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Finding 2012th term

  1. Oct 26, 2013 #1
    Hello,
    This is not a homework, just trying to solve the previous exams and i'm stuck at one of them :
    1. The problem statement, all variables and given/known data


    Consider
    Fn(x)=1/(2*pi) * [ sin(n*x/2) / sin(x/2) ]
    Calculate:

    ∫ F2012(x) from -pi to pi


    2. Relevant equations



    3. The attempt at a solution
    I've tryed replacing the sine by the exponential expression, but it doesn't really help !
    Thanks
     
  2. jcsd
  3. Oct 27, 2013 #2

    Zondrina

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    If I'm reading correctly this is your integral:

    ##\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{sin(1006x)}{sin(\frac{x}{2})} dx##

    I think a Taylor expansion would help:

    $$\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ \sum_{n=0}^{∞} (-1)^n \frac{ (1006x)^{2n+1} }{ (2n+1)! } }{ \sum_{n=0}^{∞} (-1)^n \frac{ (\frac{x^{2n+1}}{2^{2n+1}}) }{ (2n+1)! }} dx$$

    I believe you can integrate that.
     
  4. Oct 27, 2013 #3
    I completly forgot the properties of Ʃ :blushing:
    Can we simplify by eliminatig the same terms in the numerator and denominator ?
     
  5. Oct 27, 2013 #4

    LCKurtz

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    Does ##\frac {1+3+5}{2+3+6} =\frac {1+5}{2+6}##?
     
  6. Oct 27, 2013 #5
    Oh, of course not.
    So, what are the steps to calculate the integral ?
    I don't know how calculate that
     
  7. Oct 27, 2013 #6

    haruspex

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    From the structure of the problem, I would try obtaining a recurrence relation between the integrals of successive Fn. Integration by parts should help.
     
  8. Oct 28, 2013 #7

    D H

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    I would try looking at the difference between successive *even* terms of Fn and using trigonometric relationships. The integration becomes trivial.

    Hint: ##\frac{\sin(nx) - \sin((n-1)x)} {\sin(x/2)}## can be expressed in a very simple form.
     
  9. Oct 28, 2013 #8
    Hello,

    F2N=sin²(nx)/sin²(x/2)
    F2N+2=sin²((n+1)x)/sin²(x/2)

    F2N-F2N+2=[sin²(nx)-sin²((n+1)x)]/sin²(x/2) = [ cos(2nx)-cos(2n+2)x)]/(cos(x)-1)

    But I don't get the idea, how is it supposed to help me calculate F2012 ?
     
  10. Oct 28, 2013 #9

    D H

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    No! Successive terms of FN means developing FN in terms of FN-1.
     
  11. Oct 28, 2013 #10

    Dick

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    Ah, I see what you are doing. But isn't it the difference between successive odd terms that comes out to be REALLY simple? The difference between even terms is somewhat simple, but you have to sum an awful lot of those simple things to get up to 2012.
     
  12. Oct 28, 2013 #11
    You said "even" terms I took two even successive terms
    Difference betweend successive odd terms gives :
    [cos((2n+1)x)-cos((2n+3)x)]/(cos(x)-1)
     
    Last edited: Oct 28, 2013
  13. Oct 28, 2013 #12

    D H

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    Since 2012 is even, that's the problem you have to address. Yes, you'll get a sum of 1006 terms. That sum can be expressed very simply.
     
  14. Oct 28, 2013 #13
    Apparently I didn't understand the exersise your way.
    I think Fn IS a function that depends of the n we choose. But are you saying that Fn is a serie and that F2012 is the sum from n=0 to n=2012 ?
     
  15. Oct 28, 2013 #14

    D H

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    Fn is exactly what you specified in the original post: ##F_n(x) = \frac 1 {2\pi} \frac {\sin(nx/2)}{ \sin(x/2) }##. That's nice and compact, but how are you going to integrate it? One answer to solve this is to express Fn a different way, for example, as a series, each element of which is easily integrated.
     
  16. Oct 28, 2013 #15
    I made a mistake in my initial post it is Fn(x)=1/(2*pi) * [ sin(n*x/2) / sin(x/2) ]²
    I just don't get it.
    I am supposed to calculate the integral of the difference of the two even successive terms for the first 1006 terms ?

    Ʃ [ cos(2nx)-cos(2n+2)x)]/(cos(x)-1) dx

    The integral from -pi to pi
    The sum starts at n=0 to n=1006
     
  17. Oct 28, 2013 #16

    D H

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    Grrrr. Do not do that!

    We have been helping you with Fn(x) as defined in the OP. How were we supposed to know you omitted that little superscript 2? This is a very different problem.
     
  18. Oct 28, 2013 #17

    Dick

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    Ok, while I'm discarding the work I did on the original problem since it's now changed, the only way to express that simply is using sigma notation. I was actually originally wondering if somebody didn't take this question from a test in 2011 to make a new test in 2012, and then decided to 'update' the question by changing '2011' to '2012', without realizing what a difference it would make.
     
  19. Oct 28, 2013 #18
    sigma notation ?
    Even with the " ² " we can write it simply as I did or is it false ?
    I was looking at my course notes
    Can we simplify it like Fejer's Kernel ?
    Kn(x)= 1/(n+1) * [ sin((n+1)*x/2) / sin(x/2) ]² = Ʃ (n+1-|r|)/(n+1) eirx sum from -n to n
     
    Last edited: Oct 28, 2013
  20. Oct 28, 2013 #19

    Dick

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    No, I should stop chirping in here. I was talking to DH and that's just confusing you. I was talking about the originally (wrongly!) stated problem. For this one you should be able to actually get a simple numerical answer for ##F_{2012}## without having any series notation at all. Keep working.
     
  21. Oct 28, 2013 #20

    D H

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    NO!

    We were helping you with ##\int_{-\pi}^{\pi} \sin(nx/2) / \sin(x/2)\, dx## because that is what you asked us to help you with in the original post. You never corrected your mistake in your original post until just recently.

    You do not know how annoying this is. Most of us who help with homework solve the problem before we post anything.

    You need to forget everything that has been said so far because we were solving that original problem. Your new problem is *very* different. This problem is *easy*.
     
    Last edited: Oct 28, 2013
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