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Help to find limit

  1. Nov 8, 2007 #1
    Please, help to find limit:
    [tex]\lim_{n \rightarrow \infty} na(n)[/tex], where
    [tex]a(1)=1;
    a(n+1)=\frac{a(n)}{1+\left|sin(a(n))\right|}[/tex]
    Thanks for any ideas!
     
  2. jcsd
  3. Nov 8, 2007 #2

    Avodyne

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    If the limit exists, then a(n) is going to zero at large n, which can be used to simplify the right-hand side.

    Also, at large n, a finite-difference equation can be well approximated by a differential equation.
     
  4. Nov 8, 2007 #3
    Thanks. With the help of your post, Avodyne, I found that limit equals 1. Is it correct?
    Sorry, but I'm not sure.
     
    Last edited: Nov 8, 2007
  5. Nov 8, 2007 #4

    Avodyne

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    Yes, this is correct!
     
  6. Nov 8, 2007 #5
    But actually, I didn't use differential equation :)
    [tex]\frac{1}{a_{n+1}}=\frac{1+\left|sin(a_n)\right|}{a_n}[/tex]
    It's easy to show that [tex]a_n[/tex] is going to zero at large n, but remains postive. So:
    [tex]\frac{1}{a_{n+1}}=\frac{1+a_n-\frac{1}{6}a_n^3+o(a_n^3)}{a_n} =\frac{1}{a_n}+1+b_n[/tex], where [tex]b_n \rightarrow 0[/tex], when [tex]n \rightarrow \infty[/tex].
    Use this we can obtain:
    [tex]\frac{1}{a_{n+1}}=\frac{1}{a_1}+n+b_1+b_2+...+b_n[/tex]
    [tex]\frac{1}{na_{n+1}}=\frac{1}{na_1}+1+\frac{b_1+b_2+...+b_n}{n}[/tex]
    When n is going to infinity, we have: (using well-known [tex]\lim_{n \rightarrow \infty}\frac{b_1+b_2+...+b_n}{n}=0[/tex], where each of b is going to zero with large n)
    [tex]\lim_{n \rightarrow \infty} \frac{1}{na_{n+1}}=0+1+0=1[/tex]

    In any case, thank you very much, Avodyne. :)
     
  7. Nov 8, 2007 #6

    Avodyne

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    Very nice. Your analysis is more rigorous than mine was.
     
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