A recurrence relation

  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 this link. 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.
     
  2. jcsd
  3. Mentallic

    Mentallic 3,701
    Homework Helper

    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.

    [tex]g_n=\frac{a_1a_2...a_n}{a_n^2-2}[/tex]

    [tex]=\frac{a_n^2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)-2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}[/tex]

    [tex]=\frac{\left(a_n^2-2\right)\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)+2\left(\frac{a_1a_2...a_{n-1}}{a_n}\right)}{a_n^2-2}[/tex]

    [tex]=\frac{a_1a_2...a_{n-1}}{a_n}+2\frac{a_1a_2...a_{n-1}}{a_na_{n+1}}[/tex]

    [tex]=g_{n-1}+\frac{2}{a_n^2}g_n[/tex]

    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. 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. Mentallic

    Mentallic 3,701
    Homework Helper

    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

    [tex]g_n=\frac{a_n^2}{a_n^2-2}g_{n-1}[/tex]

    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. tiny-tim

    tiny-tim 26,053
    Science Advisor
    Homework Helper

    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. 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. Mentallic

    Mentallic 3,701
    Homework Helper

    Sorry about wasting your time asmani, I should've left it to the big guys :biggrin:
     
  9. No, I appreciate your participation in this thread. :smile:
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?