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Math Help - k-th convergent

  1. #1
    Senior Member bkarpuz's Avatar
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    Question [SOLVED] k-th convergent

    Dear friends, I have a problem with the following question.
    I will denote
    \langle a_{0},a_{1},\ldots,a_{k}\rangle:=a_{0}+\dfrac{1}{a  _{1}+\frac{1}{\frac{\vdots}{a_{k-1}+\frac{1}{a_{k}}}}}
    and call by k-th convergent.
    And define
    <br />
p_{k}:=<br />
\begin{cases}<br />
1,&k=-1\\<br />
a_{0},&k=0\\<br />
a_{k}p_{k-1}+p_{k-2},&k\in\mathbb{N}<br />
\end{cases}<br />
    and
    <br />
q_{k}:=<br />
\begin{cases}<br />
0,&k=-1\\<br />
1,&k=0\\<br />
a_{k}q_{k-1}+q_{k-2},&k\in\mathbb{N}.<br />
\end{cases}<br />
    It is easy to show that
    \langle a_{0},a_{1},\ldots,a_{k}\rangle=\frac{p_{k}}{q_{k}  } is true.
    We will apply induction.
    Clearly, the claim holds for k=2.
    Suppose that
    \langle a_{0},a_{1},\ldots,a_{k}\rangle=\frac{p_{k}}{q_{k}  }=\frac{a_{k}p_{k-1}+p_{k-2}}{a_{k}q_{k-1}+q_{k-2}} holds for k\geq 2.
    Then, we have
    \underset{k+1-\text{terms}}{\underbrace{\langle a_{0},a_{1},\ldots,a_{k+1}\rangle}}=\underset{k-\text{terms}}{\underbrace{\langle a_{0},a_{1},\ldots,a_{k}+1/a_{k+1}\rangle}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{p_{k-1}(a_{k}+1/a_{k+1})+p_{k-2}}{q_{k-1}(a_{k}+1/a_{k+1})+q_{k-2}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{a_{k+1}(a  _{k}p_{k-1}+p_{k-2})+p_{k-1}}{a_{k+1}(a_{k}q_{k-1}+q_{k-2})+q_{k-1}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{a_{k+1}p_  {k}+p_{k-1}}{a_{k+1}q_{k}+q_{k-1}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{p_{k+1}}{  q_{k+1}}.
    Thus, the claim is true.

    Now my question comes.
    Let x be an irrational number, and set
    b_{k}:=<br />
\begin{cases}<br />
x,&k=0\\<br />
\dfrac{1}{b_{k-1}-\lfloor b_{k-1}\rfloor},&k\in\mathbb{N},<br />
\end{cases}<br />
    where \lfloor\cdot\rfloor is the least integer function.
    How to show that x=\frac{b_{k+1}p_{k}+p_{k-1}}{b_{k+1}q_{k}+q_{k-1}} holds, where p_{k} and q_{k} are defined as above with a_{k}=\lfloor b_{k}\rfloor.
    Last edited by bkarpuz; January 10th 2009 at 05:27 AM. Reason: solved
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by bkarpuz View Post
    Dear friends, I have a problem with the following question.
    I will denote
    \langle a_{0},a_{1},\ldots,a_{k}\rangle:=a_{0}+\dfrac{1}{a  _{1}+\frac{1}{\frac{\vdots}{a_{k-1}+\frac{1}{a_{k}}}}}
    and call by k-th convergent.
    And define
    <br />
p_{k}:=<br />
\begin{cases}<br />
1,&k=-1\\<br />
a_{0},&k=0\\<br />
a_{k}p_{k-1}+p_{k-2},&k\in\mathbb{N}<br />
\end{cases}<br />
    and
    <br />
q_{k}:=<br />
\begin{cases}<br />
0,&k=-1\\<br />
1,&k=0\\<br />
a_{k}q_{k-1}+q_{k-2},&k\in\mathbb{N}.<br />
\end{cases}<br />
    It is easy to show that
    \langle a_{0},a_{1},\ldots,a_{k}\rangle=\frac{p_{k}}{q_{k}  } is true.
    We will apply induction.
    Clearly, the claim holds for k=2.
    Suppose that
    \langle a_{0},a_{1},\ldots,a_{k}\rangle=\frac{p_{k}}{q_{k}  }=\frac{a_{k}p_{k-1}+p_{k-2}}{a_{k}q_{k-1}+q_{k-2}} holds for k\geq 2.
    Then, we have
    \underset{k+1-\text{terms}}{\underbrace{\langle a_{0},a_{1},\ldots,a_{k+1}\rangle}}=\underset{k-\text{terms}}{\underbrace{\langle a_{0},a_{1},\ldots,a_{k}+1/a_{k+1}\rangle}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{p_{k-1}(a_{k}+1/a_{k+1})+p_{k-2}}{q_{k-1}(a_{k}+1/a_{k+1})+q_{k-2}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{a_{k+1}(a  _{k}p_{k-1}+p_{k-2})+p_{k-1}}{a_{k+1}(a_{k}q_{k-1}+q_{k-2})+q_{k-1}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{a_{k+1}p_  {k}+p_{k-1}}{a_{k+1}q_{k}+q_{k-1}}
    {\color{white}{\rule{3cm}{0.3cm}}}=\frac{p_{k+1}}{  q_{k+1}}.
    Thus, the claim is true.

    Now my question comes.
    Let x be an irrational number, and set
    b_{k}:=<br />
\begin{cases}<br />
x,&k=0\\<br />
\dfrac{1}{b_{k-1}-\lfloor b_{k-1}\rfloor},&k\in\mathbb{N},<br />
\end{cases}<br />
    where \lfloor\cdot\rfloor is the least integer function.
    How to show that x=\frac{b_{k+1}p_{k}+p_{k-1}}{b_{k+1}q_{k}+q_{k-1}} holds, where p_{k} and q_{k} are defined as above with a_{k}=\lfloor b_{k}\rfloor.
    I dont know how it happened but, after I typed the question here, I easily solved it even I spend some time on it before.
    I explain it below, may be someone may need it.
    Using the definition of b_{k}, we have
    x=\langle\lfloor b_{0}\rfloor,\ldots,\lfloor b_{k-1}\rfloor, b_{k}\rangle=\langle a_{0},\ldots,a_{k-1}, b_{k}\rangle
    and thus
    x=\underset{k+1-\text{terms}}{\underbrace{\langle a_{0},\ldots,a_{k}, b_{k+1}\rangle}}=\underset{k-\text{terms}}{\underbrace{\langle a_{0},\ldots,a_{k}+1/b_{k+1}\rangle}}.
    Thus, following the similar arguments in the proof above, we get the desited result.
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