text:
\section{Introduction}
Let  $s$ be a positive integer and $0\leq x < 1$ be a rational number.
In this article, we study the linear independence of values of the $s$-th Lerch function defined by
$$
\Phi_s(x,z) =\sum_{k=0}^{\infty}\frac{z^{k+1}}{{(k+x+1)}^s}~, \,\,\,\,\,\, z\in \mathbb{C}, \,\,\,  \vert z\vert <1\enspace.
$$
Note that in the case of $x=0$, we have $\Phi_s(0,z)={\rm{Li}}_s(z)$ where

\begin{equation*}
\textrm{Li}_s(z) =\sum_{k=0}^{\infty}\frac{z^{k+1}}{{(k+1)}^s}~, \,\,\,\,\,\,z\in \mathbb{C}, \,\,\,  \vert z\vert <1.
\end{equation*}
is the $s$-th polylogarithm function.

Let $r$ be a positive integer. We consider $r$ Lerch functions $\Phi_s(x,z), \,\,1\leq s \leq r$.

The linear independence of  $\textrm{Li}_s(\alpha)$ at one rational number $\alpha $, with $1\leq s \leq r$, was studied by
E.~M.~Niki\v{s}in \cite{N} in 1979.
It was generalized to the Lerch function by Kawashima \cite{Ka} and to algebraic cases by M. Hirose, M. Kawashima and N. Sato \cite{H-K-S}.
In 1990, M. Hata \cite{Ha} adapted generalized Legendre polynomials modifying Pad\'e type
constructions of G.~V.~Chudnovsky  developed in the series of articles
 \cite{ch2}, \cite{ch3}, \cite{ch9}, to obtain the linear independence of  $\textrm{Li}_s(\alpha)$
  (indeed of the Lerch transcendent function)
for different $s$ but at one rational number $\alpha$. His result implies the irrationality of $\textrm{Li}_2(1/q)$
with $q$ integer $q\geq 12$ whereas G.~V.~Chudnovsky's announced in  \cite{ch2} the irrationality of $\textrm{Li}_2(1/q)$
with $q\geq 14$.
Later, Hata gave in 1993 the irrationality of the value of $\textrm{Li}_2(1/q)$ in \cite{Ha1993}
with $q$ integer $q\geq 7$ or $q\leq -5$.

In 2005, Rhin and C. Viola \cite{RV1} adapted their permutation group method, established in 1996 \cite{RV0},
to get the irrationality of $\textrm{Li}_2(\alpha)$ for certain $\alpha\in\mathbb{Q}$, involving
the irrationality $\textrm{Li}_2(1/q)$ with $q\geq 6$, $q\in\Z$ in a qualitative and a quantitative forms. More recently in 2018,
Viola and W.~Zudilin \cite{VZ1} extended the permutation group method
with constructions
to establish
the linear independence of  $1, \, \textrm{Li}_1(1/q)$, $\textrm{Li}_2(1/q)$, $\textrm{Li}_2(1/(1-q))$ over $\mathbb{Q}$
with an integer $q\geq 9$ or $q\leq -8$ and more generally,  that of
$1, \, \textrm{Li}_1(\alpha)$, $\textrm{Li}_2(\alpha)$, $\textrm{Li}_2(\alpha/(\alpha-1))$ for certain $\alpha\in\mathbb{Q}$.

With respect to the linear independence of {\it distinct}  rational points,
G.~Rhin and P.~Toffin \cite{R-T} created a system of
Pad\'e approximants to show the linear independence
of the natural logarithms, at pairwise distinct $\alpha_1, \cdots, \alpha_m$,
either mainly rational or quadratic imaginary numbers, under a {\it metric} condition requiring the points to be very closed to
the origin $0$.

This proof opened a new path, albeit unexplored systematically during the next decades  to show the linear independence of logarithms over $\mathbb{Q}$ at
{\it distinct} $\alpha\in\mathbb{Q}$ relying only on Pad\'e approximations.
It is worthwhile noting that there is up to now no systematic method to check the linear independence of
polylogarithms over $\mathbb{Q}$

Since $\textrm{Li}_1(z)$ coincides with the usual natural logarithm, the Rhin-Toffin method suggests how to adapt
Pad\'e approximations to deal with the linear independence of polylogarithms at different points
$\alpha_1, \cdots, \alpha_m$.

Now, let $\alpha_1, \cdots, \alpha_m\in\mathbb{Q}$ be rational numbers, pairwise distinct, supposed to be
sufficiently closed to $0$ that we will precise later.

In this article, we give a criterion  as well as an outline of its proof to show the linear
independence of all the $rm+1$ numbers~:~ $\Phi_1(x,\alpha_1)$, \,$\Phi_2(x,\alpha_1), \,$
$\cdots , \ , \Phi_r(x,\alpha_1)$, \,\,$\Phi_1(x,\alpha_2)$, \,$\Phi_2(x,\alpha_2), \,$
$\cdots , \, \Phi_r(x,\alpha_2), \cdots \cdots, \Phi_1(x,\alpha_m)$,
~$\Phi_2(x,\alpha_m)$, ~$\cdots , \Phi_r(x,\alpha_m)$ and $1$, over an algebraic number field,
if the numbers
$\alpha_1, \cdots, \alpha_m$ are sufficiently closed to $0$. As far as we know, this linear independence is the first result concerning with
even polylogarithms at different points.

The new ingredient in the article relies on a few points. First and foremost, we introduce a systematic construction of Pad\'e approximants, which heavily relies  on the computations made by past authors.
Our modifications and generalizations of the method of Niki\v{s}in developed in \cite{N0} \cite{N} as well as of the Rhin-Toffin method \cite{R-T},
supply a formally regulated  construction of Pad\'e  approximants. Secondly an irrationality criterion, combined with the metric property provided for by Pad\'e approximation, leads to the irrationality for the values of the Lerch functions at
points sufficiently closed to the origin (the precise sufficient condition, that we explain later, comes from the coupling of the criteria with Pad\'e approximation). This strategy works only if one can ensure the injectivity of evaluation maps defined by systems of Pad\'e approximation, which can be now interpreted as a non-vanishing property of an Hermite-type determinant,
that we succeed in proving.

\section{Notations and Main results}
We fix an algebraic closure of $\mathbb{Q}$ and denote it by $\overline{\mathbb{Q}}$. For a subset $S\subset \overline{\mathbb{Q}}$, we define the denominator of $S$ by
$${\rm{den}}(S):=\min\{0 < n\in\Z | \ n\alpha \ \text{is an algebraic integer for any} \ \alpha\in S\}\,.$$

\medskip

Let $\mathbb{N}$ be the set of positive integers. Let $m, r\in \mathbb{N}$ and $K$ be an algebraic number field of finite degree over $\mathbb{Q}$. We denote the ring of integers of $K$ by $\mathcal{O}_K$ and the completion of $K$ with respect to the fixed embedding $\iota_{\infty}:K\hookrightarrow \mathbb{C}$ by $K_{\infty}$.
Then $[K_{\infty}:\mathbb{Q}]=1$ if $K_{\infty}\subset \mathbb{R}$, and
$[K_{\infty}:\mathbb{Q}]=2$ otherwise.

\medskip

Let $x\in \mathbb{Q}\cap [0,1)$. Put $$\mu(x):={\rm{den}}(x)\prod_{q:{\rm{prime}}, q| {\rm{den}}(x)}q^{1/(q-1)}\enspace.$$

Consider
$\boldsymbol{\alpha}:=(\alpha_1,\ldots,\alpha_m)\in (K\setminus\{0\})^m$ with $\alpha_i\neq \alpha_j$ for all $1\leq i< j \leq m$.
For $1\le g\le [K:\mathbb{Q}]$, we write  $\alpha^{(g)}$ the $g$-th conjugate
of $\alpha\in K$ over $\mathbb{Q}$.

\medskip

Let $\beta\in K\setminus\{0\}$ with $\max_{1\leq i \leq m} (|\alpha_i| )<|\beta|$.
We put
$$D(\boldsymbol{\alpha},\beta):={\rm{den}}(\alpha_1,\ldots,\alpha_m,\beta)\enspace.$$

\noindent
We also define
\begin{align*}
&\mathbb{A}(\boldsymbol{\alpha},\beta,x):={\rm{log\,}}|\beta|-(rm+1){\rm{log\,}}\max_i(|\alpha_i|)-\{rm({\rm{log\,}}D(\boldsymbol{\alpha},\beta)+r[{\rm{den}}(x)+{\rm{log\,}}(5/2)])+r({\rm{log\,}}3+{\rm{log\,}}\mu(x))\}\enspace,\\
&\mathcal{A}^{(g)}(\boldsymbol{\alpha},\beta,x):=rm\left({\rm{log\,}}D(\boldsymbol{\alpha},\beta)+{\rm{log\,}}\max(1, \min(|\alpha^{(g)}_i|)^{-1}\cdot|\beta^{(g)}|)+r[{\rm{den}}(x)-{\rm{log\,}}2]\right)\\
&\ \ \ \ \ \ \ \ \ \ +r\left({\rm{log\,}}\mu(x)+\sum_{i=1}^m{\rm{log\,}}(2^r|\alpha_i|+3^r\max(|\alpha^{(g)}_{i}|,|\beta^{(g)}|))\right)+{\rm{log\,}}3
 \ \ \ \ \text{for} \ \  1\le g\le [K:\mathbb{Q}]\enspace,
\end{align*}
and
$$
V(\boldsymbol{\alpha},\beta,x):=\mathbb{A}(\boldsymbol{\alpha},\beta,x)+\mathcal{A}^{(1)}(\boldsymbol{\alpha},\beta,x)-\dfrac{\sum_{g=1}^{[K:\mathbb{Q}]}\mathcal{A}^{(g)}(\boldsymbol{\alpha},\beta,x)}{[K_{\infty}:\mathbb{Q}]}\enspace.
$$

\bigskip

We then obtain the following statement.
{{\begin{theorem} \label{Lerch}
Assume $V(\boldsymbol{\alpha},\beta,x)>0$.
Then the $rm+1$ numbers~$:$~
$$1,\Phi_1(x,\alpha_1/\beta),\ldots,\Phi_r(x,\alpha_1/\beta),\ldots, \Phi_1(x,\alpha_m/\beta),\ldots,\Phi_r(x,\alpha_m/\beta)\enspace,$$ are linearly independent over $K$.
\end{theorem}}}

We note that it is the first result to give the linear independence of the values of the Lerch function, even in the case of
polylogarithms, at distinct algebraic numbers.

\section{Construction of Pad\'{e} approximants}
We now explain how we construct Pad\'{e} approximants of the Lerch functions. Since the full proof is long, then with the relevant details, it will be provided for in the forthcoming articles \cite{DHK1},  \cite{DHK2},  with a $p$-adic analogue as well as quantitative measures of linear independence.

First we recall the definition of Pad\'{e} approximants of formal Laurent series. In the following of this section, we denote by  $L$ a unique factorization domain of characteristic $0$.
We define the order function ${\rm{ord}}_{\infty}$ at ``$z=\infty$" by
\begin{align*}
{\rm{ord}}_{\infty}:L[z][[1/z]]\rightarrow \Z\cup \{\infty\} , \ \ \sum_{k}a_k\cdot\dfrac{1}{z^k}\mapsto \min\{k\in \Z\mid a_k\neq 0\}\enspace.
\end{align*}
\begin{lemma} \label{pade}
Let $r$ be a positive integer, $f_1(z),\ldots,f_r(z)\in 1/z\cdot L[[1/z]]$ and $\bold{n}:=(n_1,\ldots,n_r)\in \mathbb{N}^{r}$.
Put $N:=\sum_{i=1}^rn_i$.
Let $M$ be a positive integer with $M\ge N$. Then there exists a family of polynomials
$(P_0(z),P_{1}(z),\ldots,P_r(z))\in L[z]^{r+1}\setminus\{\bold{0}\}$ satisfying the following conditions~$:$~
\begin{align*}
&({\rm{i}}) \ {\rm{deg}}P_{0}(z)\le M\enspace,\\
&({\rm{ii}}) \ {\rm{ord}}_{\infty} P_{0}(z)f_j(z)-P_j(z)\ge n_j+1 \ \text{for} \ 1\le j \le r\enspace.
\end{align*}
\end{lemma}
\begin{definition}
We use the same notations as those in Lemma $\ref{pade}$.
We call a family of polynomials $(P_0(z),P_{1}(z),\ldots,P_r(z)) \in L[z]^{r+1}$ satisfying the properties $({\rm{i}})$ and $({\rm{ii}})$,
{\it Pad\'{e} type approximants of} $(f_1,\ldots,f_r)$,  {\it  of weight} $\bold{n}$ {\it and of degree} $M$.

For the Pad\'{e} type approximants $(P_0(z),P_{1}(z),\ldots,P_r(z))$, of $(f_1,\ldots,f_r)$ of weight $\bold{n}$,
we call the family of formal Laurent series $(P_{0}(z)f_j(z)-P_{j}(z))_{1\le j \le r}$,
{\it Pad\'{e} type approximations  system of }$(f_1,\ldots,f_r)$,  {\it  of weight} $\bold{n}$ {\it and of degree} $M$.
\end{definition}

In the sequel, we take $x\in L\setminus \Z_{<0}$ and assume $x+k$ are invertible in $L$ for any $k\in \mathbb{N}$.

We now introduce notations for formal primitive, derivation, and evaluation maps. Let $I$ be a finite set, we assume that $L$ contains $K[X_i,1/X_i]_{X_i\in I}$ where $K$ is a number field. In the sequel, it will be convenient to work formally and thus to treat as many quantities as variables as useful, and we shall freely extend the set $I$ as need arises.

\begin{notation} \label{notationderiprim}
\begin{itemize}
\item[(i)]  For $\alpha\in L$, We denote by ${{\rm Eval}}_{\alpha}$ the linear evaluation map $L[{T}]\longrightarrow L$, $P\longmapsto P(\alpha)$.
\item[(ii)] For $P\in L[T]$, we denote by $[P]$ the multiplication by $P$ ($Q\longmapsto PQ$).
\item[(iii)] We also denote by ${\rm Prim}_x$ the linear operator $L[T]\longrightarrow L[T]$, defined by $P\longmapsto \frac{1}{T^{1+x}}\int_{0}^{T}{\xi}^xP(\xi)d\xi$ (formal primitive).
\item[(iv)] We denote by ${\rm Deri}_x$ the derivative map $P\longmapsto T^{-x}\tfrac{d}{dT}(T^{x+1}P(T))$, and for $n\geq 1$, by $S_{n,x}$ the map taking $T^k$ to $\tfrac{(k+x+1)_n}{n!}T^k$ where
$(k+x+1)_n:=(k+x+1)\ldots(k+x+n)$
(the divided derivative $P\longmapsto \frac{1}{n!}T^{-x}\frac{d^n}{dT^n}(T^{n+x}P)=\tfrac{1}{n!}\left(\tfrac{d}{dT}+x/T\right)^nT^n(P)$), so that ${\rm Deri}_x=S_{1,x}$.
\item[(v)] If $\varphi$ is an $L$-automorphism of an $L$-module $M$ and $k$ an integer, we denote
$$\varphi^{(k)}:=\begin{cases}
\overbrace{\varphi\circ\cdots\circ\varphi}^{k-\text{times}} & \ \text{if} \ k>0,\\
{\rm{id}}_M &  \ \text{if} \ k=0,\\
\overbrace{\varphi^{-1}\circ\cdots\circ\varphi^{-1}}^{-k-\text{times}} & \ \text{if} \ k<0.
\end{cases}
$$
\end{itemize}

\end{notation}

For a given $l\in \Z$, we define the linear map $\varphi_{\alpha,x,l}$ as follows.
\begin{notation}
$$\varphi_{\alpha,x,l}:=[\alpha]\circ{\rm Eval}_{\alpha}\circ {\rm Prim}^{(l)}_x\enspace.$$
\end{notation}

\

\

For any non-negative integers $k$, note that $\varphi_{\alpha,x,s}(T^k)$ is a formal analogue of
$$\dfrac{1}{(s-1)!}\int^{\alpha}_{0}T^{k+x} {\rm{log\,}}^{s-1}\dfrac{1}{T}dT\enspace.$$

\

\

For the convenience, we collect below the following elementary facts.

\

\begin{facts}
\label{faitselem}
\begin{itemize}
\item[(i)] The map ${\rm Prim}_x$ is an isomorphism and its inverse is ${\rm Deri}_x$ for $x\in L\setminus\Z_{<0}$.
Hence $\varphi_{\alpha,x,s}$ is well-defined for $s\leq -1$.
\item[(ii)] For any integers $n_1\geq 0, n_2\geq 0$ and $x\in L\setminus\Z_{<0}$ with $x+k$ invertible in $L$ for any $k\in \mathbb{N}$, the divided derivatives $S_{n_1,x}$ and $S_{n_2,x}$ commute, namely $S_{n_1,x}\circ S_{n_2,x}=S_{n_2,x}\circ S_{n_1,x}$.
\item[(iii)]  For any integer $s\in{\Z}$ and any $\alpha\in L$, we have $\varphi_{\alpha,x,s}\circ{\rm Deri}_x=\varphi_{\alpha,x,s-1}$.
\item[(iv)] By continuity, all the above mentioned maps extend to $L[[T]]$ with respect to the natural valuation.
\item[(v)] The kernel of the map $\varphi_{\alpha,x,0}$ is the ideal $(t-\alpha)$ for any $x\in L\setminus\Z_{<0}$.
\end{itemize}
\end{facts}

\

Using Fact~\ref{faitselem} (iv), the classical Lerch function is indeed expressed as a natural image by $\varphi_{\alpha,x,s}$ with $s\geq 1$, as follows~:
\begin{align} \label{Lerch integral rep}
\varphi_{\alpha,x,s}\left(\dfrac{1}{z-T}\right)=\Phi_s(x,\alpha/z)\enspace.
\end{align}

\

Let $x\in L\setminus \Z_{<0}$ with $x+k$  invertible for any $k\in \mathbb{N}$. Consider
$\boldsymbol{\alpha}:=(\alpha_1,\ldots,\alpha_m)\in (L\setminus\{0\})^m$ with $\alpha_i\neq \alpha_j$ for $i\neq j$.
We study Pad\'{e} approximants of type  I\hspace{-.1em}I of the functions $(\Phi_{s}(x,\alpha_i/z))_{\substack{1\le i \le m \\ 1\le s \le r}}$.

Let $l$ be a non-negative integer with $0\le l \le rm$. For  a positive integer $n$, we define a family of polynomials~:~
\begin{align}
&P_{n,l}(\boldsymbol{\alpha},x|z):={\rm Eval}_z\circ S^{(r)}_{n,x}\left(T^l\prod_{i=1}^m(T-\alpha_i)^{rn}\right), \label{Qnl}\\
&P_{n,l,i,s}(\boldsymbol{\alpha},x|z):=\varphi_{\alpha_i,x,s}\left(\dfrac{P_{n,l}(\boldsymbol{\alpha},x|z)-P_{n,l}(\boldsymbol{\alpha},x|T)}{z-T}\right) \ \text{for} \ 1\le i \le m, 1\le s \le r\enspace.\label{Qnlijsj}
\end{align}

\bigskip

Under the notations above, we obtain the following theorem.

\

\begin{theorem} \label{Pade appro Lerch}
For each $0\leq l\leq rm$, the family of polynomials $(P_{n,l}(\boldsymbol{\alpha},x|z),P_{n,l,i,s}(\boldsymbol{\alpha},x|z))_{\substack{1\le i \le m \\ 1\le s \le r}}$ forms Pad\'{e} type approximants system of $(\Phi_{s}(x,\alpha_i/z))_{\substack{1\le i \le m \\ 1\le s \le r}}$,
of
weight $(n,\ldots,n)\in \mathbb{N}^{rm}$ and of degree $rmn+l$.

\end{theorem}
\begin{proof}
By definition of $P_{n,l}(\boldsymbol{\alpha},x|z)$, we have
$$
{\rm{deg}}P_{n,l}(\boldsymbol{\alpha},x|z)=rmn+l\,.
$$
Hence the condition on the  degree is verified. We only need  to check the condition on the valuation.

Put $R_{n,l,i,s}(\boldsymbol{\alpha},x|z):=P_{n,l}(\boldsymbol{\alpha},x|z)\Phi_{s}(x,\alpha_i/z)-P_{n,l,i,s}(\boldsymbol{\alpha},x|z)$.
Then, by  definition of $R_{n,l,i,s}(\boldsymbol{\alpha},x|z)$ with the property $(\ref{Lerch integral rep})$, we obtain
$$
\begin{array}{lcl}
\displaystyle
R_{n,l,i,s}(\boldsymbol{\alpha},x|z)&= & \displaystyle P_{n,l}(\boldsymbol{\alpha},x|z)\varphi_{\alpha_i,x,s}\left(\dfrac{1}{z-T}\right)-P_{n,l,i,s}(\boldsymbol{\alpha},x|z) \label{residual term decompose}\\
                                                       &= & \displaystyle\varphi_{\alpha_i,x,s}\left(\dfrac{P_{n,l}(\boldsymbol{\alpha},x|T)}{z-T}\right)
                                                       =\sum_{k=0}^{\infty}\varphi_{\alpha_i,x,s}(T^kP_{n,l}(\boldsymbol{\alpha},x|T))\dfrac{1}{z^{k+1}}\enspace. \nonumber \end{array}
$$
We note that we have the following identities in ${\rm{End}}_K(K[T])$~:~
\begin{align*}
&S_{n,x}=S_{1,x}\circ\ldots \circ (S_{1,x}+n-1) \ \text{for} \ n\in \mathbb{N}\enspace,\\
&[T^k]\circ S_{1,x}=(S_{1,x}-k)\circ [T^k] \ \text{for} \ k\in \Z_{\ge0}\enspace.
\end{align*}
By definition of $P_{n,l}(\boldsymbol{\alpha},x|T)$ and the  identities above, for each $1\le s \le r, \,\,0\le k \le n-1$, there exists a polynomial $U_{s,k}(X)\in \mathbb{Q}[X]$ of\,\, ${\rm{deg}}U_{s,k}=nr-s$, satisfying
$$T^kP_{n,l}(\boldsymbol{\alpha},x|T)=S^{(s)}_{1,x}\circ U_{s,k}(S_{1,x})\left(T^{k+l}\prod_{i=1}^m(T-\alpha_i)^{rn}\right).$$
By Leibniz rule, we obtain that  $U_{s,k}(S_{1,x})\left(T^{k+l}\prod_{i=1}^m(T-\alpha_i)^{rn}\right)$
belongs to the ideal $ (t-\alpha_i)$ for each $1\le i \le m$.
Hence we get
\begin{align*}
\varphi_{\alpha_i,x,s}(T^kP_{n,l}(\boldsymbol{\alpha},x|T))
                                                                          &=\varphi_{\alpha_i,x,0}\circ U_{s,k}(S_{1,x})\left(T^{k+l}\prod_{i=1}^m(T-\alpha_i)^{rn}\right)=0\enspace,
\end{align*}
for $1\le i \le m, 1\le s \le r$ and $0\le k \le n-1$.

Consequently, by the expansion above of $R_{n,l,i,s}(\boldsymbol{\alpha},x|z)$, we obtain
$$
{\rm{ord}}_{\infty}R_{n,l,i,s}(\boldsymbol{\alpha},x|z)\ge n+1 \ \text{for} \ 1\le i \le m, 1\le s \le r\enspace.
$$
Then Theorem $\ref{Pade appro Lerch}$ follows.
\end{proof}

\section{Metric approximations and linear independence criteria}

We now give a few of  estimates associated with the Pad\'e approximation we just constructed. They do not need involved arguments to be proven; however due to the technical nature of the construction, computations are somewhat heavy and we do skip them to keep in line with the spirit of this article.

The estimates in Lemma \ref{upper coefficients}
can be combined with an appropriate linear independence criterion to provide for a measure.
\begin{lemma} \label{upper coefficients}
Let $n$ be a positive integer, $x$ be a rational number with $0\le x<1$ and $\beta\in K\setminus\{0\}$. Then for any $1\le g \le [K:\mathbb{Q}]$, we have

\begin{align*}
\max_{\substack{1\le i \le m \\ 0\le l \le rm \\ 1\le s \le r}}|P^{(g)}_{n,l,i,s}(\boldsymbol{\alpha},x|\beta)|&\le
\max(|\alpha^{(g)}_i|)^{rm}\left(\dfrac{3}{2}\right)^{r^2m+r}\left(\dfrac{3}{2^{rm}}\prod_{j=1}^m\left[2^r|\alpha^{(g)}_{j
}|+3^r\max(|\alpha^{(g)}_{i}|)\right]\right)^{rn} \\
&\times\begin{cases}
\dfrac{\left(\min(|\alpha^{(g)}_i|)^{-1}|\beta^{(g)}|\right)^{rm(n+1)}}{\min(|\alpha^{(g)}_i|)^{-1}|\beta^{(g)}|-1} & \ \text{if} \ \min(|\alpha^{(g)}_i|)^{-1}|\beta^{(g)}|>1\enspace,\\
rm(n+1)  & \ \text{if} \ \min(|\alpha^{(g)}_i|)^{-1}|\beta^{(g)}|\le 1\enspace.
\end{cases}
\end{align*}
for $1\le i \le m$.

For the error term, we have~$:$~
\begin{align*}
\max_{0\le l \le rm}|R_{n,l,i,s}(\boldsymbol{\alpha},x|\beta)|\le \dfrac{\max_{1\le i \le m}(1,|\alpha_i|)^{rm+1}}{|\beta|-\max_{
j
}(|\alpha_{
j
}|)}\left(\dfrac{3}{2}\right)^{r^2m+r}\left(\dfrac{\max_{

}(|\alpha_{
j
}|)^{rm+1}}{|\beta|}\right)^n
 \left(3\left(\dfrac{5}{2}\right)^{rm}\right)^{rn}\enspace.
\end{align*}
\end{lemma}

\bigskip

We then state a general linear independence criterion\,:~
\begin{proposition} \label{critere version II}
Let $K$ be an algebraic number field of finite degree over $\mathbb{Q}$. We denote the completion of $K$ with respect to the fixed embedding $\iota_{\infty}$ by $K_{\infty}$.
Let $m\in\mathbb{N}$ and $\theta_0:=1,\theta_1,\ldots,\theta_{m}\in \mathbb{C}\setminus\{0\}$.
\\
Suppose that there exists a set of matrices $$\{(A_{n,l,j})_{0\le l,j\le m}\}_{n\in \mathbb{N}}\subset {\rm{M}}_{m+1}(\mathcal{O}_K)\cap {\rm{GL}}_{m+1}(K)\,.$$
Assume further there exist
positive real numbers
$$
\{\mathcal{A}^{(g)}\}_{\substack{1\le g \le [K:\mathbb{Q}]}}$$
and a positive real number $\mathbb{A},
$
satisfying the following conditions~$:$~

$$\max_{0\le l,j \le m}|A^{(g)}_{n,l,j}|\le e^{\mathcal{A}^{(g)}\cdot n+o(n)} \ \text{for} \ 1 \le g \le [K:\mathbb{Q}]\  \  \
(n\rightarrow \infty) \  \  \text{and}$$

$$\max_{\substack{0\le l \le m \\ 1\le j \le m}} |A_{n,l,0}\cdot \theta_j-A_{n,l,j}|\le e^{-\mathbb{A}\cdot n+o(n)}\  \  \
(n\rightarrow \infty).
$$
We put
$$
V:=\mathbb{A}+\mathcal{A}^{(1)}-\dfrac{\sum_{g=1}^{[K:\mathbb{Q}]}\mathcal{A}^{(g)}}{[K_{\infty}:\mathbb{R}]}\,.
$$
If  $V>0$, then the numbers ${\theta_0},\ldots,\theta_m$  are linearly independent over $K$.
\end{proposition}

\

\

Theorem~\ref{Pade appro Lerch} gives us the sequence of  matrices. The growth control of the size of the matrices to carry out the approximations is provided for in Lemma~\ref{upper coefficients}.  However, the matrices do
not always
have algebraic integer entries. This is not a big deal. The defect of integrality comes from our operators ${\rm Prim}_x,{\rm Deri}_x$ and it  is corrected by multiplying by a suitable power of ${\rm lcm}(1,\ldots,n)$ which is standard in the theory.

Just plugging in these estimates in Proposition \ref{critere version II}, it leads us to the proof of the
main theorem. The metric condition requiring the numbers to be sufficiently closed to the origin, is translated to the condition $V>0$ in  Proposition \ref{critere version II}, the linear independence criterion.

However, there is still a significant step to be performed. Now we need to prove that the matrices coming from the Pad\'e  approximation are indeed invertible. We describe this main step in the next section.

\section{Non-vanishing of a determinant and the final step of the proof}
In this section, we use the following notations.
Let $m, r$ be positive integers and $K$ be  a field with characteristic $0$. We assume that $\alpha_1,\ldots,\alpha_m, z,t$ all belong to the set of variables $I$, so our ring $L$ contains $K[\alpha_i,z,t,1/\alpha_i,1/z,1/t]$. Put  $\boldsymbol{\alpha}:=(\alpha_1,\ldots,\alpha_m)$.

For a positive integer $l$ with $0\le l \le rm$, and for $x\in K$, we put
\begin{align*}
&P_{n,l}(z):=P_{n,l}(\boldsymbol{\alpha},x|z)\enspace,\\
&P_{n,l,i,s}(z):=P_{n,l,i,s}(\boldsymbol{\alpha},x|z) \ \text{for} \ 1\le i \le m, 1\le s \le r\enspace.
\end{align*}
The polynomials in the right-hand sides above have been already defined in $(\ref{Qnl})$ and in $(\ref{Qnlijsj})$ respectively.

We define a column vector $\vec{p}_{n,l}(z)\in K[z]^{rm+1}$ by
\begin{align*}
&\vec{p}_{n,l}(z):={}^t\Biggl(P_{n,l}(z),
{P_{n,l,1,1}(z),\ldots, P_{n,l,1,r}(z)}, \ldots, {P_{n,l,m,1}(z),\ldots, P_{n,l,m,r}(z)}\Biggr).
\end{align*}

\

\begin{proposition} \label{non zero det}
We use the same notations as above. For any positive integer $n$, we have
$$
\Delta_n(z):=
{\rm{det}}  {\begin{pmatrix}
\vec{p}_{n,0}(z) \ \cdots \ \vec{p}_{n,rm}(z)
\end{pmatrix}}
\in K{{(\alpha_1,\alpha_2,\ldots,\alpha_m)}}\setminus\{0\}\enspace.
$$
\end{proposition}

\

\

To prove this statement, we firstly prove that the determinant $\Delta_n=\Delta_n(z)$ is a constant,  is independent of $z$.
Secondly, we regard $\Delta_n$ as an element of $K(\alpha_1,\ldots,\alpha_m)$ viewing  $\alpha_1,\ldots,\alpha_m$ as
 indeterminates, and factor it up to a constant depending only on $n, m, r$. We finally show that this absolute constant
 $\Delta_n$ is non-zero.
For this last step, we  identify this determinant with certain integral  to show that it does not vanish.

\

We shall prove~:
$$
\Delta_n(z)\in K{{(\alpha_1,\ldots,\alpha_m)}} \ \text{for all} \ n\in \mathbb{N}\enspace.
$$

\

For this, denote $P_{n,l}(z)\Phi_{s}(x,\alpha_i/z)-P_{n,l,i,s}(z)$ by $R_{n,l,i,s}(z)$  as above ($0\le l \le rm$, $1\le i \le m$, $1\le s \le r$).

In the matrix giving the determinant $\Delta_n(z)$, we add, the first row multiplied by the $\Phi_{s}(x,\alpha_i/z)$,
to the $(i-1)r+s+1$-th row ($1\le i \le m$, $1\le s \le r$), to obtain
                     $$
                     \Delta_n(z)=(-1)^{rm}{\rm{det}}
                     {\begin{pmatrix}
                     P_{n,0}(z) & \dots &P_{n,rm}(z)\\
                     R_{n,0,1,1}(z) & \dots & R_{n,rm,1,1}(z)\\
                     \vdots    & \ddots & \vdots  \\
                     R_{n,0,1,r}(z) & \dots & R_{n,rm,1,r}(z)\\
                     \vdots & \ddots & \vdots\\
                     R_{n,0,m,1}(z) & \dots & R_{n,rm,m,1}(z)\\
                     \vdots    & \ddots & \vdots  \\
                     R_{n,0,m,r}(z) & \dots & R_{n,rm,m,r}(z)\\
                     \end{pmatrix}}\enspace.
                     $$
We denote   by $\Delta_{n,s,t}(z)$, the $(s,t)$-th cofactor of the matrix in the right-hand side of the identity above.
Then we have, developing along the first row
\begin{align} \label{formal power series rep delta}
\Delta_n(z)=(-1)^{rm}\left(\sum_{l=0}^{rm}P_{n,l}(z)\Delta_{n,1,l+1}(z)\right)\enspace.
\end{align}
Since we have
\begin{align*}
{\rm{ord}}_{\infty} R_{n,l,i,s}(z)\ge n+1 \ \text{for} \ 0\le l \le rm, \ 1\le i\le m \ \text{and} \ 1\le s \le r\enspace,
\end{align*}
we get
$$
{\rm{ord}}_{\infty}\Delta_{n,1,l+1}(z)\ge (n+1)rm\enspace.
$$
Combining the fact ${\rm{deg}}P_{n,l}(z)=rmn+l$ with the lower bound of ${\rm{ord}}_{\infty}\Delta_{n,1,l+1}(z)$ above, we obtain
$$
P_{n,l}(z)\Delta_{n,1,l+1}(z)\in 1/z\cdot K[[1/z]] \ \text{for} \ 0\le l \le rm-1\,,
$$
and
$$
P_{n,rm}(z)\Delta_{n,1,rm+1}(z)\in K[[1/z]].
$$
Note that in the  relation above, the constant term of $P_{n,rm}(z)\Delta_{n,1,rm+1}(z)$ is
$$
``\text{Coefficient of} \ z^{rm(n+1)} \ \text{of} \ P_{n,rm}(z)"\times ``\text{Coefficient of} \ 1/z^{rm(n+1)} \ \text{of} \ \Delta_{n,1,rm+1}(z)".
$$
Thus by $(\ref{formal power series rep delta})$, the determinant $\Delta_n(z)$ is a polynomial in $z$ with non-positive valuation with respect to ${\rm{ord}}_{\infty}$, consequently it turns to be a constant.
Moreover, the terms of  strictly negative valuation, they should be canceled out. Hence we have
\begin{align}\label{cal delta n}
\Delta_n=\Delta_n(z)=(-1)^{rm}\kern-3pt\times\kern-4pt\left(\sum_{l=0}^{rm}P_{n,l}(z)\Delta_{n,1,l+1}(z)\right)=(-1)^{rm}\kern-3pt\times\kern-3pt ``\text{constant term of}\,  P_{n,rm}(z)\Delta_{n,1,rm+1}(z)" \in K.\enspace
\end{align}

\vspace{\baselineskip}

We now need to rewrite $\Delta_n$ as a rational function of $\alpha_1,\ldots,\alpha_m$ in a workable way.
We further extend the set of variables and assume that the set $I$ contains the $rm$ variables $t_{i,s}, 1\leq i\leq m, 1\leq s\leq r$, so that $L$ contains
$$K[\alpha_1,\ldots,\alpha_m, z,t,1/\alpha_1,\ldots,1/\alpha_m,1/z,1/t][t_{i,s}].$$

For each variable $t_{i,s}$ and any integer $l$, we have a well-defined map for $\alpha\in L$~:
\begin{equation*}
\varphi_{\alpha,t_{i,s}, l}:L[t_{i,s}]_{1\leq i\leq m,1\leq s\leq r}\longrightarrow L[t_{i',s'}]_{(i',s')\neq (i,s)}
\end{equation*}
since  $L[t_{i,s}]_{1\leq i\leq m,1\leq s\leq r}$ can be regarded as a polynomial ring in one variable $L'[t_{i,s}]$ over $L'=L[t_{i',s'}]_{(i',s')\neq (i,s)}$.

\

Now for a positive integer $n$ and an integer $l$ with $ 0\le l \le rm$, we put
$$
A_{n,l}(T):=T^l\prod_{i=1}^m(T-\alpha_i)^{rn}\enspace.
$$
By the definition of $A_{n,l}(T)$, we have $P_{n,l}(z)={\rm Eval}_z\circ S^{(r)}_{n,x}A_{n,l}(T).$

\

Let us define a column vector $\vec{r}_{n,l}\in L^{rm}$ by
{\footnotesize{
\begin{align*}
&\vec{r}_{n,l}:=\\
&{}^t\Biggl(\varphi_{\alpha_1,t_{1,1},1}(t^n_{1,1}A_{n,l}(t_{1,1})), \ldots, \varphi_{\alpha_1,t_{1,r},r}(t^n_{1,r}A_{n,l}(t_{1,r})), \ldots, \varphi_{\alpha_m,t_{m,1},1}(t^n_{m,1}A_{n,l}(t_{m,1})), \ldots, \varphi_{\alpha_m,t_{m,r},r}(t^n_{m,r}A_{n,l}(t_{m,r}))\Biggr)\enspace.
\end{align*}}}
\begin{lemma} \label{another presentation}
Under the notations above, we obtain the following identity~$:$~
$$
\Delta_n=(-1)^{rmn}{\left(\dfrac{(1+rmn+rm+x)_n}{n!}\right)^r }
{\rm{det}}  {\begin{pmatrix}
\vec{r}_{n,0} \
\cdots \
\vec{r}_{n,rm-1}
\end{pmatrix}}\enspace.
$$
\end{lemma}
\begin{proof}
Using $(\ref{cal delta n})$, we calculate constant term of $P_{n,rm}(z)\Delta_{n,1,rm+1}(z)\in K[[1/z]]$.

We need to deal with the non-commutativity of the multiplication by $[T]$ and the morphisms $S_{n,x}^{(k)}$. The defect of
the commutativity is given by the following identity~:~
there exists a set of rational numbers $\{e_{n,k}\}_{0\le k \le rn}\subset {\Q}$ with $e_{n,0}=(-1)^{rn}$ and
$$
[T^n]\circ S^{(r)}_{n,x}=\sum_{k=0}^{rn}e_{n,k}S^{(k)}_{1,x}\circ [T^n]\enspace.
$$
Then we obtain
\begin{align*}
\varphi_{\alpha_i,x,s}(T^nP_{n,l}(T))&=\sum_{k=0}^{rn}e_{n,k}\varphi_{\alpha_i,x,s}\circ S^{(k)}_{1,x}\circ[T^n](A_{n,l}(T))\\
                                         &=\sum_{k=0}^{s-1}e_{n,k}\varphi_{\alpha_{i,x,s-k}}\circ[T^n](A_{n,l}(T))+\sum_{k=s}^{rn}e_{n,k}\varphi_{\alpha_i,x,0}\circ S^{(k-s)}_{1,x}\circ[T^n](A_{n,l}(T))\nonumber\\
                                         &=\sum_{k=0}^{s-1}e_{n,k}\varphi_{\alpha_{i,x,s-k}}(T^nA_{n,l}(T)),
\end{align*}
for $1\le i \le m$ and $1\le s \le r$, the conclusion follows, interpreting the above relations as linear manipulations of lines, columns leaving the determinant unchanged.
\end{proof}

Now, for non-negative integers $u, n$, we put\,:
$$P_{u,n}(t_{i,s})=\prod_{i=1}^{m}\prod_{s=1}^r\left[ t_{i,s}^u\prod_{j=1}^m(t_{i,s}-\alpha_j)^{rn}\right]\prod_{(i_1,s_1)<(i_2,s_2)}(t_{i_2,s_2}-t_{i_1,s_1}),
$$
where the order $(i_1,s_{1})<(i_2,s_{2})$ follows lexicographically.

\

By $\bigcirc$, we denote the composite of morphisms.
When no confusion is deemed to occur, we omit the subscripts $\underline{\alpha}=(\alpha_1,\ldots,\alpha_m)$ and write
$$\psi=\psi_{\underline{\alpha}}:=\bigcirc_{i=1}^{m}\bigcirc_{s=1}^r\varphi_{\alpha_i,t_{i,s},x,s}\enspace.$$

Note that, by definition of
$
{\rm{det}}  {\begin{pmatrix}
\vec{r}_{n,0} \
\cdots \
\vec{r}_{n,rm-1}
\end{pmatrix}},
$
we have
$$
{\rm{det}}  {\begin{pmatrix}
\vec{r}_{n,0} \
\cdots \
\vec{r}_{n,rm-1}
\end{pmatrix}}=\psi(P_{n,n})\enspace.
$$
Let $u$ be a non-negative integer. We are going to study the value
$$
C_{n,u,m}:=\psi(P_{u,n})\,.
$$
By induction, we obtain the following proposition.
\begin{proposition} \label{decompose Cnum}
There exists a non-zero constant $c_{n,u,m}\in K$  satisfying
$$
C_{n,u,m}=c_{n,u,m}\prod_{i=1}^m\alpha^{r(u+1)+r^2n+\binom{r}{2}}_i \prod_{1\le i_1<i_2\le m}(\alpha_{i_2}-\alpha_{i_1})^{(2n+1)r^2},
$$
with $\displaystyle\binom{r}{2}=0$ if $r=1$.
\end{proposition}

\

We write the detail of the proof of the proposition in a forthcoming articles \cite{DHK1} \cite{DHK2}, however, we describe here our basic idea.
Indeed, we prove the proposition by the proceedures
\begin{itemize}
\item[(i)] Show that $C_{n,u,2}$ is homogeneous of degree $2r(u+1)+2r^2n+2{r\choose 2}+(2n+1)r^2$.
\item[(ii)] Show that $(\alpha\beta)^{r(u+1)+r^2n+{r\choose 2}}$ divides $C_{n,u,2}$.
\item[(iii)] Show that $(\alpha-\beta)^{(2n+1)r^2}$ divides $C_{n,u,2}$.
\end{itemize}

\

Here, we explain how the constant $c_{n,u,m}$ in Proposition $\ref{decompose Cnum}$ becomes non-zero.
Whenever it is shown, then the determinant does not vanish.
\

We use the same notations as those in Proposition $\ref{decompose Cnum}$.
Define $$D_{n,u,m}:=\dfrac{C_{n,u,m}}{\prod_{i=1}^m\alpha^{r(u+1)+r^2n+\binom{r}{2}}_i}=c_{n,u,m}\times\prod_{1\le i_1<i_2\le m}(\alpha_{i_2}-\alpha_{i_1})^{(2n+1)r^2}.$$
A straightforward calculation of an integral gives us
\begin{align*}
D_{n,u,m}=&\bigcirc_{i^{\prime}=1}^{m}\bigcirc_{s^{\prime}=1}^r\varphi_{1_{i^{\prime}},x,s^{\prime}}
\Biggl(\prod_{i=1}^m\prod_{s=1}^{r} \left[t^{u}_{i,s} \cdot (t_{i,s}-1)^{rn} \cdot \prod_{\substack{\tilde{i}\neq i\\ 1\le \tilde{i} \le m}}(\alpha_i t_{i,s}-\alpha_{\tilde{i}})^{rn}\right]\times\\
&\prod_{i=1}^m\left(\prod_{s_{1}<s_{2}}(t_{i,s_{2}}-t_{i,s_{1}})\right)\times \prod_{\substack{(i_1,s_{1})<(i_2,s_{2}) \\ i_1<i_2}}
(\alpha_{i_2}t_{i_2,s_{2}}-\alpha_{i_1}t_{i_1,s_{1}})\Biggr).
\end{align*}

We substitute $\alpha_m=0$ in $D_{n,u,m}$, then we have
{\small{
\begin{align*}
&\left. D_{n,u,m}\right|_{\alpha_m=0}=c_{n,u,m}\prod_{i=1}^{m-1}(-\alpha_i)^{(2n+1)r^2}\prod_{1\le i_1<i_2\le m-1}(\alpha_{i_2}-\alpha_{i_1})^{(2n+1)r^2}\\
&=\pm\prod_{i=1}^{m-1}\alpha^{(2n+1)r^2}_i \bigcirc_{s^{\prime}=1}^{r}\varphi_{1,x,s^{\prime}}\left({\prod_{s=1}^{r}} \left[t^{u}_{m,s} \cdot (t_{m,s}-1)^{rn}\right] \times\prod_{s_{1}<s_{2}}(t_{m,s_{2}}-t_{m,s_{1}})\right) \\
&\times \bigcirc_{i^{\prime}=1}^{m-1}\bigcirc_{s^{\prime}=1}^{r}\varphi_{1_{i^{\prime},x,s^{\prime}}}\Biggl(\prod_{i=1}^{m-1}{\prod_{s=1}^{r}} \left[t^{u+r(n+1)}_{i,s} \cdot (t_{i,s}-1)^{rn} \cdot \prod_{\substack{\tilde{i}\neq i\\ 1\le \tilde{i} \le m-1}}(\alpha_i t_{i,s}-\alpha_{\tilde{i}})^{rn}\right]\times\\
&\prod_{i=1}^{m-1}\left(\prod_{s_{1}<s_{2}}(t_{i,,s_{2}}-t_{i,s_{1}})\right)\times \prod_{\substack{(i_1,s_{1})<(i_2,s_{2}) \\ i_1<i_2\le m-1}}
(\alpha_{i_2}t_{i_2,s_{2}}-\alpha_{i_1}t_{i_1,s_{1}})\Biggr)\\
&=\pm\prod_{i=1}^{m-1}\alpha^{(2n+1)r^2}_i \bigcirc_{s^{\prime}=1}^{r}\varphi_{1,x,s^{\prime}}\left({\prod_{s=1}^{r}} \left[t^{u}_{m,s} \cdot (t_{m,s}-1)^{rn}\right] \times\prod_{s_{1}<s_{2}}(t_{m,s_{2}}-t_{m,s_{1}})\right) \\
&\times c_{n,u+r(n+1),m-1}\prod_{1\le i_1<i_2\le m-1}(\alpha_{i_2}-\alpha_{i_1})^{(2n+1)r^2}.
\end{align*}}}
Thus we obtain
\begin{align*}
c_{n,u,m}&=\pm\bigcirc_{s^{\prime}=1}^{r}\varphi_{1,x,s^{\prime}}\left({\prod_{s=1}^{r}} \left[t^{u}_{m,s} \cdot (t_{m,s}-1)^{rn}\right] \times\prod_{s_{1}<s_{2}}(t_{m,s_{2}}-t_{m,s_{1}})\right)c_{n,u+r(n+1),m-1}\\
&=\pm \prod_{i=1}^{m}\left(\bigcirc_{s^{\prime}=1}^{r}\varphi_{1,x,s^{\prime}}\left(\prod_{s=1}^{r}
 \left[t^{u+(i-1)r(n+1)}_{s} \cdot (t_{s}-1)^{rn}\right] \times\prod_{s_{1}<s_{2}}(t_{s_{2}}-t_{s_{1}})
\right)\right).
\end{align*}
We are then in a position to conclude.
Indeed, using the definition of the operators $\varphi_{1,x,s}$, the composition of these operators is nothing but an integral over $[0,1]^r$, more precisely~:~
\begin{align*}
&\bigcirc_{s^{\prime}=1}^{r}\varphi_{1,x,s^{\prime}}\left({\prod_{s=1}^{r}} \left[t^{u}_{s} \cdot (t_{s}-1)^{rn}\right] \cdot \prod_{s_{1}<s_{2}}(t_{s_{2}}-t_{s_{1}})\right)=\\
&\prod_{s^{\prime}=1}^{r}\dfrac{1}{(s^{\prime}-1)!}\int^1_{0}\cdots\int^1_{0}{\prod_{s=1}^{r}} \left[t^{u+x}_{s} (t_{s}-1)^{rn}{\rm{log\,}}^{s-1}\dfrac{1}{t_{s}}\right] \cdot \prod_{s_{1}<s_{2}}(t_{s_{2}}-t_{s_{1}})\prod_{s=1}^{r}dt_{s}\,
\end{align*}
then a direct computation enables us to show that this last integral does not vanish,
which yields the Proposition $\ref{non zero det}$.

The statement of Theorem \ref{Lerch} now follows from the fact that the determinant   is a non-vanishing algebraic constant.
We multiply  the determinant by the denominator to obtain that it is an algebraic integer. Then the absolute value of the norm $\geq 1$,
that gives a linear independence measure.

\section{Examples}

We show here two examples of linear independent polylogarithms, which are  shown by our criterion.
\begin{example}
Put $r=m=10$ and $x=0$. Take $\boldsymbol{\alpha}:=(1,1/2,\ldots,1/10)$ and $\beta=b$ with $|b|\ge e^{2715}$. Then we have $D(\boldsymbol{\alpha},b)=d_{10}=2520$.
Since we have the inequalities:
\begin{align*}
{\rm{log\,}}2520<7.84, \ {\rm{log\,}}3<1.10, \ {\rm{log\,}}(5/2)<0.92,\,\,\,
\end{align*}
we have
$$\,\,\,{\rm{log\,}}|b|>100(10+{\rm{log\,}}2520+10\,{\rm{log\,}}(5/2))+10\,{\rm{log\,}}3\,.$$
Then the $10^{2}+1$ numbers
\begin{align*}
1, {\rm{Li}}_1(1/b),\ldots, {\rm{Li}}_{10}(1/b),\ldots, {\rm{Li}}_1(1/(10b)),\ldots, {\rm{Li}}_{10}(1/(10b))
\end{align*}
are linearly independent over $\mathbb{Q}$.
\end{example}
\begin{example}
Let $k\ge 2$ be an integer, set $r=m=10^k$, $x=0$. Take $\boldsymbol{\alpha}:=(j)_{1\le j\le 10^k}$ and $\beta=b\in \Z$ with $|b|\ge {\rm{exp}}(2\cdot 10^{3k})$.
Since $k\ge 2$, we can easily verify
\begin{align*}
{\rm{log\,}}|b|&>(rm+1){\rm{log\,}}10^k+\left(r^2m(1+{\rm{log\,}}(5/2))+r{\rm{log\,}}3\right)\\
              &=k(10^{2k}+1){\rm{log\,}}10+\left(10^{3k}(1+{\rm{log\,}}(5/2))+10^k{\rm{log\,}}3\right)\enspace.
\end{align*}
Then
the $10^{2k}+1$ numbers
\begin{align*}
1, {\rm{Li}}_1(1/b),\ldots, {\rm{Li}}_{10^k}(1/b),\ldots, {\rm{Li}}_1(10^k/b),\ldots, {\rm{Li}}_{10^k}(10^k/b)\enspace,
\end{align*}
are linearly independent over $\mathbb{Q}$. For instance, we take $r=m=10^4$ and $b=3^{2\cdot 10^{12}}$ then the $10^{8}+1$ numbers
\begin{align*}
1, {\rm{Li}}_1(1/3^{2\cdot 10^{12}}),\ldots, {\rm{Li}}_{10^4}(1/3^{2\cdot 10^{12}}),\ldots, {\rm{Li}}_1(10^4/3^{2\cdot 10^{12}}),\ldots, {\rm{Li}}_{10^4}(10^4/3^{2\cdot 10^{12}})
\end{align*}
are all linearly independent over $\mathbb{Q}$.
\end{example}
\vspace{-5mm}

redpajama_set_name: RedPajamaArXiv


text:
#include "stdafx.h"
#include "WebScript.h"

CWebScript::CWebScript()
{
	pDoc	= NULL;
	m_vp.Clear();
}

CWebScript::~CWebScript()
{
	Close();
}

BOOL CWebScript::SetWeb(CHtmlView *pView)
{
	if ( pDoc )	return false;
	LPDISPATCH pdisp = pView->GetHtmlDocument();

	HRESULT hr = pdisp->QueryInterface(IID_IHTMLDocument2,(void**)&pDoc);

	if	(FAILED(hr))	return FALSE;

	Load();
	pdisp->Release();
	return true;
}

void CWebScript::Close()
{
	m_vp.Clear();
	pDoc	= NULL;
}

BOOL CWebScript::GetJScript(CComPtr<IDispatch>& spDisp)
{
	HRESULT hr = pDoc->get_Script(&spDisp);
	ATLASSERT(SUCCEEDED(hr));
	return SUCCEEDED(hr);
}

BOOL CWebScript::GetJScripts(CComPtr<IHTMLElementCollection>& spColl)
{
	HRESULT hr = pDoc->get_scripts(&spColl);
	ATLASSERT(SUCCEEDED(hr));
	return SUCCEEDED(hr);
}

BOOL CWebScript::Fire(CString stFunc,CComVariant* pVarResult)
{
	CStringArray paramArray;
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CComVariant* pVarResult)
{
	CStringArray paramArray;
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	paramArray.Add(strArg2);
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	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
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	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CString strArg7,CComVariant* pVarResult)
{
	CStringArray paramArray;
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	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	paramArray.Add(strArg6);
	paramArray.Add(strArg7);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CString strArg7,CString strArg8,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	paramArray.Add(strArg6);
	paramArray.Add(strArg7);
	paramArray.Add(strArg8);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CString strArg7,CString strArg8,CString strArg9,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	paramArray.Add(strArg6);
	paramArray.Add(strArg7);
	paramArray.Add(strArg8);
	paramArray.Add(strArg9);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CString strArg7,CString strArg8,CString strArg9,CString strArg10,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	paramArray.Add(strArg6);
	paramArray.Add(strArg7);
	paramArray.Add(strArg8);
	paramArray.Add(strArg9);
	paramArray.Add(strArg10);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CString strArg7,CString strArg8,CString strArg9,CString strArg10,CString strArg11,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	paramArray.Add(strArg6);
	paramArray.Add(strArg7);
	paramArray.Add(strArg8);
	paramArray.Add(strArg9);
	paramArray.Add(strArg10);
	paramArray.Add(strArg11);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,CString strArg1,CString strArg2,CString strArg3,CString strArg4,CString strArg5,CString strArg6,CString strArg7,CString strArg8,CString strArg9,CString strArg10,CString strArg11,CString strArg12,CComVariant* pVarResult)
{
	CStringArray paramArray;
	paramArray.Add(strArg1);
	paramArray.Add(strArg2);
	paramArray.Add(strArg3);
	paramArray.Add(strArg4);
	paramArray.Add(strArg5);
	paramArray.Add(strArg6);
	paramArray.Add(strArg7);
	paramArray.Add(strArg8);
	paramArray.Add(strArg9);
	paramArray.Add(strArg10);
	paramArray.Add(strArg11);
	paramArray.Add(strArg12);
	return Fire(stFunc,paramArray,pVarResult);
}

BOOL CWebScript::Fire(CString stFunc,int f,int s,int t,CComVariant* pVarResult)
{
	CComPtr<IDispatch> spScript;
	if(!GetJScript(spScript))
		return false;

	CComBSTR bstrMember(stFunc);
	DISPID dispid = NULL;
	HRESULT hr = spScript->GetIDsOfNames(IID_NULL,&bstrMember,1,
											LOCALE_SYSTEM_DEFAULT,&dispid);
	if(FAILED(hr))
		return false;

	int arraySize = 3;

	DISPPARAMS dispparams;
	memset(&dispparams, 0, sizeof dispparams);
	dispparams.cArgs = 3;
	dispparams.rgvarg = new VARIANT[3];
	dispparams.rgvarg[0].intVal	= t;
	dispparams.rgvarg[1].intVal	= s;
	dispparams.rgvarg[2].intVal	= f;
	dispparams.rgvarg[0].vt	= VT_INT;
	dispparams.rgvarg[1].vt	= VT_INT;
	dispparams.rgvarg[2].vt	= VT_INT;
	dispparams.cNamedArgs = 0;

	EXCEPINFO excepInfo;
	memset(&excepInfo, 0, sizeof excepInfo);
   	CComVariant vaResult;
	UINT nArgErr = (UINT)-1;  // initialize to invalid arg

	hr = spScript->Invoke(dispid,IID_NULL,0,
							DISPATCH_METHOD,&dispparams,&vaResult,&excepInfo,&nArgErr);

	delete [] dispparams.rgvarg;
	if(FAILED(hr))
		return false;

	if(pVarResult)
		*pVarResult = vaResult;

	return true;
}

BOOL CWebScript::Fire(CString stFunc,int f,int s,CComVariant* pVarResult)
{
	CComPtr<IDispatch> spScript;
	if(!GetJScript(spScript))
		return false;

	CComBSTR bstrMember(stFunc);
	DISPID dispid = NULL;
	HRESULT hr = spScript->GetIDsOfNames(IID_NULL,&bstrMember,1,
											LOCALE_SYSTEM_DEFAULT,&dispid);
	if(FAILED(hr))
		return false;

	int arraySize = 2;

	DISPPARAMS dispparams;
	memset(&dispparams, 0, sizeof dispparams);
	dispparams.cArgs = 2;
	dispparams.rgvarg = new VARIANT[2];
	dispparams.rgvarg[0].intVal	= s;
	dispparams.rgvarg[1].intVal	= f;
	dispparams.rgvarg[0].vt	= VT_INT;
	dispparams.rgvarg[1].vt	= VT_INT;
	dispparams.cNamedArgs = 0;

	EXCEPINFO excepInfo;
	memset(&excepInfo, 0, sizeof excepInfo);
   	CComVariant vaResult;
	UINT nArgErr = (UINT)-1;  // initialize to invalid arg

	hr = spScript->Invoke(dispid,IID_NULL,0,
							DISPATCH_METHOD,&dispparams,&vaResult,&excepInfo,&nArgErr);

	delete [] dispparams.rgvarg;
	if(FAILED(hr))
		return false;

	if(pVarResult)
		*pVarResult = vaResult;

	return true;
}

BOOL CWebScript::Fire(CString stFunc,int f,CString starg,CComVariant* pVarResult)
{
	CComPtr<IDispatch> spScript;
	if(!GetJScript(spScript))
		return false;

	CComBSTR bstrMember(stFunc);
	DISPID dispid = NULL;
	HRESULT hr = spScript->GetIDsOfNames(IID_NULL,&bstrMember,1,
											LOCALE_SYSTEM_DEFAULT,&dispid);
	if(FAILED(hr))
		return false;

	int arraySize = 2;

	DISPPARAMS dispparams;
	memset(&dispparams, 0, sizeof dispparams);
	dispparams.cArgs = 2;
	dispparams.rgvarg = new VARIANT[2];
	CComBSTR bstr = starg;
	bstr.CopyTo( &dispparams.rgvarg[0].bstrVal );
	dispparams.rgvarg[1].intVal	= f;
	dispparams.rgvarg[0].vt	= VT_BSTR;
	dispparams.rgvarg[1].vt	= VT_INT;
	dispparams.cNamedArgs = 0;

	EXCEPINFO excepInfo;
	memset(&excepInfo, 0, sizeof excepInfo);
   	CComVariant vaResult;
	UINT nArgErr = (UINT)-1;  // initialize to invalid arg

	hr = spScript->Invoke(dispid,IID_NULL,0,
							DISPATCH_METHOD,&dispparams,&vaResult,&excepInfo,&nArgErr);

	delete [] dispparams.rgvarg;
	if(FAILED(hr))
		return false;

	if(pVarResult)
		*pVarResult = vaResult;

	return true;
}

BOOL CWebScript::Fire( CString stFunc,int n)
{
	CComPtr<IDispatch> spScript;
	if(!GetJScript(spScript))
		return false;

	CComBSTR bstrMember(stFunc);
	DISPID dispid = NULL;
	HRESULT hr = spScript->GetIDsOfNames(IID_NULL,&bstrMember,1,
											LOCALE_SYSTEM_DEFAULT,&dispid);
	if (FAILED(hr))	return false;

	DISPPARAMS dispparams;
	memset(&dispparams, 0, sizeof dispparams);
	dispparams.cArgs = 1;
	dispparams.rgvarg = new VARIANT;
	dispparams.rgvarg->intVal	= n;
	dispparams.rgvarg->vt		= VT_INT;
	dispparams.cNamedArgs = 0;

	EXCEPINFO excepInfo;
	memset(&excepInfo, 0, sizeof excepInfo);
   	CComVariant vaResult;
	UINT nArgErr = (UINT)-1;  // initialize to invalid arg

	hr = spScript->Invoke(dispid,IID_NULL,0,
							DISPATCH_METHOD,&dispparams,&vaResult,&excepInfo,&nArgErr);

	delete [] dispparams.rgvarg;
	if(FAILED(hr))
		return false;

	return true;
}

BOOL CWebScript::Fire(CString stFunc, CStringArray& paramArray,CComVariant* pVarResult)
{
	CComPtr<IDispatch> spScript;
	if(!GetJScript(spScript))
		return false;

	CComBSTR bstrMember(stFunc);
	DISPID dispid = NULL;
	HRESULT hr = spScript->GetIDsOfNames(IID_NULL,&bstrMember,1,
											LOCALE_SYSTEM_DEFAULT,&dispid);
	if(FAILED(hr))
		return false;

	int arraySize = paramArray.GetSize();

	DISPPARAMS dispparams;
	memset(&dispparams, 0, sizeof dispparams);
	dispparams.cArgs = arraySize;
	dispparams.rgvarg = new VARIANT[dispparams.cArgs];

	for( int i = 0; i < arraySize; i++)
	{
		CComBSTR bstr = paramArray.GetAt(arraySize - 1 - i); // back reading
		bstr.CopyTo(&dispparams.rgvarg[i].bstrVal);
		dispparams.rgvarg[i].vt = VT_BSTR;
	}
	dispparams.cNamedArgs = 0;

	EXCEPINFO excepInfo;
	memset(&excepInfo, 0, sizeof excepInfo);
   	CComVariant vaResult;
	UINT nArgErr = (UINT)-1;  // initialize to invalid arg

	hr = spScript->Invoke(dispid,IID_NULL,0,
							DISPATCH_METHOD,&dispparams,&vaResult,&excepInfo,&nArgErr);

	delete [] dispparams.rgvarg;
	if(FAILED(hr))
		return false;

	if(pVarResult)
		*pVarResult = vaResult;

	return true;
}

void CWebScript::Load()
{
	m_vp.Clear();
	m_vp.pRoot->New("JScript");

	CComPtr<IHTMLElementCollection> spScripts;
	if(!GetJScripts(spScripts))
		return;

	long count = 0;
	HRESULT hr = spScripts->get_length(&count);
	ATLASSERT(SUCCEEDED(hr));

	long idx = -1;
	while(count--)
	{
		idx++;

		CComPtr<IDispatch> spDispatch;
		CComVariant name(idx),index(idx);
		hr = spScripts->item(name,index,(IDispatch**)&spDispatch);
		ATLASSERT(SUCCEEDED(hr));
		if(FAILED(hr))
			continue;

		CComPtr<IHTMLScriptElement> spScriptObj;
		hr = spDispatch->QueryInterface(IID_IHTMLScriptElement,(void**)&spScriptObj);
		ATLASSERT(SUCCEEDED(hr));
		if(FAILED(hr))
			continue;

		CComBSTR bstr;
		hr = spScriptObj->get_text(&bstr);
		ATLASSERT(SUCCEEDED(hr));

		CString strText(bstr);
		CString stScript;
		vToken tok(strText);
		tok.SetSeparator("\r\n");

		vTag *pCur;
		int n=0;
		for (int i=0;i<tok.GetSize();i++)
		{
			vToken tk(tok[i]);
			tk.SetSeparator(" \t");

			BOOL bFunction=FALSE;
			for (int j=0;j<tk.GetSize();j++)
			if ( tk[j]=="function")
			{
				stScript.Empty();
				vString tmp = tk[j+1];

				vToken tkTag(tmp);
				tkTag.SetSeparator("()");

				pCur	= m_vp.pRoot->Add((char*)tkTag[0]);	//
				tmp	= tok[i];	//
				pCur->AddScript(tmp);	//
				bFunction	= TRUE;
			}

			if (!bFunction)	pCur->AddScript(tok[i]);	//
		}

	}

	/*
	for (int i=0;i<3;i++)
	{
		vTag *p	= m_vp.pRoot->Search(i);
		CString st = p->stScript;
	}
	*/
}

redpajama_set_name: RedPajamaGithub


Muzio & Rovero Insurance Services provides insurance to businesses in Mill Valley, San Francisco, San Rafael, Redwood City, Healdsburg, Napa, and surrounding areas.
No matter what business you're in, each workday is dedicated to doing your job well and providing your customers with the best service possible. Muzio & Rovero Insurance Services would like to return the favor by giving you the coverage you need to ensure the security of your business.
At Muzio & Rovero Insurance Services, we're committed to keeping your business protected – so you can focus on your day-to-day operations. And with the right insurance plan, your business will be covered from top to bottom.


When sixteen-year-old Mia returns home from swim practice, she discovers her mother attempting suicide and rushes her to the hospital just in time. But once they return home, there is a suffocating silence between them. Mia keeps the event a secret and continues to train for an upcoming swim meet, suppressing her pain and shame – until the bubble bursts at the pivotal moment, forcing mother and daughter to confront their unspoken trauma.
DRYLAND trailer #2 from Liv Colliander on Vimeo.


Established in 2010, there are 23 handcrafted artisan ales on tap in Arcata, and now at a new location in Eureka. Shuffleboard and board games available. Kombucha on tap. There's an excellent food truck, but feel free to bring in your own food as well. Happy hour 2pm-6pm M-W.
Favorite local hang- out where many of the beer logos have been designed by the kinetic sculpture artist Duane Flatmo. Happy Hour Mon.-Fri. 4pm-6pm. Lunch and dinner.
Schedule a free 30 minute tour of their state-of-the-art facilities to learn a bit about the Lost Coast story and brewing process. 12-6pm, call for availability.
Adjacent to the Mad River in Blue Lake. The inside/outside tap room is frequented by regulars who enjoy the warm afternoons on the patio. Live music Tue-Sat. Hoppy Hour from 3pm-6pm M-F. Lunch and dinner.
Declared America's first certified organic brewery in 1999. The Taproom is located on the banks of the Eel River. Lunch and dinner. Happy hour 3pm-6:30pm M-F.


Responsible for improving a company's organic search results. Identifies areas of improvement and attempts to improve the clients' sites' rankings in major search engines.
Deliver international SEO projects for large websites.
Undertake online sector and competitor analysis.
Perform keyword research using industry standard tools.
Develop and implement innovative link building strategies.
Be a key point of knowledge on SEO, constantly keeping up-to-date on new trends.
To research and test new SEO techniques and to feed these back into the organization's SEO best practices.
To keep up to date and report on search engine updates.
Proven track record of evolving strategies to continually improve site and search performance to agreed targets.
Since 1999, MetricFox has been catering to various Online Marketing requests of companies from varying industries. As the name goes, we specialize in strategically placing the client's marketing plans.
© 2016 MetricFox. All Rights Reserved.


Conference "Organic production and food security"
On April, 23, 2015 III International Scientific and Practical Conference "Organic production and food security" took place in National Agroecological University in Zhytomyr.
The goal of the conference was to review the results of researches and practical experience, exchange views on solving urgent problems in the organic production. The participants of the conference visited the largest in Ukraine organic milk processing plant of the "Organic Milk" LLC.
The event was supported by the Swiss-Ukrainian project "Development of the organic market in Ukraine" (2012-2016), implemented by the Research Institute of Organic Agriculture (FiBL, Switzerland) and funded by the Swiss Confederation through the State Secretariat for Economic Affairs (SECO).


It's a good time for seeing planets. Venus moved past the Pleiades and Hyades star clusters in early April. Now it's Mercury's turn. In this image, you see the speedy little planet near the two famous star clusters on May 1, 2015. The Hyades star cluster is tangled in the branches at left. For the next week, Mercury makes its best appearance this year in the western sky after sunset. Venus is much higher above the horizon after sunset, and Jupiter higher still. Saturn rises in the east before midnight, its rings tilted dramatically, as it moves to its closest approach to Earth later in May.


text:
Deane Cameron
Music Biz Headlines, May 27, 2019
A look at superstar record producer Bob Ezrin (pictured), Drake is dissed for buffoonery, and John and Yoko's bed-in remembered. Others in the headlines include Deane Cameron, Lydia Persaud, Snotty Nose Rez Kids, Bookie, Ron Hawkins, Loud, Higher Brothers, Inuit pop, Aretha, Moby, Beastie Boys, Ultra Music Festival, and Rudimental.
Funeral For A Friend: Today We Bid Farewell To Deane Cameron
by FYI Staff
Canada's music industry and inbound foreign dignitaries, along with longstanding friends and family, gather this afternoon at Roy Thomson Hall to pay tribute to Deane Cameron, CM, affectionately known to many as Captain Canada.
Deane Cameron's Family Selects MusiCounts For Donations In His Memory
by Karen Bliss
The beloved Canadian music industry leader can be remembered by donations to the music education charity that was close to his heart.
The Measure Of The Man, As Told By Friends Of Deane Cameron
Updated June 13: -- If one is to be measured, let it be by one's actions, not just the words.
Words Of Kindness Regarding The Passing Of Deane Cameron
The outpouring of condolences yesterday from many was extraordinary by any stretch of the imagination.
RIP: Deane Cameron, Known To Many As 'Captain Canada'
by David Farrell
One of Canada's best-known ambassadors of its homegrown music scene died Thursday at age 65. Longtime friend and former bandmate Tom Cochrane called his loss a "sad day" and said he was devastated by the news, describing his friend as "one of the most generous people I know, as well as forever being one of Canada's musicians' staunchest supporters."
RIP: Paul White, Capitol Canada's 'Creative Pioneer'
Perhaps best known for being the first in North America to release recordings by the Beatles, his greatest contribution as head of Capitol Canada's A&R department was his enthusiasm for developing a roster of successful Canadian artists that included Anne Murray.

redpajama_set_name: RedPajamaCommonCrawl


text:
Some of the exclusive merchandise and items for sale at Dinsey's Star Wars Galaxy's Edge theme park will soon be available online on the ShopDisney website. The site will sell clothes and iconic lightsabers, and eventually even droids and toys. Bungie revealed plans to overhaul the Ghost system in Destiny 2. Starting with the Beyond Light expansion, which has a release date of Nov. 10, players can customize their Ghosts like they do with armor. PlayStation CEO Jim Ryan clarified whether the newest PlayStation console, the PlayStation 5, will be able to run the company's extensive back catalog of PlayStation 4 games. But how will it actually work? Dates, times and TV broadcast information for the 2020 Stanley Cup Final. The annual convention has gone all digital this year, celebrating the work of Black and Latinx developers, and developers of other marginalized racial backgrounds. The Expo will offer online panels, new game announcements, playable demos, and even a Steam sal…
In honor of the upcoming NCAA tournament and Ben Affleck's new drama "The Way Back," we rank the top 20 basketball movies.
Economy 14/03/2020 watinet
Ben Affleck stars as a reluctant basketball coach who struggles with addiction in director Gavin O'Connor's "The Way Back."
Shaquille O'Neal and Sean Connery will probably never share an alley-oop on a fast break, but in the sports movie subgenre of basketball flicks, they're both big stars.
The upcoming NCAA tournament is our annual reminder how much hoops and Hollywood intersect, from Jack Nicholson and Spike Lee being total devotees to the Lakers and Knicks, respectively, to Kobe Bryant winning an Oscar. On the big screen, underdog basketball stories are like comfort food, and there's pretty much a hall of fame of inspirational coaches: The latest is Ben Affleck's troubled former star player in need of redemption in the drama "The Way Back."
In honor of a new roundball flick as March Madness is on the way, let's rank the top 20 basketball movies of all time. (Apologies to "The Air Up There," "Eddie," "Juwanna Mann," "Like Mike" and others that just don't have the right stuff to make our starting lineup.)
Competing coaches Pete Bell (Nick Nolte, right) and Bobby Knight (himself) shake hands in "Blue Chips."
(Photo: BOB GREENE/PARAMOUNT)
Review: Ben Affleck soars in sobering 'The Way Back,' an addiction drama with underdog hoop dreams
Ben Affleck: 'The Way Back' star found playing basketball harder than playing an addict in his new movie
20. 'Blue Chips' (1994)
Bugaboos of college basketball, from gambling to corruption to illegal payments for athletes, are part of the playbook for the drama from director William Friedkin (The Exorcist). Nick Nolte is the irascible Western University coach based on Bobby Knight (who also makes an appearance) while co-stars ONeal and Penny Hardaway give the film some serious rim-rattling cred.
19. 'Cornbread, Earl and Me' (1975)
Its a trip to see a teenage Laurence Fishburne make his film debut in this drama starring NBA standout Jamaal Wilkes as Cornbread, a good-hearted and gawky talent whos the popular big man of his neighborhood on the way to a college career. Tragedy strikes involving cops and an innocent man, and residents want justice in a retro movie that still feels timely.
Derek Luke (right) stars as Texas Western point guard Bobby Joe Hill in the period drama "Glory Road."
(Photo: FRANK CONNOR/DISNEY)
18. 'Glory Road' (2006)
The story of how Texas Western coach Don Haskins (Josh Lucas) took an all-black starting lineup the first in NCAA history to a championship title in 1966 against mighty all-white Kentucky is an obvious inspiration for a big-screen underdog tale. Sure, it leans predictable but still works thanks to standout basketball action and strong performances by Lucas and Derek Luke.
Billy Crystal stars as quick-witted NBA referee Mickey Gordon in the romantic comedy "Forget Paris."
(Photo: BRUCE MCBROOM/CASTLE ROCK ENTERTAINMENT)
17. 'Forget Paris' (1995)
Billy Crystal directs and stars in the decently sweet romantic comedy as an NBA referee who takes his recently deceased fathers ashes to France and falls for an airline employee (Debra Winger). Their relationship goes through its ups and downs, but the best stuff is on the court, where Crystal gets to riff with an assortment of basketball icons, including Kareem Abdul-Jabbar and Isiah Thomas.
Leonardo DiCaprio plays a high school basketball player who gets involved in drugs and crime in "The Basketball Diaries."
(Photo: JOHN CLIFFORD/NEW LINE CINEMA)
16. 'The Basketball Diaries' (1995)
A scrawny Leonardo DiCaprio leads the biopic as high school hoopster and fledgling writer Jim Carroll whose life goes south when he gets mixed up in drugs and crime. The melodrama is pretty overwrought, though the cast is top notch, including Mark Wahlberg as one of Jims ill-behaved buddies and Ernie Hudson as a kindly sort who tries to get Jim right.
15. 'Above the Rim' (1994)
A whole lot of drama follows Duane Martin as a New York City high school senior who, while waiting to find out if hell play at Georgetown, gets caught up in the goings-on between a former star (Leon) and a drug dealer (Tupac Shakur). The basketball scenes are solid, and Shakur is especially good as the films vicious heavy.
James Bond III (far left) is the ballboy, Julius Irving's the star and Stockard Channing's the astrologer behind a successful bunch of misfits in "The Fish That Saved Pittsburgh."
(Photo: EVERETT COLLECTION)
14. 'The Fish That Saved Pittsburgh' (1979)
Philadelphia 76ers legend Dr. J Julius Erving headlines the goofy disco-era delight as Moses Guthrie, star of the lowly Pittsburgh Pythons. The team goes on a championship run when an astrologer (Stockard Channing) pitches the wacky idea to the owner (Jonathan Winters) that Moses should have only Pisces teammates. Only in the movies, folks.
Mekhi Phifer is a high school star who becomes the target of jealous vengeance by his teammate and best friend in "O," a modern reimagining of Shakespeare's "Othello."
(Photo: BOB BASHA/LIONSGATE FILMS)
13. 'O' (2001)
Tim Blake Nelsons intense teen drama is a decently effective modern reimagining of Shakespeares Othello. Mekhi Phifer is talented star player Odin whose bright future and relationship with young Desi (Julia Stiles) are at risk thanks to the shady and jealous machinations of the Iago-esque Hugo (Josh Hartnett).
Kyrie Irving stars as an old streetball legend who schools the new generation in "Uncle Drew."
(Photo: QUANTRELL D. COLBERT)
12. 'Uncle Drew' (2018)
It turns out putting a bunch of pro players in gray hair and old-age makeup is sort of hilarious. Kyrie Irving shines in the soulful comedy as the philosophical title character (whos got some ankle-breaking moves), Chris Webbers a power-forward preacher, Reggie Millers a nearly blind shooter and ONeal is the scene-stealing kung fu center.
Ken Carter (Samuel L. Jackson, center) gets real with his players during a game in "Coach Carter."
(Photo: TRACY BENNETT/PARAMOUNT PICTURES)
11. 'Coach Carter' (2005)
Samuel L. Jackson brings the coolness and loads of tough love as a coach who returns to his old high school, gives the youngsters life lessons and locks out his undefeated team and causes a ruckus when they dont make the grade in the classroom as well as on the court. (Plus, its worth a watch for the film debut of Channing Tatum and an Ashanti appearance.)
Bill Murray, Bugs Bunny and Michael Jordan draw up a play in "Space Jam."
(Photo: WARNER BROS.)
10. 'Space Jam' (1996)
Whether you grew up with Bugs Bunnys antics or His Airness, theres plenty to enjoy in the live-action/animated hybrid in which Michael Jordan and his golf partner Bill Murray are recruited by Bugs and the Looney Tunes gang to take on the Monstars, a super-duper squad of aliens with the abilities of NBA legends such as Patrick Ewing and Charles Barkley.
Sean Connery, left, is a reclusive novelist, and Rob Brown plays a scholarly athlete who balances basketball and writing in "Finding Forrester."
(Photo: GEORGE KRAYCHYK/COLUMBIA PICTURES)
9. 'Finding Forrester' (2000)
Gus Van Sants satisfying albeit schmaltzy drama centers on an African American star (Rob Brown) at an upper-crust private school whos as gifted a writer as he is a baller. The source of inspiration in this story isnt a coach but rather a reclusive Pulitzer-winning novelist (Sean Connery), who brings out the best in the youngster and vice versa.
Jack Cunningham (Ben Affleck) coaches a team of scrappy youngsters at his Catholic school alma mater in "The Way Back."
(Photo: RICHARD FOREMAN)
8. 'The Way Back' (2020)
Starring as an alcoholic onetime hoops star who returns to his alma mater to coach a losing team, Ben Affleck turns in one of his best performances in a drama that manages to buck the expectations, avoid formulaic hokum and deliver on a gripping addiction narrative.
Michael J. Fox gets hairy when hooping it up in "Teen Wolf."
(Photo: MGM/EVERETT COLLECTION)
7. 'Teen Wolf' (1985)
Michael J. Fox not only time-traveled in a DeLorean back in 85, he also transformed into a slam-dunking, van-surfing werewolf. The high school comedy stars Fox as a teen whose basketball skills get way better when he gets furry, though his new success brings extra coming-of-age problems.
Omar Epps and Sanaa Lathan are longtime friends and talented hoops players who take their relationship to the next level in "Love & Basketball."
(Photo: SHANE HARVEY/NEW LINE CINEMA)
6. 'Love & Basketball' (2000)
Theres some definite truth in advertising with that title: Sanaa Lathan and Omar Epps sizzle as the central couple of next-door neighbors who dream of becoming professional basketball players, reach their dreams and you guessed it fall for each other along the way.
Basketball star Kevin Garnett, left, stars as himself alongside Lakeith Stanfield and Adam Sandler in "Uncut Gems."
(Photo: WALLY McGRADY)
5. 'Uncut Gems' (2019)
If Die Hard is a Christmas movie, then Adam Sandlers two-hour anxiety fest is a hoops flick. Real NBA all-star Kevin Garnett plays a version of himself obsessed with a rare opal, Sandler is the gambling jeweler trying to make serious bank off the rock, and a close playoff game fuels the tension-filled last act.
Woody Harrelson and Wesley Snipes are a pair of rival hustlers-turned-teammates in "White Men Can't Jump."
(Photo: NEIL LEIFER/20TH CENTURY FOX)
4. 'White Men Cant Jump' (1992)
Pickup games, yo mama jokes and incandescent star chemistry fuel the comedy featuring a slick Wesley Snipes and geeky Woody Harrelson teaming up as a pair of trash-talking rival hustlers. The movie tackles relationships and race but allows for serious fun, as in the iconic scene in which Harrelson tries to dunk. As a great man once said, Its hard work being this good.
Gene Hackman, center, stars as the coach of small-town Hickory High in the underdog classic "Hoosiers."
(Photo: ORION PICTURES)
3. 'Hoosiers' (1986)
A film thats in the annals of most folks favorite sports movies, Hoosiers has an undeniable feel-good spirit with the 1950s-set story of disgraced college coach Norman Dale (a fabulous Gene Hackman), whos kind of an abusive jerk, taking small-town Hickory High on one heck of a run for the Indiana state championship. Its no Rudy, yet still an underdog classic.
"Hoop Dreams" followed high school star Arthur Agee and his dreams of making it to the NBA.
(Photo: FINE LINE FEATURES)
2. 'Hoop Dreams' (1994)
One of the best documentaries ever made definitely deserves to make this list. The film spectacularly uses the story of two black teens in Chicago, William Gates and Arthur Agee, and their pie-in-the-sky goals of playing in the NBA to highlight issues of class, race and education in America.
Jake Shuttlesworth (Denzel Washington, left) is a former hoops star and convicted felon trying to connect with his estranged son Jesus (Ray Allen) in Spike Lee's "He Got Game."
(Photo: DAVID LEE/TOUCHSTONE PICTURES)
1. 'He Got Game' (1998)
Spike Lees stupendous work features a fantastic Denzel Washington as Jake Shuttlesworth, a convict at Attica whos paroled in order to get his estranged son Jesus (Ray Allen), a top-ranked recruit, to go to the governors alma mater. Game offers great performances, a gripping narrative and an intriguing vibe of modern Americana given Lees deft use of Aaron Copland music.
Read or Share this story: https://www.usatoday.com/story/entertainment/movies/2020/03/09/march-madness-best-basketball-movies-ranked-even-the-way-back/4981841002/
Gaming on demand just got a whole lot better with Stadia from Google and Verizon internet. Get the Stadia Premier bundle with qualifying Fios internet plans.
Commentary: The coronavirus outbreak may persuade employers to adopt more flexible policies that allow remote work from home.
Bungie revealed plans to overhaul the Ghost system in Destiny 2. Starting with the Beyond Light expansion, which has a release date of Nov. 10, players can customize their Ghosts like they do with armor.
The United Nations celebrates the first-ever International Equal Pay Day—but women of color still aren't close.
More about The Last Word on MSNBC. Join The Last Word Lawrence O Donnell community for more news, videos and articles about politics & global change.


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Culture and heritage Toggle secondary navigation
North British Mine
The North British Mine loop walk is closed until further notice. at North British Mine, Thursday 28 March, 2019Archive
View all changed conditions for Maldon Historic Area
Gold was first found at Maldon in 1853, resulting in a rush of some 20,000 diggers to the alluvial gullies in the area. By the following year only 2000 remained. After the initial rush, alluvial mining continued using puddling machines powered by horse or steam, and companies were set up to mine the quartz reefs.
In the 1860s the amount of quartz reef gold produced at Maldon almost rivalled Bendigo, Victoria's richest quartz mining field, and the town grew rapidly with many shops, houses and other buildings. But this prosperity was short-lived, and from 1870 to 1900 Maldon remained the poor cousin to Bendigo. As gold production declined, the mines closed and the population dwindled.
The last mine, the North British, closed in 1926. Maldon grew and changed little through the 20th century and this preserved the historic buildings and features that remain today.
North British Mine Aluvial gold was discovered in the district by Captain John Mechosk, a German prospector, in 1853. Three years later, with the opening up of quartz reefs containing some of the…
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heritageSite buildingStructureRuins


Yeah! The Space Cowboys will be back on the Playa this year! They are hosting a nautical-themed fundraiser on Sunday, August 2nd. Ahh, taste the multi-layered irony as burners and breakbeat heads rock a yacht club while wearing sweaters around their necks. So put your dancing pants on and support the crew with the coolest art car around, the Unimog!


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Tom's of Maine to Help Green Sheryl Crow's "100 Miles From Memphis" Spring Tour and Hold Volunteer Events to Restore Natural Beauty in Key Cities
Tom's of Maine plans to bring back the natural beauty of smiles with its new, natural whitening toothpaste Simply White, and a fun partnership with Sheryl Crow's "100 Miles From Memphis" concert tour kicking off May 7. The project involves greening the concert tour, beautification events in key markets, a live chat with Sheryl Crow and an exclusive free song download for fans while supplies last.
"We wanted to find a memorable way to introduce Simply White and celebrate natural beauty," said Ellen Saksen, Oral Care Associate Brand Manager at Tom's of Maine. "Sheryl Crow is an American icon who shares our passion for the environment and healthy living. As we take Simply White on tour, we hope people will join us on the ground to bring back natural beauty and think about ways they can take care of their own natural beauty."
Bringing Back Natural Beauty with Simply White
Now available nationwide, Simply White toothpaste is clinically proven to whiten teeth and contains no bleaching ingredients. It provides whitening, cavity protection and fresh breath all in one and combines naturally sourced silicas to gently remove surface stains. As with all Tom's of Maine products, there are no artificial flavors or preservatives and no animal testing or animal ingredients.
Greening Crow's "100 Miles From Memphis" Tour
Through support from Tom's of Maine and Reverb, the music industry's leading environmental organization, Sheryl Crow's spring tour will feature sustainability changes such as biodiesel fuel, recycling bins at concert venues and a commitment to natural cleaning products, local and organic catering and reusable/biodegradable hospitality products, among other initiatives. At her tour venues, concertgoers can visit Tom's of Maine at the on-site Eco-Village to learn more and to receive a free sample of Simply White toothpaste while supplies last.
Working Together To Bring Back Natural Beauty
To help celebrate the Simply White launch, Tom's of Maine will be partnering with non-profits in select cities to help restore parks, rivers and gardens with volunteer beautification efforts. The events are scheduled for Boulder, CO (May 14), Eugene, OR (May 14), Boise, ID (May 14), and Salt Lake City (June 24). Local residents can join in the fun and help with projects fit for all ages. Events are open to the public space permitting or on a first come, first serve basis. Sign-up information can be obtained by calling the non-profit listed on the Tom's of Maine Facebook page.Tom's of Maine is providing funding for trees, flowers, supplies and equipment and will offer free Simply White samples as a thank you to participants while supplies last. Details, including time and location, are available atfacebook.com/tomsofmaine.
http://multivu.prnewswire.com/mnr/mnr_lib/201002/players/player-single.swf?job=49937
Enjoy A Free Song Download
Fans who can't join Tom's of Maine and Sheryl Crow on the road can visit the Tom's of Maine Facebook page for an exclusive free download of her live recording "Long Road Home – Live from the Pantages", while supplies last.
Join For A Live Chat on May 19
To discuss greening her tour and her passion for the environment, Crow will be doing a live, online video chat from the road on May 19 at 7 p.m. ET. She will take questions from fans who submit questions in advance or tune in for the chat live. For more information on the chat, visit the Tom's of Maine Facebook page.
Going Across Country: Sheryl Crow Tour Schedule
The "100 Miles From Memphis" tour, in partnership with Tom's of Maine, includes the following dates. For ticket information visitwww.sherylcrow.com/events.
May 8th Rochester, MN Taylor Arena
May 9th Duluth, MN Decc Auditorium
May 11th Waukegan, IL Genesee Theater
May 12th Omaha, NE Orpheum Theater
May 15th Boulder, CO Boulder Theater
May 19th Boise, ID Idaho Botanical Gardens
May 20th Eugene, OR Hult Center
May 21st Sonoma, CA Field of Dreams Park
June 24th Salt Lake City, UT Red Butte Garden Amphitheater
June 25th Aspen, CO Aspen Music Festival
June 27th Beaver Creek, CO Vilar Performing Arts Center
June 29th Sioux City, IA Orpheum Theater
June 30th Joliet, IL Rialto Square Theater
| Tags: Conservation, Eco Friendly, Environment, environmental programs, going green, green business, green celebrities, green living, green living news, sustainability
Cyclone Power Technologies' Licensee in China Completes Initial Milestones in Prototyping Stage


Bandol is considered as the only region where Mourvèdre reaches its full potential, thanks to its sun-drenched terroir. Mourvèdre, a late ripening grape variety, needs as much sun as it can get, but can at the same time be prone to drought stress, making is not always easy to find the right conditions. Bandol however, is perfect: 3,000 hours of sunshine, clay soils that retain water, cooling ocean breezes and shelter provided by an amphitheater formed by the Gros Cerveau in the south, Sainte-Baume to the North and Caume to the East. This does not mean however that it is a uniform region: clay and limestone around Le Castellet, sand and limestone around Sainte Anne d'Evenos (where Dupuy de Lôme makes fantastic wines, or red clay near Ollioulles (home of Terrebrune, which I'll talk about down the road), all these soil types lead to unique wines.
At its core, Mourvèdre is all about fruit. It may be hidden by layers of game, spice, wood or meat but when you peel these away with time, there is a brimming, shiny core of juicy plums and freshly picked blackberries. Vinification can go a long way in constructing these layers, or it can help to bring the fruit out in a much more direct way. In Bandol, they clearly take the long way around and the best estates can strike just the right balance between structure and nuance.
So if it is fruit you're after, more often than not you need to be patient. Youthful Bandol may sometimes be deceptively fruity on the nose, but I think that it is the other grape varieties luring you in only to smack you against a wall of tannic power for being foolish enough to tackle the wine too soon. The higher the percentage of Mourvèdre, the rawer the wine, almost begrudging you for stirring it from its slumber and knowing how much Mourvèdre was use can generally help you along in discovering when you can start approaching a wine.
Take the wines of the Bunan family for instance. Paul and Pierre Bunan entered the wine business in the sixties, and the family has since expanded it tobecome one of the largest independent producers of Bandol, yet not without continuously focusing on quality and character. They actually own two estates: Domaine Bunan up in the hills near La Cadière d'Azur (almost pure clay), and La Rouvière on the way up to Castellet (chalk and limestone running through). Aside from soil composition, the blend of the two wines differs slightly, but vinification is the same.
A comparison of 2013 proved to be enlightening: Bunan Moulin de Costes (70% Mourvèdre, 20% Cinsault, 10% Syrah) is black olives, red berries and black cherries with the tannins giving a nudge but not dominating, versus Chateau de la Rouvière, (90% Mourvèdre, 10% Syrah) which goes more on game, cloves and liquorice with a tight impression on the palate. The Moulin de Costes was the more approachable one, but that does not mean that it would not benefit from additional ageing, as proven by the 2002 I enjoyed a few weeks ago: blackcurrant, plum, olives and a bit of dust, yet juicy and fleshed out on the palate.
Of course, Mourvèdre is not the only factor in the blend that accounts for a wine's character and attitude in its youth. Grenache can play quite the important role as well, as it does not have the compact structure that necessarily requires a lot of time to get ready. I had the pleasure of enjoying Jean-Pierre Gaussen's 2008 (70% Mourvèdre, 30% Grenache) on two separate occasions over the summer, and the wine delivered on both. A little bit of reduction and stink that blows off, followed by liquorice, coffee, even a bit of chocolate. The palate is quite fleshed out, with juicy yet elegant fruit impressions. Tannins are present, but integrated and as a whole the wine has hit a high point on its maturity curve, great for yielding to temptation already, but equally rewarding for those who will be able to wait.
There is not a lot of Carignan planted in Bandol but what is there possesses the best quality you can find in the grape variety: age. Young Carignan has too much of everything: acidity, sugar (alcohol) and tannins, but as the vines grow older and yields go down, the true qualities surface, and a delightful combination of Middle-Eastern spices and earthiness is formed. There are only a couple of producers who still maintain the vines and actually use it in their blends, and my favorite by far is La Tour du Bon, run by the lovely Agnes Henry.
In vintages where yields are ok and where there is enough material to create a good entry level Bandol, a separate cuvee 'Saint Ferréol' containing only the best Mourvèdre and what little Carignan there is will be produced as well. If it's not possible, the Carignan will find its way into the entry Bandol as was the case n 2002 or 2015 for example. Like Mourvèdre, Carignan is a variety that needs to go through a calm phase, so it should not come as a surprise that the wines need a lot of time.
The 1998 Saint Ferréol (85% Mourvèdre, 15% Carignan) has just hit its prime: liquorice, cinnamon and a hint of nutmeg, followed by cherries and blueberries. Focused, more straight in character than most Bandol on the palate, with well-integrated tannins and generous fleshy fruit, only to end on a note of freshness, with everything being simply so incredibly balanced. A wine that will stay at its peak for at least another five years!
Now with all that talk on the red wines of Bandol, I forewent an economic reality: 73% of all wine produced is rosé (based on 2016). These are however not the pale and bland wines that have become so ubiquitous in the Provence, but wines of character, maturity and intensity. I love the ripe, succulent strawberries and pomegranate seeds that you can encounter in any decent Bandol rosé, combined with a whiff of garrigue. Add a rather tight acidity and a firm structure on the palate, and it should come as no surprise that particularly in their youth, these are food wines.
When it comes to rosé, Mourvèdre often plays second fiddle to Cinsault or (in increasingly rare cases, Grenache). Cinsault is a seriously underestimated variety, and its combination of earthiness with deliciously crunchy red fruit harmonizes perfectly with the structure and firmness that Mourvèdre brings to the table. Domaine Bunan's Moulin des Costes rosé is a wonderful example of graceful Cinsault (60%) with an edge to it. Appellation rules allow the inclusion of white varieties, namely Clairette, and Domaine de la Tour du Bon makes perhaps the sunniest expression of rosé that I have encountered in the region, even with a couple of years behind it.
There are estates that rely almost exclusively on Mourvèdre for their rosé production though, and it is with these wines, as with the reds, that the ageing potential of the variety really shines through. Pibarnon's rosé (65% Mourvèdre) is for me personally the most perfect yet stereotypical Bandol rosé with garrigue just jumping out of the glass regardless of the wine's age (as proven by the remarkable 2000 mentioned before).
The most atypical rosé wines that I have encountered in Bandol are those from Domaine de la Bégude. The estate itself is magnificent, set at the highest point of the appellation close to the Massif de Sainte Baume. 500 hectares, of which only 22 planted with vines, with a magnificent view all the way to the Mediterranean Sea, if ever there was a bias-inducing environment, this would be it! The estate itself is owned by Guillaume Tari, of Chateau Giscours fame, and the current president of the Syndicat des Vignerons.
The vinification style is decidedly modern, whereas all the vineyards are organically certified. The two rosés produced at the estate immediately stand out because of their colour: you would be forgiven if you thought you were in Bordeaux! The Estate rosé (80% Mourvèdre, 15% Grenache, 5% Cinsault) is dense, fruit driven but bright, whereas the Cuvée Irréductible (90% Mourvèdre & 10% Grenache) goes even further, fleshed out, more on the spicy side. It is a bit too much in its youth to be honest, so hopefully it will come around with a bit of bottle time.
I've loved Bandol for many years. The wine that got me interested was actually a Bunan, initially Mas de la Rouvière some time in the 1980s (I only drank the Château de la Rouvière after). I always wanted to visit, but I never have.
I've drunk a few nice Bandols this year, Pibarnon and some of the Tempier cuvées, but I only own some of Tempier's 2005. I very much enjoyed reading this piece.
Thanks for your kind comment! Coincidentally, it was Chateau de la Rouviere for me as well, tasted about six or seven years ago at the Real Wine Fair and stopping me in my tracks.
The big names find their way to the export markets, smaller ones are not always easy to find unfortunately, also as they have quite a strong domestic market.


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Q: How to add time to the current timestamp in Ansible? I am attempting to automate a few tasks in Ansible and while I've gotten everything else to work, I need to have a start_time and end_time variable that adds time to the current timestamp as the ServiceNow system does not accept a current timestamp when creating a change request.
For example, I have a variable start_time as "{{ lookup('pipe', date +\"%Y-%m-%d %r\"') }}" but I would need that to be the current time +5 minutes for example.  Likewise on the end_date, but something like +15 minutes.

A: Something like?
{{ ansible_date_time.date }}
{{ ansible_date_time.hour|int +1|int }}
{{ ansible_date_time.minute|int +15|int }}

A: If you are going to use lookup_plugins and pipe, you may just add 5 mins by using date +"%Y-%m-%d %r" -d "5 mins".
Thanks to

*

*Incrementing current date by 5 minutes
Nevertheless it is recommend to use Ansible variables like ansible_date_time from facts.
date_time:
  date: '2021-11-23'
  day: '23'
  epoch: '1637678908'
  hour: '15'
  iso8601: '2021-11-23T14:48:28Z'
  iso8601_basic: 20211123T154828773386
  iso8601_basic_short: 20211123T154828
  iso8601_micro: '2021-11-23T14:48:28.773386Z'
  minute: '48'
  month: '11'
  second: '28'
  time: '15:48:28'
  tz: CET
  tz_offset: '+0100'
  weekday: Tuesday
  weekday_number: '2'
  weeknumber: '47'
  year: '2021'

Form there you could use epoch, minute or what would fit to your use case.

A: You can use the strftime filter to format times, Ansible's built-in fact gathering to get the current time, and a little bit of arithmetic to add an offset.
- hosts: localhost
  tasks:
    - debug:
        msg:
          - "{{ '%Y-%m-%d %r' | strftime(ansible_facts.date_time.epoch | int + 300 ) }}"
          - "{{ '%Y-%m-%d %r' | strftime(ansible_facts.date_time.epoch | int + 900 ) }}"

TASK [debug] *******************************************************************
ok: [localhost] => {
    "msg": [
        "2021-11-23 01:10:45 PM",
        "2021-11-23 01:20:45 PM"
    ]
}

redpajama_set_name: RedPajamaStackExchange


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The Kalamazoo Symphony Orchestra Invites Second Music Director Candidate, Edwin Outwater, to the Podium on November 11, in Brahms & Rachmaninoff, Featuring Pianist, Anna Vinnitskaya
Oct 31, 2017 | Entertainment
In recognition of Veteran's Day, and in appreciation of their service, the KSO has offered all servicemen, and a guest, free tickets to attend this symphonic concert.
KALAMAZOO, MI, October 31, 2017 /24-7PressRelease/ — The Kalamazoo Symphony Orchestra, with guest conductor and Music Director Candidate, Edwin Outwater, presents a program of symphonic masterpieces by Brahms and Rachmaninoff. Anna Vinnitskaya, winner of the 2007 Queen Elisabeth Contest, solos with Rachmaninoff's brilliant Piano Concerto No. 3–a piece ideally suited to her interpretive gifts. Maestro Outwater rounds out the program with Brahms pastoral Symphony No. 2.
"I'm thrilled to return to Kalamazoo as a music director candidate," states Outwater. "I really enjoyed my last appearance with them, and hope to go even further musically together this time around. The works on this program by Brahms and Rachmaninoff are among the greatest in the orchestral repertoire, and I can't wait to bring them to life with this wonderful orchestra."
In recognition of Veteran's Day, the KSO has invited all servicemen, and a guest, to attend this concert at no cost. The evening will begin with a special dedication of Service Men on Parade, led by KSO's Resident Conductor/Interim Music Director, Daniel Brier, and will feature WWMT3's news anchor, Andy Dominianni. A reception will be held for the Veterans in the Grand Tier Lobby at intermission. Tickets can be reserved at the Miller Auditorium Box Office the night of the concert or by calling 269.387.2300.
The Kalamazoo Symphony Orchestra is currently in the midst of a search for its next Music Director, to follow the long tenure of Maestro Raymond Harvey. Edwin Outwater is the second of five candidates to visit the city and work with the orchestra. Mr. Outwater will spend a full week in Kalamazoo meeting with various groups from the KSO, rehearsing with the orchestra and talking with musicians before the culminating event of the concert on Saturday.
As a candidate for the KSO's Music Director position, Edwin Outwater adds, "I also look forward to getting to know this wonderful community better during my week there."
Edwin Outwater
Edwin Outwater is Director of Summer Concerts at the San Francisco Symphony (SFS), and was recently appointed Artistic Director of the Eastern Sierra Symphony, a summer festival in Mammoth Lakes, California. After a remarkable decade as Music Director of the Kitchener-Waterloo Symphony (KWS), he begins as their Conductor Laureate in 2017. Equally adept at interpreting canonical masterworks, premiering new commissions, and connecting audiences with repertoire beyond the mainstream, the American conductor is, as San Francisco Classical Voice notes, "headed for a top-tier future."
Outwater's final season as Music Director of the KWS continued their groundbreaking Intersections concerts, exploring the relationship between music and science in a collaboration with the Institute for Quantum Computing and cellist Johannes Moser. In his third year with SFS, Outwater led the orchestra in performances including Beethoven Symphony No. 9 and a multi-concert collaboration with jazz pianist Makoto Ozone. He also curated and conducted performances in the trailblazing series SoundBox in SFS's new nightclub-style performance space.
Recent guest appearances featured returns to the New World and Chicago Symphonies, and the BBC National Orchestra of Wales. He also debuted with the Brussels Philharmonic, the Grant Park Festival, and the Montreux Jazz Festival, and appeared at the Kennedy Center in their new music series led by composer Mason Bates. Outwater has conducted the New York and Los Angeles Philharmonic Orchestras, as well as symphony orchestras including Baltimore, Detroit, Houston, and Seattle. In Canada, he has led the National Arts Centre Orchestra and the symphonies of Toronto, Calgary, Edmonton, Winnipeg, and Victoria. International appearances include the Tokyo Metropolitan Orchestra, Tokyo Symphony, Kyoto Symphony, BBCNOW, the New Zealand Symphony, among many others.
A native of Santa Monica, California, Edwin Outwater graduated cum laude in English literature from Harvard University, where he was music director of the Bach Society Orchestra and the a cappella group Harvard Din and Tonics. He received his master's in conducting from UC Santa Barbara, where he studied with Heiichiro Ohyama and Paul Polivnick, besides studying music theory and composition with John Stewart, Joel Feigin, and Leonard Stein.
Anna Vinnitskaya
Hailed by the Washington Post as a "lioness at the keyboard," Anna Vinnitskaya has emerged as one of the most exciting pianists of her generation. She made her U.S. recital debut in 2010 at the Kennedy Center to great critical acclaim, and subsequently made her U.S. orchestral debut with the Indianapolis Symphony under Krzysztof Urbanski. In 2017, she debuted with the Cincinnati Symphony playing Prokofiev's Concerto No. 2 under the direction of Gilbert Varga. She returns to Indianapolis in 2017/2018 as one of three world-renowned pianists playing all five Prokofiev concerti in one weekend.
Worldwide, Ms. Vinnitskaya's performances include the Athens State Orchestra, City of Birmingham Symphony, Brno Philharmonic/Czech Republic, Deutsche Kammerphilharmonie/Bremen, Israel Philharmonic Orchestra, Minas Gerais Philharmonic/Brazil, Munich Philharmonic, the radio symphony orchestras of Berlin, Hamburg, and Stuttgart, National Orchestra of Russia, New Japan Philharmonic, Oslo Philharmonic, the Tchaikovsky Symphony Orchestra/Moscow, and the KBS Symphony Orchestra/Seoul. Conductors with whom she has collaborated include Marek Janowski, Dimitri Jurowski, Louis Langree, Kyrill Petrenko, Krzysztof Urbanski, and Gilbert Varga, among others.
Anna Vinnitskaya is also an active recording artist and chamber musician. Among her numerous awards and prizes, she was Gold Medalist at the 2007 Queen Elisabeth Competition in Brussels and winner of the Leonard Bernstein Award at the 2008 Schleswig-Holstein Music Festival. From 2011-2014, she was a regular participant in the concert series "Junge Wilde" at the Konzerthaus in Dortmund. Her latest project will be recording the complete Bartok concertos with Jukka-Pekka Saraste and the WDR Symphony/Cologne where she is artist- in-residence during the 2016/2017 season.
Born in 1983 in Novorossiysk, Russia, Ms. Vinnitskaya played her first orchestral concert at the age of eight. After studies in Russia with Sergej Osipenko, she became a protege of Evgeni Koroliov at the Hamburg University of Music and Theater in 2002, where she herself was appointed Professor of Piano in 2009.
CALENDAR LISTING
Kalamazoo Symphony Orchestra presents Symphonic Series Concert:
Brahms & Rachmaninoff
Saturday, November 11, 2017 – Miller Auditorium at 8pm (Wiser Financial Prelude begins at 7pm)
Edwin Outwater, Conductor
Anna Vinnitskaya, Piano
HAMLISCH Service Men on Parade
Daniel Brier, Conductor & featuring Andy Dominianni
RACHMANINOFF Piano Concerto No. 3 in D minor, Op. 30
The Kalamazoo Symphony Orchestra, with guest conductor & Music Director Candidate, Edwin Outwater, presents a program of symphonic masterpieces by Brahms and Rachmaninoff. Anna Vinnitskaya, winner of the 2007 Queen Elisabeth Contest, solos with Rachmaninoff's brilliant Piano Concerto No. 3–a piece ideally suited to her interpretive gifts. Maestro Outwater rounds out the program with Brahms pastoral Symphony No. 2. In recognition of Veteran's Day, the KSO has invited all servicemen, and a guest, to attend this concert at no cost. The evening will begin with a special dedication of Service Men on Parade. A reception will be held for the Veterans in the Grand Tier Lobby at intermission. Tickets can be reserved at the Miller Auditorium Box Office the night of the concert or by calling 269.387.2300.
Tickets: $60 – $24
Veteran tickets can be reserved at the Miller Box Office the night of or in advance at 269.387.2300
Student discounts available.
$5 Student Rush tickets with valid ID, and $10 Harvey's Hideaway seating available at the box office the night of the concert.
For tickets, visit www.kalamazoosymphony.com, call the KSO Box Office at (269) 349-7759 or call Miller Auditorium Ticket Office at (269) 387-2300.
Visit www.kalamazoosymphony.com for up-to-date information, details and schedules. Prices, artists, dates, time and program are subject to change without notice.
The Kalamazoo Symphony Orchestra receives major support from the Irving S. Gilmore Foundation, the Kalamazoo Community Foundation, the National Endowment for the Arts, and the Michigan Council for Arts and Cultural Affairs. The Kalamazoo Symphony Orchestra also receives generous support from other local, state and national foundations, as well as private and corporate support. For more information, visit www.kalamazoosymphony.com.
ABOUT THE KALAMAZOO SYMPHONY ORCHESTRA
Founded in 1921, the Kalamazoo Symphony Orchestra is Southwest Michigan's premier musical organization, providing musical enrichment to over 80,000 adults and youth per year. The third-largest professional orchestra in the state, the KSO has won numerous awards and grants, including the Met Life Award for Arts Access in Underserved Communities, the National Endowment for the Arts for its extensive education programs, and a major Ford Foundation grant to found its innovative Artist-in-Residence program.
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A line forms at the front door (usually) and everyone keeps their own place in line. Some people will walk around and talk once they have established a place in line and this is generally accepted. Leaving the property usually means you have abandoned your place in line and will have to go to the end of the line if you return. Many an argument has started over this situation. When you get in line remember you are responsible for knowing who you are behind and even how many are in front of you.
This is the practice of gathering a bunch of items in a pile and taking your time looking and making a decision on whether to buy or not. We do not allow this type of practice and require you to buy what you pile. Hoarding effectively removes the items from the sale during a time when the sale might have the most potential buyers. This is not fair to the estate sale company or others attending the sale.
Some companies allow you to mark items sold that you do not want to carry around the sale with you or are to large to carry. However, we only allow our workers to mark an item sold. Once an item is marked sold you are obligated to purchase the item because you have effectively removed it from the sale. It is not fair to the company and others to hold an item by marking it sold and then changing your mind. Someone else may have been ready to purchase the item and then left because it was marked sold. Also removing a sold sticker is not fair to the person who was first to decide to purchase the item.
Please try to leave your large purse at home when attending estate sales. Companies are leery of small items finding their way into a large purse or bag. Please try to understand the company's position on this. They may know you but they do not know everyone and have to treat all customers the same.
It is not a good idea to bring children to estate sales. There are several reasons for this and most important is the safety of the child. There are huge crowds present and a child may get hurt accidentally. Also, it is most company's policy that if an item is broken that the customer pay for it. Small hands and breakable items do not go well together. Please think about a few things before bringing a child to a sale: child safety, company policy, other customers. We have even seen people trying to bring a stroller into a sale, try to imagine pushing a stroller around a crowded house.
© 2018 Nicks Estate Sales LLC. All Rights Reserved.


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Some travelers assume fall is a difficult time to travel. Kids are back in school, calendars are full of events, temperatures are falling and the excitement of summer getaways has passed. However, fall also means less crowded destinations, lower prices and comfortable temperatures for hiking, mountain biking and other outdoor adventures.
Fall is one of the best times of year to get out and explore places near and far. The following eight tips will help you plan the perfect fall getaway, so you can take some time to relax and have an adventure in this otherwise hectic season.