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Another infinite series

  1. Aug 28, 2010 #1
    I am having a difficult time working with some of these infinite series. I studied them in calc 2, but that was a few years ago.

    Could someone help me figure out how to find what the following sum converges to:

    [tex]
    \sum_{n=-N}^N{cos(n \theta)}
    [/tex]

    Shouldn't there be some property or convergence test I could do to find out what it actually converges to?
     
    Last edited: Aug 28, 2010
  2. jcsd
  3. Aug 28, 2010 #2
    This expands to:

    [tex]
    cos(-N\theta)+ ... +cos(-4\theta)+ cos(-3\theta)+ cos(-2\theta) + cos(-\theta) + cos(0)
    + cos(\theta) + cos(2\theta) + cos(3\theta)+ cos(4\theta)+ ... + cos(N \theta)
    [/tex]


    And since [tex] cos(-\alpha) = cos(\alpha) [/tex] the above expansion boils down to:

    [tex]
    cos(N\theta)+ ... +cos(4\theta)+ cos(3\theta)+ cos(2\theta) + cos(\theta) + cos(0)
    + cos(\theta) + cos(2\theta) + cos(3\theta)+ cos(4\theta)+ ... + cos(N \theta)
    [/tex]

    Which can be simplified as:

    [tex]
    cos(0) + 2cos(\theta) + 2cos(2\theta) + 2cos(3\theta)+ 2cos(4\theta)+ ... + 2cos(N \theta)
    [/tex]
     
    Last edited: Aug 28, 2010
  4. Aug 28, 2010 #3
    So I am actually looking at the case where [tex] \theta = \frac{\pi}{2} [/tex]

    Therefore:

    [tex]
    2cos(0) = 2
    [/tex]

    [tex]
    2cos( \frac{\pi}{2} ) = 0
    [/tex]

    [tex]
    2cos(\pi) = -2
    [/tex]

    [tex]
    2cos( \frac{3\pi}{2} ) = 0
    [/tex]

    [tex]
    2cos( 2\pi ) = 2
    [/tex]

    [tex] . [/tex]

    [tex] . [/tex]

    [tex] . [/tex]

    [tex]
    2cos(N \frac{\pi}{2})
    [/tex]

    So the sequence would be:
    [tex]
    \{2, 0, -2, 0, 2, ... , 2cos(N \frac{\pi}{2}) \}
    [/tex]

    Is my logic correct?
    And if so, how do I sum the sequence to find what it converges to?
     
    Last edited: Aug 28, 2010
  5. Aug 28, 2010 #4

    mathman

    User Avatar
    Science Advisor

    Your logic is correct. The conclusion is that it doesn't converge.
     
  6. Aug 29, 2010 #5
    Okay. Well this is part of a homework problem, so if an admin could move this to the homework section in "calculus and beyond" that would be great.

    Here is the whole problem:

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{|cos(n \frac{\pi}{4})|^2}
    [/tex]


    Here is how far I have evaluated it:

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{cos^2(n \frac{\pi}{4})}
    [/tex]

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{\frac{1+cos(2n \frac{\pi}{4})}{2}}
    [/tex]

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    (\sum_{n=-N}^N{\frac{1}{2}} +
    \sum_{n=-N}^N{\frac{1}{2}cos(n \frac{\pi}{2})}})
    [/tex]

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{\frac{1}{2}} +
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{\frac{1}{2}cos(n \frac{\pi}{2})}}
    [/tex]

    [tex]
    \frac{1}{2}
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}(2N+1)+
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    (\frac{1}{2})
    \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
    [/tex]

    [tex]
    \frac{1}{2}
    \lim_{N \rightarrow \infty}(1)+
    \frac{1}{2}
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
    [/tex]

    [tex]
    \frac{1}{2}+
    \frac{1}{2}
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
    [/tex]
     
    Last edited: Aug 29, 2010
  7. Aug 29, 2010 #6
    The final answer is
    [tex]
    \frac{1}{2}
    [/tex]

    So as long as I have done everything else correct so far this part should go to zero:

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
    [/tex]

    Is that correct?
    And if so, how does that go to zero?
     
  8. Aug 29, 2010 #7
    Maybe that is as far as I can simplify:

    [tex]
    \sum_{n=-N}^N{cos(n \frac{\pi}{2})}}
    [/tex]

    So I should evaluate the limit on the outside. That would make:
    [tex]

    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}=0
    [/tex]


    Therefore making the entire term:

    [tex]
    \lim_{N \rightarrow \infty}
    \frac{1}{2N+1}
    \sum_{n=-N}^N{cos(n \frac{\pi}{2})}} = 0
    [/tex]

    Is that correct logic?
     
  9. Aug 29, 2010 #8

    mathman

    User Avatar
    Science Advisor

    I suggest you go back to start.
    cos2(nπ/4) =1 for n=0, 1/2 for n=1, 0 for n=2, 1/2 for n=3, and cycles from then on. For each set of 4 terms, the sum is 1, so the limit of the original sum is 1/4.
     
  10. Aug 29, 2010 #9
    Could you please elaborate. I don't understand.
     
  11. Aug 30, 2010 #10

    mathman

    User Avatar
    Science Advisor

    I'm not sure how to elaborate, except to say that I made an error in arithmetic. The sum of four terms (1 + 1/2 +0 + 1/2) is 2 (not 1), so the limit = 1/2. This agrees with the result you got.
     
  12. Sep 2, 2010 #11
    Actually, you can find a close form for your summation expression. Please refer to http://www.voofie.com/concept/Mathematics/" [Broken] for details:

    http://www.voofie.com/content/156/how-to-sum-sinn-q-and-cosn-q/" [Broken]

    In short,

    [tex]\sum _{n=-N}^N \cos (n \theta )=\cos (N \theta )+\cot \left(\frac{\theta }{2}\right) \sin (N \theta )[/tex]

    With this close form, you can solve your limit problem easily.
     
    Last edited by a moderator: May 4, 2017
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