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Math Help - continued fraction

  1. #1
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    israel
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    continued fraction

    I try to understand why by definition

    $[c0,c1,...cn]=[c0,[c1,...cn]]$ and also
    $[c0,c1,...cn]=[c0,c1,...c(n-2),[c(n-1),cn]]$ .

    But not
    $[c0,c1,...cn]=[c0,c1,[c2,...cn]]$ for example.
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  2. #2
    MHF Contributor

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    Re: continued fraction

    I presume that your "[c0, c1, ..., cn]" means the continued fraction \frac{c0}{1+ \frac{c1}{1+ \cdot\cdot\cdot}+ cn"}}. It should be clear that you can factor out that top "c0" to get c0\frac{1}{1+ \frac{c1}{1+ \cdot\cdot\cdot+ cn}} which is your "[c0, [c1, ..., cn]]".

    Try doing that with n= 2 or 3 to see what is happening.
    Last edited by HallsofIvy; March 31st 2013 at 05:21 AM.
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