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

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

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

    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 $\displaystyle q_1$. If $\displaystyle q_k\geq 2^{k/2}$ was true then $\displaystyle q_1\geq 2^{1/2} = \sqrt{2}$. But this is false because consider $\displaystyle [1;1,...]$ then $\displaystyle p_1/q_1 = 1 + \frac{1}{1} = \frac{2}{1}$.
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