k-th convergent

**bkarpuz** · January 10th 2009, 03:24 AM

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
$ p_{k}:= \begin{cases} 1,&k=-1\\ a_{0},&k=0\\ a_{k}p_{k-1}+p_{k-2},&k\in\mathbb{N} \end{cases} $
and
$ q_{k}:= \begin{cases} 0,&k=-1\\ 1,&k=0\\ a_{k}q_{k-1}+q_{k-2},&k\in\mathbb{N}. \end{cases} $
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}:= \begin{cases} x,&k=0\\ \dfrac{1}{b_{k-1}-\lfloor b_{k-1}\rfloor},&k\in\mathbb{N}, \end{cases} $
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 $a_{k}=\lfloor b_{k}\rfloor$ .

**bkarpuz** · January 10th 2009, 03:48 AM

Originally Posted by bkarpuz

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
$ p_{k}:= \begin{cases} 1,&k=-1\\ a_{0},&k=0\\ a_{k}p_{k-1}+p_{k-2},&k\in\mathbb{N} \end{cases} $
and
$ q_{k}:= \begin{cases} 0,&k=-1\\ 1,&k=0\\ a_{k}q_{k-1}+q_{k-2},&k\in\mathbb{N}. \end{cases} $
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}:= \begin{cases} x,&k=0\\ \dfrac{1}{b_{k-1}-\lfloor b_{k-1}\rfloor},&k\in\mathbb{N}, \end{cases} $
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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