2}^{\pi}1\cdot dx[/tex]

    What do you get if you integrate this??
     
  11. Jun 14, 2012 #10
    For [itex]n=0[/itex] you have

    [itex]\dfrac{1}{\pi}\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi} \cos (0 \cdot x) dx = \dfrac{1}{\pi}\displaystyle\int\limits_{\frac{\pi}{2}}^{\pi} 1 dx = \dfrac{1}{\pi} \cdot \dfrac{\pi}{2}[/itex]

    Edit: Oh, Infinitum was faster :) Sorry
     
  12. Jun 14, 2012 #11
    [tex] \frac{1}{\pi} \int_{\pi/2}^\pi \cos(0x) dx = \frac{1}{\pi} \int_{\pi/2}^\pi dx = \frac{1}{\pi}(\pi - \frac{\pi}{2}) [/tex]

    Alternatively, you can take the limit:
    [tex] \lim_{n\rightarrow 0} \frac{\sin(n \pi/2)}{n \pi} = \frac{1}{2}\lim_{n\rightarrow 0} \frac{\sin(n \pi/2)}{(n \pi/ 2)} = \frac{1}{2} [/tex]
     
  13. Jun 14, 2012 #12
    i understand now
     
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