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Sequence Bound

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to show that series cos(n) for n=0,1,...,inf is bounded.


    2. Relevant equations



    3. The attempt at a solution

    I understand that each of the terms of the sequence is bounded by 1. However, I cannot show the bound for the entire sequence? Could someone give some hints?
     
    Last edited: Mar 16, 2012
  2. jcsd
  3. Mar 16, 2012 #2

    Dick

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    So what's your bound for the individual terms?
     
  4. Mar 16, 2012 #3
    Well ,obviously, -1 <= cos(n) <= 1, so |cos(n)| <=1
     
  5. Mar 16, 2012 #4

    Dick

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    That's a bound for the sequence as well, isn't it?
     
  6. Mar 16, 2012 #5
    How so? Thats not quite obvious to me.
     
  7. Mar 16, 2012 #6

    Dick

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    If you call the general term of the sequence [itex]a_n=\cos n[/itex] then [itex]|a_n| \le 1[/itex]. Isn't that what you mean by a sequence bound?
     
  8. Mar 16, 2012 #7
    Actually, I meant to say series, not sequence. I need to show bound of the series. My apologies.

    Thus, for some n=N, the partial sum is bounded by N. I am struggling to show that the entire sum is bounded.
     
  9. Mar 16, 2012 #8

    Dick

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    That's different! Try using [itex]e^{i n}=\cos n+i \sin n[/itex].
     
  10. Mar 16, 2012 #9
    Do I need to express cos(n) as e[itex]^{in}[/itex] and then use geometric series?
     
  11. Mar 16, 2012 #10

    Dick

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    Find a partial sum of the geometric series [itex]e^{in}[/itex]. Then to find the partial sum of cos(n) just find the real part of that.
     
  12. Mar 16, 2012 #11
    I undestand. But isn't it easier to just find the total sum from 1 to infinity of the geometric series and essentially say that both cos(n) and sin(n) converge to the obtained number's real and imaginary parts respectively?
     
  13. Mar 16, 2012 #12

    Dick

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    The series doesn't converge. But it is bounded.
     
  14. Mar 16, 2012 #13
    But if you can obtain a finite number for geometric series of e[itex]^{in}[/itex], doesnt it mean that its real part is finite and cos(n) thus converges to a finite number? I think i am misunderstanding something.
     
  15. Mar 16, 2012 #14

    Dick

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    The geometric series formula gives you (1-r^(n+1))/(1-r) for a partial sum. The r^(n+1) part doesn't converge. You have to try to bound the partial sum.
     
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