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Limit problem

  1. Jun 16, 2009 #1
    Evaluate the following limit:
    [tex]
    \lim_{n\rightarrow \infty}\frac{1+\cos(\frac{x}{n})+...+\cos(\frac{n-1}{n}x)}{n}
    [/tex]
     
  2. jcsd
  3. Jun 17, 2009 #2

    tiny-tim

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    Homework Helper

    Hi sit.think.solve! :smile:

    Hint: learn your trigonometric identities …

    what is cosA + cosB? :wink:
     
  4. Jun 17, 2009 #3
    You can write the sum on closed form if you think about cosine as the real part of the complex exponential.
     
  5. Jun 17, 2009 #4
    There might be a problem here. The sum is discrete while the limit is continuous. Is there any practical application here or is this just math?
     
  6. Jun 17, 2009 #5
    Thanks for the tips,

    having some trouble trying to reduce the sums though:

    Presumably I should add,

    [tex]
    \cos(\frac{0}{n}x)+\cos(\frac{n-1}{n}x)
    [/tex]
    [tex]
    \cos(\frac{1}{n}x)+\cos(\frac{n-2}{n}x)
    [/tex]

    and so on, but then what would be the last term in such sum? In fact, can I do sums like this?
     
  7. Jun 18, 2009 #6
    Like this
    [tex]
    \sum_{k=0}^{n-1} \cos{\frac{kx}{n}}=\Re{\left(\sum_{k=0}^{n-1} \exp{i\frac{kx}{n}}\right)}
    [/tex]

    which is a geometric series. I don't think it converges to anything particular as n->infinity, but it is bounded when x is not equal to zero...
     
    Last edited: Jun 18, 2009
  8. Jun 18, 2009 #7

    tiny-tim

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    Hi sit.think.solve! :smile:
    Yes, of course you can …

    and you should get a common factor which you can then put outside a bracket.

    Alternatively, use daudaudaudau's :smile: method, and remember that each term inside the ∑ is (eix/n)k :wink:
     
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