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Real analysis, subsequences

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that for any L[tex]\in[/tex][-1,1] there exists a subsequence of cos(n) such that that subsequence converges to L

    3. The attempt at a solution
    I have no idea.
    I suppose that the ultimate goal would be to find a subsequence nk so that nk converges to x, where x = cos-1(L) + 2[tex]\pi[/tex]k
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 9, 2008 #2

    Mark44

    Staff: Mentor

    Think about the graph of y = cos(x) and the parts of this graph that lie between the horizontal lines
    [tex]y = L + \epsilon [/tex]
    and
    [tex]y = L - \epsilon [/tex].
    Some of the values of cos(n) will lie in this band.
     
  4. Nov 9, 2008 #3
    how can you guarantee that there is a value of n in that band? n is a natural number so i'm not sure how you can say that it is in there.
     
  5. Nov 9, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The easiest way is to show that n mod 2*pi is dense in [0,2pi]. Then use that cos is continuous. To show the density use that pi is irrational. If r is irrational then n*r mod 1 is dense in [0,1]. I KNOW this is true. For some reason the proof doesn't stick in my head. I've looked it up more than once. Do I have to do it again, or can you figure it out? I know it involves the pigeonhole principle.
     
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