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A recurrence relation

  1. Aug 8, 2011 #1
    Hi all

    Suppose that [itex]a_1=\sqrt5[/itex], [itex]a_{n+1}=a_n^2-2[/itex] and [itex]g_n=\frac{a_1a_2...a_n}{a_{n+1}}[/itex].
    Evaluate [itex]\lim_{n\rightarrow \infty } g_n[/itex].

    I have seen some information in http://oeis.org/search?q=3,7,47,2207&sort=&language=english&go=Search". Besides, the sequence gn seems as a good rational approximation for [itex]\sqrt5[/itex]. I know that the answer is 1, But I can't find any nice solution. Any hint is strongly appreciated.
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Aug 8, 2011 #2


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    I have yet to study recurrence relations for myself, but I played around with it for a bit and it seems like it was headed in the right direction.






    Hence we can express gn in terms of gn-1, and the factor multiplying gn-1 seems to say a lot about what gn is as n gets very large.
  4. Aug 8, 2011 #3
    Thanks. I can't see what does it say about gn. Can you help me on this?

    I found out how to show that:
    [tex]a_n=\frac{\varphi ^{2^n}-\varphi ^{-2^n}}{\varphi ^{2^{n-1}}-\varphi ^{-2^{n-1}}}[/tex]
    And then the rest is easy. But I'm still looking for a more elegant solution.
  5. Aug 8, 2011 #4


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    Hmm, after looking at it a bit more I realize it's really not telling us as much as I originally thought.

    After solving for gn in

    [tex]g_n=g_{n-1}+\frac{2}{a^2_n-2}g_n[/tex] to obtain


    I was simply arguing that the factor approaches 1 as n gets large, but this doesn't really tell us about the limit of gn...
  6. Aug 8, 2011 #5


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    hi asmani! :smile:
    reverse-engineering that expression, we get

    an = sinh(C2n)/sinh(C2n-1) = 2cosh(C2n-1)​

    sooo … try defining an = 2coshbn :wink:
  7. Aug 8, 2011 #6
    Thanks. I think it's better not to simplify the fraction to get:
    [tex]g_n=a_1a_2\cdots a_n\frac{1}{a_{n+1}}=\frac{\sinh(C2^1) }{\sinh(C2^0)}\frac{\sinh(C2^2) }{\sinh(C2^1)}\cdots \frac{\sinh(C2^n) }{\sinh(C2^{n-1})}\frac{\sinh(C2^n) }{\sinh(C2^{n+1})}[/tex]
    [tex]=\frac{\sinh^2(C2^n) }{\sinh(C2^0)\sinh(C2^{n+1})}[/tex]
    Where C=Log(φ).
    And now calculating the limit is easy.
  8. Aug 8, 2011 #7


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    Sorry about wasting your time asmani, I should've left it to the big guys :biggrin:
  9. Aug 9, 2011 #8
    No, I appreciate your participation in this thread. :smile:
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