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Math Help - Need help on continued farction

  1. #1
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    Need help on continued farction

    I'm trying to show the following argument but have a problem

    If \frac{p_k}{q_k} is the kth convergent of [a_0;a_1,...,a_n], show that for 0 \leq k\leq n, q_k \geq 2^{\frac {k-1}{2}}

    I use that fact that
    q_k = a_kq_{k-1}+q_{k-2} \geq 2q_{k-2} \geq 2, and
    2q_{k-2} = a_{k-2}q_{k-3}+q_{k-4} \geq 2^2q_{k-4} \geq 2^2 etc,
    we can see that q_k is decreasing in 2n steps.

    But why q_k \geq 2^{\frac {k-1}{2}} instead of 2^{\frac{k}{2}}?

    Can anyone explain it to me? Thank you.
    Last edited by kleenex; May 12th 2008 at 05:05 PM. Reason: fix formula
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  2. #2
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    This comes down to the inductive case q_1. If q_k\geq 2^{k/2} was true then q_1\geq 2^{1/2} = \sqrt{2}. But this is false because consider [1;1,...] then p_1/q_1 = 1 + \frac{1}{1} = \frac{2}{1}.
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