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Math Help - Recurrence relation

  1. #1
    Senior Member Dinkydoe's Avatar
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    Recurrence relation

    Hee, I got a really simple recurrence relation:

    \lambda_n=2-\frac{1}{\lambda_{n-1}} with \lambda_1=2.

    Calculating a few values we easily notice \lambda_n = \frac{n+1}{n},

    but how is this formally derived?

    Can someone offer a quick insight?
    Last edited by Dinkydoe; October 15th 2011 at 11:40 AM. Reason: Moo pointed out typing error
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Re: Recurrence relation

    Hello,

    I guess you're a little wrong with your few values, for example, I get \lambda_3=-1
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Re: Recurrence relation

    Yes, that's quite annoying...

    It shouldve been  \lambda_n=2-\frac{1}{\lambda_{n-1}}
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  4. #4
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    Re: Recurrence relation

    Quote Originally Posted by Dinkydoe View Post
    \lambda_n = \frac{n+1}{n}
    Prove this by induction on n.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: Recurrence relation

    Quote Originally Posted by Dinkydoe View Post
    Hee, I got a really simple recurrence relation:

    \lambda_n=2-\frac{1}{\lambda_{n-1}} with \lambda_1=2.

    Calculating a few values we easily notice \lambda_n = \frac{n+1}{n},

    but how is this formally derived?

    Can someone offer a quick insight?
    The difference equation...

    \lambda_{n+1}= 2-\frac{1}{\lambda_{n}}\ ,\ \lambda_{0}=2 (1)

    ... is non-linear and in most cases like that an ad hoc solving procedure has to be found. In this particular case it is easy to see that the solution is a continued fraction...

    \lambda_{n}= a_{0} - \frac{1}{a_{1}-\frac{1}{a_{2}-...-\frac{1}{a_{n}}}} (2)

    ... where a_{0}=a_{1}= ... = a_{n}=2. Now if You use the standard algorithm to write the rational number \frac{n+1}{n} in term of continued fraction You obtain exactly the expression (2) so that is...

    \lambda_{n}=\frac{n+1}{n} (3)

    Kind regards

    \chi \sigma
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