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Math Help - Convergent in Continued fraction

  1. #1
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    Convergent in Continued fraction

    If \frac{p_k}{q_k} is the kth convergent of [a_0;a_1, ... ,a_n] and a_0 >0, show

    \frac{q_k}{q_{k-1}} = [a_k:a_{k-1}, ... , a_2, a_1].

    I think i have to use recurrence relation of the continued fraction: q_k=a_kq_{k-1}+q_{k-2} but I don't know how to.

    Can anyone help me on the question. Thank you.
    Last edited by kleenex; May 7th 2008 at 09:31 PM. Reason: Input wrong recurrence relation formula
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  2. #2
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    Quote Originally Posted by kleenex View Post
    If \frac{p_k}{q_k} is the kth convergent of [a_0;a_1, ... ,a_n] and a_0 >0, show

    \frac{q_k}{q_{k-1}} = [a_k:a_{k-1}, ... , a_2, a_1].

    I think i have to use recurrence relation of the continued fraction: q_k=a_kq_{k-1}+q_{k-2} but I don't know how to.

    Can anyone help me on the question. Thank you.
    You do this by induction. Note that, \frac{q_k}{q_{k-1}} = \frac{a_kq_{k-1}+q_{k-2}}{q_{k-1}} = a_k + \frac{q_{k-2}}{q_{k-1}}.
    But \frac{q_{k-1}}{q_{k-2}} = [a_{k-1};a_{k-2},...,a_1] thus, \frac{q_{k-2}}{q_{k-2}} = [0;a_{k-1},a_{k-2},...,a_1]. And so a_k + \frac{q_{k-2}}{q_{k-1}} = a_k + [0;a_{k-1},...,a_1] = [a_k;a_{k-1},...,a_1].
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