text:
\section{Introduction}

Several phenomena may be modelled using systems of differential equations and a key notion to get their solutions is the exponential of (finite) square matrices. Therefore, the problem of computing these exponentials is relevant. Moreover, some crucial developments of Linear Algebra have been motivated by the study of the matrix exponential. We recommend \cite{mol-van} for a survey on this topic.

Furthermore, the matrix exponential lies in the core of Lie Theory and connects the natural Lie algebra of all $n\times n $ matrices with the general linear group $GL(n, {\Bbb K})$. We think that \cite{Rossmann} is a very good introductory text for this point of view.

By allowing concepts as manifolds modelled in infinite dimensional spaces (e.g., Banach spaces, Frechet spaces etc.), the theory of Lie groups has been extended to the infinite dimensional framework \cite{Milnor}. See \cite{H-N} for some coherence relationships between infinite dimensional Lie group structures and pro-Lie group structures when both are shared by a group. This is the case of the Riordan group. The previous facts motivated us  to introduce and develop in \cite{Lie} a structure of infinite dimensional Lie group on the Riordan group.

In the following section we recall some basic facts about the Riordan group and its Lie group and pro-Lie group structures described in \cite{Lie}. Notice that although this group appeared under this name more or less recently, the group structure and many of its elements are latent in many developments of classical mathematics (special sequences of numbers and polynomials, Umbral Calculus and much more). It is clear that,  historically,  the first (and surely the best) known Riordan matrix is Pascal's Triangle. Besides this, now, there is a lot of historical names of mathematicians related to some Riordan matrices.

The aim of this note is to describe the matrix exponential, for some infinite lower triangular matrices, from the solution of certain partial differential equations, which we find using the method of characteristics (see \cite{john}). We do all of this within the framework of the Riordan Lie group and the corresponding Lie algebra. Then, motivated by general symmetry properties of dynamical systems (\cite{Lam-Rob} and \cite{ofarrel2}), we describe some of those systems in ${\Bbb K}[[x]]$ that come from the Riordan group with a very special Riordan involution as a symmetry or time-reversal symmetry. We consider a sequence of linear ordinary differential equations in Euclidean spaces as problems approaching certain partial differential equations, using for that the pro-structures of both: the Riordan group and the corresponding Lie algebra. We propose a geometric analysis of these problems and we enumerate symmetric properties of a special sequence related to the Pascal Triangle.

Apart from the general description of the Riordan group  contained in Section $2$, we also need, along the paper, some specific results from \cite{Lie}. This is the reason why in Section $3$ and $4$ we recall some particularly related results from \cite{Lie} to make this note as self-contained as possible.

Since our first approach to the Riordan group, in  \cite{teo}, is different from the usual one in the literature, we have also different notation. This is the reason why we recommend  our previous works \cite{teo}, \cite{2ways}, \cite{finitas} and, of course, \cite{Lie} (ordered chronologically) for information about basic results and notation used herein. We still maintain the notation ${\Bbb K}$ for a field although in this paper we are only considering ${\Bbb K}$ as the real numbers. This is because many of the results and ideas can be translated, at least, to the case ${\Bbb K}={\Bbb C}$, the field of complex numbers.

\section{Basic facts on the differentiable structure of the Riordan group} The results of this section can be found in \cite{teo},  \cite{2ways}, \cite{finitas} and  \cite{Lie}.
\subsection{Riordan matrices and the Riordan group}
The definition of {\it Riordan matrix} and the related concept  of {\it Riordan group} appeared in the foundational paper \cite{Sha91} due to Shapiro, Getu, Woan, and Woodson. The original definition of a Riordan matrix given in \cite{Sha91} is more restrictive than that used currently in the literature, which is precisely the one that we are going to use herein.

\textit{A Riordan matrix is an infinite matrix $D=(d_{i,j})_{i,j\in{\Bbb N}}$   whose columns are the coefficients of successive terms of a geometric progression in ${\Bbb K}[[x]]$ where the initial term is a formal power series of order $0$ and the common ratio is a formal power series of order $1$ (so $D$ needs to be lower triangular and its diagonal needs to be a geometric progression in $\mathbb{K}$). }

We represent a Riordan matrix $D$ by  $T(f\mid
g)$, where $f(x)=\sum_{k=0}^{\infty}f_k x^k$ and $g(x)=\sum_{k=0}^{\infty}g_k x^k$  are formal power series in ${\Bbb K}[[x]]$ with $f(0)\neq0$ and $g(0)\neq0$, so that
$\displaystyle{d_{i,j}=[x^i]\frac{x^jf(x)}{g^{j+1}(x)}}$.
Consequently, the first term of the geometric progression is $\displaystyle{\frac{f(x)}{g(x)}}$
and the common ratio is  $\displaystyle{\frac{x}{g(x)}}$. In this terms, Pascal's triangle is $T(1\mid 1-x)$.

The above definition can be reinterpreted saying that the generating function of the $j$-th column (starting at $j=0$) of $D$ is the formal power series $\frac{x^jf(x)}{g^{j+1}(x)}$, which makes sense because $g(0)\neq0$. Hence, $D$ is a lower triangular matrix and it is invertible because $f(0)\neq0$.

In \cite{Sha91} it was stated one of the main results about Riordan matrices. Currently many authors call it the
{\it Fundamental Theorem for Riordan matrices (FTRM)}. \textit{Let $D=T(f\mid g)$ be a Riordan matrix and let $\gamma(x)=\sum_{k=0}^{\infty}
\gamma_k x^k$ be a power series in ${\Bbb K}[[x]]$. Consider the column vector ${\bf c}=(\gamma_0, \gamma_1, \gamma_2, \cdots)^T$. Then, the generating function of the matrix product $D{\bf c}$ is $\frac{f(x)}{g(x)}\gamma(\frac{x}{g(x)})$.  }
This fact is represented by $T(f\mid g)(\gamma)=\frac{f(x)}{g(x)}\gamma(\frac{x}{g(x)})$. A proof of this result, using a special ultrametric space $({\Bbb K}[[x]], d)$ can be found  in \cite[Proposition 19]{teo}.

The Riordan group (i.e., the set of all Riordan matrices with the usual product of matrices), denoted by ${\mathcal{R}}({\Bbb K})$  or shortly ${\mathcal{R}}$, is a subgroup of the group of invertible infinite lower triangular matrices with the usual product of matrices as the
operation. The product is given by
\[T(f\mid g)T(l\mid m)=T\left(fl\left(\frac{x}{g}\right)\big|
gm\left(\frac{x}{g}\right)\right),\] where
$\displaystyle{fl\left(\frac{x}{g}\right)\equiv f(x)\cdot
l\left(\frac{x}{g(x)}\right)}$ and analogously for the second term, and the inverse is given by
\[(T(f\mid g))^{-1}\equiv T^{-1}(f\mid g)=T\left(\frac{1}{f(\frac{x}{A})}\Big| A \right),\]
where
$\displaystyle{\left(\frac{x}{A}\right)\circ\left(\frac{x}{g}\right)=\left(\frac{x}{g}\right)\circ\left(\frac{x}{A}\right)=x}$. See \cite[Proposition 20]{teo} for more details.

The sequence of the coefficients of the previous formal power series, denoted by $A$, is the so-called
$A$-sequence of $T(f\mid g)$. Obviously, the A-sequence of $T(f\mid g)$ depends only on the power series $g$. Moreover, if $A=\sum_{k\geq0}a_{k}x^{k}$, then
\[d_{i,j}=\sum_{k=0}^{i-j}a_kd_{i-1,j-1+k} \qquad i,j\geq1.\]

\subsection{The Lie and pro-Lie group structures on the Riordan group}
Suppose that ${\Bbb K}$ is the field of real or complex numbers, denoted
by ${\Bbb R}$ and ${\Bbb C}$ respectively. Let us consider a natural way to give a completely metrizable topology  in ${\Bbb K}[[x]]$, by means of the identification ${\Bbb K}^{{\Bbb N}}\equiv{\Bbb K}[[x]]$ obtained by passing from sequences to ordinary generating functions and vice versa.

The topology considered  in ${\Bbb K}^{{\Bbb N}}$ is always the product topology for the usual topology in ${\Bbb K}$. Therefore, we convert ${\Bbb K}[[x]]$ into a Frechet space, that is, a completely metrizable locally convex linear topological space.

This is the starting point to describe a natural Frechet Lie group structure on the Riordan group. Beside this, the Riordan group can be described as the inverse limit of an inverse sequence of groups of finite matrices obtaining a pro-Lie group structure on the Riordan group.

It is well known that any Riordan matrix is completely determined by its first column and
its $A$-sequence. In this way, any Riordan matrix $D=(d_{i,j})_{i,j\in{\Bbb N}}$ is defined
by a sequence $\textbf{u}=(u_k)_{k\in{\Bbb N}}$ with $u_0\neq0$,
$u_1\neq0$, and $u_{2k}=x_k$, $u_{2k+1}=a_k$, being $x_k=d_{k,0}$, $A(x)=\sum_{n\geq0}a_nx^n$ and $d_{i,j}=\sum_{k=0}^{i-j}a_kd_{i-1,j-1+k}$ for $j\geq1$. We denote the matrix $D$ described above by
$\varphi_{\infty}(\textbf{u})$.

Let us consider ${\Bbb K}$ with the usual Euclidean topology, the product topology in ${\Bbb K}^{{\Bbb N}}$ and the basic open set
\[
\mathcal{U}_{\infty}=\left\{\textbf{u}=(u_k)_{k\in{\Bbb N}}\in{\Bbb K}^{{\Bbb N}} \ |
\ \ u_0\neq0, \ u_1\neq0\right\}
\]
in ${\Bbb K}^{{\Bbb N}}$. Set
\[
\begin{matrix}
\varphi_{\infty}& : & \mathcal{U}_{\infty} &  \longrightarrow & {\mathcal{R}}({\Bbb K})\\
&& \textbf{u}& \longmapsto & \varphi_{\infty}(\textbf{u}).
\end{matrix}
\]
Obviously, $\varphi_{\infty}$ is a bijective function. So, we consider the unique topology on ${\mathcal{R}}({\Bbb K})$, that makes
 $\varphi_{\infty}$ a homeomorphism. Note that the topological space ${\mathcal{R}}({\Bbb K})$, the locally convex vector space ${\Bbb K}^{{\Bbb N}}$ and the map $\varphi_{\infty} : \mathcal{U}_{\infty}\rightarrow  {\mathcal{R}}({\Bbb K})$ fit all conditions to get:
 \begin{thm}
$({\mathcal{R}}({\Bbb K}), (\mathcal{U}_{\infty}, \varphi_{\infty}) )$ is a smooth
manifold modelled on the locally convex vector space ${\Bbb K}^{{\Bbb N}}$. Moreover, ${\mathcal{R}}({\Bbb K})$ with this smooth structure is a Lie group.
\end{thm}

One of the main tools that we have used to get results on the Riordan group is to consider it as the inverse limit of an inverse sequence of groups of finite matrices. A natural way to do this is as follows. For every $n\in{\Bbb N}$, consider the general linear group $GL(n+1,{\Bbb K})$
formed by all $(n+1)\times(n+1)$ invertible matrices with coefficients in ${\Bbb K}$. Since every Riordan matrix is lower triangular, we can define a natural homomorphism $\Pi_n: \mathcal{R}\rightarrow GL(n+1,{\Bbb K})$ given by
\[
\Pi_n((d_{i,j})_{i,j\in{\Bbb N}})=(d_{i,j})_{i,j=0,\cdots,n}.
\]
For obvious reasons, we will refer to this homomorphism
as the projection of the corresponding Riordan matrix.

To describe the Riordan group as an inverse limit of an inverse
sequence of groups of finite matrices we use the results of \cite{finitas}. We first consider the
subgroup of $GL(n+1,{\Bbb K})$ defined by
$\mathcal{R}_n=\Pi_n(\mathcal{R})$.

\begin{defi}
Let $D=(d_{i,j})_{i,j=0,\cdots,n+1}\in{\mathcal{R}}_{n+1}$. We define
$P_n:{\mathcal{R}}_{n+1}\rightarrow{\mathcal{R}}_n$ by
\[
 P_{n}((d_{i,j})_{i,j=0,1,\cdots,n+1})=(d_{i,j})_{i,j=0,\cdots,n}.
\]
\end{defi}
$P_n(D)$ is obtained from $D$ by deleting its last row and its
last column. $P_n$ is a group homomorphism for every $n$ because
the matrices are lower triangular. Moreover, the diagram below is
commutative
\begin{displaymath}
\xymatrix{
&{\mathcal{R}} \ar[d]_{\Pi_n} \ar[rd]^{\Pi_{n+1}}& \\
&{\mathcal{R}}_{n}  & \ar[l]^{P_n} {\mathcal{R}}_{n+1} .}
\end{displaymath}
From this, we get
\begin{thm} The  Riordan group ${\mathcal{R}}$ is isomorphic to
$\underleftarrow{\lim}\{(\mathcal{R}_n)_{n\in{\Bbb N}},(P_n)_{n\in{\Bbb N}}\}$. Consequently, ${\mathcal{R}}$ is a pro-Lie group.
\end{thm}
See \cite{finitas} for more details and notation.
\subsection{The Lie Algebra of the Riordan group}

Again, using the results of \cite{Lie} we obtain a full and faithful representation of the Lie algebra
$\mathcal{L}({\mathcal{R}}({\Bbb K}))$ of the Lie group ${\mathcal{R}}({\Bbb K})$. We have that $\displaystyle{L=(\ell_{i,j})_{i,j\in{\Bbb N}}}\in \mathcal{L}({\mathcal{R}}({\Bbb K}))$
if and only if $L$ is lower triangular and there are two sequences
$(\chi_i)_{i\in{\Bbb N}}$ and $({\alpha}_i)_{i\in{\Bbb N}}$ such that

{\small
\[
L=\left(
                                \begin{array}{ccccccc}
                                  \chi_0 &  &  &  &  &  &  \\
                                  \chi_1 & \chi_0+{\alpha}_0 &  &  &  &  &  \\
                                  \chi_2 & \chi_1+{\alpha}_1 & \chi_0+2{\alpha}_0 &  &  &  &  \\
                                   \vdots & \vdots & \vdots & \ddots &  &  \\
                                  \chi_{n-1} & \chi_{n-2}+{\alpha}_{n-2} & \chi_{n-3}+2{\alpha}_{n-3}&  \cdots & \chi_0+(n-1){\alpha}_0  &  \\
                                  \chi_n & \chi_{n-1}+{\alpha}_{n-1} & \chi_{n-2}+2{\alpha}_{n-2} & \cdots & \chi_1+(n-1){\alpha}_1 &  \chi_0+n{\alpha}_0\\
                                    \vdots & \vdots & \vdots & \cdots& \vdots&\vdots&\ddots  \\
                                \end{array}
                              \right)
\]} with the usual sum of matrices and the usual product by scalars in
${\Bbb K}$. The Lie bracket is
\[
[L_1,L_2]=L_1L_2-L_2L_1
.\]
We denote the matrix $L$ above as $L(\chi,\alpha)$, where $\displaystyle{\chi(x)=\sum_{n\geq0}\chi_nx^n, \
{\alpha}(x)=\sum_{n\geq0}{\alpha}_nx^n}$, and

\[
\mathcal{L}({\mathcal{R}}({\Bbb K}))=\left\{L(\chi,{\alpha}) \  |  \ \chi,
{\alpha}\in{\Bbb K}[[x]]\right\}.
\]

\begin{prop}
Any $L=L(\chi, \alpha)\in\mathcal{L}({\mathcal{R}}({\Bbb K}))$ induces a linear continuous map,
denoted again by $L:{\Bbb K}^{{\Bbb N}}\rightarrow{\Bbb K}^{{\Bbb N}}$, given by $L(h)=
\chi(x)h(x)+x{\alpha}(x)h'(x)$ where
$h(x)=\sum_{n\geq0}h_nx^n$.
\end{prop}
The continuous linear map $L(\chi,{\alpha})$ can be viewed as a
$C^{\infty}$ vector field in the Frechet space ${\Bbb K}^{{\Bbb N}}$ under the
canonical identification $T_{h}{\Bbb K}^{{\Bbb N}}={\Bbb K}^{{\Bbb N}}$ in the tangent space
at any $h$. From this point of view we have the following
proposition.
\begin{prop}\label{prop:initialValue}
The initial value problem
\[
\left\{
  \begin{array}{ll}
    \gamma'(t)=L(\chi,{\alpha})(\gamma(t)) \\
    \gamma(0)=h
  \end{array}
\right.
\]
in ${\Bbb K}^{{\Bbb N}}$ has a unique solution given by
\[
\gamma(t)=e^{tL(\chi,{\alpha})}(h).
\]
Consequently, there is a one-parameter group of Riordan matrices
$T(f(x,t)\mid g(x,t))$ such that
\[
\gamma(t)=T(f(x,t)\mid g(x,t))(h) =\frac{f(x,t)}{g(x,t)}h\left(\frac{x}{g(x,t)}\right).
\] In particular, $\{T(f(x,t)\mid g(x,t))\}_{t\in{\Bbb R}} $ is an abelian subgroup of the Riordan group, because it defines a continuous dynamical system (or flow) in ${\Bbb K}[[x]]$.
\end{prop}
Due to the special patterns followed by the matrices in the Lie Algebra and those in the Riordan group, we named the corresponding section in \cite{Lie} as {\it Arithmetic vector fields and geometric flows.}
\subsection{The partial differential equation induced by an element in the Lie Algebra.}
For any $t\in {\Bbb R}$ and  $L\in
\mathcal{L}({\mathcal{R}}({\Bbb K}))$ we
have a well-defined Riordan matrix $e^{tL}$. Note that for any $t\in {\Bbb R}$, $\gamma(t)\in {\Bbb K}[[x]]$, where $\gamma$ satisfies the conditions of Proposition \ref{prop:initialValue}. With this in mind, another way to interpret
the above proposition is as follows.
\begin{cor}
Let $\chi,{\alpha}\in{\Bbb K}[[x]]$. Then the unique solution of the initial value
problem
\[
\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}=\chi(x)u(x,t)+x{\alpha}(x)\frac{\partial u}{\partial x} \\
    u(x,0)=h(x)
  \end{array}
\right.
\]
in ${\Bbb K}[[x,t]]$ is given by
\[
u(x,t)=e^{tL(\chi,{\alpha})}(h(x))=\frac{f(x,t)}{g(x,t)}h\left(\frac{x}{g(x,t)}\right).
\]
\end{cor}

\section{From the matrix exponential $e^{tL}$ to the solutions of the PDE $\frac{\partial u}{\partial t}=\chi(x)u(x,t)+x{\alpha}(x)\frac{\partial u}{\partial x}$ and back}
Once recalled some basic results from \cite{Lie}, we propose the following \textit{strategy}.

Given any element $L=L(\chi, \alpha)$ in the Lie Algebra of the Riordan group, we associate to it the partial differential equation $\frac{\partial u}{\partial t}=\chi(x)u(x,t)+x{\alpha}(x)\frac{\partial u}{\partial x}$. At this point, we run into the following {\it dichotomy}.
\begin{itemize}
\item[(a)] If we are able to compute the one-parameter group $e^{tL}$ and to recognize $e^{tL}=T(f(x,t)\mid g(x,t))$ as a Riordan matrix for any $t\in {\Bbb R}$, then this allows us to solve in ${\Bbb K}[[x,t]]$ the initial value problem \[
\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}=\chi(x)u(x,t)+x{\alpha}(x)\frac{\partial u}{\partial x} \\
    u(x,0)=h(x)
  \end{array}
\right.
\]
 for any $h\in {\Bbb K}[[x]]$  because the solution is given by \[
u(x,t)=e^{tL(\chi(x),{\alpha}(x))}(h(x))=\frac{f(x,t)}{g(x,t)}h\left(\frac{x}{g(x,t)}\right).
\]
\item[(b)] If, on the contrary, we are able to solve in ${\Bbb K}[[x,t]]$ the initial value problem \[
\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}=\chi(x)u(x,t)+x{\alpha}(x)\frac{\partial u}{\partial x} \\
    u(x,0)=h(x)
  \end{array}
\right.
\]
 for any $h\in {\Bbb K}[[x]]$, then we get the one-parameter subgroup   $e^{tL(\chi(x),{\alpha}(x))}=T(f(x,t)\mid g(x,t))$. We can compute both parameters $f(x,t)$ and $g(x,t)$, because $\frac{f(x,t)}{g(x,t)}$ is the unique solution for the initial condition $h(x)\equiv 1$ and $\frac{f(x,t)}{g(x,t)}\frac{x}{g(x,t)}$ is the unique solution for the initial condition $h(x)=x$. Evaluating now at $t=1$, we have the corresponding matrix exponential.
\end{itemize}

\subsection{From the matrix exponential to the solution of the PDE}
\begin{ejem}
\begin{itemize}
\item[(i)] Consider the matrix

\[
D=\left(
  \begin{array}{ccccc}
    1 & 0 & 0 & 0 & \cdots \\
    0 & 2 & 0 & 0 & \cdots \\
    0 & 0 & 3 & 0 & \cdots \\
    0 & 0 & 0 & 4 & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right).
\]
Obviously, $D=L(1,1)$ is in $\mathcal{L}({\mathcal{R}}({\Bbb K}))$. The partial differential equation associated to the matrix $D$ is \[\frac{\partial u}{\partial t}=u(x,t)+x\frac{\partial u}{\partial x}\]
The solution of the corresponding initial value problem \[
\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}=u(x,t)+x\frac{\partial u}{\partial x} \\
    u(x,0)=h(x)
  \end{array}
\right.
\] is given by $u(x,t)=e^{tD}(h)$. It is clear, by definition,  that
\[
e^{tD}=\left(
  \begin{array}{ccccc}
    e^{t} & 0 & 0 & 0 & \cdots \\
    0 & e^{2t} & 0 & 0 & \cdots \\
    0 & 0 & e^{3t} & 0 & \cdots \\
    0 & 0 & 0 & e^{4t} & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right).
\] For any $t\in {\Bbb R}$, we have that $e^{tD}=T(1\mid e^{-t})$ as a Riordan matrix. Therefore, the solution is $u(x,t)=T(1\mid e^{-t})(h)=e^{t}h(xe^{t})$.

\item[(ii)] In this more interesting example, we consider the matrix

\[
H =\left(
  \begin{array}{ccccc}
    0 & 0 & 0 & 0 & \cdots \\
    1 & 0 & 0 & 0 & \cdots \\
    0 & 2 & 0 & 0 & \cdots \\
    0 & 0 & 3 & 0 & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right).
\] Note that $H=L(x,x)$ and so $H\in \mathcal{L}({\mathcal{R}}({\Bbb K}))$.  The PDE induced by $H$ is
\[ \frac{\partial u}{\partial t}-x^{2}\frac{\partial u}{\partial x}=xu(x,t) \]

The solution of the corresponding initial value problem
\[
\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}=xu(x,t)+x^{2}\frac{\partial u}{\partial x} \\
    u(x,0)=h(x)
  \end{array}
\right.
\] is given by $u(x,t)=e^{tH}(h)$. To compute this solution we are going to take advantage of some previous work of another authors. Particularly, we are going to follow \cite{Aceto2}, where the matrix $H$ is called the {\it creation matrix}. Formula $(9)$ in \cite[Page 233]{Aceto2} computes $e^{t\Lambda_{n-1}(H)}$. Moreover, it is obvious   that $P_{n-1}(e^{t\Lambda_{n}(H)})=e^{t\Lambda_{n-1}(H)}$ for any $t\in{\Bbb R}$ and any integer $n\geq2$ (see \cite[Page 542]{Lie} for the notation concerning $\Lambda_n$). Hence, in our Riordan matrix notation, we get $e^{tH}=T(1 \mid 1-xt)$ and the solution of the corresponding initial value problem is given by \[u(x,t)=e^{tH}(h)=\frac{1}{1-xt}h\left(\frac{x}{1-xt}\right).\]

Evaluating at $t=1$, we obtain $e^{H}=T(1\mid 1-x)$, which is the {\it Pascal triangle}.

\end{itemize}

\end{ejem}
\subsection{From the solution of a PDE to the matrix exponential: the Method of Characteristics}

 To point out how to go from the solution of a PDE to the exponential map of the Lie algebra to the Riordan group, we are going to deal with the following family of significant examples.

 For any couple of real numbers $a$, $b$ and any non-negative integer number $n$, consider the infinite lower triangular matrix $L^{a,b}_{n}=(d_{i,j}^{(n)})_{i,j\in \mathbb{N}}$, where $d_{i,j}^{(n)}=0$ if $i-j\neq n$ and $d_{i,j}^{(n)}=a +jb$ if $i-j= n$. How can we compute or describe the exponential of each of the matrices $L^{a,b}_{n}$?

Note that $L^{a,b}_{n}=L(ax^n , bx^n)$. This means that they belong to the Lie algebra of the Riordan group. Moreover,

\[
L^{a,b}_{0}=\left(
  \begin{array}{ccccc}
    a & 0 & 0 & 0 & \cdots \\
    0 & a+b & 0 & 0 & \cdots \\
    0 & 0 & a+2b & 0 & \cdots \\
    0 & 0 & 0 & a+3b & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right),
\]

\[
L^{a,b}_1 =\left(
  \begin{array}{ccccc}
    0 & 0 & 0 & 0 & \cdots \\
    a & 0 & 0 & 0 & \cdots \\
    0 & a+b & 0 & 0 & \cdots \\
    0 & 0 & a+2b & 0 & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right),
\] and so on. Observe that  $D=L^{1,1}_0$ and $H=L^{1,1}_1$, where $D$ and $H$ are the matrices considered in the previous examples.
The PDE induced by $L^{a,b}_{n}$ is
\[ \frac{\partial u}{\partial t}-bx^{n+1}\frac{\partial u}{\partial x}=ax^{n}u(x,t) .\]
Note that if $b=0$; then, $e^{tL^{a,0}_{n}}=T(e^{atx^{n}} \mid 1)$ is a Toeplitz matrix (in the Riordan group) for any $t\in{\Bbb R}$. Consequently,  the solution  of the corresponding initial value problem
\[
\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}=ax^{n}u(x,t) \\
    u(x,0)=h(x)
  \end{array}
\right.
\] is given by $u(x,t)=e^{tL^{a,0}_{n}}(h)=e^{atx^{n}}h(x)$.

From now on, assume $b\neq 0$. Consider the matrix $L^{a,b}_{n}$ and the corresponding  initial value problem
\begin{equation}\label{eq:1}
 \quad\left\{
  \begin{array}{ll}
    \frac{\partial u}{\partial t}-bx^{n+1}\frac{\partial u}{\partial x}=ax^{n}u(x,t) \\
    u(x,0)=h(x).
  \end{array}
\right.
\end{equation}
Let us use the Method of Characteristics to solve it (see \cite{john}, Chapter I). Suppose $t=t(r,s)$, $x=x(r,s)$ and $z=z(r,s)$. So we consider the following system of ODEs

\[
\left\{
  \begin{array}{lll}
    \frac{dt}{ds}=1 \\
    \frac{dx}{ds}=-bx^{n+1} \\
    \frac{dz}{ds}=ax^{n}z
  \end{array}
\right.
\]
with initial conditions
\[
\left\{
  \begin{array}{lll}
  t(r,0)=0 \\
  x(r,0)=r \\
  z(r,0)=h(r)
  \end{array}
\right.
\]
Integrating the system and imposing the initial conditions we get
  \[
\left\{
  \begin{array}{lll}
t=s\\
 \frac{1}{x^n}=\frac{1+nbsr^n}{r^n}\\
 z=(1+nbsr^{n})^{\frac{a}{nb}}h(r)
  \end{array}
\right.
\]
and
 \[
\left\{
  \begin{array}{lll}
s=t\\
r=\frac{x}{\sqrt[n]{1-nbtx^{n}}}\\
z=\left(\frac{1}{1-bntx^{n}}\right)^{\frac{a}{nb}}h\left(\frac{x}{\sqrt[n]{1-nbtx^{n}}}\right)
  \end{array}
\right. ,
\]
which yields that the solution of (\ref{eq:1}) is given by \[ u(x,t)=\left(\frac{1}{1-bntx^{n}}\right)^{\frac{a}{nb}}h\left(\frac{x}{\sqrt[n]{1-nbtx^{n}}}\right)\] or
\[ u(x,t)=\sqrt[n]{\left(\frac{1}{1-bntx^{n}}\right)^{\frac{a}{b}}}h\left(\frac{x}{\sqrt[n]{1-nbtx^{n}}}\right).\]
All above in this subsection may be summarized as
\begin{thm} Suppose that  $a$ and  $b$ are two real numbers and that $n$ is a non-negative integer number. Consider the  infinite lower triangular matrix  $L^{a,b}_{n}=(d_{i,j}^{(n)})_{i,j\in \mathbb{N}}$, where $d_{i,j}^{(n)}=0$ if $i-j\neq n$ and $d_{i,j}^{(n)}=a +jb$ if $i-j= n$. Then, $L^{a,b}_{n}$ is in the Lie algebra of the Riordan group and for any $t\in{\Bbb R}$ the Riordan matrix $e^{tL^{a,b}_{n}}$ is given by \[e^{tL^{a,b}_{n}}= \left\{
  \begin{array}{ll}
    T\left(\sqrt[n]{(1-bntx^{n})^{\frac{b-a}{b}}}\mid   \sqrt[n]{1-bntx^{n}}\right)   &     \text { if   $b\neq0$} \\
    T\left(e^{atx^{n}}\mid 1\right) & \text{if b=0}.
  \end{array}
\right.
\]
\end{thm}
\section{A special involution as symmetry or time-reversal symmetry for flows in ${\Bbb K}[[x]]$ related to the Riordan group.}
There is an involution $M$ in the Riordan group playing a special role in such group. This matrix is
\[
M=T(-1|-1)=\left(
  \begin{array}{ccccc}
    1 & 0 & 0 & 0 & \cdots \\
    0 & -1 & 0 & 0 & \cdots \\
    0 & 0 & 1 & 0 & \cdots \\
    0 & 0 & 0 & -1 & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right).
\]

Considered as an endomorphism in ${\Bbb K}[[x]]$, $M$ is continuous and has exactly two eigenvalues $1$ and $-1$. The corresponding eigenspaces are $E(1)=\{h\in {\Bbb K}[x]] /   h(x)=h(-x)\}$ and $E(-1)=\{h\in {\Bbb K}[x]] /  -h(x)=h(-x)\}$, i.e., they are the linear subspaces of even, respectively odd, formal power series. Both of them are closed subspaces in ${\Bbb K}[[x]]$ and we get ${\Bbb K}[[x]]=E(1)\oplus E(-1)$ and $(M-I)\circ (M+I)=(M+I)\circ (M-I)\equiv 0$, where $I=T(1\mid 1)$ is the identity.

The reason of the interest about $M$ is because it is used for defining what is known as a  pseudo-involution in the Riordan group. Following \cite{Nkwanta1}, we say that a Riordan matrix $R$ is a {\it pseudo-involution} if the product $RM$ is an involution, that is, $RMRM=I$.

In Group Theory we have the definitions of {\it reversible} and {\it strongly reversible} elements (see \cite{ofarrel2}). We recall it here for completeness.
\begin{defi}
\begin{itemize}
\item[(i)] An element $g$ of a group $G$ is said to be reversible in $G$ if there is another element $h$ of $G$ such that \[hgh^{-1} = g^{-1}.\] In this situation we also say that $h$ reverses $g$ and $h$ is a reverser of $g$.
    \item[(ii)] An element $g$ of a group $G$ is said to be strongly reversible in $G$ if there is an involution  $h$ of $G$ such that \[hgh^{-1} = g^{-1}.\]
\end{itemize}
\end{defi}

\begin{prop}If $R$ is a pseudo-involution, then $R$ is strongly reversible in the Riordan group and is the product of two involutions.
\end{prop}
\begin{proof}
Since $RMRM=I$ we directly obtain, multiplying  on the left by the inverse $R^{-1}$, that $MRM=R^{-1}$ so $R$ is strongly reversible. Moreover, $R^{-1}M$ is an involution because $(R^{-1}M)^{-1}=MR=MMR^{-1}M=R^{-1}M$. Consequently, we have that $R=MR^{-1}M$.
\end{proof}

The above proposition  tells us that the pseudo-involutions are particular examples of {\it strongly reversible} elements in the Riordan group and $M$ is a reverser for any of them.

Consider the left and right translations in ${\mathcal{R}}({\Bbb K})$ given,
respectively, by
\[
\begin{matrix}
L_D& : & {\mathcal{R}}({\Bbb K}) &  \longrightarrow & {\mathcal{R}}({\Bbb K})\\
&& X & \longmapsto & L_D(X)=DX,
\end{matrix}\qquad \qquad \begin{matrix}
R_D& : & {\mathcal{R}}({\Bbb K}) &  \longrightarrow & {\mathcal{R}}({\Bbb K})\\
&& X & \longmapsto & R_D(X)=XD.
\end{matrix}
\]
Since the product is a $C^{\infty}$-function  both $L_D$ and $R_D$
are diffeomorphisms. We need to recall the following facts from \cite{Lie}.
\begin{prop}\label{T: conj}
Let $T(f\mid g)$ be a Riordan matrix. The tangent space $T_{T(f\mid
g)}{\mathcal{R}}({\Bbb K})$ to ${\mathcal{R}}({\Bbb K})$ at $T(f\mid g)$ is given by
\[
T_{T(f\mid g)}{\mathcal{R}}({\Bbb K})=\{T(f\mid g)L(\chi,{\alpha})| \chi,
{\alpha}\in{\Bbb K}[[x]]\}= \{L(\chi,{\alpha})T(f\mid g)| \chi, {\alpha}\in{\Bbb K}[[x]]\}.
\]
Moreover, the conjugation by $T(f\mid g)$ defined by
\[
\begin{matrix}
\text{conj}_{T(f\mid g)}& : & {\mathcal{R}}({\Bbb K}) &  \longrightarrow & {\mathcal{R}}({\Bbb K})\\
&& X & \longmapsto & T(f\mid g)XT^{-1}(f\mid g)
\end{matrix}
\]
is a $C^{\infty}$-diffeomorphism and its tangent (or differential) map at the identity
\[
\begin{matrix}
D\text{conj}_{T(f\mid g)}(I)& : & \mathcal{L}({\mathcal{R}}({\Bbb K})) &
\longrightarrow & \mathcal{L}({\mathcal{R}}({\Bbb K}))
\end{matrix}
\]
is given by
\[
D\text{conj}_{T(f\mid g)}(I)(L(\chi,{\alpha}))=T(f\mid
g)L(\chi,{\alpha})T^{-1}(f\mid g).
\]
Finally, for any $t\in{\Bbb K}$ we have \large{
\[
e^{tD\text{conj}_{T(f\mid g)}(I)(L(\chi,{\alpha}))}=\text{conj}_{T(f\mid
g)}(e^{tL(\chi,{\alpha})}).
\]}
\end{prop}
From this, we deduce
\begin{cor}
Given $T(f\mid g)\in{\mathcal{R}}({\Bbb K})$ and
$L(\chi,{\alpha})\in\mathcal{L}({\mathcal{R}}({\Bbb K}))$, there exists a unique
$L(\tilde{\chi},\tilde{{\alpha}})\in\mathcal{L}({\mathcal{R}}({\Bbb K}))$ such that
\[
T(f\mid g)L(\chi,{\alpha})=L(\tilde{\chi},\tilde{{\alpha}})T(f\mid g)
\]
or equivalently
\[
D\text{conj}_{T(f\mid
g)}(I)(L(\chi,{\alpha}))=L(\tilde{\chi},\tilde{{\alpha}}).
\]
Moreover,
\begin{equation}\label{eq:chialpha}
\tilde{\chi}=\frac{\chi\left(\frac{x}{g}\right)(g-xg')f-x{\alpha}\left(\frac{x}{g}\right)(f'g-g'f)}{f(g-xg')},
\qquad \tilde{{\alpha}}=\frac{g{\alpha}\left(\frac{x}{g}\right)}{g-xg'}.
\end{equation}
\end{cor}

As a consequence, we obtain
\begin{prop}\label{T: eigen} Consider the involution $M$ and the $C^{\infty}$-diffeomorphism \[\begin{matrix}
\text{conj}_{M}& : & {\mathcal{R}}({\Bbb K}) &  \longrightarrow & {\mathcal{R}}({\Bbb K})\\
&& X & \longmapsto & MXM.
\end{matrix}
\]
Then
\begin{itemize}
\item[(i)] the tangent (or differential) map at identity
\[
\begin{matrix}
D\text{conj}_{M}(I)& : & \mathcal{L}({\mathcal{R}}({\Bbb K})) &
\longrightarrow & \mathcal{L}({\mathcal{R}}({\Bbb K}))
\end{matrix}
\]
is a linear involution
\item[(ii)] for any couple  of real numbers $a$ and  $b$ such that $a\neq 0$ and $b\neq 0$ the  matrix $L^{a,b}_{n}$ is an eigenvector of $D\text{conj}_{M}(I)$ corresponding to the eigenvalue $-1$ if $n$ is odd and an eigenvector of $D\text{conj}_{M}(I)$ corresponding to the eigenvalue $1$ if $n$ is even (including $n=0$).
\end{itemize}
\end{prop}
\begin{proof} It is clear that $D\text{conj}_{M}(I)\circ D\text{conj}_{M}(I)=I_{\mathcal{L}({\mathcal{R}}({\Bbb K}))}$ because $M$ is an involution in ${\mathcal{R}}({\Bbb K})$, where $I_{\mathcal{L}({\mathcal{R}}({\Bbb K}))}$ represents the identity map in $\mathcal{L}({\mathcal{R}}({\Bbb K})) .$ To prove (ii), recall that $M=T(-1\mid -1)$ and that $L^{a,b}_{n}=L(ax^n , bx^n)$. From the previous corollary we get \[
D\text{conj}_{M}(I)(L(\chi, {\alpha}))=L(\tilde{\chi},\tilde{{\alpha}})\] where
$\tilde{\chi}$ and $\tilde{{\alpha}}$ are given by (\ref{eq:chialpha}). In this case, $f=g=-1$ and $\chi(x)=ax^{n}$, ${\alpha}(x)=bx^{n}.$ For the case that $n$ even, the announced result is obvious. The same also holds for the case $n$ is odd because $ -L(\chi,{\alpha})=L(-\chi,-{\alpha})$ for any $\chi$, ${\alpha}$ in ${\Bbb K}[[x]].$
\end{proof}

Finally, we obtain the following (here we use \cite{Lam-Rob} for some of the definitions appearing below).
\begin{thm}\label{T: TR} Let  $a$ and $b$ two real numbers such that $a=b=0$ does not hold. Suppose also that  $n$ is a non-negative integer number. Consider the infinite lower triangular matrix $L^{a,b}_{n}$. Then we have the following.
\begin{itemize}
\item[(i)] If $n$ is even, the one-parameter subgroup $\{e^{tL^{a,b}_{n}}\}_{t\in {\Bbb R}}$ is contained in the centralizer of the involution $M$. In other words, $M$ is a symmetry for the flow  $\{e^{tL^{a,b}_{n}}\}_{t\in {\Bbb R}}$ in ${\Bbb K}[[x]]$.
    \item[(ii)] If $n$ is odd, any element in the one-parameter subgroup $\{e^{tL^{a,b}_{n}}\}_{t\in {\Bbb R}}$ is a pseudo-involution and the involution $M$ is a time-reversal symmetry of the flow $\{e^{tL^{a,b}_{n}}\}_{t\in {\Bbb R}}$ in ${\Bbb K}[[x]]$.
\end{itemize}
\end{thm}
\begin{proof} Suppose $n$ is even. Using Proposition \ref{T: eigen}, we have $D\text{conj}_{M}(I)(L^{a,b}_{n})=L^{a,b}_{n}$. Hence \[e^{tL^{a,b}_{n}}=e^{tD\text{conj}_{M}(I)(L^{a,b}_{n})}=Me^{tL^{a,b}_{n}}M.\] The last equality above is a consequence of Proposition \ref{T: conj}. Consequently, $e^{tL^{a,b}_{n}}$ is in the centralizer of the involution $M$ for every $ t\in {\Bbb R}$.

Suppose now that $n$ is odd. By analogous arguments, we first have that $D\text{conj}_{M}(I)(L^{a,b}_{n})=-L^{a,b}_{n}$ and then
\[e^{-tL^{a,b}_{n}}=e^{tD\text{conj}_{M}(I)(L^{a,b}_{n})}=Me^{tL^{a,b}_{n}}M.\] Therefore, the Riordan matrix $e^{tL^{a,b}_{n}}$ is reversible for any $ t\in {\Bbb R} $ and the involution $M$ is a reverser for all of them. This fact implies that $e^{tL^{a,b}_{n}}$ is strongly reversible and that  $e^{tL^{a,b}_{n}}$ is a pseudo-involution because $e^{tL^{a,b}_{n}}M$ is an involution ($e^{tL^{a,b}_{n}}Me^{tL^{a,b}_{n}}M=e^{tL^{a,b}_{n}}e^{-tL^{a,b}_{n}}=I)$.
\end{proof}
\begin{cor}\label{T: CTR}The same as in the above theorem is true, word by word, changing the involution $M$ by the involution $-M=T(1 \mid -1).$
\end{cor}

\begin{section}{On some  dynamical properties  of Pascal Triangle}

Using some of the results obtained herein we can get some new information about the Riordan group. For example, in \cite{Lie}  we recognized as a subgroup of the Riordan group, the {\it substitution group of formal power series}. This  group was introduced in \cite{Jenn} (see also \cite{Babenko} for a good survey about it) where  results about topological generation are established. One can notice at once that some of our one-parameter groups $\{e^{tL^{a,b}_{n}}\}_{t \in {\Bbb R}}$ are involved in \cite[Section 3]{Jenn}. In this section, we focus only on the Pascal Triangle and  assign to it others of the many properties related to patterns and symmetries that it has.

We will use the pro-Lie group structure of the Riordan group. So, we will consider elements of the Riordan group and of the Lie algebra as approximated by the components  of the corresponding points in the inverse limit interpretation of both ${\mathcal{R}}({\Bbb K})$ and $\mathcal{L}({\mathcal{R}}({\Bbb K}))$ (see \cite{Lie} for the pro-structure of $\mathcal{L}({\mathcal{R}}({\Bbb K}))$).

To clarify the above approaching process recall that related to any problem \begin{equation}\label{E_: GEN}
 \gamma'(t)=L(\gamma(t))
\end{equation} with $L\in\mathcal{L}({\mathcal{R}}({\Bbb K}))$ or, equivalently \[\frac{\partial u}{\partial t}=\chi(x)u(x,t)+x{\alpha}(x)\frac{\partial u}{\partial x},\] if $L=L(\chi(x), {\alpha}(x))$,  we have a sequence of finite dimensional problems, denoted by $\{(\ref{E_: GEN})_{n}\}_{n\in {\Bbb N}}$. We call this sequence as the sequence of {\it approaching}  problems of  $(\ref{E_: GEN})$.

{\bf Problem $(\ref{E_: GEN})_{n}$: Approaching problems.}
\textit{
Consider the Euclidean space ${\Bbb R}^{n+1}$. For any $x\in {\Bbb R}^{n+1}$ denote by $x=(x_{0}, x_{1}, \cdots, x_{n})$ to its usual components. Suppose $x^{T}$ represents the transpose matrix of $x.$ Let $x:{\Bbb R}\longrightarrow {\Bbb R}^{n+1}$ be any derivable curve given by $x(t)=(x_{0}(t), x_{1}(t), \cdots, x_{n}(t)).$ We denote by $x'(t)=(x'_{0}(t), x'_{1}(t), \cdots, x'_{n}(t))$ the derivative of $x$  at $t$, where $x'_{i}$ is the usual derivative of a real function with real variable for any $i=0,\cdots, n$. When we refer to the {\it approaching problem} $(\ref{E_: GEN})_{n}$,  we refer to the linear differential equation \[x'^{T}= D\Pi_{n}(I)(L)x^{T}.\] In the above equation $D\Pi_{n}(I)$ represents the differential at the identity $I$ in the Riordan group ${\mathcal{R}}({\Bbb K})$ of the projection \[\Pi_{n}:{\mathcal{R}}({\Bbb K})\longrightarrow {\mathcal{R}}_{n}({\Bbb K})\] in the pro-Lie group structure of ${\mathcal{R}}({\Bbb K})$, see again \cite{Lie} if needed.}

\subsection{The approaching problems related to Pascal Triangle}

Recall that, Pascal Triangle, is the time $1$ map of the dynamical system generated by the linear differential equation in ${\Bbb K}[[x]]$
\begin{equation}\label{E_:ECU}
 \gamma'(t)=L^{1,1}_1(\gamma(t)),
\end{equation} where $L^{1,1}_1=L(x,x)$, or, equivalently, \[ \frac{\partial u}{\partial t}=xu + x^{2}\frac{\partial u}{\partial x} \] in ${\Bbb K}[[x,t]].$
Recall also that $\gamma(t)\in {\Bbb R}[[x]]$ for any $t\in {\Bbb R}$ and that $L^{1,1}_1(\gamma(t))=x\gamma(t)+x^{2}\frac{d\gamma(t)}{dx}$, where
$\frac{d}{dx}$ is the formal derivative in ${\Bbb R}[[x]]$. While $\gamma:{\Bbb R} \longrightarrow{\Bbb R}[[x]]$ is a curve and $\gamma'(t)$ is the usual derivative in $t$, when we consider ${\Bbb R}[[x]]$ identified with ${\Bbb R}^{\Bbb N}$ with the product topology. Using the pro-Lie group structure in ${\mathcal{R}}({\Bbb K})$, the corresponding pro-Lie algebra structure in $\mathcal{L}({\mathcal{R}}({\Bbb K}))$ and recalling that \[
L^{1,1}_1 =\left(
  \begin{array}{ccccc}
    0 & 0 & 0 & 0 & \cdots \\
    1 & 0 & 0 & 0 & \cdots \\
    0 & 2 & 0 & 0 & \cdots \\
    0 & 0 & 3 & 0 & \cdots \\
    \vdots  & \vdots  & \vdots  & \vdots & \ddots \\
  \end{array}
\right).
\]

We can associate to problem $(\ref{E_:ECU})$ a countably infinite  family of finite dimensional problems $(\ref{E_:ECU})_{n}$ in the euclidean space ${\Bbb R}^{n+1}$ for any non-negative integer $n$.  As we said before, we  will interpret the family of problems $\{(\ref{E_:ECU})_{n}\}_{n\in {\Bbb N}}$ as {\it approaching} the problem $(\ref{E_:ECU})$ when $n$ tends to $\infty$. For the first few values of $n$ we have:
\begin{itemize}
\item[$(\ref{E_:ECU})_{0}$] for $n=0$, we consider the differential equation $x'_{0}=0$ in the one dimensional euclidean space ${\Bbb R}$.
\item[$(\ref{E_:ECU})_{1}$] for $n=1$, we consider the differential equation in ${\Bbb R}^{2}$
\[\begin{pmatrix}x'_{0} (t) \\ x'_{1} (t) \end{pmatrix} = \begin{pmatrix}0&0\\  1& 0
\end{pmatrix}\begin{pmatrix}x_{0} (t) \\ x_{1} (t) \end{pmatrix} \]
\item[$(\ref{E_:ECU})_{2}$] for $n=2$, we consider the differential equation in ${\Bbb R}^{3}$
\[\begin{pmatrix}x'_{0} (t) \\ x'_{1} (t) \\x'_2 (t) \end{pmatrix} = \begin{pmatrix}0&0&0\\  1 & 0 & 0 \\ 0 & 2 & 0
\end{pmatrix}\begin{pmatrix}x_{0} (t) \\ x_{1} (t) \\x_{2}(t) \end{pmatrix} \]
\item[$(\ref{E_:ECU})_{3}$] for $n=3$, we consider the differential equation in ${\Bbb R}^{4}$
\[\begin{pmatrix}x'_{0} (t) \\ x'_{1} (t) \\x'_2 (t) \\x'_{3}(t) \end{pmatrix} = \begin{pmatrix}0&0&0&0\\  1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0
\end{pmatrix}\begin{pmatrix}x_{0} (t) \\ x_{1} (t) \\x_{2}(t) \\ x_{3} (t) \end{pmatrix} \]

\end{itemize}
and so on.

The problem $(\ref{E_:ECU})_{0}$ is very easy to analyze. Any point in the phase space ${\Bbb R}$ is an equilibrium point. The phase flow is trivial and nothing is moving under it.

Let us now consider $(\ref{E_:ECU})_{1}$. The equilibrium points in this case are just the points in the $x_{1}$-axe, which means that they are of the form $(0,b)$, where $b\in {\Bbb R}$. All of them are unstable in the Liapunov sense. The rest of the orbits are the straight lines $x_{0}=a$ for a fix non-null $a\in {\Bbb R}.$ Through any orbit the motion is uniform. In the semiplane $x_{0}>0$ the sense of the motion is increasing, respect to the $x_{1}$-axe as $t$ increases, i.e., the particle comes from the $-\infty$ part respect to the $x_{1}$-axe and goes to (positive) $\infty$ of such axe as $t$ increases. In the semiplane $x_{0}<0$ the sense of the motion is the opposite one. The constant speed of the motion in the orbit $x_{0}=a$  is $\mid a \mid$, the absolute value of $a.$ Particles move quickly for large values of $\mid a \mid$ and slowly for small values $\mid a \mid$ and they do not move in the $x_{1}$-axe. Therefore, there seems to be something in the $x_{1}$-axe slowing down the motion. We can deduce all above only knowing that the corresponding velocity vector field for the  equation $(\ref{E_:ECU})_{1}$, in the euclidean plane,  is given by $X(a,b)=(0,a)$. We can also compute, quickly and easily the solution of problem  $(\ref{E_:ECU})_{1}$ with initial condition  $x_{0}=a$ and $x_{1}=b$ because the corresponding matrix is nilpotent. It is the curve $x(t)=(a, at+b)$.

From  Theorem $\ref{T: TR}$ and Corollary $\ref{T: CTR}$, we get that $\Pi_1 (M)=\begin{pmatrix}1&0\\  0& -1
\end{pmatrix}$  and $\Pi_1(-M)=\begin{pmatrix}-1&0\\  0& 1
\end{pmatrix}$ are time-reversal symmetries for the flow generated by the equation $(\ref{E_:ECU})_{1}$. Consider the orbit $\theta(a,b)=\{(a,at+b), a\neq0, b, t \in {\Bbb R} \}$ of $(\ref{E_:ECU})_{1}$. Then it is symmetric respect to the involution $\Pi_1 (M)$, see Definition 4.1 in \cite{Lam-Rob}. This, in particular, means that if we transform the orbit $\theta(a,b)$ by  $\Pi_1 (M)$ we get again $\theta(a,b)$ but the parametrization obtained by means of $t$ is not a solution. However, if we finally change $t$ by $-t$ in such obtained curve we have a solution of $(\ref{E_:ECU})_{1}$; in this case the initial value at $t=0$ is $(a,-b)$. Of course, $\theta(a,b)=\theta(a,-b)$.

What is the behaviour under the action of the involution $\Pi_1(-M)$? if we transform the orbit $\theta(a,b)$ by means of $\Pi_1(-M)$, we get another {\it different } orbit of $(\ref{E_:ECU})_{1}$. In fact, we obtain  $\Pi_1(-M)(\theta(a,b))=\theta(-a,b)$ and $\theta(a,b)\neq\theta(-a,b). $ Again, the parametrization so obtained by means of $t$ is not a solution of $(\ref{E_:ECU})_{1}$. But, again, if we change $t$ by $-t$ we get another solution but with {\it different} orbit. Finally see that the composition of both time reversal symmetries is really a {\it symmetry} for $(\ref{E_:ECU})_{1}$. Then we obtain that $-I$ is a symmetry and it implies that if  $x(t)=(a, at+b)$ is a solution of the problem with initial condition $x(0)=(a, b)$, then $-x(t)=-I(x(t))$ is a solution with initial condition $(-a,-b).$

\begin{rmk} Note that any orbit $x_{0}=a\neq0$ of $(\ref{E_:ECU})_{1}$  is symmetric respect to the time reversal symmetry $\Pi_1(M)$. On the contrary, no orbit (except for equilibrium points) is symmetric respect to the time reversal symmetry $\Pi_1(-M).$
\end{rmk}

We propose to the reader the beautiful exercise of analysing the problem $(\ref{E_:ECU})_{2}$ and the role of the involutions $\Pi_2 (M)=\begin{pmatrix}1&0&0\\  0& -1&0 \\ 0&0&1
\end{pmatrix}$ and $\Pi_2 (-M)=\begin{pmatrix}-1&0&0\\  0& 1&0 \\ 0&0&-1
\end{pmatrix}$ as time reversal symmetries for $(\ref{E_:ECU})_{2}$. The geometric analysis of this problem is, obviously, richer than that of $(\ref{E_:ECU})_{1}$. In the problem $(\ref{E_:ECU})_{1}$ the involutions $\Pi_1(M)$ and $\Pi_1(-M)$ are conjugated and they represent reflections about different axes. But in $(\ref{E_:ECU})_{2}$, $\Pi_2 (M)$ is a reflection about the plane $x_{1}=0$, which is  a plane of fixed points of $\Pi_2 (M)$, and $\Pi_2 (-M)$ is a rotation of angle $\pi$ around the  $x_{1}$-axe. Of course they are not conjugated. Analysing this case,  one can see first that the equilibrium points are those of the form $(0,0,c)$ and that under the flow generated by $(\ref{E_:ECU})_{2}$, particles are moving within affine hyperplanes $x_{0}=a$. When $a=0$ we obtain the motion described in $(\ref{E_:ECU})_{1}$ in this hyperplane. For $a\neq0$ the orbits of points are parabolas. In this case, the velocity vector field is given by $X(a,b,c)=(0,a,2b)$ and the solution of $(\ref{E_:ECU})_{2}$ with initial conditions $x_{0}(0)=a$, $x_{1}(0)=b$ and $x_{2}(0)=c$ is given by $x(t)=(a, at+b, at^{2}+2bt+c)$. As in the previous case, any orbit, which is not an equilibrium point, is symmetric with respect to the involution $ \Pi_2 (M)$, but any non-trivial orbit is transformed by    $\Pi_2 (-M)$ into another different orbit. In both cases, if after the transformation, we change $t$ by $-t$ we obtain new solutions of the problem $(\ref{E_:ECU})_{2}.$ Finally $-I:{\Bbb R}^{3}\longrightarrow {\Bbb R}^{3}$ is a symmetry for the problem.

\subsection{Facts and/or conjectures and/or speculations on the problems $(\ref{E_:ECU})_{n}$ and $(\ref{E_:ECU})$ }

The flow induced by problem $(\ref{E_:ECU})$ in ${\Bbb K}[[x]]$ is given by \[\Phi:{\Bbb K}[[x]]\times {\Bbb R} \longrightarrow {\Bbb K}[[x]] \] where $\Phi(h,t)=e^{tL^{1,1}_{1}}(h)= \frac{1}{1-xt}h\left(\frac{x}{1-xt}\right)$, while the flow generated by the problem $(\ref{E_:ECU})_{n}$ in ${\Bbb R}^{n+1}$ is \[\Phi_{n}:{\Bbb R}^{n+1}\times {\Bbb R} \longrightarrow {\Bbb R}^{n+1} \] whose matrix expression is $\Phi_{n}(x,t)=\Pi_{n}(e^{tL^{1,1}_{1}})x^{T}.$

We now are going to state, without proofs, properties related to problems $(\ref{E_:ECU})_{n}$ and $(\ref{E_:ECU})$. This is the reason why we entitled this subsection as we did.
\begin{prop}{\bf (Dynamical properties related to problems \textbf{$(\ref{E_:ECU})_{n})$}}
Let $n$ be a non-negative integer number, then we have the following properties.
\begin{itemize}
\item[(i)] The orbit of any point $a=(a_{0}, a_{1}, \cdots, a_{n})$ in ${\Bbb R}^{n+1}$ is contained in the affine hyperplane $x_{0}=a_{0}.$ Moreover if $a_{0}=0$ and if one consider ${\Bbb R}^{n}=\{(x_{0}, x_{1}, \cdots, x_{n})\in {\Bbb R}^{n+1} / x_{0}=0\}$, the motion induced by $\Phi_{n}$ in ${\Bbb R}^{n}$ is $\Phi_{n-1}$.
    \item[(ii)] The equilibrium points in $(\ref{E_:ECU})_{n}$ are those in the $x_{n}$-axis and if $n>0$ all of them are unstable in the Liapunov sense.
    \item[(iii)] The non-trivial orbits in $(\ref{E_:ECU})_{n}$, i.e., those which are not equilibrium points, are related to the so called {\it moment curve} in the corresponding hyperplane. In particular the solution of $(\ref{E_:ECU})_{n}$  with initial condition $(1, 0, \cdots, 0)\in {\Bbb R}^{n+1}$ is $x(t)=(1, t,t^{2} \cdots, t^{n})$ which is a copy of the corresponding moment curve in the hyperplane $x_{0}=1.$
    \item[(iv)] Any non-trivial orbit in $(\ref{E_:ECU})_{n}$ is symmetric with respect to the time-reversal symmetry $\Pi_{n}(M)$ and no one of them is symmetric respect to $\Pi_{n}(-M).$ Anyway, if we have a non-trivial solution of $(\ref{E_:ECU})_{n}$, i.e., a non-constant one, and we transform it by any of the involutions $\Pi_{n}(M)$ or $\Pi_{n}(-M)$ and then change $t$ by $-t$ we get another solution of $(\ref{E_:ECU})_{n}.$
    \item[(v)] $-I:{\Bbb R}^{n+1}\longrightarrow {\Bbb R}^{n+1}$ is a symmetry for the equation $(\ref{E_:ECU})_{n}.$
\end{itemize}

\end{prop}

Motivated by the previous result and knowing that Pascal triangle is the time one map of the flow   $\Phi$, we can state:

\begin{prop}{\bf (Some dynamical properties of Pascal Triangle)}
\begin{itemize}
\item[(i)] The flow $\Phi$ has not equilibrium points up the null power series, that is,  all the coefficients of the power series are null.
\item [(ii)] The orbit of the formal power series constantly $1$, by means of $\Phi$, is the set of geometric progressions $\{\frac{1}{1-tx}\}_{t\in {\Bbb R}}. $ Then one can think about $1$ moving through the set of the germs of analytic functions, at $x=0$, because for any $t\in {\Bbb R}$ the function $f_{t}(x)=\frac{1}{1-tx}$ for  $x\in (-\frac{1}{\mid t \mid}, \frac{1}{\mid t \mid})$  is analytic at $x=0.$ Moreover, this orbit can be seen as the asymptotic behaviour, as $n$ goes to $\infty$, of the moment curves.
\item[(iii)] The Riordan involution $M$ and $-M$ are time-reversal symmetries for the flow $\Phi$. Any orbit of problem $(\ref{E_:ECU})$ is symmetric respect to $M$ and no orbit, except for the unique equilibrium point, is symmetric respect to $-M$. Anyway, if we transform any solution of $(\ref{E_:ECU})$ by means of $M$ or $-M$ and then change $t$ by $-t$ we get another solution of the problem.
\item[(iv)] $-I:{\Bbb K}[[x]]\longrightarrow {\Bbb K}[[x]]$ is a symmetry for the equation $(\ref{E_:ECU}).$
\end{itemize}
\end{prop}

To finish this paper note the following.

{\bf Claim:} {\it  Any non-trivial orbit of the problem  $(\ref{E_:ECU})_{n}$ has empty $\alpha$-limit set and $\omega$-limit set.} On the other hand, the problem $(\ref{E_:ECU})_{n}$ has time reversal symmetries. This fact allows us to think that there should be relationships between both limit sets. We then decided to force the corresponding flows, by means of considering a compactification of  the corresponding phase spaces, to get non-empty $\alpha$-limit and  $\omega$-limit sets and to look for relationships between them. We proceed as follows.

Consider the one point (or Alexandroff) compactification of ${\Bbb R}^{n+1}$ which is, topologically, the $n+1$-dimensional sphere $S^{n+1}.$ Let us denote by $\infty$ the added point. Note that any $t$-map of the flow $\Phi_{n}$, $\Phi^{t}_{n}$, can be continuously extended to a map $\widetilde{\Phi^{t}_{n}}:S^{n+1}\longrightarrow S^{n+1}$ defining only $\widetilde{\Phi^{t}_{n}}(\infty)=\infty.$ In this way we get a dynamical system
\[\widetilde{\Phi_{n}}:S^{n+1}\times {\Bbb R} \longrightarrow S^{n+1}. \]
For every non-negative integer $n$ we can also extend the time reversal symmetries  $\Pi_{n}(M)$ and $\Pi_{n}(-M)$ to continuous maps
$\widetilde{\Pi_{n}(M)}$  and $\widetilde{ \Pi_{n}(-M)}$  from $S^{n+1}$ onto itself imposing that the point $\infty$ is a fixed point for both of them. We can identify, topologically,   ${\Bbb R}^{n+1}$   with $S^{n+1}\backslash \{\infty\}$ (which is a dense subset of $S^{n+1}$). With all these constructions we have
\begin{prop} Let $n$ be a non-negative integer number. Then we have the following.
\begin{itemize}
\item[(i)] The maps $\widetilde{\Pi_{n}(M)}$ and $\widetilde{\Pi_{n}(-M)}$  are continuous involutions in  $S^{n+1}$ and they are time-reversal symmetries for the dynamical system $\widetilde{\Phi_{n}}.$
\item[(ii)] The orbits of the dynamical system $\widetilde{\Phi_{n}}$ are those of $\Phi_{n}$ (after the mentioned identification) plus  $\{\infty\}$ which is an equilibrium point. Moreover, every non-trivial orbit of $\widetilde{\Phi_{n}}$ is {\it homoclinic}, being the point $\infty$ an attractor and a repeller of all of them.

\end{itemize}
Consequently, the $\alpha$-limit  and the $\omega$-limit  sets of any non-trivial orbit coincide.
\end{prop}

\end{section}

redpajama_set_name: RedPajamaArXiv


text:
Vietnam resolved to realise UN 2030 Agenda: Deputy PM
The Vietnamese Government is always resolved to strive for the successful realisation of the 2030 Agenda for Sustainable Development, Deputy Prime Minister Vu Duc Dam said in Hanoi on December 11th.
Deputy PM Vu Duc Dam receives Cuban guest
Receiving Resident Representative of the United Nations Development Programme (UNDP) in Vietnam Caitlin Weisen, the official further said the Government advocates bringing the sustainable development goals into every national development strategy and programme.
Deputy Prime Minister Vu Duc Dam receives Resident Representative of the United Nations Development Programme (UNDP) in Vietnam Caitlin Weisen. (Photo: VNA)
He also added that in the time to come, Vietnam will continue boosting the awareness from the whole society and mobilise the participation of all the relevant sides into the work.
As Vietnam is going to shoulder the roles of the ASEAN Chair 2020 and a non-permanent member of the United Nations Security Council, besides the issue of climate change, the Government proposes the UNDP support in studying and implementing a number of activities to push up Vietnam's priorities at the UN.
Deputy PM Vu Duc Dam also spoke highly of the cooperation and assistance from the UN in general and the UNDP in particular to Vietnam over the past more than 40 years, stressing that the assistance programmes are described as well implemented and effective, making positive contributions to the building and implementation of the reform policy in Vietnam.
The UNDP official, for her part, briefed her host on a number of contents of the UNDP global human development report. As for Vietnam, the human development index increases every year, standing at 0.693 last year, placing the country the 118th out of the 193 countries and territories, the life expectancy is 75.3 years, the schooling is 12.7 years, and the income gap is 18.1 percent – the lowest in the group./.
TỪ KHÓA: UN 2030 Agenda Vietnam Vu Duc Dam

redpajama_set_name: RedPajamaCommonCrawl


Congratulations, Yuki Chen! You're the lucky winner! Please check your inbox. --------------- As a show of appreciation to my fans and friends for their support of GSD, my team put together a GIVEAWAY! Win a #GongShouDao movie poster signed by Jack Ma and me! Here's how to enter: 1) Like this post 2) Comment "GSD" below 3) Follow Gong Shou Dao One winner will be randomly selected. WInner will be announced on this post and contacted through DM on February 15.


text:
Henry Leck (born c. 1946) is the founder and previous Artistic Director of the Indianapolis Children's Choir and the Indianapolis Youth Chorale. He is a choral clinician and a specialist on boys' changing voices.

In 1986, Leck formed the Indianapolis Children's Choir which has toured internationally with much acclaim.

Leck has conducted festival choirs in nearly every state in the U.S., and has led many international children's choir festivals, including the Musica Mundi Tuscany Children's Choir Festival in Italy, the Central European International Children's Choral Festival, the International Children's Choir Festival in Beijing, China, the Vienna Children's & Boys Choir Festival with the Vienna Sangerknaben, and the KI Concert Spain Festival. He recently conducted the Tokyo International HS Honor Choir in Japan. Leck founded the ARCI São Paulo Children's Honor Choir in Brazil, where he conducts bi-annually.

Until his retirement in 2016, Leck served as Artistic Director of the Indianapolis Children's Choir for 30 years, where he built the legacy of one of the most respected Children's Choirs in the United States. Leck also served on the Choral Faculty at Butler University for 27 years. He retired from his position as Associate Professor of Music at Butler
in 2015, and now serves as Professor Emeritus. He is also recognized as an Honorary Associate Professor at the Hong Kong Institute for Teacher Education.

Works

Leck has published many choral arrangements with Hal Leonard Corporation and Colla Voce Music Publishing.  He also authored the choral section of Silver Burdett music textbooks for grades 4, 5 and 6.

He has produced four teaching videos entitled The Boy's Expanding Voice: Take the High Road, Vocal Techniques for the Young Singer, Creating Artistry Through Movement, Dalcroze Eurhythmics and Creating Artistry with the Male Maturing Voice.

Leck is also the editor of two nationally known choral series published by Hal Leonard and Colla Voce, from which many of his own compositions can be purchased.

In 2012, Leck's arrangement of "The National Anthem" was performed by Kelly Clarkson, and Indianapolis Children's Choir, at Super Bowl XLVI.

Creating Artistry

Each year Leck teaches his conductor's workshop, Creating Artistry, at Butler University. In this progressive three level workshop, Leck mentors conductors and teachers of all ages, working towards building musicality through expressive conducting gesture. Leck recently released a textbook called Creating Artistry Through Choral Excellence.

Education

Leck received his bachelor's degree in Music from the University of Wisconsin–Stevens Point and a master's degree in Choral Conducting from Indiana University

Awards
 Chorus America's Distinguished Service Award (2017)
 NUVO's Cultural Vision Award - Lifetime Achievement Award (2016)
 National Association for Music Education's Lowell Mason Fellows Award (2015)
 Indiana Governor's Art Award (2007)
 Indiana Music Education Association "Outstanding University Music Educator of the Year" (1992)

References

Living people
Indiana University alumni
University of Wisconsin–Stevens Point alumni
1946 births

redpajama_set_name: RedPajamaWikipedia


text:
Пааво Матті Вяюрюнен (; 2 вересня 1946, Кемінмаа, Фінляндія) — фінський політик; в минулому Міністр закордонних справ Фінляндії; кандидат на виборах президента Фінляндії 2012 року від партії Фінляндський центр, почесний голова цієї партії. Кандидат в президенти Фінляндії на виборах 2018 року, зайняв на них 4-е місце.

Життєпис
Народився 2 вересня 1946 року в Кемінмаа в родині агронома Юхо Еемелі Вяюрюнена і Анни-Лійса Кайянкоскі. У 1970 р. Здобув ступінь магістра соціальних наук в Гельсінському університеті., а в 1988 р. Здобув ступінь доктора соціальних наук в Академії Ебо в Турку. У 1996 році він став доцентом Лапландського університету в Рованіємі.

У 1970—1995 роках він мав мандат депутата в Едускунті, представляючи округи Лапландії та Уусімаа. З 1970 по 1971 рік був секретарем прем'єр-міністра. У 1972—1980 рр. він був заступником голови Партії Центру, потім до 1990 року керував цим угрупуванням. Він неодноразово обіймав урядові посади в кабінетах Марті Міеттунена, Калеві Сорса, Мауно Койвісто та Еско Аго. Був міністром освіти (1975—1976), роботи (1976—1977) та закордонних справ (1977—1982, 1983—1987 та 1991—1993), а в 1983—1987 рр. — також віце-прем'єр-міністром. У 1988 та 1994 роках він невдало балотувався на президентських виборах.

У 1995 році він зайняв посаду члена Європарламенту (після вступу Фінляндії до Європейського Союзу), а тому відмовився від засідання в національному парламенті (пізніше він був членом Едускунті протягом тижня в 1999 році). У 1996, 1999 та 2004 роках він був переобраний на європейських виборах. У ЄП з 1997 по 2004 рік був заступником голови групи Партії європейських лібералів, демократів та реформаторів.

На національних виборах 2007 року його знову обрали до Едускунті. У другому уряді Матті Ванханена він обійняв посаду міністра зовнішньої торгівлі та розвитку. Він зберігав свою посаду і в уряді, призначеному 22 червня 2010 року Марі Ківініемі. Він покинув національний парламент, а потім уряд у 2011 році..

У 2012 році він був кандидатом партії Центр на виборах президента (у першому турі голосування набрав 17,5 % голосів, посівши 3 місце). У 2014 році він був переобраний до Європейського парламенту. У 2016 році він покинув партію Центр, до 2017 року очолив новостворену групу Kansalaispuolue. У 2017 році християнськими демократами його обрали до Гельсінської міської ради. У 2018 році як незалежний кандидат він знову взяв участь у виборах президента, посівши 4 місце з результатом 6,2 % голосів.

У 2018 році він подав у відставку з Європейського Парламенту, взявши на себе мандат в Едускунті, який він отримав у 2015 році.

Примітки

Посилання

 Офіційна сторінка Пааво Вяюрюнена
 Офіційна сторінка Пааво Вяюрюнена на сайті Парламенту Фінляндії

Міністри закордонних справ Фінляндії
Міністри освіти Фінляндії
Депутати парламенту Фінляндії
Депутати Європарламенту від Фінляндії
Кандидати в президенти Фінляндії


Happy Tuesday! The day started with time outside on the patio. We enjoyed the fresh air and seeing our beautiful flowers. Before lunch, we all had a good laugh watching funny and cute babies. After lunch and some rest, a group had a good time playing community scrabble. What a great job they all did! The day ended with videos of gorgeous scenery along to relaxing music. It was a great day!


ITC, India's leading FMCG major today unveiled 'Farmland' - its new brand for fresh Fruits & Vegetables. Farmland makes a debut in the Indian market with an exciting range of variants in potatoes such as Natural Low Sugar Potatoes, Potatoes with Natural Antioxidant, Baby Potatoes and French Fry Potatoes.
In line with ITC's uncompromising commitment to provide superior quality, safe, hygienic and nutritious offerings to consumers, 'Farmland' has been crafted with a mission to provide Safe and Nutritious potatoes to consumers and is an outcome of years of research and rich experience in working closely with farmers.
Farmland Potatoes come with the promise of low sugar, low fat and great taste. These potatoes are naturally low in fat and sugar with fat content below 0.2% and sugar level below 2% respectively. Farmland potatoes have been tested and endorsed by ICAR-CPRI, Shimla (Indian Council for Agricultural Research - Central Potato Research Institute), the premier institute for potato research and development in India. ITC along with ICAR-CPRI intend informing consumers of the nutritional benefits of potatoes and help dispel some of the myths around it.
With a consumer spend of approximately Rs. 500,000 crores, the Indian Fruits & Vegetables market is pegged at approximately Rs. 300,000 crores for the producers. ITC's brand Farmland will play in the fresh fruits and vegetables segment catering to the emerging consumer demand for safe and nutritious fruits and vegetables.
Announcing the launch, Mr. S Sivakumar, Group Head - ITC's Agri Business and Information Technology, said, "At ITC we are guided by the twin philosophies of creating value through world class consumer brands while serving national priorities. Fresh Fruits & Vegetables are a fundamental part of the consumer's food basket and is also key to providing enhanced incomes to farmers who are an integral part of ITC's farm-to-fork value chain. Our brand Farmland will provide yet another win-win for the consumers by serving their emerging need for safe and nutritious fruits and vegetables as well as open up a new value-added market segment for the Indian farmers".
Potatoes are the largest consumed vegetable in the world and in India. We believe that Farmland potatoes with their superior quality & taste in different variants will resonate well with Indian consumers who today have distinct tastes and preferences. We are introducing them at highly competitive prices", he added.
The total market for table potatoes in India in consumer spend terms is valued at Rs. 70,000 crores with consumption in metros and mini-metros alone close to Rs. 5000 crore, which is what ITC's Farmland Potatoes range will address in the first phase.
Over the past several years, ITC's agri business has acquired unparalleled strengths in the potato category, managing close to 250 varieties of potatoes, access to the best potato technology in the world and global supply linkages. The ITC group is today the Number 1 seed potato company in India and by entering this segment is providing a value added market access to its farmers. This depth along with the scale of ITC's storage, handling and distribution infrastructure provides a compelling driver for the company to enter this fresh fruits and vegetables segment.


text:
Q: Is there a way to determine the number of PCIe lanes my system has and uses With a core i9-9900K and an ASUS MPG Z390 GAMING PRO CARBON AC/ MPG Z390 GAMING PRO CARBON motherboard, is there a way to determine how many PCIe lanes my system has?
With an ASUS Geforce RTX 2080 installed and allegedly running in x16 mode and an ASUS XG-C100C 10G PCI-E Network Adapter fitted in, is there a way to determine how many lanes are being used?
Lane usage solutions are welcome for both Windows and Fedora.

A: Information about computer hardware can be found in the DMI table, which is read and parsed with dmidecode utility:
 # dmidecode
 ...
 Handle 0x0900, DMI type 9, 17 bytes
 System Slot Information
        Designation: PCIe Slot 1
        Type: x16 PCI Express 3
        Current Usage: Available
        Length: Long
        ID: 1
        Characteristics:
                3.3 V is provided
                PME signal is supported
...

All slots in this server are empty, so I can't show, but for "In Use" slot there will be a Bus Address entry.
This command also shows which memory this system has, which memory slots are occupied and so on. The dmidecode package has other useful utilities, for example, biosdecode.
To find out which device has which PCI bus address, we closely look onto lspci output from pci-utils. This address is in the first column:
# dmidecode
...
Handle 0x2902, DMI type 41, 11 bytes
Onboard Device
        Reference Designation: Embedded NIC 3
        Type: Ethernet
        Status: Enabled
        Type Instance: 3
        Bus Address: 0000:03:00.0
...

# lspci
...
03:00.0 Ethernet controller: Broadcom Limited NetXtreme BCM5720 Gigabit Ethernet PCIe
...

redpajama_set_name: RedPajamaStackExchange


text:
Wells and Sturt's tree walking track
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Learn more about why this park is special
Wells and Sturt's tree walking track is in Sturt National Park. Here are just some of the reasons why this park is special:
A vast and varied precious landscape
Tibooburra means 'heaps of rocks' in the local Aboriginal language, and you can't miss the rocks - ancient granite tors that surround Tibooburra and line the road on the way to the park. This contrasts with the red sand of the desert on the western side of the park and with the 'Jump Ups' that rise from the plains in the central part of the park. Wherever you go in Sturt National Park, you're sure to be inspired by the dramatic changes in scenery and amazed by the true scale of the vast, arid expanse of outback.
Jump-Up Loop Road drive A fantastic self-guided car tour of the outback country of Sturt National Park, Jump-Up Loop Road drive offers scenic desert views, historic heritage and excellent birdwatching.
The Granites walking track Explore the geological history of Sturt National Park along The Granites walking track. See wildflowers, kangaroos and lizards as you walk over the ancient boulders.
A feral predator-free area
Locally extinct mammals are being reintroduced to Sturt National Park as part of the NSW National Parks and Wildlife Service (NPWS) feral predator-free areas project, in partnership with Wild Deserts. Over 10 years this project will reintroduce the greater bilby, crest-tailed mulgara, western barred bandicoot, greater stick-nest rat, golden bandicoot, burrowing bettong and western quoll.
Reintroducing these species, some of which have been absent for over 100 years, will play an important role in restoring the desert ecosystem in Sturt National Park. The project, funded by the NSW government, is a collaboration between NPWS and Wild Deserts (University of New South Wales and Ecological Horizons).
Talpero lookout Visit Talpero lookout for expansive desert views and see 'the big bandicoot' sculpture. It's located 2 hours from Tibooburra, near Fort Grey campground in Sturt National Park.
Wells and Sturt's tree walking track Retrace the footsteps of a famous explorer in the stunning and remote outback landscape of NSW's far north-west. You'll rediscover history, see majestic red river gums and cross a dry lakebed, near Ca...
Ancient connections
Sturt National Park is the traditional land of the Wangkumara People, whose Country extended from what is now southwest Queensland and northeast South Australia down through Tibooburra to Milparinka. The Wangkumara People travelled widely throughout this large and arid land to make the most of waterholes, permanent soaks, useful plants and animals. Today there is much evidence of the Wangkumara People's connection with this land - throughout the park you might notice middens and stone relics; reminders of the role this landscape played as the giver and sustainer of life.
Jump-Up walking track Discover the landscape of Sturt National Park along the Jump-Up walking track; you're likely to see Aboriginal sites, wildflowers and kangaroos along the way.
A waterbird oasis
Lake Pinaroo is around 80 km north west of Tibooburra and 24km south east of Cameron Corner. It's an ephemeral lake, meaning it floods for short periods of time and then may not hold water for several years, depending on rainfall. In 1996, it was listed as an internationally important wetland under the Ramsar Convention because of its retention of water for long periods, and the rarity of wetlands in arid NSW. Lake Pinaroo plays a crucial role in the survival of many plants and animal species, and supports large numbers of waterbirds and waders, including international migratory species and threatened species. When full, Lake Pinaroo is a stunning contrast to the dry landscapes of Sturt National Park, and you can see waterbirds like the freckled and blue-billed ducks, as well as brolgas, grey falcons and budgerigars.
Heritage values of the homestead
Historic Mount Wood Homestead is located on the oldest sheep station in northwest NSW, taken up around 1881. Listed on the State Heritage Register, it's one of the most complete examples of a self-reliant sheep station in the region, spanning 368,385 acres. It was a hub for washing sheep wool on the long journey by camel train or cart to Wilcannia, prior to shipping. Today, the woolscour is a rare example of a complete set of wool washing equipment, and the only 19th century station-based scour in NSW to survive intact. Surviving the harsh outback, you can still see the original stone hut built in 1890, a stone homestead (1897), and an art deco-styled homestead (1935). There's also a woolshed, shearers' quarters, woolscour, blacksmith shop, stables, windmills and outstations. The buildings provide a fascinating window into pastoral life and changing technology over almost 100 years.
Plants and animals you may see
Wedge-tailed eagle (Aquila audax)
With a wingspan of up to 2.5m, the wedge-tailed eagle is Australia's largest bird of prey. These Australian animals are found in woodlands across NSW, and have the ability to soar to heights of over 2km. If you're bird watching, look out for the distinctive diamond-shaped tail of the eagle.
Emu (Dromaius novaehollandiae)
The largest of Australian birds, the emu stands up to 2m high and is the second largest bird in the world, after the ostrich. Emus live in pairs or family groups. The male emu incubates and rears the young, which will stay with the adult emus for up to 2 years.
Red kangaroo (Macropus rufus)
The red kangaroo is one of the most iconic Australian animals and the largest marsupial in the world. Large males have reddish fur and can reach a height of 2m, while females are considerably smaller and have blue-grey fur. Red kangaroos are herbivores and mainly eat grass.
Sturt's desert pea (Swainsona formosa)
One of Australia's most famous desert wildflowers, Sturt's desert pea is found across inland arid regions of Australia, including far west NSW. One of the most easily-recognised Australian native plants, Sturt's desert pea thrives in red sandy soil, or loam, and has vibrant red leaf-shaped flowers with a black centre, known as a 'boss'.
Saltbush (Atriplex nummularia)
A hardy Australian native plant, the saltbush is a small spreading shrub that can withstand dry salty soils such as those found in the desert plains of western NSW. It is grey-white in colour and has small spear-shaped succulent leaves. It flowers from December to April.
Look out for...
Dromaius novaehollandiae
Environments in this park
Desert and arid shrubland environments
Woodland environments
Education resources (1)
Who cares about national parks?, Stage 2 (Years 3-4), Geography
Naturescapes e‑newsletter
Subscribe to Naturescapes
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China's leaders say they are done devaluing the yuan. But wise investors shouldn't necessarily take them at their word.
Many fear that China's fragile stock market could fall further if the government again surprises the world by devaluing the yuan in an effort to boost the country's economy. This could hurt many investors with Chinese exposure.
Those seeking a bit of downside protection might consider purchasing puts on the Deutsche X-Trackers Harvest CSI 300 Index ETF (ticker: ASHR ), a popular proxy for mainland Chinese stocks.
Puts increase in value when associated security prices decline.
"It's hard to trust data from China," Scott Fullman, Revere Securities' chief strategist, said.
With ASHR at $40, investors can buy the ETF's October $37 put for $2. The exchange-traded fund tracks mainland China's stock prices.
If ASHR is at $30 at expiration, the put is worth $7. If ASHR advances, investors lose the $2 put premium.
"From a macro point-of-view, I believe that hedges for China are valid, since the economy appears to be slowing and confidence from Europe and the U.S. appears to be waning," Fullman said.
ASHR's October $39 put is another hedge investors can consider. Because the put is just below ASHR's current price, the October $39 will increase in value faster if ASHR declines. The put was recently trading at $2.70.
So many investors are wagering Chinese stocks will decline that ASHR is listed on "hard to borrow" stock lists, according to Trade Alert, which analyzes trading data.
When a security is hard to borrow, it means brokerage firms have effectively lent their "long inventory" to bearish investors who think the stock will collapse. Investors sell the borrowed securities in the market in anticipation the shares can be repurchased at a lower price.
To be sure, investors have reason to be bearish about China's stock market. The yuan was devalued last week after a report showed China's exports declined 8% in July. Exports were hurt by a strong yuan.
Now, investors are increasingly concerned about the trajectory of China's economic growth and the yuan's impact.
Ma Jun, the chief economist of the People's Bank of China, has tried allaying investor concerns. He recently said the yuan will probably move in both directions as the economy stabilizes, an item Credit Suisse reminded clients of in a Monday trading advisory.
Jun's statement is reminiscent of J.P. Morgan's famously non-committal line when asked for his view of the stock market. The financier's response: it will fluctuate.
My colleague, Randy Forsyth, has a similar view of China's yuan. He believes it may fall another 10% or more since the yuan's value increased so much in recent years. Since the yuan's strength exacerbated China's economic weakness, China may need to further devalue its currency to support economic growth.
Meanwhile, other major Asian economies are struggling, raising concerns Asia is facing an economic crisis.
This view gained credence Monday when Japan reported the economy contracted 1.6% in the second quarter.
Fullman said Japan's gross domestic product report could indicate that economic weakness in Asia is more widespread than understood. If Japan, a developed nation is struggling, emerging markets like China may face even greater difficulties.
And that is exactly why it makes sense to hedge China's stock market with puts.
Correction: An earlier version of this story misstated the result if the ASHR fund advances. In that event, investors would lose the $2 put premium.
China's leaders say they are done devaluing the yuan.


Troma Entertainment proudly presents: The World Premiere of Return to Return to Nuke 'Em High aka Vol.2 at the Arcades 3 Cinema Cannes Film Festival, May 23rd.
World Premiere of Lloyd Kaufman's highly anticipated second installment of his 2-part event film Return to Return to Nuke 'Em High AKA Vol.2 at the 2017 Cannes Film Festival in the prestigious Arcades 3 Cinema, May 23rd at 17:30 (5:30pm), it was announced today by Lloyd Kaufman, President of Troma Entertainment and creator of The Toxic Avenger.
"I couldn't be happier with Return to Return to Nuke 'Em High AKA Vol.2 . The Troma Team and I have been working on this film for the past 7 years. We put our sweat, blood and slime into the production and can't wait for the world to see it! Vol.2 truly is as my partner and VP of Troma Entertainment, Michael Herz says, "The best Troma movie ever made!" stated Lloyd Kaufman, Director of Return to Return to Nuke 'Em High AKA Vol.2 .
Quentin Tarantino has been a fan and friend of Troma for years, and when he made Kill Bill in two volumes to create an "Event Film," Lloyd Kaufman took note. "Michael Herz and I decided that to mark Troma's 40th year we, too, would produce an 'Event Film' in two volumes just like Quentin did," says Kaufman, "except slightly less lavish in the budget area." Return to Nuke 'Em High: Vol. 1 & 2, directed by Lloyd Kaufman, is a hilarious, thoughtful sci-fi "Event Film" with themes ripped straight from today's headlines: the contamination and degradation of the world's food supply, rampant bullying, and LGBTQ love triumphing over prejudice and violence.


text:
This environment is composed on the following elements:
 * the main development will be performed in a VirtualBox VM running XUbuntu 14.10
    - the local versions of some services, like HBase and Kafka, will run in XUbuntu, as these services are difficul to run in windows
    - due to problems communicating the Windows host with the guest XUbuntu, finally development with Eclipse and Anaconda Spyder will also be performed in the VM. Anyway a Linux environment is more confortable for developing

 * a EMR cluster launched by some Fabric scripts from the XUbuntu VM will be used to run the jobs in a cluster, for the final acceptance and performance tests

## Components and versions:
    * Spark 1.3
    * Scala 2.10: required by Spark
    * Python 2.7: required by Spark
    * Apache Kafka kafka_2.10_0.8.1.1: required by Spark

## Network Setup for VirtualBox
First of all install the ssh server (package openssh-server) in the VM. Then following http://stackoverflow.com/questions/5906441/how-to-ssh-to-a-virtualbox-guest-externally-through-a-host I take the option of using the NAT network configuration for VirtualBox, and then in Configuration -> Networking -> Port forwarding we add a new rule with protocol = TCP, host port = 3022 and guess port = 22. This forwards petitions to host:3022 to the VM:22. Hence we can connect to the VM by ssh from the Windows host with:

```bash
C:\Users\bc24u_000\Sistemas>ssh -p 3022 juanrh@localhost
The authenticity of host '[localhost]:3022 ([127.0.0.1]:3022)' can't be established.
...
juanrh@localhost's password:
Welcome to Ubuntu 14.10 (GNU/Linux 3.16.0-33-generic x86_64)
```

This is enough as most of the time we'll use the VM directly, and ssh is enough for moving data from the host to the guest, and viceversa

## Scripts for the development environment
Fabric is used for automation
 * Run `install_prerequisites.sh` to setup Fabric
 * Run `install_devenv.sh` to install the development environment

 For now all the Fabric tasks are defined in the single file `fabfile.py`

## Scala Environment for Spark
### Version of Scala
Spark is compiled for Scala 2.10, but 1) Scala IDE 4 is distributed with Scala 2.11, and 2) Scala IDE can only
have a single version of Scala installed (http://scala-ide.org/docs/current-user-doc/faq/index.html "Currently,
it is not possible to install more than one Scala IDE plugin within the same Eclipse installation"), and 3)
we cannot run a binary for Scala 2.10 in the Scala 2.11 runtime (https://groups.google.com/forum/#!topic/scala-user/mUZSYgCLigA).
Hence we should use Scala IDE 3.0.3 (http://scala-ide.org/download/prev-stable.html careful, Scala IDE 3.0.4 is
for Scala 2.11) which is the last version of the plugin for Scala 2.10. This works for Eclipse Kepler

### Configure Scala IDE
- Import project as a Maven project
- FIX for Scala IDE 4: this is not needed for Scala IDE 3.0.3, but for Scala IDE 4 we cannot execute the
code because the runtime is Scala 2.11. In project properties -> Scala Compiler select "use project settings" and "Scala Installation" to
"Fixed Scala Installation: 2.10.4 (built-in)"

### Configure Eclipse for Sun Java 7
In Ubuntu, after installing the sun JDK with the corresponding task of the fabfile in this directory we have to modify eclipse.ini as indicated [here](https://wiki.eclipse.org/Eclipse.ini) in order to use that JDK. For that we use `-vm`, that needs to occour before `-vmargs`, for example:

```bash
....
--launcher.appendVmargs
-vm
/opt/java7/jdk1.7.0_75/bin/java
-vmargs
-Dosgi.requiredJavaVersion=1.6
...
```

## Kafka maintenance for development in local mode

### Deleting a Kafka topic
Deleting a topic is not recommended, this is more for development purposes, in production the idea is letting the TTL for the topic arrive to Kafka.
For Kafka 0.8.1.1 we can use http://stackoverflow.com/questions/24287900/delete-topic-in-kafka-0-8-1-1 we can do it:

 * From the command line utils that come with Kafka:
```bash
bin/kafka-run-class.sh kafka.admin.DeleteTopicCommand --zookeeper localhost:2181 --topic test
```

 * With the Java / Scala API
```java
ZkClient zkClient = new ZkClient("localhost:2181", 10000);
zkClient.deleteRecursive(ZkUtils.getTopicPath("test2"));
```

### Purging a Kafka Installation
For development, as seen in http://stackoverflow.com/questions/16284399/purge-kafka-queue we can:
 * Stop the Apache Kafka daemon
 * Delete the topic data folder: rm -rf /tmp/kafka-logs/MyTopic-0
 * Delete the topic metadata: zkCli.sh then rmr /brokers/MyTopic
 * Start the Apache Kafka daemon

redpajama_set_name: RedPajamaGithub


The Township of Thorold was bounded by the Welland River on the south, separating it from the townships of Crowland and Wainfleet; on the west by the Township of Pelham, and Stamford Township on the east. Grantham Township sat directly to it's north.
The Township of Thorold was settled as early as 1787 when several Quaker families arrived there from the United States. By 1817 the population was estimated to be 830 with one Quaker Church in the township. A small settlement of about 200 inhabitants had formed at St. John's, along the Twelve Mile Creek where a saw mill and a grist mill, as well as a public school, several churches and shops were located.
In 1829 the Welland Canal was completed which traversed the township from north to south and brought prosperity to such communities as Port Robinson and Allanburg.
In the year 1885 the township contained 816 horses, 1,268 sheep, 1,847 horned cattle, and a population of about 2,000. There were 2,211 acres of winter wheat and 807 acres planted to orchards and gardens.
By 1800 a small settlement of about 200 inhabitants had formed at St. John's, along the Twelve Mile Creek where a saw mill and a grist mill, as well as a public school, several churches and shops were located.


The Minnesota Council of Nonprofits AmeriCorps VISTA Program began in 2014 with only seven VISTA members and six host site organizations. In 2015 we grew to 11 VISTAs, and have welcomed 19 service members this year! We are proud to have a strong network of engaged alumni. Learn more about them in the spotlight stories below and on the VISTA Blog.
Kaitlyn was the AmeriCorps VISTA for Exodus Lending during the 2017-18 service year. It was there that Kaitlyn was able to apply what she learned in school to address social inequalities. As a VISTA, she developed procedures for analyzing reports, streamlined the internal data tracking process, and even created and launched the communications and fundraising plans.
Jenna was a VISTA for HACER in the 2014-15 cohort. As a VISTA for HACER, Jenna's main project was to develop a comprehensive five-year strategic plan to guide the organization and advance its mission.
In 2014 Amalia was looking to gain experience working with indirect services. While looking around, she found MCN's VISTA program. After applying Amalia became MCN's very first VISTA for the 2014-15 service year.
Jess was the VISTA for the Minnesota Indian Women's Resource Center (MIWRC) in the 2017-18 cohort. Like many VISTAs, she decided to stay in the local nonprofit sector after service. Jess now works at the Institute for Local Self-Reliance as part of the Community Broadband Network Initiative.
After only a year, MCN VISTA alumni are already making an impact in the post-service world.
Inspired by her year at Nexus Community Partners working with local immigrant and refugee communities, VISTA alumna Giselle Efon received her Master of Social Work with an emphasis in clinical mental health from the University of Minnesota this May. In December she will return to her home country of Cameroon for the first time since moving to the United States, where she will spend a month with the goal of better understanding her own roots and identity.


text:
£248m of new mortgage lending for The West Brom
By John CorserBusinessPublished: Nov 26, 2020 Last Updated: Nov 26, 2020
The West Bromwich Building Society has reported £248 million of new mortgage lending in the six months to the end of September.
Chief executive Jonathan Westhoff
The 171-year-old society's half year pre-tax profit figure was down from £5.5 million a year before to £2.9m.
Chief executive Jonathan Westhoff said: "Set against the challenges of operating in full lockdown for half of the period, I'm pleased to report the society has delivered a robust first half performance. Throughout this period, we have been focusing on prioritising the wellbeing of our members, colleagues and communities, remaining operationally and financially resilient and ensuring our products, services and premises are safe and accessible."
The new mortgage lending was £3m less than a year before with with 48 per cent of new mortgages going to first-time buyers.
It was the first lender to launch a new mortgage product proposition specifically designed to support mortgage prisoners, who have previously been trapped in paying higher rates due to strict borrowing criteria.
The society maintained its average rate for savers at 17 per cent above the market average, rising to 52 per cent above by the period end.
The profit figure was lower because of setting aside additional provisions for credit losses in respect of the anticipated economic impact arising from the Covid-19 pandemic.
Mr Westhoff said: "We responded to the first lockdown back in March by adapting our operating model to ensure we could maintain our high service standards. We did not place any employees on the furlough scheme, and ensured all employees were paid their full salary irrespective of whether they are required to work their full contractual hours. We supported home working for a high number of employees, and reduced the number of people working in our head office and branches to enable social distancing.
"A priority was to support our most financially vulnerable borrowers, and help them access a payment deferral and other specific support measures if required. We set up a dedicated team to support these customers, and as of September 30, 14 per cent of all our residential mortgages had taken a payment deferral. For those that had reached the end of their initial payment deferral, 84 per cent had either restarted their monthly payments or redeemed their mortgage, with 16 per cent requiring an extension, representing two per cent of total residential mortgages outstanding.
"Our activity in the mortgage market, after a relatively short hiatus at the start of the first lockdown, recovered strongly. The knock on effects of a pause in the housing market mean that gross lending for the six months to September does not accurately reflect the true strength of our performance; compared with the same period last year, our application volumes were up 39 per cent to £563m.
"Although we took the responsible approach to limiting the amount of high loan to value lending, this did have an impact on the volume of first-time buyer activity. This segment of the market, however, is still at the heart of our purpose and we were able to support a further 699 first-time buyers to become homeowners in the six months to September 30."
Mr Westhoff said that with the Bank of England base rate at an all-time low of 0.10 per cent, and the prospect of the rate potentially turning negative, savers hade been hit hard by the pandemic with savers' rates at extremely low levels.
"Despite this, throughout the first six months of the year, the society has continued to help savers by paying average rates some 17 per cent above the market average; this had increased to 52 per cent by September 30. In monetary terms, this means the society has paid an additional annualised £2.6m in interest and equivalent to £5.7m using the period end rates. While this is lower than previous years, this is reflective of our duty to balance the needs of savers and borrowers.
"An area that has been hit hard by the pandemic is the third sector, with donations and fundraising levels falling due to the lockdown restrictions. As a mutual, supporting the communities in which we operate is a core part of our ethos, and we were able to channel our fundraising efforts to helping groups most in need as a result of the pandemic. This includes donations to local food banks, care packages to new mothers at Walsall Manor Hospital, and colleagues volunteering to provide hot meals to the homeless. We also raised £30,000 to support the Midlands Air Ambulance to support the vital, lifesaving work they do across the region."


The vibrant Blue shaggy rug is ideal for adding that cool and tranquil look to your family home. This Sable Two shaggy is hand tufted from 100% polyester, featuring a shaggy border and a soft glossy centre. Available in a range of colours and 3 sizes to buy online now.


Anyone can help with loading screen mod?
Hi guys, I just created a pack of replacement loading screenshots, but everytime I start the launcher (european client), it automatically downloads the default screens from internet and overwrites the ones I made.
Is there a way to prevent the loader from downloading the default screens?
I have read on Lotrointerface that is possible to modify a file to instruct the launcher to use images from a specified folder.
""I had a small problem with the pictures ... the Launcher randomly redownloaded the old ones. After some testing I found a small workaround ... I created a seperate folder with your pics and copy the adress (e.g. F:\The Lord of the Rings Online\raw\de\logo\Samoo). Then Ive opened the DownloadFileList.xml (C:\Users<<USER>>\AppData\Local\Turbine\The Lord of the Rings Online\Launcher) and copied the adress of the created folder in the DownloadURL lines (e.g. <DownloadUrl>F:\The Lord of the Ring Online\raw\de\logo\Samoo\lotro_ad_pregame_de.jpg</DownloadUrl>).""
Unfortunately the trick above doesn't work for me, since I can't find any file in my whole hard disk named DownloadFileList.xml.
PS: Something is very strange, anyway. One image out of 11 has been left untouched by the launcher. It replaced all my images except the one named "lotro_generic_teleport_screen_02".


text:
Live Decoy Birds
Raptor Persecution
Articles and Reports
Wildlife Laws
Corvid Traps
Larsen Traps
Ladder Traps
SongBird Survival Trust
Bloodsports Links
Flawed Squirrel Report
SST Accused
GWCT
Songird Survival Flawed Squirrel Research
On 28 July 2006 the SongBird Survival Trust issued a press release claiming:
The grey squirrel is having a devastating impact on our songbird population. It ranks alongside the domestic and feral cat as the top predator of farmland birds. This is just one of the findings of Professor Roy Brown of Birkbeck, University of London, in a new review carried out for the wildlife charity SongBird Survival.
Many factors have been blamed for the decline in our songbirds in recent years. The debate about raptors has raised the issue of predation but mammal predation had largely been overlooked," says Professor Brown. The review shows that the true impact of this predation is much worse than expected.
Needless to say the report was lapped up by the pro-killing brigade and the press. On the 7 November 2006 Lord Rotherwick took up the case in the House of Lords asking, "What is their (UK government) response to the finding in the report by the Songbird Survival trust that grey squirrels are one of the main predators of songbirds."
The Minister of State, Department for Environment, Food and Rural Affairs (Lord Rooker): "My Lords, it is not all the fault of squirrels; I would think that pesticides have something to do with this issue. We note the report that the Songbird Survival trust has produced and are grateful for its contribution to the debate. We would encourage the trust to share any unpublished evidence and research with the scientific community. Of the 15 species covered in the report by the trust, only three species were identified as being affected by grey squirrel predation: blackbirds, robins and whitethroats. These results are not consistent with the findings of the repeat woodland bird survey of 2005."
So much for 'devastating impact'!
Copyright 2020 - Against Corvid Traps


Enjoy an outdoor performance of a Shakespeare great in Tiverton this month.
Bring your picnic, blanket and chairs along to the Tiverton Community Arts Theatre (TCAT) on Saturday, June 30 as The Festival Players return with The Winter's Tale.
Julian Morgan of TCAT said: "We are very pleased to welcome back the Festival Players this year. This is now the fifth time they have performed for us and every time they have entertained the audience with a polished performance.
Hamlet in 2016 was a triumph following on from their previous all male cast performances. We have found that Shakespeare performed in the open air on a summer's evening is just the best way to experience the Bard's work.
The performance starts at 7.30pm (Grounds open from 6pm), Prices are £14.00 Adults, £8 Concession price for students.
Tickets available at www.tivertontheatre.com or Busy Bee Newsagents, Fore Street, Tiverton.


PowerPoint 2013 boasts updated features and new possibilities; this highly visual tutorial provides step-by-step instructions to help you learn all the capabilities of PowerPoint 2013. It covers the basics, as well as all the exciting new changes and additions in a series of easy-to-follow, full-color, two-page tutorials. Learn how to create slides, dress them up using templates and graphics, add sound and animation, and more. This book is the ideal "show me, don't tell me" guide to PowerPoint 2013.


text:
20 Costumes That Will Earn You a Halloween Beating
Single Billionaire Seeks Bubbly Female For Moon Vacation
Jeff Bennion ·
On Monday we brought you 35 insane costumes from around the world-- costumes so inexplicable and bizarre that if you actually wore one around for a night, you'd probably be greeted by one long, roving standing ovation.
What follows are 20 samples from the other end of the costume spectrum--good for inspiring a healthy mixture of douche-chills and rage. If you're wearing one of these, consider yourself warned: those guys aren't getting up from the bar to come over and shake your hand.
Spongebob Adult Man's Costume
Aside from the "Would you like to buy some Bibles?" look on the model's face, it's the white leotards that seal the wearer's fate. Or possibly the little red dick-tie.
ATM Man
If you want to look like a Muslim Darth Vader, this costume is for you. Despite the claims in the picture, don't be surprised if women dressed like money do not approach you to withdraw cash from your wiener.
Breathalyzer
The early model Terminators were easily identifiable, since Skynet's files on what penises really look like were badly damaged in the initial attack.
Cain The Vampire Tyrant
Oh shit! It's Cain! Cain the Vampire Tyrant! And he's been playing the Nintendo with his power Glove!
This dude finally gets a modeling gig that has nothing to do with his biceps or his dreadlocks, so he improvises. The woman is sad because she knows the inevitable attempt to "unlock" here will cause her entrails to go flying out of her lower back, a gruesome and undignified death via impalement on a wacky costume.
This lacks the sexual connotations of the lock and key outfit above, but we can't figure out if that makes the costume more sad or less. At least aliens won't be able to read their thoughts.
Oh, we get it. "Hung" as in hung like a horse. Like, you have a big penis. And you convey this by... attaching a stuffed horse's head to your groin? With a hangman's noose? This costume's designer has many a dead hooker in his basement.
We can think of two people off the top of our heads who haven't seen this movie: the guy who designed this costume and the guy wearing it.
I've Got a Heart On
We know. We can see it. And the children can see it. Warning: This costume is illegal in 48 states.
Just because he is wearing a hat and carrying frankincense does not mean this is not a Geisha Girl costume.
Super Jew
Whether the kid is Jewish or not, we're pretty sure this costume qualifies as some kind of hate crime.
Baby and Mommy
If you think it looks bad now, every time he walks, it looks like an 8 year old in diapers humping a babushka wearing basketball shoes.
The Munchkin
Okay, that's fucking terrifying. Is that a wig or not? Forget it, we don't want to look at it any more. We're going to wake up some night and see this bastard staring down at us, orange cheeks and all.
God's Gift to Women
"From: God, To: Women?" Well, they are going to be disappointed when they open it up and see that it's just the rest of this guy.
The good news for him is that next year when they get divorced, he can just buy a sombrero and he's got a Mexican stereotype costume. She's stuck going as an amoeba.
If you take off the Taz mask, you've got a pretty terrifying childbirth costume here. Complete with dentata!
Yes, he's dressed as a one-night stand. GET IT? These "abstract idea costumes" actually wind up being some of the most disturbing. Such as...
The Shit Hit the Fan
If you don't have this jackass to explain the joke, this looks more like maggots crawling out of a drain. Which actually makes one of the most awesome and disturbing Halloween costumes we've ever seen. Congratulations on the accidental horror, guys.
Goth Milk
There is no place on earth where this costume won't get you a vicious beating. You wouldn't even make it out of Quaker country in this thing. Goths, puns, suggested genital piercings on a child... it's like they distilled everything a good man finds offensive and expressed it in shitty costume form.
Slave Leia
My goodness what a lame costume. Take it off.
For male celebrities who go through life looking like they're wearing an old gay woman costume, check out 25 Men Who Look Like Old Lesbians. Or find out about some action stars whose careers had a less than happy ending in 5 Movie Martial Artists That Lost a Deathmatch to Dignity.
5 'Ugh' Trump Stories The News Totally Forgot About
We're so inundated with Trump news that we shrug off scandals that would tank any other president.
By E.M. Caris
4 Crapsack Things That Are Still Legal In The USA
6 Obnoxious Tourist Scams From Around The Globe
Every tourist destination has scammers looking to separate the unwary from their money.
The Craziest Movie Easter Egg That Nobody Noticed
We thing this might just be the craziest, most elaborate Easter egg in movie history.
6 Soulless Companies That Own, Like, Everything
Here are some companies we're just sorta letting take over the world.
By Maxwell Yezpitelok
5 Towns Ruined By The Movies Filmed There
We're not sure if you've noticed this, but movie fans can get a little ... obsessive.
By Steven Assarian, Zanandi Botes, Jordan Breeding, Casper Orens, Jon Kaulay