text:
\section{ Simultaneous approximation and Diophantine exponents}

In this paper we deal with simultaneous approximation to real numbers $\alpha_1,\ldots,\alpha_n$,
that is integer solutions  $(q,a_1,\ldots,a_n)\in \mathbb{Z}^{n+1}$ of the system  inequalities
\begin{equation}\label{s4}
\begin{cases}
\displaystyle{\max_{1\leqslant j \leqslant n}} \, |q\alpha_j - a_j| \leqslant Q^{-\gamma},
\cr
1\leqslant q \leqslant Q.
\end{cases}
\end{equation}
The \emph{ordinary  Diophantine exponent}  $ \omega = \omega(\pmb{\alpha})$ for real vector
$\pmb{\alpha}\in \mathbb{R}^n$ is defined as supremum  over all $\gamma$ such that the system \eqref{s4} has integer solutions for arbitrary large values of $Q$.
The {\it uniform   Diophantine exponent}  $ \hat{\omega} = \hat{\omega}(\pmb{\alpha}) $ is defined as supremum  over all $\gamma$ such that the system \eqref{s4} has integer solutions for all $Q$ large enough.
Generalizing  the results by  Jarn\'{\i}k \cite{J}, in \cite{MaMo}
it was proven that under the condition that
the numbers $1,\alpha_1,\ldots,\alpha_n$ are linearly independent over $\mathbb{Q}$ one has
\begin{equation}\label{MaMo}
\frac{\omega}{\hat{\omega}} \geqslant G_n(\hat{\omega}),
\end{equation}
where $G_n(\hat{\omega})$
is the unique root of the equation
$$
g^{n-1} = \frac{\hat{\omega}}{1-\hat{\omega}} (g^{n-2}+ \cdots +g+1).
$$
An alternative proof of this result is given in  \cite{Nguen}. This bound is known to be optimal.\\

In this paper we deal with approximations to three numbers. In this cas, $n=3$ and
$$ G(\hat{\omega}) = G_3(\hat{\omega}) =
\frac{1}{2} \left(\frac{\hat{\omega}}{1-\hat{\omega}} +
\sqrt{
\left(\frac{\hat{\omega}}{1-\hat{\omega}} \right)^2+
 \frac{4 \hat{\omega}}{1-\hat{\omega}} }
 \right)
.
$$
was obtained in  \cite{mche}. Note that in an earlier paper \cite{R} a lot of geometric observation concerning the  behavior of the corresponding solutions of the system \eqref{s4} was done. In the present paper we want to carry out a deeper analysis of the geometry of the best approximations vectors in  the case  $n=3$ and introduce stronger new lower bounds for the ratio $\frac{\omega}{\hat{\omega}}$ under further condition.

\section{Patterns of best approximation vectors}

Suppose that $\pmb{\alpha} \in \mathbb{R}^n\setminus \mathbb{Q}^n$.
We deal with the infinite  sequence of  best approximation vectors
$$\pmb{z}_\nu = (q_\nu, a_{1,\nu},\ldots,a_{n,\nu})\in \mathbb{Z}^{n+1}, \nu \geqslant 1$$
such that
$$q_1<q_2< \cdots <q_\nu<q_{\nu+1}< \cdots$$
form the increasing sequence of common denominators and the remainders of approximation
$$\xi_\nu = \max_{1\leqslant j \leqslant n} |q_\nu \alpha_j - a_{j,\nu}|$$
satisfy
$$\xi_1>\xi_2>\cdots>\xi_\nu>\xi_{\nu+1}> \cdots,$$
and
$$\xi_\nu = \min_{q\in \mathbb{Z}_+,\,\, q\leqslant q_\nu; \,\, (a_1,\ldots,a_n)\in \mathbb{Z}^n\setminus \{(a_{1,\nu},\ldots,a_{n,\nu})\}}
\,\,\,\,\,\,
\max_{1\leqslant j \leqslant n} |q \alpha_j - a_{j}|.$$

For approximations to one number, from continued fractions' theory we know that
$$
\left|
\begin{array}{cc}
q_\nu &a_{1,\nu}
\cr
q_{\nu+1} &a_{1,\nu+1}
\end{array}\right|
= (-1)^\nu,\,\,\,\forall \, \nu
$$
and in particular
each couple of consecutive  best approximations
$\pmb{z}_\nu, \pmb{z}_{\nu+1}$ consists of two linearly independent vectors.

In the case $n = 2$  when $1,\alpha_1, \alpha_2$ are linearly independent over $\mathbb{Q}$,  Jarn\'{\i}k proved that  there  exist infinitely many $\nu$ such that
three consecutive vectors $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}$ are linearly independent.

In the present paper we deal with the case $n=3$. The behavior of the best approximation vectors in this case is a little bit more complicated. From  Jarn\'{\i}k it follows again that  there  exist infinitely many $\nu$ such that three consecutive vectors $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}$ are linearly independent. However it may happen that there is no quadruples   $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}, \pmb{z}_{\nu+2}$ consisting of four consecutive independent vectors. It is the \emph{phenomenon of degeneracy of dimension}, see \cite{Che,Moor,MLMS,Mosin,Schlei}. Nevertheless, for any $\nu_0$ the sequence $ \pmb{z}_{\nu_0+j}, \, j \in \mathbb{Z}_+$ generates the whole space $\mathbb{R}^4$.

   Let us consider $\nu$ such that   three consecutive vectors $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}$ are linearly independent and define
   $$
   j = \min \{ \nu'>\nu:\,\,
    \text{vectors} \,\, \pmb{z}_{\nu'-1}, \pmb{z}_{\nu'}, \pmb{z}_{\nu'+1}\,\,\text{are linearly independent}\}.
    $$
    Now we deal with the collection
     \begin{equation}\label{first}
    \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1},
    \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1},\,\,\,\, \nu< j
    \end{equation}
    of two successive triples
    $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}$
    and
    $ \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1}$ of independent vectors.
    We distinguish two cases.

    It may happen that  triples   $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}$ and $ \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1}$ generate the same three-dimensional subspace of $\mathbb{R}^4$, that is
    $$
    \langle \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1} \rangle_\mathbb{R} =  \langle \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1}
    \rangle_\mathbb{R}.
    $$
    In this case we say the collection of vectors \eqref{first} has \emph{pattern} $A$.

    In the case when
   triples   $ \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}$ and $ \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1}$ generate different three-dimensional subspaces in $\mathbb{R}^4$, that is
    $$
    \langle \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1} \rangle_\mathbb{R} \neq  \langle \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1}
    \rangle_\mathbb{R}
    ,$$
    and so
        $$
    \langle \pmb{z}_{\nu-1}, \pmb{z}_{j-1}, \pmb{z}_{j}, \pmb{z}_{j+1} \rangle_\mathbb{R} =
    \langle \pmb{z}_{\nu-1}, \pmb{z}_\nu, \pmb{z}_{\nu+1}, \pmb{z}_{j+1} \rangle_\mathbb{R}
    =\mathbb{R}^4,
    $$
    we say the collection of vectors  \eqref{first} has \emph{pattern} $B$.

    We should note that for each collection \eqref{first} there exists the next collection of the form \eqref{first}
      \begin{equation}\label{next}
    \pmb{z}_{j-1}, \pmb{z}_j, \pmb{z}_{j+1},
    \pmb{z}_{l-1}, \pmb{z}_l, \pmb{z}_{l+1},\,\,\,\, j<l
    \end{equation}
    where the last triple  in \eqref{first} coincide with the first triple in \eqref{next}. So, by encoding the collections \eqref{first} by letters $A$ and $B$ accordingly, for each vector $\pmb{\alpha}\in \mathbb{R}^3$ with
    $1,\alpha_1,\alpha_2,\alpha_3$
    linearly independent over $\mathbb{Q}$  we define an infinite word
    \begin{equation}\label{word}
    \frak{W} (\pmb{\alpha}) =
    C_1 C_2 C_3 \ldots \, , \,\,\,\, C_j \in \{ A,B\}.
    \end{equation}
    From the discussion above we see that when
      $1,\alpha_1,\alpha_2,\alpha_3$ are
    linearly independent over $\mathbb{Q}$,
    the word $
        \frak{W} (\pmb{\alpha}) $ is infinite and contains infinitely many letters $B$.

        In the present paper  we show how we can determine a sharper lower bound for the ratio of the exponents
        $\frac{\omega (\pmb{\alpha})}{\hat{\omega}(\pmb{\alpha})}$ in terms of the word
        $
        \frak{W} (\pmb{\alpha}) $.

\section{Parameters and functions}

 We  deal with parameters
$$
\frac{1}{3}<\lambda < 1
,\,\,\,\,
 \frac{\lambda}{1-\lambda}< g < \frac{1}{1-\lambda}
 .
$$
Define
\begin{equation}\label{pere}
\theta= \theta(\lambda) = \frac{1-\lambda}{\lambda},\,\,\,\,\,
\sigma = \sigma (g,\lambda) =\frac{\frac{1}{1-\lambda}-g}{ g - \frac{1}{\theta}} >0
\end{equation}
and define functions
$$
f_k  = f_k (g,\lambda),\,\,\,\,
k \geqslant 0
$$
inductively by
\begin{equation}\label{fu}
f_0  = 1,\,\,\,\,\, f_{k} = \frac{1}{\theta -\sigma \prod_{j=0}^{k-1} f_j}, \,\ \, k  \geqslant 1.
\end{equation}
We will see that the denominator is positive in our application.
In particular for $k=1$ we have the function
$$
f_1 =\frac{1}{\theta-\sigma}=
\frac{(1-\lambda)g - \lambda}{\frac{1-\lambda}{\lambda} g +\lambda- 2 }.
$$
By induction we see that
\begin{equation}\label{india}
f_k = \frac{1}{\frac{1}{\lambda} - \theta f_{k-1}}.
\end{equation}
So
$$
f_k =
\frac{1}{\displaystyle{\frac{1}{\lambda}-\frac{\theta}{\displaystyle{\frac{1}{\lambda}-
\frac{\theta}{\displaystyle{\frac{1}{\lambda} -\dots -
\frac{\theta}{
\displaystyle{ \frac{1}{\lambda} -\frac{\theta}{\theta-\sigma}}} }}}}}}
$$
(here $\frac{1}{\lambda}$ occurs $k-1$ times).
If we set
$$
F_0 = f_0 = 1,\,\,\,\,
F_k = \prod_{j=0}^k f_j ,
\,\,\, k \geqslant 1
$$
from \eqref{fu} we get
$$
\frac{1}{F_k} =
\frac{\theta -\sigma F_{k-1}}{F_{k-1}} = \theta^k -  \sigma S_k ,\,\,\,\, \text{where}\,\,\,\,\,   S_k = \sum_{j=0}^{k-1}\theta^j,
$$
and
\begin{equation}\label{fu2}
f_k = \frac{F_k}{F_{k-1}} = \frac{ \theta^{k-1}- \sigma S_{k-1} }{\theta^{k}- \sigma S_{k} }
.
\end{equation}

\begin{lemma}\label{Lemma1}
 For any $ k\geqslant 1$ and for any $\lambda \in \left(\frac{1}{3}, 1\right)$  equation
\begin{equation}\label{fu3}
f_k(g,\lambda)=g
\end{equation}
in the interval $\left( 1,\frac{2}{\theta}\right)$
has a unique root $\frak{g}_k = \frak{g}_k (\lambda)$.
For this root, we have lower bound
\begin{equation}\label{GG}
 \frak{g}_k (\lambda) > G(\lambda).
\end{equation}
Moreover, for a fixed $\lambda$ one has
\begin{equation}\label{fu4}
 \frak{g}_k (\lambda)< \frak{g}_{k+1} (\lambda)
 \,\,\,\,\,
 \text{and}
 \,\,\,\,\,
 \overline{\frak{g}}=
 \lim_{k\to \infty}  \frak{g}_k (\lambda) =
 \begin{cases}
 \frac{1}{1-\lambda},\,\,\,\text{if}\,\,\, \lambda \in \left[\frac{1}{2}, 1\right), \cr
 \frac{2}{\theta},\,\,\,\text{if}\,\,\, \lambda \in \left(\frac{1}{3},\frac{1}{2}\right].
 \end{cases}
\end{equation}
\end{lemma}

The explicit form of   equation $
f_1 (g,\lambda) = g
$
is
\begin{equation}\label{xRoy}
(1-\lambda) g^2 - (3-2\lambda)\lambda g +\lambda^2 = 0,
\end{equation}
and
$\frak{g}_1 > G(\lambda).
$
 Here we should note that
 the equation
$
f_1 (\theta,\lambda) = \theta
$
just determines Roy's number
$$\frak{g}_1 (\theta )=
\frac{1}{2}
 \left(2+\sqrt{5}-\sqrt{7+2\sqrt{5}}\right) = 0.42451^+
 $$
 which appeared in the problem of uniform simultaneous approximation to a real number, its square and its cube \cite{R}.

\begin{proof}[Proof of Lemma \ref{Lemma1}]
By formulas \eqref{pere} \eqref{fu2} we rewrite equation \eqref{fu3} as
$$
R_k(g) = 0,\,\,\,\,\,\text{where} \,\,\,\,\, R_k(g) =
 (\theta S_k +\theta^{k+1})g^2 -\left(2\theta^k+\frac{S_k}{\lambda}+\theta S_{k-1}\right) g + \frac{S_{k-1}}{\lambda} +\theta^{k-1},
$$
and in particular
$$
R_1(g)  = (\theta+\theta^2)g^2 -\left(2\theta +\frac{1}{\lambda}\right) g +1.
$$
We see that
$$
R_1\left(\frac{1}{\theta}\right) = -1,\,\,\,\,\,
\text{and}
\,\,\,\,\,
R_1\left(\frac{2}{\theta}\right) =
\frac{3\lambda -1}{1-\lambda} >0
$$
because of $ \lambda \in \left(\frac{1}{3},1\right)$. But for every $n$ we have
\begin{equation}\label{pr}
R_{k+1}(g) -R_{k}(g)= \theta^k (\theta g -1)(\theta g -2) ,
\end{equation}
and so
$$
R_k\left(\frac{1}{\theta}\right) = R_1\left(\frac{1}{\theta}\right) <0,\,\,\,\,\,
\text{and}
\,\,\,\,\,
R_k\left(\frac{2}{\theta}\right) =  R_1\left(\frac{2}{\theta}\right)   >0
$$
for every $n$.
As $R_k(g)$ is a quadratic polynomial in $g$,
we see that
\begin{equation}\label{prw}
\frac{1}{\theta} < \frak{g}_k < \frac{2}{\theta}, \,\,\,\,k \ge1 \,
.
\end{equation}
Moreover
$$
R_k(1)  = \theta^{k} (\theta- 2)< 0
$$
because of $\lambda >\frac{1}{3}$,
 and  so
 $$
 1
  < \frak{g}_k < \frac{2}{\theta},\,\,\,\,k \geqslant 1
,
$$
as it was claimed in Lemma \ref{Lemma1}. Substituting $\frak{g}_k$ into \eqref{pr} and taking into account \eqref{prw} we see that
$$
R_{k+1}(\frak{g}_k)<0,
$$
and this shows that
$\frak{g}_{k+1}>\frak{g}_k$. The limit condition from \eqref{fu4} is clear, because the limit value $\overline{\frak{g}}$ satisfies equation
$$
(1-\lambda)g^2 -(2-\lambda)g + 1=
((1-\lambda)g-1)(g-1) = 0
\,\,\,\,\,\text{for}\,\,\,\,\, \lambda \in \left[\frac{1}{2}, 1\right),
$$
or
$$
\theta^2g^2 - 3\theta g + 2 =  (\theta g-1)(\theta g - 2) =0
\,\,\,\,\,\text{for}\,\,\,\,\, \lambda \in \left(\frac{1}{3}, \frac{1}{2}\right].
$$
Lemma \ref{Lemma1} is proven.
\end{proof}

We need some more parameters. Define
\begin{equation}\label{g11}
\frak{g}_{1,0} = \frac{1}{\theta \frak{g}_1-1 }
 = \frac{\frak{g}_1-\lambda}{(1-\lambda) \frak{g}_1 }
=
\frac{1-\lambda-\frac{\lambda}{\frak{g}_1}}
{1-(1-\lambda)\frak{g}_1}
\end{equation}
where equalities follow from the equation \eqref{xRoy}.

It is clear that for $\lambda \in [1/3,1)$ one has
\begin{equation}\label{g1}
1\leqslant
\frak{g}_{1,0} \leqslant \frak{g}_1.
\end{equation}
For $ k> 1$ we define inductively
\begin{equation}\label{anyg}
\frak{g}_{k,0} = \frac{1}{\theta\frak{g}_{k} -1},\,\,\,\,\,
\frak{g}_{k,j} = f_j(\frak{g}_{k},\lambda),\, j = 1,\ldots, k-1.
\end{equation}
Calculations show that
\begin{equation}\label{ssdd}
\max_{0\leqslant j \leqslant k-1} \frak{g}_{k,j} \leqslant \frak{g}_{k}.
\end{equation}
Equalities \eqref{india} and \eqref{anyg} give us
\begin{equation}\label{glue}
\frak{g}_k= \frac{\frak{g}_{k,0}+1}{\theta\frak{g}_{k,0}}
,\,\,\,\,\,
\frak{g}_{k,j-1}
 = \frac{\frak{g}_{k,j}-\lambda}{(1-\lambda) \frak{g}_{k,j }},\,\,\,\
 j = 1, \ldots, k-1
.
\end{equation}

\section{ Results}

We consider the quantity
$$
k = k(\pmb{\alpha}) =\max\left\{ j:\,\, \text{subword}\,\, \underbrace{A\ldots A}_{j\, \text{times}}B\,\,
\text{occurs infinitely many times in \eqref{word}}\right\}
$$

First of all, let us discuss the case $ k(\pmb{\alpha}) =0$. In this case the word  $\frak{W} (\pmb{\alpha})$ has only finitely many letters $A$ and we do not improve on the bound $\frac{\omega(\alpha)}{\hat{\omega}(\alpha)} \geqslant G(\hat{\omega})$
which is known to be optimal in this case (see \cite{R1} and the explanation from \cite{SS_MJCNT}). In the case $ k \geqslant 1$ we have the following result.

\begin{maintheorem}\label{Thm1}
For any $\pmb{\alpha}\in \mathbb{R}^3 $
with $k(\pmb{\alpha}) = k\geqslant 1$  one has
$$
\frac{\omega(\pmb{\alpha})}{\hat{\omega}(\pmb{\alpha})} \geqslant \frak{g}_k(\hat{\omega}(\alpha)),
$$
where $\frak{g}_k (\cdot)\geqslant 1$ is the root of  \eqref{fu3} defined in Lemma \ref{Lemma1}.
\end{maintheorem}

In the next statement we show the optimality of the bound from Theorem \ref{Thm1}.

\begin{maintheorem}\label{Thm2}
For any $\lambda \in \left(\frac{1}{3}, 1\right) $
there exists $\pmb{\alpha}\in \mathbb{R}^3$  such that the sequence of patterns
\eqref{word} for
 $\pmb{\alpha}$ is a periodic sequence  with the period
 $$
 \underbrace{AA\ldots A}_{k \, \text{times}} B
 $$
and
\begin{equation}\label{T2}
\hat{\omega}(\pmb{\alpha}) = \lambda,
\,\,\,\,\,
\,\,\,\,\,
\frac{\omega(\pmb{\alpha})}{\hat{\omega}(\pmb{\alpha})} = \frak{g}_k (\lambda).
\end{equation}
 \end{maintheorem}

\begin{remark}
It is possible  to show that any sequence from two  letters $A,B$ with infinitely many $B$'s can occur as the word
$\frak{W}(\pmb{\alpha})$ for a certain $\pmb{\alpha}$ with linearly independent components.
Moreover, for any $\lambda  \in \left(\frac{1}{3}, 1\right)  $ and  for any word $\frak{W}$  with infinitely many $B$'s, it is possible to construct
$\pmb{\alpha}$ with linearly independent components such that
$\frak{W}(\pmb{\alpha})=\frak{W}$  and for
 $ k =k(\pmb{\alpha})$ (determined by the word $\frak{W}$)
the equalities \eqref{T2} are valid.
\end{remark}

We  give a proof of Theorem \ref{Thm1} in Section \ref{5} below. As for Theorem \ref{Thm2}, we give a detailed proof of the case $k=1$ in Sections 6 - 8.  The proof for the case $k>1$ uses similar argument. The difference is that  in the inductive process described below we need to repeat {\bf Stage 1} $k$ times. All necessary comments for the case $k>1$ are given in Section \ref{9}.

\section{Proof of Theorem 1}\label{5}

 First of all, we should note that for any $\lambda <\hat{\omega} (\pmb{\alpha}) $ for all $\nu$ large enough one has
\begin{equation}\label{sing}
\xi_\nu \leqslant q_{\nu+1}^{-\lambda}.
\end{equation}

  Let
 \begin{equation}\label{triples}
     \pmb{z}_{\nu_k-1}, \pmb{z}_{\nu_k}, \pmb{z}_{\nu_k+1};\,\,\,
     \pmb{z}_{\nu_{k-1}-1}, \pmb{z}_{\nu_{k-1}}, \pmb{z}_{\nu_{k-1}+1};\,\,\, \ldots \,\,\, ; \,\,\,
     \pmb{z}_{\nu_1-1}, \pmb{z}_{\nu_1}, \pmb{z}_{\nu_1+1};\,\,\,
      \pmb{z}_{\nu_0-1}, \pmb{z}_{\nu_0}, \pmb{z}_{\nu_0+1};\,\,\,
     \pmb{z}_{l-1}, \pmb{z}_{l}, \pmb{z}_{l+1}
 \end{equation}
 be the triples of successive best approximation vectors corresponding to the subword
 $ \underbrace{A\ldots A}_{k\, \text{times}}B$,
 so
 $$
 \nu_k<\nu_{k-1}<\cdots<\nu_1<\nu_0< l
 $$
 and each successive  triples
  $$
     \pmb{z}_{\nu_i-1}, \pmb{z}_{\nu_i}, \pmb{z}_{\nu_i+1};\,\,\,
     \pmb{z}_{\nu_{i-1}-1}, \pmb{z}_{\nu_{i-1}}, \pmb{z}_{\nu_{i-1}+1},\,\,\,\,\, 1\leqslant i \leqslant k
     $$
     determine pattern $A$, while successive  triples
     $$
        \pmb{z}_{\nu_0-1}, \pmb{z}_{\nu_0}, \pmb{z}_{\nu_0+1};\,\,\,
     \pmb{z}_{l-1}, \pmb{z}_{l}, \pmb{z}_{l+1}
 $$
 determine pattern $B$.

 Consider two-dimensional lattice
 $$
   \Lambda = \langle \pmb{z}_{\nu_0},\pmb{z}_{\nu_{0}+1}\rangle_{\mathbb{Z}}  =
   \langle \pmb{z}_{{l}-1},\pmb{z}_{{l}}\rangle_{\mathbb{Z}}
   $$
   and three-dimensional lattices
    $$
    \Gamma_{i} = \langle \pmb{z}_{\nu_i-1}, \pmb{z}_{\nu_i},\pmb{z}_{\nu_{i}+1}\rangle_{\mathbb{Z}} ,\,\, i = 0,\ldots,k
      \,\,\,\,\,  \text{and}\,\,\,\, \,\,
  \Gamma =
   \langle \pmb{z}_{{l}-1},\pmb{z}_{{l}}\, \pmb{z}_{l+1}\rangle_{\mathbb{Z}}.
   $$
    Consider  two-dimensional subspace
  $  \frak{L}= {\rm span}\, \Lambda $.
    As for three-dimensional subspaces,
   from the definitions of patterns $A$ and $B$ we see that
   $$
   \frak{G}^*=
   {\rm span} \,\Gamma_0 =
      {\rm span} \,\Gamma_1= \cdots
      =
         {\rm span} \,\Gamma_k
   $$
   is the same three-dimensional subspace,
   $$
      \frak{G} =    {\rm span} \,\Gamma,\,\,\,\,
      {\rm span} \,(\Gamma_{i} \cup \Gamma) = \mathbb{R}^4,\,\,\,\,
      0\leqslant i \leqslant k,
      $$
      and
 $$
    ({\rm span} \,\Gamma_{i} )     \cap
      ( {\rm span} \,\Gamma)
      = \frak{G}^* \cap \frak{G} = \frak{L},\,\,\,\, 0\leqslant i \leqslant k.
      $$
      Moreover, vectors
      \begin{equation}\label{ind}
      \pmb{z}_{\nu_0-1}
      ,
        \pmb{z}_{l-1},
        ,
          \pmb{z}_{l}
          ,
            \pmb{z}_{l+1}
            \end{equation}
            are independent.
      As  vectors \eqref{ind} are independent vectors in $\mathbb{R}^4$, their coordinates form a non-zero integer determinant and
      $$
         1\ll
        \xi_{\nu_0-1}\xi_{l-1}\xi_{l} q_{l+1},
        $$
        or taking into account \eqref{sing} and the definition  of $\theta$ from \eqref{pere},
        \begin{equation}\label{tw2}
        q_{\nu_0}\ll q_{l}^{-1} q_{l+1}^{\theta}
 .
        \end{equation}
            By standard argument we have the bounds
      \begin{equation}\label{tw}
      {\rm det}\, \Lambda \asymp \xi_{\nu_l-1}q_l
      \end{equation}
      and
         \begin{equation}\label{tw1}
      {\rm det}\, \Gamma_{i} \ll \xi_{\nu_i-1}\xi_{\nu_i}q_{\nu_i+1},\,\, 0\leqslant i \leqslant k,\,\,\,\,\,
      {\rm det}\, \Gamma \ll \xi_{\nu_l-1}\xi_{\nu_l}q_{\nu_l+1}.
      \end{equation}
        Now we consider rational subspaces $\frak{G}, \frak{G}^*$ and $\frak{L}$.  For their heights
           \begin{equation}\label{tw4}
           H (\frak{G}^*) \leqslant    {\rm det}\, \Gamma_k,\,\,\,\,
             H (\frak{G}) \leqslant    {\rm det}\, \Gamma,
             \,\,\,\,
             H(\frak{L}) \asymp   {\rm det}\, \Lambda
             \end{equation}
             by Schmidt's inequality
             (see \cite{SchHeights}, Lemma 8A)
             we have the   bound
                   \begin{equation}\label{tw5}
             1\ll   H (\frak{G}^*) \cdot \frac{ H (\frak{G})  }{  H(\frak{L})}
           \end{equation}
Now \eqref{tw4},\eqref{tw5} together with \eqref{tw},\eqref{tw1}  gives us
 $$
 1\ll   \xi_{\nu_k-1}\xi_{\nu_k}q_{\nu_k+1} \cdot \frac{\xi_{l}q_{l+1}}{q_{l}},
  $$
  or by \eqref{sing},
            \begin{equation}\label{tw6}
            1\ll q_{\nu_k}^{-\lambda} q_{\nu_k+1}^{1-\lambda} \cdot q_l^{-1} q_{l+1}^{1-\lambda},
            \end{equation}

\begin{lemma}\label{Lemma2}
 Let $ k\geqslant 1$.  Suppose that
 \begin{equation}\label{gg}
 q_{l+1}\leqslant q_l^g
 \end{equation}
 with $g <\frac{1}{1-\lambda}$.
 Then
 \begin{equation}\label{kk}
 q_{\nu_k+1}\gg  q_{\nu_k}^{f_k},
 \end{equation}
 where $f_k = f_k(g,\lambda) $ is defined in \eqref{fu} and  the sign $\gg$ contains a certain positive constant depending on $k$.
 \end{lemma}

 \begin{proof}
 We proceed by induction in $k$.

 First of all we verify the base of induction for $k=1$.
 Suppose that \eqref{gg} holds and assume
  \begin{equation}\label{hh}
 q_{\nu_1+1}\leqslant q_{\nu_1}^{f}.
 \end{equation}
 From $\nu_1<\nu_0$ and \eqref{tw2},\eqref{gg} we deduce
 \begin{equation}\label{AA}
 q_{\nu_1+1}\leqslant q_{\nu_0}\leqslant q_l^{-1+\theta g},
 \end{equation}
 and in particular $ \theta g -1 >0$.
  From \eqref{tw6} with $ k = 1$ and \eqref{gg},\eqref{hh} we see that
 \begin{equation}\label{BB}
 1\ll q_{\nu_1+1}^{-\frac{\lambda}{f}+1-\lambda}\cdot q_l^{-1 +(1-\lambda)g}.
 \end{equation}
 Last two inequalities \eqref{AA},\eqref{BB} lead to
 $$
 1\ll q_l ^{(-1+\theta q)\left(-\frac{\lambda}{f}+1-\lambda\right) - 1+ (1-\lambda)g},
 $$
 or
 $$
 (-1+\theta g)\left(-\frac{\lambda}{f}+1-\lambda \right) - 1+ (1-\lambda)g \geqslant 0.
 $$
 This can be simplified to
 $$
 f \geqslant \frac{1}{\theta-\sigma} = f_1
 $$
 by the definitions of $\theta$ and $\sigma $ (see \eqref{pere}), and we have proved \eqref{kk} for $k=1$.

 Now we deal with the inductive step. We suppose that we have proved the desired inequality \eqref{kk} for all
 values of parameters up to $k$ and deal with $k+1$. Again suppose that
   \begin{equation}\label{hhk}
 q_{\nu_{k+1}+1}\leqslant q_{\nu_{k+1}}^{f}.
 \end{equation}
 By inductive assumption
 $$
 q_{\nu_i+1}\geqslant q_{\nu_1}^{f_i},\,\,\,\,\,
 1\leqslant i \leqslant k $$
 and so as
 \begin{equation}\label{extra}
 q_{\nu_i+1}\leqslant q_{\nu_{i-1}},\,\,\,1\leqslant i \leqslant k
 ,
 \end{equation}
 we get
 \begin{equation}\label{CC}
 q_{\nu_k}\leqslant q_{\nu_0 }^{\frac{1}{\prod_{j=1}^k f_j}}.
 \end{equation}
 Here we should note that in \eqref{extra} in the left hand side we have the denominator for the last vector
 $\pmb{z}_{\nu_i+1}$ of the $i$-th triple and in the right hand side we have the denominator  for the second vector
 $\pmb{z}_{\nu_{i-1}}$ of the $(i-1)$-th triple.

   From \eqref{tw6} with $k+1$ instead of $k$ and \eqref{gg},\eqref{hh} get
    \begin{equation}\label{DD}
 1\ll q_{\nu_{k+1}+1}^{-\frac{\lambda}{f}+1-\lambda}\cdot q_l^{-1 +(1-\lambda)g}.
 \end{equation}
 We take into account the inequality $q_{\nu_{k+1}+1} \leqslant q_{\nu_k}$ and deduce from
 \eqref{DD} and \eqref{CC} with \eqref{tw2} the bound
 $$
 \frac{ (-1+\theta g) \left(-\frac{\lambda}{f}+1-\lambda\right)}{\prod_{j=1}^k f_j} -1+(1-\lambda) g  \geqslant 0
 ,
 $$
 which is equivalent to
 $$f \geqslant \frac{1}{\theta - \sigma \prod_{j=1}^k f_j} =  f_{k+1}. $$
  In particular, as $f$ is finite, we see that  from the existence of subword
 $ \underbrace{A\ldots A}_{k\, \text{times}}B$
  it follows that
  $
\theta > \sigma \prod_{j=0}^{k-1} f_j
$.
   So the inequality \eqref{kk} is proven.
 \end{proof}

 Now we finish with the proof of Theorem \ref{Thm1}.
 We can suppose that \eqref{gg} holds with $g <\frac{1}{1-\lambda}$,
  otherwise $ \frac{\omega}{\hat{\omega}} \geqslant \frac{1}{1-\lambda}$ and there is nothing to prove.
By Lemma \ref{Lemma2} we  deduce \eqref{kk}.
 This means  that
 $$
 \text{either}\,\,\,\,
 q_{l+1}\gg q_l^{\frak{g}_k}\,\,\,\,\,
 \text{or}
 \,\,\,\,\,
  q_{\nu_k+1}\gg q_{\nu_k}^{\frak{g}_k}.
  $$
  This gives the bound $ \frac{\omega}{\hat{\omega}} \geqslant \frak{g}_k$ and Theorem \ref{Thm1} is proved. \qed

\section{Inductive construction for Theorem 2: $k=1$}

We construct $\pmb{\alpha}\in \mathbb{R}^3$ such that  for any  $j$, its best approximation vectors $\pmb{z}_j = (q_j, a_{1,j},a_{2,j},a_{3,j})$
satisfy
 \begin{equation}\label{iv}
 \xi_{j-1} \asymp q_{j}^{-\lambda}
 \end{equation}
 and moreover for any $\nu$, the five successive best approximation vectors
\begin{equation}\label{collek}
\pmb{z}_{2\nu-1}, \pmb{z}_{2\nu}, \pmb{z}_{2\nu+1},\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}
\end{equation}
satisfy the conditions ({\bf i}) - ({\bf vii}) below.
By
 $$
 \pmb{Z}_j = \left(
 \frac{a_{1,j}}{q_j}, \frac{a_{2,j}}{q_j},  \frac{a_{3,j}}{q_j}\right)
 \in \mathbb{Q}^3.
 $$
 we denote the corresponding rational vector.
 We say that  vectors
\begin{equation}\label{bas}
\pmb{u},\pmb{v},\pmb{w}\in \mathbb{Z}^4
\end{equation}
form a \emph{primitive triple} if
$$
\langle
\pmb{u},\pmb{v},\pmb{w}
\rangle_\mathbb{Z} =
\mathbb{Z}^4 \cap
\langle
\pmb{u},\pmb{v},\pmb{w}
\rangle_\mathbb{R}
,
$$
that is  vectors \eqref{bas} form a basis of the three-dimensional lattice
$\mathbb{Z}^4\cap
\langle
\pmb{u},\pmb{v},\pmb{w}
\rangle_\mathbb{R}
.$

 \vskip+0.3cm

\noindent
({\bf i})  vectors
$
\pmb{z}_{2\nu-1}, \pmb{z}_{2\nu}, \pmb{z}_{2\nu+1} $, as well as vectors
$
\pmb{z}_{2\nu+1}, \pmb{z}_{2\nu+2}, \pmb{z}_{2\nu+3} $
  form  primitive triples;

\noindent
({\bf ii}) with some integers $ a_\nu$ and $b_\nu$  we have
$\pmb{z}_{2\nu+2} = \pmb{z}_{2\nu-1} + a_\nu\pmb{z}_{2\nu}+ b_\nu\pmb{z}_{2\nu+1}$ and so
$
  \pmb{z}_{2\nu}, \pmb{z}_{2\nu+1},\pmb{z}_{2\nu+2} $  form a primitive triple;

\noindent
({\bf iii})   vectors
$
\pmb{z}_{2\nu-1}, \pmb{z}_{2\nu}, \pmb{z}_{2\nu+1} , \pmb{z}_{\nu+3}$
(and so  vectors
$
\pmb{z}_{2\nu}, \pmb{z}_{2\nu+1}, \pmb{z}_{2\nu+2} , \pmb{z}_{2\nu+3}$)
form a  basis in $\mathbb{Z}^4$;

\noindent
({\bf iv})  \,\, \,
 $q_{2j+1}\asymp q_{2j}^{\frak{g}_1},\,\,\,\,\,\,\, j =\nu, \nu+1$;

\noindent
({\bf v})  \,\, \,
 $q_{2j+2}\asymp q_{2j+1}^{\frak{g}_{1,0}},
 \,\,\,\, j =\nu-1, \nu
 $;

\noindent
({\bf vi})  \,\,\,  the angle  between rational three-dimensional vectors
$\pmb{Z}_{2\nu+1} - \pmb{Z}_{2\nu+3}$ and $\pmb{Z}_{2\nu+2}- \pmb{Z}_{2\nu+3}$ is
greater that $\frac{\pi}{4}$ and less than $\frac{3\pi}{4}$;

\noindent
({\bf vii})  for the fundamental volume $\Delta$  of three-dimensional lattice
$$
\Gamma=  \langle\pmb{z}_{2\nu+1},\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}
\rangle_\mathbb{Z}  = \mathbb{Z}^4 \cap  \langle\pmb{z}_{2\nu+1},\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}
\rangle_\mathbb{R}
$$
we have
$$
\Delta \asymp
 \xi_{2\nu +1}\xi_{2\nu+2} q_{2\nu +3}
 .
 $$

  \vskip+0.3cm

If we construct such an $\pmb{\alpha}$, we prove Theorem \ref{Thm2} for $ k=1$. Indeed, from  \eqref{iv} it follows that
$\hat{\omega}(\alpha)= \lambda $ and  from the conditions  ({\bf iv},{\bf v})
we see that $ q_{2j+1}\asymp q_{2j}^{\frak{g}_1}, q_{2j+2}\asymp q_{2j+1}^{\frak{g}_{1,0}}$ for every $j$, and the inequality $ \frak{g}_{1,0}\leqslant \frak{g}_1$ gives us
$\frac{\omega(\alpha)}{\hat{\omega}(\alpha)} = \frak{g}_1$. Meanwhile, the conditions ({\bf i},{\bf ii}) and ({\bf iii})   show that the sequence of patterns for $\pmb{\alpha}$ is just the periodic sequence $ABABABABAB\ldots$ .

Vector $\pmb{\alpha}$ is constructed as a result of a certain inductive procedure and conditions ({\bf vi}, {\bf vii}) are necessary to proceed with inductive step.

We need some further objects.
 Suppose that  $\pmb{Z}_{j-2}, \pmb{Z}_{j-1}$ and $\pmb{Z}_j$ do not belong to the same one-dimensional affine subspace and define the unique  two-dimensional  affine subspace $ \frak{Z}_j$  containing three points
  $\pmb{Z}_{j-2}, \pmb{Z}_{j-1},\pmb{Z}_j$.
  Now we  define the \emph{angular neighborhood}
 $$
 \mathcal{U}_j =
 \{\pmb{X}\in \mathbb{R}^3:\,\,\,\,
 |\pmb{X}-\pmb{Z}_j|\leqslant  q_j^{-u_j},\,\,\,\,
 $$
 $$
 \text{angle between vectors}\,\,\pmb{Z}_{j-1}- \pmb{Z}_j\,\,\text{and}\,\,
 \pmb{X}- \pmb{Z}_j \,\,\text{is greater  than}\,\, \frac{\pi}{4} \,\,\text{and less than}\,\, \frac{\pi}{2}\},
 $$
 where
 $$
 u_j =
 \begin{cases}\label{uj}
 1+ \lambda \frak{g}_{1,0},\,\,\,\,\,\,\, j\,\, \text{odd},
 \cr
 1+ \lambda \frak{g}_1\,\,\,\,\,\, j\,\, \text{even}.
 \end{cases}
 $$
 For a point $\pmb{Y}\in \mathbb{R}^3$ we consider the ball
 $$\mathcal{W}_j (\pmb{Y}) =  \{\pmb{X}\in \mathbb{R}^3:\,\,\,\,
 |\pmb{X}-\pmb{Y}|\leqslant \frac{1}{10} \cdot q_j^{-u_j}\}.$$
The neighborhoods under consideration are shown on the Fig. 1 below.

\begin{figure}[h]
  \centering

\begin{tikzpicture}[scale=1.25]

  \node[draw=black,fill=white,circle,inner sep=50pt] at (-5, 0) {};
  \node[draw=black,fill=black,circle,inner sep=1pt] at (-5, 0) {};

  \draw [pattern=north east lines, fill opacity=0.25] (-5,0) -- (-5,-2) arc (-90:-45:2) -- cycle;
  \draw [pattern=north east lines, fill opacity=0.25] (-5,0) -- (-5,2) arc (90:45:2) -- cycle;

  \node[draw=black,fill=black,circle,inner sep=1pt] at (-1.5, 0) {};

  \draw (-7,0) -- (-1.5,0);
  \draw (-5,-2) -- (-5,2);
  \draw (-5,0) -- (-3.6,1.4);
  \draw (-5,0) -- (-3.6,-1.4);

  \draw (-5.2, -0.25)   node { $\pmb{Z}_j$};
  \draw (-1.45, -0.25)   node { $\pmb{Z}_{j-1}$};

  \node[draw=black,fill=gray,circle,inner sep=7pt] at (-4.5,1.2) {};

  \draw (-4.5,1.2) -- (-1.5,0);
  \draw (-4.5,1.2) -- (-5,0);

  \node[draw=black,fill=black,circle,inner sep=1pt] at (-4.5,1.2) {};

  \draw (-4.6, 1.7)   node { $\mathcal{W}_j$};
  \draw (-4.03, 1.48)   node { $\pmb{X}_j$};
  \draw (-4.45, 0.2)   node { $\pi/4$};
  \draw (-4.45, -0.2)   node { $\pi/4$};
  \draw (-2.5, -1.8)   node { \begin{Large}$\mathcal{U}_j\cap S$\end{Large}};

   \draw (2.73, 1.4)   node { $\pmb{X}_j$};
   \draw (2.6, 0.2)   node { $\pi/4$};
   \draw (2.3, -0.25)   node { $\pmb{Z}_j$};
   \draw (5.45, -0.25)   node { $\pmb{Z}_{j-1}$};
   \node[draw=black,fill=black,circle,inner sep=1pt] at (2, 0) {};
   \node[draw=black,fill=black,circle,inner sep=1pt] at (5.5, 0) {};
   \draw (1,0) -- (5.5,0);

   \draw (1.6,1.56) -- (5.5,0);
   \draw (2.5,1.2) -- (2,0);
   \node[draw=black,fill=black,circle,inner sep=1pt] at (2.5,1.2) {};
   \draw (2,-0.5) -- (2,2);
   \draw (1.7,-0.3) -- (3.4,1.4);
   \draw (1.6, -0.5)   node { $\pmb{Z}'$};
   \draw (1.5, 1.7)   node { $\pmb{X}''$};
   \draw (3.4, 1.1)   node { $\pmb{X}'$};
    \draw (3.4, -1.2)   node { \begin{Large}\text{inequalities for angles visualized}\end{Large}};

\end{tikzpicture}

      \caption{Angular domain $\mathcal{U}_j$ and the neighborhood $\mathcal{W}_j$ inside, section by subspace $S$; inequality for angles
      $ \frac{\pi}{2}-o(1) =\angle \pmb{Z}_{j-1}\pmb{X}''\pmb{Z}_j <
      \angle \pmb{Z}_{j-1}\pmb{X}j\pmb{Z}_j  < \angle \pmb{Z}_{j-1}\pmb{X}'\pmb{Z}_j
      <
      \angle \pmb{Z}_{j-1}\pmb{Z}_j\pmb{Z}'  = \frac{3\pi}{4}.
      $
       }

\end{figure}

\begin{lemma}\label{Lemma3}
For any two-dimensional affine subspace $S\subset\mathbb{R}^3$ such that $\pmb{Z}_{j-1}, \pmb{Z}_j\in S$,
 there exists $ \pmb{X}_j \in \mathcal{U}_{j}\cap S$ such that
 \begin{equation}\label{AW}
 \mathcal{W}_j = \mathcal{W}_j(\pmb{X}_j) \subset \mathcal{U}_{j}
 \end{equation}
 and
  \begin{equation}\label{AW1}
 \mathcal{W}_j  \cap \mathcal{W}_j(\pmb{Z}_j) = \varnothing.
 \end{equation}
 Moreover
 if we suppose that
 $j$ is large enough and  $\mathcal{W}_{j-1}$ satisfies \eqref{AW1} with $j$ replaced by $j-1$ and
   \begin{equation}\label{AW2}
   \pmb{Z}_j\in \mathcal{W}_{j-1},
 \end{equation}
 then
 for any $\pmb{X} \in  \mathcal{W}_j $
 the angle between    vectors
$\pmb{Z}_{j} - \pmb{X}$ and $\pmb{Z}_{j-1}- \pmb{X}$ is
greater that $\frac{\pi}{4}$ and less than $\frac{3\pi}{4}$.
\end{lemma}

\begin{proof}
To get \eqref{AW} we simply need to take a unit vector $\pmb{e}$ orthogonal to  the unit vector
  $\pmb{e}' = \frac{\pmb{Z}_{j-1}-\pmb{Z}_j}{|\pmb{Z}_{j-1}-\pmb{Z}_j|}$ and such that
  $\pmb{Z}_j + \pmb{e}\in S$
   and take
  $$
  \pmb{X}_j = \frac{1}{2}\cdot q_j^{-u_j}  \pmb{e} +  \frac{1}{4}\cdot q_j^{-u_j}  \pmb{e}'.
  $$
  Then  \eqref{AW}  and \eqref{AW1} follow from triangle inequality.
  The condition on the angle between $\pmb{Z}_{j} - \pmb{X}$ and $\pmb{Z}_{j-1}- \pmb{X}$
can be established as follows. From \eqref{AW2} and the growth conditions ({\bf iv},{\bf v}) we see that in the triangle
$\pmb{X}\pmb{Z}_j\pmb{Z}_{j-1}$ the length of the side $\pmb{Z}_j\pmb{Z}_{j-1}$ is much larger than the length of the
side $\pmb{X}\pmb{Z}_j$ and so the angle between  $\pmb{Z}_{j} - \pmb{X}$ and $\pmb{Z}_{j-1}- \pmb{X}$
is in the interval between $ \frac{\pi}{2} -o(1)$ and $\frac{3\pi}{4}$ (see Fig. 1).
\end{proof}

We need further notation. Consider the affine subspace
 $$
 \mathbb{R}_1^3  =\{ (x_0,x_1,x_2,x_3)\in \mathbb{R}^4:\, x_0 = 1\}.
 $$
 For a point $ \pmb{X}= (X_1,X_2,X_3)\in \mathbb{R}^3$ we define the point
 $ \overline{\pmb{X}}= (1,X_1,X_2,X_3)\in \mathbb{R}^3_1$. For any set
 $\mathcal{A}\subset\mathbb{R}^3$ we define
 $$
 \overline{\mathcal{A} } = \{ \overline{\pmb{X}}:\,\, \pmb{X} \in \mathcal{A}\} \subset \mathbb{R}^3_1.
 $$
 Of course we can apply Lemma \ref{Lemma3} to the similar objects in affine subspace $\mathbb{R}^3_1$ instead of $\mathbb{R}^3$.

  The proof of Theorem \ref{Thm2} is based on an inductive construction.
 We construct a sequence of integer vectors $\pmb{z}_\nu$ with \eqref{iv} such that  any collection \eqref{collek} of successive vectors satisfy conditions  ({\bf i}) - ({\bf vii}), neighborhoods form a nested sequence
 $$
  \mathcal{U}_1\supset \mathcal{W}_1\supset \cdots \supset  \mathcal{U}_{\nu-1}\supset\mathcal{W}_{\nu-1}\supset   \mathcal{U}_{\nu}\supset \mathcal{W}_\nu\supset  \cdots
  $$
  with non-empty intersection and for every $\pmb{\eta}\in  \mathcal{U}_{2\nu+3}$ the  collection
  \eqref{collek} will be a set of five successive best approximation vectors. Then the unique point of the intersection $ \cap_\nu  \, \mathcal{U}_{\nu}$ is  the vector $\pmb{\alpha}$ satisfying  the desired conditions.

 The base of induction is clear as we do not take care in the absolute  constants in signs $\asymp$.
 We may suppose that $q_1$ is large enough.
 So it is enough to explain the inductive step, namely how for given collection
  \eqref{collek} satisfying  ({\bf i}) - ({\bf vii}) and for given  neighborhood $\mathcal{U}_{2\nu+3}$ satisfying the inductive assumption
one can construct the collection for the next value of $\nu$ satisfying the similar properties and the next neighborhoods   satisfying all the necessary conditions.
For this purpose it is sufficient to construct
vectors $\pmb{z}_{2\nu+4}$ and $\pmb{z}_{2\nu+5} $  and to verify all the properties of these vectors and the corresponding neighborhoods
$\mathcal{U}_{2\nu+4}\supset \mathcal{U}_{2\nu+5}$. This procedure corresponds to a general concept of \emph{going down} and \emph{going up}.
This will be done in Section \ref{8}.

\section{Two- and three-dimensional lattices}\label{7}

We consider the two-dimensional subspace
$$
 L = \langle\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}
\rangle_\mathbb{R}
$$
the two-dimensional lattice
$$
 \Lambda = \langle\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}
\rangle_\mathbb{Z}  = \mathbb{Z}^4 \cap L\subset L
$$
as well as the three-dimensional subspace
$$
G=  \langle\pmb{z}_{2\nu+1},\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}
\rangle_\mathbb{R}
$$
and  the lattice
$
\Gamma $ defined in ({\bf vii}).
Let $ d$ be the two-dimensional fundamental volume of lattice $\Lambda$ and $\Delta$ be  the three-dimensional fundamental volume of lattice $\Gamma$ defined in condition ({\bf vii}). Then
\begin{equation}\label{dD}
d\asymp \xi_{2\nu+2}q_{2\nu+3}\asymp q_{2\nu+3}^{1-\lambda},\,\,\,\,\,\,
\Delta \asymp \xi_{2\nu +1}\xi_{2\nu+2} q_{2\nu +3} \asymp q_{2\nu+2}^{-\lambda}q_{2\nu+3}^{1-\lambda}
\asymp q_{2\nu+3}^{1-\lambda-\frac{\lambda}{g_1}},\,\,\,\,\,\,
\frac{\Delta}{d} \asymp q_{2\nu+2}^{-\lambda}.
\end{equation}
We need to consider the two-dimensional affine subspace
 $$
 L_{1}
=  L+\pmb{z}_{2\nu+1} \subset G
$$
and the affine lattice
 $$
 \Lambda_{1}
=  \Lambda+\pmb{z}_{2\nu+1} \subset \Gamma.
$$
We should note that the Euclidean distance between subspaces $L$ and $L_1$ is equal to
$\frac{\Delta}{d}.
$

Also we should deal with three-dimensional affine subspace
 $$
 G_{1}
=  G+\pmb{z}_{2\nu}  \subset \mathbb{R}^4
$$
and the affine lattice
 $$
 \Gamma_{1}
=  \Gamma+\pmb{z}_{2\nu} \subset \mathbb{Z}^4.
$$
The Euclidean distance between  $G$ and $G_1$ is equal to
$\frac{1}{\Delta}.
$

The parallelogram
$$
\mathcal{P} = \{ \pmb{z} =(x,y_1,y_2,y_3)\in L :
0\leqslant x \leqslant q_{2\nu+2}+q_{2\nu+3},\,\, \sqrt{y_1^2+y_2^2+y_3^2}\leqslant \xi_{2\nu+2}\} \supset
$$
$$
\supset
\{ \pmb{z} =(x,y_1,y_2,y_3)= \lambda_1 \pmb{z}_{2\nu+1}+ \lambda_2 \pmb{z}_{2\nu+2},\,\, 0\leqslant \lambda_1,\lambda_2\leqslant 1\}
$$
contains a fundamental domain of the lattice $\Lambda$,  while the cylinder
$$
\mathcal{C}= \{ \pmb{z} =(x,y_1,y_2,y_3)\in G :
0\leqslant x \leqslant q_{2\nu+1}+q_{2\nu +2}+q_{2\nu+3}, \,\,\sqrt{y_1^2+y_2^2+y_3^2}\leqslant 3\xi_{2\nu+1}\} \supset
$$
$$\supset
\{ \pmb{z} =(x,y_1,y_2,y_3)= \lambda_1 \pmb{z}_{2\nu+1}+ \lambda_2 \pmb{z}_{2\nu+2}
+ \lambda_3 \pmb{z}_{2\nu+3}, \,\, 0\leqslant \lambda_1,\lambda_2,\lambda_3\leqslant 1\}
$$
contains a fundamental domain of the lattice $\Gamma$.

 \section{Step of induction}\label{8}

 In the proof of Theorem \ref{Thm2} for case $k=1$, conditions
({\bf vi},{\bf v}) and the exponents $\frak{g}_1$ and $\frak{g}_{1,0}$ defined in Lemma \ref{Lemma1} and \eqref{g11} play a crucial role.

 Let the collection  \eqref{collek} satisfy ({\bf i}) - ({\bf vi}) and  suppose that for every $\pmb{\xi} \in \mathcal{U}_{2\nu+3}$
 vectors are successive five best approximation. We should construct vector   $\pmb{z}_{2\nu+4}$ ({\bf Stage 1}) and
 vector  $\pmb{z}_{2\nu+5}$ ({\bf Stage 2}).

\subsection{Stage 1}

 First, we construct integer vector $\pmb{z}_{2\nu+4}$.
 For  $\pmb{z}_{2\nu+3}$ Lemma 3
 applied in $\mathbb{R}_1^3$ with $ S = \mathbb{R}_1^3\cap G$
  gives us the point $X_{2\nu+3}$ and the neighbourhood $\mathcal{W}_{2\nu+3}$.

  Then we consider central projection $\frak{P}_1$  from affine subspace $\mathbb{R}^3_1$ onto affine subspace $L_1$, that is for $\pmb{x}\in \mathbb{R}_1^3$ its image (if it exists)  is defined as intersection of the line ${\rm span} \, \pmb{x}$ with subspace $L_1$.

  \begin{figure}[h]
  \centering

\begin{tikzpicture}[scale=1.25]

\draw[rotate=-10] (-2.2,2) -- (-5,2);
\draw[rotate=-10] (-1.5,2) -- (6,2);

\draw[rotate=-10] (-7,0) -- (-4.5,0);
\draw[rotate=-10] (-1.5,0) -- (4,0);

\draw[rotate=-10] (-5,2) -- (-7,0);
\draw[rotate=-10] (4,0) -- (6,2);

    \draw (2,1) [rotate=-10] ellipse (2cm and 0.3cm);

       \draw (-3.4,0.02) [rotate=-10] ellipse (0.08cm and 0.16cm);

    \draw[rotate=-10,dashed] (-5,0.5) -- (4.5,0.5);
     \draw[rotate=-10] (6,0.5) -- (4.5,0.5);

\draw[rotate=-10] (-7,-1.5) -- (2,-1.5);

\draw[rotate=-10] (4,-1.5) -- (2.7,-1.5);

\draw[rotate=-10] (-5.5,0) -- (-7,-1.5);
\draw[rotate=-10] (4,-1.5) -- (6,0.5);

   \draw[rotate=-10,dashed] (-5.5,0) -- (-5,0.5);
\draw[rotate=-10] (-6,-0.5) -- (-3.5,-0.5);

\draw[rotate=-10] (-6,-0.5) -- (0.3,1.16);
\draw[rotate=-10] (-6,-0.5) -- (3.4,0.785);

 \node[draw=black,fill=black,circle,inner sep=1pt] at (-6, 0.55) {};

 \node[draw=black,fill=black,circle,inner sep=1pt] at (-3.5, 0.11) {};

 \node[draw=black,fill=black,circle,inner sep=1pt] at (1.5, 0.76) {};

 \draw[rotate=-10] (1.2,0.86) -- (1.5,1.16);

 \draw[rotate=-10] (2.7,0.86) -- (3,1.16);

 \draw[rotate=-10]  (1.5,1.16) -- (3,1.16);

  \draw[rotate=-10] (1.2,0.86) -- (2.7,0.86);

         \draw[rotate=-10,dashed] (-6,-0.5) -- (1.3,1);

           \node[draw=black,fill=black,circle,inner sep=0.5pt] at (-3.35, 0.62) {};

 \draw (-6.2, 0.7)   node { $0$};

  \draw (2, -1.85)   node { $L$};

    \draw (-1.5, 2.3)   node { $L_1$};

       \draw (-3.3, -0.2)   node { $\pmb{Z}_{2\nu+3}$};

            \draw(2.23, 0.63) node  [rotate=-12]   { $\pmb{z} + \mathcal{P}$} ;

\end{tikzpicture}

      \caption{To Lemma 4: small ellipsis at the left side close to the point $\pmb{Z}_{2\nu+3}$ is the neighbourhood $\overline{\mathcal{W}}_{2\nu+3}\subset \mathbb{R}_1^3$  with center $\overline{\pmb{X}}_{2\nu+3}$ and  the large ellipsis at the right side is the image  $\frak{P}_1(\overline{\mathcal{W}}_{2\nu+3})  \subset L_1$.}

\end{figure}

 \begin{lemma}\label{Lemma4}
 The image $ \frak{P}_1( \overline{\mathcal{W}_{2\nu+3 }})$
 contains a shift of  a fundamental  domain of lattice $\Lambda$.
 \end{lemma}

 \begin{proof}
 The geometry of the proof is illustrated on Fig. 2.\\

Consider the unit vector
 $\pmb{e}_0=
 \frac{\pmb{z}_{2\nu+3}}{|\pmb{z}_{2\nu+3}|}
 $
 and the unit vector
 $\pmb{e}_1\in   G, |\pmb{e}_1| = 1$ which is orthogonal to $L$
 such that    $\frac{\Delta}{d}\pmb{e}_1\in L_1$. We also need a unit vector $ \pmb{e}_2 = (0,u,v,w)\in L$ which belongs to subspace $L$,  and  such that both vectors $\pmb{e}_2$ and $\pmb{z}_{2\nu+2}$ lie in the same half-subspace of $L$ with respect to the one-dimensional subspace $ \langle\pmb{z}_{2\nu+3}\rangle_\mathbb{R} \subset L$.
 Vectors $\pmb{e}_j, j =0,1,2$ are defined uniquely by the conditions above.

    As $ \overline{\pmb{X}}_{2\nu+3} \in S\subset G$ the image
    $\pmb{z} =\frak{P} (\overline{\pmb{X}}_{2\nu+3})\in L_1$ is well defined and satisfies
    $$
    \pmb{z} =  r \pmb{e}_0+
 s\pmb{e}_1+t \pmb{e}_2.
    $$
Calculations show that
$$
 r
 \asymp  \frac{\Delta}{d}\cdot q_{2\nu+3}^{1+\lambda \frak{g}_{1,0}}
 ,\,\,\,\,\,\,
 s =\frac{\Delta}{d},
 \,\,\,\,\,\,
 t\asymp  \frac{\Delta}{d},
$$
and the constant in  sign $\asymp$ are absolute.
 From \eqref{dD} and  ({\bf iv})  and the definition \eqref{g11}  of $\frak{g}_{1,0}$ we see that
  \begin{equation}\label{rst1}
  r\asymp  q_{2\nu+3}^{\frak{g}_{1,0}},
\,\,\,\,\,\text{
   meanwhile}\,\,\,\,\,
  t  \asymp q_{2\nu+2}^{-\lambda}.
   \end{equation}
   The image $ \frak{P}_1 ( \overline{\mathcal{W}_{2\nu+3 }}) $ is an ellipsis.
The large major axis  of the ellipsis is parallel to $\pmb{z}_{2\nu+3}$ and  has length
$\asymp r$, the section $ \{ x_0 =r\}$ of the ellipsis  has length $\asymp t$.
 From \eqref{rst1} we see (as $q_{2\nu+3}>q_{2\nu+2}>q_1$ are large enough),
 that the large axis is $ \geqslant 4q_{2\nu+3}$ and the section  $ \{ x_0 =r\}$  has length $ \geqslant 4\xi_{2\nu+2}$.
Then, by convexity
 $$
 \pmb{z}+ \mathcal{P}\subset  \frak{P}_1 ( \overline{\mathcal{W}_{2\nu+3 }}).
 $$
 Lemma is proven.
 \end{proof}

 As $ \frak{P}_1 ( \overline{\mathcal{W}_{2\nu+3 }})$ contains a  shift of a fundamental domain of lattice $\Lambda$ and lattice $\Lambda_1$ is congruent to $\Lambda$, the image
  $ \frak{P}_1 ( \overline{\mathcal{W}_{2\nu+3 }})$ contains an integer point. This is just the point $\pmb{z}_{2\nu+4} = (q_{2\nu+4} , a_{1,2\nu+4}, a_{2,2\nu+4}a_{3,2\nu+4})$ which we are constructing.
 First of all we should note that, as $ \pmb{z}_{2\nu+4}  \in \Lambda_1$, it is of the form
 \begin{equation}\label{newform}
  \pmb{z}_{2\nu+4}  =  \pmb{z}_{2\nu+1}  + a_{\nu+1}\pmb{z}_{2\nu+2}  +b_{\nu+1}\pmb{z}_{2\nu+3}
 \end{equation}
 with integers
 $ a_{\nu+1},b_{\nu+1}$
 and we have ({\bf ii})  for the next step.
 From  \eqref{rst1} we see that
 $$
 q_{2\nu+4}\asymp q_{2\nu+3}^{\frak{g}_{1,0}}
 $$
 and this is just the new (second) inequality from the condition ({\bf v}) for the next step.

 Now we consider neighborhood $\mathcal{U}_{2\nu+4}$. It is clear from triangle inequality   and
 the asymptotics for parameters \eqref{rst1} that
 $\mathcal{U}_{2\nu+4}\subset\mathcal{W}_{2\nu+3} $ and that
 for every $\pmb{\eta}\in \mathcal{U}_{2\nu+4}$  vectors $\pmb{z}_{2\nu+3}$ and $\pmb{z}_{2\nu+4}$ are successive best approximation vectors. Indeed, there is no best approximation vectors $\pmb{z}= (q,a_1,a_2, a_3)$
 with $ q_{2\nu+3}\leqslant q\leqslant q_{2\nu+4}$ outside $G$ because   for
 the corresponding integer vectors
  $ (q,a_1,a_2,a_3)\in \mathbb{Z}^4 \setminus G$ one has
 $$
 \sqrt{|q\eta_1-a_1|^2 +|q\eta_2-a_2|^2 +|q\eta_3-a_3|^2}
 \geqslant
 \min_{\pmb{\eta}_1\in G,\, \pmb{\eta}_2\in \mathbb{Z}^4\setminus G} |\pmb{\eta}_1-\pmb{\eta}_2|
 - q\times  \text{diameter of} \,\, \mathcal{U}_{2\nu+3} ,
 $$
 $$
 \min_{\pmb{\eta}_1\in G,\, \pmb{\eta}_2\in \mathbb{Z}^4\setminus G} |\pmb{\eta}_1-\pmb{\eta}_2|
 =
 \frac{1}{\Delta}\asymp \xi_{2\nu} \asymp q_{2\nu+1}^{-\lambda},
 \,\,\,
 q\times
 \text{diameter of} \,\, \mathcal{U}_{2\nu+3} \ll  q_{2\nu+4} q_{2\nu+3}^{-1-\lambda \frak{g}_{1,0}} \asymp
 q_{2\nu+3}^{-\lambda/\frak{g}_1} \asymp q_{2\nu+2}^{-\lambda} = o(q_{2\nu+1}^{-\lambda}),
 $$
 and so
  $$
 \sqrt{|q\eta_1-a_1|^2 +|q\eta_2-a_2|^2 +|q\eta_3-a_3|^2}
 >
 \max_{\pmb{\eta} = (\eta_1,\eta_2,\eta_3) \in \mathcal{U}_{2\nu+3}} \sqrt{||q_{2\nu+3}\eta_1||^2 +||q_{2\nu+3}\eta_2||^2  +||q_{2\nu+3}\eta_3||^2}.
 $$
 Meanwhile in $G$ the only candidates to be  new best approximation  vectors  in the range $ q_{2\nu+3}\leqslant q\leqslant q_{2\nu+4}$ are $\pmb{z}_{2\nu+4}$ and $\pmb{z}_{2\nu+4}-\pmb{z}_{2\nu+3}$
 (there is no integer points in $G$ in the strip between $L$ and $L_1$).
 Vector $\pmb{z}_{2\nu+4}$ is a best approximation vector indeed, and $\pmb{z}_{2\nu+4}-\pmb{z}_{2\nu+3}$ is not a best approximation vector as $\pmb{\eta}\in \mathcal{U}_{2\nu+4}$ and this angular neighbourhood just separates  $\pmb{\eta}$ from
  $\pmb{z}_{2\nu+4}-\pmb{z}_{2\nu+3}$.

  We have established all the approximation properties for all the vectors $\pmb{\eta}\in \mathcal{U}_{2\nu+4}$.

\subsection{Stage 2}

 Now we should construct vector $\pmb{z}_{2\nu+5}$ and  establish all the necessary properties of points
  $\pmb{\eta}$ in the new neighborhood $\mathcal{U}_{2\nu+5}$.

      First of all we should note that by ({\bf{i}}) for the second triple and \eqref{newform} the triple
  $\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}
 $
 form a basis of the lattice $\Gamma$, meanwhile the quadruple
 $\pmb{z}_{2\nu},\pmb{z}_{2\nu+2},\pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}
 $
 is a basis of $\mathbb{Z}^4$ (we take into account ({\bf iii}).

 We should note that $\pmb{z}_{2\nu-1}\not \in G$.
 Together with three-dimensional linear subspace $G$ we consider affine subspace
 $ G_1 = G+ \pmb{z}_{2\nu-1}$.
 The Euclidean distance between parallel subspaces $G$ and $G_1$ is equal to $\frac{1}{\Delta}$.
 We want to construct a new integer point $\pmb{z}_{2\nu+5}$  which belongs the the rational affine subspace $
 G_1$. Subspace $G_1$ contains a three-dimentional integer lattice  $G_1\cap\mathbb{Z}^4$. This gives us certain freedom to choose the point $\pmb{z}_{2\nu+5}$. However it would be enough  to localise the new point $\pmb{z}_{2\nu+5}$
close to a three-dimensional linear subspace $S$
 defined below.

We should apply Lemma \ref{Lemma3}. We use it for $ j = 2\nu +4$ and for points $\pmb{Z}_{\nu+4},\pmb{Z}_{\nu+3}$. We choose $S$ as follows.
    Let $G'\subset \mathbb{R}^4$ be three dimensional subspace with basis
  $ \pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}, \pmb{f}_1$,
  where   $\pmb{f}_1, |\pmb{f}_1| = 1$  is the unit which is orthogonal to $G$ and
 such that    $\frac{1}{\Delta}\pmb{f}_1\in G_1$.
 We define $ S = G'\cap \mathbb{R}_1^3$. Then Lemma \ref{Lemma3} gives point $\pmb{X}_{2\nu+4}$ and neighborhood $\mathcal{W}_{2\nu+4}$.

  We consider the central projection $\frak{P}_2$  from affine subspace $\mathbb{R}^3_1$ onto affine subspace $G_1$.

 \begin{lemma} \label{Lemma5}
  The image $ \frak{P}_2( \overline{\mathcal{W}_{2\nu+4 }})$
 contains a shift of  a fundamental  domain of lattice $\Gamma$.
 \end{lemma}

 \begin{proof}
 We consider the unit vector
 $\pmb{f}_0=
 \frac{\pmb{z}_{2\nu+4}}{|\pmb{z}_{2\nu+4}|}
 $
 and the unit vector
 $\pmb{f}_1, |\pmb{f}_1| = 1$ which is orthogonal to $G$
defined before.
 Again
 we need a unit vector $ \pmb{f}_2 = (0,u,v,w)\in  \langle \pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}  \rangle_{\mathbb{R}}$  such that both vectors $\pmb{f}_2$ and $\pmb{z}_{2\nu+3}$ lie in the same half-subspace of the two-dimensional subspace $\langle \pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}  \rangle_{\mathbb{R}}$ with respect to the one-dimensional subspace $ \langle\pmb{z}_{2\nu+4}\rangle_\mathbb{R} $.
In particular, vectors $\pmb{f}_0$ and $\pmb{f}_2$ generate the same two-dimensional linear subspace
as the vectors $\pmb{z}_{2\nu+3}$  and $\pmb{z}_{2\nu+4}$. This is convenient for our purposes.
 Vectors $\pmb{f}_j, j =0,1,2$ are defined uniquely by these conditions.

 We consider the image $\pmb{z}' = \frak{P}_2( \overline{\pmb{X}}_{2\nu+4})$ which is well-defined and for which we have
 $$
\pmb{z}' =  r' \pmb{f}_0+
 s'\pmb{f}_1+t' \pmb{f}_2 ,
 $$
where
 \begin{equation}\label{rst1w}
 r'\asymp \frac{q_{2\nu+4}^{1+\lambda \frak{g}_1}}{\Delta} ,\,\,\,\,\,\,
 s'= \frac{1}{\Delta},
 \,\,\,\,\,\,
 t' \asymp \frac{1}{\Delta}.
 \end{equation}
 Note that  as $\overline{\pmb{X}}_{2\nu+4}\in S$, so does $ \pmb{z}' \in S$. From the last equality \eqref{g11} we see that
   \begin{equation}\label{rst2w}
   r'\asymp q_{2\nu+4}^{\frak{g}_1}.
   \end{equation}
   The image $ \frak{P}_2( \overline{\mathcal{W}_{2\nu+4 }})$ is an ellipsoid. Its larger major axis is
   $\asymp r' $ and is larger than $ 6q_{2\nu+4}$. The section $ \{ x_0 = r'\}$ of the ellipsoid is a ball of radius
   of order
   $\asymp s' =\frac{1}{\Delta}\asymp q_{2\nu+3}^{-1+\lambda+\frac{\lambda}{\frak{g}_1}} $. From the right inequality from \eqref{prw} we see that $\xi_{2\nu+1} \leqslant q_{2\nu+2}^{-\lambda} = o(s)$.
   So  the radius  of the section $ \{x_0 = r' \} $ of the ellipsoid is $ \geqslant 6\xi_{2\nu+1}$. By convexity
   $$
   \frak{P}_2( \overline{\mathcal{W}_{2\nu+4 }})\supset \pmb{z}' + \mathcal{C}
   ,$$
   and the lemma is proven as  $\mathcal{C}$ contains a fundamental domain of lattice $\Gamma$.
   \end{proof}

From Lemma \ref{Lemma5} we see that  $ \frak{P}_2( \overline{\mathcal{W}_{2\nu+4 }})$ contains a shift of a fundamental domain of the lattice $\Gamma$ and lattice $\Gamma_1$ is congruent to $\Gamma$. So $ \frak{P}_2( \overline{\mathcal{W}_{2\nu+4 }})$
 contains an integer point. This is just the point $\pmb{z}_{2\nu+5} = (q_{2\nu+5} , a_{1,2\nu+5}, a_{2,2\nu+5}a_{3,2\nu+5})$ which we are constructing.

 As we defined $\pmb{z}_{2\nu+5}$, we may consider angular neighborhood $\mathcal{U}_{2\nu+5}$.
 From the  inequalities on  parameters \eqref{rst1w} and triangle inequality we see that
 $\mathcal{U}_{2\nu+5}\subset \mathcal{W}_{2\nu+4}$. Moreover, it is clear from the definition of the angular neighborhood
  $\mathcal{U}_{2\nu+5}$ that for all $\pmb{\eta} \in \mathcal{U}_{2\nu+5}$ vectors $\pmb{z}_{2\nu+4}$ and $\pmb{z}_{2\nu+5}$ are successive best approximation vectors to $\pmb{\eta}$.

 \vskip+0.3cm
  Now we will verify the rest of conditions ({\bf i}) - ({\bf vii}).

 \vskip+0.3cm

It is clear that the triple $\pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}, \pmb{z}_{2\nu+5}$ is primitive and both quadruples
$$\pmb{z}_{2\nu+1}, \pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}, \pmb{z}_{2\nu+5}\textrm{ and } \pmb{z}_{2\nu+2}, \pmb{z}_{2\nu+3}, \pmb{z}_{2\nu+4}, \pmb{z}_{2\nu+5}$$
 form a basis of $\mathbb{Z}^4$.

 As for the value of $r'$ we have inequality \eqref{rst2w} and $\pmb{z}_{2\nu+5}\in \pmb{z}'+ \mathcal{C}$, we see that
 $$
 q_{2\nu+5}\asymp q_{2\nu+4}^{\frak{g}_1},
 $$
 and we get (second) inequality from the condition ({\bf iv}) for the next step.

  So we have constructed two vectors $\pmb{z}_{2\nu+4}, \pmb{z}_{2\nu+5}$ which determine the next collection
  $$
   \pmb{z}_{2\nu+1}, \pmb{z}_{2\nu+2}, \pmb{z}_{2\nu+3},
  \pmb{z}_{2\nu+4}, \pmb{z}_{2\nu+5},
  $$
  and verified  conditions ({\bf i}) - ({\bf v}).

  Geometric  condition ({\bf vi})  for the angle between vectors
  $\pmb{Z}_{2\nu+3}-\pmb{Z}_{2\nu+5}$ and
  $\pmb{Z}_{2\nu+4}-\pmb{Z}_{2\nu+5}$
  is valid because of Lemma \ref{Lemma3} as we take $ \pmb{Z}_{2\nu+5} \in \mathcal{W}_{2\nu+4}$.

  As for ({\bf vii}), it immediately follows from ({\bf vi}) by means of the following auxiliary statement.

 \begin{lemma} \label{Lemma6}
  Consider the three-dimensional parallelepiped
  $$
  \Pi = \{ \pmb{z} = \lambda_1 \frak{z}_1+ \lambda_2  \frak{z}_2 + \lambda_3  \frak{z}_3,\,\,\,
  0\leqslant \lambda_j \leqslant 1 \}
  $$
  generated by  linearly independent vectors $ \frak{z}_j = (q_j,a_{1,j}, a_{2,j},a_{3,j})
   \in \mathbb{R}^4 , \, j =1,2,3$.
  Let $\frak{Z}_j = \left(\frac{a_{1,j}}{q_1},\frac{a_{2,j}}{q_2}, \frac{a_{3,}j}{q_3}\right)
  $ be the corresponding points on $\mathbb{R}^3$.
  Then
  $$
  {\rm Vol} \, \Pi \gg q_1q_2q_3\cdot |\frak{Z}_1- \frak{Z}_3|\cdot |\frak{Z}_2-\frak{Z}_3|\cdot \sin \varphi
  $$
  where $\varphi $ is the angle between $\frak{Z}_1- \frak{Z}_3$ and $\frak{Z}_2- \frak{Z}_3$.
  \end{lemma}

\begin{proof}We consider linear transformation $\mathcal{G}$  such that
$$
\mathcal{G}\frak{z}_j =  \overline{\frak{Z}}_j = \left(1, \frac{a_{1,j}}{q_1},\frac{a_{2,j}}{q_2}, \frac{a_{3,}j}{q_3}\right),
\,\,\,\,
j = 1,2,3.
$$
Then
$$
 {\rm Vol} \, \mathcal{G} \Pi = \frac{1}{q_1q_2q_3} {\rm Vol} \, \mathcal{G} \Pi .
 $$
 But  $\mathcal{G} \Pi$ contains simplex with vertices $\pmb{0},  \overline{\frak{Z}}_1, \overline{\frak{Z}}_2, \overline{\frak{Z}}_3$ whose volume is greater than
 $$ \frac{ |\frak{Z}_1- \frak{Z}_3|\cdot |\frak{Z}_2-\frak{Z}_3|\cdot \sin \varphi}{6}.$$
 Lemma is proven.
 \end{proof}

    \vskip+0.3cm
    Now we finish with ({\bf vii}).
  Indeed, let
  $\Delta_*$  be the fundamental volume
  of the lattice
  $$
  \Gamma_* =
   \langle\pmb{z}_{2\nu+3},\pmb{z}_{2\nu+4},\pmb{z}_{2\nu+5}
\rangle_\mathbb{Z}  = \mathbb{Z}^4 \cap  \langle\pmb{z}_{2\nu+3},\pmb{z}_{2\nu+4},\pmb{z}_{2\nu+5}
\rangle_\mathbb{R}
  .
 $$
 We apply Lemma \ref{Lemma6} to the vectors $\frak{z}_j =\pmb{z}_{2\nu+2+j}$ and take into account   ({\bf vi})  together with the inequalities
 $$
  |\frak{Z}_j- \frak{Z}_3| =
  |\pmb{Z}_{2\nu+2+j} - \pmb{Z}_{2\nu+5}| \asymp \frac{ \xi_{2\nu+2+j}}{q_{2\nu+2+j}},\,\,\,\, j = 1,2
  $$
  to obtain
  $$
  \Delta_* \gg q_{2\nu+2}q_{2\nu+4}q_{2\nu+5}  \cdot  \frac{ \xi_{2\nu+3}}{q_{2\nu+3}}  \cdot \frac{ \xi_{2\nu+4}}{q_{2\nu+4}} = \xi_{2\nu+3}\xi_{2\nu+4}q_{2\nu+5}.
  $$
  The upper bound
   $$
  \Delta_* \ll  \xi_{2\nu+3}\xi_{2\nu+4} q_{2\nu+5}
  $$
  is standard.

  We have checked all the conditions of the next step of indiction. Thus, Theorem \ref{Thm2} in the case $k=1$ is proven.$\Box$

\section{Comments on the case $ k>1$}\label{9}

In the case $k=1$, we constructed vector $\pmb{\alpha}$ where best approximations form a periodic sequence of patterns $AB$, where growth of denominators $q_\nu$
 is determined by conditions ({\bf iv}) and ({\bf v}). That is, on the even step  ({\bf Stage 1}) the exponent of growth was equal to $\frak{g}_1$ and on the odd step  ({\bf Stage 2}) it is $\frak{g}_{1,0}$ defined in Lemma \ref{Lemma1} and \eqref{g11} respectively. The possibility of \emph{gluing}  pattern $A$ and pattern $B$ is enabled by equalities in  \eqref{g11} which comes from the equation \eqref{xRoy}. This alternance is enclosed in the sequence $u_j$ defining the sets $\mathcal{U}_j$, which is periodic with period of length $2$.
  To carry out the construction for arbitrary $k>1$ we simply need to construct periodic patterns
   $ \underbrace{A\ldots A}_{k\, \text{times}}B$
   corresponding to triples \eqref{triples} with $ \nu_{j-1} =\nu_j +1,  1\leqslant j \leqslant k$. For this, we consider the sets $\mathcal{U}_j$ defined by a periodic sequence $u_j$ with period of length $k+1$, defined by
    $$
 u_j =
 \begin{cases}\label{uj}
 1+ \lambda \frak{g}_{k,l},\,\,\,\,\,\,\, j \mod k+1 = l
  \cr
 1+ \lambda \frak{g}_k, \,\,\,\,\,\, j \mod k+1 =k
 \end{cases}
 $$
   where $\frak{g}_k$ is defined in Lemma \ref{Lemma1} while $\frak{g}_{k,l}$ for $0\leqslant l \leqslant k-1$ are defined in \eqref{anyg}. In the inductive construction, at step $j$ we apply ({\bf Stage 2}) if $ j \mod k+1 =k$ and $({\bf Stage 1})$ otherwise. In this construction the last two elements of each triple
 $ \pmb{z}_{\nu_j-1}, \pmb{z}_{\nu_j}, \pmb{z}_{\nu_j+1}$
 are just the first two elements of the triple
  $ \pmb{z}_{\nu_{j-1}-1}, \pmb{z}_{\nu_{j-1}}, \pmb{z}_{\nu_{j-1}+1}$,
  that is $\nu_{j-1} = \nu_j +1$.
 The possibility of gluing together separate  patterns $A$ and $B$ into
 $ \underbrace{A\ldots A}_{k\, \text{times}}B$  follows from \eqref{glue} because of the identities
 $$
 q_{\nu_j+1}^{(1-\lambda)\frak{g}_{k,j}-1} \asymp \frac{\Delta}{d_j}
, $$
where
$
\Delta
$ is the height of the three-dimensional rational subspace containing
all the triples $ \pmb{z}_{\nu_j-1}, \pmb{z}_{\nu_j}, \pmb{z}_{\nu_j+1},
0\leqslant j \leqslant k$ and
$
d_j \asymp
 q_{\nu_j+1}^{1-\lambda}
 $
 is the height of  the two-dimensional subspace  generated by vectors
 $ \pmb{z}_{\nu_j}, \pmb{z}_{\nu_j+1}$.

 The conclusion $ \frac{\omega(\pmb{\alpha})}{\hat{\omega}(\pmb{\alpha}) } =\frak{g}_k
$ follows from \eqref{ssdd}.\\

 {\bf Acknowledgment}
 The authors would like to thank the \emph{Centro Internazionale per la Ricerca Matematico} (CIRM) in Trento for the opportunity of a \emph{Research in Pairs} stay from May 19th to June 1st 2019. Many ideas of this paper appeared during this fruitful time.

redpajama_set_name: RedPajamaArXiv


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text:
Forces Children's Trust funds boosted by RAF Valley
2nd August 2014 FCT Admin News
WHEN ex-RAF Valley Master Aircrewman Antz Males lost his life in tragic circumstances in 2012, the Forces Children's Trust was a great help to his widow Bridget and two young children Jack and Macey. The Trust provides for a range of activities to support children of bereaved service personnel, from Outward Bound expeditions to counselling and emotional support.
Friends and colleagues at the Valley-based Search & Rescue Training Unit (SARTU) – which trains helicopter crews from all three armed services in vital search and rescue techniques – rallied round to organise a charity boxing night at the Anglesey base to raise money for the trust. Master Aircrewman Simon Allen explains:
"Antz was heavily into his boxing and had organised a charity boxing night with some SARTU pals in 2011 which was a great success. So to honour his life and raise funds for the Forces Children's Trust we decided to do another one. We had help from the RAF Boxing Association, and over sixty local businesses generously put up auction prizes, as well as Brietling and the London Boxing Club. The event raised £3750 which will be a real help to the FCT which does some tremendous work. I'd like to thank all my friends and colleagues from SARTU, RAF Valley, the Llandudno Boxing Club and our catering and leisure contractors ISS who worked so hard to make this event such a success"
The Forces Children's Trust is devoted to helping children whose father or mother has died, or has sustained life changing injuries whilst serving as a member of the armed forces. The charity, by reason of its aims and size, has the flexibility and ability to make decisions and offer help as needed, with minimum fuss. More details are available on its website: www.forceschildrenstrust.org
Image: Organisers of the Charity Boxing event in front of two Griffin helicopters of RAF Valley's Search & Rescue Training Unit (SARTU)
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Home / News / Moffett Wins Lakeland Stages Rally Despite Last Minute Dramas!
Moffett Wins Lakeland Stages Rally Despite Last Minute Dramas!
Updated: 2 September 2018
(Image: PhilPics.net) There was drama after the final stage of the Trailer Parts & Spares Lakeland Stages Rally as the long-time leader, Josh Moffett, struggled to complete the final road section when his Fiesta ground to a halt…
"It could have been temperature related but we don't really know yet," Moffett said. "We got through the final stage without any trouble but about halfway through the road section to the finish, the car just stopped. Then it went, and then it stopped again."
Moffett and navigator Stephen Thornton had to push their car to the final control and were four minutes late, incurring 40 seconds worth of road penalties. However, luck was on their side as they had extended their lead to 41 seconds after the final stage. That meant the crew won the rally in their Fiesta R5 by just one second!
"Unfortunately, Barry McKenna and Andrew Purcell dropped out early on so it made our day a bit easier," Moffett said. "It's always good to have a race with those guys but it was good to get the win!"
In a final twist, event organisers received a protest alleging that the winning crew had received 'outside assistance' to get to the final time control – something that isn't permitted within rally regulations. However, after much deliberations, Moffett and Thornton were eventually confirmed as the winners around 11pm.
The event hosted round five of the McGrady Insurance N.I. Rally Championship and it was Desi Henry who arrived home as the series' top competitor in second overall. He and Liam Moynihan nabbed second overall after a push on the final stage saw them displace Vivian Hamill and Andrew Grennan from the runner-up position.
"We had a problem on the first loop that we couldn't figure out until we got to service," Henry explained. "One of the sensors wasn't working so we were down on power. We got that fixed and I think we were second quickest over the last three stages."
Henry's increased pace didn't come without its dramas as a heavy landing over one of the jumps resulted in some technical troubles.
"We had a bit of a scare when the car reset itself and turned off stage mode," Henry explained. "We didn't know what had happened but as soon as we switched 'stage mode' back on we were fine."
Vivian Hamill and Andrew Grennan were disappointed to drop a position on the final stage but it was still a fine display from the crew who were only contesting their second event with their new Ford Fiesta.
Cathan McCourt and Barry McNulty finished fourth in their Fiesta on a day where McCourt was also trying to get used to his car, while Adrian Hetherington and Gary Nolan successfully brought their battle-scarred Corolla WRC to the finish in fifth.
"We changed our tyres for the second loop and that gave us a lot more grip," Hetherington said. "I really enjoyed that!"
Derek McGarrity and Paddy Robinson survived a stage four puncture, losing around six seconds, to come home in sixth overall, picking up plenty of valuable N.I. Championship points.
Alan Smyth and Mac Kierans secured seventh overall in their Evo X while N.I. Championship hopeful, Alan Carmichael, ended the day with a smile on his face after pushing hard over the big jumps on stage six.
"We had a lot of problems on the second loops but it wasn't as bad as this morning," Carmichael said. "We had a bit of fun on the last stage and hit the big jumps as hard as we could!"
Stephen McCann and Kaine Trainor sealed ninth overall in their Fiesta R5 after a "much cleaner run" over the second loop of stages while Niall Henry and Damien Duffin took their Citroen DS3 R5 into 10th overall despite breaking his power-steering on the final stage.
Two-Wheel Drive
In the battle among the two-wheel drive crews, David Crossen and Aileen Kelly secured the win, as well as 11th overall, despite a scare during the closing stages.
"We broke the shock on stage five," Crossen explained. "We lost about 10 seconds to Frank (Kelly) but we still managed to win the category. We were on a wing and a prayer over some of the bumps!"
Frank Kelly, who finished second in the two-wheel drive category and 12th overall, was also on a "wing and a prayer" over one particular section…
"We were on the grass at very inappropriate speeds!" Kelly laughed. "I tried to take a corner flat in sixth gear and I realised when I got there that it wasn't really flat! So I caught the grass, and we ended up on the grass the whole way around the corner. The car didn't want to come out of it but luckily there was nothing solid to hit."
Father and son, Jordan and Paul Hone, completed the top three 2WD finishers after a fantastic drive in their Opel Adam R2. They fended off the hard-charging Oran Donnelly and Mo Downey who were lucky to escape unscathed after a stage one brush with the scenery.
"I put slightly narrower tyres on the back of the car for the second loop," Donnelly explained. "I think it was a combination of the tyres and a cleaner road that made it feel really nice."
Group N
In the Production category, Liam Regan and Gary McElhinney claimed a well-deserved win after early leader, Niall McGonigle, retired during the final stage with a suspected driveshaft failure. In the end, Regan held off Conor McCourt and Caolan McKenna to seal the win, as well as 13th overall, by 25 seconds.
Other retirements during the final loop of stages included Paul McCann who had two punctures and not enough spares. The same drama befell Colm McFall who got a puncture on stage five and then another one on stage six, putting him out of the rally.
Top two-wheel drive competitor, Shane McGirr, also retired during the final loop of stages while Barry McKenna, who rejoined the event under 'Rally 2' after his stage one excursion, lost an argument with a tree stump during the second loop but was lucky to escape any major damage.
The penultimate round of the McGrady Insurance MSA N.I. Rally Championship takes competitors to Omagh for the Bushwhacker Rally on 22 September. More details can be found on www.nirallychampionship.com.
Lakeland Stages Rally Results (Top 10)
Full results available from www.rallyscore.net.
1 Josh Moffett/Stephen Thornton (Fiesta R5) 32m17s
2 Desi Henry/Liam Moynihan (Fabia R5) 32m18s
3 Vivian Hamill/Andrew Grennan (Fiesta R5) 32m20s
4 Cathan McCourt/Barry McNulty (Fiesta R5) 32m34s
5 Adrian Hetherington/Gary Nolan (Corolla WRC) 32m35s
6 Derek McGarrity/Paddy Robinson (Focus WRC) 32m55s
7 Alan Smyth/Mac Kierans (Evo X) 33m19s
8 Alan Carmichael/Ivor Lamont (Mini WRC) 33m25s
9 Stephen McCann/Kaine Treanor (Fiesta R5) 33m42s
10 Niall Henry/Damien Duffin (DS3 R5) 33m48s
← Previous Story SS3: Moffett Surges Into 8 Second Lakeland Lead
Next Story → Bushwhacker Rally: Regulations and Entry Form


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Wells Fargo must face litigation on defective mortgages: US judge
Published Fri, Mar 31 20177:07 AM EDT Updated Fri, Mar 31 20177:08 AM EDT
Rick Wilking | Reuters
A federal judge on Thursday said Wells Fargo
must face litigation seeking to hold it responsible for billions of dollars of claimed investor losses stemming from its alleged failures as a trustee overseeing risky residential mortgage-backed securities.
U.S. District Judge Katherine Polk Failla in Manhattan said the plaintiffs, including a few dozen funds from BlackRock
, Pacific Investment Management Co., Prudential Financial
, and TIAA-CREF, may pursue breach of contract and conflict of interest claims related to 53 trusts.
Failla also said the investors may pursue some claims alleging breaches of fiduciary duty and due care, but she dismissed claims alleging general negligence and the violation of a New York law governing mortgage trusts. Failla also denied Wells Fargo's bid to dismiss claims by Germany's Commerzbank.
Ancel Martinez, a Wells Fargo spokesman, declined to comment. A lawyer for many of the plaintiffs did not immediately respond to requests for comment.
Failla's 80-page decision covers five lawsuits, which comprise one of the largest remaining pieces of U.S. litigation seeking to hold banks liable for risky mortgage securities that were a major cause of the 2008 global financial crisis.
Much of this litigation targeted lenders, but some targeted trustees that oversaw the securities' performance.
Investors accused Wells Fargo of having taken "virtually no action" to require lenders to buy back or fix defaulted or poorly underwritten loans that backed their securities, despite knowing of shortfalls.
They said the San Francisco-based bank's resistance stemmed from concern that acting would have exposed its own "misconduct" in other residential mortgage-backed securities trusts, and jeopardized its business dealings with lenders and servicers, court papers show.
Failla said the plaintiffs "more than met" the legal standard for letting the breach of contract claims proceed, having pointed to internal Wells Fargo documents to suggest the bank knew about many loan defects but did nothing.
"It is plaintiffs' contention that such allegations go far beyond many other RMBS trustee complaints, which themselves have been found sufficient to state a claim," Failla wrote, without ruling on the merits. "The court agrees."
Failla also said the National Credit Union Administration may pursue various claims against Wells Fargo on behalf of five failed credit unions.
The NCUA has already recouped roughly $4.3 billion in litigation against many banks over securities that the credit unions bought. A spokesman could not be reached for comment.
The cases in the U.S. District Court, Southern District of New York are: BlackRock Allocation Target Shares Series S Portfolio et al v. Wells Fargo Bank NA et al, No. 14-09371; Royal Park Investments SA/NV et al v. Wells Fargo Bank NA et al, No. 14-09764; National Credit Union Administration Board v Wells Fargo Bank NA, No. 14-10067; Phoenix Light SF Ltd et al v. Wells Fargo Bank NA, No. 14-10102; and Commerzbank AG v. Wells Fargo Bank NA, No. 15-10033.


California BSIS Licensed security officers are required to complete an 8 hour annual "refresher" training course to maintain your license.
BSIS can require proof of completion of this training from you at any time. Employees of Monument Security are required to submit proof of completion annually to maintain employment. This class is open to all BSIS licensed guards regardless of employer. Fees are non-refundable and must be paid in advance to hold your seat. Walk-ins are accepted if space is available on the day of the class. Completion certificates are sent within 10 days of class completion.
Please see the Calendar below for scheduled classes. Click on the event for detailed information and map information. Classes are updated monthly and classes are added as they are filled so please check the calendar regularly.


text:
Inquiry - Freedom of Information Amendment (Administrative Arrangements) Bill 2014
For the following reasons, Liberty Victoria is opposed to this Bill:
1. The Bill seeks to repeal entirely the scheme for Freedom of Information (FOI) Review set in place in 2010. It is a return, therefore, to a system of external review that had been found wanting only four years ago. The 2010 administrative reforms sought to achieve a number of very important objectives, each of which were designed to facilitate the public's right of access to governmental information. In summary these objectives were the following:
a. To strengthen the publication requirements of the FOI Act so as to make access to non-exempt documentation easier by dispensing with the need for individuals to apply for it in accordance with the FOI Act's legal requirements.
b. To make external review of departmental and agency decisions more expert and informed by creating the Office of Information Commissioner with terms reference confined to FOI and Privacy alone.
c. To reduce the cost to the public of applying for departmental and agency documentation by, among other things, reducing processing charges and excluding application fees for requests for personal information.
d. To provide for the systematic monitoring and reporting of the FOI regime by conferring powers on the Information Commissioner to undertake reviews of adverse FOI decisions; to audit departments' and agencies' systems for FOI implementation; to establish guidelines for the interpretation of the FOI Act's provisions; and to provide independent annual reports on the Act's implementation across government.
e. On the basis of this activity to make independent and impartial recommendations for the legislation's improvement, legislatively and administratively.
2. The present Bill repeals the administrative arrangements sought to achieve these worthwhile objectives. The Bill:
a. Abolishes the office of the Australian Information Commissioner while leaving the Office of the Privacy Commissioner intact.
b. External review of decisions to refuse access to governmental information will be returned to the Administrative Appeals Tribunal (AAT).
c. The Commonwealth Ombudsman will once again become responsible for investigating complaints about the way in which departments and agencies handle FOI requests. That is, the Ombudsman will assume responsibility for determining whether FOI procedures have been implemented appropriately.
d. The Attorney-General will take responsibility for the monitoring and reporting functions presently residing with the Information Commissioner. These include, for example, the drafting of FOI guidelines for agencies, the collection of statistics disclosing trends in requests for access, refusals of access, applications for external review and other similar matters, and the presentation of annual reports.
3. The defects in the return to the old system are readily apparent. These are the principal ones:
a. The abolition of the Office of Information Commissioner will increase the costs of external merits review. Instead of costs being minimized, applications for governmental information will now be required to pay the normal application costs for AAT review. The base application fee for such reviews is in the vicinity of $800. The fee is high enough to deter most ordinary members of the public from pursuing external review. This is, plainly, a blow to open and transparent government.
b. The AAT is a generalist mechanism for the external merits review of governmental decisions. While Liberty has every confidence in the capacity and impartiality of the AAT's membership, it is inevitable that the legal and administrative expertise that presently resides in the Office of the Information Commissioner will be lost as cases are referred to and heard by AAT members who, speaking generally, will be far less informed as to the law and practice of FOI in government than the Commissioner's office is. Again, 'freedom' of information will be compromised.
c. The return of the complaints jurisdiction to the Office of the Ombudsman will also have the effect of diluting the specialist administrative expertise that the implementation of FOI legislation requires. One of the supposed benefits of the return to the old system is that it may reduce complexity and duplication. On the contrary, the coordinated administration of the substantive and procedural review of governmental decisions situated within the Information Commissioner's office is to be bisected and dismantled. Jurisdictional complexity will increase. Liberty notes also that the Office of the Ombudsman is heavily over-worked and very substantially under-funded. The additional responsibility of handling complaints will only exacerbate this situation.
d. Under the proposals, the Attorney-General will take over responsibility for monitoring FOI's implementation and reporting upon its legal and administrative adequacy. The problem with this is readily apparent. The Office of the Information Commissioner was established to ensure that these functions are undertaken independently and impartially by an Ombudsman type body free from the political influence of politicians; and decisions taken in the interest of the administrative convenience of departments and agencies. Such independence and impartiality is critical to the effective operation of open government legislation. It is now to be sacrificed to the very entities in whose interests the limitation of access to governmental information will, from time to time, be prevalent. FOI guidelines can be politicised. FOI statistics can be massaged. FOI reports can be rendered anodyne. These are consequences that ought not to be contemplated. Each will now be distinctly possible should the old regime return.
4. We note that the comprehensive review of the FOI Act, conducted very recently by Allan Hawke AO, concluded that the Office of the Information Commissioner had been a valuable addition to the manner in which the legislation was implemented. He concluded that:
"The review considers that the establishment of the OAIC has been a very valuable and positive development in the oversight and promotion of the FOI Act."
"In essence the review found that the recent reforms were working well and have had a favourable impact in accordance with their intent".
Given these findings by an individual who has occupied the highest positions in Commonwealth administration, it becomes even more perplexing as to why it is that the present Government has chosen to return to an old system, that was sorely in need of reform, that was reformed, and that has by common consent operated efficiently and effectively.
Regrettably, the only feasible rationale for this backward Bill is that it is designed to return the control of access to governmental information, and in particular politically and administratively sensitive information, to those in whose interests secrecy lies.
Should you wish to discuss any aspect of this submission further please contact Prof Spencer Zifcak.


Choosing among stencils for tattoos in Fort Lauderdale can be quite a fun activity, especially if you have yet to get a tattoo. You can select from thousands of stencils or customize your own design.
In order to select tattoos in Ft. Lauderdale, you need to find an area tattoo artist who is well-versed in tattooing and comes highly recommended. You will minimize the pain and nervousness that comes with getting a first tattoo by planning ahead and using the services of a reputed artist.
Indeed, you can obtain awesome tattoos in Fort Lauderdale and the surrounding area by choosing such shops as Miami's Salvation Tattoo Lounge. The tattoo parlor features skin art for anyone seeking tattoos in Fort Lauderdale and beyond. You can also obtain body piercing or permanent makeup at the Miami shop, too. So, if you wish to obtain one of the other services at a later date, it may be to your benefit to choose an all-inclusive shop like Salvation.
With that being said, you are probably wondering about how the process will proceed once you choose from one of the tattoos in Fort Lauderdale that are featured in area shops. A good tattoo artist is adroit at handling a needle. He should never draw an excess amount of blood. Therefore, find an artist who has a lighter touch when he operates the machine.
That's because the ink is applied to the area of skin known as the dermis – a layer that is a tier above the nerves and vessels in the skin. The ink is never applied to the epidermis, or outer layer, as the skin's top layer sheds. Therefore, it is not a good practice to apply tattoo ink to an area where it can easily fade.
If this is your first venture in choosing tattoos in Fort Lauderdale, you will need to arm yourself with some information about the shop where you want to obtain your tattoo and get some details about the artist. What are some of the things people say about him or her? Does he have a good in-shop reputation? Don't ever permit any artist who makes you feel uncomfortable proceed and outline the stencil or ink in the image for your tattoo.
Therefore, when shopping for tattoos in Fort Lauderdale and beyond, you will want to take a careful look at the shop you are considering. Again, check out such tattoo facilities as the Salvation Tattoo Lounge. See what the shop features among its services, and find out if it is well-respected in the community. By doing so, you can make obtaining a tattoo a less taxing experience.


Bone ash is not common in glazes. When employed it can cause the slurry to flocculate and thicken (and produce a very thick layer on the ware surface which cracks during drying). People often react to this by adding more water, producing a glaze that shrinks even more on drying and eventually thickens again. A better way is to add a little deflocculant to the glaze slurry (like Darvan).
Bone ash or calcium phosphate are used to opacify opal glass (1-3%) because the P2O5 content forms colorless compounds with iron impurities.


text:
NewsPositively Hampton Roads
Photos: US Women's soccer team celebrates fourth World Cup win
The US Women's National Team won its second consecutive FIFA Women's World Cup Sunday, beating the Netherlands 2-0 and securing its fourth World Cup overall.
On Wednesday, the soccer team was celebrated with a ticker tape parade in New York City.
The last ticker tape parade was held in July 2015 after the women' national team defeated Japan 5-2 to win the World Cup, according to the Alliance for Downtown New York, a commercial advocacy, research, and information group. The team was the first female athletes to be honored with a ticker tape parade at the time, the Alliance said.
The US Women's National team also won the World Cup in 1991 and 1999.
Related: US women's soccer team celebrates World Cup success with parade in NYC


text:
Q: Uninstall Berkeley DB in ubuntu I installed Berkeley DB 6.2.23 form this tutorial. Also before it, I installed libdb5.3 from ubuntu repositories. Now when I'm running a Perl script in command line, I got this error:
DB_File was build with libdb version 6.2.23,
but you are attempting to run it with libdb version 5.3.28
Compilation failed in require at ./ROUGE-1.5.5.pl line 174.

I guess installation of these two packages is the problem, but during removing libdb5.3 there are many dependency error :
The following packages have unmet dependencies:
 adduser : Depends: passwd (>= 1:4.1.5.1-1.1ubuntu6)
 libkf5service5 : Depends: libkf5dbusaddons5 (>= 4.97.0) but it is not going to be installed
                  Recommends: libkf5service-bin (= 5.18.0-0ubuntu1) but it is not going to be installed
 libpam-modules : PreDepends: libdb5.3 but it is not going to be installed
                  Recommends: update-motd but it is not going to be installed
 libqt5svg5 : Depends: libqt5widgets5 (>= 5.3.0) but it is not going to be installed
 libqtgui4:i386 : Depends: libsm6:i386 but it is not going to be installed
 libsm6 : Depends: libuuid1 (>= 2.16) but it is not going to be installed
 libxt6:i386 : Depends: libsm6:i386 but it is not going to be installed
 openssh-client : Depends: passwd
 procps : Depends: initscripts
 python3 : Depends: python3.5 (>= 3.5.1-2~) but it is not going to be installed
           Depends: libpython3-stdlib (= 3.5.1-3) but it is not going to be installed
           Depends: dh-python but it is not going to be installed
 python3-requests : Depends: python3-urllib3 (>= 1.13.1) but it is not going to be installed
                    Depends: python3-chardet but it is not going to be installed
                    Depends: python3-urllib3 (< 1.13.2) but it is not going to be installed
 wget : Depends: libuuid1 (>= 2.16) but it is not going to be installed
E: Error, pkgProblemResolver::Resolve generated breaks, this may be caused by held packages.

Because of that, I want to uninstall version 6.2.23 from my system, and want to know how can I do it.

A: I has solved the problem that you met. Because the database has some dependencies, it is hard to install the database with a usual way. You need to install the database according to the following guidelines.
1. Install berkeley-db
Please install the berkeley-db with the commands as follows :
$ wget http://download.oracle.com/berkeley-db/db-4.8.30.zip
$ unzip db-4.8.30.zip
$ cd db-4.8.30/
$ cd build_unix/
$ ../dist/configure --prefix=/usr/local --enable-cxx
$ make -j12
$ make install
$ sudo make install

The above commands enable to install the libraries that the following bsddb3 needs.
2. Install bsddb3-6.26 :
$ wget https://files.pythonhosted.org/packages/e9/fc/ebfbd4de236b493f9ece156f816c21df0ae87ccc22604c5f9b664efef1b9/bsddb3-6.2.6.tar.gz
$ tar -xvzf bsddb3-6.2.6.tar.gz
$ cd bsddb3-6.2.6/
$ python setup.py install --berkeley-db=/usr/local/

After completing the installation, it will show the following message.
Processing dependencies for bsddb3==6.2.6
Finished processing dependencies for bsddb3==6.2.6.
Cheers,
Mike

redpajama_set_name: RedPajamaStackExchange


Today I am sharing my story of how I preserved my husband's family history at Heidi Swapp's blog! You won't want to miss out on reading the steps I took to get it done!
One way to make sure this big project got done was to delegate a family historian to be in charge of communicating with the entire family. Cousin Susan was a huge help and my number one contact. When I had a question, I would go to her and she would try and find the answer. Here are the steps in which I had her help.
1. Email the Family Questionnaire with a deadline of three weeks to have them returned (three weeks seemed long enough and a month it was too easy to forget about).
2. Have the family email them to her when they were finished.
3. She proofed the questions before forwarding them on to me, just in case she saw something that needed to be clarified.
4. I used quotes from the questionnaire to match with the correct family photos I had scanned and created an album. She proofed the album to make sure my work was correct.
There of course wasn't enough room in the album to include everyone's life stories. So, I started a Blurb book, placed everyone's stories inside, and printed a paperback. Just like that, everyone's story has been preserved!
- Have a back up plan for those family members that do not type well on the computer. Delegate a family member to type up their written answers for you.
- Have a friend that is an editor? Send them the full copy text via email to proof for you. Reading an email is different than holding a book in your hand. The words and sentences need to flow better in a book. It is also better for a non family member to proof the text so that generations later will understand the abbreviations, nicknames, and family humor.
- Set a deadline for your final copy to be finished.
- Be sensitive to those family members that might have hard feelings toward other family members. Keep your questions "neutral" while also giving an open ended question for the family to share more if they want to. If you know of family members that might be offended, remove questions that might bring up bad memories for them. Have your delegated family member proof your questions first before emailing them out to the family. When you receive the answers back, if there is more you would like to know about something specific, ask that family member directly.
Here are the questions I asked the family. Feel free to use these questions to kick off your own family questionnaire!
How are you related? Please write an auto biography of YOU, your marriage, children, grandchildren.
Please identify and write about the other siblings and write specific memories of them, IE. Gussie, Dorothy, Golda, Florence etc. No memory is too small, writing something about them all would be wonderful, just identify who you are writing about.
What do you remember about your family relationship, Christmas' traditions that were passed down through the family, stories of things you did because they were tradition?
Do you have a piece of clothing or jewelry that holds significance to you?
Do you remember hearing stories how it was like for them to live through the Depression?
Do you remember stories of their first car ride, electrical appliance etc?
Do you remember what their early married life was like/challenges?
Describe a typical family dinner, what was a favorite family meal?
Recipes that could be passed down?
What is the first family picnic you remember?
What picnic or picnics are the most memorable?
Did your family go every year, if not why?
What is the last picnic you attended?
Write anything you feel would be of interest to the history of this family.
Create Well: Preserving your family's history doesn't just bless the family you are doing it for. It will bless your children and their children. It is something you will do in your lifetime that will never be forgotten.
Thanks so much for sharing the process with me, What an awesome treasure. I'm in the middle of scanning all of my Grandmother's photos and needed some inspiration to push forward onto the next step. I am so grateful for your sharing this and yes, what a treasure it is to have this preserved for you and generations to come. Thanks again.
Thanks for sharing your process and questions, this was an amazing thing to do with photos that were no so familiar, what a lucky family you have that you were able to complete such a huge task.