Let $\omega=(\omega\_1,\ldots,\omega\_{m})$ be an $m$-tuple of real numbers. Let $|\omega|\_{m}:=\sup\limits\_{1 \leq j \leq m}|\omega\_j|\_{1}$ be a metric on flat torus $\mathbb{T}^{m}=\mathbb{R}^{m}/\mathbb{Z}^m$, i. e. $|\theta|\_{1}$ is the distance from $\theta$ to nearest integer.


We say that an $m$-tuple $\omega$ satisfies *the Diophantine condition of order* $\nu \geq 0$ if there is a constant $C>0$ such that for all natural $q$ the inequality
$$|\omega q|\_{m} \geq C \left(\frac{1}{q}\right)^{\frac{1+\nu}{m}}$$
is satisfied.


Suppose we have some algorithm that provides a sequence of convergents (fractions which approximate $\omega$) with denominators $\{ q\_{k} \}$. I want to find such an algorithm, which satisfyes the following properties


**Property 1.** If $\omega$ satisfies the Diophantine condition, then $q\_{k+1}=O\left(q^{1+\nu}\_{k}\right)$.


**Property 2.** For many (in some sense) $\omega$ there is a constant $\hat{C}>0$ such that the estimate
$$\sum\limits\_{k=N}^{\infty}\frac{1}{q\_{k}} \leq \hat{C}\frac{1}{q\_{N}}$$
holds for all large enough $N$.


The classic continued fraction algorithm for $m=1$ provides the required denominators: property 1 is equivalent to the Diophantine condition and Property 2 holds for all real numbers.


Are there known multi-dimensional algorithms, satisfying the above properties? I will be grateful for any help.



Simultaneous Diophantine Condition and Growth Rate of Convergents Denominators
**Theorem.** Suppose that an $m$-tuple $\omega$ satisfies the Diophantine condition of order $\nu \geq 0$. Then there exists a sequence $\{q\_{k}\}$ of natural numbers, $k=1,2,\ldots$, and a constant $\hat{C}=\hat{C}(\omega)>0$ such that


* $|\omega q\_{k}|\_{m} \leq \hat{C} \cdot
 \left(\frac{1}{q\_{k+1}}\right)^{1/m}$.
* $q\_{k} \leq q\_{k+1}$, $q\_{k+1}=O\left(q^{1+\nu}\_{k}\right)$ and
there exists constants $\gamma\_{2}>\gamma\_{1}>1$ and $A\_{1},A\_{2}>0$
such that $$A\_{1} \gamma^{k}\_{1}\leq q\_{k} \leq A\_{2}
\gamma^{k}\_{2}.$$
* $\sum\limits\_{k=N}^{\infty}\frac{1}{q\_{k}} \leq
\hat{C}\frac{N}{q\_{N}}$.


**Proof.** 


* By the Dirichlet theorem, for all $Q>0$ there exists a natural number
$1 \leq q \leq Q$ such that $$|\omega q|\_{m} \leq
 \left(\frac{1}{Q}\right)^{1/m}.$$ Let $\alpha>1$ be fixed and let
$q\_{k}$ be a number $q$ from the Dirichlet theorem for $Q =
 \alpha^{k}$. If we get $q\_{k+1}<q\_{k}$ for some $k$, then we can put
$q\_{k}:=q\_{k+1}$ and so on.
* Now for some $C>0$ we have $$C
 \left(\frac{1}{q^{1+\nu}\_{k}}\right)^{1/m} \leq |\omega q\_{k} | <
 \left(\frac{1}{\alpha^{k}}\right)^{1/m} \leq
 \left(\frac{1}{q\_{k}}\right)^{1/m}$$ and, consequently,
$$C^{\frac{m}{1+\nu}}\alpha^{\frac{k}{1+\nu}} \leq q\_{k} \leq
 \alpha^{k}.$$ It is easy too see that $$q\_{k+1} \leq
 \alpha^{k+1}=\alpha\cdot\alpha^{k}=\alpha
 C^{-m}\left(C^{\frac{m}{1+\nu}}
 \alpha^{\frac{k}{1+\nu}}\right)^{1+\nu}\leq \alpha C^{-m}
 q^{1+\nu}\_{k}.$$ Thus, $q\_{k+1}=O(q^{1+\nu}\_{k})$. One can put
$\gamma\_{1}:=\alpha^{\frac{1}{1+\nu}}$ with
$A\_{1}:=C^{\frac{m}{1+\nu}}$ and
$\gamma\_{2}:=\alpha$ with $A\_{2}:=1$. Also note that
$$|\omega q\_{k}| <
 \left(\frac{1}{\alpha^{k}}\right)^{1/m}=\alpha^{1/m}\left(\frac{1}{\alpha^{k+1}}\right)^{1/m}
 \leq \alpha^{1/m}\left(\frac{1}{q\_{k+1}}\right)^{1/m}.$$
* Now we will estimate
$\sum\limits\_{k=N}^{\infty}\frac{1}{q\_{k}}=\frac{1}{q\_{N}}\sum\limits\_{k=0}^{\infty}\frac{q\_{N}}{q\_{N+k}}.$
From the previous inequalities we have $$\frac{q\_{N}}{q\_{N+k}} \leq
 \frac{A\_2}{A\_1}\frac{\gamma^{N}\_{2}}{\gamma^{N+k}\_{1}}=\frac{A\_2}{A\_1}\alpha^{\frac{N\nu
 - k}{1+\nu}}.$$ We use this estimate only for $k \geq 2\nu N$, for low values of $k$ we will use $\frac{q\_{N}}{q\_{N+k}} \leq 1$. So,
$$\frac{1}{q\_{N}}\sum\limits\_{k=0}^{\infty}\frac{q\_{N}}{q\_{N+k}} \leq
 \frac{1}{q\_{N}} 2\nu N + \frac{1}{q\_{N}}\sum\limits\_{k \geq 2\nu
 N}\alpha^{\frac{N\nu - k}{1+\nu}} \leq \hat{C}\cdot\frac{N}{q\_{N}},$$ where
$\hat{C}>0$ is an appropriate constant.


